Shear strength in soils–Introduction–Definitions

Mohr-Coulomb criterion–Introduction–Lab tests for getting the shear strength

Direct shear test–Introduction–Procedure & calculation–Critical void ratio

Triaxial shear test

The strength of a material is the greatest stress it can sustain;

So that the unit of strength is the same as stress (Pa in SI unit system);

The shear strength of a soil is its resistance to shearing stresses.

It is a measure of the soil resistance to deformation by continuous displacement of its individual soil particles

Shear strength in soils depends primarily on interactions between particles

Shear failure occurs when the stresses between the particles are such that they slide or roll past each other

Soil derives its shear strength from two sources:

1. Cohesion between particles (stress independent component)

o Cementation between sand grains

o Electrostatic attraction between clay particles

2. Frictional resistance between particles (stress dependent component)

Cohesion (c), is a measure of the forces that cement particles of soils

Dry sand with no cementation

Dry sand with some cementation

Soft clay

Stiff clay

Internal Friction angle (φ), is the measure of the shear strength of soils due to friction

Cohesionless soils – If the soil do not have physico-chemical interaction (i.e. c value = 0). The shear strength is derived from internal friction only. Example – sand, gravel

Cohesive soils – These soils exhibit cohesion but the angle of friction is Φ = 0. Example –saturated clays and silts

Cohesive-frictional soils – Composite soils or mixture of fine grained and coarse grained soils. These soils are also called c-Φ soils. Example natural soils, sandy clay, clayey sand, silty sand etc.

Failure happens if the shear stress along the failure surface reaches the soil’s shear strength.

Soil generally fails in shear

failure surface mobilised shear resistance

At failure, shear stress along the failure surface () reaches the shear strength (

f).

This theory states that a material fails because of a critical combination of normal stress and shear stress, and not from their either maximum normal or shear stress alone.

Charles MohrCharles Coulomb

The shear strength parameters of a soil are determined in the lab primarily with two types of tests: 1) Direct Shear Test; and 2) Triaxial Shear Test

Direct shear test is Quick and Inexpensive

Shortcoming is that it fails the soil on a designated plane which may not be the weakest one

Used to determine the shear strength of both cohesive as well as non-cohesive soils

ASTM D 3080

The test equipment consists of a metal box in which the soil specimen is placed

The box is split horizontally into two halves

Vertical force (normal stress) is applied through a metal platen

Shear force is applied by moving one half of the box relative to the other to cause failure in the soil specimen

1.Measure inner side or diameter of shear box and find the area2.Make sure top and bottom halves of shear box are in contact and fixed together.3.Weigh out 150 g of sand.4.Place the soil in three layers in the mold using the funnel. Compact the soil with 20 blows per layer.5.Place cover on top of sand6.Place shear box in machine.7.Apply normal force. The weights to use for the three runs are 2 kg, 4 kg, and 6 kg if the load is applied through a lever arm, or 10 kg, 20 kg, and 30 kg, if the load is applied directly.

Note: Lever arm loading ratio 1:10 (2kg weight = 20 kg)

8. Start the motor with selected speed (0.1 in/min) so that the rate of shearing is at a selected constant rate9. Take the horizontal displacement gauge, vertical Displacement gage and shear load gage readings. Record the readings on the data sheet. 10.Continue taking readings until the horizontal shear load peaks and then falls, or the horizontal displacement reaches 15% of the diameter.

Specimen preparation is easy, the test is simple and convenient

As the thickness of the specimen is relatively small, the drainage is immediate and the pore pressure dissipates very rapidly. Consequently, the consolidated-undrained and the consolidated-drained tests take long period of time

It is ideally suited for conducting drained tests on cohesionless soils

The test apparatus is relatively cheap Samples may be sheared along predetermined

planes. This is useful when the shear strengths along fissures or other selected planes are required.

The stress condition is only known at failure, thus Mohr circle cannot be drawn

The stress distribution on the failure plane is not uniform

The area under shear gradually decreases as the test progresses.

The failure plane is fixed- this plane may not be the weakest plane

Controlled over drainage condition is very difficult – only permeable specimen can be tested on drained conditions

The measurement of pore water pressure is not possible

The side walls of the specimen holder caused lateral restraint on the specimen and do not allow it to deformed laterally.

Since development of the much better triaxial test, the used of the direct shear test has decreased

Developed by Casagrande in an attempt to overcome some of the serious disadvantages of the direct shear test.

Advantages over DST

More Versatile

Drainage can be well controlled

There is no rotation of the principal stresses like the direct shear test

Also the failure plane can occur anywhere

The triaxial compression test is used to measure the shear strength of a soil under controlled drainage conditions

A cylindrical specimen of soil is encased and subjected to a confining fluid/air pressure and then loaded axially to failure.

The test is called "triaxial" because the three principal stresses are assumed to be known and are controlled.

During shear, the major principal stress, σ1 is equal to the applied axial stress (∆σ= P/A) plus the chamber (confining) pressure,σ3

The applied axial stress, σ1-σ3 is termed the "principal stress difference" or sometimes the "deviatory stress“

The intermediate principal stress, σ2 and the minor principal stress, σ3 are identical in the test, and are equal to the confining or chamber pressure

Drained conditions occur when rate at which loads are applied are slow compared to rates at which soil material can drain (k -dependent)

Sands drain fast; therefore under most loading conditions drained conditions exist in sands

Exceptions: pile driving, earthquake loading in fine sands

In clays, drainage does not occur quickly; therefore excess pore water pressure does not dissipate quickly

Therefore, in clays the short-term shear strength may correspond to undrainedconditions

Even in clays, long-term shear strength is estimated assuming drained conditions

There are three types of tests:

1. Unconsolidated-Undrained (UU or Q) Test

2. Consolidated-Undrained (CU or R) Test

3. Consolidated-Drained (CD or S) Test

Complete control over the drainage condition Pore pressure changes and volumetric

changes can be measured directly The stress distribution on the failure plane is

uniform The specimen is free to fail in the weakest

plane The state of stress at all intermediate stages up

to failure is known, Mohr circle can be drawn at any stage during the test

The complete stress-strain-strength behaviorcan be investigated

The apparatus is elaborate, costly and bulky The drained test takes longer period of time as

compared to that in direct shear test The strain condition at both ends of the

specimen are not uniform due to frictional restraint produce by the loading cap and the pedestal disc.

