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Essentials of Soil Mechanics and Foundations Mccarthy 7e

ISBN 978-1-29203-939-8

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Essentials of Soil Mechanics and Foundations: Basic Geotechnics David F. McCarthy Seventh Edition

Pearson New International Edition Essentials of Soil Mechanics and Foundations: Basic Geotechnics David F. McCarthy Seventh Edition

Pearson Education Limited Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companies throughout the world Visit us on the World Wide Web at: www.pearsoned.co.uk © Pearson Education Limited 2014 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without either the prior written permission of the publisher or a licence permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, Saffron House, 6–10 Kirby Street, London EC1N 8TS. All trademarks used herein are the property of their respective owners. The use of any trademark in this text does not vest in the author or publisher any trademark ownership rights in such trademarks, nor does the use of such trademarks imply any affiliation with or endorsement of this book by such owners.

ISBN 10: 1-292-03939-6 ISBN 10: 1-269-37450-8 ISBN 13: 978-1-292-03939-8 ISBN 13: 978-1-269-37450-7

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library Printed in the United States of America

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Table of Contents

1. The Soil and Rock of Planet Earth: Geologic Overview David F. McCarthy

1

2. Soil Types and Soil Structure David F. McCarthy

39

3. Soil Composition: Terminology and Definitions David F. McCarthy

61

4. Index Properties and Classification Tests, and Soil Classification Systems David F. McCarthy

79

5. Site Investigations: Purpose and Methods, Information and Procedures Available David F. McCarthy

121

6. Movement of Water Through Soil: Basic Hydrogeology, Subsurface Flow, Permeability, Capillarity David F. McCarthy

201

7. Movement of Water Through Soil: Practical Effects: Seepage, Drainage, Frost Heave, Contamination David F. McCarthy

239

8. Combined Stresses in Soil Masses: Stress at a Point and Mohr's Circle David F. McCarthy

303

9. Subsurface Stresses David F. McCarthy

319

10. Settlement: Soil Compression, Volume Distortion, Consolidation David F. McCarthy

343

11. Shear Strength Theory David F. McCarthy

397

12. Earthquakes and the Effects David F. McCarthy

445

13. Foundations: Introductory Concepts David F. McCarthy

475

I

14. Foundations: Design Considerations and Methods David F. McCarthy

505

15. Site Improvement: Earth Moving, Compaction, and Stabilization David F. McCarthy

615

16. Stability of Unsupported Slopes David F. McCarthy

679

17. Lateral Pressures and Retaining Structures David F. McCarthy

743

18. Appendix: Laboratory Procedure to Determine Coefficient of Consolidation

II

David F. McCarthy

827

Index

833

The Soil and Rock of Planet Earth Geologic Overview

It has long been known that the earth is a dynamic constantly changing planet, but facts uncovered by scientists throughout the last half-century indicate that these ongoing changes involve factors of nature that are more profound than ever imagined previously. The major constituents of the earth’s outer zone—the soil, rock, and water—are continually subjected to forces that incite change. The changes in rock and soil deposits are often slow and subtle, and may not be recognized over short geologic periods, such as the human lifetime. However, rapid (sometimes instantaneous) changes also occur because of natural phenomena or human activities (mass movements during earthquakes, landslides, and tidal waves, or effects of war and large construction projects). Related to the struggle to survive and the drive to improve, humans have learned about many of the factors responsible for changes that occur in the earth’s rock and soil deposits and the factors that affect the (geologically) short-term behavior. Humans have always realized that soil influences their survival. Soil is the ground on which we stand. We build with soil. We have relied on soil to support structures and the paths of transportation. We depend on soil to grow our food and provide the products we use for living, protection, and comfort. We have trusted the soil to be stable and permanent. In some of these presumptions, we have been incorrect. This text studies soil as a material that is used to build with or on, but also as a material of the environment that may act in combination with other forces of nature or of civilization to affect landforms, our structures, and the state of our environment. This study is the field of geotechnics. This chapter presents a concise overview of important geologic-based information, to provide an understanding of the factors responsible for the formation and behavior of today’s rock and soil deposits and the factors that will cause changes in the future. In this introductory study of geotechnics, the agricultural aspects of soil are not considered, with the exception of

From Essentials of Soil Mechanics and Foundations: Basic Geotechnics, Seventh Edition. David F. McCarthy. Copyright © 2007 by Pearson Education, Inc. Published by Pearson Prentice Hall. All rights reserved.

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The Soil and Rock of Planet Earth

discussions relating to the vegetation used as a means of erosion control and as a factor influencing the creation of present-day soil types. The earth’s crust is composed of soil and rock. Rock can be defined as a natural aggregate of minerals that are connected by strong bonding or attractive forces; for this reason, rock is often considered a consolidated material. Soil may be defined as the unconsolidated sediments and deposits of solid particles that have resulted from the disintegration of rock. To the construction industry and the engineering profession, however, soil is also assumed to include the residue of vegetable and animal life, including civilization’s buried trash, garbage, and industrial wastes. Soil is a particulate material, which means that a soil mass consists of an accumulation of individual particles that are bonded together by mechanical or attractive means, though not as strongly as with rock. In soil (and in most rock), voids exist between particles, and the voids may be filled with a liquid (usually water) or a gas (usually air). As a result, soil deposits are often referred to as a three-phase material or system (solids plus liquid plus gas).

1

Rock: The Source of Soils Most of the nonorganic materials that are identified as soil originated from rock as the parent material. Rock types are grouped into three major classes—igneous, sedimentary, and metamorphic—determined by their origin or method of formation. The type of soil that subsequently develops relates to the rock type, its mineral components, and the climatic regime of the area. Igneous rock resulted from the cooling and hardening of molten rock called magma, which originated deep within the earth. Molten magma that escaped to and near the surface of the earth through volcanoes and fissures in the earth’s crust (termed lava) cooled quickly. As a result of rapid cooling, the mineral components solidified into small crystals and possessed a fine, interlocking texture. In some situations, the cooling was so rapid that a crystal-free, glassy texture resulted. The molten materials (lava) that cooled rapidly at or near the earth’s surface are called extrusive or volcanic rock types and include the basalts, rhyolites, and andesites. Molten rock trapped deep below the surface of the earth (magma) cooled slowly. The mineral components formed in large interlocking crystals, and coarse-textured rocks resulted. These rocks are classified as intrusive or plutonic types and include the granites (the most common) as well as the syenites, diorites, and gabbros. Many of the mineral combinations in igneous rocks are unstable in the environment existing at the earth’s surface. Upon exposure to air, water, chemicals in solution in water, freezing temperatures, varying temperatures, and erosive factors, the rock minerals break down to the soil types existing today. Rock whose chief mineral is quartz or orthoclase (potassium feldspar), minerals with high silica content, decomposes to predominantly sandy or gravelly soil with little clay.1 Granites, syenites, and rhyolites are in this category. Because of the high silica content, these rocks are classified as acidic. 1According

to the engineering definition, coarser soils include sands and gravels and are particles larger than 0.074 mm. Particles that comprise the fine-textured silt and clay category of soils are smaller than 0.074 mm. The 0.074 mm size is close to the smallest particle size observable with the unaided eye under normal conditions. Most clay particles are smaller than 0.002 mm.

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The Soil and Rock of Planet Earth

Rocks whose minerals contain iron, magnesium, calcium, or sodium, but little silica, such as the gabbros, diabases, and basalts, are classified as basic rocks. These rocks decompose to the fine-textured silt and clay soils. Generally, the acidic rocks are light-colored, while the basic rocks are very dark. Intermediate colors reflect an intermediate chemical composition. Rock types that are intermediate between acidic and basic include the trachytes, diorites, and andesites. Because of their mineral components, diorite and andesite easily break down into the fine-textured soils. The clay portion of fine-textured soil is the result of primary rock minerals decomposing to form secondary minerals. The clays are not small fragments of the original minerals that existed in the parent rock. Because of this change, the properties and behavior of clay soils are different from those of the gravel, sand, and silt soils, which are still composed of the primary rock minerals. Rocks that are acidic (not basic), such as the granites, are considered to be good construction materials. Sedimentary rocks are formed from accumulated deposits of soil particles or remains of certain organisms that have become hardened by pressure or cemented by minerals. Pressure sufficient to harden or solidify a soil deposit results from the weight of great thicknesses of overlying material. Under this pressure, the deposit is compacted and consolidated, and strong attractive bonds are established. Cementing minerals such as silica, calcium carbonate, and the iron oxides are abundant in soil as a result of rock weathering, and when dissolved in the water circulating through a soil deposit, they precipitate out onto the soil particles. Other cementing may be obtained from within the mass by solution or chemical change of materials. Sedimentary-type rocks include the limestones and dolostone (dolomites),2 shale (claystone, siltstone, mudstone), sandstone, conglomerate, and breccia. Geologic conditions in past historic times have had a very significant effect on the location and type of sedimentary rocks that exist across North America today. In early prehistoric times, most of what is now the United States was under water (Figure 1). Gradually, much of the land rose. Accumulated sediments in these shallow seas eventually became the limestones, shales, and sandstones of today. As a consequence of the layered manner of soil deposition, many sedimentary rocks are easily recognized today because of their stratified appearance (Figure 2). Shales are predominantly formed from deposited clay and silt particles. The degree of hardness varies, depending on the type of minerals, the bonding that developed, and the presence of foreign materials. The hardness is generally due to external pressures and the particle bonds that resulted, not to cementing minerals. Many shales are relatively stable when exposed to the environment, but some expand or delaminate (the layers separate) after contact with water or air. Weathering breaks down shale to fragments of varying sizes. These fragments, in turn, may be quickly reduced back to clay particle sizes. The properties of shale are quite important to the construction industry, for it is estimated that shale represents approximately 50 percent of the rock that is exposed at the earth’s surface or closest to the surface under the soil cover. Sound shale can provide a good foundation material. Its use as a construction material is questionable, however, because of its tendency to break down under handling, abuse, and weathering.

2Historically, the term dolomite has referred to both the rock-forming mineral CaMg(CO ) and sedimentary 3 2 rock. Dolostone is a recent proposal for designating the rock material, to avoid confusion.

3

The Soil and Rock of Planet Earth

Volcanoes

Land area

Swamp areas

Land area

Shallow sea

High mountains Volcanoes

Shallow sea

Low hills Plains

Figure 1 Generalized geographic map of the United States for period approximately 100 million years ago. Sediments deposited in shallow seas became sedimentary rock that exists in such areas today. (Source: U.S. Department of the Interior, Geological Survey)

Figure 2 Sedimentary rock in central New York State showing stratified formation.

Sandstone is predominantly quartz cemented together with mostly silica, but also calcium carbonate, or iron compounds. Sandstones are generally considered good construction materials. Conglomerate (cemented sand and gravel) and breccia (cemented angular rock fragments) are similar to sandstone. Limestone is predominantly crystalline calcium carbonate (calcite) formed under water. This rock material forms as a result of chemicals precipitating from solution, and from the remains of marine organisms and action of plant life. Because of the sedimentary nature, limestone formations can be heterogeneous and include alien materials such as

4

The Soil and Rock of Planet Earth

clays and organic materials (including trapped gases). (Limestone, and limestonedolomite, formations may be referenced as Karst or Karstic terrain.) The degree of hardness and durability of limestones will vary. Some rock formations are very sound, but some are relatively soluble to ingredients carried by nature’s groundwaters, with the result being the development of cavities. If the rock cap for a cavity located near the surface of the limestone formation deteriorates, a typical result is that the overlying earth material collapses into the void (referred to as a sinkhole) (see Figure 3). Dolostone is a variety of limestone, but harder and more durable. Marl and chalk are softer forms of limestone. Weathering of limestones can produce a soil that includes a large range of particle sizes, but the fine-grained soils predominate. Limestone is a good foundation material, provided that the formation is sound and free of cavities. Sound limestone is considered a good construction material. Frequently, because of the manner of deposition of the original rock-forming sediments, sandstone and shale, or limestone and shale, and sometimes sandstone, limestone, and shale are interbedded. Metamorphic rock results when any type of existing rock is subject to metamorphism, the changes brought about by combinations of heat, pressure, and plastic flow so that the original rock structure and mineral composition are changed. Plastic flow for rock refers to slow viscous movement and rearrangement within the rock mass as it changes and adjusts to the pressures created by external forces. Under these conditions, limestone is changed to marble and sandstone to quartzite. When subject to metamorphism, shale is transformed to slate or phyllite. Higher levels of these factors change the shale, or slate, to schist. Because

Figure 3 Sinkhole occurrence in area underlain by carbonaceous (limestone family) rock. (Source: USGS photo)

5

The Soil and Rock of Planet Earth

of the processes affecting their formation, slates and schists become foliated rocks (that is, layered as in a folio). Gneiss is a foliated rock with distinctive banding that results from the metamorphosis of sedimentary rock or basalt or granite. Despite their possibly different origin, a distinction between gneisses and schists is not always clear. These two rock types on occasion appear to be gradational. Metamorphic rocks formed from sound igneous or sedimentary rocks can be good materials for construction. However, schist, gneiss, and slate are questionable construction materials, because the foliated or banded structure can act to originate planes of weakness that affect strength and durability. Upon weathering, some metamorphic rocks break down to soil types comparable to that which would be derived from the original igneous or sedimentary-type rock. Others reflect the changes brought about by metamorphism. Gneiss and schist decompose to silt–sand mixtures with mica. Soils from slates and phyllites are more clayey. Soils derived from marble are similar to those resulting from limestone. Decomposition of quartzite generally produces sands and gravels. The processes of rock changing to soil, soil changing to rock, and alteration of rock are continuous and occur simultaneously. The process of change or alteration takes place over long periods of time, and there is no set sequence in which changes occur. There are many rock types. To establish a proper perspective, the construction industry’s concern is generally not with rocks’ names but with their properties. In-place properties, such as hardness, and possible presence of fractures or fissures affect drilling, blasting, and excavation operations. The suitability for use as a foundation for structures is related to strength, durability, and possible presence of cavities or of fractures and fissures. The commercial value of excavated and crushed rock for fill and as an ingredient of concrete is influenced by soundness and durability. In a general way, desirable and undesirable properties have been associated with the different rock types.

2

Soil Categories: Transported, Residual Soil deposits can be grouped into one of two broad categories—residual or transported— based on the process responsible for the formation. Residual Soil formations are created from the weathering of rock or the accumulation of organic materials where the material remains at the location of origin. Transported soils are those materials that have been moved from the place of origin; transportation may have resulted from the effects of gravity, wind, water, glaciers, or human activity—either singularly or in combination. The presence of a soil formation that blankets an area is a result of global position, geologic history, climate, and topography; but as a simplified generalization, residual formations are more apt to be found in humid and tropic regions of the planet (reflecting the strong influence of that type of climate), whereas transported soils are more prevalent for the temperate and cold climatic regions. Though the engineering properties of all soil deposits (the properties important for designing and constructing structural foundations, earthworks projects, construction site work, etc.) are affected by the method responsible for the development and accumulation, some important properties of the two categories differ because of the different formation processes. The general distribution of residual soil formations and transported soil deposits across the United States is indicated by Figure 4.

6

The Soil and Rock of Planet Earth

A tla

ntic

Oce

an

c Ocean Pacifi

Canada

Gulf of Mexico Cuba Mexico

Puerto Rico

20° N Hawaii 155° W

Key Areas with residual soil formations Predominantly transported soil deposits

Figure 4 Distribution of transported soil and residual soil formations for the United States. (Compiled from several sources)

Transported Soils Transported soils are those materials that have been moved from their place of origin. Soil particles are often segregated according to size by, or during, the transportation process. The process of transplantation and deposition has a significant effect on the properties of the resulting soil mass; various details are discussed in subsequent sections of this chapter. Much of the knowledge about the engineering-related properties and behavior of soil formations is based on the vast experience gained from construction projects in areas of the world where transported soil deposits exist (which relate to the geography of the many established population and industrial areas of the world, such as locations near bodies of water, in temperate climates, etc.). Gravity- and Wind-Transported Soils. Gravity is generally capable of transporting aggregate particles only limited distances, such as down a hill or mountain slope,

7

The Soil and Rock of Planet Earth

with the result that little change in the soil material is brought about by the transportation procedure. Wind can move small particles by rolling or carrying them. Soils carried by wind and subsequently deposited are designated aeolian deposits. Particles of small sand sizes can be rolled and carried short distances. Accumulations of such wind-deposited sands often form dunes. Dunes are typically characterized by low hill and ridge formations.They generally occur in sandy desert areas and on the down-wind side of bodies of water having sandy beaches. Dune material is a good source of sand for some construction purposes, but if the particles are of uniform size and very weathered and rounded, the sand may not be highly suitable for all construction purposes. Fine-textured soils, the silts and clays, can be carried great distances by wind. Silt soils in arid regions have no moisture to bond the particles together and are very susceptible to the effects of wind. Clay, however, has sufficient bonding or cohesion to withstand the eroding effects of wind. Deposits of wind-blown silts laid down in a loose condition that has been retained because of particle-bonding or cementing minerals is classified as loess. Significant loess deposits are found in North America, Europe, and Asia (Figure 5(a)). In the United States, great thicknesses of loess exist in the vicinity of the Mississippi and Missouri rivers. With these materials, accumulations have built up slowly, and grasses growing at the surface could keep pace with the rate of deposit. The resulting rootholes and grass channels that remain have created a soil that has a high porosity and cleavage in the vertical direction. Natural and human-made cuts in this material will stand with nearly vertical slopes, as illustrated in Figure 5(b). However, if the soil is exposed to excessive water (becomes saturated or inundated), or is subject to severe ground vibrations, the soil’s stable structure can be broken down. Subsidence or settlement results. Consequently, loess formations should be considered as poor foundation soils unless they can be protected from the effects of water and vibrations. Volcanic eruptions have also produced “wind-transported” soils. The volcanic ash carried into the air with the escaping gases consists of small fragments of igneous rock. The soil type expected to result will be related to the mineral characteristics of the igneous rock, as discussed earlier. Generally, remains of volcanic ash deposits are limited. Because of surface deposition, they are quickly affected by weathering agents. Glacial Deposits. Much of Canada and the northern United States, as well as northern Europe and Asia, have been subjected to the past effects of massive moving sheets of ice, the continental glaciers (see Figure 6). The most recent geological period of glaciation is referred to as the Great Ice Age, and scientists generally estimate that it covered the span of time extending from about 2 million years ago to about 10,000 years ago. Recent theories supported by evidence consider that extended glaciation occurred periodically even earlier throughout prehistoric times, in cycles apparently related to the significant variation that occurs in the earth’s orbit (which gradually varies from nearly circular to elliptical, with the related change in distance between the sun and the earth being approximately 11 million miles), to variations in the tilt of the earth’s axis with respect to the plane of the orbit (from about 21.5˚ to about 24.5˚), and to a wobble about the axis of rotation. Conditions conducive to glaciation apparently develop on approximately 100,000-year cycles. The past periods of glaciation resulted in some present-day areas having been covered once (Figure 6), some areas several times. Glaciers expanded and advanced over the land

8

The Soil and Rock of Planet Earth S. Dakota Minn. Wis. Nebraska

Iowa Ill. Ind.

Col.

Okla.

Texas

Mo.

Ky. Tenn.

Ark. La.

Miss.

Kansas

Ala.

Gulf of Mexico (a)

Figure 5 (a) Major loess deposits of the world (shaded areas); enlargement shows area of loess deposits in central United States. (b) Highway cut through loess deposit in Iowa shows natural vertical cleavage.

when climatic conditions permitted or contributed to the formation of ice. Glacial advance ceased when melting at its limits equaled the rate of expansion. When, because of climatic or other changes, the rate of melting exceeded the rate of growth, the glaciers receded or shrank. Generally, glacier expansion or shrinkage and movement were slow. At present, only the polar regions of the planet remain covered by glacial ice, and for the present the condition is considered stable. Considerable quantities of soil have been moved and deposited by or because of glacial action. Such deposits are referred to as glacial drift. However, although glaciers moved vast quantities of soil and created a surface topography, the major topographical features such as mountain ranges or plains areas are not the direct result of glaciers. Indirectly, however, major topographical features have very likely been affected by the continental glaciers. The glaciers were several thousand feet thick in many areas, and this had two very significant effects: The tremendous amount of water taken to form the glaciers lowered the level of the

9

The Soil and Rock of Planet Earth

Figure 6 Areas of the Northern Hemisphere that have been covered by glacial ice (shaded areas). (Source: After Ernst Antevs, “Maps of the Pleistocene Glaciations,” Geological Society of America, Bulletin 40, 1929)

sea by some 125–150 m (400–500 ft), and the tremendous weight of the glaciers caused the land beneath them to depress. As a result of the lowering of the sea level, the continental limits of North America extended beyond their current limits. It is probable that much glacially eroded soil was dropped or washed to areas that are now under the sea. As a result of the land’s being depressed, areas of low elevation were subsequently flooded. After the glaciers retreated and the areas were relieved of their weight, the land rose in elevation. Rock and soil deposits that had been carried to the low flooded areas rose to create new land surfaces. Many lake areas became diminished in size or disappeared entirely. The Great Salt Lake in Utah, for instance, is the shrunken remains of a 50,000-km2 (20,000-square-mile) glacial lake that once had a depth in excess of 300 m (1,000 ft). As a glacier grew or advanced, it gathered and pushed soil ahead of it, or enveloped and gathered the soil into itself. All sizes of particles were picked up and mixed together, with no sorting according to size. Some of the material picked up was subsequently dropped during the advance, either under the glacier or in front of it, and then overrun by the continued movement. When a glacier advance stopped, soil being pushed by the glacier and soil being freed by the melting process accumulated in front of the glacier. When the glacier receded, all soil trapped in the melting ice was dropped. Such direct glacial deposits are a heterogeneous mixture of all soil sizes and are termed glacial till. The land form or topographic surface resulting after a glacier receded is called a ground moraine or till plain. The hills and ridges of till that formed at the front of the glacier and marked its farthest advance are terminal moraines (Figure 7). Recessional

10

The Soil and Rock of Planet Earth Developing terminal moraine

Glacier

Streams (from melting glacier and glacial rivers or streams)

Glacial till

Glacial outwash

Pre-glacial soil or rock (a)

Glacier

Moraine

Glacial outwash

(b)

Figure 7 (a) The development of a terminal moraine and outwash plain in front of a glacier. (b) Aerial photo of glacier showing developing moraine and outwash area.

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The Soil and Rock of Planet Earth

moraines are hills or ridges that represent deposits along the front of a glacier where it made temporary stops during the recession process. (Long Island, located in the northeastern United States, illustrates such a glacial landform of geographical importance;3 the western segment provides approximately half the land area for New York City, and the island houses one of the world’s highest population densities.) Debris dropped along the side of a glacier as it moved through a valley is termed a lateral moraine. Long low hills of till that extend in the direction of the glacial movement are called drumlins. Where the till material was dropped under the glacier and overrun, it became very dense and compact, and can provide excellent foundation support. The suitability of a till material for construction purposes, such as for a compacted earth fill, depends on the quantity and range of sizes of the soil particles; till deposits that have a preponderance of coarse particles are good construction materials, whereas deposits containing large percentages of silt and clay materials are, generally, relatively difficult to handle and compact. Where an area was subjected to repeated glacial action, early deposits could be overlain by the more recent glacial till, or the original deposits may have been moved and redeposited by the more recent glacier as a new landform. As a result, the original source of a material may be difficult to determine. In some situations, however, the color may provide information about its source. With reference to glacial deposits in the northern United States, material gathered by glaciers from the Hudson Bay area in Canada is gray, whereas soils picked up from glaciers originating in the area northwest of the Great Lakes are red in color as a result of the high iron content in the original soils of that area. Even while the glaciers covered a land area, there were streams and rivers of water flowing on the surface of the glaciers and in subterranean tunnels eroded within the glaciers. These flowing bodies of water carried material picked up from the land surface or Valley wall (soil and rock)

Meandering, intertwining streams River

Subterranean stream

Lake

Stagnant (stationary) ice

Kame terrace Esker

Kames

Kettles

Figure 8 Flowing water on a glacier and resulting effect on landform.

3Present-day Long Island is actually the result of two glacial actions, one forming the eastern or outer section (Ronkonkoma Ridge) and the other forming the western or inner section (Harbor Hill Ridge).

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The Soil and Rock of Planet Earth

eroded from the glaciers (Figure 8). Some such soil was ultimately carried to the front of the glacier, but much was also dropped along the routes of flow and where the water became trapped within the glacier or between a glacier and a valley wall. Soils deposited by the surface and subsurface glacial rivers, to remain in the form of long winding ridges (Figure 8), are called eskers. These deposits, sorted and stratified, are usually mostly coarse grained (sands and gravels). The range of particle size can vary considerably in cross section or over short distances, however. Nevertheless, eskers can provide a good source of coarse-grained soils for construction purposes, but the uniformity of gradation should not be expected to be consistent at different locations within the esker. Eskers frequently followed along locations of low ground elevation, such as river valleys. As a result, they often represent paths of good foundation soils across areas where poor soil deposits, such as fine-grained soil sediments or marshes, have subsequently accumulated. Such paths of coarse-grained soils ideally suit the requirements for highway subgrades because of their good drainage characteristics, low susceptibility to frost heave, and ease of handling during the construction process. Kames provide soil deposits somewhat similar to eskers; they are the remains of material dropped along the boundaries of a glacier and a valley wall (kame terrace) or in holes in a glacier. The resulting landform has the shape of knobs or small hills. As with eskers, the gradation of soils from kames should be expected to be variable. In some areas where kame terraces and outwash plains exist, the surface topography is further affected by depressions called kettle holes. Kettle holes apparently formed when great blocks of ice remained after a glacier receded and became buried by the glacial soil. When the ice melted, a hole or depression in the soil surface resulted. When water-filled, the depressions are kettle lakes. River Deposits. Flowing bodies of water are capable of moving considerable volumes of soil by carrying the particles in suspension or by rolling, sliding, and skipping them along the river bottom. The largest-diameter particle that can be carried in suspension is related to the square of the velocity of the flowing water. Coarser particles being carried in suspension are dropped when a decrease in the water velocity occurs, as when the river deepens, widens, or changes direction. Finer particles remain in suspension to be deposited in quieter waters. Thus, river deposits are segregated according to size. For most rivers, the volume of water that flowed and the volume of soil that was transported were variables that changed as land drainage forms changed or as seasonal variations in precipitation occurred. All soils carried and deposited by rivers are classified as alluvial deposits. However, glacial soils carried by rivers created from melting glacial waters and subsequently sorted and dropped according to size to create deposits of stratified glacial drift are referred to as glaciofluvial deposits or stratified drift. Glacially created rivers were developed from water escaping at the edge of a glacier or from trapped water breaking through a recessional moraine left by a receding glacier. Upon leaving the glacier or moraine area, the flowing water rapidly fanned out over a broad area of land, temporarily flooding it. With the resulting decrease in the flow velocity, the larger soil particles dropped out, forming fan- or delta-shaped flat beds of predominantly sand and gravel soils. Overlapping deltas of coarse soils spread over broad areas

13

The Soil and Rock of Planet Earth

created land forms classified as outwash plains (see Figure 7). The finer soil particles remained in suspension in the escaping water. Subsequently, these too settled out where the velocity of flow slowed or the water became ponded. At locations where a heavily loaded natural or glacial river broadened or encountered flatter terrain so that its velocity decreased, coarse soil particles dropped out to form submerged spreading triangular-shaped deposits termed alluvial fans. The alluvial fans are good sources of sand and gravel for construction purposes. Rivers flowing through broad flat valleys have often overflowed their banks during periods of flooding. When this occurred, the overflow velocity quickly diminished, and the heavier gravel and sand particles dropped out in the vicinity of the bank, forming low ridges termed natural levees. The broad lowland areas on either side of the river were also flooded over, but the materials dropped in these areas were the finer-grained soils. These are floodplain deposits (Figure 9). Where rivers bend or curve to change direction, the velocity of the flowing water can vary considerably between the inside and outside edges of the curve. These natural curves are called meander bends. Erosion may take place along the outside, whereas deposition takes place on the inside. The deposits are the coarser soils, generally sorted according to size. Constant erosion along the outside of a bend while the inside is being built up with sediments causes the river to migrate laterally. The old river locations provide good sources of coarse soils for construction and are good foundation sites. When a river shortcuts a large bend when eroding a new route, the old channel left behind is cut off from new flow, and the trapped water forms an oxbow lake (Figure 10). Such lakes eventually fill with predominantly fine-grained soils carried by low-velocity floodwaters or surface runoff. These areas become poor foundation sites. Lake areas were created in natural basins in the topography or, in glacial areas, in depressed reservoirs created between a terminal (or recessional) moraine and a retreating glacier. Natural lakes and glacial lakes often covered vast areas of land. The coarsest soil particles (sands) carried by rivers feeding into the lake would fall out of suspension quickly after entering the lake area because of the sharp decrease in velocity. Such coarse soil deposits, termed lake deltas because of the resulting shape of the deposit, are good sources of sand and gravel for construction purposes. Fine-grained particles remained in suspension and were carried to the body of the lake, where they eventually settled out. The larger particles (fine sand and silt) settled out first, whereas the smaller particles (clay) continued to remain in suspension. After the waters in the lake quieted, as in periods of little or no flow into the lake, the clay settled out. Alternating layers of these fine-grained deposits continually built up as the variation of water flowing into the lake area continued

Floodplain area (silt, clay soil)

Natural levee (sand, gravel soil)

River

Original valley soils (variable)

Figure 9 Cross section of floodplain deposit.

14

Fine-grained (silt, clay) soil

The Soil and Rock of Planet Earth

Figure 10

Aerial photo of meandering river forming oxbow lakes.

(Figure 11). If the lake area were extremely large, the coarser silt particles might settle out in areas close to the shore and the clays settle out in the quiet central areas of the lake. Frequently, the basin eventually filled with soil, or drained and left the lake deposit if the trapped water found or eroded an outlet. Soil formations remaining at the locations of

15

The Soil and Rock of Planet Earth Stream or river flow feeding lake Alternate layering of silt and clay soil (varied clays) in quiet lake area

Clays

Silts

Gravel and coarse sand deposited in river bed Silt and fine sand Sands (deposited close to lake shore)

Figure 11 Cross section of soil deposits in a lake area, indicating typical alternate layering of silt and clay soil.

former lake areas are termed lacustrine deposits. When layers are less than 1 cm in thickness, the sediments are described as laminated. Deposits of such fine sand–silt and clay layering are termed laminated clay. Laminations that have been deposited over a one-year period are varves. Unless such lake sediments were subject to the weight of a new glacier during a new glacial advance or to other overburden pressures, they are weak and compressible and make poor foundations. However, the deltas of coarse soils, dropped when the flowing waters entered the lake area, provide good foundation support and soils for construction use. At some former glacial lake locations, the silt–clay soil near the surface has become firm because of drying (desiccation). This can give the illusion of a strong deposit with good foundation capabilities. Such an area may be suitable for carrying roadways and light structures, but foundations for larger and heavier structures generally cannot be satisfactorily supported. In some areas where flowing waters carried fine-grained soils to ocean or seawater areas, some types of clay particles flocculated in the presence of the saltwater.4 The silt and clay then settled out of suspension at about the same rate, creating deposits of marine clays. Despite their classification, marine clays may actually consist of more silt than clay. The marine clays are typically gray or blue-gray in color. Frequently, these deposits contain the shells or remains of shells from marine life. Much of the seaboard area of the northeastern United States and southeastern Canada has such deposits. Because of the method of deposition, marine clay deposits are generally weak and compressible, and are therefore poor foundation materials. Where the land covered by the marine clays has subsequently elevated to be above sea level, there is an additional danger that the sodium (from the seawater) that reacted with the clays to cause them to flocculate may have been leached from the soil by percolating rainwater. The result is that the clays suffer a loss in strength and become sensitive to disturbance, and land areas that had been stable become unstable. Beach Deposits. Ocean beach deposits are predominantly sand materials and are constantly being changed by the erosive and redistributing effects of currents and wave action. These same currents and wave actions keep silt and clay particles in suspension and carry them to the deeper, quieter, offshore areas, where they eventually settle out. Long ridges of sand that form slightly offshore are termed bars (Figure 12). When the formation that 4Flocculation may be defined as the development of an attraction between, and bonding of, individual particles to

form larger particles.

16

The Soil and Rock of Planet Earth

Beach Lagoon Spit Barrier bar Sand Submerged sand bar

Open ocean

Figure 12

Types of sandbar deposits along a shoreline.

develops includes two or more submerged ridges, they are longshore bars. Such deposits have been built up by the breaking waves. When a small exposed ridge forms offshore from a gently sloping beach, the formation is termed an offshore bar or island bar (also called barrier beach or barrier island). A barrier bar is a deposit that almost completely blocks the entrance to a bay. A sand or sand–gravel accumulation that is connected to the shore and extends into open water like a finger is a spit. Marine sands are somewhat rounded and smooth, particularly the larger particles, and the deposit at a particular location may consist of particles of uniform size. Such factors, plus the corrosion potential due to salinity, may affect their usefulness for certain construction purposes, such as making quality concrete. Generally, however, such sands are potentially good sources of materials for construction purposes, particularly for waterfront and marine structures, because excavation is generally uncomplicated and transportation will be economical if barges or hydraulic pumping techniques (through pipelines) can be used. Swamp and Marsh Deposits. Swamps and marshes develop in stagnated areas where limited depths of water accumulate, or where periodic inundation and drying occur because of fluctuations in the groundwater level, and vegetation therefore has the chance to grow. The soils that subsequently form on the surface of swamp areas are generally of high organic content (from the decaying vegetation) and soft and odoriferous. Accumulations of decomposed or partially decomposed aquatic plants in swamp or marsh areas are termed muck or peat. Muck, geologically older than peat, is almost fully decomposed vegetation and is relatively dense. Peat includes partially decomposed vegetation and is normally spongy and relatively light. These materials are generally weak and highly compressible. Muck or peat deposits may be buried beneath the ground surface if the marsh area is subsequently overlain by glacial or alluvial materials. Peat has been used as a fuel in many areas of the world. In an early geologic period, when much of the present United States was covered over by shallow seas, an eastern section of the continent was generally a low swampy area where heavy vegetation flourished (Figure 13). Because of subsequent geologic changes, these lush forest areas slowly became flooded and covered over. While they were being destroyed, however, vast accumulations of partially decomposed forest vegetation were also being deposited. Subsequently, heat and pressure due to the depth of burial changed these deposits to the extensive coal beds found throughout Pennsylvania, West Virginia, Tennessee, Illinois, and Kentucky.

17

The Soil and Rock of Planet Earth

Deep sea

Shallow sea

Plains

Hills High mountains Swamp area Swamp areas Hills

Low hills

Figure 13 Generalized paleogeographic map of the United States approximately 200 million years ago, indicating extensive swamp areas that subsequently formed into coal deposits. (Source: U.S. Department of the Interior, Geological Survey)

Sanitary (Solid Waste) Landfills. The term sanitary, or solid waste, landfill relates to the technique of using burial methods for disposing of solid waste resulting from human activities. In recent times, many municipalities have accepted the sanitary landfill as a relatively economical and manageable method for handling the large volumes of trash being generated. Landfill sites typically have involved relatively large areas and often thicknesses equivalent to the height of multistory buildings; a result has been a surficial deposit of material equivalent to, or greater than, many of nature’s soil deposits. The reliance on the buried landfill method for disposing of solid waste has diminished in recent years as a result of national and state environmental regulations and the perceived dangers to the surrounding environment that will occur if thorough planning and careful operational procedures are not followed. Nevertheless, new landfills are being opened while some older landfills remain in operation, and numerous filled or completed landfill sites are already in place. The solid waste in the modern landfill is commonly the throw-away material generated by commercial business and the private or residential sector (such as garbage and trash), but may also include disposal from industry. Usually, but not always, the solid wastes accepted for burial at the present time are the materials rated nonhazardous and, preferably, degradable. Materials in the hazardous category, such as radioactive waste, medical waste, and petroleum or chemical wastes, are usually not permitted. Typically, modern solid waste landfills have each day’s delivery of waste covered by one or more layers of soil to help reduce odors, occurrence of flyaway materials, development of fires, and attraction to scavengers, rodents, and insects, and to improve the general appearance of the area being worked. The completed landfill then consists of multiple zones or cells encapsulated by earth (Figure 14). When a landfill site has been

18

Filled and covered trench section

Daily earth cover

(a) Trench method for solid waste landfill

Earth cover obtained from excavation in trench This area not yet excavated

Buried zones or cells of solid waste

(b) Alternate procedure, trench method Covered trench section Active trench (open for solid waste)

Cover soil

Excavated trench soil is stockpiled to use as soil cover Filled trenches (covered solid waste) (c) Slope and area method for solid waste landfill

Cover soil (obtained from on-site or off-site)

Zones or cells of covered solid waste

Figure 14

Methods for burying and covering solid waste in sanitary (solid waste) landfill.

19

The Soil and Rock of Planet Earth

filled (totally utilized), the new land area may be recycled to other use; commonly, recreational facilities requiring large surface areas, such as playgrounds and golf courses, are selected. Buried solid waste landfills do create some problems which should be recognized. Some can be controlled by proper planning and careful operation. The more significant issues relate to understanding several factors. Surface settlements will vary as the buried waste decomposes (the decomposition and settlement process generally takes many years). In addition, the decomposition of organic materials results in the generation of gases (some of which are combustible and some of which can be hazardous to animal and plant life). Also, surface and subsurface water passing through the buried waste can place into the solution undesirable particle matter, bacteria, and chemicals (the resulting polluted liquid is termed leachate) which might migrate and contaminate nearby areas. Present-day landfill designs commonly include an installed barrier liner for the base of the landfill area to prevent underground escape of the leachate (which is collected for treatment and proper disposal). At completion, the landfill area is sealed or capped to reduce future infiltration of surface water before receiving the topping layer of soil for the final vegetative cover. New and recently opened solid waste landfills typically serve a municipality or geographical area, and are controlled by a related governmental agency that is assigned the responsibility to have the design and operation comply with the environmental regulations established for the area. Municipal solid waste (MSW) typically has been mostly organicbased material subject to biological decomposition, but the inclusion of throw-away synthetics and nondisposable materials has been increasing. For a MSW composed primarily of degradable material, compression of the mass and resulting settlement is attributed to mainly two features: (1) early-term shifting and reorientation of materials and compression due to placement and (2) long-term decomposition resulting in volume decrease. When considering the property maintenance responsibilities associated with MSW sites, settlement is a major issue. Municipal sites are being monitored to determine the settlement versus time behavior, and as data is accumulated methods for long-range predictions are being developed. Such information is important for planning and budgeting the expected needs for expensive maintenance improvements, such as site service or public roads, and the reconstruction of the protective environmental earth cap cover. Most of the information used has been limited to sites existing for less than 20 years. The accumulated and published information (e.g., Ref. 268) indicates that graphical presentations of settlement versus time using a linear settlement scale and an accelerated (compressed) logarithmic time scale plot as a shallow slope line for the initial phase of the MSW accumulation (earlyterm settlement) followed by a long-term steeper slope line appears to permit extrapolation into future years (see Figure 15). The application of the accelerated logarithmic time compression scale reflects the decreasing rate of the decomposition process. Information as shown in Figure 15(a) and (b) implies that an eventual total settlement may be considerable.

20

The Soil and Rock of Planet Earth Time (days)

Time (days) 1 0

100

10,000 1,000,000

1 0

100

20

40

60

Site A

Vertical strain (%)

Vertical strain (%)

Site A

10,000 1,000,000

20

40

Measured (L-1)

Measured (L-1)

Predicted (hyperbolic function method)

Predicted (hyperbolic function method)

60 (a) (b) Figure 15 Long-term time vs. settlement data for a MSW landfill site—actual vs. predicted by different methods (hyperbolic function method and logarithmic function method). (Source: American Society of Civil Engineers, The Journal of Geotechnical & Geoenvironmental Engineering, Vol. 128, #2 February 2002; “Evaluation of Decomposition Effect on Long Term Settlement Predictions . . . ” by I. H. Park. With permission from ASCE.)

The long-term settlement data represented by Figure 15 are from existing sites where the daily procedure to place, compact, and bury the solid waste has not been well documented. Current methods for placement include the use of spreading-and-compaction procedures that will break down the new waste materials to minimize the volume occupied during placement so to maximize the utilization life of the site, recognizing that permitted landfill locations are designed and authorized only for an identified final volume (i.e., the final elevations and topography established for the site). Figure 16 shows compaction equipment designed to chop, tear, and shred the type of materials typically included in solid waste.

Residual Soils Residual soils (or sedentary soils) have principally formed from the weathering of rock or accumulation of organic material, and remain at the location of their origin. The weathering process may be attributed to mechanical weathering or chemical and solution weathering but does also involve the effects of biologic activity. Mechanical weathering refers to physical disintegration resulting from the effects of wind, rain, running water, ice and frost wedging, and tectonic forces. Chemical and solution weathering is rock decomposition due to chemical reactions in the rock materials that results from exposure to the atmosphere, temperature changes, water and water-based solutions, or other materials. Climate, topography, drainage, and vegetative cover have great influence on the chemical–solution weathering process. The accumulated depth of most residual formations is the result of chemical–solution weathering. Across the continental United States and offshore locations, some geographic areas are known for extensive deposits of residual soils—the Piedmont region of the Southeast, the Pacific Northwest, the central states south of the Ohio and Missouri rivers, and Hawaii

21

The Soil and Rock of Planet Earth

Figure 16 Heavy compaction equipment designed for MSW landfills includes large diameter wheels with projecting cleats intended to chop and shred solid waste materials prior to being covered with soil. (Photo courtesy of Terex Corporation)

and Puerto Rico—but residual formations also exist in other areas. Figure 17 outlines the areal extent of residual soil accumulations across the continental United States. The residual soils of the southeastern United States (Figure 18), commonly referred to as the Piedmont region but actually geographically larger and with geologic differences, cover a large region of population and industrial growth where the related construction and agricultural activity is directed by properties of residual soils (an area which includes the cities of Atlanta, Charlotte, Baltimore-Washington, D.C., and Philadelphia). Residual soils can include particles having a wide range of sizes, shapes, and composition, depending on the amount and type of weathering as well as the minerals in the parent rock. However, aged formations typically include a preponderance of fine-grained particles (silt or clay sizes) whose mineral composition and behavioral properties relate to the mineral composition of the parent rock. For example, the clays of the Piedmont have derived from deeply weathered metamorphic rock, whereas the residual clays of the Pacific Northwest and the areas south of the Ohio and Missouri rivers have evolved from deeply weathered sedimentary and volcanic rock. The thickness of residual soil existing at any particular location is affected by the rate of rock weathering and the presence or absence of erosive forces to move the accumulation of particles. A profile of residual material lying above the unweathered rock often shows the transitional stages of soil formation; downward from the surficial zone of soil (a zone that may bear little mineralogical resemblance to the parent rock) progressively lesser degrees of rock weathering and soil-type materials exist until finally the unaltered rock is encountered. Figure 19(a) is representative of residual soil formations in tropical regions, and Figure 19(b)

22

The Soil and Rock of Planet Earth Canada

c Ocean Pacifi

12

N

5

11 6

4

Oce

1

9

ntic

3

A tl a

10

7

an

2

8

Gulf of Mexico Mexico

Map Number

Area

1

Southern Piedmont and Blue Ridge

2

Appalachian Ridge and Valley

3

Appalachian Plateau

4 5 6 7 8 9 10 11

Springfield Plateau Missouri Plateau High Plains Colorado Plateau Edwards Plateau Osage Plains Plains Border Cascade Mountains

12

Columbia Plateau

13

Puerto Rico

14

Hawaii

Figure 17

Parent Rocks Gneiss, schist, granite, gabbro (igneous Sedimentary-metamorphic) Limestone, sandstone, shale (Sedimentary) Limestone, sandstone, shale (Sedimentary) Limestone (Sedimentary) Shale (Sedimentary) Shale (Sedimentary) Shale, sandstone (Sedimentary) Limestone (Sedimentary) Shale (Sedimentary) Limestone (Sedimentary) Volcanic ash, lava (Sedimentary-igneous) Lava, volcanic ash (igneous-Sedimentary) Volcanics, limestone (igneous-Sedimentary) Volcanic ash, lava (Sedimentary, igneous)

Residual Soil Profiles 0 to 30 m continuous

0 to 20 m limestone 0 to 3 m shale, sandstone continuous 0 to 10 m continuous 0 to 3 m irregular 0 to 2 m 0 to 2 m 0 to 10 m 0 to 2 m 0 to 10 m 0 to 20 m scattered 0 to 5 m scattered 0 to 20 m local, scattered 0 to 20 m

Principal residual soil regions for the United States [341].

23

PENN.

N.J. MD.

Appalachian Valley and Ridge Physiographic Province

DEL.

Piedmont Physiographic Province Blue Ridge Physiographic Province

W. VA. VA.

N.C. TENN.

S.C.

GA.

ALA.

Figure 18

Residual soil regions of the southeastern United States [68]. Topsoil–humus (highly organic) Surficial soil zone (oldest soil material; mineralogical composition may be altered and different than the deeper, newer soil)

Leaching soil accumulating

Zone of completely weathered rock (virtually all soil material) Zone of highly weathered rock (mostly soil material) Partially weathered rock Relatively sound rock zones

(a)

Zone of moderately weathered rock (decomposed or decayed rock)(rock 50% to 90%) Zone of slightly weathered rock Gneiss to Schist

Unweathered rock; some fractures and fissures in upper zone

Granite to Gabbro

(b)

Figure 19 Representative subsurface profiles for residual soil areas, indicating stages of transition from rock to soil. (a) Weathering profile of crystalline rocks in humid temperate regions [341]. (b) Weathering profile representative of formations in humid, tropical regions [230].

24

Saprolite

Saprolite

The Soil and Rock of Planet Earth

represents a cross section of conditions encountered in the temperate southeastern portion of the United States. The term saprolite refers to material that is essentially soil but which includes physical features representative of the parent rock (the “fabric”), as well as fragments of the parent rock. A feature of residual soil formations is that the buried soil zone of completely weathered rock (Figure 19, the “newer” soil zone directly beneath the surficial layer of well-aged soil), as well as the deeper zones of highly weathered and moderately weathered rock, retain some of the strength-related characteristics of the parent rock (such as the structural appearance or fabric, the particle bonding or cohesion, the resistance to deformation, etc.); such features are not found in deposits of transported soils. The rate of weathering is generally greater in warm, humid regions than in cool, dry regions. In the tropics, well-drained regions produce lateritic soils (soils with an absence of silica and alkalines but enriched with iron and aluminum compounds), whereas poorly drained areas show a prevalence of dark expansive montmorillonite clay; andosols (allophanic soil, a soil with an indefinite mineral composition but rich in silica) develop over volcanic ash and rock. Most residual soils can be used for construction projects, but typically some care and special treatment is required. Of significance, the behavioral properties of disturbed or reworked, wetted then dried, soil may be considerably different from those of the same soil that has not been disturbed. For a summary and overview of the interrelationships between rock and soil formations, see Figure 20.

Soil Gases and Liquids Soil deposits, being composed of irregularly shaped particles which have a range of sizes, will include open pore spaces or voids. Typically the void spaces include liquid— commonly water, but not always—and air (gas), whose properties are close to those found in the surface atmosphere. However, gases other than air or liquids other than water may be present as a result of natural conditions or the activities of humankind. Soil overlying deposits of petroleum and natural gas, or radioactive compounds, can include these other liquid or gaseous materials as a result of upward leakage. Radon, a naturally occurring gas, results from the decay of uranium, radium, and polonium and from radioactive waste. As part of the natural decomposition process occurring within masses of buried vegetation, gases such as methane, carbon dioxide, ammonia, hydrogen sulfide, and nitrous oxide are generated, as well as acidic fluids. Foreign liquids and gases find their way into soil deposits as a result of accidents and poorly planned human activities, including the establishment of waste dumps and landfills, the application of agricultural fertilizers and pesticides, the discharge of inadequately treated sewage, and the accidental spills and discharges of industrial and commercial compounds. Frequently, gases or liquids in low concentrations are accepted into the surrounding environment without significant effect and subsequently reduced to a less noxious material or dispersed. However, high concentrations will pollute the soil, the groundwater, and possibly the surface zone of the atmosphere. In sufficient concentrations, methane gas is explosive, hydrogen sulfide is toxic, carbon dioxide causes asphyxiation, and radon is carcinogenic.

25

IGNEOUS ROCK The original rock type for the planet; results from cooling of molten rock (magma, lava).

Some rock undergoes change caused by heat, pressure, plastic flow

Some rock breaks down, deteriorates, decomposes, to become soil

METAMORPHIC ROCK Results from igneous or sedimentary rock undergoing change, generally to a harder, stronger state

Some rock breaks down, decomposes, to become soil

Some rock undergoes change caused by heat, pressure, plastic flow Some rock breaks down to become soil SOIL DEPOSITS

Residual deposits: Remain in place at location formed

Deposits from transported soils: Soil materials transported, then deposited to create new soil formation

Waterborne Glacial action Ability to carry particles - Soil materials pushed relates to flow velocity: in front of glacial Effects can result in a advance, mixing occurs; "sorted" soil deposit at eventual deposit is a locations where flow heterogeneous mixture velocity decreases (larger of all soil particle sizes sand and gravel particles - Soil materials picked up and mixed into the glacial drop out of suspension, ice; a direct deposit of but smaller silt and clay these soils results in a particles remain in heterogeneous mixture suspension); fine grained of all particle sizes (if silts and clays eventually settle from suspension in the laid down soil is then a quiet body of water overrun by the glacier, (lakes, ponds, etc.) the soil materials will be compacted by the weight of the ice) Wind Humans - Rolls particles - Some projects have - Pick up and carry created new land - Glacial meltwaters and small particles; forms and/or streams/rivers flowing sand (sand dunes), extensively changed on and within the glacial silt (loess) surface topography ice results in deposition of (landfill and reclamsoil materials; some waters ation sites, dam and breach natural dams of reservoir projects, lakes to further transport open-pit or strip then deposit soil particles mines, etc.)

Gravity - Rolls particles downhill, etc. - Effects of tectonic forces - Volcanic ash

- From remains of marine life

- From breakdown and decomposition of parent rock

SEDIMENTARY ROCK

Hardening of soil deposits by cementing or pressure and strong particle bonding to become sedimentary rock

Some soil deposits Some soil deposits harden to rock remain soil

SOIL

Figure 20 Summary and overview: interrelationships between rock and soil formations. (Arrangement presentation after C. A. Matrosic)

26

The Soil and Rock of Planet Earth

3

Plate Tectonics The concept of plate tectonics deals with the development, formation, and changes occurring to the earth’s land and ocean areas on a long-term, large-scale basis. Plate tectonics relates to the recent recognition that the outer shell of the earth is made up of a small number of large, thick plates that mesh together like pieces of a gigantic puzzle (see Figure 21), but that also move somewhat independently of each other. Continents and oceans are carried (float) on these plates. The phenomenon of plate movement has had a profound effect on what constitutes today’s continents, including the type and location of rock and soil materials, mineral deposits, and animal and plant life. Very significantly, areas near plate boundaries coincide with the most active earthquake and volcanic regions of the world. In macroscopic terms, the earth’s interior consists of a core, mantle, and outer crust, concentrically zoned, as depicted in Figure 22. The uppermost region of the mantle possesses properties important to the plate tectonics theory. The outer zone, ranging from about 70 to 150 km thick and termed the lithosphere (rock sphere), consists of dense and hard brittle rock materials. Generally, the lithosphere is thickest beneath continental mountain masses. Underlying the lithosphere is a 200- to 250-km-thick zone of dense, semisolid, or plastic rock material termed the asthenosphere (weak rock). The rock materials making up the lithosphere and asthenosphere actually are similar, with the demarcation between zones being established on the basis of a difference in condition. Material in the lower asthenosphere increases in hardness and rigidity, grading to become the more solid material characteristics of the deeper mantle. Continental and ocean crusts of relatively light-density rock material (in comparison to the rock material of the mantle) top the lithosphere. The crust underlying continental areas is thicker and less dense than oceanic crust; continental crust ranges from about 20 to 60 km in thickness (thickest beneath mountain areas), whereas ocean crust is generally on the order of 10 km thick. The areal boundaries of the shell-like plates are the result of discontinuities or fractures that have developed in the lithosphere. At some locations, these plates of lithosphere are moving apart (diverging); at other regions the plates are moving toward each other (converging) or sliding past each other. The plates are rigid and tend to not change shape with time. Changes result only at zones where convergence or divergence occurs; the process of plate convergence and divergence will be described in later paragraphs. The causes of plate movement are not known, but it has been hypothesized that regions of the asthenosphere and underlying mantle are in motion because of a temperature difference between the earth’s core zone and the outer mantle (similar to the way heated air moves by convection through a cool room), with the lithosphere being carried along by the asthenosphere. What is established is that the plates of lithosphere have been in motion for a long period of the earth’s history and that the plates have been moving relative to each other. Continental and ocean crusts atop the lithosphere move with the lithosphere. The position and orientation of continent-supporting plates have been changing with respect to other continental plates and with respect to location on the earth’s surface. Similarly, ocean areas have been changing; some of these changes have been so extensive that oceans have been

27

A iccif Pa

ic rct nta

r is

Kermadec-Tonga trench

Macquarie ridge

Fiji plate e

Ridge axis Subduction zone

Antarctic plate

li ris

Chi

South American plate

Alps

t lan At

African plate

Iran plate

Eurasian plate

In ic-

n di a

r id

ge

Carlsberg ridge

Uncertain plate boundary

Antarctic plate

e

ridg

Turkish plate

Adriatic plate Hellenic plate

Transform Direction of plate motion

lan tic

-At

Mid

s

Figure 21 Major plates of the lithosphere and general direction of movement. (Source: After J. F. Dewey, “Plate Tectonics,” copyright May 1972 by Scientific American, Inc. All rights reserved.)

Solomon's plate

Peru-Chile trench

South East Indian rise

Australian plate

Nazca plate

Co pla cos te

Caribbean plate

North American plate

Mexico trench

ies ck

Philippine Pacific plate plate Bismark plate New Hebrides trench

e

San Andreas fault

Marianas trench

Kuril trench Japan trench

Ro

Java trench

Eurasian plate

ch tren tian Aleu Gorda plate

ns ia

rise

h at rp

East pacif ic

Ke yk ja

ne ge

rid

28 n ia ab te Ar pla

es

Ca

s

And

u Ca

su ca

The Soil and Rock of Planet Earth Continental crust 0 Lithosphere

Oceanic crust

100

Mantle

Asthenosphere

200

Kilometers

Sea level

100 km (±)

Cr us

t

350 km (±)

Lithosphere 288

Asthenosphere

0k

Mantle Core

514 0

m

km

Liquid outer core

6370 km Solid inner core

Figure 22 Cross section depicting composition of the earth (numerical distances indicate approximate depths from the earth’s surface).

both created and destroyed. Figure 23 illustrates the relative position of continental areas and oceans through the past 200-million-year period, and an expectation for the future. A rift develops where plates of lithosphere are moving apart, creating the condition where molten rock from the asthenosphere rises to create new lithosphere and crust; volcanism is involved. Plate divergence and the creation of new lithosphere in ocean ridge zones may be referred to as sea-floor spreading. Plate boundaries identified as ridges and rises (Figure 21) represent locations where plates are spreading apart. Most of the currently known locations of major plate divergence occur in what is now ocean area. It is likely that early in the earth’s history the spreading of some plates began under continental areas, but the widening separations filled with shallow crust, had low surface elevation, and were flooded over. The surface area of the earth remains approximately constant with time. The creation of new lithosphere where plates diverge must be balanced by an equivalent loss of lithosphere at other plate boundaries. Such losses occur at locations where plates converge, by having one plate subduct (dip) below the other. The subducted section of lithosphere passes into or through the asthenosphere, where it is absorbed. Boundaries where subduction occurs are identified as trenches in Figure 21. Ocean crust will be carried along when its supporting plate of lithosphere is being subducted, but continental crust does not subduct because of its light density; ocean crust is thus destroyed by subduction and created where plate divergence occurs, but continental crusts tend to be permanent.

29

80°

Sinus Borealis

60° 40°

P 20° Panthalassa

A

80°

40°

40°

N

Tet hys

G

20°

80°

120°

Se a

Sinus Australis

40° A

E

A

60° 80° (a)

Figure 23 Location and position of continents and oceans over the past 200-million-year period, on the basis of plate movements. (Source: After R. S. Dietz and J. C. Holden, “The Breakup of Pangaea,” copyright October 1970 by Scientific American, Inc. All rights reserved.) (a) Universal land mass Pangaea may have looked like this 200 million years ago. Panthalassa was the ancestral Pacific Ocean. The Tethys Sea (the ancestral Mediterranean) formed a large bay separating Africa and Eurasia. The Appalachian, Alps, and Ural mountain ranges were already ancient by this time. The Rockies and Andes were in existence but were relatively young. These ranges probably developed as a result of collisions between continental plates predating Pangaea (plates are shown lightly shaded).

60° L A U R A S I A 40° 20° 40°

80°

G

40°

O

N

80°

120°

20°

D

W 40°

A

N

A 80° (b)

(b) After 20 million years of drift (180 million years ago), the northern group of continents, collectively known as Laurasia, had split away from the southern group, known as Gondwana. The latter had started to break up: India had been set free by a Y-shaped rift, which had also begun to isolate the Africa–South America land mass from Antarctica–Australia. The Tethyan trench (hatched lines in black), a zone of crustal uptake, runs from Gibraltar to the general area of Borneo. Black lines and black arrows denote megashears, zones of slippage along plate boundaries. The white arrows indicate the vector motions of the continents since drift began.

(continued)

30

80°

40° 20° 80°

40°

40°

80°

120°

20° 40°

80° (c)

(c) After 65 million years of drift (135 million years ago), the North Atlantic and Indian oceans had opened considerably. The birth of the South Atlantic had been initiated by a rift. The rotation of the Eurasian land mass had begun to close the eastern end of the Tethys Sea. The Indian plate continued its northerly movement.

80° 60° 40°

80°

40°

40°

80°

120°

20° 40° 60° 80° (d)

(d) After 135 million years of drift (65 million years ago), the South Atlantic had widened into a major ocean. A new rift had carved Madagascar away from Africa. The rift in the North Atlantic had switched from the west side to the east side of Greenland. The Mediterranean Sea is clearly recognizable. Australia still remained attached to Antarctica. An extensive north–south trench (not shown) must also have existed in the Pacific to absorb the westward drift of the North American and South American plates.

(continued)

31

The Soil and Rock of Planet Earth

20° 160°

120°

80°

40°

40°

80°

160°

200°

20° 40° 60° 80°

(e)

(e) The world as it looks today was produced in the past 65 million years. Nearly half of the Atlantic Ocean floor was created in this geologically brief period. India completed its flight northward by colliding with Asia, and a rift separated Australia from Antarctica. The North Atlantic rift finally entered the Arctic Ocean, fissioning Laurasia. The Antilles and Scotia area now occupy their proper positions with respect to neighboring land masses. Extrapolation indicates that, 50 million years into the future, the Atlantic (particularly the South Atlantic) and Indian oceans will continue to grow at the expense of the Pacific. Australia will drift northward and begin rubbing against the Eurasian plate. The eastern portion of Africa will split off. Baja California and a sliver of California west of the San Andreas fault will be severed from North America, to begin drifting to the northwest.

Boundaries where plate edges slide past each other are classified as transform faults. Such locations are identified as faults on Figure 21. Surface area is neither created nor destroyed at such plate edges. Small “filler” plates between the boundaries of main transforming plates apparently rotate in place. When two plates carrying continental crust converge, a subduction of one lithosphere plate occurs, but its continental crust is too lightweight and thick to be carried along. The result is that the two continental masses stay afloat but undergo a condition of collision over a sustained period, causing the crust material to fold slowly, accordion style, and giving rise to a mountain range (for instance, the Himalayas). Regions near the boundary where two continent-supporting plates converge are earthquake zones but are not volcanically active. Earthquakes relate to the quick release of strains (deformations) that build up within the lithosphere and crust as a result of the plate sections moving past each other. Where two ocean plates converge, or where an ocean plate dips below a plate carrying continental crust, the boundary region becomes an area of active volcanism and earthquake activity (for instance, the Andes Mountains of South America). Apparently the volcanic

32

The Soil and Rock of Planet Earth Basaltic lava rises from asthenosphere into mid-oceanic ridge-rift valley Spreading edge (new lithosphere) Ocean

Rising molten rock Subducted lithosphere and ocean crust

Arrows shown on lithosphere indicate direction of plate movement

nt

Ocean

ne nti

M ou nt Co ai ns nt in en t

an Oce

Co

Subduction zone exists at ocean trench (lithosphere destroyed)

Oceanic crust (basaltic, heavy) Continental crust (silicic, light)

Magma from remelting Lithosphere (solid, rigid, lithosphere and crust rock material) rises to feed volcanic Lithosphere chain Asthenosphere (semi-solid, plastic rock material)

Figure 24

Movement of lithospheric plates—plate divergence and plate subduction.

activity relates to the surface breakthrough of a rising, relatively low-density magma, which is created when ocean crust, carried into the asthenosphere on the lithosphere plate being subducted, melts (Figure 24). Mountain ranges near the boundary where a continental plate overrides a subducting oceanic plate, such as the Andes, may be the product of volcanism or the result of horizontal compression and folding that continental crust experiences because of the plate convergence. The folding and distortion of rock crust due to plate movements can occur in regions considerably distant from the plate boundaries as well as at the near locations (see Figure 25). Transform faults occur at plate boundaries originally created by divergence or convergence. The sliding of one plate past another causes earthquake activity but no volcanism. Transform faults develop where a section of plate boundary is aligned with the general direction of the plate’s movement. At some locations the fault is of great length— for example, the San Andreas fault near the west coast of the United States. More prevalently, however, numerous limited segments of transform fault are incorporated into the plate boundaries of the ocean ridges and rises identified in Figure 21, a condition brought about by the irregular fracturing of lithosphere where plate spreading is taking place. In recent times, the planet has been experiencing about 80,000 earthquakes a year, but most are small. Historically, the worse quakes have usually occurred in regions near subduction zones where the converging plates are in direct collision. This type of convergence apparently causes the descending plate to deflect downward at a shallow angle; resulting high friction developed between the two plates restricts continuous movement but subsequent sudden slippage releases the great energy which accumulated as the shifting plates deformed, triggering the level of earthquake event responsible for massive ground movements.

33

The Soil and Rock of Planet Earth

Figure 25 (a) Distortion and fracturing of a rock formation, indicating influence of plate tectonics at a location in central Pennsylvania. Rock face shown is about 25-m high. (b) Stratified sandstone in the Colorado foothills bordering the easterly side of the Rocky Mountains. Formed in the area of an ancient shallow sea (see Figure 1), this sedimentary rock (continental crust) has been lifted to be almost vertical by the plate tectonics activity responsible for the Rocky Mountain range.

Recent evidence suggests that plate movement does not occur at a steady rate, as was earlier believed. New crust formed along rift boundaries assumes the north–south magnetism in effect at the time molten magma cools. It has been established that, for reasons as yet unexplained, the magnetism of the earth has reversed numerous times during the planet’s existence—possibly 170 times in the last 76 million years. (Recent information hints that the solid core of the earth rotates slightly faster than the surrounding molten outer core and mantle zones, and an association of fluid and electrical currents between the inner core and outer core is related to the generation of the planet’s natural magnetic field.

34

The Soil and Rock of Planet Earth

If the inner core revolves slightly faster than the outer core, the alignment of the magnetic field will be slowly changed, which, with time, results in periodic reversals of magnetic north–south.) Variations found in the width and directional magnetism of the bands of new crust adjacent to ocean rifts provide a reference to the past rate of plate movement. The causes of plate acceleration and deceleration are not yet understood. The movements of the major plates are related, whereby the plates speed up or slow down together. Of significance, volcanic activity and the mountain building process increases during the periods of plate acceleration. In the past, apparently each period of acceleration and deceleration extended over several millions of years. The earth is probably in a session of accelerated plate movement at the present time. The plate tectonics theory assumes that the various geologic processes associated with the formation of rock and soil materials as discussed in the early part of this chapter have been ongoing simultaneously with plate movements. If we recognize that the global position of a continent influences the conditions and forces of nature that the land becomes subjected to, such as climate (including temperature and moisture), effects of oceans if nearby (including chances of inundation), and exposure to glacial, volcanic, and earthquake activity, it becomes evident that much of the earth’s currently existing rock and soil represents the influence of plate movements.

4

Effect on Design and Construction Many large land areas have been formed with soils primarily deposited by one of the transportation methods described in this chapter. One area may be underlain by glacially deposited soils, another by lake or marine deposits, another by river deposits. It is not unusual, however, to have stratification of soils that have been deposited by different methods. For instance, a glacial till may be covered with a glacial outwash or other alluvium. Soft lake deposits are known to exist over compact glacial till or river deposits of sands and gravels. Coarse soil alluvium may overlie soft lake deposits or marine clays. In areas that have been subjected to repeated glaciation, stratification of different types of glacially transported soils having different properties can exist. The result is that loose or soft soils may overlie compact or firm soils, and vice versa. In areas blanketed by residual soil, the character of the material will vary with depth, and both depth and properties can vary across limited horizontal distances. A residual soil formation may be covered by a transported soil, or reduced by erosive and other transportive forces. The type and condition of soil deposits underlying any proposed construction site must be an important consideration to the engineering and construction personnel concerned with the project, for it is the soil, or rock, that provides the support for the structure. Proper design for a structure includes investigation and evaluation of soil conditions underlying the proposed structure, as well as probability and extent of significant natural occurrences, such as seismic activity, site flooding, and so on. However, knowledge of geology and the manner in which land forms and the soils in them have been created, as discussed throughout this chapter, can frequently serve to provide a preliminary evaluation of an area and the potential advantages or problems.

35

The Soil and Rock of Planet Earth

Problems 1 Assemble a listing of all the types of occurrences where serious problems for humans and their developments have resulted when soil deposits (read “solid ground”) did not perform as expected. 2 Name the three main classes of rock and describe how they originated.

16 What effect does the shape of a channel have on stream flow and the related carried or deposited soil sediments?

3 What factor of formation most influences the texture of igneous rocks?

17 How do natural levees originate? What is the major soil type expected in a natural levee?

4 Provide generalized comments on the hardness, soundness, and durability associated with the three basic categories of rock.

18 What soil types would be expected in a river or stream delta?

5 What visual properties can frequently be used to distinguish between acidic rocks and basic rocks? 6 Describe the two typical processes that occur to transform soil sediments into sedimentary rock. 7 The shales represent what percentage (approximately) of rock materials closest to the ground surface? Give probable reasons why this condition exists. 8 What processes occur to cause metamorphosis of rocks? 9 What type of bedrock formations would be thought to offer good foundation support for structures? What types would be more suspect of being poor bedrock materials? 10 If sinkhole topography is related to the underground erosion of rock which results in cavities, what type of rock formations would be considered most susceptible to having the condition develop? 11 What two major geologic changes resulted across the North American continent during, and because of, the planet’s last period of heavy glaciation (the last “ice age”) when massive sheets of ice extended across Canada and the northern half of the United States, as well as other continental areas of the northern hemisphere. 12 What are loess soils, and what is the potential danger of loss of stability in loess deposits? 13 Indicate the soil types to be expected in a glacial terminal moraine. 14 Provide an outline indicating the relationship between soil deposit, glaciation, and sedimentary rock formations.

36

15 Indicate why sand or gravel deposits are frequently found along old river and stream locations.

19 Describe the process by which a glacial lake is formed. 20 How can lakes exist at elevations above the groundwater table? 21 What type of transported soil deposit commonly is found at the location of filled-in ancient lakes and basin areas? 22 Why are naturally filled-in lake locations often thought of as areas that offer poor support for building foundations? 23 How do eskers differ from drumlins with regard to formation and soil types? 24 In glacially affected areas, what types of glacial formation represent possible good sources of sands and gravels for the construction industry? 25 Does an area covered by glacial till represent a location of advantage or disadvantage to the building and construction industry? 26 What is the potential danger to stability in areas where the land is formed from marine clays? 27 Do soils typical of beach deposits represent any advantage or disadvantage to the building and construction industry? 28 (a) Briefly explain the geologic difference between a transported soil deposit and a residual soil deposit. (b) What combination of geographic, geologic, topographic, and climatic conditions are associated with, or responsible for, transported soil deposits? (c) What combination of geographic, geologic, topographic, and climatic conditions are associated with the presence of residual soil formations?

The Soil and Rock of Planet Earth 29 Formations of residual soils typically have strength-related characteristics distinctly different from deposits of transported soils. Briefly describe the significant differences. 30 On the basis of known lithosphere plate boundaries: (a) Identify the major continental areas likely to experience earthquakes in the future. (b) Identify the major continental areas likely to experience active volcanic activity in the future. 31 Indicate the present-day beliefs used to explain the movements of plates of the lithosphere. 32 Develop a listing that indicates comparative values of specific gravity (or heaviness) for rock

materials in the lithosphere, the asthenosphere, the continental crust, and the ocean crust. 33 Explain how the boundary regions for the plates of lithosphere relate to the presence of the earth’s major mountain ranges. 34 Explain how the boundary regions for plates of lithosphere related to the presence of presentday active volcanoes. 35 Speculate on possible reasons for the development of the fractures in the lithosphere (the plate boundaries). 36 Explain the reason why fossils of cold-region or polar life have been discovered in tropical areas in recent times.

37

38

Soil Types and Soil Structure

From Essentials of Soil Mechanics and Foundations: Basic Geotechnics, Seventh Edition. David F. McCarthy. Copyright © 2007 by Pearson Education, Inc. Published by Pearson Prentice Hall. All rights reserved.

39

Soil Types and Soil Structure

The term soil, as generally used, refers to the accumulation of particles of disintegrated rock and, frequently, also human-made materials. Because of the experienced wide variation in their characteristics and behavior, soils have been subdivided into categories based on the materials’ physical properties. In nature, soils are made up of particles of varying size and shape. Particle size and shape are factors that have been found to be related to, or to affect, the material behavior of soil to some degree. Historically, the soil categories or types that have been developed are basically referenced to size. To distinguish between soils where size cannot be visually discerned (the particles are too small), an additional property, plasticity (or nonplasticity), is used as a criterion. Experience and study have proved that a soil’s important behavioral properties are not always controlled by particle size and plasticity. Soil structure and mineralogical composition, and the intereffect with water, may also have significant influence on the properties and behavior deemed important for design and construction. Under certain conditions, simple typing of the soil provides adequate information for design and construction, whereas certain other conditions require that detailed information about the soil’s composition and structure be determined.

1

Major Soil Types The major engineering categories of soil are gravel, sand, silt, and clay. There is not unanimous agreement on the exact division between each of these major soil types, but gravel and sand are universally considered coarse-grained soil, for the individual particles are large enough to be distinguished without magnification. Silts and clays are considered fine-grained soil because of their small particles—too small, for the most part, to be seen unaided.

40

Soil Types and Soil Structure Table 1 Size Range for Soil Types Soil Type

Upper Size Limit

Lower Size Limit

Gravel

Varies from 80 mm up to about 200 mm (3 in. to 8 in.)

Sand

4.76 mm or 2.00 mm

Silt and clay

0.074 mm or 0.05 mm

4.76 mm (about 0.20 in.) (as determined by a #4 U.S. Standard sieve) or 2.00 mm (#10 U.S. Standard sieve) 0.074 mm (#200 U.S. Standard sieve) or 0.050 mm (#270 U.S. Standard sieve) None

The most commonly used divisions for classifying soils for engineering and construction purposes are shown in Table 1. On a comparative basis, the division sizes between gravel and sand (4.76 mm or 2.00 mm) and between sand and silt–clay (0.074 mm or 0.05 mm) are actually quite close. As a result, lack of agreement on these division sizes normally does not cause serious problems. Particles larger than gravel are commonly referred to as cobbles or boulders. Again, no unanimous agreement exists on the range of sizes. When gravel extends up to the 200 mm (8 in.) size, anything larger would be termed a boulder. Where the 80 mm (3 in.) size, or thereabouts, is taken as the upper size for gravel, the sizes between 80 mm and 200 mm may be designated as cobbles, and anything larger than 200 mm (8 in.) as boulders. However, 150 mm or 300 mm (6 in. or 12 in.) may also be taken as the division between cobbles and boulders. As for sands and gravels, these discrepancies usually do not cause serious problems. Conventionally, when a construction project requires a particular material, it has become standard practice to indicate the soil or aggregate requirements on the basis of size instead of, or in addition to, classification. In conclusion, particle size serves as the basis for classification of sands, gravels, cobbles, and boulders. The classification of a fine-grained soil as either a silt or a clay is not done on the basis of particle size but, rather, is based on the plasticity or nonplasticity of the material. Clay soil is plastic over a range of water content; that is, the soil can be remolded or deformed without causing cracking, breaking, or change in volume, and will retain the remolded shape. The clays are frequently “sticky.” When dried, a clay soil possesses very high strength (resistance to crushing). A silt soil possesses little or no plasticity and, when dried, has little strength. If a small sample of moist silt is shaken easily but rapidly in the palm of the hand, water will appear on the surface of the sample but disappear when the shaking stops. This is referred to as dilatancy. When a sample of moist clay is similarly shaken, the surface will not become wetted. The reason for the difference in behavior between clay and silt relates to the difference in mineralogical composition of the soil types and particle shape. Silt soils are very small particles of disintegrated rock, as are sands and gravels, and possess the same general shape and mineralogical composition as sands and gravels (which are nonplastic). The clay minerals, however, represent chemical changes that have resulted from decomposition and alteration of the original rock minerals. The effect is that their size and shape are significantly different from those of other types of soil particles. This is discussed further in a following section.

41

Soil Types and Soil Structure

Naturally occurring soil deposits most generally include more than one soil type. When they are classified, all the soil types actually present should be indicated, but the major constituent soil type should dominate the description, while the soils of lesser percentage are used as modifying terms; for example, a material that is mostly sand but includes silt would be classified as a silty sand, whereas a silt–clay mixture with mostly clay would be termed a silty clay. Although a soil may be predominantly coarse-grained, the presence of silt or clay can have a significant effect on the properties of the mixture. Where the amount of finegrained material exceeds about one-third of the total soil, the mixture behaves more like a fine-grained soil than a coarse-grained soil. The condition also exists where small fragments of decomposed vegetation are mixed with the soil, particularly fine-grained soils. Organic material mixed with the nonorganic soil can have striking detrimental effects on the strength and compressibility properties of the material. The presence of organic material should be carefully considered. A foul odor is characteristically though not always associated with such soils, as is a blackish or dark gray color. Soils in this category are designated as organic (e.g., organic silt or organic clay) in comparison to a nonorganic designation for soil free of decomposed vegetation.

2

Particle Shapes and Sizes Particles in the sand, gravel, and boulder categories are considered as “bulky grain,” indicating that particle dimensions are approximately equal; that is, the dimensions in the length, width, and thickness directions would be of the same order of magnitude (commonly, one dimension is no more than five times larger or smaller than another dimension). Individual particles are frequently very irregular in shape, depending somewhat on the rock they were derived from, their age, and exposure to weathering and transporting processes. Generally, a new particle is “angular” and rough-surfaced, and is then modified with time and exposure to become more smooth-surfaced and rounder. The various stages of transition—angular, subangular, rounded—are illustrated in Figure 1. Generally, the angular particles possess better engineering properties, such as higher shear strength, than do weathered and smooth particles. Particles in the silt category, though classified as “fines” along with the clays, are still angular or bulky in shape, and of the same mineralogical composition as the coarsegrained soils. Because of the mineralogical composition, such particles rarely break down to less than 2 µ or 0.002 mm in size (1 µ, i.e., micron, equals 0.001 mm). The mineralogical composition of true clay is distinctly different from the mineral components of the other soil types, inviting the distinctions clay minerals and nonclay minerals. Clay minerals typically result from the alteration of rock minerals (which, unaltered, constitute the coarser soil particles). Almost all clay minerals are crystalline minerals (minerals consisting of an orderly and repetitious arrangement of molecules to produce a sheetlike structure) that are capable of developing cohesion (because of an attraction and bonding between soil sheets and water) and plasticity. Clay particles may be made of many sheets on top of one another. Clay particles are mostly found in sizes less than 0.002 mm

42

Soil Types and Soil Structure Edges distinct but fairly well rounded

Subangular

Sharp edges as in crushed stone

Angular

Angular

Very round

Rounded

Subrounded

Figure 1 Shapes of granular soil particles [1].

(or 2 ␮) or easily break down to this size. However, it is the mineral type and not the small size that is primarily responsible for the high cohesion and plasticity that clays possess. Where particles of nonclay minerals have been broken down to smaller than 0.002 mm, the clay properties are not developed. Because of the sedimentary origin of fine-grained soil deposits and the overlap in sizes of the clay and nonclay minerals, it is unusual to find natural deposits of pure clay mineral soils. Very frequently, so-called clay deposits are actually a mixture of clay minerals and nonclay minerals. Because of this, the term clay material has been used to prevent confusion when one is designating a naturally occurring soil deposit consisting of finegrained soils that have the general properties of cohesion and plasticity. The clay minerals themselves can and do vary in their composition and, therefore, in their behavior properties. The building blocks, or constituent sheets, that combine to form most of the different types of clay minerals are the silica tetrahedral sheet (Figure 2) and alumina octahedral sheet (Figure 3). The silica tetrahedron consists of four oxygen ions and one silicon ion. The molecular arrangement is such that the four oxygens are spaced and located at what would be the corners and tip of a three-dimensional, three-sided pyramid, with the silicon located within the pyramid. Oxygen ions at the base are shared by adjacent tetrahedrons, thus combining and forming the sheet. The thickness of a silica sheet is 5 * 10-7 mm , or 5 Angstrom units (one Angstrom unit or 1 Å = 1 * 10-7 mm ). The alumina octahedron consists of six oxygens and one aluminum. Three of these oxygens are in the top plane of the octahedron, and three are in the bottom plane. The aluminum is within the oxygen grouping. It is possible that the aluminum ion may be replaced with magnesium, iron, or other neutral ions. To obtain a valence balance, some of

43

Soil Types and Soil Structure and and

= Oxygens = Silicons

In perspective Diagrammatic sketch showing sheet structure of silica tetrahedrons arranged in a hexagonal network.

Single silica tetrahedron

Projected on the plane of the base of the tetrahedrons.

Figure 2 Assembly of atoms forming the basic clay mineral sheet silica tetrahedron [136, 137].

Single octahedral unit

Diagrammatic sketch showing the sheet structure of the octahedral units. and

= Hydroxyls

Aluminums, magnesiums, etc.

Figure 3 Assembly of atoms forming the basic clay mineral sheet alumina octahedron [136, 137].

the oxygens may also carry a hydrogen ion (resulting in a hydroxyl at some oxygen locations). The alumina sheet is also 5 * 10-7 mm or 5 Å units thick. It also happens that oxygens from the tip of a silica tetrahedron can share in an alumina sheet, thus layering sheets. Different arrangements of sheets can then combine to form the different clay minerals. The composition and typical properties of the more commonly occurring clays are summarized in Table 2. Although the thickness of a clay mineral sheet is limited, the dimensions in the length and width direction are not. As a result, the clay minerals have a flat, platelike shape (like an irregular sheet of paper), where the length and width can be several tens or several hundreds times the thickness.

44

Soil Types and Soil Structure Table 2 Basic Properties of Some Typical Clays Clay Mineral

Composition

Layer Thickness

Kaolinite

One silica, one alumina sheet. Very strongly bonded together.

7.5 Å

Halloysite

One silica, one alumina sheet make up the layer. Has sheet of water molecules between layers. (Similar to kaolinite except for sheet of water.)

10 Å

Illite

Alumina sheet sandwiched between two silica sheets. Potassium provides the bond between layers.

10 Å

Montmorillonite (also identified as smecite)

Alumina sheet sandwiched between two silica sheets. Iron or magnesium may replace the alumina in the alumina sheet; aluminum may replace some silicons in the silica sheet (isomorphous substitution). Weak bond between layers. Alumina sheet sandwiched between two silica sheets, but layers are bonded together with an alumina sheet.

Chlorite

9.5 Å (variable)

14.1 Å

Shape of Mineral, General Properties, and Comments The most prevalent clay mineral. Very stable, with little tendency for volume change when exposed to water. Kaolinite layers stack together to form relatively thick particles. Particles are plate-shaped. Form from crystalline rocks in humid climates. Sheets of halloysite curl into tubes. Strength and plasticity are significantly affected by drying and removal of the water. After drying, the clay mineral will not reinstate a water layer if again exposed to water. Caution is required in identifying this mineral and in using remolded (and rewetted) samples in laboratory testing to determine properties. Dried halloysite has characteristics of kaolinite. Rewetted samples appear stronger and less plastic than naturally wetted halloysite. Irregular flake shape. Generally more plastic than kaolinite. Does not expand when exposed to water unless a deficiency in potassium exists. Illite clays seem most prevalent in marine deposits and soil derived from micaceous rock (schists, etc.). Irregular plate shapes or fibrous. Because of the weak bond between layers and the negative charge resulting because of isomorphous substitution, the clay readily absorbs water between layers. Has a great tendency for large volume change because of this property. Forms mostly from ferromagnesium rock and develops mostly in semi-arid and temperate climates; also from decomposition of volcanic ash. Irregular plate shapes. Nonexpanding. Formed from well-drained soils and micaceous rocks in humid areas.

Note: 1 Å = 1 * 10-7 mm.

3

Clay and Water The surfaces of clay mineral particles have a net electrical charge that is negative, whereas the edges have positive and negative charges. This results from the molecular grouping and arrangement of ions. However, the charges are not uniform, but vary in intensity at different locations on the particle. Because of the extremely small size of clay particles and the very high ratio of particle surface to particle mass, the forces of electrical charge have a profound effect on the behavior of particles coming in association with other particles and water (or other fluids) present in the soil. And because of the manner of development or deposition, clay deposits almost always exist in the presence of some water. In contrast, the engineering behavior of coarse particles is not significantly affected by surface electrical charges because of a comparatively low ratio of particle surface to mass.

45

Soil Types and Soil Structure

The water molecule has a somewhat peculiar arrangement, as illustrated in Figure 4. The electron cloud configuration shown in Figure 4(e) is responsible for a behavior that is almost unique (and actually responsible for life on earth as we know it). The term electron cloud refers to the varying zone of space traveled by the orbiting electrons. Referring to Figure 4(c) and (e), note that of the six electrons in the oxygen atom’s outer orbit, two are shared or bonded to the hydrogen atoms and two pairs of electrons are unshared in the arms of the electron clouds opposite the hydrogens. The center of gravity of the positive and negative electrical charges do not coincide, and the relative positions of the negatively charged electrons result in a molecule that possesses an electrostatic dipole moment.1 Because the effect is an assembly that has a positive charge at one end and a negative charge at the other, similar to a bar magnet, water molecules are represented as polar molecules (Figure 4(f)). The two electron cloud arms possess a negative charge that can attract the positive partial charges of the hydrogen atom in a nearby water molecule; water molecules then prefer to be bonded together. Water has been labeled nature’s universal solvent, and this feature (of water) can affect earth materials. Also, groundwater is rarely pure, instead containing dissolved gases, minerals, and other compounds in solution or suspension. As a result, the groundwater for an area may be acidic (pH less than 7) or alkaline (pH greater than 7) instead of neutral (pH of 7). In the presence of groundwater, then, various rock, soil, and other minerals included underground will disassociate (break down) into the component cations (positively charged ions or groups of atoms) and anions (negatively charged ions), which are then transported by the groundwater (i.e., as a solution) to react with minerals at other locations to form new minerals. Clay mineral particles would, by themselves, tend to repel each other because of the net negative charges present on the surface, unless edge-to-surface contact were made (positive to negative would attract). Because of the net negative charge, however, the particles will attract cations (positive ions) such as potassium, sodium, calcium, and aluminum present in the soil moisture (very typically present as a result of solutions from rock weathering), so as to obtain an electrically balanced or equilibrium condition. Further, because of the net positive charge of the cations, they in turn can also attract negative charges. As a result of this phenomenon, water becomes bonded to the cations. The negative tips of water molecules are attracted and held to the cation, which in turn is held by the clay particle. The resulting effect is that significant water (significant with respect to the size and weight of the clay particle) becomes “bonded” to the clay. Water molecules are also held to the particle surface, where they become attracted directly to a location of negative charge. Figure 5 depicts the various features of the attraction-bonding. Additional water molecules also become attracted to the clay particle because of a chainlike arrangement of negative ends to positive ends of molecules and by hydrogen bonding (the condition where hydrogen atoms in water are shared with hydrogen atoms in the clay).

1Without

this dipole moment and the negatively charged electron cloud arms, water molecules would not stick together in liquid form but instead would exist in the gaseous state. These features also explain why water remains liquid over a comparatively wide range of temperatures and why the surface tension property important to capillary movement is so great.

46

Oxygen

O

0.

97

104.5° H

H ° 1.54 A

Hydrogen

1 proton (+) 1 electron (−) 8 protons (+) 8 electrons (−)

Atomic number = 1 Atomic weight = 2

(c) Water molecule - two hydrogen and one oxygen combine (note shared electrons in outer orbits)

Atomic number = 8 Atomic weight = 16

(a) Representation of hydrogen atom

(b) Representation of oxygen atom

(−) X X

X

X X

O

O

H

H

(−)

X

X

H

X

X

Electron cloud (from Horne, ref. 152)

H

(d) Chemistry representation of water molecule

Oxygen X

X

H

H

X

X

X

X

H

X

(+)

H

(+)

Hydrogens

(e) Relative positions of hydrogens and oxygen (−) X X X X

(−) (+)

(−)

(−)

(−)

X

X X

(−) (+)

(+)

X X

(or) (+) X

(f) Simplified representation of water molecule

H

(+)

H

X

X

X

X

H

X

(−)

X X

(−) X

X X

H (+)

(−) (+)

X

(+) H

X

H

X

X

(−) X X X X

X

X

(−) X

X

(+) H

H (+)

(g) Hydrogen bonding of water molecules (positive hydrogen atom in one water molecule is attracted to the unshared negatively charged electron pairs in another water molecule)

Figure 4 The water molecule.

47

Soil Types and Soil Structure

+ −

+

+ +

+

+

− +

+

+

+

+

+ +

+

+

+

+

− +

+

+

Random water

Cation (positively charged atom or group of atoms) Water molecule

+ + + +

+

+ + +

+

+ −

+

+

+

+

+

− −

+ +

+

+

+

+ +

+

+

+ + + −

+

+

+ + +

+ +

+

+

+ − + +

+

Clay particle (not to scale)

Limits of diffuse double layer +

+

+ −

+

+

+

+

+ +

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+ − + +

+

+

+

+ −

+ − +

+

+ +

Random water

+

+

+

+

Water in this zone is adsorbed water (water attracted and bound to clay particle)

+

+

Figure 5 Concept: Adsorbed water and cations in diffuse double layer surrounding clay particle.

The state or nature of the water immediately surrounding a clay particle is not clearly understood by soil scientists, but it is generally known to possess properties different from liquid water. It may be in a very dense and viscous state. It is certain, however, that this water is very strongly attracted to the clay particle. The attraction for cations necessary to balance the negative charge of the clay extends beyond the surface layer of molecules surrounding the particle. The further from the particle surface, though, the weaker the attraction becomes. Therefore, the concentration of cations becomes lessened. The resulting effect is that water molecules are still attracted to the clay particles, indirectly, but the further from the particle, the weaker the attraction. At a distance beyond where cations and, therefore, water molecules are attracted to a clay particle, water in the soil is considered “loose” or “normal” pore water. The distance from the clay particle surface to the limit of attraction is termed the diffuse double layer (see Figure 5), relating to the negative charges on the particle surface and the distribution of attracted (held) cation charges close to the particle. An effect is that immediately surrounding the particle a thin, very tightly held layer of water, perhaps 1 * 10-6 mm 110 Å2 thick, exists, and a second diffused, more mobile zone extends beyond this first layer to the limit of attraction. Molecular movement in the outer layer (or outer zone), and probably also in the tightly held layer, continually occurs, however. The water that is held in the diffuse double layer is frequently termed adsorbed water or oriented water, to differentiate it from normal pore water, which is not oriented.

48

Soil Types and Soil Structure

The plasticity that clay soils possess is attributed to the attracted and held water. And, as a restatement, water molecules are attracted because of their dipole structure. The unusual properties of plasticity possessed by clays occur because of the unusual molecular structure and the common presence of water in soil deposits. Experiments performed with clay using nonpolar liquid in place of water have resulted in a “no-plasticity” condition similar to that noted for coarse-grained sandy soil.

4

Soil Structure The particle arrangement of the equidimensional particles—gravel, sand, and silt—has been likened to arrangements that can be obtained by stacking marbles or oranges. For similar-sized spherical particles, a loose condition (condition with a high void ratio) is obtained from an arrangement as shown in Figure 6(a). A dense condition (condition with a low void ratio) is obtained from an arrangement as in Figure 6(b). Actual soil deposits are made of accumulations of soil particles having at least some variation, but more frequently great variation, in particle size. As a result, the soil structure is not quite like that presented in Figure 6. Generally, the greater the range of particle sizes, the smaller the total volume of void spaces there will be. For a given soil deposit, however, a range of conditions between loose and dense is possible. Typical values for different types of soil mixtures are tabulated in Table 3. In relating the volume of void spaces to properties desirable for building construction purposes, it is generally anticipated that the smaller the void ratio (or the denser the material), the higher the strength and the lower the compressibility will be. Coarse soil in an initially loose condition may be prone to quick volume reductions and loss of strength if subjected to shock or vibrations, unless there is some cementing at points of particle contact or “cohesive strength” provided by moisture menisci. Experience indicates that it is possible for sands or silts to be deposited in such a manner that an unusually loose or honeycomb structure results. Grains settling slowly in quiet waters, or a loosely dumped moist soil, can develop a particle-to-particle contact that bridges over relatively large void spaces between the aggregates and carries the weight of the overlying material. A possible particle arrangement is shown in Figure 7.

(a) Loose

(b) Dense

Figure 6 Schematic diagram of grain arrangement for loose and dense granular soils.

49

Soil Types and Soil Structure Table 3 Typical Void Ratios and Unit Weights for Cohesionless Soils Range of Void Ratio Soil Description Well-graded fine to coarse sand Uniform fine to medium sand Silty sand and gravel Micaceous sand with silt

Range of Unit Weight

emax (loose)

emin (dense)

Condition (moisture)

0.70

0.35

0.85

0.50

0.80

0.25

1.25

0.75

Saturated Dry Saturated Dry Saturated Dry Saturated Dry

gmin

kN

m3 (loose) 19.5 15 19 14 18 14 17 12

gmax

kN

m3 (dense)

γmin pcf (loose)

γmax pcf (dense)

22 19 20.5 17.5 22.5 17 19.5 15

125 95 120 85 115 90 110 75

140 120 130 110 145 130 125 95

Figure 7 Honeycomb structure in a granular soil.

The presence of flake-shaped particles, such as mica flakes, in a coarse soil has a significant effect on the void ratio, density, and compressibility of a deposit. The flake-shaped particles are capable of bridging over open spaces so that relatively large void ratios develop. When subject to external loading, however, the flakes are incapable of providing great support, bending or breaking and rearranging under load. For coarse-grained soils and silts, the mass of an individual particle is relatively great compared to the surface area. Therefore, the effect of gravity has the most influence over the arrangement of deposits of such soils. (The effect of electrical charges on the particle surface is negligible.) Conversely, clay particles, because of their large surface-to-mass ratio, are more affected by the electrical forces acting on their surfaces than by gravity forces. Clay deposits developed from clay particles that have settled out of suspension in a freshwater or saltwater environment tend toward a flocculated structure, whereby the attraction and contact between many of the clay particles is through an edge-to-face arrangement. Clays settling out in a saltwater solution tend to a structure more flocculent than do clays settling out in a freshwater situation. Saltwater (oceans are approximately a 3.5 percent saline solution) acts as an electrolyte in which the repulsion between particles is reduced. With respect to other particles, particle sedimentation then occurs with a random orientation, creating the flocculent structure. Sedimentation in a weaker electrolyte, such as freshwater (fresh surface or subsurface water is not “pure”), produces a structure

50

Soil Types and Soil Structure

where some parallel orientation of settled particles occurs, but the overall structure is flocculent; see Figure 8(a) and (b). Clay particle sedimentation occurring in ponds and wetlands where organic decay is taking place will result in highly flocculent structures. Clay deposits with flocculent structures will have high void ratios, low density, and probably high water contents. The structure, however, is quite strong and resistant to external forces because of the attraction between particles. But if the environment surrounding the clays is changed, such as by having the salts leached from the deposit (as has occurred where ocean deposits later rose above sea level and freshwater subsequently percolated through the soil, leaching out the salts with it), the attraction and strength between particles can be markedly decreased. Clays that have been further transported after being deposited (as from glacial action, or when used in human-made earth fills) are reworked or remolded by the transportation process. The particle structure that develops from remolding is a more parallel arrangement or orientation of particles than existed in the flocculent condition, as shown in Figure 9. Such a particle arrangement is considered a dispersed or oriented structure. When used for construction, clays that were in a flocculent condition before use generally lose some strength as a result of remolding. Subsequent to remolding, and with the passage of time, however, the strength increases, though not back to the strength of the originally undisturbed clay. Reasons for this increase appear to be related to a time-dependent rebuilding from the remolded, dispersed structure toward a less dispersed, more flocculent

Clay particle Silt particle Undisturbed saltwater deposit (b)

Undisturbed freshwater deposit (a) Flocculated-type structure (edge-to-face contact)

Figure 8 Schematic diagram of types of particle orientations [201].

Dispersed-type structure (face-to-face contact)

Figure 9 Schematic diagram of particle orientation.

51

Soil Types and Soil Structure

structure. In the dispersed condition, the equilibrium of forces between particles is disturbed. With time, the particles become sufficiently reoriented (only very small movements are necessary) to reacquire a structure in which the forces between particles are again in equilibrium (but not as originally structured; see Figure 10). This phenomenon of strength loss–strength gain, with no changes in volume or water content, is termed thixotropy. Thixotropy has been defined as a “process of softening caused by remolding, followed by a time-dependent return to the (original) harder state.” The degree of difference between the undisturbed strength and remolded strength, and the extent of strength gain after remolding, are affected by the type of clay minerals in the soil. Generally, the clay types that adsorb large quantities of water, such as the montmorillonites, experience greater thixotropic effects than do the more stable clay types, such as kaolinite. For many construction situations, thixotropy is considered a beneficial phenomenon, since with the passing of time soil structures (dams, highway embankments, etc.) and disturbed foundation soils get stronger and presumably safer. However, the phenomenon also causes its problems: Construction sites may be quickly transformed into a mire of mud when construction equipment travels across the area, making handling of equipment and materials very difficult. Thixotropic influences have also affected piles driven in clay soils when the driving operation caused remolding and weakening of the clay surrounding the pile.

Shaded area represents adsorbed water layer

(a) Structure immediately after remolding or compaction

Attraction > > repulsion Water in high-energy structure

Attraction > repulsion (b) Structure after thixotropic hardening partially complete

Clay particle Silt particle

Strength

Aged strength

(c) Final structure at end of thixotropic hardening

Attraction = repulsion Water in low-energy structure

ing

ing

Ag

Ag

Remolding

Remolded strength Time (d) Thixotropic change in strength following remolding or disturbance

Figure 10 Schematic diagram of thixotropic structure change in a fine-grained soil [242].

52

Soil Types and Soil Structure

The soil strength is sometimes recovered relatively quickly. This has been experienced where a pile has initially been driven part of its length, and attempts at continued driving after a one- or two-day wait have met with considerable resistance due to the increasing adhesion (directly related to a clay’s strength) along the pile surface. This is one reason that piles embedded in cohesive soils should be fully driven whenever possible.

5

Some Special Soil Categories To those in the construction industry, the term special soil implies a soil type with a property or behavior that is considered unusual and capable of causing problems which therefore requires special treatment. Collapsible soils are one of the groupings of special soils. The loess soils are in this collapsible category, for the in-place structure of these aeolin deposits makes them susceptible to significant volume reduction when in contact with water. Collapsible gravel is also included in this grouping, due to the significant volume decrease that occurs upon inundation or exposure to large amounts of water. The design and construction of facilities in areas where collapsible deposits exist should include means for either protecting the foundation soil from water or otherwise compensating for its adverse effects. Serious ground settlement can also occur in areas underlain by saturated sand in a loose condition; such deposits can lose much of their shear strength when significant vibration or seismic shock occurs. The actual occurrence is termed liquefaction. (Note: Liquefaction more accurately represents a soil condition rather than a soil type.) Some clays are prone to large volume changes directly related to changes in water content, such as shrinking in dry seasons and swelling in wet seasons or when otherwise in contact with water. Such soils are termed expansive clays or swelling clays. Clays including the montmorillonite mineral are particularly noted for their high volume change characteristic. The expansive force of a swelling montmorillonite clay can lift pavements and structures, while volume decrease due to reduction of water content can be responsible for significant ground settlement. A common construction practice in areas where expansive clays exist is to keep the soil beneath and surrounding a structure stable by providing protection against changes in the moisture content. Residual soil formations may not consist of special or problem-type soils, but do represent a special condition for soil deposits that should be understood. Residual formations typically show a gradational variation with depth, whereby the surficial zone comprised of the accumulation of soil size particles (therefore, classified as “true soil”) transitions into the underlying weathered rock (saprolite, or mixture of soil particles and assortment of weathered rock fragments) that extends to where deeper unweathered sound rock exists. As for the collapsible soils, sensitive clay, and loose sands susceptible to liquefaction, the strength-related properties of a disturbed (or lab-tested) residual soil-weathered rock material can be significantly different from the properties of the in-place material. Field procedures for in-place testing usually are required when properties for the undisturbed condition are needed (such as, to plan excavation work and design foundations). Collapsible soils, liquefaction, and expansive clays are discussed further in the sections following. Dispersive clays and lateritic soils similarly represent materials that require consideration, and are also discussed. Permafrost, the permanently frozen ground that exists in the vicinity of the earth’s polar regions.

53

Soil Types and Soil Structure

Collapsible Soils Collapsible soils refer to the category of soil deposits that experience significant decrease in volume when exposed to water. Collapsible deposits are typically found in arid regions. The loess soil deposits (predominantly silt-size particles) discussed in Chapter 1 are in this category, but granular deposits that include considerable gravel can also be collapsible. One feature of collapsible soils is the geologic process responsible for the deposition. Wind-transported silts that formed the present-day loess deposits typically were laid down slowly in a dry environment. These deposits commonly retain the original loose structure but also include vertical rootholes and grass channels that function to create a high permeability in the vertical direction (as well as establishing a natural plane of cleavage for earth slopes and excavation cuts). Most collapsible gravels are water transported and deposited formations such as alluvial fans, mud flow or slope wash deposits, or torrential stream deposits, where rapid deposition was followed by drying. A property of the collapsible gravels is the presence of silt or clay fine-grained materials (typically 25 to 45 percent of the soil weight, but as low as 10 percent) which act as a binder to produce a fragile metastable structure vulnerable to breakdown in the presence of water. Granular deposits that are prone to collapse can be difficult to detect with standard subsurface investigation procedures involving soil sampling because high resistance to penetration by the sampling equipment can develop. In addition, high density or unit weight can exist. Collapsible soils will often indicate a low moisture content, however. A procedure for evaluating collapse potential is to compare the volume and strength for soil samples in their natural condition and after exposure to water; the review of the geologic history for an area of proposed construction should include checking for conditions indicative of collapsing soils at neighboring projects (unexpected large settlements, sinkholes, etc.). Sites underlain by collapsible deposits have been improved by flooding and by using mechanical processes that resulted in collapsing the soil to achieve a stable condition prior to construction. After construction, efforts to prevent structural damage concentrate on preventing water from reaching the collapsible strata. Case histories of construction projects located in areas of collapsible soils, including details of conditions and remedies, are discussed by Rollins et al. [304] and Mitchell [243].

Liquefaction Liquefaction is a condition that can occur when saturated cohesionless sand deposits exist in a relatively loose condition. If subject to vibration or shock waves, as can result during earthquake and from explosion or operation of some types of machinery, the soil grains attempt to quickly move (shake) into a denser or more compact arrangement, but the presence of the void space water (i.e., pore water) interferes, and particle-to-particle contact is prevented. Temporarily, much of the soil shear strength is lost, with the result that the deposit assumes the properties of a viscous liquid with little strength. The sand will flow or displace if supporting any type of loading when the event occurs. Structures underlain by sands that undergo liquefaction may experience significant vertical or lateral movements, whereas unsupported earth slopes tend to slide. The sand deposits susceptible to liquefaction can be identified from soil boring or penetration resistance testing, which indicates the relatively loose condition, or from

54

Soil Types and Soil Structure

relative density testing. Generally, deposits of uniform sands are considered more susceptible than well-graded sands, and fine sands are considered more susceptible than coarse sands. Probability of liquefaction also relates to the severity of shock vibration passing through the deposit; for example, a medium-dense sand deposit may not be affected by minor events but seriously affected by a strong event. Site improvements that can be accomplished to reduce or eliminate the chance of liquefaction occurrence include drainage to remove the saturation condition or a soil densification-compaction procedure.

