Shear strength of soil and stability of slopes

Shear strength is the maximum resistance a soil can mobilise against sliding along a plane. Almost every geotechnical failure — bearing failure of foundations, sliding of retaining walls, landslides on Nepal's hill roads — is a shear failure. This note explains stresses on planes using principal stresses and Mohr's circle, the Mohr–Coulomb failure criterion, the laboratory shear tests and the basic methods of slope stability analysis.

Key concepts

Principal planes and principal stresses

• At any point there are three mutually perpendicular planes with zero shear stress — the principal planes. Normal stresses on them are the major ($\sigma_1$), intermediate ($\sigma_2$) and minor ($\sigma_3$) principal stresses.
• Maximum shear stress $\tau_{max} = (\sigma_1 - \sigma_3)/2$ acts on planes at 45° to the principal planes.

Stresses on an inclined plane

For a plane inclined at $\theta$ to the major principal plane: $\sigma_\theta = \frac{\sigma_1+\sigma_3}{2} + \frac{\sigma_1-\sigma_3}{2}\cos2\theta$ $\tau_\theta = \frac{\sigma_1-\sigma_3}{2}\sin2\theta$

• Mohr's circle: centre $(\sigma_1+\sigma_3)/2$, radius $(\sigma_1-\sigma_3)/2$; a plane at $\theta$ in reality is at $2\theta$ on the circle. The pole (origin of planes) method locates any plane.

Mohr–Coulomb theory

$\tau_f = c + \sigma\tan\phi\quad\text{(total)};\qquad \tau_f = c' + \sigma'\tan\phi'\quad\text{(effective)}$

• $c$ = cohesion (kPa), $\phi$ = angle of internal friction, $\sigma$ = normal stress on failure plane.
• Failure occurs when Mohr's circle touches the failure envelope.
• Failure plane is inclined at $\theta_f = 45^\circ + \phi/2$ to the major principal plane.
• Typical: clean sand $c = 0$, $\phi$ = 28–40°; saturated clay in undrained condition $\phi_u = 0$, $\tau_f = c_u$.

Principal stress relation at failure

$\sigma_1 = \sigma_3\tan^2\!\left(45^\circ+\frac{\phi}{2}\right) + 2c\tan\!\left(45^\circ+\frac{\phi}{2}\right)$ For $c = 0$: $\sigma_1/\sigma_3 = \tan^2(45^\circ+\phi/2) = \dfrac{1+\sin\phi}{1-\sin\phi}$.

Types of shear tests

Test	Features
Direct shear (box)	Failure plane forced (horizontal); quick, simple; drainage poorly controlled; best for sands
Triaxial	Cylinder in cell; controlled drainage and pore-pressure measurement; UU, CU, CD tests; most versatile
Unconfined compression (UCS)	Triaxial with $\sigma_3 = 0$; saturated clays; $c_u = q_u/2$
Vane shear	Field/lab, soft sensitive clays; $c_u = \dfrac{T}{\pi D^2(H/2 + D/6)}$
Ring shear	Residual strength at large strain

Drainage conditions:

• UU (quick): no drainage; $\phi_u = 0$ for saturated clay — short-term (end of construction) stability.
• CU: consolidated, then sheared undrained; with pore pressure gives $c'$, $\phi'$.
• CD (slow): drained throughout; long-term stability.
• Sensitivity $S_t$ = undisturbed strength ÷ remoulded strength (quick clays > 16).

