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:
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:
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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