24-Bld-A6 Geotechnical Materials and Analysis, December 2016, Q5

Question 5 (20 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 drained direct-shear results on the same silty sand: $(\sigma_n,\tau_f)=(0,25)$ kPa and $(\sigma_n,\tau_f)=(200,130)$ kPa.

Find. (a) φ′; (b) σ₁ and σ₃ from a Mohr circle tangent to the strength envelope at the given point; (c) the stresses on, and orientation of, the corresponding failure plane.

Check: the two data points define an apparent cohesion intercept c′ = 25 kPa for a "silty sand," which is unusual for a clean drained sand (normally $c'\approx0$); this is taken at face value as printed, most plausibly reflecting the silt fraction, dilation at low confinement, or dataset rounding, but the strength parameters are used exactly as the two given points define them.

Approach. Fit the linear Mohr–Coulomb envelope through the two shear-strength data points to get c′ and φ′, then construct the Mohr circle that is tangent to that envelope exactly at the second test point (the geometric condition that the radius to the tangent point is perpendicular to the envelope), and finally locate the theoretical failure-plane orientation relative to the major principal plane and confirm it reproduces the given stress point.

(a) Friction angle from the two-point envelope. The first test (zero normal stress) gives the cohesion intercept directly, $c'=25$ kPa. The slope through the two points gives $$\tan\phi' = \frac{\tau_f-c'}{\sigma_n} = \frac{130-25}{200}=0.525 \quad\Rightarrow\quad \boxed{\phi' = 27.7^{\circ}}$$
(b) Locate the Mohr circle tangent at (200, 130). The circle's centre lies on the σ-axis, and the radius drawn to the tangent point must be perpendicular to the envelope $\tau=c'+\sigma\tan\phi'$. Requiring the vector from the centre $(\sigma_c,0)$ to $(200,130)$ to be normal to the envelope direction gives $$\sigma_c = \sigma_f+\tau_f\tan\phi' = 200+130(0.525) = 268.3\ \text{kPa},$$ $$R = \sqrt{(\sigma_f-\sigma_c)^2+\tau_f^2} = \sqrt{(-68.3)^2+130^2} = 146.8\ \text{kPa}.$$ $$\boxed{\sigma_1 = \sigma_c+R = 415.1\ \text{kPa},\qquad \sigma_3 = \sigma_c-R = 121.4\ \text{kPa}}$$
(c) Orientation and stresses on the failure plane. The failure plane makes the classical angle with the major principal plane, $$\theta = 45^{\circ}+\frac{\phi'}{2} = 45+13.85 = 58.8^{\circ}.$$ Substituting $\sigma_1$, $\sigma_3$ and θ back into the plane-stress transformation, $$\sigma=\frac{\sigma_1+\sigma_3}{2}+\frac{\sigma_1-\sigma_3}{2}\cos2\theta,\qquad \tau=\frac{\sigma_1-\sigma_3}{2}\sin2\theta,$$ reproduces $\sigma=200.0$ kPa and $\tau=130.0$ kPa exactly — confirming that the tangent point used to build the circle in part (b) IS the stress state on the theoretically-critical plane, oriented at 58.8° to the major principal plane (31.2° to the minor principal plane).

Mohr circle for the 200 kPa test, tangent to the Mohr–Coulomb envelope (c′ = 25 kPa, φ′ = 27.7°) exactly at the measured failure point (200, 130).

Question 5 — final results
Quantity	Value
Cohesion intercept, c′	25 kPa
Friction angle, φ′	27.7°
Major principal stress, σ₁	415.1 kPa
Minor principal stress, σ₃	121.4 kPa
Failure plane orientation (from major principal plane)	58.8°
Stress on failure plane (check)	σ = 200 kPa, τ = 130 kPa

Concept & Theory

Topic. Recovering principal stresses and failure-plane orientation from direct-shear data via the Mohr circle.

Key relations. Mohr–Coulomb envelope $\tau_f=c'+\sigma_n'\tan\phi'$; tangency condition (radius perpendicular to envelope at the failure point); failure-plane orientation $\theta=45^{\circ}+\phi'/2$ from the major principal plane; plane-stress transformation to recover $(\sigma,\tau)$ on any plane from $\sigma_1,\sigma_3,\theta$.

Why this works. A direct shear box physically forces the specimen to fail on a predetermined horizontal plane, which is not generally the plane a triaxial specimen would choose for itself; however, IF that forced plane happens to be the critical one (as is assumed whenever a direct-shear result is used to define a point ON the Mohr–Coulomb envelope, which is the standard interpretation of the test), then a unique Mohr circle tangent to the envelope at that exact point exists, and it correctly recovers the equivalent triaxial principal-stress state. The closing check in step (c) — that transforming back from $\sigma_1,\sigma_3,\theta$ reproduces the original (200, 130) — is what confirms the construction is self-consistent.

Common pitfall. Assuming the failure plane in a direct shear test is automatically "properly oriented" without checking; also, sign errors locating the circle centre when the tangent point does not lie directly above the centre (it rarely does, unless $c'=0$).

Source. Das, Principles of Geotechnical Engineering, Ch. 12 (shear strength, Mohr circle construction); Craig's Soil Mechanics, Ch. 6.

Question 5 of 5

Question 5 (20 marks)