The cross-section of a bar is subjected to a uni-axial tensile stress “p”. The shear stress on a plane inclined at θ to the cross-section of the bar would be
- A(p sin 2θ) / 2Correct
- Bp sin 2θ
- Cp cos 2θ
- D(p cos 2θ) / 2
Solution
Concept
Shear Stress on Inclined Plane
Formula
τ = (p / 2) × sin(2θ)
Step-by-step solution
- Identify given stress components: σx = p, σy = 0, and τxy = 0 - Apply the shear stress transformation formula: τ = ((σx - σy) / 2) \* sin(2θ) - Substitute the values into the formula to obtain τ = (p / 2) \* sin(2θ)
Explanation
When a bar is subjected to a uni-axial tensile stress p, the shear stress on an inclined plane at an angle θ to the cross-section is given by τ = (p/2) sin(2θ), which simplifies to (p sin 2θ) / 2. This represents the component of the applied tensile load that acts parallel to the inclined plane divided by the inclined area.
Shortcut
To easily remember: Shear stress in uni-axial loading is always half the normal stress multiplied by sin(2θ) [τ = (p/2) sin(2θ)]. Remember that shear is zero at θ = 0° and maximum (p/2) at θ = 45°, which matches the behavior of sin(2θ).
Common mistake
A common mistake is selecting Option B (p sin 2θ) by forgetting to divide by 2, or confusing the shear stress formula τ = (p/2) sin 2θ with the normal stress formula σ_θ = p cos²θ (or (p/2)(1 + cos 2θ)), leading to choosing Option C or D.
Correct answer: A) (p sin 2θ) / 2