The consolidated–drained (CD) triaxial test is used to obtain the effective strength parameters of soils. First, a soil specimen is saturated by circulating deaired water
through the specimen, from bottom to top, utilizing the two drainage tubes shown in Figure 5.9. After the specimen is fully saturated, the triaxial test is done in two stages: a stress initialization stage and a shearing stage. In the first stage a confining pressure is applied via the confining fluid. Because the soil specimen is fully saturated, excess pore water pressure will be generated (≈ confining pressure).
The soil specimen is allowed to consolidate by opening the two drainage valves throughout this stage. That will allow the excess pore water pressure to dissipate gradually and the specimen to consolidate. The volume of the dissipated water can be measured using a graduated flask. The volume of the dissipated water is equal to the volume change of the specimen because the specimen is fully saturated. The volumetric strain can be calculated by dividing the volume change by the initial volume of the specimen. The consolidation curve for this stage can be plotted in the time versus volumetric strain (εv) plane, as shown in Figure 5.11.
In the shearing stage of the CD test a deviator stressσd = σ1− σ3 is applied very slowly while the drainage valves are opened, to ensure that no excess pore
σ3
σ3
σ3
σ3
σ3
σ3
Time u
Time
Compression εv
Time
Time
Consolidation
σ′3
FIGURE 5.11 Stage I of a consolidated drained triaxial test: consolidation stage.
water pressure is generated. Consequently, the effective stresses are equal to the total stresses during this stage of the CD test. Because of its stringent loading requirements, the CD test may take days to carry out, making it an expensive test.
Figure 5.12a shows stress–strain behavior typical of loose sand and normally con-solidated clay. Note that εa is the axial strain and σ1− σ3 is the deviator stress.
The soil shows smooth nonlinear stress–stress behavior reaching an ultimate shear strength (σ1− σ3)f. Figure 5.12b shows the deformation behavior in the axial strain
Compression Dilation
Compression Dilation
εa εa
εa
εa
εv
εv
∆u = 0
∆σd= σ1− σ3
∆σd= σ1− σ3 σ3
(d ) (b)
(c) (a)
σ3
σ3
σ3
(σ1− σ3)f
(σ1− σ3)f
(σ1− σ3)ult
(σ1− σ3) (σ1− σ3)
FIGURE 5.12 Stage II of a consolidated drained triaxial test: shearing stage (a, b) loose sand; (c, d) dense sand.
(εa) versus volumetric strain (εv) plane typical of loose sand and normally consol-idated clay. In this figure the soil compresses as the shearing stress is increased.
When the ultimate shear strength is approached, the curve levels out, indicating that the volumetric strain is constant (i.e., no further volume change). From the measured deviator stress at failure (σ1− σ3)f, one can plot Mohr’s circle, which describes the stress state of the soil specimen at failure. Such a plot is shown in Figure 5.13. The Mohr–Coulomb failure criterion can be obtained by drawing a line that is tangent to Mohr’s circle and passing through the origin. This is appli-cable to sands and normally consolidated clays since these soils havec= 0. The slope of the Mohr–Coulomb failure criterion is the effective (or drained) friction angleφ of the soil.
Figure 5.12c shows typical stress–strain behavior of dense sand and overcon-solidated clay in the axial strain (εa) versus deviator stress plane. The soil shows nonlinear stress–stress behavior reaching a peak shear strength (σ1− σ3)f at an early stage of shearing. After the peak strength is reached, a sharp decrease in strength is noted. Then the stress–strain curve levels out, approaching an ultimate strength (σ1− σ3)ult. Note that this ultimate strength is the same as the ultimate strength of the loose sand if the two sands were identical (not in terms of their relative density). Figure 5.12d shows typical deformation behavior of dense sand and overconsolidated clay in the axial strain (εa) versus volumetric strain (εv) plane. Initially, the soil compresses as the shearing stress is increased. But shortly
τ = σ′ tanφ′
τ (kPa)
σ′ (kPa) 0
50
0 50 100
A
P
τA
φ′
φ′
90°
θ θ
σ′1= σ1
σ′3= σ3
σ′A
σ′3 σ′1
(σ′1− σ′3)/2
(σ′1+ σ′3)/2
FIGURE 5.13 Determination of the Mohr–Coulomb failure criterion for sand (CD triaxial test).
after that the soil starts expanding (dilating) as the axial strain increases. When the ultimate shear strength is approached, the curve levels out, indicating that the volumetric strain is constant.
We need only one CD test to determine the strength parameters of dense sand (becausec= 0). From the peak deviator stress measured (σ1− σ3)f, we can plot Mohr’s circle, which describes the stress state of the soil specimen at failure as shown in Figure 5.13. The Mohr–Coulomb failure criterion can be obtained by drawing a line that is tangent to Mohr’s circle and passing through the origin. The slope of the Mohr–Coulomb failure criterion is the effective (or drained) friction angleφ of the soil.
For overconsolidated clays the cohesion intercept,c, is not equal to zero. There-fore, we will need to have the results of at least two CD triaxial tests on two identical specimens subjected to two different confining pressures. From the mea-sured peak deviator stress (σ1− σ3)f of these tests, we can plot Mohr’s circles that describe the stress states of the soil specimens at failure, as shown in Figure 5.14.
The Mohr–Coulomb failure criterion in this case can be obtained by drawing a line that is tangent to the two Mohr’s circles. The Mohr-Coulomb failure crite-rion will intersect with the shear stress axis at τ = c, as shown in Figure 5.14.
This is the effective cohesion intercept of the overconsolidated clay. The slope of the Mohr–Coulomb failure criterion is the effective friction angle φ of the overconsolidated clay.
