Notes 1_Shear Strength(8)

Geotechnical Engineering 367 Dr Mohamed Shahin Curtin University Page 1

Shear Strength of Soils


Introduction

Shear strength of soils is the soil ability to resist sliding along internal surfaces within a soil mass. These internal surfaces are called slip surfaces or failure surfaces. The shear strength of soils is thus one of the most important aspects that geotechnical engineers must be able to calculate properly because if the applied stress exceeds the shear strength of soils, failure occurs. The principal design situations addressed by geotechnical engineers regarding the shear strength of soils are the bearing (or load-carrying) capacity of foundations, earth pressures on retaining walls and stability of slopes. The failure modes that govern these situations are shown in Figure 1.

Figure 1: Modes of failure for different geotechnical engineering structures (Wesley, 2010)


Concept of Shear Strength The shear strength (or resistance) of soils is the result of two main components:

frictional strength and cohesive strength.

The frictional strength is similar to the classic sliding friction between solid bodies from basic statics or physics. In Figure 2(a), consider two blocks sliding over each other on a horizontal plane. The coefficient of friction between the two blocks is . The upper block is subjected to a constant normal force, N, and the force, F, is producing sliding that will be resisted by a force T that is equal to the normal force, N, multiplied by the friction coefficient , i.e. T = N . If A is the overall contact area of the two blocks, the shear strength, s, may be written as:

nA

N

A

Ts (1)

where, n is the normal stress.

If the two blocks are made of the same material, as the case with soils (see

Figure 2b), the coefficient of friction, , may be replaced by the soil friction

angle, , and Equation (1) can be re-written as:

tanns (2)


Figure 2: Illustration of friction resistance

F or

N or n

T or s

(a)

n

s

(b)

The basic concept of friction as explained previously can be applied to soils

that are purely granular in nature (e.g. sands and gravels). The friction resistance

in such soils is produced by the interlocking of the soil particles.

In cohesive soils, however, there is very little, if any, interlocking between

particles and the resistance to shear is developed primarily due to cohesion or

cohesive strength. Cohesion is the molecular forces of attraction between soil

particles and is analogous to that which would be exhibited between two sticky

surfaces. Cohesion is usually described by the variable c.

In mixed grained soils, the resistance is a combination of the separate resistance

of cohesion and friction, and the Mohr-Coulomb general expression, which is

world widely used, can give the relationship between the soil shear strength at

failure and the two resisting components of friction and cohesion, as follows:


tanncs (3)

where:

s = shear strength of soil;

c = soil cohesion;

= soil friction angle; and n = normal stress on the plane of sliding.

If the shear stress on any plane within a soil mass exceeds the value given in Equation (3), movement (or yield) will occur on that plane. The parameters c and are called the shear strength parameters.

If a series of laboratory tests are carried out in which a number of soil specimens are subjected to different values of normal stresses (e.g. n1, n2, n3), the specimens will fail at different values of shear stresses (1, 2, 3) and the results can be represented graphically as shown in Figure 3. The straight line in Figure 3 is called the Mohr-Coulomb failure envelope and is mathematically represented by the Mohr-Coulomb failure criterion expressed in Equation (3).


n

s

n

tanncs

(n1, 1)

(n2, 2)

Figure 3: Mohr-Coulomb failure envelope and criterion

tanncs c

tann

c

(n3, 3)


The significance of the Mohr-Coulomb failure envelope shown in Figure 3 can be explained as follows (see Figure 4):

If the normal stress and shear stress on a plane in a soil mass are such they plot as point A, shear failure will not occur along that plane.

At point B, shear failure will occur along the plane.

A state of stress on a plane represented by point C cannot exist because it plots above the failure envelope and shear failure in a soil would have occurred already.

n

A

B

C

Figure 4: Significance of Mohr-Coulomb failure envelope


In general, the shear strength for different kinds of soils, as determined by the Mohr-Coulomb failure criterion, can be illustrated graphically as in Figure 5.

Figure 5: Mohr-Coulomb failure envelopes for different types of soil

n

c

cs For pure clays and silts

tanns

For clean sands or gravels

For mixed-grained soils

tanncs


The previous discussion has been made on consideration of total stresses where the pore water pressures are not considered. As described in Geotechnical Engineering 268, soil behaviour is rather governed by the effective stresses. This is to say that at any plane passing through a fully saturated soil, the total normal stress, n, applied to the plane is, in general, the sum of two components: stress carried by the solid particles, n, and pore water pressure in the voids, u, that is:

unn ' (4)

This implies that, in terms of effective stresses, Equation (3) of the Mohr-Coulomb failure criterion should be rewritten as follows:

where:

s = shear strength of soil;

c = soil cohesion in terms of effective stress;

= soil friction angle in terms of effective stress; n = effective normal stress on the plane of sliding; and

u = pore water pressure on the plane of sliding.

