Soil Mechanics and Foundations

• • SOILM AND F5(j}!ANICS DATIONS R A U C SE A FO SOIL MECHANICS AND FOUNDATIONS MUNI BUDHU JOHN WilEY & SONS, I...

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• •

SOILM AND F5(j}!ANICS DATIONS

R A U C SE A

FO

SOIL MECHANICS AND FOUNDATIONS MUNI BUDHU

JOHN WilEY & SONS, INC. "lew York / Chichestcr IlVeillheim I Brisballe I 5111gopore I TorOIlfO

Editor Wayne Anderson Marketing Manager Katherine Hepburn Semor Production Manager Lucille Buonocore Production Editor Leslie Surovick Cover Designer Lynn Rogan Illustration Editor Sigmund Malinowski Illustration Studio Radianl Illustration II:- Design Cover Photo CORBISlRogu Wood Th is book was set in 10112 Times Ten by UG I GGS Informati pnnted and bound by RR DonnelJey/Wiliard . The cover was p Corporation. This book is printed on acid-free paper.

e

The paper in this book was manufactu red by a include sustained yield harvesting of tts timbedan principles ensure that the numbers of of new growth.

tf'J''t';'

NOTES FOR INSTRUCTORS

ix

Chapter 11 is about slope stability. Here stability conditions are described based on drained or undrained conditions. An appendix (Appendix A) allows easy access to frequently used typical soil parameters and correlations.

F

CHAPTER LAYOUT

dvent 0 ersonal computers, learning has become more visual. Some , studies hav epo[ ed that visual images have improved learning by as much as 400% . This tex-t ok is accompanied by a CD ROM that contains text, interactive animation, images, a glossary, notation, quizzes, notepads, and interactive problem solving. It should appeal, particularly, to visual learners. A quiz is included in appropriate chapters on the CD ROM to elicit performance and provide feedback on key concepts. Interactive problem solving is used to help students solve problems similar to the problem-solving exercises. When an interactive problem is repeated, new values are automatically generated. Sounds are used to a limited extent. The CD ROM contains a virtual soils laboratory for the students to conduct geotechnical tests. These virtual tests are not intended to replace the necessary hands-on experience in a soil laboratory. Rather , they complement the hands-on experience, prepare the students for the real experience, test relevant prior knowledge of basic concepts for the interpre-

X

NOTES FOR INSTRUCTORS

tation of the test results, guide them through the evaluation and interpretation of the results, allow them to conduct tests that cannot otherwise be done during laboratory sessions, and allow them to use the results of their tests in practical applications.

ABET REQUIREMENTS The United States Accreditation Board for Engi (ABET) has introduced new criteria for accreditation p this book has the author's judgment on how it satisfies ence (ES) and engineering design (ED) crit mended percentages allocated to ES and

FO

COURSE MATERIAL

NOTES for Students and Instructors

FO

PURPOSES OF THIS BOOK echanics and its appli-

technical engineering. The goals of this te as follows:

o characterize soil properties, cs to analyze and design simple geo-

ng this textbook you should be able to: • Descr be soi and determine their physical characteristics such as, grain size, wate ~ ontent, and void ratio • Classify soils • Determine compaction of soils • Understand the importance of soil investigations and be able to plan a soil investigation • Understand the concept of effective stress • Determine total and effective stresses and pore water pressures • Determine soil permeability • Determine how surface stresses are distributed within a soil mass • Specify, conduct, and interpret soil tests to characterize soils

xi

xii

NOTES FOR STUDENTS AND INSTRUCTORS

• Determine soil strength and deformation parameters from soil tests, for example, Young's modulus, friction angle and undrained shear strength • Discriminate between "drained" and "undrained" conditions • Understand the effects of seepage on the stability of strucrure Estimate the bearing capacity and settlement of structures unded on soils Analyze and design simple foundations • Determine the stability of earth structures, for examFi slopes

et '

in~

Distribution of Main Topi

Foundation and earth structures Description

6

7

8

9

10

11

0

0

0

0

0

0

FO

0

• •

Stresses in soils

D

Drained and undrained conditions

U

Settlement and deformation

C

Shear strength

T

Bearing capacity and settlement of foundations

0

Stability of earth structures

N

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

• • 0

0

0 0

0

0

0

• • • • • • • 0

0

0

O.

0

0

•

Seepage

0

0

• •

0

0

0

•

0

0

0 0

• •

NOTE S FOR STUDENTS AND INSTRUCTORS

xiii

ASSESSMENT You will be assessed aD how well you absorb and use the fundamentals of soil mechanics. Three areas of assessment are incorporated in the Exercise sections of this textbook. The first area called "Theory" is intended fog y-ou to demon· strate your knowledge of the theory and extend it to uncovec1!Cwt elat ionships. The questions under "Theory" will help you later in your cf.~eec to address unconven tional issues using fundamental principles. The selo n a cea ca~ d " Problem Solving" requires you to apply the fundamenta prin i les and--concepts to a wide variety of problems. These problems willtes o ur un erstandi ng and use of the fundamental principles and concep ts. The thirtl ea caUl d " Practical" is intended to create practical scenarios for you to use "a t 'o.n~y the subject matter in the specific chapter but prior mate rial/ i"h ya.u "havl' encountered. These problems try to mimic some aspects of r~tuatl \fls and give you a feel for how the materials you have studied so far t"~ be pplied 'o practice. Communications ese "Pract ica l" are. at leas!. as important as the technic detail~ 1n many 0 prob lems you are placed in a sit atlOn I co iv' ce stakeholders f you r technical competence. A q uiz (multip) choice) on each chapter' • d ud~d in the CD \0 tcst you r general knowledge of t subj ct matter in that chapter: h,e questions on the quiz are re lated to the sec Io n Question.!J6 Guide Your eading,"' included in each chapter. ~ ~

SUGGESTIONS F R PRO.B EM

SOL\lING~

Enginee ring is, ~ ,a bout propl em 5"' I~ n ~ r most engineering problems, there i~~ nique method or procedure for fi nding solutions. Often, there is no unique .s~ u Ion to an engineering ~;~bl~~A suggested problem-solving pro~ cedure is ~t1ined be low.

,

1. Rea/ the

prObleJ.1l...c a ref~ UY; ate or wrile down what is given and what you

~ required to find.

2. Draw clear diagrams or e tches wherever possible. 3. Devise trategy tOtn the solution. De termine what principles, concepts, and.rq ua tl ns a e needed to solve the problem. 4. p.lrfocm calcuJations making sure that you are using the correct units. 5. Chcc 'hethl r your results are reasonable. The units of measure ment used in this textbook follow the SI system. Engineering calculations are approximations and do not result in exact numbe rs. A ll cal1.0 1.0 to 10- 3 10- 3 to 10- 1 r2 , h1' h2' and qv (flow rate of the pump), k can be calculated from Eq. (2.41). This test is only practical for coarse-grained soils. Pumping tests lower the groundwater, which then causes stress changes in the soil. Since the groundwater is not lowered uniformly as shown by the drawdown curve in Fig. 2.22, the stress changes in the soil will not be even. Consequently, pumping tests near existing structures can cause them to settle unevenly. You should consider the possibility of differential settlement on existing structures when you plan a pumping test. Also, it is sometimes necessary to temporarily lower the groundwater level for construction. The process of lowering the groundwater is called dewatering.

2.11 DETERMINATION OF THE COEFFICIENT OF PERMEABILITY

I

Observation wells

57

Pumping well

---- - -~----e--- --~- ----------------!

!-15m~

~30m

+

15 m

Impervious

FIGURE E2. 14

soil bed of thickness ate of pumping was 10.6 cated at 15 m and 30 m from the

EXAMPLE 15 m and the center

.4

FO

ketc

Step 1: Step 2:

e measurements to directly apply Eq. (2.41) to f the pump test to identify the values to be

tch of the pump test with the appropriate dimensions-see Substitute given values in Eq. (2.41) to find k. 1'2 =

hI

k

30 m ,

1'1

= 15 m,

= 15 - (1.9 + 1.6) =

qu In(rzlr1) 1T( h~ - hD

=

=

15 - (1.9 + 1.4)

h2 =

=

11.7 m,

11.5 m 3

10.6 X 10- In(30/15) = 50 X 10-2 I 1T(1l.72 - 11.5 2) 10 4 ' em s

•

What's next . .. Water, although regarded as the "foe" in geotechnical engineering, can be used to improve soil strength, reduce soil deformations under loads, and reduce the permeability. Next, we will study how water can assist in the improvement of soils.