You recall from the strength of materials laboratory that when a concrete cylin-drical specimen was crushed between the jaws of a compression machine, there existed a failure plane making an angle θ with the horizontal. Soils exhibit sim-ilar behavior—they also have a failure plane that occurs at failure in a triaxial
τ =c′ + σ′ tan φ′
c′
σ′(kPa)
τ (kPa)
0 50
0 50
Test 1
Test 2
φ′
FIGURE 5.14 Determination of the Mohr–Coulomb failure criterion for overconsolidated clay (CD triaxial test).
compression test. Let us calculate, both graphically and analytically, the orienta-tion of the failure plane. In reference to Figure 5.13, a sandy soil will have a zero cohesion intercept, and the Mohr–Coulomb failure criterion is tangent to Mohr’s circle. The tangency point (point A in Figure 5.13) represents the stress state on the failure plane at failure. Thex and y coordinates of point A give, respectively, the normal and shear stresses exerted on the failure plane. We can measure the x and y coordinates at point A directly from the figure provided that our graph is drawn to scale. We can determine the failure plane orientation using the pole method. For a triaxial compression test, the pole (the origin of planes) is located at point P in Figure 5.13. The orientation of the failure plane can be obtained by connecting point P with point A. The orientation of the failure plane is the angle between line PA and the horizontal. We can use a protractor to measure the angle θ. This is the graphical solution.
The analytical solution for the angleθ can be obtained easily from Figure 5.13.
Using simple trigonometry, you can show that θ = 45◦+φ
2 (5.2)
As mentioned above, the x and y coordinates of point A give, respectively, the normal stress (σA) and shear stress (τA) exerted on the failure plane at failure.
Let’s calculate those analytically. In reference to Figure 5.13 we can write σA=σ1+ σ3
2 −σ1− σ3
2 sinφ (5.3)
τA=σ1− σ3
2 cosφ (5.4)
The friction angleφ can be calculated from sinφ= σ1− σ3
σ1+ σ3 (5.5)
Example 5.3 A CD triaxial compression test was conducted on a sand specimen using a confining pressure of 42 kPa. Failure occurred at a deviator stress of 53 kPa.
Calculate the normal and shear stresses on the failure plane at failure. Also calculate the angle made by the failure plane with the horizontal. (a) Solve the problem graphically and (b) confirm your solution analytically.
SOLUTION: (a) Graphical solution Given: σ3f = 42 kPa and (σ1 − σ3)f = 53 kPa. We need to calculate the major principal stress in order to draw Mohr’s circle:
σ1f = σ3f + 53 kPa = 42 kPa + 53 kPa = 95 kPa The radius of Mohr’s circle is
(σ1− σ3)f
2 = 53
2 = 26.5 kPa
Thex-coordinate of the center of Mohr’s circle is (σ1+ σ3)f
2 = 95+42
2 = 68.5 kPa
Now we can plot Mohr’s circle with its center located at (68.5 kPa, 0) and radius
= 26.5 kP a, as shown in Figure 5.15.
Since the soil is sand, we havec= 0. The Mohr–Coulomb failure criterion is drawn as a straight line passing through (0,0) and tangent to the circle as shown in the figure. Using the same scale as that used in the drawing, we can now measure the coordinates of the tangency point: The normal stress at point A isσA≈ 57 kPa, and the shear stress at point A isτA≈ 26 kPa. The orientation of the failure plane can be obtained by connecting pole P with point A. The orientation of the failure plane is the angle between line PA and the horizontal. Using a protractor, we can measure the angleθ ≈ 56◦.
(b) Analytical solution First, we calculate the friction angle of the sand using (5.5):
φ= sin−1σ1− σ3
σ1+ σ3 = sin−1 53
95+ 42 = 22.75◦ Therefore,
θ = 45◦+φ
2 = 45◦+22.75◦
2 = 56.38◦
σ′ (kPa)
τ (kPa)
τ = σ′ tan φ′
φ′ = 22.75°
θ = 56.38°
0 50
0 50 100
A
P 68.5 kPa
σ′1= 95 kPa σ′ = 42 kPa3
σ′ = 42 kPa3
σ′ = 95 kPa1
φ′
90° τA
τA
σ′A
σ′A
θ
FIGURE 5.15 Mohr’s circle for a CD triaxial test on sand.
In reference to Figure 5.15, we can write
Note that there is a relatively good agreement between the graphical and analytical solutions. Solving problems graphically first can reveal better ways of solving them analytically. Also, sometimes it is more efficient to solve problems graphically.
Example 5.4 Three CD triaxial compression tests were conducted on three over-consolidated clay specimens using three confining pressures: 4, 20, and 35 kPa.
Failure occurred at the deviator stresses of 19, 36 and 54 kPa, respectively. Deter-mine the shear strength parameters of the soil.
SOLUTION: Let’s calculate the major principal stress at failure for each test. Then let’s determine thex-coordinate of the center of each Mohr’s circle and its radius:
σ3f (kPa) (σ1− σ3)f (kPa) σ1f (kPa) (σ1+ σ3)f/2 (kPa) (σ1− σ3)f/2 (kPa)
4 19 23 13.5 9.5
20 36 56 38 18
35 54 89 62 27
Using the results from the fourth and fifth columns, we can draw three Mohr’s circles as shown in Figure 5.16. The failure envelope (Mohr–Coulomb failure crite-rion) is then established. The friction angle and the cohesion intercept are measured from the figure asφ≈ 23◦ andc≈ 5 kPa, respectively.
FIGURE 5.16 CD triaxial compression tests on three overconsolidated clays.