'tan'' ncs (5)

unn 'or


Drained versus Undrained Shear Strength

To obtain the shear strength on a plane within a soil mass in terms of the effective stress, we simply need the shear strength parameters c and , as can be seen in Equation (5). This shear strength is also the drained shear strength on this plane, provided that the soil is sheared sufficiently slow that any pore water pressures generated by the shearing process is dissipated as shearing occurs.

The situation with respect to the undrained shear strength on the same plane is more complicated. The undrained shear strength is by definition the strength available on the plane when it is sheared to failure under undrained conditions, as the case in which an embankment is built quickly above a fully saturated ground clay. In this case, the total stress condition applies and Equation (3) can be used to obtain the shear strength, provided that the shear strength parameters c and are obtained. However, the values of c and are not of practical relevance because in undrained condition, the soil behaves as if it has no frictional component of shear strength and the values of c and become:

c (or cu) = su (undrained shear strength of soil)

(or u) = 0

In reality, however, the soil still has a frictional component of shear strength dependent on its value.


Measurement of Shear Strength Parameters

There are several methods for determining the shear strength parameters of soils, discussed here are the following:

1. Direct shear test;

2. Triaxial compression test;

3. Unconfined compression test; and

4. Vane shear test.

The first three tests are conducted in the laboratory, whereas the fourth test can be carried out either in the laboratory or in the field. The above tests are made on representative samples of soil with loading and drainage conditions approximating those in the field, where possible.

Great care should be taken in obtaining, packaging and transporting the soil samples from the site to the laboratory so that the in-situ structure, density and moisture content can be preserved.


Direct Shear Test

The direct shear test is probably the oldest and simplest shear strength test which is normally used for cohesionless soils, but can also be used for cohesive soils.

The apparatus is shown in Figure 7(a) and basically consists of a metal specimen container, or shear box (60 mm square), which is separated horizontally in two halves. One half is fixed, and the other is either pulled or pushed. The soil specimen (generally is about 25 to 30 mm high) is placed into the shear box surrounded by porous stones at the top and bottom to allow for drainage. A normal (vertical) load, P, is applied to the specimen in the shear box through a rigid loading cap. The specimen is allowed to consolidate under this load, which is kept constant during the test. Once the soil has been fully consolidated, a horizontal force, T, is applied by moving one half of the box relative to the other until the specimen fails under shear. During deformation, the horizontal shearing force, T, is measured as well as the lateral deformation, L, and vertical deformation, h. The test is carried out for a number of applied normal stresses, n (where n = P/A and A is the initial cross-sectional area of the sheared specimen) and the shear stresses, , at each normal stress are measured. The shear stress is calculated as the shearing force, T, divided by the corrected cross-sectional area of the sheared specimen, Ac = L (L L) and L is the initial length of sheared specimen.

The following plots are drawn from the results of the test:

(a) shear stress, = T/Ac, versus shear strain, s = L /L (Figure 7b);

(b) vertical displacement, h, versus lateral displacement, L (Figure 7c); and

(c) shear stress, , versus normal stress, n (Figure 7d).

Geotechnical Engineering 367 Dr Mohamed Shahin Curtin University Page 13 Figure 7: Direct shear test: (a) apparatus; (b) and (c) test results; (d) Mohr-Coulomb diagram

(a)

= T/Ac

1

3

2

Test 1 (at n1 = P1/A)



n1

n2 n3

L

Com

pre

ssio

n

Dil

atio

n

(c)

h

P

T

h

L

Photo by ELE International

Schematic diagram

n

Test 1

Test 2

Test 3

(d)

c

s = L/L

(b)

Soil

sample Porous

stones

E

1

1

3 2

n1 n3 n2


From Figure 7b, Youngs modulus of elasticity, E, at different normal stresses can be obtained, and from the results of the maximum shear stresses, the shear strength parameters are determined, as shown in Figure 7d.

Depending on the equipment used, the shear test can be either stress-controlled or strain-controlled. In stress controlled tests, the shear force is applied in equal increments until the specimen fails. After the application of each incremental load, the shear displacement is measured from the horizontal dial gauge. In strain-controlled tests, a constant rate of shear displacement is applied to one half of the box by a motor that acts through gears. The resisting shear force of the soil corresponding to any shear displacement can be measured by a horizontal proving ring or load cell.

Usually drained tests are performed in the direct shear box apparatus. For cohesionless soils, the test can be performed quickly since water does not significantly affect the strength. For cohesive soils, the rate of shearing must be chosen to prevent the build up of excess pore water pressures. Since there are no excess pore water pressure and the pore pressure is approximately zero, the total and effective stresses will be identical and thus, the effective shear strength parameters c and are obtained.


