In a uniaxial loading case, a plane is taken at an angle θ to the direction of the applied force. What is the ratio of the tangential (shear) stress to the normal stress on that plane?

Introduction / Context:
Stress transformation on an inclined plane is a classic topic in strength of materials. For a uniaxial direct stress, the normal and shear components on a plane at angle θ have standard relations. This question asks for the ratio of shear to normal stress on that plane.

Given Data / Assumptions:

• Uniaxial direct stress σ acts along a reference direction.
• An inclined plane is cut at angle θ to this stress direction.
• Plane stress transformation relations apply.

Concept / Approach:
For a uniaxial stress σ, the stresses on an inclined plane at angle θ (measured between the plane and the stress direction) are:
Normal stress: σ_n = σ * cos^2(θ)Shear stress: τ = σ * sin(θ) * cos(θ)Therefore the desired ratio is τ / σ_n = tan(θ).

Step-by-Step Solution:
1) Write σ_n = σ cos^2 θ.2) Write τ = σ sin θ cos θ.3) Compute τ/σ_n = (σ sin θ cos θ) / (σ cos^2 θ) = tan θ.

Verification / Alternative check:
Using Mohr’s circle, the radius is σ/2 and the coordinates on the circle at angle 2θ from the σ-axis give the same relations, yielding τ/σ_n = tan θ.

Why Other Options Are Wrong:
sin θ, cos θ, sec θ, cot θ: do not match the derived ratio from the standard transformation equations for uniaxial stress.

Common Pitfalls:

• Confusing the angle definition (plane angle vs. normal angle).
• Using σ_n = σ cos θ instead of σ cos^2 θ.
• Dropping the common factor σ when forming the ratio.

Final Answer:
tan θ