58

CHAPTER 2

PHVSlCAl CHARACTERISTICS Of SOILS ANO SOlllNVESTlGATIONS

2.12 DRY UNIT WEIGHT- WATER CONTENT RELATIONSHIP 2 . 12.1 Basic Concept Let 'S examine Eq . (2.12) for dry unit weight, that IS , '1" =

(~)'Y. = C+ ~.'G~S)'Yw

(2.42)

The extreme right· hand side term was obtained by rc la ci n~ b~ e = wGjS. How can we increase the dry unit weight? Examination iE . (2~) reveals [hat we have to reduce the void ratio; tha i is, wlS mllsLbe r du .' T he theoret iea! maximum dry unit weight is ob tained 1 tha is

W 7.:.

e~

(2. 43 )

2 . 12.2 Proctor Compact) n Te A laboratory test . ca lled the Proci was developed to delive landa rd amou nt of mechanical ener (comp~c i e effort) t i:lelermine th maximum d ry unit weight of a soi l. In the fi and,V'd Proctor te • a r soi l s~im en is mixed with water and com acted a c:tlmdrical mold vOlu 044 x 10-· m) (sta ndard Procto mold) by r Beated blows fro m th ass a hammer, 2.5 kg, falling freely (fa a height jpf 05 mm (Fig.....2. ). . IS compacted in th ree layers, each of w his su iected to 2~s. A nfodified Pr test was de eloped or c paction of airfields to support hea aire f I ads. In the modifi ed Pro tor tesl, a hamme r with a mass 4.54 kg falls eely om a height of 4 mm. e SOil is compacted in five layers with 25 blo per layer in l)Atandard Proctor mold. Fou r or ore tests a r~o ~ duct ed on the soil using different water con tents. T h last test is idenl lf\e w e additional water causes the bulk uni t weight of e soil fO decrease fThe resu l ared,lo!led as dry unit weight (ordinate) versus wa r'content (ab ssa ). T ieal unit weighl- water content plots are shown in f ig. 2.24. Clays usually yjeld l=shaped curves. Sands do nOl often Yield a clear bell ;

dry

Yo) Moltl

(b) Hamme.

FIGURE 2.23 Compaction apparatus . (photo courtesy of Geotesr.)

2.12 DRY UNIT WEIGHT-WATER CONTENT RELATIONSHIP

59

12 10L---------~L---------~----~~~~~~~----~

4

6

FO

R

FIGURE 2 .24

un aturated at the maximum dry unit weight, that is, can det mine the degree of saturation at the maximum dry unit weight using q. ( -A2). We. know 'Yd = ('Yd)max and W = WopI from our Proctor test results. If Gs is know w. an solve Eq. (2.42) for 5. If G s is not known, you can substitute a value of 2.7 with little resulting error in most cases. Equation (2.42) can be used to plot a series of theoretical curves of dry unit weight versus water content for different degrees of saturation (Jines of constant degree of saturation) as shown in Fig. 2.24 for 5 = 100% and 5 = 80 % . You plot these curves as follows:

1. Assume a fixed value of 5, say, 5 = 1 (100 % saturation). 2. Substitute arbitrarily chosen values of w, approximately within the range of water content on your graph. 3. With the fixed value of 5 and either an estimated value of Gs (= 2.7) or a known value, find 'Yd for each value of W using Eq . (2.42) and plot the results of 'Yd versus w. 4. Repeat for a different value of 5.

60

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOilS AND SOil INVESTIGATIONS

OJ

"

Increasing compaction

1 Water content

Effect of increasing compaction efforts on water content relationship . FIGURE 2.25

FO

The curve corresponding to S = 1 is k voids line. This line represents the mini ... _-.: ...._ ._ water content [Eq. (2.43)]. The achievement of zero Proctor test, using higher leve.l mum dry unit weight at a lower 0 (Fig. 2.25). The degree 0 aturation i than the standard compactIDlnii"'t,~I"""".i·

a uration line or zero air ble at a given

,

2.12.5 Field Compaction A variety of mechanical equipment is used to compact soils in the field. You may have seen various types of rollers being used in road construction . Each type of roller has special mechanical systems to effectively compact a particular soil type. For example, a sheepsfoot roller (Fig. 2.26a) is generally used to compact finegrained soils while a drum type roller (Fig. 2.26b) is generally used to compact coarse-grained soils.

2.12.6 Compaction Quality Control A geotechnical engineer needs to check that field compaction meets specifications. Various types of equipment are available to check the amount of compac-

61

FO

R


Two types of machinery for field compaction. (Photos courtesy of Vibromax America, Inc.)

FIGURE 2.26

tion achieved in the field . Three popular apparatuses are (1) the sand cone, (2) the balloon, and (3) nuclear density meters.

2.12.6.1 Sand Cone A sand cone apparatus is shown in Fig. 2.27. It consists of a glass or plastic jar with a funnel attached to the neck of the jar. The procedure for a sand cone test is as follows: 1. Fill the jar with a standard sand-a sand with known density-and determine the weight of the sand cone apparatus with the jar filled with sand

CHAPTER 2

PHYSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

Jar

Ottawa sand

FIGURE 2 .27

STM) recom2. 3.

FO

62

D ry umt . welg . h t:

'fd

W" = 11

2 . 12.6.2 Balloon Test The balloon test apparatus (Fig. 2.28) consists of a graduated cylinder with a centrally placed balloon. The cylinder is filled with water. The procedure for the balloon test is as follows : 1. Fill the cylinder with water and record its volume, VI' 2. Excavate a small hole in the soil and determine the weight of the excavated soil (W). 3. Determine the water content of the excavated soil (w).


63

Air re lease va lve

Ba lloon Pump

FIGURE 2.28

Balloon test device.

4.

5. Record the volume of wat

6. Calculate the unit weigb;

FO

nt of a soil at a particular site.

FIGURE 2 .29

Nuclear density meter. (Photo courtesy of Seaman Nuclear Corp.)

CHAPTER 2

1. 2.

3. 4. 5.

PHYSICAL CHARACTERISTICS OF SOILS AND SOIL INVESTIGATIONS

The essential points are: Compaction is the densijication of a soil by the expulsion of air and the rearrangement of soil particles. The Proctor test is used to determine the maximum dry unit weight and the optimum water content and serves as the reference for field specijications of compaction. Higher compactive effort increases the maximum dry unit weight and reduces the optimum water content. Compaction increases strength, lowers compressibility'; and reduces the permeability of soils. A variety offield equipment is used to check the dry unit weights achieved in the field Popular fiel!1 equiwnent includes the sand cone apparatus, the balloon apparatus.., an the nu lear density eter.

EXAMPLE 2.15 The results of a standard

5 % standard compaction?

n plot the results of 'Yd versus

FO

64

w

(%) . Then

Zero air voids

Water content (%)

Bulk unit weight (kN/m3)

Dry unit weight (kN/m 3)

Y 'Yd=1+w

Dry unit weight (kN/m3) Water content

Yd = (, +

~Gs/5)YW;

(%)

5=1

22.8

6.2

16.9

15.9

6

8.1

18.7

17.3

8

21 .8

9.8

19.5

17.8

10

20.8

11.5

20.5

18.4

12

20.0

12.3

20.4

18.2

14

19.2

13.2

20.1

17.8

2.13 SOil INVESTIGATION

--1~ 23

.I

22

~ J

19

,

~ 18

.

Zero .Ir I'OIds

~ \ _ 18.4 kNim

M .... Mum unit...,

95% =pachon

On

.~

I

I' 15 6

65

I

,

7

8

9

10

11115 12 ~ 3

i

r-.

Wale< conlen! (%)

FIGURE E2.1S

Step 2: Step 3:

Compaction test results.

Extract the desired

Step 4:

.7

What~
.. .W e have

(18.419.8) _ 0 7 ~ % 18.419.8 . 1 71

•

discu ss~

s,p.ll s w it h the tacit assum ption that we have obof sOils from th e field on which we conducted tests s h ave b e ~ escribed. So' Is are o bserved and recove red during a soil investigation of a pro oseCl site. A s9-i invest igati n'is an essential part of the design and construct ion of a p ro posed structu al sy te (bu ildings, dams, roads and highways, etc.) . You i!!...B.P
What's next . , .I n the next two sections, we will write the genera l expression for Hooke's law, wh ich is t he fu ndamenta l law for linear elastic materials, and then conside r two loading cases appropriate to soils.

88

STRESSES, STRAINS, AND ELASTIC DEFORMATIONS OF SOILS

HOOKE'S LAW 3.5.1 General State of Stress Stresses and strains for a linear, isotropic, elastic soil are related t law. For a general state of stress (Fig. 3.9), Hooke's law is 1 -v -v

ex

-v

ey ez

1 -v

-v -v

0

0

0

0

1

0

0

0

o 2(1

+ v)

'{yz

0

0

0

0

'{zx

0

0

0

'{xy

1 E

(3.10)

(3.11 )

where

FO

3.5

CHAPTER 3

(3.12)

shown in Table 3.2.

z y

~--------L-----+X

FIGURE 3 .9

General state of stress.

3.5 HOOKE'S LAW

TABLE 3 .2

Typical Values of E and G Description

E-,MPal

G- (MPal

Clay

Soft M edium Stiff

Sand

loose Medium

1-15 15-30 30- 100 10-20 20- 40 40- 80

0.5-5 5-15 15-40 5- 10 10-15 15-35

Soil type

89

Dense

"These ale .velage seCln! elu!;c mod>! li for !he d .. ined condition (~ee discussions letel in Ihis c~ep!e., Sections 3.12 and 3.131.

3.5.2 Principal Stresses If the stresses applied to a soil are

iifclOa l stre e , then Hooke': (w reduces

to

(3.13)

The matrix on the Thc invcrse of Eq.

-calledoillhe- compl iance

matrix.

(3.14)

matrix on the righ led the stiffness matrix. If you know the s'8nu the mate I 1 parameter E and v, you can usc Eq. (3.13) to calcula te U:!'e st ains; or if Y9 know:ne s~ il1 s. E, and v. you can use Eq, (3 .14) to calcu late t e stresses. 'V'" tre

...._"

3.5.3 Displacements from Strains and Forces from Stresses The displacements and forces are obtained by in tegration. For example, the vertical displacement. 6.z. is (3.15)

and the axial force is (3.16)

where dl. is the heigh t or thickness of the element and dA is the elemental area.