Expansive Clays The generic name for clay minerals that swell (expand) or shrink (collapse) as changes in soil water occur is smectite. Montmorillonite is one of the prominent smectite clay minerals found worldwide. Clays containing the montmorillonite mineral expand in volume if the soil water content is below a stability value2 when water becomes available. Conversely, these clays experience considerable shrinkage when water content is reduced. The volume change is related to the thickness and mobility of the water film adsorbed onto or surrounding the montmorillonite particle, being increased or decreased relatively easily during natural wetting and drying conditions. The combination of silica–alumina–silica sheets that comprise the montmorillonite mineral remains intact (refer to Figures 2 and 3 and Table 2), but the sheets tend to associate and layer with one another, and where grouped form substacks. Connecting substacks of varying thickness and linear dimension (assemblages) are termed quasicrystals (substantial monovalent cations such as sodium in the absorbed water) or tactoids (substantial divalent cations such as calcium in the absorbed water), and form the skeletal microstructure of the particle. Water can and usually does exist between substacks, and between quasicrystals or tactoids. Probably most of the swelling or shrinkage that occurs is due to water moving in or out of the pore spaces (channels of non-solid, or voids, space) between the quasicrystals (or tactoids) and not between the layers [139]. A requirement for significant shrinking-swelling is that the soil have many fine pores in the 0.001-mm to 0.002-mm range; pores of this size permit rapid acceptance or release of water. The montmorillonite mineral carries a large net negative charge on the surface, and has a high capacity for ion exchange (the effects resulting from changing adsorbed cations are greater where the ion exchange capacity is high). Cation exchange occurs outside the structure of the silica–alumina–silica sheet combination, and the resulting most-prevalent ion tends to be the most abundant or available. Where calcium ions are available and replace ions such as sodium but also increase the ion concentration, a lower ion exchange capacity results; an effect is that the tendency for expansion or swelling is reduced. The expansive force created by a clay undergoing an increase in water content and volume can be considerable, being capable of lifting heavy structures and imposing lateral pressures that can move retaining walls and basement walls. Swelling pressures in excess of 500 kPa or 10,000 psf have been measured. High swelling pressures can occur even in already high-moisture clays if additional water is adsorbed. Swelled clays can experience

2Stability

value of water content for clays prone to expanding refers to the condition in which additional water will not be attracted to the particle and (further) expansion does not occur.

55

Soil Types and Soil Structure

shrinkage equal to the volume increase if reduction in water content (drying) occurs. Damage to structures can result when a clay swells or shrinks; the more notorious problems have been associated with the swelling condition, but shrinkage-related ground settlement can result in serious damage to many types of structures. For structures such as roadway and airfield pavements subject to cycles of clay expansion and shrinkage, effects can be more damaging than those resulting from conventional loadings. The existence of clay soils that can develop large swelling pressures or undergo shrinkage damaging to structures is apparently a condition recognized only during the twentieth century. But the presence of expansive clays has now been documented for virtually all continental areas of the planet (North, Central, and South America, Hawaii, Europe, Asia, Africa, and Australia). A generalized distribution of expansive clays in the United States is shown in Figure 11, with the most severe conditions reported for Colorado, Texas, and Wyoming. In Canada, expansive clays have been identified in deposits found across the western sector (Canadian prairie regions, Saskatchewan, Alberta, and Manitoba). West-central India and the generally inland regions of south-central China have well-known deposits of expansive soils. The countries of North Africa and Eurasia that border the Mediterranean Sea similarly have identified deposits of expansive soils. The large volume change property of expansive clay formations is typically due to the presence of montmorillonite clay minerals (although significant shrinkage-related problems reported for China occur with clays dominant with illite). Montmorillonites form from the basic igneous rocks (gabbros and basalts, volcanic glass) and some sedimentary rocks (shales containing montmorillonite clay particles but also limestones and marls rich

Montmorillonite generally abundant and in continuous geologic formations Montmorillonite distribution sporadic

Figure 11 General distribution of montmorillonite expansive clay across the United States (Courtesy of Association of Engineering Geologists, Denver, CO) [365].

56

Soil Types and Soil Structure

in magnesium). However, climate is an important factor; the montmorillonites are typically found in semiarid tropical and temperate regions. Conditions associated with the formation of montmorillonites are extreme disintegration of the parent material and an alkaline environment, coupled with a semiarid climate, which results in an availability of some but limited water; an absence of water leaching through the material permits magnesium, calcium, iron, and sodium ions—particularly magnesium—to accumulate. (Where leaching has occurred and the majority of ions have been flushed, the clay mineral remaining is typically kaolinite, a “stable” clay.) Relating to construction practice, various methods have been used in the attempt to stabilize expansive clays against swelling and shrinkage. Protecting a soil formation from variations in moisture content responsible for volume changes represents a sensible concept but it is often difficult to accomplish or maintain in practical situations. Chemical stabilization using additives that reduce the inclination to attract or lose moisture can be successful if the additives are adequately mixed or distributed throughout the accumulation of clay particles; lime slurry mixing is a widely used procedure relating to the availability and economy of lime (calcium) materials. A method in use where only a limited area or depth requires protection against soil volume changes is replacement (e.g., excavate the clay material from the zone susceptible to moisture change and replace it with a stable granular soil). Site investigation that includes obtaining soil samples for testing is important in identifying the presence of expansive clay in an area of proposed construction, so that procedures for preventing problems can be implemented during the design-planning stage of the project.

Dispersive Clays Fine-grained soils that will deflocculate in still water and erode if exposed to low-velocity water are termed dispersive clays.3 Ordinary clays typically do not erode in the presence of water unless the flow velocity is relatively high (more than about 1 m/sec or 3 to 4 ft/sec). Areas blanketed by natural deposits of dispersive clays characteristically show steep erosion gullies and eroded tunnels, though not always. Embankments constructed with dispersive clays similarly experience the development of gullies and tunnels. Critically, the presence of dispersive clay in earth dams, engineered and otherwise, has been the reason for piping (leakage through eroded tunnels) of such great magnitude that major repair became necessary to keep the structures functional; frequently, the piping condition developed very soon after the initial pooling of the reservoir. A clay’s susceptibility to dispersion has been found to be related to the presence of cations (sodium, calcium, magnesium, potassium) in the soil pore water. Generally, repulsive forces between clay particles that act to cause deflocculation decrease as the concentration of

3Note that the terms dispersive clay and dispersed structure for clay refer to different properties. The similarity of

terms is unfortunate, but when each term is used in the proper context the description will be understood. Simply stated, dispersive clay refers to an easily eroded clay, whereas dispersed structure refers to a positional relationship between particles; a clay having a dispersed structure may or may not be dispersive (easily eroded).

57

Soil Types and Soil Structure

Note: Position of curves is affected by type of clay mineral and pH of the pore water, i.e.,

ed

ers

p Dis

ition

ns Tra

d

late Floccu

(+)

pH increasing

Sodium absorption ratio, Na SAR = Ca + Mg 2

(+)

Kaolinite Illite Montmorillonite

Ion concentration (milliequivalents per liter)

Figure 12

Qualitative relationship between ion concentration and dispersivity.

ions increases. However, repulsion increases as the quantity of sodium ions increases; the thickness of the diffuse double layer of water adsorbed to the soil particles is relatively great when monovalent sodium ions are present, and the distance-related attractive forces that exist between particles are thus weakened. Usually, in a clay–water system, the repulsive forces and the tendency toward dispersion increase as the concentration of ions increases along the series calcium–magnesium–potassium–sodium–lithium (e.g., a clay high in sodium is more dispersive than one having a high concentration of calcium and little sodium). Figure 12 expresses a qualitative relationship between cations present in a clay deposit and the tendency for dispersion. The piping that occurs in compacted earth dams that have been constructed of dispersive clay may originate at locations where water enters minuscule cracks, such as those that would develop from settlement or hydraulic fracturing. In the presence of water, the exposed clay particles quickly go into suspension, and a progressive erosion rapidly results. Where the presence of a dispersive clay has not been recognized during construction, dams have experienced piping as the first reservoir filling was underway, or soon after filling has been completed. Relating to the effect of ion concentration, earth dams that have satisfactorily retained reservoirs of saltwater for long periods have been known to develop piping problems quickly when freshwater replaced the saltwater, a result of the dilution of the ion concentration in the dam’s clay. The erodibility of dispersive clay can be reduced through the use of hydrated lime or aluminum sulfate admixtures, 1 to 2 percent by weight. Where core zones of dams must be constructed of dispersive clays, soil transition–filter zones that border the dispersive clay can be designed to control erosion and seal concentrated leaks. The use of geofabrics (geotextiles) can also be investigated as an erosion control.

Laterites The term laterite or description lateritic soil refers to a category of residual soil formed from the weathering of igneous rock under conditions of high temperature and high rainfall such as those typically occurring in tropical regions, where the decomposition process

58

Soil Types and Soil Structure

results in a soil leached of silica (SiO2) and calcium carbonate, but retaining high concentrations of iron and aluminum sesquioxides (Fe2O3, Al2O3). An Al2O3/SiO2 ratio of onehalf or greater is one criterion used to classify a soil as lateritic. Calcium, magnesium, and potassium contents are very low. Laterites are frequently reddish in color but not always, and all reddish tropical soils are not lateritic. Extensive areas of Asia, South America, and Africa are blanketed with lateritic soils. Concerning their engineering behavior, these soils have been found to possess some unusual properties when compared to soils existing in the temperate regions of the world. Agriculturally, most lateritic soils typically contain only meager concentrations of the nutrients considered necessary for productive farming. Deposits of lateritic soils may be found in a hard or cemented state, particularly in areas where vegetation is sparse or has been removed. The cementation, attributed to the presence of free iron oxide in the soil, is such that the hardened soil can be quarried and used as a bricklike material for building construction purposes. Generally, with a tropical residual soil the weathering process is not complete, and material can exist at all stages between fresh rock and soil. However, lateritic soils seem to be typified by a gap grading, existing with a prevalence of gravel-sized and clay materials but having very limited sand- and silt-sized particles. When used as a construction material, these weathered soils often have been found to be unstable, subject to further breakdown during handling and testing or excavation and placement. It appears that the soils from a damp regime are more apt to change physical properties than are soils from an arid region; soil from the damp environment has never been dehydrated but becomes so after exposure to air, an occurrence that has a weakening effect. Further, there is a great possibility of significant changes in properties if a laterite is completely dried and then wetted. Reworking a lateritic soil at a given water content frequently alters properties such as plasticity and compaction characteristics. Widely used correlations between soil classification index properties and the subsequent engineering behavior, established primarily from experiences with soils formed in the temperate regions of the world, cannot be relied on to predict behavior of soils formed in tropical areas. In their native regions, lateritic soils have, of economic necessity, been used as a construction material for road bases and embankment structures, and to support foundations. Experience has proved that these materials can serve successfully when protected from percolating or migrating water and the effects of heavy repetitive loadings. Properties can be improved by the use of common admixtures, such as cement and lime.

Problems 1 List the soil types included in the coarse-grain category and the fine-grain category and the reason for the division into categories.

3 What is the essential reason for the difference in behavior of natural clays and other soil types such as silts and sands?

2 Clay is a soil material that possesses plasticity in the presence of water. What does the term plastic mean in relation to clay soils?

4 Comment on the difference between the shape and size of clay particles compared to other soil types such as silts and sands.

59

Soil Types and Soil Structure 5 What are the “building blocks” of most clay minerals? Comment on the comparative length– width–thickness dimensions of a typical clay particle. 6 Referring to the attraction that typically exists between water and clay particles, what is adsorbed water?

14

7 Describe what is meant by the dipole nature of a water molecule. How is this related to adsorbed water and plasticity in a clay soil?

15

8 Relating to particles of clay and adsorbed water, what is the diffuse double layer?

16

9 Why does the presence of water in a soil typically have a much greater effect on clays than on sand or gravel materials? 10 Expansive (swelling) clays exist in various areas across North America. What clay mineral is typically associated with the swelling condition, and what is the explanation for the expansion? Also, identify some of the problems caused to structures by expanding clays. 11 Clays that expand or swell when in contact with water (such as the montmorillonite clays) can cause problems for roads and structures supported on such soil. The expansive-swelling property can also make this type of clay almost impervious (acting as a barrier to the passage or flow of water), a feature that can be used to advantage for handling the water and seepage conditions associated with some construction projects. Briefly, indicate the type of situation where this waterbarrier feature could be utilized. 12 In a general way, the strength of a coarse-grained soil is related to the soil deposit’s structure (or

60

13

particle orientation) and void ratio (or density). What is this relationship? With reference to coarse-grained soils, what type of particle structure is susceptible to having large volume changes occur? Provide a probable description for the soil deposit that could become a collapsible gravel. Briefly describe the phenomenon of liquefaction in soil deposits, indicating type and condition of soil required and occurrences responsible. (a) Briefly describe the difference between a flocculent structure and a dispersed structure in clay soils. (b) What type of structure is most expected in a clay deposit that has formed in a naturally occurring underwater environment? (c) What type of soil particle alignment (or soil structure) would be expected where a clay had been used for a compacted fill on a construction project?

17 With regard to the thixotropy phenomenon in clay soils, what is it and what causes it? 18 Provide a brief description of dispersive clays and the typical construction-related problems known to develop. 19 What are laterites (or lateritic soils), and why are such soils considered in the category of requiring special consideration on construction projects? 20 For planning the design associated with the construction of a new structure, indicate the importance and benefit of learning that the subsurface material at the building site is included in the category described in this chapter as a special (or problem) type soil.

Soil Composition Terminology and Definitions

Soil deposits comprise the accumulated solid particles of soil or other materials plus the void spaces that exist between the particles. The void spaces are partially or completely filled with water or other liquid. Void spaces not occupied by fluid are filled with air or other gas. Since the volume occupied by a soil bulk may generally be expected to include material in the three states of matter—solid, liquid, and gas—soil deposits are referred to as three-phase systems. Significant engineering properties of a soil deposit, such as strength and compressibility, are directly related to or at least affected by basic factors such as how much volume or weight of a bulk soil is solid particles or water or air. Information such as soil density or unit weight,1 water content, void ratio, degree of saturation—terms defined in the following sections—is used in calculations to determine the bearing capacity for foundations, to estimate foundation settlement, and to determine the stability of earth slopes. In other words, such information helps to define the condition of a soil deposit for its suitability as a foundation or construction material. For this reason, an understanding of the terminology and definitions relating to soil composition is fundamental to the study of soil mechanics.

1

Soil Composition: Analytical Representation Bulk soil as it typically exists in nature is more or less random accumulations of soil particles, water, and air space, as shown in Figure 1(a). For purposes of study and analysis, it is convenient to represent this soil mass by a phase or block diagram, with part of the 1In

soil mechanics, the term density has frequently been used (or misused) to indicate unit weight. Unit weight should be expressed in pounds per cubic foot or kilonewtons per cubic meter. Most commonly, density is expressed in kilograms per cubic meter. From Essentials of Soil Mechanics and Foundations: Basic Geotechnics, Seventh Edition. David F. McCarthy. Copyright © 2007 by Pearson Education, Inc. Published by Pearson Prentice Hall. All rights reserved.

61

Soil Composition Soil particles

Air Water Solid particles (soil)

Water surrounding particles and at points of contact between particles, and filling small void spaces

Air in irregular spaces between soil particles

(a)

(b)

Figure 1 (a) Actual soil bulk consisting of soil particles, water, and air. (b) Phase diagram representation of soil bulk.

diagram representing the solid particles, part representing water or other liquid, and another part air or other gas, as shown in Figure 1(b).

Weight–Volume, Mass–Volume Relationship On the phase diagram, the interrelationships of weight and mass to the volume that make up the soil system being analyzed can be shown. The relationships are summarized in Figure 2. As the diagram shows, the total weight WT of the soil volume is taken as the sum of the weights of solids Ws plus water Ww (or other ground liquid). In practical problems, all weighings are made in air, and the weight of air (in the voids) measured in air (the earth’s atmosphere) is zero. If gas other than air is present, it may have a measurable weight, but it would normally be very small compared to the total weight of soil plus water (liquid) and therefore can be neglected without causing serious error:

Air, gas

Volume of air = Va

Weight of water = Ww Mass of water = Mw

Water, Liquid

Volume of water = Vw

Weight of solids = Ws Mass of solids = Ms

Solid particles (soil)

Volume of solids = Vs

Total mass, MT

Total weight, WT

Weight of air = 0

Figure 2 Relationship between volume and weight/mass of a soil bulk.

62

Volume of voids = Vv Total volume, VT

Soil Composition

WT = Ws + Ww

(1a)

MT = Ms + Mw

(1b)

Similarly, for measurement of mass:

The total volume of the soil bulk includes the volume occupied by solids plus water (or liquid) plus air (or other gas). The total space occupied by water and air may collectively be indicated as the volume of voids: VT = Vs + Vw + Va = Vs + Vv

(2)

where VT Vs Vw Va Vv

= = = = =

total volume volume of soil solids volume of water volume of air volume of voids

The relationship between weight and volume, for any material (x), is: Wx = VxGxγw

(3a)

while that between mass and volume is: Mx = VxGxρw

(3b)

where Wx Vx Gx γw

weight of the material (solid, liquid, or gas); lb, N, dyne volume occupied by the material; m3, ft3 specific gravity of the material; a dimensionless value unit weight of water at the temperature to which the problem refers. (In most soils work, γw is taken as 62.4 pcf, 980 dynes/cm3, or 9.81 kN/m3 regardless of temperature. In experimental work, accurate values are used.)2 Mx = mass of the material; kg, slug ␳w = density of water; 1 gm/cm3, 1 Mg/m3 (103kg/m3), or 1.95 slugs/ft3 (note 1 slug = 1 lb-sec2/ft) = = = =

2The

gram is a unit of mass, whereas the pound is a unit of weight or force. A value of grams per cubic centimeter is density (by definition, density is mass per unit volume and is given the symbol ρ). Unit weight is force or weight per unit volume. Unit weight and density are therefore not identical, but they are related by the expression γ = ρg, where g is the acceleration of gravity. In soil mechanics work, the terms are often taken to have synonymous meaning. Since comparative values of unit weight or density are normally sought to relate to other properties, this practice does not cause complications. To overcome this discrepancy between γ and ρ, unit weights based on a gram measurement can be expressed as gram-force (gf ) per unit volume.

63

Soil Composition

For soil mechanics problems, then: Ws = VsGsγw

(4a)

Ms = VsGsρw

(4b)

Ww = VwGwγw = Vwγw since Gw = 1

(5a)

Mw = VwGwρw = Vwρw

(5b)

The specific gravity of most commonly occurring rock or soil materials is between 2.30 and 3.10. For many deposits, the specific gravity of soil solids lies within the range of 2.60 to 2.75.

2

Basic Terms Relating to Soil Composition and Condition The unit weight of a soil γ is conventionally expressed as pounds per cubic foot, or kilonewtons per cubic meter (unit weight is weight per one unit of volume). Unit weights are reported as wet unit weight γwet or dry unit weight γdry: Wet unit weight

gwet =

WT 1kN>m3, pcf2 VT

(6a)

Dry unit weight

gdry =

Ws 1kN>m3, pcf2 VT

(6b)

From the definition, it can be seen that the wet unit weight includes the weight of water as well as soil particles in a soil bulk. The dry unit weight is based on only the weight of soil solids in the accumulation. Soil density, ρ, expressed in kilograms per cubic meter or grams per cubic centimeter (or other terms of mass per unit volume), can also be in terms of a wet or dry value, whereby:

Wet density

Dry density

rwet =

rdry =

MT 1kg>m3, gm>cm32 VT Ms 1kg>m3, gm>cm32 VT

(7a)

(7b)

In some applications, it is necessary to relate density and unit weight; basic equivalencies as shown in Table 1 can be used to complete conversions (see Illustration 1).

64

Soil Composition Table 1 Equivalencies between Density and Unit Weight for Water Densitya (metric)

Density (SI)

Unit Wt.a (U.S. customary)

Unit Wt. (SI)

1 gram/cm3 (1 g/cm3)

1 Mg/m3

62.4 pcf

9.81 kN/m3

gm/cm3 is the equivalent of 62.4 lb/ft3. Since W = mg, the weight for one gram of mass is: W = 11 gm21980 cm>sec22 = 980 dynes (where 1 dyne = 10-5 N = 0.2248 * 10-5 lb, and 980 dynes = 0.002203 lb). Therefore, the weight for 1 cm3 of water is 980 dynes = 0.002203 lb>cm3, and the weight for 1 ft3 of water = 10.002203 lb>cm32128,317 cm3>ft32 = 62.4 lb>ft3. a1

By definition, water content w is the ratio of the weight of water in a soil volume to the weight of soil solids, or of the mass of water in a soil volume to the mass of solids, but typically is expressed as a percent: w% =

Ww * 100% Ws

or

Mw * 100% Ms

(8)

where w% Ww Ws Mw Ms

= = = = =

water content expressed as a percentage weight of water weight of dry soil mass of water mass of dry soil

(Cautionary note: Water content is not the weight of water divided by the total weight WT, nor the mass of water divided by the total mass MT .) The relationship of water content and weight of dry soil to total wet weight of a soil volume is as follows: WT = Ws + Ww

(from Eq. 1)

and since: Ww = ¢

w% ≤ Ws 100%

WT = Ws + ¢ = Ws ¢ 1 +

(from Eq. 8)

w% ≤ Ws 100% w% ≤ 100%

then, by rearranging, we obtain: Ws =

WT w% 1 + 100%

(9a)

65

Soil Composition

In a similar manner: Ms =

MT w% 1 + 100%

(9b)

These equations enable the dry soil weight (or mass) to be easily determined when the wet weight or mass of a large soil sample is known, and the water content is determined from a small representative portion taken from the sample. This procedure for determining dry soil weight or mass is frequently used in laboratory and construction work. Two terms, void ratio and porosity, express a relationship between the volumes in a soil material occupied by solids and nonsolids. Void ratio e is: e =

Vv Vol. of voids = Vol. of solids Vs

(10)

Void ratio is expressed as a decimal. Porosity n, conventionally expressed as a percentage, is:

n% =

Vv Vol. of voids * 100% = * 100% Total vol. VT

(11)

The relationship between void ratio and porosity is: n% 100% e = n% 1 100%

n% =

e * 100% 1 + e

(12a)

(12b)

The term degree of saturation, S, indicates the portion of the void spaces in a soil material (bulk) that is filled with water. Degree of saturation is expressed as a percentage: S% =

Vw * 100% Vv

(13)

Full saturation, or 100 percent saturation, indicates that all voids are filled with water. A soil can remain 100 percent saturated even though its water content is changed if the soil experiences compression or expansion (since compression or expansion indicates a decrease or increase in void spaces).

66

Soil Composition

Vv = eVs VT

Va

Air

Vw

Water

Vs

Vs

Wa = 0 Mw = wMs

Soil

Ms

Ww = wWS

WT or MT

Ws

Note: In the expressions Mw = wMs and W = wWs the w term is water content expressed in decimal (not percentage) form.

Figure 3 Phase diagram used to develop equations applicable to weight–volume relationship.

With reference to a phase diagram, as shown in Figure 3, other useful weight– volume relationships can be developed. Since: VT = Vs + Vv and with e = Vv / Vs, we obtain: VT = Vs + eVs = Vs11 + e2

or:

Vs =

VT 1 + e

(14a)

(14b)

Also, since: γdry =

Ws VT

(from Eq. 6), we obtain: γdry =

VsGsγw VsGsγw Gsγw = = VT Vs11 + e2 11 + e2

(15)

For a fully saturated soil, the unit weight becomes: γsat =

=

=

WT Ws + Ww = VT Vs11 + e2

(from Eq. 6)

VsGsγw + Vwγw VsGsγw + eVsγw = Vs11 + e2 Vs11 + e2 1Gs + e2γw 11 + e2

(16a)

67

Soil Composition

For a partially saturated soil, the wet unit weight, γwet (or γtotal), becomes: gwet1or gtotal2 =

=

VsGsgw + 1SVv2gw WT VsGsgw + Vwgw = = VT Vs11 + e2 Vs11 + e2 VsGsgw + S1eVs2gw Vs11 + e2

=

1Gs + Se2gw 1 + e

(16b)

A very informative relationship can be obtained by proper substitution of terms into Equation 5a: Ww = VwGwgw = Vwgw

(from Eq. 5a)

or: Vw =

Ww gw

Dividing both sides by Vs gives: Vw Ww = Vs Vsgw Multiplying the left term by

Vv gives: Vv Vw Vv Ww * = Vs Vv Vsgw

and: Vw Vv wWs * = Vv Vs Vsgw S * e = w¢

Ws ≤ Vsgw

Se = wGs

(17)

In this equation, both the S and w terms are expressed as a percent. For a given soil and unit weight, this equation shows the relationship between void ratio and water content, and the limiting water content that can be obtained. A summary of the weight–volume and mass–volume equations most frequently used in solving problems is presented in Figure 4. In performing the analysis to determine weight or volume properties, it is extremely helpful to sketch the phase diagram and

68

Soil Composition

Va

Wa = 0

Air

Vv = eVs

w% Ws 100% WsSe = SeVs γw = Gs

Ww = Vw γw =

Vw = SeVs = SVv

Water

VT = Vs (1 + e) Vs =

VT 1+e

e=

=

Ms Gsρw

=

Ws

Vv

Ms = VsGs ρw = MT − Mw Mw

w% =

Vs

MT = Ms + Mw

Ws = Vs Gs γw = WT − Ww

Soil

Gsγw

WT = Ws + Ww

Ms

=

Ww Ws

S% =

× 100%

Vw Vv

× 100%

× 100%

Figure 4 Summary of weight–volume and mass–volume equations.

indicate on the diagram the information that is known (from given data or as it is developed). This procedure helps to guide the analyst through the proper steps or equations necessary for a complete solution and will eliminate unnecessary computations. Illustration 1 A sample of soil obtained from a test pit 0.0283 m3 (or 1 ft3) in volume has a mass of 63.56 kg (weight of 140 lb). The entire sample is dried in an oven and found to have a dry mass of 56.75 kg (weight of 125 lb). Calculate the water content, wet density, and dry density.

Air Water VT = .0283 m3

MT = 63.56 kg Soil

MS = 56.75 kg

Solution Mass of water Mw = 63.56 - 56.75 = 6.81 kg 1Weight of water Ww = 140 - 125 = 15 lb2 Mass of dry soil Ms = 56.75 kg 1Weight of dry soil Ws = 125 lb2 Total volume of sample VT = 0.0283 m3 = 1.0 ft3 Wet density rwet =

63.56 kg MT = = 2.246 Mg>m3 VT 0.0283 m3

69

Soil Composition

¢ Wet unit weight gwet =

WT 140 lb = = 140 pcf, or, VT 1.0 ft3

2.246 Mg>m3 * 9.81

Dry density rdry =

kN>m3 1 Mg>m3

= 22.03 kN>m3 ≤

56.76 kg Ms = = 2.005 Mg/m3 VT 0.283 m3

¢ Dry unit weight gdry =

Ws 125 lb = = 125 pcf, or, VT 1.0 ft3

2.005 Mg/m3 * 9.81

Water content w% =

kN>m3 1 Mg>m3

= 19.65 kN>m3 ≤

Mw Ww 1100%2 = 1100%2 = 12% Ms Ws

Illustration 2 Determine the wet density, dry unit weight, void ratio, water content, and degree of saturation for a sample of moist soil that has a mass of 18.18 kg and occupies a total volume of 0.009 m3. When dried in an oven, the dry mass is 16.13 kg. The specific gravity of the soil solids is 2.70. (Acceleration of gravity, g = 9.81 m>sec2.)

Air

Va

Water

Vw

Vv MT = 18.18 kg Ms = 16.13 kg

Soil

VT = 0.009 m3 Vs

Solution Wet density rwet =

18.18 kg MT = = 2,020 kg>m3 VT 0.009 m3

Dry unit weight gdry =

Water content w% = Void Radio e =

70

116.13 kg219.81 m>sec22 Ms g Ws 158.235 N = = = = 17.58 kN>m3 3 VT VT 0.009 m 0.009 m3

118.18 - 16.132kg Mw * 100% = * 100% = 12.7% Ms 16.13 kg

Vv .0031 m3 = = 0.53 Vs .0059 m3

Soil Composition

B where Vs =

Ms Gsρw 16.13 kg

= 12.702 ¢ 1.0

gm cm3

16.13 kg = 3

2.70 * 10 and

kg

* 106

cm3 m3

* 0.001 kg>gm ≤

= 0.0059 m3

m3

Vv = VT - Vs = 0.009 m3 - 0.0059 m3 = 0.0031 m3 R

Degree of saturation

S% =

112.7%212.702 w%Gs = = 64.7% e 10.532

Illustration 3 A 150-cm3 sample of wet soil has a mass of 250 g when 100 percent saturated. When oven-dried, the mass is 162 g. Calculate the dry density, dry unit weight, water content, void ratio, and Gs.

Water

MW

Soil

MS = 162 g

MT = 250 g

VT = 150 cm3

Solution ρdry =

162 g g Ms = = 108 3 3 VT 150 cm cm

gdry = 11.08 g>cm32 ¢ 62.4 w% = ¢

Vw =

lb>ft3 g>cm3

≤ = 67.5 pcf 1refer to Table 12

250 g - 162 g Mw ≤ 1100%2 = ¢ ≤ 1100%2 = 54.3% Ms 162 g

250g - 162g Mw = 88 cm3 = Vv for this problem, since for 100% saturation = Gwrw 11.0211.0 g>cm32 all voids are filled with water

71

Soil Composition Vs = VT - Vv = 150 cm3 - 88 cm3 = 62 cm3 e =

Gs =

Vv 88 cm3 = = 1.42 Vs 62 cm3 162 g Ms = 2.61 = 3 Vsρw 162 cm 211.0 g>cm32

Illustration 4 Laboratory test data on a sample of saturated soil show that the void ratio is 0.45 and the specific gravity of soil solids is 2.65. For these conditions, determine the wet unit weight of the soil and its water content.

WW

Water

VV = eVS

WS

Soil

VS = 1.0 (assumed)

Solution This sample is saturated; thus all voids are filled with water: e =

Vv = 0.45 Vs

but Vv and Vs are not known. On the phase diagram, assume that Vs is unity (i.e., Vs = 1.0 ft3). Therefore: VT = Vs + eVs = 1.0 ft3 + 0.45 ft3 = 1.45 ft3 As a result:

and:

from which:

Ws = VsGsgw = 11.0 ft3212.652162.4 pcf2 = 165 lb Ww = Vwgw = 1.45 ft32162.4 pcf2 = 28 lb WT = Ws + Ww = 165 lb + 28 lb = 193 lb

gwet =

w% =

72

WT 193 lb = = 133 pcf VT 1.45 ft3

Ww 28 lb 1100%2 = ¢ ≤ 1100%2 = 17% Ws 165 lb

Soil Composition

3

Submerged Soil In many soil mechanics problems, it is necessary to determine the net intergranular weight, or effective weight, of a soil when it is below the groundwater table. (In this context, intergranular refers to the weight or force that acts at the point, or on the surfaces, where soil particles are in contact.) Effective soil weight is used to determine effective stress in a soil deposit, a value that influences factors such as soil shear strength, soil compressibility and settlement, and slope stability—topics discussed in later chapters. For this “underwater” condition, the soil solids are buoyed up by the pressure of the surrounding body of water, and the submerged soil weight becomes less than for the same soil above water. The effective soil weight then becomes the unit weight of the soil material when it is weighed under water. The water in the voids has zero weight (when submerged, all voids can be assumed to be filled with water), and the weight of the soil solids is reduced by the weight of the volume of water they displace. Therefore, a submerged soil weight (Wsub) equals the soil weight above water minus the weight of water displaced, or: Wsub = VsGsgw - VsGwgw = Vsgw1Gs - Gw2 Wsub = Vsgw1Gs - 12

(18)

Since unit weight is total weight divided by total volume: gsub =

Vsgw1Gs - 12 Wsub = VT Vs11 + e2 gsub =

1Gs - 12

gw

(19a)

ρsub =

1Gs - 12

ρw

(19b)

Similarly, in terms of density:

11 + e2

11 + e2

For a given soil (the subject soil) the effective unit weight when submerged (or the submerged effective density) will be the same regardless of depth below the water surface.3 For the condition where the soil is 100 percent saturated and the wet unit weight is known, the equations for submerged soil unit weight reflect that the weight (or mass) of both the soil particles and the voids water are for the buoyant condition, to become: gsub = gsat - gw 3The hydrostatic pressure p developed within a body of water relates to the depth below the water surface (that w is, pw = gw * depth), and acts equally in all directions. For a unit volume of soil particles submerged in a body of water, such as one cu ft or one cu meter, the difference between the upward water pressure on the base of the unit volume and the downward pressure on the top of the unit volume (the reason for buoyancy) is always the product of the height dimension for the unit volume and the unit weight for water (a calculation that is independent of depth).

73

Soil Composition

or: gsub = gsat - 62.4

= gsat - 9.81

1in pcf2

(20a)

1in kN>m32

(20b)

Equation 194 indicates that an accurate determination of the submerged unit weight requires that the specific gravity of the soil solids and the void ratio be known. Unfortunately, in terms of time and expense, some testing or physical analysis is required to determine the specific gravity, which is then used to calculate the void ratio. Also, soil properties such as particle size distribution, void ratios, and soil weights vary somewhat over even relatively limited distances (including areas identified as “uniform deposits”). Commonly for foundation studies at building sites, representative unit weights are selected on the basis of values determined from tested soil samples (samples obtained from borings, test pits, etc.) and assumed for areas between locations of known conditions (borings, etc.). Because of the various practical aspects, the effort to make highly accurate determinations of submerged soil weights is seldom undertaken when studies and designs are done. Instead satisfactory estimates, which can be made from knowing a wet weight, are frequently utilized. For many soils, and fortunately for ease of computation, the submerged unit weight is on the order of half the wet soil unit weight above the water table (related to the relatively limited range of specific gravity values for particles and typical void ratios), see Eq 21; a notable exception to this condition is soils containing significant decomposed vegetation or organic material. For many practical applications, the effects from adopting the simplification are negligible (but where accuracy is required, Equation 19 should be used). gsub L

1 2 gwet 1approximately2

(21a)

rsub L

1 2 rwet

1approximately2

(21b)

Illustration 5 Undisturbed soil obtained from a test pit 0.0283 m3 (1 ft3) in volume is found to have a wet weight of 0.459 kN (103.2 lb). The dry weight of the sample is 0.376 kN (84.5 lb). What would be the effective unit weight of such a soil if it were submerged below the groundwater table? The specific gravity of the soil is determined to be 2.70.

Solution By Equation 21, the effective submerged weight is approximately: gsub ⬵

1 1 1 0.0459 kN g or 1103.2 pcf2 ⬵ 52 pcf ⬵ ¢ ≤ ⬵ 8.11 kN>m3 2 wet 2 2 0.0283 m3

a word of caution, it is pointed out that γsub is not equal to γdry, the dry unit weight of soil. The term γdry conventionally is restricted to indicate dry soil and does not include the effect of buoyancy as caused by submergence. 4As

74

Soil Composition An accurate determination is as follows:

Vs =

Ws 0.376 kN = = 0.0142 m3 Gsgw 12.70219.81 kN>m32

Vv = VT - Vs = 0.0283 m3 - 0.0142 m3 = 0.0141 m31or 0.50 ft32 e =

Vv 0.0141 = ⬵ 1.00 Vs 0.0142

gsub = ¢ = ¢

Gs - 1 ≤ gw 1 + e

(from Eq. 19)

2.70 - 1 ≤ 19.81 kN>m32 = 8.34 kN>m31or 53.2 pcf2 1 + 1.0

Illustration 6 Assume that a sample of soil similar to the soil from the preceding illustration is excavated from a 1 ft3 test hole at a location below the water table. The soil analysis is now for the 100 percent saturated condition. What saturated weight would be expected?

Solution gsub = gsat - gw or:

gsat = gsub + gw = 53.2 pcf + 62.4 pcf = 115.6 pcf

(from Eq. 21a)

Problems 1 A sample of soil taken from a borrow pit has a wet mass of 14.56 kg. Completely dried, the soil dry mass is 11.78 kg. Determine the water content of this soil. 2 A sample of soil obtained from a construction site is found to have a wet weight of 29.4 lb. When completely dried, the soil weighs 25.9 lb. What is the water content of the soil sample? 3 A large soil sample obtained from a borrow pit has a wet mass of 26.50 kg. The in-place volume occupied by the sample is 0.013 m3. A small portion of the sample is used to determine the water content; the wet mass is 135 g, and after drying in an oven, the mass is 117 g. (a) Determine the soil’s water content.

(b) Determine the soil wet and dry density for conditions at the borrow pit. 4 A sample of soil obtained from a borrow pit has a wet weight of 42 lb. The total volume occupied by the sample when in the ground was 0.34 ft3. A small portion of the sample is used to determine the water content. When wet, the sample mass is 150 g; after drying, 125 g. (a) What is the water content of the sample? (b) Determine the wet and the dry unit weights of the soil in the borrow pit. 5 The mass of a dried soil sample is 250 g, as determined on a laboratory balance (scale). When immersed in water, the soil particles displace 95 cm3, and this is then the volume of soil solids.

75

Soil Composition Using this data, determine the specific gravity of soil solids. 6 The following data apply to a soil sample taken from a construction site: Specific gravity, Gs = 2.70 Void ratio, e = 0.80 For this soil, determine the dry density (kg/m3) and dry unit weight (in both kN/m3 and pcf). 7 An undisturbed soil sample has a dry mass of 59 kg and an in situ (in-ground) volume of 0.035 m3. The specific gravity of the soil particles is 2.65. Determine the void ratio, e. 8 A 1-ft3 sample of undisturbed soil is found to have a dry weight of 107 lb. If the specific gravity of soil solids is 2.70, what is the void ratio of the sample? What is the porosity? 9 A laboratory test container holds a volume of 1/30 ft3 (0.000943 m3). The container is carefully filled with a dry sand, 1.36 kg, to determine a value for the minimum density. The specific gravity of the soil solids is 2.70. Determine the dry density (kg/m3), the dry unit weight (in both kN/m3 and pcf ), and the void ratio. 10 A dry sand is placed in a container having a volume of 1⁄4 ft3. The dry soil weight is 27 lb. If the specific gravity of soil solids is 2.75, determine the void ratio of the sand in the container. Also, calculate the dry unit weight in kN/m3 and pcf. 11 A dry sand is placed in a container having a volume of 0.30 ft3. The dry weight of the sample is 31 lb. Water is carefully added to the container so as not to disturb the condition of the sand. When the container is filled, the combined weight of soil plus water is 38.2 lb. From these data, compute the void ratio of the soil in the container and the specific gravity of the soil particles. Also, determine the wet and dry unit weights, in kN/m3 and pcf. 12 What will be the dry unit weight of a soil (kN/m3 and pcf ) whose void ratio is 1.20, where the specific gravity of soil solids is 2.72? 13 A volume of undisturbed soil, 0.015 m3, obtained from a construction site has a wet mass of 31 kg and a dry mass of 27.5 kg. Gs of soil solids is 2.71. Determine the void ratio, water content, and degree of saturation for the in-place condition.

76

14 A sand is densified by compaction at a construction site so that the void ratio changes from 0.80 to 0.50. If the specific gravity of solids is 2.70, what is the increase in the dry unit weight of the sand (in kN/m3 and pcf)? 15 Demonstrate that e = n>11 - n2 when n is used as a decimal. 16 A sample of clay soil taken from a construction site has a wet mass of 45.6 kg and a related inplace volume of 0.021 m3. After being dried in an oven, the dry soil mass is 39.1 kg. The specific gravity of the soil particles is 2.73. Determine the soil water content, void ratio, and degree of saturation. 17 An undisturbed sample of clay is found to have a wet weight of 63 lb, a dry weight of 51 lb, and a total volume of 0.50 ft3. If the specific gravity of soil solids is 2.65, determine the water content, void ratio, and degree of saturation. 18 A soil sample is taken from a small test hole in a soil borrow pit. The volume of the test hole is determined to be 0.028 m3. The wet mass of the soil is 56 kg and the dry mass is 49 kg. The specific gravity of soil solids is 2.72. Determine wet and dry density, wet and dry unit weight, water content, void ratio, and degree of saturation. 19 A clay sample has a wet mass of 417 g and occupies a total volume of 276 cm3. When oven dried, the mass is 225 g. If the specific gravity of soil solids is 2.70, calculate the water content, void ratio, and degree of saturation. 20 Given the following data for an undisturbed soil sample:

Gs = 2.69, e = 0.65, w = 10% determine wet unit weight and dry unit weight (in kN/m3 and pcf), wet and dry density (mg/m3), and degree of saturation. 21 A saturated sample of undisturbed clay has a wet mass of 700 g. The sample has a total volume of 425 cm3. When dry, the soil mass is 450 g. What is the specific gravity of the soil solids? 22 A sample of soil obtained from below the groundwater table is found to have a water content of 20 percent. If it is assumed that the specific gravity of most soil particles is within the

Soil Composition range of 2.60 to 2.75, what is the approximate void ratio of the soil sample? 23 Construction projects may involve working with soil deposits that are under a body of water or below the groundwater table that exists at a land area. (a) For soil submerged below water, explain what is meant by the term effective unit weight. (b) What is the importance of knowing, or being able to determine, an effective unit weight for a submerged soil? 24 When a soil is submerged, the solid particles are subject to buoyancy caused by the water pressure existing at the “underwater” depth position being analyzed. The result is that the effective intergranular weight (that is, the magnitude of soil weight transmitted at the points where the particles are in contact) of the submerged soil bulk is less than for the condition where the soil bulk is not submerged. Why is the difference between the unit weight of a submerged soil and the same soil above water (the reduction due to buoyancy) independent of the depth below the water surface?

25 In an undisturbed soil formation, it is known that the dry unit weight is 18.06 kN/m3. The specific gravity of the soil particles is 2.75. (a) What is the saturated wet unit weight of the soil, in kN/m3 and pcf? (b) What is the submerged effective weight of the soil, in kN/m3 and pcf? (c) What is the submerged effective density (Mg/m3)? 26 An undisturbed soil sample has a wet unit weight of 120 pcf when the water content w is 15 percent. The specific gravity of the soil particles is 2.65. (a) Approximate the submerged effective weight of this soil. (b) Determine the submerged effective weight, using the formula for the exact value. 27 An undisturbed soil sample has a wet density of 2.22 Mg/m3 when the water content is 10 percent. The specific gravity of the soil particles is 2.69. (a) Approximate the submerged effective density for this soil. (b) Determine the submerged effective density using the formula for an exact value.

77

Answers to Selected Problems

1. 2. 3. 4. 5. 7. 8. 9. 10. 11.

12. 13. 14.

78

w = 23.6% w = 13.5% w = 15.4%, rwet = 2038.5 kg>m3, rdry = 1766.5 kg>m3 w = 20%, ␥wet = 123.5 pcf, ␥dry = 103 pcf Gs = 2.63 e = 0.57 e = 0.57, n = 36% rdry = .1442 kg>m3, ␥dry = 90 pcf = 14.13 kN>m3, e = 0.87 e = 0.58, ␥dry = 108 pcf = 16.96 kN>m3 e = 0.62, Gs = 2.68, ␥dry = 103.3 pcf = 16.22 kN>m3 ␥wet = 127.3 pcf = 19.99 kN>m3 ␥dry = 77 pcf = 12.09 kN>m3 e = 0.48, w = 12.7%, Sat ⬵ 72% increase = 18.5 pcf = 2.90 kN>m3

16. 17. 18.

19. 20.

21. 22. 26. 27.

w = 16.6%, Sat = 97.3%, e = 0.467 w = 23.5%, e = 0.62, Sat = 100% rwet = 2.0 Mg>m3 = 124.8 pcf rdry = 1.75 Mg>m3 = 109.2 pcf w = 14.3%, e = 0.556, Sat = 70% w = 85%, e = 2.31, Sat = 100% ␥wet = 112.2 pcf = 17.62 kN>m3 ␥dry = 102 pcf = 16.01 kN>m3 rwet = 1792 kg>m3 rdry = 1630 kg>m3 Sat = 42% Gs = 2.57 e = 0.541;2 (a) ␥sub L 60 pcf (b) ␥sub = 65 pcf (a) rsub = 1.11 Mg>m3 (b) rsub = 1.27 Mg>m3

Index Properties and Classification Tests, and Soil Classification Systems

From Essentials of Soil Mechanics and Foundations: Basic Geotechnics, Seventh Edition. David F. McCarthy. Copyright © 2007 by Pearson Education, Inc. Published by Pearson Prentice Hall. All rights reserved.

79

Index Properties and Classification Tests, and Soil Classification Systems To aid the engineering profession and the design–construction field, soils have been divided into basic categories or classifications based on certain physical characteristics. But because of the range of characteristics for the different soil variations that exist in nature, classification categories have been, of necessity, relatively broad in scope. All of a soil’s properties are not checked to obtain a classification. Consequently, for proper evaluation of a soil’s suitability for construction or foundation use, information about its properties, in addition to classification, is frequently necessary. Those properties that do help to define a soil’s engineering qualities and that are used to assist in determining accurate classification are termed index properties. The tests necessary to determine index properties are classification tests. Index properties include those characteristics that can be determined relatively quickly and easily and that will have bearing on items of engineering importance, such as strength or load-supporting ability, tendency to settle or expand, and effect of water and freezing conditions. Standard Test Procedures. For a test-determined property or identification to be meaningful, the method used for testing needs to be standardized in regard to procedure and to the equipment or apparatus unique to the procedure. When soil and rock properties are to be determined for construction projects in the United States, the testing procedures presented in the American Society for Testing and Materials (ASTM) publications1 and American Society of State Highway and Transportation Officials (AASHTO) publications2 are probably the most frequently referenced; ASTM standards are considered 1Annual

Book of ASTM Standards, Volume 04.08, Soil and Rock (I) and 04.09, Soil and Rock (II). ASTM, West Conshohocken, PA 19428. 2Standard Specifications for Transportation Materials and Methods of Sampling and Testing. AASHTO, Washington, D.C. 20090.

80

Index Properties and Classification Tests, and Soil Classification Systems

universal, whereas the AASHTO standards typically relate to transportation projects. Many of the individual procedures for testing presented by these two organizations are identical. Typically, a proposed procedure is adopted as a standard by ASTM or AASHTO only after a thorough review and consensus by appropriate researchers, educators, and practitioners. Because of the applicability, ASTM and AASHTO standards are also referenced internationally. Additionally, testing procedures established by some governmental agencies associated with construction work, as well as organizations outside the United States, often carry the designation of the individual agency or organization, but the test methods are identical to the ASTM and AASHTO procedures.

1

Index Properties Index properties refer to those properties of a soil that indicate the type and condition of the soil, and provide a relationship to structural properties, such as the strength and the compressibility or tendency for swelling and permeability. Generally, for coarse-grained soils, properties of the particles and the relative state of compaction are most significant. For fine-grained soils, the consistency (firm or soft) and plasticity are particularly important. The index properties that provide the desired information for coarse-grained and fine-grained soils are summarized in Table 1. In studies and analyses for construction projects it frequently is not necessary to determine all the index properties for the soil. Properties to be determined relate to the information

Table 1 Index Properties and Related Classification Tests Soil Type Coarse-grained

Index Property

Classification Test

Range of particle sizes and distribution of sizes Shape of particles Presence of fine-grained particles

Particle-size distribution (mechanical analysis) by sieving, or sedimentation test Visual From mechanical analysis (usually from use of a fine-mesh sieve) In situ density determination, and relative density test From mechanical analysis, or visual identification based on grain size Field or laboratory evaluation of unconfined compressive strength or shear strength (cohesion) Unconfined compressive strength or cohesion for the remolded soil Water content Atterberg limits (liquid limit and plastic limit) From visual identification and Atterberg

In-place density and relative state of compaction Classification Fine-grained

Consistency (strength and type of structure in the undisturbed state) Change in consistency due to remolding Water content Plasticity Classification

limits Presence and type of clay

Indirectly from determination of plasticity and change in consistency, and/or directly from a clay mineral analysis

81

Index Properties and Classification Tests, and Soil Classification Systems

that is needed and how such information is eventually to be used. For example, a clay mineral analysis requires very specialized equipment and is not performed in studies for foundation designs, unless the conditions are unusual. For organic soils, it is important to know of the presence and, at least the approximate, amount of organic material because of its influence on compressibility and strength. For all soils, the description should include the color. Color may have bearing on the mineralogical composition and is also extremely useful for determining homogeneity of a soil deposit and as an aid for identification and correlation during field construction.

2

Classification Tests Particle Size Distribution (Mechanical Analysis) This classification test determines the range of size of particles in the soil and the percentage of particles in each of the sizes between the maximum and the minimum. Two methods are in common use for obtaining the necessary information. Sieving is generally used for coarse-grained soils, and a sedimentation procedure is used for analyzing fine-grained soils. Sieving is the most direct method for determining particle sizes, but there are practical lower limits to sieve openings that can be used for soils. This lower limit is approximately at the smallest size attributed to sand particles. Information on sieves in common use is shown in Table 2. In the sieve analysis, a series of sieves (screens) having different-sized openings are stacked with the larger sizes over the smaller (Figure 1). The soil sample being tested is dried, clumps are broken, and the sample is passed through the series of sieves by shaking. Larger particles are caught on the upper sieves, and the smaller particles filter through to be caught on one of the smaller underlying sieves. The weight of material retained on each sieve is converted to a percentage of the total sample. The resulting data are conventionally presented as a grain- or particle-size distribution curve plotted on semilog coordinates, where the sieve size opening is on a horizontal logarithmic scale, and the percentage

Table 2 Common Sieve Types and Mesh Openings Sieve Size Designation #4 #8 #10 #20 #40 #60 #100 #200 #270 #400

82

U.S. Standard

Tyler Standard

British Standard

in.

mm

in.

mm

in.

mm

0.187 0.0937 0.0787 0.0331 0.0106 0.0098 0.0059 0.0029 0.0021 0.0015

4.76 2.38 2.00 0.84 0.42 0.25 0.149 0.074 0.053 0.037

0.185 0.093 0.065 0.0328 — 0.0097 0.0058 0.0029 0.0021 0.0015

4.70 2.362 1.651 0.833 — 0.246 0.147 0.074 0.053 0.038

— 0.081 0.0661 — — 0.0099 0.0060 0.0030 — —

— 2.057 1.676 — — 0.251 0.152 0.076 — —

Index Properties and Classification Tests, and Soil Classification Systems Figure 1 Set of sieves being assembled. Sieves shown are U.S. Standard sieves, typically used in laboratories.

(by weight) of the size smaller than a particular sieve opening is on a vertical arithmetic scale. Results may also be presented in tabular form. A detailed procedure for performing the sieve analysis is presented in the ASTM (American Society for Testing and Materials) Testing Manual, under ASTM Test Designation D-422. A typical presentation is shown in Figure 2. Note the “reversal” of the logarithmic scale (size increases from right to left). Most soil grains are not of an equal dimension in all directions. Hence, the size of a sieve opening will represent neither the largest nor the smallest dimension of a particle, but some intermediate dimension. As an illustration, assume a brick-shaped particle whose length, width, and thickness are different. The dimension that controls whether or not the

Log cycle

Log cycle

Partial log cycle

Log cycle

100

Percent finer by weight

80

60

40

20 0 10.0

1.0

0.1

0.01

Particle diameter, mm (logarithmic scale)

Figure 2 Particle-size or grain-size distribution curve.

83

Index Properties and Classification Tests, and Soil Classification Systems

particle passes through the sieve opening is the intermediate dimension, assuming that the particle is aligned so that the greatest dimension is perpendicular to the sieve opening. Sieve tests can be performed in a laboratory or in the field (at the area being explored, such as a proposed borrow pit or construction site). The appearance of the particle-size distribution plot depends on the range and amounts of the various sizes of particles in the soil sample. These, in turn, have been affected by the soil’s origin or the method of deposition. Well-graded soils (a distribution of particles over a relatively large range of sizes) produce a longish straight curve (Figure 3(a)). A uniform soil (soil having most of the particles of approximately similar size) plots as shown in Figure 3(b). A gap-graded soil (an absence of intermediate sizes) plots as in Figure 3(c). The grain-size plot can provide an indication of a soil’s history. A residual deposit has its particle sizes constantly changing with time as the particles continue to break down, and typically produces grain-size curves as shown in Figure 4. The curves shown in Figure 5(a) and (b) represent glacial and glacial–alluvial deposits. River deposits may be well-graded, uniform, or gap-graded, depending on the water velocity, the volume of suspended solids, and the river area where deposition occurred. Certain properties of clean sands have been related to particle diameters. The effective size of a sand is taken as the particle size corresponding to the 10 percent passing size from the grain-size curve, and is indicated as D10. It is this size that is related to permeability and capillarity. It has also been found that the D15 particle size can be related to soil permeability. Another relationship, the ratio D60 /D10, is termed the uniformity coefficient, Cu, and provides a comparative indication of the range of particle sizes in the soil (Figure 6).

% finer

100

% finer

100

0 10

1.0

0.1

0.01

0 10

0.001

1.0

0.1

0.01

Particle diameter, mm (log scale)

Particle diameter, mm (log scale)

(a) Well-graded soils

(b) Uniform soil

% finer

100

0 10

1.0

0.1

0.01

0.001

Particle diameter, mm (log scale) (c) Gap-graded soil (absence of intermediate-size particles)

Figure 3 Shape of particle-size distribution curve expected for described soil type.

84

0.001

Index Properties and Classification Tests, and Soil Classification Systems

% finer

100

% finer

100

0 10

1.0

0.1

0.01

0 10

0.001

1.0

0.1

0.01

0.001

Particle diameter, mm (log scale)

Particle diameter, mm (log scale)

(a) Young residual

(b) Intermediate maturing

% finer

100

0 10

1.0

0.1

0.01

0.001

Particle diameter, mm (log scale) (c) Fully maturing

Figure 4 Typical particle-size curves for residual soils.

% finer

100

% finer

100

0 10

1.0

0.1

0.01

0.001

0 10

1.0

0.1

0.01

Particle diameter, mm (log scale)

Particle diameter, mm (log scale)

(a) Glacial

(b) Glacial–alluvial

0.001

Figure 5 Typical particle-size curves for transported soils.

Sand having a wide range of particle sizes is considered well-graded and has Cu values greater than 10. A uniform soil has Cu values less than about 5. The procedure commonly used for obtaining particle-size distribution information for silts and clays is the sedimentation method. In this method, the soil is placed into solution with distilled water, and the soil particles are permitted to settle out of the solution. As settling occurs, the average specific gravity of the solution decreases. Readings of specific gravity or a related property by use of a hydrometer, made at different time intervals, provide an indication of the weight of the soil remaining in solution and also information on the sizes of particles that have settled out of the solution (Figure 7). Most conventionally, the test data are reduced to provide particle diameters and the percentage (weight) that is finer than a particular size by using the Stokes equation for spheres falling freely in a fluid of known properties. This application is not absolutely correct, since most fine-grained soil

85

Index Properties and Classification Tests, and Soil Classification Systems

Log cycle

Log cycle

Partial log cycle

Log cycle

100

Percent finer by weight

80

D60 = 0.7 mm D10 = 0.12 mm D Cu = 60 = 0.70 = 5.8 D10 0.12

60

40

20

0 10.0

5.0

2.0

1.0

0.5

0.2

0.1

0.05

0.02

0.01

Particle diameter, mm (log scale)

Figure 6 Determining uniformity coefficient from particle-size curve.

d1

Start

d2

t1

d3

t2

d4

t3

t4

Time increasing from start of test (b)

Figure 7 (a) Hydrometer used to determine specific gravity of a soil–water solution being inserted into the solution; (b) diagram showing the change in hydrometer depth as soil particles settle out of solution.

particles are not round; in fact, most clay particles are flat or plate shaped. The method is, therefore, more applicable to silts than to clays. Nevertheless, the method is in wide use, for it is felt to be a practical way to obtain reasonable approximations of the particle-size distribution for fine-grained soils. Resulting effects are not serious, since particle-size distribution is not used for evaluation of the significant engineering properties of fine-grained soils. Details on the method to determine the particle-size distribution by using the sedimentation test (hydrometer method) are described by ASTM Test Designation D-422.

86

Index Properties and Classification Tests, and Soil Classification Systems

In-Place Density, In-Place Unit Weight The term in-place density refers to the volumetric weight, usually expressed as kilonewtons per cubic meter (kN/m3), megagrams per cubic meter (Mg/m3), or pounds per cubic foot (pcf) of a soil in the undisturbed (or in situ) condition or in a compacted fill. (Actually, pcf and kN/m3 indicate unit weight, whereas Mg/m3 indicates density. The term in-place density has been widely used for years to refer to the field procedure for determining volumetric weight, but with the advent of the SI system, improper usage of weight and mass terms has become more conspicuous; correct usage is evolving slowly. For reference, the density of water is 1 Mg/m3 (that is, 1,000,000 g/m3). Generally, for the coarse-grained soils, the greater the density, or unit weight, the better the shear strength and the lesser the tendency for compression (settlement). In-place density or unit weight determinations are made of borrow pit soils so as to estimate the change in volume (the space occupied by the soil particles and voids) because of shrinkage or expansion (“swell”) that will occur as the soil is transported or compacted in place at a fill location. Where compacted earth fills are being constructed, it is standard practice to determine the in-place density or unit weight of the soil after it is placed to establish whether the compaction effort has been adequate or if more compaction is required. Of the equipment and methods used for making in-place density or unit weight determinations, the sand-cone method and the rubber-balloon method have a long history of use (Figure 8). With these methods, a small test area is selected and a volume of compacted soil is dug up and weighed. The sand-cone equipment or the balloon equipment is used to determine the volume of the dug hole. Knowing soil weight (or mass) and corresponding volume permits the unit weight (or density) to be calculated. A modern development for making in-place density and unit weight determinations involves the use of nuclear equipment (Figure 9). Through controlled use of a nuclear material, gamma rays (photons) are emitted into the tested soil. These photons collide with electrons in the soil materials, some being scattered and some being absorbed. The quantity of photons reaching a detection device (part of the test equipment) relates to the soil

Glass or plastic jar holds sand

Density plate (metal) Sand-cone (metal)

Figure 8 Sand-cone and balloon apparatus for determining in-place density. (Photo courtesy of ELE International)

87

Index Properties and Classification Tests, and Soil Classification Systems

Figure 9 (Left) Representative in-place density equipment (left to right): nuclear moisture-density gauge, standard sand-cone, balloon apparatus, and NYS-DOT sand-cone apparatus. (Right) Nuclear moisture–density meter in use. (Courtesy of Troxler Electronic Laboratories, Inc.)

density. To determine water content, a neutron-emitting material and detector are used. Compared to the above-mentioned methods of determining in-place density, a significant advantage with the nuclear method is the rapid speed with which results are obtained. Figure 9 shows a grouping of in-place density or unit weight equipment, for the purpose of comparison.

Relative Density For a granular soil, the shear strength and resistance to compression are related to the density (or unit weight) of the soil; higher shear strength and more resistance to compression are developed by the soil when it is in a dense or compact condition (high density) than when it is in a loose condition (low density). In a dense condition, the void ratio is low; in a loose condition, the void ratio is high. To evaluate the relative condition of a granular soil, the in-place void ratio can be determined and compared to the void ratio when the soil is in its densest condition and when it is in the loosest condition (Figure 10). This comparison is the relative density DR. Relative density is expressed as a percentage. High values indicate a dense or compact material; low values represent a loose material. DR% =

88

emax - e0 * 100% emax - emin

(1)

Index Properties and Classification Tests, and Soil Classification Systems

Vv

Voids

Soil Solids

Vs

Loosest condition void ratio = emax

Figure 10

Vv

Voids

Vs

Voids

Vv

Soil Solids

Soil Solids

Vs

Intermediate condition void ratio = e0

Densest condition void ratio = emin

Relative conditions of a granular soil.

where emax = void ratio of the soil in its loosest condition emin = void ratio of the soil in its densest condition e0 = void ratio of the soil in the natural condition or condition in question The maximum density (or minimum void ratio) is determined in the laboratory by compacting the soil in thin layers in a container of known volume and subsequently weighing the soil. The compaction is achieved by applying a vibratory and compressive force simultaneously. The compressive force needs to be sufficient to compact the soil without breaking the individual particles of soil. Because of the irregular size and shape of granular particles, it is not possible to obtain a zero volume of void spaces. Practically, there will always be voids in a soil mass. The minimum density (or maximum void rate) can be determined in the laboratory by carefully letting the soil slowly flow into the test container through a funnel. When this task is carefully performed, the soil will be deposited and remain in a loose condition, from whence the loose density and void ratio can be calculated. In terms of dry unit weight and dry density (which are more convenient to work with), relative density is: 1 1 gmin g0 DR% = 1100%2 1 1 gmin gmax 1 rmin r0 1100%2 DR% = 1 1 rmin rmax

(2a)

1

-

(2b)

where gmin1rmin2 = dry unit weight (density) for the loosest condition gmax1rmax2 = dry unit weight (density) for the densest condition g01r02 = dry unit weight (density) for the condition in question Typical relative density values are presented in Table 3.

89

Index Properties and Classification Tests, and Soil Classification Systems Table 3 Representative Values of Relative Density Descriptive Condition

Typical Range of Unit Weight and Density

Relative Density, %

pcf

kN/m3

Mg/m3

635 35–65 65–85 785

690 90–110 110–130 7130

614 14–17 17–20 720

61.4 1.4–1.7 1.7–2.0 72.0

Loose Medium dense Dense Very dense

Illustration 1 An undisturbed sample of fine sand is tested in the laboratory and found to have a dry mass of 3.63 kg (dry weight of 8 lb), a total volume of 0.00198 m3 (0.07 ft3), and a specific gravity Gs of 2.70. Other laboratory tests are performed to determine the maximum and minimum density for the sand. At the maximum density, it is determined that the void ratio is 0.35; at the minimum density, the void ratio is 0.95. Determine the relative density of the undisturbed sample.

Solution SI Units Void ratio of undisturbed sample e0: VT = .00198 m3 3.63 kg

MS = .00134 m3 = GSrW 12.7021103 kg>m32

VS =

VV = VT - VS = 0.00198 m3 - 0.00134 m3 = 0.000636 m3 e0 =

VV 0.636 * 10-3 m3 = = .475 VS 1.34 * 10-3 m3

10.95 - 0.4752 emax - e0 1100%2 = 1100%2 = 79% emax - emin 10.95 - 0.352

DR% =

Dry density =

3.63 kg MS = = 1.83 Mg>m3 VT 0.00198 m3

U.S. Customary Units Void ratio of undisturbed sample e0: VT = 0.07 ft3 VS =

WS 8 lb = = 0.0474 ft3 GSgW 12.702 162.4 pcf2

VV = VT - VS = 0.070 - 0.0474 = 0.0226 ft3 e0 =

90

VV 0.0226 ft3 = = 0.476 VS 0.0474 ft3

Index Properties and Classification Tests, and Soil Classification Systems

DR% = ¢

emax - e0 0.95 - 0.476 ≤ 1100%2 = ¢ ≤ 1100%2 = 79% emax - emin 0.95 - 0.35

Dry unit weight of soil =

WS 8.0 lb = M 115 pcf VT 0.07 ft3

In its natural condition, the soil is probably dense (see Table 3).

Water Content For coarse- and fine-grained soils, water content can have a significant effect on the soils’ behavioral properties when used for construction purposes. By definition and as previously described, water content (w) is the ratio of the weight of water in a soil to the dry weight of the material. As a result, in laboratory and field work the weight of water in the test sample has to be determined. Conventionally this is done by drying the original wet sample, recording the wet and dry weights, and performing the required calculation. Various other methods are available for drying soil samples to determine water content, such as by using gas pressure extractors or infrared-based equipment. Another device, the Speedy Moisture Tester, bases its operation on the reaction that occurs between a carbide reagent and soil moisture to determine a soil’s water content. The wet soil sample is placed in a sealed container with calcium carbide, and the pressure generated by the vaporized moisture is then related to water content. This method provides a rapid procedure for determining moisture content and is finding increased usage in laboratory and field work.3

Consistency of Clays Consistency refers to the texture and firmness of a soil and is often directly related to the strength. Consistency is conventionally described as soft, medium stiff (or medium firm), stiff (or firm), or hard. These terms, unfortunately, are relative and have different meanings to different observers. For standardization, it is reasonable and practical to relate consistency to strength. With clays, shear strength is discussed in terms of cohesion and unconfined compressive strength. The unconfined compressive strength is obtained by imposing an axial load to the ends of an unsupported cylindrical sample of clay and determining the load that causes shear failure in the sample. Unconfined compression tests can be performed in the laboratory or in the field. The unconfined compressive strength is, under practical conditions, twice the cohesion (or shear strength) of a clay soil. The cohesion can be determined in a laboratory strength test. For a consistency classification in the field or laboratory, special soil testing equipment, such as a vane-shear device or pocket penetrometer,

3Somewhat

unfortunately, the Speedy Moisture reading is expressed as a percentage of the soil’s wet weight. Conversion to moisture content by dry weight can be performed using the relationship: w% =

wsp 1 - wsp

* 100%

where wsp is the moisture content indicated by the Speedy device, expressed as a decimal.

91

Index Properties and Classification Tests, and Soil Classification Systems

6"

Pocket penetrometer for estimating strength of field or lab samples Vane

Vane-shear device for determining shear strength of undisturbed soil in borings or lab sample

1" Torvane device for determining shear strength of lab or field samples

Figure 11

Testing equipment for making strength or consistency determinations on cohesive soils.

Table 4 Consistency and Strength for Cohesive Soils

Shear Strength, t/ft2 or Kg/cm2 (kN/m2 or kPa)

Unconfined Compressive Strength, t/ft2 or Kg/cm2 (kN/m2 or kPa)

60.25 ( 624)

60.5 ( 648)

Medium (medium stiff or medium firm) Stiff (firm)

0.25–0.50 (24–48)

0.50–1.0 (48–96)

0.50–1.0 (48–96)

1.0–2.0 (96–190)

Very stiff (very firm)

1.0–2.0 (96–190)

2.0–4.0 (190–380)

72.0 ( 7190)

74.0 ( 7380)

Consistency Soft

Hard

Feel or Touch Blunt end of pencil-sized object makes deep penetration easily. Blunt end of pencil-sized object makes 12 mm or half-inch penetration with moderate effort. Blunt end of pencil-sized object makes moderate penetration (about 6 mm or about 1冫4 in.). Blunt end of pencil-sized object makes slight indentation; fingernail easily penetrates. Blunt end of pencil-sized object makes no indentation; fingernail barely penetrates.

provides a means of making quick and easy determinations (Figure 11). A tabulation of strength values for various consistency terms is shown in Table 4. The strength of a clay soil is related to its structure. If the original structure is altered because of changes in particle arrangement (from reworking or remolding) or chemical changes, the strength of the altered clay is less than the original strength. Sensitivity is the term that provides an indication of remolded strength related to original strength. When remolded, the strength of a clay is affected by the water content. At lower water contents,

92

Index Properties and Classification Tests, and Soil Classification Systems

strength is generally greater. However, sensitivity should be based on the comparison of remolded to undisturbed soil strength at an identical water content.

Sensitivity St =

Unconfined compressive strength, undisturbed clay Unconfined compressive strength, remolded clay

(3)

For most clays, sensitivities range between 2 and 4. Clays considered sensitive have St values between 4 and 8. Clays classified as extrasensitive have values between 8 and 16. Clays with sensitivity values greater than 16 are classified as quick clays. Such clays are very unstable. Normally, clays with a high degree of sensitivity possess a very flocculent structure in the undisturbed condition.

Consistency in the Remolded State and Plasticity In the remolded state, the consistency of a clay soil varies in proportion to the water content. At a higher water content, the soil–water mixture possesses the properties of a liquid; at lesser water contents, the volume of the mixture is decreased and the material exhibits the properties of a plastic; at still lesser water contents, the mixture behaves as a semisolid and finally as a solid. The water content indicating the division between the liquid and plastic state has been designated the liquid limit. The water content at the division between the plastic and semisolid state is the plastic limit. The water content at the division between the semisolid and the solid state is the shrinkage limit. At water contents above the shrinkage limit, the total volume of the soil–water mixture changes in proportion to change in the water content. Below the shrinkage limit, there is little or no change in volume as water content varies (Figure 12). Definition of the various states of consistency and the establishment of criteria to determine these various states were first formally proposed by A. Atterberg, a Swedish soil scientist, in the early twentieth century. Initially intended for agricultural use, the method has been adapted for engineering use in classifying soils. Although the limit values (water contents) have little direct meaning insofar as engineering properties of soils are concerned, correlations between liquid or plastic limits and engineering properties have been established over the years to aid in evaluating a soil for use as structural fill (dams, embankments, landfills), for highway construction, and for building support. The liquid limit is taken as the water content at which the soil “flows” (i.e., the condition where a very viscous liquid shears). Special equipment and procedures, defined in ASTM Test Designation D-4318, are required for determining the liquid limit. Briefly, the test procedure involves placing the soil sample in the liquid limit device cup (Figure 13), creating a thin groove (or split) in the center of the sample with a special tool, then subjecting the cup-sample to a series of small drops (or taps) to create vibrations that will cause the soil to “flow” and close the groove; the vibrations cause “miniature landslides” in the two separated portions of the soil sample adjacent to the groove. The typical test procedure to obtain the liquid limit value (i.e., the water content) consists of performing a series of trials, with the soil at a different water content for each trial, to determine the water content where the groove will “close” (soil from both sides of the open groove flows

93

Index Properties and Classification Tests, and Soil Classification Systems Semisolid state Solid state

Plastic state SL

PL

Liquid state LL

Total soil volume

Total volume

Water

Volume of water (varies)

Solids

Vol. solids

LL = liquid limit PL = plastic limit SL = shrinkage limit

Note relationship between total volume and volume of water indicates soil material is 100 percent saturated when water content is greater than SL.

0 Volume of water

Figure 12 Variation of total soil volume with change in water content for a fine-grained soil (for soil in the remolded state).

Brass 54 mm

27 mm 60 mm

2.0 mm 50 mm

Hard rubber (b)

(a) (c)

Figure 13 (a) Liquid limit trial being performed; (b) details of liquid limit device; (c) soil thread being rolled to determine plastic limit.

to establish contact) when the tap count is 25 blows. The testing should be performed using a moist sample that has not been allowed to dry, with a new (unused) portion of the sample for each test trial. Relating to the test procedure, the shearing resistance of all soils (free of organic content) is similar when at the liquid limit (near 2 kPa). The plastic limit is the water

94

Index Properties and Classification Tests, and Soil Classification Systems

content when the soil can just be rolled into a 18 -inch-diameter thread before crumbling, as described in ASTM Test Designation D-4318. Both test procedures use the portion of soil sample finer than 0.42 mm (#40 sieve). Various designations have been used to indicate the liquid and plastic limits. Using LL and PL for liquid limit and plastic limit, respectively, provides an easily understood terminology that has little chance of misinterpretation. The plasticity index (PI) is the numerical difference of the liquid and plastic limits, and indicates the range of water content through which the soil remains plastic. For proper evaluation of a soil’s plasticity properties, it has been found desirable to use both the liquid limit and the plasticity index values. Engineering soil classification systems use these values as a basis for classifying the fine-grained soils. For fine-grained soils, determining the natural water content (the water content of a soil in an undisturbed condition in the ground) and relating it to the plastic and liquid limits can provide an indication of the soil’s consistency and/or sensitivity potential. One such relationship is the liquidity index, LI: LI =

w% - PL% w% - PL% = LL% - PL% PI%

(4)

where w is the natural water content of the soil. A value less than 1 indicates that the natural water content is less than the liquid limit. A very low value for the LI, or a value near zero, indicates that the water content is near the plastic limit, where experience has shown that the sensitivity will be low and the cohesive strength relatively high (a stiff or hard consistency). As the natural water content approaches or exceeds the liquid limit, the sensitivity increases. The undisturbed strength still very much depends on the undisturbed structure. Negative values of the LI are possible and normally indicate a desiccated (dried), hard soil.

Presence of Clay Minerals The presence of even small amounts of certain clay minerals in a soil mass can have a significant effect on the properties of the soil. Identifying the type and amount of clay minerals may be necessary in order to predict the soil’s behavior or to develop methods for minimizing detrimental effects. The identification of clay minerals requires special techniques and equipment and trained personnel. Many different techniques are available. Some are useful for identifying only a particular type of clay, whereas other methods are suitable for identifying several types of minerals. These techniques include microscopic examinations, X-ray diffraction, differential thermal analysis, infrared absorption, optical property determination, and electron micrography. Description of these techniques and equipment is beyond the scope of this text; further information is available in publications dealing with clay mineralogy. Even with the techniques available to today’s soil scientists, the accurate determination of some clay minerals is not possible. Generally, however, qualitative if not quantitative identifications can be made that are adequate for many engineering problems. A somewhat indirect method of obtaining information on the type and effect of clay minerals in a soil is to relate plasticity to the quantity of clay-sized particles. For a given

95

Index Properties and Classification Tests, and Soil Classification Systems

amount of clay mineral the plasticity resulting in a soil will vary for the different types of clays. One relationship is activity, defined as: Activity =

Plasticity index, % Percentage of clay sizes, %

(5)

For this analysis, the percentage of clay sizes is that portion of the soil, by weight, consisting of particle sizes below 0.002 mm. Such information is available from the conventional hydrometer analysis used to determine a particle-size distribution. Activity, therefore, can be determined from standard laboratory tests. Clay materials with kaolinite, a stable clay mineral, will have low activity, whereas those soils with montmorillonite, known to be a type subject to large volume changes depending on available water, will have a high activity value. A relative activity classification is as follows: Activity 60.75 0.75–1.25 71.25

Classification Inactive clays Normal clays Active clays

The general relationship between clay’s potential for swelling or expansion (high, medium, low) and information about activity, clay particle sizes, and plasticity can be indicated diagrammatically as shown in Figure 14. The volume expansion associated with the various categories on the diagram is, crudely, as follows: Category Very high High Medium Low

Potential Volume Expansion 710% 5–10% 2–5% 62%

Testing for Dispersive Clays On the basis of testing many clay soils for dispersion (dispersive clays), the relationship between the presence of ions and susceptibility to deflocculation shown in Figure 15 has been proposed. Atomic absorption spectroscopy or flame photometry is used to determine the ion concentrations. Unfortunately, a clay’s tendency to disperse is not indicated by conventional soil classification tests such as the Atterberg limits. The atomic absorption spectroscope and flame photometer are not ordinarily found in soil mechanics laboratories; simple qualitative tests, using standard laboratory equipment, capable of being performed in the field have been developed and are outlined next. Judgment is involved when evaluating some of the results, but generally the tests are good indicators. 1. Soil Conservation Service Dispersion Test (ASTM D4221): This procedure, also known as the double hydrometer test, compares the percentage of soil sample particles smaller than 0.005 mm as determined by a conventional hydrometer analysis (where a deflocculating agent is used and the soil sample is mechanically agitated prior to test)

96

Index Properties and Classification Tests, and Soil Classification Systems

Ac

tiv

ity

tivi ty 2 Ac

Plasticity index (PI) of whole sample

.0

1. 0

100

75 Very high 50 Low

High

.5

ty 0

tivi

Ac

25 Medium Low 0

25

50

75

100

Percent of clay (particles smaller than .002 m) in whole sample

Percent sodium =

Percent Sodium Na(100) Na(100) = TDS Ca + Mg + Na + K milliequivalents per liter of saturation extract

Figure 14 Volume change potential classification for clay soils (Van der Merwe, or South African, method). (Modified from NAVFAC DM-7.01, 1986 [378])

100

80

Zone A – Dispersive (most soils)

60 Zone C – Intermediate (nondispersive for most soils) 40

20

0 0.1

Zone B – Nondispersive (most soils)

0.2

0.5

1.0

5.0

10

50

100 200 300

Total dissolved salts in saturation extract in milliequivalents per liter (TDS = Ca + Mg + Na + K)

Figure 15 Relationship of pore water salts and soil dispersion. (American Society of Civil Engineers, The Journal of Geotechnical Engineering, Vol. 102, #4, April 1976; “Identification and Nature of Dispersive Soils” by Sherard, Dunnigan, and Decker. With permission from ASCE.)

and as determined when a similar sample is exposed to a plain distilled water environment without being agitated, whence: % dispersion =

% finer than 0.005 mm in plain water test % finer than 0.005 mm in conventional hydrometer test

A percentage dispersion greater than 35 percent indicates a dispersive clay soil.

97

Index Properties and Classification Tests, and Soil Classification Systems

2. Crumb Test: A small sample of soil (10 mm or less in diameter) preserved at the natural water content is placed in a beaker of distilled water. The reaction in terms of a colloidal cloud around the crumb is an indication of the soil’s tendency to disperse; no cloud indicates a nondispersive material. 3. Pinhole Test (ASTM D4647): The test soil is compacted, and distilled water is then made to flow through a 1-mm-diameter hole in the sample. The test can be set up in the permeameter apparatus used for permeability determinations, using pea gravel filters at the ends of the test sample. With a dispersive clay, the hole erodes and the flowing water discharge is colored, whereas for a nondispersive soil the hole does not enlarge and the water discharge remains clear.

Other Properties Where a soil is to be used to support building foundations or other types of buried structures (such as pipelines, tunnels, or storage vaults), it is good practice to have pH and soluble sulfate determinations made, so as to obtain information on the corrosion potential of buried metal (piles or piping) or the potential deteriorating effect on concrete foundations.

3

ASTM and AASHTO Test Procedure Designations A summary of ASTM and AASHTO test procedure designations is presented in Table 5. Most commercial, agency, and educational program laboratories perform soil testing in accordance with these procedures.

Table 5 ASTM and AASHTO Test Procedure Designation Soil Property to be Determined or Test Identification Name 1. Particle sizes, distribution of particle sizes 2. Presence of fine-grain particles 3. Classification, classification systems 4. Shape of particles 5. Water content 6. Specific gravity 7. Plasticity, liquid limit, plastic limit 8. Presence of clay, type or behavior of clay 9. Soil shear strength, consistency 10. Change in consistency due to remolding 11. In-place density, relative density, compaction

12. Permeability, hydraulic conductivity 13. Compressibility, consolidation, soil collapse

98

Related ASTM Designations

Related AASHTO Designations

D422 D422, D1140 D2487, D2488, D3282 D2488 D2216, D3017, D4643, D4944, D4959 D854, D5550 D4318 D4221, D4647, D4829 D2166, D2573, D2587, D2850, D3080, D4767, D5315 D2166 D698, D1556, D1557, D2167, D2922, D2937, D4253, D4254, D5080 D2434, D3385, D5126 D2435, D4186, D4546, D5333

T11, T27, T88 T11 M145 — T93, T217, T239, T265 T84, T85, T100 T89, T90 — T208, T223, T234, T236, T296, T297 — T99, T180, T191, T204, T205, T233, T238, T239 T215 T216

Index Properties and Classification Tests, and Soil Classification Systems

4

Classification Systems Soil classification systems have been primarily devised to facilitate the transfer of information between interested parties. In the engineering and construction field, the broad general properties of concern relate to the performance or usefulness for supporting structures and the handling or working qualities of a soil. Because of the wide variation in properties and behavior of soils, classification systems generally group together in relatively broad categories the soils that have similar features or properties that are considered to be of importance. As a result, a classification system is not necessarily an identification system in which all pertinent engineering properties of a material are determined. Because of this, soil classification should not be used as the sole basis for design or construction planning. Historically, the most widely used method of classifying soils has been through visual identification, the size of soil grains, and the plasticity of the soil being used as the basis for indicating the soil type. To a great degree, the refined and more recently developed classification systems that rely on laboratory-determined properties for accurate identification still use these two criteria as the basis for indicating soil type. The desirable requirements for a satisfactory engineering soil classification system include the following: 1. There should be a limited number of different groupings, so that the system is easy to remember and use. Groupings should be on the basis of only a few similar properties and generally similar behavioral characteristics. 2. The properties and behavioral characteristics should have meaning for the engineering and construction profession. Generally, the properties should, at least crudely, relate to a soil’s handling characteristics, shear strength, volume change characteristics, and permeability. 3. Descriptions used for each grouping should be in terms that are easily understood and are in common use for indicating the soil type and its properties. Symbols or shorthand used to describe the grouping should easily relate to the soil type. Coded symbols are not desirable. 4. Classification into any grouping should be possible on the basis of visual identification (generally limited to differentiating between particle sizes, coarse- and finegrained soils, and plasticity) without special tests or equipment being necessary. The Unified Soil Classification System satisfies the above requirements of a classification system and is the system that is coming into prevalent use in the engineering and construction fields. This system is shown in Figure 16. Classifications are on the basis of coarse- and fine-grained soils, and retain the four common groupings of soil—gravel, sand, silt, and clay. The symbols are easily associated with the classification, being simply the first letter of the soil type (except for silt, which has the designation M, from mo, the Swedish word for silt). Refinement in grouping is based on a coarse soil’s being well or poorly graded and a fine-grained soil’s being of a high or low plasticity. What at first appear to be a large number of groupings are in fact very logical categories. Experience has proved that previously untrained personnel very quickly learn the system and use it with accuracy. The Unified System includes the use of a plasticity chart for aiding the classification of fine-grained soils.

99

Soil Classification Chart Soil Classification Criteria for Assigning Group Symbols and Group Names Using Laboratory TestsA COARSEGRAINED SOILS More than 50% retained on No. 200 sieve

FINE-GRAINED SOILS 50% or more pass the No. 200 sieve

Group Symbol

Group NameB

Gravels More than 50% of coarse fraction retained on No. 4 sieve Sands 50% or more of coarse fraction passes No. 4 sieve

Clean Gravels Less than 5% finesC

Cu ⱖ 4 and 1 ⱕ Cc ⱕ 3 Cu 6 4 and兾or Cc 7 3E

GW GP

Well-graded gravelF Poorly graded gravelF

Gravels with Fines

Fines classify as ML or MH

GM

Silty gravelF,G,H

More than 12% finesC

Fines classify as CL or CH

GC

Clayey gravelF,G,H

Clean Sands

Cu ⱖ 6 and 1 ⱕ Cc ⱕ 3E

SW

Well-graded sandI

Less than 5% finesD

Cu 6 6 and兾or 1 7 Cc 7 3E

SP

Poorly graded sandI

Sands with Fines

Fines classify as ML or MH

SM

Silty sandG,H,I

More than 12% finesD

Fines classify as CL or CH

SC

Clayey sandG,H,I

Silts and Clays Liquid limit less than 50

inorganic

PI ⬎ 7 and plots on or above “A” lineJ

CL

Lean clayK,L,M

PI ⬍ 4 or plots below “A” lineJ

ML

SiltK,L,M

Liquid limit - oven dried

OL

Organic clayK,L,M,N

PI plots on or above “A” line

CH

Organic siltK,L,M,O Fat clayK,L,M

PI plots below “A” line

MH

Elastic siltK,L,M

Liquid limit - oven dried

OH

Organic clayK,L,M,P

PT

Organic siltK,L,M,Q Peat

E

organic

Liquid limit - not dried inorganic

Silts and Clays Liquid limit 50 or more

organic

Liquid limit - not dried HIGHLY ORGANIC SOILS

6 0.75

6 0.75

Primarily organic matter, dark in color, and organic odor

A

Based on the material passing the 3-in. (75-mm) sieve. B If field sample contained cobbles or boulders, or both, add “with cobbles or boulders, or both” to group name. C Gravels with 5 to 12% fines require dual symbols: GW-GM well-graded gravel with silt GW-GC well-graded gravel with clay GP-GM poorly graded gravel with silt GP-GC poorly graded gravel with clay D Sands with 5 to 12% fines require dual symbols: SW-SM well-graded sand with silt SW-SC well-graded sand with clay

SP-SM poorly graded sand with silt SP-SC poorly graded sand with clay (D30)2 E Cu = D60兾D10 Cc = D10 * D60 F If soil contains ⱖ 15% sand, add “with sand” to group name. G If fines classify as CL-ML, use dual symbol GC-GM, or SC-SM. H If fines are organic, add “with organic fines” to group name. I If soil contains ⱖ 15% gravel, add “with gravel” to group name. J If Atterberg limits plot in hatched area, soil is a CL-ML, silty clay.

K

If soil contains 15 to 29% plus No. 200, add “with sand” or “with gravel,” whichever is predominant. L If soil contains ⱖ 30% plus No. 200, predominantly sand, add “sandy” to group name. M If soil contains ⱖ 30% plus No. 200, predominantly gravel, add “gravelly” to group name. N PI ⱖ 4 and plots on or above “A” line. O PI ⬍ 4 or plots below “A” line. P PI plots on or above “A” line. Q PI plots below “A” line.

Figure 16(a) Unified Soil Classification System (ASTM Designation D-2487. Copyright ASTM INTERNATIONAL. Reprinted with permission. 60 E

IN

L U"

Plasticity index, PI

50 "

CH

40

or

" "A

H

O

NE

LI

30 20 10 7 4 0

CL CL − ML 0

10 16 20

or

L

O

MH or OH

ML or OL 30

40

50

60

70

80

90

100

110

Liquid limit, LL

Figure 16(b) Plasticity chart for classification of fine-grained soils and fine-grained fraction of coarse-grained soils. (Test results relate to soil fraction finer than 0.42 mm.) (Copyright ASTM INTERNATIONAL. Reprinted with permission.)

100

101

>12% fines

5–12% fines

12% fines

5–12% fines

Cu < 6 and/or 1 > Cc > 3

Cu ≥ 6 and 1 ≤ Cc ≤ 3

Cu < 6 and/or 1 > Cc > 3

Cu ≥ 6 and 1 ≤ Cc ≤ 3

Cu < 4 and/or 1 > Cc > 3

Cu ≥ 4 and 1 ≤ Cc ≤ 3

Cu < 4 and/or 1 > Cc > 3

GP-GC

GM GC GC-GM

fines = CL, CH, (or CL-ML) fines = ML or MH fines = CL or CH fines = CL-ML

SP-SM SP-SC

SM SC SC-SM

fines = CL, CH, (or CL-ML) fines = ML or MH fines = CL or CH fines = CL-ML

SW-SC

fines = CL, CH, (or CL-ML) fines = ML or MH

SW-SM

fines = ML or MH

SP

SW

GP-GM

GW-GC

fines = CL, CH, (or CL-ML) fines = ML or MH

GW-GM

fines = ML or MH

GP

GW

GROUP SYMBOL

11 - m2, where E is the soil modulus of elasticity and µ is the Poisson’s ratio. Material index, ID = ¢p>1p0 - u02, where u0 is the pore water pressure at test depth. Horizontal stress index, KD = 1p0 - u02>sv, where svo is the vertical effective stress. Relationships using these indices to obtain soil properties important for design have evolved from the developers and users of the equipment. Table 3 presents a summary of soil parameters that can be obtained from the dilatometer procedure. Experiences indicate that

Depth, m

Material Z index (m) .1 5 10 0 .5 1 0 0 CLAY SILT SAND 2 2

Constrained Undrained Z Z modulus (m) shear strength (m) 400 800 0 .5 1 1.5 2 2.5 0 0 0 2

2

4

4

4

4

6

6

6

6

8

8

8

8

10

10

10

10

12

12

12

12

14

14

14

14

16

16

16

16

18

18

18

18

20

20

20

20

22

22

22

22

24

24

24

24

.6 1.8 ID

400 800 M (bar)*

* 1 bar = 1 kgf/cm2 ≈ 100 kPa.

Figure 32

170

Illustrations, DMT results [365, 220].

0 .5 1 1.5 2 2.5 0 Cu (bar)*

Horizontal stress index 3 6 9 12

3

6 9 12 KD

171

Corrected first reading (DMT list)

Corrected second reading (DMT list) Material index Horizontal stress index Dilatometer modulus

Coefficient, earth pressure in situ

Overconsolidation ratio

Undrained shear strength

Friction angle

Coefficient of consolidation Coefficient of permeability Unit weight and description Vertical drained constrained modulus

Equilibrium pore pressure

p0

p1 ID KD ED

K0

OCR

cu

ϕ

ch kh γ M

u0

RM = 0.14 + 2.36 log KD RM = 0.5 + 2 log KD

u0 L p2 L PC - ZM + ¢A

if 0.6 6 ID 6 3 RM = RM,0 + 12.5 - RM,02 log KD with RM,0 = 0.14 + 0.151lD - 0.62 if KD 7 10 RM = 0.32 + 2.18 log KD if RM 6 0.85 set RM = 0.85

if ID … 0.6 if ID Ú 3

ChDMTA = 7 cm2>Tflex k h = c hgw>Mh1Mh L K0MDMT2 (see Figure 33) MDMT = RMED

wsafe,DMT = 28° + 14.6 log KD - 2.1 log2 KD

cu, DMT = 0.22 svo10.5

KD21.25

OCRDMT = 10.5 KD21.56

K0,DMT = 1KD>1.520.47 - 0.6

p1 = B - ZM ¢B ID = 1p1 - p02>1p0 - u02 (dimensionless) KD = 1p0 - u02>svo (dimensionless) ED = 34.71p1 - p02 (units as for stress)

p0 = 1.051PA - ZM + ¢A2 - 0.051PB - ZM - ¢B2

In freely draining soils

Tflex from A-log t DMTA-decay curve

for ID 7 1.8

for ID 6 1.2

for ID 6 1.2 (not for sand)*.

u0 = pre-insertion pore pressure svo = pre-insertion overburden effective stress ED is not a Young’s modulus E. ED should be used only after combining it with KD (stress history). First obtain MDMT = RMED [can estimate E L 0.8 MDMT] for ID 6 1.2 (not for sand)*.

ZM = Gauge reading when vented to atm. If ΔA and ΔB are measured with the same gauge used for current readings PA, PB then set ZM = 0 (ZM is compensated)

Basic DMT Reduction Formulae

*For

sands (except cemented sands), approximate values for K0 and OCR can be obtained from K0 = 0.376 + 0.095 KD - 0.00461qc >svo2 where qc is the soil resistance value from the cone y K0 1 d where y = 1>10.8 sinax2 fax = fps - c 1fps - 322 d . Note that φax is the drained penetrometer test (CPT) and svo is the effective vertical stress at the test depth, OCR = c 1 - sin fax 3 axisymmetrical angle of internal friction determined from the plane-strain dilatometer test value for angle of internal friction, φps, as obtained for the direct shear test, refer to Chapter 11.

Description (test property, soil property)

Symbol

Table 3 Soil Parameters and Related DMT Formulae [365, 220]

Site Investigations

the best applications of the dilatometer are to define the M and cu profiles of a soil formation, predict foundation settlements, identify soil type (classification), differentiate permeable from low-permeability zones, and identify thin layers of weak material embedded within otherwise firm formations. Figure 33 is used to relate dilatometer results and soil classification. p

p

Clay

Sand

p1

p0

p1

p0

Comparitive results of DMT related to Id: [ID = (pi – p0)/(p0 – u0)] Resulting graphical scale developed on basis of experiences (ID is a very sensitive parameter) ID = 0.1

0.6

1.8

Clay

10 (log scale)

Silt

Sand

(a) Basis for soil classification chart.

2000

2000 Sand 5

2.1

1000

500

2

Dilatometer modulus ED (bar) (1 bar = 1 kgf/cm2 ≈ 100 kPa)

5

1.9

1.8

1.8

1.8

1.7

1.7

*

1.6

1.9

5

2.0

200

200

1.9

100

50

500

2.1

Clay Silty

1000

Silty

Sandy

Clayey

Silt

100

D

Indicates approximate value for γ soil /γ water; can use to estimate γ soil.

C

1.7 B

*

1.6

20

20

A

Mud and/or peat

12 10

0.33 Mud

0.8

0.2

3.3

0.5

1

Material index ID (b) Soil classification chart.

Figure 33

172

10

1.8

(*) If Pl>50, reduce γ ratio by 0.1

1.5 5 0.1

1.2

0.6

Soil type related to DMT parameters [220].

2

5

5

Site Investigations

Figure 34 Menard pressuremeter equipment (borehole probe and surface control unit). (Courtesy of Roctest Ltd., St. Lambert, QC, Canada)

The pressuremeter is a device that includes a cylindrical probe that can be pressurized and expanded after it has been inserted to test depth in a borehole (see Figure 34). This equipment differs from the flat-blade dilatometer in that usually a borehole must be provided before the pressuremeter probe can be inserted (but some pressuremeters are a selfboring type with capability for advancing through fine-grained soils); the dilatometer can be advanced to the test depth by driving or jacking. Probe pressures are increased in measured increments, and corresponding volume changes are recorded. Correlations between pressure and volume changes are related to soil parameters.

6

Groundwater, Soil Water, and Soil Gas Sampling Obtaining samples of groundwater and soil water is an important phase of any program to monitor or protect a subsurface environment. If a suspected liquid seeps at the ground surface, say from the face of an earth slope or from an artesian condition, or from an open pit,

173

Site Investigations

samples can easily be obtained. Although any procedure that permits a sampling of ground liquids can be appropriate, monitoring and observation wells are a commonly used method to obtain water samples that permit checking for the presence of contaminants (Figure 35(a)). Monitoring and observation wells are typically constructed using one of the procedures for making subsurface borings and installing water supply wells, whereby manual excavation procedures or procedures involving specialty boring or drilling equipment are used (Figure 35(b)). Additionally, the well may require a protective casing to protect the hole from cave-in. The selection of materials must be such that reactions between the groundwater and a well casing, or subsequently used sampling equipment, are prevented. A difference between a monitoring well (or observation well) and a conventional cased-boring or water supply well is the necessity that a perforated casing (perforated with screening or thin slots) be used at the depth where the groundwater sample is to enter into the casing (Figure 35(c)). Water samples can be obtained from an open well or open excavation such as a test pit, but the depth that is the source of the contaminated water may not be identified. If a water sample from a particular depth in the well is desired, methods to seal the well below and above the test zone will be necessary. Water samples can be obtained from different depths by using a cluster of individual wells with each extending to the desired depth, or by installing a multiple of different-length casings in a single large-diameter boring (Figure 36). Water samples may be obtained from locations below the groundwater table or from the vadose zone (above the water table). Procedures and equipment needed for obtaining samples above the water table are different from the procedures and equipment for sampling below the water table. Below the water table, groundwater tends to flow into the well opening because of gravity effects, but above the water table it is common to use suction or vacuum methods to draw water held in soil voids by capillary and adsorption forces. Wells typically range between 50 mm and 150 mm (2 in. and 6 in.) in diameter. The smaller-diameter wells are preferred where the groundwater flow rate is limited, because a smaller volume of water inflow is necessary to obtain adequate samples. The minimum well size is usually controlled by the size of pumping and sampling equipment planned for the well. The installation of a monitoring well usually represents an intrusion into previously undisturbed natural earth. The wall area of the well excavation is disturbed because of the smearing disturbance created during the installation, a factor that can later interfere with the flow of liquids into the completed well. The related passage of contaminants into the well could be restricted, an undesirable condition if the presence and concentration is to be determined. The performance of a well, and information obtained from water sampling, can be affected by the skill and care of the well installers. Following the installation, it is usually necessary to clear the new well of the water used for the construction and of the residue sediment accumulated during the installation, to ensure that subsequently obtained water samples are true groundwater samples unmixed with other liquids. A surging and flushing process is used whereby water is forced into the well, then removed; this procedure clears sediment and accumulated water and also helps clear the soil smear along the well wall. Where the well is to be used to obtain groundwater samples over a period of time, as is typical, the top of the well is capped and the surficial soil zone surrounding the well is sealed, to prevent surface water from entering the well and surrounding earth.

174

Upper protective steel casing and hinged cover (lockable)

Well cap

Concrete Ground surface

Outside diameter of drilled hole Cement grout or low-permeability sealer surrounds pipe casing

(b) Installation of observation-monitoring well at a gasoline station, to check for leaks from buried storage tanks

Well casing has flush joints

Riser pipe casing (steel, PVC, etc.) Clay bentonite seal Fine sand (seal)

Well screen Coarse sand or sand-gravel packing around well pipe casing screen section

(c) Perforated well casing; small slotted openings in lower sections are not visible in photograph

Cap or plate covering end (a) Schematic—typical monitoring well, screened over one vertical interval (compiled from various sources)

(d) Removable cover for observation-monitoring well, to permit periodic sampling

Figure 35

Observation-monitoring well information.

175

Site Investigations Well casings (tops capped)

Well casings (tops capped) Ground surface

Clay bentonite or other impermeable sealer (typical)

Soil backfill or cement grout (low permeability)

Sand or sand-gravel packing around riser screen (typical)

(a) Individual wells each to desired different depth

(b) Single large-diameter drilled hole, but individual well casings to desired depths

Figure 36 Well cluster to monitor different depths.

The locations selected for monitoring wells are usually adjacent to an area suspected of, or being checked for, escaping contaminants (for example, monitoring the boundary area of a waste dump) or are surrounding a location being protected (such as water supply sources or wells) (Figure 37). Tracers placed into suspected sources of contamination can expedite the process to determine if nearby water well supplies are being affected. Where more than one location is the possible source of groundwater contamination, tracers are most appropriate; a different tracer material is added at each suspected source, and the type of tracer(s) subsequently found at the monitor wells establishes the link to the proper source. Tracers can be any type of matter—liquid, solid, or gaseous—which can be transferred into the ground at or near a suspected contamination source and flow with characteristics similar to groundwater and the suspected contaminant. The tracer should not be a hazardous material or pose a danger to the environment (e.g., a radioactive material). The methods required to recover samples of the tracer reaching the monitoring points and then to perform analysis to indicate a presence and concentration should be simple. Dyes are commonly used. Water at an elevated temperature can be used if flow distance and time are not great. Solid particles, yeast, some types of bacteria, ionic compounds, liquids that retain their identity, and inert gases can also be used. Radionuclides (radioactive isotopes)

176

Site Investigations Waste dump, landfill, etc.

Ground surface

Landfill site

Monitoring wells for surface water supply Water supply, surface body of water

Groundwater table Groundwater flow Monitoring wells, typically to different depths

(a) Monitoring wells in and surrounding location of contaminant source, to determine if contaminants leak into the surrounding earth

Figure 37

(b) Monitoring wells (observation wells) located in region surrounding a water supply, to detect movement of contaminants into the water supply

Concept of locations for monitoring wells.

function well as tracers, but their use is questionable and may be regulated because of the potential dangers. Water sampling equipment, the equipment used to “grab” a sample of water from a known depth inside the well casing, ranges from very basic, manually operated items to sophisticated, electrical-mechanical-pneumatically operated equipment. The simplest, bailers, are used for open wells that are screened over one vertical interval (Figure 38(a)). Bailers are basically open cups or sections of open tubing or piping provided with ball-type check valves. The bailer is lowered to the desired water level depth in the well using a cord line, then withdrawn to the surface. The cup bailer is a container with the top open. Check-valve bailers have a ball-and-seat check valve at the bottom of the bailer tube, or at both top and bottom. With the single, bottom-check-valve-type bailer, the check valve remains open as the bailer is lowered into the water, but closes when the bailer is withdrawn because of the downward weight of the water inside the bailer. The doublecheck-valve bailer (check valve at both bottom and top of the bailer) functions much the same; both check valves are in the open position as the bailer is being lowered, but both close as the water sample is being raised to the surface. The syringe-type sampler illustrated in Figure 38(b) obtains the water sample after being lowered to the desired depth in the well by having a vacuum or negative pressure induced in the container via a tube connected to a pump at the ground surface. Water pressure in the well, in excess of the reduced pressure inside the syringe container, forces water through the syringe needle. The syringe sample is brought up to the surface, where the tip is sealed so the unit can be used to store and transport the water sample. Wellwater samples can also be obtained by using negative pressure displacement (suction lift) and positive displacement methods. The negative (suction) pressure method, depicted by the schematic diagram of Figure 38(c), involves use of pump equipment to create a reduced pressure (vacuum pressure) in a container which has a tube extending into

177

To pressure-vacuum pump at ground surface

Line for lowering and lifting

Bailer open at top

Flexible vinyl tubing

Tubing of inert material

Ballast (weight to help sink unit) Threaded connecting tube

Sealing washers Syringe container

Syringe plunger (moves upward as vacuum is applied) Check valve—ball-and-seat type Syringe needle (water sample enters when vacuum is applied)

(a) Single check-valve-type bailer (b) Vacuum-syringe-type sample

Gas entry tubes to pressurize container when installing Flexible tubing Reduced (suction) pressure created in tubing

Gas supply tube when water sample is being forced to surface

Tubing

Outlet

Well casing

Vacuum pump Atmospheric and hydrostatic pressure on water forces sample up tubing

Sample container Tube used to accept and carry water sample to ground surface Ball-and-seat check valve

Sample collection container Slotted well screen (water entry)

(c) Concept of negative (suction) pressure lift-type sampler (d) Simple gas-driven water sampler

Figure 38 Illustrations of common monitor wellwater samplers.

178

Site Investigations

the wellwater at the desired sampling depth. Atmospheric and hydrostatic pressure acts on water in the well, then forces the water sample up the tube into the collection container. The height of lift relates to the vacuum pressure which the pump can develop, but typically is limited to about 7 m, or 20 ft. This equipment can also be used to obtain samples of gas that lies in the well above the water level. The positive displacement methods involve use of conventional-type well-pumping equipment (piston-type, submersible, venturi-jet, etc.), either placed within the well or operating from the ground surface. Gas-driven water samplers are represented by the simple-in-principle device illustrated in Figure 38(d). More sophisticated variations exist. To operate the sampler shown, a positive gas pressure is first created in the sampler container using a pressurized supply at the surface, to hold the check valve closed. At the desired sampling depth, the gas pressure is released, and water flows into the sampler. A positive gas pressure is again imposed, a procedure which both causes the check valve to shut and forces the water sample up the discharge tube to be collected at the ground surface. To obtain samples of soil water in the zone above the water table (the vadose zone), suction equipment is necessary to free water held in the void spaces by capillary and adsorption forces (this soil moisture will not flow into a sampler by the pull of gravity). The concept of a sampler commonly used for the vadose zone, the pressure-vacuum lysimeter, is illustrated in Figure 39. This type of sample collector consists of a long container, usually of a rigid plastic material such as PVC, having a cup base made of porous ceramic. In use,

2-way pump

Sample bottle

Ground surface Soil backfill

Pressurevacuum inlet tube

Bentonite seal

Sample collector (plastic pipe 24 in. or 608 mm long, 2 in. or 51 mm dia)

Porous cup (typically ceramic)

Figure 39

Discharge tube (copper or flexible plastic)

Borehole (6 in. or 150 mm)

Sand backfill

Bentonite seal

Schematic representation of pressure-vacuum lysimeter.

