Combined Stress — Maximum Principal (Normal) Stress A body is under a direct tensile stress of 300 MPa in one principal plane and a simple shear stress of 200 MPa. Determine the maximum normal (principal) stress.

Difficulty: Medium

Correct Answer: 400 MPa

Explanation:


Introduction:
This problem tests Mohr’s circle or principal stress formulas for a state with one normal stress and in-plane shear. We need the maximum principal (normal) stress.


Given Data / Assumptions:

  • σx = 300 MPa (tension), σy = 0 MPa (no perpendicular direct stress stated).
  • τxy = 200 MPa (simple shear).
  • Linear elasticity; plane stress condition.


Concept / Approach:
Principal stresses for plane stress: σ1,2 = (σx + σy)/2 ± sqrt( ((σx - σy)/2)^2 + τxy^2 )


Step-by-Step Solution:
Compute the average: (σx + σy)/2 = (300 + 0)/2 = 150 MPa.Compute the radius: sqrt( ((σx - σy)/2)^2 + τxy^2 ) = sqrt( (300/2)^2 + 200^2 ) = sqrt(150^2 + 200^2 ) = sqrt(22500 + 40000) = sqrt(62500) = 250 MPa.Therefore, σ1 = 150 + 250 = 400 MPa (maximum); σ2 = 150 - 250 = -100 MPa (minimum).


Verification / Alternative check:
The computed pair (400 MPa, -100 MPa) preserves the invariant σ1 + σ2 = σx + σy = 300 MPa, confirming consistency.


Why Other Options Are Wrong:

  • -100 MPa: This is the minimum principal stress, not the maximum.
  • 250 MPa: Not a principal stress here; it is the Mohr circle radius.
  • 300 MPa: The applied direct stress, not accounting for shear transformation.


Common Pitfalls:
Forgetting σy = 0, misusing average vs radius, or adding magnitudes incorrectly leads to wrong answers.


Final Answer:
400 MPa

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