Axial tension and inclined plane: A prismatic member carries tensile force P over a normal cross-sectional area A (uniform). What is the normal stress on a plane inclined at angle θ to the transverse (normal) cross-section?

Introduction / Context:Stress transformation on inclined planes is a key topic in mechanics of materials. Even under simple axial tension, planes cut at an angle experience different normal and shear components of stress than the basic P/A on the transverse plane.

Given Data / Assumptions:

• Member under axial tensile force P.
• Uniform normal cross-sectional area A.
• Plane considered is inclined at angle θ to the transverse (perpendicular-to-axis) plane.

Concept / Approach:The force resultant across the inclined section equals P (equilibrium). The area of the inclined section increases as A_inclined = A / cos θ. The traction vector on the plane resolves into normal and shear components governed by geometry. The well-known results for axial tension are: σ_n = (P/A) * cos^2 θ and τ = (P/A) * sin θ * cos θ.

Step-by-Step Solution:Compute inclined area: A_inclined = A / cos θ.Traction magnitude on the plane: p = P / A_inclined = (P/A) * cos θ.Normal component: σ_n = p * cos θ = (P/A) * cos^2 θ.

Verification / Alternative check:At θ = 0°, σ_n = P/A (as expected). At θ = 90°, σ_n = 0 (the plane parallels the load and carries no normal stress).

Why Other Options Are Wrong:(P/A) * sin^2 θ: gives zero at θ = 0°, contradicting basic axial stress.(P/A) * tan θ or * sec^2 θ: not dimensionally or physically consistent for σ_n.Zero: only at θ = 90°, not general.

Common Pitfalls:Confusing angle definitions (to the axis vs to the transverse plane) and mixing normal stress with shear stress expressions.

Final Answer:(P / A) * cos^2 θ.