179

Site Investigations

the sampler is sealed in place at the desired depth in a boring, so a vacuum will be effective for drawing soil water inward through the porous cup. The lysimeter is provided with two tubes that will extend to the ground surface for connection to a pressure-vacuum pump and a sample holder bottle. To obtain a soil water sample, the discharge tube is closed (or clamped) and the pump is used to create a vacuum in the sampler. The difference between liquid and air pressure in the soil voids and the reduced (vacuum) pressure in the sampler forces a flow through the porous ceramic cup. The collected water sample is then delivered to the surface sample bottle by the procedure of opening (unclamping) the discharge tube and using the pump to deliver air or gas into the sampler at a pressure sufficient to force an upward flow. The method is capable of obtaining soil water samples from locations relatively deep in a monitoring well, but the pressures induced to recover the collected water sample cannot be too great because of the possibility of forcing the water back out through the ceramic cup into the surrounding earth. Soil gas samples can be obtained using grab sampling techniques or by passive sampling. Passive sampling involves placing (burying) a sorbent material, such as sorbent charcoal, in the soil zone being monitored for a period of time (days or weeks, typically). Contaminants diffused through the soil gases are sorbed (trapped) on the charcoal. The retrieved charcoal sampler can then be delivered to the laboratory, where desorption and chemical analysis are performed. Grab sampling refers to the procedure for obtaining a small volume of the soil gas that is present at the time of sampling. Grab samples are classified as static (the sample is obtained from a more or less immobile body of gas) or dynamic (the sample is obtained from an actively moving volume). The principle for obtaining grab samples is relatively simple—a small probe is inserted to the depth or the location zone to be tested, then suction equipment, such as a hand pump or bellows or a mechanical vacuum pump, extracts the desired air–gas sample. The penetrating probe may be as small as a hypodermic needle (for shallow depth penetrations) or made of heavy tubing (necessary for deeper penetration). Soil gas samples can also be obtained from borings or wells using the lysimeter apparatus shown in Figure 39 as well as by working with soil samples extracted from borings (Figure 40(a), (b)). It is possible to obtain samples of soil gases that reach the ground surface where open observation wells exist (for example, where solid waste fill sites are being monitored). Figure 40(c) shows one type of soil gas sampling probe that can be installed by driving or insertion in a bored hole. The BAT probe (after Bengt Arne Torstensson, developer) represents another type of apparatus used to obtain samples of soil water/gases in boreholes. The probe type illustrated in Figure 41 was developed for use in offshore exploration boreholes, drilled in deep waters as undertaken for commercial drilling ventures where the detection of potentially explosive gases in the marine sediments and underlying material is important to planning for the safety of personnel and equipment. Soil gas analysis methods are particularly effective for determining the presence of volatile organic compounds (VOCs), products commonly escaping from industrial waste fill and solid waste landfills, industrial sites, and so on. Grab samples are often analyzed at the job site in mobile laboratories or with portable instruments (some small enough to be hand held), but the gas samples can also be stored in small glass or stainless steel containers for transportation to a conventional laboratory for testing.

180

Site Investigations

(a) Installation of monitoring well at an industrial site. Drilling crews are wearing protective suits

(b) Soil sampling at monitoring well installation, to check for contaminants

Syringe (to extract gas within tubing) Valve to deliver collected gas to holding container Tenax GC trap Pounding plate (to install pipe)

(d) PID instrument to determine presence of soil gas as monitoring well is being installed Driving pipe Inside tubing Coupling Air holes Driving point

(c) Soil gas sampling probe

Figure 40 Observation-monitoring well installation and sampling equipment. (Photographs courtesy of Atlantic Testing Laboratories, Ltd., Canton, New York)

The handheld analyzers are typically limited to indicating gross levels of gases, or total volatiles. One type, the photoionization detector (PID) equipment, is small, reads quickly, and is commonly used to indicate or check for certain components of the volatile compounds (Figure 40(d)). For identification of individual components, field or fixed

181

Site Investigations Connection to surface equipment (barge, work platform, etc.)

Motor, transmitter, and thermometer wires

∅30 mm

Motor Gear-box transmitter

∅42 mm

Thermometer chip Ball-screw

1350 mm

Transmitter wires

Pressure transmitter

Rubber discs Spring and double-ended needle unit

Main body

40 mm

Container retains water/gas sample (capacity 36 cm3 or 155 cm3)

∅22 mm

40 mm

Variable

Filter

Figure 41

∅30 mm

Conical tip

Offshore BAT probe [286].

laboratory gas chromatography (GC) equipment is necessary, whereas flame ionization detection (FID) is used to indicate concentrations. The electron capture detector (ECD) is used to analyze chlorinated hydrocarbons. Field GC and FID information is usually not as detailed as the results from a conventional laboratory.

182

Site Investigations

A practical field procedure available for determining the presence of particular gaseous compounds involves the use of gas detector tubes. A large number of different detector tubes, each for covering a different kind of gas, are available from the marketplace. The detector tube is filled with an appropriate solid reagent. When a measured volume of soil gas is passed through the tube, a color change indicates the presence of a particular compound. The types of detector tubes selected relate to the types of gases that need to be detected or investigated. If the tubes are graduated, the concentration is also indicated.

7

Geophysical Methods The determination of subsurface materials through the use of borings and test pits can be time consuming and expensive. Considerable interpolation between checked locations is normally required to arrive at an areawide indication of conditions. Geophysical methods involve the technique of determining underground materials by measuring some physical property of the material and, through correlations, using the obtained values for identification. Most geophysical methods determine conditions over a sizable distance. Frequently, this is an advantage over the “point” checking accomplished by borings and test pits. Most geophysical measurements can be rapidly obtained. Thus, the methods lend themselves well to the checking of large areas. In the engineering–construction profession, several types of geophysical investigation have been found useful: the seismic refraction method, the electrical resistivity method, and ground-penetrating radar. Though these methods have proven to be reliable, there are also certain limitations as to the data that can be obtained. Thus, at the present time, subsurface investigations can rely heavily on geophysical methods, but conditions should at least be spot-checked with borings or test pits. Typically, when a thorough investigation is made, a number of borings will be required in order to obtain test samples to make accurate determinations of soil properties, such as strength and compressibility. It is these borings that can provide the detail required to check and complement the geophysical data.

Seismic Refraction When a shock or impact is made at a point on or in the earth, the resulting seismic (shock or sound) waves travel through the surrounding soil and rock at speeds relating to the density and bonding characteristics of the material. In refraction seismology, the velocity of seismic waves passing through subsurface soil or rock materials is determined, and the magnitude of the velocity is then utilized to identify the material. A seismograph, the instrument used to make a seismic refraction study, consists of a shock- or impact-inducing mechanism, such as an impact hammer or small explosive, plus a receiver to indicate when the seismic wave reaches a point at a particular known distance from impact as well as a timing instrument for measuring the time for the wave to travel the distance from the point of impact to the point of measurement. In shallow refraction seismology, as used for determining subsurface conditions for construction purposes, the shock impact is created with a

183

Site Investigations

Figure 42

Seismic refraction study being performed in field.

sledgehammer hitting a striking plate placed on the ground (Figure 42). The seismic wave is then picked up by a sensitive geophone. (The geophone is actually a transducer, an electromechanical device that detects vibrations and converts them into electric signals that can be measured.) The field survey involves obtaining a series of geophone readings at different distances along a straight line directed from the impact point. For geophone spacings close to the strike plate, the vibrations picked up by the geophone will be from those direct waves traveling through the upper layer of earth material (Figure 43). For direct waves traveling through the upper layer, the time to reach the geophone is proportional to the distance from the point of impact.

Hammer Earth surface

Figure 43

184

Method of imparting sound waves into a soil.

Geophone

Site Investigations Ground surface

Soil (lower velocity) Rock or other hard material (higher velocity)

Figure 44

Soil Rock

Travel of sound waves through different subsurface materials.

Time

Continuation of direct wave, not observed because of arrival of earlier refracted wave.

Refracted wave, with slope equal to rock velocity.

Direct wave, with slope equal to soil velocity. Horizontal or surface distance, X

Figure 45

Typical travel-time graph for soil overlying rock.

When the surficial layer is underlain by a harder layer, the seismic waves from the strike plate also progress downward and enter the harder layer. The seismic velocity will be greater in the harder material. Waves traveling through the upper portion of the harder layer transfer energy back into the upper layer through the surface of contact. This energy becomes a refracted wave. For large strike-plate-to-geophone distances, the refracted wave will reach the geophone more quickly than the direct wave, even though the path of travel is longer. This occurs because part of the path is through the harder, high-velocity material (Figure 44). Seismic velocities for the earth materials in the upper and lower strata would be obtained from plotting the measured values of geophone distance and time on a travel-time graph (both coordinate values use an arithmetic scale); see Figure 45. The slope of the first segment of the plot represents the seismic velocity for the material in the upper stratum. The slope of the second segment is the seismic velocity for the deeper layer. To obtain data that properly define the plotted segments, it is recommended that the maximum geophone distance be about five times the depth of investigation. Figure 46(a) illustrates a soil layer–rock layer subsurface and indicates the type of seismic wave (direct or refracted) picked up at the various geophone positions. The resulting travel-time graph appears in Figure 46(b). Seismic velocity values representative of different earth materials and conditions are shown in Table 4.

185

Shock point Point of intersection Refracted waves arrive first x

Primary waves arrive first X1 50

100

150

200

250

300

Seismometers (geophones)

400

Soil overburden H1

Bedrock

(a)

Point of intersection X1

x

t

0.060

0.050

Velocity, rock

0.030 T

Time, seconds

0.040

0.020 Velocity, soil overburden 0.010

50

100

150

200

250

300

350

400

Distance, feet or meters (b)

Figure 46 Seismic refraction conditions and data: (a) subsurface conditions showing direct and refracted sound waves; (b) travel-time plot for conditions in (a). (Courtesy of Acker Drill Company)

186

187

Below 3000 800–1500 1500–2000 5000 800–2000 1500–4000 1500–3500 1500–3000 2000–4000

Below 900 250–450 450–600 1500 250–600 450–1200 450–1100 450–900 600–1200

m/sec

Note: Occasionally, formations may yield velocities that lie outside of these ranges. Source: Courtesy of ELE International

Most unconsolidated materials Soil—normal —hard-packed Water Loose sand—above water table —below water table Loose mixed sand and gravel, wet Loose gravel, wet Hard clay

ft/sec

Soil—Unconsolidated Material

Most hard rocks Shale—soft —hard Sandstone—soft —hard Limestone—weathered —hard Basalt Granite and unweathered gneiss Compacted glacial tills, hardpan, cemented gravels Frozen soil Pure ice

Table 4 Representative Seismic Velocity Values (Velocity in ft/sec and m/sec)

Above 8000 4000–7000 6000–10,000 5000–7000 6000–10,000 As low as 4000? 8000–18,000 8000–13,000 10,000–20,000 4000–7000 4000–7000 10,000–12,000

ft/sec

Above 2400 1200–2100 1800–3000 1500–2100 1800–3000 1200? 2400–5500 2400–4000 3000–6000 1200–2100 1200–2100 3000–3700

m/sec

Rock—Consolidated Material

Site Investigations

Where a two-layer condition exists, the thickness or depth of the upper layer can be determined from: H1 =

X1

V2 - V1

2 A V2 + V1

(4)

where H1 = depth or thickness of the upper layer X1 = distance, taken from the travel-time graph, where the two plotted slopes intersect V1,V2 = seismic velocities in the upper and lower layer, respectively For the condition of three successively harder layers existing in an area, the travel-time graph will show three different slopes. The seismic velocity for each of the materials is the slope of the respective segment of the plot. The thickness of the upper layer can be calculated from Equation 4. The thickness of the intermediate layer, H2, can be determined from: H2 = 0.85H1 +

X2

V3 - V2

2 A V3 + V2

(5)

where H1 = thickness of the upper layer X2 = distance from the travel-time graph, where plotted segments 2 and 3 intersect V2,V3 = seismic velocities of layers 2 and 3 as determined from the travel-time graph The use of the seismic refraction method for determining subsurface conditions has certain significant limitations: 1. The method should not be used where a hard layer overlies a softer layer, because there will be no measurable refraction from a deeper soft layer. Refraction seismic test data from such an area would tend to give a single-slope line on the travel-time graph, indicating a deeper layer of uniform material. 2. The method should not be used on an area covered by concrete or asphalt pavement, because these materials will represent a condition of a hard surface over a softer stratum. 3. A frozen surface layer may give results similar to those obtained where a hard layer is over a soft layer because of the velocity increase resulting from the better wave transmission through the more “solid” frozen material. Further, some topographic and underground features will give seismic data that are difficult to interpret fully and correctly. Such situations include the condition of an irregular or dipping underground rock surface, the condition where discontinuities such as rock faults or earth cuts or banks exist, the condition where layers having gradual changes in their velocity values occur, and the condition of thin layers of varying materials. Because of the possibility for misinterpretation of data with such occurrences, the seismic analysis should be performed by trained personnel, and, as a minimum, spot-checks should be made with borings or test pits.

188

Site Investigations Velocity in feet per second ⫻ 1000 0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Topsoil Clay Glacial till Igneous rocks Granite Basalt Trap rock Sedimentary rocks Shale Sandstone Siltstone Claystone Conglomerate Breccia Caliche Limestone Metamorphic rocks Schist Slate Minerals & ores Coal Iron ore

Rippable

Marginal

Non-rippable

Figure 47 Seismic rippability chart developed by Caterpillar Tractor Company for D9 with moulded No. 9 ripper. (Courtesy of ELE International)

Where rock exists at or close to the surface at a construction site, predetermining the methods necessary for removal (to lower the surface elevation) or for excavation becomes important for estimating, bidding, and scheduling purposes. Seismic velocity data have been used to determine when the rock material is capable of being ripped with dozer rippers (or other similar equipment) and when drilling and blasting are required. Such seismic rippability information, as developed by one major construction equipment manufacturer, is shown in Figure 47.

Electrical Resistivity Resistivity is a property possessed by all materials. The electrical resistivity method for determining subsurface conditions utilizes the knowledge that in soil and rock materials, the resistivity values differ sufficiently to permit that property to be used for identification purposes. To determine resistivity at a site, electrical currents are induced into the ground through the use of electrodes. Soil or rock resistivity can then be determined by measuring the change in electrical potential (voltage) between known horizontal distances within the electrical field created by the current electrodes (Figure 48).

189

Site Investigations

Ground surface

Figure 48 Path of current flow (solid lines) through a soil in the electrical resistivity study (dotted lines are equipotential surfaces).

Current meter

Battery

Volt meter

Spacing, A

Figure 49

V Spacing, A

Spacing, A

Wenner configurations for arrangement of voltage and current electrodes.

A schematic diagram of equipment for resistivity testing, including electrode arrangements (the Wenner configuration), is shown in Figure 49. With four electrodes equally spaced along a line, as indicated, the resistivity is calculated from: V b = 2pSa b I

(6)

where b S V I

= resistivity of the earth material in ohm-feet = electrode spacing = difference in potential (volts) between the inner electrodes = current flowing between the outer electrodes

The calculated resistivity value is an apparent resistivity. This is a weighted average of all the earth material within the zone created by the electrodes’ electric field. The depth of material included in the measurement (depth of penetration) is approximately the same as the spacing between electrodes. Two different field procedures are in use for obtaining information on subsurface conditions. One method, electrical profiling, is well suited for establishing boundaries between different underground materials and has a practical application in prospecting for sand and gravel deposits or ore deposits. The second method, electrical sounding, can

190

Site Investigations

Figure 50

Electrical resistivity field study being performed.

provide information on the variation of subsurface conditions with depth and has practical application in indicating layered conditions and approximate thicknesses. It can also provide information on depth to a water table or water-bearing stratum. Figure 50 shows fieldwork for an electrical resistivity study being performed. In the electrical profiling method, an electrode spacing is selected, and this same spacing is used in running different “profile” lines across an area (Figure 51). The information obtained applies for the particular location of the line and is sometimes referred to as an “electrical trench,” implying that the subsurface data revealed are similar to the information that would be obtained by examining an open-trench excavation. The information resulting from a profile line can most simply be plotted on arithmetic coordinates, as shown in Figure 52. A change in the plotted curve indicates a change in

Surface of the earth

A

A

A A 1 Stations:

Figure 51

A A

2

A A A

3

A A A

4

Electrode arrangement in a profiling survey.

191

Site Investigations Apparent resistivity, ρ (ohm-feet) 400

a = 20'

300 200 a = 50'

100

100

200

300

400

500

600

700

800

Position of center of spread (feet)

Figure 52 Apparent resistivity versus position of the center of electrode spread for two values of electrode separation. (Courtesy of ELE International)

The position of the center of the spread is fixed a1 a1

a2

a3

Figure 53

a2

a3

a1

a2

a3

Representative electrode position during a sequence of sounding measurements.

the underground material. From the series of profile lines, boundaries of areas underlain by different materials can be established on a map of the area (areal map). In the electrical sounding method, a center location for the electrodes is selected and a series of resistivity readings is obtained by systematically increasing the electrode spacing, as indicated by Figure 53. Since the depth of information recovered is directly related to electrode spacing, the series of resistivity data obtained from successively increased electrode spacings will indicate changes of resistivity with depth and, hence, information on layering of materials. This method is capable of indicating subsurface variations where a hard layer underlies a soft layer and also the condition of a soft layer underlying a hard layer. Data can be presented on Cartesian (arithmetic-scale) coordinates or logarithmic coordinates, as indicated in Figure 54. Logarithmic coordinates are the most popular. The spacing of electrodes is important. To obtain data that give a reliable indication of conditions, the closest electrode spacing should be no more than half the estimated thickness of the upper stratum. Three feet is the minimum recommended spacing, however. The largest spacing is between 5 and 10 times the total depth of interest. The number of intermediate

192

Site Investigations Apparent resistivity, ρ (ohm-feet)

1000

750

500

250

50

100

150

200

Electrode separation, a (feet) (a) Apparent resistivity, ρ (ohm-feet) 800 600 500 400 300 200

100

2

3

4

5 6 8 10

20

30

40 50 60 80 100

200

Electrode separation, a (feet) (b)

Figure 54 Illustration of two methods for graphing sounding data: (a) Cartesian (arithmetic) coordinates; (b) logarithmic coordinates.

spacings selected should be adequate to provide sufficient points to plot a well-defined curve. Representative sounding curves for some typical subsurface conditions appear in Figure 55.

Correlation between Resistivity and Earth Materials In earth materials, resistivity decreases with increasing water content and increasing salt concentration. Increasing degrees of water content and salinity make it easier for an electrical current to flow through the material. Consequently, nonporous materials (holding

193

Site Investigations Key

Ground surface Low Designates relative High resistivity of layer compared to other layers

Resistivity curve Apparent resistivity or App. resis. Ratio Resis. surface layer

Subsurface condition

ρ

Ground surface Log scales a

Electrode spacing or Elect. spcg. Ratio Thickness top layer

Medium

Layer 1

Low

Layer 2

High

Layer 3 a (5)

Ground surface Higher

Lower

Ground surface

ρ

Layer 1 Layer 2

a

High

Layer 1

Low

Layer 2

Medium

Layer 3

(1)

Higher

a

Ground surface

ρ

Layer 1

ρ

(6)

Ground surface Lower

ρ

Low

Layer 1

Medium

Layer 2

High

Layer 3

ρ

Layer 2 a (2)

Ground surface

Ground surface Medium

Layer 1

ρ

Layer 2 Low

Layer 3

High

Layer 1

Medium

Layer 2

Low

Layer 3

a

(3)

Layer 2

Medium

Layer 3

ρ

Low

Layer 1

High

Layer 2

Low

Layer 3

ρ

a (4)

Figure 55

a

Ground surface

Layer 1

High

ρ

(8)

Ground surface Low

a (7)

a (9)

Representative resistivity curves for differing subsurface conditions.

little water) will have high resistivity values. Such materials include most igneous and metamorphic rock plus some dense sedimentary rock, such as dense limestone and sandstone. In soil materials, clean gravel and sand have a relatively high resistivity value. Silts, clays, and coarse-grained–fine-grained soil mixtures have comparatively low resistivity

194

Site Investigations Table 5 Representative Resistivity Values Types of Materials Wet-to-moist clayey soils Wet-to-moist silty clay and silty soils Moist-to-dry silty and sandy soils Well-fractured to slightly fractured bedrock with moist soil-filled cracks Sand and gravel with silt Slightly fractured bedrock with dry soil-filled cracks; sand and gravel with layers of silt Massive bedded and hard bedrock; coarse dry sand and gravel deposits

Resistivity (ohm-ft) 5–10 10–50 50–500 500–1,000 1,000 1,000–8,000 8,000 +

Source: Courtesy of ELE International

values. Soil formations in nonglaciated areas typically have lower resistivity values than soils in glacial areas. Representative values of resistivity for commonly occurring earth materials are presented in Table 5.

Thickness of Layers A method of obtaining a reliable and accurate determination of the depth to, or thickness of, soil layers by using resistivity data is currently not available. Approximation methods exist, but they are cumbersome to handle and not particularly suited for use in fieldwork. With present knowledge, depth or thickness information is best obtained from borings or by seismic methods.

Ground-Penetrating Radar Ground-penetrating radar (also identified as ground-probing radar) is capable of defining the shallow zones of soil and rock materials that underlie an area. The method relies on the penetration and reflection of high-frequency radio waves (UHF/VHF frequencies). The field equipment commonly includes a radar control unit, transmitting and receiving antennas, and data storage or display devices; for implementation the portable radar and antenna module is moved across the ground surface along a path whose depth profile is to be outlined (Figure 56(a)). Short pulses of high frequency radio waves are transmitted continuously into the earth, and some of the wavelength energy is reflected back to the equipment receiving antenna in accord with the electrical properties of the materials penetrated and encountered. The two-way travel time for the wave transmission and reflection is recorded and converted to a distance (depth), see Figure 56(b). Notable variations in the reflected signal occur where a change (interface) of material exists. Data presentation is typically in the form of a quasi-image of the subsurface (Figure 56(c)). Practically, the method is commonly limited to evaluating depths of 15 meters or less although greater penetrations can be obtained; the deepest readings have been obtained in dry coarse-textured materials such as granular soils, and concrete and dry

195

Site Investigations

rock materials such as granite and limestone. When electrically conductive soils such as wet, fine-grain silt and clay deposits are present, the evaluation is restricted to shallow depths. The most reliable applications of the method have been for delineating sharp contrasts in subsurface conditions (such as between soil and rock) but the method also has been found appropriate for identifying the presence of buried tanks, piping, foundations, and ground cavities (sinkholes).

(a)

Mobile Radar and Antenna Module Radio frequency Transmitter

Receiver

Dipole antenna—converts electrical current to electromagnetic waves

Direction of travel Air gap

Reflected wave Transmitted (refracted) wave Stratum A

Interface depth

Ground surface

Reflector surface (Electrical interface) Stratum B

Two way travel time for electromagnetic (radio frequency) wave from transmitter to be reflected back to receiver antenna is measured and used to calculate interface depth. (b)

Figure 56 Ground-penetrating radar equipment and results: (a) Field equipment. (b) Basic principle of operation. (Ref: Bevan, David, “Cone Penetration Tests (CPT) and Radar Scanning (GPR),” Proceedings of the International Symposium on Cone Penetration Testing, CPT ’95, Swedish Geotechnical Society, SGF Report 3:95, 1995)

196

Site Investigations 0.0

5.00

10.0

Ground Surface

Trench Wall

15.0

20.0

Pipes 25.0

30.0

35.0

40.0 ns

(c)

Figure 56 (continued ) (c) Photo of radar wave data on field CRT monitor, indicating presence of buried metal storage tanks (monitor screen is in color; waves shown here in black and white are actually different colors to enhance recognition of different materials). Top of tanks is about 4 ft (1.3 m) deep. (Courtesy of Geophysical Survey Systems, Inc.)

Problems 1 What publicly available maps serve as good references for obtaining information about surface conditions in an area? 2 List and describe some of the advantages of learning about land surface features from aerial photographs. 3 Suggest a list of procedures that would be practical to implement in order to obtain information on surface and subsurface conditions appropriate for construction planning at a proposed project site that extends over a large area that is presently in a natural, undeveloped state. 4 Outline the general procedure used to drill soil borings and obtain soil samples for classification and testing. 5 In soil borings, what is the standard penetration test? 6 Soil borings are drilled to investigate subsurface conditions at a planned construction site. Soil samples are obtained using the standard

penetration test procedure. The drilling equipment utilizes an automatic hammer for driving the soil sampler. Resistance blow-counts are N = 18 , N = 22 , N = 23 , and N = 19 for samples obtained from the topmost soil stratum. Assume that the blow-count data will be applied to preliminary foundation design using foundation bearing capacity information based on a sample drive hammer efficiency of 60 percent. Modify the blow-count values obtained with the automatic hammer to relate to the N60 data. 7 Compare advantages and disadvantages of test pits versus soil borings for obtaining information on subsurface conditions. 8 Make a comparison between the static cone penetrometer and standard penetration test methods for determining subsurface soil conditions. 9 What are the practical limitations to information provided by soil borings and cone penetration tests?

197

Site Investigations 10 Prepare a borings log from the following infor1 mation, as obtained from a22 -in. boring wher standard split spoon soil samples were taken. Sample Depth (ft)

Blow-Count N

Soil Classification

12–132

3-4-4 7-8-7 8-9-10

17–18 22–23

9-12-11 9-25-27

1 2 1 21 27–282

29-35-39

Brown fine to medium sand Brown fine to medium sand Brown fine to medium sand in upper part of sample; red-brown clay in lower part of sample Red-brown clay Gray fine to coarse sand, silt and clay, occasionally gravel (compact glacial till) Gray fine to coarse sand, silt, clay with gravel (compact glacial till) Refusal (no sample recovered)

1 2–32 1 7–92

1

1

31–312

100

Water level encountered at 21 ft.

11 Referring to the borings log data from Problem 10: (a) Indicate the blow-count value, N, for each of the SPT samples obtained. (b) Determine the depth correction factor, CN, for each of the SPT samples (as determined from text Figure 18), assuming the soil at the site has a unit weight of 110 pcf, then indicate the corrected blow-count for each sample. 12 For coring rock in borings, three types of core barrels are available—the single wall barrel, the fixed double barrel, and the swivel-type double wall barrel. Describe the advantage that the swivel-type double wall barrel offers for obtaining cores from weathered rock that probably includes cracks, fissures, and seams. 13 Conventional borings will be made for a planned industrial building in an area where it is known that a relatively limited thickness of soil (estimated to range between 10 ft and 25 ft deep) overlies rock. For the purpose of planning excavation work for the new structure and estimating the cost, it is important that the depth of the soil overburden be defined, the zone of underlying weathered rock be defined

198

(material that probably could be excavated with conventional digging equipment), and the depth where “sound” rock begins be determined (material excavation that will require special equipment or procedures). The standard split spoon sampler and SPT procedure will be used to obtain samples of the subsurface materials. What SPT blow-count criteria can be applied to define soil, weathered rock, and “sound” rock, and the depths where a change of material occurs? 14 (a) A CPT sounding for a tested location near a planned building location and obtained at a depth of approximately 10 m provides the data indicated. The sounding is 3 m below the groundwater table for the site. Total vertical overburden stress at the test depth is about 160 kPa. The cone “a” factor is 0.80. qc = 7450 kPa fs = 150 kPa cone pore pressure, u2 = 36 kPa Referring to Figure 22, determine the SBT for the test depth. (b) For this same CPT data, use Figure 23 to determine the soil identification type, and indicate the final values for qc,1 and fs,1, including the related exponential components. 15 List and briefly discuss the major advantages and disadvantages of in-place shear tests, such as those performed by the in-place vane shear. 16 Piezometers are installed into a buried stratum of compressible clay to monitor excess pore water pressures during the placement of earth fill in an area where construction of a new highway is taking place. The top of the buried clay layer and the area groundwater table are at a depth of 10 m below the natural ground surface. Initially, the water level reading in the piezometers is at a depth of 10 m. When a 5-m thickness of soil fill is placed over the area, the water level reading in the piezometers rises to an elevation 9 m above the top of the clay layer. What is the stress increase at the level of the clay layer in kPa (i.e., what is the excess pore water pressure that develops)? 17 Dilatometer testing is performed at a planned construction site as part of the subsurface investigation. The dilatometer instrument gauge

Site Investigations indicates pressure in bars. At one location and depth, the corrected dilatometer test pressure readings are: p0 = 5.30 bar 1530 kPa2 , p1 = 11.8 bar 11180 kPa2 , p2 = 0.15 bar 115 kPa2 . For this test location and data, indicate the soil classification and approximate soil unit weight in pcf. (To determine ED, ID, etc., perform calculations using bar as the unit for pressures.) 18 Dilatometer test results obtained as part of a subsurface investigation for a planned construction project indicate the following data for one of the tested locations (note the dilatometer instrument gauge indicates pressure in bars): p0 = 3.0 bars 1300 kPa2 p1 = 8.76 bars 1876 kPa2 p2 = u0 = 0.20 bars 120 kPa2 The soil overburden pressure, svo , at the test depth is 53 kPa (0.53 bar). From this information, determine the soil classification, approximate soil unit weight in pcf, and estimate the soil shear strength angle of internal friction f. (To determine ED, ID, KD, etc. perform calculations using bar as the unit for pressures.) 19 How does obtaining samples of soil groundwater and soil gas relate to diagnosing and monitoring subsurface contamination? 20 For groundwater sampling in monitoring wells, what procedures for installing the well are necessary to ensure that the depth indicated for the water sample is accurate? 21 What is the main difference in the procedure required for obtaining a sample of groundwater from below the groundwater table and from above the water table in a monitoring well?

22 How might samples of soil gas be obtained from a desired depth below the ground surface? 23 (a) Briefly describe the principles on which seismic refraction studies for subsurface explorations are based. (b) Briefly describe the principles on which electrical resistivity studies for subsurface explorations are based. 24 List and briefly describe the type of subsurface information that seismic refraction studies can provide and the limitations on information that can be obtained. 25 List and briefly describe the type of subsurface information that electrical resistivity studies can provide and the limitations to data that can be obtained. 26 A seismic refraction study made for an area provides the following field data: Distance from Impact Point to Geophone (ft) 50 100 200 300 400

Time to Receive Sound Wave (sec) 0.025 0.05 0.10 0.11 0.12

(a) Graph the travel-time data and determine the seismic velocity for the surface layer and the underlying layer. (b) Determine the thickness of the upper layer. (c) Using the seismic velocity information, give the probable earth materials in the two layers.

199

Answers to Selected Problems

14. 17.

200

(a) Soil type 6, clean sand to silty sand (b) Sand mixture ID = 1.26, Sandy silt, ␥ L 120 pcf

18. 26.

ID = 2.06, KD = 5.28, Silty sand, ␥ L 120 pcf (b) 82 ft.

Movement of Water Through Soil Basic Hydrogeology, Subsurface Flow, Permeability, Capillarity

Water is one of the most abundant substances found on this planet. Water can be found in any of the three phases of matter—gas, liquid, or solid—in what are considered normal conditions of nature. Water is a relatively stable material, but it also combines easily with many other substances and permits other solids, liquids, and gases to mix easily into solution. Water has had a profound effect on the development of the planet, and it is essential for many forms of life. Water is present on the surface of the planet, below the surface, and in the atmosphere. Close to three-fourths of the earth’s surface is covered by water. Of the planet’s total quantity of water, approximately 97 percent is saltwater, as found in the oceans. The remaining water is classified as “fresh water,” even though other materials may be present in solution. Almost three-fourths of the fresh water volume is frozen, locked up in the planet’s polar ice caps and glacial ice. Less than 1 percent of the planet’s water is mobile, fresh water, and of that amount more than half is very deep subsurface water (not readily accessible for use at and near the surface). Remarkably, then, much of today’s continental topography and plant and animal life, including some marine life, are the result of functioning with approximately one-half of 1 percent of the planet’s supply of water. Nevertheless, the volume of fresh water at, near, and above the earth’s surface—that volume influential and accessible to humans—is considerable (see Table 1). Most earth scientists believe that the total volume of water on the planet has been relatively constant from the time the earth reached its present form (i.e., having solid land masses capable of supporting life, the various ocean areas, and an atmosphere). Some of the present-day water supply is lost through the atmosphere when photodissociation occurs (the separation of hydrogen and oxygen molecules, with the hydrogens escaping into space), but new water (juvenile water) is also created when gases and liquids escape from the depths of the earth. Surface and near-surface water can be very mobile, however, in form and in location, and the volume in any area is very susceptible to change, even over From Essentials of Soil Mechanics and Foundations: Basic Geotechnics, Seventh Edition. David F. McCarthy. Copyright © 2007 by Pearson Education, Inc. Published by Pearson Prentice Hall. All rights reserved.

201

Movement of Water Through Soil Table 1 Earth’s Water: Estimated Quantity and Distribution Location and Type

Volume (km3)

Percent of Total

Oceans (saltwater) Ice caps, glaciers (fresh water) Lakes, rivers, other land surface bodies, fresh Groundwater, fresh (shallow, zone less than 0.8 km deep) Groundwater, fresh (deeper than 0.8 km) Atmosphere

1,300,000,000 29,000,000 125,000 4,000,000 5,000,000 13,000

97 + 2+

less than 1

short periods of time. Natural forces, including the activities of animal and plant life, cause water to move. Humans use large volumes of water in their activities, but as with the effects of other animal and plant life, the usage of such water is typically a “borrowing” or “a relocation” instead of a “using up.” An important issue is that moving water affects the properties and behavior of soil and therefore can influence both construction operations and the performance of completed construction. Since groundwater conditions are frequently encountered on construction projects, those in the construction profession have found it necessary to understand how movement of water through soil can occur and its possible effects. The discussion on permeability (hydraulic conductivity) and capillary action in soil in this chapter relates to both the type and manner of water movement.

1

Basic Hydrogeology Subsurface Flow, Basic Facts The cycle of changes and movements that surface water and shallow groundwater repeatedly passes through is referred to as the hydrologic cycle. The major features of this cycle, indicating typical phases of movement, use, and recycling, are depicted by Figure 1. Hydrogeology is the study of the groundwater phase of the hydrologic cycle (both surface and subsurface) related to the effects of geophysical features of soil and rock formations. Precipitation reaching the earth’s surface tends to either flow over the ground surface as dictated by gravity or to infiltrate into the ground. Precipitation may infiltrate after it has traveled some distance as surface water. The features of nature that influence the rate of infiltration include the slope of the land, the presence or lack of plant life, and the porosity of the soil or rock. Where soil is present, water seeping into the ground may be prevented from continuing movement by factors of attraction, such as capillarity and adsorption, or may be taken into plant root systems. Infiltrating water that does continue to migrate due to gravity effects is drawn to the underground zone where theoretically all of the pore or void spaces in the soil are filled with water. The surface or upper boundary of this saturatedwith-water zone is identified as the water table or phreatic surface. The water present below the phreatic surface creates a positive hydrostatic pressure so that a total of air plus water pressure is greater than atmospheric pressure. The soil zone above the phreatic

202

Movement of Water Through Soil

Precipitation (rain, snow)

Transpiration (water converted or evaporated by plant life)

Su In

rfac e flow wate

lan

Water infiltration or percolation (becomes part of subsurface groundwater supply)

r

Evaporation

ds

urf ace

Groundwater absorbed by plant roots

Evaporation

Creek, surface water taken for low volume Rivers, streams (natural supply collectors of water and subsequent routes of flow)

Subsurface flow

Evaporation

Well Waste water returned to ground

Reservoirs, oceans, lakes (large, collectors and holders of water)

Infiltration Well water supply Infiltration

Figure 1 Hydrologic cycle.

surface includes air or other gas and is partially saturated with water classified as soil moisture (water held by capillary and adsorption forces); this zone is designated the vadose zone. The pore water held in this partially saturated zone above the phreatic surface is at negative pressure so that a total pressure from air plus water is less than atmospheric. In scientific terms, the identification of the water table or phreatic surface is where the groundwater pressure changes from a negative value to a positive value; in other words, where pore water pressure is zero. In general, the position of the water table tends to reflect the surface topography of the area, being at a higher elevation where the ground surface is high and at a lower elevation where the ground surface is low (Figure 2). (Commonly, for some geographical areas, such as glacially affected ones, the relative elevations of a soil surface are also indicative of the boundary of the underlying rock surface.) However, it is also known that the depth or elevation of the groundwater table will vary, even over relatively short horizontal distances. The depth to groundwater can vary seasonally, and from year to year. The presence of groundwater in a soil deposit is affected by factors such as the characteristics of the various soil strata in the soil profile; the depth and type of underlying rock; the elevation of the soil surface related to the surrounding terrain; surface water coming to the area; and the presence of subsurface water in the adjacent soil and rock deposits. The groundwater in an area may be either relatively stationary or mobile (that is, underground flow is occurring). The case of a stationary volume of groundwater occurs

203

Movement of Water Through Soil Vadose zone above water table, voids partially filled with water Ground surface

er Wat

Marsh area (water table at or near soil surface)

table

Stream Below water table all pores or voids between solid particles filled with water

Figure 2 The position of the water table generally conforms to the land surface.

where an equilibrium condition has developed, such as where the phreatic surface is at similar elevation over a large area or when adjacent to a large body of water, or where the top boundary of a subsurface soil or rock layer restricts the entry of water and creates a buried basin in which the collected groundwater is prevented from escaping via movement through the buried soil or rock. (This condition creates what is classified as a perched water table; the phreatic surface for the groundwater trapped in the buried basin is positioned or “perched” above the region’s true water table.) In contrast, where strata of soil or rock have porous characteristics that permit the flow of liquid, the groundwater tends to be mobile, moving primarily under the effect of gravity forces. Where the elevation of the groundwater table measured at two different locations is different and the earth material between the two locations is porous (that is, permits the passage of water), flow occurs in the direction from the high elevation toward the lower. Physical conditions responsible for such flow are indicated by the Bernoulli Theorem for a steady condition of laminar flow involving an incompressible liquid such as water. Mathematically, the Bernoulli Theorem expresses the energy head possessed by a body of water at any position as:

h =

p n2 + + z g 2g

(1a)

where h = energy head, based on total or absolute hydraulic head, possessed by a body of water at an elevated location (units are length, L) n = flow velocity 1L>T2 g = acceleration of gravity 1L>T 22 p = fluid pressure at any point in the body of water 1F>L22 g = unit weight of the water 1F>L32 z = a gravitational position potential head indicating the vertical distance between an arbitrary reference datum and the point where pressure p is measured (L) The flow of water between two positions through soil and rock occurs because of the energy head difference between those positions. For flow through soil, where velocity typically is low and can be neglected in computations without incurring significant error, the

204

Movement of Water Through Soil

energy head difference responsible for flow between location 1 and location 2 becomes (Figure 3): h1 - h2 = a

p1 g

+ z1 b - a

p2

+ z2 b

g

(1b)1

The pressure difference, Δp, causing flow to continue between the two locations, becomes, approximately: ¢p = g1h1 - h221approximately2

(2)

The underground flow of water through a soil deposit can occur only where the soil material is porous enough to permit continuous passage. Some soil types, such as coarse granular materials, are more porous than other types, such as the fine-grained silt and clay soils (see Section 2 on permeability, a property also referred to as hydraulic conductivity). Where a porous stratum of soil (typically a coarse-grained or granular soil) lies below an area’s groundwater table or is in contact with an elevated source of water, such as a surface body (lake, river, etc.), so that flow in significant quantity can or does occur, that stratum is classified as an aquifer or transmitter of water. Aquifers typically are considered to be a good source for obtaining subsurface water (when installing water supply wells, etc.). Soil strata that restrict or prevent the flow of water, usually because of the fine-grained composition and related absence of large voids necessary to achieve rapid passage, are Ground surface

Total distance = h1

Energy head due to p1/γ 1

H2O particle, subject to total energy head, h1

Energy head due to z1

Wate rt

Flow

able (p

hreatic

h1-h2 surface)

Total distance = h2

Direction of flow

Energy head due to p2/γ H2O particle, subject to total energy head, h2 Energy head due to z2 2

Arbitrary elevation selected as reference datum

Figure 3 Description of conditions for the Bernoulli Theorem of flow.

1The

Bernoulli Theorem was developed from the concept that the total energy (kinetic plus potential energy) of a unit of flowing water remains constant between position 1 and position 2 (i.e., assuming no energy losses occur). For water flowing through a soil or rock formation, however, energy is used in overcoming resistance to flow caused by friction, etc., which is one reason a steady state of flow between locations while having water levels at different elevations can represent a stable condition.

205

Movement of Water Through Soil

classified as aquitards or preventors of flow. Clay-type soils and, to a lesser extent, silt–soil mixtures function as aquitards. Aquifers can be classified as unconfined, confined, or artesian (Figure 4). The unconfined aquifer typically exists where the porous soil (a granular soil) constitutes the surface stratum. The position for the water table is within the stratum. However, flow through the stratum would not be prevented from reaching or breaking through to the surface should volume and pressure become great enough. The confined aquifer condition results where the porous stratum is bounded above and below by an aquitard. The confinement created by the aquitards prevents flow within the aquifer from escaping, unless an outlet exists. The position of the water table may be within the aquifer, or the pressure (energy head) developed by the flow may be great enough to force water entering a well that is tapped into the aquifer to rise above its top boundary. Where the energy head for water flowing within a confined aquifer is sufficient to force water to rise above the top boundary, an artesian condition exists. The artesian well, identified by the phenomenon of water flowing from the top of the well without a need for mechanical pumping, actually illustrates the condition where the energy head for flow in a confined aquifer is great enough to raise water to a height above the ground surface. As described previously, the position of an area’s groundwater table can fluctuate seasonally, and from year to year, as influenced by precipitation and weather, the volume of groundwater drawn off by users, etc. It is also recognized that changes in topography, natural or related to humans, can be responsible for changes in groundwater, as illustrated by Figure 5.

2

Permeability (Hydraulic Conductivity) Soil, being a particulate material, has many pore or void spaces existing between the solid grains because of the irregular shape of the individual particles. In a mass of particles that are rounded and roughly equidimensional in shape, such as the gravels, sands, and silts, or are platey or flakelike, such as clays, the pore spaces are interconnected. Fluids (and gases) can travel or flow through the pore spaces in the soil. Thus, soil deposits are porous, and the material is considered a permeable material. It should be realized that flow is occurring through the void spaces between particles and not actually through the particles themselves (Figure 6).

Factors Affecting Flow The actual path taken by a fluid particle as it flows through void spaces from one point toward another is a tortuous and erratic one in most soils because of the random arrangement of the soil grains and the range of sizes. It is highly probable that the direction of flow and velocity of flow vary considerably. The factors that can affect the flow of a fluid through soil are known, but the influence of all factors has not been clearly established. These factors include: 1. 2. 3. 4.

206

The pressure difference existing between the two points where flow is occurring The density and viscosity of the fluid The size, shape, and number of pore openings The mineralogical, electrochemical, or other pertinent properties of the fluid and the soil particles, which affect the attraction between the two materials

Surface body of water

Ground surface Perched water trapped above true water table because of discontinuous aquitard

Water ta

ble

Aquifer (pervious soil)

Aquitard

Ground

Aquitard (impervious soil or rock)

water fl

ow

(a) Groundwater flow in unconfined aquifer

Surface body of water

Ground surface Aquit

ard (

Flow Aqu

itard

(imp

ervio

us)

impe

rviou

s)

Burie d (per aquifer viou s)

Water ta

ble Flow

(b) Groundwater flow in confined aquifer

Surface body of water

Note: buried water table is at top of confined aquifer

Ground surface

Artesian well; water gushes from surface because of high pressure in aquifer (no pump required) Elevation of energy Aq head indicated by uit ard Bernoulli equation (im pe rvi Natural artesian ou s) Aqu condition ifer (pe rvio us) Aquit Flow ard ( impe Flow rviou s)

Quicksand area

Upward seepage forces. Soil particles suspended in the upward flow (surface appears to be solid but is actually liquid or "quick") (c) Artesian condition for groundwater flow

Figure 4 Illustrations of groundwater flow in different types of aquifers.

207

Shallow well

Deep well

Ground surface

Original position of water table

Shallow well may lose its source of water because of drawdown

Drawdown position of water table due to heavy pumping from deep well (a) Permanent lowering of water table because of heavy well pumping

Surface body of water

Original position of water table

Aquife (pervio r us

)

Construction excavation intersects aquifer

Water supply well

Flow

After excavation, aquifer flows to newly exposed soil slope, probably collected in drainage ditches

Aquifer flow intercepted at excavation – water supply to well is lost

(b) Excavation intersects aquifer, cuts off or lowers groundwater flow to well

New dam and reservoir Old valley stream

New position for water table because of reservoir (higher than originally)

Aquifer (per

vious)

Aquitard

(c) Reservoir responsible for higher water table in area downstream from dam

Figure 5 Conditions responsible for changed elevation of groundwater table.

208

Original water table

Movement of Water Through Soil

Water

Figure 6 Schematic diagram indicating manner in which water flows through soil.

The effects of items 3 and 4 on flow are the most difficult to evaluate, partly because of the tremendous variation that occurs in natural deposits (even in homogeneous soils) and partly because of insufficient knowledge. In many engineering and construction problems, the concern is primarily with the quantity of fluid, usually but not always water, that is flowing through or out of a soil mass. The seepage velocity is frequently sufficiently low that no problems result because of this factor, and it may not even require consideration. An exception is where the velocity is great enough to cause movement of the soil particles, or erosion; this problem is discussed in Section 7.1 on flow nets. (But for geoenvironmental problems where concerns relate to movement of a contaminant fluid including escape from containment, even low velocity and small volumes might be hazardous.) For the situation where the quantity of flow is to be determined, an average discharge velocity is assumed. This is a fictitious velocity compared to the actual velocity of flow. This discharge velocity is simply the volume of fluid flow per unit of time divided by the total area (soil plus voids) measured normal to the direction of flow. If an average seepage velocity (average actual velocity of flow) is desired, it can be obtained by dividing the average discharge velocity by n, the porosity of the soil (recall that porosity n equals Vv /VT). Experimental studies have shown that fluid flow is affected by the shape and dimensions of the channel through which flow is occurring and properties of the fluid such as the viscosity. To relate these effects to flow velocity, the terms hydraulic radius and shape factor have been developed. Hydraulic radius, RH, provides a relation between the crosssectional area of flow and the channel walls that are in contact with the fluid: RH =

Area of flow Wetted perimeter

(3)

The coefficient that reflects the shape factor (i.e., the influence of the channel crosssection) is CS. For example, with a circular tube flowing full (Figure 7), the hydraulic radius is one-half the tube radius, and CS = 12 . To indicate how permeable a porous material will be to any flowing fluid (liquid or gas), the term K, absolute or intrinsic permeability, is used. Mathematically: K = CS R2H n

(4)

209

Movement of Water Through Soil

. ia

=

2r

D

Cross-section of flow area = πr2 Wetted perimeter = 2πr 2 RH = πr = r = 1 r 2πr 2 2

Figure 7 Hydraulic radius for pipe flowing full.

where n = porosity of the material. This term applies only to the material through which the flow could occur. It reflects the effect of the size, shape, and number of flow channels, and is completely independent of any fluid properties. In the dimensional (units) analysis, the shape factor and porosity are dimensionless terms, and therefore K has a unit of L2, or area. The theoretical basis for determining absolute permeability K of a porous material, the Kozeny-Carman equation, and the concepts for flow through soil deposits based on Darcy’s law for flow, are presented in the following sections of the chapter. For porous materials such as soil deposits, where the void space channels through which fluids (and gases) flow vary significantly in size and are of irregular cross section, the value for K would typically be determined experimentally. The value of K for soils is expressed in darcys (one darcy equals 9.88 * 10 - 13 m2 or 1.062 * 10 - 11 ft2). This term is little used in practical problems for relating the flow of water in soil deposits but does have application when other fluids are studied or comparisons are made. To indicate the ease or difficulty with which a particular fluid will flow through a permeable material2, the properties of the fluid are incorporated with K, the properties of the permeable material, to provide a coefficient of permeability, k (the lowercase letter is commonly used for this term) or khyd, where: k = khyd = K

g g = K had rhk

(5)

where g = had = hk = r = 2The

unit weight of the fluid 1e.g., in kN>m3, pcf2 absolute or dynamic viscosity of the fluid3 kinematic viscosity density of the fluid

letter “k” historically has been used as the symbol to represent coefficient of permeability (hydraulic conductivity) and continually appears in engineering literature and references for that property. Somewhat unfortunately, k is also used to represent other terms or properties associated with soil deposits. To help prevent misinterpretation of symbols or the application of equations, the modification khyd is used in this text to indicate soil coefficient of permeability and hydraulic conductivity (the “hyd” subscript adapted as an abbreviation for and reference to hydraulic). 3Units for absolute or dynamic viscosity are FT/L2; units for kinematic viscosity are L2/T. In the SI system, absolute-dynamic viscosity is expressed in Pa # sec (where 1 Pascal = 1 N>m2); in traditional metric units, the absolute-dynamic viscosity is expressed in poises or centipoise (cp) (where 1 poise = 1 g>cm # sec or 0.1 Pa # sec; and in U.S. Customary units, the absolute-dynamic viscosity is expressed in lb # sec/ft2. For kinematic viscosity, the SI units are cm2/sec, and the U.S. Customary units are ft2/sec. For water at 20°C (68°F), absolute viscosity is 1 cp, or 0.001 Pa # sec, or 2.083 * 10-5 lb # sec/ft2; the kinematic viscosity is 0.0112 cm2/sec, or 1.076 * 10-5 ft2/sec. (The SI unit of cm2/sec for kinematic viscosity is also called a Stoke; that is, 1 Stoke = 1 cm2>sec.)

210

Movement of Water Through Soil

When the values of unit weight, viscosity, and density for a fluid are not known, the information can often be found listed in reference manuals that tabulate physical properties; it is important to recognize that values for unit weight and viscosity usually vary with temperature. Historically, the phrase coefficient of permeability has been an engineering term. For this same property, geologists, environmentalists, hydrologists, and groundwater specialists use the term hydraulic conductivity. This latter term is descriptive of a physical property associated with the transmission of matter or energy, in the family of terms such as electrical conductivity and thermal conductivity. Geotechnical personnel have become familiar with this dual usage of terms; both are commonly found (used) in engineering literature. To avoid confusion between the terms permeability and coefficient of permeability, the hydraulic conductivity phrase will eventually be adopted by everyone. Illustration 1 The relationship expressed by Equation 5 permits a soil’s coefficient of permeability (hydraulic conductivity) to be determined with fluids other than water if the value of K, or k for water, is known. For example, when compared to water 1unit weight equal to 62.4 pcf or 9.8 * 103 N>m3, absolute viscosity at 20°C equal to 1 * 10-3 Pa # sec = 11 * 10-321N>m221sec22, the coefficient of permeability (hydraulic conductivity) for a gasoline (assume calculations are performed to study the effects of a spill or underground leak) whose specific gravity is 0.72 and whose absolute viscosity is 0.337 * 10-3 Pa # sec will be approximately two times as great, since for water: kwater = aK

g 9.8 * 103 N>m3 b = K¢ ≤ = K19.8 * 1062m>sec had w 1 * 10-3 Pa # sec

and for the gasoline: kgas = aK

0.72 * 9.8 * 103 N>m3 g b = K¢ ≤ = K120.96 * 1062m>sec had g 0.337 Pa # sec * 10-3

and the ratio: kgas>kwater =

K120.96 * 1062 K19.81 * 1062

⬵ 2.14

Darcy’s Law for Flow In the mid-eighteenth century, H. Darcy performed experiments to study the flow of water through sands. With an arrangement represented by Figure 8, it was found that the quantity of water flowing through the soil in a given period was proportional to the soil area normal to the direction of flow and the difference in water levels indicated in the piezometers (open standpipes), and inversely proportional to the length of soil between piezometers through which flow took place. Mathematically: 1¢h2A 1¢h2A Q r = 1a constant2 t L L

211

Movement of Water Through Soil Δh

L

Soil

q=Q t

h2

h1

q=Q t Reference datum

Area A

Figure 8 Darcy’s sand filtration experiment.

where Q t ¢h A L

= = = = =

volume of water flowing through the soil in time t time period for the volume Q to flow h1 - h2 cross-sectional area of the soil sample length of soil through which flow occurs, between points h1 and h2

The factors A and L relate to the volume occupied by the soil but not to its properties. The value of Δh relates to the pressure acting to force the water to flow through the soil. The constant of proportionality, a factor that indicates if the volume of flow is to be great or small, relates to the ease or difficulty with which the water moves through the soil. This constant of proportionality is khyd, Darcy’s hydraulic coefficient of permeability (or hydraulic conductivity) for water. This is the same coefficient of permeability indicated in Equation 5, but attributed to Darcy, who first established it. Consequently: 1¢h21A2 Q = khyd t L

(6)

which is Darcy’s law. The ratio of Δh/L is termed the hydraulic gradient, i, and therefore: Q = q = khydiA t

(7)

where q = volume of flow per unit time. The units of k are length per unit time (L/T); i is dimensionless (length divided by length). For steady flow, the volume of flow q passing a point is equal to the product of flow velocity v and the cross-sectional area A through which flow occurs: q = Av where units of v are length or distance per unit of time.

212

(8)

Movement of Water Through Soil

From Equations 7 and 8, an expression for velocity of flow is obtained. Since: q = khydiA then: v = khydi

(9)

This is a theoretical average velocity, and will be lower than an actual average velocity. Its determination is of use in practical problems, however, where only order of magnitude is required.

Laminar and Turbulent Flow The movement of a fluid through a channel or pore space can be described as laminar or turbulent flow, depending on the path followed by the flowing water particles. Laminar (layered) flow indicates that adjacent paths of water particles are parallel, even when changing direction, and the paths never cross. This is an orderly flow with no mixing. Turbulent flow indicates a disorderly random path for moving water particles, with lines of movement crossing and frequently moving at an angle with or contrary to the general direction of flow. A high degree of mixing occurs. Velocity has direct bearing on whether a flow is laminar or turbulent. Darcy’s law for fluid flow applies provided that the flow is laminar. In soils, the velocity of flow is affected by the size of the void opening as well as the hydraulic gradient i. Studies show that for soils in the coarse sand and finer range, and frequently for small gravel, laminar flow occurs provided that i is 5 or less. In practical soil mechanics work, Darcy’s law thus has a wide range of application.

Effect of Soil Type The volume of water that can flow through a soil bulk is related more to the size of the void openings than to the number or total volume of voids. This is shown by observing that the values of k for coarse soils are greater than for fine-grained soils (even though void ratios are frequently greater for the fine-grained soils), along with the knowledge that voids in a soil mass can range up to the size of the particles. This phenomenon of higher permeability in coarse-grained soil can be explained, at least in part, by the manner in which water flows through a conduit. The fluid flow measured at increments of distance extending between the walls of the conduit indicates that the velocity varies from a very low value adjacent to the wall of the conduit (or against the soil particle) to a maximum at the center of the conduit, as shown in Figure 9. This variation in flow is caused by the friction developed at the conduit wall and the viscous friction developed in the moving fluid. For fine-grained soil, where void spaces are very small, all lines of flow are physically close to the “wall of the conduit,” and therefore only lowvelocity flows occur. In clays, flow in already small “flow channels” is further hampered because some of the water in the voids is held, or adsorbed, to the clay particles, reducing the flow area and further restricting flow. Typical ranges for coefficient of permeability (hydraulic conductivity) for different soil types and resulting drainage characteristics are listed in Table 2.

213

Movement of Water Through Soil Velocity at wall = Vmin. Velocity at center = Vmax.

CL

Figure 9 Variation of flow velocity across the cross section of a tube.

Table 2 Typical Ranges for Coefficient of Permeability (Hydraulic Conductivity): Water and Different Soil Types*

Soil Type Clean gravel Clean sand, sand and gravel mixtures Fine sands, silts Sand–silt–clay mixtures, glacial tills Homogeneous clays

Relative Degree of Permeability

khyd, Coefficient of Permeability or Hydraulic Conductivity (mm/sec)**

Drainage Properties

High Medium

10 to 100 10 to 10−2

Good Good

Low Very low

10-2 to 10-4 10-3 to 10-6

Very low to practically impermeable

10-3 to 10-6 10-6 to 10-10

Fair to poor Poor to practically impervious Poor to practically impervious

*For other fluids, values of khyd are expected to vary from those shown, refer to Equation 5. **To convert, use 1 mm>sec = 0.2 ft>min = 86.4 m>day.

Kozeny-Carman Equation for Permeability Theoretical and experimental studies of fluid flow in a porous material such as water moving through a soil deposit typically are related to Poiseuille’s law for flow in circular tubes, wherby an average flow velocity, vavg, is determined from: vavg =

gR2 i 8h h

where h g R ih

= = = =

the dynamic viscosity of the fluid unit weight of the fluid radius of the circular passage or channel hydraulic gradient as defined by Darcy’s law for flow (refer to Figure 8)

Because of the variation for the size and shape of void spaces in soil deposits and therefore the flow passages, the hydraulic radius term, RH, is substituted for the tube radius, R (or R = 2RH, refer to Figure 7), giving the Poiseuille’s equation for a circular tube flowing full: q = 1vavg21a2 = B

g g 1 12RH22ih R 1a2 = B a b a bR2Hih R 1a2 h 2 8h

where a is the cross-sectional area of the tube.

214

Movement of Water Through Soil

For a flow channel having a cross-sectional shape other than circular, the generalized shape factor CS replaces the coefficient 1⁄2 used to define the hydraulic radius for circular tubes, and: g q = CS B R2H ih R 1a2 h For a group of tubes of different size but where each tube retains a constant cross section and letting A equal the total cross-sectional area of the tube openings plus the tube material (i.e., wall thicknesses), the cross-sectional area where flow occurs, Af , is: Af = A S n where S = degree of saturation

n = porosity of the total cross section, A (for soils, n = Vv>Vt). The hydraulic radius for this condition is:

RH = 1volume available for flow2>1wetted surface area in contact with fluid2 = Af 1length of tubes2>31wetted perimeter21length of tubes24

and: from:

RH = Vw>1Vs So2 Af = S n Atotal = 1Vw>Vv21Vv>Vt2Vt = Vw

where So is the wetted surface area per unit volume of particle material, or: 1wetted surface area2>Vs to establish the relationship: Vs So = Vs31wetted surface area2>Vs4 = wetted surface area Since Vw = SVv = S1eVs2, and n = e>11 + e2, obtain: g e q = CS B R2H i R SAa b h 1 + e g 1SeVs22 e = CS B b R 1SAih2a h 1VsSo22 1 + e = CS B

g 1S2e2V2s 2 e S 1SAih2a b h 1V2s S2o2 1 + e

g 1 e3 = CS B 1S32 ¢ 2 ≤ ¢ ≤ R Aih h So 1 + e

215

Movement of Water Through Soil

Referring to Darcy’s law for flow, q = khyd ih A, obtain: g e3 1 kh = CS B 1S32 ¢ ≤¢ ≤R h 1 + e S2o For full saturation, S = 100% or one. Fluid flowing within a soil deposit travels an erratic and tortuous path through void spaces that vary in size and shape. The general effect is to impede flow, and the shape factor CS is used to represent the influence on the actual volume of flow that occurs. Using ko as a configuration factor for the shape of the soil void channel and Ω as a tortuosity factor so that 1/(koΩ2) defines the overall shape factor CS, get: kh = B ¢

g 1 1 e3 ≤ ¢ 2≤ ¢ ≤R 2 h ko Æ So 1 + e

(10a)

This is the Kozeny-Carman equation developed to study the flow of fluids through porous materials. The group of terms within the brackets define the intrinsic or absolute permeability, K, of a material (Equation 4). For porous materials having equal-size voids channels and the related values for koΩ2, Cs becomes approximately 1/5. At 20°C, the unit weight for water is 9.81 kN/m3 and the dynamic viscosity, n, is close to 10-3 Pa-sec (or 10-3 N-sec/m2), and: g = 19.93 * 1042>1cm-sec2 L 11052>1cm-sec2 h

1for water at 20°C2

The viscosity of water is affected by temperature, decreasing for lower temperatures and increasing for higher temperatures. For water at 20°C flowing through soil voids, obtain: g 1 1 e3 kh = B ¢ ≤ R ¢ ≤ ¢ ≤ h ko Æ 2 S2o 1 + e ⬵ B

105>cm # sec 1 e3 R ¢ 2≤ ¢ ≤ 5 So 1 + e

and when units for So are in 1cm2-1: khyd1cm>sec2 ⬵ B

2 * 104 1 e3 R ¢ 2≤ ¢ ≤ # cm sec So 1 + e

(10b)

If a material consists of equal-size spheres, the value of So for spheres having a diameter D is: So = 1wetted surface area for the unit of volume2>1unit volume of the particle2 = 1surface area of a sphere having a unit volume2>1volume of the sphere2 = 1pD22>1pD3>62 = 6>D

216

Movement of Water Through Soil

For a porous material comprised of same-size spheres, the permeability equation becomes: khyd1cm>sec2 L 12 * 1042cm-1 sec - 1 31D22cm2>136243e3>11 + e24 L 555 D23e3>11 + e24

(10c)

Direct application of the above equation does not provide highly accurate results for most soil deposits because of the assortment of particle sizes and shapes. Nevertheless, the equation is commonly used to approximate soil permeability. For example, if the D50 size obtained from a particle-size distribution analysis is assumed to be the representative size for indicating the permeability of a uniform sand soil deposit (i.e., the Cu or D60/D10 value, as obtained from a particle size distribution analysis, is numerically small, refer to Chapter 4), the hydraulic conductivity would be estimated as: khyd 1cm/sec2 L 15552 1D50 cm22 3e3>11 + e24

(10d)

Predictions of permeability based on the Kozeny-Carman equation improve for commonly encountered soil deposits if values assigned to terms such as So, ko, and Ω match conditions occurring in the soil deposit. The variations of the sizes and shapes for void space openings and the related greater tortuosity are impediments to flow, resulting in a lesser value for Cs than indicated for a material of same-size spheres. Referring to the definition for So, the total surface area per unit volume increases as particle sizes decrease (visualize a brick-shaped particle having a volume of one unit that is then fractured to become two halves; for the same total unit volume, two additional end surface areas are created at the fracture). So for an irregular shape particle is greater than So for a spherical particle having the same volume. Similarly, the So value for deposits increase where the soil is comprised of irregular shaped particles that are of varying size. The significance is that kh decreases as So increases. A solution for the Kozeny-Carman equation shown by Equation 10c involves an arbitrary selection of a particle diameter to represent the permeability characteristics of the deposit. Because the smaller particles have significant influence on permeability, a particle size from the smaller size range is an appropriate choice. For example, when the D10 particle size is used, the Kozeny-Carman equation provides order-of-magnitude agreement with values obtained from other methods such as Figure 10(b) and Equation 11b. The Kozeny-Carman equation is considered appropriate for predicting flow through sand and silt deposits comprised of similar-size particles, and the use is commonly extended to obtain approximations for well-graded sandy soils. However, for some categories of application, particularly for projects that involve a large area, the Kozeny-Carman equation should be viewed as providing values that are approximations or order-of-magnitude only (with natural soil deposits, the properties that relate to terms in the equation typically vary across the area of the deposit). Predictions are expected to be best for constructed earth projects where the soil materials placed for the project have similarity and homogeneity (projects to provide separation or containment susch as dam and levee structures, geoenvironmental projects including holding basins, sanitary landfills, etc.). Studies and experience have established that the Kozeny-Carman equation does not have application for clay-type soils (a consequence relating to the complex and variant nature of clay deposits). Properties of clay soil such as the mineral composition of particles

217

10

50

0.500

0.400 100

200

300 400 500 1000 Coefficient of permeability, 10−3 mm/sec. (log scale)

10

0.5

0.2 0.1 0.05

lower limit for e ≈ 0.3, Cu ≈ 12

2 1.0 0.5

Cu > 12

Cu = D60 /D10

0.02

0.1

(a) 0.01 0.1

0.2

0.5

1

2

5

0.05 10

Effective particle size, D10 (mm) (b)

101

10

Coefficient of permeability, khyd(mm/sec)

1 1.0

2

3 6

5

7

0.1

8 10

0.01 12

4

10–1

9 10–2

11

10–3

13

0.001

100

10–4

0.0001

0.00001 0

20

40

60

80

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.

coarse Sand coarse to med Sand coarse to med Sand med to fine Sand cse to med Sand, some med to fine Gravel cse to med Sand, some med to fine Gravel fine Sand cse to fine Sand, some fine Gravel cse to fine Sand, some fine Gravel, trace Silt cse to fine Sand, some fine Gravel, trace Silt fine Sand and cse Silt cse Silt cse to fine Sand, some cse to fine Silt, some cse to fine Gravel

10–5 100

Relative density, DR (%) (c)

Figure 10 Coefficient of permeability for sandy soils—affect of various properties. (a) Void ratio versus permeability plotted on semilog coordinates. (b) Coefficient of permeability related to void ratio, uniformity coefficient, and effective particle size (band limits based on an assortment of field tests and laboratory tests) [378]. (c) Coefficient of permeability versus particle size and relative density. [198]

218

(mm/sec)

5

1

e, lo wC u e, h igh C u

0.600

2

low

Void ratio, e

0.700

20

upper limit for e ≈ 0.75 Cu ≈ 2

high

Coefficient of permeability, khyd(ft /min)

5

Movement of Water Through Soil

and the shape and arrangement of particles that affect the configuration and size of voids, and therefore the ability for water to flow, do not compare well with the properties assumed for porous materials.

Empirical Relationships Considerable information on the flow of fluids through porous media has been obtained from studies of flow through tubes and conduits. From such information, attempts have been made to relate permeability to a soil’s grain size. Practically, such a relationship appears more possible for sands and silts than for clays, because of the particle size, shape, and overall soil structure. One of the more widely known relationships is: khyd L 101D1022

(11a)

where khyd is given in millimeters per second and D10 is the 10 percent particle size, expressed in millimeters, from the grain-size distribution analysis (from the curve resulting from plotting percent finer by weight versus particle diameters). This relationship was developed from the work of Hazen on sands.4 This expression applies only to uniform sands in a relatively loose condition. More recent studies to evaluate the permeability and filter properties of sands [326] have determined a relationship between the D15 size and the khyd value for dense or compacted sands. A close approximation for coefficient of permeability based on the D15 size is: khyd L 3.51D1522

(11b)

when D15 is expressed in millimeters and the khyd value is in millimeters per second. The permeability for loose sands will be greater than indicated by Equation 11b. For sands where the uniformity coefficient Cu value is less than about 10, the khyd value for the loose condition can be estimated as 1.5 to 3 times the equation value. When the Cu value is more than 10, the permeability for a loose sand can be considerably greater (5 to 15 times greater) than the value indicated by Equation 11b. For sand deposits and for silt deposits, the typical values for the coefficient of permeability are, practically speaking, within the relatively narrow ranges indicated in Table 2. By comparison, the values associated with clay deposits extend across a significantly broad range (i.e., 10-3 to 10-10 mm/sec) because of the complex and variable geometry of void channels (or flow channels) that result with different clay minerals and the tendency for particles to flocculate or cluster, open fissures and seams or other discontinuities that develop in deposits, effects of stress history, etc. Relating to such practical factors, theorybased methods that will closely predict permeability of clay deposits have not been developed. Accordingly, for a construction project where the permeability characteristics of a 4Originally

included in “Some Physical Properties of Sands and Gravels with Special Reference to their Use in Filtration,” Allen Hazen, 24th annual report of the State Board of Health of Massachusetts, 1892, and numerous subsequent engineering manuals, including reference [378].

219

Movement of Water Through Soil

Undisturbed flocculent structure

Remolded or dispersed structure (a)

2.8 1.25

Void ratio, e

Curve values = PI + CF (both decimal)

1.00

2.0 0.75

PI = plasticity index, (LL – PL) CF = clay fraction (particles < 0.002mm)

1.2 0.50

0.4 1 × 10–8

1 × 10–7

1 × 10–6

5 × 10–5

k v (mm/sec) Vertical coefficient of permeability for clay (b)

Figure 11 Conditions and properties affecting coefficient of permeability of clay soils. (a) Cohesive soil with flocculent structure will have higher permeability than soil with dispersed structure. (b) Coefficient of permeability related to soil properties. [198]

naturally occurring clay deposit are important such data usually is determined from field permeability tests or laboratory tests on undisturbed samples to complement the subsurface information obtained from the site investigation process (borings, test pits, etc.) and review of the area’s geology. Clay soils that have been excavated, transported, and placed at a new location are disturbed and remolded by the process. Projects in this category include dredging and deposition projects, and constructed earth fill projects including dams and levees. The condition of deposited fill clays will be affected by the method of placement but also from behavior influenced by the properties of clay minerals associated with soil plasticity. Developed from studies on remolded clays, Equation 12 below relates the permeability of remolded clays to easily determined soil plasticity index values (Carrier and Beckman, ref. 53). Limited data implies that the application can be extended to approximating the coefficient of permeability for normally consolidated deposits of marine clays. khyd1m>sec2 L

220

0.0174 e - 0.02731PL2 - 0.2421PI24 4.29 b r 11 + e2 1PI2

(12)

Movement of Water Through Soil

where PL and PI are the plastic limit and plasticity index values (percent water contents) expressed as numeric values (e.g., indicate 25 pct water content as 25).

Permeability Tests Much of the available information from studies of flow through uniform porous media is not directly applicable to soils because of the extent of variation for properties, such as the size and shape of void spaces in a soil deposit. For soil, it has been found to be practical and more appropriate to evaluate flow directly through use of laboratory or field tests on the soil in question. Experience has shown that for a given soil, a relationship exists between permeability and void ratio. Generally, a semilog plot (e plotted on an arithmetic scale, khyd plotted on a logarithmic scale) produces approximately a straight line for most soils. Thus, permeability tests can be performed on a soil at two or three widely different void ratios, and the results can then be plotted. Permeability at intermediate void ratios would be determined by interpolation (Figure 10a). The permeability of a soil deposit is affected by the in-place structure. For granular (or coarse-grained) soils, the opportunity for passage of water relates to the void ratio (the relative volume of open spaces within the total volume occupied by the soil particles), the particle sizes and size of the void openings (the largest voids tend to be similar to the size of the smaller particles), the range of particle sizes (uniform sands where all particles are close in size tend to have larger void ratios and lower Cu values than sands comprised of a wide range of particle sizes with a high Cu value), and the Relative Density DR (such as a loose condition having a high void ratio or a dense condition with low void ratio). The effect of various physical properties on the coefficient of permeability for granular soils are illustrated in Figure 10(b), (c). Illustration 2 Estimate the coefficient of permeability for a sand formation, given the following representative information. The representative void ratio value is based on calculations using results of in-place density tests and laboratory tests for the specific gravity of soil particles. The particle size data from grain-size-distribution analyses is: Void ratio e = 0.65, D60 = 2.0 mm, D10 = 0.40 mm

Solution

Uniformity coefficient Cu = D60>D10 = 2.0 mm>0.40 mm = 5 Effective particle size D10 = 0.40 mm Referring to Figure 10(b), and interpolating, obtain khyd L 0.35 ft>mm L 2 mm>sec

The sizes and volume of the void spaces in a clay deposit, and the resulting opportunity for passage of water, relates to the arrangement of particles (soil structure), the type of clay mineral making up the particles and the effect of contact with water (properties which also influence results of the liquid limit and plastic limit tests), and presence of silt or sand sizes included in the deposit. At similar void ratios, clay with an

221

Movement of Water Through Soil

undisturbed flocculated structure possesses larger void openings than the same clay having a remolded, dispersed, or oriented structure (see Figure 11(a)), with the result that the coefficient of permeability is many times greater for the flocculated soil. Other correlations between plasticity properties, the clay sizes content of the deposit, and coefficient of permeability are illustrated in Figure 11(b). Extremely important is the significance of stratification. In layered soils, the permeability measured for flow across layers can be greatly different from that for flow parallel to the layering. Fine-grained deposits, such as clay or alternating layers of silt and clay, have a permeability in the direction parallel to bedding of the layers (often, such bedding is approximately horizontal) many times the permeability of cross-bedding flow. Thus, the use of undisturbed test samples, where the sample has retained its original structure and is tested so that flow through the sample is in the correct direction (to correspond to horizontal or vertical flow, as will actually occur in the field), is important if reliable results are to be obtained. This also points out an important consideration regarding the use of test data: Data from remolded samples may not apply to field conditions if the natural soil remains undisturbed, or test data from undisturbed samples may not apply if, in the field, the soil is to be disturbed and rehandled (as in placing a compacted fill). Conditions other than the size and number of voids will affect the quantity of flow through a soil deposit. Trapped air or gases prevent flow, whereas seams, cracks, fissures, and cavities that exist in a soil deposit increase the opportunity for fluid movement. Field investigations need to provide information on the presence of these conditions if reliable measures of flow through the soil deposit at a construction site are to be obtained. For this reason, field permeability tests are felt to provide more accurate information than laboratory tests. Field testing has its disadvantages, however. Costs and time involved are usually greater than for a laboratory test, and the field test provides only information on conditions in the limited proximity of the test location. In practical applications, determining permeability from field or laboratory tests or from indirect analytical methods is for order-of-magnitude use only. This is realistic when it is realized that subsurface conditions and soil properties most probably vary over even short horizontal distances, that soil conditions and properties are generally not known in all areas of influence at a site, and that, frequently, external factors causing or affecting flow are not accurately known during planning and design.

Laboratory Permeability Tests Two of the more conventional laboratory permeability tests are the constant-head test and the falling-head test. Schematic diagrams showing each of these methods, and the mathematics to calculate the coefficient of permeability, are shown in Figures 12 and 13. The constant-head permeability apparatus (permeameter) is in wide use for testing the coarse-grained soils, such as sands, where the volume of flow through the soil will be relatively large. For fine-grained soils, such as silt and clay, the falling-head permeameter is generally used. In the constant-head test, permeability is computed on the basis of fluid that passes through the soil sample. In the falling-head test, permeability is computed on the basis of fluid flowing into the sample. The reason for the distinction is simple. In a fine-grained soil, a very limited volume of fluid will flow through the sample. With the constant-head test, time is required to accumulate the fluid volume necessary to perform

222

Constant head

Note: t is time for volume of water, Q, to flow through sample.

Porous filters

Δh Soil sample

L

Cross-sectional π area = A = 4 D2

Q

Spring

Sample diameter = D Constant-head test: Q = kiA t khyd = Q ⋅ 1 = Q t iA t

Figure 12

L A(Δh)

Constant-head permeability test: Representation of permeameter and definition of related terms.

(t 1)

Time when water level is at h1 is t1

a = area of standpipe Time when level is at h2 is t2

h1 (t 2)

A = cross-sectional area of soil sample

h2

L

Soil sample

Falling-head test: k = or

Figure 13

khyd =

L (t2 − t1)

(2.303)L (t2 − t1)

a A

⋅ a A

In

h1 h2

log10

h1 h2

Falling-head permeability test: Representation of permeameter and definition of related terms.

223

Movement of Water Through Soil

computations. Extreme care would be required to prevent leaks in the apparatus and evaporation of discharged water. With the falling-head method, the duration of the test is shortened, and there is no concern about the volume of discharge. Care is required, however, to prevent evaporation of water in the inlet tube. Darcy’s coefficient of permeability is the factor for a condition of steady flow through a soil. In performing laboratory permeability tests, it is essential that volumes be measured only after steady flow has been occurring for some period. It is important to ensure that no air or other gases are trapped within the soil to interfere with flow. A vacuum may be required to remove trapped air. In general, the constant-head test is easier to perform and requires less skill and experience than the falling-head test. Care is required during testing of fine granular soils (such as in the fine sand range) to prevent the particles from being carried along with the discharging water. Details for performing permeability tests are presented in the ASTM Procedures for Testing Soils (D2434).

Illustration 3 In a laboratory, a constant-head permeability test is performed on a sample of granular soil. The test setup is as indicated in Figure 12. The length of the soil sample is 15 cm, and the cross-sectional area is 10 cm2. If a 24 ml (or 24 cm3) volume of water passes through the soil sample in a 3-minute period, when Δh is 30 cm, compute the coefficient of permeability.

Solution Use: khyd = a

Q L ba b t A¢h

where Q t L A ¢h khyd

24 cm3 3 minutes 15 cm 10 cm2 30 cm 24 cm3 15 cm = a ba b = 0.4 cm>min = 0.067 mm>sec 3 min 10 cm2 * 30 cm = = = = =

Illustration 4 A falling-head permeability test is performed on a silty soil. The test setup is as shown in Figure 13. For the test data summarized below, what is the coefficient of permeability for this sample? Sample length = 8 cm Cross-sectional area of sample = 10 cm2 Area of standpipe = 1.5 cm2 Height of water in standpipe at start of test period h1 = 100 cm Height of water in standpipe at end of test period h2 = 90 cm Time for change from h1 to h2 = 60 min

224

Movement of Water Through Soil

Solution Use: khyd = khyd =

h1 12.3032L a a blog t2 - t1 A h2

12.303218 cm2 1.5 cm2 100 cm a b log 60 min 90 cm 10 cm2

= 0.00212 cm>min = 3.5 * 10-4 mm/sec

Where consolidation tests are performed on soil samples, the permeability test can be adapted to determine the rate of flow through the consolidation sample, and the coefficient of permeability can be computed from these data. Alternatively, the coefficient can be determined from the consolidation data obtained to calculate the rate of consolidation of the soil.

Field Permeability Tests Field permeability tests offer the advantage of testing undisturbed soil in the natural location with respect to the ground surface, water table, and other factors that could influence the rate of flow. Various methods for determining permeability are available, depending, among other things, on the soil’s being above or below the groundwater table. The methods described herein are of the type where a cased boring is made into the soil that is to be tested. The casing and related equipment necessary are of the type normally used by soilboring contractors. Whenever possible, it has been found expedient to perform field permeability tests during the investigative stage of planning a project, at the time that the subsurface investigation (soil borings) is being made. Essentially, the field permeability test involves obtaining a record of the time that it takes for a volume of water to flow out of, or into, the boring casing. A schematic presentation of different conditions and the related equations for calculating the coefficient of permeability are shown in Figure 14. On a practical note, simple field permeability tests of the types depicted in Figure 14 are considered appropriate for evaluating granular and silt soils. The low coefficient of permeability associated with clays and silt–clay mixtures would involve a long time period for the field test and require consideration of various additional factors, such as evaporation and the like.

3

Capillarity The groundwater table (or phreatic surface) is the level to which underground water will rise in an observation well, pit, or other open excavation in the earth. All voids or pores in soil located below the groundwater table would be filled with water (except possibly for small isolated pockets of trapped air or gases). In addition, however, soil voids for a certain height above the water table will also be completely filled with water (full saturation). Even above this zone of full saturation, a condition of partial saturation will exist. Any water in soil located above the water table is referred to as soil moisture. The phenomenon in which water rises above the groundwater table against the pull of gravity but

225

Movement of Water Through Soil Q Cased hole, soil flush with bottom.

GWL

Q khyd = 0.024 Q Rh where Q = gal/min R = feet dw = feet k = ft/min

dw Ground surface

At least 10R

dw

GWL

2R

2R

Used for permeability determinations when water is above or below bottom of casing. Q is quantity of water to keep casing filled.

2R GWL Casing Cased hole, soil flush with bottom.

h2 D

h1

h 2πR khyd = In 1 11(t2 − t1) h2 For D from 6" to 60" (150 mm to 1.5 m)

Used for permeability determination at shallow depths below the water table. May yield unreliable results in falling-head test with silting of bottom of hole.

2R GWL Cased hole, uncased or perforated extension of length L.

h2

Casing

h1

R2 2L(t2 − t1) h In L In 1 R h2 For L > 8 R

khyd =

Used for permeability determinations at greater depths below water table.

( ) ( )

L

2R Cased hole, column of soil inside casing to height L.

Figure 14

Casing

h2

h1

h1 khyd = 2πR + 11L In 11(t2 − t1) h2

L

( )

Principal use is for permeability in vertical direction in anisotropic soils.

Methods for performing field permeability tests.

is in contact with the water table as its source is referred to as capillary rise. The water associated with capillary rise is capillary moisture. The soil region directly above the water table and wetted by capillary moisture is designated the vadose zone.

Water in Capillary Tubes The basic principles of capillary rise in soils can be related to the rise of water in glass capillary tubes (tubes with very small diameters) under laboratory conditions. When the end of a vertical capillary tube is put in contact with a source of water, the water rises up in the

226

Movement of Water Through Soil

tube and remains there. The rise is attributed to the attraction between the water and the glass and to a surface tension5 which develops at the air–water interface at the top of the water column in the capillary tube. This surface tension can be thought of as an infinitely thin but tough film, such as a stretched membrane. (The surface tension phenomenon is one of the reasons that small insects can “walk” on water.) The water is “pulled up” in the capillary tube, to a height regulated by the diameter of the tube, the magnitude of the surface tension, and the density of the water. The attraction between the water and the capillary tube affects the shape of the air–water interface at the top of the column of water. For water and glass, the shape is concave downward; that is, the water surface is lower at the center of the column than at the walls of the tube. The resulting curved liquid surface is termed the meniscus (Figure 15). The column of water in the capillary tube depicted by Figure 15 has risen above the surface of the water supply and against the pull of gravity. For a condition of equilibrium, the effect of the downward pull of gravity on the capillary column of water has to be resisted by the ability of the surface film to adhere to the wall of the capillary tube and hold the column of water. This wall adhesion must equal the surface tension of the fluid. If Ts is Surface tension, Ts Meniscus contact with tube wall Glass tube h Free water surface

Figure 15

Capillary rise and water meniscus in a glass tube.

5The kinetic theory explains the surface tension phenomenon (all molecules are in motion and the speed or kinetic energy increases as temperature increases, and molecules attract each other at close range). Within the body of a liquid, a molecule of water will be subject to attractions that are equal in all directions because of the surrounding molecules. At the surface of a body of water (the air–water interface), the water molecules are subject to an unequal force in the direction of the body of water, since in the air (gaseous state) the molecules are widely dispersed and, in total, exert a lesser attractive force. The normal molecular motion of the water molecules is thereby restricted by the unbalanced pull (attraction). The resulting net effect between the natural molecular motion and unbalanced attraction is that the surface zone of the water acts as a stretched membrane (i.e., a membrane in tension). If the attractions internal to the body of water (cohesion) are less than the attractions between the water and a solid (adhesion), then the water surface curves outward (from the body of water) where the water is in contact with the solid, producing a curved meniscus. Since glass materials (and soil particles) are solids that possess a polar surface, the result is an attraction or adhesion between the water molecules and the surface of the glass (or soil particles).

227

Movement of Water Through Soil

the value of surface tension, expressed in units of force per unit length, the vertical loading that can be supported is: 1Tube circumference2 * 1Ts2 * 1cos a2 where α is the angle formed between a tangent to the meniscus and the capillary wall. With water and glass, the meniscus at the wall of the capillary tube is tangent to the wall surface, and the angle α is zero degrees. Therefore, cos α is 1, and the column weight that is capable of being supported because of surface tension is: 12pr21Ts2 where r is the radius of the capillary tube. The weight of the column of water in the capillary tube is: 1pr22 * 1h2 * 1r2 * 1g2 where h = height of the column of water r = density of water, mass per volume, taken as 1 gm/cm3, 1 mg/m3, or 1.95 slugs/ft3 g = acceleration of gravity, 9.81 m/sec2, 981 cm/sec2, or 32.2 ft/sec2 If hc is the maximum height of capillary rise that can occur for prevailing physical conditions (such as for a particular temperature), equilibrium requires that: 12pr21Ts2 = 1pr2hc21r21g2 = pr2hc

1weight2 6 1volume2

= pr2hcgw where gw ⫽ unit weight of water, taken as 62.4 pcf, 9.81 kN/m3, or 981 dynes/cm3.7 The maximum height of capillary rise is then: hc = hc =

12pr21Ts2

1pr221r21g2 2Ts

r1gw2

=

=

4Ts

d1gw2

2Ts

1r21r21g2

(13a) (13b)

where d = the diameter of the capillary tube.

weight equals mass times acceleration of gravity ( W = mg ), the product of ρg equals: weight * g, or . 1volume2 volume g-cm 7Recall that the dyne is a unit of force; units are . sec2 6Since

1mass2

228

Movement of Water Through Soil

The value of Ts for water varies according to temperature. At normal room temperatures, Ts is close to 0.005 lb/ft, 0.064 N/m, or 73 dynes/cm. In applying the development of capillary rise in tubes to capillary rise in soils, these values for Ts are sufficiently accurate for many practical problems. Thus, the equation for capillary rise can be expressed as: hc ⬵

31 mm d

(14)

provided that d is in millimeters. Illustration 5 Compute the height of capillary rise for water in a tube having a diameter of 0.05 mm (in SI units).

Solution hc =

4Ts dgw

=

14210.064 N>m2

15 * 10-5 m219.81 kN>m32

= 0.52 m

There are situations, however, where the temperature effects should be considered. Generally, as temperature increases, the value of Ts decreases, indicating a lessening height of capillary rise under warm conditions or an increasing height of capillary rise for conditions of falling temperatures. At freezing, Ts for water is about 0.067 N/m. Note from the terms of Equation 13 that the height of capillary rise is not affected by a slope or inclination in the direction of the capillary tube, or by variations in the shape and size of the tube at levels below the meniscus (Figure 16). But for water migrating up a capillary tube, a large opening can prevent further upward movement in an otherwise small-diameter tube. The determining factor is the relationship between the size of the opening (tube diameter) and the particular height of its occurrence above the water supply. In the capillary tube, at the level equal to the free surface of the water in the supply pan, the hydrostatic pressure in the water is zero. Hydrostatic pressures increase below that free water surface according to the relationship: pw = gw z

(15)

Atmospheric pressure, Patm, just above meniscus surface

Point c, just beneath meniscus h Patm, atmospheric pressure at free surface of water supply

Figure 16 Capillary heights of capillary tubes of various shapes are the same if their menisci diameters are the same.

229

Movement of Water Through Soil

where z = depth below the water surface pw = water pressure at depth z Conversely, hydrostatic pressure measured in the capillary column above that free water surface is negative, according to the relationship: pw = -gwh

(16)

where h is the distance measured upwards, from the free water surface. The negative sign indicates that water pressure in the capillary tube is at less than atmospheric pressure. As a result, water in the capillary column is said to be in tension. The maximum negative pressure, or capillary tension, exists at the maximum height of rise hc. If the pressure at the water surface in the supply pan depicted in Figure 16 is atmospheric pressure, patm, the pressure in the capillary water just beneath the meniscus, point c, is patm - hc ␳g. By substitution of terms: patm - hc rg = patm - a

2Ts rrg

b1rg2 = patm -

2Ts r

or 2Ts /r less than the pressure existing just above the meniscus. (The combination of atmospheric and water pressures is used to define the position of the phreatic surface in a soil deposit; a net pressure greater than patm indicates gravitational water while a net pressure less than patm indicates capillary water above the ground water table.) As the column of water stands in the capillary tube, “hanging” by the surface tension at the meniscus, the weight of the column is transferred to the walls of the capillary tube, creating a compressive force in the walls. An effect of this occurrence within soil deposits is discussed in a later subsection, “Effects of Surface Tension.” Capillary rise is not limited to tube, or enclosed, shapes. If two vertical glass plates are placed so that they touch along one end and form a V, a wedge of water will rise up in the V because of the capillary phenomenon (Figure 17). The height of rise relates to the Meniscus

Capillary rise

Capillary "wedge" Water surface Glass plates (Elevation)

Figure 17

230

Capillary rise in corner formed by glass plates.

(Plan)

Movement of Water Through Soil

attraction between the water and plates and the physical properties of the water, as in tubes, but also to the angle formed by the V. The significance of this type of capillary rise is discussed in the following subsection, “Capillary Rise in Soil.”

Capillary Rise in Soil In soils, the shapes of void spaces between solid particles are unlike those in capillary tubes. The voids are of irregular and varying shape and size, and interconnect in all directions, not only the vertical. These properties make the accurate prediction of the height of capillary rise in soil almost impossible. However, the features of capillary rise in tubes are applicable to soils insofar as they facilitate an understanding of factors affecting capillarity, and help to establish an order of magnitude for capillary rise in the different types of soils. Illustration 6 Limited laboratory studies indicate that for a certain silt soil, the effective pore size for height of capillary rise is 15 of D10, where D10 is the 10 percent particle size from the grain-size distribution curve. If the D10 size for such a soil is 0.02 mm, estimate the height of capillary rise.

Solution d = effective capillary diameter =

1 1 D = 10.02 mm2 = 0.004 mm 5 10 5

31 1from Equation 142 d (mm) 31 L L 7750 mm L 7.75 m ⬵ 25 ft 0.004 mm

hc (mm) L

Review of Equation 13 indicates that even relatively large voids will be filled with capillary water if the soil is close to the source of water, that is, the groundwater table. As the distance from the water table increases, only the smaller voids would be expected to be filled with capillary water. The larger voids represent interference to upward capillary flow and would not be filled. Consequently, the soil closest to the water table but above it is fully saturated as a result of capillarity. Above this zone of full saturation lies a zone of partial saturation due to capillarity (Figure 18). The upper limit (or height) to which capillary water rises in the capillary zone is termed capillary fringe. In the upper zone of the capillary fringe, where partial saturation exists, capillary movement occurs through the capillary tube phenomenon but may also occur in the wedges of the capillary V formed where soil particles are in contact (similar to the V formed by vertical plates discussed in the preceding section; see Figure 19). Since the largest void spaces in soil are of the same order of magnitude as the particle sizes, it follows that the height of capillary rise would be greater in fine-grained soils than in coarse-grained soils. For orders of magnitude, relative values of capillary rise for different types of soils are presented in Table 3. (Though attempts at establishing a relationship between height of capillary rise and some aspect of soil’s grain size or void ratio have been made, the general applicability and accuracy of proposed relationships currently are limited to providing estimates of order of magnitude.)

231

Movement of Water Through Soil Well

Ground surface

Downward percolation (may be held in Zone of partial saturation due to suspension by capillary capillary rise, downward percolation forces) of water, and adsorbed water. Capillary rise Vadose zone

Zone of 100% saturation due to capillary rise and adsorbed water. Zone of 100% saturation (below water table).

Groundwater table

Figure 18

Distribution of soil moisture in a soil profile. Capillary water in wedge formed by soil particles

Soil particle

Figure 19

Soil particle

Wedge of capillary water at point of contact between particles. Table 3 Representative Heights of Capillary Rise: Water in Soil Soil Type Small gravel Coarse sand Fine sand Silt Clay

m

ft

0.02–0.1 0.15 0.3 to 1 1 to 10 10 to 30

0.1–0.4 0.5 1–3 3–30 30–90

Temperature plays an important practical role in capillary rise in soil. The height of rise is greater at lower temperatures than at high temperatures. Demonstrations of this fact have frequently been observed in construction performed in a cool season following a warm season (i.e., in the fall months at locations in the northern hemisphere). Cool temperatures cause an increase in migration of water toward the surface. Frequently the additional moisture increases the difficulty in handling or working with the soil, or creates a muddy surface, which makes the moving of equipment a problem. Horizontal migration of water can also occur near a heated building. Water may move from beneath a structure in cold seasons if the presence of the structure affords some protection from cold temperatures. If the building itself contributes to a lowering of the ground temperature, as in a

232

Movement of Water Through Soil

poorly insulated cold-storage warehouse, frozen foods processing plant, or ice skating rink, capillary flow will be induced toward the building.

Time Rate of Capillary Rise The time necessary for the expected maximum height of capillary rise to occur requires consideration in some practical problems, such as where a structural fill has been placed for highways, buildings, or other purposes. On the basis of typical void sizes, clay and fine silt soils will have significant heights of capillary rise. However, the time period required for the rise to occur may be so great that other influences, such as evaporation and change in groundwater level, also have to be considered. The term indicating the rate of capillary rise is capillary conductivity or capillary permeability, kcap. Factors known to affect a soil’s capillary conductivity are void sizes, moisture content, and temperature in the soil. Generally, capillary conductivity is greater at higher moisture contents and lower temperatures. Absolute values of capillary conductivity are not available. Though not identical to Darcy’s coefficient of permeability, the relative rates of capillary conductivity can be thought of in terms of the comparative values for permeability—that is, rapid for coarse soils, low for silts and clays.

Suspended Capillaries Water attempting to percolate downward through a soil (such as infiltrating surface water from rain or melting snow or pore water resulting from a formerly higher water table) can be held suspended in the soil voids because of the same surface tension phenomenon responsible for capillary rise. For this, the column of water would have a meniscus at both ends of the suspended column. Each meniscus would be in tension. The maximum length of a column would be controlled by the same factors affecting height of capillary rise.

Elimination of Capillary Water in Soil For capillary rise to occur, the existence of an air–water interface is required. Capillary water will not exist in soil at a level below the groundwater table. It follows that a condition of capillarity will cease where submergence of a soil zone occurs. Capillary water can be removed from a soil by evaporation. As a result, capillary water can represent a very mobile category of water where the evaporation is continually replaced by new capillary water.

Effects of Surface Tension At the location of the meniscus, the surface tension imposes a compressive force onto the soil grains in contact with the meniscus that is equal in magnitude to the weight of the water in the capillary column, as indicated in the earlier discussion for capillary tubes. This effect applies to both water resulting from capillary rise and pore water suspended above a capillary zone. The compressive force is imposed throughout the soil particles in contact with the “held column” of water, and serves to compress or shrink the soil.

233

Movement of Water Through Soil Table 4 Representative Values of Compressive Stress Resulting from Capillary Forces psf

kg-f/cm2

kN/m2

200–2000 2000–6000

0.1–1.0 1.0–3.0

10–100 100–300

Soil Type Silt Clay

In clays, this phenomenon results where the groundwater table drops subsequent to the time the clay deposits were formed. Such “drying” of an area can cause compressive forces within the clay mass of such great magnitude that the affected soil zone becomes firm and strong. This is referred to as drying by desiccation, and the clays are referred to as desiccated clays. A land area formed by clay deposits that were originally soft and weak may have a firm crust or upper layer overlying the still soft and weak deeper clays because of desiccation. The thickness and strength of a desiccated zone has frequently been found sufficiently adequate that roads and light building structures could be satisfactorily supported by it. Typical magnitudes of compressive stress that can result in fine-grained soils are tabulated in Table 4. In partially saturated granular (coarse-grained) soils, the surface tension phenomenon also contributes to the strength of the mass. At partial saturation, all voids are not filled with water. The available water collects in the interstices adjacent to where soil particles touch, forming a wedge of moisture but leaving the center portion of the void filled with air. Thus, an air–water interface, or meniscus, is formed. The surface tension in the meniscus imposes a compressive force onto the soil particles, increasing the friction between particles and thus the shearing strength. The strength gain in granular soil due to partial saturation and the surface tension phenomenon is termed apparent cohesion. The strength gain can be quite significant, as the firm condition of wet sandy beach surfaces, which easily support the weight of vehicles, indicates. The menisci and surface tension, along with the apparent cohesion, will disappear when the soil is fully saturated or dries.

Problems 1 Of the earth’s total volumetric supply of water, approximately how much is fresh water (quantity and percentage)? How much is underground water? 2 Define hydrogeology. 3 Discuss the hydrologic cycle in terms of water on land areas (that is, the source of land-area surface and subsurface water, movement patterns, and change of location).

234

6 Explain the difference between a soil stratum that is an aquifer and one that is an aquitard. 7 Referring to Figure 3, if the elevation difference between h1 and h2 is 10 m, what is the difference in energy head pressure (approximately) between the groundwater surface at point 1 and point 2?

4 What is the groundwater table, and what defines its position?

8 Referring to Figure 3, if the elevation difference between h1 and h2 is 15 ft, what is the energy head pressure difference (approximately) between the groundwater surface at point 1 and point 2?

5 Outline practical reasons why the elevation (depth) of the groundwater table in an area may fluctuate seasonally or annually.

9 Explain how the process of pumping large volumes of water from a well can alter the position of a groundwater table.

Movement of Water Through Soil 10 Outline examples of how construction projects can be responsible for altering the elevation of the groundwater table for an area.

19 What effect does the presence of adsorbed water in clay have on the coefficient of permeability for this type of soil?

11 On the basis of general soil types that form an aquifer and an aquitard, indicate a probable range for the coefficient of permeability in aquifers and in aquitards.

20 In the flow of water through soil, what are the conditions necessary in order for Darcy’s law to apply?

12 In an area of hilly topography, a section of property has a ground surface that is a long hill extending for a 400 m distance at an average slope of 20 percent. An aquifer is buried beneath the hill as depicted by Figure 4(c) and intersects a large pond at the top of the hill slope and a flowing creek at the base of the hill. For these conditions, determine if a well installed midway up the hill and penetrating to the aquifer will flow as an artesian well; for this problem, assume the ground surface adjacent to the well is 45 m below the top of the hill. 13 Compute the value of the hydraulic radius for a circular pipe flowing half-full. 14 Calculate the value of the hydraulic radius for flow in an open channel where the bottom of the channel is 3 ft wide, side slopes are 1 on 1 (slope of 1 ft horizontal to 1 ft vertical), and the depth of flow is 2 ft. 15 A thick fluid having a high viscosity and a thin fluid having a low viscosity are to be passed through a permeable material. Indicate which fluid will have the greatest resulting coefficient of permeability and briefly explain why. For which fluid (thick or thin) is the quantity of flow through the material expected to be greatest? 16 Outline the general relationship between soil particle sizes and size of void spaces between the particles and the corresponding effect on a soil deposit’s coefficient of permeability. 17 The coefficient of permeability is generally greater for coarse soils (sands and gravels) than for fine-grained soils (silts and clays). What effect does particle size have on permeability? 18 The void ratio for clay soils commonly is greater than the void ratio for the coarse-grained soils, but comparisons of the coefficient of permeability indicate considerably lower values for clay. Explain.

21 (a) Estimate the coefficient of permeability for a uniform sand where a sieve analysis indicates that the D10 size is 0.15 mm (by Hazen formula, Eq. 11a). (b) The particle-size distribution curve for a sand indicates that the D15 size is 0.19 mm. If the sand deposit is in the dense or compact state, estimate the coefficient of permeability. (c) Apply Kozeny-Carman equation (Equation 10c), using both the D10 and D15 sizes, to compare calculated values with those values obtained for (a) and (b). Use the answer for (a) and Figure 10(b) to estimate a value for e. 22 (a) Estimate the coefficient of permeability for a sand deposit where field testing indicates the in-place void ratio is about 0.50, and representative laboratory grain-size analyses indicate the D10 size is 0.55 mm and the uniformity coefficient Cu is eight. (Use Fig. 10.) (b) Apply the Kozeny-Carman equation (Equation 10c) to estimate the coefficient of permeability, then compare the result to the value obtained for (a). (c) Using the above information, and referring to Table 2, indicate the probable soil type (soil classification). 23 (a) Estimate the coefficient of permeability for a sand deposit where field testing indicates the in-place void ratio is about 0.70, and representative laboratory grain-size analysis indicate the D10 size is 0.20 mm and the uniformity coefficient Cu is three. (Use Fig. 10.) (b) Using the above information, determine the D60 particle size, and then by referring to Table 2 indicate the probable soil type (soil classification). (c) Apply the Kozeny-Carman equation (Equation 10c) to estimate the coefficient of permeability, then compare the result to the value obtained for (a). 24 Based on grain-size-distribution analyses of samples obtained from a soil deposit, the soil is

235

Movement of Water Through Soil described as predominantly coarse-to-fine sand, with some coarse-to-fine silt and some coarseto-fine gravel sizes. The relative density of the in-place soil is about 50 percent. (a) Using Figure 10(c) as a reference, approximate the coefficient of permeability for the in-place soil. (b) What new coefficient of permeability would be expected after the deposit is densified by compaction equipment to a relative density of 90 percent, and what is the extent of the change from part (a). (c) Provide the probable reason (or explanation) for the comparatively low coefficient of permeability for this soil, as judged against a typical value for “clean” coarse-to-medium sand.

30

31

25 Why might the permeability in fine-grained soil deposits be expected to be greater for horizontal flow than for vertical flow? 26 Briefly give reasons why the coefficient of permeability in an undisturbed clay deposit possessing a flocculent structure would be expected to be greater than if the same clay had a dispersed (remolded) structure. 27 Estimate the vertical coefficient of permeability for a uniform clay deposit where representative test results are as shown: LL = 92, PL = 34 percent of clay size particles in the soil = 40 void ratio 1approximate2 = 1.75 28 Assume the conditions described in problem 27 refer to a compacted clay soil at an earth fill project. Estimate the coefficient of permeability by applying Equation 12. 29 A constant-head permeability test is performed, and the information below indicates the test conditions and results. From the given data, provide a sketch representing the test arrangement, and calculate the value for the coefficient of permeability. On the basis of the computed value for khyd, indicate the probable soil type. • Water flows horizontally through the soil sample. • The height of the reservoir supply source is 2 m above the elevation of the outlet reservoir. • The soil sample length in the direction of water flow is 200 mm.

236

32

33

• The cross-sectional area of the soil sample is 1,950 mm2. • The measured volume of flow for the steadystate condition is 500 ml in a 10-minute time period (note 1 ml = 1,000 mm3). A constant-head permeability test is performed in a laboratory where the soil sample is 25 cm in length and 6 cm2 in cross section. The height of water is maintained at 2 ft at the inflow end and 6 in. at the outlet end. The quantity of water flowing through the sample is 200 ml in 2 minutes. (a) Make a sketch of the described conditions. (b) What is the coefficient of permeability in millimeters per minute? A constant-head permeability test is performed where the hydraulic gradient is 0.75. The crosssectional area of the sample is 0.25 ft2. The quantity of water flowing through the sample is measured to be 0.004 ft3/minute. (a) What is the coefficient of permeability in feet per minute? (b) What is the coefficient of permeability in millimeters per second? How much water will flow through a soil mass in a 5-minute period when the sample length is 150 mm, the cross section is 20 mm by 20 mm, and a constant head of 2 ft is maintained? The soil has a khyd value of 0.1 mm/sec. The coefficient of permeability for a fine-grained soil is determined in a laboratory by use of a falling-head test. Test conditions and results are as indicated below. Determine the coefficient of permeability, then indicate the probable soil classification. • Length of soil sample = 150 mm • Cross-sectional area of sample = 1,140 mm2 • Cross-sectional area of standpipe = 200 mm2 • At the start of the test, the water level in the supply standpipe is 1 m above the top of the permeameter. • One hour after start, the water level in the supply standpipe is 0.95 m above the top of the permeameter.

34 A falling-head permeability test is performed on a fine-grained soil. The soil sample has a length of 120 mm and a cross-sectional area of 600 mm2. The water in the standpipe flowing into the soil is 0.60 m above the top of the sample at the start of the test. It falls 50 mm in 30 minutes.

