Shear Strength of Soil
Shear Strength of soil
Q: Explain drainage conditions for determination of shear strength parameters
A different drainage situation affects the shear strength of soil. Hence in the laboratory shear test, the drainage condition required to be simulated for correct prediction of shear strength parameters. There are three different types of drainage condition used in laboratory shear tests.
Unconsolidated Undrained Test (UU test)
Consolidated Undrained Test (CU test)
Consolidated Drained Test (CD test)
1. Unconsolidated Undrained Test (UU)
This is also called quick test. In this test, drainage is not permitted throughout the test. In the case of direct shear test drainage is not permitted during the application of both normal stress and shear stress. In the case of triaxial compression test drainage is not permitted during the application of both cell pressure and deviator stress. The test is conducted fast allowing no time for either consolidation of sample initially or dissipation of pore pressure in later stage.
2. Consolidated Undrained Test (CU)
In this type of shear test the soil specimen is allowed to consolidate fully under initially applied stress and then sheared quickly without allowing dissipation of pore pressure. In the case of direct shear test the specimen is allowed to consolidate fully under applied normal stress and then sheared at high rate of strain to prevent dissipation of pore pressure during shearing. In the case of triaxial compression test the specimen is allowed to consolidate fully under the applied cell pressure and then the pore water outlet is closed and the specimen is subjected to increasing deviator stress at higher rate of strain.
3. Consolidated Drained Test (CD)
In this type of shear test drainage is allowed throughout the test. The specimen is allowed to consolidate fully under the applied initial stress and then sheared at low rate of strain giving sufficient time for the pore water to drain out at all stages. The test may continue for several hours to several days (till required degree of consolidation).
Q: Draw shear strength envelope for different types of soils (C, C-φ, φ soils)
Q:Explain shear strength, Coulomb’s theory and failure envelope
Q:Explain Mohr–Coulomb failure theory
The shear strength of soil is basically made up of the following components:
Frictional component: It is mainly due to interlocking of soil particles and the friction between them.
Cohesion component: It is due to mutual attraction that exists between the fine particles of some soils. The shear strength of cohesive soil results both from friction as well as cohesion whereas for cohesion-less soil it results from friction alone.
The Coulomb Equation:
The shear strength equation was first proposed by French engineer Coulomb. He expressed the shear strength’s’ as a linear function of total normal stress ‘σ’ on the potential surface of sliding
τ = c + σ tan (ɸ)
Where τ = shear strength
c = apparent cohesion
α = total normal stress
φ= angle of internal friction
The coulomb strength envelope is shown in the figures. It is clear from the plot that the frictional component (tan φ) of shear strength increases with normal stress whereas the cohesion component (c) remains constant. Experiments have shown that the shear strength in a soil is developed only by solid particles, because the water and air phases have no shear strength.
Q: Explain Direct shear test with merits and demerits
Merits of direct shear test
The sample preparation is easy. The test is simple and convenient.
As the thickness of the sample is relatively small, the drainage is quick and the pore pressure dissipates very rapidly. Consequently, the consolidated-drained and the consolidated-undrained tests take relatively small period.
Direct shear test is ideally suited for conducting drained tests on cohesion-less soils.
The apparatus is relatively cheap.
Demerits of direct shear test
The stress condition is known only at failure. The conditions prior to failure are indeterminate and, therefore, the Mohr circle cannot be drawn.
In direct shear test, the stress distribution on the failure plane (horizontal plane) is not uniform. The stresses are more at the edges and lead to the progressive failure, like tearing of a paper. Consequently, the full strength of the soil is not mobilized simultaneously on the entire failure plane.
The area under shear gradually decreases as the test progresses. But the corrected area cannot be determined and therefore, the original area is taken for the computation of stresses.
The orientation of failure plane is fixed. This plane may not be the weakest plane.
Control on the drainage conditions is very difficult. So, only drained tests can be conducted on highly permeable soils.
The measurement of pore water pressure is not possible in direct shear test.
The side walls of the shear box cause lateral restraint on the specimen and do not allow it to deform laterally.
Q: Explain Unconfined compression test with merits and demerits
Merits
This is the simplest test for determining the undrained shear strength of soil.
A quick test and field apparatus gives a fast determination of undrained strength.
The test is inexpensive
Demerits
Fissured samples may not give reliable results because there is no confining pressure to close the fissures.
Samples with high sand content give unreliable results.
Q: Explain triaxial shear test with merits and demerits
Merits
There is complete control over the drainage conditions
Pore pressure changes and volumetric changes can be measured directly
The stress distribution in the failure plane is uniform
The specimen is free to fail on the weakest plane
The state of stress at all intermediate stages up to failure is known
The test is suitable for accurate research work
Demerits
The apparatus is elaborate, costly and bulky
The drained test takes a longer period as compared with that in a direct shear test
The strain condition in the specimen are not uniform
It is not possible to find out the cross sectional area of the specimen accurately under large strains
The test simulates only axis symmetric problems
The consolidation of the specimen in the test is isotropic whereas in the field, consolidation is generally anisotropic.
Q: Explain Vane shear test with merits and demerits
Merits
The test is simple and quick
It is ideally suited for determination of the in-situ undrained shear strength of non fissured, fully saturated clay
The test can be conveniently used to determine the sensitivity of the soil
Demerits
The test cannot be conducted on the fissured clay or the clay containing silt or sand laminations
The test does not give accurate results when the failure envelope is not horizontal
Sample disturbance not systematically accounted for shear strength identification.
Q: Derive following equation or show that stress at a point in soil mass is
1= 3 + 2 c ∙tanα
The stresses acting on the soil specimen during Triaxial testing are shown in figure. In this case minor principle stresses is equal to the cell pressure (fluid pressure) σ3. The major principle σ1 stress is equal to deviator stress plus the cell pressure.
Let the stress components on the failure plane be σ and τ and the failure plane is inclined at an angle α to the major principle plane.
Let the envelop DF cut the abscissa at angle φ , K be the centre of the Mohr’s circle.
From ΔADK and ΔFDK we get,
=45+ 2
Principle stresses relationships at failure
BK =DK = radius of Mohr’s circle =12 1-3
OK= 12 1+3
OF= ccot
From ΔFDK,
sin = DKFK= radius of Mohr’s circle OF+OK
sin = DKFK= 12 1-312 1+3+ccot
Rearranging and solving we get
sin = 1-31+3+2ccot
sin × 1+3+2ccot = 1-3
1sin + 3sin + 2ctan = 1-3
1sin - 1= -3- [+ 3sin + 2ctan ]
1 )= 31+ sin +2ctan
1= 31+ sin )+2ctan )
1= 3tan2α+2c tanα
1= 3 N+2c N
From the Mohr’s circle drawn, the straight line forms the failure plane and the intercept created is the effective strength parameter c of the soil. The angle made by the plane ø forms the angle of friction.
Hence, effective shear strength is given by, τ = c + σ tan (ɸ)
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