Download - Soil Mechanics II

Addis Ababa University Faculty of Technology Department of Civil Engineering

1 Lecture Note (Instructor: Amsalu Gashaye) Academic Year: 2001E.C

CHAPTER ONE

STRESS DISTRIBUTION IN SOIL 1.1 Introduction

Stress in soil is caused by one or both of the following:

a) Self weight of soil b) External loading (unloading) = structural loading

The vertical stress in soil owing to self weight is given by: 'v = z,

where 'v = the vertical stress in the soil at depth z below the surface = unit weight of soil

Foundations are designed with an adequate factor of safety against shear failure of soil; thus it is safe to presume that the operating stresses in soil are small enough to assume stress-strain proportionality. Fortunately, the order of magnitudes of stress transmitted in to soil from structural loading are also small and hence the application of elastic theory for the determination of stress distribution in soil gives reasonably valid results. Hence stresses due to applied structural loadings are determined by using elastic theory. The resultant stress can be obtained by superposition of structural load stress and overburden pressure. 1.2 Stress Distribution for a Point Load Bossinesq (1885) has given the solution for the stresses caused by the application of a point load at the surface of a homogenous, elastic, isotropic and semi-infinite medium.

zy

A (x,y,z)

Q

Yz

X

zx

zyxz

xyx

yzyx

Z

r

Q

R

z

y

xr

R

rz

r

Fig. 1.1 Point Load on the surface

With reference to Fig. 1.1, the expressions for the increase in stress at point A due to a point load Q at its surface are

ratiospoissonvRz

Rr

zyxzrR

yxr

'

cos;sin

22222

22

===

++=+=+=



In rectangular coordinates

)6.1(23

)5.1(23

)4.1()(

)2(321

23

)3.1()(

)2()(

1321

23

)2.1()(

)2()(

1321

23

)1.1(23

5

2

5

2

235

323

2

5

2

323

2

5

2

5

3

RyzQRxzQ

zRRxyzRv

RxyzQ

Rz

zRRyzR

zRRv

RzyQ

Rz

zRRxzR

zRRv

RzxQ

RzQ

yz

xz

xy

y

x

z

=

=

+

+=

+++

+=

+++

+=

=

In cylindrical coordinates

)10.1(23

)9.1()(

1)21(2

)8.1()(

2132

)7.1(23

5

2

3

5

2

5

3

RrzQ

Rz

zRRvQ

zRRv

RzrQ

RzQ

rz

r

z

=

+=

+

=

=

In most foundation problems it is very necessary to be acquainted with the increase in vertical stresses (for settlements) and the increase in shear stresses (for shear strength analysis). The equation for the vertical stress can be rewritten as follows

25

2

2

25

222522

3

5

3

)(11

23

)11.1(infdim

)(11

23

)(23

23

+=

==

+=+==

zrI

factorluenceensionalnonIwherezQI

zrzQ

zrzQ

RzQ

z

zz

z

Values of Iz for different values of r/z ratio can be tabulated (Table 1.1) or plotted as Iz versus r/z ratio (Fig 1.2) and hence can be used for routine stress calculation. Note that the maximum vertical stress is observed directly below the load (r=0).



Table 1.1r/z I z0 0.4775

0.2 0.43290.4 0.32950.6 0.22140.8 0.13861.0 0.08441.2 0.05131.4 0.03171.6 0.02001.8 0.01292.0 0.00852.2 0.00582.4 0.0042.6 0.00282.8 0.00213.0 0.0015 Fig 1.2 Non dimensional influence factor

00.050.1

0.150.2

0.250.3

0.350.4

0.450.5

0 0.5 1 1.5 2 2.5 3 3.5r/z

Iz

The equation for the shear stress can be rewritten as follows

)11.1(

23

2

5

2

abovegivenfactorInfluenceIwherezQ

zrI

zr

RrzQ

zzrz

zrz

==

==

Pressure distributions Graphical vertical stress distribution on a horizontal plane at any depth z below the ground surface can be drawn as shown here under. The vertical stress on a horizontal plane at a depth z is given by

depthspecifiedabeingzz

QzIz 2

=

For several assumed values of r, r/z is calculated and Iz is found for each, the value of z is then computed. As an example consider the following cases Case 1: Let z =c (where c= constant number) and r be varied as 0, 0.25c, 0.5c, 0.75c, c, 1.25c, 1.5c, 1.75c, 2c, etc

Case 2: Let z=2c and r be varied as case 1 Case 3: Let z=4c and r be varied as case 1

Now the stresses are calculated (tabulated) and plotted together below.



r r/z I z (zc2)/Q r r/z I z (zc2)/Q r r/z I z (zc2)/Q0 0 0.4775 0.4775 0 0 0.4775 0.1194 0 0 0.4775 0.0298

0.25c 0.25 0.4103 0.4103 0.25c 0.125 0.4593 0.1148 0.25c 0.0625 0.4728 0.02960.5c 0.50 0.2733 0.2733 0.5c 0.250 0.4103 0.1026 0.5c 0.125 0.4593 0.0287

0.75c 0.75 0.1565 0.1565 0.75c 0.375 0.3436 0.0859 0.75c 0.1875 0.4380 0.0274c 1.00 0.0844 0.0844 c 0.500 0.2733 0.0683 c 0.25 0.4103 0.0256

1.25c 1.25 0.0454 0.0454 1.25c 0.625 0.2094 0.0523 1.25c 0.3125 0.3782 0.02361.5c 1.50 0.0251 0.0251 1.5c 0.750 0.1565 0.0391 1.5c 0.375 0.3436 0.0215

1.75c 1.75 0.0144 0.0144 1.75c 0.875 0.1153 0.0288 1.75c 0.4375 0.3082 0.01932c 2.00 0.0085 0.0085 2c 1.000 0.0844 0.0211 2c 0.5 0.2733 0.0171

Case 1 (z=c) Case 2 (z=2c) Case 3 (z=4c)

0.000.050.100.150.200.250.300.350.400.450.50

-3 -2 -1 0 1 2 3r/c

zc2 /Q

z=cz=2cz=4c

From the above curves one can see that the stress diminishes as we move down from the ground surface and also as we move away from the point of load application. Similarly vertical stress distribution on a vertical plane at any radial distance r from the load can be drawn as shown here under. For several assumed values of z, r/z is calculated and Iz is found for each, the value of z is then computed. As an example consider the following cases Case 1: Let r=a (where a= constant number) and z be varied as 0, 0.5a, a, 2a,

5a, 10a, etc Case 2: Let r=2a and z be varied as case 1 Case 3: Let r=4a and z be varied as case 1

Now the stresses are calculated and plotted below.

z r/z I z (za2)/Q z r/z I z (za2)/Q z r/z I z (za2)/Q0 - - indet. 0 - - indet. 0 - - indet.

0.5a 2.00 0.0085 0.0342 0.5a 4.0 0.0004 0.0016 0.5a 8 0.0000 0.0001a 1.00 0.0844 0.0844 a 2.0 0.0085 0.0085 a 4 0.0004 0.0004

2a 0.50 0.2733 0.0683 2a 1.0 0.0844 0.0211 2a 2 0.0085 0.00215a 0.20 0.4329 0.0173 5a 0.4 0.3295 0.0132 5a 0.8 0.1386 0.005510a 0.10 0.4657 0.0047 10a 0.2 0.4329 0.0043 10a 0.4 0.3295 0.0033

Case 1 (r=a) Case 2 (r=2a) Case 3 (r=4a)

0

2

4

6

8

10

12

0.00 0.02 0.04 0.06 0.08 0.10

z/a

za2/Q

r=ar=2ar=4a



Stress isobar or Pressure Bulb An isobar is a stress contour or a line which connects all points below the ground surface at which the vertical pressure is the same. In fact, an isobar is a spatial curved surface and resembles a bulb in shape; this is because the vertical pressure at all points in a horizontal plane at equal radial distances from the load is the same. Thus, the stress isobar is also called the bulb of pressure or simply the pressure bulb. The vertical pressure at each point on the pressure bulb is the same. An isobar diagram, consisting of a system of isobars appears as shown in Fig 1.3.