Movement of Water Through Soil The standpipe has a cross-sectional area of 200 mm2. (a) Make a sketch of the described conditions. (b) What is the coefficient of permeability in millimeters per second? (c) What is the coefficient of permeability in feet per minute? (d) On the basis of the calculated value for khyd, what is the probable soil type? 35 A field permeability test is performed by measuring the quantity of water necessary to keep a boring casing (pipe) filled. The distance from the top of the casing to the bottom (in the ground) is 10 ft. The groundwater table is below the bottom of the casing. The casing has an inside diameter of 6 in. In a 10-minute period, 1 gallon of water was used to keep the casing filled. What is the coefficient of permeability for the soil at the bottom of the casing? 36 A field permeability test indicates that the coefficient of permeability for a certain soil is 2 * 10-2 mm>sec . Is this a relatively high or low coefficient of permeability? What type of soil would this probably be? 37 (a) Outline the practical reasons why field permeability tests to obtain a value for coefficient of permeability might be preferred over the laboratory testing method. (b) Outline reasons why the coefficient of permeability determined by a laboratory test procedure might be preferred over that from a field test procedure. 38 To what height would water rise in a glass capillary tube that is 0.01 mm in diameter?

39 What is the water pressure just under the meniscus in a capillary tube where the water has risen to a height of 6 ft? 40 A glass capillary tube is 0.001 mm in diameter. (a) What is the theoretical maximum height of capillary rise for a tube of this size? (b) What compressive pressure results in the capillary water just under the meniscus? 41 In a silt soil, the D10 size is 0.01 mm. If the effective pore size for estimating capillary rise is taken as 15 of D10, approximately what height of capillary rise will occur? 42 Why is it expected that the maximum height of capillary rise is greater for fine-grained soils than for coarse-grained soils? 43 How is the height of maximum capillary rise of water in soil affected by temperature? 44 What is the vadose zone in an area’s subsurface region, and what is the explanation for the condition? 45 The effects of soil capillary water are often cited as the reason why deposits of fine-grained soils (silts and clays) that generally are in relatively weak condition because of high water content can have a drier, firmer surface zone. Explain. 46 Explain how capillarity is related to the dried and firm condition frequently observed to exist in the surface zone of fine-grained soil deposits. 47 Indicate the ways that capillary water and the effects of capillarity can be removed from a soil.

237

Answers to Selected Problems

13. 14. 21.

22.

23. 24. 27.

238

RH = 0.5 r RH = 1.2 (a) khyd = 0.225 mm>sec (b) khyd = 0.13 mm>sec (c) khyd L 0.17 mm>sec 1using D102 khyd L 0.27 mm>sec 1using D152 (a) khyd L 0.5 ft>min L 2.5 mm>sec (b) khyd L 1.35 mm>sec (c) probably sand and gravel khyd L 0.15 ft>min L 0.7 mm>sec probably clean sand (a) khyd L 0.001 mm>sec (b) khyd L 0.0001 mm>sec khyd L 7 * 10-7 mm>sec

28. 29. 30.

khyd L 0.4 * 10-6 mm>sec khyd = 2.56 mm>min = 0.043 mm>sec khyd = 1.5 mm>sec

31. 32. 33. 34.

khyd = 0.021 ft>min L 0.11 mm>sec 80 ml = 0.08 liter khyd = 3.75 * 10-4 mm>sec khyd = 1.93 * 10-3 mm>sec L 4 * 10-4 ft>min khyd = 9.6 * 10-4 ft>min hc = 3.1 m pw = -375 psf pw = -304 kPa hc = 15 m

35. 38. 39. 40. 41.

Movement of Water Through Soil

From Essentials of Soil Mechanics and Foundations: Basic Geotechnics, Seventh Edition. David F. McCarthy. Copyright © 2007 by Pearson Education, Inc. Published by Pearson Prentice Hall. All rights reserved.

239

Movement of Water Through Soil Practical Effects: Seepage, Drainage, Frost Heave, Contamination

The handling of mobile and stationary underground water during construction operations and making provisions so that the effects of its presence will not interfere with the function of completed structures are of vital concern to the construction profession; this chapter’s discussion on flow nets and seepage, drainage, and frost heave relates to the practical aspects of controlling groundwater during and after construction. The section on soil and groundwater contamination and problem soil gases discusses factors responsible for the development of such conditions and methods for prevention or accomplishing remediation.

1

Flow Nets and Seepage Flow of Subsurface Water The flow of water beneath the ground surface through all soils except coarse gravel and larger materials occurs as laminar flow; that is, the path of flow will follow a regular pattern, with adjacent paths of water particles all flowing parallel. For this condition, Darcy’s law for water traveling through soils can be applied to determine the rate and quantity of flow and the seepage forces that result from this flow. In its most direct form, Darcy’s law is: q = khydiA

(1)

(as developed in Chapter 6, Equation 6–7), where q is the quantity of flow in a unit time period (or rate of flow), khyd is the coefficient of permeability for the soil, A is the crosssectional area of the soil through which flow is occurring (normal to the direction of flow), and i is the hydraulic gradient (the difference in the energy head of water between two points divided by the distance between the same two points).

240

Movement of Water Through Soil: Practical Effects River surface elevation A

Horizontal distance, L

Impermeable soil (clay, etc.) Subsurface flow between river and canal occurs through permeable stratum.

Impermeable soil (clay, etc.)

Canal surface elev. B

Δh = El. A − El. B

Permeable soil

Figure 1 Instance in which direct application of Darcy’s law can be used to determine underground flow.

When underground water is flowing over a relatively long distance and within a soil zone having well-defined boundaries, such as that shown in Figure 1, the quantity of flow can be determined by using the foregoing expression directly. Illustration 1 A river and a canal run parallel to each other but at different elevations, as indicated by Figure 1. If the difference in the water surface elevations is 5 m, the horizontal separation is 200 m, and the thickness of the permeable stratum is 2 m, compute the seepage loss between the river and canal, per km of river–canal length. Coefficient of permeability khyd = 1 m/day.

Solution ¢h 5m A = 11.0 m>day2a b12 m * 1,000 m2 hL 200 m = 50 m3>day>km of length

q = khydiA = k

The Need for Flow Nets and Flow Net Theory Where the zones of flow or directions of flow are irregular, where water enters and escapes from a permeable zone of soil by traveling a short distance, or where the flow boundaries are not well-defined (a boundary being the separation between where flow does and does not occur), it may be necessary to use flow nets to evaluate flow. Flow nets are a pictorial method of studying the path that moving water follows. Darcy’s law can be applied to flow nets to evaluate the effects of flow.1 For a condition of laminar flow, the path that the water follows can be represented by flow lines. In moving between two points, water tends to travel the shortest distance. If changes in direction occur, the changes take place along smooth curved paths. A series of flow lines to represent flow through a soil mass would be parallel except where a change in the size of an area through which flow occurs takes place. Figure 2 shows flow lines for water seeping through a soil in a simple laboratory model. Flow of water occurs between two points because of a difference in energy (usually expressed as head of energy or as pressure). Since it is the value of energy difference between two points, energy head can be referenced to an arbitrary datum for convenience. The total head of any point is that resulting from the sum of velocity head, potential head, 1A scale

drawing is necessary where actual quantities of flow are to be calculated or the effects of flow are to be determined.

241

Movement of Water Through Soil: Practical Effects

Figure 2 Laboratory seepage tank showing flow lines indicating path of water flowing through the soil.

Pressure head at B

Pressure head at A

Piezometer (open tube)

B Reference datum

Potential head at A

Potential head at B

A

(a)

Equipotential lines (b)

Figure 3 Points of similar energy head indicated on flow lines, and drawing of the flow net by addition of equipotential lines.

and pressure head.2 For water flowing through soil, velocity head is neglected, since it is small compared to potential and pressure head.3 In Figure 3(a), piezometers (open standpipes) inserted at points A and B would have the water level rise to the elevations shown. This figure shows the pressure and potential energy heads. upon the Bernoulli equation for steady-state laminar flow, total head h = v2>2g + p>g + z, where n is the velocity of flow, g is the acceleration of gravity, p is pressure, g is the unit weight of the fluid flowing, and z is the vertical distance between the point where the pressure p is determined and a reference elevation or datum. 3Because the velocity head can be assumed negligible, the total head difference between two points along a path of flow is equal to the difference in height that water in open tubes would rise—for example, ¢h1 = h1 - h2. Thus, flow occurs because of differences in total energy head. 2Based

242

Movement of Water Through Soil: Practical Effects

At certain points on different flow lines, the total energy head will be the same, as seen in Figure 3(a). Lines connecting points of equal total energy head can be drawn, and are termed equipotential lines (Figure 3(b)). It becomes apparent that there is no gradient to cause flow between points on the same equipotential line. On a flow net, equipotential lines must cross flow lines at right angles. (See Appendix: Application of the LaPlace Equation to Flow Nets, for mathematical proof.) The flow lines and equipotential lines together are the mesh strands which form the flow net (see the box below for a discussion of the development of the flow net), and are used to determine the quantities and other effects of flow through soils. When seepage analyses are made, flow nets can be drawn with as many flow lines as desired. The number of equipotential lines will be determined by the number of flow lines selected. A flow net therefore can be a fine-mesh net or a coarse-mesh net. A general recommendation is to use the fewest flow lines that still permit reasonable depiction of the flow path along the boundaries and within the soil mass. For many problems, three or four flow channels (a channel being the path between adjacent flow lines) are sufficient. The expression that results from the flow analysis to calculate the volume of water flowing or seeping through a soil mass (refer to the flow net development discussion below) is: q = khydhw ¢

Nf Nd

≤ 1width2

3for isotropic soil conditions4

(2)

where khyd = coefficient of permeability appropriate to the soil mass hw = elevation difference between energy head at upstream and downstream limits of the flow net Nf = number of flow channels for the flow net, a channel being the zone between adjacent flow lines Nd = number of equipotential drops for the flow net, a drop being the zone between adjacent equipotential lines width = width of structure or soil zone involved, a distance perpendicular to the cross section or perpendicular to the plane of the flow net Development of the Flow Net—Isotropic Soil Diagram 1 shows a condition in which flow occurs through a soil mass possessing isotropic properties (applied to fluid flow, the term isotropic implies that soil permeability is equal in all directions). Flow occurs because of the difference in total energy head caused by the unequal heights of water, ht - h0. 1. To begin the flow net, establish the boundaries of the soil mass through which the flow occurs and conditions at the boundaries. In Diagram 1, lines a–b and c–e–f–d are drawn to represent flow lines at the boundary. Line c–a is an equipotential line with total head equal to ht. Line b–d is an equipotential line with total head equal to h0. These four lines establish the boundaries for flow through the soil in this problem. Note the line that indicates the variation in head between the ends of the soil volume.

243

Movement of Water Through Soil: Practical Effects

Line indicating variation in energy head between a and b c

a b

ht

d

e

ho

f Reference datum

Diagram 1

2. Next, sketch in intermediate flow lines parallel to those flow lines along the boundaries. Start by spacing the flow lines an equal distance apart in the sections where it is known that the flow lines will be parallel to each other (where the direction of flow is in a straight line and not curved) (Diagram 2).

c

a b

ht

d

Flow lines e

ho

f Reference datum

Diagram 2

3. After the flow lines have been selected, locate a point of equal total energy head on each flow line. A line connecting these points will be an equipotential line (Diagram 3). Draw additional equipotential lines, selecting a spacing so that the distances y1, y2, and y3 permit the proportion: y2 y3 y1 = = l1 l2 l3 to be maintained. Note the corresponding changes in total head at each equipotential line (Diagram 4).

c

a b

ht

y1

y2 y3

e

Diagram 3

244

l1

l2

d f

Reference datum

ho

Movement of Water Through Soil: Practical Effects

4. Apply Darcy’s law, q = khydiA . Since a volume (of flow) is to be determined, width as well as length and height of soil mass must be considered. For convenience, assume a thickness or width (perpendicular to the page) of unity. Let the flow past y1 be q1, the flow past y2 be q2, and the flow past y3 be q3. Referring to Diagram 4, and noting that i (the hydraulic gradient) is the difference in head divided by the distance between the points where the head is measured, or Dh/DL (where l1, l2, are DL), obtain: q1 = 1khyd2

¢h1 1y12112 l1

q2 = 1khyd2

¢h2 1y22112 l2

q3 = 1khyd2

¢h2 1y32112 l2

Since y1 and y2 are bounded by the same flow lines, q1 and q2 must be equal, because flow is continuous. Also, since the ratios of the sides of each mesh block are equal, y3 is equal to y2 and therefore q3 equals q2.

Δh1 c

b

ht

y1

y2 y3

e

Diagram 4

hw = ht − ho

Δh2

a

l1

l2

d f

ho

Reference datum

If all the mesh blocks are proportional, the quantity of flow between flow lines is equal, and it follows that Dh1 equals Dh2. In flow net solutions, the mesh blocks are usually drawn as squares, although actually only the proportionality of distances between the two sides of a block must be maintained. By using squares, however, the ratio of the sides (height/length ratio) is 1, and the expression for flow becomes: q1 = khyd ¢h where Δh is the total head divided by the number of equipotential line pressure drops ᏺd, or: ¢h =

hw ᏺd

Total flow per unit of width is the summation of the flow in each flow channel, or the sum q = q1 + q2 + q3 + Á , which is also q1 times the number of flow channels ᏺf , or q1ᏺf . When analyzing a condition having a known width, total flow q is:

245

Movement of Water Through Soil: Practical Effects

q = q1ᏺf 1width2 = khyd ¢hᏺf 1width2 = khyd

hw ᏺ 1width2 ᏺd f

or: q = khydhw a

ᏺf ᏺd

b # 1width2

3for isotropic soil4

In constructing flow nets for most problems, it is improbable that the figures resulting from the assumed flow lines and equipotential lines will all be squares. The requirement for this condition is that, for each block, the distance across the center of the block between flow lines must equal the distance between the two equipotential lines, and the equipotential lines must cross the flow lines at right angles.

The combination of flow lines and equipotential lines create the mesh (or individual blocks) of the flow net. For most problems, the flow lines and equipotential lines are partially or totally curvilinear, and the mesh resulting will not be uniform throughout the net, but instead will vary in shape and size. The requirement for this situation is that, for each mesh block, the distance across the center of the block between flow lines must equal the distance between the two equipotential lines, and the equipotential lines must cross or meet the flow lines at right angles (refer to Appendix: Application of LaPlace Equation to Flow Nets). Figure 4 traces the steps used in drawing a flow net for a practical problem, and shows the computations to determine seepage.

Boundaries for the Flow Net Figure 5 provides illustrations of typical flow net problems. These illustrations can serve as a guide to establish the boundaries of flow and the general pattern of a flow net for many types of problems. For problems like that shown in Figure 4, the upper boundary is the flow line following along the base of the structure. The lower boundary is a flow line that, for application to the flow net, is a line following along the surface of the impermeable stratum. The locations and directions of other flow lines within these boundaries are selected by using the methods described earlier. Earth dams represent the condition in which the upper limit of flow is not defined by a natural boundary.4 In some cases, such as homogeneous earth dams situated on an impervious stratum (see Figure 6), seepage will break out on the dam’s downstream face. The flow line representing the upper limit of seepage moving through the dam is approximately parabolic in shape. The point where the seepage breaks out onto the downstream face can be established from: d d2 h2 cos u A cos2 u sin2 u 5 1for slope angles, u, less than 30°2 a =

4The

method for studying seepage through earth dams is presented in Arthur Casagrandes’ paper “Seepage through Dams,” which appeared in the Journal of the New England Water Works Association, June 1937. This paper has been reprinted in Contributions to Soil Mechanics, 1925–1940, Boston Society of Civil Engineers, Boston, Mass. 5Earth dams typically have slopes flatter than 2 horizontal to 1 vertical—that is, flatter than 30°.

246

L = 15 m

hw = 6 m Permeable stratum

khyd = 1 m/day

Impermeable stratum (a)

(b)

(c)

hw = 6 m

( ) (width) 3 = (1m/day)(6m)( ) (1m of width) 9

Seepage = q = khyd

Nf Nd

= 2m3/day per m of width (d)

Figure 4 Steps in drawing a flow net: (a) scale drawing of conditions; (b) trial flow lines; (c) trial equipotential lines; (d) final flow net and related seepage computation.

247

Filter Permeable stratum

Permeable stratum Surface of impervious stratum

Surface of impervious stratum

(a)

(b) Filter

Permeable stratum Surface of impervious stratum (c)

Surface of impervious stratum (d) Filter

Impervious blanket

Filter Permeable stratum Surface of impervious stratum (e)

Rock toe

Surface of impervious stratum (f)

Figure 5 Flow nets for representative seepage problems: (a) masonry dam with sheetpile cutoff at heel; (b) sheetpile; (c) masonry dam with sheetpile cutoff and filter at toe; (d) earth dam; (e) masonry dam with upstream impervious blanket and downstream filter; (f) earth dam with downstream rock toe filter.

248

Movement of Water Through Soil: Practical Effects L

d L/3 Point where seepage outbreak occurs A

B

Upper flow line for seepage

h

s

a θ

Impervious stratum Point A: theoretical beginning point for parabolic-shaped upper flow line. Point B: actual point for water entry on upper flow line.

Figure 6 Seepage through an earth dam; definition of terms necessary to locate seepage outbreak.

where values of a, d, and h are defined in Figure 6. The method used to establish the shape and position of the upper line of seepage through the dam is outlined in Figure 7. After the upper flow line is established, the remainder of the flow net can be sketched. The quantity of seepage q can be estimated from: q ⬵ khyd A 2d2 + h2 - d B 1width2

(3)

The effect of seepage forces is discussed in a later section of the chapter.

Flow Nets for Nonisotropic Soils The method of flow net analysis discussed above applies for the frequently found condition in which the soil permeability in the horizontal and vertical directions is similar. In stratified soil deposits, however, the horizontal and vertical coefficients of permeability may differ; usually the horizontal permeability is greater than the vertical. In such instances, the methods for drawing the flow net need to be modified. Use of a transformed section is an easily applied method that accounts for the different rates of permeability. Vertical dimensions are selected in accordance with the scale desired for the drawing. Horizontal dimensions, however, are modified by multiplying all horizontal lengths by the factor 2kver>khor, where kver and khor are the vertical and horizontal coefficients of permeability, respectively. A distorted diagram results, with shortened horizontal dimensions, as illustrated in Figure 8. The conventional flow net is then drawn on the transformed section using the procedures presented previously. For flow through the nonisotropic soil, the seepage equation becomes: q = hw ¢

Nf Nd

≤ 2kver>khor 1width2

3for nonisotropic soils4

The values for ᏺf and ᏺd in this equation are taken directly from the transformed section.

249

A

n

m

n1

m1

B

s n

m

C O p

1. Establish point A and swing an arc OB having the radius AO (point O will be the focus of the parabola); 2. Draw vertical line BC (this will be the directrix of the parabola); 3. Locate point p midway between O and C; 4. Draw the vertical line m-m in the general location indicated, and determine the horizontal distance from line BC to line m-m; 5. Using point O as a center, swing an arc with radius m1 B to intersect line m-m. This intersection establishes a point on the parabolic flow line; 6. Draw the vertical line n-n in the general location indicated, and determine the distance n1 B; 7. Using point O as a center, swing an arc with radius n1 B to intersect line n-n. This intersection establishes another point on the parabolic flow line; 8. Continuing the procedure just outlined, establish as many additional points as necessary to sketch the total length of flow line between A and s; 9. Modify the beginning section of the flow line freehand. The freehand line must intersect the upstream face of the dam at a right angle; 10. The parabolic flow line is assumed to follow the downstream face of the dam below point s; 11. Other flow lines are sketched in, assuming a shape generally similar to the upper seepage line.

Figure 7 Procedure for locating the upper seepage line for flow through an earth dam. 1 3

Actual length, La

Nonisotropic soil; kver = 1, khor = 9 or khor = 9 kver Impermeable stratum (a)

La

Permeable stratum

Horizontal distance = natural scale distance x = natural scale distance x = natural scale distance

x 13

(b)

Figure 8 Method of drawing the transformed section and flow net for nonisotropic soil conditions: (a) actual section (to scale); (b) transformed section (distorted scale).

250

kver khor 1 9

Movement of Water Through Soil: Practical Effects

Uplift Forces For the problem of seepage beneath a structure such as a dam, the uplift force acting on the base of the structure because of this seepage can be evaluated from the flow net. The pressure acting at the upstream and downstream ends of the structure (and at selected inbetween points if necessary) is determined and a pressure diagram is made. The total uplift force (per unit of width perpendicular to the drawing) is then the area of the pressure diagram (Figure 9).

Other Seepage Forces Water seepage beneath a structure and then escaping by flowing upward at the downstream end imposes an upward force on the soil. If the upward force is sufficiently great, soil particles will be carried away, eroded by the escaping water. There follows an illustration of how the flow net can be used to determine if the pressure of the escaping water is great enough to erode the soil. The submerged unit weight soil is: gsub = a

Gs - 1 bgw 1 + e

(4)

If the value of Gs (specific gravity of soil solids) is about 2.7 and the void ratio is about 0.70 (both reasonable values), the submerged weight of the soil would be about equal to the weight of water. The hydraulic gradient between the last two equipotential lines is the difference in pressure head across these two points divided by the distance between these Note: In this figure, the h values on the equipotential lines indicate the height above the ground surface to which the water level in a piezometer would rise. Pressure head drop across adjacent equipotential lines = hW 6m = = 0.67 m/drop 9 drops Nd L = 15 m

b

a h=6m h = 5.33 m

hw = 6 m h = 0.67 m h = 1.33 m

h = 4.67 m

For reference, use elev. 0 to indicate soil surface. Base of dam for this problem is at elev. −1 meter Permeable stratum Impermeable stratum

Pb = (0.67 m + 1 m) 9.81 kN/m3 = 16.4 kN/m2 = 16.4 kPa Pa = (5.33 m + 1 m) 9.81 kN/m3 = 62.1 kN/m2 = 62.1 kPa

Uplift force = 62.1 + 16.4 kPa (15 m length) 2 = 589 kN per meter of width perpendicular to page

(

)

Figure 9 Procedure used to compute uplift pressure acting on the base of a masonry dam.

251

Movement of Water Through Soil: Practical Effects

equipotential lines, or Δh/l. The pressure difference between the two equipotential lines is Δh times the unit weight of water, γw If the hydraulic gradient is unity, the upward force due to the moving water is equal to the unit weight of water. There then exists an upward pressure of 9.81 kPa or 62.4 psf acting on a soil where gravity effects a downward pressure of 9.81 kPa or 62.4 psf. This is a fully buoyant condition in the soil. A slightly greater uplift pressure would carry away soil particles. For this reason, an escape or exit hydraulic gradient of about unity or greater is usually considered an indication that erosion may occur. To prevent erosion, the pattern of flow could be modified to reduce the pressure of the escape gradient (increasing the length of flow either by making the structure larger or by embedding sheeting beneath the structure are two alternatives), or a coarse material of carefully graded particle sizes that is not susceptible to erosion by the escaping water could be placed beneath the tip of the structure (a soil filter). Illustration 2 A masonry dam having a sheetpiling cutoff at the upstream end is located at a reservoir site, as indicated by the sketch. Draw a flow net for the subsurface flow and compute the seepage. Also calculate the uplift force acting on the base and the escape gradient of the water at the downstream tip of the dam. 48' El. +20' El. 0'

El. 0' Elev. −5' El. −30' Soil

Ft. khyd = 0.01 Min. El. −60'

Rock

48' El. +20' h = 20'

Elev. −5'

l = 6'

Elev. 0'

hw = 20' h = 0'

B A Δh

=2

khyd = 0.01

'

h = 18' h = 16'

h = 8'

h = 14' h = 12'

h = 6'

h = 4'

Ft. Min.

h = 2'

h = 10'

Solution

The number of flow channels, ᏺf = 4. The number of pressure drops, ᏺd = 10 (note feet of head acting at each equipotential line, where differences Δh are calculated from): ¢h =

252

hw 20.0 ft ft = = 2.0 ᏺd 10 drop

Movement of Water Through Soil: Practical Effects

Seepage q = khydhw q = a 0.01 = 0.8

Nf Nd

1width2, or

ft 4 b120 ft2 a b11.0 ft wide2 min 10

ft3 per foot of width min

Uplift Force on Base pA = 15 ft + 7 ft2gw

pB = 15 ft + 2 ft2 gw Uplift = a

112 ft + 7 ft2162.4 lb>ft32148 ft2 PA + PB b1L2 = 2 2

= 29,200 lb per foot of width 1perpendicular to page2 L = 48 ft

pA

pB

Escape Gradient at Downstream Tip ¢h between last two equipotential lines = 2.0 ft, and l = 6 ft ¢h 2 = = 0.33 1therefore erosion is not expected2 l 6 (Figure 10 shows a dam where foundation erosion did occur.) i =

Practical Considerations The correct drawing of a flow net can be a tedious operation, particularly in the case of nonuniform soil conditions. Fortunately, even crudely drawn flow nets can provide reliable information on quantities of seepage and uplift pressures. Proper evaluation of an escape gradient requires a correct and carefully drawn flow net, however. For flow nets, as well as for all seepage studies, it should be expected that there will be a variation in the coefficient of permeability for the soil in the seepage zone under study. For practical problems, answers obtained from a seepage analysis should be considered as approximate, and are not to be believed as being precise.

Quicksand The dreaded quicksand condition occurs at locations where a sand or cohesionless silt deposit is subjected to the seepage force caused by an upward flow of groundwater. This condition can occur in depressions of areas where the water table is high (close to the ground surface) or where artesian conditions exist. The upward

253

Movement of Water Through Soil: Practical Effects

gradient of the water flow is sufficient to hold the soil particles in suspension, in effect creating a material with the properties of a heavy liquid. A no-support condition exists even though the soil gives the appearance of being firm ground. Other effects of upward seepage forces were described previously (refer to Figure 9 and Illustration 2). The suction or pull attributed to quicksand is in reality gravity exerting its normal effect but in a heavy liquid environment whose viscous properties also exert “drag” on a thrashing body. Theoretically, a person falling into a quicksand area can float as in water.

Figure 10 Small concrete dam (approximately 7-m or 21-feet high) that experienced underdam erosion of the foundation soils due to seepage and piping [226]. (a) Dam viewed from downstream side. (b) Depiction of dramatic blowout geyser at downstream side (when foundation piping channel broke through), as reported by witnesses. (c) Closeup of downstream face and toe of dam.

254

Movement of Water Through Soil: Practical Effects

Figure 10 (d) Foundation cavity tunnel approximately 0.7-m or 2-ft deep that developed as a result of piping. Light area near lower-right portion of photo is daylight from back (upstream) side of the dam. (e) Eroded foundation area at rear (upstream) side of dam, associated with the piping channel viewed in photo (d).

Since the quicksand condition is caused by the forces of seepage and not some mystical property of the soil, elimination of the seepage pressure will return the soil to a normal condition capable of providing support.

2

Drainage Among the more common problems in construction work is the need to handle subsurface water encountered during the construction sequence, and to handle subsurface water after construction so that the completed facility is not damaged nor its usefulness impaired.

255

Movement of Water Through Soil: Practical Effects

During construction, lowering of the groundwater table and removal of water from working areas is desirable from the standpoint of better working conditions for workers and equipment. In some cases, dewatering may be necessary to ensure that proper construction can be achieved. Conditions at some locations require that the structure be protected from erosive effects of flowing groundwater to prevent the loss of foundation stability. Additionally, usable interior portions of structures located below groundwater elevation are normally required to be free from serious seepage or leakage.

Conditions Requiring Drainage In planning excavations for construction projects that extend below the water table in soils where the permeability is greater than about 1 * 10-5 ft/min or 5 * 10-5 mm/sec, it is generally anticipated that at least some continuous drainage or dewatering procedures will be required if the work area is to be kept dry for construction. Excavations in the more impermeable soil types may remain dry after an initial dewatering, especially if the excavation is to remain open for only a limited time. (For all soil conditions, the actual amount of seepage expected would be affected by the depth below the water table and by the period of time that the excavation is to be open, as well as by the soil properties.)

Dewatering Shallow Excavations For construction of shallow foundations and for other excavations of limited depth made in coarse soil, open drainage or interceptor ditches can be an expedient and relatively inexpensive method for lowering the groundwater table a slight distance. The interceptor ditch has to penetrate deeper than the elevation of the work area because of the pattern that the underground water surface takes (termed the drawdown curve) in the area surrounding the interceptor ditch. Water collecting in such ditches normally has to be pumped out of the ditch for disposal. Since gravity flow is relied on to bring the water to the ditch, the continued inflow is dependent on the water level in the ditch being kept low. With this method, it is common to construct small pits in the ditch, termed sumps, for locating the necessary pumps (sump pumps). Where shallow foundations are to be installed at just about the water table elevation, it has frequently been found possible to obtain sufficient lowering of the water level to permit working in the dry area by locating sump pits within, or immediately adjacent to, the foundation excavation. The drawing down of the water table can also be accomplished by constructing a series of sump pits (Figure 11), or, if greater depth is required, some type of drainage wells around the construction area and pumping the water from these pits or wells. Subsurface water that flows in an upward direction into an excavation area that is being dewatered imparts a seepage force that tends to loosen the soil, reducing the soil strength. The change in strength should be considered in designing excavation bracing and foundations.

Dewatering Intermediate Depths Where excavations in coarse-grained and silty soils are to extend more than one meter or a few feet below groundwater level, open ditches or pits typically are not practical, and more advanced methods are used. Discussion of several methods follows.

256

Movement of Water Through Soil: Practical Effects Excavated area Original elevation of water table

Interceptor ditch Drawn-down water table (depth and configuration of drawdown curve is related to soil type and properties)

Figure 11

Sump pit in interceptor ditch

Shallow drawdown of water table by use of interceptor ditches and sump pit in excavation.

Well Points. For dewatering to intermediate depths (to about 9 m or 30 ft, but more if sufficient area is available for installing the necessary equipment), well-point systems are normally used. Basically, a well point is a closed-end pipe or tube, having perforations along its lower end, that is installed to the desired depth below the water table (Figure 12). Groundwater entering the well point through the perforations is pumped to the surface through a riser pipe connected to the well point (Figure 13). For construction dewatering, the well-point perforations are provided with a protective screen or filter to prevent soil particles from clogging the perforations. Well points are conventionally installed in drilled holes or, most usually, by jetting. In jetting, water is pumped through (into) the riser and discharges through special openings at the tip of the well point, an action that displaces the soil below the tip of the point. This procedure is continued until the desired penetration is achieved. To dewater an area, a series of well points is installed around the perimeter of the area. The groundwater level within the perimeter will be lowered when the well-point system is put in operation. The spacing of the well points varies according to the soil type and depth of dewatering. Spacings conventionally vary between 1 to 3 meters (or 3 to 10 ft). Each Riser

Screen Suction tube

Suction port

Jet point

External

Figure 12

Ball valve, closed on suction

Cross section

A well-point assembly.

257

Movement of Water Through Soil: Practical Effects Valve Union Riser pipe

Header pipe (carries collected water to point of disposal)

Top soil Clay Original

Depressed water table

water table

Sand

Well point

Figure 13

Installed well point.

well-point riser is connected by hose or pipe to a horizontal header or collector pipe, which is connected to a pump. Usually, many risers are connected to a common header. Wellpoint systems are usually rented from specialty contractors who also install the system. With most contractors, the riser pipes are about 50 mm (2 in.) in diameter, and headers are 150 to 300 mm (6 to 12 in.), depending on the flow to be handled. Normally, the screen provided with the well point is sufficient to keep medium sands and coarser materials from clogging the perforations. Where finer soils are to be dewatered, it is necessary also to provide a sand filter around the well point. The sand filter has the added benefit of acting to increase the size of the well point, permitting easier and greater flow into the point. To obtain a filter, it is necessary to create a hole having a larger diameter than the well point, prior to or when installing the well point and riser. The sand filter and backfill are then placed around the installed riser. A large-diameter hole can be formed by providing an attachment to the well point so that it can still be jetted to the desired depth, or by installing a casing by jetting or other method, after which the well point and surrounding sand filter are installed and the casing is pulled. For normal requirements, the sand filter material consists of a uniform graded medium-to-coarse sand. The horizontal extent to the soil zone surrounding a well point that will show lowering of the groundwater table is affected by the depth of the well point and the permeability of the soil. Generally, the horizontal distance that is affected increases as permeability of the soil increases. The drawdown curve values shown in Table 1 can be used to estimate the horizontal distance that will be influenced by well-point dewatering. Table 1 Approximate Slope of Drawdown Curve Due to a Single Row of Well Points Soil Being Dewatered Coarse sand Medium sand Fine sand Silt–clay

Slope of Drawdown Curve, %a 1–3 2–5 5–20 20–35

aRefer to Figure 14 and recall that percentage slope is vertical change (ft or m) per 100 units of horizontal length (ft or m).

258

Movement of Water Through Soil: Practical Effects

With the type of pumping equipment conventionally used for well points, the depth of dewatering that can be achieved by a single line of well points located around the perimeter of an excavation is usually about 6 to 7 m (or about 20 ft). This is due to the limit on the practical lifting, or suction, capacity of the pumping equipment. Lowering the water table through a greater distance may require the use of a two (or more) -stage (multistage) installation. Where a two-stage installation is required, the well points for the first stage of drawdown are located near the extreme perimeter limits of the area that can be excavated, and are put into operation. Well points for the second stage are subsequently located within the area that has been excavated, near to the bottom elevation that has been dewatered by the first stage. The second-stage well points then lower the water table to the additional depth necessary to complete the excavation dry (see Figures 14 and 15). Although the principle on which well-point dewatering is based is a simple one, care and skill are required during installation and operation to obtain the intended results. Proper assembly of the piping system is particularly crucial, because if air leaks develop at connections, the efficiency of the pump suction is reduced. Vacuum Dewatering. Where well points are in use, flow to the well point is through gravity. In the coarser soils (soils with less than 25 percent of particles smaller than 0.05 mm), Water table before well points Water level after 1st stage dewatering

Slope of drawdown curve

Well points

Water levels with 2nd stage well points in operation

Figure 14

Multistage well-point operation.

Figure 15 Installed two-stage well-point system in operation. (Courtesy of Moretrench American Corporation)

259

Movement of Water Through Soil: Practical Effects

gravity flow is normally adequate to achieve dewatering. For silts, gravity flow is restricted because of capillary forces that tend to hold the pore water. However, by applying a vacuum to the piping system, satisfactory dewatering of silty soils can be achieved. For maximum efficiency, the vacuum dewatering system requires that the well point and riser be surrounded to within a few feet of the ground surface with filter sand and that the top few feet be sealed or capped with an impervious soil or other material. By having the pumps maintain a vacuum pressure, the hydraulic gradient for flow to the well points is increased. With this system, closer well-point spacings are required than for the conventional system. Where suction pumps are used to draw the collected groundwater from the well points, as is common, the practical maximum height of lift is about 5 to 6 m (15 to 18 ft). Where an excavation is to extend more than this distance below the groundwater table, it will be necessary to dewater the area in two or more stages. With multistage dewatering, the well points for the first stage are installed on a perimeter line outside of the actual excavation area required for the construction and put in operation. Excavation proceeds within the perimeter formed by the well points to the depth where groundwater is encountered. Well points for the second stage are next installed within this excavation to further lower the water table for the actual construction area. Electro-osmosis. Sand and the coarser silt soils usually can be dewatered by using gravity draining or vacuum-assisted well points. In fine silts, clay, or coarse-grained– fine-grained mixtures, the effective permeability may be too low to obtain a successful dewatering by using well-point methods. Drainage of such low-permeability soils may be achieved by electro-osmosis, provided that no more than 25 percent of the soil particles are smaller than 0.002 mm. The basic principle of electro-osmosis is as follows: If a flow of direct-current electricity is induced through a saturated soil between a positive and negative electrode (anode and cathode, respectively), pore water will migrate toward the negative electrode. If a well or well point is made the cathode, collected water drained from the soil can be removed by pumping as with a conventional well-point system. Movement of water toward an electrically negative terminal occurs because of the attraction of the cathode for positive ions (cations) that are present in groundwater. Cations in pore water are the result of dissolved minerals going into solution with the groundwater. The cations concentrate around the negatively charged surface of clay particles to satisfy the electrical charge on the particle. These cations, in turn, attract the negative “end” of “dipole” water molecules. As the cations are drawn to the cathode, water molecules held to the cation follow. In a field installation, it is considered most beneficial to locate the anode close to the excavation perimeter and place the cathode further back, so that the direction of drainage is away from the excavation. This arrangement eliminates the danger that seepage forces caused by the draining water could act to cause the exposed slopes of the excavation to slide or slough in toward the excavation (construction area). Because of the techniques involved, the required special equipment, and the high electrical consumption, drainage by electro-osmosis is expensive compared to gravity drainage methods. Consequently, the method is used only when other methods cannot be applied.

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Movement of Water Through Soil: Practical Effects

Deep Drainage If excavations are to extend deep below the groundwater table, or to penetrate through a deep permeable stratum, well points may not be applicable because of the limits to which well points can raise water. For this situation, deep wells and deep well pumps (jet or venturi pumps) can be used. Deep wells are frequently relatively large-diameter (on the order of 600 mm (2 ft) drilled holes). A perforated protective casing is installed, and the deep well pump is then placed inside the casing near the bottom. Coarse filter material is placed between the outside of the casing and the walls of the drilled hole.

Consolidation Drainage Consolidation refers to the process that occurs in clay deposits when the development of compressive stresses in the soil mass from external loading causes the soil to undergo a reduction in volume (due to a reduction in void spaces) that is simultaneously accompanied by water in the soil pores being expelled to permit the decrease in voids. Thus, subsurface drainage of fine-grained soils by consolidation can be achieved by imposing a surface load onto an area. In general, a reduction in water content also results in an increase in the shear strength of the clay. Unfortunately, the rate of drainage is quite slow in clay soils because of the low coefficient of permeability. Sand drains represent one method that has been used with success to accelerate the rate at which consolidation drainage occurs in fine-grained materials. The method is based on reducing the distance that pore water has to travel to escape from the consolidating soil. Sand drains are vertical columns of freely draining sand installed so as to penetrate the soil strata to be consolidated. In clays, the permeability in the horizontal direction is frequently many times the vertical permeability because of the stratified manner in which clay deposits have been formed. The presence of relatively closely spaced sand drains thus reduces the distance that pore water must travel to escape and also permits an easier (horizontal) drainage path. In draining from the clay, water flowing to the sand drains is under pressure as a result of the external loading that is applied to cause the consolidation (usually an earth fill to act as a surcharge). Therefore, the water can flow upward as well as downward in the sand drain.

Summary: Techniques for Subsurface Drainage A graphical summary relating soil categories and appropriate techniques for accomplishing subsurface drainage is presented in Figure 16.

Drainage after Construction Preventing groundwater from seeping into a structure or through a structure after it has been built may be necessary in order to obtain proper use of the structure or to protect it from damage. In some cases, the dewatering methods used during construction can be kept in use to protect the structure. Normally, however, this procedure is not followed; the equipment or

261

Movement of Water Through Soil: Practical Effects Gravel Fine Coarse

Sand Medium

Silt

Fine

Clay

Subaqueous excavation – heavy pumping yield Well points – vacuum system needed if quick conditions exist Well points – theoretical limits for gravity drainage Well points – gravity drainage very slow Well-point vacuum system Electro-osmosis possible

10.0

1.0

0.1

0.01

0.001

Grain size in millimeters (log scale)

Figure 16 Summary: Relationship between method for dewatering soil and range of particle sizes [8].

methods may interfere with the usefulness of the finished project, may be too expensive for long-term use, or may be impractical in some other way. When it is known that a usable part of a structure will be located below the groundwater table, it is desirable to build the facility using waterproof design and construction techniques. One of the desirable features for a subaqueous structure is to have all seams and joints provided with water stops and/or to have the structure built with no seams or joints wherever possible. For example, in small conventional buildings, concrete basement walls and floors can be poured monolithically. Admixtures are also available to make concrete more resistant to water penetration; similarly, chemical coatings are available to waterseal the surface of concrete.

Foundation Drains Where groundwater will be flowing in the vicinity of the structure, provision can be made so that such water will be quickly carried away from the building area, and at worst only a limited height of groundwater buildup against the exterior sides of the substructure will occur. With adequate provisions, the large hydrostatic pressures that tend to force seepage entry thus will not develop. A method for achieving this control when the depth below the water table is not too great is the use of foundation drains, conventionally placed around the building exterior at footing level and adjacent to the footing (Figure 17). Such drains should not be lower than the bottom of the footing. Where there is concern over the ability of the exterior drain to handle the expected groundwater, the foundation drains may be located on the interior side of the foundation as well. With a high water table, it may also be necessary to place interceptor drains at some short distance from the building and at an elevation higher than the footings so that the water table is lowered in stages. The installed drain normally consists of pipe or a synthetic composite provided with perforations or installed with open joints so that groundwater can enter into the pipe. Gravel and/or sand filter material, or a geosynthetic fabric, must surround the drain pipe so that soil particles tending to be carried along by the inflowing water are prevented from clogging the drain pipe openings or causing erosion. The topic of geosynthetics and criteria for filter material are presented in subsequent sections of this chapter.

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Movement of Water Through Soil: Practical Effects

Masonry basement wall (exterior waterproofed)

Ground surface (sloped away from wall)

Soil backfill

Below-grade basement area

Original elevation of groundwater table Drawn-down water table Select coarse filter material to prevent erosion of natural soil (may be wrapped with a pervious geosynthetic to prevent intrusion of boundary soils)

Perforated footing drain (slope to area of low elevation,for gravity flow of collected water)

Figure 17

Footing drain tile installation for disposing of groundwater against a basement wall.

An outlet for collected water is a necessity; preferably disposal will be by gravity flow to a storm drain system or other drainage facility, such as a ditch, dry well, or pool located at an area of lower elevation than the building. If disposal by gravity flow is not possible, drainage water will have to be directed to a sump pit or other collector and pumped to a disposal.

Blanket Drains Where a basement is to be located below groundwater level, a blanket or layer of filter material can be placed beneath the floor slab to provide a highly permeable drainage path for the removal of groundwater acting against the bottom of the slab (Figure 18). Providing an escape path serves to reduce uplift pressures and the possibility for seepage to occur through the floor. The blanket connects to a sump where collected water is pumped out, or to drainage pipes where disposal occurs by gravity flow. The filter blanket normally consists of a sand layer placed over the natural subgrade soils and a coarser, small-gravel or crushed rock layer through which most of the horizontal flow is intended to take place.

Vertical limits of soil excavation for the foundation construction Interior basement area Concrete floor

Wall

Soil backfill

Original water table elevation

Waterproof joint

Gravel or crushed rock

Footing Movement of water to sump

Coarse sand

Sump-pit for collecting and disposing of water (screened openings)

Figure 18

Water level drawdown curve

Natural soil

Details of a blanket drain beneath a concrete floor.

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Movement of Water Through Soil: Practical Effects

Interceptor Drains In paved highways and airfield runways, the design frequently calls for interceptor trench drains located parallel to the shoulder, as illustrated in Figure 19. The purpose of the drains is to lower the groundwater table to a level beneath the pavement and to permit easy lateral drainage (escape) for water finding its way into the coarse base material provided beneath the pavement (water resulting from upward capillary migration and from infiltration through cracks, joints, or voids in the pavement). The intent of the drainage facility is to keep the base and subgrade soils dry so that they will maintain high strength and stability. With coarser soils, the intent is probably achieved quite successfully. In fine-grained soils, the degree of achievement is questionable. Most probably, a groundwater table in such soils is not significantly lowered because of the material’s low permeability. However, the drainage ditches do serve to prevent excess pore water pressures from building up in the subgrade soils (excess pore water pressures have the effect of weakening the soil). The drains also provide a means for disposal

Construction cut Original ground Seal

Original ground

Seal

Base and subbase

Cut Water level Longitudinal drain with drainage (pipe conduit or geosynthetic)

Longitudinal drain (pipe conduit or geosynthetic) (a)

(b) CL

175 ft Original ground line

2 ft minimum ditch bottom

Original water level

CL

Pavement

Original water level Water level with drainage

Pavement

Runway paving surface

75 ft Impervious backfill 1% slope

Subsurface draincontrolled gradation backfill

75 ft

175 ft

1% slope

Granular subbase course

1.5% slope

Base Place drain on both sides where needed

(c) Road pavement or wearing surface

Open side ditch

Granular base Movement of subsurface water towards ditch

(d)

Figure 19 Highway and airfield drains: (a) interceptor drain for highway constructed in a sidehill (after Cedergren [60]); (b) interceptor drain for highway in flat terrain [60]; (c) typical airport runway interceptor drains (Courtesy of Civil Aeronautics Administration); (d) typical open-shoulder ditch for roads.

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Movement of Water Through Soil: Practical Effects

of surface and near-surface water, and will intercept underground flow trying to enter the roadway from the side areas. Similarly, paved and unpaved roads will benefit somewhat from open drainage ditches located adjacent to shoulder areas (Figure 19(d)). Such ditches handle surface water, provide a lateral drainage outlet for soil near the level of the road surface that helps prevent the development of excess pore water pressures, intercept surface and near-surface water flowing toward the roadway area from the sides, and provide storage for plowed snow and a flow channel for melting snow so that it does not run onto the roadway. Currently, the common methods for achieving highway and airfield drainage involve state-of-the-art technology and geosynthetics as shown in Figure 22.

Flow Through a Structure In some types of structures, such as earth dams and dikes, it is known that water will be flowing through the structure after it is put in use. The design of the structure, therefore, considers the effect that such flow will have on the stability of the structure. Flow is not prevented but, rather, is permitted and directed so as not to create a stability problem. For instance, in a homogeneous earth dam or dike, flow is typically as indicated in Figure 20. If the velocity of the water flowing through the structure is too great, erosion and instability of the downstream side will result. The placement of a toe drain or an underdrain blanket that consists of coarse materials helps direct the flow to those areas that will not erode or lose stability because of the flowing water (Figure 21). There are basically similar situations encountered in building construction, where localized underground flow occurs through seams or a stratum of pervious soil (frequently referred to as underground springs) and presents a potential seepage problem for the structure. Providing some type of drainage conduits for the flow to pass around or through the

Earth dam

Impermeable stratum

Figure 20

Pattern of flow through a homogeneous earth dam.

Earth dam Rock-fill toe

Drain

Earth dam Impervious stratum (a) Earth dam with rock fill toe

Figure 21

Impervious stratum (b) Earth dam with downstream drain

Flow through an earth dam modified by a toe drain or an underdrain.

265

Movement of Water Through Soil: Practical Effects

structure is frequently technically more desirable and less expensive than attempting to cut off or stop the flow. The main precaution to be observed is that the flow conduit be designed so that seepage forces do not affect the building and that soil erosion does not occur.

Filter Design (Aggregate Filters) Water flowing toward a drainage structure must be able to enter the drain easily and quickly but must not be permitted to carry with it small soil particles that could eventually clog the drain or cause erosion or cavitation in the areas surrounding the drain. Currently, the desirable separation is achieved by use of the geosynthetics discussed in the section, “Drainage, Filtration, Separation, and Protection with Geosynthetics (Synthetic Fabrics).” But for the situation where a drainage structure is open to the soil, it is necessary to provide a drainage filter. Sand and gravel blends are typically used for such filters. To permit quick, easy flow into the drain facility, and to “draw” water to the drain, it is necessary to have the permeability of the filter material considerably greater than the permeability of the surrounding natural soil. To satisfy this criterion, the following ratio between particle sizes for the filter material and natural soil has been established: D15 of filter material Ú 4 d15 of natural soil To prevent particles of natural sandy soils from migrating into the filter material and eventually clogging it, the following ratios of particle sizes should be used as a design criteria: D15 of filter material 6 5 d85 of natural soil D50 of filter material 6 25 d50 of natural soil D15 of filter material 6 20 d15 of natural soil Of these criteria, the D15/d85 ratio is considered to be the most important and indicative of filter performance. In these ratios, the D15, d15, and other, values refer to the particle dimensions obtained from a particle-size distribution curve (e.g., the D15 or d15 refers to the particle size for which 15 percent of the soil material, by weight, is smaller). In the past, filter design criteria have suggested that natural and filter materials have parallel but transposed particlesize distribution curves, but this feature of filters has been found to be unnecessary and perhaps even undesirable. Where a sand and gravel filter is to be in contact with a fine-grained soil, some additional criteria beyond those just outlined should be applied to the selection of filter materials. For fine clays (a d85 range of 0.03–0.10 mm), the D15 for the filter should not exceed 0.5 mm. For fine-grained silt (a d85 range of 0.03–0.10 mm) with low cohesive strength

266

Movement of Water Through Soil: Practical Effects

and low plasticity (liquid limit less than about 30), the D15 for the filter should not exceed about 0.3 mm. For sandy silts and clays having noticeable sand content (a d85 range of 0.1–0.5 mm), a sand filter material is adequate when the D15/d85 ratio is less than 5. The D50/d50 and D15/d15 ratios should not be considered when selecting a sand or a sand–gravel filter that will be in contact with the fine-grained soils described. To ensure that filter particles will not enter and clog the drain structure where holes or slots are provided for the entry of water, the following criteria should be followed: For circular holes: D85 of filter material 7 1.2 Diameter of opening For slotted openings: D85 of filter material 7 1.4 Width of slot For proper performance, the filter material must surround the drain, under as well as over. The filter material should also be compacted so that it does not settle. To help minimize any tendency for fine particles to migrate through the filter, and to drain as much water as possible, the holes or slots in the drain structure should be placed facing downward, or as near to downward as possible, so that the flow into the drain facility is not downward but instead is lateral or slightly upward. If the natural soil is fine-grained, it may be necessary to provide a filter that consists of two or more layers, each having a different gradation. The material closest to the drain would be coarsest, with the outer material filter that is next to the natural soil being less coarse. For the situation in which two or more filter materials are used, the criteria presented earlier for filter and natural soil would also apply between any two adjacent filter materials. The filtering function accomplished by selection of soil materials as described in this section can also be achieved using the new technology of geosynthetics described in the section following. However, the use of soil filters is not a discontinued practice. Much of the information used to design filter systems with geosynthetics is based on soil filter technology.

Drainage, Filtration, Separation, and Protection with Geosynthetics (Synthetic Fabrics) The successful performance of subsurface and surface construction, and earth embankment structures, often relates to the proper handling of surface and subsurface water, and to preventing the otherwise detrimental effects of its presence. Underground, usual factors requiring consideration include a means for accomplishing the drainage of subsurface water that might cause problems, a means to achieve filtration where groundwater movement can be tolerated but must occur without migration of soil particles, and a means to maintain separation where differing soil materials in contact must stay zoned. For some surface areas, the potential for soil erosion caused by water flow must be considered (for example, to determine if protection against loss of a foundation’s supporting soil that

267

Movement of Water Through Soil: Practical Effects

would result in undermining is required, to prevent problems that result where an eroded soil is eventually deposited, etc.). The subsurface drainage, filtration, and separation functions can be accomplished particularly well by the geosynthetic products when compared to accomplishing these functions in the “conventional” manner with earth materials (see representative uses illustrated in Figure 22). It is expected that in the near future the geosynthetics will be the first choice for all drainage, filtration, and separation designs (except where soil mass is required to aid stability or necessary to provide volume for other necessary functions, such as aerobic and anaerobic decompositions). Though the use of the geosynthetics is relatively new, these materials are rapidly becoming popular with designers and construction contractors because of their ability to perform necessary functions while offering practical advantages, such as a wide availability of products from the marketplace, the relative ease of shipping and field handling, rapid installation, durability and long life when properly selected, and general environmental safety (will not degrade). Various details relating to the geosynthetic products are presented in the two boxes, “The Role of Geosynthetics in Construction” and “The Materials, Manufacture, and Properties of Geosynthetics Used for Construction Projects.” Most of the geosynthetic materials used for filtration and separation are from the geotextile grouping described in the insets. Geosynthetics used because of their capability of accomplishing the drainage function are from the geotextile, geonet, and geocomposite groupings. Many of the geosynthetics are clothlike or sheetlike and available in wide sheets or rolls; application involves spreading the sheet over the earth zone being improved, and large areas can be covered quickly. Since they are manufactured, the synthetics can be produced to possess different properties; for example, geotextiles are available in a range of strengths, in a range of coarse to fine textures, and in a range of permeability. The concern for separation of soil materials is significant in works such as zoned earth dams, roadways, airfields, and railroads. With projects of this nature, one zone of material is designed to be coarser than an adjacent zone (e.g., an earth dam may consist of a finegrained soil core with shell zones of coarse material, whereas a roadway base of coarse granular material rests on a subbase or subgrade of finer soil). It is necessary to prevent migration of fine material into the coarse material, or intrusion of coarse material into the fine material. Geotextiles can be used at the boundary between such different soil materials to maintain a separation but still permit the movement and passage of water. Geotextiles and geocomposites similarly have application with foundation drains and underdrains. (It has also been proven that the presence of geotextiles or geogrids incorporated into a structure such as an earth dam or a road pavement foundation will act as reinforcement, helping support and redistribute applied loads because of the material’s tensile and shearing strength). Geosynthetics with low permeability or impermeability (geomembranes and geocomposites) can be used as buried or exposed linings to prevent seepage or infiltration in reservoir and impounding areas. The movement of water through an earth dike can be restricted by installing a geomembrane lining on the dike’s upstream side. Seepage beneath a dam can be reduced by placing an impervious geomembrane or geocomposite to act as an upstream blanket. Conversely, geotextiles and geocomposites with high planar permeability (ability to conduct flow within the thickness of the fabric) can be used as a passageway for water (and gases). They have applications as protective drains where it is necessary to

268

Core zone

Separation, filtration, and drain geotextile

Road pavement or railroad ties and rails

Separation and filtration fabric (geotextile or geogrid)

Drain fabric

Earth dam Shell zone

Granular base Subgrade (b) Geosynthetic to reinforce roadbase and to maintain separation from subgrade

(a) Geosynthetic to maintain separation of different soils and carry seepage Ground surface

Road pavement (concrete or bituminous)

Backfilled trench

Pervious enveloping fabric (separation and filtration)

Coarse soil; high permeability

Protective soil cover Impervious geosynthetic (geomembrane)

Earth dike

Soil subgrade Geocomposite edge drain (d) Highway and airfield pavement edge drains (geocomposite drains replace conventional drains of perforated pipe and crushed stone)

(c) Arrangement for foundation drain or roadway underdrain Fabric mattress (erosion-resistant)

Crushed rock or gravel base Paved shoulder

Geosynthetic (geotextile or geocomposite) with high planar permeability (drainage and filtration)

Backfill soil

Natural soil Collector drain

(e) Geosynthetic to reduce seepage through or beneath an impounding structure

(f) Geosynthetic to drain water from soil behind a retaining wall Drive mandrel (mandrel withdraws after installing wick)

Granular soil working pad (surcharge load added after wicks are installed)

Wick drains in place Dashed arrows represent path of flow for draining water

Surcharge load (soil fill, etc.) Fine-grained or organic soil to be drained Firm-compact soils (g) Geocomposite wick drains used to hasten consolidation drainage of weak organic and fine-grained soil deposits

Figure 22

Representative uses of geosynthetics for drainage, filtration, and separation.

269

Gases discharge to atmosphere through vent stack outlet above roof level Basement walls

Geosynthetic snow fence or silt fence (snow fence is open texture to permit passage of some snow; silt fence material permits passage of water but few soil particles)

Basement floor slab

Roadway Footing drain disposal by gravity flow Footing drain

Geosynthetic silt fence (prevents silt in surface runoff from clogging Surface creek) water runoff Creek

Crushed rock or clean gravel base (6 inches to 8 inches thick) Drainage geosynthetic to permit entry and passage of soil gases (radon, etc.)—can connect into foundation footing drains provided outlet is external

(i) Silt fencing and snow fencing to protect road areas, creeks, etc.

(h) Soil gas venting for building substructure or slab-on-grade construction Turf surface Topsoil-loam layer 75–150 mm thick Sand layer 150–200 mm thick Small gravel-coarse sand layer 100–200 mm thick or geocomposite with high planar permeability

Separation-filtration geotextile (pervious)

Separation-filtration geotextile (pervious)

Subgrade Water collection trenches or perforated conduit (j) Athletic playing field (section, design for field drainage)

(k) Completed geomembrane liner in place for waste liquid retention pond

Figure 22 (continued)

270

(l) Seaming operation, for joining geotextile sections

(m) Installing geotextile and geogrid as part of a road construction project to help stabilize an area where weak subgrade soils exist (to be covered by crushed stone base layer)

(n) Highway shoulder subsurface drainage system utilized to intercept groundwater flowing from the adjacent hillside

(o) Pavement edge drain installation (perforated flexible drain conduit surrounded by small crushed rock, wrapped in filter-separation geotextile)

(p) Silt fencing of geofabric used to protect natural landscape area adjacent to an earth works project

(q) Foundation system for a new river-crossing bridge consisting of cast-in-place reinforced concrete piles, geotextiles and rock riprap (temporary bridge shows in upper background)

Figure 22 (continued)

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Movement of Water Through Soil: Practical Effects

collect and carry seepage, such as within earth dams and solid waste landfills and behind retaining walls, and as road and airfield pavement drains. Geocomposites with high planar permeability are being used for wick drains, to accelerate the consolidation process for thick deposits of weak compressible soil, and in marsh areas. Manufacturers have developed geotextiles and geocomposites for venting radon and other gases that are present in the earth under and surrounding buildings and under waste liquid retention ponds and landfills. It is expected that the development of special products involving the geosynthetics will continue as the opportunity to compete with conventional procedures or materials, or to solve problems differently, is recognized. The use of geosynthetic fabrics as a means to control soil deposition that results after a surface is eroded by storm water runoff is appropriate to both the temporary and permanent categories of construction. Erosion control is particularly an issue for projects involving earthwork; see Figure 22(p) illustrating use of silt fencing to contain soil carried by runoff. The means to provide protection against the erosive effects of rapidly flowing water can also be achieved by designs that combine “old” and “modern” techniques; Figure 22(q) illustrates a now-standard design concept for bridge foundations that consists of long piles plus use of geosynthetics and heavy rock rip-rap protection at the river bed’s soil surface to prevent erosion scour and undermining from starting. Further information on the use of geosynthetics to prevent groundwater contamination appears in Section 4.

The Role of Geosynthetics in Construction Of the new engineering materials developed within the last quarter century, the greatest impact on the construction industry has been made by those grouped within the geosynthetics classification. The large-scale usage of such synthetics for construction projects is a relatively new development, but use is becoming more common because of the performance experience and confidence gained by designers and contractors. The geosynthetics—made of synthetics instead of natural materials—will perform some function of a construction assemblage that is in contact with the ground (hence the prefix “geo-”). The geosynthetics typically have been produced in the form of fabrics, creating clothlike or sheet materials with which to work. Special assemblages will have other forms, however. The variety of fabric types and the wide range of applications, the general availability, the quality of production, improved methods for field handling and seaming, and the continuing development of specialty products are all factors contributing to the increasing popularity of the construction synthetics. The basic component materials of the geosynthetics are manufactured by the petrochemicals and plastics industry and mainly consist of polymers such as the polyesters, polypropylenes, polyethylenes, nylon, and the polyvinylchlorides, although other durable materials, such as fiberglass and rubber, are also used. The synthetics are selected because of their resistance to degradation or decomposition in the presence of moisture and the atmosphere and when buried (although degradation may occur when exposed to some chemicals); fabrics of natural fibers, such as cotton and wool, will deteriorate when buried and exposed to moisture. The geosynthetics are permitting many important aspects of construction to be accomplished better, faster, or more economically (such as buried drainage facilities, maintaining separation

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Movement of Water Through Soil: Practical Effects

of dissimilar materials, providing ground reinforcement, accomplishing containment of liquids and solids, and providing erosion control), and are also permitting some accomplishments that previously were not practical (such as providing reinforcement, containment, sealing, or capping). Items comprising construction geosynthetics may be grouped into the geotextiles (or geofabrics), the geogrids and geonets, geomembranes, and geocomposites (see Figure 23). The geotextiles consist of synthetic fibers assembled by either a woven, nonwoven, or knitting process to form a porous fabric sheet. The geogrids consist of plastic materials where the arrangement of heavy strands outline relatively large and thick gridlike openings in the resulting sheet. The geonets consist of plastic or polymer sheets that have been pressed or extruded to produce an open net or weblike form. The geomembranes are solid sheets of impervious plastic or synthetic rubber material. The geocomposites represent any combination of geotextile, geogrid, geonet, or geomembrane. The popularity of geosynthetics relates to their being particularly appropriate for functions where immunity to degradation is crucial and the fact that properties such as strength, weight, and flexibility can be controlled or selected because the base materials are selected, sized, and assembled to provide particular features. Geosynthetics are used for construction projects to provide one or more of the following functions (in other words, the function is achieved because of the presence of the geosynthetic): 1. Separation 2. Filtration 3. Drainage 4. Reinforcement 5. Barrier or containment 6. Protection 7. Erosion control The separation function is important where it is necessary to prevent mixing of dissimilar materials that are in contact, such as two different soil types (for example, the coarse granular base for a road pavement and a fine-grained subgrade soil, or different soil materials in a zoned earth dam). In the filtration function, the geosynthetic permits the passage of water but prevents the migration of soil particles from one side of the fabric to the other. The drainage function relates to providing a passageway within the thickness of the geosynthetic for controlling the movement or flow of liquids and gases, to encourage flow to occur along a desired path, or to direct flow to a selected location for collection, discharge, or escape. The reinforcement function involves placing geosynthetics on a surface such as a road subgrade or within a soil fill to improve the ability of the affected soil mass to carry loads by increasing the resistance to tensile and shear stresses. The barrier or containment function involves use of an impervious geosynthetic for situations where structures require a waterproofing membrane, or to function as a no-leak ground lining for liquid and solid waste disposal sites and the top capping seal. The protection function relates to including a protective geosynthetic for strength or resistance to surrounding conditions as part of a geocomposite in a situation where the material used to provide a major function, say drainage, is vulnerable to conditions present in the surrounding environment. The erosion control function is concerned with holding soil surfaces in place and preventing erosion. Some geosynthetics permit protective vegetation to

273

Figure 23

274

Representative samples of various geosynthetics.

Movement of Water Through Soil: Practical Effects

grow through the texture of the fabric so that a natural (rooted) resistance to erosion develops (the geosynthetic may be designed to gradually decompose or degrade). Materials in the geotextiles group are capable of providing any of the functions described above, as long as a proper synthetic and composition are selected. The synthetics used are durable, but the possibility for degradation if exposed to chemicals and sunlight requires consideration. The geogrids have been developed to fulfill the reinforcement function but also may have application for special cases where the separation function is required. The geonets are almost exclusively used for the drainage and erosion control functions. Geomembranes are used for their containment or barrier properties as protective liners and covers.

Land Drainage One of the earliest methods for reclaiming flat expanses of marginal wetland areas for use consisted of constructing a network of deep drainage ditches to lower the water table and obtain a stable surface suitable for agriculture and for supporting light structures. With a network of drainage ditches, drainage is by gravity, and best results are obtained in soils with relatively high coefficients of permeability. In clays or fine silts, where permeability is low and capillary forces holding the pore water are high, drainage from gravity effects will be limited. Water can be replaced by normal precipitation as quickly as drainage occurs. However, where the subsurface soils include strata or seams of coarse soils, as frequently exist in coastal and lake areas, predominantly fine-grained soil deposits have been drained. The drainage ditch network, generally laid out in some form of grid pattern, requires outleting to areas of elevation rather than the draining area.

Effects of Drainage Water flowing toward a drain also produces seepage forces that act on the soil particles. These forces are capable of moving soil particles, particularly small particles. Unless the migration of particles by a properly functioning filter is prevented, the small particles will be washed into the drain, eventually eroding soil from the area surrounding the drain. This occurrence could cause the drainage structure to collapse because of inadequate support from the surrounding soil. This process is referred to as internal erosion. If an open drainage system (e.g., open ditches) causes seepage forces to act toward an open excavation, there is the possibility that the forces will precipitate a sloughing-in of the soil embankment. Normally, removing water from soil (particularly a fine-grained soil) will increase the shear strength of the soil, in effect making the soil mass stronger. Lowering of a water table, however, also results in an increased effective vertical pressure acting on the soil mass, because the soil weight changes from a submerged weight to a saturated but unbuoyed weight. The usual result of an additional loading to the area within the zone of drawdown is settlement. The magnitude of the settlement depends on the amount of water table lowering and the strength and compressibility of the soil. If an undeveloped area (one

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The Materials, Manufacture, and Properties of Geosynthetics Used for Construction Projects Most of the geosynthetics used for construction projects are manufactured from polymers such as the polyesters, polypropylenes, polyethylenes, nylons, and polyvinylchlorides (polymers refer to the products obtained by the joining of two or more like molecules to form a more complex molecule whose physical properties are different). Textured fabrics (the geotextiles) developed from filaments of polyester or polypropylene are widely used. Some fabrics use bicomponent fibers (fibers of two or more polymer types). Compared to natural fibers, the polymer synthetics offer long-term durability in the presence of elements commonly encountered in construction (e.g., moisture and other types of waste fluids). The fabric filaments (fibers or strands) may be obtained by an extrusion process or by the slit-film process. Extruded filaments result when a heated and liquefied polymer is forced through small openings, such as the holes in a spinnerette. The extruded strands cool and solidify and are then ready to be combined to form fabric. In the slit-film process, the polymer is melted and forced (extruded) into a thin sheet, or film. When cooled and solidified, the polymer sheet is cut (slit) into thin strips, which are then ready to be combined to form fabric. Filaments are assembled into the geotextile fabric using either a woven or a nonwoven procedure. Woven fabric is obtained by the conventional technique of weaving warp yarn (filaments extending in the long direction of the weave) and filler yarn (filaments in the transverse or cross direction). Properties of the woven fabric are controlled or affected by the spacing of the warp and filler yarns, as well as by the size and properties of the filaments. Generally, strength-related properties of the completed woven fabric (tensile strength, resistance to distortion and tearing, etc.) tend to be bidirectional. The nonwoven procedures include the spun-bonded technique and the dry-laid-nonwoven process. In the spun-bonded process, continuous filaments resulting from extrusion of polymer through a spinnerette are commonly laid down in a circular pattern on a laydown belt to form the fabric web. The web strands can be bonded together using a needle-punching procedure, by heat bonding, or by resin (chemical) bonding. The needle-punching procedure involves use of a needle loom containing a number of barbed needles. The barbed needles repeatedly punch through the fabric web, forcing the randomly laid-down strands to interlock mechanically. The needles’ barbs are designed so strands are grabbed only on the downward stroke (picture the reverse of fishhook barbs). The resulting fabric texture is soft, with a noticeable thickness. To heat-bond a fabric, the unbonded web is passed through heated rollers in order to cause strand surfaces to melt and fuse together at points of contact. The fuse bonding is retained after the fabric cools. With resin bonding, the unbonded web is treated with a liquidcementing resin that adheres strands at their points of contact when the fabric passes through heated rollers. Excess resin is subsequently flushed from the fabric. Dry-laid-nonwoven fabrics are manufactured from staple fibers, which are fibers with limited length (typically, fiber lengths range between 1.5 and 6 in., or 4 to 15 cm). The fibers are laid down on a moving belt by a cross-lapper, to create a crisscross layering of short strands (staple). The staple is bonded by either needle punching, heat, or resin.

276

Movement of Water Through Soil: Practical Effects The manufactured fabric is wound into rolls for storage and shipping. Roll widths vary according to manufacturer and product, but many products are in the range of 4 to 5 m or 12 to 16 ft wide. At the construction site, the fabric is unrolled and cut if necessary. Large areas are covered using multiple rolls. The edge and end seams are mechanically sewed, glued, fused, or simply overlapped. Most geogrids are made from high-density polyethylene, polyester, or polypropylene. One method of manufacture involves taking solid sheets of relatively thick or heavy-gauge material, then, by hole-punching and control-stretching the sheet, producing a grid skeleton that has the desired strength strands and size openings. A second method of manufacture involves arranging strips or bundled fiber strands into the desired grid pattern, then bonding the intersections. At present, most geonets are manufactured from polyethylene by an extrusion and stretching process. Heated liquefied polyethylene is extruded through a rotating die called a stenter to create a netlike pattern of closely spaced ribs or strands. The newly formed netted sheet is forced over a cylindrical mandrel which has an increasing diameter, a process which stretches the still-plastic material to achieve the desired-size net strand and net opening. The cooled and stabilized cylindrical-shaped sheet is cut in the direction of the long axis of the mandrel in the desired length, then placed into rolls. Most of the geomembranes are made from modified forms of the synthetic polymers— polyethylene, polyvinylchloride, and polyvinylidine-chloride materials that will result in obtaining virtually impervious sheets. The appropriate polymer resin is blended with additives, such as plasticizers, vulcanizers, and other processing aids, then extruded into sheets. Sheets can also be obtained by extruding the liquid onto a cylinder or blowing it into the shape of a long giant bubble; the cylindrical-shaped sheeting is then cut to the desired width and length. Sheets may also be further processed by passing through rollers (calendering) to improve the properties. Geomembranes are also produced by saturating an otherwise porous geotextile with polymer resin to obtain an impervious sheet. Multiple geomembranes or geocomposites can be obtained by laminating the single-ply geomembrane with other geosynthetics. Some generalities exist concerning properties, performance, and suggested uses for synthetic fabrics. (Note: Some degree of caution is warranted with generalities; exceptions are to be expected. Furthermore, not all of a fabric’s properties are a consideration for use.) Needlepunched fabrics are generally considered more flexible than chemically and heat-bonded fabrics. Nonwoven fabrics are generally considered to have similar strength properties across all directions in the plane of the fabric, whereas woven fabrics are probably strongest in the direction of fabric strands. Fabrics manufactured from continuous filaments generally are considered stronger than fabrics made with staple fibers of similar material. Polyester filaments have good resistance to ultraviolet rays (sunlight), but unprotected polypropylene is subject to deterioration. Polyesters are not expected to be damaged by exposure to diesel fuels, but polypropylenes could be. Polyester fabrics generally retain better flexibility at low temperatures than do polypropylene fabrics. Geotextiles woven from filament fibers and used for the drainage function in cohesionless soils have less tendency to clog than do the nonwoven and slit-film wovens. Geomembranes to be used for liners or capping seals at liquid and solid waste disposal sites should be tested for durability when exposed to the chemicals to be contained. Descriptions of representative physical properties commonly used as measures to determine fabric suitability are presented in Table 2; testing to determine the properties is usually performed by the manufacturer.

277

278 — — —

Friction (surface friction)

D-4355 Corps of Engineers Test CW-02215 —

D-4716

D-4491

D-3786 D-3787, D-4833

D-4533

D-4632, D-4595

D-4751

ASTM Test Procedures

Seam strength Creep

Fatigue strength

Permitivity (also referred to as cross-plane permeability) Transmissivity (also referred to as planar permeability) UV resistance Soil clogging

Mullen Burst Test Puncture resistance

Trapezoidal tear

Grab tensile strength and elongation %

AOS or EOS (apparent opening size, or equivalent opening size)

Property Designation General Description of Property Test, Purpose, etc.

Property intended to relate ability of fabric to retain or permit passage of solid particles (e.g., soil). Test procedure involves use of a series of equal-diameter glass beads; the smallest-diameter beads for which 95% by weight is retained (or 5% passes) is taken as the equivalent size opening (expressed in mm opening or the equivalent sieve size). Maximum tensile strength (resistance for one unit of width and an indicated fabric thickness), and percent elongation (stretch) at maximum tensile load. Tensile type test performed on fabric which is placed in a skewed position (on a bias) in test apparatus so that a progressive tearing or breaking of individual fibers occurs; an indication of resistance to tearing. Fabric is stretched (distorted) into a hemispheric shape until tearing or bursting occurs. Resistance to puncture by pointed objects such as sharp stones. Test is performed by pushing an 8.0-mm-dia rod through a stretched fabric (the concept of the “elbow through the sleeve”); indicates resistance to potential damage by stones, pieces of timber and roots, etc. The cross-plane permeability or hydraulic conductivity; indicates ease or difficulty for liquid to flow across (or through) the fabric. A measure of the liquid volume that can flow within the thickness of the fabric; a significant property for a product intended to facilitate drainage or transmission of liquid. Ability of the fabric to retain original strength property after exposure to ultraviolet (sun) rays. Indication of extent to which fabric will become clogged because of trapped soil particles; clogging can interfere with the cross-plane and the planar permeability. Strength of the fabrics after being subject to repetitive loading. Fatigue strength is expected to be less than that for a single application loading, and is affected by the number of load applications. Ability of tensile or shear force to be transmitted across the seam where fabrics are joined. Measurement of the stretching or deformation that occurs over a period of time under the effect of a constant, long-term load. An important property for fabrics where sliding along the surface of contact is possible, as when a fabric is placed on a slope. The friction value is determined by applying a normal load to the fabric surface and measuring the force required to move/slide the fabric (a process similar to the basic shear test to determine a coefficient of friction).

Table 2 Summary Tabulation: Commonly Listed Physical Properties of Geosynthetics

Movement of Water Through Soil: Practical Effects

with no buildings, etc.) is dewatered and the surface settles, usually no problems result. If developed areas are in the zone of drawdown, however, settlement of buildings and utilities in the area could occur. If settlement of structures adjacent to a construction site being dewatered is anticipated, effects can be minimized by the process of recharging. Recharging consists of pumping water back into the ground, usually with well points, in the area between where the endangered structures are located and where the dewatering takes place. Careful monitoring of groundwater levels is required for proper overall results.

Soil Percolation Rate Soil percolation rate refers to the ease or difficulty with which soil accepts fluid, generally water, into it. It is a condition frequently of interest for planning private underground sewage disposal systems, such as the leach fields and leach pits for septic tanks. With a septic tank system, all sewage passes into the septic tank. The solids are retained, but the liquid is passed on for absorption into the earth. The percolation rate can be related to the coefficient of permeability of a soil: the coarser, permeable soils are capable of accepting larger volumes of water, and at a faster rate, than fine-grained soils having low permeability. However, percolation and permeability do not refer to the same property of a soil. For example, a coarse soil having a high coefficient of permeability may have a low rate of percolation if the groundwater table is high and the soil is not capable of accepting additional water. Procedures to check the percolation rate for a soil generally stipulate testing of the area and depth where the septic tank leach field or leach pit is to be located. For the common requirement that leach field drain lines be installed about 0.7 m (or 2 ft) below the ground surface, percolation tests are performed about 0.75 m (or 30 in.) below the surface. A small test hole, usually 150 to 300 mm (or 6 to 12 in.) in diameter, is dug at least 150 mm (or 6 in.) deep. The test hole area is saturated by filling the hole several times with water and letting it drain completely before beginning the actual test. For the test, the hole is filled with water and the time for the water level to drop a specified distance (e.g., from 150 mm down to 125 mm, or from 6 in. deep down to 5 in. deep) is recorded. A fast rate of drop indicates “good” percolation. A slow rate or no drop indicates poor percolation. Areas having poor percolation rates should not be used as leach fields. Details of the applicable methods for performing percolation tests should be obtained from local health department authorities to ensure that procedures are in accord with their requirements.

3

Frost Heave in Soils When soil is exposed to a freezing climate (i.e., conditions below the freezing temperature of water) for a sufficient period of time, soil temperatures will eventually depress to below freezing. Temperatures will vary within the soil mass, being lowest where contact with the freezing source is made (usually this would be the ground surface in contact with the freezing air temperatures, but the source of a freezing temperature could also be the floor of a frozen storage warehouse or ice rink) and gradually increasing with distance

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Movement of Water Through Soil: Practical Effects

from the freezing source to reach the stable, above-freezing temperature that exists underground. When freezing temperatures develop in a soil mass, most of the pore water in the soil is also subject to freezing.6 As water crystallizes, its volume expands approximately 9 percent. In considering normal void ratios and the degree of saturation for soils in general, expansion of a soil material as a result of freezing might be expected to be on the order of 3 or 4 percent of the original volume. This may mean a vertical expansion on the order of 25 to 50 mm (1 to 2 in.) for an area with the climate that exists in the northern half of the United States. Experience indicates, however, that in some soil types, the volume of expansion resulting during freezing periods (a condition termed frost heave) is frequently considerably greater than could be expected from expansion of normal pore water. Frost heaves exceeding 0.3 m (1 ft) are not uncommon. Investigation of heaved soils indicates that pore water indeed has frozen, but additionally, much of the frozen water has been segregated in discontinuous layers, or lenses, throughout the frozen soil, and the volume of the water in the ice lenses is considerably greater than the volume of the original pore water. Further, the magnitude of ground surface heave has been found to approximately equal the combined thickness of ice lenses existing in the soil at the heave location (Figure 24). Studies conducted at field sites and in the laboratory to investigate the frost heave phenomenon have been able to identify conditions and occurrences associated with ice lens formation, but all factors involved are not yet understood. Water in soil freezes at temperatures lower than 0°C (32°F) because of ions from dissolved minerals present in the water, soil particle surface forces, and the existence of negative pore water pressures if the soil is not saturated. The concentration of ions will affect the temperature at which water begins to freeze (nucleate), and ion concentrations will likely vary across the soil deposit. The

Depth of frost penetration

Magnitude of frost heave approximately equals thickness of ice lenses underlying this location

Unfrozen zone

Figure 24

6This

Frost-heaved elevation of ground surface

Original elevation of ground surface Ice lenses

Frost line or frost depth

Ice lens formation in a frozen soil.

explains the “frozen-solid” condition noted for soil ground in cold weather. The frozen water is responsible for the strong bonding between soil particles; the soil particles generally experience no change as “freezing” temperatures occur, for they are already in the frozen or solid state.

280

Movement of Water Through Soil: Practical Effects

temperature for nucleation also varies across the soil void space, being lowest in the diffuse double layer adsorbed to the particle surface and increasing as distance from the particle increases. The thickness to this film of unfrozen water surrounding the soil particle in an otherwise frozen soil is also affected by temperature, decreasing as temperatures decreases. The unfrozen film of water surrounding particle surfaces can be several degrees lower than the conventional freezing temperature for water. Ice, therefore, tends to originate in the zone of soil void space farthest away from the particle surfaces. Additionally, the surface of an ice zone consists of a film of adsorbed water; the ice zone increases by attracting additional water molecules (from adjacent soil voids and particle surfaces) into the adsorbed film. The energy gradients responsible for the attraction of new water to the growing ice zone (the forming ice lens) are complex but probably involve temperature and pressure differences, the phenomenon of capillarity, and osmotic flow.7 It is probable that migration of water molecules toward an expanding ice lens occurs in a zone within or bordering the films that constitute the diffuse double layer on particle surfaces. But if the temperature drops too low, the mobility of this water is restricted. The process of expanding ice lenses will cease when the lens can no longer attract water because the required energy gradient becomes too great or the available sources are diverted to other lenses. The development of ice lens formations significant enough to cause large heaves requires that a source of water be sufficiently close to the freezing zone of soil so that it can be drawn into this zone. In the normal frost heave occurrence, the source of water is the groundwater table, as illustrated by Figure 25 (but other conditions, such as leaking underground pipes, can also be a source). Upward movement from a water table to the freezing zone relates to a potential for migration (i.e., capillary rise). Height of capillary rise is quite limited in clean, coarse-grained soil, and unless a source of water is close to the zone of freezing soil, frost heave problems are normally not expected in such soils. Ground surface

Frost depth

Upward migration of water by capillarity Groundwater level

Below freezing

Above freezing 0°C 32° F

Variation of soil temperature with depth, F

Figure 25

Ice lens formation by capillary movement of groundwater into zone of freezing.

7Osmotic flow relates to the movement between liquid solutions having different ion concentrations. Simply stated, there is the tendency for the more dilute solution to flow into, and mix, with the highly concentrated solution. Water in soil normally has some concentration of cations because of dissolved minerals. Applied to the ice lens phenomenon, crystallization of soil pore water may increase the ionic concentration; thus the more dilute free water in the soil would be attracted to the ice lens.

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Movement of Water Through Soil: Practical Effects

Additionally, sands tend to freeze uniformly throughout the frozen zone and ice lenses do not form. The theoretical height of migratory rise is greatest in the fine-grained silt and clay soils. Practically speaking, however, under typical seasonal conditions of limited cold periods, ice lens growth is affected by the rate of capillary rise in soil. In clays, this rate is so limited that under normal conditions sufficient water cannot be obtained to result in ice lens growth. Silts, however, have a relatively high rate of capillary rise, and water can migrate easily to the zone where ice lenses are forming. Silts, therefore, because of their potential for high capillary rise (distance) coincident with the potential for rapid rise, are conducive to ice lens formation and frost heave. As a result, silts are termed frost-susceptible soils. Silt mixtures, such as silty sands and silty clays, are similarly regarded as frost susceptible. Because of the practical limitations on ice lens growth within them, coarsegrained and true clay soils are normally considered as non-frost-susceptible soils. In summary, several simultaneously occurring conditions are required for the notorious frost heave problem to occur: 1. 2. 3. 4.

Presence of a frost-susceptible soil Presence of temperatures lower than the freezing temperature of water Presence of a supply or source of water to help form and feed the ice lenses A time period sufficient for ice lenses to form and enlarge

The rate at which freezing temperatures penetrate the frost-susceptible soil has an effect on the total ice lens formation and, consequently, on the frost heave of the ground surface. The maximum effects result when there is a gradual decrease in temperature as opposed to a rapid or flash decrease. The longer and the colder the freezing period, the greater will be the depth of frost penetration and the greater will be the frost heave. Damages from frost heave result in structures supported on the soils being heaved. Commonly, this includes roadways, pavements, building foundations, and exposed groundlevel building slabs. The heaving force can be considerable; it has the ability to lift pavements, slabs, and building foundations. Because the magnitude of heave is rarely uniform over even short horizontal distances, cracking of pavements and slabs frequently occurs. In buildings, walls or floors may crack or distort so that doors, windows, and equipment supported on the walls may not operate properly. If heave is great enough, utility pipelines passing through or over the walls may be ruptured because of the movement. If the frozen soil adheres to the exterior side of a masonry foundation wall, the soil heave results in an uplift force on the upper section of the wall, which, for light loading conditions, may cause wall mortar joints to crack. Damage is not restricted to that resulting from the heaving. Much of the ice lens volume represents new water introduced to the soil, and when the soil thaws from the surface downward in a warm period, the free water cannot drain through the still-frozen underlying soil. The effect is that the melting ice lenses significantly increase the water content of the soil, and loss of soil strength occurs. Loss of soil bearing and settlement follow. For building foundations not located below frost depths, the cycle of heaving in winter and settling in spring is an annual occurrence that is capable of causing progressive damage. Similar effects can occur in pavements and unprotected ground-level building slabs. In roadway surfaces, the effects are compounded by the repeated loadings applied to an area

282

Movement of Water Through Soil: Practical Effects

by vehicle movement. Under the action of traffic, the roadway surface will be broken up because of the loss of the underlying soil support. This is generally referred to as the spring breakup. Further, the soft soil will be displaced from the road base, causing holes and depressions, which are referred to as potholes or chuck holes. Frost heave also may have its effect where new construction takes place during freezing periods. For example, should the foundation soil for planned structural components such as footings, floor slabs, or pavements be in a frozen and heaved condition at the time of construction, the eventual thawing and settling of the soil can cause the component to experience settlement, distortion, or structural damage. To prevent the effects of frost heave damage to buildings, it is common practice to construct foundations to a depth at least equal to the depth of frost penetration for the area. Such procedures might be thought to be necessary only in a frost-susceptible situation. If non-frost-susceptible soils exist at a building site, or if there is no source of water, locating foundations below frost depth could be unnecessary unless required for other reasons. However, in natural soil deposits, there should also be some suspicion that even “non-frost-susceptible” soils will have some properties that could make them susceptible. For example, many naturally occurring coarse-grained soils include some fines, or include seams or strata of finer, more susceptible material. Clay soils frequently have considerable silt content, or the clay deposit may include fissures and hairline cracks that can act as capillary tubes for the migration of water. Also during the life of a structure, the depth of the groundwater table is subject to change. In considering the poor performance and damage that can result if frost heave should affect foundations, most designers and constructors are of the opinion that it is worth the relatively small extra expense to take precaution at the time of construction to install foundations for exterior elements below frost depth. Figure 26 shows maximum frost depth data prepared from information obtained across the United States. For grade-level structures such as building slabs and highway pavements (no deep foundations), other approaches to the control of the frost heave problem are taken. Control usually requires achieving at least one of the following: 1. Removing existing frost-susceptible soils to about frost depth 2. Removing or cutting off the water source that could feed ice lens growth 3. Protecting the susceptible soil from freezing temperatures For road or airfield pavements, the effort for control or protection has frequently included removing the frost-susceptible soils to frost depth and replacing them with a nonfrost-susceptible material or providing drainage facilities to keep the water table low in the vicinity of the pavement. For building slabs, replacing the frost-susceptible soil is frequently a practical solution. Non-frost-susceptible soils for replacement purposes include coarse-grained soil (sand and sand–gravel mixtures) and clays. Coarse-grained soils are preferred because they are easier to place and compact, are less difficult to work with in inclement weather, and have other desirable properties when in place, such as good drainage. To be non-frostsusceptible, studies and experience indicate that a well-graded granular soil should contain

283

Movement of Water Through Soil: Practical Effects

18-35 72-108 54-71 72-108 72-108 54-71 54-71

54-71

36-53

0-5 18-35

36-53 36-53 36-53

36-53

18-35 18-35

6-17 18-35 0-5

6-17 6-17

0-5

0-5

Figure 26 Maximum depth of frost penetration in inches. (From Special Report 1, HRB, National Resource Council, Washington, D.C., 1952, Figure 112, page 125.)

not more than 3 percent of particle sizes smaller than 0.02 mm,8 while uniform granular soils can have up to about 10 percent of particle sizes smaller than 0.02 mm. Some ice lens formation may still result with these soils, but such development will generally be of limited extent and tolerable. Other methods for control of frost heave have included installation of a barrier at about frost depth to prevent upward capillary movement that could feed ice lenses. The barrier can be a very coarse soil of a thickness through which capillary rise cannot occur, an impermeable material such as a densely compacted clay layer, or a membrane or other fabric material through which water cannot penetrate. These methods appear most suitable for protecting limited areas, such as building slabs, or for highway fill sections. The methods are not considered practical for highway construction in areas of cut section or at-grade construction because of the expense of excavating and replacing subgrade soils that would be associated with installation of a barrier. Further, frost-susceptible subgrade soils may be difficult to replace and compact, especially in wet weather. Successful protection has also been achieved by wrapping pavement subgrade soils within an impermeable membrane after they have been properly compacted, so as to keep

8To retain a perspective on sizes, it is noted that the #200 sieve has openings of 0.074 mm; a #400 sieve has open-

ings of 0.037 mm.

284

Movement of Water Through Soil: Practical Effects

the soil at a desired optimum moisture content and to prohibit ice lens formation. A disadvantage of this method of control relates to costs of installing the membrane and placing the soil. If the soil beneath a slab or pavement can be protected so that it does not freeze, there will be no frost heave problem. In the construction of frozen-storage warehouses, ice rinks, and similar facilities, insulation is placed between the soil subgrade and the floor of the facility to protect the soil from the cold temperatures prevailing in the structure. Similar methods are adaptable for protecting exposed ground-level slabs for buildings. Various methods that have been used are illustrated in Figure 27. For highways, some work has been done with providing insulating material beneath the pavement.

Permafrost Permanently frozen ground, termed permafrost, exists in the northern areas of North America, Europe, and Asia, and across most of Greenland (Figure 28), a result of the severe climate in those areas. It is estimated that one-fifth of the earth’s northern hemisphere is underlain by permafrost. The depth of frozen soil is on the order of 300 m (1,000 ft) in regions corresponding to Alaska’s North Slope, and probably extends to more than 1,000 m (3,300 ft) at the Arctic Circle. The surface zone of soil in the permafrost regions thaws in summer (a condition that also occurs where permafrost is located beneath a structure that enables lost heat to penetrate the ground), but the deeper material stays frozen. The zone thawed by a summer climate refreezes the next winter. The soil depth that is subjected to seasonal cycles of thawing and freezing is termed the active zone. Thawed soil in the active zone is usually weak and compressible because of the trapped water content. Structures built in permafrost areas should have their foundations installed in the permanently frozen earth that exists below the active zone, to protect the structure against seasonal movements that would otherwise result. These foundations must

Area temperature at 0° C (32° F) Concrete floor Insulation aggregate Subgrade

Area temp. at 0° C to −12° C (32° F to 10° F) Concrete floor Plastic foam or cork insulation Coarse aggregate Concrete foundation slab Subgrade

Area temp. at 0° C to −18° C (32° F to 0° F) Concrete floor

Air space

Concrete floor

Concrete

Coarse aggregate or insulation aggregate

Concrete foundation slab Subgrade

Figure 27

Support posts

Area temp. at −9° C to −24° C (15° to −10° F)

Foam or cork insulation

Subgrade Heating ducts for warm air heating

Some methods that have been used to insulate floor systems for refrigerated areas.

285

Movement of Water Through Soil: Practical Effects

Asia

Europe

90°W 0° 80°

70°

60°

50°

180° 90°E

Area underlain by continuous permafrost Areas of discontinuous, intermittent or isolated masses of permafrost

Figure 28

North America

Extent of permafrost zones in the Northern Hemisphere [15].

be designed to ensure that heat is not conducted through or by them to the permafrost that provides the support; insulating materials could be necessary along the vertical perimeter of the foundation and at the base. Similarly, to prevent building heat losses from thawing the permafrost, the floors and subsurface walls that would be in contact with the ground must be insulated, artificially cooled, or separated (with small buildings, the first floor can be elevated above the ground surface). The subarctic plains have a characteristic vegetative cover of mosses and low shrubs underlain by mucklike soil, a combination termed tundra. This surficial blanket acts as an insulating material limiting the depth of summer thawing and therefore the active zone. A result of recognizing this beneficial effect of the tundra is that roadways for permafrost areas are now constructed on embankments of gravel, which act as a non-frost-susceptible

286

Movement of Water Through Soil: Practical Effects

insulating pad to keep the underlying soil frozen and stable. Equally significant, the presence of tundra has proved to be helpful in keeping sloped and hilly areas stable against slides.

4

Soil and Groundwater Contamination Soil contamination and groundwater contamination can be broadly defined as the inclusion of matter or substances which affect normally expected use. Contamination of soil and groundwater involves degradation due to the presence of physical, biological, chemical, or radiological materials in undesirable concentrations or concentrations considered abnormal. Most natural soil deposits and “normal” groundwater typically include minerals, as well as biological and chemical substances, but in low concentrations considered tolerable. Concentrations considered acceptable for one type of use may be unacceptable for another, however. Pollution, such as water pollution, refers to the loss of purity through contamination. Polluted water is water unfit for consumption and agricultural and commercial use. The presence of contaminants can be the result of natural occurrences (natural contamination), but is commonly considered to be a condition associated with human activities (artificial contamination). Most government environmental protection regulations reflect intentions to prevent or control the release of contaminants as related to human activities. Table 3 lists the more well-known types of soil and groundwater contamination sources. Figure 29 illustrates similar information. Most groundwater contaminants are included in one of three broad groups: biological organisms, organic chemicals, and inorganic chemicals. Knowledge of a contaminant general grouping can be helpful in developing a remedial process which achieves decontamination. The source of biological organisms is animal, human, and vegetative waste, frequently resulting where heavy concentrations occur such as at inadequate sewage treatment operations, cesspools and septic systems, sewage treatment settling ponds or lagoons, and from poorly controlled land spreading of dried sewage sludge. Organic chemicals result from materials containing carbon, hydrogen, and oxygen. Petroleum products and petrochemicals are included in this group, as are the byproducts of decomposing natural and synthetic organics. Commonly, these materials enter the groundwater supply from waste disposal sites, from processing and manufacturing locations, and from accidental spills. Inorganic chemical contaminants are of mineral origin, being materials that exist naturally but that become concentrated to dangerous levels at mining and storage operations, or that are a result of accumulated waste products from industrial processing. Surface and near-surface contaminants can be transported from the source location by natural occurrences such as the movement of water and air, or by intentional and unintentional human activities. Land surface and near-surface contaminants can cause pollution of surface bodies of water (lakes, streams, etc.) via overland flow. Subsurface contaminants are subject to the same forces responsible for the movement of underground water, and for many types of contaminants the water will be the conveying medium (the groundwater transports the contaminant as part of the natural flow). A special case of contaminated subsurface liquid termed leachate results when clean surface and subsurface water seeps

287

Movement of Water Through Soil: Practical Effects Table 3 Sources of Groundwater Contamination Category I—Sources designed to discharge substances Subsurface percolation (e.g., septic tanks and cesspools) Injection wells: Hazardous waste Nonhazardous waste (e.g., brine disposal and drainage) Nonwaste (e.g., enhanced recovery, artificial recharge, solution mining, and in situ mining) Land application: Wastewater (e.g., spray irrigation) Wastewater byproducts (e.g., sludge) Hazardous waste Nonhazardous waste

Open burning and detonation sites Radioactive disposal sites Category III—Sources designed to retain substances during transport or transmission Pipelines: Hazardous waste Nonhazardous waste Nonwaste Materials transport and transfer operations: Hazardous waste Nonhazardous waste Nonwaste

Category II—Sources designed to store, treat, and/or dispose of substances; discharge through unplanned release

Category IV—Sources discharging substances as consequence of other planned activities

Landfills: Industrial hazardous waste Industrial nonhazardous waste Municipal sanitary Open dumps, including illegal dumping (waste) Residential (or local) disposal (waste) Surface impoundments: Hazardous waste Nonhazardous waste Waste tailings Waste piles: Hazardous waste Nonhazardous waste Materials stockpiles (nonwaste) Graveyards Animal burial Aboveground storage tanks: Hazardous waste Nonhazardous waste Nonwaste Underground storage tanks: Hazardous waste Nonhazardous waste Nonwaste Containers: Hazardous waste Nonhazardous waste Nonwaste

Irrigation practices (e.g., return flow) Pesticide applications Fertilizer applications Animal feeding operations De-icing salts applications Urban runoff Percolation of atmospheric pollutants Mining and mine drainage: Surface mine–related Underground mine–related Category V—Sources providing conduit or inducing discharge through altered flow patterns Production wells: Oil (and gas) wells Geothermal and heat recovery wells Water supply wells Other wells (nonwaste): Monitoring wells Exploration wells Construction excavation Category VI—Naturally occurring sources whose discharge is created and/or exacerbated by human activity Groundwater–surface water interactions Natural leaching Saltwater intrusion/brackish water upconing (or intrusion of other poor-quality natural water)

Source: Office of Technology Assessment [262].

through landfill or disposal site material and carries along soluble matter responsible for the contamination; leachates are commonly acidic, which increases the capacity for dissolution of toxic materials (usually a benefit) but also the transport of toxic materials (usually a detrimental effect). Solid contaminants are prone to remain at the source location unless transported by a moving medium such as water, air, or humans. Liquid contaminants,

288

Movement of Water Through Soil: Practical Effects

Precipitation Evapotranspiration Injection or disposal well Pumping well Land spreading Septic tank/ or irrigation cesspool Sewer

Pumping well Landfill, dump or refuse pile

Lagoon, pit or basin Water table

Percolation

Stream Leakage

Discharge Leakage

Percolation

Water table aquifer

Confining zone Artesian aquifer (fresh)

Leakage

Ground water movement Confining zone Artesian aquifer (saline)

Intentional input Unintentional input

Discharge or injection

Figure 29 How waste disposal practices may contaminate groundwater (from EPA Journal, Vol. 10, No. 6, “Sources of Groundwater Contamination,” July-August 1984).

however, have different behavioral characteristics, possessing the tendency to flow without requiring the presence of another medium such as groundwater. Liquids having a fluid viscosity and density less than water (for example, gasoline) will be able to move through a soil deposit more easily than water; soils relatively impervious to the movement of water will be pervious to the flow of low-viscosity fluids (refer to Equation 6-5, relating the coefficient of permeability for soil and viscosity of the flowing liquid). In some instances, the contaminants present at a surface or subsurface source location may pass into solution as water moves through, or particles of matter may be physically transported with the flowing water. Some types of contaminants, however, are subject to chemical reactions and changes, or to biological breakdown, or become absorbed (bonded) to soil particle surfaces; in effect, a filtration occurs. These three factors are often categorized as retardation or attenuation mechanisms. An important result is that some types of contaminants can travel underground long distances without significant change; some types of contaminants will be transported underground but at a retarded rate; some will break down or dilute as part of the transportation process (possibly being sufficiently diluted to become relatively harmless); and some will move only a limited distance. The extent to which the various mechanisms actually function influences the degree to which contaminated groundwater will spread, and affects remedial measures. The bulk of subsurface contaminant migrations take place in the presence of moving groundwater—both the flowing category and the more or less stationary groundwater

289

Movement of Water Through Soil: Practical Effects

condition—due to the process of advection and hydrodynamic dispersion. Advection identifies the movement of contaminants suspended or carried within the volume of the groundwater flow. Hydrodynamic dispersion is associated with the spreading and dilution that occurs as moving water travels through the irregular orientation of channel paths formed by the interconnected void spaces between soil particles; as flow occurs, portions are diverted or spread out from the main direction of movement. Some osmotic mixing is probably involved (referring to the tendency for liquid solutions that have different ion concentrations to mix). The zone of water carrying contaminants is therefore enlarged along the general direction of the underground flow, although a less-concentrated or more-diluted volume results. The subsurface zone affected by the spread of contaminated groundwater is referred to as the contaminant plume, or simply plume. Figure 30 illustrates a variety of subsurface soil and groundwater conditions and contaminant plumes that tend to result. The configuration of a contaminated plume is influenced by an array of factors, such as the physical properties of the soil stratum being polluted (the thickness, depth, and slope, the mineral composition of the soil particles, and the permeability or hydraulic conductivity); the presence of groundwater; the properties of the contaminant, including the concentration, solubility, and density and flow (viscosity) characteristics; and time. A liquid contaminant may initially be, or become, mixed with a groundwater supply, but then separates if it is of a significantly lighter or heavier density. For example, gasoline escaping into the earth will accumulate or travel along the top of a groundwater supply and tend to flow faster, because of the lighter density and lower viscosity. Of practical interest, cased water supply wells have continued to provide unpolluted water even with gasoline present in the ground, as long as the intake section of the well casing remained below the level of the gasoline phase. The migration of contaminants to pollute a water supply is not the full extent of the environmental problem. Contaminants that enter the earth zones important to plant life (either a surface, subsurface, or above-surface region) and are then absorbed may have harmful effects on the vegetation and may enter the food chain to harm higher species. Two situations of groundwater contamination due to factors other than a reaction with waste materials and chemicals are the occurrence of saltwater intrusion into the subsurface supply of fresh water and the presence of trihalomethanes in drinking water. The saltwater intrusion condition develops in seacoast regions when fresh groundwater is removed in sufficient quantity to lower the land area groundwater level to or near the level of the bordering body of seawater. The heavier saltwater displaces, or intrudes into, space previously occupied by the fresh groundwater. Water supply wells situated in the zones of such seawater encroachment will then deliver saltwater instead of fresh water. The presence of trihalomethanes is the rather unusual phenomenon of contamination because of a disinfection effort. When a source of water containing organic matter and bromide salts is treated with chlorine, certain halogenated organic compounds, such as chloroform, bromoform, bromodichloromethane, or chlorodibromomethane (the halogen-substituted methanes, or trihalomethanes), are produced. Organic particulate matter and the bromide ions can be found in many natural waters and saltwater. Surface waters tend to contain greater amounts of organic particles in fall and winter because of natural factors such as fallen tree leaves. Trihalomethanes such as chloroform are suspected carcinogens, and desirably the concentration in drinking water supplies will be limited to levels considered tolerable (less than 100 micrograms per liter).

290

Precipitation

Ground surface

Landfill or waste disposal

Precipitation

Leachate

Leachate

"Stationary" groundwater

Pervious soil

Pervious soil

wta wta

Contaminant plume

Aquitard, impervious (b) Flowing groundwater, uniform soil, groundwater in single layer

(a) "Stationary" groundwater (limited subsurface flow), uniform soil conditions (groundwater in single layer)

Landfill or waste disposal Leachate

Pervious soil

Contaminant plume

Groundwater flow

Aquitard, impervious

Precipitation

Landfill or waste disposal

Precipitation

Landfill or waste disposal Stream or lake

wta

Impervious soil layer

wta Groundwater flow

Contaminant plume Groundwater flow

Aquita rd, imp erviou s

Aquitard (impervious)

Pervious soil

(d) Flowing groundwater, contaminant plume affects surface body of water

(c) Flowing groundwater, pervious and impervious soil layers

Injection or disposal well (deep well injection disposal)

Contaminant plume

Auto service - gasoline -

Ground surface

Pervious soil wta

wta

Groundwater flow

Contaminant plume Aquitard, impervious

(e) Flowing groundwater, deep well disposal affects groundwater at greater depths

Figure 30

Street

Leaking buried storage tank

Groundwater flow

Sewer

Contaminant plume Pervious soil (f) Contaminant plume infiltrates buried sewer, spreading petroleum liquid and vapor

Contaminant plume relating to site conditions and contaminant source.

291

Movement of Water Through Soil: Practical Effects Storage tank

Storage tank

Tank leakage

Tank leakage wta

wta Contaminant plume

Pervious soil

Groundwater flow

Groundwater flow

Pervious soil

Aquitard (impervious)

Contaminant plume

Aquitard

(g) Contaminant plume where liquid is less dense than water

(h) Contaminant plume where liquid is more dense than water Storage tank

Tank leakage wta Pervious soil Groundwater flow Aquitard (impervious soil or rock)

(i) Contaminant plume, liquid more dense than water

Figure 30 (continued)

Protecting a subsurface region including the groundwater against pollution involves procedures to monitor for, or detect, the presence of contaminants and, where pollution has occurred, the implementation of remedial processes (that is, become aware of the problem; identify the contaminants and source if possible; then undertake remedial measures). The monitoring process conventionally is used where a known source of contaminants exists, such as at a landfill or waste dump site. The air, surface, and subsurface materials surrounding the site are monitored (commonly by analysis of air, soil, vegetation, or groundwater samples) to check for the presence (migration) of contaminants. The typical detection process is a protective undertaking for a commodity such as a water supply, and involves checking the air, soil, surface, or groundwater for the arrival of polluting contaminants and the need to provide remediation. The monitoring and detection techniques are similar, using surface and subsurface sensor procedures to indicate change, and samplinganalysis procedures to identify the presence and concentration of contaminants. Commonly, subsurface monitoring or detection programs include the use of wells which permit samples of soil, groundwater, or gases to be obtained for analysis and evaluation. Surface monitoring–detection procedures include physical observation of surface conditions (such as change in vegetation and presence of odors which could indicate escaping soil gases) and air sample analysis. Comparison of aerial photographs taken at different time periods is useful in identifying the changes in surface vegetation that indicate the presence of underground contaminants.

