Plane stress under pure shear: for a rectangular element under pure shear of intensity q, the principal planes and principal stresses are what values?

Introduction / Context:
Mohr’s circle analysis shows that pure shear stress states produce principal normal stresses oriented at ±45° to the original element faces. This result is widely used in mechanics of materials and failure criteria.

Given Data / Assumptions:

• State of stress: τ_xy = q, σ_x = σ_y = 0 (pure shear).
• We seek principal planes (angles) and principal stresses (σ1, σ2).

Concept / Approach:
Principal planes are those with zero shear stress and extremal normal stress. For pure shear, rotation by 45° eliminates shear on the rotated faces, yielding normal stresses equal in magnitude and opposite in sign.

Step-by-Step Solution:
Mohr’s circle center at (σ_avg, 0) = (0, 0) with radius R = q.Principal stresses: σ1 = +q, σ2 = -q.Angle from original x-plane to principal plane: 2θ_p = 90° → θ_p = 45°.Thus principal planes at 45° and 135° to the original faces, with ±q as principal stresses.

Verification / Alternative check:
Direct transformation equations also give σ_θ = ±q at θ = 45° and τ_θ = 0.

Why Other Options Are Wrong:

• 0°/90° or 30°/120° do not nullify shear under pure shear.
• Option with missing magnitudes is incomplete; correct values are ±q.

Common Pitfalls:

• Confusing the physical angle θ with 2θ on Mohr’s circle.

Final Answer:
45°, 135°; + q and - q