Shear Strength / Slope Stability: 18-Geol-A6 Soil Mechanics, May 2015, Q3

Question 3: Shear Strength / Slope Stability (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. Cohesionless sand at failure, $\phi'=30^{\circ}$, $c'=0$. On a plane $\theta=30^{\circ}$ from horizontal: $\sigma_n=50$ kPa, $\tau=10$ kPa. Two Mohr circles are tangent to the failure envelope and pass through $(\sigma_n,\tau)$.

Find. (a) $\sigma_1$, $\sigma_3$ and the orientation of the major principal plane, using the smaller circle. (b) $\sigma_n$, $\tau$ on the actual failure plane.

Approach. For $c'=0$, a Mohr circle of centre $C$ and radius $R$ is at failure when $R=C\sin\phi'$. Writing the stress-transformation equations for the known point in terms of $C$ and the angle $\psi$ between the given plane and the major principal plane gives two equations in two unknowns; the quadratic in $C$ has two roots — the smaller is the required circle.

Figure — both candidate Mohr circles (tangent to the $\phi'=30^{\circ}$ envelope) through the given point $(50,10)$ kPa; the smaller circle governs Part (a).

Set up the two circle equations. With $\sigma_n=C+R\cos2\psi$, $\tau=R\sin2\psi$, and tangency $R=C\sin\phi'$, eliminating $\psi$ gives a quadratic in $C$: $$(\sigma_n-C)^2+\tau^2=(C\sin\phi')^2\ \Rightarrow\ (1-\sin^2\phi')C^2-2\sigma_nC+(\sigma_n^2+\tau^2)=0$$ Substituting $\sigma_n=50$, $\tau=10$, $\phi'=30^{\circ}$ ($\sin\phi'=0.5$) gives $0.75\,C^2-100\,C+2600=0$, with roots $$C=\boxed{35.40\text{ kPa (smaller)}}\quad\text{and}\quad C=97.94\text{ kPa (larger)}$$
Part (a) — principal stresses, smaller circle. $R=C\sin\phi'=35.40(0.5)=17.70$ kPa. $$\sigma_1=C+R=35.40+17.70=\boxed{53.10\text{ kPa}},\qquad \sigma_3=C-R=35.40-17.70=\boxed{17.70\text{ kPa}}$$
Part (a) — orientation. From $\cos2\psi=(\sigma_n-C)/R=(50-35.40)/17.70=0.825$ and $\sin2\psi=\tau/R=10/17.70=0.565$, both consistent at $2\psi=34.4^{\circ}$, so $\psi=17.2^{\circ}$ is the angle between the given plane and the major principal plane. Since the given plane is at $\theta=30^{\circ}$ from horizontal, $$\alpha=\theta-\psi=30-17.2=\boxed{12.8^{\circ}\text{ from horizontal}}$$ is the orientation of the major principal plane ($\sigma_1$ acts normal to it, i.e.\ along a line $12.8^{\circ}+90^{\circ}=102.8^{\circ}$ from horizontal).
Part (b) — stresses on the actual failure plane. At the Mohr-circle tangent point, $2\psi_f=90^{\circ}+\phi'$, which reduces (using $R=C\sin\phi'$) to $$\sigma_{n,f}=C\cos^2\phi'=35.40(0.866)^2=\boxed{26.55\text{ kPa}},\qquad \tau_f=C\sin\phi'\cos\phi'=35.40(0.5)(0.866)=\boxed{15.33\text{ kPa}}$$ (check: $\tau_f/\sigma_{n,f}=0.577=\tan30^{\circ}$ ✓). The failure plane itself is oriented $45^{\circ}+\phi'/2=60^{\circ}$ from the major principal plane, i.e.\ at $\approx12.8^{\circ}+60^{\circ}=72.8^{\circ}$ from horizontal (with a conjugate plane at $12.8^{\circ}-60^{\circ}$).

Quantity	Value
Circle centre $C$ / radius $R$ (smaller)	35.40 / 17.70 kPa
$\sigma_1$, $\sigma_3$	53.10 kPa, 17.70 kPa
Major principal plane, from horizontal	12.8°
Failure-plane $\sigma_n$, $\tau$	26.55 kPa, 15.33 kPa
Failure plane, from horizontal	≈72.8° (conjugate at −47.2°)

Part (2). All limit-equilibrium slope-stability methods share the same underlying logic, regardless of how they subdivide the sliding mass:

A trial slip surface (circular, log-spiral, or an arbitrary composite surface) is assumed, and the soil mass above it is treated as a free body, or divided into a series of vertical slices when the surface or the stratigraphy is irregular. The factor of safety is defined as the ratio of the total resisting shear strength available along the surface to the total driving shear stress actually mobilised to maintain equilibrium:

$$FS=\frac{\sum \tau_f\,\Delta\ell}{\sum \tau_{mobilised}\,\Delta\ell}=\frac{\sum\big[c'+(\sigma_n-u)\tan\phi'\big]\Delta\ell}{\sum \tau_{mobilised}\,\Delta\ell}$$

Static equilibrium (force and/or moment) of the sliding mass, or of each slice, is enforced to solve for the mobilised shear and normal stresses along the surface; different methods (Fellenius, Bishop, Janbu, Spencer, Morgenstern–Price) differ only in which equilibrium equations they satisfy and what simplifying assumption they make about the inter-slice forces, which is why they can give slightly different FS values for the same geometry. The process is then repeated for many trial surfaces, and the surface giving the minimum FS is taken as the critical (most likely) failure surface. A slope is judged adequate when the minimum FS exceeds a code-specified target (typically 1.3–1.5 for static long-term conditions).

Concept & Theory

Topic. Mohr-circle representation of stress at failure (Part 1) and the limit-equilibrium framework for slope stability (Part 2).

Key relations. Failure envelope $\tau_f=c'+\sigma_n'\tan\phi'$; tangency condition $R=C\sin\phi'$ (cohesionless); failure-plane orientation $\theta_f=45^{\circ}+\phi'/2$ from the major principal plane; $FS=\sum\tau_f\Delta\ell/\sum\tau_{mob}\Delta\ell$.

Why this works. A Mohr circle tangent to the strength envelope represents the unique stress state at which the soil is on the verge of shearing on some plane; because two different orientations of the same physical stress state can produce the identical $(\sigma_n,\tau)$ pair on a given plane, two distinct circles (two different absolute stress magnitudes) can both satisfy one measured point, which is exactly the ambiguity this question exploits. Limit equilibrium works because, at incipient failure, the mobilised shear strength along the trial surface exactly balances gravity-driven demand; searching over many trial surfaces for the minimum FS approximates finding the true, unknown failure surface without solving the full stress field.

Common pitfall. Assuming the "smaller" and "larger" circles are simply arithmetic mistakes of each other — both are mathematically valid tangent circles through the same point; the exam's own wording ("two Mohr circles satisfy these conditions") is deliberately testing whether the candidate solves the full quadratic rather than assuming a single solution.

Source. Craig, Craig's Soil Mechanics, Ch. 5 (Shear Strength) and Ch. 9 (Stability of Slopes).

Question 3 of 6: Shear Strength / Slope Stability

Question 3: Shear Strength / Slope Stability (20 marks)