Uniaxial loading and oblique planes:\nA prismatic body is subjected to a direct tensile stress σ in one plane. What is the tangential (shear) stress acting on an oblique section whose normal makes an angle θ with the direction of σ?

Introduction / Context:
When a member is subjected to a direct uniaxial tensile stress, the stresses on any oblique section can be obtained by resolving the applied stress into normal and shear components on that plane. This concept underpins strength-of-materials topics such as failure criteria and Mohr’s circle.

Given Data / Assumptions:

• State of stress: uniaxial tension of magnitude σ.
• Plane of interest is an oblique section whose normal is inclined at angle θ to the loading direction.
• Material is linearly elastic; small deformation; classic continuum assumptions.

Concept / Approach:
The stress transformation relations for a plane making an angle θ with the load direction give the normal and shear components on that plane. For uniaxial stress, the shear component simplifies to a compact trigonometric form involving sin 2θ. The maximum shear on an oblique plane occurs at θ = 45° to the load direction, consistent with Mohr’s circle predictions.

Step-by-Step Solution:

Verification / Alternative check:
From Mohr’s circle, the shear stress ordinate at an angle 2θ from the point representing the original stress equals τ = (σ/2) sin 2θ. At θ = 45°, sin 90° = 1, so τ = σ/2, which is the well-known maximum in-plane shear under uniaxial loading.

Why Other Options Are Wrong:

• σ sin 2θ and σ cos 2θ miss the factor 1/2.
• σ/2 cos 2θ corresponds to transformed normal stress deviation, not shear.
• σ tan θ is not a correct transformation expression for shear on the oblique plane.

Common Pitfalls:
Confusing the angle of the plane with the angle to the plane’s normal; omitting the 1/2 factor; mixing normal and shear formulas from Mohr’s circle.

Final Answer:
σ/2 sin 2θ