Z

Isobars

Q

Fig 1.3 Isobar diagram

The procedure for plotting an isobar is as follows. Let it be required to plot an isobar for which z=0.2Q per unit area (20% isobar) From 22.0

22.0

2

2z

Q

zQzI

Q

zz

zIz

QzIz ====

Assuming various values for z, the corresponding Iz- values are computed; for these values of Iz, the corresponding r/z-values are obtained; and, for the assumed values of z, r-values are got. It is obvious that, for the same value of r on any side of the z-axis, or line of action of the point load, the value of z is the same; hence the isobar is symmetrical with respect to this axis. When r=0, Iz=0.4775; the isobar crosses the line of action of the load at a depth of: unitszIzzIzzzI 545.12.0/4775.02.0/2.0/2.0

22 ===== The calculations are best performed in the form of a table as given below. The plot is shown too.

z Iz r/z r (unit) z0.5 0.0500 1.211 0.605 0.2Q1.0 0.2000 0.645 0.645 0.2Q1.5 0.4500 0.155 0.232 0.2Q

1.5452 0.4775 0.0 0.000 0.2Q

0

0.4

0.8

1.2

1.6

2

-1.0 -0.5 0.0 0.5 1.0r

z

Westergaard (1938) has obtained an elastic solution for stress distribution in soil under a point load. He has assumed the soil to be laterally reinforced by numerous, closely spaced horizontal sheets of negligible thickness but of infinite rigidity, which prevent it from



undergoing lateral strain. The vertical stress z caused by a point load, as obtained by Westergaard, is given by (all notations as above): HL\CH1.doc

)12.1(232

2221

2221

21

2

+

=

zr

vv

vv

z

Qz

1.3 Stress Due to a Line Load a) Infinite Length: The increases in stresses at point A (Fig 1.4) due to an infinite line load of q/unit length are:

x

Ax

z

q/unit length

Z Fig 1.4: Line load of infinite length b) Finite Length: The increases in stresses at point A (Fig 1.4-a) due to finite line load of q/unit length are:

Fig 1.4a: Line load of finite length 1.4 Stress Due to Uniform load on an Infinite Strip Let a uniform load of intensity q/unit area be acting on a strip of infinite length and a constant width B (=2b) as shown in Fig 1.5. The increase in stress at point A due to the strip load q are given with reference to Fig 1.5 as

x

zA

B=2b

z

q/unit area

x

Fig 1.5: Uniform load The vertical stresses at different depths below the center of a uniform load of intensity q and width B are as follows:

Depth z 0.1B 0.2B 0.5B B 2B 5B 10B z 0.997q 0.977q 0.818q 0.550q 0.306q 0.126q 0.064q

[ ][ ]

)18.1()2sin(sin

)17.1()2cos(sin

)16.1()2cos(sin

+=

+=

++=

qxz

qx

qz

)15.1(2)22(

22

)14.1(2)22(

22

)13.1(2)22(

32

xzxzq

xz

xzzxq

x

xzzq

z

+=

+=

+=

(1.13d)3

12m2n

n

12m2n

3n2

12m 2

1I where,

(1.13c)I z

qz or

(1.13b)3

12m2n

n

12m2n

3n2

12m z 2

qz

becomes(1.13a)equationy/znandx/zmngsubstitutiand gRearrangin

(1.13a)

3

2z2y2x

y31

2z2y2x

y2)2x22(

3z3qz

++

++

+=

=

++

++

+=

==

++

+++=

q/unit length

x

B

o

z

xzA

y

xyy

-z

(z2



A few typical pressure bulbs for this case of strip loading are shown in Fig 1.6.

q/unit area

/q=1/8 /q=1/2

/q=1/4

z

z

z

B=2b

Fig 1.6: Pressure bubs for strip load of infinite length

1.5 Stress Due to Triangular load on an Infinite Strip When the applied load increases linearly across the width of the strip to give a triangular distribution as in Fig 1.7, the increases in stress at point A are given by

q/unit area

A

z

B=2b

x

xz

R1 R2

Fig 1.7: Triangular load Results of section 1.4 and 1.5 above can be superimposed in order to estimate the stress changes that result from the construction of embankments or the formation of cuttings in a soil mass. 1.6 Stress Due to Circular loaded Area Let a uniform load of intensity q/unit area be acting on a circular area of radius R. The increase in vertical stress at a depth z below the center of a flexible circular area (Fig 1.8 (a)) is given by

q/unit area2R

R

A

z

q/unit area

2R

A

z R

q/unit area

r

(a) (b)

z z

Fig 1.8: Circular load However, for points other than those under the center, the solutions have an extremely complex form and are generally presented in the form of charts or tables. With reference to point A in Fig 1.8 (b), the increase in total vertical stress is given as

)23.1( qIz =

)21.1(22cos12

)20.1(2sin21

22

21ln

)19.1(2sin21

+=

+=

=

Bzq

xz

R

R

Bz

Bxq

x

Bxq

z

)22.1(2/3

2)/(1

11

+=

zRqz



where the influence factor I depends on R, z and r. Values of I as a function of the parameters z/R and r/R can be obtained from Fig 1.9. or Table 1.2

Fig 1.9: Values of influence factor I.

Table 1.2: Influence coefficients I for vertical stress due to uniform load on a circular area r/R z/R

0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0.5 0.911 0.840 0.418 0.060 0.010 0.003 0.000 0.000 0.000 1.0 0.646 0.560 0.335 0.125 0.043 0.016 0.007 0.003 0.000 1.5 0.424 0.374 0.256 0.137 0.064 0.029 0.013 0.007 0.002 2.0 0.284 0.258 0.194 0.127 0.073 0.041 0.022 0.012 0.006 2.5 0.200 0.186 0.150 0.109 0.073 0.044 0.028 0.017 0.011 3.0 0.146 0.137 0.117 0.091 0.066 0.045 0.031 0.022 0.015 4.0 0.087 0.083 0.076 0.061 0.052 0.041 0.031 0.024 0.018 5.0 0.057 0.056 0.052 0.045 0.039 0.033 0.027 0.022 0.018 10.0 0.015 0.014 0.014 0.013 0.013 0.013 0.012 0.012 0.011

Pressure bulbs or isobar patterns for vertical stress is shown in Fig 1.10

q/unit areaB=2R

0

R

2R

3R

4R

0.9q0.8q0.7q0.6q0.5q0.4q0.3q0.2q

0.15q

0.1q0.05q

Fig 1.10: Pressure bubs for uniform circular load



1.7 Stress Due to Uniformly Loaded Rectangular Area The more common shape of a loaded area in foundation engineering practice is a rectangle, especially in the case of buildings. Based on Bossinesqs theory, Newmark (1935) has given an expression for the vertical stress at a point below the corner of a uniformly loaded rectangular area (Fig 1.11) as

q/unit

area

m=B/zn=L/z

Bz

A

L

Fig 1.11: Vertical stress at the corner of a uniformly loaded rectangular area

)24.1(22122)1222(1tan

)122(

)222(22122)1222(

4

++

+++++++

+++++=

nmnm

nmmn

nm

nm

nmnm

nmmnqz

Since this equation is symmetrical in m and n, the values of m and n are interchangeable. The above equation can be rewritten in the form:

)25.1( qIz = Where I = Influence value

++

+++++++

+++++= 22122

)1222(1tan)122(

)222(22122)1222(

41

nmnm

nmmn

nm

nm

nmnm

nmmn

Based on this equation, Fadum (1941) has prepared a chart for the influence values for sets of values for m and n, as shown in Fig 1.12.



Fig 1.12 Values of influence factor I for calculating vertical stress at the corner By superposition we can determine stress at any point within or out side the loaded area. For instance the stresses at point A and point B are

I IV

IIIII

S

T

P

U B

R V

WQ

A

PWBT:I QWBU:IIISVBT:II RVBU:IV

+=

+++=

IVI

IIII

III

IIqzBpoAt

IVI

IIII

III

IIqzApoAt

:int

:int



1.8 Stress Due to Uniformly Loaded Irregular Areas- Newmarks Chart Newmark (1942) devised a simple, graphical procedure for computing the vertical stress in the interior of a soil medium (semi-infinite, homogenous, isotropic and elastic), loaded by uniformly distributed, vertical load at the surface of a flexible area. The chart devised is called an influence Chart. The vertical stress underneath the center of a uniformly loaded circular area has been shown to be

)22.1.(2/3

2)/(1

11 EqSeezR

qz

+= .