It is not possible to measure the cross sectional area of the specimen at large strains/pressures

The consolidation test in the triaxial is isotropic whereas, the in the field consolidation are generally anistropic

Also called slow test. Drainage valves OPEN during consolidation as well as

shearing phases. Complete sample drainage is achieved prior to

application of the vertical load. The load is applied at such a slow strain rate that

particle readjustments in the specimen do not induce any excess pore pressure. (can take up to 2 weeks)

Since there is no excess pore pressure total stresses will equal effective stresses.

This test simulates long term shear strength for cohesive soils.

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Conduct a CD test on sand.

Soil specimens will be loaded to failure under 3 different confining pressures; 15, 30 and 45 psi

Failure will be defined as the peak or 3 maximum value of principal stress difference reached.

ASTM D 2850

Apply σ3 and wait until the soil consolidates Drainage valves open during consolidation phase but

closed during the shearing phase (Drainage and consolidation is allowed to take place

during the application of the confining pressure σ3) Loading does not commence until the sample ceases to

drain (or consolidate). This test can simulates long term as well as short term shear

strength for cohesive soils if pore water pressure is measured during the shearing phase

For this Test, cT≠c’and φT= φ’ From this test we obtain; c’, φ’ and u (Effective stress) cT, φT (Total stress)

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�′= ������� − ��

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This test is also called the quick test. σ3 and ∆σ are applied fast so the soil does not have time to settle or consolidate.

The test is performed with the drain valve closed for all phases of the test. (Water is not allowed to drain)

UU test simulates short term shear strength for cohesive soils.

For this test, φ =φ’ = 0

s=cu= Su = (σ1-σ3)/2 = (σ’1-σ’3)/2

A special type of UU test that is commonly used for clay specimen.

σ3 = 0

Axial load rapidly applied to the specimens to cause failure.

At failure the total minor principal stress is zero and total major stress is σ1

Vane shear apparatus : consists of four thin, equal sized steel plates welded to a steel torque rod

This test is used to determine shear strength of cohesive soils

Normally done in-situ on a soft clayey soil (cohesive soil)

The vane is pushed into the soil Torque, T is applied at the top of the torque rod to

rotate the vane at a uniform speed A cylinder of soil (height h & diameter d) will resist the

torque until the soil fails

where = coefficient of shear strength mobilization at the end of soil cylinder.

=1/2 @ 2/3 @ 3/5

Bjerrum (1974) showed that as the plasticity of soils increases, cu obtained from VST may give results that are unsafe for foundation design. Therefore, he suggested that some correction should be made :

More recently, Morris and Williams (1994) gave the correlations of λ as :

cu (design) = cu (VST)

where = correction factor = 1.7 - 0.54log(PI)

= 1.18e-0.08(PI) + 0.57 (for PI 5) = 7.01e-0.08(LL) + 0.57 (for LL 20)

1. Ring shear test

2. Cone penetration test

(ASTM D3441 and D5778)

3. Standard penetration test

(ASTM D1586)

4. Penetrometer test

Fieldtesting

Laboratory testing

A convenient way of plotting triaxial test data is through diagrams called stress paths.

Stress path = line that connects a series of point, each of which represents a successive stress state experienced by a soil specimen during the process of a test

Lambe (1964) suggested a new coordinate system of p’ versus q’ where,

Results of triaxial tests can be represented by stress path

Consider a normally consolidated clay subjected to an isotropicallyconsolidated-drained triaxial test.

At the beginning of the test, the deviatoric stress σ1’ = σ3’ = σ3, so

and the p’ and q’ will plot as point I in the figure on the next slide. As the load is increased, the deviatoric stress σ1’ = σ3’ + Δσd and σ3’ = σ3, and the Mohr circle A corresponds to that state.

These two equations plot as point D’ at the top of the Mohr circle A.

Each successive plot of p’ and q’ will continue this straight line that is at an angle of 45° with respect to the p’ axis.

Eventually we will reach the point D on the Mohr circle B, and the sample fails (it has reached the modified failure envelope).

From this, q ' = p ' tan α

Sensitivity = the reduction in unconfined compression strength of clay soils due to remolding assuming no changes in the water content occurred

Degree of sensitivity, St =

)(

)(

remoldedu

undistubedu

q

q

Thixotrophy = is a condition where a soil specimen that is kept in undisturbed condition (after remolding) will experienced a continuous increased in strength with time

The loss of strength is gradually regained with time when the materials are allowed to rest

However most soils are partially thixotropic (part of the strength loss caused by remoldingis never regained with time)

The soils strength will never achieved the initial strength although the specimen were prepared undisturbed

Factors affecting overburden pressure

i) Depth of the point of interest

deeper the point, larger the pressure

ii) Groundwater level

same soil profile will give different pressure value due to the water level

iii) Types of soil

soil with higher unit weight value (dense soil) give larger overburden pressure