Normal stress on an inclined plane under biaxial normal stresses: A prismatic bar is subjected to a longitudinal normal stress σ1 and a transverse normal stress σ2. What is the normal component of stress on a plane inclined at an angle θ to the longitudinal (σ1) direction? Express the answer using only σ1, σ2, and θ.

Introduction / Context:
Stress transformation is fundamental in solid mechanics. When a member carries normal stresses in two perpendicular directions (σ1 longitudinal, σ2 transverse), the stress on an arbitrarily inclined plane changes with the angle. Engineers need the transformed normal stress to assess failure on critical planes.

Given Data / Assumptions:

• Only normal stresses σ1 (longitudinal) and σ2 (transverse) act; shear stress τ12 is zero.
• The plane of interest is inclined at angle θ to the longitudinal axis of σ1.
• Small deformation, linear elasticity, continuum assumptions.

Concept / Approach:

Using plane stress transformation, if the plane itself is at angle θ to the σ1 direction, its normal is at (90° − θ). The general normal stress transformation (with τ = 0) reduces to σ_n = σ1 cos^2 φ + σ2 sin^2 φ where φ is the angle between the plane's normal and σ1. Substituting φ = 90° − θ gives σ_n = σ1 sin^2 θ + σ2 cos^2 θ.

Step-by-Step Solution:

Verification / Alternative check:

At θ = 0°, the plane is parallel to σ1 (normal is along σ2), giving σ_n = σ2, which matches the formula. At θ = 90°, σ_n = σ1, again consistent.

Why Other Options Are Wrong:

• σ1 sin θ * σ2 cos θ mixes stresses multiplicatively and has wrong dimensions.
• σ1 cos^2 θ + σ2 sin^2 θ corresponds to the case when θ is the angle between normal and σ1, not the plane and σ1.
• (σ1+σ2)/2 + ... cos 2θ is correct if θ denotes the normal's angle to σ1; the problem defines θ for the plane, not the normal.
• σ1 cos θ + σ2 sin θ is a simple projection, not the transformation law.

Common Pitfalls:

• Confusing the plane angle with the normal angle.
• Forgetting that shear is zero only in the original axes, not necessarily on the inclined plane.

Final Answer:

σ1 sin^2 θ + σ2 cos^2 θ.