The tri-axial shear test is most versatile of all the shear test testing methods for getting shear strength of soil i.e. Cohesion (C) and Angle of Internal Friction (Ø), though it is bit complicated. This test can measure the total as well as effective stress parameters both. These two parameters are required for design of slopes, calculation of bearing capacity of any strata, calculation of consolidation parameters and in many other analyses. This test can be conducted on any type of soil, drainage conditions can be controlled, pore water pressure measurements can be made accurately and volume changes can be measured. In this test, the failure plane is not forced, the stress distribution of failure plane is fairly uniform and specimen can fail on any weak plane or can simply bulge.
Fig. 1: Triaxial Shear Test Apparatus
Fig. 2: Triaxial Shear Test Setup
Fig. 3: 3.8 cm (1.5 inch) internal diameter 12.5 cm (5 inches) long sample tubes.
Fig. 4: Rubber Ring
Fig. 5: Open ended cylindrical section
Fig. 6: Weighing balance
In consolidated-drained and consolidated-undrained tests, the consolidation of the specimen takes place during the first stage. As the volume of the specimen decreases, its post-consolidation dimensions are different from the initial dimensions. The post consolidation dimensions can be determined approximately assuming that the sample remains cylindrical and it behaves isotropically. Let Li, Di and Vi be the length, diameter and the volume of the specimen before consolidation. Let L0, D0 and V0 be the corresponding quantities after consolidation.
Volumetric change, Δ Vi = Vi – V0
Volumetric change is measured with the help of burette.
Volumetric strain, Єv = Δ Vi / Vi
For isotropic consolidation, the volumetric strain is three times the linear strain (Єl), thus:
Єl = Єv / 3
L0 = Li – Δ Li = Li – Li x Єl
L0 = Li (1 - Єl) = Li (1 – Єv/3)
Similarly, D0 = Di (1 – Єv/3)
The post consolidation diameter D0
can also be computed after L0 has been determined from the relation:
(ϖ/4. D02) x L0 = V0
Or D0 = √ V0 / [(ϖ/4) x L0]
As the sample is sheared, its length decreases and the diameter increases. The cross-sectional area A at any stage during shear can be determined assuming that the sample remains cylindrical in shape. Let ∆ L0 be the change in length and ∆V0 be the change in volume. The volume of the specimen at any stage is given by V0 ± ∆V0.
Therefore, A (L0 - ∆L0) = V0 ± ∆V0
Or, A = V0 ± ∆V0 / L0 - ∆L0 = V0 (1 ± ∆V0 /V0) / L0 (1 - ∆L0/L0)
This is the general equation which gives the cross-sectional area of the specimen and it can be written as:
A = A0 L0 (1 ± ∆V0/V0) / L0 (1 - Єl)
A = A0 (1 ± ∆V0/V0) / (1 - Єl)
Where, Єl is the axial strain in the sample.
For an undrained test, the volumetric change (∆V0) is zero and then the equation above becomes:
A = A0 / (1 - Єl)
The stresses in the specimen at various stages of shear should be calculated using the cross-sectional area A as found above.
Deviator Stress: The deviator stress (σd) acting on the
specimen when the axial load applied by the loading
machine is P can be obtained as:
σd = P/A
The deviator stress (σd) is equal to (σ1 – σ3).
It may be noted that the load indicated by the proving ring is slightly more than P because of friction on the ram and the upward thrust on the ram due to pressure of the water in the cell. The correction can be determined separately.
A more convenient procedure is to lift the ram above the specimen when the cell pressure has been applied. The machine is started keeping the strain rate the same as to be used in the actual test. The proving ring records the load. To account for correction, the dial gauge on the proving ring is set to zero to indicator zero load. This automatically compensates the ram friction and the upward thrust on the ram due to cell pressure. Thus the load indicated by the proving ring during shear would be equal to the load P applied to the specimen.
The minor principal stress (σ3) is equal to the cell pressure (σc). The major principal stress (σ1) is equal to the sum of the cell pressure and the deviator stress.
Thus, σ1 = σ3 + (σ1 - σ3) = σ3 + σd
The deviator stress at failure (σ1 - σ3) is known as the compressive strength of the soil.