This equation can be rewritten in the form:

)26.1(132

1

=

qz

zR

If a series of values is assigned for the ratio z/q, such as 0, 0.1, 0.2, ., 0.9 and 1.0, a corresponding set of values for the relative radii, R/z, may be obtained. If a particular depth is specified, then a series of concentric circles can be drawn. Since the first has zero radius and the 11th has infinite radius, in practice only 10 circles are drawn. Each ring or annular space causes a stress of q/10 at a point beneath the center at the specified depth z, since the number of annular spaces (m) is 10. The relative radii can be tabulated as shown below:

S. No of circle z/q Relative Radii (R/z)1 0 0.0002 0.1 0.2703 0.2 0.4004 0.3 0.5185 0.4 0.6376 0.5 0.7667 0.6 0.9188 0.7 1.1109 0.8 1.387

Now lets assume that a set of equally spaced rays, say n in number, is drawn emanating from the center of the circles, thus dividing each annular area in to n sectors, and the total area in to m*n sectors. If the usual value of 20 is adopted for n, the total number of sectors in this case will be 10x20=200. Each sector will cause a vertical stress of 1/200th of the total value at the center at the specified depth and is referred to as a mesh or an influence unit. The value 1/200 or 0.005 is said to be the influence value (or influence factor) for the chart. Construction of the Newmarks chart

1. For the specified depth (say 20m), calculate the radii of the circles as: R/z 0 0.270 0.4 0.518 0.637 0.766 0.918 1.11 1.387 1.908

R (m) 0 5.40 8.00 10.36 12.74 15.32 18.36 22.20 27.74 38.16 2. Draw circles to a convenient scale (say 1:100) 3. Draw a suitable number of uniformly spaced rays usually 20 emanating from the

center of the circles. 4. The resulting diagram will appear as shown in Fig 1.13; on it draw a vertical line

AB, representing the depth z to the scale used in drawing the circles. Thus here AB=20cm.



Influence value =0.005

20cm

A

B

One influence unit or mesh

Fig 1.13: Newmarks influence chart

This diagram can be used for other values of z by considering the scale to which it is drawn alters. So if z is to be 10m, line AB now represents z in scale of 1:50. Use of the Newmarks chart The chart can be used for any uniformly loaded foundation of any shape. The procedure is as follows

1. Draw the loaded area on a tracing paper, using the same scale to which the distance AB on the chart represents the specified depth.

2. Place the plan drawn above in such a way that the point under which the vertical stress z is required coincides with the center of the circles on the chart.

3. Count the number of elements on the chart enclosed by loaded area including fractional units, if any; let this total equivalent number be N.

4. Determine the vertical stress z at the specified depth as z =INq, where I= influence value of the chart.

1.9 Approximate Methods

Approximate methods are used to determine the stress distribution in soil under the influence of the complex loadings and/or shapes of loaded areas, saving time and labour without sacrificing accuracy to any significant degree. The following are some. a) Equivalent Point Load Method

Q1

Q2 Q3

Q4

A

zB

R1

R2 R3R4

Fig 1.14: Equivalent Point Load Method

Referring to Fig 1.14, if the influence values are I1, I2, I3, . for the point loads Q1, Q2, Q3,, then z is found as: z = (Q1I1+ Q2I2+ Q3I3+..) (1.27) This method gives a good result if z/B> 3.



b) Two is to One Method

Fig 1.15: Two to One Method

The average vertical stress at depth z is obtained as:

)27.1())((

..zLzB

LBqz ++=

c) The 30o or 60o Method

Fig 1.16: 30o or 60o Method

The average vertical stress at depth z is obtained as:

)28.1()5.1)(5.1(

..zLzB

LBqz ++=

1.10 Initial Settlement or Elastic Compression

This is also referred to as the 'immediate or distortion or contact settlement' and it is usually taken to occur immediately on application of the foundation load (within about 7 days). Immediate settlement computation The settlement of the corner of a rectangular base (flexible) of dimensions B' X L' on the surface of an elastic half-space can be computed from an equation from the Theory of Elasticity [e.g., Timoshenko and Goodier (1951)] as follows:

FSs

oi IIEvBqS

=

21'

qo = intensity of contact pressure in units of Es B' = least lateral dimension of contributing base area in units of S. Es, v = elastic soil parameters

Ii = influence factors, which depend on L'/B', thickness of stratum H, Poisson's ratio v,



and base embedment depth D. The influence factor Is (see Figure 3.1 for identification of terms) can be computed using equations given by Steinbrenner (1934) as follows:

1 2 1 21 2 with and as follows:1S

vI I I I Iv

= +

+++++++

++++++=

1

)1)(1(ln)11(

))(11(ln122

22

22

222

1NMM

NMMNMM

NMMMI

1 12 2 2

tan tan in radians2 1N MI

N M N = + +

B'HN

B'L'Mwhere; == and

ii cornerforLL'andcenterfor2LL';corner forBB'andcenterfor

2BB' II ====

IF = influence factor from the Fox (1948b) equations, which suggest that the settlement is reduced when it is placed at some depth in the ground, depending on Poissons ratio and L/B. Figure 3.1 can be used to approximate IF. Note: if your base is "rigid" you should reduce the Is factor by about 7 percent (that is, Is, rigid = O.931 Is, flexible)

Figure 1.17: Influence factor IF for footing at a depth D. Use actual footing width and

depth dimension for this D/B ratio.



CHAPTER TWO

SHEAR STRENGTH OF SOILS 2.1 Introduction HL\CH2.doc The shear strength of a soil is its maximum resistance to shear stresses and a consequent tendency for shear deformation. Its value determines such factors as the stability of slopes (cuts, earth dams), the allowable bearing capacity of a foundation and the thrust of a soil against retaining walls. Knowledge of the shear strength is thus an essential prerequisite to any analysis concerned with the stability of a soil mass. Basically, a soil drives its shearing strength from the following:

1) Resistance due to the interlocking of particles. 2) Frictional resistance between individual soil grains, which may be sliding

friction, rolling friction, or both. 3) Adhesion between soil particles or cohesion

Granular soils or sands may derive their shear strength from the first of the two sources, while cohesive soils or clays may derive their shear strength from the second and third source. Highly plastic clays, however, may exhibit the third source alone for their shearing strength. Most natural soil deposits are partly cohesive and partly granular and as such, may fail in to the second of the three categories just mentioned, from the point of view of shearing strength. 2.2 Basic Concepts of Shearing Resistance

a) Friction between Solid Bodies Consider a prismatic block resting on the surface MN (Fig 2.1). The block is subjected to two forces: force Pn acting at right angle to surface MN and force Fa acting tangentially to the plane MN.

Fig 2.1 Friction between bodies As Fa increases, Fr also increases such that Fa=Fr. The block will start sliding

when angle of obliquity reaches a maximum value, m. o If the block and surface MN are the same material, m= (=angle of

internal friction) and thus tan = ( =coefficient of friction) o If the block and surface MN are different materials, then m= angle of skin

(wall) friction.

b) Internal Friction within Soil Masses In granular or cohesionless soil masses, the resistance to sliding on any plane through the point with in the mass involves the movement of one particle relative to another.

Let Pn remain constant, Fa increases gradually until sliding starts.

If Fa is relatively small, the block will remain at rest and the resisting force can be written as Fr=Pn tan. This resisting force is due to surface roughness between block and plane.



The components giving rise to movement include: 1) Sliding or rolling 2) Removal of interlocking between particles

The angle of internal friction, which is a limiting angle of obliquity and hence the primary criterion for slip or failure to occur in a certain plane, varies appreciably for given sand with the density, since the degree of interlocking is known to be directly dependent upon the density. This angle also varies somewhat with the normal stress. However, the angle of internal friction is mostly considered constant, sine it is almost so for a given sand at a given density. Since failure or slip within a soil mass cannot be restricted to any specific plane, it is necessary to understand the relationships that exist between the stresses on different planes passing through a point, as a prerequisite for further consideration of shearing strength of soils. 2.3 Principal Planes and principal Stresses-Mohrs Circle Consider a solid mass acted on by a series of forces F1, F2, , Fn in a 2D Plane (Fig 2.2 (a)). These forces produce stresses at a point with in the body. And every plane passing through that point will be subjected, in general, to a normal or direct stress and a shearing stress. In geotechnical engineering, compressive normal stress and shear producing counter-clockwise couple/moment on the element or clockwise moment about a point just out side the element face are usually considered positive (Fig 2.2 (b)).

Fig 2.2: a) Solid mass acted by forces b) Sign convention Now lets consider an element of soil whose sides are chosen as the principal planes (planes which do not have shear stresses), the major and the minor, as shown in Fig 2.3 (a). Let it be required to determine the stress conditions on a plane inclined at angle from the major principal plane measured CCW (Fig 2.3(b)). Considering the equilibrium of this element redrawn as Fig 2.3 (c), we have

)2.2(2sin2

)31(

)1.2(2cos2

)31(

2

)31(

2cos)(2sin2cos 31331

=

+=+=+=



Fig 2.3 Stresses on a plane If we square and add equations 2.1 and 2.2, we have

)3.2(2

2312

2

231

2

23122cos

231

2312

2

2312

2

2312cos

231

2312

2

23122

=+

+

=+

+

+

+

++

+=+

This is an equation of the circle with radius 2

31 and center (2

31 + , 0). This circle is drawn in , space (Fig 2.4) and is known as Mohrs circle of stress.