90

CHAPTER 3


The essential points are: 1. Hooke's law applies to a linearly elastic material. 2. As a first approximation, you can use Hooke's law to calctJlO:le stresses, strains, and elastic settlement of soils. 3. For nonlinear materials, Hooke's law is used with an.ap oximate elastic modulus (tangent modulus or secant modul s an he cal~ula tions are done for incremental increases in stres

What's next , ..The stresses and strains in three when applied to real problems. For practical puq~ tion, we will discuss two conditions that sim

3.6 PLANE STRAIN AND A"".~... SYMMETRIC CONDITIONS

FO

(3.17) (3.18)

Reta ining wall

Z (1 )

r

;

(2)

~X(3) FIGURE 3 .10 Plane strain condition in a soil element behind a retaining wall.

3.6 PLANE $TRAIN ANO AXIAL SYMMETRIC CONOITIONS

and 10'1 .., t.{0'1 + a,}

I

91

(3.19)

In matnx form, Eqs. (3.17) and (3.18) become

{,,} . ~ [' -, E

tl

- ],/

-,

1 -],/

]{"'} O'J

(3.20)

The inverse of Eq. (3.20) gives ~~----~--~~

{::J -

(I +

,)~

2,) [' : '

(3.2l)

3.6.2 Axisymmetric Condition T he other co ndition that occurs in practica l ble is axial symmetry or the axisymmerric condi tion where two Slres a re u61. Let us consi er a water tank or an oil tank founded on a mass ;11 trated in Fi 3.U . The rad ial stresses (0",) and clccumfer ntial stresses (cr,) on a lindrical element of soil directly und the ce n t~ qt-the ta nk are equal because of ax ial symmetry . The oil tank will pply 8 unifo rm vertic· J ( xial) stress at the soil surface and the soi l element w~1I be subjected to an merea in aJial stress, 6 0, "" 60"10 and an incre€e" in radial stress, 60 , "" 6 0, "" crl' ill a soil element under the edge of j\e tank bj under an axis): e. ndition? The answer is no, since the stres ' s at th., edge of the ta nk c...e all differen t; the re is no "" symmetry. Hooke' la 'fo r the axisymmetric condition is (3.22) (3.23)

or, in p

tri x form , (3.2 4)

z

, Tank

,I

1 FIGURE 3.11 tank.

I ""

Axisymmetric condition on

a soil element u nder the center o f a

CHAPTER 3

STRESSES. STRAINS. AND ELASTIC DEFORMATIONS OF SOILS

The inverse of Eg. (3.24) gives

t:}

= (1 +

v~1 _ 2v) [1 ~ v~v]{::}

The essential points are:

(3.25) i

1. A plane strain condition is one in which the strain in one or more ai-

rections is zero or small enough to be neglected. 2. An axisymmetric condition is one in which two stre$Ses are qual. EXAMPLE 3.1

A retaining wall moves outward causing a strain of 0.05% on a soil element loc e(:l

FO

. e stliess condition and write the appropriate

Step 2: ~(Jl

~(J3

= 9615.41(0.7 x 0.0005) + [0.3 x (-0.001)]1 = 0.5 kPa = 9615.41(0.3 x 0.0005) + [0.7 x (-0.001)]! = -5.3 kPa; the negative sign means reduction

Step 3:

Calculate the lateral force per unit length. ~cr3

=

~Px =

~(Jx

f ~(Jx

dA = - (

5.3 (dx

x 1) =

-[5.3x]8 = -31.8 kN/m

•

EXAMPLE 3.2

An oil tank is founded on a layer of medium sand 5 m thick underlain by a deep deposit of dense sand . The geotechnical engineer assumed, based on experience,

93

3.6 PlAlIIE STRAIN AND AXIAL SYMMETRIC CONDITIONS

Tan k

.ok"" 2OkPa _

T

2.5m

.1

--

5m

Med,um Sind

FIGURE E3. 2

/'.

that the sertlement oC the lank would occurArom ~ ttlem t in the mcdium sand. The vertical and lateral stresses al the middle f em dlum sand directly under nd a, re ~. Ivcly. The valu s of E and the center of the tank are 50 k Pa ~ 1/ are 20 M"Pa and 0.3, respectively ssuming a li.pear, isotropi elas c material vertical behavior, calculate the strains im ed on t e medium sana and I settlement. Strategy You have to decide the stress condition on the il element directly under the c~ter of th tan . Once you mak e yo deCision, use the appropriate equatioils to find th~ trains and then integrate e vertical strains to ca lculate the set emen!. D a.Jr a diagram i6us~ng the roblem.

Solution 3.