Advantages and disadvantages of direct shear test

Whilst the direct shear test is very simple, it suffers from a number of limitations, including:

Drainage cannot be controlled and pore water pressures cannot be measured, thus, undrained tests are not possible;

Failure is forced along a predetermined plane;

Shear stress along the failure plane is not uniform, stress concentrations exist, and failure occurs progressively from the edges towards the centre; and

Despite these limitations, the test is useful because of its simplicity, relatively low cost and ease of sample preparation.


A series of three direct shear tests has been conducted on a saturated soil using a

shear box of cross-sectional area of 100 cm2. The tests were performed slowly

enough to produce drained conditions under applied vertical loads of 1, 2 and 4

kN. Failure occurred at shear loads of 0.7, 1.1 and 2.0 kN. Determine the shear

strength parameters of this soil. [Answers: c = 30 kPa and = 22o]

Worked Example (1)


Triaxial Compression Test A more useful test than the direct shear is the triaxial test, which is the most reliable and

widely used shear strength test that is suitable for almost all types of soils. As shown in Figure 8(a), the test consists of a cylindrical soil specimen (about 38 mm in diameter and 76 mm high), and on special occasions, larger diameter samples are used. The specimen is wrapped with an impermeable rubber membrane, and o-rings are used at the top and bottom to provide a watertight seal, thus allowing drainage from only the top and bottom sides of the soil sample. The membrane prevents the cell water from penetrating the pores of the soil. The sample is placed on a metal pedestal (with provisions for drainage and pore water pressure measurements) and is encased inside a plastic cylindrical chamber that is filled with water.

The specimen is first saturated and then subjected to equal all-round confining (cell) pressure, 3 (see Figure 8b) by the compression of water in the chamber and this cell pressure is usually held constant during the test. To cause shear failure, an axial load P (see Figure 8a) is steadily applied to the specimen through a vertical loading ram until failure occurs. The axial load P causes axial stress d = P/Ac (Ac is the corrected cross-sectional area of soil specimen) known as the principal stress difference or deviator stress (see Figure 8b). The stresses at failure applied to the soil specimen will then be the horizontal cell (or confining) pressure 3 and vertical stress 1. These stresses are principal stresses since there is no shear stress applied on the planes of 1 and 3. In calculating the deviator stress, d, the change in the cross-sectional area of soil specimen during the test must be taken into account as follows:

where, A is the initial cross-sectional area of the soil specimen; a = L/L is the vertical

)1(

)1(

a

voc

AA

(6)


strain; and v = V/V or (a + 2l) is the volumetric strain. l = D/D is the lateral strain, and L, L, V, V, D and D are, respectively, the vertical deformation, initial length, volume change, initial volume, lateral deformation and initial diameter of the soil specimen (see Fig. 8c).

The test is performed a number of times on several specimens of the same soil using different initial cell pressures, 3, and the corresponding deviator stresses, d, are determined (see Figure 9a) so that the axial stresses at failure are calculated, i.e. 1 = 3 + d. The results are used to obtain the shear strength parameters using the Mohrs circles of stress, as will be explained later. The relationship between the axial strain, a, and volumetric strain, v, are also used to get an idea of whether the soil is contracting or dilating during shearing (an example of a contracting soil is shown in Figure 9b).

Drainage can be permitted via porous stones placed on top and bottom of the soil specimen and through tubes connected to the top and base of the specimen. The pore water pressure can be measured via pressure gauges connected to the tubes. There are basically three types of triaxial test which are commonly assigned by a two-letter symbol. The first letter refers to what happens before shearing (i.e. whether the specimen is consolidated or not), and the second letter to the drainage during shearing. These tests include:

ConsolidatedDrained CD

ConsolidatedUndrained CU

UnconsolidatedUndrained UU

CD or CU test is carried out to determine the effective shear strength parameters c and , while the UU test is carried out to determine the undrained shear strength of soil, su or cu. The three tests will be explained in detail later.

Geotechnical Engineering 367 Dr Mohamed Shahin Curtin University Page 19 Figure 8: Triaxial compression (a) test apparatus; (b) stress system; (c) change in sample dimensions at failure

Photo by ELE International

3

d 3

d

3

1

3

1 (b)

Drainage and

back pressure Cell pressure

Rubber

membrane

Cell water

Porous stones

Plastic

cylindrical

chamber

Axial load, P

Soil

(a) A Ac

L

L

(c)

O-ring

D/2 D D/2


Figure 9: Triaxial compression test results (a) deviator stress versus axial strain; (b) volumetric

strain versus axial strain

Test 1

Test 2

Test 3

Com

pre

ssio

n (

+ve)

