Equal normal and shear components on an inclined plane in uniaxial loading A member is subjected to a single axial force. On a plane inclined at angle θ to the direction of the force, the normal stress component equals the tangential (shear) stress component. What must θ be?

Introduction / Context:
Stress transformation in a uniaxially loaded member shows how normal and shear components vary with the orientation of the cut plane. Recognizing special angles simplifies analysis and helps in failure predictions using Mohr’s circle or transformation equations.

Given Data / Assumptions:

• Uniaxial normal stress σ acts along the member axis.
• Plane is inclined at angle θ to the force direction (equivalently, to the axis of σ).
• Linear elastic continuum; plane stress transformation applies.

Concept / Approach:
For uniaxial stress, the normal and shear components on a plane at angle θ to the stress direction are: σ_n = σ * cos^2(θ) and τ = σ * sin(θ) * cos(θ). Equating σ_n and τ gives the characteristic angle where normal and shear magnitudes match.

Step-by-Step Solution:
Write equality: σ * cos^2(θ) = σ * sin(θ) * cos(θ).Cancel σ and cos(θ) (for θ not equal to 90°): cos(θ) = sin(θ).Therefore, tan(θ) = 1 ⇒ θ = 45°.

Verification / Alternative check:
Using Mohr’s circle with center at σ/2 and radius σ/2, the plane where σ_n = τ lies at 45° from the stress axis, consistent with the algebraic result.

Why Other Options Are Wrong:
30°/60°: These produce unequal components because cos^2(θ) ≠ sin(θ)cos(θ) at those angles.90°: On a plane normal to the stress direction, σ_n = 0 and τ = 0, not equal nonzero values.15°: No special equality occurs at this angle.

Common Pitfalls:

• Mixing the angle from the plane normal vs. plane itself; the formula above uses the angle between the plane and the stress direction.
• Forgetting that both components scale with σ; equality depends only on angle, not magnitude.

Final Answer:
45°