,

Y

Ora a diagram of the proble ~g . E3.2. Decid ~.Jm a stress cop dition. The element is dir t U r he center of the tank , so Ihe .... ' _'axi ymmclric conditio

~~~~~.s the2~:pe { E)

== 20

llE) = 20

~

t()l [1

= 20

~

101

.6.£3

WI~06l(50}

~ to' [-~.3

0.7

20

e gel

Using alg bra

Step 4:

u.lion, ,nd

x SO - 0.6 x 201 = 1.9 x 10 - 3

[- 0.3 x SO + 0.7 )( 20] - -5 )( lO-s

Calculate vertical displacemcnt. 6£) = 6(, 6z

=

f

6£., dz - [1.9

X

10- 3

z15 = 9.5

)( 10- 3

m=

9.5

mm

•

What's next . ..We have used the elastic equations to ca lcul ate stresses, strains, and displacements in soils assuming that soils are linear, isotropic, elastic materials.

94

CHAPTER 3

STRESSES. STRAINS. AND ELASTIC DEFORMATIONS OF SOilS

Soils, in general, are not linear, isotropic, elastic materials . We will briefly discuss anisotropic, elastic materials in the next section.

F

3.7

ANISOTROPIC ELASTIC STATES Anisotropic materials have different elastic parameters' Anisotropy in soils results from essentially two causes

1.

2. The difference in stresses in induced anisotropy.

s Poisson's ratio determined from the ratio of the strain (X direction) to the strain in the vertical direction (Z direction) with ad applied in the vertical direction (Z direction). In the laboratory, the direction of loading of soil samples taken from the field is invariably vertical. Consequently, we cannot determine the five desired elastic parameters from conventional laboratory tests. Graham and Houlsby (1983) suggested a method to overcome the lack of knowledge of the five desired elastic parameters in solving problems on transverse anisotropy. However, their method is beyond the scope of this book. For axisymmetric conditions, the transverse anisotropic, elastic equations are irecti

(3.26)

95

3.1 ANISOTROPIC ELASTIC STATES

where the subscript z de notes vertical and , denotes rad ial. By su perposit ion, \1,/ "'" = EJ E•.

The essential poinls are: J. Two forms of allisotropy an praenl ill soils. One is struc/ural anlsolropy, which is nlatuJ to the history of 10000l11g and enfJlronmental conditiolls during chposilion, 4IId Ihe other is slrtSS-induced onisolropy, which raults from differmcrs in strrsses in different directions. 2. The pNwulenl form of structural anisQtropy In soils is /ralU ven 'e anisolropy; 'he soil properties and the soil rtSponse ;n tire la/eral directions are the slime but lire different/rom those in the vertical direction. 3. You need 10 Ji.d Ihe elastic parameters In different directions of a soil mass to dele,..",/ne elaJlic strrsses, strains, and displacements.

EXAMPLE 3 .3 Redo .Example 3.2 but now the soil un ~e 0 11 ta nk is an anisotropIc maten al Wit h E~ = 20 MPa, E~ MPa v'r == 0.15, v" "" 0.\

Strategy The (326)

SOI~f this problen~ IS a stralA~wara app"

aSllc

tiOn of Eq.

,

Solution 3 .3 Step 1:

Determine

( by superpositio

1-2 0.,,]

Find the ~!~~

use' . ~3.,

{""J At,

...

to-

,[

x

20 -0.12 (l

25

20

The solu tion is E~ 0.03%.

Step 3:

=

~50.3) {~}

2.26 X 10- 3 = 0.23% and €, = 0.26 x 10-' -

De termine vert ical displacement. 4z =

fE,

dz -

(2.26 x lO-lzl&

=

11.3 X 10 -) m

=

11 .3 mm

The ve rt ical d isplacemenl in the anisotropic case is about 19% more than in the isotropic case (Example 3.2). Also. the radial strain is te nsile fo r the isotropic case but compressive in the anisotropic case for this problem. •

96

CHAPTER 3


What's next . . .We now know how to calculate stresses and strains in soils if we assume soils are elastic, homogeneous materials. One of the important tasks for engineering works is to determine strength or failure of materials. We can draw an analogy of the strength of materials with the strength of a chain. The cain is only as strong as its weakest link. For soils, failure may be initiated at a poi w ithin a soil mass and then propagate through it; this is known as progressive fai reo The tress state at a point in a soil mass due to applied boundary forces y equal t o the strength of the soil, thereby initiating failure. Therefore, as e g1 ed to know the stress state at a point due to applied loads. We will using your knowledge in strength of materials.

3.8

STRESS AND STRAIN STATES

FO

1

rY,

T

~

0

M

'~ 1-0,

Plane on wh ich the

Txz

(J

II'

(rY,. (a)

FIGURE 3 .12

-T,) (b)

Stresses on a two-dimensional element and Mohr's circle.

l .8 STRESS AND STRAIN STATES

97

and 3. The stresses at these points are the major principal stress, (TI! and the minor principal stress, (Tl' The principal stresses are related to the stress componen ts (1. . (T... 'T u by cr, +

2

CT~ =

a, + )(CT, -2 aA) ~ + r.

,~

+

0-, a~ -, --

J(" -2 ':

x0

6U1 -]0 _

AU~

0) as illustrated by the inset fig ure labeled "3" in Fig. 3.30. The increases in stress invariants are

a1 consta nt (lla l = 0) and then increase aJ (lla )

6.

Pl -

0 + 2611)

3

=

260"]

3

130

CHAPTER 3


The stress invariants at the end of loading "3" are P3

= P2 + q2 +

q3 =

D.P3

= 5D. Ul + ~D.U3

D.q 3 =

~Ul -

~U3

The stress path for loading "3 " is shown as BC in Fig. 3.30. The slope of BC is ilq3

-~U3

3

~P3

~~U3

2

F O /

"'4¥10 I

FIGURE 4.4 (al Soit sample at a depth z below ground surface. (b) Expected onedimensional response.

4.4 CALCULATION OF PRIMARY CONSOLIDATION SETTLEMENT

re e in vertical stress due to pie, is Au z ' (Recall that you 3.11.) The final vertical stress

FO

R

to construct a the building a can find Au z USI

153

OCR

=

1

(4.14)

4.4.3 Primary Consolidation Settlement of Overconsolidated Fine ~ Grained Soils If the soil is overconsolidated, we have to consider two cases depending on the magnitude of AO'z . We will approximate the curve in the log u~ versus e space as two straight lines, as shown in Fig. 4.5. In Case 1, the increase in Au z is such that O' ~ in = u~ o + AO'z is less than u ~c (Fig. 4.5a) . In this case, consolidation occurs along the URL and (4.15)

154

CHAPTER 4

ONE·DIMENSIONAL CONSOlIDATION SETTLEMENT OF FINE·GRAINED SOilS

,

,

'"

)

'0 C,

C,

URe

'"

0';.

O" bn

Ioi a;

0';,

"'";&

(4.16)

o- ; , ~

> 0-;
0

a grid .

Divide' ' he d7 ,'O five lay"" ti..z = - =lm 5 C~ = 8 x 10-· cm 2/s = 8 X 10- 8 m2/s == 2.52 m2 /y r

Assume 6.t

=

0.1 yr. = C~ 61 _ 0:

Step 2:

6 z2

2.52 x 0.1 _ 025 12

.

05

-,undary, is

Let us calculate the excess par node located at row 2, colu 2. U 2. I+l

= 7

.~_..~·~ ng

a spreadsheet program are ults are plotted in Fig. E4.7 .

FO

5

30.0

61 47 39

0.20

0.30

0.50

0.0 46.3 65.0 60.0 48.5 43.0

0.0 39.4 59 .1 58 .4 50 .0 45.8

0.0 34.5 54.0 56.5 51 .0 47 .9

Excess pore water pressure (kPa )

00

FIGURE E4.7

20

40

60

•

What's next . . .We have only described primary consolidation settlement. The other part of the total consolidation settlement is secondary compression, which will be discussed next.

4.6 SECONDARY COMPRESSION SETTLEMENT

SECONDARY COMPRESSION SETTLEMENT You will recall from our experiment in Section 4.3 that consolidation settlement consisted of two parts. The first part is primary consolidation, which occurs at early times. The second part is secondary compression, or creep, w ., takes place under a constant vertical effective stress. The physical reason for secondary compression in soils are not fully understood. One plausible planatiom is the expulsion of water from micropores; another is viscous fon of he soil structure. We can make a plot of void ratio versus the logan experimental data in Section 4.3, as shown in Fig. 4.11. Pri assumed to end at the intersection of the projectio of the the curve. The secondary compression index is (4.41)

(4.42)

F

4.6

171

Primary consolidat ion

Secondary Gompression

/

ep

e,

I

A

Slope = C" : I I

' - - - -- -- - - - - - L --

--L----+-Iog I

Ip

FIGURE 4 . 11

Secondary compression.

172

ONE-DIMENSIONAL CONSOLIDATION SETTLEMENT OF FINE-GRAINED SOILS

4.7 ONE-DIMENSIONAL CONSOLIDATION LABORATORY TEST 4.7.1 Oedometer Test The one-dimensional consolidation test, called the oedome

FO

2

CHAPTER 4

Displacement gauge

Porous sto ne (b ) Fixed ring ce ll

Displacement gauge

Ring

Soi l sample

Porou s stone (c) Fl oating ring ce ll

(a) A typical consolidation apparatus (Photo courtesy of Geotest.) (b) a fixed ring cell and (c) a floating ring cell.

FIGURE 4.12

4.7 ONE·DIMENSIONAL CONSOLIDATION LABORATORY TEST

173

consolidation load completely, a negative excess pore water pressure that equals the final consolidation pressure would develop. This negative excess pore water pressure can cause water to flow into the soil and increase the soil's water content. Consequently, the tinal void ratio calculated from the final water content would be erroneous. The data obtained from the one-dimensional consolidation est are as follows: 1. Initial height of the soil, Ho, which is fixed by the hei fl. t 0 2. Current height of the soil at various time intervals un settlement data) . 3. Water content at the beginning and at the en of the soil at the end of the test. You now have to use these data to deter We will start with finding Cu'

FO

4.7.2 Determination of the Coefficient of Consoli

-':,i----->-YT:

(ff) A1 1'-------'--

1.15( ff)A

FIGURE 4.13

Correction of laboratory early time response to determine Co.

CHAPTER 4

ONE-DIMENSIONAL CONSOLIDA liON SETTLEMENT OF FINE-GRAINED SOILS

