Uniaxial tension — shear stress on an oblique plane A prismatic member with a normal cross-sectional area A is subjected to a tensile force P (uniaxial tension). What is the shear stress τ on a plane inclined at an angle θ to the normal (transverse) plane of the member?

Introduction / Context:
Stress transformation under uniaxial loading is a core topic in strength of materials. On planes inclined to the axis of loading, both normal and shear components exist. This question asks for the shear stress on such an oblique plane under a simple uniaxial tensile state.

Given Data / Assumptions:

• Uniaxial stress σ = P/A acts along the member axis.
• Plane of interest is inclined at angle θ to the normal cross-section (i.e., to the transverse plane).
• Continuum, small deformation, standard Cauchy stress transformation applies.

Concept / Approach:

For uniaxial stress σ, the stress components on a plane at angle θ to the normal plane are: normal stress σ_n = σcos²θ and shear stress τ = σ(sin 2θ)/2. These arise from resolving σ along the plane’s normal and tangential directions or via Mohr’s circle (uniaxial case has centre at σ/2 and radius σ/2).

Step-by-Step Solution:

Verification / Alternative check:

From Mohr’s circle for uniaxial stress, the point corresponding to plane angle θ from the transverse plane is located at an angular position 2θ on the circle; the ordinate is τ = (σ/2) sin 2θ, confirming the result.

Why Other Options Are Wrong:

• (b) and (c) miss the correct trigonometric dependence.
• (d) gives a normal component form, not shear.
• (e) is dimensionally incorrect across the full range of θ.

Common Pitfalls:

• Mixing up the “plane angle” with the “Mohr’s circle angle”; remember the factor of 2.

Final Answer:

τ = (P / A) * (sin 2θ) / 2.