Fig 2.4 Mohrs Circle of stress

Once this circle is drawn, we can determine normal shear stresses at any plane through a point. The procedure is based on a unique point called pole or origin of planes. Once the pole is known the stress on any plane can readily be found by drawing a line from the pole parallel to the plane where we want stresses. The coordinates of the point of intersection with the Mohrs circle determine the stresses on that pale. 2.4 Strength Theories for Soils a) Coulomb Failure Equation In 1976 Coulomb observed that if the thrust of a soil against a retaining wall caused the wall to move forward slightly, an essentially straight slip plane formed in the retained soil. He postulated that the maximum resistance to shear, f, on the failure plane is given by f = c + f tan (2.4) Where: is the total stress normal to the failure plane is the angle of internal friction c is the cohesion of the soil

Coulombs Envelope f = c + f tan c



The use of Coulombs equation didnt always result in successful design of soil structures. This is due to the fact that water cannot sustain shear stress; the shear resistance of a soil must result solely from the frictional resistance arising at the particle contact points, the magnitude of which depends solely on the magnitude of the effective stress carried by the soil skeleton. Hence, expressing Coulombs equation in terms of effective stress: f = c' + 'f tan' (2.5) in which the parameters c' and ' are properties of the soil skeleton, referred to respectively as the effective cohesion and the effective angle of friction. b) Mohr-Coulomb Theory of Shear strength The Mohr- Coulomb theory of shearing strength of soil, first postulated by Coulomb (1976) and later generalized by Mohr, is the most commonly used concept. Mohr stated that when shear stress on the failure plane reaches a unique function of the normal stress on the failure plane, failure takes place. Thus if we conduct several tests and obtain principal stresses at failure, the Mohrs circles at failure can be constructed as in Fig 2.5. A line drawn tangent to the failure circles is called an envelope (Mohr failure envelope). Therefore, it can be said that the Mohrs circle of stress relating to a given stress condition would represent, incipient failure condition if it just touches or is tangent to the strength or failure envelope (circles C, D, E); otherwise, it would wholly lie below the envelope as shown in circle A. circles lying above the Mohr failure envelope (circle B) cannot exist.

2.5 Determination of Shearing Strength of Soils Determination of shearing strength of a soil involves the plotting of failure envelopes and evaluation of the shear strength parameters for the necessary conditions. The following tests are available for this purpose: Laboratory tests:

1. Direct Shear Test 2. Triaxial Compression Test 3. Unconfined Compression Test 4. Laboratory Vane Shear Test

1) Direct Shear Test The direct shear device, also called the shear box apparatus, essentially consists of a brass box, split horizontally at mid height of the soil specimen, as schematically shown in Fig 2.6. The soil sample is gripped in perforated metal grills, behind which a porous discs

Field tests: 1. Vane Shear Test 2. Penetration Tests



can be placed if required to allow the specimen to drain. For undrained tests, metal plates and solid metal grilles may be used.

Fig 2.6: Schematic representation of Direct Shear Test

After the sample to be tested is placed in the apparatus or shear box, a normal load which is vertical is applied to the top of the sample by means of a loading yoke and weights. Since the shear plane is predetermined as the horizontal plane, this becomes the normal stress on the failure plane, which is kept constant through out the test. A shearing force is applied to the upper-half of the box, which is zero initially and is increased until the specimen fails. Two types of application of shear are possible- one in which the shear stress is controlled and the other in which the shear strain is controlled. In the stress controlled type, the shear stress which is controlled variable, may be applied at a constant rate or more commonly in equal increments by means of calibrated weights hung from a hanger attached to a wire passing over a pulley. Each increments of shearing force is applied and held constant, until the shearing deformation ceases. The shear displacement is measured with the aid of a dial gauge attached to the side of the box. In the strain-controlled type, the shear displacement is applied at a constant rate by means of a screw operated manually or by motor. With this type of test the shearing force necessary to overcome the resistance within the soil is automatically developed. The shearing force is measured with the aid of a proving ring-a steel ring that has been carefully machined, balanced and calibrated. The shear displacement is measured with the aid of another dial gauge attached to the side of the box. In both cases, a dial gauge attached to the plunger, through which the normal load is applied, will enable one to determine the changes in the thickness of the soil sample which will help in the computation of volume changes of the sample, if any. HL\CH2.doc For the readings taken, we can determine the following 1) Normal stress: i = Pi/A 2) Shear stress: i = Fi/A 3) Vertical Strain (Volume change) n = h/h 3) Shear Strain h = A/A And the following plots can be produced (Example Fig 2.7 for overconsolidated clay) 1) Normal stress vs. shear stress ( vs. ) 2) Shear strain vs. shear stress (h vs. )



3) Normal stress vs. volume change ( vs. h) 4) Mohr circle at failure

Fig 2.7 Different plots

The direct shear test is relatively simple and quick drainage is possible as the thickness of the sample is small. However, it has some draw backs like

1. The failure plane is predetermined and this may not be the weakest plane.* 2. Drainage conditions cannot be controlled 3. The normal and shear stresses in a plane are not uniform 4. The area of the sliding surface is not constant better to account it 5. The effect of the lateral restraints by the side walls of the shear box is likely to

affect the results. 2) Triaxial Compression Test Apparatus and procedure Triaxial HL\CH2.doc Fig 2.8 shows the essential details of the triaxial cell in which a soil sample is sealed with in water tight rubber membrane and enclosed in a cell filled with water through which a confining pressure is applied to the sample. Drainage facility is provided by strips of filter paper placed vertically around the sample. These connect with a pours disc in the top platen, take-off being through a nylon tube which passes out of the cell through its base. The pore water pressure in the sample is measured through a saturated porous disc sealed flush in the base pedestal and connected via a water-filled duct to an electrical pressure transducer. The sample is sheared by a vertical piston load applied through the top platen. To minimize the friction forces at the top and bottom of the sample and allow an unrestricted lateral deformation during shear, greased rubber discs are placed between the sample and the end caps. The form of the test may either be strain controlled, in which case the vertical piston load is applied by a motorized loading frame greased to deform the sample vertically at a constant rate of strain, or stress controlled, in which case the sample is allowed to strain freely under a vertical piston load applied by dead weights.



Fig 2.8 Triaxial cell

By eliminating the shear stresses on the top and bottom boundaries of the sample, and since no shear stress can act between the cell water and the vertical surface of the sample, it follows that the axial stress and the ambient cell pressure are both principal stresses. Under the conditions of the triaxial test, the major principal stress 1 is the axial stress and the intermediate and minor principal stresses, 2 and 3, are both equal to the all-round cell pressure. This of course acts not only on the vertical surface of the sample but also on the end faces. Thus, if at some stage of the shear test the vertical piston load is P and the current cross-sectional area of the sample is A then

3131 =+= AP

AP

The applied piston stress P/A is therefore in actual fact equal to the difference between the major and minor principal total stresses, 1-3, which is termed the deviator stress and denoted as . The application of the all-round cell pressure and application of the deviator stress form two separate stages of the test. Whether drainage of the sample is permitted during each stage depends upon the soil type and nature of the field problem being investigated. A soil element in the field (notably an element of clay soil) may fail either undrained, partially drained or drained. We thus have three basic types of tests: Undrained, consolidated-undrained and drained. Area correction for the Determination of Deviator Stress During the application of the axial load, the specimen undergoes axial compression and horizontal expansion to some extent. Little error is introduced if the volume (or area) is assumed constant, although the area of the x-section varies as axial strain increases. This assumption is perfectly valid if the test is conducted under undrained conditions, but, for drained conditions, the exact relationship is somewhat different and given below. If Ao, ho, and Vo are the initial area of cross section, height and volume of the soil specimen respectively, and if A, h, V are the corresponding values at any stage of the test, the corresponding changes in the values being designated A, h, and V, then



V=A (ho - h) = Vo - V a

oVV

oA

ohh

oh

oVV

oV

hoh

VoVA

=

=

=1

1

1

1

For undrained tests, V = 0 a

oAA = 1 . And the ratio a11 is called the correction

factor. Once the corrected area is determined, the additional axial stress or the deviator stress, , is obtained as

areacorrectedreadingringprovingfromLoadAxial )(

31 == Mohrs Circle for Triaxial Test (Relationship between principal stresses and shear strength Parameters) The stress condition in a triaxial test may be represented by a Mohrs circle, at any stage of the test, as well as at failure as shown in Fig 2.8 (a). The cell pressure, c, which is also the minor principal stress 3 is constant and 11, 22, 13, , 1f are the major principal stresses at different stages of loading and at failure. The Mohr circle at failure will be tangential to the Mohr-Coulomb strength envelope, while those at intermediate stages will be lying wholly below it. The Mohrs circle at failure will be as shown in Fig 2.8 (d).

