A circular shaft in torsion is a basic model for many real machine parts, including drive shafts, axles, drill bits, and screwdrivers. When equal and opposite torques act at the ends of a shaft, the material resists by developing internal shear stress. Understanding torsion helps engineers choose safe diameters, materials, and allowable loads.
It also predicts how much a shaft twists during operation, which matters for alignment and control.
Key Facts
- Torsion shear stress in a circular shaft is tau = Tr/J.
- Maximum shear stress occurs at the outer surface, where r = c, so tau_max = Tc/J.
- Angle of twist for a uniform circular shaft is theta = TL/JG.
- Polar moment of inertia for a solid circular shaft is J = pi d^4/32.
- Polar moment of inertia for a hollow circular shaft is J = pi(D^4 - d^4)/32.
- Shear stress varies linearly with radius, so tau = 0 at the center and increases to tau_max at the surface.
Vocabulary
- Torque
- Torque is a twisting moment that tends to rotate a shaft about its long axis.
- Shear stress
- Shear stress is the internal force per unit area acting parallel to a material plane.
- Polar moment of inertia
- Polar moment of inertia is a geometric measure of how strongly a circular cross-section resists twisting.
- Angle of twist
- Angle of twist is the angular rotation between two cross-sections of a shaft caused by applied torque.
- Shear modulus
- Shear modulus is a material property that measures resistance to elastic shear deformation.
Common Mistakes to Avoid
- Using area moment of inertia I instead of polar moment of inertia J is wrong because torsion of circular shafts depends on resistance to twisting about the axis, not bending about a centroidal axis.
- Assuming shear stress is uniform across the circular cross-section is wrong because tau = Tr/J shows that stress increases linearly with radius.
- Forgetting to use the outer radius c when finding tau_max is wrong because the maximum stress occurs at the outer surface, not at the diameter or the center.
- Mixing units such as N mm for torque with meters for length is wrong because torsion equations require consistent units to produce correct stress and twist values.
Practice Questions
- 1 A solid steel shaft has diameter 40 mm and carries a torque of 600 N m. Using J = pi d^4/32, find the maximum shear stress at the outer surface.
- 2 A uniform solid circular shaft is 1.2 m long, has diameter 30 mm, carries a torque of 250 N m, and has shear modulus G = 80 GPa. Find the angle of twist in radians using theta = TL/JG.
- 3 A solid shaft and a hollow shaft have the same material, length, mass, and applied torque. Explain which design can usually resist torsion more efficiently and why the radial distribution of material matters.