Drained Test: Fig. 7 shows the stress-strain curve for a drained test. The y-axis shows the deviator stress (σ1 - σ3) and the x-axis, the axial strain (Єl). For dense sand (and over-consolidated clay), the deviator stress reaches a peak value and then it decreases and becomes almost constant, equal to the ultimate stress, at large stains. For loose sand (and normally consolidated clay), the deviator stress increases gradually till the ultimate stress is reached.
The volumetric strain is shown in Fig. 8. In dense sand (and over-consolidated clay), there is a decrease in the volume at low strains, but at large strains, there is an increase in the volume. In loose sand (and normally consolidated clay), the volume decreases at all strains. (For some loose sands, there is a slight tendency to increase in the volume at large strains).
Consolidated-undrained test: Fig. 9 shows the stress-strain curve for a consolidated-undrained test. The shape of the curves is similar to that obtained in a consolidated-drained test. In a consolidated-undrained test, there is an increase in the pore water pressure throughout for loose sand (and normally consolidated clay), as shown in Fig.10. However, in the case of dense sands (and over-consolidated clay), the pore water pressure increases at low strains but at large strains it becomes negative (below atmospheric pressure).
Fig. 11 shows the failure envelope for a normally consolidated
clay in terms of effective stresses obtained from a consolidated drained
test. The failure envelope has an angle of shearing resistance of Փ and
passes through origin. First the Mohr circles for the three tests are
drawn in terms of effective stresses corresponding to failure conditions.
Then the best common tangent is drawn to the three circles. The common
tangent is the failure envelope. As each circle represents a failure,
there must be at least one point on it which gives the stresses
satisfying the failure criterion. Obviously, the common tangent joins
all such points of the three circles. Thus for normally consolidated
clays, shear strength is:
s = σ’ tan ø’
Fig. 12 shows the failure envelope for over-consolidated clay in
terms of effective stresses. The failure envelope is slightly curved
in the initial portion, but, for convenience, it is approximated as a
straight line. The failure envelope has an intercept c’ on the Y-axis,
the angle of shearing resistance is Փ’. In the case of over-
consolidated clays, shear strength is:
s = c’ + σ’ tan ø’
The failure envelopes in terms of effective stresses can also be drawn from the results of a consolidated-undrained test (CU test) when the pore water pressure measurements are also taken. The shear strength parameter c’ and Փ’ obtained from the consolidated-undrained tests and that from consolidated-drained test are approximately equal. Drained tests on dense sands and over-consolidated clays give slightly higher values of Փ’ due to extra work required during dilation (increase in volume), but the difference is small, and therefore, usually neglected.
The failure envelope in terms of total stresses can be drawn from the test results of a consolidated-undrained test.
The failure envelopes are similar in shape to that in terms of
effective stresses but the values of the strength parameters are
quite different. Fig. 11 shows the failure envelopes for effective
stresses and also for total stresses for normally consolidated clay.
The angle of shearing resistance in terms of total stresses (Փcu)
is much smaller than the angle (Փ’).
In the case of normally consolidated clays, shear strength is:
S = σ tan Փcu
Fig. 12 shows the failure envelope for an over consolidated clay
in terms of total stresses. The angle of shearing resistance (Փcu)
is much smaller than the angle Փ’ obtained in terms of effective
stresses. In the case of over consolidated clays, shear strength is:
S = ccu + σ tan Փcu
The angle of shearing resistance Փcu obtained from the total stress envelopes is also known as apparent angle of shearing resistance.
Fig. 13 shows the failure envelope in terms of total stress obtained from an unconsolidated-undrained test on normally consolidated clay. The failure envelope is horizontal (Փ = 0) and has a cohesion intercept of cu. In this case, shear strength is S = cu. The failure envelope for over-consolidated clay is also horizontal, but the value of cu will be more, depending upon the degree of over-consolidation.
For an unconsolidated-undrained test, the failure envelope cannot be drawn in terms of effective stresses. In all the tests conducted at different confining pressures, the effective stress remains the same. This is due to the fact that an increase in confining pressure results in an equal increase in pore water pressure for a saturated soil under undrained conditions. Thus only one Mohr circle in terms of effective stresses is obtained from all the three tests. It may be noted that the deviator stress at failure is the same for all specimens.