Combined plane stress: A body has tensile stresses of 1200 MPa and 600 MPa on mutually perpendicular planes, with a shear stress of 400 MPa on the same planes. What is the maximum in-plane shear stress?

Difficulty: Medium

Correct Answer: 500 MPa

Explanation:


Introduction / Context:
In a general plane stress state, the maximum in-plane shear governs yielding for Tresca and influences fatigue assessments. It equals the radius of Mohr's circle constructed from the normal and shear components.



Given Data / Assumptions:

  • σx = 1200 MPa (tension), σy = 600 MPa (tension).
  • τxy = 400 MPa (shear on the same planes).
  • Plane stress; linear elasticity.


Concept / Approach:
Maximum in-plane shear stress is given by:tau_max = √[ ((σx − σy)/2)^2 + τxy^2 ]Geometrically, this is the radius of Mohr's circle.



Step-by-Step Solution:

Compute half-difference: (σx − σy)/2 = (1200 − 600)/2 = 300 MPa.Square and add: 300^2 + 400^2 = 90000 + 160000 = 250000.Take square root: tau_max = √250000 = 500 MPa.


Verification / Alternative check:
Plot Mohr’s circle with points (σx, τxy) and (σy, −τxy). The radius measured horizontally equals 500 MPa, consistent with the computation.



Why Other Options Are Wrong:
400 MPa: ignores the normal stress difference.900 MPa and 1400 MPa: overestimates inconsistent with the circle radius.300 MPa: equals (σx − σy)/2 only, not including shear.



Common Pitfalls:
Using τmax = |σx − σy| / 2 (only valid when τxy = 0); arithmetic errors when squaring and adding.



Final Answer:

500 MPa

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