292

Movement of Water Through Soil: Practical Effects

Where a soil or groundwater contamination condition exists, it may be necessary to undertake remediation, or at least containment or other mitigation, for health, safety, environmental protection, or other reasons. Factors influencing the selection of a corrective program relate to the type and extent of contamination, the procedures that could be successful and their cost, the time involved to implement and complete the program, and the hazards associated with implementation. Table 4 outlines a variety of corrective actions that have been applied to contamination–pollution problems. It is not unusual to have more than one corrective procedure involved where a complex condition exists or where a large area is affected. Undoubtedly, the refinement of present procedures and the development of new technology will result in more choices being available.

Soil Gas Recognizing the possibility of a soil gas presence other than normal air has become extremely important for personnel dealing with any aspects of underground construction or environmental safety and protection. Some soil gases are hazardous, being toxic or carcinogenic or presenting the danger of fire or explosion. Significantly, the presence of soil gas in near-surface soils has been found to be a reliable indicator of deeper groundwater and soil contamination, and rapid gas detection procedures are being used that can efficiently check regions for underground pollution and outline affected areas. The gases distributed throughout the voids of a soil deposit would normally be expected to have properties similar to those of the earth’s atmosphere. However, displacement by contaminant gases can occur because of pressures or thermal energy generated at a buried source or because of heavier density. Buried sources of gas producers can be present in an area because of natural occurrences or because of human activities. The types of gases found in a soil deposit relate to the types of buried materials responsible for their formation, and the possibilities for such gas-producing reactions are numerous. But conditions responsible for the development of soil gas volumes that possess the capacity for fire or explosion are most associated with the presence of buried organic materials and organic chemicals. Seeping mature natural petroleum–gas deposits and juvenile gases generated within still-decomposing masses of buried vegetation (e.g., in peat or marsh areas) represent the typical natural occurrence. Waste dumps, areas of buried vegetation resulting from land clearance operations, and garbage-trash landfills, rich with organic throwaway materials, represent the more active gas producers associated with human activities. Radioactive gas, such as radon, can result from natural sources of uranium and from processed or waste materials containing uranium, radium, or polonium. On the elemental level, decomposition of buried organic matter occurs by reactions classified as oxidation procedures and reduction procedures. Oxidation, the removal of electrons from atoms in the organic molecule, occurs where oxygen is present. Reduction, the addition or gain of electrons, occurs where oxygen is lacking but organic matter is plentiful. Where the decomposition process involves oxidation and reduction (the electrons gained equal those being removed or, in other words, being transferred so as to change the involved molecules), the process is termed a redox reaction. With organic matter, oxidations tend to occur before reductions if oxygen is available. With buried organic materials,

293

Movement of Water Through Soil: Practical Effects Table 4 Corrective Action Alternatives: Techniques and Descriptions I. Containment: This category consists of geotechnical methods that act to limit the mobility and prevent the further spreading of contaminants. Contaminants are not actually removed from the subsurface but are contained or isolated from the rest of the environment—e.g., via physical barriers or hydrodynamic pressures. Techniques are applied in relation to either the contaminants or their source. 1. Slurry wall: Consists of a material (slurry) barrier wall constructed in place; is usually located below the water table and surrounding a site to limit the horizontal migration of contaminants in the saturated zone; is also used to reduce hydraulic gradients, facilitate withdrawal, or channelize groundwater flow. 2. Sheet pile: Consists of a material (e.g., concrete, steel, or wood) barrier wall inserted into place by driving or vibration; is usually located below the water table and around a site to limit the horizontal migration of contaminants in the saturated zone. 3. Groutinga: Consists of a material cutoff injected into voids of water-bearing strata either to cover, bottomseal, or bind together the subsurface materials at a site. 4. Geomembrane cutoff: Involves the insertion of synthetic sheeting into an open trench (combining aspects of both the slurry wall and sheet pile) to form a barrier wall; is used primarily to limit the horizontal migration of contaminants in the saturated zone. 5. Clay (or other) cutoff b: Clay (or other material, e.g., concrete) barrier wall; normally is constructed above the water table and downgradient of a site to limit the horizontal migration of contaminants in the unsaturated zone (which is commonly negligible). 6. Linerc: Consists of a material (e.g., clay or synthetic) barrier constructed or emplaced to isolate (e.g., cover or seal) contaminating sources in order to limit the vertical migration of contaminants; is often a facility design component. 7. Natural containment: Involves limitation of contaminant mobility by naturally occurring geochemical, geologic, and/or hydrologic conditions; is evaluated by analytical and/or empirical methods. 8. Surface sealingc: Is used as an infiltration control measure to limit the vertical migration of contaminants by reducing leachate production and/or by recharge. 9. Diversion ditchc: Is used as an infiltration control measure to limit surface runoff into a contamination management area (e.g., a slurry-walled area) by channelizing and diverting surface drainage. 10. Hydrodynamic control: Limits the horizontal migration of contaminants in the saturated zone through selective pumping and the subsequent creation of pressure troughs or pressure ridges.

294

II. Withdrawal: Withdrawal options include methods for either directly removing or facilitating the removal of contaminated groundwater and/or contaminated soils from the subsurface. Techniques are principally applied in direct relation to the contaminants. 1. Pumping: Involves the removal of contaminated groundwater by pumping from wells or drains; controls the lateral (and in some cases, vertical) migration of contaminants; can be used for flushing (via artificial recharge). 2. Gravity drainage: Involves the removal of groundwater from the subsurface using the force of gravity (e.g., using sumps or French drains) instead of pumps; controls the lateral (and in some cases, vertical) migration of contaminants. 3. Withdrawal enhancement: Enhances the ability to withdraw either groundwater or contaminants, typically by increasing contaminant solubility in water (e.g., by injecting steam or heat, bacteria or nutrients, or surfactants). 4. Gas venting: Removes gases associated with contamination (e.g., methane and petroleum-related products). 5. Excavation: Involves the direct removal of contaminated soil and/or groundwater resulting from source leakage. III. Treatment: This category includes physical and chemical/biological treatment methods for detoxifying contaminants found in groundwater. These methods presume that contaminants have already been withdrawn from the subsurface (e.g., via withdrawal methods) in the form of contaminated groundwater or contaminated soils. Treatment can be applied at the source, at the site of contamination (e.g., in on-site treatment units), prior to the distribution of groundwater for use (e.g., in municipal wastewater treatment facilities), and at the point of end use (e.g., at the tap). A. Physical treatment: 1. Skimming: Involves the removal of floating contaminants (e.g., oil, grease, and hydrocarbons) in a multilayer solution. 2. Filtration: Involves the physical retention and subsequent removal of contaminants present as suspended solids. 3. Ultrafiltration: Involves the physical filtration, through semipermeable membranes, of suspended and dissolved metals, emulsified hydrocarbons, and substances of high molecular weight. 4. Reverse osmosis: Involves the osmotic filtration, through semipermeable membranes, of contaminants (e.g., metals and radioactive wastes) present as dissolved solids; operates at high pressures (up to 1,500 psig).

Movement of Water Through Soil: Practical Effects Table 4 (continued) 5. Air stripping: Uses air injection to facilitate the volatilization and removal to the atmosphere of contaminants (e.g., volatile organics and hydrogen sulfide) that are present in water as dissolved solids. 6. Steam stripping: Involves the fractional distillation of volatile organics or gases by heating. B. Chemical/biological treatment: 7. Precipitation/clarification/coagulation: Removes contaminants (e.g., suspended and colloidal solids, phosphates, and heavy metals) through the use of chemical additives, such as coagulants and coagulant aids. 8. Ion exchange: Removes selected ions (primarily inorganic) via the exchange of ions between an insoluble solid salt (“ion exchange”) and a solution containing the ion(s) to be removed. 9. Adsorption: Removes contaminants (primarily organics) via their tendency to condense, concentrate, or adhere on the surface of another substance (e.g., granular activated carbon and synthetic resins) with which they come into contact. 10. Electrodialysis: Separates and removes positive or negative ions under the action of an electrical field. 11. Chemical transformation: Involves oxidationreduction reactions for the chemical conversion of contaminants to less toxic substances (e.g., by ozone treatment, hydrogen peroxide treatment, ultraviolet photolysis, and chlorination). 12. Biological transformation: Involves the transformation and removal by microorganisms of dissolved and colloidal biodegradable contaminants; includes both aerobic and anaerobic processes. 13. Incineration: Involves the high-temperature transformation of contaminants into constituent components; many types of thermal destruction systems are included.

IV. In situ rehabilitation: In situ rehabilitation techniques are directed at immobilizing or otherwise detoxifying contaminants in place. 1. Biological degradation: Involves either stimulating the growth of native microflora or injecting specific organisms to consume or otherwise alter contaminants. 2. Chemical degradation: Involves the injection of specific chemicals that react with or otherwise alter contaminants. 3. Water table adjustment: Involves either the isolation of the contaminated zone (and creation of a detoxifying unsaturated environment) by lowering the water table or the artificial inducement of increased flushing action by raising the water table. 4. Rehabilitation via natural processes: Involves the natural degradation, dispersion, or detoxification of contaminated groundwater; is evaluated by analytical and/or empirical methods. V. Management options: Management options are usually applied to prevent further contamination or to protect potential exposure points from contaminated groundwater. These methods thus focus on sources and exposure points rather than on contaminants per se. The methods also tend to be institutionally based rather than technology based. 1. Limiting/terminating aquifer use: Limits access or exposure of receptors to contaminated groundwater. 2. Development of alternative water supply: Involves the substitution of alternative supplies (e.g., surface water diversions and/or storage, desalination, and new wells) for contaminated groundwater. 3. Purchase of alternative water supply: Includes bottled water and water imports. 4. Source removald: Involves the physical removal of the source of contamination and includes measures to eliminate, remove, or otherwise terminate source activities; could also include modification of a source’s features (e.g., operations, location, or product) to reduce, eliminate, or otherwise prevent contamination. 5. Monitoring: Involves an action evaluation program with a “wait and see” orientation. 6. Health advisories: Involves the issuance of notifications about groundwater contamination to potential receptors.

7. Accepting increased risk: Involves the decision to accept increased risk; is usually a “no action” alternative. aCan

be considered a form of chemical immobilization if injected directly into the plume of contamination. barriers located above the water table will not affect the horizontal migration of contaminants in the saturated zone. cMost often used in the context of either “source removal” or the prevention of recharge to the groundwater system, rather than as a containment option per se. dModification of a source’s features is often an important element of corrective action in the context of preventing future groundwater contamination (i.e., reducing the need for future corrective action). Source: Office of Technology Assessment [262]. bPhysical

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Movement of Water Through Soil: Practical Effects

the early stages of decomposition usually occur by oxidation because of the presence of trapped air and oxygen in groundwater. As the available oxygen is consumed, the reduction process prevails. Oxidation processes include the breakdown of organic matter to carbon dioxide, ammonia to nitrate, methane to carbon dioxide, and sulfide to sulfate. Reduction processes occurring in the presence of anaerobic acid-forming microorganisms include the change of nitrate to nitrogen gas,9 nitrate to nitrite, nitrite to ammonia, sulfate to sulfide (such as hydrogen sulfide, the “rotten egg” odorous gas), carbon dioxide to methane, and nitrogen gas to ammonia. Methane is a volatile, explosive gas. Additionally, other gases can be carried along with methane (for example, vinyl chloride, a known carcinogen which results from degradation of the solvent trichlorethylene, as well as benzene and xylene, have been found to be included when methane was detected). Hydrogen sulfide gas possesses an obnoxious odor and is toxic in high concentrations. Otherwise harmless gases such as carbon dioxide can, in large volumes, cause asphyxiation. Nitrates and sulfates can pollute groundwater.10 Nitrate is considered toxic to infants. Ammonia is a pungent gas that can affect respiration. Recognizing the possibility of problem gases, deep construction work such as tunneling and even shallow rock excavation (particularly in sedimentary formations) includes the use of gas detection (sniffer) equipment, while solid waste landfill projects will build into the protective containment system a means for venting gases that are expected to be generated. Radon is a radioactive gas whose presence is related to rock, soil, and waste material whose composition includes uranium, radium, or polonium. Radium and polonium, products of decay (the progeny) from naturally occurring uranium, in turn decays into radon. Radon gas is invisible, odorless, and tasteless, and is soluble in water. As radon gas decays into solid particles of lead and bismuth, it emits small amounts of radiation. Naturally occurring radon gas exists in low concentration in many areas of the earth without causing harm. On the basis of information obtained from experiences with highly radioactive materials, however, medical and health specialists fear that inhaling air that includes high concentrations of radon for extended periods could cause the type of serious health problems associated with excessive exposure to radiation and carcinogens. Where waterborne, the dissolved radon apparently represents a direct hazard only after it separates and becomes airborne. Unusual concentrations of radon have been encountered in and near uranium mines, at locations where uranium or phosphate tailings (i.e., waste materials and byproducts of mining operations) exist, at disposal sites for manufacturing processes that included uranium–radium materials, in some geologic areas underlain by extensive deposits of granite and sillimanite (aluminum silicates), and from granular soil deposits resulting from glacially transported granite debris. In open space at such locations, health may not be affected because dangerous concentrations do not accumulate. But a significant hazard to health is anticipated where high concentrations accumulate indoors (in houses and at places of work) when a building is constructed over uranium-bearing rock, soil, or waste material and where

9The

loss of nitrates from soil, termed denitrification, is undesirable in the eyes of the agricultural community, where nitrate is considered to be a plant nutrient. 10Denitrification is often part of the treatment process in municipal wastewater (sewage) treatment plants.

296

Movement of Water Through Soil: Practical Effects

radioactive materials have been used inadvertently to manufacture material used for the building construction (e.g., masonry building blocks made with radioactive phosphate slag). A concentration of airborne radon exceeding about 4 picocuries per liter of air (pCi/l) typically is taken as the condition requiring a remedial action, but some agencies recommend 2 pCi/l for the threshold.11 The radon gas works its way into the building through construction joints or other openings in contact with the earth. Domestic water will be the source of radioactive material if the surface or subsurface supply has absorbed radon; for example, the radon becomes airborne when the water is sprayed. A concentration of 10,000 pCi per liter of water is taken as the threshold level requiring a remedial action. If the presence of radon in limited concentrations is known or suspected at the site of a planned building, measures for having the radon vented harmlessly to the atmosphere (instead of becoming dispersed or trapped within the building) can be incorporated into the building design and construction phases (Figure 22(h)). When radon in limited concentration is detected from groundwater or subsoil conditions at the location of an existing building, adequate mitigation methods consisting of gas interception and venting procedures often can be achieved at moderate expense. If a planned-for development is known as a location of high radon concentrations, the fill or soil or rock responsible can be removed (and possibly replaced with an unoffending material) if practical. Where the problem source cannot be identified or worked, realistic concerns relating to liability and health issues would probably dictate that the site be abandoned.

Use of Geosynthetics to Prevent Ground Contamination Though some soil and groundwater contamination occurs because of natural phenomena, the waste disposal locations associated with human activities—the solid waste (trash) landfills, sewage treatment and industrial waste liquid retention ponds, and mining waste dumps—are potentially the greatest problem because of the large volume and localized heavy concentrations. Recent efforts toward protecting the environment have dealt with preventing contaminants present at such waste disposal locations from escaping into the surrounding earth. Practical reasons involve factors such as ready identification of the contaminant source and of the property owner, which may be a municipality. In the case of a new facility, the disposal site can be properly prepared before being put into operation, or, for an older, in-use site, a modification can be accomplished to limit contaminant migration. The relatively recent availability of the geosynthetics discussed earlier in the chapter have improved on the opportunity to achieve protective measures that are reliable and practical. The use of geosynthetics for environmental protection against ground contamination almost always involves containment. Where possible, such as where planning identifies a waste disposal site before it is put into service, the geosynthetics are used to create a bottom seal for the ground surface

11The

curie (Ci) is the unit to indicate radioactivity. Radioactivity is the release of particles or radiation from the nucleus of an unstable atom as it is in the process of transforming to a stable atom. The release can be alpha particles, beta particles, or gamma radiation. One curie represents 3.7 * 1010 disintegrations per second; that is, the release of 3.7 * 1010 alpha particles or beta particles per second, or 3.7 * 1010 photons per second of gamma radiation. One picocurie represents 1 * 10-12 Ci, or 13.7 * 10102 11 * 10-122 * 1602 = 2.2 disintegrations per minute.

297

Movement of Water Through Soil: Practical Effects

across the basin area so the waste material will be isolated from the natural surroundings. When a top cover seal is eventually added, the waste material ends up being enveloped within the protective geosynthetic system. As is typical for present state-of-the-art systems, the protective geosynthetics will be multilayered and include zones where filtered liquids and gases can be collected for safe disposal. Representative designs for solid waste disposal and waste liquid retention ponds are shown in Figure 31(a), (b), and (c). Liquid chemical and petroleum storage tanks also have been a source of ground contamination (leaks, spills). As protection against the possibility of leaking, above-ground and buried tanks can be multishell, or installed where a geosynthetic envelope is also provided to ensure separation and confinement in the event of tank failure (Figure 31(d) and (e)). Soil cover layer (operating layer) Geotextile LCR (primary) Geotextile

Solid waste

FML (primary) Clay matting LCR (secondary) Anchor trench

FML (secondary) Legend:

2 to 3 ft compacted clay

LCR − leachate collection and removal (geonet) FML − flexible membrane liner (geomembrane) LCR + FML can be a geosynthetic composite (a) Double-liner system for solid waste landfills (geotextile, drainage composite and geomembrane with clay underlayer constitute each liner) Note: Amendment to the 1976 Federal Government Resource Conservation and Recovery Act (1984) requires use of a double liner for hazardous waste disposal facilities Optional geotextile and soil layer over geomembrane Geomembrane Geotextile underliner with optional drainage geosynthetic, for venting soil gases

Liquid waste Anchor trench Natural subgrade or compacted soil fill (b) Single-liner system for liquid waste retention ponds (geomembrane for containment overlies geotextile with high planar permeability for gas venting; optional design includes soil layer plus geotextile above geomembrane)

Figure 31 Use of geosynthetics to protect against ground contamination.

298

Movement of Water Through Soil: Practical Effects Gases generated by decomposing waste discharge to atmosphere or are collected

Small, low vegetation (small, shallow root system) Capping soil layer for vegetative surface root system

Gas vent

Filter geotextile

x

Cap drainage (pervious soil)

xx

x

x x xxx

xx

xxx x

x xxxxx x x

x x x x xx x xx xxx x x x x x x

Geomembrane to prevent infiltration of surface water Compacted clay layer

x x xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Gas venting geotextile Operational cover (soil) for solid waste provides operating surface for placement of gas-venting geotextile

Solid waste

Infiltrating surface water flows through cap drainage layer, to perimeter collection system

Base liner system (c) Detail of closure cap for solid waste landfill

Geomembrane liner

Geomembrane liner Buried storage tanks

Perimeter containment dike (e) Ground protection liner for above-ground storage tanks

(d) Liner system for underground storage tanks and related piping

Figure 31

(continued)

Problems 1 A canal is located in an area where the alignment parallels that of a nearby river. The difference in the water surface elevation of the canal and the river is 10 m. A 3-m-thick buried stratum of granular soil (permeable) extending between the canal and river is bounded by layers of clay (impermeable). The horizontal distance

between the two bodies of water is 400 m. The coefficient of permeability for the granular soil is 0.75 m per day. (a) Make a sketch of the described conditions. (b) Calculate the volume of seepage that occurs from the canal to the river, per kilometer of canal length.

299

Movement of Water Through Soil: Practical Effects 2 A river is located in an area where its route parallels that of a nearby canal. The elevation of the water surface in the river is +375 ft. In the canal, the water surface is at elevation +345 ft. A stratum of coarse-grained soil 10 ft thick is sandwiched between relatively impermeable fine-grained soils, and extends between the canal and the river. The distance between the river and the canal is 500 ft, and the coefficient of permeability of the buried coarse stratum is 0.01 ft per hour. (a) Make a sketch of the described conditions. (b) Calculate the seepage loss from canal to river, in gallons per day per mile of river canal length. 3 What are flow nets and why are they used? 4 Flow lines and equipotential lines constitute flow nets used to study the subsurface movement of fluids. Explain the physical factor that the flow lines represent and what the equipotential lines indicate. 5 Are flow nets to study subsurface flow appropriate only to coarse-grained soil with a high coefficient of permeability? Explain. 6 A concrete dam is 30 m long from upstream side to downstream end. The dam is situated on a 15-m-thick stratum of soil that has a coefficient of permeability equal to 0.5 m per day. Firm, hard clay and rock underlies the described soil stratum. The base of the dam is installed at a level 2 m below the natural soil surface. Draw the flow net for seepage occurring beneath the dam, and compute the seepage in cubic meters per day per meter of dam width (perpendicular to the paper), for the condition where the reservoir water against the upstream face of the dam is 18 m above the soil surface and the water level at the downstream end is at the soil surface. 7 A concrete gravity dam is 100 ft long from upstream to downstream end. It is constructed on a stratum of sand–silt soil 40 ft thick. The coefficient of permeability for this soil is 0.1 ft per day. Rock underlies the soil stratum. The bottom of the dam is 6 ft below the soil surface. (a) For the condition where the height of water against the upstream face of the dam is 45 ft above the soil surface and the water level at the downstream end is just at the soil

300

surface, draw the flow net for seepage beneath the dam and calculate the quantity of seepage in gallons per day per foot of width of dam. (b) For the condition where the height of water against the upstream face of the dam is 35 ft above the soil surface and the water level at the downstream end is 5 ft above the soil surface, draw the flow net for seepage beneath the dam and compute the seepage in gallons per day per foot of width of dam. 8 The vertically upward flow of groundwater (an artesian condition) occurs through a sand deposit where the void ratio is 0.60 and the specific gravity of soil particles is 2.65. What hydraulic gradient is necessary for a quicksand condition to develop? 9 The hydraulic gradient for an occurrence where there is a vertically upward flow of water through a sand mass is 0.95. If the specific gravity of the soil particles is 2.75 and the void ratio is 0.65, determine if a quicksand or erosion condition could develop. 10 Explain why a pumped well results in a coneshaped, depressed, or drawn-down water table in the region surrounding the well location. 11 Why is the vacuum well-point dewatering system expected to withdraw a greater quantity of groundwater than the conventional well-point dewatering system? 12 Well points are used to dewater an excavation that extends 5 m below the surface of the water table. The soil being dewatered is a fine sand. At what distance (approximately) outside of the well-point system is the original groundwater table unchanged (not drawn down)? 13 Well points are to be used to dewater a large trench excavation that extends 35 ft below the surface of the water table. Assume that the bottom width of the trench excavation is 70 ft. (a) Develop a simple sketch of the necessary two-stage well-point system. Use separate diagrams to show the progress of the necessary stages of installation. (b) Assume that half the necessary lowering of the water table elevation is achieved with the outer row of well points. If the soil being dewatered is a fine-to-medium sand,

Movement of Water Through Soil: Practical Effects approximately at what distance beyond the well-point location is the original groundwater table not affected by well-point pumping?

22 Why is it expected that a period of rapidly developed extreme cold (weather) will not produce severe soil frost heave?

14 Indicate the intended functions for foundation drains and the major requirements for design and installation so as to achieve a properly operating system.

23 For extensive freezing-related soil heave to occur, an important contributing natural feature must continue. What is this feature and how does it occur? (Hint: Consider the extent of ice lens formations.)

15 What are the intended functions and limitations of interceptor drains as conventionally used for highway and airfield construction?

24 What methods are commonly used to prevent the occurrence of, or to offset the effects of, frost heave?

16 What are the primary reasons for having limitations on the sizes of particles used to construct aggregate-type drainage filters?

25 In regions of permafrost, why is it expected that the thickness of the frozen soil zone tends to be relatively constant?

17 Relating to interceptor drains for highway and airfield pavements: (a) Outline the traditional method for installing pipe drains. (b) Outline techniques and materials for installations using geosynthetics.

26 For construction work in permafrost regions, what type of foundation construction and protection is necessary to maintain structural stability? 27 (a) Briefly define groundwater contamination.

18 Relating to the use of geosynthetics on construction projects, briefly describe what is meant by (a) the separation function, (b) the filtration function, (c) the drainage function, (d) the reinforcement function, and (e) the containment function. 19 Why is it important that the geotextiles (geofabrics) consist of synthetics and not natural fibers? 20 Relating to geotextiles (geofabrics), what is a significant difference in appearance and texture between the woven and nonwoven materials? 21 Briefly describe the conditions necessary and the resulting sequence of occurrences that typically take place in a frost heaving situation. Specify the soil type considered most susceptible to large frost heaves.

(b) What is leachate? 28 Briefly describe how the direction of groundwater flow and the density properties of a contaminating liquid affect the configuration of a contaminant plume in an aquifer. 29 Relating to contaminant materials buried or accumulated underground, list the categories (general procedures) available to correct or control problems that tend to result, and briefly describe the procedures involved. 30 List the types of soil gases expected to be generated at a solid waste landfill site and the typical problem associated with each. 31 Briefly describe what radon is and the reasons for being concerned about its presence. 32 Indicate briefly, using words or sketches, how geosynthetics have been used to prevent groundwater contamination.

301

Answers to Selected Problems

1. 2. 6. 7. 8.

302

Q = 56.25 m3>day per km length Q = 32 ft3>hr = 5700 gal>day 2.7 m3兾day per m wide (a) 8.5 gal兾day per ft wide (b) 5.6 gal兾day per ft wide i = 1.03

9. 12. 13.

i␥w = 0.95 ␥w (no quicksand, close) 33 meters distance approx. 350 ft

Combined Stresses in Soil Masses

From Essentials of Soil Mechanics and Foundations: Basic Geotechnics, Seventh Edition. David F. McCarthy. Copyright © 2007 by Pearson Education, Inc. Published by Pearson Prentice Hall. All rights reserved.

303

Combined Stresses in Soil Masses Stress at a Point and Mohr’s Circle

When a body or mass is subjected to external loading, various combinations of internal normal and shear stresses are developed at the different points within the body or mass. Generally, data concerning internal stress conditions are used to determine deformations (in soils work, deformation is most frequently associated with settlement) and to check for the possibility of a material (soil) failure occurring because the strength is exceeded. In performing a stress study, it is convenient to use methods from engineering mechanics for analyzing stress at a point. With these methods, stresses acting on any plane passing through the point can be determined. The combinations of normal and shear stress that develop will vary, depending on the plane being analyzed (Figure 1) and the magnitude of the external loading. The “stress at a point” analysis is also applicable for determining the

Loading applied to surface causes stresses within the soil deposit Soil deposit

One combination of normal and shear stress act on this plane passing through the point Point under analysis

Different combination of normal and shear stress act on this plane passing through the point

Figure 1 Concept for analysis of stress at a point where external loading causes different combinations of normal and shear stress within the material.

304

Combined Stresses in Soil Masses

weakest plane or potential plane of failure in a material, which is not always easily evident, and for indicating the magnitude of the stresses that act on this plane. This type of analysis has particular application for soil and foundation studies, since stability failures in soil masses are the result of the shear strength of the soil being exceeded.

1

Stress at a Point: Analytical Development The stresses acting on any plane passed through a point within a material consist of a normal stress (compression or tension) and a shearing stress. Depending on the type of external loading causing the stress condition, it is possible for the shear or the normal stress, or both, to be zero on some planes. In soil problems, most external loadings are compression. (The downward weight of a structure supported by a soil mass would be a compressive loading.) For a situation where the loading is compressive, normal stresses that develop on any plane would almost always have a value other than zero; some shear stress would act on all planes with the exception of two planes, where it will be zero (discussed in the following paragraphs). If the combination of normal and shearing stresses acting on any two mutually perpendicular planes (orthogonal planes) is known, the combination of stresses acting on any other plane through the same point can be determined. In analyzing stress at a point, it is convenient to assume an incremental element that represents the stress conditions at the point and to show the known stresses acting on it, as indicated in Figure 2(a). For equilibrium, the sum of forces (not stresses) acting in any direction must equate to zero, and the rotational moments about any axis caused by forces similarly must equal zero. To satisfy this latter requirement, shear stresses acting on orthogonal planes must be equal in magnitude.

σa, σb = normal stresses (compression) τ1 = shear stress τ1

σa

τ1

σb

σc

τ1 σa

τ1

τ2

σb

τ2

σd τ2

σd τ2

β

σc, σd = normal stresses τ2 = shear stress β = angle of rotation between element of Figure 1 (a) and this figure.

σc (a)

(b)

Figure 2 Basic representation for stress-at-a-point analysis, indicating stress combinations acting on an incremental element. (a) Incremental element with representative stresses assumed for analysis of “stress at a point”; (b) method of rotating incremental element to determine stresses on different planes of study.

305

Combined Stresses in Soil Masses

At the same point, but on a differently oriented element (actually a different plane passing through the identical location), the combination of normal and shearing stresses that act will be different (Figure 2(b)). As will be shown, however, there is a relationship between the normal and shear stresses acting on all planes (or orientation of elements) that pass through the same point. Consider all the planes that can be passed through a point. On one particular plane, not yet defined, the shear stress will be zero, whereas the normal stress will be the maximum possible value of all the normal stresses acting on the various planes through that point. On a plane perpendicular to the plane just referred to, the shear stress will also be zero, but the normal stress that acts will be the least, or the minimum, of all the normal stresses acting on different planes through the point. These maximum and minimum normal stresses are called principal stresses. The planes on which they act are principal planes. The shear stress on a principal plane is always zero. In many practical soil problems, the principal stresses act in the vertical and horizontal directions (or on horizontal and vertical planes) and are easily calculated. For instance, where the incremental element represents a point within a soil mass where the ground surface is horizontal, the vertical stress is due to the weight of the soil overburden at that point. This would be the effective unit weight of soil multiplied by the depth of the point below the ground surface. The horizontal stress is proportionate to the vertical stress. Thus, the magnitudes of the principal stresses are known, as are the orientations of planes on which the stresses act. If the major and minor principal stresses are ␴1 and ␴3, respectively, the magnitude of a normal stress ␴n and shear stress ␶n on any other plane can be determined. Referring to the stressed element in Figure 3(a) and letting the area on the cut plane be unity (Figure 3(b)), it is apparent that the area of the plane on which ␴1 acts becomes 112 * 1cos ␪2, and the area on which ␴3 acts becomes 112 * 1sin ␪2. The total normal force on the cut plane is 1␴n2 * 112. The total force on the vertical surface is 1␴32 * 1sin ␪2, and the total force on the horizontal surface is 1␴12 * 1cos ␪2.

σ1 rea

Fn

= σn

×A

= σn

×1

= σn

Assign value of one square unit to the area of this plane. Therefore, area = 1 Area = 1 × sin θ

σ3

σ3

θ

τn

= τn

×1

F3 = σ3 × Area = σ3 sin θ

θ

Area = 1 × cos θ σ1

F1 = σ1 × Area = σ1 cos θ

(a) (b)

Figure 3 Basic step in determining stresses on a random plane through a point in terms of known principal stresses.

306

Combined Stresses in Soil Masses σn os

σ3

sin

θc

θ

σ3 sin θ sin θ

θ

τn

σ3 sin θ

θ θ

σ1 cos θ cos θ

σ1 cos θ σ1 cos θ sin θ

Figure 4 Resolution or principal stresses into components parallel and perpendicular to a random plane. This step is in preparation for determining stresses on the plane in terms of principal stresses.

To determine ␴n and ␶n in terms of ␴1 and ␴3, the forces acting on the horizontal and vertical planes are resolved into components parallel and perpendicular to the cut plane (Figure 4). Summing forces parallel to ␴n (normal to the cut plane) gives: ␴n = ␴1 cos ␪ cos ␪ + ␴3 sin ␪ sin ␪ = ␴1 cos2 ␪ + ␴3 sin2 ␪ and by trigonometric identity: ␴n = ␴1 cos2 ␪ + ␴3 sin2 ␪ =

␴1 + ␴3 ␴1 - ␴3 + cos 2␪ 2 2

(1)

Summing forces parallel to the cut plane in the direction of ␶n gives:

␶n = ␴1 cos ␪ sin ␪ - ␴3 sin ␪ cos ␪ = 1␴1 - ␴32 sin ␪ cos ␪

and by trigonometric identity: ␶n = 1␴1 - ␴32 sin ␪ cos ␪ = a

␴1 - ␴3 b sin 2␪ 2

(2)

When applying these equations, it is necessary to use some method that will prevent compressive stress from being confused with tensile stress; for geotechnical problems, it is typical to identify compressive stress as a positive value and tensile stress as a negative value. From examination of Equation 2 it should be recognized that the maximum shear stress will occur on a plane that is 45° from the major principal plane 1␪ = 45°2 and will have a magnitude equal to 121␴1 - ␴32. On this plane, the normal stress is always 1 2 1␴1 + ␴32. The previous developments apply to a two-dimensional stress analysis (stresses in a plane). In actuality, for soil problems, the study of stress at a point involves a three-dimensional analysis. An incremental element with three principal stresses acting—␴1, ␴2, and ␴3—is shown in Figure 5. For many practical conditions of loading, the intermediate stress, ␴2, is equal to ␴1, or to ␴3. For some situations, however, ␴2 may be a value intermediate

307

Combined Stresses in Soil Masses σ1

σ2 σ3

Figure 5 Three-dimensional incremental element showing principal stresses in mutually perpendicular directions.

between ␴1 and ␴3. As far as practical effects of the intermediate stress are concerned, it appears that there is some influence on the strength and stress–strain properties of the material, but its effect is not clearly understood. To keep a proper perspective, however, it is pointed out that the methods available for making determinations of the stresses within a soil mass that result from external loading are not highly refined. As a result, for most practical problems, the degree of accuracy does not appear to be significantly diminished by neglecting the effect of the intermediate stress and working with the simpler twodimensional condition (␴1 and ␴3 only).

2

Mohr’s Circle The varying values of normal stress and shear stress corresponding to differing values of ␪ can be determined by using Equations 1 and 2. If the combination of normal and shear stress resulting from each value of ␪ is plotted as a point on a coordinate system where the horizontal axis represents normal stress and the vertical axis represents shear stress (Figure 6(a)), the locus of many plotted points will form a circle (Figure 6(b)). This fact can be used to advantage. By working with simple properties of a circle, a graphical or pictorial method of solving for normal and shear stresses on any plane is easily developed once the stresses on orthogonal planes are known. The method simplifies Shear stress, τ

τ

σ

τ

σ

τ

Normal stress, σ

σ

0,0 (origin)

(a)

(b)

Figure 6 Method of representing the combination of normal and shear stresses acting on any plane through an incremental element on shear stress–normal stress coordinates: (a) single point represents the stress combination on one plane; (b) stress combination occuring on different planes through an incremental element provides points that form a circle.

308

Combined Stresses in Soil Masses σ1 τ

τ

σ3

Stresses shown on horizontal plane: positive normal stress positive shear stress

σ3 τ τ σ1

Stresses shown on vertical plane: negative normal stress negative shear stress

Figure 7 Sign convention assigned to stresses for the Mohr’s circle analysis.

the calculation necessary for determining stresses and eliminates the need to work with cumbersome equations. The method is referred to as the Mohr’s circle for determining stresses, after Otto Mohr (1835–1918), who is credited as the developer of the method. In using Mohr’s circle, a sign convention is required. For soil problems, compressive stresses are conventionally assumed to be positive, and shearing stresses that provide a clockwise couple are also considered positive (Figure 7). For a stress combination like that shown in Figure 8(a), ␴1 and ␴3 are the major and minor principal stresses. There is no shear stress acting on these major and minor principal planes. To construct the Mohr’s circle for this combination (Figure 8(b)), locate ␴1 (or 2.5 kPa) and ␴3 (or 0.50 kPa) on the horizontal axis of the coordinate system (since the shear is zero). Next, establish the location (or value) for the center of the Mohr’s circle, knowing that the diameter of the circle has a value equal to ␴1 minus ␴3. (This numerical difference between ␴1 and ␴3 is called the deviator stress.) For this illustration, the deviator stress is 2.50 kPa minus 0.50 kPa, or 2.0 kPa. The radius of the circle is then 1.0 kPa, and the center of the circle will plot at 1.5 kPa, which is the value of ␴3 plus the radius. After the center of the circle and the diameter are established, the circle itself can be constructed; thereafter, the stress combination on any plane can be determined. In working with the Mohr’s circle, it is convenient to reference the plane under study to the major principal plane whenever possible. Remember that the major principal stress τ

2.50

σ1 = 2.5 kPa

σ3 = 0.5 kPa

σ3

σ1

0.5

Deviator = 2.0

σ3

Center = 1.5

σ1

0.50 + 1.0 = 1.5 (a)

(b)

Figure 8 Representation of principal stresses acting at a point and the related Mohr’s circle plot: (a) stresses acting on incremental element; (b) Mohr’s circle plot.

309

Combined Stresses in Soil Masses

acts on the major principal plane; the angular measurement to the plane in question is made on the Mohr’s circle by starting from the point representing the major principal stress. If the angle to be measured is formed by two radii, the angle is measured at the center of the circle. Because of the properties of a circle, the central angle on the circle must be twice the value of the angle ␪ measured on the original element (Figure 9). The direction of measurement from the principal plane to the cut plane on the element, clockwise or counterclockwise, is the same direction to be used on the Mohr’s circle (Figure 9). By reference to the Mohr’s circle diagram in Figure 10, the expressions for shear and normal stress for any point (representing any plane) can be reaffirmed. The normal stress coordinate is the value of the center of the Mohr’s circle plus or minus the horizontal projection of the radius. As shown in Figure 10, the normal stress would then be: ␴n =

␴1 + ␴3 ␴1 - ␴3 + cos 2␪ 2 2

τ

τ σn, τ θ

σ3

σ3

σ

σ1

σ1

θ

σ

2θ σn, τ

σ1 σn τ

θ

σ3

σ3

θ

τ

σn σ1

(b)

(a)

Figure 9 Relation of positions of planes on the incremental element to points on the Mohr’s circle: (a) reference angle measured counterclockwise; (b) reference angle measured clockwise. τ σn =

σ1 + σ3 σ1 − σ3 + cos 2θ 2 2

P R τ

σ3

R cos 2θ σ − σ3 σ3 + 1 2 σ1 + σ3 = 2

(

)

τn = R sin 2θ = σ1 R=

σ1 − σ3 sin 2θ 2 σ 1 − σ3 2

σ1 − σ3 2

Figure 10 Establishment of the general equations for normal stress and shear stress from the Mohr’s circle.

310

Combined Stresses in Soil Masses τ σn = σ3 + radius = 1.50

σ1 = 2.50 kPa θ = 45°

R = 1.0

σ3 = 0.50 kPa

σ3 θ = 45°

2θ = 90°

1.50 σ3 = 0.50

τ = 1.0 = τmax σ σ1 = 2.50

σ1

Figure 11 Use of the Mohr’s circle to determine the shear and normal stress acting on a plane 45° from the major principal plane.

which agrees with Equation 1. The shear stress coordinate is the vertical projection of the radius, or: ␶n =

␴1 - ␴3 sin 2␪ 2

which agrees with Equation 2. As an illustration, to determine the magnitude of the normal and shear stress acting on a plane 45° from the major principal plane (Figure 11), the angle measured at the center of the Mohr’s circle would have to be twice 45°, or 90°. Note that, on the element, the angle ␪ is 45° measured counterclockwise, regardless of whether the top or bottom of the original element is used as the reference plane. On the Mohr’s circle the central angle, 2␪ or 90°, is also measured counterclockwise. If the principal stress equations previously developed are used, or by applying simple mathematics if the Mohr’s circle is used pictorially, or by scaling if the graphical method is used, the value of the normal stress is found to be 1.5 kPa and the shear stress is 1.0 kPa. From examination of the Mohr’s circle, it should be evident that the maximum shear stress (in this case 1.0 kPa) always acts on the plane that is 45° from the major principal plane. Further, the maximum shear stress always has a magnitude equal to the radius of the Mohr’s circle. As a second illustration, with the same principal stresses previously used, assume that it is desired to find the stresses acting on a plane that is 60° clockwise from the major principal plane (Figure 12). On the Mohr’s circle, the central angle to be measured is 120°. Mathematically, from a pictorial Mohr’s circle, the shear stress ␶n equals 1R2 * 1sin 60°2, or 11.0 kPa2 * 10.8662, which is 0.866 kPa. The normal stress ␴n is 1.5 kPa – (R) * 1cos 60°2 , or 1.0 kPa, as shown in Figure 12. The 1.5 kPa value locates the center of the circle. On the desired plane, the stresses are as shown in Figure 13. If the stress combination is to be determined on a plane measured with reference to the minor principal plane, the related angular measurement on the Mohr’s circle is made from the minor principal stress, ␴3, as shown in Figure 14. The direction of the angle will be the same on the Mohr’s circle as on the incremental element.

311

Combined Stresses in Soil Masses τ σ1 = 2.5 kPa

R = 1.0

R cos 60° θ = 60°

σ3 = 0.50 σ3 = 0.5 kPa

σ3

σ1 = 2.50

τ= R sin 60° = 0.866

60°

2θ = 120°

P σ1 σn = 1.50 − R cos 60° = 1.0

Figure 12 Use of the Mohr’s circle to determine stresses on a plane 60° clockwise from the major principal plane. τ = .866 kPa σn = 1.0 kPa τ = .866 kPa

σn = 1.0 kPa

Figure 13 Magnitude and direction of stresses acting on the plane studied in the Mohr’s circle analysis of Figure 12. τ σ1 σ, τ

σ3

σ1

σ1

σ3

θ

σ3

σ1

σ

τ σ3

θ

σn

Figure 14 Method of locating the stress–combination point on the Mohr’s circle when the minor principal plane is used as a reference.

The Mohr’s circle pictorial or graphical method can similarly be used for determining principal stresses if the stress conditions for any two orthogonal planes other than principal planes are known. For an element where the normal shear stresses are as shown in Figure 15 (note that the shear stresses on the mutually perpendicular planes are equal in magnitude), the Mohr’s circle construction proceeds as follows (Figure 16):

312

σa = 3.0 kPa τ = 0.50 kPa

σb = 0.6 kPa

σb

τ = 0.5 kPa

Figure 15 planes.

An incremental element with stresses that are not principal stresses acting on principal

τ

τ R = 1.22 + 0.52 = 1.30

3.0 − 0.60 = 2.40 1.20

.30

(3.0, 0.5) 1.20

σ

0.50

1.2 + 0.6 = 1.80

(0.60, −0.50)

σ3 = 1.8 − R = 0.50

=1

0.50

1.20 .30

σ

1

(a) τ

R

(b) τ

1.80

2θ 1.20

σ σ1 = 1.80 + R = 1.80 + 1.30 = 3.10

0.50

σ

Tan 2θ = .500 1.20 = 0.416 2θ = 22.6° θ = 11.3°

(d)

(c) σ1 θ = 11.3° σ3 = 0.50 kPa

σ3

σ1 = 3.1 kPa (e)

Figure 16

Procedure for applying the Mohr’s circle analysis to solve for principal stresses.

313

Combined Stresses in Soil Masses

1. On the Mohr’s circle coordinates, establish the two points corresponding to ␴a, ␶, and ␴b, ␶. 2. Connect these two points, thereby establishing the diameter of the Mohr’s circle (Figure 16(a)), and determine the coordinate for the center of the circle. 3. Calculate the value for the radius of the Mohr’s circle (Figure 16(b)). 4. The value of ␴1 (the major principal stress) is the circle radius added to the value established for the circle center. The value of ␴3 (the minor principal stress) is the radius subtracted from the coordinate for the circle center (Figure 16(c)). 5. The angle 2␪ on the Mohr’s circle is obtained from simple geometry, as shown in Figure 16(d). 6. The orientations of the principal planes with respect to the original element are shown in Figure 16(e). 7. Since the complete Mohr’s circle is established, stresses on any other plane can be determined by using the methods covered previously. Illustration 1 The major and minor principal stresses acting at a point are 50 Pa compression and 10 Pa compression, respectively. Draw the Mohr’s circle for this stress combination, and determine the magnitude of shear and normal stress on one of the planes where shear is a maximum.

Solution τ

Element showing principal stresses.

σ1 = 50 Pa σn = 30 2θ = 90°

σ3

σ3 = 10 Pa

τ max = 20 30

σ1 = 50

σ3 = 10 τmax = radius = 20 Pa 2θ = 90° θ = 45°

σ

σ1 σn = 30 Pa

τ = 20 Pa θ = 45° τ σn

Element showing stresses on plane where shear is a maximum, and the orientation relating to principal planes.

Illustration 2 At one point in a soil mass, the minor principal stress is 1,000 psf. What is the maximum value possible for the major principal stress at this point if the shear stress cannot exceed 2,000 psf?

314

Combined Stresses in Soil Masses

Solution τ

τ max = R σ3

σ3 = 1000

σ1

σ

2R = 4,000

τ

The maximum shear stress at a point is the value of the radius of the Mohr’s circle; that is: Radius = 2,000 psf Since the major principal stress s1 = s3 + 2R: s1 = 1,000 + 212,0002 = 5,000 psf

Problems 1 In terms of the stress-at-a-point analysis, define principal stress. Indicate how it is different from other normal stresses (tensile or compressive) that also act at the point. 2 At a point in a stressed material, the major principal stress is 50 kPa compression and the minor principal stress is 20 kPa compression. Use Equations 1 and 2 to determine the maximum shear stress that acts through the point and the normal stress on the plane of this maximum shear. 3 At a point in a stressed material, the major principal stress is 44 kPa tension and the minor principal stress is 18 kPa tension. Using Equations 1 and 2, determine the maximum shear stress that acts through the point and the normal stress on the plane of maximum shear. 4 At a point in a stressed material, the major principal stress is 7.2 ksi compression and the minor principal stress is 3.6 ksi compression. Using Equations 1 and 2, determine the maximum shear stress that acts at the point and the normal

stress that acts on the plane of the maximum shear stress. 5 Using the equations for determination of stress at a point, calculate the value for the maximum shear stress that develops at a point if the major principal stress is 36 kPa compression and the minor principal stress is 14 kPa tension. Also indicate the value of normal stress acting on the plane of maximum shear. 6 Using the equations for determination of stress at a point, calculate the value for the maximum shear stress that develops if the major principal stress is 7.2 ksi compression and the minor principal stress is 3.6 ksi tension. 7 At a point in a stressed material, the major principal stress is 120 kPa compression and the minor principal stress is 40 kPa compression. (a) Draw the Mohr’s circle for this stress combination. (b) Determine the maximum shear stress. Also indicate the normal stress on the plane where shear is a maximum.

315

Combined Stresses in Soil Masses (c) Determine the value of normal and shear stress acting on the plane 30° counterclockwise (ccw) from the major principal plane. 8 At a point in a stressed material, the major principal stress is 5,000 psf compression and the minor principal stress is 2,000 psf compression. (a) Draw the Mohr’s circle for this stress combination. (b) Find the maximum shear stress, and also indicate the value of normal stress acting on this same plane. (c) Determine the values of the normal and shear stresses acting on a plane that is 60° ccw from the major principal plane. 9 At a point in a stressed material, the major principal stress is 180 kPa compression and the minor principal stress is 40 kPa tension. (a) Determine the Mohr’s circle for this stress combination. (b) What is the maximum shear stress that acts at the point, and what is the value for the normal stress acting on the plane of maximum shear? (c) Determine the value of the normal and shear stress acting on the plane 22.5° ccw from the major principal plane. 10 At a point in a stressed material, the major principal stress is 80 psi compression and the minor principal stress is 20 psi compression. (a) Draw the Mohr’s circle for this stress combination. (b) What is the maximum shear stress acting at the point, and what value of normal stress acts on this same plane? (c) Determine the value of the normal and shear stress acting on a plane that is 60° clockwise (cw) from the minor principal plane. 11 At a point in a stressed material, the major principal stress is 66 kPa compression and the minor principal stress is 34 kPa tension. (a) Use the Mohr’s circle to determine the shear and normal stress on a plane 45° ccw from the major principal plane. (b) Determine the stress combination acting on the plane 75° ccw from the major principal plane. 12 The major and minor principal stresses acting at a point in a stressed material are 80 psi compression and 20 psi tension.

316

(a) Draw the Mohr’s circle for this stress combination. (b) What is the maximum shear stress acting at the point, and what value of normal stress acts on the plane? (c) Determine the value of the shear and normal stresses acting on a plane that is 30° ccw from the major principal plane. 13 At a point in a stressed material, the stress combination acting on a plane is a 90 kPa compressive stress and a 30 kPa shear stress, while on a perpendicular plane, the stress combination is 30 kPa compression and 30 kPa shear. Use the Mohr’s circle for this stress analysis. (a) Determine the values of the major and minor principal stresses. (b) Determine the angle between the major principal plane and the plane where the 90 kPa compressive stress acts. 14 At a point in a stressed material, the stress combination acting on a plane is 60 kPa compression and 20 kPa shear, while on an orthogonal plane, the stress combination is 40 kPa tension and 20 kPa shear. Use the Mohr’s circle for this stress analysis. (a) Determine the values for the major and minor principal stresses. (b) Determine the value of the maximum shear stress and the value of the normal stress on the plane of maximum shear. 15 At a point, the normal and shear stresses acting on one plane are 8,000 psf compression and 2,000 psf, respectively. On a perpendicular plane, the normal and shear stresses are 2,000 psf compression and 2,000 psf, respectively. (a) Draw the Mohr’s circle. (b) Determine the value of the principal stresses. (c) Find the angle between the plane on which the 8,000 psi compressive stress acts and the major principal plane. (d) What is the maximum shear stress acting at the point? 16 The normal and shear stresses acting on one plane passing through a point in a soil mass are 120 psi compression and 25 psi, respectively. On an orthogonal plane, the respective stresses are 40 psi compression and 25 psi. (a) Draw the Mohr’s circle for this stress condition. (b) What are the principal stresses?

Combined Stresses in Soil Masses (c) Determine the angle between the plane on which the 120 psi stress acts and the major principal plane. (d) What is the maximum shear stress acting at the point? 17 The major principal stress and deviator stress at a point are 100 kPa and 60 kPa, respectively. Indicate the magnitude of the minor principal stress. What is the value of the maximum shear stress and of the normal stress acting on the same plane? (Deviator stress is the algebraic difference between the values for the principal stresses.) 18 For a minor principal stress of 50 kPa, what is the maximum value possible for the major principal stress if the maximum shear stress is not to exceed 40 kPa? 19 At a location where the ground surface is horizontal, the vertical stress at one point in the underlying soil deposit is 190 kPa and the horizontal stress is 76 kPa. If these stresses represent principal stresses, what is the maximum shear stress passing through the point? What is the value of the normal stress on the plane of maximum shear? 20 In a soil where the ground surface is horizontal, the vertical stress acting at a point is 2,000 psf and the horizontal stress is 1,000 psf. If these stresses represent the principal stresses, what is the maximum shear stress acting at the point? 21 A level ground surface exists in an area where soil in the underlying stratum has a wet unit weight of 19 kN/m3. The vertical pressure at any depth represents the major principal stress. The vertical pressure (stress) is the product of the soil unit weight and depth. The lateral pressure at a corresponding depth is 0.45 times the vertical pressure and represents the minor principal stress. (a) Determine the principal stresses acting at a depth 4 m below the surface, and draw the Mohr’s circle for the condition. (b) What is the maximum shear stress acting at this depth? 22 Assume a soil mass with a level ground surface and a material having a wet unit weight of

120 pcf. The vertical pressure at any depth represents the major principal stress at that particular point. The vertical pressure (stress) is the product of the soil unit weight and depth. The lateral pressure acting at any point is one-half the vertical pressure and represents the minor principal stress. (a) Determine the principal stresses acting at a depth of 10 ft below the ground surface, and draw the Mohr’s circle for these conditions. (b) What is the value of the maximum shear stress at this depth? 23 In an area where the ground surface is level, the underlying soil has a unit weight of 16 kN/m3. Determine the principal stresses acting at a depth of 5 m if the lateral pressure is one-half the vertical. The vertical pressure (stress) is the product of the soil unit weight and depth, and is the major principal stress. 24 The soil in a deposit underlying an area where a level ground surface exists has a unit weight of 21 kN/m3. (a) Determine the principal stresses at a depth of 5 m below the soil surface if the lateral pressure is one-half the vertical pressure. The vertical pressure (stress) is the product of the soil unit weight and depth. (b) Determine the value of the maximum shear stress at the 5-m depth. 25 A soil deposit exists with a level ground surface. The saturated unit weight of soil in the deposit is 125 pcf. (a) Determine the principal stresses acting at a depth 15 ft below the ground surface if the lateral pressure is 0.45 times the vertical pressure. The vertical pressure (stress) is the product of the soil unit weight and depth. (b) What is the magnitude of the maximum shear stress at this depth? 26 A soil deposit exists in an area where the ground surface is level. The unit weight for the soil is 22 kN/m3. The lateral pressure is 0.5 times the vertical pressure. The vertical and lateral pressures represent principal stresses. The vertical pressure (stress) is the product of the soil unit weight and depth. Determine the value for the major and minor principal stresses at a depth of 10 m.

317

Answers to Selected Problems

2. 3. 4. 5. 6. 7. 8. 10. 12. 15.

318

␶max = 15 kPa, ␴n = 35 kPa 1C2 ␶max = 13 kPa, ␴n = 31 kPa 1T2 ␶max = 1.8 ksi, ␴n = 5.4 ksi 1C2 ␶max = 25 kPa, ␴n = 11 kPa 1C2 ␶max = 5.4 ksi, ␴n = 1.8 ksi 1C2 (b) ␶max = 40 kPa, ␴n = 80 kPa (c) ␶ = 34.64 kPa, ␴n = 100 kPa (b) ␶max = 1500 psf, ␴n = 3500 psf (c) ␴n = 2750 psf, ␶ = 1300 psf (b) ␶max = 30 psi, ␴n = 50 psi 1C2 (c) ␶ = 26 psi, ␴n = 65 psi 1C2 (b) ␶max = 50 psi, ␴n = 30 psi (c) ␶ = 43 psi, ␴n = 55 psi (b) ␴1 = 8600 psf 1C2, ␴3 = 1400 psf 1C2 (c) ␪ = 16.9° (d) ␶max = 3600 psf

16.

17. 18. 20. 21. 22. 23. 24. 26.

(b) ␴1 = 127 psi 1C2, ␴3 = 33 psi 1C2 (c) ␪ = 16° (d) ␶max = 47 psi ␶max = 30 kPa, ␴n = 70 kPa ␴1 = 130 kPa ␶max = 500 psf ␴1 = 76 kPa, ␴2 = 34.2 kPa ␶ = 20.9 kPa (a) ␴1 = 1200 psf, ␴3 = 600 psf (b) ␶max = 300 psf ␴1 = 80 kPa, ␴3 = 40 kPa ␴1 = 105 kPa, ␴3 = 52.5 kPa ␶ = 26.25 kPa ␴1 = 220 kPa, ␴3 = 110 kPa

Subsurface Stresses

From Essentials of Soil Mechanics and Foundations: Basic Geotechnics, Seventh Edition. David F. McCarthy. Copyright © 2007 by Pearson Education, Inc. Published by Pearson Prentice Hall. All rights reserved.

319

Subsurface Stresses

At a point within a soil mass, stresses will be developed as a result of the soil lying above the point and by any structural or other loading imposed onto that soil mass. The magnitude of the subsurface stress at a point is affected by the groundwater table if it extends to an elevation above the point. In most foundation design problems, the safe bearing capacity of the soil (the ability to support structural load) and the settlement (the soil volume change resulting from loading) that will develop under a given intensity of structural loading are major items of concern. For such analysis, the significant stresses are considered to be those acting in the vertical direction. In the design of vertical structures such as retaining walls, sheeting for braced excavations and waterfront structures, and some types of pile foundations, the soil stresses acting in the horizontal or lateral direction are the most significant.

1

Stresses Caused by the Soil Mass Vertical Stresses In a soil mass having a horizontal surface, the vertical stress caused by the soil at a point below the surface is equal to the weight of the soil lying directly above the point. Vertical stress thus increases as the depth of the soil overburden increases. The vertical stress can be calculated as the weight of a “column” of soil extending above a unit area (Figure 1(a)). For a homogeneous soil having a wet unit volumetric weight of gt (normally expressed as kilonewtons per cubic meter or pounds per cubic foot), the stress s␯ (normally in kilonewtons per square meter or kilopascals, or pounds per square foot) at a depth Z m or ft below the ground surface is: sn = gt Z

320

(1)

Subsurface Stresses Ground surface

Ground surface γ t = unit weight of soil, homogeneous from ground surface to depth Z

Z

σv

Stratum A

Za

γ a = unit soil weight

Stratum B

Zb

γb

Stratum C

Zc

γc

σv

Unit area σv = γ t Z

σv = γ aZa + γ bZb + γ cZc

(a)

(b)

Figure 1 Vertical subsurface stress resulting from the soil mass.

If the soil mass is made up of strata of different soil types and the unit weights of the soil in each stratum are different, the vertical stress at a depth Z will be equal to the total weight of the different segments of the soil “column,” as indicated by Figure 1(b).

Effect of Groundwater Table When a soil exists below the groundwater table, the submerged soil particles are subject to a buoyant force resulting from the hydrostatic water pressure, the same phenomenon that acts on any submerged solid. The submerged weight of the soil, gsub, is termed the effective soil weight, and the subsurface stress that results is termed the effective stress. Effective stress represents the actual intergranular pressure that occurs between soil particles. This effective stress is the stress that influences shear strength of the soil and volume changes or settlements. If a condition exists where the water table is at the ground surface and the soil mass is homogeneous, the effective stress sn at a depth Z is: sn = gsubZ

(2)

If the total weight of soil as it exists above the water table is gt (the wet soil weight before the buoyant effects of submergence are considered), the effective stress is: sn = gt Z - gw Z

(3)

where gw is the unit weight of water. (It is conventional to assume 62.4 pcf or 9.81 kN/m3 for gw.) The last term of this equation is the total water pressure at the depth Z. Total water pressure at a point is termed the neutral stress, u, for it acts equally in all directions. Neutral stress refers to any water pressure that develops at a point as caused by hydrostatic conditions. It is important to recognize that the neutral stress acts to reduce the intergranular stress that develops between soil particles. This condition frequently has an adverse effect on the strength of a soil. The effective stress for the conditions just described can be expressed as: sn = gt Z - u

(4)

321

Subsurface Stresses

To compute the effective stress for a condition where the groundwater table lies below the ground surface, and for the condition where strata of soils of different types and weights exist, two different approaches are possible. One approach involves determining the total soil pressure (disregarding buoyancy effects) and then subtracting the hydrostatic pressure (the neutral pressure) at the point being analyzed. The neutral pressure u is the unit weight of water gw , multiplied by the depth below the water table. A second approach is to determine directly the effective stress of the column of soil above the point by using the effective or submerged weight of all soil in the “column.” Above the water table, the effective soil weight is the total soil weight, including the pore water; below the water table, the effective soil weight is the submerged or buoyant weight (Figure 2(a)). Where the soil surface is below water (such as in oceans and lakes), the effective stress should be computed by using the submerged or effective soil weight multiplied by the depth measured from the soil surface (Figure 2(b)). A summary of the analytical procedures for computing subsurface stresses resulting from various soil and water table conditions, and numerical illustrations, are presented in Figure 3.

Horizontal (Lateral) Stresses The magnitude of vertical stress is relatively simple to determine when the ground surface is level. When this condition does exist, it is also convenient to indicate horizontal (lateral) stresses that exist in a soil mass in terms of the soil vertical stress. The ratio of lateral stress to vertical stress, K, is termed the coefficient of lateral earth pressure. Mathematically, K =

Horizontal soil pressure sh

(5)

Vertical soil pressure sn

The use of effective pressures for determining the lateral pressure coefficient accounts for the influence of submergence where the soil deposit is wholly or partially below water or the water table (i.e., submerged unit weights are used when appropriate to determine sn). In a horizontal, uniform soil mass of infinite extent, the lateral movement of the soil at any depth is not possible, because the confining pressure is equal in all horizontal directions. Water surface

Ground surface Za

Zb

γ ta

γ tb

Stratum A Stratum B Zw γ subb = γ t b − γ w

γ sub

Zs

σv

σv = γ subZs

σv = γ t aZa + γ t bZb − γ w Zw (or) σv = γ t aZa + γ t b(Zb − Zw ) + γsubbZw

(Note: γ sub is unit weight of submerged soil γt is total wet unit weight of soil.)

(a)

(b)

Figure 2 Subsurface stress below water.

322

Soil surface

Water table

σv

Subsurface Stresses Ground surface Z=4m

γ t = 18 kN3 (≈ 115 pcf) m

σv

σv = γ t z kN kN (4 m) = 72 2 = 72 kPa m3 m = (115 pcf)(13.1 ft) = 1506 psf

(

= 18

(a)

)

Groundwater table deeper than Z distance shown

Ground surface Z = 3.65 m

σv

Water table γ saturated = 19.5 kN3 m (124.2 pcf)

Ground surface Za = 1.5 m

σv

Zb = 3 m

(b)

γ t = 18 kN (115 pcf) m3 Water table kN γ sat = 20 3 m (127.4 pcf)

σv = γ subZ = (γ saturated − γ w)Z kN kN = 19.5 3 − 9.8 3 (3.65 m) = 35.4 kPa m m = (124.2 pcf − 62.4 pcf)(12 ft) = 742 psf

(

)

σv = γ t Za + γ sub Zb kN kN kN = 18 3 (1.5 m) + 20 3 − 9.8 3 (3 m) m m m = 57.6 kPa = (115 pcf)(4.9 ft) + (127.4 pcf − 62.4 pcf)(9.84 ft) = 1203 psf

(

(c)

)

(

)

Water surface Zw = 5 m Zs = 2 m

σv

Soil surface kN m3 (108.3 pcf)

γ sat = 17

(d)

σv = γ subZs = (γ sat − γw)Zs kN kN = 17 3 − 9.8 3 (2 m) = 14.4 kPa m m = (108.3 pcf − 62.4 pcf)(6.56 ft) = 301 psf and μ, neutral (water) pressure = (5 m + 2 m) γ w = 68.6 kPa

(

)

kN kN = 20.89 psf; 1 3 = 6.36 pcf m2 m kN kN 1 psf = 0.048 2 ; 1 pcf = 0.157 3 m m

1 kPa = 1

Figure 3 Summary: Method to compute subsurface stress in a soil mass.

Thus a state of static equilibrium exists and the soil is in the at-rest condition. The coefficient of lateral pressure for the at-rest condition is indicated by K0. The magnitude of K0 for a given soil mass is affected by the soil deposit’s stress history. Soils that have been subjected to heavy loading at some time in their history, such as now-dense granular (sand or gravel) deposits and hard, overconsolidated 1 clays, would have had to develop resistance to high lateral stress in order to maintain stability. Deposits that have not been exposed to heavy loading, such as loose granular soils and

1Clays that at some past time were subject to loading greater than the weight of all currently existing overlying soil.

323

Subsurface Stresses Table 1 Typical Values of K0 Soil Type

K0

Granular, loose Granular, dense Clay, soft Clay, hard

0.5–0.6 0.3–0.5 0.9–1.1 (undrained) 0.8–0.9 (undrained)

soft, normally consolidated or underconsolidated clays, would not have developed high lateral strength. Typically, then, dense granular soils and hard clays end up having lower values of K0 than do loose granular soils and soft clays. Such typical values are presented in Table 1. Illustration 1 The unit weight of the soil in a uniform deposit of loose sand is 100 pcf. Determine the horizontal stress that acts within the soil mass at a depth of 10 ft. Ground surface Z=3m kN m3 Ko = 0.5 γ = 16

σv σh

Solution sn = sZ = 116 kN>m3213 m2 = 48 kN>m2 = 48 kPa sh = K0sn = 10.52148 kPa2 = 24 kPa where K0 is obtained from Table 1.

In soils below the water table, determination of the total lateral pressure requires that the hydrostatic pressure due to the water be added to the effective at-rest soil pressure computed by using a value from Table 1 and the effective soil weight. Illustration 2 A concrete basement wall for a structure extends below the groundwater table. For conditions indicated by the sketch, calculate the total lateral pressure acting against the wall at a point 8 ft below the ground surface. Ground surface γ = 120 pcf Medium dense sand, Zw = 4 ft ko = 0.4 γ sub = 60 pcf

4 ft 8 ft σh

324

Subsurface Stresses

Solution Soil pressure, sh, at 8 ft = K0gZ

= 10.42314 ft * 120 pcf2 + 14 ft * 60 pcf24 = 288 pcf

Total lateral pressure = K0gZ + gwZw = 288 psf + 14 ft2162.4 pcf2 = 538 pcf

2 Stress Within the Soil Mass Resulting from Vertical Surface Loading Uniform Homogeneous Soils When a vertical loading from a structure or other body is applied at the surface of a soil mass, new stresses are created within the mass. Because of shearing resistance developed within the soil, loading transferred to the soil mass will be spread laterally with increasing depth from the point or area of application (Figure 4). With increasing depth, the area over which new stresses develop will increase, but the magnitude of the stresses will decrease. For an equilibrium condition, the sum of the new vertical stresses developed in the soil mass on any horizontal plane must equal the weight or force of the surface loading. The manner in which the stresses become distributed throughout the soil mass is affected by the properties of the soil, including modulus of elasticity and Poisson’s ratio and any stratification of different soil types.

Boussinesq Stress Distribution One of the classical methods in common use for calculating stresses that result in a soil mass from a surface loading is based upon the work of Boussinesq, a nineteenth-century French mathematician. Boussinesq assumed a homogeneous, isotropic material (properties the same in all directions) of semi-infinite extent (unlimited depth) and developed equations for the stress distribution resulting from a point load. Adapted to soil masses, the described conditions are as indicated by Figure 5. The vertical stress increase ⌬sn

Surface loading

Z3

Z2

Z1

Figure 4 Variation of vertical stress at different depths Z.

325

Subsurface Stresses Q

Δσv

Z

r

Figure 5 Definition of terms applicable to Boussinesq and Westergaard equations.

resulting at a depth Z and a distance r, measured horizontally from where a point loading Q is applied, becomes: ¢sn =

3Q Q Z3 = 2 2 2 5>2 2p 1r + Z 2 Z

3 r 2 5>2 2p B 1 + a b R Z

(6a)

This equation indicates that as the depth increases the stress decreases. Similarly, the stress decreases as the horizontal distance from the point of loading is increased. For a given depth, the intensity of stress is greatest directly beneath the point of load application. With the Boussinesq equation for vertical stress, the modulus of elasticity and Poisson’s ratio are not required, which indicates that the stress is independent of these properties as long as the material is homogeneous and isotropic. The suitability of using the Boussinesq equation for determining subsurface stresses in a foundation analysis depends on how closely the actual soil conditions and properties resemble the theory’s original assumptions. For practical problems, conditions of a homogeneous and isotropic material are commonly assumed for homogeneous clay deposits, for human-made fills where the soil fill has been placed and compacted in thin layers, and for limited thicknesses of uniform granular soil deposits.

Westergaard Stress Distribution Some sedimentary soil deposits consist of alternating thin layers of sandy soil (coarse, relatively incompressible material) and fine-grained silt–clay soils (compressible material)—for example, stratified deposits such as laminated clays. For such conditions, the Westergaard equations provide a better means of evaluating the subsurface stresses. In his development, Westergaard assumed that thin layers of a homogeneous and anisotropic material were sandwiched between closely spaced, infinitely thin sheets of rigid material that would permit compression but no lateral deformation. For the case where Poisson’s ratio is zero, the equation for subsurface stress resulting from a concentrated point loading reduces to: Q ¢sn = Z 2p B 1

r 2 3>2 + 2a b R Z

(7a)

The terms Q, r, and Z are as defined for the Boussinesq equation and as shown in Figure 5.

326

Subsurface Stresses 0.5

0.48

0.4 0.32 0.3

IB

IB, IW 0.2 IW

0.1 0

0.5

1.0

1.5

2.0

2.5

3.0

r/Z

Figure 6 Values of IB and Iw for calculating vertical stress resulting from surface load Q.

Computational Aids With both the Boussinesq and Westergaard equations, the subsurface stress resulting from a given point load will be related to the Z and r distances, or the r/ Z ratio. If the Boussinesq equation is written in terms of a stress influence factor IB, which is related to r/ Z as follows: ¢sn =

Q Z2

3 = r 2 5>2 2p B 1 + a b R Z

Q I Z2 B

(6b)

calculations for values of IB for different r/ Z ratios can be computed and presented as shown in Figure 6. The effort to determine a subsurface stress then becomes greatly simplified. Similarly, the Westergaard equation can be written in terms of an influence factor Iw as follows: Q ¢sn =

= Z 2p B 1

r 2 3>2 + 2a b R Z

Q I Z2 w

(7b)

and values of Iw versus r/ Z ratios can be developed as presented in Figure 6. Illustration 3 For Boussinesq conditions, what subsurface stress will result at a point 3 m below where a 45 kN point load is applied? Q = 45 kN

Z=3m

Δσv

327

Subsurface Stresses

Solution r = 0 ft,

Z = 3 m,

r>Z = 0

Q I Z2 B 45 kN 10.482 = 2.4 kN>m2 = 2.4 kPa. = 13 m * 3 m2

For r>Z = 0, obtain IB = 0.48 (from Figure 6) and ¢sn =

Illustration 4 For the Westergaard conditions, what subsurface stress will result 10 ft below and 10 ft horizontally from where a 10,000-lb concentrated load is applied?

Q = 10,000 lb

Z = 10 ft

Δσv

r = 10 ft

Solution r = 10 ft,

Z = 10 ft,

r>Z = 1.0

For r>Z = 1.0, obtain Iw = 0.065 (from Figure 6) and ¢sn =

Q Z2

=

Iw

10,000 lb 10.0652 = 6.5 psf. 110 ft * 10 ft2

Application for Foundation Loading In construction practice, the condition of a concentrated point loading is rarely encountered. More commonly, building loads are supported on foundations that cover a finite area (square, rectangular, or round footing), or the structure itself directly imposes loading over a finite area (earth structures such as dams and dikes). The subsurface stresses that result from loading acting over an area can be determined by integration of Equation 6 or 7, where the loading on infinitely small increments of the foundation area can be assumed as point loads. The subsurface stress at a point is the summation of the effects resulting from all of the applied point loadings. Such integrations have been accomplished for uniform loads acting on square, rectangular, strip, and circular areas, and also for uniformly varying loads such as those developing from the weight of an earth structure having a sloped cross section (dams and dikes, for instance). Results are available in the form of charts, tables, and graphs in order to provide generalized solutions. In these presentations, the subsurface stresses are expressed as a percentage of the foundation loading intensity. Measurements

328

Subsurface Stresses

of the depth and horizontal distance at which a subsurface stress acts are expressed in terms of the dimensions of the loaded foundation area. Subsurface stress conditions indicated by Boussinesq and Westergaard equations for commonly occurring foundation loadings are presented in Figures 7 and 8. The following illustration problems serve to demonstrate the use of these figures. The stress curves in Figures 7 and 8 show that the stresses beneath the center of a loaded foundation area will be greater than stresses beneath the edge of the foundation until a depth of about twice the foundation width is reached. Below this level, the stresses beneath the center and edge become practically equal. Consequently, in making determinations of the stresses resulting from a foundation loading, it is suitable to assume a concentrated point loading, if convenient, where the subsurface depth is greater than twice the foundation width. Δσv in percent of uniform footing pressure, q 20

40

60

rc ul a ed r e ge dg

r

nte

are

ce

2.0

B q = uniform footing pressure

2.5 Z

Δσv Δσv Edge

3.0 Center

Ratio of depth to footing width (Z/B)

1.5

Lo

ng

(s

tri

p)

ce

nt

u

Sq

er

re ua Sq

1.0

100

cen

ular

Circ

Ci

0.5

80

ter

e

3.5

4.0

4.5

Note: Z is depth from base of footing to point in underlying soil

5.0

Figure 7 Variation of vertical stress beneath a foundation: Boussinesq analysis.

329

Subsurface Stresses Δσv in percent of uniform footing pressure, q 0

20

40

60

re Cir cul a

0.5

dg

e

re

er

ent

ec

r qua

100

Cir

ed

a qu

S

80

enter cular c

ge

S

p

tri

ng

(s

er

nt

e )c

Lo

1.0

2.0

B q = uniform footing pressure

2.5 Z

Δσv Δσv Edge

3.0 Center

Ratio of depth to footing width (Z/B)

1.5

3.5

4.0

Note: Z is depth from base of footing to point in underlying soil

4.5

5.0

Figure 8 Variation of vertical stress beneath a foundation: Westergaard analysis.

Illustration 5 A foundation supported on the surface of a uniform, homogeneous soil is 2 m square and carries a loading of 500 kN. What subsurface stress increase occurs beneath the center of the foundation at a depth of 2 m? Q = 500 kN B=2m

Z=2m

330

Δσv

Subsurface Stresses

Solution For uniform homogeneous soil, use Boussinesq conditions.

q =

Q 500 kN = = 125 kPa at base of foundation A 12 m * 2 m2 Depth 2m = = 1.0. Width 2m

From Figure 7, Δsv in terms of q is 34 percent for Therefore:

¢sv = 0.341125 kPa2 = 42.5 kPa

Illustration 6 A circular storage tank is supported on soils that satisfy the Westergaard assumptions. What subsurface stress increase develops 10 ft beneath the edge of the tank? The tank is 20 ft in diameter, and the stored fluid material causes a pressure of 1000 psf at the tank base.

Solution 20 ft dia.

z = 10 ft

Δσv

Surface load q = 1,000 psf Ratio of depth>width =

10 ft = 0.5 20 ft

From Figure 8, ¢sn = 23 percent of q from “circular edge” curve. Therefore, ¢sn = 0.2311,000 psf2 = 230 psf.

Sixty-Degree Approximation A method in wide use for making rough estimates of subsurface stresses resulting from a loaded foundation area is the so-called 60° approximation. In this method, it is assumed that the subsurface stresses spread out uniformly with depth, the stressed area increasing at a slope of 1 m or ft horizontally for each 2 m or ft of depth as measured from the edges of the foundation. These assumed conditions are sketched in Figure 9. At a given depth, the subsurface stress Δsn is assumed to be uniform over the area stressed; the method

331

Subsurface Stresses Q

Q

W 1 2

B Δσv

B Depth, Z W

Δσv

+ Z

B+Z

B+Z

Figure 9 Method for approximating vertical stress increase resulting beneath a loaded foundation (60° approximation).

differs from the Boussinesq and Westergaard theories in this respect. The stress at a depth Z then becomes:

¢sn =

Q 1B + Z21W + Z2

(8)

This approximation method is incorrect in representing subsurface stresses as being uniform across a plane area. In a homogeneous soil, the computed subsurface stresses directly beneath the foundation will be less than indicated by the Boussinesq analysis, whereas at distances beyond the edges of the foundation the computed stress will be greater. This method’s best application may be estimating stress conditions in deep layers below a foundation and for determining an order of magnitude in a preliminary analysis.

Layered Soils Having Different Properties The condition of soil deposits having uniform properties over a great vertical distance is not always found in practice. The situation is frequently encountered where, within the zone (depth) that will be affected by structural loads, two or more layers of significantly different soils exist. For the condition of two different strata, the possibilities of relative properties include a firm upper layer overlying a soft layer, and the reverse. For the condition where a firm layer overlies a soft layer (a dense sand above a compressible clay, or a surface zone of firm desiccated clay overlying the still-saturated, softer and compressible lower zone of the clay), the firm layer tends to bridge over the soft layer, spreading out the area over which stresses are transferred into the softer layer. This results in creating lower stresses in the soft material than would result for a homogeneous soil. With a soft layer overlying a firm layer, the stresses reaching the deeper layer in the area directly beneath the foundation will be greater than those indicated for a uniform homogeneous deposit.

332

Subsurface Stresses

A diagram for determining how the variation in subsurface stress beneath a uniformly loaded circular area is affected by a two-layer soil mass appears in Figure 10. This diagram is for the special case where the thickness of the upper soil layer is equal to the radius of the loaded area. The curves of Figure 10 can be used for approximating stress conditions beneath a square area as well as the circular area by assuming foundation dimensions giving the same area as the circular foundation. The illustration presented below indicates the use of Figure 10.

2R

H

E1

Profile—two-layer subsurface condition

Δσv

Z

E2

X E = modulus of elasticity for soil Δσv in percent of uniform foundation pressure 0

50

100

50

100

50

100

K = 1000 K = 1000

K = 1000

10

10

10 1 2 Ratio of Z/H

1

1

2

1

0.001

0.001 0.001 2 k=

E1 E2

k=

E1 E2

k=

E1 E2

3 Beneath center (X = 0)

For X = R/2

Beneath edge (X = R)

Figure 10 Variation of vertical stress beneath a circular foundation. Two-layer subsurface condition for condition where R = H.

333

Subsurface Stresses

Illustration 7 A circular storage tank 10 m in diameter is located in an area where soil conditions are as shown by the sketch. If the tank pressure at the soil surface is 15 kPa, what stress results beneath the center of the tank at a depth of 7.5 m?

Solution 10 m dia.

H=5m Z = 7.5 m

Firm stiff clay E1 (E1 = 10 E2) Layer 1 Δσv

Medium soft clay E2 Layer 2

H = radius = 5 m 7.5 m Z = = 1.5 Ratio of H 5m E1 = 10 k = E2 From Figure 10, Δsn is 20 percent of surface load q. Therefore, ¢sn = 0.20115 kPa2 = 3 kPa.

Effect of Foundation Installation Below Finished Grade In the practice of building design and construction, foundations are seldom placed directly on the ground surface. More often, foundations are placed at some depth below the ground surface as protection against frost and seasonal soil volume changes, erosion, or other factors. In other words, some excavation is performed before the foundation can be constructed. The subsurface stresses developed in the soil below the foundation are determined by using the previously described methods, and the depth for computing the stress increase is measured from the base of the foundation. The total stress at a point in the soil below the foundation is the sum of the stress caused by the soil overburden plus the stress due to the foundation loading. In many cases, a foundation excavation is backfilled to the ground surface or interior floor level after the foundation is constructed. For practical purposes, then, the stress caused by the soil remains equal to the original soil overburden. If an excavation is made and not backfilled, the effect is to reduce the subsurface stresses. The reduction in stress can be computed by assuming that the excavation is a “foundation” whose loading is the weight of the excavated soil. The size of the “foundation” is the plan area of the excavation. The subsurface stress is determined as for a real foundation but is handled as a negative stress to be subtracted from the stress caused by the actual foundation that is ultimately installed in the excavation. Illustration 8 A heavy materials storage building 30 ft by 30 ft is constructed with a basement that extends 6 ft below the ground surface. A foundation for an interior supporting column is constructed at floor level in the middle of the basement. The column footing is 4 ft square and carries a loading of 50 kips.

334

Subsurface Stresses What is the net stress change in the soil 5 ft below the middle of the basement floor? (For this example, neglect the weight of the concrete footing and floor.) Excavated area = 30 ft × 30 ft 50 k B = 4 ft 6 ft 5 ft

Δσv γsoil = 120 pcf (Boussinesq conditions)

Solution A “negative” foundation loading results from the basement excavation. Negative load is 6 ft deep * 120 pcf = 720 psf . For a foundation 30 ft * 30 ft placed at a depth 5 ft below the floor: Depth 5 ft = = 0.17 Width 30 ft and the stress decrease ≤sn is 95 percent * 720 psf = -685 psf (percentage as determined from Figure 7). Stress increase from column footing, where: q =

50 kips = 3.14 ksf = 3,140 psf 4 ft * 4 ft

is + ¢sn = 125%213,140 psf2 = +780 psf , for the ratio of: Depth 5 ft = = 125 (ref. Figure 7). Width 4 ft Therefore, the actual stress increase is: ¢svnet = +780 - 685 = 95 psf

Effect of Changing the Surface Grade On large construction sites, the surface grade may be changed to improve the topography and to obtain a surface more suitable for the needs of the project. If the surface grade is lowered, this results in a reduction in the subsurface stresses. Where earth fill is placed to raise the surface level, an increase in the subsurface stresses results. When a uniform thickness of material is placed or removed over a large area, the subsurface stress increase or decrease is the product of the unit weight of the soil added or removed and the thickness involved. This subsurface stress change is constant with depth; it does not diminish with increasing depth, as for foundation loadings. Where the thickness of fill or soil removal is variable, as is frequently the case, a limited area can be analyzed and the average thickness of soil added or removed can then be used to calculate the subsurface stress change for the area.

335

Subsurface Stresses

Illustration 9 A compacted earth fill for a shopping center development is placed over an area where soil conditions are as shown by the sketch. What stress increase results in the middle of the clay layer from the weight of this soil fill? Final ground surface

Compacted soil fill γ = 19.5 kN/m3

3m

Sand

2m 1m 2m

Original ground surface

Δσv

Clay

Sand and gravel

Solution Since this is an areal fill (fill placed over a large area), the stress increase in the clay layer is: ¢sn = 1gfill21height of fill2 = 19.5 kN>m3 * 3 m = 58.5 kPa

Subsurface stresses caused by foundations subsequently installed in fill or cut areas are computed by using the methods already presented. It is important to remember that the final total subsurface stress must reflect the changes caused by filling or lowering the surface of the site as well as the foundation loading. Illustration 10 Assume that a building foundation is constructed at a location where an 8 ft soil fill has been placed. If the footing is 5 ft square (5 ft by 5 ft) and carries a loading of 75 kips, what net stress results in the middle of the clay layer beneath the center of the footing for conditions indicated by the sketch? (Consider effects of fill and foundation loading.)

Solution 75 k Footing 5 ft × 5 ft Footing installed 2' below grade

2' 8'

Compacted fill, γ = 125 pcf

6 ft

Sand

6 ft

Clay Sand and gravel

336

15 ft Δσv

Subsurface Stresses

qfooting =

75 kips = 3 ksf = 3,000 psf 5 ft * 5 ft

Depth 15 ft = = 3 Width 5 ft Estimate stress increase by either Boussinesq or Westergaard, ¢sn L 5% * q (percentage determined from Figure 7 or Figure 8), therefore: ¢sn = 0.0513,000 psf2 = 150 psf from foundation Stress increase from fill = 125 pcf * 8 ft = 1,000 psf Total stress increase = 150 + 1000 psf = 1,150 psf

Problems 1 Relating to soil deposits, provide a definition of subsurface effective stress and total stress, and explain the relationship between the two. 2 At a planned construction site, subsurface sampling indicates that the soil wet unit weight is 19.5 kN/m3. (a) Determine the effective vertical stress at a depth of 4 m if the groundwater table is deep. (b) Determine the effective vertical stress if the groundwater table is at the ground surface. (c) Determine the effective vertical stress at the 4 m depth if the groundwater table is 2 m below the surface. 3 At a planned construction site, subsurface sampling indicates that the wet unit weight of the soil is 123 pcf. (a) Determine the effective vertical stress at the 12-ft depth if the water table is deep. (b) Determine the effective vertical stress and the neutral stress at the 12-ft depth if the water table rises to within 6 ft of the ground surface. (c) Determine the effective vertical stress at the 12-ft depth if the water table is at the ground surface. 4 At a lake location, the soil surface is 6 m under the water surface. Samples from the soil deposit underlying the lake indicate a wet unit weight of 18.8 kN/m3. Determine the effective vertical stress in the deposit at a depth 12 m below the soil surface. Also indicate the total vertical stress at the same point.

5 At an offshore location, the soil surface is 30 ft below the water surface. Weighing of soil samples obtained in the subsurface investigation indicates the saturated unit weight to be 118 pcf. Determine the effective vertical pressure at a depth 55 ft below the water surface. 6 The soil in a deposit underlying an area where a level ground surface exists has a saturated unit weight of 21 kN/m3. The groundwater table is at the soil surface. (a) Determine the principal stresses for the total stress condition at a depth of 5 m below the soil surface if the lateral pressure is onehalf the vertical pressure. Recall that the subsurface vertical and horizontal pressures or stresses are the principal stresses when the site surface is horizontal.) (b) Determine the principal stresses for the effective stress condition at the same 5-m depth. (c) Determine the value of the maximum shear stress for the effective stress condition at the 5-m depth. 7 A soil deposit exists with a level ground surface. The saturated unit weight of the soil is 125 pcf, and the water table is at the ground surface. (a) Determine the principal stresses acting at a depth 15 ft below the ground surface that results from the effective (submerged) soil weight if the lateral pressure is 0.45 times

337

Subsurface Stresses the vertical pressure. (Recall that the subsurface vertical and horizontal pressures or stresses are the principal stresses when the site surface is horizontal.) (b) What is the magnitude of the maximum shear stress relating to the effective stresses at this depth? 8 A soil exists in an area where the ground surface is level and the groundwater table is 3 m below the surface. The saturated unit weight for the soil is 22 kN/m3, and the wet weight for the soil above the water table is 18 kN/m3. The lateral pressure is 0.5 times the vertical pressure. The vertical and lateral pressures represent principal stresses. Determine the value for the major and minor principal stresses at a depth of 10 m (determine values for both the total stress and effective stress condition), and the maximum shear stress that develops because of the effective principal stresses. 9 A building is to be constructed in an area where the ground surface is horizontal. The groundwater table at the site is very deep. For a point within the soil mass, 10 feet below the surface, the preconstruction vertical effective stress is 1,120 psf and the horizontal effective stress is 560 psf (consider these stresses to be principal stresses). After construction, the vertical stress at the same point will increase by 440 psf because of foundation loading and building use. Assume the horizontal stress will be unchanged. (a) Indicate the change to the maximum shear stress at the described point that will occur because of the construction. (b) Assume that after the building is completed and occupied, the groundwater table at the site raises to within 6 ft of the ground surface as a result of area rainfall. The vertical effective stress in the soil mass changes because of the high water table. Determine the maximum shear stress (effective stress conditions) at the analyzed point in the soil mass for the changed situation if the horizontal effective stress is unchanged (from part a) by the rising water table. 10 A building is constructed in an area where the ground surface is horizontal. The groundwater table at the site is very deep. For a point within

338

the soil mass, 3 m below the surface, the preconstruction effective vertical stress was 52 kPa and the effective horizontal stress was 26 kPa (based on a lateral pressure coefficient K equal to 0.5); consider these stresses to be principal stresses. The vertical stress at the same point is increased by 20 kPa because of foundation loading and building use, but assume the horizontal stress is not changed. Shortly after the building is completed and occupied, the groundwater table at the site raises to within 1.5 m of the ground surface. The vertical effective stress in the soil mass changes because of the raised water table. Determine the maximum shear stress at the analyzed point in the soil mass for the changed condition if: (a) the horizontal effective stress is not changed by the rising water table (b) the horizontal effective stress is reduced to a value equal to K times the submerged effective vertical stress (based on effective soil unit weight only and not including stress due to construction loading). 11 Estimate the lateral earth pressure developed at a depth 5 m below the soil surface of a loose sand deposit (wet unit weight of soil is 17.5 kN/m3). Also estimate the lateral pressure for the same depth if the sand is in a dense condition (unit weight is 20.5 kN/m3). What would be the total lateral pressure for the loose sand condition described if the water table were at the ground surface? Use Table 1 for reference. 12 (a) Estimate the lateral earth pressure at a depth 20 ft (6.1 m) below the ground surface in a loose sand deposit. The wet unit weight of the sand is 115 pcf (19 kN/m3). What would the total lateral pressure be if the water table rose to the ground surface (in psf or kPa)? Refer to Table 1. (b) What difference in lateral pressure is expected when the soil is in a dense condition (unit weight is 130 pcf or 20.4 kN/m3)? 13 A basement wall for a commercial building extends 3.1 m below the ground surface. Assume the soil unit weight is 18.5 kN/m3. For a point 3 m below the ground surface, compare the lateral soil pressure acting against the wall for the condition (a) when the soil is a sand–gravel material and (b) when the soil is clay.

Subsurface Stresses 14 A deep basement for a building is constructed in dense granular soil whose unit weight is 126 pcf (19.8 kN/m3). Assume KO ⫽ 0.40. For a point 10 ft (3 m) below the ground surface, estimate the increase in total lateral pressure that results by having the water table change from a location below the basement level to the ground surface (in psf and kPa). Assume the submerged unit weight for the soil is half the indicated unit weight (the given unit weight is a wet but not saturated unit weight). 15 A concrete tank structure is constructed below ground level for a large commercial swimming pool in an area where dense granular soil exists (20 kN/m3 wet unit weight). The diving board end of the tank-pool is 6 m deep (below the ground surface). For the design, it is assumed that the groundwater table could rise close to the ground surface at various times. For the condition where the pool is empty, estimate the total lateral pressures (due to soil and to water) acting against the exterior side of the end wall at points 2 m, 4 m, and 6 m below the ground surface for the high water table condition. Assume the submerged unit weight for the soil is half the indicated unit weight (the given unit weight is a wet but not saturated unit weight). 16 A 200-kN concentrated (point) load acts on the surface of a soil mass. Determine the vertical stress 3 m below the ground surface at locations directly beneath the point of load application, 3 meters horizontally from the load, and 6 meters horizontally from the load. (a) Assume the Boussinesq conditions apply. (b) Assume the Westergaard conditions apply. 17 A 20-kip concentrated (point) load acts on a surface of a soil mass. Determine the vertical stress 10 ft below the ground surface at locations directly beneath the load, 10 ft horizontally from the load, and 20 ft horizontally from the load, for: (a) Boussinesq conditions. (b) Westergaard conditions. 18 A 30-kN concentrated load is imposed at the surface of a granular soil deposit. Determine the subsurface vertical stress created at depths of 5 m and 10 m directly beneath the loading and

at a horizontal distance 5 m from the line of application (5 m and 10 m depths) for: (a) Boussinesq conditions. (b) Westergaard conditions. 19 A square footing foundation, 3 m by 3 m, and positioned on the ground surface of a soil deposit, supports a column load of 1,350 kN. Determine the vertical stress resulting from the foundation loading at a depth 3 m below the base of the footing for locations beneath the center and beneath the edge, assuming: (a) Boussinesq conditions apply. (b) Westergaard conditions apply. 20 A square foundation, 4 ft by 4 ft, located on the surface of a soil mass, supports a column load of 96 kips. Determine the vertical stress resulting from the foundation loading at a point 6 ft below the ground surface for locations beneath the center of the footing and beneath the edge of the footing, assuming: (a) Boussinesq conditions apply. (b) Westergaard conditions apply. 21 A 300-kN loading is supported on a 1.5 m by 1.5 m square foundation constructed on the ground surface of a homogeneous soil deposit (soil unit weight is 19 kN/m3). Determine the values of the preconstruction and the postconstruction vertical stress imposed on the soil at a point 2 m below the center of the foundation area (assume the Westergaard conditions apply). 22 A long strip-footing foundation will be used to support a masonry bearing wall for a commercial building. The footing will be 1.5 m wide and 45 m long. The wall loading imposed onto the foundation will be 200 kN per meter of wall length. Determine the vertical stress increase for points 1 m, 2 m, and 4 m below the footing center, assuming: (a) Boussinesq conditions apply. (b) Westergaard conditions apply. 23 A 4-ft-wide long (strip) footing carries a wall loading of 20,000 lb per foot of wall length. What vertical stress increase results below the center of the footing at depths of 4 ft, 8 ft, and 12 ft, assuming: (a) Boussinesq conditions apply? (b) Westergaard conditions apply?

339

Subsurface Stresses 24 A 2 m by 2 m square foundation will be used to support an item of machinery weighing 800 kN. The foundation will be placed on the soil surface. Determine the values of preconstruction and postconstruction vertical stress in the soil zone 1.5 m below the base of the foundation, under the center, and under the edge. Soil unit weight is 18 kN/m3. (a) Assume Boussinesq conditions apply. (b) Assume Westergaard conditions apply. 25 A 1-m-wide surface-supported strip footing carries a loading of 100 kN per meter of wall length. Determine the total vertical stress acting at a depth of 1 m below the center of the foundation width (note that the total stress will be the sum of the original vertical stress due to the soil mass plus the increase due to the foundation loading). Assume the Boussinesq conditions apply. Use a soil unit weight equal to 18 kN/m3.

30

31

32

26 A steel storage tank 20 m in diameter will be used to hold a liquid petroleum product. When filled, the tank causes a bearing pressure of 150 kPa. The soil underlying the tank has a unit weight of 18.5 kN/m3. For a depth 10 m below the base of the tank, compute the effective vertical stress when the tank is empty and when full, for points beneath the center and beneath the edge. (a) Assume the Boussinesq conditions apply. (b) Assume the Westergaard conditions apply. 27 An oil storage tank 50 ft in diameter imposes a maximum loading of 2,500 psf onto the ground surface where the tank is supported. The soil underlying the tank has a unit weight of 125 pcf. For a point 25 ft below the ground surface, compute the effective vertical stress when the tank is empty and when the tank is full, at locations beneath the center and the edge of the tank. Assume that the Boussinesq conditions apply. 28 Compare the stress increase occurring 2 m below the center of a 3 m by 3 m square foundation imposing a bearing pressure of 145 kPa (145 kN/m2) when: (a) the Boussinesq stress distribution is assumed. (b) the 60° approximation is assumed. 29 Compare the stress increase resulting 8 ft below the center of a 10-ft square foundation imposing 500 psf when the 60° approximation is assumed

340

33

34

and when the Boussinesq conditions are assumed. A 5-m-diameter tank supported on the surface of a soil deposit imposes a bearing pressure of 225 kPa (225 kN/m2). For a point 4 m below the tank base, compare the vertical stress increase due to the tank loading when: (a) the Westergaard conditions are assumed. (b) the 60° approximation is assumed. A circular foundation 12 ft in diameter imposes a pressure of 8,000 psf onto the soil. At the 12-ft depth, determine the vertical stress increase beneath the center and the edge of the loaded area, assuming: (a) the Westergaard conditions apply. (b) the 60° approximation. An 8-m-diameter storage tank is located on the ground surface of an area where a 4-m thickness of desiccated (firm) fine-grained soil is underlain by a very thick layer of similar fine-grained soil that retains a high water content and, as a result, is relatively weak and compressible. The modulus of elasticity for the firm upper zone is 10 times the modulus for the underlying layer. The tank bearing pressure is 125 kPa. (a) Determine the increase in vertical stress beneath the center of the tank at a depth of 6 m. (b) Compare the result from (a) to the value of stress increase if homogeneous conditions (Westergaard) are assumed. A 30-ft-diameter storage tank is supported on the ground surface at a site where a 15-ft-thick layer of dense sand overlies a very thick clay layer. The modulus of elasticity for the upper layer is ten times the modulus for the lower layer. The tank imposes a pressure of 3 ksf. (a) Determine the increase in vertical stress below the center of the tank and at a depth of 30 ft. (b) Determine the stress increase at the 30-ft depth if subsurface conditions were homogeneous (assume Boussinesq), and compare the results with (a). The site for a commercial building project has the surface elevation lowered 2 m to improve area grading for building access, parking, and drainage. Foundations for interior building columns will be constructed at the level of the

Subsurface Stresses new ground surface. Each foundation is 1.5 m by 1.5 m square and imposes a bearing pressure of 200 kPa. What net stress increase results in the soil 1.25 m below the center of the foundation? Assume Boussinesq conditions apply and soil unit weight is 18 kN/m3. 35 For a construction project, the ground surface at a building location is lowered 5 ft. A 4-ft square footing (4 ft by 4 ft) imposing a bearing pressure of 4,000 psf is then constructed at the level of the new surface. What net stress increase results in the soil mass 4 ft below the center of the foundation? Assume that the soil unit weight is 120 pcf. 36 A 3-m thickness of compacted earth fill is placed across an area as preparation for constructing an industrial building. An item of heavy industrial equipment will be placed at a location within the building and supported on a foundation installed

at the surface of the compacted soil fill. The equipment has a total weight of 5,000 kN. The supporting foundation will be 5 m by 5 m. What net stress increase results beneath the center of the foundation in the original soil mass 2 m below the natural ground surface? Use 19 kN/m2 for the soil unit weight. Assume the Boussinesq conditions apply. 37 For a construction project, 10 ft of compacted earth fill is placed in preparation for erecting a building. At one of the building’s interior locations, a machine foundation 20 ft square will be installed at floor level to support equipment weighing 2,000 kips. What net stress increase results below the center of the machine in the original soil mass, 5 ft under the natural ground surface? Assume a unit weight of 125 pcf for the compacted fill. Assume the Boussinesq conditions apply.

341

Answers to Selected Problems

2.

3.

4. 5. 6.

8.

12.

13.

(a) ␴v = 78 kPa (b) ␴v = 39 kPa (c) ␴v = 58.5 kPa (a) ␴v = 1476 psf (b) ␴v = 1110 psf (c) ␴v = 738 psf ␴v = 107.9 kPa, ␴v 1total2 = 284.5 kPa ␴v = 1390 psf (a) ␴1 = 105 kPa, ␴3 = 52.5 kPa (b) ␴1 = 56 kPa, ␴3 = 28 kPa (c) ␶ = 14 kPa

18.

¢␴v = 0.144 kPa at r = 5 m ¢␴v = 0.108 kPa at CL, z = 10 m ¢␴v = 0.087 kPa at r = 5 m 20.

21. 23.

342

(a) ¢␴v = 2.7 ksf at -4 ft ¢␴v = 1.5 ksf at -8 ft ¢␴v = 1.05 ksf at -12 ft

25.

␴3 = 69.7 kPa

27.

␴v1total2 = 73 kPa

␴v = 3125 psf (empty) ␴v = 4730 psf (center, full), 4025 (edge, full)

(b) Ko = 0.4, ␴h = 1040 psf,

29.

¢␴v = 1.54 ksf and 2.20 ksf

¢␴h = 110 psf (deep wta), 60 psf (high wta)

31.

(a) ¢␴v = 2400 psf (center), 1520 psf (edge) (b) ¢␴v = 1580 psf

granular Ko = 0.5, ␴h = 27.8 kPa

33.

(a) ¢␴v = 540 psf (b) ¢␴v = 840 psf

clay Ko L 1.0, ␴h = 55.5 kPa 14.

␴v0 = 36 kPa, ¢␴v = 24 kPa, total ␴v = 60 kPa

Eff. stress ␴1 = 139.3 kPa,

␴h = 1828 psf 1high wta2

(a) ¢␴v = 1.08 ksf (center), 0.90 ksf (edge) (b) ¢␴v = 0.72 ksf (center), 0.54 ksf (edge)

Total stress ␴1 = 208 kPa, ␴3 = 104 kPa

(a) Ko = 0.5, ␴h = 1150 psf 1deep wta2,

(a) ¢␴v = 0.576 kPa at CL, z = 5 m

¢␴h = 364 psf = 17.43 kPa

35.

¢␴v = +760 psf

37.

¢␴v = 3650 psf

Settlement

From Essentials of Soil Mechanics and Foundations: Basic Geotechnics, Seventh Edition. David F. McCarthy. Copyright © 2007 by Pearson Education, Inc. Published by Pearson Prentice Hall. All rights reserved.

343

Settlement Soil Compression, Volume Distortion, Consolidation

The weight of any structure on the earth will result in stresses being imposed on the soils below the level of the base or foundation of that structure. The deformations that develop in the soil because of these stresses cause dimensional changes in the soil volume, with the result that the structure undergoes settlement. The extent of foundation settlement that will actually occur is related to the bearing pressures (stresses) imposed on the soils and the stress–strain properties of the soil.

