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16-Civ-A4 Geotechnical Materials and Analysis · December 2016

Question 5 of 5: Drained direct-shear strength of a silty sand

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Examination 98-Civ-A4 Geotechnical Materials and Analysis — December 2016. Closed book; 3 hours; total 100 marks; answer ALL five questions. Newmark / m–n influence charts and a formula sheet are supplied with the paper.

Reference texts: B.M. Das & K. Sobhan, Principles of Geotechnical Engineering, 9th ed. (Cengage); R.F. Craig, Craig's Soil Mechanics, 8th ed. (Spon Press); Holtz, Kovacs & Sheahan, An Introduction to Geotechnical Engineering, 2nd ed. (Pearson).

Question 5: Drained direct-shear strength of a silty sand (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. Drained direct-shear tests on identical silty-sand specimens: at zero normal stress the shear strength is $\tau_f = 25$ kPa; at a normal stress of 200 kPa the specimen fails at $\tau_f = 130$ kPa.

Find. (a) the friction angle; (b) the principal stresses at failure under $\sigma_n=200$ kPa; (c) the normal and shear stresses on the failure plane.

failure plane (200, 130)s3=121s1=415tau = c + sigma tan(phi)c=25normal stress (kPa)shear (kPa)
Mohr circle at failure, tangent to the strength envelope $\tau=c+\sigma\tan\phi$ at the failure-plane point (200, 130).

Approach. The two test points define the Mohr–Coulomb envelope $\tau_f=c+\sigma_n\tan\phi$; the failure Mohr circle is tangent to this envelope at the direct-shear point, from which the principal stresses follow by geometry.

  1. (a) Cohesion and friction angle. The zero-normal-load test gives the cohesion intercept $c = 25$ kPa. From the second point, $$\tan\phi = \frac{\tau_f - c}{\sigma_n} = \frac{130-25}{200} = 0.525 \;\Rightarrow\; \boxed{\phi = 27.7^\circ,\quad c = 25\ \text{kPa}}.$$
  2. (b) Failure Mohr circle geometry. The direct-shear point $(\sigma_n,\tau_f)=(200,130)$ is the tangent point, where the radius meets the envelope at right angles. The radius is $$r = \frac{\tau_f}{\cos\phi} = \frac{130}{\cos 27.7^\circ} = 146.8\ \text{kPa},$$ and the circle centre is $s=\tfrac12(\sigma_1+\sigma_3)=\sigma_n + r\sin\phi = 200 + 146.8\sin27.7^\circ = 268.2$ kPa.
  3. Principal stresses. $$\sigma_1 = s + r = 268.2 + 146.8 = \boxed{415\ \text{kPa}},\qquad \sigma_3 = s - r = 268.2 - 146.8 = \boxed{121\ \text{kPa}}.$$ Cross-check with $\sigma_1 = \sigma_3\tan^2(45^\circ+\phi/2) + 2c\tan(45^\circ+\phi/2) = 121.4(2.73)+2(25)(1.65) = 415$ kPa. ✓
  4. (c) Stresses on the failure plane. In the direct-shear apparatus failure occurs on the horizontal plane on which the load is applied, and this is exactly the tangent point of the Mohr circle. Hence on the failure plane $\sigma_{ff}=\sigma_n=200$ kPa and $\tau_{ff}=130$ kPa. (Equivalently, $\sigma_{ff}=s-r\sin\phi=200$ kPa and $\tau_{ff}=r\cos\phi=130$ kPa, and the plane is inclined at $45^\circ+\phi/2 = 58.9^\circ$ to the major principal plane.)
Final results — Question 5
QuantityValue
Cohesion $c$ / friction angle $\phi$25 kPa / $27.7^\circ$
Major principal stress $\sigma_1$415 kPa
Minor principal stress $\sigma_3$121 kPa
Failure plane $(\sigma_{ff},\tau_{ff})$(200 kPa, 130 kPa)
Failure plane orientation$58.9^\circ$ from major principal plane
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