Triaxial test pros, cons, procedure

The triaxial compression test applies an all-round cell pressure $\sigma_3$ to a cylindrical specimen and then increases the axial stress until the sample fails.

Advantages

• Drainage conditions can be controlled and pore pressure measured.
• Failure takes place on the weakest plane; stresses are fairly uniform.
• Suitable for all soil types and for different stress paths.
• Effective and total stress parameters can both be obtained.
• Volume change and strain can be recorded.

Disadvantages

• Equipment is costly, and the test needs a skilled operator.
• CD and CU tests take a long time (days for clays).
• Stress state is axisymmetric ($\sigma_2 = \sigma_3$), which is not the real condition for plane-strain problems such as strip footings and slopes.
• Sample disturbance during preparation, and end restraint from the platens.
• Preparation of an undisturbed sample is difficult in sand.

Procedure

1. Prepare a cylindrical specimen (typically 38 mm diameter, 76 mm long), cover it with a rubber membrane and place it between porous discs on the pedestal.
2. Fill the cell with water and apply the cell pressure $\sigma_3$.
3. For CU and CD, allow consolidation under $\sigma_3$ with drainage open; for UU, keep drainage closed.
4. Apply axial load at a constant rate of strain until failure (or 15 to 20% strain), recording load, deformation and, if required, pore pressure $u$.
5. Repeat on at least three identical specimens at different cell pressures.

Computing the shear parameters

For each test, the deviator stress at failure is

$$\sigma_d = \sigma_1 - \sigma_3 = \frac{P}{A_c}, \qquad A_c = \frac{A_0}{1-\varepsilon}$$

where $P$ is the axial load and $A_c$ is the corrected area.

1. Calculate $\sigma_1 = \sigma_3 + \sigma_d$ for each test (use $\sigma' = \sigma - u$ for effective stress).
2. Draw a Mohr circle for each test on the $\sigma$-$\tau$ plane with diameter from $\sigma_3$ to $\sigma_1$.
3. Draw the common tangent to the circles, which is the failure envelope.
4. Read the intercept on the $\tau$-axis as $c$ and the slope as $\phi$ (or $c'$, $\phi'$ for effective stress).

Alternatively, $\sin\phi = \dfrac{\sigma_1 - \sigma_3}{\sigma_1+\sigma_3+2c\cot\phi}$, and for $c=0$, $\sin\phi = \dfrac{\sigma_1-\sigma_3}{\sigma_1+\sigma_3}$.