A circular shaft resists both bending and torsion. Where does the maximum normal stress occur?
- AAt the neutral axis
- BAt 45° to the axis
- CAt the outermost fibreCorrect
- DUniformly distributed
Solution
Concept
Maximum normal stress in combined loading
Formula
σ1 = σ/2 + √((σ/2)2 + τ2)
Step-by-step solution
1. Under combined bending and torsion, normal stress σ = M·y/I and shear stress τ = T·r/J act on the shaft cross-section. 2. Bending stress is zero at the neutral axis and maximum at the outermost fiber (y = R). 3. Torsional shear stress is zero at the center and maximum at the outermost surface (r = R). 4. Since both component stresses peak at the outer periphery, the maximum principal normal stress occurs at the outermost fibre. Answer: C
Explanation
Under combined bending moment and torque, normal bending stress and torsional shear stress simultaneously reach their maximum values at the outer boundary of the shaft. Consequently, the principal normal stress calculated by combining these components is maximized at the extreme surface fibers.
Why the other options are wrong
A) Bending stress vanishes at the neutral axis. B) 45 degrees is the inclination angle of the principal plane, not the physical cross-sectional location. D) Stresses vary linearly with radial distance and are not uniform.
Shortcut
Both bending and torsional shear peak at the outer surface, so max normal stress is at the outermost fibre.
Common mistake
Confusing the 45-degree principal plane angle in pure torsion with the spatial physical position in the cross section.
Correct answer: C) At the outermost fibre