Shear strength and failure planes — which statement about shear stress on the failure plane at limiting equilibrium is correct?

Introduction / Context:
In Mohr–Coulomb failure, a soil element reaches limiting equilibrium when the Mohr circle touches the strength line tau = c + sigma' * tan(phi). The plane of failure is not generally the plane of maximum shear stress; rather, it is oriented such that the shear and normal stresses satisfy the failure criterion simultaneously.

Given Data / Assumptions:

• Mohr–Coulomb behavior with friction angle phi and cohesion c.
• Failure occurs when stresses reach the envelope.
• Principal planes carry zero shear; maximum shear occurs on planes at 45 degrees to principal planes for phi = 0.

Concept / Approach:

The plane of maximum shear stress is oriented at 45 degrees to the principal planes when phi = 0 and near 45 degrees for small phi, but the failure plane for Mohr–Coulomb is inclined at 45 + phi/2 to the major principal plane. Consequently, the shear stress on the failure plane is not the absolute maximum in the element; it is the critical combination of shear and normal stress that first meets the envelope.

Step-by-Step Solution:

Verification / Alternative check:

Plot the Mohr circle and draw the failure line; the tangency point corresponds to a plane different from the one at which tau is maximized when phi > 0.

Why Other Options Are Wrong:

(a) contradicts Mohr–Coulomb geometry. (c) is also incorrect; at phi = 0, failure plane is at 45 degrees but the criterion reduces to tau = c, not necessarily the absolute max-shear plane in general stress states. (e) false because principal planes have zero shear.

Common Pitfalls:

Assuming failure occurs on the plane of maximum shear stress; forgetting the role of normal stress in frictional resistance.

Final Answer:

the failure plane does not carry the maximum shear stress