Torsion Calculator
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Torsion Calculator
Calculator
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Formula: tau_max = Tc/J | theta = TL/(GJ) | J = pi*R^4/2 (solid)
Worked example โ tau_max = 25.46 MPa, angle of twist = 0.73 degrees
Formula
tau_max = Tc/J | theta = TL/(GJ) | J = pi*R^4/2 (solid)
Maximum shear stress equals the applied torque times the outer radius divided by the polar moment of inertia. The angle of twist equals the torque times the length divided by the product of shear modulus and polar moment of inertia. For solid circular sections, J equals pi times the radius to the fourth power divided by 2.
Worked Examples
Example 1: Solid Steel Shaft
Problem:Find the maximum shear stress and angle of twist for a solid 100mm diameter steel shaft, 2m long, under 5000 N-m torque. G = 80 GPa.
Solution:J = pi * 50^4 / 2 = 9,817,477 mm4 tau_max = T*c/J = 5,000,000 * 50 / 9,817,477 = 25.46 MPa theta = TL/(GJ) = 5,000,000*2000/(80,000*9,817,477) = 0.0127 rad = 0.73 deg
Result:tau_max = 25.46 MPa, angle of twist = 0.73 degrees
Example 2: Hollow Shaft Comparison
Problem:Same conditions but with a hollow shaft (100mm outer, 60mm inner diameter).
Solution:J = pi*(50^4 - 30^4)/2 = 8,545,132 mm4 tau_max = 5,000,000 * 50 / 8,545,132 = 29.27 MPa theta = 5,000,000*2000/(80,000*8,545,132) = 0.0146 rad = 0.84 deg
Result:tau_max = 29.27 MPa, twist = 0.84 deg (15% more stress, 36% less material)
Frequently Asked Questions
What is torsion in structural and mechanical engineering?
Torsion is the twisting of a structural member when it is loaded by torques (moments) that produce rotation about the longitudinal axis. The resulting shear stresses vary linearly from zero at the center to a maximum at the outer surface for circular sections. Torsion is common in shafts, beams loaded eccentrically, and spandrel beams in concrete frames where floor loads apply twisting moments.
What is the polar moment of inertia J?
The polar moment of inertia J is a geometric property that quantifies a cross-section's resistance to torsional deformation. For a solid circular section, J equals pi times the radius to the fourth power divided by 2. For hollow circular sections, J is the difference between the outer and inner values. A larger J means less shear stress and less angular twist for a given applied torque, making the member stiffer in torsion.
Why are hollow shafts more efficient than solid shafts?
Hollow shafts use material more efficiently because the core of a solid shaft contributes very little to torsional resistance while adding significant weight. The shear stress in torsion is zero at the center and maximum at the surface, so removing the lightly stressed core has minimal impact on strength. A hollow shaft with the same outer diameter and 60% inner diameter retains about 87% of the torsional strength while reducing weight by 36%.
How do you calculate the angle of twist?
The angle of twist theta equals the applied torque T times the member length L divided by the product of the shear modulus G and the polar moment of inertia J. The formula is theta = TL/(GJ) in radians. For steel with G around 80 GPa, a 100mm diameter shaft 2 meters long carrying 5 N-m of torque will twist about 0.032 degrees. Longer shafts and softer materials produce more twist for the same torque.
What is the torsion section modulus Zp?
The torsion section modulus Zp (also called the polar section modulus) equals the polar moment of inertia J divided by the outer radius c. It relates the applied torque directly to the maximum shear stress through the formula tau_max = T/Zp. This is analogous to the flexural section modulus S that relates bending moment to bending stress. A larger Zp means a lower maximum shear stress for a given torque.
How is the shear modulus G related to Young's modulus E?
For an isotropic material the two elastic constants are linked through Poisson's ratio by G = E / (2(1 + nu)). Structural steel with E near 200 GPa and nu near 0.3 therefore has G of about 77 GPa, which is why 80 GPa is a common design value. Aluminium alloys sit near 26 GPa and copper near 45 GPa. Use G, not E, in the twist equation: substituting Young's modulus understates the rotation by a factor of roughly 2.6 for steel.
Why can't these formulas be used for a square or I-shaped section?
The Tc/J and TL/GJ results depend on axisymmetry, which guarantees that cross-sections stay plane and radii stay straight. Non-circular prisms warp out of plane when twisted, so Saint-Venant's torsion constant replaces the polar moment of inertia. For an open thin-walled shape such as an I-beam or channel the constant is roughly the sum of b t cubed over three for each rectangular element, a value far smaller than the polar moment of inertia, which is why open sections are notoriously weak in torsion.
What is the difference between an open and a closed thin-walled section in torsion?
A closed tube carries torque as a continuous shear flow that circulates around the wall, and the Bredt-Batho relation gives the stress as T divided by twice the enclosed mid-line area times the wall thickness. An open section of identical material and wall thickness cannot form that loop and must resist by through-thickness shear couples instead, so it is often one or two orders of magnitude less stiff. Slitting a tube lengthwise destroys most of its torsional capacity while barely changing its bending strength.
How does a keyway or shoulder fillet affect the calculated shear stress?
The linear Tc/J distribution assumes a smooth prismatic shaft. Any abrupt change in geometry, such as a keyway, cross-hole, splined end or a step with a small fillet radius, concentrates stress locally by a factor typically in the range of about 1.5 to 3 depending on how sharp the transition is. Multiply the nominal stress by the appropriate stress-concentration factor when checking fatigue, and increase fillet radii rather than material grade to reduce the peak.
How do I convert transmitted power into the torque input for Torsion Calculator?
Torque and power are linked through rotational speed: P = T times omega, where omega is angular velocity in radians per second, equal to 2 pi N over 60 for N revolutions per minute. So a shaft transmitting 15 kW at 1500 rpm carries omega = 157.1 rad/s and T = 15000 / 157.1, about 95.5 N-m. Slower shafts carry proportionally more torque for the same power, which is why the low-speed output shaft of a gearbox is always the larger diameter.
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Reviewed for accuracy by Abdullah, Technical Content Specialist ยท Editorial policy
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