1

Basic Considerations For the typical situation, loadings imposed by a structure at foundation level act in a downward vertical direction. The resulting stresses developed within the soil mass beneath the structure act vertically but in other directions as well. When the stresses that develop are well below the ultimate strength of the soil (a usual condition of foundation designs), settlement estimates of practical accuracy can be obtained by considering only the vertical stresses. Load-induced settlement of foundations is due to volume distortion in the foundation soil caused by shear strains (volume distortions occurring without change in volume) and by soil compression resulting from decrease in soil void ratio (soil particles are forced into a closer arrangement). Volume distortion tends to occur upon application of loading and results in an almost immediate initial settlement. Primary soil compression occurs at the rate in which foundation stresses are transferred onto the soil particles (or the soil skeleton). Primary compression occurs almost instantly in coarse-grained soil, but is a timerelated condition in fine-grained soil deposits having a high degree of saturation because of the lag period associated with pore water escaping (or being squeezed) from the soil void spaces; such time-related compression is termed consolidation compression. An additional time-related compression, termed secondary compression, is volume decrease attributed to

344

Settlement

creep or rearrangement of soil particle positions when a condition of constant stress exists over an extended period. Total foundation settlement results from the combination of immediate settlement plus primary (or consolidation) compression settlement plus secondary compression settlement, or: St = Svd + Spc + Ssc

(1)

where St Svd Spc Ssc

= = = =

total foundation settlement settlement due to volume distortion settlement due to primary compression settlement due to secondary compression

For foundations bearing on coarse-grained soils free of organic content (sands, sand mixed with silt or gravel), most of an expected total settlement occurs upon application of load. The settlement due to volume distortion and primary compression is immediate; the time-related transfer of foundation stress onto the soil skeleton—that is, primary (consolidation) compression—occurs rapidly even in fully saturated coarse-grained deposits because of the high coefficient of permeability (or hydraulic conductivity) and rapid draining characteristics. Secondary compression or creep effects are typically small. For foundations supported on fully saturated and almost-saturated fine-grained soil deposits free of organic content (the soils possessing cohesion such as clays and silt–clay mixtures), some volume distortion (immediate) settlement occurs, but primary (consolidation) compression settlement which extends over a period of time is expected to be the cause of most of the total settlement. A major exception is where foundations bear on firm soil (high shear strength) and the foundation pressures do not exceed the preconsolidation stress for the soil (preconsolidation pressure is discussed in Section 3); for this situation, immediate settlements may be on the order of half the total settlement. Long-term secondary compression settlement also takes place with fine-grained soil deposits, but typically is limited for soils free of organic matter. Where foundations bear on soils with significant organic content, the immediate settlement, primary (consolidation) settlement, and secondary compression settlement can each be significant. Typically, soils containing organic material are expected to undergo greater total settlement than nonorganic soil deposits. Primary (consolidation) compression and secondary compression may begin when foundation loads are first imposed onto a soil deposit, occurring concurrently with the processes responsible for immediate or initial settlement. For determining volume distortion settlements, calculation methods based on elastic theory and knowledge of soil properties are in use. Primary and secondary compression properties of fine-grained soils, relating to volume changes occurring in the soil deposit as void spaces between particles decrease when the particles rearrange to develop resistance to new (foundation) loading, are typically determined from laboratory compression tests on undisturbed soil samples. Laboratory compression tests are seldom performed on cohesionless soils, relating to the practical difficulty associated with obtaining truly undisturbed samples of granular soil from the construction site. Also, because the rate of settlement for granular

345

Settlement

soils is rapid (so that settlements often occur during the construction period), there may not be the problems of drawn-out postconstruction settlement as occur where the fine-grained cohesive soils are present; an exception for coarse-grained soil deposits is where much of the anticipated loading is from postconstruction activity, such as live loads and repetitive loads. Where compression characteristics of a granular soil are needed, the information is usually determined from results of in-place testing such as the dilatometer and pressuremeter procedures, or obtained indirectly, such as from an in-place density determination, relative density testing, or correlations from the results of procedures utilized in subsurface investigations such as penetration resistance or soil sampling in borings. Table 1 provides a summary of the relationship between foundation settlements and soil conditions. In practice, determination of the settlement characteristics of a soil deposit is frequently attempted by one of, or combination of, several methods: 1. Working backwards from observations of the behavior of structures in the area near the planned new building site. 2. Prior to construction, performing large-scale field load tests at the actual building site. 3. Performing laboratory tests (such as compression tests) on soil samples obtained from borings or test pits made at the planned construction site. 4. Estimating compressibility or volume change characteristics on the basis of index property tests performed on soil samples from borings or test pits. (Such tests typically include classification, moisture density determinations, liquid limit and plastic limit determinations, and relative density determinations.) 5. Using data from field exploration results (such as standard penetration tests used to obtain boring samples and cone penetrometer tests) that have been correlated empirically to design values for allowable foundation bearing pressures and foundation settlements. 6. Performing in situ tests in the field (in drilled boreholes, etc.) that relate to soil compressibility or expected settlement for a planned foundation. Methods 1 and 2 have inherent shortcomings. With method 1, it can be difficult to accurately ascertain the actual extent of settlement that an existing structure has experienced, and, perhaps more importantly, soil conditions may not be identical at different sites. However, checking the behavior or performance of existing structures in an area is good practice, provided that the procedures discussed hereafter are also followed. For method 2, field load tests seldom truly represent the loading conditions that will be developed by a structure, and the resulting settlement or compression data may be seriously misleading. Data from field load tests may be grossly in error when cohesive soil strata underlie the test area (relating to the time lag for consolidation settlement). Full-scale load tests are relatively expensive to perform. Evaluation of a soil’s compressibility characteristics can be accurately determined from laboratory tests if the tested samples represent the natural or undisturbed condition (method 3). With soil borings or other form of subsurface investigation being an accepted necessity for construction projects, it becomes practical to also obtain soil samples for purposes of performing laboratory tests. Compression tests can provide accurate information

346

347

Yes. Occurs quickly but for typical soils is expected to be a small part of total settlement. Probably yes. Expected to occur quickly.

Cohesive or fine-grained soils (clay, silt–clay mixtures)

Organic soils (soils including high organic content, swamp deposits, peats, etc.)

Yes. Occurs quickly (upon application of load or soon after)

Svd

Granular or coarse-grained soils (sands, sands with limited silt)

Soil Type

Yes. Occurs gradually, relating to consolidation properties of the soil deposit. Probably yes. Expected to occur quickly.

Yes. Occurs quickly (upon application of load or soon after)

Spc

Ssc

Yes. Can be considerable and extend over a very long period.

Yes. Typically expected to be a small part of total settlement.

Yes. Magnitude typically is small, but foundations subjected to repetitive or fluctuating loads are expected to expericence greater time-related settlement than foundations supporting static loading.

Table 1 Summary: Type and Rate of Foundation Settlement that Occurs, Related to Soil Classification

Most of the expected total settlement occurs quickly upon application of loading. For many types of structures, settlement may be almost completed during the construction period. But structures subjected to repetitive or fluctuating loads will experience postconstruction settlement. In overconsolidated or prestressed soil deposits, total settlement expected is considerably less than where normally compressed soil exists. In overconsolidated soils, settlement expected is considerably less than for normally consolidated and underconsolidated soil deposits. Highly organic and highly fibrous materials typically have very low shear strength and very low bearing capacity.

Additional Comments

Settlement

(method 3), but index property tests (method 4) may also provide adequate information (particularly if great accuracy is not required) and are quicker and less costly to perform. Correlating foundation settlement to soil conditions data obtained from cone penetrometer results and standard penetration tests in soil borings (method 5) represents a practical approach to designing foundations, since borings and penetrometer tests are typically included in the subsurface investigation phase of planning a project. This procedure also represents a necessary approach for conditions where undisturbed soil samples suitable for strength-deformation tests cannot be obtained (i.e., it is considered virtually impossible to obtain samples of cohesionless soil without affecting the soil particle structure which influences settlement properties). In situ field testing procedures to directly determine the volume change characteristics of a soil zone, such as accomplished with the pressuremeter apparatus and flat-blade dilatometer, are not yet in widespread use but are increasing as the required specialty equipment is becoming more available through boring contractors and commercial testing laboratories. Procedures for calculating the settlement of foundations supported on sand deposits and those supported on clay deposits are discussed separately in the following sections. If the subsurface conditions consist of a cohesionless soil stratum underlain by a cohesive layer, it may be necessary to calculate the settlement of the granular layer and of the clay layer separately then sum these separate values to obtain the total expected settlement.

2

Settlement of Foundations on Sand The settlement of foundations supported on sand relates to volume distortion and primary compression occurring in the stressed soil zones, and is expected to occur rapidly after application of loading (almost instantaneously). For construction where the major loading is from dead load due to the weight or mass of the structure, such settlement commonly is completed within the timeframe of the construction period, with the result that postconstruction settlement tends to be very limited. Conversely, the structures whose function involves significant live loading will experience postconstruction settlement. Limited documentation [44] further indicates that settlement attributed to the secondary compression or creep properties of the sand deposit should be anticipated where the live loading is fluctuating or variable, such as from wind effects and vibration.

Schmertmann Method Immediate settlement of cohesionless soil will be due to the combined effects of volume distortion and primary compression. The process of determining soil properties related to these effects through the use of laboratory testing has been hampered by the practical difficulty associated with obtaining undisturbed samples; important physical relationships such as soil particle orientations, effect of past compressive loadings and degree of compaction, and influence of in-ground confinement are altered by the sample-taking procedure. However, equations from elastic theory and model studies indicate distributions of vertical strains in the soil zone stressed by a foundation loading that are similar (Figure 1) and provide the basis for obtaining settlement information that is in agreement with field observations.

348

Settlement Center line vertical strain, %

Relative depth below surface footing

0.2

0.4

0.6

0.8

Range for long foundations

2B Range for square foundations

3B

4B

Figure 1 Vertical strain in sands below foundations (representative results: theory and model studies).

The Schmertmann method [312, 314], utilizing empirical relationships between subsurface investigation data and soil properties to approximate the pattern for vertical strains in the stressed soil zone, offers a procedure to calculate settlement resulting from the combined effects of volume distortion and compression in sand deposits that have not been precompressed (previously loaded). Strain influence factors for the Schmertmann method are shown in Figure 2. From elastic theory, the expression for vertical strain, ev, at a point in the stressed soil zone is: ev =

¢q I Es v

(2)

where ¢q = net foundation bearing pressure imposed onto soil Es = sand modulus of elasticity (also referred to as Young’s Modulus) Iv = strain influence factor for the soil zone beneath the foundation The value for settlement is obtained by summing the strains occurring throughout the depth being stressed by the foundation loading. Since the values of Δq, Es, and Iv are expected to vary with depth below the base of the foundation area, the calculation procedure to determine settlement involves subdividing the stressed zone into finite sublayers and assigning a representative or average value of Δq, Es, and Iv to each sublayer. The settlement for each

349

Settlement Rigid footing vertical strain influence factor, Iv 0

0.1

0.2

0.3

0.4

B/2

0.6

Calculate Ivp, use Δq σ'vp

Ivp = 0.5 + 0.1

B Relative depth below footing level

0.5

B Δq = q − q'vo 2B

Square foundations (L/B = 1)

q q'vo

B/2 (square) B (long)

3B Long foundations (L/B > 10)

σ'vp

Depth to Ivp

4B

B = least width of foundation L = length of foundation

q = foundation bearing pressure q'vo = soil overburden pressure at foundation level σ'vp = soil overburden pressure at depth for Ivp (B or B/2).

Figure 2 Strain influence factors for use with Schmertmann method to estimate settlement of foundations on sand.

sublayer is calculated and the results summed to obtain total settlement, or ¢s1 + ¢s2 + ¢s3 + . . . , where Δs values are the settlements of the respective sublayers. The settlement for any sublayer Δsn is determined as the product of the average strain value within the layer, evavg, and the related sublayer thickness, Δzn, or: ¢sn = 1evavg2 1¢zn2 = ¢q1Iv>Es2 1¢zn2

(3)

The strain influence factors, Iv, can be approximated from the diagram of Figure 2. A correction factor C1 is applied to compensate for the effect of foundation depth (or embedment), where: C1 = 1 - 0.51svo>¢q2

(4)

where ¢svo = soil overburden pressure at the level of the foundation base (corresponds to q¿ vo shown on Figure 2) The minimum value assigned to C1 is 0.5. A second correction factor C2 can be applied to consider effects of creep and other long-term compression factors for sand deposits, which include thin zones of fine-grained soil (but a factor of unity is suggested

350

Settlement

where it is known that the sand deposit is free of thin fine-grained soil layers). Where appropriate, C2 is determined from: C2 = 1 + 0.21log 10t2 where t = elapsed time in years Total settlement St then becomes:

St = C1C2 ©1evavg21¢zn2

= C1C2 ©1¢q21Iv>Es21¢zn2

(5)

Relating to the impractical aspects associated with obtaining values of Es through the testing of soil samples, correlations have been developed to relate this soil modulus to cone penetrometer resistance results, qc, determined as part of the subsurface investigation for a construction site. Values suggested for Es extend between 1.5qc and 4qc; values from the lower range are appropriate where sand and silt mixtures underlie square-shaped foundations, while values from the higher range are applicable where dense sand or sand–gravel mixtures underlie long foundations. Subsurface investigations frequently consist of drilled borings where the soil sampling involves using the standard penetration test (SPT) procedure and where the blow count N associated with obtaining each sample is recorded. The SPT method (a dynamic method involving use of a drop hammer procedure) is not considered an appropriate method for determining values of the soil modulus Es. But for sites where cone penetrometer data are not available, crude correlations between the SPT blow count N and the cone penetrometer resistance qc provide estimations of Es for use in settlement studies. Results considered representative of the correlations between qc and N are shown in Figure 3. Clay Clayey silts Sandy silt & silty clays & silt

Silty sand

Sand

10 9 8

Ratio, qc /N

7 6 5 4 3 qc in bars (1 bar = 100 kPa) N, blows/ft

2 1 0 0.001

0.01

0.1

1.0

Mean grain size, D50, mm

Figure 3 Ratio qc/N related to mean grain size. (Range based on standard penetration test N, resulting from typical field procedure having approximately 60 percent efficiency.) [300]

351

Settlement Table 2 Values of Es Related to Soil Type Approximate Value for Es (kgf/cm2, 0.1 MPa, ton/ft2) Soil Type Sand–silt mixture Fine-to-medium sands, fine–medium–coarse sands Sand–gravel mixtures

In terms of N

In terms of qc

4N 7N–10N (relating to density and compactness) 12N

1.5qc 2qc–3qc (relating to density and compactness) 4qc

A summary of the relationships between qc, the SPT blow count N, and the soil modulus Es is presented in Table 2. Illustration 1 Using the Schmertmann method, calculate the settlement for a foundation being supported on a sand stratum, given the following conditions: Square foundation, 3 m by 3 m; foundation bearing pressure q = 165 kPa; foundation installation depth Df = 1.5 m; soil unit weight g = 17.5 kN>m3

Solution Net foundation pressure Δq: ¢q = 165 kPa - 117.5 kN>m3 * 1.5 m2 = 138.75 kPa

Rigid footing vertical strain influence factor Iv 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 (0.83)

(0.33)

Depth below footing level

Δq B/2

Ivp = 0.5 + 0.1 (0.67)

B (1.08)

(1.67) 2B

352

σvp

= 0.66

Settlement Vertical pressure at depth B/2 below foundation σvp (prior to foundation construction) svp = 117.5 kN>m3 * 1.5 m2 + a17.5 kN>m3 *

B b = 52.5 kN>m2 = 52 kPa 2

Maximum strain influence factor Ivp (refer to Figure 2): Ivp = 0.5 + 0.1 2¢q>svp = 0.5 + 0.1 2138.75>52.5 = 0.66

C1 factor = 1 - 0.5a

svo 17.5 * 1.5 b = 1 - 0.5 a b = 0.905 ¢q 138.75 C2 factor = 1.0

The value for (Iv /Es) Δz is obtained using the tabulation procedure following. Table values shown are for a site where cone penetrometer resistance data are available. Layer Identification and Soil Type (below fdtn level) (1) f-c sand (2) f-m sand (3) f-m sand (4) f-c sand (5) f-c sand

aUse bUse

Layer Thickness Δz (m) (below fdtn level)

Distance to Center of Layer (m) (below fdtn level)

0.5 1.0 1.0 1.5 2.0

0.25 (or 0.083B) 1.0 (or 0.33B) 2.0 (or 0.67B) 3.25 (or 1.08B) 5.0 (or 1.67B)

qc(or N)a (kgf/cm2) 30 25 27.5 35 30

L L L L L

3 MPa 2.5 MPa 2.75 MPa 3.5 MPa 3.0 MPa

Es (MPa)

Ivb 0.13 0.47 0.59 0.40 0.15

2.5qc 2.5qc 2.5qc 2.5qc 2.5qc

= = = = =

7.5 6.25 6.88 8.75 7.5

(Iv /Es)Δz (m/MPa) 0.009 0.075 0.086 0.069 0.040 © = 0.274

average value for the layer thickness. Iv values to depth equal to 2B for square foundation (ref. Figure 2).

Settlement St = 1C1C22¢q ©1Iv>Es2¢z

= 10.905 * 121138.75 kPa210.274 m>MPa2 = 34 * 10-3 m = 0.034 m = 34 mm

The Schmertmann procedure applies to normally-loaded sand deposits and will overestimate foundation settlement if the sand has been precompressed or densified by construction activity such as compaction. Where geologic or other information indicates precompressed sands, the settlement to expect can be estimated as roughly half of the value calculated by the Schmertmann method.

Dilatometer Method The Marchetti flat-blade dilatometers is used to determine in situ soil properties. The results of dilatometer readings can be used to estimate the settlement of spread footing foundations by calculating the compression that results from foundation load-induced stresses within each of the incremental layers assumed to represent part of the overall stressed soil zone underlying the foundation (similar to the process

353

Settlement

used with the Schmertmann method). The foundation settlement is obtained by totaling the increment of compression occurring in each of the incremental layers, or: Settlement S = ©

¢sv 1¢z2 MDMT

where ¢sv = the subsurface stress resulting from the net pressure imposed on the soil at the base of the foundation (the net pressure Δqv being the pressure of the applied foundation load, q, minus the soil overburden pressure at the elevation of the foundation base); Δσv can be determined from the Boussinesq equations for stress distribution. ¢z = thickness of the incremental layer under study MDMT = Vertical Drained Constrained Modulus, a dilatometer parameter related to the Dilatometer Modulus ED and Horizontal Stress Index KD (which represents the influence of the lateral pressure condition, or OCR, in the soil). This value is equivalent to the laboratory compression test value for tangent modulus used to indicate the slope for the plot of effective stress versus strain for a desired load range. Illustration 2 Estimate the foundation settlement attributed to soil compressions occurring in the soil formation under a 2 m by 2 m square footing foundation constructed one meter below the soil surface, for the following conditions: Foundation loading q = 250 kPa = 250 kN>m2 Soil unit weight g = 17.5 kN>m3 Dilatometer procedure MDMT values obtained at different depths are indicated in the tabulation below.

Solution

Net foundation bearing pressure ¢q = 250 kN>m2 - 117.5 kN>m3211 m2 = 232.5 kN>m2 . Use three incremental soil layers to represent the soil formation under the foundation, with thickness of 1 m, 2 m, and 2 m (measured downward from the base of the foundation).

Layer (m) a b c

Layer Thickness Δz(m)

Distance to Center of Layer (m, below fdtn base)

MDMT (kPa, kN/m2)

Δ␴v (kPa, kN/m2)

¢sv 1¢z2 MDMT

1 2 2

0.50 (or 2.25B) 2.0 (or 1.0B) 4.0 (or 2.0B)

34,000 39,000 41,000

10.9821¢q2 = 227.8 10.3521¢q2 = 81.4 10.1221¢q2 = 27.9

0.0067 0.0042 0.0014

∴ Estimated settlement S = 0.0123 m L 12 mm (or about 0.5 inches)

354

a = 0.0123 m

Settlement

Approximations for Estimating Settlement of Foundations on Sand The planning associated with construction projects may require that estimates of foundation settlements be obtained before subsurface conditions and soil properties are fully identified.

Burland Data. The following information permits a probable order-of-magnitude settlement to be calculated. The data represented by Figure 4, developed from a large assortment of case study observations [48], can be used to perform rapid calculations to indicate a probable magnitude of settlement. If the soil conditions are identified and uniform, the ratio of settlement to applied pressure can be taken from the range between one-half and three-fourths of the upper limit value indicated by Figure 4 (but a value from near the upper limit is suggested for long, strip foundations).

Illustration 3 The preliminary design for a structure indicates that a foundation having a width of 3 m would impose a soil bearing pressure of 250 kPa (or 250 kN/m2). Site data from maps and preliminary field studies indicate that the construction area is underlain by a uniform deposit of dense sand. Estimate the foundation settlement using Figure 4.

SPT criteria

r li ppe

U

ense

0.1 0.08 0.06

it,

er lim

Upp

0.04

.d med

it, er lim

s

sand

s

and

se s

den

Upp

0.02 0.01 0.1

0.2

0.4 0.6 0.8 1

2

4

6 8 10

20

Range observed, loose sands

loos

mit,

0.2

ive)

ntat

s (te

nd e sa

0.4

Range observed med. dense sands

1.0 0.8 0.6

dense sands

2.0

Loose sand N < 10 Medium 10 < N < 30 dense sand Dense sand N > 30

Range observed

s 2 q , settlement/applied pressure, mm/(kN/m )

4.0

40 60 80 100 200

Width B, m

Figure 4 Data for estimating settlement of foundations on sands. Ranges of s/q related to foundation width as derived from case study observations. (After J. B. Burland and M. C. Burbridge, “Settlement of Foundations on Sand and Gravel.” Proceedings, Institution of Civil Engineers 78, 1985. Courtesy of Thomas Telford Journals)

355

Settlement

Solution For B equal to 3 m, select an s/q ratio equal to 0.05 (from Figure 4, upper limit value for dense sand). Then, estimate settlement as: S = 0.051q2 1342 = 0.051250 kN>m221342 = L 10 mm 1or less than 12 in.2

Subgrade Reaction Method. In uniform, cohesionless soil deposits, the soil rigidity (or modulus of elasticity) is expected to increase with depth because the overburden pressure and confining pressure, which affect the rigidity, increase with depth. For the condition where the increase in rigidity is uniform with depth, the following expression can be used to estimate settlements, provided that the depth of the foundation below the ground surface is less than the width of the foundation and that the foundation width is not greater than about 6 m (20 ft): Settlement, S 1in feet2 =

4qB2

Kv1B + 122

Settlement, S 1in meters2 =

1All terms in Customary Units2

13.932qB2

Kv1B + 0.30522

1All terms in SI units2

(6a)

(6b)

where B = width of foundation in meters or feet q = pressure imposed by the foundation in kN/m2, kPa, or kips/ft2 Kv = modulus of vertical subgrade reaction for 0.3 m square plate (1 ft square) bearing on the ground surface, in kN/m3 or kips/ft3 The modulus of vertical subgrade reaction Kv can be determined directly from a platebearing test performed in the field at the planned location for the structure. In this test, loads are applied to the plate in increments, and the plate settlement is recorded for each load. The results are plotted as shown in Figure 5.

Pressure, kips/ft2 or kN/m2

Settlement, ft or m

0 0.02 0.04 0.06 0.08 0.10 0.12

5

10

15

Slope = Kv

Figure 5 Plot of load settlement data from a plate-bearing test, to determine modulus of vertical subgrade reaction.

356

Settlement Table 3 Typical Values of Kv for Differing Sand Densities Representative Values of Dry Unit Weight Condition Loose Medium dense Dense Very dense

Values of Kva

Relative Density, %

pcf

kN/m3

kips/ft3

6 35 35–65 65–85 7 85

6 90 90–110 110–130 7 130

6 14 14–17 17–20 7 20

100 150–300 350–550 7 600 – 700

kN/m3 15 * 125 – 502 * 155 – 852 * 195 – 1102 *

103 103 103 103

aFor the condition where the groundwater table is at a depth greater than 1.5B. If the water table is at the base of the foundation, use 12 K v. Interpolate for intermediate locations of the water table.

The tangent to the initial straight-line portion of the plot is Kv, the modulus of vertical subgrade reaction. Mathematically: Kv =

Pressure kN kips = 3 or 3 Settlement m ft

Field plate-bearing tests are time consuming and costly. For purposes of estimating settlements, representative values of Kv that have been correlated with other soil properties can be used; see Table 3. Illustration 4 A foundation footing is to support a total column loading of 2,250 kN. Half of this will be live load. The building site is underlain by a thick stratum of sand. Tests on soil samples obtained from the site indicate dry unit weights for the sand between 15 and 17 kN/m3. The design calls for the footing to be located 1 m below the ground floor level of the building. Preliminary design data indicate that a soil bearing pressure of 250 kN/m2 can be used. Estimate the foundation settlement.

Solution

Assume a square footing shape. The required footing area becomes 12,250 kN2> 1250 kN>m22 = 9 m2 and for a square footing, B = 29m2 = 3 m. From Table 3, estimate a value for Kv; select Kv = 45 * 103 kN>m3. Estimated settlement, S, is: S =

13.932qB2

Kv1B + 0.30522

=

13.9321250 kN>m2213 m * 3 m2

145 * 103 kN>m3213.305 m * 3.305 m2

= 0.018 m = 18 mm

The expected postconstruction settlement (settlement due to live loading only) would be in the ratio of the live load to the total load. Since half of the total load is live load, the postconstruction settlement will be about 12118 mm2 = 9 mm (or less than 12 in.).

3

Settlement of Foundations on Clay Soils The total settlement expected for foundations bearing on clay soils is the sum of volume distortion (immediate) settlement plus primary compression (consolidation) settlement plus secondary compression settlement, or Sr = Svd + Spc + Ssc.

357

Settlement

Volume Distortion Settlement for Clays The settlement due to soil volume distortion, Svd, is typically a small proportion of the total settlement where foundations are supported on cohesive soil deposits. However, for firm, stiff clays (soil possessing high shear strength), the initial settlement may approach half the total settlement if the foundation pressures do not exceed the clay’s preconsolidation stress (but total settlement for this situation is usually small). Expressions from the linear theory of elasticity are in use for estimating the magnitude of volume distortion settlement; Equation 7 is based on a foundation bearing on a cohesive soil deposit possessing homogeneous and isotropic properties, and of infinite horizontal extent. The expression provides reliable results for static loading conditions when the foundation stresses are low, well within the limits of the ultimate soil bearing capacity (a condition satisfied where the factor of safety applied to the ultimate soil bearing capacity described in Chapter 14 is on the order of 3, a typical value). Svd = CsqB ¢

1 - v2 ≤ Eu

(7)

where Cs = shape and foundation rigidity factor q = magnitude of equivalent distributed load acting on the foundation area (total load/foundation area) B = plan width or diameter of the foundation Eu = undrained clay elastic modulus (Young’s Modulus or modulus of elasticity) v = Poisson’s ratio for the applied stress range (assume 0.5 for saturated clays, slightly less for partially saturated) The shape factors, Cs, are based on the design function of the foundation, rigid or flexible, because foundation bearing pressure distributions vary, as shown by Figure 6. Tables 4 and 5 present values, based on rigid or flexible category, shape, and position beneath the footing, where the settlement calculation applies.

q', applied loading

q', applied loading

Foundation–soil contact pressure (a) Rigid foundation

(b) Flexible foundation

Figure 6 Variation of foundation–soil contact pressures for rigid and flexible foundations on cohesive soil, and related settlement pattern.

358

Settlement Table 4 Values of Cs for Foundations on Clay Soil of Infinite Deptha Shape Flexible foundation: Circular Square Rectangular: L>B = 2 L>B = 5 L>B = 10 Rigid foundation: Circular Square Rectangular: L>B = 2 L>B = 5 L>B = 10 aSoil

Center

Corner

Edge at Middle of Long Side

Average

1.00 1.12

— 0.56

0.64 0.76

0.85 0.95

1.53 2.10 2.56

0.76 1.05 1.28

1.12 1.68 2.10

1.30 1.82 2.24

0.79 0.82

— 0.82

0.79 0.82

0.79 0.82

1.12 1.60 2.00

1.12 1.60 2.00

1.12 1.60 2.00

1.12 1.60 2.00

depth extends greater than 10B.

Table 5 Values of Cs for Foundation on Clay Soil of Limited Depth (D) above a Rigid Substratum (Rock) Cs Under Center of Rigid Circular Foundation

Cs, Under Corner of Flexible Rectangular Foundationa

Depth to Width Ratio (D/B)

L>B = 1

L>B = 2

L>B = 5

L>B = 10

L>B = q

1 2 5 10 q

0.15 0.29 0.44 0.48 0.56

0.12 0.29 0.52 0.64 0.76

0.10 0.27 0.55 0.76 1.05

0.04 0.26 0.54 0.77 1.28

0.04 0.26 0.52 0.73 —

0.35 0.54 0.69 0.74 0.79

aTo determine C for center of a foundation area, divide foundation shape into four equal subrectangles, then assign the B s dimension based on the size of one of the subrectangles. Multiply the selected value of Cs by 4 to use with Equation 7.

The elastic modulus, Eu, can be determined from the results of undrained triaxial compression tests performed on undisturbed soil samples, as indicated by Figure 7. Values for Eu frequently lie in the range between 500cu and 1,500cu, where cu is the soil cohesion shear strength as determined from undrained tests. The lower range is for clays of high plasticity and where foundation loads are large, while the higher range is for clays of low plasticity and where foundation loading is low. An estimated value for Eu, based on the consistency of the soil, can be selected from Table 6. Illustration 5 Calculate the immediate settlement (due to volume distortion) expected beneath the center of a rigid rectangular-shaped foundation bearing on a deep clay deposit for the following conditions: Foundation length = 6 m Foundation width = 3 m L>B ratio = 2

359

Settlement Axial stress, σ1

Soil sample

σ3

Deviator stress σ1 − σ3

Eu = slope Test confining pressure, σ3

σ1

Axial strain, ε x

(a) Representation of triaxial compression test

(b) Representative results, stress vs. strain, and determination of Eu

Figure 7 Method used to determine Eu from triaxial compression tests.

Table 6 Values of Eu Related to Soil Consistency Values for Eu Clay Consistency

Cohesion (Shear Strength)

MPa

ksf

Soft Medium to stiff Very stiff to hard

624 kPa 16500 psf2 25–100 kPa (500 to 2,000 psf) 7100 kPa 172,000 psf2

2.5–15 15–50 50–200

50–300 300–1,000 1,000–4,000

Solution Foundation bearing intensity, q = 200 kPa Clay cohesion (shear strength), c = 90 kPa Estimated value for elastic modulus, Eu = 45 MPa. Use Poisson’s ratio, v = 0.5 Shape and foundation rigidity factor, Cs = 1.12

Svd = CsqB ¢

1 - v2 ≤ Eu

= 11.1221200 kPa213 m2 ¢

1 - 0.52 ≤ 45 MPa

= 11.1221200 kPa213 m2 ¢

45 * 103 kPa

0.75

≤ = 0.011 m = 11 mm

Primary Compression Settlement for Clays In primary compression, decreases in soil volume are attributed principally to decreases in void spaces between soil particles as the particles rearrange so as to develop resistance to the new external loading; practically, there is no decrease in the actual volume of soil

360

Settlement

particles. The term compressibility is used to indicate one-dimensional volume changes that occur in a soil deposit when compressive loading is applied and the particles rearrange as a reaction. Although applicable to all soil types, the term is more typically associated with the study of fine-grained, cohesive soils (clay, clay–silt mixtures). The compression properties of a fine-grained soil can be determined directly by performing a laboratory compression test, frequently called a consolidation test. In this test, an undisturbed sample is fit into the ring of a consolidometer apparatus so that the sample is confined against lateral displacement, and compressive loading is imposed on the soil. [Basically, the consolidometer apparatus consists of a heavy brass metal ring to hold the soil sample being tested, and includes porous stone discs to cover the top surface and bottom surface of the sample to enable pore water to escape from the soil as compression is taking place (ref. Figure 8); an extensometer or micrometer dial gauge is used to measure changes in soil sample thickness.] For known magnitudes of load, the amount of compression and also the time required for compression to occur are recorded. The test is usually performed by Compressive loading

Compressive loading Ring

Loading cover

Ring Standpipe

Porous stones

Soil sample Base

Soil sample

Base

Fixed ring consolidometer

Floating ring consolidometer (a)

(b)

Figure 8 (a) Schematic representation of conventional types of laboratory consolidometers; (b) laboratory consolidation equipment (mechanical-lever type).

361

Settlement

imposing a series of increasing compressive loadings and determining time-rate-ofcompression data for each increment of loading. The entire body of data permits the compressive stress–strain characteristics of the soil to be determined (discussed more fully later in this chapter). The speed or rate of time that is required for volume changes to occur differs significantly for the coarse-grained (cohesionless) soils and the fine-grained (cohesive) soils. The cohesionless soils experience compression relatively quickly—frequently instantaneously— after loading is imposed. Conversely, the fully saturated and almost fully saturated clay soils (common conditions for many naturally occurring clay deposits) generally require a significant period before full compression under an applied loading results. Relating compression with the time period necessary for the compression to occur (that is, the time rate of compression) includes the process termed consolidation. Consolidation and its effects are discussed in Section 4. Presentation and Analysis of Laboratory Compression Test Data. Compression test data are presented in any of several ways, usually depending on the preferences and experiences of the individual using the data. Because compression is due to changes in the void spaces in the soil, methods in common use frequently indicate compression as a change in the void ratio. Using arithmetic coordinates, a typical test result for change in void ratio versus the increase in loading pressure is as shown in Figure 9. The slope of the curve at any point is av, the coefficient of compressibility. Mathematically, av = de>d sv. The dimensional units for av are square feet per pound (or kip or ton) or square meter per kilonewton. Because of the constantly changing slope of the curve, it is somewhat difficult to use av in a mathematical analysis, as is desired in order to make settlement calculations. When semilog coordinates are used and the void ratio is plotted versus the logarithm of pressure, the data will plot approximately as a straight line (or, as described later, a series of straight lines) (Figure 10). In this form, the test data are more adaptable to analytical use. For all plots, the imposed compressive test pressure is the intergranular or effective pressure (stress), sv. Another method of presentation shows unit change in sample thickness (or strain) versus the logarithm of pressure (Figure 11). This method gives essentially the same results as

2.2 2.0 Void ratio, e

1.8 slope a v

1.6 1.4 1.2 1.0 0.8

1

2

3

4

5

Imposed test pressure σv (tsf, ksf, kN/m2, kPa, etc.) (arithmetic scale)

Figure 9 Presentation of compression test data on arithmetic coordinates.

362

Settlement

Void ratio, e (arithmetic scale)

2.5

2.0

1.5

1.0 0.5 0.1

0.2

0.5

1.0

2

3

4

5 6

8 10

Imposed test pressure σv (tsf, ksf, kN/m2, kPa, etc.) (log scale)

Figure 10

Presentation of compression test data on semilog coordinates.

Compression in inches per inch or % (arithmetic scale)

0 Cc' 0.25 0.50 0.75 0.1

0.2

0.5

1.0

2

3

4

5

10.0

Imposed test pressure, σv (tsf, ksf, kN/m2, kPa, etc.) (log scale)

Figure 11

Method of indicating soil compression as strain instead of void ratio.

the method of Figure 10, but it has the advantage of minimizing the amount of work associated with reducing the test data. It may also make for simpler settlement calculations. With this method, no conversion to void ratio is necessary. The compression data for the sample are converted to strain by dividing total compression by the original sample thickness, and strain is then plotted against the logarithm of pressure. If, during the compression test, the pressures on a sample are increased to a certain magnitude, unloaded to a lesser value, and then reloaded and increased to magnitudes greater than previously, results like those shown in Figure 12 are obtained. Note that the soil does not expand to its original volume when pressure is removed. Some of the volume change due to external loading is permanent. Soil, therefore, is not an elastic material. Upon reloading, the resulting slope of the compression curve is less steep than the original slope. These factors of soil behavior have a significant effect on the settlement of structures. Consider a soil sample obtained from a site where conditions are as shown in Figure 13(a). The ground surface overlying the sample has never been above the existing surface, and there never was extra external loading action on the area. For this condition, the maximum vertical pressure ever imposed on the sample being considered is the current weight of overlying soil, svo. The result of a compression test performed on the sample is as shown in Figure 13(b). For laboratory loading less than svo, the slope of the compression curve is

363

Settlement

Void ratio, e (arithmetic scale)

2.5

2.0 Cc 1.5

1.0 0.5 0.1

0.2

0.5

1.0

2

3

4

5

10

Imposed test pressure, σv (tsf, ksf, kPa, kN/m2, etc.) (log scale)

Results of loading–unloading–reloading cycle applied to a soil.

Ground surface

Uniform soil has a unit weight of γ pcf

Z = Depth σvo

Cr Void ratio e or compression in inches per inch (arithmetic scale)

Figure 12

Cc

σvo = γZ

Test sample

Pressure, σv (log scale)

(a)

(b)

Figure 13 Description of conditions applying to compression test sample: (a) location of soil sample obtained for compression test; (b) result of compression test.

less steep than it is for loads greater than svo since, to the soil, values less than svo are in the “reloading range.” The slope of the curve at loading greater than svo is termed the virgin compression curve, and the slope of the curve is the compression index, Cc. The slope of the curve at values less than svo is the recompression slope Cr. Mathematically, the slope of the curve, for either Cc or Cr, is Δe/Δlog sv, Cc and Cr are terms without dimensional units (that is, they will be indicated as numerical values that have no dimensional units). From examination of Figures 12 and 13, it should be concluded that a change in the slope of the compression curve results when the previous maximum pressure ever imposed onto the soil is exceeded. If the ground surface had at some time in past history been above the existing surface and eroded away, or if the weight of a glacier had been imposed on the area at some time in history, svo (the existing overburden) is not the maximum pressure that has been imposed on the soil sample. The greatest pressure that previously existed on the soil is svmax, which would be the total pressure that developed from the existing soil overburden and the weight of the eroded soil or glacier. Compression test results for this occurrence would be as shown in Figure 14.

364

Settlement

Void ratio e or compression in inches per inch (arithmetic scale)

“Break” in slope of curve

σvo = γZ σvmax Pressure σv, (log scale)

Figure 14 pressure.

Compression test results where past pressure on soil has exceeded existing overburden

The effect of this soil property on the settlement of structures is shown by a comparison of conditions from Figures 13(b) and 14. Assume that the weight of the structure would cause an additional stress equal to ¢sv to act on the sample being studied. The total soil compression that will occur within the soil deposit, and therefore the settlement that the structure experiences, is related to the compression occurring in the test sample. Comparative results are shown by Figure 15(a) and (b). It is seen that, for the same magnitude of structural loading, soil conditions of Figure 15(a) cause more compression and, therefore, more building settlement than do the conditions of Figure 15(b). This indicates that the stress history of a soil may be more significant than other soil properties insofar as settlement of structures is concerned. Soil deposits whose condition is represented by Figures 13(b) and 15(a) are termed normally loaded or normally consolidated, meaning that the present overburden pressure is the greatest pressure that has ever been imposed on the soils. Soils whose condition is represented by Figures 14 and 15(b) are termed precompressed, preconsolidated, or overconsolidated, meaning that at some time in past history there were imposed pressures greater than those that currently exist.

Δσv Δe or Δ strain

σvo

σvf

Δe or Δ strain

Void ratio e or compression in inches per inch

Void ratio e or compression in inches per inch (arithmetic scale)

Δσv

σvf σvo

σvf

σvmax

Pressure σv, (log scale)

Pressure σv, (log scale)

(a)

(b)

Figure 15 Comparison of compression that occurs for (a) normally consolidated and (b) overconsolidated soils at loading above the overburden pressure.

365

Settlement

If the virgin curve of a laboratory compression test begins at a loading less than svo, the soil is termed underconsolidated, which means that the soil has not fully adjusted or stabilized under the current overburden pressures. (This would represent a “new” soil deposit.) The ratio of svmax to svo, indicated as the overconsolidation ratio, OCR (i.e., svmax/ svo), is a summary term used to help reflect the stress history for a soil deposit. Important strength and deformation properties of a soil, such as shear strength, τ, modulus of elasticity, Es or Eu, lateral pressure coefficient, K, as well as compressibility, are related to the OCR (generally, strength properties and resistance to deformation increase as the OCR increases, while values for lateral pressure coefficients and compressibility decrease as the OCR increases). Numerically, the OCR will be greater than one for overconsolidated or preconsolidated deposits, and less than one for underconsolidated soils. Originally conceived for describing deposits of clay and fine-grained soils, the term is also applied to sand deposits. For design purposes, the OCR for clays and fine-grained soils can be determined from the laboratory compression test procedure described earlier in this section. Empirical relationships that associate the OCR to the results of laboratory tests and results of field exploration techniques such as the blow count N60 from the standard penetration test (SPT) used in borings or the cone tip resistance qc from the cone penetrometer test (CPT), as shown below, can be used for estimating purposes. From lab test index properties, cohesive soil:

OCR L 1patm>svo21011.11 - 1.62 LI2

where LI = Liquidity Index = 1wnat - PL2>1LL - PL2, see Chapter 4

patm atmospheric pressure 101 kPa 2.1 ksf OCR L 0.58 N60 1patm>svo2 For SPT, cohesive soil: For CPT, cohesive soil:

OCR L 0.32 1qc - svo2>svo

The OCR for sands can be estimated from the results of laboratory shear strength tests or determined from the results of in-place testing using equipment that measures soil resistance to deformation such as the pressuremeter and dilatometer types (relating to the practical difficulties associated with obtaining undisturbed samples from sand deposits for laboratory testing), e.g., For DMT, sand: or where

OCR L 10.75 KD21.70, see Chapter 5

OCR L B

K0 R 1 - sin fax

y

y = 1>10.8 sin fax2

fax = fps - B 13 - 1fps - 322 R and φax is the drained axisymmetrical angle of internal friction determined from the plane–strain dilatometer test value for angle of internal friction, φps (refer to chapter 11 for definitions of axisymmetrical strain and plane strain).

366

Settlement

Illustration 6 A laboratory compression test performed on a clay sample provides results as shown by Figure 14. The test value indicated for svmax is 380 kPa. The soil sample was obtained from a depth of 5 m in a clay deposit known to be overconsolidated. Results of tests on other samples obtained at the site indicate the representative soil unit weight is 18.5 kN/m3. What is the OCR for this clay deposit?

Solution The current soil overburden pressure at the depth where the laboratory test sample was obtained is: svo = 118.5 kN>m3215 m2 = 92.5 kPa The overconsolidation ratio, OCR = svmax>svo = 380 kPa>92.5 kPa = 4

For reliable results, undisturbed soil samples are required for the compression test. If the soil sample to be tested is disturbed because of the soil boring techniques used to obtain the sample or during transportation and handling, the test data may be in error. Typical test results occurring from disturbed samples are shown in Figure 16. Estimating Compressibility from Index Properties. Laboratory compression tests on fine-grained soils normally require days to complete (about 2 weeks is the period conventionally used). It is frequently necessary to obtain information about the compressibility of a soil in as short a time as possible. Or, in the interest of job economy, it may be desirable to limit the number of compression tests but still to evaluate the compressibility characteristics of many soil boring samples. For these situations, it is possible to use correlations that have been established between compression properties and some more easily or more quickly determined properties of soils. These correlations, which permit estimates of the compressibility of silts and clays to be made, are: (8a)

Cc = 0.541e0 - 0.352 where e0 is the in-place void ratio:

Void ratio e or compression in inches per inch (arithmetic scale)

Cc = 0.005412.6w - 352

(8b)

Undisturbed sample Disturbed sample

Note: For disturbed sample there is no distinct “break” in the curve.

Pressure, σv, (log scale)

Figure 16

Typical compression test curve shapes for disturbed and undisturbed test samples.

367

Settlement

where w is the in-place water content, and: (8c)

Cc = 0.0091LL - 102

where LL is the liquid limit. Equation 8c should be applied only for normally consolidated clays. In a normally consolidated clay, the natural water content is about at the liquid limit. If a clay has its natural water content significantly less than the liquid limit, it is preconsolidated. The shear strength of a clay soil can also be used to indicate if the soil is normally consolidated or overconsolidated. The magnitude of the maximum pressure imposed on a soil is from 3 to 4 times its shear strength. By determining the shear strength and then svmax, the magnitude of svmax can be compared with the overburden pressure that acted on the sample. Close agreement would indicate a normally consolidated soil. Settlement Due to Primary Compression of Clay. Assume that a building is to be constructed at a location where soil conditions are as indicated by the sketch of Figure 17(a). The results of a laboratory compression test performed on a sample obtained from the center of the clay stratum are shown in Figure 17(b). The value of svo, the overburden pressure, is equal to ygsand + 1H0 > 22gclay , where γsand and γclay are the unit weights (in pcf or kN/m3) of the sand and clay soils, respectively, and the y and H distances are in units (m, ft) compatible with the soil unit weight values. The foundation load imposed by the new structure results in an additional pressure, Δσv, acting on the clay. Figure 18 uses the phase diagram method to illustrate changes that the clay sample experiences under the increased loading. The change in volume, ΔV, is equal to the change in void ratio. The relationship of the change to the original volume is: ¢e ¢V or VT 1 + e0

Ground surface

y

Ho

H/2

σvo

Very compact sand and gravel (assume as incompressible) Clay (compressible)

e (arithmetic scale)

Cr Cc

σv

Very compact sand and gravel

max

Pressure, σv (log scale) (b) (a)

Figure 17 Description of conditions applying to compression test sample: (a) subsurface profile indicating source of compression test sample; (b) compression test results.

368

Settlement σvo

σvo + Δσv = σvf

Vv = eo

Voids

Vs = 1

Solids

ΔV

VT

Δe = e0 − e1 e1 = Vv

Solids

Vs = 1

VT

Final volume after Δσv is applied

Total volume when σvo acts

Figure 18

Voids

Phase diagram illustrating change in soil volume that occurs with increase in loading.

If the clay sample represents the average volume change that occurs throughout the clay stratum of Figure 17(a), the following proportion will apply: Spc H0

=

¢e 1 + e0

where H0 is the original thickness of the clay layer and Spc is the compression that this layer will experience. The settlement of a foundation supported above the clay layer will be equal to the soil compression that occurs, or: Settlement = Spc = H0 ¢

¢e ≤ 1 + e0

(9)

As previously noted, the slope of the curve from the compression test for loadings greater than any previous maximum overburden pressure is Cc, where: Cc =

¢e ¢log sv

Rearranging gives: ¢e = Cc1¢ log sv2 Substituting in the equation for settlement gives: H0 C 1¢log sv2 1 + e0 c

(10a)

H0 H0 Cc1log svf - logsvo2 = 1C 2log 1svf >svo2 1 + e0 1 + e0 c

(10b)

Spc = or: Spc =

369

Settlement

In this equation, svo is the overburden soil pressure, and svf is the sum of the overburden soil pressure svn and the pressure caused by the weight of the structure ¢sv. If the compression test data are presented in the form shown by Figure 11, the settlement equation is: Spc = H0 * 1change in percentage of compression for the load change ¢sv2 or: Spc = H0C¿ c1log svf - log svn2

(11)

where C′c is the slope of the compression versus log sv plot (see Figure 11). Illustration 7 Referring to the conditions shown by Figure 19, assume that y is 3.66 m (12 ft), H0 is 2.44 m (8 ft), and γsand and γclay are 21.2 kN/m3 (135 pcf) and 17.27 kN/m3 (100 pcf ), respectively. The clay stratum is normally consolidated [e.g., svo = svmax; e0 is 1.20 and the compression index Cc is 0.20 (both dimensionless values)]. The weight of the structure causes a stress of 28.8 kN/m2 (600 psf) at midheight of the clay layer. [Therefore, ¢sv = 28.8 kN>m2 (600 psf).]

Solution The settlement would be: Spc =

H0 C 1log svf - log svo2 1 + e0 c

(a) Problem worked in SI units: Since: svo = 13.66 m * 21.2 kN>m32 + ¢

2.44 m kN kN * 17.27 3 ≤ = 98.9 2 2 m m

and: svf = svo + ¢sv = 198.9 + 28.82

kN m2

= 127.7

kN m2

Ground surface Very compact sand and gravel (assumed incompressible)

Thickness = y

Thickness = Ho

Ho /2

σvo = pressure of existing overburden Clay (compressible layer) Very compact sand and gravel

Figure 19

370

Soil conditions for problem of settlement due to compression in a buried clay layer.

Settlement therefore: 12.44 m2 12.202

Spc =

10.2021log 127.7 - log 98.921 = 0.0246 m = 25 mm

If the groundwater table is at the soil surface, the effective soil overburden pressure, svo, is due to the submerged (or effective) weight of the soil. If the unit weights of 21.2 kN/m3 and 17.27 kN/m3 represent saturated unit weights, the submerged unit weights will be (using gw = 9.81 kN>m3): For the sand: gsub = gsat - gw = 21.2 kN>m3 - 9.81 kN>m3 = 11.39 kN>m3 For the clay: gsub = gsat - gw = 17.27 kN>m3 - 9.81 kN>m3 = 7.46 kN>m3 If the soils are not fully saturated, it is generally sufficiently accurate to assume that a submerged effective soil weight is about half its weight when not submerged. The stress increase ¢sv is not affected by submergence. Using the values calculated, the overburden pressure svo is then: svo = 13.66 m2 ¢ 21.2

kN m3

= 141.69 + 9.102 svf = svo

- 9.81

kN m3

≤ + a

2.44 kN kN mb ¢ 17.27 3 - 9.81 3 ≤ 2 m m

kN

kN = 50.79 2 m2 m + ¢sv = 150.79 + 28.82 kN>m2 = 79.59 kN>m2

and: Spc =

12.44 m2 12.202

10.2021log 79.59 - log 50.792 = 0.043 m = 43 mm

Thus, the effect of a high water table is to cause more settlement. (b) Problem worked in U.S. customary units: Since:

and:

therefore:

svo = 112 ft * 135 pcf2 + 14 ft * 110 pcf2 = 2,060 psf svf = svo + ¢sv = 12,060 + 6002 psf = 2,660 psf 8 ft 10.2021log 2,660 - log 2,0602 1 + 1.20 18 ft210.202 = 13.426 - 3.3152 = 0.0806 ft = 1 in. 1; 2 2.20

Spc =

1From

mathematics, recall that the process of subtraction the logarithms of numerical values is actually accomplishing a division of those values. Hence the units for the involved terms, such as psf or kN/m2, will cancel, and the result of the logarithmic subtraction is dimensionless (without units).

371

Settlement If the goundwater table is at the soil surface, the effective soil overburden pressure, svo, is due to the submerged (or effective) weight of the soil. If the unit weights of 135 pcf and 110 pcf represent saturated unit weights, the submerged unit weights will be: For the sand: gsub = gsat - gw = 135 pcf - 62.4 pcf = 72.6 pcf For the clay: gsub = gsat - gw = 100 pcf - 62.4 pcf = 47.6 pcf If the soils are not fully saturated, it is generally sufficiently accurate to assume that a submerged effective soil weight is about half its weight when not submerged. Using the values calculated, we obtain: svo = 172.6 pcf * 12 ft2 + 147.6 pcf * 4 ft2 = 1,062 psf and: svf = svo + ¢sv = 1,062 pcf + 600 pcf = 1,662 psf (Note that ¢sv is not affected by the water table or submergence.) Spc =

=

18 ft210.202 2.20

H0 C 1log svf - logsvo2 1 + e0 c

13.222 - 3.0282 = 0.14 ft = 1.7 in. 1; 2

If the soil is an overconsolidated soil, the equation for settlement calculations is modified to: Spc =

H0 C 1log svf - logsvo2 1 + e0 r

(12)

provided that svmax 7 svf (Recall that svmax is the pressure where the slope of the compression test plot changes from Cr to Cc).2 For an overconsolidated soil where svf is greater than svmax, the settlement equation should be used as: Spc =

H0 C 1log svmax - logsvo2 1 + e0 r

+

H0 C 1log svf - logsvmax2 1 + e0 c

2To obtain a value of C from the laboratory compression test for use in this equation, the recommended practice r is as follows: Impose loading on the sample in increments to above the precompression stress, unload or rebound to at least the existing overburden, then reload the sample, and use the last recompression slope for determining Cr.

372

Settlement

or Spc =

H0 3C 1log svmax - log svo2 + Cc1log svf - log svmax24 1 + e0 r

(13)

Illustration 8 Assume that a buried stratum of clay 1.83 m (6 ft) thick will be subjected to a stress increase of 33.6 kPa (700 psf) at the center of the layer. The magnitude of the preconstruction soil overburden pressure is 48 kPa (1,000 psf) at the center of the layer. A laboratory compression test indicates that the clay is overconsolidated, with svmax equal to 72 kPa (1,500 psf). The value of Cc is 0.30, and the value of Cr is 0.05. What change in thickness results in the clay layer due to the stated conditions? Ground surface

eo = 1.50

1.83 m

σvo = 48 kPa Δσv = 33.6 kPa

σvo = 48 kPa

Clay layer

Cr = 0.05 Void ratio, e

Incompressible soil

Cc = 0.30 σvmax = 72 kPa (1,500 psf) σvf = 81.6 kPa (1,700 psf) σvo

σvmax σvf

Pressure, σv (log scale)

Solution The change in thickness of the clay layer, from Equation 13, is: Spc =

H0 3C 1log svmax - logsvo2 + Cc1log svf - log svmax24 1 + e0 r

(a) Problem worked in SI units: 1.83 m 310.0521log 72 - log 482 + 10.3021log 81.6 - log 7224 2.50 = 0.0184 m = 18.4 mm

Spc =

(b) Problem worked in U.S. customary units: 72 in 310.0521log 1,500 - log 1,0002 + 10.3021log 1,700 - log 1,50024 2.50 = 0.73 in.

Spc =

If the thickness of the compressible soil layer is great, it should be divided into thinner layers for purposes of making the settlement calculations. The increment of settlement for each layer should be handled by using the methods discussed. The value of svo is the overburden pressure at midheight of each layer, and ¢sv is the increase in pressure from the weight of the structure, also calculated for the midpoint of each layer. The total settlement will be the sum of increments occurring in all the layers. The depth of soil that is considered to be affected by foundation loading is limited to where ¢sv is approximately one-tenth of the soil overburden pressure.

373

Settlement

Illustration 9 A settlement calculation is to be made for a structure that is planned for a site underlain by a thick deposit of normally consolidated clay soil. Soil properties are as indicated on the sketch. What building settlement is expected? Ground surface Layer 1

eo = 1.30

1.22 m (4')

σvo = 10.56 kPa (220 psf) Δσv = 48 kPa (1000 psf)

Layer 2

eo = 1.26

1.22 m (4')

σvo = 31.7 (660) Δσv = 28.8 (600)

Layer 3

eo = 1.20

1.83 m (6')

σvo = 58.1 (1210) Δσv = 16.8 (350)

Layer 4

eo = 1.16

2.44 m (8')

σvo = 94.1 (1960) Δσv = 9.6 (200)

Void ratio, e

Cc = 0.20

e = 1.26

(Representative) σvo = 31.7 kPa (660 psf)

Pressure, σv (kPa, psf) (log scale)

Solution For this condition, the settlement calculation should be performed by dividing the thick clay deposit into sublayers and calculating the compression that occurs in each sublayer. The overburden pressure is calculated for the center of each layer, and the stress increase due to the structure is also calculated for the center of each layer. Results of these computations are shown on the sketch. The related settlement calculations follow. (a) Problem worked in SI units: S1 =

1.22 m 58.56 10.202 log a b = 0.0789 m = 79 mm 2.30 10.56

S2 =

1.22 m 60.48 10.202 log a b = 0.0303 m = 30 mm 2.26 31.67

S3 =

1.83 m 74.9 10.202 log a b = 0.0184 m = 18.4 mm 2.20 58.1

S4 =

103.7 2.44 m 10.202 log a b = 0.0095 m = 9.5 mm 2.16 94.1 Total settlement = 137 mm

(b) Problem worked in U.S. customary units: 48 in. 10.2021log 1,220 - log 2202 2.30 48 in. 10.2021log 1,260 - log 6602 S2 = 2.26 72 in. 10.2021log 1,560 - log 1,2102 S3 = 2.20 96 in. S4 = 10.2021log 2,160 - log 1,9602 2.16 Total settlement S1 =

374

= 3.11 in. = 1.19 in. = 0.72 in. = 0.38 in. = 5.4 in.

Settlement

Secondary Compression of Clay Soil Primary consolidation compression, discussed in the preceding section, refers to the process where soil particles in the stressed zone of saturated soil are rearranging into a more compact or tighter configuration as void water is squeezed out (expelled). The effective stress acting on the soil skeleton (the soil particles) is gradually changed from the initial (before-load) condition to the final condition (where the full magnitude of the construction loading is imposed onto the soil particles), while simultaneously, the excess void water (pore water) pressure gradually diminishes to zero (i.e., theoretical full consolidation has occurred). Secondary compression is the additional compression that occurs at a constant value of effective stress after excess pore water has been dissipated. The full nature of secondary compression is not yet completely understood but does relate to the slowly occurring adjustment of the particle arrangement under the effect of newer imposed loads. The process typically continues slowly over a long period of time. For most inorganic deposits, secondary compression is usually small compared to consolidation compression. On a practical basis, the procedure to analyze the effects of secondary compression in a soil layer subject to usual types of construction loading is complicated because the process of secondary compression can be occurring simultaneously with the process of consolidation compression. (In Section 5 it is more fully explained that in a consolidating soil layer the degree of consolidation varies across the depth or thickness of the layer until full consolidation for the entire layer is reached; the soil zones adjacent to the drainage surface for the escaping water consolidate faster than do the soil zones farthest from the drainage surface.) A compression index for secondary compression, Cα, can be determined from the plot of void ratio versus logarithm of time (data obtained from the laboratory consolidation test, refer to Appendix: Laboratory Procedure to Determine Coefficient of Consolidation) and the relationship: Ca =

¢e ¢e = ¢ log t log t2 - log t1

(14a)

The value for Δe is the change in test sample void ratio that occurs over the time period between t1 and t2 that follows the time when full theoretical consolidation (i.e., full primary compression) has been reached. The value of Cα should be determined for the consolidation test stress increment range comparable to that which will develop at the construction project. Typically, the values of Ca are small compared to the value of Cc (the compression index for consolidation compression). Test-determined values of Cα and Cc imply that a relatively limited range of the ratio Ca/Cc exists for naturally occurring soil deposits. For soft inorganic clays and highly organic, plastic silt–clays, the range is: Ca>Cc = 0.03 to 0.06

(14b)

Values from the lower end of this range are expected for the inorganic clays, whereas the higher values are expected for the organic, highly plastic silt–clay soils. These values do not apply to the highly organic deposits such as peat.

375

Settlement

After a value of Cα has been obtained, secondary compression settlement can be estimated from the following expression: Ssc = Ca ¢ = Ca ¢

H0 ≤ 1log tf - log tp2 1 + e0 H0 ≤ log 1tf >tp2 1 + e0

(15)

where tp is the time at the end of consolidation compression and tf is the time at the end of the time period (a final time) over which the settlement occurs (or is to be determined for). Illustration 10 The following data are applicable to settlement studies for a construction project. The construction site is underlain by a buried stratum of normally consolidated, inorganic clay 5 m thick. Laboratory tests indicate Cc = 0.20 and e0 = 1.15. Calculations indicate that consolidation settlement (primary compression) will be 30 mm for the expected construction loading. Full consolidation settlement (primary compression settlement) will require approximately 8 years. Estimate the total settlement to be expected over a 20-year time span, considering the effects of secondary compression.

Solution Estimate Ca = 0.04Cc = 0.0410.202 = 0.008 Ssc = Ca ¢

H0 5m 20 b log = 0.0074 m = 7.4 mm ≤ log 1tf>tp2 = 0.008 a 1 + e0 1 + 1.15 8

Total settlement = Spc + Ssc = 30 mm + 7.4 mm = 37.4 mm

4

Settlement Resulting from Earth Fill Volume changes in subsurface soils should be expected as a result of any new loading imposed on the soil. Filling of construction sites with compacted earth so as to raise the grade of a low area or make an uneven site level is now a relatively common undertaking in the construction industry. Areal settlements due to fills should always be considered. And where structures are to be supported on the compacted earth fill, it should be recognized that the total settlement for the structure could well be the sum of the volume changes occurring in the natural soil from the weight of the fill plus the volume changes in the fill and natural soil caused by the weight of the structure itself. Only if the completed fill is left in place a sufficient time so that full compression of the natural soils occurs before the structure is built can the effect of settlement due to fill weight be neglected. Settlement of granular soils will still, of course, occur almost instantaneously. However, compression of fine-grained soils does usually require a time period. Areal settlement due to the placement of fill can be computed by using methods already presented. However, the increase in stress Δsv caused by the weight of the fill is

376

Settlement

assumed to remain constant throughout the full depth of the natural or original soil. This is in contrast to the condition that occurs beneath a foundation area, where the value Δsv decreases with increasing depth below the foundation. The settlement of fill can be assumed to result from compression that occurs in the original soils underlying the fill. For cohesive soils, the necessary value of Cc or Cr is determined or estimated from laboratory tests. For granular soils, the expression for soil compression is also used. Appropriate values of Cc can be estimated from index properties, or from the following: For loose uniform sands (DR from 25% to 40%): Cc = 0.05 to 0.06

(16a)

For dense or compact sands (DR from 60% to 80%): Cc = 0.02 to 0.03

(16b)

Conversely, removal of soil from an area results in reducing the vertical stress imposed on the remaining underlying soil, and some expansion will occur because of the unloading. If new loading subsequently occurs (effect of structures, etc.) the soil undergoes re-compression (refer to Figure 12).

5

Consolidation The compression of clay deposits usually occurs gradually when new loading is applied (in contrast to coarse-grained deposits where compression related to loading typically is rapid). The causes for this are related to the low permeability of these fine-grained soils and the condition of their common occurrence in nature. Many clay deposits were formed underwater, an exception being the clays in residual and volcanic soil deposits. In the underwater deposit, all void spaces were occupied with water. Even with the passage of time, these sedimentary materials remained fully or almost fully saturated. In soil, volume decreases are due to reductions in the void spaces between the soil particles as the particles rearrange in order to support the additional pressure resulting from external loadings. For compression to occur in a saturated soil, water in the voids must be expelled to permit the decrease in void spaces to occur. This behavior may be visualized as similar to that of a saturated sponge having the water squeezed out of it by a pressing device. When external load is applied to a saturated clay, the water in the voids (or pores) becomes subject to an immediate increase in pressure equal to the subsurface stress caused by that loading. This is the same principle that is in effect for any hydraulic loading system, where an external force results in pressure developing in the fluid throughout the system. The soil’s trapped pore water then starts to flow toward regions of lesser pressure, moving away from the area where loading is occurring or toward bordering soil layers where a lesser pore water pressure exists. But because of the low permeability of finegrained soil, the movement or escape of this pore water takes place slowly. Only as water escapes from the stressed zone can soil particle rearrangement and decrease in void spaces

377

Settlement

take place. The result is that, for saturated clay soils, compression is a gradual occurrence. As voids space water escapes, the pressure caused by the external loading is transferred from this (previously present) water to the soil particles. The process of load transfer to the soil as pore water escapes is the consolidation process.3 The rate of consolidation for a stratum of clay soil is affected by several factors, including: 1. The permeability of the soil 2. The extent or thickness of the compressible soil and the distance that pore water in the soil must travel to escape from the zone where pressures due to foundation loading exist 3. The in-place void ratio of the compressible soil 4. The ratio of new loading to the original loading 5. The compression properties of the soil The consolidation properties of a soil (not the soil stratum) are dependent on items 1, 3, 4, and 5. The effect of these factors can be grouped together to obtain a property termed the coefficient of consolidation, cv, a value that indicates how rapidly or slowly the process of consolidation takes place. The coefficient of consolidation is: cv =

k11 + e2 avgw

¢ Note: Units for cv are

length2 .≤ time

(17)

where e khyd av gw

= = = =

void ratio (dimensionless value) coefficient of permeability (mm/sec, cm/min, m/day, ft/min, ft/day, etc.) coefficient of compressibility (m2/kN, ft2/lb) (ref. Figure 9) unit weight of water (kN/m3 or lb/ft3)

Values for k, e, and av can be determined separately and substituted into the equation to determine cv. It is common practice, however, to determine cv directly from a laboratory consolidation test. The consolidation test is a compression test but with the provision that time readings also are obtained (amount of compression versus time, for each applied loading increment). Appendix: Laboratory Procedure to Determine Coefficient of Consolidation provides details of the procedures used to compute values for cv from laboratory consolidation test results. As previously identified, the consolidation process requires two interrelated occurrences: the soil particles will be shifting to a denser configuration so that the particle 3A

concept that compares to the dissipation of soil pore water pressures as associated with soil consolidation refers to the rate of heat loss or temperature decrease occurring in a heated metal plate that is surrounded by a material at lower energy level or temperature. An energy gradient exists between the plate and the surrounding material. The outer zones of the plate cool fastest because of the short escape path, but simultaneously the high energy core region is losing heat energy to the low energy regions at the plate boundaries. The process of energy transfer continues until energy equilibrium is reached (no temperature differential within the plate material or between the plate and surrounding material).

378

Settlement

arrangement (that is, the soil structure or soil skeleton) develops the additional resistance necessary to support the increased loading, while simultaneously pore water in the void spaces is escaping so that the required decrease in voids can occur. Referring to Figures 20 and 21, note that initially the stress increase due to structural loading Δsv is imposed fully on the trapped pore water. This means that, at first, none of the new loading is carried by the soil particles. Only as water is being squeezed from the voids does the new loading start transferring to the soil particles; the rate at which pore water escapes from soil voids and the newly imposed construction-related loads are shifted to the soil particles is affected by the soil coefficient of permeability. Consolidation theory indicates that the end of the consolidation process has been reached when the water pressure in the soil voids drops to the hydrostatic pressure normal for the location so that the draining of pore water ceases. Full consolidation (100 percent) is assumed to have occurred when the full loading of the structure is carried by the soil particles (no excess pressure exists in the pore water). Values of consolidation less than

Open standpipe (piezometer) to measure height to which water, now under pressure, will rise

Foundation loading

Sand

GWL Height above GWL Δσv

Stress caused by foundation = Δσv

Clay Note: Δσv = pw = γ w × height of water column above GWL

Sand

Hydrostatic pressure in clay layer when foundation loading is first applied (a)

GWL Δσv

Note: water level in standpipe same as groundwater level (zero pw above GWL)

Hydrostatic pressure in clay layer when 100% consolidation has occurred (b)

Figure 20 Change in pore water pressure occurring as clay soil consolidates: (a) hydrostatic pressure in clay layer when foundation loading is first applied; (b) hydrostatic pressure in clay layer when 100% consolidation has occurred.

379

Settlement

σv

Fine-grained soil deposit; soil deposit is 100% saturated. Before construction, a compressive stress due to the soil overburden exists (is imposed onto the soil particles); σvo = γz. Water pressure in soil void spaces is normal hydrostatic pressure.

σv + Δσv

σv + Δσv

Construction loading causes a stress increase in the soil deposit. Due to effect of the new stress Δσv, particles attempt to shift into a more compact or tighter arrangement (so as to develop more compressive resistance as necessary for stability). However, water occupies the void spaces and prevents the soil particles from moving into those void spaces. The compressive stress increase, Δσv, therefore is imposed onto the water trapped in the void spaces, and not onto the soil particles. The result is that the void water is under an “excess” pressure (pressure greater than normal hydrostatic pressure expected from the position of the groundwater table).

Because void water is subject to excess pressure created by Δσv, the water gradually flows to locations of lesser water pressure (toward location of “normal” water pressure). The rate of flow depends on soil permeability. As void water pressure is dissipated, the soil particles gradually shift and the Δσv stress is slowly transferred from the void water over to the soil particles (that is, onto the “soil skeleton” for the soil deposit). The process of stress being slowly transferred onto the soil skeleton as void water is being expelled or squeezed from the stressed region of the soil deposit, and related gradual compression of the soil deposit, is “consolidation.”

Figure 21 Sequence of foundation loading being transferred onto soil particles (the soil skeleton) by the consolidation process.

100 percent indicate that, for the applied loading, soil compression (and settlement of the structure overlying the consolidation stratum) is still in process. If the term Uavg% represents the average percent consolidation throughout the thickness of a compressing soil layer, Uavg% and excess pore water pressure are related by the expression: Uavg% = ¢ 1 -

ut ≤ 1100%2 ui

(18)

where ut = average excess pore water pressure in the consolidating soil layer, corresponding to the time when the percent consolidation is being determined ui = initial excess pore water pressure in the consolidating soil layer Until the time the consolidation process is completed, excess pore water pressures in the consolidating soil vary in magnitude across the thickness of the compressing layer, being lower near the boundaries of the layer and greatest at the locations farthest from the

380

Settlement Top of stratum

Stratum thickness, H

Excess pore pressure dissipated at time t Drainage direction Excess pore pressure ut remaining at time t

Excess pressure variation for low percent consolidation

Excess pressure variation for high percent consolidation

Drainage direction

Initial excess pore pressure, ui Bottom of stratum Excess pore water pressure, u

u=0 (this condition represents 100% consolidation)

Figure 22

ui = Δσv, stress increase due to construction loading (this condition represents 0% consolidation)

Variation of excess pore water pressure across stratum thickness, doubly drained stratum.

boundaries because of the time lag associated with travel distance (similar to the manner in which a heated metal plate cools from the surface inward) (Figure 22). A time factor T is used to relate the rate at which excess pore water pressures dissipate to the period required for an average percentage consolidation to occur (Figure 23). Tv is used for the condition where the escaping pore water travels vertically (i.e., vertical drainage, Figure 24), while Th will apply to the condition where horizontal drainage of pore water takes place. The solid curve of Figure 23 applies to the condition of vertical double drainage and also single drainage of the consolidating layer for the occurrence where the initial excess pore pressure ui is uniform throughout the full thickness of the layer (note from Figure 24 that for double drainage of a layer, the drainage distance is half 0

U%, percent consolidation

10 u1

20 30 40

u α = u1 2

α=1

50

u2

α=5

60 70 80

α=∞

90 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.1

1.2

Time factor, Tv (vertical drainage)

Figure 23

Variation of time factor Tv with percentage of consolidation U.

381

Settlement Foundation loading

Foundation loading

Sand

Ht = Hdr

u1 (top of consolidating layer)

Sand Hdr

Ht = 2Hdr

Hdr

u2 (bottom of consolidating layer)

Impervious

Figure 24

u α = u1 2

One-way drainage (single drainage)

Two-way drainage (double drainage)

Distribution of initial excess pore pressure

Maximum distance water must travel to escape from clay layer equals the layer thickness (Hdr = Ht)

Maximum distance water must travel to escape from clay layer is half the layer thickness (Hdr = Ht /2)

(Method for determining α to use with Figure 10-23)

Vertical drainage conditions in consolidation theory.

the layer thickness, while for single drainage, the drainage distance is the full layer thickness). The situation of uniform initial excess pore pressure can be assumed where the construction loading responsible for the development of excess pore pressure is of large areal extent (such as results from placement of an earth fill over a large area or from the loading of a structure that covers a large plan area). For the case where construction loading acts over a limited area (such as with individual foundations for a building), the increase in subsurface stress due to the structural loading diminishes with depth; since the initial excess pore water pressure is expected to equal the increase in construction-caused subsurface stress, the excess pore water pressure will therefore vary across the thickness of the consolidating layer (greatest at the top, least at the bottom). If the ratio of the initial pore pressure at the top of the layer to the pore pressure at the bottom of the layer is a (i.e., a = Ui top>Ui bot ), the dashed line curves on Figure 23 can be used to select a time factor Tv appropriate for the expected variation in excess pore pressure distribution. For given field conditions, where the distance that trapped pore water must travel to escape from the compressible layer (Hdr in Figure 24) and cv are known, the time period t required for a given percentage of consolidation to occur is: t =

TvHdr2 cv

(19)

where Tv = time factor for consolidation due to vertical drainage (a dimensionless value) Hdr = longest drainage distance for escaping pore water cv = coefficient of consolidation The accuracy of the time rate of settlement calculation relates to how well the features of the consolidation theory and actual subsurface conditions match. Equation 19 is a solution for the case of one-dimensional diffusion resulting as a fluid is being forced through a

382

Settlement

porous media. The expression assumes that rate of soil compression is governed only by the dissipation of soil pore water. Other practical factors that can affect computed predications include variations in the thickness of the consolidating layer, variation in the soil properties, and occurrence of horizontal drainage in the perimeter regions of the consolidating layer. Compression of a clay layer resulting because trapped pore water under pressure has escaped is consolidation compression or primary compression. Some additional long-term compression from what is considered secondary compression also takes place as soil particles continue to adjust under the applied loading. However, for inorganic soil deposits, the effects of secondary compression are expected to be small; where the calculated primary compression settlements are small, the effects of secondary compression are commonly ignored. The time rate for secondary compression settlement is estimated from the plot of consolidation test results for the appropriate load increment, using the ratio of time for full compression to time for the theoretical 100 percent consolidation compression. Illustration 11 Referring to Illustration 5 and Figure 19, assume that laboratory consolidation test data for the clay indicate that, for the range of loading applied to the soil, cv is 0.016 m2 per month. (Recall that the settlement computed in Illustration 5 is 25 mm ±.) (a) How long will it take for half of the estimated settlement to occur? (b) How much settlement will occur in one year?

Solution (a) Half the estimated settlement is a U of 50 percent. Referring to Figure 23, assuming the solid curve applies, for U = 50%, obtain the value Tv = 0.20:

t =

TvHdr2 cv

10.22a

=

2.44 m 2 b 2

10.016 m2>month2

= 18.6 months

(Note that for this problem, Hdr is one-half the total thickness of the clay layer.) t = 1 yr = 12 months

(b) Tv =

tcv

= 2

Hdr

112 months210.016 m2>month2 11.22 m22

= 0.13

From Figure 23, for Tv = 0.13, we obtain U = 42%, and therefore settlement, Spc, is: Spc = 0.42 * 25 mm 1from III. 109–952 ⬵ 11 mm

6

Surcharging The situation may exist where, because of compressible or weak subsoils, a proposed structure would undergo settlements that would exceed tolerable limits. Alternatively, stresses resulting in the weak soil from the weight of the structure could cause a weak soil

383

Settlement

(increasing)

Shear strength, τ

to fail. If the site is to be used, the problem soil somehow must be circumvented (by removing it if possible, or by using pile foundations penetrating to below the poor soil) or improved so that the strength is adequate and the compressibility reduced. For construction projects extending over large areas, such as highways, improvement of poor subsoils is always given consideration because of possible cost savings. The improvement of poor soils may possibly be accomplished by using a surcharge program. Surcharging of a soil, in its simplest intent, merely involves imposing an external loading for a long enough duration to cause desirable changes in the soil before the structure planned for the area is erected and supported on the soil. Under the weight of a surcharge load, a poor soil will compress and increase in strength. The surcharge is made to the desired weight or loading and continued for a sufficient period to achieve results satisfactory for the need of the planned structure. Generally, this might include having the surcharge cause the magnitude of soil compression (settlement) that the planned structure would have caused, so that after the surcharge is removed and the structure is built, its settlement will be negligible, or having the surcharge cause a sufficient increase in the strength of the poor soil so that it can safely support the weight of the planned structure. Any material that will impose pressure onto the subsoils that require improvement can be used for the surcharge loading. Most conventionally, the material used is soil (borrow or fill). Soil is used because of its general availability and low cost, its ease of handling, and the absence of problems with deterioration. Surcharging for improvement is most commonly applied for cohesive and organic soils—soil types that require a time period for compression (consolidation) and strength gain to occur. Surcharging programs are applicable for improving loose, granular soil deposits, but other methods, frequently quicker and more economical, can be used for these soils (methods that do not have application with cohesive soils). Surcharges for improving the strength of weak soils have to be applied slowly or in increments so that the weak soils are not overstressed (failed) before they have time to improve. With cohesive soils, the compression–consolidation process involves a readjustment of soil particles as water is squeezed from the voids of the soil. A gain in strength

Sheer strength increases as consolidation (compression and reduction in water content) occurs

(increasing) Water content, w

Figure 25

384

Change in soil strength and water content as consolidation under surcharge occurs.

Settlement

occurs because of the consolidation, but the strength gain results only as rapidly as water is squeezed from the soil (an implication from the strength curve shown in Figure 25). Consequently, a properly performed surcharge program requires field monitoring to establish a safe rate for increasing the surcharge and to determine how long it should remain in place. Monitoring normally includes instruments such as piezometers to measure excess pore water pressures developed within the weak soil, to learn of the rate at which consolidation is occurring, and to indicate when additional increments of the surcharge can be safely added. Settlement plates located in or on the surface of the original soil are used to monitor the rate and amount of settlement caused by a surcharge. For accurate records, any reference benchmark to measure settlement must be located outside of the area that is settling. In some cases, the structure itself can be used to apply a surcharge load in increments. An example of this is with ground-level storage tanks, such as those used to hold petroleum or other fluids. For many situations, the tolerable settlement of storage tanks is relatively large, provided that large differential settlements that might cause the tank to rupture do not occur. A fluid, possibly the material to be stored but more frequently water, is used to partially fill the tank. The tank is left partially filled until the subsoil achieves the necessary increase in strength. Then additional liquid is pumped in, and the next waiting period is allowed to elapse. This procedure continues until the subsoils supporting the tank reach the desired degree of stability. The tank can then be put into permanent use. Permanent piping connections should not be made until all, or most of, the settlement has occurred. With this method, a laboratory test program should be performed before field operations are begun to ensure that the weak subsurface soils will be able to reach the strength required to support the tank and to obtain an estimate of the time required to complete the consolidation process. A laboratory program involves consolidating samples of the weak subsoil under the same vertical pressures that the tank will impose onto the soil and then performing shear tests to determine if adequate shear strength is being obtained.

Acceleration of the Soil Improvement Process Use of a surcharge program requires that the surcharge be applied and left in place for long enough to obtain the desired compression or strength gain. Depending on the properties and thickness of the poor soils, a surcharge may remain in place for months or longer before it accomplishes its purpose and can be removed. Construction cannot begin until the surcharge is removed, of course. If time is an important factor, means of accelerating the consolidation period can be undertaken. One method to reduce the time for a desired amount of compression (or settlement) to result is to use an excessive surcharge, provided that the subsurface soils have sufficient strength. A given amount of settlement will occur more quickly under a heavy surcharge than under a light surcharge. Illustration 12 Assume that an area is underlain by a stratum of compressible soil 5.3 m thick. The weight of the proposed structure, say an earth embankment for a new highway, will result in an increase of 28.2 kPa pressure to the compressible soil. The settlement under this loading would be 0.30 m. The

385

Settlement compressible layer is doubly drained, and the soil has a cv value of 0.4 m2/month. Calculations indicate that it would take approximately 22 months for the settlement to occur under the weight of the proposed embankment. How long would be required for a surcharge imposing a pressure of 56.4 kPa to cause the 0.30 m of settlement?

Solution The first step of the solution requires that the total settlement to be expected from the 56.4 kPa surcharge be computed. Using Equation 10, the settlement is computed to be 0.51 m. To obtain but 0.30 m of settlement, about 60 percent consolidation 1i.e., 0.3 m>0.51 m * 100%2 must occur. The Tv value for 60 percent consolidation is 0.3, and the time required is: t =

10.3212.65 m * 2.65 m2 TvH2dr ⬵ 5.3 months = cv 0.4 m2>month

Horizontal Drainage When a surcharge is used, another method for accelerating the consolidation process is to shorten the length of the drainage path for the pore water escaping from the consolidating soil. This can be accomplished by providing vertical drains or drain wells at spacings closer than the drainage distance for vertical flow so that horizontal drainage flow will occur. Frequently, too, the horizontal coefficient of permeability for cohesive soil deposits is many times greater than the coefficient of permeability for the vertical direction (because of the manner of deposition), which further increases the speed of consolidation. Sand Drains. Vertical drain wells can be holes that are drilled into the compressible soil and filled with sand (frequently called sand drains). Since the sand is very permeable material, pore water under pressure because of the surcharge will flow to the drain wells and be forced upward (or downward, too, if the well extends to a deeper permeable stratum underlying the compressible soil). After flowing upward to the surface, the water is carried away from the surcharge area (Figure 26(a)). If the drain wells are located as indicated in Figure 26(b), the zone of drainage can be assumed to be circular. The expression for the time period necessary to achieve consolidation when this condition exists is:

t =

Th1gwav2

kh11 + e02

=

Thdd2 ch

(20)

where t Th ch dd

= = = =

elapsed time time factor for horizontal drainage (dimensionless) horizontal coefficient of consolidation (note: units are length2/time) diameter of the zone draining to a drain well (ft or m)

The value of Th varies according to the ratio of the draining area to the escape (drain well) area. If dd is the diameter of the draining area and dw is the drain well diameter, n can

386

Settlement Sand layer to drain water from surcharge area

Surcharge

Drain wells Firm layer

Compressible layer

Water flow

Water flow

(a) Cross section of installed drain wells and surcharge Drain wells

S

S

S

Hexagon pattern indicates zone of influence for one well (S indicates actual well spacing)

dd

Equivalent diameter, dd

dw, well diameter (b) Drain-well pattern

Figure 26 Cross section and plan of typical sand drain installation: (a) cross section of installed drain wells and surcharge; (b) drain well pattern [24].

be the ratio of dd to dw. Curves of Th, similar to those for Tv used for vertical consolidation, are available for different ratios of n (Figure 27). The drain-well spacing affects the time for consolidation more than does the drainwell size. Quicker results will be obtained by using a small drain-well spacing. However, there is the practical problem of constructing the wells close together without collapsing or otherwise damaging them. There is also the potential problem that the sand drains may shear along their length if the compressible soils experience large settlement or lateral movement during compression, with the result that water flowing from the consolidating soils to the drain well cannot escape to the surface, thus invalidating the drain well system. Because of the specialized techniques and equipment required for a sand drain installation, it is generally not economical to use the method for improving small projects. Illustration 13 The settlement problem of Illustration 10 examined the effects of using a surcharge to improve the subsurface soils for a planned highway embankment. The surcharge program involves the placement and removal of an amount of soil greater than that required for the embankment, with a related expense. It is desired also to study the effect of using a sand drain installation, on the possibility that

387

Settlement 0 10

Consolidation (%), Ur

20 30 10

40

50

10

60

40

5

n = ratio

70

dd dw

V

u al

es

of

n

80 90 100

0.004

0.01

0.04 0.10 Th, time factor for horizontal drainage

0.40

1.0

Figure 27 Variation of time factor Th with percentage of consolidation U for sand drain installations [24].

it may be less costly than surcharging (because less surcharge is required) and that faster results may be provided. Laboratory testing of samples indicates that the coefficient of permeability for horizontal flow ranges from 5 to 15 times the coefficient for vertical flow; the coefficient of consolidation for horizontal drainage would be similarly affected. For a preliminary analysis, use a value of ch that is 7 times the value of cv. Since cv was 0.4 m2/month, ch will become about 2.8 m2/month. It is planned to install sand drains that are 0.7 m in diameter and on a center-to-center spacing of 3.5 m in a pattern as indicated in Figure 26(b). Settlement calculations indicate that a 37.6 kPa surcharge would cause a settlement of 0.34 m. Related to this, the 0.30 m settlement expected from the necessary embankment fill represents 0.3 m/0.34 m, or 89 percent consolidation. The ratio of dd to dw is 3.5/0.7, or 5. With reference to Figure 27, the value of Th for 89 percent consolidation and n equal to 5 gives a time factor of 0.27. Substitution of appropriate values into Equation 20 gives: t =

Th1dd22 ch

=

10.27213.5 m * 3.5 m2 2.8 m2>month

⬵ 1.2 months

Wick Drains. A practical development of the 1970s was the prefabricated band drain consisting of synthetic fabrics and including a permeable core. The typical band drain has a small cross section (e.g., 4 in. wide by 0.25 in. thick, or 100 mm by 7 mm), is extremely flexible, and can be used on a job site while being unwound from a storage or dispensing reel. The fabric drain can be rapidly installed with mobile crane or mast-equipped rigs by using a needle-punching technique (a nondrilling, soil displacement procedure; see Figure 28). A long, thin mandrel (shaft) on the installation rig is vibrated or hydraulically pushed into the soil to be drained. The fabric drain, attached at the tip of the mandrel, is pulled into the earth as the mandrel advances. At the desired penetration, the drain is disengaged from

388

Settlement

Band drain feed

Band drain on reel

Drive mandrel-hydraulic or vibratory advance

Granular soil working pad

Fine-grained or organic soil to be drained – consolidated

Mandrel with wick drain penetrates soil; mandrel is withdrawn, leaving drain in place

Firm-compact soils

Figure 28

Typical wick drain installation equipment and procedure.

the mandrel and also cut from the band reel at the ground surface. The mandrel is withdrawn, but the fabric drain remains in place. This vertical drain or wick provides the escape path for excess pore water in the penetrated soil. The rate of drain installation can be on the order of 1 m or 3 ft per second. Installation depths up to about 30 m or 100 ft have been used. In the typical application, the wick drains are installed at closely spaced intervals (on the order of 1.5 m or 5 ft or less) in a staggered pattern similar to that indicated in Figure 26(b). A surcharge material is applied to the area after the drains have been installed to create an increase in subsurface stress and pore water pressure, as for the sand drain installation. As with sand drains, a granular soil layer is placed on the site surface under or as part of the surcharge, to provide a drainage pad for water wicked upward. The close spacing of drains is conducive to rapid consolidation because of the small lateral distance that squeezed pore water must travel to reach a drain. The fabric drains offer some distinct advantages over sand drains, such as lower materials cost, easier and more rapid installation, and less danger of damage to the drain as the treated soil consolidates and settles. Accordingly, wick drains generally have supplanted the use of sand drains. The theory previously outlined for sand drains can be applied where analytical studies involving wick drains are necessary. As an additional item of information, installed columns of gravel-size material are still being selected for projects, for reinforcing primarily (to increase the shear strength or bearing capability of a weak soil formation) but to simultaneously function as paths of drainage for the consolidation process.

389

Settlement

Whether or not the use of drains is more economical than a surcharge program without drains (where a heavier surcharge or longer time period would be involved) depends on the relative cost of the drain installation and price of embankment fill. Factors to make a drain installation more attractive would include the possibility for the surcharge to be adequately compacted during placement so that it can be left in place as the structural embankment when the desired settlement has been reached.

Problems 1 The settlement of foundations is typically the result of three separate occurrences that take place in the soil zones providing support. List these factors and give a brief description of each. 2 Indicate the typical time period associated with foundation settlement for the condition where granular soil provides the support and for the condition where saturated clay provides the support. Discuss the reasons for the difference. 3 The major portion of settlement typically resulting for a foundation bearing on a sand deposit is different than for a foundation bearing on a saturated clay deposit. Indicate and explain the cause for the major portion of settlement for the two soil categories. 4 Compare the relative degree of settlement typically expected where a foundation bears on a soil deposit with significant organic content to the settlement expected where a foundation bears on soil free of organic material. 5 For structures supported on foundations that bear on sandy soils, it is presumed that the major portion of expected settlement occurs upon application of loading; the practical effect is that total settlement can take place by the end of the time period required to complete the construction. However, there is a major exception to this statement; indicate this condition. 6 Provide reasons why field plate-loading tests performed at a planned construction site may give erroneous settlement information if the area is underlain by cohesive soils. (Hint: Field loading tests are frequently performed with bearing plates smaller in area than the foundation footings will be. Also consider the effects of consolidation.)

390

7 Use the Schmertmann method to estimate the settlement for a foundation bearing on a sand deposit and where the following conditions apply: • Square foundation, 2 m by 2 m, installed at a depth of 1 m • Foundation bearing pressure, q = 200 kPa • Soil unit weight, g = 18 kN>m3, deep water table Use the tabulation procedure as shown in Illustration 1. Analyze for conditions extending to a depth 2B below foundation level, using four 1-m-thick sublayers. Cone penetration resistance values (in MPa) qc are 3.0, 3.5, 5.0, and 3.5, respectively, for the four sublayers. Assume that the C2 factor equals 1. 8 Use the Schmertmann method to estimate the settlement of a long strip-footing foundation bearing on a sand deposit where the following conditions apply: • Footing width = 1.5 m, installed at a depth of 1 m • Foundation bearing pressure, q = 150 kPa • Soil unit weight, g = 18.5 kN>m3 , deep water table Use the tabulation procedure as shown in Illustration 1. Analyze for conditions extending to a depth 4B below foundation level, using sublayers 1 m, 1 m, 2 m, and 2 m thick. Subsurface samples have been obtained using the standard penetration test (SPT), where N values of 24, 22, 29, and 36 can be used as the respective average values for the sublayers. Assume Es = 10N (Table 2), C2 = 1. 9 Use the Schmertmann method to estimate the settlement of a foundation bearing on a sand deposit and where the following conditions apply:

Settlement • Square foundation, 4.5 ft by 4.5 ft, installed at a depth of 3 ft • Foundation bearing pressure, q= 4 ksf = 2 tsf • Soil unit weight, g = 118 pcf above water table Use the tabulation procedure as shown in Illustration 1. Analyze for conditions extending to a depth 2B below foundation level, using three sublayers each 3 ft thick. Standard penetration test (SPT) values for the three layers are Navg = 36 , 27, and 32. Assume Es = 9N and C2 = 1. The groundwater table is at the elevation of the base of the foundation (use submerged unit weight for soil zone below foundation level when computing σvp). 10 Apply the dilatometer method to predict the foundation settlement expected for the following conditions: • Spread footing foundation, 1.75 m by 1.75 m (square shaped) • Loading pressure imposed at the base of the foundation, q = 275 kPa • Foundation installed at depth 1.5 m below the soil surface • Unit weight for soil at the site = 17 kN>m3 • MDMT = 30,000 kPa for the 5-m zone under the foundation Limit the settlement analysis to the 5-m zone underlying the base of the foundation, using three incremental soil layers (use layers 1-m, 2-m, and 2-m thick). 11 Apply the dilatometer method to predict the foundation settlement expected for the following conditions: • Spread footing foundation, 6 ft by 6 ft (square shaped) • Foundation installed at depth 4 ft below the soil surface • Loading pressure imposed at the base of the foundation, q = 5 ksf • Unit weight for soil at the site = 110 pcf Limit the settlement analysis to the 12-ft zone underlying the base of the foundation, using three incremental soil layers (use layers 2-ft, 4-ft, and 6-ft thick). Apply MDMT values of 700 ksf, 800 ksf, and 850 ksf, respectively, to these layers. 12 Referring to the foundation settlement information summarized by the Burland studies (Figure 4),

estimate the settlements expected for the following foundations, assuming that the supporting sand deposit is medium-dense to loose: (a) Square foundation 2 m by 2 m, q = 200 kPa (b) Long foundation 1.5 m wide, q = 150 kPa (c) Square foundation 4.5 ft by 4.5 ft, q = 4 ksf 13 A plate-bearing test is performed to determine a value for the modulus of subgrade reaction at a construction site underlain by granular soil. The load versus settlement data graphs as a straight line for the initial range of applied loads. The graphical plot indicates a settlement of 1.4 cm when the loading is 450 kN/m2. What is the value for the modulus of subgrade reaction? 14 A plate-bearing test is performed on soil near the surface of a granular stratum in order to determine the modulus of vertical subgrade reaction. The load versus settlement information appears as a straight line on a graphical plot. At a point where the load is 5 tons/ft2, the indicated settlement is 0.5 in. What is the value for the modulus of vertical subgrade reaction? 15 A square footing 3 m by 3 m supports a building column load that results in a foundation bearing pressure of 200 kPa. The deep sand stratum underlying the foundation has an average unit weight of 17.0 kN/m3 and is classified as medium-dense. Use the subgrade reaction method to estimate the foundation settlement. 16 A square footing 5 ft by 5 ft supported on a sandy stratum is designed to carry a column loading of 250,000 lb. The unit weight of the sands underlying the footing location vary between 120 and 125 pcf. Estimate the foundation settlement to be expected using the subgrade reaction method. (Neglect the weight of the footing.) 17 A square foundation is to be supported on a thick clay soil deposit. Calculate the foundation settlement due to volume distortions occurring in the supporting soil zone, given the following conditions: • Square foundation (flexible category), 2 m by 2m • Total loading on foundation is 700 kN • Clay shear strength cohesion, c = 100 kPa; saturated clay

391

Settlement 18 A square footing foundation (flexible category), 5 ft by 5 ft, imposes a total loading of 75 kips. The foundation bears on a thick stratum of saturated clay where the shear strength c is 1,500 psf. Estimate settlement due to volume distortions occurring in the clay. 19 A long strip-footing foundation 1.5 m wide will be used to support an exterior bearing wall for a commercial building. The foundation will be installed near the soil surface and impose a loading of 200 kN per meter of wall length. The soil is saturated clay, with cohesion c (shear strength) equal to 50 kPa. (a) Calculate the settlement under the center of the foundation due to volume distortion in the soil, assuming a flexible foundation. (b) Calculate the volume distortion settlement, assuming the foundation is rigid. 20 Calculate the settlement under the center of a flexible foundation 4 m by 4 m due to volume distortions occurring in a saturated clay stratum but where rock exists at a depth 8 m below the foundation. The clay shear strength c is 60 kPa. The total foundation loading imposed onto the soil is 2,400 kN. 21 Calculate the settlement for the center of a flexible foundation due to volume distortions in the underlying clay stratum. The clay soil has a cohesion shear strength c equal to 1,000 psf. The foundation is 6 ft by 6 ft and imposes a total loading of 120 kips. The clay stratum is 12 ft thick; rock underlies the clay. 22 Calculate the soil volume distortion settlement under the center of a 2-m-wide strip foundation (flexible type) which carries a loading of 400 kN per meter of wall length. The foundation bears on a saturated clay stratum (cohesion c is 75 kPa) that is 10 m deep. Rock underlies the clay. 23 Calculate the settlement due to volume distortion in a clay where a rectangular-shape (flexible category) foundation 6 ft by 12 ft imposes a total loading of 220 kips. The layer of saturated clay (cohesion c is 1,500 psf) extends 12 ft deep and is underlain by rock. 24 Indicate advantages related to performing laboratory compression tests on clay samples obtained from strata underlying a building site for determining soil compression data.

392

25 Briefly review the manner in which laboratory compression tests are conventionally performed so as to obtain load deformation data for a soil. 26 In the conventional laboratory compression test, what is the cause of soil volume decrease? 27 In a laboratory compression test, the void ratio in a soil sample changes from 1.46 to 1.32 when the compressive loading increases from 100 kN/m2 to 200 kN/m2. Calculate the value for the coefficient of compressibility and the compression index for this load range. 28 In a laboratory compression test, the void ratio changes from 1.55 to 1.36 as loading increases through the range from 2,000 to 4,000 psf. (a) What is the value of the coefficient of compressibility for these conditions? (b) What is the compression index for this loading range? 29 Referring to occurrences observed in a typical field or laboratory compression test, explain why a soil mass does not expand to its original volume when loading is released. 30 Explain the difference between soil compression and consolidation. 31 Briefly indicate the difference between a normally consolidated and an overconsolidated soil deposit. 32 (a) A completed laboratory compression test performed on a clay sample obtained during the soil borings phase of a subsurface investigation provides results similar to that illustrated by chapter Figure 14. The “break” in the plotted curve svmax occurs at a compressive pressure of 5,000 psf. The soil sample was obtained at a depth of 15 ft in the soil deposit, and the representative unit weight for the in-place soil is 118 pcf. From this information, determine the OCR for the clay in this deposit. (b) A completed laboratory compression test on a clay sample indicates svmax is 50 kPa. The sample was recovered from a depth of 4 m in a clay deposit where the representative soil unit weight is 16.5 kN/m3. Indicate the OCR ratio for the deposit and the category of its “consolidated age.” 33 A cohesive soil sample obtained from a known normally consolidated clay deposit is found to

Settlement have a liquid limit of 80 percent. Approximately what would be the compression index for this soil? 34 At a planned construction site, a 2-m-thick stratum of normally consolidated clay underlies a surface layer of compact granular soil 3 m deep. The unit weight for the compact granular soil is 20.2 kN/m3. The clay material has a unit weight of 17.6 kN/m3. The groundwater table is very deep. Laboratory testing of the clay indicates an in-place void ratio of 1.35 and a compression index of 0.42. The building planned for the site will create a stress increase of 24.5 kN/m2 at the center of the clay layer. (a) Assume that the foundations for the building will be situated near the surface of the upper compact granular soil layer. Determine the foundation settlement due to primary compression occurring in the clay layer because of the stress increase. (b) Calculate the settlement to be expected if the groundwater table were at the soil surface, and compare to (a). 35 For the foundation loading and soil conditions described in problem 34(a), determine the compression occurring in the clay layer if the clay is overconsolidated, Cr equals 0.09, and: (a) the preconsolidation pressure is 120 kPa. (b) the preconsolidation pressure is 90 kPa. 36 A 7-ft layer of clay is buried beneath a 10-ft stratum of very compact granular soil. Compact sand underlies the clay. The layer of granular soil is composed of material having a unit weight of 130 psf. The clay unit weight is 105 pcf. A laboratory compression test on a sample of the clay indicates a compression index of 0.40 and a natural void ratio of 1.30. A planned building loading will cause a 550-psf stress increase at the middle of the clay layer. (a) What amount of primary compression occurs in the clay layer for the indicated conditions? (b) How much primary compression of the clay layer would result if the groundwater table was at the ground surface (all other conditions remain the same)? (c) How much clay layer compression would occur if the clay was an overconsolidated material, the past maximum pressure was

2,000 psf, and the Cr value was 0.10? Assume a deep water table. 37 Calculate the foundation settlement due to primary compression occurring in a buried clay layer where the following conditions exist: • The groundwater table is very deep. • The thickness of buried clay layer is 6 ft. • An 8-ft-thick layer of compact sand overlies the clay layer and extends to the ground surface. • A deep zone of compact sand underlies the clay layer. • The unit weight of the compact sand is 135 pcf, and the unit weight for the clay is 100 pcf. • The foundation size and loading causes a stress increase of 700 psf at the center of the clay layer. • Laboratory testing of the clay indicates the clay is normally consolidated, the soil void ratio is 1.28, and the compression index is 0.33. 38 A building foundation 3 m by 3 m will impose 1,500 kN total loading onto the surface of a thick stratum of normally consolidated (beneath center) clay soil. Determine the foundation settlement due to primary compression in the clay, using layers that are 1 m, 1 m, 2 m, and 2 m thick, respectively, from foundation level downward. For simplification, use a soil unit weight of 17.5 kN/m3 constant with depth, an in-place void ratio of 1.20, and a compression index equal to 0.38 for each layer. Assume Boussinesq conditions apply. 39 A planned construction site is underlain by a thick deposit of normally consolidated clay soil. A building foundation 6 ft square will be located on the ground surface and carry a total loading of 180,000 lb. Determine the foundation settlement (beneath center) by analyzing the volume changes due to primary compression in layers that are 2, 4, and 6 ft thick, respectively, from the foundation level downward. For simplification, assume a soil unit weight of 115 pcf constant with depth, an inplace void ratio of 1.05, and a compression index of 0.35 for each layer analyzed. Use Boussinesq conditions. 40 Estimate the settlement due to effects of primary compression plus secondary compression

393

Settlement

41

42

43 44

45

394

in a clay stratum 3 m thick where the compression index Cc is 0.33 and the value for e0 is 1.25. Settlement due to primary compression (or consolidation) is computed to be 45 mm and will require approximately 6 years to complete. Calculate the settlement expected after a 15-year time period by summing the primary compression and secondary compression (select a ratio for Cα/Cc from the lower values of the typical range). Approximate the settlement expected after a 25-year period from the effects of secondary compression in a clay layer 10 ft thick where the primary compression settlement is calculated to be 2 in. and will require a period of about 12 years for completion. The compression index for the clay is 0.28, and e0 is 1.15. (Refer to Equation 15.) Describe the events that take place when a saturated clay soil undergoes consolidation (the process of load transferring onto the soil skeleton, etc.). What is the physical meaning of the coefficient of consolidation, cv? At a planned construction site, a 2-m-thick buried clay layer lies beneath a surficial stratum of free-draining granular soil. Free-draining granular soil also underlies the clay layer. Double drainage from the clay layer can therefore occur when construction loads cause consolidation. The coefficient of consolidation for the clay is 0.001 m2/day. Settlement calculations indicate that the clay layer will eventually compress 4 cm (primary compression or consolidation) due to the effect of building loads. (a) How long a time period is required for 90 percent of the estimated settlement to occur? (b) How much settlement occurs in the first 12 months? (c) What time period is required for a settlement of 2 cm? A buried clay layer 10 ft thick is sandwiched between strata of free-draining granular soil so that double drainage during consolidation can occur. Calculations indicate that, because of a planned building loading, an ultimate settlement of 2 in. is expected as a result of primary compression in the clay layer. The coefficient of consolidation for the clay is 0.01 ft2/day.

(a) How long will it take for 90 percent of the estimated settlement to take place? (b) How much settlement will occur in one year? (c) How long will it take for 1 in. of settlement to take place? 46 At a building site, a 3-m-thick clay layer is overlain by a stratum of sand and bounded on the bottom by rock. Calculations indicate that foundation settlement due to primary compression (consolidation) will be 40 mm. The coefficient of consolidation for the clay is 0.055 m2/month. (a) Determine the time period for 90 percent of the primary compression settlement to occur. (b) Indicate the settlement expected one year after construction. 47 A buried saturated clay layer 8 ft thick is bounded on top by a layer of granular soil having a high coefficient of permeability and along the bottom by impervious rock. Calculations for a foundation indicate that 2.5 in. of settlement will occur due to primary compression in the clay. The coefficient of consolidation (vertical drainage) for the clay is 0.02 ft2/day. (a) Determine the period of time for 90 percent of the primary compression settlement to occur. (b) Indicate the settlement (inches) that will occur in the first year after construction. 48 A surcharge and sand drain installation is proposed for an airport project to accelerate the consolidation of a thick deposit of fine-grained soil underlying the construction area. The plan is to install sand drains 1.5 ft in diameter at a staggered spacing so that each sand drain will handle a plan area 15 ft in diameter. Laboratory tests indicate that the coefficient of consolidation for horizontal drainage of the soil is 0.5 ft2/day. Computations indicate that the surcharge load will cause an eventual settlement of 12 in. How long a period will be required for 90 percent of this settlement to take place? 49 A surcharge and wick drain installation is planned as part of a surcharge program for a highway project undertaken to accelerate the consolidation of a fine-grained soil deposit underlying the construction area. The wick drains will be approximately 10 mm by 15 mm

Settlement in cross section. Wicks will be installed in a hexagonal pattern so that each drains an area having an equivalent diameter of 1.25 m (ref. Figure 26); assume an equivalent n ratio to be about 20 for the conditions described. Laboratory tests indicate that the coefficient

of consolidation for horizontal drainage is 2.5 m2/month. Calculations indicate that the surcharge loading will cause an eventual total consolidation settlement of 1 m. How long a period will be required for 90 percent of this settlement to occur?

395

Answers to Selected Problems

7. 10. 12.

13. 14. 16. 17. 20. 27. 32.

396

S L 25 mm S L 15 mm (a) estimated S = 20 to 30 mm (b) estimated S = 14 to 20 mm (c) estimated S = 18 to 25 mm Kv = 32.14 kN>m3 Kv = 240 kcf S = 0.75 in. Svd L 6 mm Svd L 12 mm av = .0014 m2>kN, Cc = 0.465 (a) OCR = 2.8 (b) OCR = 0.76 (since OCR < 1.0, soil is unconsolidated, consolidated age is young)

33. 36.

39. 44.

48.

(a) Spc = 42 mm (b) Spc = 75 mm. (a) Spc = 1.9 in. (b) Spc = 3.2 in. (c) Spc L 1.0 in. Spc = 17 in. (a) t = 860 days (b) S = 27 mm (c) t = 200 days t L 200 days

Shear Strength Theory

From Essentials of Soil Mechanics and Foundations: Basic Geotechnics, Seventh Edition. David F. McCarthy. Copyright © 2007 by Pearson Education, Inc. Published by Pearson Prentice Hall. All rights reserved.