D

iala

tion

(-v

e)

a v

(b)

d

a

Test 1 at 3-1

Test 3 at 3-3

Test 2 at 3-2

(a)

d-1

d-2

d-3


Mohrs Circle of Stress To obtain the Mohr-Coulomb failure line using the triaxial test results (and thus the values

of the shear strength parameters), it is normal practice to make use of a graphical solution known as the Mohr circle of stress. Figure 10 shows state of stress on a soil element within a soil mass where the major and minor principal stresses are acting in the vertical and horizontal directions, respectively. Generally, the principal stress with the largest magnitude is denoted 1 and called the major principal stress, and the smallest principal stress is denoted 3 and called the minor principal stress. The stress in the third dimension is the intermediate principal stress, 2 , and is neglected in our analysis since most problems in geotechnical engineering exhibit two-dimensional (plane strain) conditions. The state of stress in Figure 10 is similar to that of a soil specimen in a triaxial test. We now need to know the stress state on any plane making an angle to the horizontal since failure cannot occur on either the vertical or horizontal plane. The stresses acting on the plane EE making an angle to the horizontal consist of two components: a stress perpendicular to the plane, known as the normal stress, n, and a stress acting parallel to the plane, known as the shear stress, . These stresses can be obtained by analysing the equilibrium condition of the soil element and considering the stresses acting on the faces of the triangle FHG and applying simple statics, as follows:

Figure 10: Illustration of stress state at a point in a soil mass

1

E

3 3

E

1

B

A D

C

F

G

n

F H

G

3n

1

H


Resolving forces normal to the plane FG gives:

Dividing through FG gives:

Therefore:

Rearranging gives:

sincos 31 GHFHFGn

sincos 31FG

GH

FG

FHn

232

1 sincos n

2cos22

3131

n (7)

Now:

1cos22cos 2 and 2sin212cos


Resolving forces parallel to the plane FG gives:

Dividing through FG will give:

cossin 31 GHFHFG

cossin 31FG

GH

FG

FH

cossin)( 31

2sin2

31

(8)

or

Since sin2 = 2 sin cos, therefore:

In Equation (9), when = 45o, = max =1/2(13)


A simple graphical solution using the Mohrs circle of stress can be used to obtain the stresses of the analytical Equations 7 and 8, as follows (refer to Figure 11):

Draw the perpendicular axes x and y to represent n and , respectively, and mark off OA = 3 (the minor principal stress); and OB = 1 (the major principal stress).

Draw a circle of diameter AB whose centre lies on the x-axis at C (only half of the circle is needed because of symmetry around the horizontal axis).

Draw a line making an angle to the horizontal from Point A to intersect the circle at Point P. We will examine the magnitudes of OD and PD, as follows:

That is:

Also:

2sin2

2sinOAOB

CPPD

(9)

2cos2

CPOBOA

CDOCOD

2cos22

3131

n


Figure 11: Graphical representation of stress using Mohrs circle

n

3

max

n

B D A C

P

O

Soil element

1

3

3

1

Mohr Circle

1

2

Thus:

Equations 9 and 10 are identical to those obtain from the analytical derivation of

Equations 7 and 8. Thus the distance OD gives the value of n and DP gives the

value of on a plane inclined at an angle from the horizontal. Note that the

maximum shear stress, max, can be also obtained from the Mohrs circle, as

indicated in Figure 11.

(10)

n

2sin2

31


Mohr circles of stress as explained previously can be used for plotting the triaxial test results from which the shear strength parameters can be obtained. By doing a series of triaxial tests at different cell pressures, we can plot a series of Mohr circles as shown in Figure 12. The tangent to these circles defines the Mohr-Coulomb failure envelope. No state of stress can exist which would be represented by a circle that crosses this line as failure would occur before this could happen. The intercept of the failure envelope to y-axis gives the soil cohesion, c, and the inclination of the failure envelope to x-axis gives the soil friction angle, .

Figure 12: Graphical representation of triaxial test results using Mohrs circles of stress

n 3 1 3 3 1 1

Test 1

Test 2

Test 3

c


There are situations where it is useful to express the Mohr-Coulomb failure criterion in terms of principal stresses. This involves obtaining a relationship between the principal stresses at failure, 1 and 3, and shear strength parameters c and . This relationship can be obtained by referring to the Mohr circle in Figure 13. In the triangle ACD:

however:

Substituting the values of DC and AC gives:

Rearranging yields:

Which from the trigonomeric identities leads to:

AC

DCsin

2

31 DC2

cot 31

cBCABACand

2/)(cot

2/)(sin

31

31

c

2/)(cot

2/)(sin

31

31

c

)2/45tan(2)2/45(tan231 c (11)


Figure 13: Mohr circle and failure envelope for an element of soil

3 1

C

D

A

c

B n

Also, from Figure 13, the sum of the interior angles of triangle ACD is 180o

and thus: (180o 2f) + 90o + = 180o, therefore, the angle of failure plane

to the horizontal can be expressed as:

(12)

