What are the engineering assumptions of the Torsional Shear Stress in Circular Shafts?

The bar has a circular cross-section (solid or hollow). The material is homogeneous, isotropic, and obeys Hooke's Law in shear. Cross-sections remain plane and do not warp during twisting.

What are the design limitations of the Torsional Shear Stress in Circular Shafts?

Not applicable to non-circular cross-sections (which undergo warping, requiring Prandtl stress function analysis). Underestimates stress if twisting exceeds the elastic limit of the material.

Engineering DisclaimerEngineering calculations are provided for preliminary design and educational/reference purposes only. Users must verify all results according to applicable engineering standards, supplier data, manufacturing requirements, and professional engineering judgment.

Strength of Materials

Standard Equation

Torsional Shear Stress in Circular Shafts

Calculates the shear stress at any radial location in a circular solid or hollow shaft under torsional load.

Primary Mathematical Expression

\tau = T * r / J

Design Schematic

Nomenclature & Variables

Symbol	Variable Name	Metric Unit	Imperial Unit	Description
T	Applied torque	N·m	in·lb	The torsional moment load twisting the bar.
r	Radial distance	mm	in	Radial distance from center axis to the point of interest.
J	Polar moment of inertia	mm⁴	in⁴	The geometric torsional resistance property of the cross-section.
\tau	Torsional shear stress	MPa	psi	The resulting shear stress at radius r.

Step-by-Step Derivation

1
A circular bar is subjected to pure torsion, causing cross-sections to rotate relative to each other about the longitudinal axis.
2
Shear strain \gamma is proportional to radial distance r: \gamma = r * \phi, where \phi is the twist rate per unit length.
3
Assuming elastic behavior, Hooke's Law for shear applies: \tau = G * \gamma = G * r * \phi.
4
The resisting internal torque must balance the external torque: T = \int \tau * r * dA = \int (G * r² * \phi) * dA = (G * \phi) * J.
5
Substituting G * \phi = T / J back into the stress equation yields the torsional shear stress formula: \tau = T * r / J.

Worked Example Calculation

Problem Statement

A solid circular shaft of radius 15 mm is subjected to a torque of 450 N·m. The polar moment of inertia J is 7.95 \times 10^4 mm⁴. Calculate the maximum torsional shear stress.

Calculation Steps

•Identify input parameters: T = 450 N·m = 450,000 N·mm, r = 15 mm, J = 79,500 mm⁴.
•Apply the Torsional Shear Stress Formula: \tau = T * r / J.
•Compute: \tau = (450,000 * 15) / 79,500 = 6,750,000 / 79,500 = 84.9 MPa.

Final ResultShear Stress = 84.9 MPa

Engineering Assumptions

•The bar has a circular cross-section (solid or hollow).
•The material is homogeneous, isotropic, and obeys Hooke's Law in shear.
•Cross-sections remain plane and do not warp during twisting.

Design Limitations

•Not applicable to non-circular cross-sections (which undergo warping, requiring Prandtl stress function analysis).
•Underestimates stress if twisting exceeds the elastic limit of the material.

Academic References & Standards

Shigley Ch 3textbook

Shigley's Mechanical Engineering Design, 11th Edition

Textbook covering torsional stresses and shafts.