Uniaxial tension – Resultant (traction) on an inclined plane\nA body is subjected to a direct tensile stress σ. On a plane inclined at an angle θ to the normal cross-section, what is the magnitude of the resultant stress (traction) on that plane?

Introduction / Context:
In strength of materials, the traction vector on an inclined plane has a normal component (normal stress) and a tangential component (shear stress). For uniaxial tension, their combination yields a simple expression for the resultant magnitude.

Given Data / Assumptions:

• Uniform direct tensile stress σ along one axis.
• A plane is taken at angle θ to the normal cross-section (i.e., to the plane perpendicular to the load).
• Continuum mechanics relations for plane stress apply.

Concept / Approach:
Resolve traction into normal and shear components using standard transformation equations: σ_n = σ cos^2 θ and τ = σ sin θ cos θ. The resultant traction magnitude is the vector sum of these components.

Step-by-Step Solution:
Normal stress on plane: σ_n = σ cos^2 θ.Shear stress on plane: τ = σ sin θ * cos θ = 0.5 * σ * sin 2θ.Resultant: t = sqrt(σ_n^2 + τ^2).t = sqrt(σ^2 cos^4 θ + σ^2 sin^2 θ cos^2 θ) = σ * sqrt(cos^4 θ + sin^2 θ cos^2 θ).Factor cos^2 θ: t = σ * cos θ * sqrt(cos^2 θ + sin^2 θ) = σ * cos θ.

Verification / Alternative check:
At θ = 0°, the plane is normal to the load, so t = σ, and σ cos 0° = σ. At θ = 90°, the plane is parallel to load, so t = 0, matching σ cos 90° = 0.

Why Other Options Are Wrong:
σ sin θ understates traction and is zero at θ = 0°, which is incorrect. σ sin 2θ and σ cos 2θ relate to transformed shear/normal components, not the resultant magnitude.

Common Pitfalls:
Confusing θ as the angle with the axis instead of the normal section, or mixing up individual components with the resultant.

Final Answer:
σ cos θ