245

f

f f2

Soil element

3

3

1

f

1


In the figure below, determine the effective shear and normal stresses applied on

the failure surface of a soil element located at point (A), and check the stability of

the soil element against shear failure assuming that the applied vertical and

horizontal stresses are principal stresses. [Answer: FoSagianst shear failure = 2.4]

Worked Example (2)

Potential shear surface

A

50o

4m

5m c = 20 Kpa

= 30o

d = 15 kN/m3

sat = 18 kN/m3

Ko = 0.5


This test is used to obtain the effective shear strength parameters c and . In this test, the soil specimen is first fully saturated by applying simultaneous increments of cell pressure (by compression of cell water) and back pressure (by compression of water inside the soil sample), while the drainage valves are open. This process will ensure that the air which may be present inside the voids is dissolved. The effectiveness of the saturation process shall be checked by closing all drainage valves, increasing the cell pressure by a small amount, c, and measuring the consequent excess pore pressure, uc. Skemptons parameter B is then determined (B = uc/ c), and if the specimen is fully saturated, B should be close to 1. A value of B 0.95 is accepted and when B < 0.95, the above saturation process shall be repeated until B 0.95 is reached.

With all drainage valves open, the required confining pressure, 3, is applied and the consequent excess pore water pressure, u3, is allowed to dissipate over time until consolidation occurs. The full consolidation will take place when u3 becomes zero (Figure 14a), and then the C part of the CD test is complete.

The next phase of the test is shearing the soil specimen to failure. This phase starts with keeping the drainage valves open, and applying an increasing axial deviator stress d = 1 3 until the specimen fails. d is calculated by dividing the axial deviator load by the corrected area of soil sample obtained from Eq. (6). The axial deviator load should be applied very slowly so that no excess pore water pressure develops during shearing. The vertical and horizontal stresses on the sample are then the effective major and minor principal stresses (i.e. 1 = 1 and 3 = 3), as shown in Figure 14b. Once shearing is done, the D part of the CD test is complete.

Consolidated-Drained Test (CD)


Changing 3 allows several tests of this type to be conducted on samples of the same soil and the effective shear strength parameters (c and ) can be determined by plotting the Mohr circles at failure, as shown in Figure 14c.

The CD test simulates the condition where a soil has been consolidated due to some prior loading, and then loaded to failure so slowly so that pore water pressures will not develop or will dissipate quickly. In this case, the soil strength parameters needed for design are c and , and the analysis is referred to as the effective stress analysis or long-term stress analysis. Practical example is an embankment constructed slowly subsequent to consolidation under its original weight.

The CD test on clay takes several days which leads to practical problems in the laboratory such as leakage of valves, seals, and the membrane that surrounds the sample. An alternative method is the CU test where the induced pore water pressure can be measured.


(a) Part C

3

3

3 3 u3 = 0

Figure 14: Schematic diagram of the stress conditions of the CD test with typical Mohr-Coulomb failure envelope

n c

3 1 1 3 3 1

(b) Part D

d

d

1 = 1

1 = 1

ud = 0 3 = 3

3

3

3 = 3

(c)


This test is an alternative quicker way of obtaining the effective shear strength parameters, c and . In this test, the consolidation phase (part C of the test) is done as in the CD test. The shearing phase is started by applying an increased deviator stress, d, until failure occurs. During this phase, the drainage from the specimen is prevented and an excess pore water pressure, ud, is developed and measured, thus, both the total and effective stresses at failure can be determined as follows (see Figure 15b):

Minor total principal stress = 3 Major total principal stress = 1 = 3 + d

Minor effective principal stress = 3 = 3 ud Major effective principal stress = 1 = 1 ud Note that : d = 1 3 = 1 3

Consolidated-Undrained Test (CU)

Figure 15: Schematic diagram of the stress conditions of the CU test

(a) Part C (b) Part U

d

d

1 1 3

3 1 1

ud 0

3

3

3 3 u3= 0 3 3 3 3


The deviator stress, d , at failure and corresponding excess pore water pressure, ud , are used to determine the Skemptons pore water pressure parameter A as follows:

A = ud/d

The Skemptons parameter A indicates whether the soil specimen during shearing will compress or dilate. Normally consolidated clays and loose sands tend to contract and give A value close to 1, whereas overconsolidated clays and dense sands tend to dilate and give very low or negative A value. The following A values can be used as a guide:

A > 0.5 (soil that contracts during shearing)

A < 0.5 (soil that dilates during shearing).

Several tests of this type are to be conducted on samples of the same soil and the effective shear strength parameters, c and , are determined by plotting Mohrs circles at failure, as shown in Figure 16.