Fig 2.8: Mohrs Circle at Failure The Mohrs circles at failure for one particular cell pressure are shown for the three typical cases of a -soil, a c-soil and a general c- soil in Figs. 2.8 (b), (c), and (d) respectively. With reference to Fig 2.8 (d), the relation ship between the major and the minor principal stresses at failure may be established from the geometry of the Mohrs circle, as follows: From AOF, 2 = 90o + = 45o+/2 the failure plane is inclined at an angle of 45o+/2 to the major principal plane. Again from AOF, sin = FO/AO= FO/(AB+ BO)

Since axial strain, a=h/h



/2)(45tantanN where2aswritten also is This

)6.2(tan2tan)2/45tan(2)2/45(tan)sin1(

cos2)sin1()sin1(

cos2)sin1()sin1(sin)(cos2)(

)31(cot2

)31(

2/)31(cot

2/)31(sin

o2231

23

231

31

313131

+==+=+=+++=

++=

++=++=++

=++=

NcN

cc

ccc

cc

These state of stress at failure is defined as plastic equilibrium condition, when failure is imminent. From one test, a set of 1 and 3 is known; but it can be seen from equation 2.6 that at least two such sets are necessary to evaluate the parameters c and . Conventionally, three or more such sets are used from a corresponding number of tests. However, the usual procedure is to plot the Mohrs circles for a number of tests and take the best common tangent to the circles as the strength envelope. A small curvature occurs in the strength envelope of most soils, but since the effect is slight, the envelope for all practical purposes, may be taken as a straight line. The intercept of the strength envelope on the -axis gives cohesion and the angle of the slope of this line with the -axis gives the angle of internal friction. In derivation of equation 2.6, we have sin)(cos2)( 3131 ++= c ( )

sincossin2/)(cos2/)( 3131

pcqc

+=++=

which plots as a straight line as q against p (Fig 2.9), the slope of the line defining sin and the intercept ccos, and thus provide an alternate method for determining shear strength parameters from the results of triaxial tests.

Fig 2.9: Alternative Procedure of Evaluating shear strength parameters

The application of the all-round cell pressure and application of the deviator stress form two separate stages of the test. Whether drainage of the sample is permitted during each stage depends upon the soil type and nature of the field problem being investigated.

i) Undrained Test (UU test) HL\CH2.doc Drainage is not permitted at any stage of the test, that is, either before the test during the application of the normal stress or during the test when the shear stress is applied. Hence no time is allowed for dissipation of pore water pressure and consequent consolidation of the soil; also, no significant volume changes are expected. Usually, 5 to 10 minutes may be adequate for the whole test (called quick test or Q or Qu test), because of the shortness of drainage path. However, undrained tests are often performed only on soils of low permeability.



This test is applicable to the short term stability analysis of works formed on or in clay deposits where it is considered that insufficient time has elapsed for any dissipation of excess porewater pressure to have occurred by the end of construction. Such works generally include small embankments, cuttings, retaining walls and foundation of buildings. Test on Saturated Clay If the clay sample is saturated, the increase in cell pressure is carried entirely by the porewater as an excess porewater pressure with no change in effective stress in the sample and therefore no change in shear strength. Thus, the deviator stress required to fail the sample is independent of the cell pressure at which the test is run. Fig 2.10 shows the corresponding Mohr stress circles, the common tangent to which defines the failure envelope for the soil, which in this case is horizontal giving u=0; the intercept on the vertical shear stress axis defines the undrained cohesion Cu. The undrained shear strength, u, in terms of the total stress is thus u= Cu + tanu, with u =0,

fu = Cu. (2.7) The undrained cohesion thus defines the undrained shear strength. And Cu is more generally referred to as the undrained shear strength.

Fig 2.10: Failure envelope for undrained tests on Saturated Clay Also, Cu = (1 - 3)/2 = f /2 , where the deviator stress at failure, f, defines the compression strength of the sample. Here it is clear that the deviator stress, (1 - 3) =constant for any values of cell-pressure, 3, used. Thus the parameters Cu and u are unique and independent of the test procedure used or the stress path followed. This is particularly important result for it means that soil elements in the field where the stress path conditions to failure are more complex with 1 and 3 both varying, will all have the same value of Cu and u=0. This would suggest that where the stress change likely to cause failure occurs under undrained conditions, the stability analysis may be carried out in terms of total stresses. This particular form of analysis, known as the u=0 analysis, leads to a very simple and rapid assessment of stability since the undrained shear strength at any point is defined solely by the undrained cohesion Cu. If the porewater pressure is measured in a series of undrained tests on saturated samples it is found that the same effective stress failure circle is obtained in each case, and hence the effective stress parameters C' and ' cannot be determined from such tests. Recourse is made to consolidated-undrained or drained tests for this purpose.



Unconfined Compression Test This is a special case of triaxial test with 3=0 or c=0. The test is carried out in saturated cohesive soil sample which can stand without lateral support (rubber membrane). This test may be conducted on undisturbed or remolded cohesive soils. It cannot be conducted on coarse grained soils such as sands and gravels as theses cannot stand with out lateral support. From above,

fu = Cu = (1 - 3)/2 = f /2 fu = Cu = 1 /2 Usually 1 is designated in this case as qu and is referred as unconfined compressive strength. Thus,

fu = Cu = qu /2 (2.8) The Mohrs circle for this case is shown in Fig 2.10 and designated as circle A. The value of the undrained shear strength of clay may be used to indicate its consistency state (Table 2.1).

Table: 2.1 Consistency of Clays

Consistency BS5930:1981 Terzaghi and Peck: 1967Very soft 200

Undrained shear strength Cu(Kpa)

correlationHL\CH2.doc

Test on Partially saturated Clay Undrained tests may also be carried out on partially saturated soils, for example, samples of a rolled clay fill or laboratory-compacted samples of earth fill. In this case the increase in cell pressure is carried partly by the pore water and partly by the soil skeleton. Thus, the greater 3 the greater the increase in the effective confining pressure, and therefore the greater the increase in shear strength and consequently the greater the increase in deviator stress required to cause failure. The increase in the deviator stress becomes progressively smaller and finally ceases when the applied stresses are large enough to force all the air in the voids in to solution in the pore water, when the sample behaves as a saturated soil with u=0. The failure envelope with respect to the total stress is curved (Fig 2.11(a)) and values of Cu and u vary with the magnitude of the normal stress on the failure plane. If the pore water pressure is measured during each test, Mohr circles of effective stress can be drawn and it is found that the failure envelope in terms of the effective stress is linear over a wide range of (Fig 2.11(b)).

Fig 2.11 Failure envelopes for undrained tests on partially saturated clay w.r.t (a) total stress (b) effective stress

Curved failure envelope



ii) Consolidated Undrained Test (CU test) In the previous section we have considered the relevance of the undrained test to model a field situation where a clay deposit is subjected to a stress change which is rapid in relation to the time for the dissipation of excess porewater pressure so that a potential failure would occur under undrained conditions. If, however, the construction period extends over several seasons (for example, in the case of an earth dam) it is reasonable to assume that some consolidation of the soil mass will have occurred by the end of construction. If at this stage the shear stresses induced in the soil are of sufficient magnitude to cause failure, this would occur quickly under conditions of no further drainage. This behavior is modeled in the consolidated-undrained test in which a soil sample is fully consolidated under the cell pressure and then sheared by the deviator stress under undrained conditions. Fig 2.12 shows a typical deviator stress-axial strain and porewater pressure-axial strain curves obtained for normally consolidated and over consolidated clays. For a normally consolidated soil the pore water pressure rises to failure, reflecting the volume decrease that would occur if drainage were to be allowed. For a heavily over consolidated clay the porewater pressure decreases during shear, reflecting the dilatancy that would occur if the sample were free to undergo volume change.

Fig 2.12: Stress-strain-pore pressure relationships for consolidated-undrained tests on clay

The greater the pressure to which a sample is consolidated, the greater the deviator stress required to cause failure. Fig 2.13(a) shows typical Mohr circles of total stress. The intercept and slope of the failure envelope define the total stress shear strength parameters for the soil, which for a consolidated-undrained test are denoted by Ccu and cu respectively. If the porewater pressure is measured during the test, as is the usual practice, then the corresponding Mohr circles of effective stress may be drawn (Fig 2.13(b)), the failure envelope now defining the effective stress shear strength parameters C and '. The effective stress circles may lie either to the left or right of their respective total stress circles, depending on whether the porewater pressure is positive or negative.

Fig 2.12: Failure envelopes for consolidated-undrained tests on clay w. r. t (a) total stress, (b) effective stress.



