Combined biaxial normal stresses with in-plane shear: For stresses σx and σy on perpendicular planes with shear τxy, what is the maximum shear stress τ_max in the material?

Introduction / Context:
State of stress at a point in a plane generally consists of two perpendicular normal stresses and an in-plane shear. Determining the maximum shear stress is key for yielding criteria (e.g., Tresca) and failure analysis. This is a staple result derived conveniently via Mohr’s circle or stress transformation equations.

Given Data / Assumptions:

• Plane stress state with normal stresses σx and σy, and in-plane shear τxy.
• Material is homogeneous and isotropic; linear elastic relations used for transformation.
• Sign convention consistent with standard mechanics texts.

Concept / Approach:
Using Mohr’s circle for plane stress, the circle center is at (σx + σy)/2 and radius R equals 0.5 * sqrt( (σx − σy)^2 + 4 * τxy^2 ). The maximum in-plane shear stress equals the radius of Mohr’s circle. Out-of-plane considerations are neglected here (plane stress).

Step-by-Step Solution:

Verification / Alternative check:
Special cases: if τxy = 0 → τ_max = |σx − σy| / 2, matching the uniaxial/biaxial no-shear case. If σx = σy → τ_max = |τxy|, as expected for pure shear superposed on hydrostatic stress.

Why Other Options Are Wrong:

• (σx + σy)/2 is the circle center, not τ_max.
• |σx − σy|/2 applies only when τxy = 0.
• sqrt( σx^2 + σy^2 + τxy^2 ) is not a recognized invariant for τ_max.
• 2 * τxy is incorrect except under nonstandard conditions.

Common Pitfalls:
Confusing radius with diameter; forgetting the factor 0.5; neglecting the 4 * τxy^2 term, which is critical when shear is significant.

Final Answer:
0.5 * sqrt( (σx − σy)^2 + 4 * τxy^2 )