397

Shear Strength Theory

The ability of a soil deposit to support an imposed loading or of a soil mass to support itself is governed by the shear strength of the soil. As a result, the shearing strength of the soil becomes of primary importance in foundation design, highway and airfield design, slope stability problems, and lateral earth pressure problems that deal with forces exerted on underground walls, retaining walls, bulkheads, and excavation bracing.1 In the study of the shear strength of soils, it is common to consider the two major categories of soil types—cohesionless and cohesive—separately. Overall, the factors that can affect the shearing strength of both soil types are the same. Practically, however, the factors that have the most influence on the shear strength that is or will be developed by each soil type are different. The shearing strength and related deformations (or stress–strain relationship) of a foundation or construction soil is conventionally studied in the laboratory by testing soil samples obtained from the construction site, using established testing procedures. Additionally, field test methods have been developed for determining the shear strength of soil in its natural location, for reasons of expediency and economy and sometimes necessity, if samples for testing cannot be obtained. Typically, field determinations of shear strength are quick procedures, so that many soil samples (from different borings or test pits and different depths) can be checked easily and economically. Commonly with field tests, strength values only (no deformation data) are obtained. 1The shearing strength of soil deposits (which we recognize as consisting of an accumulation of discrete particles,

possibly with some degree of interparticle bonding present) almost always refers to the resistance along a plane that passes between or along particle surfaces (but not through the particle). The mineral compositions of soil particles produce materials that have relatively high resistance to compressive and shear forces and that rarely are fractured or sheared when a soil mass “shears.” A plane of resistance passing between or along particle surfaces is weaker than a plane of resistance through the solid particles.

398

Shear Strength Theory

1

Laboratory Tests The most widely used laboratory tests for studying the shear strength and related deformations of soils include the direct shear test (single and double shear), the triaxial compression test, and the unconfined compression test. For cohesive soils, vane shear tests can be used in the laboratory and the field to determine the cohesive strength, or cohesion, of soils (cohesion is related to shear strength). Currently, the most used or preferred laboratory type of strength test is the triaxial test. The unconfined compression test is a category of triaxial test that is appropriate only for cohesive soils. Though the use of the direct shear test has decreased since the development of the triaxial test, it still represents the basic approach for studying the stress–strain characteristics of a soil during shearing and possesses merit as a learning tool. Direct shear tests may represent soil behavior better than triaxial tests for some applications (e.g., some types of foundation designs, retaining wall designs, slope stability analyses). See later discussions in this chapter on plane strain and axisymmetrical strain.

Direct Shear Test The direct shear apparatus for performing single shear is essentially a rectangular or circular box having separated lower and upper halves (Figure 1(a)). After the sample to be tested is placed in the apparatus, a normal (compressive) loading is applied to compress the soil. The upper half of the apparatus is then moved laterally by a recorded shearing force, forcing the sample to shear across the plane between the two halves of the apparatus. The normal force is kept constant during the test. The shearing force starts at zero and increases until the sample fails (is sheared). Usually, a record of the magnitude of the shearing force and the resulting lateral movement is kept so that shearing resistance stress versus shearing displacement can be computed and plotted graphically. Changes in sample thickness that occur during the shearing process are also recorded so that volume change versus shearing stress or shearing strain can be studied. Typically, results of shearing stress versus shearing strain are as shown in Figure 2. The initial portion of the diagram is curvilinear (constantly changing slope), which continues until the maximum shear is reached, after which continuing deformation occurs with no increase in loading—that is, no increase in resistance. Failure is considered to have occurred at the maximum shearing value. In direct shear testing, shearing can be accomplished by either controlling the rate of strain or the rate of stress. For the strain-controlled test, the shearing deformation (lateral movement) occurs at a controlled rate, usually continuously and at a constant speed. With this type of test, the shearing force necessary to overcome the resistance within the soil is automatically developed. In the stress-controlled test, the magnitude of the shearing force is the controlled variable. The force (and therefore the stress) is increased at either a uniform rate or in established increments. For each increment of shearing force, it is applied and held constant until the shearing deformation ceases. The strain-controlled shear test appears to be the most widely used of the two methods, probably because a mechanically operated straincontrolled apparatus is the simplest to devise. Double shear testing is similar to the single shear test, except that two parallel surfaces are sheared. This type of test is usually performed on soil samples obtained from test

399

Dial gauge to measure sample compression or expansion

N, Normal (compressive) loading (constant)

Loading plate

Porous stone Soil sample

Shearing plane Shearing force, S (variable)

Porous stone Dial gauge to measure shearing displacement, δ (a)

Dial gauge for measuring change in sample thickness

Proving ring for measuring shearing force

Soil sample in shear box

Gearbox and crank for applying shearing force

Weights for imposing normal force

(b) Figure 1 (a) Schematic diagram of direct shear apparatus; (b) laboratory direct shear equipment (manually operated). (Courtesy of ELE International)

400

Shear Strength Theory

(c) Figure 1 (continued ) (c) State-of-the art electronic direct shear apparatus. (Courtesy of ELE International)

Shear stress, τ τ = Shear force = S Sample area A

+

+ Shearing displacement, δ

Figure 2 Typical plotted representation of shearing stress versus shearing displacement data.

borings that have utilized special tube equipment designed to fit directly into a shearing apparatus, so that a minimum of sample handling is required. As an illustration of one method, the soil-sampling apparatus used to extract samples from test borings is such that the soil is obtained in a tube whose interior lining consists of a series of rings. In the laboratory, three rings of soil are taken for the shear test. The shearing takes place as indicated in Figure 3. The stress–strain data from the double shear test are similar to the data from the single shear test. A variation of the direct shear test is a torsional test, where one section of the tested sample is twisted relative to the other. The torsional resistance developed on the failure plane is related to the shear strength of the soil. This test is not in common use. Similarly, some past studies have been made using an extrusion-type test for cohesive soils, but such tests are not in common use.

Triaxial Compression Test The triaxial test is currently the most popular test for determining the shearing strength of soils. Though not as simple a test as the direct shear test, it has several advantages that are of practical importance: (1) the loading conditions in the triaxial test can be made to simulate

401

Shear Strength Theory Shear force, S, applied to middle ring

Normal (compressive) force held constant during shearing

S

Soil ring

Shearing plane

Soil ring

Outer rings clamped to prevent movement Soil ring

N = normal force

Shearing plane

Dial gauge to measure shearing displacement, δ

Figure 3 Schematic diagram of double-ring shear.

more accurately the loading conditions that the soil was (or will be) subjected to in its natural state; and (2) failure is not forced to occur across a predetermined plane, as occurs for the direct shear test. In the triaxial test, a cylindrical soil sample (as conventionally obtained in soil borings) is wrapped in an impervious membrane for protection and placed in a chamber where an all-around, or confining, pressure can be applied. The confining pressure is usually applied through water or air introduced into the sealed chamber. The sample sits on a fixed pedestal, and a cap attached to a vertical piston rests on the top of the sample. In testing, the confining pressure is applied all around and to the top of the sample and, usually, held constant. An axial (vertical) load is subsequently applied to the sample through the piston, which passes through the top of the chamber. The axial load is steadily increased until failure of the sample occurs. Figure 4(a) shows the triaxial testing arrangement schematically. Analysis of a three-dimensional element within the tested triaxial specimen shows that a lateral normal stress equal to the confining pressure pc acts on orthogonal vertical planes, and a vertical stress acts on the horizontal plane. The vertical stress is equal to the sum of the confining pressure pc plus that pressure increase resulting from the axial loading, Δp (Figure 5). In practical problems, the confining pressure is frequently selected to be equal to the confining or lateral pressure that acted on the soil in its natural location within the soil mass, and the axial pressure that eventually causes the sample to fail is assumed to be representative of the maximum vertical pressure that can be imposed on the soil at the depth at which the confining pressure acts. By relating test pressures to subsurface stresses, the confining pressure represents s3, the minor principal stress, while the total axial stress represents s1, the major principal stress. In the triaxial test setup, porous discs or stones are placed at the top surface and bottom surface of the sample so that pore water in the sample can be drained, if and when desired. The pedestal and loading cap are provided with drainage tubes for the water to escape and from which the pore water pressure can be determined. If the problem under study requires, water can also be introduced into the sample to determine the effect of increased water content on the strength. The numerical difference between the total axial pressure s1 and the confining pressure s3 in the triaxial test is termed the deviator stress. When stress–strain data are

402

Shear Strength Theory Loading piston for applying vertical force To pressure gauge Pressure supply (air or water)

Air or fluid in chamber, under pressure

Top loading plate Porous stone Tie rod

Flexible, water-tight membrane around soil sample

Soil sample

Cylinder (lucite, or metal for high pressure testing) Porous stone

Base

(a)

Triaxial chamber for soil sample

Motor and mechanism for applying axial load

Control panel for confining pressure and soil pore water pressure determinations

(b) Figure 4 (a) Schematic diagram of triaxial compression test apparatus; (b) laboratory triaxial equipment (dual-chamber, motorized). (Courtesy of ELE International)

graphed, it is the deviator stress that is conventionally plotted against strain. Typical qualitative results are shown in Figure 6. For axially loaded test specimens, the related slope of the stress–strain curve is the modulus of elasticity E. The modulus of elasticity is an indication of the stiffness or resistance

403

Shear Strength Theory Pc + Δ p

Pc

Pc Pc + Δ p

Deviator stress σ1 − σ3

Deviator stress σ1 − σ3

Figure 5 Stress combination acting on incremental element of soil subjected to triaxial testing.

Axial strain εx

Axial strain εx

(a)

(b)

+

Ei = tangent modulus of elasticity (initial)

Deviator stress σ1 − σ 3

Et = tangent modulus (general) Es = secant modulus (general)

x

+ Axial strain εx (c)

Figure 6 Representative triaxial test data: (a) typical result for loose sands and normally consolidated clays; (b) typical result for dense sands and overconsolidated clays; (c) indication of common terms used to relate stress and strain.

to deformation of the material; the higher the value of E, the stiffer or stronger the material. A tangent modulus is the slope of a line drawn tangent to a point on the curve. A secant modulus is the slope of a line connecting any two points on the stress–strain curve. By reference to Figure 6, it should be clear that a tangent modulus or a secant modulus will not be constant for all parts of the curve. Both the tangent modulus and secant modulus are used in problem analysis.

Plane Strain and Axisymmetrical Strain The condition of plain strain is closely approximated by the typical laboratory direct shear test where strain during shearing can occur only in the axial direction (direction of the test compressive loading) and one lateral direction (the direction of the applied shearing force)

404

Shear Strength Theory

but not in the transverse lateral direction. Geometrically, plane strain is considered a twodimensional condition (no deformation occurs in the direction extending at a right angle to the displacement). Practical applications of the plane-strain condition include the analysis of soil bearing capacity for long wall footing foundations, determining the lateral force of the soil mass behind long retaining walls, and studying the landslide condition for long sections of ground having a sloped topography. The triaxial test permits the soil particles freedom to move in three directions as loading is applied, and represents the axisymmetrical-strain condition. The axisymmetrical-strain condition applies where the shear-related movement in a stressed soil zone is geometrically three-dimensional, such as occurs during development of the bearing capacity for foundations which have similar plan-view crosssectional dimensions (i.e., square, rectangular, and circular shape footing foundations and pile foundations), or where a landslide occurs across a short section of a sloped area. Application of test data to field projects should consider that laboratory-determined values of shear strength from plain-strain tests are typically slightly greater than obtained with axisymmetrical types of tests (reflecting the greater liberty for particle movements and lesser particle interlocking that results when directional strains are not restricted).

Unconfined Compression Test The unconfined compression test (Figure 7) is a triaxial compression test but a type where the all-around confining pressure is zero. The axial force represents the only source of external pressure imposed onto the soil. Because it is necessary that the soil sample under test be capable of standing in the testing apparatus under its own internal strength, the test is limited to use for soils possessing cohesion. Test results are generally similar to conventional triaxial test results.

Vane-Shear Test Vane-shear tests are coming into wider use for determining the in situ strength of cohesive soils. The vane-shear apparatus consists of thin-bladed vanes that can be pushed into the soil with a minimum of disturbance. A torque applied to rotate the vanes is related to Axial stress =

Axial load, P (variable)

P Axial load = Sample area A

Top loading plate Soil sample

Unsupported soil sample

Lateral stress = 0

Bottom loading plate or base

Dial gauge for measuring axial deformation (optional)

P=0

P=0 Axial stress =

(a)

P A

(b)

Figure 7 Representation of unconfined compression test: (a) test arrangement; (b) stresses acting on incremental element.

405

Shear Strength Theory

(c)

Figure 7 (continued ) (c) Unconfined compression apparatus. (Photo courtesy of ELE International) Torque

L = 10D

H = 2D

D

Field test procedure (refer ASTM D2573) 1. Insert vane into undisturbed soil at bottom of borehole. 2. Rotate vane to record torque related to determining the value for the undisturbed undrained shear strength, su or cu; rotate at speed of 0.1 degrees per sec. 3. Obtain remolded shear strength by rotating vane about 10 times, then record a final torque value (used to calculate the remolded shear strength). Calculation Procedure Undrained shear strength, vane shear test ⫽ su or cu ⫽ T/K where T is the vane test torque, in N-m or lb-ft and K ⫽ ␲(D2H/2) [1 ⫹ D/(3H)] . . . for D, H in meters or K ⫽ (␲/1,728)(D2H/2) [1 ⫹ D/(3H)] . . . for D,H in inches Sensitivity, St ⫽ (undisturbed shear strength)/(remolded shear strength).

Figure 8 (a) Vane of vane-shear apparatus, for determining in-place shear or cohesive strength (also ref. Figure 5-27); (b) vane geometry and procedure to determine in-place shear strength.

the shear strength of the soil. A vane-shear apparatus, as pictured in Figure 8, can be attached to a long vertical rod and inserted into borings, so that the in-place strength of the soil can be determined without removing the soil from its natural state in the ground. Similar equipment is also used in the laboratory. It is specifically useful for determining

406

Shear Strength Theory

.25 mm/ (1 inch) φ

Figure 9 Torvane for determining shear or cohesion of soil in field or laboratory.

the cohesive strength of soft or sensitive clays that could possibly be affected by the handling that the conventional laboratory shear tests require. The procedure for performing a vane-shear test to determine the in-place soil shear strength is outlined in Figure 8. The shear strength calculated by the basic equation shown typically will overestimate the actual shear strength available. An adaptation of the vane-shear apparatus is the torvane (Figure 9), used for determining the cohesive strength of samples in the field or laboratory.

Shear Test Results Plotted on Mohr’s Circle Coordinates Results of the direct shear and triaxial tests can be plotted graphically (or semigraphically) on a shear stress versus normal stress coordinate system. They are the same coordinates used for Mohr’s circle plots. The stress combination that represents the condition at the maximum shearing strength or the ultimate shearing strength (at failure) is generally used for making the plot. For the direct shear test (Figure 10(a)), the results of each test would plot as a point, as indicated in Figure 10(b).

τf

τf

Stress–strain result when normal load is N1

Shear strain, Ω (a)

Shear stress, τ

Shear stress, τ

2

σ1 =

N1 Area of sample

σ1 Normal stress, σ (b)

Figure 10 Direct shear results for one test: (a) stress–strain plot; (b) result of test on normal stress–shear stress coordinates.

407

Shear Strength Theory

σ3 is known (held constant) for the test; σ1 varies (increases) until sample fails

Mohr's circle is drawn knowing σ3 and σ1 − σ3

Shear stress, τ

Deviator stress σ1 − σ3

Maximum deviator stress

σ1 σ3

Axial strain, εx

(a)

Normal stress, σ

(b)

Figure 11 Triaxial results for one test: (a) stress–strain plot; (b) result of test on normal stress–shear stress coordinates.

With triaxial test results, the values of s3 and s1, or the principal stresses, are known, and plotting these points permits the Mohr’s circle for these stress conditions to be drawn (Figure 11(b)). If identical homogeneous soil samples were tested under identical conditions in a direct shear test and a triaxial test, the same strength information, for practical purposes, would be obtained. Where the results from a direct shear test are superimposed on the results from a triaxial test, the plotted point of τ versus s appears as shown in Figure 12(a) or (b).

Strength or Failure Envelope

Result from direct shear test

σ3

Mohr’s circle from triaxial test σ1

Shear stress, τ

Shear stress, τ

If a series of direct shear tests were performed on different samples of homogeneous soil and the normal loading were different for each test, the results plotted on tf versus s coordinates could be as shown in Figure 13. A curve passing through the plotted points establishes what is referred to as the strength or failure envelope curve. This envelope indicates the limiting shear strength that a material will develop for a given normal stress. Since the strength or failure envelope is a boundary established from test results, it should be evident that it is physically impossible to achieve a stress combination whose coordinates plot above the boundary. The same information can be obtained from a series of triaxial tests performed on a homogeneous soil. In each test the confining pressure s3 is different. This consequently Result from direct shear test

σ3

Mohr’s circle from triaxial test σ1

Normal stress, σ

Normal stress, σ

(a)

(b)

Figure 12 Result of direct shear and triaxial test plotted on same coordinates: (a) general location of τ versus σ point from direct shear test on cohesionless soil; (b) general location of τ versus σ point from direct shear test on cohesive soil.

408

Shear Strength Theory "Strength (or failure) envelope" Shear stress τf

(4) (3)

Test number

(2) (1)

Normal stress, σ σ=

Results of direct shear tests on samples of a homogeneous soil.

Shear stress τ

Figure 13

Normal loading during test Area of sample

"Strength (or failure) envelope"

Test 3 Test 1 σ3

σ1

Test 2 σ3

σ3

σ1

σ1

Normal stress, σ

Figure 14

Results of triaxial test on samples of a homogeneous soil.

affects the maximum axial stress s1 that develops at failure. The Mohr’s circle test results would be as shown in Figure 14. A curve that is tangent to all the Mohr’s circles establishes the strength or failure envelope curve. The point of tangency to each circle establishes the combination of shear and normal stresses that act on the failure plane in the tested soil sample. These should be the same combinations of shear and normal stresses obtained from direct shear tests. From reference to the Mohr’s circle plot, it should be understood that the plane of failure does not necessarily occur on the plane where the shear stress is a maximum. Rather, it is the combination of shear and normal stress developing within the material that is critical. The combination that is critical relates to the slope or angle of the failure envelope curve. If the strength (failure) envelope is horizontal (parallel to the normal stress axis), the maximum shear stress indicated by the Mohr’s circle is the shear stress on the failure plane. Where the envelope curve makes an angle with the horizontal axis, the shear stress on the failure plane is less than the maximum shear stress represented by a Mohr’s circle. In soils, the condition of both the horizontal and the sloped strength–failure envelope exists. This is discussed further in Section 3. Quite frequently in soil analysis, the Mohr’s circle plot requires use of only the upper half of the circle, the lower half being considered redundant because it is a mirror image of the upper half. However, the entire circle is best shown when strength–failure envelopes are discussed so as to clearly establish the concept of boundaries for the nonfail and fail combination of stresses (see Figure 15). For strength conditions indicated by the Mohr’s circle of Figure 15, it should be seen that an incipient failure situation develops on two different planes simultaneously. However, when it is desired only to learn the combination of principal stresses that will

409

Shear Strength Theory τ Stress combinations which theoretically plot outside the strength envelope boundaries represent a failure condition

Stress combinations which plot within boundaries of the strength envelope represent a stable or safe (not failed) condition

σ

τ

Boundaries of strengthfailure envelope

σ

Incipient failure

Figure 15 Concept of strength–failure envelope showing combinations of stress representing failure and stable conditions.

cause failure or the magnitude of shear and normal stresses acting on the failure planes, as is usual, using only half of the Mohr’s circle is sufficient.

3

Shearing Strength Basic understanding of the shearing strength of cohesive and cohesionless soils can be obtained by reference to the results from triaxial or direct shear tests. A discussion relating to the more simple direct shear test is easiest to follow, and is presented first. The behavior described, and the explanation of the factors affecting the behavior, apply to triaxial conditions as well as to direct shear conditions. A discussion of triaxial testing and related results is presented later in this chapter.

Shearing Strength of Cohesionless Soil The results of a direct shear test performed on a dry cohesionless soil (e.g., a dry sand), where the normal load is held constant during the shearing process, are presented on stress–strain coordinates. Typical results are indicated in Figure 2. The actual configuration of the stress–strain curve will be affected by the size and shape of the soil particles and by the density of the sample at the beginning of the test, as well as by the magnitude of the normal loading. A sample that is initially loose will develop a stress–strain curve as indicated in Figure 16(a), whereas an initially dense sample will present a curve as shown in Figure 16(b). The ultimate strength for samples whose only difference is the initial density will be the same. The difference in the behavior is explained by the factors that contribute to the shearing strength of dry cohesionless soil. Resistance to movement across the failure plane, and hence the shearing strength, is developed from friction that occurs between particle surfaces under the applied normal loading and to interlocking between particles. The extra shearing resistance for the initially dense sample (difference between peak strength and ultimate strength shown in Figure 16(b)) is attributed to a greater degree of interlocking. This greater interlocking is overcome as shearing displacements increase, indicating that the initially dense sample loosens during

410

Shear Strength Theory Peak strength

Same ultimate strength Shear stress, τ

τp τf

τf Ultimate strength

Ultimate strength

Shear strain

Shear strain

(a)

(b)

Figure 16 Comparison of shear test results as affected by soil density: (a) cohesionless soil, sample initially loose; (b) cohesionless soil, sample initially dense.

shearing. This is verified by recording the thickness (or volume) of the sample during testing and observing that a thickness increase occurs. In other words, the void ratio of the initially dense sample increases during shearing. Conversely, the thickness of the initially loose sample decreases during testing, indicating a decrease in the void ratio. At the ultimate strength, after shearing movement, the final void ratio will be similar for the initially dense and loose materials, regardless of the initial void ratio. This final void ratio is referred to as the critical void ratio. Soils whose natural void ratios are above the critical void ratio will attempt to decrease in volume during shearing, whereas the reverse is true for soils whose natural void ratio is below the critical ratio (Figure 17). If a series of shearing tests are performed on identical samples, but the normal loading that acts during shearing is different, the ultimate shearing resistances (and peak resistances if the soil is initially dense) will also be different. Greater shearing resistances, or shearing strengths, are obtained at higher normal loadings (Figure 18). If the results of the series of direct shear tests are plotted on shear stress versus normal stress coordinates, as for a Mohr’s circle analysis, a strength–failure envelope curve for the soil is obtained (Figure 19). Through the range of normal stresses usually encountered in foundation problems, the plotted points of t versus s establish a slightly curved line that for many practical applications can be satisfactorily represented as a straight line. The angle that the strength–failure envelope creates with a horizontal line is indicated as f, the angle of internal friction for

Void ratio, e

Initially loose soil

ecr = critical void ratio

ecr

Initially dense soil

Shearing strain

Figure 17

Effect of initial density on change in void ratio during shearing.

411

Shear Strength Theory Na

Test a

Nb

Sa

Nc

Sb

Test b

Test c

Sc

Shear stress, τ

Nc > Nb > Na (Sc, Sb, Sa are ultimate shearing forces) τf c

τfb

τfa τfa =

Sa A

τfb =

Shear displacement

Figure 18

Sb A

τfc =

Shear displacement

Sc A

Note: Area, A, is cross-sectional area of the shear sample

Shear displacement

Effect of normal loading on shear strength of dry cohesionless soil.

Shear stress, τf

Strength–failure envelope curve τc τb τa φ σa

σb

σc

Normal load, N Normal stress, σ = Area of sample, A

Figure 19 Summary representing maximum shear strength versus applied normal stress for dry cohesionless soil.

the soil. For dry cohesionless soils, the strength–failure envelope obtained from ultimate shear strength values is assumed to pass through the origin of coordinates. Thus, the relation of shearing strength to normal stress can be calculated from: tan f =

t s

or: t = s tan f

(1a)

If the peak values of shear strength are plotted, values of f somewhat greater than obtained from the ultimate shear strength values will be obtained. Representative values for f are presented in Table 1. The value of f (ultimate or peak) selected for use in practical soil or foundation problems should be related to the soil strains that are expected. If soil deformation will be limited, using the peak value for f would be justified. Where deformations might be relatively great, ultimate values of f should be used. The typical laboratory direct shear procedure represents a plane-strain (two-dimensional) condition of testing, and the triaxial test procedure represents a three-dimensional or axisymmetrical-strain condition. The difference in the conditions of constraint (or freedom for particle movements to occur) during testing does affect results. Values for the angle of

412

Shear Strength Theory Table 1 Representative Values of f for Cohesionless Soils Angle f (degrees) Soil Type

Ultimate

Peak

33–36 32–35 29–32 27–32 26–30

40–50 40–50 32–35 30–33 30–35

Sand and gravel mixture Well-graded sand Fine-to-medium sand Silty sand Silt (nonplastic)

internal friction determined from direct shear tests commonly are 1 degree to 4 degrees greater than indicated by triaxial testing but larger differences can occur. Illustration 1 A sample of dry sand is tested in direct shear. A normal load equivalent to 96 kPa is imposed for the test. The shearing force applied to fail the sample is increased until shearing does occur. The shear stress at failure is 65 kPa. What is the angle of internal friction f for the sand?

Solution 65 kPa = 0.677 96 kPa s = 34° ;

τ, shearing stress, kPa

tan s =

65 N = 96 kPa

Test results on stress-displacement coordinates

τ, shearing stress, kPa

Shearing displacement

65

Test information plotted on Mohr's circle coordinates

φ 96 Normal stress, σ, kPa

Illustration 2 A dry cohesionless soil is tested in a triaxial test to determine the angle of internal friction f. A confining pressure equal to 1,000 psf is used. The sample fails when the axial load causes a stress of 3,200 psf. What is the value of f?

Solution 3200 - 1000 = 1100 1psf2 2 Center of circle = 1000 + 1100 = 2100 1psf2

Radius of circle =

413

Shear Strength Theory Stress values plotted on Mohr's circle coordinates Point where strength– Shear failure envelope stress, τ (psf) is tangent to Mohr circle

σ1 = 3,200 psf

σ3 = 1,000 psf

R= 1,100

φ σ3

σ1

Normal stress, σ (psf)

2,100 σ3 = 1,000 σ1 = 3,200

From the Mohr’s circle plot: 1,100 psf = 0.525 2,100 psf f = 31.5° ;

sin f =

Shear stress, τ

If moisture is present in the soil samples during shearing, the subsequent plotting of ultimate shearing strength versus the applied normal stress does not present a failure envelope curve that passes through the origin. Rather, at zero normal stress, the failure envelope intersects the shear stress coordinate, as shown in Figure 20. This indicated “no load” shearing strength is referred to as apparent cohesion, a shear strength value attributed to factors other than friction developed from the normal stress. With damp cohesionless soils, the extra strength is due to compressive forces exerted on soil particles as a result of surface tensions where water menisci have formed between soil particles (as in capillary water). The extra shearing strength exists as long as the soil retains some moisture. The extra strength would be lost if the soil were to dry or to become saturated or submerged. For the saturated and submerged case, all voids are filled with water, and all water menisci are lost. For these reasons, the extra shear strength attributed to apparent cohesion generally is neglected in foundation studies. Values of f are not significantly affected when the soil is below a groundwater table or otherwise submerged. However, any hydrostatic pressure u that acts on a plane under analysis reduces the intergranular compressive stress (and will affect the soil shear strength). The resulting effective stress, s (equal to ␴t - u ) Strength–failure envelope Value of apparent cohesion

Normal stress, σ

Figure 20 Results of shear tests on moist cohesionless soils indicate an “apparent cohesion” value. [Note: The slope of the strength—failure envelope line as illustrated is different than the result from testing the soil in the dry condition.]

414

Shear Strength Theory

is utilized to obtain a preferred general expression for the shear strength of a cohesionless soil, whereby: t = 1st - u2 tan f = s tan f

(1b)

where st is the total stress acting at a point. When u is zero, s equals st and Equation 1a, derived for the case of a dry cohesionless soil, results. The angle of internal friction determined from effective stress analysis or testing is often indicated as fdrained or f′. Illustration 3 (a) Samples taken from a uniform deposit of granular soil are found to have a unit weight of 19.6 kN/m3 and an angle of internal friction of 35°. What is the shearing strength of the soil on a horizontal plane at a point 4 m below the ground surface?

Solution At a depth of 4 m, the soil overburden pressure, or normal stress, is: 119.6 kN>m3214 m2 = 78.4 kN>m2 = s on horizontal plane The shearing resistance that can be developed is: t = s tan f = 178.4 kN>m221tan 35°2 = 54.9 kN>m2 (b) A proposed structure will cause the vertical stress to increase by 60 kN/m2 at the 4 m depth. Assume that the weight of the structure also causes the shearing stress to increase to 52 kN/m2 on a horizontal plane at this depth. Does this shearing stress exceed the shearing strength of the soil?

Solution The total vertical pressure due to the structure and soil overburden is: 60 kN>m2 + 78.4 kN>m2 overburden pressure = 138.4 kN>m2 The shearing strength that can be developed by the soil at this depth is: t = 1138.4 kN>m221tan 35°2 = 96.9 kN>m2 This would indicate that the shear strength of the soil is greater than the imposed shear stress; therefore, a shear failure does not occur 196.9 kN>m2 7 52 kN>m22. If the water table rose to the ground surface, the effective soil overburden pressure would be reduced to about: 112 * 19.6 kN>m3214 m2 = 39.2 kN>m2

[This value represents the effective vertical stress, i.e., s = st - u = gsubZ = 1gt - gw2Z.] The total vertical stress would be: 39.2 kN>m2 + 60 kN>m2 = 99.2 kN>m2 The shear strength available is: t = 199.2 kN>m221tan 35°2 = 69.46 kN>m2 This is still greater than the shear stress resulting from the loading conditions; that is, 69.46 kN>m2 7 52 kN>m2.

415

Shear Strength Theory

The preceding discussion has indicated that the shearing strength that can develop is directly proportional to the effective normal stress that acts on the plane under analysis. For most practical problems, the effective stress in cohesionless soil is calculated by using the effective overburden weight plus any stress increase created by structural loading. However, if the soil is saturated, or nearly so, there is the possibility that the stress resulting from loads newly applied onto the soil mass will not result in a related increase to the effective stress acting within the soil. This is the situation of an excess hydrostatic pressure condition developing (pore water in the soil voids becomes subject to a pressure greater than the normal hydrostatic). With this situation, there is the potential danger that the shearing stress resulting from the external loading may increase faster than the soil shearing strength (which is controlled by the effective stress), and a shear failure may occur. Fortunately, cohesionless soils have relatively high rates of permeability that under most conditions permit rapid drainage of pore water when new loadings and stresses are imposed onto the soil mass. The danger of excess hydrostatic, or excess pore pressure, conditions in cohesionless soils and possible shear failures are generally limited to the situation where loose saturated material is exposed to vibratory, instantaneous, or shock loading, such as from explosives, earthquakes, and traveling trains. The occurrence of loss of strength under these conditions is called liquefaction, for the soil momentarily liquefies and tends to behave as a dense fluid. Generally, for sandy soils, the number of vibrations or shock waves required for liquefaction to occur increases as the soil density increases, but the presence of fine-grained material decreases the susceptibility to liquefaction. The soils found to be most susceptible to liquefaction are the saturated and loose fine-to-medium sands having a uniform particle-size range.

Approximating Values of f from Site Investigation Data (SPT, CPT, DMT) Soil borings for subsurface investigations are frequently sized so that soil samples can be obtained with a “standard” two-inch- or 51-mm-diameter split-spoon soil sampler. With the split-spoon sampler, the blow count required to drive the sampler into undisturbed soil is recorded for all samples obtained. The samples recovered with this equipment are considered disturbed and are unsuitable for performing strength tests. Obtaining undisturbed samples of cohesionless soil is generally difficult under many existing subsurface conditions and may require the use of large-diameter borings and special soil samplers. However, as a result of the widespread usage of the split-spoon sampler (partially because of the relative economy) and related logging of soil conditions from the recovered samples, broad correlations between the blow count N60 (the standard penetration test) and the angle of internal friction f and relative density for sands have developed. Similarly, correlations between cone penetrometer soundings and soil properties have been developed. Generalized correlations are presented in Table 2; because of the broad range of the reference data, the soil properties and conditions should be considered only as order-of-magnitude values. Equations relating N60 blow count values (SPT) to soil properties have been developed (empirically derived, using field and lab test data as the basis), as have expressions that

416

Shear Strength Theory Table 2 Approximate Relationship between Condition of Cohesionless (sand) Soil and Results of Subsurface Investigation Procedures* Range for Relative Density, DR (pct)

General Description

Value SPT (N)**

Approximate Value for f (degrees)

Value CPT (qc/patm)

620 20–35 35–65 65–80 780

Very loose Loose Medium dense to dense Dense Dense to very dense

64 4–10 10–30 30–50 750

630° 30–35 34–40 38–45 745

620 20–40 40–120 120–200 7200

*Data

compiled from various sources. This data is for estimates or order-of-magnitude only, and should not be used for designs (for design work, use results from in-place or laboratory tests). **In the so-called standard penetration test, N is the number of blows required to drive a standard 2-in.-outside-diameter (51 mm) split-barrel soil sampler 12 inches (.3 m) into undisturbed soil with a 140-lb weight falling 30 inches (or 63.5 kg with a 0.76 m drop). Values in this table refer to soil sampling procedures where the efficiency of the drop hammer is approximately 60%. For further information on the standard penetration test, refer to Chapter 5.

relate qc values (CPT) and soil properties. Generally, relationships developed for sandy soils as shown below are rated to predict values that are of approximate nature only. For SPT:

f L tan-151N602>312.2 + 20.31svo>patm2460.34

(2a)

For CPT:

f L 17.6 + 11 log31qc>patm2>1svo>patm20.54

(2b)

Associations between N60 values and qc values also exist, and provide a method to translate soil property information in terms of qc values to N60 values (and the reverse); the equations shown require that soil particle size distribution information be known: 1qc>patm2>N60 L 4.25 - 1F>41.32

(3a)

where F = the percentage of soil particles smaller than 0.075-mm (the Number 200 sieve) patm = one atmosphere, 101 kPa, or 2.1 ksf and 1qc>patm2>N60 L 5.44 D50 0.26

(3b)

where D50 corresponds to the 50 percent finer particle size. The Marchetti flat-blade dilatometer is used to determine in situ soil properties. The angle of internal friction, f, has been correlated to the dilatometer parameter KD (the Horizontal Stress Index value), see Figure 21. The presentation of a band range on this diagram reflects the influence of lateral earth pressure conditions existing in the soil deposit (refer to Chapters 9 and 17 for information discussing the coefficient of lateral earth pressure);

417

Shear Strength Theory

Value for ␾°, angle of internal friction

45

40 Expected range for ␾ (affected by Ko value) 35 Conservative value for ␾ (use for preliminary designs, estimates) (␾° = 28 + 14.6 log KD – 2.1 log2 KD)

30

25 0.5

Figure 21

1

5 10 2 Value of horizontal stress index, KD (for uncemented sands)

20

Values of f related to dilatometer test (DMT) results [220].

typically, the higher values for f are obtained where the overconsolidation ratio (OCR) and lateral pressure are high, as for dense and well-compacted sands. The shaded band also indicates that the KD - f relationship is not established for the lower end of the range. However, the solid line curve on the diagram represents the lower boundary for the general KD - f relationship, and can be used for conservative analysis and estimations. After the value for f has been obtained, K0 and the OCR can be calculated.

Shearing Strength of Clay Soils The shearing strength that a clay deposit possesses is related to the type of clay mineral and the water content but, very importantly, also to the effective stress or consolidation pressure to which the soil has been subjected in its past (the soil’s stress history). The change that is possible in a clay’s shear strength (such as if loading conditions change because of the weight of a new structure) is affected by these just-mentioned factors but also by the pore water drainage that can occur as shearing deformations tend to occur. Consideration of drainage is of practical importance, because many clays in their natural condition are close to full saturation, and the low permeability of these soils tends to inhibit changes in pore water content that try to occur during shearing.2 2Strictly

speaking, stress history and drainage factors can also affect the shearing properties of a cohesionless soil, although shape and size distribution of particles have very considerable influence. A granular soil that had been subjected to high effective stresses or high overburden pressures in its past would be dense and capable of developing a peak strength, as shown in Figure 16(b). The prevention of pore water drainage during shearing of a saturated soil would lessen the shearing strength, but under most practical conditions there are no restrictions on drainage because of the relatively high permeability; a major exception relates to effects of seismic waves as occur during earthquakes and explosions.

418

Shear Strength Theory Ground surface

Uniform deposit of clay soil, homogeneous and isotropic Unit weight = γ Overburden pressure on soil samples = γZ

Figure 22

Depth = Z

Source of all test samples

This clay deposit is normally consolidated. The existing soil overburden pressure represents the maximum pressure that the soil has been subjected to in its history.

Source of samples for analysis of shear strength of clay soils.

The significance of effective stress and drainage during shearing can be explained by reference to a series of shearing tests made on fully saturated clays. Practically, the discussion also applies to nearly saturated clays. Assume that the soil for all the test samples is obtained from one location in a homogeneous, isotropic, normally consolidated clay deposit. All properties of all samples to be tested are identical (Figure 22). Samples are to be tested in direct shear. In each series of direct shear tests, each test sample will have a different normal load applied during shearing to determine the effect. The normal loads are greater than the effective overburden (or consolidation) pressure. For the first series of tests, no drainage of pore water occurs, either when the normal load is applied or during shearing (relating to clay’s very low coefficient of permeability). This “no drainage” control can be visualized by assuming the sample is wrapped in a flexible impermeable membrane when in the shearing apparatus. Actually, the normal load is applied just prior to beginning the shearing process and shearing is then completed relatively quickly. These test conditions are referred to as unconsolidated–undrained (U–U) shear tests, because the samples are not permitted the time or the drainage necessary to consolidate to the pressure exerted by the normal loading, and practically, no drainage or volume change takes place during the shearing process. For this condition, excess pore water pressures develop in the samples during shearing. Results for this series of tests are shown in Figure 23. For the U–U conditions, the shearing strengths for all samples are the same, since the effective stress within each sample has not changed from the condition that existed when the soil was in the ground. Even though different normal loads are applied during shearing, consolidation to the normal load pressure can only occur if pore water is permitted to drain from the sample. Since drainage and consolidation does not occur, the internal effective stresses are the same for each of the samples, and therefore, the shearing strengths will remain the same. For the second series of tests, different normal loads are applied to each test sample and full consolidation to the respective new normal load is permitted. After full consolidation has taken place so that the effective stress acting is equal to the applied normal loading, the samples are placed in the shearing apparatus and shearing is performed quickly. These test conditions are referred to as consolidated–undrained (C–U), indicating that the samples are consolidated to their respective normal load prior to shearing but that during shearing no drainage or volume change occurs. For this condition, excess pore water pressures develop in the sample during shearing, but they are not as great as those that occur in the U–U test. Results for such a series of tests are shown in Figure 24. The increase in

419

Shear Strength Theory Na

Nb

Nc

Sb

Sa

Sc

Shear intercept is the value of cohesion, c

Failure envelope

S A

Area, A, is crosssectional area of sample upon which the normal load, N, acts.

τ=

Shear stress, τ

Nc > Nb > Na (Sc, Sb, Sa represent ultimate shearing forces)

σa

σb

σc

Normal stress, σ N σ= A

Figure 23

Shear strength results for unconsolidated–undrained (U–U) tests. Na

Nb

Nc

Sa

Sb

Sc

S A τ=

Shear stress, τ

Nc > Nb > Na (Sc, Sb, Sa represent ultimate shearing forces) Failure envelope

Sc A Sb A Sa A

Area, A, is crosssectional area of shear sample.

σa

σb

σc

Normal stress, σ σ= N A

Figure 24

Shear strength results for consolidated–undrained (C–U) tests.

shearing strength (compared to the U–U conditions) is the result of the increased effective pressure to which the test samples have been consolidated. For the third series of tests, the samples are fully consolidated to the normal loads that will be applied during the shearing test. However, in this series, shearing will take place very slowly, and drainage and volume changes are permitted during the shearing process. These test conditions are referred to as consolidated–drained (C–D) conditions. No excess pore water pressures develop in the soil during shearing. Results for such a series of tests are shown in Figure 25. The results from the three series of tests are presented together in Figure 26. A difference in the C–D and C–U curves results because of a normally consolidated soil’s tendency to undergo volume decrease during shearing. If volume change is prevented, as in the undrained test, slight excess pore pressures are developed, with a subsequent reduction in the effective stress and shear strength.

420

Shear Strength Theory Na

Nb

Nc

Sa

Sb

Sc

Failure envelope

Sc A

Area, A, is crosssectional area of shear sample.

S A

Sb A

τ=

Shearing stress, τ

Nc > Nb > Na (Sa, Sb, Sc represent ultimate shearing forces)

Sa A σb

σa

σc

Normal stress, σ N σ= A

Figure 25

Shear strength results for consolidated–drained (C–D) tests.

Shear stress, τ

φCD φCU

D

C–

U C–

U–U

c A σa

σb

σc

Normal stress, σ

Figure 26

Qualitative comparison of shear strength results for U–U, C–U, and C–D tests.

As for tests on cohesionless soils, the slope of the t versus s curve is designated by the angle f. The stress represented by point A in Figure 26 is approximately the value of the effective stress svc that acted on the soil in its natural subsurface location. If the soil has been normally consolidated, svc is equal to the soil overburden pressure (svc = gZ; see Figure 22). If C–U or C–D tests were performed and the applied normal loading was less than svc, the plotted shear strengths would fall close to the values for the failure envelope curve obtained for the U–U test series. The strength envelope curves of Figure 23, 24, 25, and 26 indicate the total stress condition existing on the shear failure plane (i.e., the total normal stress and related soil shearing resistance). Recall that the total normal stress represents the combined influence of effective stress s and neutral stress u (the latter term indicating stress resulting from pore water pressure), or: st = s + u

421

Shear Strength Theory

A rearrangement gives: s = st - u

Shear stress, τ

Values of the strength envelope from the consolidated–drained (C–D) test series indicate effective stress as well as total stress, since no excess pore water pressure existed during the shearing process. For any C–U- or U–U-type test, the magnitude of excess pore water pressure existing in the soil at shear failure is indicated from the relationship between the C–D and C–U envelopes, or the C–D and U–U envelopes; at points of similar shear strength, the difference in the normal pressures shown on the C–D and C–U curves, or the C–D and U–U curves, represents the excess pore water pressure (Figure 27).

φCD

es

s res

l st

ta To

σt

ffe s ult res

=e D

C–

tr es ctiv

σt = σ (u = 0)

Normal stress, σ

Shear stress, τ

(a) s, σ es

tr es ctiv

e Eff

ore

p ess

wat

φCU

Exc

s ult

es

Dr

C–

φCD e, u sur res er p

, σt tress

s Total

s

sult –U re

σ = σt − u

C

Normal stress, σ (b)

ss,

Shear stress, τ

tre

σ = σt − u

es ctiv

C–

φCD

e Eff

Total stress, σt

s

ult

es Dr

σ

U–U results Excess pore water pressure, u

φUU = 0

Normal stress, σ (c)

Figure 27 Relationship between effective stress, total stress, and pore water pressure: (a) C–D shear test results; (b) C–U shear test results; (c) U–U shear test results.

422

Shear Strength Theory

Illustration 4 To determine the strength properties of a clay soil, a series of consolidated–undrained direct shear tests is performed. For the first test, the normal pressure is 36 kPa and the sample fails when the shear stress is 21.6 kPa. The second sample is tested under a normal load of 72 kPa, and failure occurs when the shear stress is 24 kPa. A third sample is tested under a normal loading of 120 kPa, and failure occurs when the shear stress is 38.4 kPa. From these data, estimate the cohesion of the soil in the in situ condition and the value of fCU.

Solution A problem of this type frequently is best solved graphically. 1. Locate test data on a scaled plot. 2. Draw a sloped failure envelope line to estimate fCU. 3. For a test having a low normal (compressive) pressure, assume that the pressure is less than svc, and draw a horizontal failure envelope line through it to obtain the value of in situ cohesion. 4. The intersection of sloped and horizontal failure envelope lines gives a normal pressure value that is approximately the value of svc (the maximum past overburden stress).

Shear stress, τ, kPa

48 φCU

c = 21.6 kPa 24 σVc = 62.5 0

24

48

72

96

120

Normal stress, σ, kPa

From the scaled plot, c = 21.6 kPa, fCU = 17° ;

The value of the angle of internal friction for clay soil is important for design and analysis studies involving effective stress, and reliable values can be obtained from test results. The angle of internal friction of clay soils for the drained or undrained shear strength condition, fCD and fCU, is influenced by a number of factors, including overburden pressure, stress history of the deposit (i.e., normally consolidated or overconsolidated), and category of clay minerals present. Expected values of fCD are between 20° and about 40°. The value of fCD for normally consolidated clays can be approximated from a crude relationship to the plasticity index (PI), using: sin fCD = 0.8 - 0.094 1ln PI2 Where ln PI is the natural log of the plasticity index expressed as a numeric value (i.e., if the PI is 40 percent, use 40 in the equation to obtain fCD as, approximately, 27°). In many practical soils problems involving clays, applied structural loads result in excess pore pressures being developed. Some drainage does occur, but it is restricted because of the low permeability of the soil. Consequently, the actual shearing conditions that are developed fall somewhere between the U–U and the C–D case. In many soil and foundation design problems, the U–U strength is used where the initial period of loading is

423

Shear Strength Theory

Ultimate shear strength, τ

+

(Increasing) Water content, w, percent

Figure 28

Change in strength for a cohesive soil as water content changes.

critical, since it is realized that the shear strength will become greater (increase) and conditions safer with time as excess pore water drainage occurs. These shear test results also indicate that it is possible to improve the shear strength of clay soils by consolidation, provided that time is available for permitting the necessary pore water drainage to take place. In effect, consolidation results in decreasing the water content of the clay and obtaining a related increase in shear strength, as indicated by Figure 28.

Shear Strength of Clay Related to Triaxial Testing Although the direct shear test is a convenient reference for introducing the factors affecting the shearing strength of clay soils, the triaxial compression test is the method most typically used in the laboratory for determining soil shearing strength properties. Recall from the early sections of this chapter that the triaxial compression test subjects the soil sample to a three-dimensional state of compression (i.e., compressive stresses act in the three orthogonal directions), thereby causing shear stresses to develop within the sample (Figure 29; also ref. Figure 4). The following discussion relates to clay soil as defined by Figure 22; that is, the clay soil is fully saturated and the deposit from which the test sample is obtained is a normally consolidated deposit (i.e., the existing soil overburden and lateral pressures represent the maximum pressures the soil has been subjected to in its history as part of the deposit). The strength–failure envelope for a soil can be determined by performing the triaxial test on separate but identical samples; each test uses a different confining pressure (Figure 30). It has been found adequate to work with only the s3 and s1 values on the Mohr’s circle coordinates. In Figure 30, the value of s3 is the all-around confining pressure or stress acting on the soil sample at the start of the test, and s1 is the measured vertical (axial) stress that subsequently causes the soil sample to shear. If s3c 7 s3b 7 s3a, it will be found that s1c 7 s1b 7 s1a. The typical triaxial equipment depicted by Figure 29(a) includes tubing lines connected to the porous stones positioned on the top and bottom of the soil sample; these lines can be utilized to drain pore (void) water from the sample or to measure, control, or vary the pore water pressure existing within the soil sample during the test procedure. Note the soil sample is enveloped by a flexible impervious membrane. In the following descriptions, the magnitudes of the s3 and s1 test pressures are greater than the s3 and s1 stresses that acted on the soil sample in situ (in the ground).

424

Shear Strength Theory Loading piston for applying vertical force Pressure gauge Pressure supply (air or water)

Air or fluid in chamber, under pressure

Top loading plate Porous stone

Flexible tubing Soil sample

Tie rod

Impervious membrane around soil sample Cylinder (lucite, or metal for high pressure testing) Porous stone and base

Soil sample drainage or pore pressure measurement lines (a)

σ1 = σ3 + Δσaxial

σ1

σ3 σ3

Soil sample

σ2 = σ3

σ3 = σ2

σ1 (b)

Figure 29 Triaxial compression test representations: (a) schematic diagram of test apparatus; (b) state of stress on an incremental element in the soil sample.

For the consolidated–drained (C–D) test, the confining pressure is imposed onto the test sample and the porous stone drain lines are left open to enable soil pore water to escape from the soil as the confining pressure squeezes the sample (i.e., consolidates it). After the sample is fully consolidated by the confining pressure (identified as the condition when equilibrium for the imposed loading is reached, whereby no further pore water drains), the axial force is slowly applied. The drain lines remain open as the axial force increases, permitting additional drainage of sample pore water to occur. Since additional compression (i.e., volume decrease) of the sample tends to occur because of the increasing (new) vertical stress, the continuous escape of pore water must be permitted if the fully

425

Shear Strength Theory Sample b σ1b

Sample a σ1a

a

σ3a

b

Sample c σ1c

σ3b

c

σ3c

Shear stress, τ

Note: σ3c > σ3b > σ3a

Line defining strength-failure envelope for tested soil is tangent to series of Mohr’s circles determined from series of triaxial tests

σ3a

σ1a σ3b

σ1c

σ1b σ3c Normal stress, σ

Figure 30 Results of a series of triaxial tests performed on identical soil samples, as used for determining the strength envelope for the soil. (Representative of results for cohesive soils subject to consolidated-type tests and for cohesionless soils.)

consolidated (no excess pore water pressure) condition is to be satisfied. The axial and confining pressure acting when the soil sample shears provide the values to plot one Mohr’s circle for the C–D case. A series of C–D triaxial tests on identical soil samples provide the data for determining the effective stress, s, envelope curve, as indicated in Figure 31. In the consolidated–undrained (C–U) test, the soil sample in the triaxial apparatus is fully consolidated to the all-around confining pressure, s3, by permitting the pore pressure drain lines to remain open (as for the C–D test). The drain lines are then closed and the axial load is imposed. Pressure gauges on the drain lines permit measurements of the pore

Shear stress, τ

Angle of internal friction for consolidated–drained conditions

φCD

σ envelope (strength envelope for C–D conditions)

Result from sample 3

Sample 1

Result from sample 2 Normal stress, σ

Figure 31

426

Triaxial results from consolidated–drained (C–D) tests and related strength envelope.

Shear Strength Theory

water pressure developing in the sample as the axial load is applied. The increasing axial force acting on the soil sample tries to cause sample compression (i.e., further consolidation), but since the pore water is now prevented from escaping because the drain lines are closed, an “excess pore water pressure” develops. The pore water pressure continues to increase in relation to the increase in axial loading. When the soil sample shears, the axial load, the internal pore water pressure, and the confining pressure are known (measured and recorded). The axial and confining pressures indicated by the triaxial gauges represent the total axial and confining stresses acting on the soil sample. A total stress minus the measured pore water pressure provides the value of interparticle, or effective, stress acting within the soil sample at the time of failure:3 s = st - u The test results are presented on the Mohr’s circle coordinates, as shown by Figure 32. As for the C–D test, a series of tests performed on identical samples but at different confining pressures permits a strength envelope to be obtained. In the performance of the unconsolidated–undrained (U–U) test, the soil sample is assembled in the triaxial apparatus and the pore water drain lines are closed. The confining pressure, s3, is applied and quickly followed by application of the axial load. The imposed pressures act to compress the soil sample (i.e., force the soil particles closer together), but because the pore water drains are closed, the trapped pore water prevents a decrease in sample volume from occurring. The trapped pore water has prevented the sample from consolidating. The pressure gauges on the pore water drain lines will show increasing pore water pressure as the axial force increases. The values indicated by the triaxial gauges for the confining pressure, s3, and axial stress, s1, represent the total pressure or stress internal

φCD

Shear stress, τ

σ envelope (C–D tests)

σ (t)

lope enve

–U for C

itions

cond

φCU

Effective stress circle Total stress circle

σ3

σ3(t) u

σ1

σ1(t) u

Note: σ1(t) − u = σ1 σ3(t) − u = σ3

Normal stress, σ

Figure 32

Results from consolidated–undrained (C–U) triaxial test and related strength envelope.

3This

equation helps clarify the phrase “excess pore water pressure.” That term refers to the magnitude of pressure developed in the void space water that is above the magnitude of the interparticle pressure that acts. The occurrence is similar to the undergraduate strength of materials problem, where a structural member composed of two different materials carries a compressive load; the portion of the total load carried by each material relates to the stress–strain properties or rigidity of the material (the least deformable or more rigid material supports more than its share of the total load).

427

Shear Strength Theory

to the soil sample. The effective stress, or intergranular pressure, when the sample shears is determined by deducting the pore water pressure from the total pressure (as for the C–U test), as shown in Figure 33(a), or: s = st - u

Shear stress, τ

A series of tests on identical samples, but at differing confining pressures, s3, provide the data to obtain the strength envelope for the U–U case as shown in Figure 33(b). Note that the Mohr’s circles will be the same for all samples but will be positioned horizontally in accord with the s3 value used to perform the test. Even though the s3 and s1 values that cause the sample to shear are different for each of the tests, the internal shear strength of the soil remains the same because the internal conditions remained the same (the presence and quantity of pore water, the sample void ratio, and the particles’ spacing will not change when trapped pore water prevents a volume decrease). Somewhat different occurrences result when the confining pressure applied in the triaxial test is less than the confining pressure that existed when the soil was in situ (for this

φCD

σ envelope

Effective stress circle Total stress circle σ = σt − u σ3

σ3(t)

σ1

u

σ1(t) u

Normal stress, σ (a)

φCD Shear stress, τ

σ envelope U–U envelope (φUU = 0°)

Sample 1

Sample 2

Sample 3

Normal stress, σ (b)

Figure 33 Unconsolidated–undrained (U–U) triaxial test data: (a) relationship between total stress, effective stress, and pore water pressure; (b) results of a series of tests on identical soil samples provide Mohr’s circle of equal size but transposed laterally.

428

Shear Strength Theory

occurrence, the soil is considered overconsolidated in regard to the test conditions). When the test confining pressures exceed the in situ pressures, the sample volume tends to reduce during testing; conversely, the soil sample tends to expand during testing when the confining pressure is less than the in situ pressure. If an overconsolidated sample is prevented from taking in water during testing as the sample attempts to expand (the triaxial drain lines are closed), a negative pore water pressure develops in the soil void spaces (a condition similar to the effects of capillary tension) and the relationship between total and effective stress becomes: s = st - 1-u2 = st + u

The results plotted on the Mohr’s circle coordinates are shown in Figure 34. In the natural (in situ) condition, overconsolidated clay soils will tend to undergo expansion during some types of construction activity, such as excavation work, and negative pore water pressures will develop within the soil. A high soil shear strength results, as indicated by the Mohr’s circle for the initial strength condition (Figure 34). However, natural water, which is or becomes present, will tend to be drawn into the clay voids. Eventually, as the negative pore pressures vanish, the soil shear strength decreases toward the value indicated by the effective stress envelope (Figure 35). Where the U–U strength is required to be known for analysis of a soil or foundation study, it is convenient to determine the shear strength from unconfined compression tests. As explained earlier in this chapter, the unconfined compression test is a triaxial test where the confining pressure acting on the sample is zero. This test is quick and easy to perform. The axial stress required to fail the tested soil is designated qu. The result of an unconfined compression test can be plotted on τ versus s coordinates, as shown in Figure 36. The special case of cohesion, or undrained shear strength, is simply one-half the unconfined axial stress: cu =

1 q 2 u

(4)

Some clay deposits have developed cracks or fissures (sometimes called slickensides). This condition may be the result of desiccation, seismic effects, or other factors. If tested in

Shear stress, τ

σ envelope

φCD

Total stress circle Effective stress circle

σ3(t)

σ3 u

σ1(t)

σ1 u

Note: σ = σt + u, when the sample is overconsolidated in reference to the test confining pressure, σ3

Normal stress, σ

Figure 34

Data from U–U triaxial test on overconsolidated clay soil.

429

Shear Strength Theory

σ envelope Shear stress, τ

Initial maximum shear strength Final maximum shear strength

φCD

Mohr’s circle based on initial strength possessed because of the overconsolidated condition Mohr’s circle for expected final strength

σ3

σ1(final)

σ1(initial) Normal stress, σ

Shear stress, τ

Figure 35 Mohr’s circles illustrating expected change in strength for overconsolidated clay when soil water content increases.

P = axial force when soil shears (fails) qu = P/A

c Cohesion, c = 1 σ1 2

1σ 2 1 σ1 = axial stress = qu

σ3 = 0

Axial stress to fail sample = qu = σ1

Normal stress, σ

Figure 36

Unconfined compression test data used to determine cohesion c.

unconfined compression, shear failure along the slickensides could give a low and misleading indication of the strength of soil confined in the natural deposit. For such material, a triaxial test is preferred.

Approximating Shear Strength of Cohesive Soil from Site Investigation Data (SPT, CPT, DMT, VST) The in-place shearing strength of cohesive soil represents an undrained shear strength condition that is commonly identified as the soil cohesion, and reflects a value influenced by the soil water content but also properties of the deposit (in-place shear strength of cohesive soil is known to be affected by a combination of factors, including the overburden pressure, the pore water pressure, the overconsolidation ratio, and the sensitivity). No definitive and inclusive relationship has been established between the in-place shearing strength of cohesive soil and results of the standard penetration test (SPT) or cone penetrometer soundings (CPT). The information shown in Table 3, determined from a popular relationship that undrained shear strength is approximately equal to (0.06) (N60) ( patm), is expected to be conservative for many combinations of in-place conditions, but not all (such as soils showing a high plasticity index, PI). The strength values shown in the tabulation should be for general reference and preliminary estimates only. For applications, shear

430

Shear Strength Theory Table 3 Approximate Relationship between Condition of Cohesive Soil (Clay Type) and Results of Subsurface Investigation Procedures* Approximate Value of Cohesion (undrained shear strength), cu General Description Very soft Soft Medium Stiff/firm Very Stiff (very firm) Hard

psf

kN/m2(kPa)

Value SPT (N)**

6250 250–500 500–1000 1000–2000 2000–4000 74000

612 12–24 24–48 49–96 96–190 7190

62 2–4 4–8 8–15 15–30 730

Value CPT (qc/patm)

Consistency Index, 1CI = 1.0 - LI2***

65 5 to 15

60.5 0.5 to 0.75

15–30 30–60 760

0.75–1.0 1.0–1.5 71.5

*Data

compiled from various sources. This data is for estimates or order-of-magnitude only, and should not be used for designs (for design work, use results from in-place or laboratory tests). **Values refer to sampling procedures where the efficiency of the drive hammer is approximately 60 percent; refer to Table 2. ***Consistency Index, CI = 1LL - w2>1LL - PL2 = 1 - LI, where LI = 1w - PL2>1LL - PL2, refer to Eq. 4-4.

strength data should be determined from in-place testing (such as the vane-shear test discussed below or the dilatometer procedure, DMT), or from laboratory tests using undisturbed samples. The vane-shear test (VST) used to obtain the in-place shear resistance for a cohesive soil (refer to Figure 8) typically overvalues the actual shear strength that will develop within the soil deposit. The actual shear strength available can be determined by modifying the result of the field test procedure by a factor that includes the effect of the test strain rate on the shear resistance that develops, and soil anisotropy. A simplified method that provides a close approximation is presented below. (Reference [198] discusses detailed procedures for obtaining shear strength values from the results of vane-shear tests.) 1actual available shear strength2 L

1soil strain modification factor21vane-shear strength2 L 0.22 svmax

(5)

where: the strain modification factor 1approximate, for PI 7 202 = 1.0 - 0.0041PI2 PI = soil plasticity index svmax = vertical preconsolidation stress for the soil deposit

Shear Strength of Mixed Soils Mixtures of clay and granular soils in nature are not unusual. Material that is predominantly clay, in which all granular particles are surrounded by clay materials, will behave essentially as a clay. Mixtures that are predominantly granular soil with limited clay will present a sloped failure envelope curve on a τ versus s plot, but the intercept is on the

431

Shear stress τ

Shear Strength Theory

φ

Cohesion, c c Effective normal stress, σ

Figure 37

Typical plot of t versus s for mixed soils.

t axis, as shown in Figure 37. The relationship of shear strength to normal stress can be expressed as: t = c + s tan f

(6)

For such soils, the limitations on permeability and drainage should not be overlooked when considering new loadings that are applied to the soil mass. Excess pore pressures may develop and only part of the newly applied stress may represent effective stress, with a subsequent lag in the development of shear strength.

Position of Failure Plane Related to Angle f Homogeneous soils stressed to failure during a triaxial test or unconfined compression test typically develop a distinct plane of failure, as indicated in Figure 38(a). Applying the analysis for stress at a point (Figure 38(b) and (c)), a practical relationship between the σ1 Failure plane σ3

θ

θ

σ3

σ1 (a)

(b)

Failure envelope

90° 2θ φ σ3

(180 − 2θ)

σ1

(c)

Figure 38 Mohr’s circle analysis to related f and position of failure plane: (a) failure plane on tested sample; (b) incremental element at failure plane; (c) Mohr’s circle.

432

Shear Strength Theory

angle of internal friction, f, and the position of the failure plane is obtained: f + 90° + 1180 - 2u2 = 180°, the sum of the interior angles in a triangle f + 90° = 2u u =

90° + f 2

u = a45 +

(7)

Stress Paths Stress paths refer to the series of progressive changes in shear and normal stress that develop within a soil mass as a result of construction loading being applied (such as resulting from loading due to structures and earth fill embankments that cause stress increases) or load being reduced (e.g., excavations). Section 3 explained the method of relating shear and normal stresses on the s-t Mohr’s circle coordinates (or s-t space). An alternative method for indicating the relationship is to use a p and q coordinate system, and indicate combinations of shear and normal stress in the related p-q space (Figure 39). The values of p and q represent the coordinates for the maximum shear stress on the Mohr’s circle and the related normal stress. The p and q coordinates, using total stress conditions, are identified as:

p =

1 1s + s32 2 1

q =

1 1s - s32 2 1

τ

q q = q' = 1 (σ1 − σ3) 2

Shear stress, τ

4

f bdeg 2

Maximum τ has coordinates p, q

A B (σ3 or σ3)A (σ1 or σ1)A (σ1 or σ1)B Normal stress, σ or σ (a) σ-τ coordinates

(p, q)A or (p, q)A

(p, q)B or (p, q)B

σ or σ

p p = 1 (σ1 + σ3) or p = 1 (σ1 + σ3) 2 2 (b) p-q coordinates

Figure 39 Comparison of results of triaxial tests plotted on the σ-τ coordinates and p-q coordinates.

433

Shear Strength Theory

In terms of effective stress conditions, the coordinate values p and q are identified as: 1 1s + s32 2 1 1 q = 1s1 - s32 2 p =

Since total stress is represented as s1 and s3, and effective stress is total stress less pore water pressure u (or neutral stress), the values for effective stresses s1 and s3 become: s1 = s1 - u s3 = s3 - u and: q =

1 1 1 1s - s32 = 31s1 - u2 - 1s3 - u24 = 1s1 - s32 2 1 2 2

so therefore q = q (i.e., since the q term represents shear stress, its value is not affected by pore water pressure). Correspondingly: p =

1 1 1 1s + s32 = 31s1 - u2 + 1s3 - u24 = 3s1 + s3 - 2u4 = p - u 2 1 2 2

The stress path analysis has practical application for the condition where construction loads are responsible for the development of excess pore water pressures in the soil foundation zone, such as with clay and silt–clay soils having a low coefficient of permeability. A concern is that shear stresses developing in the soil as a result of foundation loads will exceed the shear strength possessed by the soil. The problem can be studied using the s-t coordinate system to compare construction-induced stresses to the soil strength, but the necessary comparisons become cumbersome for the typical field condition where construction-related loadings are changing (e.g., increasing from structural loads, decreasing for excavations) while, simultaneously, excess pore water pressures are changing (e.g., increasing as structural loads increase but then decreasing as drainage occurs). The p-q stress path method allows easier comparisons between construction-induced shear stresses and soil shear strength.

Stress Path for Tests in Consolidated–Drained Conditions Understanding of the p-q stress path may be most easily grasped by referring to a triaxial test performed on a normally consolidated clay under consolidated–drained (C–D) conditions and where the applied test stresses exceed the in situ stress condition for the soil sample. As part of the sample preparation, the soil is fully consolidated in the triaxial chamber to an all-around confining pressure s3. Since the sample is consolidated to the confining pressure s3, the total stress and effective stress are similar, or s3 = s3. The confining pressure acts in the vertical (axial) direction as well as horizontal, so s3 = s1. The test

434

Shear Strength Theory

will be performed slowly so that drainage of pore water in the sample can occur as the axial load increases (s1 increases); referring to Figure 29(b), s1 = s3 + ¢saxial. Therefore, at the start of the test: 1 1 1s1 + s32 = 1s3 + s32 = s3 2 2 1 1 q = 1s1 - s32 = 1s3 - s32 = 0 2 2 p =

As the test progresses, s1 1i.e., s3 + ¢saxial2 increases, and: ¢saxial 1 1s3 + ¢saxial + s32 = s3 + 2 2 ¢saxial 1 q = 1s3 + ¢saxial - s32 = 2 2 p =

The stress path for axial load increasing until the sample fails in shear is shown in Figure 40. Shear failure of a soil occurs when the p-q stress combination reaches the k′f strength (or failure) envelope line; this k′f line defines the limiting strength for the soil (a strength envelope is similarly used for the Mohr’s circle stress analysis). The relationship between the angles f and a defining the strength (failure) envelope lines shown on Figure 40 can be determined, recalling that:

sin fCD

1 1s - s32 2 1 = 1 1s + s32 2 1

τ

q φCD α

Stress path

σ3 and σ1 at start

Strength (failure) envelope line for p - q plot, k'f

σ1 when sample shears σ1 for increasing Δσaxial σ1 for small Δσaxial

(a) Series of σ -τ Mohr’s circles resulting from sequence of increasing axial stress during the triaxial test.

σ

q = 1 (σ1 − σ3) 2

Strength (failure) envelope line, kf

Stress path (total stress)

45°

σ3

p

p = 1 (σ1 + σ3) 2 (b) Related stress path plotted on p-q coordinates.

Figure 40 Results of consolidated–drained (C–D) triaxial test relating s-t and p-q values to develop the stress path for isotropically consolidated soil sample, s1 = s3 at start.

435

Shear Strength Theory

and noting that: 1 1s - s32 2 1 tan a = 1 1s + s32 2 1 Both expressions are the same, and therefore: sin fCD = tan a The stress path indicated for the C–D triaxial test is also representative of effects when load is applied to granular soil where the high coefficient of permeability permits rapid drainage of pore water so that excess pore water pressures from the presence of groundwater do not develop as the construction loads are imposed.

Stress Path for Tests in Consolidated–Undrained Conditions The stress path for soil subjected to triaxial compression under consolidated–undrained (C–U) conditions is significantly different than the stress path for soil under drained conditions. On a practical basis, the C–U condition is representative of the effects of construction loading applied to a normally consolidated clay deposit, where the low soil coefficient of permeability retards drainage so that excess pore water pressures develop. Consider a soil sample of a saturated clay, normally consolidated, in a triaxial compression test under C–U conditions. The clay sample is fully consolidated in the triaxial chamber to the test value all-around confining pressure. When the sample is fully consolidated, the total lateral stress equals the confining pressure s3. Since excess pore water has been allowed to drain from the soil sample and the full triaxial chamber pressure is imposed onto the soil skeleton, s3 equals s3. At this stage, the axial stress (vertical stress) also equals the chamber confining pressure so that s1 = s1 = s3. To fail the soil in the shear, axial loading will be increased while lateral pressure is held constant. The triaxial equipment drain lines are closed so that no additional pore water can drain from the sample. Pore water pressures are measured independently from measurement of the applied loading (direct pore pressure can be determined by use of a pore pressure needle inserted into the sample or from a pore water supply line connected at the base of the sample). The applied axial load results in a stress increase, Δs1. However, because the equipment drain lines are closed to prevent drainage during the axial loading procedure, an excess pore water pressure, Δu, is created. Because of the volume strains that occur during application of the axial load, the excess pore pressure value that develops is different than the stress increase Δs1. The relationship can be expressed as ¢u = A1¢s12 where the A factor is the Skempton pore pressure coefficient. The excess pore water pressure is exerted equally in all directions, and the resulting effective stresses become: s1 = s3 + ¢s1 - ¢u = s11start2 + ¢s1 - A1¢s12 s3 = s3 - ¢u = s3 - A1¢s12

436

Shear Strength Theory

where ¢s1 s3 ¢u A

= = = =

increase in axial stress all-around confining pressure (stress) at start of test, held constant excess pore water pressure resulting from application of Δs1 Skempton pore pressure coefficient

The concept of Δu on total and effective stress conditions in a foundation soil is illustrated by Figure 41. If the ratio Δu/Δs1 were to remain constant for the range of applied loading, the effective stress path would follow a sloping straight line. However, in soil the rate of volume strains changes as applied stresses increase; the result is that the A term or related Δu/Δs1 are not constant values but increase with increases in applied stress. The value for Af or Δu/Δs1 when the sample reaches the failure shear stress has been found to range between 0.5 and 1.0 for normally consolidated clays. The Af value ranges from near zero to 0.5 for overconsolidated clays, and is between 1.0 and 1.25 for sensitive clays with highly flocculent structures. The effective stress path curves representative of normally consolidated clays are as illustrated in Figure 42.

Applications On a practical basis, the concern is when the effective stress path resulting because of construction loading extends to intersect the strength envelope line k′f for the soil, a condition indicating the start of shear failure. Continuing, as drainage of pore water in the foundation soil occurs, the excess pore water pressures dissipate. The effective stress path (ESP) curve gradually shifts toward the total stress path (TSP) curve, a stress condition where the foundation soil possesses greater shear strength (i.e., the soil can resist greater foundation-induced shear stress before a failure condition is reached).

α τ

TSP = Total stress path ESP = Effective stress path

α, for effective stress, from consolidated−drained tests

q α Δu = 1.2 Δσ1

ESP for Δu = 1.2 Δσ1

Δu = .7 Δσ1

ESP for Δu = 0.7 Δσ1 TSP

kf' line

Δσ1 (concept)

(a) Stress paths on σ-τ coordinates, indicating TSP and ESP for constant ratio of Δu/Δσ1 (concept only)

TS

P

σ σ3 = σ3 = σ1 @ start

P

ES

kf' line

p k0 = 1 line

(b) Related stress paths on p-q coordinates.

Figure 41 Stress paths for consolidated–undrained triaxial tests for condition of constant Δu/Δσ1 ratio (concept of effect of Du only) for isotropically consolidated soil samples, where s1 = s3 at start.

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Shear Strength Theory Figure 42 Stress paths (ESP and TSP) for consolidated–undrained triaxial test where ratio Δu/Δσ1 increases (for isotropically consolidated soil sample, s1 = s3 at start). To establish TSP from TX Test: Use s1 and s3 to indicate conditions at start of testing, then: p =

+ ¢s2 + s34

Shear failure occurs when stress path intersects kf' line

+ ¢s2 - s34

To establish ESP, use: p = q =

1 2 31s1 1 2 31s1

k f'

e

lin

P

TS

Δuf

P ES

q =

1 2 31s1 1 2 31s1

q

ko = 1

σ3 = σ3 = σ1 @ start

+ ¢s - A # ¢s2 + 1s3 - A # ¢s24

p

+ ¢s - A # ¢s2 - 1s3 - A # ¢s24

The triaxial test conditions described in the preceding paragraphs and Figure 42 represent the situation where isotropic stress conditions exist in the foundation soil (that is, sh = sv, and the lateral pressure coefficient K0 value of 1 is located on the horizontal axis of the s-τ and p-q coordinate systems. For a soil deposit where anisotropic stress conditions exist and K0 is less than 1 (as is common in normally consolidated and underconsolidated deposits), the influence can be shown as a sloping K0 line on the p-q coordinates. The start point for the stress path is from the sloping K0 line instead of from the horizontal axis (Figure 43). Where a saturated, normally consolidated clay is excavated, the vertical load on the remaining foundation soil zone decreases, resulting in progressive decreases in p and q values. The in-place soil tends to expand because of the stress relief. With a clay that was originally saturated, the expanding void spaces result in a partially saturated clay and the development of negative pore pressures, - ¢u (sometimes referred to as suction). The total and effective stress paths for the excavation condition extend downward, instead of upward as occurs when construction load increases (Figure 44). The effective stress path is to the right of the total stress path, indicating that the soil shear strength is greatest at the time of excavation but then decreases with time as the negative pore pressures diminish.

Figure 43 Stress paths for consolidated–undrained triaxial test (for anisotropically consolidated soil sample, s1 7 s3). Note stress path begins at K0 line (K0 = s3>s1 = 0.6 for this diagram). To establish TSP from TX Test: Use s1 and s3 to indicate conditions at start of testing, then: p = 12 31s1 + ¢s2 + s34

q = 12 31s1 + ¢s2 - s34 To establish ESP, use:

p = 12 31s1 + ¢s - A # ¢s2 + 1s3 - A # ¢s24

q = 12 31s1 + ¢s - A # ¢s2 - 1s3 - A # ¢s24

438

q Shear failure occurs when stress path intersects kf' line

e

lin

k f'

P

TS

Δuf

1 k0 < ) = 0.6 k 0 , . (i.e

ESP k0 = 1 σ3 @ start (σ3 = 0.6 σ1)

σ1 @ start

p

Shear Strength Theory q

q

1 k0 < ) = 0.6 k 0 ., e (i. TSP

p

TS P

k0 = 1

ESP (shifts left as negative pore −Δu pressures dissapate) Shear failure α if TSP or ESP intersects kf' line kf'

(a) Stress paths for isotropic stress conditions: σv = σh and k0 = 1

ESP

−Δu

α

p Safe condition if final excavation results in this p-q coordinate.

−Δu kf'

Unsafe condition if excavation results in this p-q coordinate (shifting ESP will intersect kf' line before reaching TSP when negative pore pressures dissapate)

(b) Stress paths for anisotropic stress conditions: σv > σh and k0 < 1

Figure 44 Stress paths (ESP and TSP) for rapid excavation in normally consolidated clay (representing consolidated–undrained conditions).

Problems 1 (a) A dry sand sample is tested in direct shear. The test procedure includes having a normal (compressive) stress of 200 kPa imposed while the sample undergoes shearing. The sample fails when the shear stress reaches 135 kPa. From this data, determine the angle of internal friction f for the soil. (b) A second sample of the same sand is also to be tested in direct shear, but the applied normal (compressive) stress will be 145 kPa. What shear stress is expected to cause the sample to fail? 2 A sample of dry sand is tested in direct shear. The shear box holding the sample has a circular cross section with a diameter of 50 mm. The normal (compressive) load imposed is 200 N. The sample shears when the shear force is 130 N. (a) Determine the test normal stress and shear stress at failure. (b) Determine the angle of internal friction f for this soil. (c) What is the probable condition of the tested sample (dense, loose, etc.)? 3 (a) A sample of dry sand is tested in direct shear. Under an applied normal stress of

4,800 psf, the sample falls when the shear stress reaches 3,100 psf. What is the angle of internal friction for this soil? (b) A second sample of the same sand material is also to be tested in direct shear, but the applied compressive loading will be 3,500 psf instead of 4,800 psf. What shear stress is expected to fail the sample? 4 Samples of damp sand are tested in direct shear (samples are tested at the in situ water content). The procedure involves two separate tests, with different normal loads for each test. In test one, using a normal loading which causes a compressive stress equal to 50 kPa, the sample is failed at a shear stress equal to 33 kPa. For test two, the normal stress is 80 kPa and the sample shears at a stress of 51 kPa. Use this test data to determine the angle of internal friction and the value of apparent cohesion for this sand. 5 Determine the values of apparent cohesion and angle of internal friction for a damp sand using the results of direct shear tests performed on two identical samples of the soil. In test one, the sample shears at a stress of 71 kPa when the compressive (normal) stress is 95 kPa. In test

439

Shear Strength Theory two, the sample shears at a stress of 104 kPa when the normal stress is 150 kPa. 6 A sand sample is subjected to direct shear testing at its normal (in situ) water content. Two tests are performed. For one of the tests, the sample shears at a stress of 3,000 psf when the normal stress is 4,000 psf. In the second test, the sample shears at a stress of 4,000 psf when the normal stress is 6,000 psf. From these data, determine the value of apparent cohesion and the corresponding angle of internal friction. 7 A sample of dry sand in relatively loose condition is subject to a triaxial test. The sample fails when the confining stress (minor principal stress) is 50 kPa and the axial stress (major principal stress) is 170 kPa. What is the angle of internal friction for the soil?

14

15

8 A dry sand is known to have an angle of internal friction equal to 36°. A triaxial test is planned, where the confining pressure will be 40 kPa. What maximum axial stress (major principal stress) should be predicted? 9 A dry sand sample in a triaxial test failed when the confining stress (minor principal stress) was 1,000 psf and the axial stress (major principal stress) was 4,000 psf. What is the angle of internal friction for this soil?

16

10 A dry sand is known to have an angle of internal friction equal to 35°. What is the maximum major principal stress that the soil can withstand when the minor principal stress is 20 psf? 11 It is known that the angle of internal friction for the soil comprising a granular deposit is 37°. At one depth in the deposit, the lateral pressure is 45 kPa, and this is considered the value of the minor principal stress. Use the Mohr’s circle analysis to determine the maximum vertical pressure (major principal stress) that can be applied (i.e., the vertical pressure for incipient shear).

17

12 The soil in a dry granular deposit has an angle of internal friction equal to 35°. At a point in the soil mass where the lateral pressure (minor principal stress) is 1,000 psf, what maximum vertical pressure (major principal stress) can be imposed?

18

13 Design studies for a planned building indicate that the foundation loading will cause the

440

principal stresses in the soil at one point below the building area to increase to 195 kPa and 70 kPa (major and minor principal stress values, respectively). If the soil is a dry sand having an angle of internal friction equal to 34°, determine if the indicated stresses would cause shear failure to occur. Computations indicate that a planned building loading will cause the principal stresses in the soil at a point beneath the building to increase to total values of 4,500 psf and 1,500 psf (major and minor principal stresses, respectively). If the soil is a dry sand with an angle of internal friction equal to 35°, will the indicated stresses cause a shear failure at the point? A sample of dry sand is tested in direct shear and a separate sample of the same soil is tested in a triaxial compression test. In the triaxial test, the confining pressure (or minor principal stress value) is held at 15 kPa, and the sample shears (fails) when the axial pressure reaches 51 kPa. What shear strength is expected for the sample tested in direct shear when the test normal (compressive) stress is 125 kPa? Samples of a dry sand are to be tested in a direct shear test and a triaxial test. In the triaxial test, the sample fails when the major and minor principal stresses are 140 psi and 40 psi, respectively. What shear resistance is to be expected in the direct shear test when the normal load stress equals 5,000 psf? Soil borings for the subsurface investigation of a site indicate a deep deposit of sandy soils. At one boring location, the blow count logged for the standard penetration test, N60, while sampling at a depth of 25 feet was 22 blows/ft. The sand has a unit weight of 115 pcf. Using criteria that relate SPT results and the soil angle of internal friction, estimate the value of f for the sand at the sample depth. Cone penetrometer soundings were used for the subsurface investigation of a planned construction site. At one sounding location, the value recorded for qc at a depth of 21 feet was 150 ksf. The soil has a unit weight of 115 pcf. Using criteria that relate CPT results and the soil angle of internal friction, estimate the value of f for the sand at the test depth (known from borings).

Shear Strength Theory 19 Cone penetrometer soundings were used for the subsurface investigation of a planned construction site. At one sounding location, the value recorded for qc at a depth of 8 m was 8,000 kPa. The soil has a unit weight of 16 kN/m3. Using criteria that relate CPT results and the soil angle of internal friction, estimate the value of f for the sand at the test depth. 20 Soil borings completed for a subsurface investigation indicate the site is underlain by a deposit of sandy soils. At one boring location, the blow count logged for the standard penetration test, N60, while sampling at a depth of 7 m was 24 blows. The sand has a unit weight of 16.5 kN/m3. Using criteria that relates SPT results and the soil angle of internal friction, estimate the value of f for the sand at the sample depth. 21 Dilatometer test results obtained as part of a subsurface investigation for a planned construction project indicated the following data for one of the tested locations (note the dilatometer instrument gauge indicates pressure in bars): p0 = 3.0 bars 1300 kPa2 ,

p1 = 12.76 bars 11276 kPa2 ,

p2 = u = 0.20 bars 120 kPa2 . Supplementary soil borings at the site verify that the soil is sand. The soil overburden pressure, svo, at the test depth is 53 kPa (0.53 bar). From this information, estimate the angle of internal friction f for the sand at the test depth. (To determine ED, ID, KD, etc., perform calculations using bar as the unit for pressure.) 22 A triaxial test performed on a clay sample, under unconsolidated–undrained conditions, reaches a maximum (failure) axial stress (major principal stress) of 106 kPa when the confining stress (minor principal stress) is 40 kPa. What is the value of cohesion, cu (undrained shear strength) for this sample? If another sample of the same soil is tested in an unconfined compression test, what axial load is expected to cause failure? 23 A clay soil is subjected to a triaxial test under unconsolidated–undrained conditions. At failure, the major and minor principal stresses are 3,000 psf and 1,000 psf, respectively.

(a) What is the value for cohesion, cu, for this soil? (b) If this soil were subjected to an unconfined compression test, what axial load would result in failure? 24 A clay soil has an undrained shear strength (cohesion, cu) of 800 psf. How could the strength of this soil be increased? 25 A deposit of homogeneous mixed soils (soil including both coarse-grained and fine-grained sizes) possesses shear strength due to both cohesion and internal friction 1c = 22 kPa, f = 28°2. What shear strength is expected on a plane where the normal stress is 70 kPa? 26 A mixed soil is found to possess a unit cohesion of 500 psf and an angle of internal friction of 30° (results of CU tests). What shear strength is expected at a point in the soil deposit where the normal stress is 1,500 psf? 27 Separate triaxial compression tests are performed on identical soil samples. The soil is a saturated mixed soil. Both tests follow the procedure for consolidated–undrained conditions. At failure, the respective minor and major principal stress values are 20 kPa and 46 kPa for sample one and 40 kPa and 80 kPa for sample two. From these data, determine the value for fCU and the cohesion intercept c on the Mohr’s circle shear stress coordinate. (This problem can be solved graphically using the Mohr’s circle.) 28 It is desired to obtain information about the in situ strength properties, c and f, for a saturated clay soil, and also the preconsolidation pressure. Direct shear tests under consolidated–undrained conditions are performed on three identical samples of the clay. For sample one, the applied normal pressure is 30 kPa and the sample fails when the shear stress is 19 kPa; for sample two, the applied normal stress is 70 kPa and the sample shears at 28 kPa; for sample three, the applied normal pressure is 120 kPa and the sample shears at 45 kPa. From these data, estimate the values of the in situ cohesion c and fCU, and the preconsolidation stress svc. 29 A triaxial test is performed on a normally consolidated clay. The sample is further consolidated by the test of all-around confining pressure

441

Shear Strength Theory prior to application of the axial load. The axial force is then applied very slowly. The equipment pore pressure lines remain open so that drainage of soil pore water can occur during the test. (The described conditions outline the procedure for a consolidated–drained-type test.) The sample fails (shears) when the σ3 pressure is 21 kPa and the s1 value is 61 kPa. From these data, determine the angle of internal friction for the effective stress strength–failure envelope (that is, determine fCD). 30 A triaxial test is performed on a clay sample as described in the preceding problem, but the sample fails when the axial stress (major principal stress) is 30 psi and the confining stress (minor principal stress) is 10 psi. Determine the angle of internal friction for the effective stress strength envelope, fCD. 31 A normally consolidated clay sample is subject to a triaxial test where pore water pressure measurements are made. A consolidated–undrained type of test is performed. The sample fails (shears) when the total all-around confining pressure is 45 kPa and the total axial pressure is 97 kPa. At failure, the recorded pore water pressure is 20 kPa. Determine the angle of internal friction fCD for the effective stress strength envelope and also the value for fcu. 32 A vane-shear test was performed at the bottom of a borehole drilled in a fine-grained (clay) deposit, using a 2.0-in. vane ( D = 2.0 in. or 50.8 mm, H = 4.0 in. or 101.6 mm). In the undisturbed condition, the measured torque (soil resistance) was 36 N-m. A torque resistance of 9.5 N-m was recorded for the subsequent rotation used to determine the remolded shear strength. (a) Indicate the value for the test-determined (uncorrected) undisturbed, undrained soil shear strength (or cohesion), the remolded shear strength, and the clay sensitivity. (b) Samples of the soil tested in a laboratory indicate the plasticity index, PI, is 62. Indicate the corrected undisturbed, undrained shear strength value to use for analysis and

442

design purposes. The test depth was 7 m, and the preconsolidation stress was determined to be 260 kPa. 33 Provide a brief description of the term stress path as related to stress conditions in a soil mass. 34 The results of consolidated–drained triaxial compression tests on samples of saturated clay indicate that fCD is 24° for this soil. Use this information to determine the slope of the K′f line to use for a stress path analysis. 35 A consolidated–undrained triaxial compression test is performed on a sample of saturated clay. The sample has been isotropically consolidated so that at the start of the test, s1 = s3 = 25 kPa. The slope of the K′f line is 30°. Use the stress path method to determine the value of the axial stress (s1 at start plus Δσv) when the effective stress path intersects the K′f line to cause shearing (failure) of the sample. Assume that the Skempton pore pressure coefficient at failure, Af , is 1.0. (Hint: Refer to Figure 40.) 36 A consolidated–undrained triaxial compression test is performed on a sample of saturated clay. The clay is anisotropically consolidated, and s3 = 0.6 s1. The slope of the K′f line is 28°. At the start of the test, s3 is 24 kPa and s1 is 40 kPa. Use the stress path method to determine the value of axial pressure when the effective stress path intersects the K′f line and the sample shears. Assume that the Skempton pore pressure coefficient at failure Af is 1. 37 Stress path diagrams for applications of increasing loading as illustrated in Figure 41 indicate that a clay soil shear strength will increase as consolidation progresses and pore water pressures dissipate (that is, the effective stress eventually equals the total stress). Excavation and related effects on clay soil represent a different condition, however. Referring to Figure 43, briefly describe in general terms what concerns exist relating to construction safety when excavations are undertaken in clay deposits.

Answers to Selected Problems

1. (a) ␾ = 34° (b) ␶ = 97.9 kPa 2. (a) ␴ = 102 kPa, ␶ = 68.8 kPa (b) ␾ = 34° (c) med. dense 3. ␾ L 33° 5. ␾ L 31°, apparent c = 14 kPa 6. ␾ L 27°, apparent c = 1000 psf 7. ␾ = 33° 8. ␴1 = 154 kPa 9. ␾ L 37.5° 10. ␴1 = 74 psi 12. ␴1 L 3690 psf 15. ␶ = 81.3 kPa 16. ␾ = 33.7°, ␶ = 3340 psf

␾ L 39° ␾ L 38° c = 33 kPa, qu = 66 kPa c = 1000 psf, qu = 2000 psf ␶ = 1366 psf ␾CD = 29° ␾CD = 30° ␾CD = 30.6°, ␾CU = 21.5° 1a2 undisturbed c = 75.55 kPa remolded c = 19.94 kPa sensitivity = 3.8 34. ␣ = 22.1° 35. start ␴1 + ¢␴1 = 43.3 kPa 17. 19. 22. 23. 26. 29. 30. 31. 32.

443

444

Earthquakes and the Effects

Earthquakes as events of nature have probably been occurring since the early periods of the planet’s existence; they will probably continue to occur in the future. Earthquakes unleash awesome amounts of uncontrolled energy, without warning and with typical suddenness. Great destruction is associated with earthquakes; for the human race, fear is the learned response. In recent decades, scientists have made significant advances in understanding earthquakes and have gained some insight into methods of reliable protection; much remains to be discovered, however.

1

Causes and Effects An earthquake is the phenomenon of vibration, shaking, or movement of the ground that occurs when a release of energy into the earth results in the transmission of shock, or seismic, waves. Typically, such release of energy is caused by the sudden fracture of rock in the lithosphere or overlying crust, or by a sudden rupture (shift) at the boundaries where plates of lithosphere meet, or along a fault plane (that is, along a pre-existing fracture in the crust); this type of event is classified as a tectonic earthquake. Energy release capable of causing earthquakes can also occur during volcanic eruptions (volcanic earthquakes) and from humaninduced explosions such as the detonation of nuclear devices (explosion earthquakes). With tectonic earthquakes, the release of energy may be attributed to the elastic rebound property of rock materials located in the zone responsible for the earthquake. Elastic rebound results where rock that is strained and has been deformed (because tectonic movement or thermal or other change occurred) suddenly releases the associated energy of deformation when the rock fractures or slips (analogous to the process of a compressed steel coil springing back toward the original shape when the confining force is released).

From Essentials of Soil Mechanics and Foundations: Basic Geotechnics, Seventh Edition. David F. McCarthy. Copyright © 2007 by Pearson Education, Inc. Published by Pearson Prentice Hall. All rights reserved.

445

Earthquakes and the Effects

Tectonic movements are relative movements between rock zones, usually as adjacent plates of lithosphere or overlying crust shift independently. Deep-rock tectonic movements are also associated with the presence of water, when the micro-cracks which develop in the rock as movement and strain occurs become filled with water. The expanded (dilated) rock cannot close to the original condition as the movement continues (the micro-cracks do not close completely because of the presence of water). Eventually, the loss of strength within the rock mass (because of the loss of internal bonding and inability of the micro-cracks to rebond, and the loss of friction along rough surfaces) results in fracture. (Interestingly, large increases in the volume of deep rock may be reflected by changes in the ground surface level and also by ground tilting, physical features that when monitored may be useful in forecasting the advent of an earthquake.) The location within the earth where the rock fracture or rupture develops to trigger an earthquake is identified as the focus or hypocenter. The location on the earth’s surface directly above the hypocenter is the epicenter (Figure 1). The majority of earthquakes are in the category of shallow-focus events, where the focus is less than 70 km deep (roughly 45 miles). Most of the planet’s shallow earthquakes originate in the trench regions where plates of lithosphere meet (see Figure 1-20). The energy released into the earth zone surrounding suddenly ruptured or slipped rock, or the zone of a volcanic or human-induced explosion, radiates outward in a wavelike manner, producing seismic waves (i.e., shock waves) and heat. The type of wave movement, and the velocity, depend on the elastic properties of the rock, soil, and water or other liquid material being traversed. Two categories of waves, body waves and surface waves, are used to describe the wave transmission (Figure 1 and Table 1). For a

Love wave

Epicenter X

Rayleigh wave Fault

Shear wave

X Focus or hypocenter

Compressional wave (P-wave)

Wave fronts

Figure 1 Schematic illustration of the directions of vibration caused by body and surface seismic waves generated during an earthquake. When a fault ruptures, seismic waves are propagated in all directions, causing the ground to vibrate at frequencies ranging from about 0.1 to 30 Hz. Compressional and shear waves mainly cause high-frequency (greater than 1 Hertz) vibrations. Rayleigh and Love waves mainly cause low-frequency vibrations. (Source: Facing Geologic and Hydrologic Hazards, W. W. Hays (ed.), Geological Survey Professional Paper 1240-B, 1981, courtesy of USGS)

446

Earthquakes and the Effects Table 1 Earthquake-Induced Ground Motion Waves Type

Properties

Primary or P-wave (a body wave)

Fastest body wave. Longitudinal dilational and compressional wave; energy transmitted acts to cause soil and rock materials to move/pulsate in the direction of wave travel. P-waves travel through solids and liquids. Wave velocity typically about 6 km/sec in sound rock.

Secondary (aka Shear) or S-wave (a body wave)

S-waves move through solid materials only (but not liquids); cause twisting/shear type movements in the earth which are transverse to the direction of the wave propagation (such transverse type waves create side-to-side and up-and-down motions). About 50 percent slower than P-waves but usually cause larger movement.

Love wave (a surface wave)

Wave whose energy travels in the surface zone of earth material with horizontal shear-type back-and-forth motions (sideways but no vertical) that are transverse to the direction of propagation. Travel velocity is less than for body waves.

Rayleigh wave (a surface wave)

Complex rolling or ripple type of wave moving through the surface zone of earth material in the direction of the wave propagation (visualize the rolling surface motion of ocean water waves approaching a shoreline). The rippling motion of the ground surface is visually apparent during major earthquakes; advancing waves can be seen approaching an area. Slowest type wave but often largest and most destructive type of earthquake-induced ground motions.

seismic-type event (earthquake, underground explosion, etc.), the early waves that pass through an area are typically the larger of each of the different categories. Traveling within the earth, two types of waves occur. Body waves, termed primary or P-waves, consist of compression–dilation (push–pull) types of movements. Shear-type waves, termed secondary or S-waves, cause up-and-down and side-to-side movements (Figure 2). The depiction of the effect of a shear wave in Figure 2(b) is, for simplicity, shown as having two-dimensional vertical movements; actual propagation of a shear wave through the earth occurs as a three-dimensional undulation as could be represented by a combination of Figure 2b and c. P-waves travel at higher velocities than S-waves, and will move through solids and liquids. S-waves will not be transmitted through liquid materials. Typically then, when an earthquake occurs, the effects of the P-wave will be the first noted. Building occupants typically first experience vertical jolt, and then horizontal shaking. Along the earth’s surface zone the energy transmission is in the form of Love waves and Rayleigh waves (Figure 2). Love waves are the shear-type wave that cause ground motions transverse to the direction of propagation, similar to S-waves except that only horizontal movement takes place (no vertical movement). Rayleigh waves occur as a rolling or rippling type of wave motion (similar to surface ripples when a stone is dropped into water). Surface waves move at lower velocities than body waves; generally, the Love waves travel at higher velocity than Rayleigh waves. Though the slowest of earthquake ground waves, the Rayleigh waves are often the largest and most destructive. Primary waves traveling through sound rock have a typical velocity of 5 to 6 km/sec; shear waves travel at roughly half the velocity of the primary waves. Except for locations

447

Primary (P) wave

Compressions

(a)

Body Waves

Dilatations Wave direction Secondary (S) wave

(b)

Amplitude Wavelength Wave direction Love wave

Surface Waves

(c)

Wave direction Rayleigh wave

(d)

Wave direction

Figure 2 Concept of the form of the body and surface waves resulting from earthquake. (Source: Nuclear Explosions and Earthquakes: The Parted Veil by Bruce A. Bolt. © 1976 by W. H. Freeman and Company. Used with permission.)

448

Earthquakes and the Effects

close to the epicenter, the propagation frequency for body waves is typically in the range between 0.5 and 20 hertz (Hz, or cycles per second), whereas surface waves usually have a frequency less than 1 Hz. (The frequency in Hz is the reciprocal of the time period T in seconds between successive waves moving past a reference location; a higher seismic frequency indicates that a greater number of waves per second is traveling through the earth; for example, if the period between successive waves is 0.2 sec, the frequency is 5 Hz, whereas if the period is 4 sec, the frequency is 0.25 Hz.) High-frequency waves dampen or diminish more rapidly than low-frequency waves, one result being that the low-frequency (surface) waves travel further from an epicenter than the high-frequency (body) waves. The amplitude or extent of ground motion is affected by the type of earth material, usually being greater in soil than in rock. The transmission of P-waves is similar to the movement of sound waves. P-waves reaching the earth’s surface can be passed into the atmosphere as sound waves (sound wave frequencies greater than about 15 Hz can be heard by humans). However, most of the energy of both P-waves and S-waves is reflected back into the earth. The surface zone thus becomes subjected to seismic waves that tend to go in opposite directions simultaneously, a situation that can result in amplication of the movement (i.e., shaking). The vibratory motion experienced by a rock or soil formation relates to the distance from the earthquake focus and the energy released. The seismograph is an instrument developed to measure the type and extent of movement at the instrument’s point of placement. In principle, the instrument consists of a weight suspended by a wire from a frame (somewhat like a pendulum); when the frame is shaken, the inertia of the suspended weight lags behind the motion of the frame. If a pen or other tracking mechanism attached to the suspended weight continuously plots the relative movement between the vibrating frame and the suspended weight against time, the record of the up-and-down, or back-andforth, undulations of the earth is a seismogram. Typically, surface seismographs will record movements in three directions (e.g., vertical, north-south, and east-west directions). The concept of the seismograph is shown in Figure 3, along with an example of a seismogram or accelogram (the accelograph is a strong-motion seismograph that operates only when large, earthquake-size ground vibrations occur). Earthquake magnitude refers to a numerical scaling system to rate the severity of a seismic event on the basis of the total strain energy released. The magnitude scaling system permits evaluation of earthquakes regardless of the location of the focus or epicenter. The concept of earthquake magnitude was presented by Charles Richter in 1935 and included a numerical scale which ranged upward from zero, with no upper limits (but the severest known earthquake was approximately a magnitude 9). The Richter magnitude remains in widespread usage because of the general population’s familiarity with the term, but is now identified as local magnitude, ML. The events having a Richter magnitude of 8 of greater are classified as Great earthquakes; events having a Richter magnitude between 7 and 8 are classified as Major earthquakes. The local magnitude value is defined as the logarithm (base 10) of the maximum seismic wave amplitude (expressed in thousandths of a mm) recorded on a torsional-type seismograph at a distance of 100 km from an earthquake epicenter. The type of seismic wave (P-, S-, or surface wave) selected to compute magnitude is the one creating the largest amplitude. The ground motion data recorded by a seismograph will be influenced by its distance from the epicenter, of course; the nomograph shown in Figure 4 was developed

449

Earthquakes and the Effects (a)

(b)

North-south component

0.3 0.2 0.1 0 −0.1 −0.2 −0.3 −0.4

Surface waves

Acceleration (g)

S-waves East-west component 0.2 0.1 0 −0.1 −0.2 −0.3 0.3 0.2 0.1 0 −0.1 −0.2 −0.3

Vertical component

P-waves

5

10 s

Figure 3 Seismograph and seismogram. (a) Concept of pendulum seismograph for recording the horizontal and vertical directions of ground motion. (Source: USGS National Earthquake Information Center) (b) Accelerogram of the 1940 Imperial Valley, California, earthquake recorded at EI Centro, California. (Source: W.W. Hays, 1980, Procedures for Estimating Earthquake Ground Motions, U.S. Geological Survey Professional Paper 1114, courtesy of USGS)

to include the factor of distance between a seismograph’s actual location and the 100-km reference location and permit simple determination of magnitude. Referring to the illustration within Figure 4, note that the time difference between the arrival of P-waves and S-waves is used to determine distance from the epicenter.

450

Earthquakes and the Effects

3 2

S-wave

P-wave

Trace amplitude = 23 mm

1

1

2

S-P time = 24 s

500

50

400 40

100 6 50

300 30 5 200

20

20 10

4 100

5

10

60 6

3

40 4

2 1

2 0.5

20 2 0-5 Distance (km) / S-P TIME (s)

1

0.2

0.1 0 Amplitude (mm) Magnitude scale

Figure 4 Determination of Richter magnitude. Using the maximum amplitude of the seismogram and the difference in arrival times of the P- and S-waves, the value of magnitude can be read from the nomograph. (Source: W. W. Hays, Procedures for Estimating Earthquake Ground Motions, U.S. Geological Survey Professional Paper 1114, 1980, courtesy of USGS)

Because of the logarithmic scale, each successive unit of Richter magnitude represents an amplitude 10 times greater; refer to Illustration 1 for a calculation showing the energy released by an event of known magnitude, and a typical difference in released energy for an event of the next higher magnitude. Since the time of Richter’s original work, additional earthquake magnitude scales have been developed in an attempt to obtain better definition of occurrences. These alternative magnitude scales are usually based on measurement of select seismic waves caused by the event.

451

Earthquakes and the Effects Table 2 Listing of Earthquake Magnitude Scales Scale Name or Identification

Definition

Description of Terms

General Comments

Local Magnitude, ML (the original Richter scale)

Logarithm (base 10) of the maximum amplitude seismic wave (either P-, S-, or surface wave)

Amplitude expressed in 0.001 mm

Seismic wave amplitudes as measured by a Wood-Anderson torsional seismograph located 100 km from epicenter of quake. Seismograph must have a natural period of 0.8 sec, magnification of 2800, with a damping coefficient of 80%.

Surface Magnitude, Ms

Ms = also: Ms = also: Ms = also: Ms =

A = horizontal component of the Rayleigh surface wave, for 20-sec period, in microns T = period of seismic wave, in seconds d = epicenter distance, in degrees L = length of fault activated by event Af = fault area 1length * depth, in km22

Originally developed to determine a magnitude for shallow-focus earthquakes at relatively long distances.

Ms = surface magnitude ML = local magnitude

Developed to measure magnitude of deep-focus earthquakes.

G = modulus of rigidity (approximately 3 * 106 ton>m2 or 3 * 1011 dyne>cm2)

Indirect measurement of earthquake energy.

Body Wave Magnitude, Mb

log1A>T2 + 1.66d + 3.3 1.7 + 0.8 ML - .01ML2 2.02 + 1.14 log L 4.15 + log Af

Mb = 0.56Ms + 2.9 where Ms = 1.7 + 0.8ML - 0.1ML2

Seismic Moment, Mo

Mo = GAf D (units are ton-m or dyne-cm)

Af = fault area (length * depth, in m2) D = longitudinal displacement of fault, in m Mw = 2冫3log Mo - 10.7

Seismic Moment Magnitude, Mw Richter Magnitude, M

Mo = seismic moment

M = ML 1for ML 6 5.92 M = MS 1for 5.9 6 MS 6 8.02 M = MW 1for 8.0 6 MW 6 8.32

Magnitude based on seismic moment. Redefinition, general usage Richter scale.

