Uniaxial loading and oblique planes (normal stress): A prismatic body is subjected to a direct tensile stress σ. What is the normal stress on an oblique section whose normal makes an angle θ with the load direction?

Introduction / Context:Stress transformation allows us to find the normal stress acting on any inclined plane under a uniaxial load. This is essential for analyzing potential failure planes and deriving Mohr’s circle relations.

Given Data / Assumptions:

• Uniaxial tensile stress σ.
• Plane with normal at angle θ to the load axis.
• Linear elasticity; small deformation; continuum assumptions.

Concept / Approach:Using transformation formulas, the normal stress σ_n on a plane whose normal is at angle θ to the stress direction is σ_n = σ cos^2 θ. Equivalently, using the double-angle identity, σ_n = σ/2 (1 + cos 2θ).

Step-by-Step Solution:

Verification / Alternative check:Mohr’s circle yields the same result: the abscissa at angle 2θ from the loading point gives σ_n = σ/2 (1 + cos 2θ).

Why Other Options Are Wrong:

• σ cos θ and σ sin θ are not correct for normal stress on the oblique plane.
• σ sin^2 θ is the complementary form for a plane normal at 90°−θ.
• While σ/2 (1 + cos 2θ) is mathematically equivalent, the explicitly squared form is the standard direct expression; both are correct, but the single best choice requested is σ cos^2 θ.

Common Pitfalls:Confusing angle to the plane with angle to the plane’s normal; mixing normal and shear stress expressions; sign errors in transformation.

Final Answer:σ cos^2 θ