For normally consolidated clays, the failure envelopes with respect to total and effective stress pass through the origin giving Ccu and C' equal to zero; for overconsolidated clays Ccu and C' usually range between 5 and 30 KPa. The value of ' is not influenced appreciably by overconsolidation and ranges from about 30o to 20o, decreasing with increasing plasticity index. The value of cu varies similarly and its relation to ' is determined by the magnitude of the porewater pressure at failure. In the standard triaxial test the sample is failed in a conventional manner by holding 3 constant and increasing 1. In a field situation the stress changes leading to a potential failure of an element are more complex than this and in general range from 3 constant and 1 increasing to 1 constant and 3 decreasing. It is important to consider the possible influence of this on the measured values of the shear strength parameters. This is illustrated in Fig 2.14 for the case of normally consolidated clay. In test 1 a soil sample is fully consolidated under a cell pressure 3.

Fig 2.14: Influence of test procedure on cu and

The drainage valve is then closed and the sample sheared to failure under undrained conditions by increasing 1. The line OX then defines the value of cu. In test 1 an identical soil sample is consolidated to the same cell pressure 3. The drainage valve is then closed and the sample failed by decreasing 3, at the same time maintaining 1 constant and equal to the initial cell consolidation pressure. Since the pressures to which the samples have been consolidated are the same, then the deviator stress at failure will be the same as in test 1. The line OY then defines the value of cu. It is thus seen that the value of cu is not unique and therefore of limited application in practice. Considering effective stress conditions, in test 1 the porewater pressure at failure is positive therefore the effective stress circle lies to the left of the total stress circle. In test 2, failure by unloading the sample results in a negative porewater pressure and consequently the effective stress circle lies to the right of the total stress circle. It is found that the two effective stress circles coincide (circle 3). The value of is therefore unique and consequently has a much wider application in practice. Thus, if any drainage of a soil mass has occurred by the end of construction, stability analysis should be carried out in terms of effective stress. This is logical since it is the effective stress in a soil that controls its shear strength.

iii) Drained Test (CU test) Where construction in the field is on sand or gravel deposit, a potential failure would occur under drained conditions. Cuttings in clay may also fail many years after construction when the initial (negative) excess porewater pressures have fully dissipated. These conditions are simulated by the drained test in which a sample is fully consolidated under the cell pressure and then sheared under drained conditions at a rate compatible with no build-up in excess porewater pressure, so that u remains equal to zero throughout



the shearing stage. The form of the stress-strain-volume change relationship for sands and clays are similar to those obtained from direct shear tests and are given in Fig 2.15.

Fig 2.15: Stress-strain-volume change relationships for drained tests (a) Sands (b) clay Since u=o through out the shearing process, ' 3=3, ' 1=1 and the Mohr circles of effective stress and total stress coincide (Fig 2.16). The failure envelope defines the effective stress parameters C' and '. These are often denoted Cd and d respectively. Generally there is little difference between the effective stress parameters obtained from drained and consolidated-undrained tests. However, for sands and heavily overconsolidated clays d is slightly higher than ', because of the work done by the sample as it expands against the confining pressure during shear.

Fig 2.16: Failure envelope for drained tests on clay

iv) Pore Pressure Parameters

In addition to determining the shear strength parameters of a soil, the triaxial test is also used to furnish data for predicting the initial excess porewater pressure set up at a point in a soil mass by a change in the total stress conditions. Such predictions are required in conjunction with an effective stress stability analysis and are made by means of experimentally determined pore pressure parameters. Pore parametersHL\CH2.doc

(a) Sand

(b) Clay



To introduce the idea of pore pressure parameters, consider the soil element shown in Fig 2.17 wherein the element is subjected to a triaxial loading in which 2=3. We may consider this stress system to be composed of an isotropic change of stress 3 plus a uniaxial change of deviator stress 1-3. This, of course, is the stress system imposed in the triaxial test and it is clear that the excess porewater pressure generated within the element will result firstly from the change in the all-round stress and secondly from the change in the deviator stress.

Fig 2.17: Soil element If Ua denotes the excess porewater pressure induced in the element by the application of 3 and Ud that by 1-3, then assuming these pore pressures to be a simple proportion of the applied stresses, we may write

Ua = B 3 and Ud = A (3-1) Where B and A are experimentally determined pore water parameters. If U denotes the total excess porewater pressure in the element, then assuming the principle of superposition, U = Ua + Ud and hence U = B3+A(1-3) (2.9) This problem was considered by Skempton (1954) who gives

U = B [3+A (1-3)] (2.10) This is of the same form as equation 2.9 where AB replaces A. The pore pressure parameters are obtained from a triaxial compression test. The parameter B is determined by measuring the increase in porewater pressure resulting from the increase in cell pressure and varies from 1 for a fully saturated soil to 0 for a dry soil. The parameter A, and hence A, is determined by measuring the porewater pressure induced in the sample by the application of the deviator stress. The value of A depends upon the initial consolidation stress system (whether isotropic or anisotropic), the stress history (as reflected by the overconsolidation ratio), the proportion of the failure stress applied (i.e, on the strain of the sample) and the type of stress change (whether loading or unloading). Typical failure values (Af) are given in Table 2.2.

Table 2.2: Values of the pore pressure parameter A at failure (Af)Soil type Af

Sensitive clay 1.5-2.5Normally consolidated clay 0.7-1.3Lightly consolidated clay 0.3-0.7Heavily consolidated clay -0.5-0

With B=1 for a saturated soil, A=A.



v) P-q Diagrams and Stress Paths In discussing about Mohrs circle for triaxial tests, we saw how to represent and plot the state of stress as stress point whose coordinates are (p, q), defined as

2q

2p 3131

=+=

Actually the p and q values represent the coordinates of a point on Mohrs circle, with p representing the center of the circle (always located on the abscissa, or axis) and q representing the maximum shear stress, equal to the radius of the circle. The locus of p-q points for a test series is known as a stress path. Such a graphical representation is known as the p-q diagram. Fig 2.18 shows the Mohr envelope ( line) developed from tangents to the circles at points 1, 2, and 3 and line Kf (Kf line), which passes through p-q points A, B, and C- points of maximum shear for the respective circles. Thus, the Kf line represents a limiting state of stress at impending failure. The following relationship between the line and the Kf line exist as illustrated in Fig 2.9. d = c cos , sin = tan (2.11) where d is the intercept on the q-axis and is the angle made by the Kf line with the p-axis.

Fig 2.18: Stress paths for a triaxial test series

We may vary 1 and 3 in various ways to obtain any number of stress paths. As an example it is shown for different cases below. For all cases assume h=v

Stress path HL\CH2.doc

Point A: initial conditions Po=(h+v)/2=v ; qo=(h-v)/2=0 final conditions Pf =(hf+vf)/2=v +; qf =0

Point B: initial conditions Po=(h+v)/2=v ; qo=(h-v)/2=0 final conditions Pf=(hf+vf)/2=v +3/4; qf=(hf-vf)/2=1/4

Note: stress paths can be developed for either total or effective stresses giving total stress path or the effective stress path. HL\Examples CH2.doc

Points v h RemarkA B 0.5C increses 0 Triaxial compressionD E 0 increases Triaxial extensionF decreases increases



Shear Strength of Cohesionless Soils The shear strength of sands is given by (C = 0), = ' sin =(-u) sin (2.12) Factors affecting shear strength HL\CH2.doc Generally the value of (hence shear strength of sand) is influenced by

1. Void ratio or relative density or state of compaction 2. Roughness , shape and angularity of the grains 3. Grain size distribution

Ultimate values of may range from 29 o to 35o and peak values from 32 o to 45o for sands. The values of selected for use in practical problems should be related to soil strains expected. If soil deformation is limited, using the peak value for would be justified. If the deformation is relatively large, ultimate value of should be used. Stress-strain Behavior The stress-strain behavior of sands is dependant to a large extent on initial density of packing, as characterized by the density index. This is represented in Fig 2.15 (a) above. There it can be seen that, the shear stress builds up gradually for loose sand, while for an initially dense sand, it reaches a peak value and decreases at grater values of shear/axial strain to an ultimate value compatible to that for an initially loose specimen. The behavior of medium-dense sand is intermediate to that of loose sand and dense sand. The hatched portion represents the additional strength due to the phenomenon of interlocking in the case of dense sands. Critical Void Ratio Volume change of a soil sample during testing depends on particle size, shape, grain-size distribution, principal stress difference, previous stress history and most significantly relative density or void ratio. At large strains initially loose sand and initially dense sand attain same void ratio, where further strain will not bring any volume change. Such a void ratio is called a critical void ratio, ecr (Fig 2.19).

Fig 2.19 Critical void ratio

ecr depends on 3 in the triaxial test and v in the direct shear test. The larger 3 and v, the smaller ecr.