for which Tv = 0.848 (Fig. 4.9). If point C were to lie on a straight line, the theoretical relationship between U and Tv would be U = 0.98~; that is, if you substitute Tv = 0.848, you get U = 90%. At early times, the theoretical relationship between U and Tv is given by Eq . (4.34); that is,

The laboratory early time response is represente 4. 13. You should note that 0 is below t e initi

tion is achieved

1. Plot 2.

at

i 'tial part of the curve intersecting and the abscissa (Vtime) at A.

it is

-VC:;.

FO

174

L -_ _~~_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _~

~

115V'i;;

FIGURE4.14

Roottime method to determine Cu '

4.7 ONE-DIMENSIONAL CONSOLIDATION LABORATORY TEST

175

6. The intersection of the line OB with the curve, point C, gives the displacement gauge reading and the time for 90% consolidation (t90)' You should note that the value read off the abscissa is ~. Now when U = 90%, Tu = 0.848 (Fig. 4.9) and from Eq. (4.31) we obtain

c u

=

a.848m, (90

where H dr is the length of the drainage path.

4.7.2.2 Log Time Method In the log time method, th lace ent gauge readings are plotted against the logarithm of time . typic I c e obtained is shown in Fig. 4.15. The theoretical early time s ttle ent ponse in a plot of logarithm of times versus displacement gau ding's a parabola (Section it a p arabola and a correc4.3) . The experimental early time curve is R'O tion is often required. The procedure, with reference dary

2.

FO

R

3.

~

ell el 2

~ ~

'"

""~

el 50

CD C

- log (3;

zo

FIGURE 4 . 18 Schmertmann's method to correct Cu for soil disturbances.


179

4.7.2.7 Determination of the Secondary Compression Index The secondary compression index, Ca , can be found by making a plot similar to Fig. 4.11. You should note that Fig. 4.11 is for a single load. The value of C", usually varies with the magnitude of the applied loads and other fac · such as the UR.

What's next . ..Three examples and their solutions are present how to find various consolidation soil parameters as discusse examples are intended to illustrate the determination of the com how to use them to make predictions. The third examg . lustr using the root time method. EXAMPLE 4.8

At a vertical stress of 200 kPa, the vo' an oedometer is 1.52 and lies on t e vertical stress of 150 kPa compresse

(3)

(4) an e versus log (J~ curve. Use

o n in Fig. E4.8.

F

-(1.43 - 1.52) log(350/200)

1. 52 1-------''1.' 1.4 5 t:=====t:::s~ 1.43

~-2..L. O O-3-"50 --50'-O-- log

FIGURE E4_8

0"; (kPa )

=

0.37

CHAPTER 4

Step 2:


Determine C. C,. is the slope of BC in Fig. E4.8. C,

Step 3:

- (1.43 - 1.45) log(3501200)

=

=

0.08

Determine the overconsolidation ratio. Preconsolidation effective stress:

Step 4: The void ratio at 500 kPa . consolidation line (Fig. E

• EXAMPLE 4.9

ertical strain. You know the in-

FO

180

6..£

6.. z

= -

z

Ho

0 .5 18

=-

= 0.Q28

te the modulus of volume recompressibility. m . v'

Step 3:

6..£, t;rr;

= -

0.028 =- = 1.9 x 10- 4 m2 IkN 150

Calculate the constrained elastic modulus. E~ = -

1

m vr

=

1 1.9 x 10-

4

= 5263 kPa

•

EXAMPLE 4.10

The following readings were tak en for an increment of vertical stress of 20 kPa in an oedometer test on a saturated clay sample, 75 mm in diameter and 20 mm thick. Drainage was permitted from the top and bottom boundaries.


Time (min)

0.25

t).H

0.12

0.23

2.25

4

9

0.33

0.43

0.59

16

068

25 0.74

36 0.76

181

24 hours

0.89

(mm)

Determine the coefficient of consolidation using the root time met

Strategy Plot the data in a graph of displacement rea follow the procedures in Section 4.7.2.1.

Solution 4.10 Step 1: Step 2:

FO

Step 3:

Make a plot of settlement (decreas shown in Fig. E4.10. Follow the procedures out From Fig. E4.10,

Calculate C

o .----,~~----~--~~--_.--__.

o

0.1

1-'~-+--;"~--I----+---+--1

0.2 f - -""''I.- - - ' - - - - - - I - - - - - + - - - + - - j

E .5

0.3 f ----I--¥.Nfm1

-"LO'WoL'-_ _ _ _ _ _ r,,\ = t 9.4 kNlm' lm F.ne·i!ai!1e4 SO\I 3m

[clay

~nd

... .. 62%

FIGURE E4. 13a

sill m..!ur..s)

F

190

CHAPTER 4


Strategy To calculate the primary consolidation settlement you need to know Cc and C, or m v , and (J~o , ~a-, and (J~C" Use the data given to find the values of these parameters. To find time for a given degree of consolidation, you need to find Cu from the data.

Solution 4.13 Step 1:

Find Cu using the root time method. Use the data from the 240 kPa load step t Vtime curve as depicted in Fig. E4.13b. ~ in section 4.7 to find Cu' From th

_ 0.2

~ 0.3 :;::; 0.4 c

E 0.5

-

.'!' 0.6

'iii

c/)

07 . 0.8 0.9 t+r+-t\t\~~H-t+lfl"'!"t"-;-:...~~~++-l....l.-j

=

59.3

X

10-6 m 2 /min

void ratio at the end of each load step. eo = wG. = 0.62 x 2.7 = 1.67

'tial v d ratio: Equation (4.6):

e

=

~z

e - - (1 - e) o Ho 0

=

~z

1.67 - - (1 20

= 1.67 - 13.35 x 10- 2

+ 1.67)

~z

The void ratio for each load step is shown in the table below.

a; (kPa) Void ratio

15 1.66

A plot of e - log

30 1.66

a- ~

60

120

1.64

1.52

240 1.38

480 1.25

versus e is shown in Fig. E4.13c.

4.12 SUMMARY

1.75

'"§

1.55

II

1.45

I

-0

g

' ,

Ii i II

1.65 o

I

U ,~

i\

I o-;c

Co = 0.45

",

1.35 1.25 10

191

1000

100

0-; (kPa)

FIGURE E4. 13c

Step 3:

Determine cr~e and Ce . Follow the procedures in Sectio

Step 4:

FO

R

Step 5:

Divide the clay layer into three sublayers of 1.0 m thick and compute the settlement for each sublayer. The primary consolidation settlement is the sum of the settlement of each sublayer. The vertical stress increase in the fine-grained soil layer is 90 - (

90 - 75) 3 Z

=

90 - 5z

where z is the depth below the top of the layer. Calculate the vertical stress increase at the center of each sublayer and then the settlement

192

CHAPTER 4


from the above equation. The table below summarizes the computation.

Layer

z

u ~o at center of

(m)

sublayer (kPa)

au z 87.5

0.5

48.7

2

1.5

54.9

82.5

3

2.5

61.1

77.5

(kPa)

Step 6:

t=

FO

26.3Tu (days)

~

80 90

0.2 0.8 1.9 3.3 5.2 7.6 10.6 14.9 22.3

20.8 61.3 81.6 102.1 122.5 142.9 163.3 183.7

0.567 0.848

Time (days)

5

E

5

c

Q)

E

50 100

Q)

:;:::

OJ 150

(/)

~

'"

200

FIGURE E4_13d

10

15

20

25

I

I

I

~ I

---

~

1

-

•

EXAMPLE 4.14 A geotechnical engineer made a preliminary settlement analysis for a foundation of an office building, which is to be constructed at a location where the soil strata contain a compressible clay layer. She calculated 50 mm of primary cons lidation settlement. The building will impose an average vertical stress of 15 a in the clay layer. As often happens in design practice, design changes are equired. n this case, the actual thickness of the clay is 30% more than the 0 ·gin soil pro Ie indicated and, during construction, the groundwater table ha be I by 2 m. Estimate the new primary consolidation settlement.

Strategy From Section 4.3, the primary consolidation se tl tional to the thickness of the soil layer and also t tl1 ·n [see Eq. (4.18)]. Use proportionality to find tile ne settlement.


Estimate the new primary con increase in thickness.

Step 2: ue to lowering of water table

FO

= 2 x 9.S = 19.6 kPa

•

IE XAMPLE 4.15 The foundations supporting two columns of a building are shown in Fig. E4.15. An extensive soil investigation was not carried out and it was assumed in the design of the footing that the clay layer has a uniform thickness of 1.2 m. Two years after construction, the building settled with a differential settlement of 10 mm. Walls of the building began to crack. The doors have not jammed but by measuring the out-of-vertical distance of the doors, it is estimated that they would become jammed if the differential settlement exceeded 24 mm. A subsequent soil investigation showed that the thickness of the clay layer was not uniform but

CHAPTER 4


780 kN

FIGURE E4.1S

varies as shown in Fig. E4.1S. The a expected total differential setJ;l:e~~·t.. become jammed.

FO

194

780 (LlO"z)B = (1.5 + 2.8)(1.5 + 2.8) = 42.2 kPa

Step 2:

Note: For a more accurate value of ~(J"z you should use the vertical stress increase due to surface loads on multilayered soils (Poulos and Davis, 1974). Calculate the primary consolidation settlement. Use Eq. (4.18), Ppc = Homu ~(J", to calculate the primary consolidation settlement. (PPC)A = 1.2 x 0.7 x 10- 3 X 30 = 25.2 X 10- 3 m = 25.2 mm (PpC)B = 2.8 x 0.7 x 10- 3 x 42.2 = 82.7 x 10- 3 m = 82.7 mm

EXERCISES

Slep 3:

195

Calculat e the differential settlement. Differential settlement: 0 - 82.7 - 25.2 """ 57.:5 mm

Slep 4:

Calculate the time for 24 mm differential settlement to occur.

Current differential settlement: 0, = 10 mm &~

10

8

57.5

Degree of consolidation: U - - = From Eg. (4.34).

4

4

•

•

=

0.17

T" '"' - Ui = - X 0.17< ., 0.037

. H_._ ' From Eg. (4.31): , - _T_

C.

" 7;::';:;~=

Therefore, in the next 10.25 ears, the total differential sell l e~ t would be 24 mm.

•

EXERCISES For all prob le

Theory cia soil of thie ness H is a e ~ rain on Ihe lOp boundary through a thin sand I er A vet . al stress of A waS'1fplied ~ the clay. The excess pore water pressure ~"'lIIIid~ ,i st~b uttOn was linear i fne'lo il laycl--~ a value of II, a{ the top boundary and u~ (Uh u,) ft'"tbe bottom bf u Q:dfl ry~ex.s:tss pore water pressure at the top boundary was n t zero because t ~na layer wa part ially blocked. Derive an equation for Ihe excess p re water pre ~ re disti but" n i h soil thickness and time. 4:1

A soil layer o~kness H o as only single drainage through the top boundary. The excess

