Mohr’s circle interpretation: Is the maximum shear stress equal to the radius of Mohr’s circle for a plane-stress state?

Introduction / Context:Mohr’s circle graphically represents stress transformation. The circle’s center, radius, and key points map directly to principal and shear stresses, providing intuitive insight into maximum values and orientations.

Given Data / Assumptions:

• Plane-stress condition with σx, σy, and τxy.
• Standard sign convention and Mohr’s circle construction.

Concept / Approach:The center of Mohr’s circle is at (σx + σy)/2. The radius equals sqrt( ((σx − σy)/2)^2 + τxy^2 ). The maximum in-plane shear stress is precisely this radius, occurring at the top and bottom of the circle (90° on Mohr’s circle, 45° in physical space from principal directions).

Step-by-Step Solution:

Verification / Alternative check:From transformation equations, τ^2 + [σ − (σx + σy)/2]^2 = R^2. The peak shear equals the circle’s radius by geometry and by differentiation of the shear-stress transformation formula.

Why Other Options Are Wrong:

• Restrictions to pure shear or σx = σy are unnecessary; the result holds generally in plane stress.
• “True for 3D only” is incorrect; a similar notion exists in 3D but here we address plane stress specifically.

Common Pitfalls:Mixing up the diameter and radius; forgetting the 2:1 mapping between physical angle and Mohr angle; sign mistakes when plotting τxy.

Final Answer:Correct