A listing of various earthquake magnitude scales is presented in Table 2. An expression to roughly approximate the energy E (in ergs) released by an earthquake of local magnitude ML or Richter magnitude M is: log10 E = 11.8 + 1.5M n

E = 10 ergs

452

(1a)

Earthquakes and the Effects

where n = 11.8 + 1.5M (Note: 1 erg = 1 dyne-cm = 7.382 * 10-8 ft lb of energy, or 1 ft lb = 13.547 * 106 ergs.) Another expression proposed for rough evaluation of energy released using the determination of surface magnitude, MS, is: E = 104.8 + 1.5Ms joules

(1b)

(Note: 1 joule = 1.36 ft lb; 1 erg = 107 joules.) Illustration 1 For a Richter magnitude 5 earthquake, an approximation of the energy released is: E ⬵ 10 n ergs ⬵ 10111.8 + 1.5 * 52 ergs E ⬵ 1019.3 ergs ⬵ 2 * 1019 ergs ⬵ 1.48 * 1012 ft lb For a magnitude 6 earthquake, an approximation of the energy released is: E ⬵ 10 20.9 ergs ⬵ 7.94 * 10 20 ergs ⬵ 58.6 * 1012 ft lb or approximately 40 times the energy of the magnitude 5 earthquake.

Illustration 2 Determine the surface magnitude, MS, value corresponding to a local (Richter) magnitude ML 6 earthquake, and approximate the earthquake energy associated with that MS value. Solution for ML = 6, estimate: MS = 1.7 + 0.8ML + 0.011ML22

(ref. Table 2)

MS = 1.7 + 0.8162 + 0.011622 = 6.14 and: E = 104.8 + 1.5MS E = 104.8 + 1.516.142 ⬵ 1014 joules or 74 * 1012 ft lb

Body wave transmission (subsurface transmission) becomes complicated at the boundaries where different earth materials meet, such as where a soil material overlies rock or where strata of rock materials join. The direction of a transmitted wave may be altered when it moves from one material into a different material (the phenomenon of refraction), and some of the wave energy can be reflected back into the source material. For either occurrence, the original wave may be converted into a new combination of P-waves and S-waves or S-waves into surface waves. The size of the seismic waves may increase or decrease as transmission occurs from sound dense rock into surficial weathered rock or soil.

453

Distance (in miles from epicenter)

Earthquakes and the Effects 0 Seismogram for location close to epicenter

1000 2000 3000

Seismogram for midrange location from epicenter Seismogram for L waves location distant from epicenter

4000 5000 6000

P-waves

S-waves

Quiet record

7000 0

5

10

15

20

25

30

35

40

45

50

55

60

Time (in minutes)

Figure 5 Representative comparison of travel-time relationships for major types of seismic waves.

A general reaction to the various combinations of seismic waves that reach an area is that surface zones tend to shake or move more than the deeper earth zones. The energy possessed by seismic waves dissipates with distance from the earthquake hypocenter. The faster, high-frequency body waves lose their energy over distance faster than the surface waves. The slow, long-period, long-wavelength, surface waves tend to propagate over long distances. For short-duration earthquakes, the P-waves and S-waves reach an area before the surface waves, but an earthquake of sustained duration can have both body and surface waves moving through a region at the same time. The effect of distance on the period of time for different types of seismic waves to reach a location is illustrated in Figure 5. Ground surface motions resulting from seismic waves are greater in the horizontal direction than in the vertical direction; the combination effects of P-waves and S-waves, or P-waves and S-waves plus surface waves, produce greater horizontal movement than vertical. On the ground surface, the effect of an earthquake is a vibratory-type motion (i.e., cyclic or oscillating). Movement occurs vertically up and down and also laterally in all directions. Factors affecting the stability of natural and human-made structures located at the earth’s surface include the direction of the seismic waves, the duration of the shaking, the distance or displacement of the ground motions, the acceleration of the ground motions, and the time period or frequency of the different seismic waves. For a seismictype event (earthquake, underground explosion, etc.), the early waves which pass through an area are typically the larger of each of the different categories. The effects of ground acceleration and the duration of seismic vibrations may be more responsible for the damage to a structure during an earthquake than the vertical movements; among other factors, the design and construction of structures typically is based on the everyday gravity (vertical) loads expected, and therefore structures have an inherent capacity for resisting vertical influence. Horizontal ground accelerations transferred into a structure cause an array of potentially damaging horizontal shear forces to cycle vertically up and down through the height of the structure, in accord with the principle of dynamics which identifies the relationship between force, mass, and acceleration (i.e., F = ma). Earthquake intensity is the scale of reference based on the effects of an earthquake felt at a particular location. The degree of damage that occurs is related to the distance from the

454

Earthquakes and the Effects Table 3 Modified Mercalli Intensity Scale, MMI (Abridged) Intensity I II

Effects Not felt except by a very few under especially favorable circumstances. Felt only by a few persons at rest, especially on upper floors of buildings. Delicately suspended objects may swing.

III

Felt quite noticeably indoors, especially on upper floors of buildings, but many people do not recognize it as an earthquake. Standing motor cars may rock slightly. Vibration like passing of truck. Duration estimated.

IV

During the day felt indoors by many, outdoors by few. At night some awakened. Dishes, windows, doors disturbed; walls make cracking sounds. Sensation like heavy truck striking building; standing motor cars rocked noticeably.

V

Felt by nearly everyone; many awakened. Some dishes, windows, etc., broken; a few instances of cracked plaster; unstable objects overturned. Disturbance of trees, poles, and other tall objects sometimes noticed. Pendulum clocks may stop.

VI

Felt by all; many frightened and run outdoors. Some heavy furniture moved; a few instances of fallen plaster or damaged chimneys. Damage slight.

VII

Everybody runs outdoors. Damage negligible in buildings of good design and construction; slight to moderate in well-built ordinary structures; considerable in poorly built or badly designed structures; some chimneys broken. Noticed by persons driving motor cars.

VIII

Damage slight in specially designed structures; considerable in ordinary substantial buildings, with partial collapse; great in poorly built structures. Panel walls thrown out of frame structures. Fall of chimneys, factory stacks, columns, monuments, walls. Heavy furniture overturned. Sand and mud ejected in small amounts. Changes in well water. Disturbs persons driving motor cars.

IX

Damage considerable in specially designed structures; well-designed frame structures thrown out of plumb; damage great in substantial buildings, with partial collapse. Buildings shifted off foundations. Ground cracked conspicuously. Underground pipes broken.

X

Some well-built wood structures destroyed; most masonry and frame structures with foundations destroyed. Ground badly cracked. Rails bent. Landslides considerable from river banks and steep slopes. Shifted sand and mud. Water splashed (slopped) over banks.

XI

Few, if any (masonry) structures remain standing. Bridges destroyed. Broad fissures in ground. Underground pipelines completely out of service. Earth slumps and land slips in soft ground. Rails bent greatly.

XII

Damage total. Waves seen on ground surfaces. Lines of sight and level distorted. Objects thrown upward into the air.

Source: Wood and Neuman, 1931, by U.S. Geological Survey, 1974, Earthquake Information Bulletin, v. 6, no. 5, p. 28.

earthquake epicenter and the magnitude, but also relates significantly to the type of facility or structure and the properties of the supporting rock or soil foundation materials. The Modified Mercalli Intensity Scale shown in Table 3 includes the experience data obtained from investigating damages and other after-effects of earthquakes. When used with geologic data and seismic probability information, intensity scales, such as the Modified Mercalli, have become a source of good experience-related information for establishing construction standards appropriate to the geographic area. Structures that will be located in areas where seismic activity can occur should be designed and constructed to resist the effects of such events. Earthquake-resisting design

455

Earthquakes and the Effects

procedures consider the influence of seismic-induced ground acceleration factors; the most basic approach has been to assign extra horizontal and vertical forces to the static design loads so as to simulate the effects resulting from an earthquake (as illustrated in Figure 6).1 The building codes for many areas of the world have evolved to include procedures for providing a level of earthquake resistance considered appropriate for the region. Building codes, as well as other technical sources, may provide maps indicating zones or levels of seismic risk, or levels of ground motion acceleration or velocity expected from earthquake activity; seismic maps indicating acceleration coefficients and velocity-related acceleration are based on consideration of ground motions, frequency of occurrence, and attenuation with distance. These seismic maps do not indicate the frequency of earthquakes but reflect the extent of ground movement anticipated from significant events which may occur in the area. (Such data are usually based on evaluation of past occurrences.) Building codes may state that the design regulations are primarily intended to achieve a reasonable level of public safety, whereby structures should withstand a moderate earthquake without experiencing significant damage and a major earthquake without collapse occurring. The seismic maps of the United States and North America imply the probability of large magnitude earthquake events is highest along the West Coast, the Pacific Northwest (including Alaska and Canadian territory), the Southeast Coastal Region of the United States, and the Central Continental Region traversing the states of Missouri and Mississippi. Though most of these locations are near known plate boundaries, a wide fault zone also exists in the Central United States, extending from the Gulf of Mexico northward then veering westward—possibly the effect of an ancient incomplete plate tectonic divergence (New Madrid Seismic Zone). Active areas in Canada include the southeast region near the US border and the Arctic margin region between Hudson Bay and Greenland (in addition to the highly active west coast region). Seismic activity in Mexico is concentrated along the Lower California peninsula (northwest Mexico) and the entire Pacific Coast region (extending along western and southern Mexico to the border of Central America). The Pacific Coast seismic zone continues into Central America but also diverges eastward into the Caribbean region beyond Cuba. For concept, the following listing represents a summary of seismic-related factors considered by most building codes as important to earthquake-resistant building design: 1. Level of seismic activity (i.e., ground motion) to assume for the building design procedure, including duration plus peak ground acceleration and peak ground velocity expected. 2. Information relating to the building’s structural assembly, including ability of the main structural system to absorb energy and sustain large deformations without major damage; the fundamental time period for the structure in order to help evaluate the tendency to oscillate when subject to ground motions; the dead load and live loadings (weights).

1The

effect of forces distributed throughout a building structure because of seismic activity can be evaluated using pseudostatic or dynamic analysis. The pseudostatic method involves assigning static-type forces to achieve effects that are considered comparable to those resulting from an earthquake. Dynamic analysis, considered more appropriate for representing a structure’s response to ground motion but analytically more rigorous than pseudostatic analysis, is based on applying cyclical loading to represent time-related variations in the magnitude and location of forces resulting from an earthquake. The effect of an earthquake on structures other than buildings (retaining walls, dams, earth structures) is often evaluated using pseudostatic analysis.

456

Earthquakes and the Effects FROOF F3 F2 F1

V Seismic-related ground motion creates lateral foundation force, V (a) Building structures

V Lateral foundation force V is distributed throughout height of the building as the F-forces

Additional force from earth pressures during earthquake Resultant force of earth pressures against wall, static condition

(b) Gravity-type retaining walls

Lateral force due to earthquake (additional to vertical force)

Direction of weight force, static condition (no earthquake) (c) Earth slopes

Figure 6 Typical procedures used to indicate effects of earthquake forces.

3. Subsurface conditions, including indication of soil and rock properties (such as classification, loose or dense condition, natural frequency or vibration period, etc.). 4. Building occupancy importance factors, relating to the need for the building to remain functional after a seismic event (e.g., buildings such as hospitals and those used to house public utility and communications operations have high importance).

457

125° 50° 120° 115° 110° 105° 100°

95°

29 10

5

15

125

35

10.5

113

7.0

35

54

29

35

31

125

10 0

34

30

60

11.4

25

21

40

25

60

100

25

18.8

16.9

40

18.7

35°

40

19.7

30

35

10.0

36

50

70

42 35

9.0

30°

30 35 40

10

40 27

40

50

110°

75

7.5

35

115°

28

20

28

25

25

120°

16.9

40

150

125

53

20

50

20

42

13.4 34

18.5

20

60

20

35

10

566 0 75

21

11.2

9.0

36

23

24

12.6

50

60

125

125

10

20

80

10

40

125

60

50 60

60

150 125

5

55 50 51 56

36

80

15

100

125

40°

37

58

48

18.1 8.0

35

40

26 50

37

54

47

15

12.9 10

100

50

35 40

150

13.5

45 10

110

47

17.2

40

60

30

150

43 40

10

90

12

20

42

100

25

70

84 113

40

30

28

45°

10.4

40

100

125

90 100

52

11.5

30

40

80

150

150

150

80

35

60

118

25 30 35

150 90 100

20 15

28

64

125 150

24

12.3

39

62

35°

20

59 50

60

40°

90

42

45°

50°

25

50

125

39

15

40

6.6

30°

Notes

10.3

15 17.7

– The acceleration values contoured are the random horizontal component. For design purposes, the reference site condition for the map is to be taken as NEHRP site class B. – Regional maps should be used when additional detail is required. – Contour values indicate %g

25° 105° 100°

95°

Figure 7 (a) 0.2 sec spectral response acceleration (5% of critical damping). Maximum considered earthquake ground motion for the conterminous United States, Site Class B [257]. Sources: (1) Building Seismic Safety Council, 1998, NEHRP Recommended Provisions for Seismic Regulations for New Buildings and Other Structures, FEMA 302; (2) Frankel, A., Mueller, C., Barnhard, T., Perkins, D., Leyendecker, E. V., Dickman, N., Hanson, S., and Hopper, M., 1996, National Seismic-Hazard Maps: Documentation June 1996: U.S. Geological Survey Open-File Report 96–532, 110 p3;

458

65°

50°

70° 75° 80° 85°

5 60 40 0 35

50°

90°

95°

30

35

46

35 40

40 30

6.0

3.4

35

70 60 50

5.5

45°

30

30

11.1 10

45°

27

5

25 30 25

33

40°

10

17.0

23

12.7

30

15

12.6

20

18.4

15

43

15

20

7.0

20

20 23

30 35 40

20

70°

40° 20 35 30

28

12.1

25

100

10

60 80

27

17 5 30 30 0 0

50

30

40

18.0

60

12 5

40 28

10 0 15 0

41

30

35°

60 80

166

20

15 16.8

30°

5

75° 25°

Scale 1:13,000,000

80° 85° 95°

100

100

200

300

400

500

600 MILES

90° 100

100

200

300

400

500

600 KILOMETERS

Figure 7 (b) 0.2 sec spectral response acceleration (5% of critical damping). Sources continued: (3) Frankel, A., Mueller, C., Barnhard, T., Perkins, D., Leyendecker, E. V., Dickman, N., Hanson, S., and Hopper, M., 1997, SeismicHazard Maps for the Conterminus United States, Map F - Horizontal Spectral Response Acceleration for 0.2. Second Period with 2% Probability of Exceedance in 50 Years: U.S. Geological Survey Open-File Report 97-131-F, scale 1:7,000,000; (4) Petersen, M., Bryant, W., Cramer, C., Cao, T., Reichle, M., Frankel, A., Lienkaemper, J., McCrory, P., and Schwartz, D., 1996, Probabilistic Seismic Hazard Assessment for the State of California: California Division of Mines and Geology Open-File Report 96–08, 66 p., and U.S. Geological Survey Open-File Report 96-706, 66 p. [257]

459

125° 50° 120° 115°

25

110° 105° 100°

95°

50°

10

9.0

40

25 4

15 3.4

60

45°

4.2

1.4

8.7 3.0

40 30

10.4

15

40

11.1

25

15

45°

10

10

40

25

30

8

30 40°

2.4

60 9.2

8.7

3.0

8.2

27 8.7 30 30

25

3.6

25

4.5

11.4

25

30

50

12.9

20

30

20

13.0

6

3.7

40

60 149

40°

6.6

8

30

60

25

20

8 6

60 60

5.5

100

13.0

60

5.4

20

8.7

8

6.3

7.4

18.5

4.3

15

15

10

13.8

35°

10

35°

3.7

25

21 19.3

3.1

8.2

30

6

40 25

15

159

7.0

9.5

15

40

19.4

3.7

25

10

40

50

60

30

18.9

32

16.4

4

4.3 8.6

12.7

15

40

24

25 30

22 22

9.5

11.4

7.8

3.3 9.7

8.3

10

30°

8

10

110°

120° 115°

2.6

Notes

30°

8.9

4

– The acceleration values contoured are the random horizontal component. For design purposes, the reference site condition for the map is to be taken as NEHRP Site Class B. – Regional maps should be used when additional detail is required. – Contour values indicate %g 25° 105° 100°

Figure 7 (continued) (c) 1.0 sec spectral response acceleration (5% of critical damping).

460

95°

65°

50°

70° 75° 80° 85° 50°

90°

95°

15

45°

10

11.0

1.5

6

2

15

11.7 2 45°

40°

4

4

5.4

8.4 70°

40°

10

8

6

9.2

15 20 25

13.0

15

60 40 30

13.9 13.0

12.5

20

30 25

13.2

35°

40

12.5

10

47 10

8

30°

4

2

75° 25°

Scale 1:13,000,000

80° 85° 95°

100

100

200

300

400

500

600 MILES

90° 100

100

200

300

400

500

600 KILOMETERS

Figure 7 (d) 1.0 sec spectral response acceleration (5% of critical damping).

461

Earthquakes and the Effects

2

Applications to Design In North America, The Provisions for Seismic Regulations for New Structures developed under the National Earthquake Hazard Reduction Program (NEHRP) (United States)2 and the National Building Code of Canada (NBCC) are representative of modern seismic design criteria being incorporated into building codes.3 Both National Codes reference spectral response seismic design maps (developed on the basis of regional geologic conditions and the seismic history for the region) to anticipate the severity of seismic events and ground motions that could occur, and present a related procedure for determining the seismic forces to apply to building design. The two seismic codes have a basic similarity but use different engineering design procedures; details of the NEHRP procedure are presented in this chapter for a general understanding of the process of seismic design. The NEHRP provisions include methods for calculating a design value for the lateral shear force V acting at the foundation or base level of a building when subject to an earthquake, and the method for distributing that lateral shear force throughout the height of the structure to complete the analysis and design. Expressions for the base-level lateral shear force V are shown herein, along with related maps and tables, to illustrate the application of seismic and subsurface information.4

NEHRP Provisions (USA) The NEHRP provisions include use of spectral response seismic design maps developed for areas of the United States (based on U.S. Geological Survey seismic hazard maps which provide values for peak ground accelerations, or PGA, expected for a design-condition earthquake); see Figure 7. The Spectral Response Acceleration (or SA) maps reflect the effect of an earthquake’s horizontal ground accelerations on a structure (the mapped SA values are not the ground accelerations, PGA, which occur during an earthquake). The design model of a structure’s response spectrum for the occurrence of ground motion is shown in Figure 8 and reflect that the early seismic waves which reach an area typically are the larger of each wave type. Figures 7(a,b) and 7 (c,d) present values of spectral response acceleration for short time periods (Ss) and at 1 sec (Sl); the map contours present the response acceleration values as a percent of g (the acceleration of gravity). [Map values for PGA and 0.2 seconds spectral acceleration relate to short-period energy that has the greatest harmful effect on short period structures such as low-rise buildings (e.g., one and two-story buildings) whereas spectral accelerations of 1.0 seconds and greater represent the degree of ground motion that has the greatest effect on longer-period structures such as tall buildings and bridges (reflections of the duration involved for shaking movements to propagate through a structure).] The values presented on the maps relate to preventing a collapse of structures when subject to the design earthquake. The mapped 2Prepared

by the Building Seismic Safety Council for the U.S. Federal Emergency Management Agency (FEMA). The full provisions are available from FEMA, Washington, D.C. as Publications 302 and 303. 3Data

relating to the Seismic Hazard assigned to areas of Canada can also be obtained by contacting Earthquakes Canada, Geological Survey of Canada, Ottawa, ON K1A0Y3, or National Research Council of Canada, Ottawa, ON K1A 0R6. 4Seismic effects predicted by building codes typically include the influence of factors such as knowledge of local geology and historic experiences. For construction projects, the designer must apply, as a minimum standard, criteria from the building code having jurisdiction at the site area.

462

Spectral Response Acceleration SA

Earthquakes and the Effects

Sds SA = Sdl /T

Sdl

To

Ts

1.0

Period T

Figure 8 Model of response spectrum for structures subject to ground motions. Table 4 Facility Seismic Use Groups and Related Importance Factors Seismic Use Group Designation

Facilities in Seismic Group (Partial Listing)

Importance Factor, I

III

Fire, rescue, and police stations; hospitals; power generating stations; communication centers; aviation control towers and air traffic control centers; emergency shelters; emergency vehicle garages; structures storing considerable toxic or explosive substances; water treatment facilities required to maintain supply for fire suppression

1.50

II

Educational structures; public assembly structures for more than 300 persons; all structures with capacity for more than 5,000 persons; jails and detention facilities; day-care centers with capacity greater than 150 persons; water treatment facilities; waste water treatment facilities

1.25

All structures not assigned to Seismic Group III or II

1.0

I

contours indicate the effect of ground motions on a structure for the maximum considered earthquake. The maximum considered earthquake is defined as the event having a 2 percent likelihood of exceedence in 50 years, and is not the strongest probable earthquake which might occur in an area (the mapped spectral response values reflect a judgment that it is not practical to require that all structures be designed to withstand the rare worse event). The procedure to establish the design value for seismic-event-base shear V requires information about the type and use of the building structure to determine the Seismic Use Group (buildings will be classed as either a group III or II or I structure, see Table 4), a related Occupancy Importance Factor (see Table 4), and knowledge of the site subsurface conditions (termed the Site Class, see Table 5). The NEHRP provisions permit structural analysis for loadings resulting from earthquakes to be determined using an equivalent lateral force procedure, where the seismic base shear V is calculated from: V = CSW

(2)

463

Table 5 Site Class Definitions A

Hard rock with measured shear wave velocity, vs 7 5,000 ft>sec (1500 m/s)

B

Rock with 2,500 ft>sec 6 vs … 5000 ft>sec (760 m>s 6 vs … 1,500 m>s)

C

Very dense soil and soft rock with 1,200 ft>sec 6 vx … 2,500 ft>sec (360 m>s 6 vs … 760 m>s) or with either N 7 50 or su 7 2,000 psf (100 kPa)

D

Stiff soil with 600 ft>sec … vs … 1,200 ft>sec (180 m>s … vs … 360 m>s) or with either 15 … N … 50 or 1,000 psf … su … 2,000 psf (50 kPa … su … 100 kPa)

E

A soil profile with vs 6 600 ft>sec (180 m/s) or with either N 6 15, su 6 1,000 psf, or any profile with more than 10 ft (3 m) of soft clay defined as soil with PI 7 20, w Ú 40 percent, and su 6 500 psf (25 kPa)

F

Soils requiring site-specific evaluations: 1. Soils vulnerable to potential failure or collapse under seismic loading, such as liquefiable soils, quick and highly sensitive clays, collapsible weakly cemented soils. 2. Peats and/or highly organic clays (H 7 10 ft [3 m] of peat and/or highly organic clay where H = thickness of soil) 3. Very high plasticity clays (H 7 25 ft [8 m] with PI 7 75) 4. Very thick soft/medium stiff clays (H 7 120 ft [36 m]) Exception: When the soil properties are not known in sufficient detail to determine the Site Class, Site Class D shall be used. Site Classes E or F need not be assumed unless the authority having jurisdiction determines that Site Classes E or F could be present at the site or in the event that Site Classes E or F are established by geotechnical data.

4.1.2.2 Steps for Classifying a Site (also see Table below): Step 1:

Check for the four categories of Site Class F requiring site-specific evaluation. If the site corresponds to any of these categories, classify the site as Site Class F and conduct a site-specific evaluation.

Step 2:

Check for the existence of a total thickness of soft clay 7 10 ft (3 m) where a soft clay layer is defined by: su 6 500 psf (25 kPa), w Ú 40 percent, and PI 7 20. If these criteria are satisfied, classify the site as Site Class E.

Step 3:

Categorize the site using one of the following three methods with vs, N , and su computed in all cases as specified by the definition in Sec. 4.1.2.2: a. vs for the top 100 ft (30 m) (vs method) b. N for the top 100 ft (30 m) (N method) c. Nck for cohesionless soil layers (PI 6 20) in the top 100 ft (30 m) and average su for cohesive soil layers (PI 7 20) in the top 100 ft (30 m) (su method) Summary Table, Site Classification Site Class E D C

vs

N or Nck

6 600 fps ( 6180 m>s) 600 to 1,200 fps (180 to 360 m/s)

615

71,200 to 2,500 fps (360 to 760 m/s)

15 to 50 750

su 61,000 psf ( 650 kPa) 1,000 to 2,000 psf (50 to 100 kPa) 72,000 ( 7100 kPa)

Note: If the su method is used and the Nck and su criteria differ, select the category with the softer soils (for example, use Site Class E instead of D).

The shear wave velocity for rock, Site Class B, shall be either measured on site or estimated for competent rock with moderate fracturing and weathering. Softer and more highly fractured and weathered rock shall either be measured on site for shear wave velocity or classified as Site Class C. The hard rock, Site Class A, category shall be supported by shear wave velocity measurements either on site or on profiles of the same rock type in the same formation with an equal or greater degree of weathering and fracturing. Where hard rock conditions are known to be continuous to a depth of 100 ft (30 m), surficial shear wave velocity measurements may be extrapolated to assess vS. The rock categories, Site Classes A and B, shall not be used if there is more than 10 ft (3 m) of soil between the rock surface and the bottom of the spread footing or mat foundation. Source: Courtesy of NEHRP, 1997.

464

Earthquakes and the Effects

where Cs = seismic response coefficient that reflects the combined effects of the structural design, building usage (the importance factor), and site conditions including geographic location [however, upper and lower limits apply: Cs1max2 7 Cs 7 Cs1min2] W = “seismic weight” of the structure, equal to the total dead load plus applicable portions of live loads, equipment, etc. The value for Cs is calculated from: CS = where

12>32Sms 12>32Fa Ss Sds = = 1R>I2 1R>I2 1R>I2

(3)

Sds = design spectral response acceleration in the short range period (the s subscripts refer to short) = 23 Sms = 23 Fa Ss Sms = maximum considered earthquake spectral response acceleration in the short range period = FaSs R = response modification factor, a value related to the type of structural framing and the behavior when subject to earthquake (see Table 6) Table 6 Building Construction Type and Response Modification Coefficient, R Building Height Limitation by Site Class (feet) Building Structure Type A. Bearing Wall Systems Ordinary steel braced frames Ordinary reinforced concrete shear walls Ordinary plain concrete shear walls Ordinary reinforced masonry shear walls B. Building Frame Systems Ordinary steel concentrically braced frames Ordinary plain concrete shear walls Ordinary reinforced concrete shear walls Ordinary plain concrete shear walls Ordinary reinforced masonry shear walls Ordinary plain masonry shear walls C. Moment Resisting Frame Systems Special steel moment frames Ordinary steel moment frames Ordinary reinforced concrete moment frames D. Structural steel systems not specifically detailed for seismic resistance

R

B

C

D

E

F

4 4H 1H 2

NL NL NL NL

NL NL NP NP

160 NP NP NP

160 NP NP NP

160 NP NP NP

5 2 5 2 2H 1H

NL NL NL NL NL NL

NL NP NP NP NP NP

160 NP NP NP NP NP

100 NP NP NP NP NP

100 NP NP NP NP NP

8 4 3 3

NL NL NL NL

NL NL NP NL

NL 35 NP NP

NL NP NP NP

NL NP NP NP

NL = no limit NP = not permitted

465

Earthquakes and the Effects

I = occupancy importance factor (see Table 4) Ss = mapped maximum considered earthquake spectral response acceleration for short range period, use contour value obtained from SA Map Figure 7(a,b) (map values are expressed as percent of g) Fa = function of the Site Class and Ss (See Table 7(a)) However, the value for Cs used to determine V need not be taken greater than Cs(max), where: Cs1max2 = with:

Sdl = 12>32Sml = 12>32FvSl

Sdl 1T21R>I2 1the l subscripts refer to long2

Sdl = design spectral response acceleration in the 1.0 sec long range period 2

2

= 3 Sml = 3 FvSl

Table 7(a) Values of Fa as a Function of Site Class and Mapped Short-Period Maximum Considered Earthquake Spectral Acceleration Mapped Maximum Considered Earthquake Spectral Response Acceleration at Short Periods Site Class

Ss … 0.25

Ss = 0.50

Ss = 0.75

Ss = 1.00

Ss Ú 1.25

A B C D E F

0.8 1.0 1.2 1.6 2.5 a

0.8 1.0 1.2 1.4 1.7 a

0.8 1.0 1.1 1.2 1.2 a

0.8 1.0 1.0 1.1 0.9 a

0.8 1.0 1.0 1.0 a a

Note: Use straight line interpolation for intermediate values of Ss. a = Site-specific geotechnical investigation and dynamic site response analyses shall be performed.

Table 7(b) Values of Fv as a Function of Site Class and Mapped 1 sec Period Maximum Considered Earthquake Spectral Acceleration Mapped Maximum Considered Earthquake Spectral Response Acceleration at 1 sec Periods Site Class

Sl … 0.1

Sl = 0.2

Sl = 0.3

Sl = 0.4

Sl Ú 0.5

A B C D E F

0.8 1.0 1.7 2.4 3.5 a

0.8 1.0 1.6 2.0 3.2 a

0.8 1.0 1.5 1.8 2.8 a

0.8 1.0 1.4 1.6 2.4 a

0.8 1.0 1.3 1.5 a a

Note: Use straight line interpolation for intermediate values of Sl. a = Site-specific geotechnical investigation and dynamic site response analyses shall be performed.

466

Earthquakes and the Effects Table 8 Coefficient Cu for Upper Limit on Calculated Period Design Spectral Response Acceleration at 1 sec, Sdl Ú0.4 0.3 0.2 0.15 0.1 0.05

Coefficient Cu 1.2 1.3 1.4 1.5 1.7 1.7

Sml = maximum considered earthquake response acceleration in the 1.0 second long range period = Fv Sl Sl = mapped maximum considered earthquake spectral response acceleration for the 1.0 second long range period, use contour value obtained from Map, Figure 7(c,d) T = fundamental period for the structure, CuTa Fv = function of the site class and Sl, see Table 7(b) Ta = approximate value for the structure’s fundamental period = 10.12(number of building stories), or = CTh10.752 n where CT = 0.030 for eccentrically braced steel frames (the metric coefficient is 0.0731), and CT = 0.020 for all other structural systems (the metric coefficient is 0.0488) hn = height above the base to the highest level of the structure (ft or m) Cu = coefficient obtained from Table 8 Also, the value for Cs need not be taken less than Cs(min) where:

Cs1min2 = 10.12(Sdl) (I), and providing the Site Class is not Seismic Design Category E or F, where the Seismic Design Category is obtained from Table 9.

Illustration 3 (Building in USA, use NEHRP Provisions) A two-story hospital building is planned for a location near longitude 80° and latitude 35°. The building will be 200 ft by 150 ft in plan, with a height of 33 ft. The structural type will be steel, momentresistant frame construction. The total seismic design weight for the building based on dead loads and design percentage of live loads is 11,000 kips. Subsurface conditions at the site indicate a stiff soil and soft weathered rock (some meeting the criteria for Site Class C and some for Site Class D; use Site Class D for design). Determine the value for the design seismic base shear using the procedure for equivalent lateral force (V = CsW ). A hospital is a seismic Group III facility; the Building Occupancy Importance Factor is l = 1.50. The Building Response Modification Factor for a steel, moment-resistant frame structure is R = 4.

467

Earthquakes and the Effects For the indicated site location (see Figure 7(a,b)), the map contour value for Ss = 0.38 g, and for Sl = 0.14 g, obtain: Cs =

12>32Sms 12>32Fa Ss Sds = = 1R>I2 1R>I2 1R>I2

Using: Fa L 1.5, 1interpolated from Table 7a2, Ss = 0.38, R = 4, and I = 1.50 get: Cs =

12>3211.5210.382 14>1.502

= 0.143

Also check Table 9 for proper Seismic Design Category based on Sds and Sdl values: Sds = 12>32Sms = 12>32FaSs = 12>3211.5210.382 = 0.38g Sdl = 12>32Sml = 12>32FvSl = 12>3212.2210.142 = 0.21g

Both terms require assignment of a Seismic Group III facility to Seismic Design Category D (see Table 9). Compare calculated Cs with values for Cs(max) and Cs(min): Cs1max2 =

10.212 Sdl = = 0.281 T1R>I2 10.28214>1.52

where: T = CuTa = 11.4210.202 = 0.28 since: Ta = 10.121N stories2 = 10.1212 stories2 = 0.20

Cs1min2 = 0.1 SdlI = 10.1210.21211.502 = 0.031

Since Cs1max2 7 Cs 7 Cs1min2 , use Cs = 0.143 . Therefore, Design Seismic Base Shear, V = CsW = 10.1432111,000 kips2 = 1573 kips.

Table 9(a) Seismic Design Category Based on Short-Period Response Accelerations Seismic Use Group

468

Value of Sds

I

II

III

Sds 6 0.167 g 0.167 g … Sds 6 0.33 g 0.33 g … Sds 6 0.50 g 0.50 g … Sds

A B C Da

A B C Da

A C D Da

Earthquakes and the Effects Table 9(b) Seismic Design Category Based on 1 sec Period Response Accelerations Seismic Use Group Value of Sdl

I

II

III

Sdl 6 0.067 g 0.067 g … Sdl 6 0.133 g 0.133 g … Sdl 6 0.20 g 0.20 g … Sdl

A B C Da

A B C Da

A C D Da

a Seismic Use Group I and II structures located on sites with mapped maximum considered earthquake spectral response acceleration at 1 second period, Sdl equal to or greater than 0.75g, shall be assigned to Seismic Design Category E, and Seismic Use Group III structures located on such sites shall be assigned to Seismic Design Category F.

Some generalized concepts have evolved from correlating the physical events that occur within the earth during a seismic incident and the effects on structure: 1. Building structures have a natural frequency where harmonic motion results (i.e., a smooth undulating movement occurs along a pattern of flow through the height and length of the structure). Seismic frequencies that are similar to the natural frequency of the foundation soil or rock and that of a structure will cause the greatest movement or vibration in the structure (a seismic frequency similar to the building’s natural frequency will amplify motion, while dissimilar frequencies tend to dampen movement). Low-rise buildings tend to have a higher natural frequency than tall buildings. 2. Low-rise buildings tend to be damaged by the high-frequency body waves, while tall buildings tend to be damaged by the low-frequency surface waves. High-frequency waves (body waves) dissipate energy over distance faster than low-frequency waves (surface waves). As a consequence, low-rise buildings are most prone to damage when located close to the earthquake epicenter, while tall buildings can be damaged at relatively great distances from the epicenter. Where seismic event velocity and acceleration information is available, the ratio of acceleration to velocity for a location can provide an indication of ground motion expected; a low ratio indicates that velocity is the dominant ground motion type (as occurs from a distant, large-magnitude earthquake) while a high ratio indicates that acceleration is the dominant ground motion type (as occurs from an earthquake close to the site). 3. Ground vibrations tend to be amplified when the seismic waves pass from rock into soil (though not always). Commonly, structural damage is greater for buildings constructed on soil than for buildings situated on rock; the potential for greatest damage exists where a building is constructed over deposits of weak soil. Though the direct effects of earthquake vibrations are a cause of destruction, destructive events greater than those resulting from the primary event may present a more serious danger to an area. Tsunamis, tectonically induced tidal waves originating in open ocean areas near the earthquake hypocenter, can swell to tens of meters (hundreds of feet) above normal sea level and travel for hundreds of km (miles); the enormous force from a huge wave of water smashing into the shoreline of a land area can be devastating. Another possible effect is that large areas of land subsidence can occur because of liquefaction (liquefying) of soil deposits

469

Earthquakes and the Effects

and other breakdown of bonds in the soil deposit. Conversely, some land masses may be lifted. Landslides and mud slides in hilly topography are not unusual. Secondary effects, such as collapsed dams releasing reservoirs to cause flooding, fires, destroyed utility conduits, destroyed bridges and transportation routes, breakdown of the communication systems, and lack of medical aid, contribute to the havoc resulting from earthquakes. Large earthquakes are often preceded by a series of smaller foreshocks and a series of aftershocks. Foreshocks can occur over a relatively lengthy period (months), while aftershocks typically occur within hours or days of the main event. Certainly, some dangers associated with earthquakes, such as loss of life, could be reduced if the ability for prediction existed. Unfortunately, no reliable methods for accurate prediction (place and time) are known at the present. Various methods and items have been proposed for use as indicators of an impending event, some of which have a scientific basis and some that do not. The more promising methods at present involve monitoring various earth conditions for unusual change; predictive methods based on behavioral changes in animals or the weather have not been reliable (although such changes may actually occur). Measurable changes in various earth property parameters have been noted in those earthquake regions of the world where conditions had been monitored in the period leading up to the actual event. The noted occurrences associated with some scientific reasoning include: 1. The velocity of normally ongoing P-waves traveling through the rock in the earthquake region decreases, but shortly prior to the event reverts to normal (a condition attributed to strains taking place in the rock, along with the development of small fractures which alter the physical properties of the rock materials). 2. The surfaces of regions underlain by rock that is undergoing strain and movement often experience some uplifting or tilt. 3. The emission of radon gas to the surface increases in the period prior to an earthquake (attributed to a change in the earth’s venting process as rock movement and fracture occurs). 4. The electrical resistivity of the rock materials may decrease prior to the event because of the increase in pressure throughout the zone being strained. 5. Large earthquakes show the tendency to be preceded by a series (swarm) of smaller quakes throughout the general region, a phenomenon attributed to the occurrence of rock ruptures on a somewhat limited scale. Unfortunately, some foreshocks have been of relatively large magnitude; at the time of occurrence uncertainty existed about whether that event was the main one or if larger events were yet to come. Relating to associations between surface changes and earthquakes, Synthetic Aperture Radar Interferometry is a developing method which offers opportunities for identifying and monitoring tectonic movements. Satellite radar measurements of the distance to the earth’s surface are obtained during the orbital travel and are used to define the relief or topography of sections of the earth’s surface. (Radar’s long-length radio waves are not affected by factors such as cloud cover, which can interfere with camera-like processes based on short-length light waves; the radar procedure becomes a more consistent source of precision data.) Radar readings of distance to the same ground point from different positions along the orbital path can be used to produce a type of stereoscopic or three-dimensional

470

Earthquakes and the Effects

image of the earth’s surface. Comparisons of radar measurements obtained from repetitive orbits can be used to indicate topographic changes (movements) occurring across large areas of the earth’s surface. It is anticipated that such topographic monitoring will enable identification of changes that just precede or lead to a seismic event, and that can then be used as a prediction tool.

Problems 1 List the different types of seismic waves produced during an earthquake event, and relate each type to the potential for causing building damage. 2 Explain the difference between earthquakes as defined by the Richter scale and by the Modified Mercalli scale. 3 Compare the energy released during the earthquake where ML is equal to 5.5 and the earthquake where ML is equal to 7.5. 4 (a) Determine the Richter magnitude for an earthquake where the following seismograph data have been obtained: • Maximum registered amplitude of 5 mm • Time separation between arrival of primary and secondary seismic waves at the seismograph of 30 sec (b) Also determine the distance separating the location of the seismograph and the earthquake epicenter. 5 Referencing Figure 7, indicate the seismicrelated values for the 0.2 sec spectral response acceleration (Ss) and 1.0 sec spectral response acceleration (Sl) for the following geographic locations in North America: (a) Atlanta, GA (approx. lat 37.5 deg, longitude 84 deg) (b) Boston, MA (approx. lat 42 deg, longitude 71 deg) (c) Dallas, TX (approx. lat 33 deg, longitude 97 deg) (d) Los Angeles, CA (approx. lat 34 deg, longitude 118 deg) (e) Miami, FL (approx. lat 26 deg, longitude 80 deg) (f) New Orleans, LA (approx. lat 30 deg, longitude 90 deg) (g) New York, NY (approx. lat 41 deg, longitude 74 deg)

(h) St. Louis, MO (approx. lat 38.5 deg, longitude 90 deg) (i) San Francisco, CA (approx. lat 37.5 deg, longitude 122 deg) (j) Seattle, WA (approx. lat 47.5 deg, longitude 122 deg) 6 Relating to NEHRP earthquake design criteria for classifying subsurface conditions, indicate the Site Classification for each of the following site location descriptions (see Table 5): (a) Area that is geologically blanketed with transported soils that overlay bedrock found at great depth. An extensive soil boring investigation indicates the construction site is underlain by sandy soils; results of the standard penetration test (SPT) blow count N recorded for soil samples obtained from the borings show N-values between 25 and 40 for all locations. (b) Area is geologically blanketed with a thick accumulation of firm fine-grain (silt-clay) soils, and bedrock is very deep. The area’s history indicates the soil deposit has been subject to the weight of overlying glaciers, and an affect is that the soil is highly consolidated and the cohesion shear strength is in excess of 2000 psf (100 kPa). (c) Area is blanketed by residual soils which grade into bedrock generally located at a depth of 40 ft (12 m). The residual soil below the foundation level planned for new structures (including the deeper zone of weathered rock) is very dense; SPT results from borings indicate N-values between 60 and 90. 7 Determine the NEHRP value for the design earthquake base shear, V, for the structure and location described: • Two-story municipal building, 30 ft high, ordinary steel moment-resistant frame construction • Building’s earthquake weight is 6,500 kips

471

Earthquakes and the Effects

8 Determine the NEHRP value of the design earthquake base shear V for the structure and conditions described in Problem 7 except that Ss = 0.60 g and Sl = 0.25 g for the new location.

• Building earthquake weight is 46,000 kN • Site located in area where Ss = 0.26 g and Sl = 0.11 g • subsurface site class C 10 Determine the NEHRP value of the design earthquake base shear V for the structure and conditions described in Problem 9 except that SS = 0.50 g and Sl = 0.30 g for the new location.

9 Determine the NEHRP value for the design earthquake base shear V for the structure and location described: • Three-story hospital building, ordinary steel moment-resistant frame construction, height from ground surface to top is 15 m

11 Determine the NEHRP value of the design earthquake base shear V for the structure and conditions described in Problem 9 except that the subsurface represents Site Class D conditions and the building structure will consist of special steel moment-resistant type framing.

• Seismic use group designation II • Site located in area where Ss = 0.45 g and Sl = 0.16 g • subsurface site class D

472

Answers to Selected Problems

3. ML 5.5; E L 8.28 * 1012 ft lb ML 7.5; E L 8.28 * 1015 ft lb 11000 * more energy2 4. (a) mag 4.2 ( ; ) (b) 275 km (approx) 6. Site classes: D, D, C

7. 9. 10. 11.

V V V V

= = L =

858 kips 3607 kN 6950 kN 2406 kN

473

474

Foundations Introductory Concepts

For many of humankind’s structures, it is the earth underlying the structure that provides the ultimate support. The soil at a building location automatically becomes a material of construction affecting the structure’s stability. Typically, soil is a material weaker than the other common materials of construction, such as steel, concrete, and wood. To carry a given loading satisfactorily, a greater area or volume of soil is necessarily involved. In order for loads carried by steel, concrete, or wood structural members to be imparted to the soil, load transfer devices—the structural foundations—are required. The major purpose of the structural foundation is the proper transmission of building loading to the earth in such a way that the supporting soil is not overstressed and does not undergo deformations that would cause serious building settlement. The type of structural foundation utilized is closely related to the properties of the supporting soils. A structural foundation performs properly only if the supporting soil behaves properly. Consequently, it is important to recognize that building support is actually being provided by a soil–foundation system, a combination that cannot be separated. Designers and constructors are aware of this relationship, but it has become common practice to consider the structural foundation separately, primarily because it is a cost item that is built or installed, while the supporting soil is usually the natural earth that “is there.” Since the soil–foundation system is responsible for providing support for the lifetime of a structure, it is important that all forces that may act over that time period are considered. For a building to endure, its foundations should be designed for the worst conditions that may develop. Typically, the foundation design always includes the effect of the structure’s dead plus live loads. It is important also to consider load effects that may result from environmental factors such as wind, ice, frost, heat, water, earthquake, and explosive blasts.

From Essentials of Soil Mechanics and Foundations: Basic Geotechnics, Seventh Edition. David F. McCarthy. Copyright © 2007 by Pearson Education, Inc. Published by Pearson Prentice Hall. All rights reserved.

475

Foundations

1

General Types of Foundations—Foundation Categories The various types of structural foundations can be grouped into two broad categories: shallow foundations and deep foundations. Generally, the classification indicates the depth of the foundation installation and the depth of the soil providing most of the support. Spread footing and mat (or raft) foundations usually fall within the shallow foundations category. Deep foundation types include piles, piers, and caissons. The floating foundation, a special category of foundation, is actually not a different type, but it does represent a special application of soil mechanics principles to a combination mat–caisson foundation.

Spread Footings Spread footing foundations are typically of plain concrete or reinforced concrete, although masonry and timber have also been used. The spread footing foundation is basically a pad used to “spread out” building column and wall loads over a sufficiently large soil area. Spread footings are constructed as close to the ground surface as the building design permits (considering requirements such as basements and the need to resist lateral forces) and as controlled by local conditions (considering factors such as frost penetration, soil shrinkage and expansion, the possibility of soil erosion, and building code stipulations). Footings for permanent structures are rarely located directly on the ground surface. To be classified as a spread footing, the foundation does not have to be at a shallow depth; spread footings will be located deep in the ground if soil conditions or the building design requires. Spread footing foundations for building columns, walls, and equipment bases commonly have the shapes of squares, rectangles, trapezoids, or long strips (Figure 1). Usually, the shape and dimensions for a footing result from having the structural loading positioned so that,

Q (kN or lb)

Q (kN per m or lb per foot of length)

(a)

Q1 (kN or lb)

(b) Q2 (kN or lb)

(c)

Q2 (kN or lb) Q1 (kN or lb)

(d)

Figure 1 Types of shallow spread footing foundations: (a) square spread footing to support column loading; (b) long (strip) footing to support wall loading; (c) rectangular footing for two columns (combined footing) or machine base; (d) trapezoidal footing for two columns (combined footing) or machine base.

476

Foundations Wall loading

Col. loading

Col. loading

Col. loading

Wall loading Ground surface

Steel-reinforced concrete slab (typical)

Figure 2 Mat foundation—for soils having low bearing capacity or where soil conditions are variable and erratic—is used to obtain low bearing pressure and reduce differential settlement. The mat foundation may also consist of a grid arrangement of closely spaced and interconnected grade beams capped with a floor slab.

theoretically, a uniform bearing pressure on the soil beneath the foundation is achieved. For the support of walls and single columns, the loading is usually centered on the footing. For foundations supporting two or more column loads, or machinery, the positioning of the loading or weight often makes a rectangular or trapezoidal shape necessary.

Mat (or Raft) Foundations The mat (or raft) foundation can be considered a large footing extending over a great area, frequently an entire building area (Figure 2). All vertical structural loadings from columns and walls are supported on the common foundation. Typically, the mat is utilized for conditions where a preliminary design indicates that individual column footings would be undesirably close together or try to overlap. The mat is frequently utilized as a method to reduce or distribute building loads in order to reduce differential settlement between adjacent areas. To function properly, the mat structure will be more rigid and thicker than the individual spread footing.

Pile and Pier Foundations Piles and piers are foundation types intended to transmit structural loads through upper zones of poor soil to a depth where the earth is capable of providing the desired support (Figure 3(a) and (b)). In this respect, where loadings developed at one level are transmitted to a lower level, piles and piers are similar to structural columns. Though considered as long, slender structural members, such foundations typically obtain adequate lateral support from the embedding soil along their length so usually there is no concern about buckling under axial load, as with conventional columns. These deep foundation types are also utilized in situations where it is necessary to provide resistance to uplift or where there is concern about possible loss of ground or erosion due to flowing water or other causes. In years past, the pile or pier category was indicative of the method used to install the foundation or of its size. Piles were slender foundation units, usually driven into place. Piers, typically larger in area than piles, were units formed in place by excavating an opening to the desired depth and pouring concrete. Often, such foundations were large enough

477

Foundations

Water

Earth

(a)

(b)

(c) Air shafts

Air lock Water

Water

Earth Compressed air in working chamber

Earth

(d)

(e)

Figure 3 Representative types of deep foundations: (a) slender driven, drilled, or cast-in-place pile; (b) drilled or cast-in-place pier with enlarged base; (c) pile-type open caisson; (d) box-type open caisson; (e) pneumatic caisson.

to permit a man to enter and inspect the exposed earth. Currently, a clear distinction between pile- and pier-type foundations is not always present because of changes and innovations in construction or installation techniques. For example, some types of cast-inplace piles are constructed by using the basic methods historically attributed to pier foundations. As a result, the developing practice is to classify all deep, slender foundation units simply as pile-type foundations, with terms such as driven, bored, or drilled and precast or cast-in-place to indicate the method of installation and construction.

Caissons A caisson is a structural box or chamber that is sunk in place or built in place by systematically excavating below the bottom of the unit, which thereby descends to the final depth. Open caissons may be box type or pile type. Usually, the top and bottom are open during installation. When in place, the bottom may be sealed with concrete if necessary to keep out water, or the bottom may be socketed into rock to obtain a high bearing capacity. Pneumatic caissons have the top and sides sealed and use compressed air to keep soil and water from entering the lower working chamber, where excavation to advance the caisson is occurring. Representative types of caissons are illustrated in Figure 3(c), (d), and (e).

478

Foundations

Rigid boxcaisson element

Figure 4 Rigid box caisson foundation utilizing the floating foundation concept.

Floating Foundations The floating foundation is a special type of foundation construction that is useful under proper conditions. Particularly, it has application in locations where deep deposits of compressible cohesive soils exist and the use of piles is impractical. The floating foundation concept requires that a building’s substructure (the below-ground structure) be assembled as a combination mat and caisson to create a rigid box, as shown in Figure 4. This foundation is installed at a depth selected so that the total weight of the soil excavated for the rigid box equals the total weight of the planned building. In theory, the soil below the structure is therefore not subjected to any change in loading. For such an occurrence, there would be no settlement. Usually, however, some settlement does occur because soils at the bottom of a deep excavation commonly expand somewhat after excavation because of stress relief, then recompress during and following construction.

2

Pile Foundation Types and Installation Procedures Pile Drivers for Driven Piles Driven piles are installed from the ground surface by hammering a ready-made unit or hollow shell, usually with special pile-driving equipment (Figure 5). Most driving is done with an impact type of hammer; a moving weight falls or is forced against the top of the pile (pile driving has been described as a brutal method for installing foundations). General categories of modern pile hammers include the single- and double-acting units (steam, compressed air, or hydraulic) and diesel hammers, illustrated in Figure 6. The term hammer refers to the entire driving unit; usually, the moving weight that strikes the pile is the ram. Single-acting hammers use steam or compressed air to raise the hammer ram to a ready-for-driving position (Figure 7). The ram is then released to enable it to drop on the top of the pile. A double-acting hammer uses steam or compressed air to raise the ram to a ready-for-driving position and also to accelerate the ram’s downward thrust. Differentialacting hammers, another category, are similar in operation to the double-acting hammers. Double-acting hammers actuated by hydraulic pressures are also available.

479

480 Pile

Hammer Ram Hammer cushion

Hammer

Pile gate (device at base of leads that close around pile to maintain its alignment)

Pile

Pile monkey (mechanical device to position pile under the helmet)

Pile cushion

Helmet (steel pile cap, aligns pile head under hammer, protects pile head during driving)

Figure 5 Main components in typical pile-driving equipment.

Detail: Typical arrangement for components at pile top, hammer strike location

Pile cushion (when used, typical for concrete piles)

Helmet

Hammer cushion

Top plate

Ram velocity

Crane

Brace (attaches bottom of lead to base of crane to position lead for a vertical or batter pile alignment)

Boom

Lead (the support structure for pile hammer and for keeping the pile and hammer aligned). Leads can be fixed or swinging type (the fixed lead is depicted). Swinging leads are hung from a boom tip with hanger straps or from a crane cable (the hammer is held by a separate cable line).

Piston

Piston Air or steam pressure

Compressed air (optional)

Winch (drop hammer) Ram Ram

Ram

Combustion Stroke, H

Hammer cushion

Hammer cushion

Anvil Hammer cushion

Pile

(a) Drop and singleacting hammers

Drive head

Drive head

Drive head

Pile

(b) Differential and doubleacting hammers

Pile

(c) Diesel hammers

Figure 6 Schematic diagram of ram-type pile-driving hammers [269].

Figure 7 Single-acting compressed air hammer. (Courtesy of FHWA)

481

Foundations

Figure 8 Diesel hammer. (Courtesy of Geoforum.com 2000)

Diesel hammers (Figure 8) are self-contained, self-activated units. The ram is located within an enclosed cylinder. Initially, the ram is mechanically raised to the top of the cylinder and released for its fall. A fuel mixture injected into the cylinder ahead of the falling ram is compressed due to the piston effect. Near the bottom of the stroke, the fuel is detonated, and the force of the resulting explosion and the ram impact is delivered to the pile, driving it. Within the cylinder, the force of the explosion raises the ram up to the top of the cylinder, where it is ready to begin another cycle. Single-acting hammers need to rely on the weight of a heavy ram for driving piles. Double-acting and diesel hammers can develop high driving energy that is equal to or greater than the energy of the falling ram in a single-acting hammer, but they achieve this through greater impact velocity. One advantage of the double-acting and diesel hammers over the single-acting hammer is the greater operating speed; a greater number of blows will be delivered per unit of time. When long or heavy piles are driven, a hammer that has a heavy ram should be used. For effective driving, the weight of the hammer ram needs to be approximately the same weight as the pile or greater. Equipment manufacturers provide data on the important characteristics of their hammers, including the weight of the ram and the stroke, hammer efficiency, driving energy per hammer blow, and the number of blows per minute (Table 1).

482

Foundations Table 1 Data on Representative Pile-driving Hammers Weight: Ram, Piston or Moving Parts

Impact Hammers Make McKiernan-Terry (MKT) (St. Louis, MO, USA) International Const. Equipment (ICE) (Matthews, NC, USA) American Pile Driving Equipment (APE) (Kent, WA, USA) Berminghammer Foundation Equipment (Hamilton, Ont., Canada) Conmaco/Rector, L.P. (Conmaco) (Belle Chase, LA, USA) Menck GmbH (Menck) (Ellerau, Germany) Vulcan Foundation Equipment (Vulcan) (Chattanooga, TN, USA)

Rated Energy per Blow

Blows per Minute

k-ft

kN-m

Model (type)

kips

kN

DE-20 (diesel, single-acting) DE-110 (diesel, single-acting) 11B3 (air/steam, double-acting) 60S (diesel, single-acting) 520 (diesel, double-acting) 115 (hydraulic impact) D30-32 (diesel, single-acting) D50-32 (diesel, single-acting) 14 (Hydraulic-impact) B-3005 Mark V (diesel, single-acting) B-5505 Mark V (diesel, single-acting) B-6505C Mark V (diesel, single-acting) 125ES (air/steam, single-acting) *6850 (air/steam, single-acting) *1750 (air/steam, single-acting) MRB-600 (air/steam, double-acting) *MRBS-8800 (air/steam, double-acting)

2.0 11.0 5.0 7.0 5.07 11.5 6.61 11 11 3.0 9.2 22.04 12.5 85 175 14.88 194

8.9 48.9 22.2 31.1 22.5 51.1 29.4 48.9 48.9 13.35 40.9 98.0 83.4 378.1 778.5

40–50 40–50 95 41.59 80.84 40 36–52 37–53 40–80 36–60 36–60 39–60 41 To 40 To 40 45 36

To 20 To 110 19.15 60 30 46 To 69.9 To 116.6 33 To 21.5 To 66.0 To 122.0 62.5 510 1,050 62.2 973

27.1 149.1 25.9 81.3 40.6 62.4 To 94.7 To 158 To 44.7 To 29.0 To 89.0 To 165.0 84.7 691.4 1,423.6 84.38 1,320

08 (air/steam, single-acting) 520 (air/steam, single-acting) 85C (air/steam, differential-acting) *400C (air/steam, differential-acting) *6200 (air/steam, single-acting)

8.0 20.0 8.525 40 200

35.6 88.9 37.9 178 890

50 42 111 102 36

26 100 26 113.5 1,200

35.2 135.6 35.25 153.9 1,627

*Offshore hammers developed for use in deep water applications; hammers typically are heavier and more sturdy than onshore hammers.

Vibratory Hammers Make

Model

MKT (USA) ICE (USA) APE (USA)

V-5C 223 300

Force at Operating Frequency

Total Weight kips 7.2 4.8 22

kN

Available HP

kips/cps

kN/hertz

32.0 20.5 97.9

185 325 625

106/28 166/38 207/25

470/28 735/38 1,842/25

Vibratory drivers rely on a principle different from the conventional hammers for installing a pile. Vibratory units are typically more contained and compact than the fallingram type of hammer. Basically, the vibratory driver (Figure 9) consists of a pair of counterrotating weights that are synchronized so that the lateral components of thrust always counteract, or cancel, each other’s effects. The vertical components of thrust are additive, however, and create up-and-down pulsations, or vibrations. For an installation, the driver is clamped to the pile. The pile is then vibrated into the earth under its own weight. Vibratory drivers may be the low-frequency type (operating range between about

483

Vibrator Power Source Diesel-electric Diesel-hydraulic Suspension of cables or hoses

Vibration isolator Motor Eccentric masses Excitor block

Power transmission Electric cable or hydraulic hoses

Hydraulic clamping device

Control panel Pile

(a) Schematic diagram of main components

(b) Diesel-electric generator (power source)

(c) Vibrator in process of driving steel H-pile (note size related to worker) Figure 9 Vibratory pile driver system and equipment.

484

Foundations

10 and 30 Hz or cps) or resonant type (capable of an operating frequency up to about 150 Hz or cps). The intent of a resonant driver is to create a condition of resonance between the driver, pile, and zone of affected soil so that penetration occurs very rapidly. Vibratory drivers are considered to be most effective when installing piles in sand and silty sand soils. An item of importance in pile installation is the factor of stresses created in the pile by the driving. For wood piles and concrete piles, excessive driving when high resistance is met (a high number of blows per inch of penetration) may result in structural damage or breakage in the pile. It is helpful for the foundation designer to impose an upper limit to the number of blows that can be safely imposed, for reference by field personnel. Breakage while driving can be difficult to detect but is often indicated by a drop in blow count following a high blow count. Metal tips or end covers can be installed on piles to help protect them in hard driving and to aid penetration.

Pile Types and Materials The common types of piles that are installed by driving include timber piles, steel H or pipe (Figure 10), reinforced concrete (Figure 11), and prestressed concrete. Generally, with these types, the pile unit is complete and ready to be installed when it is delivered to the job site. The driven steel shell (Figure 12) is a different category of pile installed by driving. Often, with this type, a hollow steel shell is installed to the desired depth by driving on a steel mandrel or steel core that fits inside the shell. Much of the force of driving thus acts at the tip of the pile, so, in effect, the pile is partially being pulled into the ground. At the desired penetration, the mandrel or core is withdrawn and plain or reinforced concrete is placed in the shell. Corrugated or fluted steel is conventionally used for the shell so that

Figure 10

Diesel hammer driving a battered (angled) steel pipe pile. (Courtesy of FHWA)

485

Foundations

Figure 11 Precast reinforced concrete pile being lifted for positioning in the driving rig leads (aka guides); lifting procedure is the “pile-up” process. (Courtesy of Geoforum.com 2000)

it will possess adequate strength to withstand the stresses of driving. This type of pile is usually considered a cast-in-place unit. Timber piles can be treated or untreated. “Treated” refers to the procedure by which the wood has been impregnated with a protective material or preservative. Pressure impregnation is a typical treatment for piles that are to be exposed to moisture but not permanently submerged in water. Untreated wood permanently below water will not decay and is considered to have unlimited life unless threatened by other effects. Preservative–protection treatment is required to protect wood piles from marine borers (if for a marine environment), from wood-infesting insects such as termites, or from decay (wet rot) if the pile is embedded in soil above the water table. Cast-in-place piles include the steel-shell and concrete type, described previously; the shell-less type, which is formed by excavating to a desired depth and then filling the opening with concrete; and the type in which a concrete unit is formed in the ground without prior excavation. Where pre-excavation is performed for a shell-less pile before the concrete is placed, the excavation may be unlined or may be provided with a temporary lining, such as a steel shell, whose purpose is to keep soil from caving into the open excavation and to seal off water from the soil walls (Figure 13). Excavation for this type of pile is typically performed by augering or by applying a wash boring technique similar to that utilized when borings are drilled. With either of these methods, a bentonite clay slurry may be used in the excavation while it is being drilled instead of a temporary steel lining to keep the soil walls from caving in. The bentonite slurry is a heavy liquid whose purpose is to exert a lateral pressure sufficient to hold the soil walls in place. Concrete for the foundation unit is tremied (placed under water through a large-diameter flexible hose or tube in order to prevent contamination) to the bottom of the excavation. The pile is formed from the base up, and the slurry is displaced from the excavation. The shell-less, cast-in-place pile is often referred to as a bored-pile or drilled-shaft foundation. Representative types are illustrated in Figure 14.

486

Foundations Mandrel

Steel shell

(a) Steel shell installed by driving solid mandrel fitting within shell. (Mandrel is withdrawn after shell reaches desired penetration)

(d) Straight CMP (corrugated metal pipe) shell

Figure 12

(c) Step-taper steel shell (corrugated)

(b) Tapered steel shell (corrugated)

(e) Tapered fluted steel shell (driven without mandrel)

Types of driven steel-shell, cast-in-place concrete piles.

In cohesive soils, the base of the excavation may be enlarged to provide a greater bearing area. The excavation for the shaft section and the belled or underreamed base is often performed by special auger-type drilling equipment, but can be accomplished manually. This type of foundation is often referred to as a drilled (drilled and belled) pier, drilled (drilled and belled) caisson, or drilled-shaft foundation (Figures 15 and 16). A different type of cast-in-place pile, the augered type, is constructed by augering into the earth without removing soil, then pumping concrete to the bottom of the drilled zone through a hollow stem in the auger and forming the pile from the bottom up as the auger withdraws and removes soil (Figure 17).

487

(a) Installation rig, ready-to-place reinforcing steel in new pile.

1

2

3

4

5

(b) For the construction-installation, a protective casing is driven and the soil then augered or washed (flushed) from within the casing. At the desired depth, tremie concrete is used to fill the hole from the base upward. Reinforcing steel is inserted as soon as the concrete placement is completed. Figure 13

488

Large-diameter bored pile equipment and procedure. (Courtesy of Geoforum.com 2000)

Q

Q

Poor or marginal soils

Poor or marginal soils

30°± Firm-compact soil or rock (a) Straight-sided

Figure 14

Firm cohesive soil (to permit bell to be formed)

(b) Belled or underreamed base

Typical drilled-shaft foundations: (a) straight-sided; (b) belled or underreamed base.

(b) Rig with auger ready to begin

(c) Auger being spun, to remove excavated soil from the flights of the cutting auger

(a) Auger drilling rig

(d) Placing concrete into the open augered shaft. Cylinder hanging from rig kelly bar is a “baler” (i.e., bucket) to remove water accumulated in the shaft excavation Figure 15

(e) Close-up of cutting auger

Drilled shaft foundation installation (equipment and procedure).

489

(a) Equipment in operation

Kelly bar

Borehole casing

Hinged toothed cutters Housing Direction of opening

(b) Schematic of operating principle Figure 16 Auger underream or belling equipment for forming enlarged base of drilled-shaft foundations (cast-in-place piers, piles, etc.). (Courtesy of Geoforum.com 2000)

490

(a) Pile installation auger rig

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(b) Installation procedure: (i, ii, iii) auger drills to desired pile depth without withdrawing auger (limited soil spoil brought to surface); (iv) at design depth, fluid concrete or grout is pumped through the hollow auger stem; (v) concrete or grout is continuously pumped as auger and soil is withdrawn, forming pile from base upward; (vi) auger continues to be withdrawn at the same rate as pile is being formed. Steel reinforcement is inserted as soon as pile is completed Figure 17 Continuous-flight auger pile (also known as Augercast pile). (Courtesy of Geoforum.com 2000)

491

Foundations Connection to pile cap or column Falling ram

Plug Casing

Bulb

(a) Plug of dry concrete in casing

(b) Falling ram drives concrete plug and casing to desired pile depth

(c) Concrete bulb is formed by anchoring casing while ram continues to drive concrete plug

(d) Casing is lifted while additional concrete is provided and rammed to form a shaft

(e) Completed pile

Figure 18 Procedure to form bulb-type uncased concrete pile.

Still another category of cast-in-place foundation is the bulb-type uncased concrete pile (Figure 18). To construct this type, a mass of very dry concrete is placed in a steel casing standing on the ground surface and formed into a plug by a falling heavy ram. High friction develops between the concrete and the inside casing wall, and continued blows of the ram force the concrete plug and casing into the ground. At the depth where it is desired to form a base for the pile, the casing is locked to the driving rig, and continued blows of the ram force the concrete plug from the bottom of the casing into the earth. An enlarged base forms as the driven concrete plug compacts the soil around this level. Additional dry concrete is then added and rammed while the casing is slowly lifted. A continuous pile shaft is formed as the rammed concrete continues to be driven from the bottom of the withdrawing casing. The term composite piles usually refers to piles that have different materials for their different sections of length. The most typical combinations include timber or steel H pile or pipe pile for the lower section and cased cast-in-place concrete for the upper section (Figure 19). Composite piles are normally used for reasons of economy (timber is less costly than steel or concrete) or to take advantage of certain structural features that one material possesses (steel will fare better than timber or concrete in hard driving; timber or cast-in-place concrete piles may not be available in the required length). Where composite piles are utilized, the splice joint between sections should be as strong as the pile materials, particularly if the pile will be subject to uplift forces. Composite pile is also the term applied to a pile cross section of more than one material, such as a concrete-filled pipe pile. Micropiles are a relatively new category of pile foundation units that were developed in Europe during the middle of the 20th century, and have been used in the past quarter century for projects in North America. Micropiles are small-diameter steel-reinforced concrete units, typically in the 125-mm to 250-mm diameter range. Micropiles are installed to perform as friction piles (relating to the small tip area) but can be end-bearing if extending to rock or into a firm-dense stratum. Historically, micropiles have been constructed with

492

Foundations Concrete and steel shell Concrete and steel shell Steel shell (concrete filled)

Wedge joint— with wedge ring forced into the head of the wood pile Timber

Tenon joint— with sealing ring

Timber

Concrete and steel shell

Concrete and steel shell Steel pipe or timber

(a) Composite pile

Tenon joint— with rod, pin, and socket; used for uplift resistance

Lower pipe section

(b)Splice connections

Figure 19 Composite pile showing splice connections: (a) typical makeup of composite pile; (b) typical splice connections.

considerable vertical reinforcing steel that is intended to carry most of the imposed loading (the reinforcing steel is commonly in the range of 25 to 50 percent of the pile cross section, a marked difference compared to conventional cast-in-place reinforced concrete piles). Friction-type and end-bearing piles commonly utilize a single large-diameter steelreinforcing bar capable of transmitting virtually the full axial loading, but multibar groups also are used (bars are kept centered in the cross section by use of centralizers), see Figure 20(a). Micropiles can be installed vertically or on a batter (at an incline or a vertical angle). When used for foundation support, micropiles have been used to provide a broad range of axial load capabilities (from the lower range of about 25 kN or 6 kips up to the 4500 kN or 1000 kips range). Micropiles are identified as a replacement type of foundation unit, whereby the pile is constructed by auger-drilling the necessary excavation, then placing the reinforcing steel and concrete grout (as compared to the displacement category term used for driven piles). The concrete grout can be cast-in-place via gravity or pressure-grouted in place (or a combination). The excavation may be cased or uncased, depending on soil conditions (commonly, at least the upper section of the excavation is protected by casing). Micropiles are being used to provide foundation support for new projects. But increasingly, micropiles are being used as underpinning retrofitted to existing facilities as part of a repair or improvement or renovation project (the basis for the original development), relating to the installation process for small-diameter sizes that includes the availability of small, portable soil and rock drilling equipment to access small and low-headroom areas (but capable of penetrating through difficult materials such as boulder and cobble layers and buried construction materials) and portable grouting equipment. Reticulated networks are being used to stabilize earth slopes and land regions where weak soils exist. Two classifications have been established to indicate the design application use: Case 1 is for piles that will receive structural load directly and function as a main load

493

Pile cap

Pile cut-off elevation

Tension connection (if required)

Bottom of pile cap Steel casing

Cement grout (4000 psi or 28 MPa compressive strength)

Augered excavation slightly larger than steel pipe casing Casing may remain in place or withdrawn during grouting With casing left in place, high pressure grouting procedure forces grout out bottom to flow around exterior side of casing

Centralizer (PVC material) (3m/10 ft spacing typical) Reinforcing bar

Mechanical coupling (as required)

Bottom of pile

(a) Representative foundation installation Figure 20 Micropiles—installation and applications.

494

Structure being supported

Compressible stratum

Bearing stratum (pile shaft support zone)

(b) Case 1 —micropile foundation (directly loaded)

Line of micropiles for reinforcing-stabilizing weak soil zone and to support structural load

Line of micropiles to stabilize-reinforce upper slope region

oil k sm a e W tratu s

Soil fill project for new roadway includes cap beam to transmit load onto micropiles

k soil

Wea

m

stratu

(c) Case 2 — Micropile Reticulated Network (micropile lines would extend along length of slope being stabilized or reinforced)

Figure 20

(continued ).

495

Foundations

support member, see Figure 20(b); and Case 2 is for piles used as components of a grouping placed to form a reticulated network that reinforces or stabilizes a soil zone (i.e., multiple vertical and battered-inclined piles are positioned in an interacting three-dimensional arrangement to form a network that includes the soil mass within the pile zone), see Figure 20(c). Classification of construction type (types A, B, C, D) is based on the method of placing the cement grout, which affects the degree of bond developed between the pile shaft and surrounding soil: Type A—cement grout placed by effect of gravity head only (not a pressurized placement); Type B—cement grout placed into drilled hole under pressure as temporary steel casing is withdrawn (typical grouting pressure in range of 0.5 to 1 MPa or 75 to 150 psi), limited lateral expansion into surrounding soil; Type C—two-step placement of grout: (i) initial placement as for Type A, then before hardening, (ii) post-grout with additional grout at pressure of 1 MPa or 150 psi into the original grout using a sleeved grout pipe (extra grout pressure causes additional lateral expansion into surrounding soil); Type D—multi-repeatable post-grouting: (i) initial placement of grout as for Type A or B (gravity or pressurized placement), (ii) after initial grout hardens, additional grout injected at pressure of 2 to 8 MPa or 150 to 600 psi at various levels via sleeved grout pipe; procedure may be repeated (procedure results in maximum lateral expansion into surrounding soil and maximum bond between pile and surrounding soil).

Representative Pile Load Capacities and Available Lengths The pile type and cross section to be selected for a project will be influenced by soil conditions, required pile length, required structural capability of the pile, and consideration of the pile installation method. Information on the maximum lengths generally available for the various pile types is summarized in Table 2. Representative structural load ranges for various pile types are shown in Table 3.

Methods to Aid Pile Installation The installation of driven piles can be aided by the use of spudding or predrilling. These techniques are frequently utilized in situations where obstacles that could damage the driven pile are buried in the soil to be penetrated, where compact or hard soil must be penetrated, or where driving vibrations may affect nearby structures. Another form of installation assistance is jetting, a procedure that eliminates some driving and hastens pile penetration. Spudding refers to the procedure of driving a steel H or similar section into the earth to break up obstacles before installing the pile. Beyond the depth where the obstacles exist, the spud is withdrawn. The pile is then installed in the hole and driven to its final depth. Predrilling consists of drilling a hole, approximately the diameter of the pile, through very hard soils to eliminate the danger of pile damage that might result if driving were

496

Foundations Table 2 Available Lengths of Various Pile Types Pile Type

Comments, Available Maximum Length

Timber

Depends on wood (tree) type. Lengths in the 15 to 18 m (50- to 60-ft) range are usually available in most areas; lengths to about 25 m (75 ft) are available but in limited quantity; lengths up to the 30 m (100-ft) range are available, but supply is very limited.

Steel H and pipe

Unlimited length; “short” sections are driven and additional sections are field-welded to obtain a desired total length.

Steel shell, cast-in-place

Typically between 30 and 40 m (100 and 125 ft), depending on shell type and manufacturer-contractor.

Precast concrete

Solid, small-cross-section piles usually extend to the 15 to 18 m (50to 60-ft) length, depending on cross-section shape, dimensions, and manufacturer. Large-diameter cylinder piles can extend to about 60 m (200 ft) long.

Drilled-shaft, cast-in-place concrete (including micropiles)

Usually in the 15 to 25 m (50- to 75-ft) range, depending on contractor equipment.

Bulb-type, cast-in-place concrete

Up to about 30 m (100 ft).

Composite

Related to available lengths of material in the different sections. If steel and thin-shell cast-in-place concrete are used, the length can be unlimited; if timber and thin-shell cast-in-place concrete are used, lengths can be on the order of 45 m (150 ft).

Table 3 Typical Capacities for Various Pile Types Typical Design Load Range Pile Type Wood Concrete, cast-in-place, steel shell and uncased Concrete, reinforced or prestressed (lower range for smaller cross sections, upper range for larger cross sections) Concrete, bulb-type Composite: wood and concrete Steel pipe (lower range for small diameter and wall thickness, upper range for larger diameter and heavy wall thickness) Steel HP (lower range for light sections, upper range for heavier sections) Micropiles

tons

kN

15–30 30–75 30–200

100–300 250–700 250–2,000

75–1,000 30–60 30–100

600–9,000 250–600 250–1,000

40–200

400–1,500

3–500

25–4500

attempted. Predrilling is frequently utilized on projects where driving effects must be minimized, as protection against possible damage from driving vibrations to nearby facilities. The procedure is also utilized when piles are installed in clay soils to prevent ground heave, which can result from pile driving. If ground heave occurs, previously driven piles may also heave up and have to be reseated. Jetting is the technique of using a powerful stream of water directed below the tip of a long pile penetrating sandy soil to wash ahead of the pile to assist it in advancing through the sand. The jetting nozzle may be temporarily attached directly to the pile. As sand is flushed from below the pile tip, the pile settles into the created void or is easily driven if the driving hammer is activated. Jetting causes the sand that eventually surrounds the pile to be

497

Foundations

loose. Typically, the jetting is stopped and the pile is driven for the last segment of the desired penetration to develop high end friction and tip bearing. For short piles, jetting is also used to wash a hole at a pile location, before driving, to make installation easier. It is also important to recognize that the process of pile driving will cause vibrations in the ground that act to densify the sand. The occurrence can create a field problem where the first piles installed for a group densify the sand and a resulting effect is that it becomes difficult to drive the subsequent piles to the desired depth. For types such as timber and concrete piles, hard driving induces the risk of damaging the pile. When the foundation design requires a particular embedded length for reasons of erosion protection, or resistance to uplift or lateral loads, or resistance to seismic effects, etc., jetting-assisted installation may be necessary for all the piles at the project.

Other Installation Considerations Pile foundations may be required to resist lateral forces instead of, or in addition to, vertical loads. Driven piles and some types of formed-in-place piles can be installed at an angle to the vertical to develop high resistance to lateral forces. Such piles are referred to as batter piles (see Figure 10). Most vertical piles are capable of resisting lateral forces also, but usually of only small magnitude. Deep foundations may be required at locations where available headroom is limited. Driven and formed-in-place piles can be installed where space is limited, but special equipment is necessary. If headroom is very limited, driven-type piles can be installed by working with very short sections and jacking. Sections are jacked and added to until the desired penetration is reached. The augered-excavation cast-in-place piles can offer the advantage of relatively quiet, vibration-free installation compared to driven piles. Such factors may be important if construction is in a highly developed area where there is concern about noise or the effect of driving vibrations on nearby structures. Comparative installed costs for different foundation types will vary, depending on soil conditions to be penetrated, required pile length, desired load capacity, geographic area and labor costs, availability of pile materials, site accessibility and site conditions, and contractor availability. Although drilled-shaft or cast-in-place piles are often considered more economical than driven piles where short lengths are required, and vice versa, each project should be considered unique and costs should be determined accordingly.

3

Relating Soil Conditions and Foundation Types A structural foundation serves as the intermediary element to transmit forces from a building’s superstructure to the supporting soil. It is necessary to know soil conditions and soil properties underlying an area, as well as the magnitude and type of building loading, before selection of a proper foundation type can be made. After an appropriate foundation has been decided upon, each unit is sized for proper carrying capacity. Individual foundations should be analyzed, as is any other structural member, to ensure that the element itself possesses adequate internal strength.

498

Soil conditions

Appropriate foundation type and location

Design comments

El.0, ground surface (1) Compact sand, (deposit to great depth)

Installation below frost depth or where erosion might occur

Spread footings most appropriate for conventional foundation needs. A deep foundation such as piles could be required if uplift or other unusual forces (e.g., seismic, effect of flood) could act.

El.0 (2)

Spread footings most appropriate for conventional foundation needs. Also see comment for (1) above.

Firm clay or firm silt and clay (to great depth) Installation depth below frost depth, or below zone where shrinkage and expansion due to change in water content could occur El.0 (3) Firm clay El.–3m(–10') Soft clay (to great depth)

Comments as for (2) above

Spread footing would be appropriate for low to medium range of loads, if not installed too close to soft clay layer. If heavy loads are to be carried, deep foundations might be required.

El.0 (4) Loose sand (to great depth)

(or) Depth greater than frost or erosion depth

Spread footing may settle excessively or require use of very low bearing pressures. Consider mat foundation, or consider compacting sand by vibroflotation or other method, then use spread footings. Driven piles could be used and would densify the sand. Also consider augered cast-in-place piles.

El.0 (5) Soft clay, (Soft) El.–8m(–25') but firmness increasing with depth (to very great (Med. firm) depth) El.–15m(–50') (Firmer)

Figure 21

(or)

Spread footings probably not appropriate. Friction piles or piers would be satisfactory if some settlement could be tolerated. Long piles would reduce settlement problems. Should also consider mat foundation or floating foundation.

Illustrations relating soil conditions and appropriate foundation types.

499

Appropriate foundation type and location

Soil conditions

Design comments

El.0 (6)

Deep foundation—piles, piers, caissons —bearing directly on/in the rock.

Soft clay

El.−20m(−65') Rock El.0' (7)

Spread footings in upper sand layer would probably experience large settlement because of underlying soft clay layer. Consider drilled piers with a bell formed in hard clay layer, or other pile foundation into hard clay layer.

El.−3m(−9')

Compact sand El.−6m(−20') Med. soft Clay Hard clay (extending deep)

El.0 (8)

Deep foundation best—castin-place piles such as auger piles or bulb piles into sand layer appear most appropriate.

(or)

Soft clay El.−6m(−20') Med. dense sand (extending deep) Auger pile

Bulb-type pile

El.0 (9)

Miscellaneous fill (soil, non-soil) El.−5m(−16') Loose sand, soft clay, organic mat'l

El.−3m(−10') El.−7m(−22')

(or)

Med. dense sand

Compact glacial till

El.−18m(−60')

Deep foundation types extending into medium dense sand, or more preferably, into compact glacial till. Strong possibility for drilled pier with bell constructed in till. Also consider cast-in-place and driven concrete pile, wood pile, pipe pile.

Rock El.0 (10)

El.−2.5m(−8') Miscellaneous fill (poor) Med. dense sand El.−12m(−40') Med. firm clay El.−30m(−100') Rock

Figure 21

500

(continued )

(or) New compacted sand fill

Deep foundations penetrating through fill are appropriate. With piles or piers, consider stopping in upper zone of sand layer so as to limit compression of clay layer. Also consider replacing poor fill with a compacted fill and then using spread footings in the new fill.

Foundations

Appropriate foundation type and location

Soil conditions

Design comments

El.0 (11) (or)

(or)

Soft clay Med. dense to dense sand Soft clay (to rock)

For light to med. heavy loading

If foundation loads are not too heavy, consider using piles or piers bearing in the upper zone of sand layer and check for settlement. If foundation loads are heavy, consider driven piles (steel) or caissons to rock. Also consider floating foundation.

For heavy loading

Rock Basement

El.0 (12)

Miscellaneous soil and non-soil fill

(or)

Loose sand and soft clay Rock

Figure 21

Subbasement

Foundations should bear directly on the rock which is relatively close to ground surface. If no basement areas are needed for the building, consider piers. If basement areas are useful, consider full excavation to rock and construction of two basement levels.

(continued )

As a guide for developing an understanding of the intereffect of soil conditions and a required type of foundation, illustrations of different subsurface conditions and related foundation considerations are presented in Figure 21. For the design comments, it was assumed that a multistory commercial structure, such as an office building, was to be supported.

Problems 1 Why are foundations, as a distinct structural element, required for buildings? 2 Typically, a foundation design will be most influenced by the properties of the supporting soil and the magnitude of structural loading to be carried. Of these two factors, soil properties represent the primary concern. Explain. 3 List the various types of foundations in use for construction projects such as buildings, including type of materials commonly chosen and general definition of the depth where soil support is achieved.

4 (a) What factors determine whether a foundation type is in the shallow or deep foundation category? (b) Piles and piers are considered deep foundation units. Why are deep foundation units typically long, slender members? 5 Generally, shallow foundations, such as poured concrete spread footings, offer definite cost advantages over deep foundation types, such as piles. Assume that you are making a presentation to a nontechnical audience; outline reasons for the usually significant cost differential between shallow and deep foundations.

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Foundations 6 Why are site factors such as a high water table or the potential for frost heave or soil shrinkage and swelling considerations important to the design of shallow foundations? 7 Provide a brief description of the mat-type foundation, and list the practical reasons why a mat foundation type could be selected for a building. 8 Outline the scientific reasoning which supports the concept of the floating foundation. 9 A planned subterranean vault for heavy power transformer equipment is required in an area where a thick deposit of weak soils exists. The vault is 8 m by 12 m in plan area and is 4 m high. The soil unit weight is 15 kN/m3. The approximate weight of the vault structure plus included equipment is 5,500 kN. It is proposed to apply the floating foundation concept to the design–installation of the vault, to eliminate the expense of deep foundation support. Does the proposal appear reasonable? 10 A multistory building with a plan area of 200 ft by 100 ft will impose an estimated total loading of 20,000 kips onto the supporting earth. If the floating foundation concept is to be utilized for this structure and the average unit weight of soil underlying the site is 120 pcf, what volume of soil excavation (and subsurface structure) is required? 11 Compile a summary list of the common types of pile foundations, including materials used, available lengths, and common range of load applications. 12 List the general types of pile drive hammers in use, and provide a brief description of the basic operating principle for each. 13 Structural designers commonly mention that steel piles are a good choice for sites where long piles are required and the depth to bearing will be variable and somewhat unpredictable. What are the probable reasons for this belief? 14 Indicate the difference between pile and pier types of foundations. 15 Outline the difference between the caisson type of foundation and the pier type of foundation. 16 Relating to the installation of driven piles, provide a brief explanation of spudding,

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predrilling, and jetting (indicate the general purpose for the procedure, and the differences in the methods). 17 The selection of pile foundations for any project requires that the designer and contractor consider various practical issues to ensure an adequate rate of progress (completed installations) and proper (quality) installations (no damage, etc.). For driven wood piles, driven steel-shell and concrete-filled piles, and augered-type castin-place piles, list some of these practical issues that probably would have influence on the type chosen (such as transporting longer piles to the job site, a possible problem in an urban setting, and onsite storage). 18 Soil conditions at a proposed construction site make it necessary for a driven pile foundation system to be utilized to support a planned building. The designer has the following design–cost estimate data: (a) Wood piles, 25-ton capacity, required length 35 ft, $18 per lineal foot in place. (b) Steel-shell, cast-in-place concrete piles, 50ton capacity, required length 45 ft, $24 per lineal foot in place. (c) Steel HP piles, 95-ton capacity, required length 75 ft, $28 per lineal foot in place. Of these types of piles, which appears to be the most economical for supporting the typical column loadings of 350 tons and 550 tons? 19 A piling contractor has been awarded the pile foundation contract for a large warehousing structure. The contractor’s fee is based on the following conditions and terms: • Building will require 120 separate pile foundation units, with the design indicating that the foundation units will consist of pile groups varying between 5 and 9 piles per group. • Design estimate calls for a total of 1100 piles, lengths to vary between 40 and 50 ft, with an average length of 46 ft expected (the estimated total footage for this project is therefore 50,600 linear feet, and this total serves as the basis for calculating final contractor installation costs). • Mobilization and demobilization for 4 pile rigs is $45,000. • Contract base payment for piles, total estimated linear foot basis, is $26.50 per foot.

Foundations • Overage, for total linear footage installed in excess of the estimated total of 50,600 ft, will be $32 per foot. • Underage, if installed linear footage is less than the estimated total of 50,600 ft, will be $20 per foot payment reduction for each foot less than the estimated total. • Installation period for the pile foundations is 30 days; contractor will be penalized $2,000 per day for each day beyond 30 days; and contractor bonus will be $1,000 per day if the installation is completed in less than 30 days. Compute the contract final (total) fee if: (a) 48,000 linear feet are actually installed and the contractor finishes four days ahead of the 30-day schedule. (b) 55,000 linear feet are installed, and the contractor finishes in 30 days. 20 An existing commercial building (a relatively new structure having a storage type basement) located in an urban area is to be renovated. The planned new use will result in heavier loadings being imposed onto the substructure including the foundations. The building structure is in good, serviceable condition, but with evidence that some settlement of the foundations has occurred; if the existing foundation system is retained, the expected new and heavier loadings would cause further settlement. The existing building foundations are shallow category spread footings located slightly deeper than the basement floor and supported in a stratum of medium-strength clay type soils. Very firm glacial till soil and rock exists from 20 to 25 feet

(6 to 7-m) below the basement floor slab and footing foundations. Why might micropiles be a practical choice for improving the foundation support provided for the renovated building? 21 A brief description of subsurface conditions at different building locations follows. Assume a commercial–industrial building is planned. Column loads to be imposed on foundations will vary, and could be up to 2,000 kN. For each location, list the type(s) of foundation which, for preliminary planning, probably could be considered suitable. (a) Thick stratum of glacially deposited sand– silt–clay mixture, relatively firm–compact, is underlain by rock; water table is deep. (b) Soil deposit across the area consists of widely varying layers of fine-grained soil and sand soil possessing low shear strength. Rock underlies the area at depths ranging from 10 m to 15 m below the soil surface; groundwater table is at a depth of 3 m. (c) Thin surface layer of predominantly sand soil (2- to 3-m thick) is underlain by a 3- to 4-m-thick stratum of weak, compressible fine-grained soil. A thick stratum of firm clay underlies the weak compressible layer. The groundwater table is at a depth of 10 m. (d) A surface layer of weak organic soil ranging from 2- to 4-m thick is underlain by a thick stratum of relatively dense–compact sandy soil with limited silt and gravel content. The area groundwater table corresponds to the surface of the sand stratum, or slightly below.

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From Essentials of Soil Mechanics and Foundations: Basic Geotechnics, Seventh Edition. David F. McCarthy. Copyright © 2007 by Pearson Education, Inc. Published by Pearson Prentice Hall. All rights reserved.

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Foundations Design Considerations and Methods

Building foundations need to be capable of carrying an imposed loading without undergoing movement that causes structural damage or affects the facility’s planned usage. These considerations require that the soil responsible for supporting a foundation not be stressed beyond its strength limits. Simultaneously, the deformations resulting within this soil because of loading and the action of natural forces cannot be excessive. The pressure that a foundation unit can impose onto the supporting earth mass without causing overstressing (or shear failure) is the soil’s bearing capacity. Deformations occurring because of foundation loading usually cause settlement, but lateral movements may also be of concern. It is also important to consider the possibility of foundation movements due to natural phenomena, such as soil expansion and shrinkage if moisture changes or freezing occurs. The magnitude and type of loading (static, live, or repetitive), the foundation performance requirements (how much settlement is permissible), and properties of the supporting soil all have influence on the type and size of foundation that will be necessary and its resulting behavior. Methods in widespread use for determining a soil’s bearing capacity include the application of bearing capacity equations, the utilization of penetration resistance data obtained during soil explorations, and the practice of relating the soil type to a presumptive bearing capacity recommended by building codes. Permissible bearing capacities determined by the equation method or from building code tables typically do not consider effects of soil compressibility and the possible influence of poorer soil layers underlying the bearing layer. Consequently, settlement determinations and other results of soil deformations must be analyzed separately. Foundation design criteria developed from boring–penetration resistance data often relate a foundation bearing pressure to settlement. Design data in this form are convenient to use, but the methods available do not cover all foundation and soil types and may be less precise than the analytical methods for determining bearing capacity and settlement.

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An additional method in use for determining the permissible or safe design loading (applied mostly for piles, less so for footing-type foundations) is the field load test performed on an in-place foundation unit. Load tests relate carrying capacity and settlement together, which is an advantage. Disadvantages include the cost and time involved. Further, test results require care in their evaluation, for it is known that a load test on a small shallow foundation may not be representative of the behavior of a large shallow foundation. With piles, the load test results on a single unit may not be indicative of the behavior of a loaded pile group. Of the common procedures for foundation design, the analytical method using soil mechanics principles (e.g., the use of bearing capacity equations with settlement analysis) and the use of penetration resistance data are preferred. Properly applied, these methods consider the effects of foundation type and size as well as the properties of soil to the depth that will have significant effect on the foundation performance. Load test data will provide reliable design information if properly related to the results of a subsurface exploration and a final foundation design. The use of presumptive bearing capacities is discouraged because of the heavy reliance on the soil-type description, with little correlation to the soil’s actual physical properties and no consideration of the possible existence of weak soil strata underlying the foundation bearing level. The procedures for foundation design that follow represent currently accepted methods. These methods generally have a history of being conservative. As in the past, foundation design procedures are subject to improvement for greater precision as new techniques are developed to better determine soil properties and behavior or as new analytical tools evolve.1 The foundations profession is still considered a state-of-the-art profession. Scientific methods and principles are utilized, but exact answers are not always expected. Final decisions concerning best foundation type, design criteria, expected behavior, and methods for construction and field control are greatly influenced by experience and intuition.

Shallow Foundations—Design and Installation Criteria 1

Basic Concepts—Long (Strip) Footings The ultimate soil-bearing capacity for foundations (the loading that will cause a shear failure in the supporting soils) is related to the properties of the soil, including the past stress history and the proximity of the groundwater table; it is also affected by the characteristics of the foundation, including size, depth, shape, and the method of construction or installation. For the case of an increasing load being imposed onto a shallow, horizontal strip footing resting on a homogeneous soil, a characteristic load-settlement curve is shown in Figure 1. Information important for predicting foundation behavior and for developing

1For

example, over the past decade, important advances in equipment capability and methods for instrumentation have improved the ability to determine in-place soil properties, increase understanding of the interaction between installed foundations and the surrounding soil, and monitor the behavior of earth masses (during seismic activity, where landslides or mass movements are being studied, etc.).

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Foundations Load, Q

Vertical load, Q Level ground surface

Settlement

Figure 1 Typical load-settlement relationship for shallow foundations.

Horizontal bearing surface

Firm and compact soils

Load, Q Q Settlement

Figure 2 Conditions for ultimate soilbearing capacity: involved soil zones and related foundation settlements [385].

Soft and loose soils

(a) General shear Load, Q

Settlement

Q

(b) Local shear Load, Q Test at greater depth

Settlement

Q

Surface test (c) Punching shear

criteria appropriate for foundation design has resulted from studies in which influential soil characteristics have been correlated with the settlement data (e.g., the soil descriptions have identified the materials as dense or firm, exhibiting brittle-type stress–strain behavior, or as loose or soft, exhibiting plastic properties). The soil zones involved in developing resistance to foundation loading (therefore, the soil zones responsible for bearing capacity) have been identified in a qualitative manner. The three principal modes of soil failure, established by the patterns of the shearing zones, are defined as general shear failure, local shear failure, and punching shear failure. Figure 2 illustrates the differences in the three modes and the foundation load-settlement curve typical to each mode. The general shear failure (Figure 2(a)), expected for soils possessing brittle-type stress–strain characteristics, is identified by a well-defined wedge beneath the foundation

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and slip surfaces extending diagonally from the side edges of the footing downward through the soil, then upward to the ground surface. The ground surface adjacent to the footing bulges upward. Soil displacement is accompanied by tilting of the foundation (unless the foundation is restrained). The load-settlement curve for the general shear case indicates that failure is abrupt. The punching shear failure (Figure 2(c)), occurring in soil possessing the stress–strain characteristics of a very plastic material, involves poorly defined shearing planes. Significant compression of a wedge-shaped soil zone beneath the foundation is accompanied by the occurrence of vertical shear beneath the edges of the foundation. The soil zones beyond the edges of the foundation are little affected, and no significant degree of surface bulging occurs. Aside from large settlement, failure is not clearly recognized. The local shear failure (Figure 2(b)) involves some of the characteristics of both the general shear and the punching shear failure modes. As for general shear, the well-defined wedge and slip surfaces are formed beneath the foundation, but the slip surfaces fade into the soil mass beyond the edges of the foundation and do not carry upward to the ground surface. Slight bulging of the ground surface adjacent to the foundation does occur. As with punching shear, significant vertical compression of soil directly beneath the foundation takes place. The local shear condition represents a transitional mode between the general shear and the punching shear failures and is expected for soils possessing somewhat plastic stress–strain characteristics. The load-settlement curves of Figure 2 are qualitative in nature. Investigators of the bearing capacity problem have found that relatively large settlements are required in order to have foundations reach the “failure” load (3 to 15 percent of the foundation width for very shallow installations and up to 25 percent of the foundation width for deeper installations). For application to shallow foundation design, it is commonly considered that the general shear case applies to dense granular soil and to firmer saturated cohesive soils subject to undrained loading (the U–U and C–U shearing conditions apply). The punching shear case is considered appropriate for compressible soil, such as sands having a low-tomedium relative density, and for cohesive soils subject to slow loading (the C–D shearing conditions apply).

2

Bearing Capacity Equations Historically, over the past one hundred years, a number of investigators have undertaken studies relating to foundation bearing capacity, typically applying the classical theories of elasticity and plasticity to soil behavior to develop equations appropriate for foundation design.2 Behavior described by the classical theory of plasticity has been widely used to obtain a solution for the case of a general shear failure. The original theoretical concepts for analyzing conditions considered applicable to foundation performance using the theory of plasticity are credited to Prandtl [283] and Reissner [291]. Prandtl studied the effect of a long, narrow metal tool bearing against the surface of a smooth metal mass that possessed

2Jumikis

(1962) [170] provides a summary of the works of early investigators.

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cohesion and internal friction but no weight. The results of Prandtl’s work were extended by Reissner to include the condition where the bearing area is located below the surface of the resisting material and a surcharge weight acts on a plane that is level with the bearing area. Terzaghi [356] applied the developments of Prandtl and Reissner to soil foundation problems, extending the theory to consider rough foundation surfaces bearing on materials that possess weight. Conditions for relating the classical theory of plasticity to the case of a general shear failure are indicated by Figure 3. The arrangement shown establishes criteria for developing the ultimate bearing capacity for a long strip foundation; because of the infinite foundation length, the analysis proceeds as for a two-dimensional or plane-strain problem. The theory assumes that the (soil) material in zones I, II, and III possesses the stress–strain characteristics of a rigid plastic body (the material shows an infinite initial modulus of elasticity extending to the point of shear failure, followed by a zero modulus; see Figure 3(b)). Applied to the soil mass providing support for the foundation, the theory assumes that no deformations occur prior to the point of shear failure but that plastic flow occurs at constant stress after shearing failure. It is also assumed that the plastic deformations are small and the geometric shapes of the failure zones remain essentially constant. The use of an equivalent surcharge to substitute for the soil mass above the level of the foundation, along with some estimations, simplifies the analysis, but the effect is to provide conservative results. When subject to a foundation loading near to the ultimate, zone I behaves as an active zone that pushes the radial zone II sideways and the passive zone III laterally upwards. Boundaries AC and DE shown on Figure 3(c) are essentially straight lines; the shape of section CD varies from circular (when the soil angle of internal friction is zero Qult Note: ratio L/B is very large σ

τ L B D

d a

φ

c b c, cohesion Soil properties: c, φ, γ

ε Definition of assumed stress−strain properties

σ

Definition of soil strength properties (b)

(a) Forces and soil zones involved at ultimate bearing capacity Q"ult L=∞ A

I C

B

Surcharge load/effect

III II

D

E Soil properties: c, φ, γ

(c) Assumption of forces and soil failure zones

Figure 3 Definition of conditions for developing the ultimate bearing capacity equation.

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degrees) to a curve intermediate between a logarithmic spiral and a circle (when is greater than zero degrees). Terzaghi developed a general bearing capacity equation for strip footings that combined the effects of soil cohesion and internal friction, foundation size, soil weight, and surcharge effects in order to simplify the calculations necessary for foundation design. His equation utilized the concept of dimensionless bearing capacity factors whose values are a function of the shear possessed by the supporting soils. Through ensuing years, the ultimate bearing capacity for shallow and deep foundations has continued to be studied in the quest for refined definition of foundation soil behavior and a generalized bearing capacity equation that agrees well with failure conditions occurring in model and large-scale foundation tests.3 Modifications to early concepts have emerged from such studies, but the general form of the Terzaghi bearing capacity equation has been retained because of its practicality. The ultimate bearing capacity equation, shown as Equation 1(a) and developed from the conditions outlined in Figure 3(c) for long footings, has been found to provide good agreement with ultimate loads observed for model and large-scale foundations. This equation shows the Terzaghi format, but the current dimensionless bearing capacity factors are the result of better definition of the soil failure zones: Qult Q–ult 1 = qult = cNc + Bg1Ng + g2Df Nq L 1B * L2 2 1B * 12

(1a)

where Qult = total loading acting at the base of the strip foundation whose width is B and length is L (kips, kN, etc.) Q–ult = load acting at base of foundation for unit length in L direction (kN/m, k/ft, etc.) qult = ultimate gross bearing capacity or soil bearing pressure (kN/m2, ksf, etc.) c = cohesion of the soil below foundation level (kN/m2, kPa, ksf, etc.) Df = depth of footing below lowest adjacent soil surface (m, ft, etc.) g1 = effective unit weight of soil below foundation level (kN/m3, pcf, kcf, etc.) g2 = effective unit weight of soil above foundation level (kN/m3, pcf, kcf, etc.) Nc, Ng, Nq = soil-bearing capacity factors, dimensionless terms, whose values relate to the angle of internal friction, , whereby: Nq = tan2145 + f>221e2ptanf where e = base of natural logarithims, 2.71828 Nc = 1Nq - 12 cot f 1for f 7 0°2 = 5.14 1for f = 0°24 Ng = 21Nq + 12 tan f

3Meyerhof

[231, 233], Hansen [141], Vesic [384], DeBeer [82]. rigorous mathematical solution for the value of Nc when equals zero is presented by Jumikis [171, pp. 626–27]. 4A

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Foundations 800

800

600

Nq = eπ tan φ tan2 (45 + φ/2) Nc = (Nq − 1) cot φ Nγ = 2(Nq + 1) tan φ

400 300

Nq

200

600 400 300 200

Bearing capacity factors, N

Nc 100 80

100 80

60

60

40 30

40 30

20

20

10

10

8 6

Nc

8 6

4 3

4 3

2

φ

Nc

Nq

0 5 10 15 20 25 30 32 34 36 38 40 42 44 46 48 50

5.14 6.5 8.3 14.0 14.8 20.7 30.1 35.5 42.2 50.6 61.4 75.3 93.7 118.4 152.1 199.3 266.9

1.0 1.6 2.5 3.9 6.4 10.7 18.4 23.2 29.4 37.7 48.9 64.2 85.4 115.3 158.5 222.3 319.1

0.0 0.5 1.2 2.6 5.4 10.8 22.4 30.2 41.1 56.3 78.0 109.4 155.6 224.6 330.4 496.0 762.9

2 Nq Nγ

1.0 0.8

1 0.8

0.6

0.6

Nq /Nc

0.4 0.3

0.4 0.3

0.2

0.2

0.1

5

10

15

20

25

30

35

40

45

0.1 50

Angle of internal friction, φ°

Figure 4 Bearing capacity factors for the bearing capacity equation.

The relationship between the bearing capacity factors, Nq, Nc, and N and the angle of internal friction is presented in Figure 4. The first and third terms in this equation represent the Prandtl–Reissner solution for the condition of a weightless soil. The second term represents the separate solution to a determination of the ultimate bearing capacity for a cohesionless soil with weight but no overburden effects (a necessary consideration because the Prandtl–Reissner terms do not include a shearing resistance from the effect of soil weight and internal friction, , in the failure zones). This method of superposition (i.e., a simple combination of three terms) to produce Equation 1a is not a theoretically correct procedure, but errors are on the safe side and small for most practical applications.

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Foundations Probable upper limit for qult Footing pressure, qult

Figure 5 Implications of the general bearing capacity equation: Allowable pressures increase with increases in footing width and depth (solid lines). Note that the actual rate of bearing capacity increase decreases as width becomes greater (dashed lines).

, m) 6 (ft , m) 4 (ft Df = , m) 2 (ft Df =

Df =

Foundation width, B

Equation 1a implies that the bearing capacity for foundations on soil possessing internal friction, , increases indefinitely with increases in foundation width, as indicated by the solid lines of Figure 5. However, physical studies have indicated that the rate of increase for bearing capacity decreases somewhat as foundation width increases, and there may be an upper limit to a soil’s bearing capacity (see dashed lines of Figure 5). Foundation widths beyond the value that develops the upper limit would not contribute to the bearing capacity. Fortunately, the width corresponding to such an upper limit on bearing capacity apparently is relatively large; as a result, the condition is not a serious restriction for commonly used sizes of shallow foundations. This “size effect” can be compensated for during design by using the value from soil testing appropriate to the general range of the bearing capacity pressure and by applying the rigidity index reduction factors discussed in the following paragraphs. Nevertheless, the effect of using higher bearing pressures on larger foundations requires careful consideration where settlement is important. Because of the manner in which stresses from foundation loadings are transferred into the earth, a large foundation tends to settle more than a small foundation when both impose the same intensity of bearing pressure, because the loaded area and total load Q are greater. Using a greater bearing pressure for a large footing further increases the tendency for greater settlement. Consequently, if lightly loaded and heavily loaded foundations for a structure are designed only on the basis of allowable bearing capacity, there is a strong possibility of differential settlement between the large and small footings. The assumption that the soil responsible for developing a foundation bearing capacity behaves as a rigid material is satisfied for the case of general shear but is not appropriate for punching shear and local shear. Comparison of the relative load-settlement curves of Figure 2 indicates that, for the punching and local shear cases, the ultimate load is less and the settlement is greater than for the condition of the general shear failure. Methods applied in the past to compensate for lower bearing capacity design values where a foundation was to be supported on soil that is compressible and that would u