Saturated Cohesionless Fine-Grain Soils The angle of internal friction of saturated sands and some inorganic silts is only slightly less than that of the soil in a dry state and of the same relative density (as the drainage is instantaneous). However the shear strength might be altered significantly by a change in the pore pressures (equation 2.12). Quite apparently, when the porewater pressure approaches , the shear strength approaches zero. When that happens, we may approach impending instability or perhaps motion (e.g., slope failures, boiling). Fluctuation in the water table is a common cause of significant variations in the pore stress and, thereby, in the shear strength of the soil.



Liquefaction of Fine Sand and Inorganic Silts If a saturated and/or inorganic silt is totally saturated and under hydrostatic neutral stress such that it is not subjected to any effective stress, the mass is in a state of liquefaction. Under such circumstances, the pore water pressure u equals the total normal stress , there by reducing the shear strength to zero, and then the soil is in a quick or boiling or flowing condition (conditions that result in impending upward movement of soil and water). If submerged fine sand undergoes a sudden decrease in the void ratio, an increase in the porewater pressure, u, may result such that the pore pressure may equal or exceed the value for . For example, pile driving, earthquakes, blasts, or other forms of vibration or shock may cause a sudden decrease in the volume, thereby increasing the pore pressure u as a result of a surge in hydrostatic excess pressure. Should the value of u reach sufficient magnitude, say u > , the shear strength of the soil may be totally lost, resulting in what is known as spontaneous liquefaction. Loose, fine silty sands are most vulnerable to such effects from shock or dynamic loads or sudden fluctuations in the water table. Compacting loose sand stratum is frequently a viable option to decrease the possibility of liquefaction. Shear Strength of Cohesive Soils HL\CH2.doc Shear strength behavior of clays is influenced by the fact whether the clay is normally consolidated or overconsolidated, by the fact whether it is undisturbed or remolded, by the drainage conditions during testing, consistency of the clay, by certain structural effects, by the type of test and by the type and rate of strain. Normally Consolidated Clays When the sample is extracted from the ground, the overburden pressure is removed and the pore pressure altered significantly, that is, negative pore pressures are developed. In order to simulate a somewhat realistic in-situ state of stress, the characteristics of saturated, normally consolidated clays extracted from a given stratum are commonly investigated via a CU triaxial test. A confining pressure, 3, and a deviator stress, , are applied for undrained conditions. A confining pressure, say o, of the in-situ value may be estimated as the overburden pressure for the depth of the stratum from which the sample was extracted. If several such tests are run for varying confining cell pressures, a Mohr envelope may be obtained as indicated in Fig 2.20 (a). If the confining pressure is less than the in-situ value o, the Mohr envelope depicts a range of preconsolidation of the soil; that is, relative to the confining pressures, the soil specimen appears overconsolidated. The shear strength of the clay specimen tested in this range is higher than that indicated by a straight line through the origin. The relationship between the shear strength and the normal stress in this range is designated by line portion ab, which is slightly curved, but frequently interpreted as a straight line. On the other hand, if the tests are run under confining cell pressures larger than o, the envelope to the rupture circles is approximately a straight line, represented by segment bc. Although the effect of porewater pressure is present for all ranges of normal stress within the Mohr envelope,



the pore pressure is larger for the case where the confining pressures are smaller than o. Generally, there is an increased drainage of water with increasing confining pressure. However, the effective stress is more significant in the region where the confining pressures are larger than o.

Fig 2.20 relationships from CU and drained tests (a) CU test (b) CD test Figure 2.20 (b) shows the relationship between the shear strength and the effective stress for a case of CD tests. These tests are appreciably more time consuming than the CU test, and are therefore not as common during investigations. The specimen is subjected to a confining pressure, and then the deviator stress is applied at a very slow rate and drainage is permitted at each end of the sample. A series of such tests run under varying confining cell pressures provides a strength envelope as shown in Fig 2.20 (b). The strength envelope for such test is again somewhat curved for the range of confining pressures less than o; for confining pressures greater than o, the Mohr envelope is approximately a straight line. Briefly, a comparison of the two cases shows the following 1. The corresponding slopes of straight-line segments of the Mohr envelopes (Fig 2.20)

are significantly different. 2. For the drained condition the neutral (pore-water) stress is virtually negligible. 3. The effective friction is significantly larger for the drained case than for the undrained

case. The relationship between angles and cu may be illustrated by means of Fig 2.20 (c). The shear strengths for the CD and CU tests are and 1 as shown. The actual strength may lie somewhere between that for a CD and that for an undrained condition since total drainage is unlikely; that, the actual strength of the sample may range between values of and 1. HL\CH2.doc Overconsolidated Clays HL\0.doc A soil being evaluated or tested may have been subjected to a great deal of precompression (i.e. overconsolidation pressure, Pc) induced by loads, which since then may have disappeared. Fig 2.21 shows the Mohr strength envelope for an overconsolidated clay. One notes that for < Pc, the Mohr envelope deviates from a straight line, that is, the shear strength is larger than that given by the dotted straight line. Generally preconsolidation causes or results in a smaller void ratio at failure than would otherwise exist, even though the specimen tends to expand as a result of extraction from an in-situ condition. Cohesion and general capillary forces tend to resist the volume increase, thereby resulting in somewhat greater shear strength, as indicated by the curved portion of the envelope. Beyond the preconsolidation pressure Pc, the effective normal stress and shear strength relationship is given by a reasonably straight line.



Fig 2.21 Shear strength vs. effective normal stress for OCC

The shear strength of clays that have fine discontinuities, hairline cracks, and the like, generally referred to as fissures, may be appreciably different from that of the typical overconsolidated clay described above. Depending on the magnitude and orientation of these fissures, test results may be particularly misleading in the overall evaluation. For example, the results from a direct shear test on a sample where fissures are parallel to the shearing force may be appreciably smaller than for one where the orientation of fissures is 90o to the shear force. On the other hand, the triaxial test yields somewhat more reliable results, improved perhaps by the lateral restraint of the confining pressure. Sensitivity of Clay If the strength of undisturbed sample of clay is measured and its strength is again measured after remolding at the same water content to the same dry density, a reduction in strength is often observed. This is an important phenomenon which is quantitatively characterized by sensitivity, defined as follows:

remolded strength,n compressio Unconfineddundisturbe strength,n compressio UnconfinedSySensitivit f =

A comparison of stress-strain curves for sensitive clay in the undisturbed and remolded sates is shown in the figure below. Sensitivity classification is given in the table below. Overconsolidated clays are rarely sensitive, although some quick clays have been found to be overconsolidated. 3) In-Situ Evaluation of Shear Strength As mentioned previously, the extraction of a soil sample and the subsequent changes induced by the extraction process, the handling, and the testing procedures may greatly alter the characteristics of the specimen and, therefore, the test results. Furthermore, it is not always feasible and practical to duplicate the in-situ conditions. Frequently, it is both practical and desirable to test the soil in the in-situ condition. A number of field tests are used to estimate the shear strength of the soil. Some of these give results that fit in to the theoretically based expressions used to designate the shear strength of the soil. Others are empirical in nature and greatly dependent on the engineers judgment and experience.

Table: Sensitivity classification of clays Sensitivity, Sf Classification

1 Insensitive 1-2 Low 2-4 Medium 4-8 Sensitive 8-16 Extra-sensitive >16 Quick (Sf can be up to 150)



The Vane shear Test Is used for the determination of shear strength of soft clays (clays which may be disturbed during the extraction and testing process). The test is performed at any given depth by first augering to the prescribed depth, cleaning the bottom of the borings, and then carefully pushing the vane instrument (Fig 2.22) in to the stratum to be tested. The torque is then applied gradually and the peak value noted. The shear strength of the soil can then be estimated by using the formulae derived below.

Fig 2.22 Vane Instrument The torque is resisted by T1 and T2. If both ends of the vane are submerged in the soil stratum, and if the maximum shear stress is Cu for all shear surfaces, then Resisting moment = cylindrical surface resistance + two circular end face resistance

T = 2r L (Cu r) + 2[r2 Cu (2/3r)] =2r2Cu (L+2/3r) ( )rLr 2

TC3

22u += ratio)used(commonlyr4Lif

r283TC 3u ==

If one end of the vane is submerged in the soil stratum, Resisting moment = cylindrical surface resistance + one circular end face resistance

T = 2r L (Cu r) + r2 Cu (2/3r) =2r2Cu (L+1/3r) ( )rLr 2

TC3

12u += r4Lif

r263TC 3u ==

This test is also made in laboratories using small vane instrument. Standard Penetration Test (SPT) The number of blows required to drive the split-spoon sampler is still another means used to estimate the soils resistance to shear (shear strength of the soil). Here a split-spoon sampler (Fig 2.23 (a)) is lowered to the bottom of the bore hole by attaching it to the drill rod and then driven by forcing it in to the soil by blows from a hammer (64Kg) falling from a height of 76cm. The sampler is initially driven 15cm below the bottom of the bore hole to exclude the disturbed soil while boring. It is then further driven 30cm in two stages (each 15cm). The number of blows required to penetrate the last 30cm is termed as the standard penetration value, N. Corrections have to be made for overburden pressure, dilatancy etc. Then corrected N value is used for correlation. HL\6.doc



Fig 2.23 Standard penetration test (SPT) The test is especially suited for cohesionless soils as a correlation has been established between the SPT value and the angle of internal friction of the soil. Terzaghi and Peck also gave the following correlation between SPT value, and Dr.