pore water pressur istri tion when a vertical stress. CT, is applied va ries parabolically with a va lue of zero at the top boundary and uh al Ihe bOllom boundary. Show thai

, 4.3

Show that. for a linear elaslic soil. m~

4.4

=

(1

+ v')(l - 2v') v')

£'(J

Show thaI, If an overconsolidflted soil behaves like a linear elastic matcnal,

K':

=

(OCR)K~C

- - ' -' -, (OCR - 1) )

-

,

196

CHAPTER 4

4.5

ONE-DIMENSIONAL CONSOLIDATION SETILEMENT OF FINE-GRAINED SOILS

The excess pore water pressure distribution in a 10 m thick clay varies linearly from 100 kPa at the top to 10 kPa at the bottom of the layer when a vertical stress was applied. Assuming drainage only at the top of the clay layer, determine the excess pore water pressure in 1 year's time using the finite difference method if Cu = 1.5 m 2 /yr.

4.6

4.7 me that the present

Problem Solving 4.8

4.9 4.10

FO

4.11

4.12

test on a soli sample for an

to a sample

4

9

16

36

64

100

0.49

0.61

0.73

0.90

0.95

0.97

urs the settlement was negligible and the void ratio was 1.20, corresponding of 18.2 mm. Determine C using the root time and the log time methods.

A sample of saturated clay of height 20 mm and water content of 30% was tested in an oedometer. Loading and unloading of the sample were carried out. The thickness H f of the sample at the end of each stress increment/decrement is shown in the table below.

cr~ (kPa)

100

200

H,(mm)

20

19.31

400 18.62

200

100

18.68

18.75

(a) Plot the results as void ratio versus log (J"~. (b) Determine Cc and C. (c) Determine mv between

(J"~

= 200 kPa and

(J"~

= 300 kPa.

EXERCISES

4.13

197

A sample of saturated clay, taken from a depth of 5 m, was tested in a conventional oedometer. The table below gives the vertical stress and the corresponding thickness recorded during the test. cr~ (kPa)

h(mm)

100 19.2

200 19.0

400 17.0

800 14.8

1600 12.6

800

400

13.1

14.3

100. A

15~

The water content at the end of the test was 40% and the initial h (a) Plot the graph of void ratio versus log cr~. (b) Determine Cc and C. (c) Determine mu between cr~ = 400 kPa and cr~ = 500 (d) Determine the relationship between e (void rat~nd (e) Determine cr~c using Casagrande's method. 4.14

The following observations were recorded in an oe in diameter and 30 mm high. Load

(N)

Displacement gauge reading (mm )

o 1.4

(a) (b) (c) Calculate t

FO

(d)

An oil tank is to be Jted 0 a soft alluvial deposit of clay. Below the soft clay is a thick layer of stiff clay. It w decided that a circular embankment with sand drains inserted into the clay would be constructed to preconsolidate the soil. The height of the embankment is 6 m and the saturated unit weight of the soil comprising the embankment is 18 kN/m 3 . The following data are available: thickness of clay = 7 m, mu = 0.2 m 2 /MN, Cu = 3.5 m 2 /yr, Ch = 6.2 m 2 /yr, diameter of drain = 300 mm. The desired degree of consolidation is 90% in 6 months. Determine the spacing of a square grid of the sand drains such that when the tank is constructed the maximum primary consolidation should not exceed 20 mm.

Practical 4.18

Fig. P4.18 shows the soil profile at a site for a proposed office building. It is expected that the vertical stress at the top of the clay will increase by 150 kPa and at the bottom by 90

198

CHAPTER 4

ONE-DIMENSIONAL CONSOLIDATION SETILEMENT OF FINE-GRAINED SOILS

4.19

'/so, =

17_5 kN/m 3

FO

lidated clay,

Vertical stress (kPa) Void ratio

50 0_945

100 0.895

200 0_815

400 0.750

800 0.705

Calculate the primary consolidation settlement. Assuming that the primary consolidation took 5 years to achieve in the field, calculate the secondary compression for a period of 10 years beyond primary consolidation. The secondary compression index is Cj6. [Hint: Determine ep for your Urin from a plot of e versus log u~.l

CHAPTER

5

SHEAR STRENGTH OF SOILS ED 15

INTRODUCTION

FO

5.0

d determine the shear strength d be able to: • Determine the shear strength Qf soils • Understand the differences between drained and undrained shear strength • Determine the typ o f shear test that best simulates field conditions • Interpret laboratory and field test results to obtain shear strength parameters principles learned from previous chapters and other

of stresses and strains, and stress paths • Friction (statics and/or physics) Sample Practical Situation You are the geotechnical engineer in charge of a soil exploration program for a dam and housing project. You are expected to specify laboratory and field tests to determine the shear strength of the soil and to recommend soil strength parameters for the design of the dam. In Fig. 5.1 a house is shown in a precarious position because the shear strength of the soil within the slope near the house was exceeded. Would you like this to be your house? The content of this chapter will help you to understand the shear behavior of soils so that you can prevent catastrophes like that shown in Fig. 5.1.

199

200

CHAPTER 5

SHEAR STRENGTH OF SOilS

1

5.1 Shears"e~~~~!oll

lstance to applied shearing

forces .

FO

shear strength of soils.

5.2

the change in volume of a soil when it is distorted by

QUESTIONS TO GUIDE YOUR READING 1. What is meant by the shear strength of soils? 2. What factors affect the shear strength?

3. How is shear strength determined? 4. What are the assumptions in the Mohr-Coloumb failure criterion? 5. Do soils fail on a plane? 6. What are the differences between peak, critical, and residual effective friction angles?

5.3 TYPICAL RESPONSE OF SOILS TO SHEARING FORCES

201

7. What are peak shear strength, critical shear strength, and residual shear

8.

9. 10. 11. 12.

strength? Are there differences between the shear strengths of dense and loose sands or normally and overconsolidated clays? What are the differences between drained and undrained Under what conditions should the drained shear strengtl or the u drained shear strength parameters be used? What laboratory and field tests are used to dete mi What are the differences between the results of tests?

5.3 TYPICAL RESPONSE OF S...--__. TO SHEARING FORCES

FO

s 0 soil by applying simple shear ample, which we call Type I, rep-

z ~===~ Expansion . ................ I LIz

to (a)

Original soil sample

"-------.L-----x (b) Simple shear defor mat ion of

Type I soils

FIGURE 5.2

(c)

Simple shear deformat ion of Type II soils

Simple shear deformation of Type I and Type II soils.

202

CHAPTER 5

SHEAR STRENGTH OF SOILS

on soil behavior. Since in simple shear Ex"" £7 "" 0, the volumetric strain is eq ual to the ve rtical strain , E. = 6 zlH o, where az is the vertical displaccment (posi tive for compression) and H o is the initial sample height. The shea r st rain is the small 's t e horizontal angular distortion expressed as 'Y:.t = tu l Ho, where 6x ~ displacement . We are going to summarize the important features of the esponses of these two groups of soils when subjected to a constant vertica l (n rm 1 effective slress and increasing shea r strain. We will consider the shear,...s ess e rsu ~ e shear strain, the volumetric strain versus the shear strain, a the v id ralio versus the shear strain responses, as illustrated in Fig. 5.3. When a shear band(s) develops in som~s 0 ov TC9nsolidated clays, the particles become o rien ted parallel to the dlr~ 1 0 the..shear band, causi ng the fina l shear stress of these clays to decrease belo the-critical state shear stress. We will call this type of soil, Type II·A . and he fin I~ h ear stress attained the residual shear stress, 'T,. Type I soils at ve mal effective slress can also exhibi t a peak shear stress duriny)'ie.aring.

Type I

Type II SOII$

1,

FIGURE 5.3 Response of soils to shearing.

5.3 TYPICAL RESPONSE OF SOilS TO SHEARING FORCES

Type I soils-loose sands, normally con clays (OCR :s 2)-are observed to:

203

overconsolidated

until a

Type II soils-d observed to:

FO

• Sho

The critical state shear stress is reached for all soils when no further volume change occurs under continued shearing. We will use the term critical state to define the stress state reached by a soil when no further change in shear stress and volume occurs under continuous shearing.

5.3.1 Effects of Increasing the Normal Effective Stress So far, we have only used a single normal effective stress in our presentation of the responses of Type I and Type II soils. What is the effect of increasing the

204

CHAPTER 5 SHEAR STRENGTH Of'SOILS

normal effective stress? For Type I soils. the amount of compression a nd t he magnitude of th e critical state shear stress will increase (Figs. S.5a.b). For Type II soils. the pea k shear stress tends to disappear, the crit ical shear stress increases and the change in volume expansion decreases (Figs. S.Sa.b). If we were to pial the peak shear stress and the critical s ~h ear stress for each constant normal effective stress for Type I and II soil we would get: 1. An approxima te straight line (OA. Fig. S.5c) that shea r stress values of Type ' and Type 11 soils. tween OA a nd the a;, axis. the critical sta te fric ti will be cal1ed the failu re envelope because any sh a critica l sta te shear stress.

all the edt.J.,1 state e wi call th~ angle bea ngle, • ..,.The line OA al" tressti'hatlies on it is

ij