It should be noted that during the undrained shearing, the sample water content remains unchanged as drainage is not allowed, thus, there would be no volume change in the soil specimen. This will exhibit zero volumetric strain v and Equation (7) for the calculation of the corrected area of sheared specimen simplifies to:

)1( ac

AA

(13)


Figure 16: Typical Mohr-Coulomb failure envelopes for the CU test

c 3 1

n

ud1 ud2

3 1

ud3

1 3 3 1 1 3 1 3

Effective stress circles

Total stress circles


In this test, the soil specimen is subjected to a cell pressure, 3, while drainage from the soil specimen is not permitted during the application of 3. Without being allowed to consolidate, the specimen is sheared to failure by the application of deviator stress, d, with no drainage is allowed (see Figure 17). Since drainage is not permitted at any stage, the test can be performed very quickly (usually takes about 10 to 20 minutes).

Unconsolidated-Undrained Test (UU)

(a) First U Part (b) Second U Part

d

d

3

3 3

3

3

3

1 1

1 1

ud 0 u3 0

Figure 17: Schematic diagram of the stress conditions of the UU test

3 3 3 3


Failure envelope u = 0.0

ud3

n 3 1 3 3 1 1

su = cu

3 1

ud2

ud1

Figure 18: Typical Mohr-Coulomb failure envelope for a UU test

The pore water pressure is not measured in this test, thus, only the total stresses are known and only a total stress plot is possible as shown in Figure 18. It can be seen that the cohesion is undrained, cu, and is equal to the undrained shear strength, su = (13)/2, and the friction angle is undrained , u , and is zero.

A special feature of this test is that an increased cell pressures, 3, will not increase the undrained shear strength, su , or the effective stresses, 1 and 3 , but rather will only increase the pore water pressures. This is because water is not permitted to dissipate from the specimens and since all samples will have the same density and moisture content during the test, all samples will have the same strength. If we were to measure the pore water pressure in each test and then calculate the effective stresses and plot the effective stress circles, we would find that these circles coincide, as shown in Figure 18. For this reason, only one test is sufficient to perform a UU test.

The UU test simulates a situation where a construction (e.g. embankment or foundation) is build on a soft deposit over a short period of time (a few weeks).

Effective stress circles

coincide in one circle

Total stress circles


The triaxial test has a number of significant advantages, including:

Drainage, that is, pore water pressure can be controlled and measured

Different combinations of cell pressure and axial stress that simulate field

conditions can be applied, and failure can occur anywhere in the specimen

Specimens are subjected to (approximately) uniform stresses and strains

This test is by far the most popular shear strength test, and whilst it overcomes

many of the limitations of the direct shear test, it is complex and requires

expensive equipment.

Advantages and Disadvantages of Triaxial Test


Worked Example (3)

The results shown in the table below were obtained at failure in a series of consolidated-undrained triaxial tests (with pore water pressure measurement) that are carried out on fully saturated soil specimens of 40 mm in diameter and 89 mm in height. Determine the values of the shear strength parameters.

[Answer s: c = 17 kPa and = 29o]

Cell pressure

(kPa)

Deviator force

(N)

Vertical deformation

(mm)

Pore pressure

(kPa)

30 128.1 3.1 6.6

80 194.7 4.9 32.3

140 254.0 7.8 72.8


Unconfined Compression Test

This test is one of the quickest and simplest shear strength tests, and is a special case of the unconsolidated-undrained test in which the all-round pressure is equal to the atmospheric pressure, i.e. 3 = 0 (see Figure 19a).

The test is an undrained test and carried out on samples that can stand without any lateral support, hence, it is applicable to cohesive soils only.

Specimens of height to diameter ratio of 2 are normally used and subjected to a rapid vertical stress, d = 1, until failure occurs. The vertical stress at any stage of loading can be obtained by dividing the vertical load, P, by the cross-sectional area of the sample.

During the test, the cross sectional area increases with the increase in compression and may be computed at any stage of loading as follows (refer to Figure 19b):

)/1( LL

AAc

(14)


Figure 19: Unconfined compression tests (a) photo of the test equipment by ELE International;

(b) sample dimensions during test; (c) stress-strain curve; (d) Mohr circle

su = cu = qu/2

n 3 = 0.0 1 = qu

Failure envelope u = 0.0

cu

(a)

(c) (d)

A Ac

L

L

(b)

a

qu

d = 1


By computing the normal stress, 1, and axial strain, a = L/L, at different stages of loading, stress-strain curve may be plotted (Figure 19c). The peak value of the stress-strain curve is taken as the unconfined compressive strength, i.e. qu= 1-failure. The corresponding Mohr-Coulomb failure envelope is illustrated in Figure 19d, where it can be seen that the slope of the failure envelope is horizontal, i.e. u = 0 and the undrained shear strength is calculated as:

2/uuu qcs (15)

Theoretically, the unconfined compression test should lead to the same value of cu as this obtained from the UU triaxial test if:

The specimen is 100% saturated, otherwise, compression of the air in the voids will occur and will cause a decrease in void ratio and increase in strength; and

The specimen is sheared rapidly to failure so that the undrained condition exists. If the time to failure is long, evaporation and surface drainage will lead to higher strength.


