Question 6 of 6: Shear-strength parameters from drained triaxial tests

Question 6: Shear-strength parameters from drained triaxial tests (30 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two consolidated-drained (CD) triaxial tests at failure; drained, so pore pressure is zero and total stresses equal effective stresses.

Find. The effective shear-strength parameters $c'$ and $\phi'$, verified graphically with Mohr’s circles and their common tangent.

Approach. Use the stress-point (modified Mohr) method from the formula sheet: plot $\tfrac12(\sigma_1'-\sigma_3')$ against $\tfrac12(\sigma_1'+\sigma_3')$, fit the $K_f$ line of slope $\tan\alpha'$ and intercept $a$, then convert with $\phi'=\sin^{-1}(\tan\alpha')$ and $c'=a/\cos\phi'$. Confirm by writing the failure criterion for each circle.

1. Stress points. With $s'=\tfrac12(\sigma_1'+\sigma_3')$ and $t=\tfrac12(\sigma_1'-\sigma_3')$: Specimen 1 gives $s'_1=\tfrac12(369+100)=234.5$, $t_1=\tfrac12(269)=134.5$; Specimen 2 gives $s'_2=\tfrac12(669+200)=434.5$, $t_2=\tfrac12(469)=234.5$ (all kN/m²).
2. Fit the $K_f$ line. The slope and intercept of the line through the two stress points are $$\tan\alpha'=\frac{t_2-t_1}{s'_2-s'_1}=\frac{234.5-134.5}{434.5-234.5}=\frac{100}{200}=0.500,\qquad a=t_1-s'_1\tan\alpha'=134.5-117.25=17.25.$$
3. Convert to strength parameters. $$\phi'=\sin^{-1}(\tan\alpha')=\sin^{-1}(0.500)=\boxed{30^\circ},\qquad c'=\frac{a}{\cos\phi'}=\frac{17.25}{\cos 30^\circ}=\boxed{19.9\ \text{kN/m}^2}$$
4. Verify with the failure criterion (Mohr circles). With $N_\phi=\tan^2(45^\circ+\phi'/2)=\tan^2 60^\circ=3.0$, the criterion $\sigma_1'=\sigma_3'N_\phi+2c'\sqrt{N_\phi}$ gives, for $\sigma_3'=100$: $\sigma_1'=100(3)+2(19.9)(1.732)=369\ \text{kN/m}^2$; for $\sigma_3'=200$: $\sigma_1'=200(3)+68.9=669\ \text{kN/m}^2$. Both reproduce the measured $\sigma_1'$, so a single line of slope $\phi'=30^\circ$ and intercept $c'=19.9\ \text{kN/m}^2$ is tangent to both circles.