Table: Correlation between N, , and Dr for SandsCondition N (degree ) Dr(%)

Very loose 0-4 50 >42 >85 Table: Correlation between N and qu for Clays

Consistency N qu(KPa)Very soft 0-2 30 >400

The correlation for clays is unreliable. Hence, vane shear test is recommended for more reliable information. Usually SPT is conducted at every 2m depth or at the change of stratum. If refusal is noticed at any stage, it should be recorded. Other Penetrometers Penetrometers are some times used to test the shear strength of the soil at the surface-lateral or vertical. Their use is primarily applicable to fine-grained soil; coarse and gravelly strata tend to give erroneous results. The procedure for using the penetrometer consist of first cleaning the surface of any loose material, pushing the pentrometer in to the stratum to the calibration mark on the head of the penetrometer, and recording the maximum reading on the penetrometer scale. This reading represents the pressure in force per unit area necessary to push the penetrometer to the designated mark. The reliability of the results must be interpreted with a view to the condition present at the time of testing (like water content).



CHAPTER THREE

LATERAL EARTH PRESSURE AND RETAINING WALLS

1. Introduction Soil is neither a solid nor a liquid, but it exhibits some of the characteristics of both. One of the characteristics similar to that of a liquid is that of its tendency to exert a lateral pressure (earth pressure) against any object in contact. Thus structures which retain or support soil like retaining walls, abutments, sheet pile walls, basement walls and under ground conduits need estimation of the lateral pressure for their design. Various types of retaining walls are shown in Fig 3.1 and are widely employed in civil engineering works ranging from their use in road and rail construction to support cuts and fills where space is limited to prevent the formation of appropriate side slopes, to the construction of marine structures such as docks, harbours and jetties.

Fig 3.1 Types of Earth retaining structures

The design of a retaining wall requires the determination of the pressures which act on it. These are influenced by:

The physical property of the soil (, and C) The time dependent nature of the soil strength (for claysdrainage) The interaction between the soil and the retaining structure at the interface () The type of wall and the degree and mode of wall movement The imposed loading (e.g. height of back fill, surcharge loads)

2. Lateral Earth Pressures

Let us consider a retaining wall which holds back a mass of soil. The soil exerts a push against the wall by virtue of its tendency to slip laterally and seek its natural slope or angle of repose, thus making the wall to move slightly away from the backfilled soil mass. This kind of pressure is known as the active earth pressure of the soil. The soil being the actuating element is considered to be active and hence the name active earth pressure. Next, let us imagine that in some manner the retaining wall is caused to move toward the soil. In such a case the retaining wall or is the actuating element and the soil



provides the resistance for maintaining stability. The pressure or resistance which soil develops in response to movement of the structure toward it is called the passive earth pressure which may be very much greater than the active earth pressure. The surface over which the sheared-off soil wedge tends to slide is referred to as the surface of sliding or rupture. Fig 3.2 depicts the relationship between the earth pressure and the wall movement. Po represents the magnitude of pressure when no movement of the retaining wall takes place; it is commonly referred to as earth pressure at rest. As the wall moves toward the backfill, the pressure increases, reaching a maximum value of Pp (passive earth pressure) at point C. On the other hand, if the wall moves away from the backfill, the force decreases, reaching a minimum value of Pa (active earth pressure) at point B.

Fig 3.2 Relationship between earth pressure and wall movement

Very little movement (about 0.5% horizontal strain) is required to mobilize the active pressure; however, relatively much larger movement (about 2% of horizontal strain for dense sands as high as 15% for loose sands) may be required to mobilize full passive resistance (Lambe and Whitman, 1969).About 50% of the passive resistance may be mobilized at a moment comparable to that required for the active case. HL\30-1.doc

The relative magnitude of the active and passive earth pressures may perhaps be better illustrated with the aid of Fig 3.3. For the sake of simplicity, several specific assumptions are made. Deviations from these will be discussed in the next sections.

1. Frictional forces between backfill and retaining wall are assumed negligible 2. The wall is vertical, and the surface of the backfill is horizontal 3. The backfill is homogenous, granular material 4. The failure surface is assumed to be a plane

Fig 3.3 Relationship between earth pressure and wall movement for cohesionless soil The magnitude of the active and passive forces Pa and Pp could be derived from the basic condition of static equilibrium as follows.



Case of Active Pressure: From Fig 3.3(a),

( )( )

-tancot2

21

aP

cot221)cot(

21W-WtanaP

H

HHHbut

====

The maximum value of Pa may be obtained when cPa /c=0. Thus, ( ) ( )

==

=+==

=+=

2452tanaK :whereaK

2H21

aP

(3.1)2

452tan2H21

aP

have wePa,for equation in theback thisngsubstituti2

45cr

02cosectan2seccot2H21

aP

Case of Passive Pressure: From Fig 3.3(b), ( )

( )

+===+=

tancot221

pP

cot221)cot(

21WWtanpP

H

HHHbut

The maximum value of Pp may be obtained when cPp /c=0. Thus, ( ) ( )

+==

+===

=+++=

245tanK :whereKH

21P 2pp

2p

)2.3(2

452tan2H21

:have we,Pfor equation in theback thisngsubstituti2

45cr

02costan2seccot2H21

pP

aP

p

ec

Ka and Kp are generally referred to as coefficients for active and passive pressure, respectively. They are constants for any given soil where = constant. It is clear that the value of Kp is significantly larger than Ka.

3. Earth Pressure at Rest Earth pressure at rest may be obtained theoretically from the theory of elasticity applied to an element of soil, remembering that the lateral strain of the element is zero. Referring to Fig 3.4 (a), the principal stresses acting on an element of soil situated at a depth z from the surface in semi-infinite, elastic, homogenous and isotropic soil mass are v, h and h as shown; v and h denoting the stresses in the vertical and horizontal directions respectively.

Fig 3.4: Stress conditions relating to earth pressure at rest



The soil deforms vertically under its self weight but is prevented from deforming laterally because of an infinite extent in all lateral directions. Let E and be the modulus of elasticity and Poissons ratio of the soil respectively.

rest.at pressureearth oft coefficien theasknown is K K 1

0E

E

E

strain, Lateral

oov

h

hvhh

wherev

v

v

==

=

+=

Ko is the ratio of the intensity of the lateral earth pressure at rest to the vertical stress at a specified depth. As the vertical pressure at any depth is v=z, then h=Ko v = Ko z. The distribution of the earth pressure at rest with depth is linear (hydrostatic nature) for constant properties such as E,, and , as shown in Fig 3.4 (b) If a structure such as a retaining wall of height H is interposed from the surface and imagined to be held without yield, the total thrust on the wall per unit length Po is given by 2

o00

o ..K 21..P HdzzKdz

H

o

H

h === . This is considered to act at H/3 above the base of the wall. Choosing an appropriate value for the Poissons ratio,, is by no means easy; this is the limitation in arriving at Ko from the equation Ko=[/(1-)]. However, various researchers proposed empirical relationships for Ko, some of which are given below:

Ko = (1-sin') [Jaky, 1944] Ko =0.9 (1-sin') [Fraser, 1957] Ko = 0.19 + 0.233logIp [Kenney, 1959] Ko = [1+ (2/3) sin'] [(1- sin)/(1+sin)] [Kezdi,1962] Ko = (0.95-sin') [Booker and Ireland, 1965]

' in these equations represents the effective angle of friction of the soil and Ip the plasticity index. HL\30-1.doc 30-2

4. Earth Pressure Theories The magnitude of earth pressure is evaluated by the application of one or the other of the so-called lateral earth pressure theories or simply earth pressure theories. Several investigators have proposed many theories of earth pressure after a lot of experimental and theoretical work. Coulombs and Rankines are perhaps the two best-known theories and are frequently referred to as the classical earth pressure theories.

Rankines Theory Rankine (1857) developed his theory of lateral earth pressure when the backfill consists of dry, cohesionless soil. He has made the following important assumptions.

1) The soil mass is semi-infinite, homogenous, dry and cohesionless 2) The ground surface is plane which may be horizontal or inclined 3) The face of the wall in contact with the backfill is