Stability of slopes

• Factor of safety $F = \tau_f/\tau_{mobilised}$.
• Infinite slope (long, uniform):
  • Dry cohesionless: $F = \dfrac{\tan\phi}{\tan\beta}$ (stable if $\beta < \phi$).
  • Submerged cohesionless: same as dry.
  • With steady seepage parallel to slope: $F = \dfrac{\gamma'}{\gamma_{sat}}\dfrac{\tan\phi}{\tan\beta}$ ≈ half the dry value.
  • Cohesive: stability number concept; critical depth $H_c = \dfrac{c}{\gamma\cos^2\beta(\tan\beta - \tan\phi)}$.
• Finite slopes: Swedish circle (φ = 0 method, $F = \dfrac{c_u L_a R}{W x}$), method of slices (Fellenius, Bishop's simplified), friction circle method, Taylor's stability number $S_n = \dfrac{c}{F\gamma H}$.
• Types of failure: face, toe, base (slope) failure; translational and rotational slides.
• Stabilisation: flattening slope, benching, drainage (surface and sub-surface — most important in monsoon Nepal), retaining structures, gabions, bioengineering, rock bolts, soil nailing.

High-yield points

• Failure plane at $45^\circ + \phi/2$ to the major principal plane; maximum shear plane at 45°.
• Mohr's circle: centre $(\sigma_1+\sigma_3)/2$, radius $(\sigma_1-\sigma_3)/2$; angles double on the circle.
• UU test on saturated clay gives $\phi_u = 0$; UCS gives $c_u = q_u/2$.
• Sensitivity = undisturbed ÷ remoulded strength.
• Infinite sand slope: $F = \tan\phi/\tan\beta$; seepage parallel to slope roughly halves it.

Formulas

• $\sigma_\theta = \dfrac{\sigma_1+\sigma_3}{2} + \dfrac{\sigma_1-\sigma_3}{2}\cos2\theta$; $\tau_\theta = \dfrac{\sigma_1-\sigma_3}{2}\sin2\theta$.
• $\tau_f = c + \sigma\tan\phi$; $\theta_f = 45^\circ + \phi/2$.
• $\sigma_1 = \sigma_3\tan^2(45^\circ+\phi/2) + 2c\tan(45^\circ+\phi/2)$.

Worked example

A dry sand ($c = 0$, $\phi = 30^\circ$) is tested in triaxial with $\sigma_3 = 100$ kPa. Find $\sigma_1$ at failure, the failure-plane angle and stresses on it.

• $\tan^2(60^\circ) = 3$ → $\sigma_1 = 300$ kPa
• $\theta_f = 45 + 15 = 60^\circ$
• $\sigma_f = 200 + 100\cos120^\circ = 200 - 50 = 150$ kPa
• $\tau_f = 100\sin120^\circ = 86.6$ kPa; check: $150\tan30^\circ = 86.6$ ✓

Exam tips

• Failure plane at $45^\circ + \phi/2$ to major principal plane; maximum shear plane at 45° (not the failure plane unless φ = 0).
• UCS: $c_u = q_u/2$; UU test of saturated clay gives $\phi_u = 0$.
• Infinite dry sand slope: $F = \tan\phi/\tan\beta$; seepage parallel to slope roughly halves $F$.
• Direct shear forces the failure plane; triaxial lets it form naturally.
• Drainage is the cheapest and most effective way to stabilise hill slopes.

Exam-style questions

From Calmon's question bank, which has 25 questions on this topic.

1. An infinite cohesionless slope has a factor of safety of 1.6 when dry. If seepage then occurs parallel to the slope with the water table at the surface ($\gamma_{sat} = 20$ kN/m³, $\gamma_w = 10$ kN/m³), the factor of safety becomes

• A. 1.6
• B. 1.2
• C. 3.2
• D. 0.8

Answer: D. With seepage parallel to the slope, $FS = \dfrac{\gamma'}{\gamma_{sat}}\dfrac{\tan\phi}{\tan\beta} = \dfrac{10}{20} \times 1.6 = 0.8$.

2. A main limitation of the direct shear test compared with the triaxial test is that

• A. It measures pore water pressure only
• B. It requires a cylindrical specimen
• C. It cannot be performed on sand
• D. The failure plane is predetermined and drainage cannot be properly controlled

Answer: D. In the direct shear box the sample is forced to fail along the horizontal plane between the two halves, and drainage and pore pressure cannot be controlled or measured as in a triaxial test.

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