Effective and Total Principal Stresses at Failure from Mohr–Coulomb Shear Parameters: 22-Agric-A2 Soil Physics and Mechanics, May 2014, Q7

Question 7: Effective and Total Principal Stresses at Failure from Mohr–Coulomb Shear Parameters (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.

Quantity	Value
Effective friction angle, φ′	$30^\circ$
Effective cohesion, c′	10 kN/m²
Stress ratio at failure	$\sigma_3' = \sigma_1'/10$
Pore pressure at failure, u	175.3 kN/m²

Find. Effective principal stresses at failure (a); total principal stresses at failure (b); angle of shear failure θ_f (c).

Approach. Combine the Mohr–Coulomb failure criterion in principal-stress form, $\sigma_1' = \sigma_3'\tan^2(45+\phi'/2) + 2c'\tan(45+\phi'/2)$, with the given stress ratio $\sigma_3' = \sigma_1'/10$ to solve simultaneously for the two effective principal stresses, add the pore pressure to get totals, and use $\theta_f = 45+\phi'/2$ for the failure-plane angle.

Mohr circle at failure: the circle through σ3′ and σ1′ is tangent to the c′–φ′ strength envelope.

Failure criterion coefficients. With $\phi' = 30^\circ$, $$N_\phi = \tan^2(45+\phi'/2) = \tan^2(60^\circ) = 3, \qquad \sqrt{N_\phi} = \sqrt{3} = 1.732$$
a) Solve for the effective principal stresses. Substituting $\sigma_1' = 10\sigma_3'$ (from the given ratio) into $\sigma_1' = N_\phi\sigma_3' + 2c'\sqrt{N_\phi}$, $$10\sigma_3' = 3\sigma_3' + 2(10)(1.732) \ \Rightarrow\ 7\sigma_3' = 34.64\ \Rightarrow\ \sigma_3' = \boxed{4.95\ \text{kN/m}^2}$$ $$\sigma_1' = 10\sigma_3' = \boxed{49.49\ \text{kN/m}^2}$$
b) Total principal stresses. Adding the given pore pressure to each effective principal stress, $$\sigma_1 = \sigma_1' + u = 49.49 + 175.3 = \boxed{224.79\ \text{kN/m}^2}, \qquad \sigma_3 = \sigma_3' + u = 4.95 + 175.3 = \boxed{180.25\ \text{kN/m}^2}$$
c) Angle of shear failure. The failure plane forms at the classical Mohr–Coulomb angle from the major principal plane, $$\theta_f = 45 + \frac{\phi'}{2} = 45 + 15 = \boxed{60^\circ}$$

Quantity	Value
Effective minor principal stress, σ3′	4.95 kN/m²
Effective major principal stress, σ1′	49.49 kN/m²
Total minor principal stress, σ3	180.25 kN/m²
Total major principal stress, σ1	224.79 kN/m²
Angle of shear failure, θ_f	60° from the major principal plane

Concept & Theory

Topic: Mohr–Coulomb failure criterion in principal-stress form.

Key relations: $\sigma_1' = \sigma_3'\tan^2(45+\phi'/2) + 2c'\tan(45+\phi'/2)$; $\theta_f = 45 + \phi'/2$ measured from the plane of the major principal stress.

Why this works: This algebraic form is exactly equivalent to constructing the Mohr circle through $\sigma_1'$ and $\sigma_3'$ and requiring it to be tangent to the straight-line envelope $\tau = c' + \sigma'\tan\phi'$ — it avoids an iterative graphical construction and, combined with any second relationship between the two principal stresses (here, the given ratio), turns the failure condition into a single linear equation.

Common pitfall: Mixing effective and total stress within the same Mohr–Coulomb equation — c′ and φ′ are effective-stress strength parameters and must be paired with effective principal stresses; only after solving for σ1′ and σ3′ should the pore pressure be added back to get the total stresses asked for in part (b).

Source: R.F. Craig, Craig's Soil Mechanics, 9th ed., Ch. 6 (shear strength); B.M. Das, Principles of Geotechnical Engineering, 9th ed., Ch. 6.

Question 7 of 7: Effective and Total Principal Stresses at Failure from Mohr–Coulomb Shear Parameters

Question 7: Effective and Total Principal Stresses at Failure from Mohr–Coulomb Shear Parameters (20 marks)