Drained triaxial tests — angle of shearing resistance: 16-Civ-A4 Geotechnical Materials and Analysis, May 2016, Q6

Question 6: Drained triaxial tests — angle of shearing resistance (20 marks)

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Given. Four drained (CD) triaxial tests on the same sand (same porosity). Cell pressure $\sigma_3$ and deviator stress $(\sigma_1-\sigma_3)$ at failure:

Given data (failure)
$\sigma_3$ (kPa)	$(\sigma_1-\sigma_3)$ (kPa)	$\sigma_1$ (kPa)
100	452	552
200	908	1108
400	1810	2210
800	3624	4424

Find. The effective angle of shearing resistance $\phi'$ (with $c'=0$ for sand), verified with Mohr circles.

Approach. For a cohesionless sand the failure envelope passes through the origin ($c'=0$), so each test satisfies $\sin\phi' = (\sigma_1-\sigma_3)/(\sigma_1+\sigma_3)$; compute it for every test and confirm the Mohr circles share one tangent through the origin.

Major principal stress. $\sigma_1 = \sigma_3 + (\sigma_1-\sigma_3)$ gives $\sigma_1 = 552,\ 1108,\ 2210,\ 4424\ \text{kPa}$ for the four tests.
Analytical $\phi'$. With $c'=0$, $$\sin\phi' = \frac{\sigma_1-\sigma_3}{\sigma_1+\sigma_3}.$$ Test 1: $452/652 = 0.6933$; test 2: $908/1308 = 0.6942$; test 3: $1810/2610 = 0.6935$; test 4: $3624/5224 = 0.6937$.
Consistency and result. All four give $\sin\phi' \approx 0.694$, i.e. essentially the same value, confirming a single linear envelope through the origin. Averaging, $\phi' = \sin^{-1}(0.694) = \boxed{43.9^\circ}$ with $c'=0$.
Mohr-circle verification. Plotting each circle (centre $(\sigma_1+\sigma_3)/2$, radius $(\sigma_1-\sigma_3)/2$) shows all four are tangent to the line through the origin inclined at $\phi'=43.9^\circ$, verifying the analytical value.

Mohr circles for the four drained tests, all tangent to the failure envelope through the origin ($c'=0$) at $\phi' \approx 43.9^\circ$.

Question 6 — results
$\sigma_3$	$\sigma_1$	$\sin\phi'$	$\phi'$
100	552	0.6933	43.89°
200	1108	0.6942	43.96°
400	2210	0.6935	43.91°
800	4424	0.6937	43.93°
Mean $\phi'$ ($c'=0$)			43.9°

Topic. Drained shear strength of sand from triaxial tests (Mohr–Coulomb, effective stress).

Key relations. $\sin\phi' = (\sigma_1-\sigma_3)/(\sigma_1+\sigma_3)$ for $c'=0$; envelope $\tau_f = \sigma'\tan\phi'$; $\sigma_1/\sigma_3 = \tan^2(45^\circ+\phi'/2)$.

Why this works. In a drained test $u=0$, so total and effective stresses coincide; a cohesionless sand has no true cohesion, so its strength envelope passes through the origin and its slope is set entirely by $\phi'$. Because every specimen was at the same porosity, all four circles are tangent to one line.

Common pitfall. Forgetting to add $\sigma_3$ to the deviator stress to get $\sigma_1$, or fitting a non-zero $c'$ to a sand (a small intercept here is only scatter, not real cohesion).

Source. Das & Sobhan, Ch. 12 (Shear Strength of Soil — triaxial test, drained strength of sand).

Question 6 of 6: Drained triaxial tests — angle of shearing resistance

Question 6: Drained triaxial tests — angle of shearing resistance (20 marks)