In a circular shaft under torsion, how does the shear stress at a point vary with its radial distance from the shaft axis?

Difficulty: Easy

Correct Answer: directly proportional to the distance from the axis

Explanation:


Introduction / Context:
Torsion of circular shafts is a core topic in mechanics of materials. Designers must know how shear stress distributes from the center to the outer surface to ensure safe torque transmission.



Given Data / Assumptions:

  • Solid or hollow circular shaft under pure torque T.
  • Linearly elastic, homogeneous, isotropic material.
  • No warping restraint beyond Saint-Venant torsion assumptions.


Concept / Approach:
The torsion formula gives shear stress at a radius r as tau(r) = T * r / J, where J is the polar moment of inertia of the cross-section. This indicates a linear variation of shear stress with radius—zero at the center and maximum at the outer surface.



Step-by-Step Solution:

Torsion relation: tau(r) = T * r / JFor a given T and J, tau is proportional to r.At r = 0 (axis), tau = 0; at r = R (outer radius), tau_max = T * R / J.


Verification / Alternative check:
Mohr’s circle for pure shear at a point yields zero shear at the centerline because r = 0; experimental strain-gage data on shafts confirms linear radial variation.



Why Other Options Are Wrong:
Inverse or constant dependence contradicts tau = T r / J; saying maximum at center is incorrect as tau(0) = 0.



Common Pitfalls:
Confusing bending stress distribution (linear with y) with torsional shear; forgetting hollow shaft still has tau proportional to r within its wall.



Final Answer:

directly proportional to the distance from the axis

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