Torsion Shear Stress Calculator
Calculate torsion shear stress accurately for your build. Get material quantities, waste allowances, and project cost breakdowns.
Torsion Shear Stress Calculator
Calculator
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Formula: tau_max = T * c / J; theta = T * L / (G * J)
Worked example โ Max shear stress: 20.37 MPa | Angle of twist: 0.584 degrees over 1 meter.
Formula
tau_max = T * c / J; theta = T * L / (G * J)
Where tau_max is the maximum shear stress, T is the applied torque, c is the outer radius, J is the polar moment of inertia, theta is the angle of twist, L is the shaft length, and G is the shear modulus of the material.
Worked Examples
Example 1: Solid Steel Drive Shaft
Problem:A solid steel shaft (D = 50mm, G = 80 GPa) transmits 500 N*m of torque over 1 meter. Find max shear stress and angle of twist.
Solution:J = pi x 50^4 / 32 = 613,592 mm^4 Max shear stress: tau = (500 x 1000 x 25) / 613,592 = 20.37 MPa Angle of twist: theta = (500,000 x 1000) / (80,000 x 613,592) = 0.01019 rad = 0.5836 degrees
Result:Max shear stress: 20.37 MPa | Angle of twist: 0.584 degrees over 1 meter.
Example 2: Hollow Aluminum Tube
Problem:A hollow aluminum tube (D = 80mm, d = 60mm, G = 26 GPa) carries 200 N*m over 0.5m. Find stress and twist.
Solution:J = pi x (80^4 - 60^4) / 32 = pi x (40,960,000 - 12,960,000) / 32 = 2,749,018 mm^4 Max shear stress: tau = (200,000 x 40) / 2,749,018 = 2.91 MPa Angle of twist: theta = (200,000 x 500) / (26,000 x 2,749,018) = 0.001399 rad = 0.0802 degrees
Result:Max shear stress: 2.91 MPa | Angle of twist: 0.080 degrees. Well within limits.
Frequently Asked Questions
What is torsion shear stress and how is it calculated?
Torsion shear stress is the internal stress developed in a shaft or structural member when a twisting moment (torque) is applied. The maximum shear stress occurs at the outermost fiber of the cross section and is calculated using the formula tau equals T times c divided by J, where T is the applied torque, c is the distance from the neutral axis to the outermost fiber (the outer radius), and J is the polar moment of inertia. For a solid circular shaft, J equals pi times the diameter to the fourth power divided by 32. The stress varies linearly from zero at the center to maximum at the surface. Understanding torsion shear stress is critical for designing drive shafts, axles, and structural connections.
What is the polar moment of inertia and why does it matter?
The polar moment of inertia J is a geometric property that quantifies a cross section's resistance to torsional deformation. For a solid circular shaft, J equals pi times D to the fourth power divided by 32. For a hollow circular shaft, J equals pi times the quantity D to the fourth minus d to the fourth divided by 32, where D is the outer diameter and d is the inner diameter. A larger polar moment of inertia means the shaft can resist more torque before reaching a given stress level. Hollow shafts are remarkably efficient because material near the center contributes very little to torsional resistance, so removing it saves weight while retaining most of the strength.
How do hollow shafts compare to solid shafts for torsion?
Hollow shafts offer superior strength-to-weight ratios compared to solid shafts because the material near the center of a solid shaft carries very little shear stress. For example, a hollow shaft with an inner diameter equal to 80 percent of the outer diameter retains about 59 percent of the polar moment of inertia while using only 36 percent of the material. This means it can handle 59 percent of the torque at roughly one-third the weight. In practice, hollow shafts are widely used in automotive drive shafts, aircraft structures, and bicycle frames where weight savings are critical. The trade-off is that hollow shafts are more susceptible to local buckling under extreme loads.
What is the angle of twist and how is it used in design?
The angle of twist is the angular deformation of a shaft under applied torque, calculated as theta equals T times L divided by G times J, where T is torque, L is shaft length, G is the shear modulus of the material, and J is the polar moment of inertia. The result is in radians and is often converted to degrees. Engineers use the angle of twist as a design criterion alongside shear stress. Many applications limit the angle of twist to 0.25 to 1.0 degrees per meter of shaft length to prevent excessive deflection that could cause misalignment in coupled machinery. In precision applications like machine tool spindles, even smaller twist angles are required for accuracy.
What shear modulus values should I use for common materials?
The shear modulus G represents a material's resistance to shear deformation and is essential for calculating the angle of twist. Common values in gigapascals include: carbon steel at 79 to 84 GPa, stainless steel at 74 to 86 GPa, aluminum alloys at 25 to 28 GPa, copper alloys at 37 to 44 GPa, titanium alloys at 41 to 45 GPa, cast iron at 32 to 41 GPa, and brass at 35 to 40 GPa. For design purposes, using 80 GPa for steel and 26 GPa for aluminum are common approximations. The shear modulus is related to the elastic modulus E and Poisson ratio nu by the relationship G equals E divided by two times the quantity one plus nu.
What is the polar section modulus and how is it used?
The polar section modulus Zp is the polar moment of inertia divided by the outer radius, J divided by c, which collapses the torsion formula to the compact form tau equals T divided by Zp. For a solid circular shaft Zp equals pi times D cubed divided by 16, so a 50 mm shaft has Zp of about 24,544 mm cubed and carries 20.4 MPa under 500 N-m. Because Zp packages the geometry into one number, it is the quickest way to compare candidate sections or to back-solve the torque a given shaft can carry at an allowable stress.
Why does the tau equals T c over J formula not work for rectangular or I-shaped sections?
The derivation assumes plane cross-sections stay plane and radii stay straight under twist, which is true only for circular and annular sections. Saint-Venant showed in 1855 that every other shape warps out of plane, so the polar moment of inertia no longer describes the resistance and the peak stress moves to the middle of the long side rather than the corner. Non-circular members need a torsion constant with its own shape coefficients, and open sections such as channels and I-beams are dramatically weaker in torsion than a closed tube of similar size.
How does the torque in this calculation relate to transmitted power and shaft speed?
Power equals torque times angular velocity, P equals T times omega, with omega equal to 2 pi N divided by 60 for speed N in rpm. A shaft carrying 500 N-m at 1500 rpm therefore transmits 500 times 157.08, about 78.5 kW or 105 horsepower, while the same torque at 100 rpm delivers only 5.2 kW. Torque, not power, sets the shear stress, so a slow-turning shaft in a mill drive is far more heavily stressed than a fast one delivering the same kilowatts.
Why do brittle and ductile shafts break differently under torsion?
Pure torsion creates a state of pure shear, and the principal stresses in that state are equal in magnitude to the shear stress but act at 45 degrees to the shaft axis, one tensile and one compressive. Ductile metals are weakest in shear, so they yield and fracture on the transverse plane, leaving a flat square break. Brittle materials such as grey cast iron or chalk are weakest in tension and split along a helical surface inclined at 45 degrees, perpendicular to the tensile principal stress.
When can a thin-walled tube buckle before it reaches its shear stress limit?
The 45-degree compressive principal stress that accompanies torsion can wrinkle a thin wall into a helical buckling pattern long before the material's shear strength is reached. Resistance depends strongly on the ratio of radius to wall thickness, so as tubes get thinner the elastic buckling stress falls away much faster than the material allowable does. For large-diameter, thin-wall tubing the governing check is stability rather than stress, and Torsion Shear Stress Calculator's tau and twist figures should be treated as an upper bound only.
References
Background & Theory
History
Reviewed for accuracy by Daniel Agrici, Founder & Lead Developer ยท Editorial policy
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