Question 3 of 6: Shear Strength

Question 3: Shear Strength (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. Test A (CD triaxial, cohesionless sand): $\sigma_3'=100$ kPa, deviator stress at failure $\Delta\sigma=250$ kPa. Test B (direct shear, same sand): $\sigma_n=60$ kPa held constant, $\tau_f=40$ kPa at failure (the "initial horizontal stress" of 30 kPa is the pre-shear at-rest condition and does not enter the failure-state calculation).

Find. $c'$, $\phi'$; stresses on the failure plane and on the plane of $\tau_{max}$, both tests; principal stresses and plane orientations for Test B; a comparison and discussion of test reliability.

Approach. Both are single-test results on a sand, so $c'=0$ is taken (one Mohr circle tangent to an envelope through the origin fixes $\phi'$ uniquely). For the CD triaxial, $\sigma_1'=\sigma_3'+\Delta\sigma$ gives the circle directly. For the direct shear box the failure plane is forced horizontal by the apparatus, so the stress point $(\sigma_n,\tau_f)$ is the point of tangency; standard Mohr-circle geometry back-calculates $\sigma_1'$, $\sigma_3'$ from it.

Test A — CD triaxial

1. Failure stresses and $\phi'$. $\sigma_1'=\sigma_3'+\Delta\sigma=100+250=350$ kPa. Taking $c'=0$ (sand), the envelope is a line through the origin tangent to the one available circle: $$\sin\phi'=\frac{\sigma_1'-\sigma_3'}{\sigma_1'+\sigma_3'}=\frac{250}{450}=0.556 \Rightarrow \boxed{\phi'=33.7^\circ,\ c'=0}$$
2. Stress on the failure plane. Using the tangent-point relations $\sigma_{ff}=\tfrac12(\sigma_{1f}+\sigma_{3f})-\tfrac12(\sigma_{1f}-\sigma_{3f})\sin\phi'$ and $\tau_{ff}=\sigma_{ff}\tan\phi'$: $$\sigma_{ff}=225-125(0.556)=\boxed{155.6\text{ kPa}}\qquad \tau_{ff}=155.6\tan(33.7^\circ)=\boxed{103.9\text{ kPa}}$$
3. Failure-plane angle. $\alpha_{ff}=45^\circ+\phi'/2=45+16.9=\boxed{61.9^\circ}$ (measured from the plane of $\sigma_3'$, i.e. from horizontal on a vertically-loaded specimen).
4. Maximum shear stress and available strength on that plane. $\tau_{max}=\tfrac12(\sigma_1'-\sigma_3')=\boxed{125\text{ kPa}}$, on the $45^\circ$-plane where $\sigma_n=\tfrac12(\sigma_1'+\sigma_3')=225$ kPa. Available strength there is $\tau_{avail}=\sigma_n\tan\phi'=225\tan(33.7^\circ)=\boxed{150.3\text{ kPa}}>\tau_{max}$, so this plane does not govern (factor of safety $150.3/125=1.20$) — consistent with failure actually occurring on the steeper $61.9^\circ$ plane found above.

Test B — direct shear

1. Friction angle. The failure plane is forced horizontal, so $(\sigma_n,\tau_f)=(60,40)$ kPa is itself the tangent point of a $c'=0$ envelope: $$\phi'=\arctan\left(\frac{\tau_f}{\sigma_n}\right)=\arctan\left(\frac{40}{60}\right)=\boxed{33.7^\circ}$$ (matching Test A to within reading precision — both tests are on the same sand).
2. (a) Principal stresses at failure. For a $c'=0$ tangent point, the circle centre and radius follow from $\sigma_n=p\cos^2\phi'$, i.e. $p=\sigma_n/\cos^2\phi'$ and $R=p\sin\phi'$: $$p=\frac{60}{\cos^2(33.7^\circ)}=86.7\text{ kPa}\qquad R=86.7\sin(33.7^\circ)=48.1\text{ kPa}$$ $$\boxed{\sigma_1'=p+R=134.7\text{ kPa}}\qquad\boxed{\sigma_3'=p-R=38.6\text{ kPa}}$$ (check: $\sin\phi'=(\sigma_1'-\sigma_3')/(\sigma_1'+\sigma_3')=96.1/173.3=0.555$ — matches.)
3. (b) Orientation of the failure plane. Horizontal — fixed by the shear-box geometry (the plane between the upper and lower halves of the box).
4. (c) Orientation of the major principal plane. The failure plane always lies at $\theta_f=45^\circ+\phi'/2=45+16.85=61.85^\circ$ to the major principal plane; since the failure plane is horizontal, the major principal plane is inclined at $\boxed{61.85^\circ\text{ to horizontal}}$ (equivalently, $\sigma_1'$ acts $28.15^\circ$ from vertical).
5. (d) Plane of $\tau_{max}$, available strength, FoS. $\tau_{max}=R=\boxed{48.1\text{ kPa}}$ on the plane at $45^\circ$ to the principal planes ($\approx16.9^\circ$ to horizontal), where $\sigma_n=p=86.7$ kPa. Available strength $\tau_{avail}=p\tan\phi'=86.7\tan(33.7^\circ)=\boxed{57.8\text{ kPa}}$; $\text{FoS}=57.8/48.1=\boxed{1.20}$ — essentially identical to Test A's $1.20$ on its own $45^\circ$ plane. This is not a coincidence: on the $45^\circ$ plane $\sigma_n=p$, $\tau=p\sin\phi'$, so $\text{FoS}=\tan\phi'/\sin\phi'=1/\cos\phi'$ regardless of the stress magnitude — a function of $\phi'$ alone, which is why both tests (same soil, same $\phi'$) agree.
6. (e) Which test to trust. The CD triaxial (Test A) is the more reliable result. It applies a uniform, known stress state to the full specimen and lets the failure plane develop wherever the soil's own strength dictates, with drainage and volume change monitored throughout so the effective stress path is known exactly. The direct shear test, in contrast, forces failure onto a single predetermined horizontal plane regardless of whether that is the true weakest orientation, concentrates stress non-uniformly near the box edges, and cannot measure pore pressure directly (drainage is only inferred from a slow shearing rate) — so while it is quick and simple to run, its $\phi'$ is a reasonable estimate rather than a rigorous one. Test A procedure: saturate the specimen under back pressure, consolidate under the cell pressure $\sigma_3'$ with drainage open until volume change ceases, then shear axially at a slow, drained rate (valve open, no excess pore pressure) while recording axial load, displacement and volume change to failure. Test B procedure: place the specimen in the split shear box, apply the vertical (normal) load $\sigma_n$ and allow consolidation, then displace the upper half horizontally at a slow rate while recording shear force and both vertical and horizontal displacement until the shear force peaks (failure).