During unconsolidated-undrained triaxial test on a soil specimen, the minor and

major principal stresses at failure were 96 kPa and 187 kPa, respectively. What

will be the deviator load at failure for a specimen of the same soil that has a

diameter = 50 mm and height = 100 mm which is tested in the unconfined

compression apparatus. The axial deformation of the soil sample at failure in the

unconfined compression test was 6.5 mm. [Answer: P = 191 N]

Worked Example (4)


Vane Shear Test

The vane shear test (Figure 20) is widely used for cohesive soils, particularly sensitive clays. The test consists of a steel rod having, at one end, four small projecting blades (or vanes) parallel to its axis, and situated at 90o around the rod (see Figure 20a). The vane is inserted into the soil and a torque, T, is applied until the soil fails, thus, generating a cylinder of soil. At failure, the applied torque will be equal to the resisting moment caused by the shear at the surface of the generating cylinder, therefore:

where, T = applied torque; d = diameter of vane; and h = length of vane (see Figure 20b).

)67.02(

432 dhd

Tsu

(16)

Bjerrum (1973) back computed a number of embankments failure on soft clays in which the vane undrained shear strength was used and concluded that the vane shear tended to be too high and should be corrected for the design purposes, as follows:

)()( vanesdesigns uu (17)

where, is a reduction factor that has been related to the plasticity index,

Ip, as in Figure (20c).


Figure 20: The vane shear test (a) photo of the vane device; (b) schematic diagram of the vane;

(c) vane shear correction curve of Bjerrum (1973)

(a)

(b)

h

d

T

(c)


At a depth of 6 m below the ground surface on a certain site, a vane shear test

gave a torque value of 6040 N.cm. The vane was 10 cm high and 7 cm across

the blades. A sample of the same soil was taken to the laboratory and was

found to have LL = 66 and PL = 16. Estimate the design shear strength of this

soil according to Bjerrum (1973). [Answer: su = 51.2 kPa]

Worked Example (5)


Shear Strength Characteristics of Sands

The characteristics of dry and saturated sands are the same provided that there is zero excess pore water pressure in the case of saturated sands.

Typical examples of the shear strength characteristics for loose and dense sands are given in Figure 21.

Figure 21: Typical shear test results for sands

n

Dense

Loose

(a) (c)

v a

Dilat

ion

Co

ntr

acti

on

Dense

Loose

(b)

e

a

Dense

Loose

(d)

d

a

Dense

Loose

Peak

Ultimate Residual


Figure 21(a) shows the deviator stress versus axial strain relationship for dense and loose sands. It can be seen that dense sand reaches a peak stress and then the stress decreases with increasing strain to an ultimate or residual value. This behaviour is known as work-softening and similar to the behaviour of brittle materials. The relatively sharp peak associated with dense sands is related to the degree of interlocking between the soil particles, which must be overcome in the shearing process. Once this interlocking is overcome, the shear stress necessary to continue the shear strain is reduced to the ultimate value. On the other hand, loose sands show a work-hardening behaviour which is similar to the behaviour of ductile materials.

Figure 21(b) shows the relationship between the volumetric strain, v, and axial strain, a. As can be seen, a soil can exhibit expansion (dilation), or contraction (densification). As shearing takes place in a loose sand, the material becomes more dense, as one would expect. Initially, upon shearing, a dense sand increases in density, or decreases in thickness, but upon continued shearing, the soil dilates. This is due to the fact that, for shearing to continue, at high densities, the soil grains must ride over the adjacent grains, causing a corresponding increase in specimen thickness.

Figure 21(c) shows typical Mohr-Coulomb failure envelopes for sands. As expected, the effective internal angle of friction, , increases with increasing density. Notice too that for both the loose and dense sands, the failure envelopes pass through the origin. This is a typical characteristic for clean sands since c = 0.

Figure 21(d) shows the relationship between the void ratio, e, and axial strain. As shearing takes place, the void ratio of loose sand decreases indicating an increase in density, as expected. Initially the dense sand increases in density slightly, but upon further shearing, dilates for the same reason mentioned previously. Notice that as shearing continues, both loose and dense sands approach the same level of void ratio. This is known as the critical void ratio, or the critical state.


Shear Strength Characteristics of Clays

Clay soils are said to be cohesive soils and unlike sands, their behaviour rely on the moisture content and more importantly soil stress history, i.e. whether the soil is normally consolidated or over-consolidated. A normally consolidated clay is the one which its current effective vertical stress is equal to its pre-consolidation pressure (i.e. the maximum effective vertical stress the soil had in its entire life). Over-consolidated clay is the one which is currently applied to an existing effective vertical stress that is less than the pre-consolidation pressure. Figure (22) shows typical results of triaxial tests carried out on normally and overconsolidated clays from which the following observations can be made:

In Figure 22(a), it can be seen that normally consolidated clays exhibit behaviour similar to that of loose sands, whereas overconsolidated clays show behaviour similar to that of dense sands.

