Plane stress transformation: A body has normal stresses of 1200 MPa (tension) and 600 MPa (tension) on two perpendicular planes, with a shear stress of 400 MPa on the same planes. What is the minimum (least) principal normal stress?

Introduction / Context:Principal stresses and maximum shear stresses in plane stress are central to failure theories and design. Given two perpendicular normal stresses and an in-plane shear, we can compute principal values using transformation equations or Mohr’s circle.

Given Data / Assumptions:

• Sigma_x = 1200 MPa (tension).
• Sigma_y = 600 MPa (tension).
• Tau_xy = 400 MPa (shear on the same planes).
• Plane stress; sign convention: tensile positive, shear as given.

Concept / Approach:Principal stresses are given by sigma_(1,2) = sigma_avg ± R, where sigma_avg = (sigma_x + sigma_y)/2 and R = sqrt( ((sigma_x − sigma_y)/2)^2 + tau_xy^2 ). The smaller of these is the minimum normal stress.

Step-by-Step Solution:

Verification / Alternative check:Mohr’s circle center at 900 MPa, radius 500 MPa; intersections at 1400 and 400 MPa confirm the computed values.

Why Other Options Are Wrong:

• 500 MPa and 900 MPa: intermediate values (average or misused components) not principal results.
• 1400 MPa: this is the maximum, not the minimum principal stress.

Common Pitfalls:Sign errors for shear; forgetting that both principal stresses can be tensile when both sigma_x and sigma_y are tensile.

Final Answer:400 MPa