Maximum shear orientation: The plane (or filament) on which maximum shear stress occurs makes what angle with the horizontal plane?

Introduction / Context:
In Mohr’s circle representation of stress, principal planes carry zero shear and planes oriented midway between principal planes carry maximum shear. For many geotechnical states (e.g., vertical overburden giving vertical major principal stress), understanding the orientation of max shear aids in interpreting failure planes and crack inclinations.

Given Data / Assumptions:

• Plane stress with well-defined principal directions.
• Major and minor principal planes are orthogonal.
• We consider the angle between the plane of maximum shear and the horizontal.

Concept / Approach:

Maximum shear occurs on planes at 45° to the principal planes. If the major principal plane is vertical (common under gravity loading where σ_v is major and σ_h is minor), then the plane of maximum shear is at 45° to the vertical and hence at 45° to the horizontal as well. This 45° rule is a cornerstone for interpreting failure plane directions in simple triaxial and direct shear contexts.

Step-by-Step Solution:

Verification / Alternative check:

From Mohr’s circle, the angular shift from the principal plane to the max-shear plane is 45° (2θ relationship in the circle).

Why Other Options Are Wrong:

30° and 60° correspond to other stress states or frictional failure inclinations; 90° is a principal plane, not a max-shear plane.

Common Pitfalls:

Confusing the inclination of failure planes in Mohr–Coulomb (often 45° + φ/2 to the major principal stress) with the pure max-shear plane; the two are not identical when frictional strength is involved.

Final Answer:

45°