Mohr’s Circle – Pure Shear at 80 N/mm^2 An element is in a state of pure shear with shear stress magnitude τ = 80 N/mm^2. What is the maximum principal stress at that point?

Introduction / Context:
Pure shear is a fundamental stress state frequently encountered near beam webs, around fastener holes, or in thin-walled torsion members. Determining the principal stresses from a known shear stress using Mohr’s circle is a core mechanics-of-materials skill.

Given Data / Assumptions:

• Stress state: pure shear (σx = σy = 0, τxy = ±80 N/mm^2).
• Sign convention: magnitude is 80 N/mm^2; only absolute values matter for principal magnitudes.
• Plane stress condition.

Concept / Approach:

For pure shear, the Mohr’s circle is centered at the origin with radius equal to |τ|. Principal stresses are located at the intersection of the circle with the σ-axis, yielding symmetric values ±|τ|. Hence the maximum principal stress equals the shear magnitude.

Step-by-Step Solution:

Verification / Alternative check:

Stress transformation equations give σ1,2 = (σx + σy)/2 ± sqrt[((σx − σy)/2)^2 + τ^2]. Substituting σx = σy = 0 directly yields ±τ, confirming the result.

Why Other Options Are Wrong:

• 113.14 and 120 N/mm^2: Not associated with pure shear principal magnitudes; 113.14 might be misapplied combination (e.g., τ * √2) which corresponds to maximum shear from principal, not maximum principal from shear.
• 56.57 N/mm^2: Equal to 80/√2, another common mix-up with maximum in-plane shear for a biaxial case.

Common Pitfalls:

Confusing maximum principal stress with maximum shear stress; misplacing the Mohr’s circle center; mixing up relations for combined normal and shear stress states.

Final Answer:

80 N/mm^2