Figure 22(b) shows that normally consolidated clays contract during shearing while overconsolidated clays expand, or dilate, during shearing.


Figure 22: Typical shear test results for clays

(a)

v a

Dilat

ion

Co

ntr

acti

on

(b)

Overconsolidated

Normally consolidated

Overconsolidated

Normally consolidated

d

a

Peak

Ultimate Residual


For some clays, the strength in a remoulded or disturbed condition is much

less than that in an undisturbed condition at the same moisture content. A

parameter known as sensitivity is used to indicate the amount of strength

lost by a clay soil as a result of disturbance. The degree of sensitivity, St,

may be defined as the ratio of the undrained shear strength in an

undisturbed state, su(undisturbed) to that in a remoulded state, su(remoulded) as

follows:

(18)

)(

)(

remouldedu

dundisturbeu

ts

sS

The degree of sensitivity of low sensitive clays ranges from 2 to 4,

however, for medium to high sensitive clays, St may range from 4 to 16.

Clays with St greater than 16 turn to viscous fluids upon remoulding and

are known as quick clays. This behaviour occurs because these clays have

a very delicate structure that is disturbed when they are remoulded.


Stress Path in Triaxial Testing The behaviour of soils in the field depends on many factors, including the

magnitude of induced stress, the way the stress changes and the history of loading (whether due to natural reasons such as erosion and sedimentation, or as a result of human changes such as excavation). It is therefore desirable to trace the stress of an element of soil throughout its loading history by applying similar stress changes during the laboratory test on a soil specimen. To keep track of loading progress of a soil element/sample, the Mohr circles can be used, however, an infinite number of Mohr circles would be required, leading to inconvenient representation diagram. A more convenient way of stress representation is called the stress path.

In the stress path, the state of stress is represented by stress points that corresponds to the highest points on the Mohr circles, and the continuous line connecting the stress points is the stress path, as shown in Figure 23. The stress path points have coordinates t and s given by:

2

31 t (19)

2

31 s (20)


Figure 23: Mohr circle of stress and corresponding stress points and stress path

n

t = (1 3)

Stress path points

3 1

s = (1 +3)

t = t

s or s

Stress path

3 1 3 1


It should be noted that t and s may be defined in terms of the total or effective stresses. If we adopt the normal convention of a prime indicating effective stress, then t = t and s = s u .

A simple case to illustrate the stress path in conventional triaxial test is when 3 remains constant as 1 is increased, as shown in Figure 24. In this case, the corresponding stress path is a straight line making an angle of 45o from the horizontal.

Figure 24: (a) successive Mohr circles; (b) corresponding stress pass for constant 3 and increasing 1

t

s

t

s


The stress path plots for a set of triaxial tests can be used to determine the shear strength parameters, without having to draw the Mohr circles. The line drawn through the set of points representing the appropriate failure criterion gives the stress path failure envelope, which it is referred to as the Kf line. The inclination of the Kf line to x-axis and its intercept with y-axis can be used to obtain the shear strength parameters. Figure 25 shows the stress path of three consolidated-undrained triaxial tests in terms of the total stress (TSP) and effective stress (ESP). For each test, the horizontal distance between the failure points represents the excess pore water pressure at failure, ud. The inclination and intercept of the ESP failure envelope can be used to obtain c and , as follows:

)(tansin' 1 (21)

'cos'

ac (22)

Figure 25: Stress path and failure envelope for a set of consolidated undrained triaxial tests

t = t'

s or sTest 1

Total stress path (TSP)

Effective stress path (ESP)

Test 2 Test 3

450

a

Kf line

ud2

ud3

ud1

ud2

ud3

ud1


Worked Example (6)

Consider the following triaxial test on a specimen of clean sand with 5 separate measurements of 1, 3 and ud at failure. Plot the total and effective stress paths and obtain the effective shear strength parameters.

[Answers: c = 0.0 and = 22.6]

1 (kPa) 3 (kPa) ud (kPa)

300 300 100

350 300 165

380 300 200

396 300 224

398 300 232


References:

Bjerrum, L. (1973). Problems of soil mechanics and construction on soft clays Proceedings of the 8th International conference on Soil Mechanics and Foundation Engineering, 3, Moscow.

Wesley, L. D. (2010). Fundamentals of soil mechanics for sedimentary and residual soils John Wiley and Sons, Inc., New Jersey.

Notes 1_Shear Strength(8)

Documents

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concept of shear strength

shear strength of soil

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