Combined stress state: A body has direct tensile stress σ = 300 MPa (one plane) and simple shear τ = 200 MPa. What is the maximum in-plane shear stress?

Introduction / Context:
When both normal and shear stresses act, the maximum shear stress must be computed using stress transformation or Mohr’s circle. This governs yielding under Tresca criterion and influences fatigue in multiaxial states.

Given Data / Assumptions:

• Plane stress with σx = 300 MPa, σy = 0 MPa, τxy = 200 MPa.
• Linear elasticity; sign conventions standard.
• We seek the maximum in-plane shear stress.

Concept / Approach:
Maximum in-plane shear is the radius of Mohr’s circle: tau_max = √[ ((σx − σy)/2)^2 + τxy^2 ]. This reduces to the pure-shear value σ/2 only when τxy = 0.

Step-by-Step Solution:

Verification / Alternative check:
Construct Mohr’s circle with points (300, 200) and (0, −200); the circle radius equals 250 MPa, confirming the calculation.

Why Other Options Are Wrong:
−100 MPa: shear magnitude cannot be negative; sign denotes sense only.300 or 400 MPa: overestimates; not aligned with the geometric radius.200 MPa: equals applied shear but ignores accompanying normal stress.

Common Pitfalls:
Using σ/2 blindly; neglecting the influence of τxy; arithmetic slips when squaring and adding.

Final Answer:

250 MPa