2. A cu rve (OBCA . Fig. SSc) that lin ~ pea hear-stress values for Type I I soils. We will call OBC (the cu~ ,p a)1..Qf ?B}?A). the peak shear stress envelope because any shea r stress thaYUes'o it is a peak s ~ear st ress, fo r 'Ii pe n soils is ap ~rs}S a poin t (point 9) located on OA (fig. 5. • r Type 11- soils, the r~ua l shear st resses would he on a lin OI>::beloVOA , We wil II the an~le ' betwee n 00 and the a~ axis. the...residuaJ iriCl n angle, ~; s th norm ' eUective slress increases. the cri 'fal void ratio decreases (Fig. 5,5dl.T hus,j)le critical void ra tio

At large normal effective resses, pel'l.k shea r sIr sup pre~d and only a criti cal sta te..; hear stress is observed an

u t of the norll'

is dependent on he rna,

eii~"r
Type II SOIls

• Type I 5O,ls

o

a. 'd

•

~

~

~ ~~;::~~~T:"'~';~" ~

I

lllCre.1$I"i nOflNI

effet'~ $InISS

l'"

T,-pe l150ils

I

notm~1

IIlClUSl"l effectIVe s[.esJ;

' - - - - - - - - :-

to,

'd)

'.

FIGURE $.5 Effects of i ncreasi ng norma l effective stresses on the response of

soils.

S.l TYPICAL RESPONSE OF SOILS TO SHEARING FORCES

205

5.3.2 Effects of Overconsolidation Ratio The init ial state of the soi l dictates the response of the soil to sheari ng forces. For exa mple, two overconsolida ted soi ls wi th different ove rconsolidation ra tios but the sa me mineralogica l composition would exhibi t different peak shear stresses and volume ex pansion as shown in Fig. 5.6. The highe ve onsolidated soil gives a higher pea k shea r strength and grea ter volume e ansion .

I.

2.

J. 4.

5.

6.

7.

The essential points are: Type I soils-Ioost sands and normally consolidlUed and lightly over· consolidated clays- strain harden to a critical Skltt slt",r stress and comp'YSS toward a cril/cal void ratio. Type II soils-dem'e sands and o.~rconsolidated days-naclt a peak shear stnss, strain soften to a critical stille sll«u stnss and txpmuJ 10ward a critical void ralio afler an i,dtial compression al low shear strains. The peak shear slress o/ Type II soils ;s suppressed and 'he volume txplJnsion dec«ases when Ihr normal effective stress is large. All soils 'YOM a criticaluate. irresptclive oj ,lte;r inilial Slate, at which continuous shearing occurs wUhoUl changes In sMar sinsS and J.·olum£. AI large strains, Ihe p articles 0/ some overconsolidaled days become orlentul parallel to the direction oj shear bands and the jinal shear stress attained is lower Ihan the cridcld st4fe sllear Slress. The critical slOIe shear stnss and Ihe critical void ralio tkptnd On the normal effective stress.. Higher nonnal effective strnsa mW/1 in Itigher critical -stale shear stlY!SJes and lower cril/call'oid ,.tios. Higller overconsolidation ratios nsull;n higher peak shear slrrsses and grealer l'oillme up4nsion.

FIGURE 5.8

Effects of OCR on peak strength and volume exp ansion.

206

CHAPTER 5


T Co

~L--_ _ _ _ _- +

FIGURE 5.7

Peak shear stress envelope for cemented s

5.3.3 Cemented Soils

FO

What's next . . .You shaul

(5.1)

H= f-lW

t of static friction between the block and the table and

(5.2)

w

,--T

/Sli P plane

H ---+

N

;\ R (a)

FIGURE 5.8

(b)

(a) Slip of a wooden block. (b) A slip plane in a soil mass.

FO

5.4 SIMPLE MODEL FOR THE SHEAR STRENGTH OF SOILS USING COULOMB'S LAW

207

where T f (= TIA, where T is the shear force at impending slip and A is the area of the plane parallel to T) is the shear stress when slip is initiated, (u:')f is the normal effective stress on the plane on which slip is initiated . The subscript f denotes failure, which according to Coulomb's law occurs when rigid body movement of one body relative to another is initiated. Failure doe ot necessarily mean collapse but the initiation of movement of one body relative to another.

that is, ' ansi on. We ar goin to use our knowledge of statics to investigate impending sliding of particles up or down a plane to assist us in interpreting the shearing behavior of soiJs using Coulomb's frictional law. The shearing of the loose array can be idealized by analogy with the sliding of our wooden block on the horizontal plane. At failure (impending motion),

(5.3)

ROW 20 _~_0

Row 1

(0)

FIGURE 5.9

loose

~ Row2 ~R owl (b)

Dense

Packing of disks representing loose and dense sand.

208

CHAPTER 5


w

z

,·,LL ,..

(,, ) S'reues on Iillhlle pl,ne

X

e/iol Simulated sheanng 01 8 dense . .

FIGURE 11.10 Simulation of failure in dense sand

Consider two particles A and B in tj1e d ~n se a e bly and draw the freebody diagram of Ihe stresses at the slidi st o tact 1ktween A and as depicted in Fig. S.lO. We now appeal to our wooden ock r an analogy describe the shearing behavior o f the dense array. Fa the d; nse array, II e woe<Jen block is placed on a plane oriented at a "gle a t~ he horizontal ( Fig .~.lOb). Our goal is to find the horizontal force to inii1ate movemenl of lhe block up e incli ne. You may have solved this f fob lem in .sties. Anywa we are gain to solve it where N is the normal t ree. Using the again. At impending mario, T. force eq ui librium eq \jons the I. and Z directib s, w e t 'LF. ... . H - Nsin ~ . 0 I.F "'0: Ncos ~. a-W - O

Solving Co

and w,-we obtain

e

(5.5)

, cos a)

(5.6)

W - N(cos a-- f.I. sin Q)

(5.7)

ding Eq. (S.6) by Eq.. .-(S ) n~ sf H W

(5.4)

+ an 1

).I.

plifying, we obtain =- tao 4l' + tan Q J Ian 4>' Ian 0.

Q

tan 0

l~emblY' we can replace H by TJ and 130$'+ 13001 (O'~)J 1 A..'

tan,+, tan

0.

::

, (O'~)J

,

tan(¢, + 0.)

W by

«(f~)J' (5.8)

Let us investigate the impl ications of Eq . (5.8). If a = 0, Eq . (S.8) reduces to Coulomb's frictional eq uation (S.2). If Cl increases, the shear strcngth , "), gets larger. For instance, assume 4l' = 30" and (a~)J is constant; th en for (II ". 0 we gel TJ = O.S8(a~ )f' but i f (II - 10" we get Tf = O.84(O'~)f' that is, an increase of 4S% in shear st rength for a 10% increase in Q. If the normal effective stress increases on our dense disk asse mbly, the amou nt of " riding up" of the disks wiU decrease. In fact. we can impose a sufficien tly high normal effective stress to suppress Ihe "riding up" tendencies of the de nse disk assembly. Therefore, the abilily of the dense disk assembly to e xpand depends on the magnitude of the normal effective stress. The lower the normal effective stress, Ihe grea ter the value of (II. The ne t effect o f a due 10 normal effective st ress increases is th at the fai lure envelope becomes curved as illustra ted

5. 4 SIMPLE MODEL FOR THE SHEAR STRENGTH OF SOILS USING COULOMB'S LAW

209

Curved CoulQrnl) lall ure envelope caused by dllal lon A

o

G·

FIGURE 5 . 11

~

Effects of dilation on Cou"lomb's failu reaflve lope

others occu r (5.9)

where the positive sign refe to soils in wfiic -rhe net ovemen t of the particles is initialed up \ plane and the " n lti e--sign e rs to net particle movement lane. down I W dl caJUhe angle 0: the I ation ang e. It is a measure of Ihe change in volumetri trair7"wilh respecl lo th change In shear strain. Soils Ihal have pos-i.tive value f a expand during shearing whi le soils with nega tive values of ~ ~ contract uring shearfng. I hr's circle of st rain (Fig. 5.12). the dilation ngleJ (5.10)

fiGURE

5.12 Mohr's circle of strain and angle of dilation.

210

CHAPTER 5 SHEAR STRENGTH OF SOilS

where L'l denotes change. The negative sign is used because we want Ct to be positive when the soil is expanding. You should recall that compression is taken as positive in soil mechanics. The angle Ct is also the tangent to the curve in a plot of volumetric strain versus shear strain as illustrated for simple shear in Fig. 5.3b. ~ If a soil mass is constrained in the lateral directions, th~ilation angle is represented as (5 .11)

Dilation is not a peculiarity of soils but ocetu:S in example, rice and wheat. The ancient trade s 0 er phenomenon of volume expansion of grai it was Osborne Reynolds (1885) who described the phenomena brought it to the attention of the scientific community. For cemented soils, Coulo (5.12)

where

FO

o

Co

is called cohesio

The essential points are' 1. Shear failu fe of soils can be modeled using Coulomb'sfrictionallaw, Tf = (u~)r tlm (cf> , ± It) where Tr is the shear stress when slip is initiated;, (u~)r is the normal effective stress on thlJ slip plane, cf>' is the friction angle and it is the dilation angle. 2. The effect of dilation is to increll$e the shear strength of the soil and cause the Coulomb'sfoilure envelope to be curved. 3. Large normal effecti e strt s$es tend to suppress dilation. 4. At the critical slatt:, the dilati(m angle is zero. S. For cemented soils, CouloJ1(b'sjrictionallaw is Tr = Co + (u~), tan(cf>' + a) where Co is called cohesion.

hat's next . . .In the next-section, we will define and describe various parameters a interpret the shear strength of soils. It is an important section, which you should read carefully, because it is an important juncture in our understanding of shear strength of soils for soil stability analyses and design considerations.

1

5.5 INTERPRETATION OF THE SHEAR STRENGTH OF SOILS The shear strength of a soil is its resistance to shearing stresses. In this book, we will interpret the shear strength of soils based on their capacity to dilate. Dense sands and overconsolidated clays (OCR > 2) tend to show peak shear stresses

5.S INTERPRETATION OF THE SHEAR STRENGTH OF SOilS

Z 11

