Planes of maximum shear under biaxial stress: For a biaxial state with principal planes defined, what is the orientation of the planes of maximum shear stress?

Difficulty: Easy

Correct Answer: True

Explanation:

Introduction / Context:Understanding the orientation of maximum shear stresses relative to principal planes is fundamental for failure analysis and for interpreting Mohr’s circle in biaxial stress states.

Given Data / Assumptions:

  • Biaxial stress condition with defined principal planes (no in-plane shear on those planes).
  • Linear elasticity; standard sign conventions.

Concept / Approach:Principal planes carry only normal stress; shear is zero there. Maximum shear planes occur midway between principal planes. In physical space, this corresponds to a 45° inclination to principal planes, and two orthogonal maximum-shear planes exist, 90° apart from each other.

Step-by-Step Solution:

On Mohr’s circle, principal planes lie at the horizontal intercepts where τ = 0.Maximum shear occurs at the top and bottom of the circle (τ = R), 90° on Mohr’s circle, which translates to 45° in the physical plane.Hence, the two planes of maximum shear are orthogonal and each is 45° to the principal planes.

Verification / Alternative check:Transforming stresses via the stress transformation equations and differentiating τ(θ) with respect to θ yields the same 45° condition for extremum shear.

Why Other Options Are Wrong:

  • Any angle other than 45° contradicts the transformation result.
  • Coincidence with principal planes is impossible because shear there is zero by definition.

Common Pitfalls:Mixing Mohr’s-circle angle (2θ) with the physical angle θ; forgetting the two orthogonal maximum-shear planes; sign convention errors.

Final Answer:True

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