Under a uniaxial direct stress, the tangential (shear) stress on a plane inclined at angle θ to the direction of the load equals the normal stress on that plane multiplied by which factor?

Introduction / Context:Stress transformation on inclined planes under uniaxial loading is a staple of Mohr’s circle and elementary mechanics of materials. The relationship between the normal and shear components on an inclined plane is frequently tested in exams.

Given Data / Assumptions:

• Uniaxial direct stress σ acts along one direction.
• Consider a plane inclined at angle θ to the load direction.

Concept / Approach:For uniaxial stress, the normal stress on the inclined plane is σ_n = σ cos^2 θ, and the shear (tangential) stress is τ = σ sin θ cos θ. Eliminating σ gives a relation between τ and σ_n: τ = σ_n * tan θ.

Step-by-Step Solution:1) σ_n = σ cos^2 θ.2) τ = σ sin θ cos θ.3) Divide: τ / σ_n = (σ sin θ cos θ) / (σ cos^2 θ) = tan θ.4) Therefore, τ = σ_n * tan θ.

Verification / Alternative check:Mohr’s circle also yields τ = (σ/2) sin 2θ and σ_n = (σ/2)(1 + cos 2θ). Their ratio simplifies to tan θ, consistent with the above derivation.

Why Other Options Are Wrong:

• sin θ, cos θ: Incomplete; they relate to σ components relative to σ, not to σ_n.
• sin θ cos θ and sin 2θ: These multiply σ directly, not the normal stress on the plane.

Common Pitfalls:

• Confusing the relation between τ and σ (gives sin θ cos θ) with the relation between τ and σ_n (gives tan θ).

Final Answer:tan θ.