and expand (positive dilat ion angle), while loose sands and normally consolid ated and lightly overconso lidated clays do not show peak shea r stresses except at very low normal effective stresses and tend to compress (negative dila tion angle). In our interpretation of shea r st rength, we will describe soils as dilating soils when they exhibit peak shear stresses at 0: > 0 and nondilating soi ~ n they exhibit no peak shear stress and attain a maximum shear stress O. However. a nondilating soil does not mean that it does not change volume (expand or contract) during shearing. The terms di lating and nondilating 0 Iy refer t particular stress stales (peak and critical) during soil deformatIon, We will refer to key soil shear strength pa eters. using the follow ing notation. The peak shear strength, Tp. is the peak he strf$Ss attained by a dilating soil (Fig. 5.3). The dilation angle a eak,..sh r strc s will be denoted as O:p. The shea r stress attained by all soils lar~ he r slla'ins (Yu > 10%), when the dilation angle is zero, is the criti I Slate sh r strength denoted by""" The void ratio corresponding to the critil\! ~ai'e sh . r strength' the critical void ratio denoted by eC$' The effccl;l'e...frictlon angle correspon 'ng to the critical state shear strength and critictvora ratio is $;,. The peak effeclive fric ('0 angle for { dila ting soi I S,_ .....

ara. =:

(5.13)

Tesl results ~' 19

J

$; - ...

.

(5,14)

W e will continuu J se Eq. (5.1.3 fb~:1 " rae icc you can make (Eq. ( -4)):ugg, sted by Bolto ~986) .

the adj ustment

T1 ical 'talues o f $~, ~; and if:l.' for

ils are shown in T able 5.1. effecti e~~scribi ng friction angle and accept it by default s c:h tha t effej {lVe critical state friction angle becomes critical state friction angle, $;'. and "Cf cit e pe'ak friction angle becomes peak fric tion angle, We will drop the ter

;,.

=

Solution 5 . 1 Step 1:

Find 4>;" From Eq. (5 .20) ,

!

sin ' = ('1;), - (O'i), _ 300 - 100 = ~ = eo (O'il, + (O'~)! 300 + 100 4 2 ~~

Step Z: Find

- 30'

e.

From Eq. (5.23),

8 - 45.....

~.

~=45 ..

2

30"

+--6O' 2

•

CHAPTER 5


EXAMPLE 5 .2

Figure E5.2 shows the soil profile at a site for a proposed building. Determine the increase in vertical effective stress at which a soil element at a depth of 3 m, under the center of the building, will fail if the increase in lateral effe tive stress is 40% of the increase in vertical effective stress. The coefficient lateral earth pressure at rest, Ko, is 0.5. Strategy You are given a uniform deposit of sand an data given to find the initial stresses and then use the to solve the problem. Since the soil element is under th c axisymmetric conditions prevail. Also, you are that fore, all you need to do is to find LlO'~.

er 0'3

b equation he building, O.4LlO'~. There-

0


Th(

The subscript

Step 2:

0

de ote

~I;; (a~)

pres.",e is =

Find LlO'l' t failure:

vertical effective stress to bring the soil to

FO

216

34.4 + Llai 17.2 + O.4Lla;

-----'-- = 3

Ground surface

'I'

--

r

FIGURE ES.2

•

5.1 UNDRAINED AND DRAINED SHEAR STRENGTH

217

What's next . . .In the next section, we will consider two rather extreme condition sdrained and undrained conditions-under which soil is loaded and the effects these load ing conditions have on the shear strength . Drained and undrained conditions are the bounds to evaluate soi l sta bi lity.

5.7 UNDRAINED AND DRAINED SHEAR STRENGTH stress paths in C hapter 3. Drained cond itio curs e excess pore water pre ssu re developed du ring loading of ~il slpat I.e., nil = O. Undrained condition occurs when th e excess pqre wa ter p e~ u re cannot dra in, at least quick ly, from the soil; that is, nu *- 0. e t ,,'Thte"n of either cW1dilion- drained or undrained-depends on the soil tYR:' ~olog1cal fo rmation ~ ssures, sa nd layers in clays, etc.), and the ratrof load{ng. The ratc of load ing unoel' the un~ rained condition'is o n uch fa ster than s re wate r pressure and the l ume change the rate of d issipation of the ex ~ e result a is suppressio is a change in tendency of the soil is upP'resseCl . excess pore water press reXfutio shearing. .xs0~1 ith a te deney 10 compress during drained I j;ling ~i e~ibil an incre~~~n e ess pore waler press ure (positive excess pore wa~# pressu re, Fig.-S.14J u~der . ndrained condit ion resulling in a d rease in ~ff ctive stress. Aso~hat expan ds during drained loading will exhibit a c ease n excess p~te r pressure (nega tive excess pore wa ter press ~ , Eig. 5.14) under undraint d conc!.i Ion ulting in an increase in effective stress~ ~~§-e.. h ges in excess P\'~e water ressure occur because the void ratio does not C'h ange during undrained ~ hat is, the volume of the soil rema ins ~ constant. Dunng t he life a ge ~c qnical structure, ca lled the long-term condition, the ex
~n-~1

I~r---~--~~----'-~-r~~

: Ol$placement

000

~

~ -0.30 li ~ -0.40

:>

-0.50 -060

,

:--..,

...0,\0

l...o20

i

0

I.,

•

I

6

,

, ,,

8

10

12

iO

12

(m ml

8

=- F ,-,' ......0. -1'\

'

r-

--

- 0.70

FIGURE ES.S

I

I

'"

10 -

mlal force and

; oo d- 1O-~ m1

5 .8 LABORATORY TESTS TO OETERMINE SHEAR STRENGTH PARAMET"ERS

Normal stress:

«'4)

176.4

'{?

300

""

~ 200

•

~"

~150

O ~OC---1r e.~~r---~~--'8----'IOO----71·' """'-

€l (%)

~. 7b

Extract 'rp and "l" p

Step 3:

= (0"; -

2

Te •.

O"~)p = 250 = 125 kPa 2

. ~

=:

(0"; - (13)"" _ 175.8 = 879 kP

2

2

.

a

Determ ine £ ' and £ 3. The initial slope of Fig. ES.7b gives E' and the slope of the line fro m {he origin to 2Tp gives £;. •

54

£ = 0.002 - 27.000 kPa

E; =

250 0.035 .. 7143 kPa

FO

234

CHAPTER 5

Step 4:


Determine ~s' The deviatoric stress and the volumetric change appear to be constant from about £1 = 10%. We can use the result at £1 = 11 % to determine ~ . (a3)cs = 100 kPa , (aDcs = 175.8 + 100 = 5.8 kPa. ,j.. '

,+, cs

=

.

SIn

- I

(45"

Loose Medium Dense Very dense

"These values correspond to

q,~.

5.11 FIELD TESTS

TABLE 5.4 Correlation of and Su for Saturated FineGrained Soils Nso

Su

N60

(kPa)

< 10 10-25 25-50 50-100 100-200 > 200

Very soft Soft Medium Stiff Very stiff Extremely stiff

FO

R

0-2 3-5 6-9 10-15 15-30 > 30

Description

249

(5.52)

where N k is a cone factor that depends on the geometry of the cone and the rate of penetration. Average values of N k as a function of plasticity index can be estimated from

I .

N = 19 k

I - 10 - p- _ .

5'

Ip > 10

(5.54)

250

CHAPTER 5

SHEAR STRENGTH Of SOILS

Cooe

'e~'5t ance

q, (MPa)

"",,'

$o.t de5l;.iption

0

2

0

Pore reS i$tMce u (MPal

310 0 25 0.50 0.75 100

Fill

-'-COnnecting

,,'"

5

10

Reda.med

~nd

uPP'" mallne ,'~ (S, . 42)

15

Cone

pte5sure measurement

i"a

20

< / \

A

i

dO',

I

v'; ,'" \

\

__

.... \,...\...............

\

\

I I

I

>

/ '\ \ /

_

II

I' I

I

\

4S' +~'12

",(

_ /

\

\

\

~

/-

II' II'

V \

Pressure cell

I

\ d O'o//-

II

Gu ard cell

'-',

t

as Ie zone

~ Ex~::tn~

-

~_____

/

EI

_----

Gas pressure to / ' , /"\ In f late guard ce ll s Dlrect lonsof prin cipal stresses;' \

,

251

;' " \

I I' \

'''''''-l I

I' \

-

I

I

v,

I ,

Guard ce ll

I

I "

-.../

I

(a) Vertical secti on

FO

R

FIGURE 5.32

Cutaway view

-

FIGURE 5.33

Wroth,1972.)

~-'---

Auger

Schematic of Cambridge Camkometer. (Redrawn from Hughes and

252

CHAPTER S SHEAR STRENGTH OF SOILS

An excellent source on the inlerpretation of the pressure meter test is Wroth (1984). The essential points art

I . Variousfield leslS are used 10 deltrmine soil sIrength parameters. 2, The most simplefil!ld Itst, and Ihe mOSI popular, j~' lh e standard penetration ust (SPT).

What's next . ..Severa l em p irical relat io nships hav b eelL!" opqse d to obtain soil strength parameters from laboratory tests, for example, h e- er6erg limits, or from statistical analyses of field and laboratory test esults. 5 .Q:.1 ~ of t hese re lationships are presented in t he next secti on. T 6.12 EMPIRICAL RELATIONS,,{,S FOR SHEAR _ ... , STRENGTH PARAMETERS ) some-suggested empirical rei X r the in Table 5.5. Thes. elationr'ips stlould only nary design cal ations. , . TABLE

~.5

Emp ~Soil

Strengt

Shea~[ength ~f so~ are shown ed as

gu'Ke and in prelimi-

Relations hips Referenee

Normally consolidated fays Overconsotidated clays

~ "

: 0. 11 + 0.0037 1"

•

Is

~I ...

.. {OCRlo,

·...u,

(sJa..:

Skemplon 11957) Ladd 61 al. (1977) Jamioll::owski et 81. (1985)

A ll clays

~ O.22 Wo. + 3D,(10 - In pil - 3, where pi is the mean effective stress at fa ilure !in kPa) and D, is rela tive density, This equation should only be used if 12 > (cI>~ - 4>~.) -:> O.

<j)~ .. <j)~.

Mesri 119751 Bolton (1986)

The strength of soi ls is in terpreted usi ng Cou lomb's frictio nal law. All soi ls regardless of their in itial state of stress will reach a critical state characterized by conti nuous shearing al constant shear stress and constant volume. The ini tial void ratio of a soil and the norma l effective stresses determine whet her the soil will dila te or not. Dilating soi ls often exhibit (1) a peak shear stress and then stra in soften to a constant shear stress, and (2) ini tia l contraction followed by expansion

5.13 SUMMARY

253

towa rd a critical void ratio. Nondilating soils (1) show a gradual increase of shea r stress, ultimately reaching a constant shea r stress, a nd (2) contract toward a critica l void ratio. The shear st rength parameters are the fri ction angles (

Soil Mechanics and Foundations

Mechanics Soil and Foundations

Soil Mechanics and Foundations

An introduction to soil mechanics and foundations