User Rating 0.0 ★★★★★

Total Usage 0 times

Category Physics & Science

Cross-Section Type

Diameter (mm) Outer diameter of the shaft

Inner Diameter (mm) Must be less than outer diameter

Mean Radius (mm)

Wall Thickness (mm) Must be < 10% of mean radius for accuracy

Material Preset

Shear Modulus G (GPa)

Applied Torque T (N·m)

Shaft Length L (mm)

Enter shaft parameters and click Calculate

Angle of Twist — —

Polar Moment of Inertia J —

Max Shear Stress τ_max —

Twist per Unit Length —

Torsional Stiffness GJ —

Is this tool helpful?

Your feedback helps us improve.

★ ★ ★ ★ ★

About

Torsional deformation in shafts is a critical failure mode in power transmission systems. A miscalculated angle of twist θ can cause misalignment in coupled machinery, fatigue cracking at keyways, or catastrophic shear failure when τ_max exceeds the material's shear yield strength. This calculator applies the classical torsion formula θ = TLGJ to compute the twist angle in both radians and degrees, the polar moment of inertia J, and the maximum shear stress τ at the outer fiber. It assumes linear-elastic, homogeneous, isotropic material behavior per Saint-Venant torsion theory. Results are invalid beyond the proportional limit or for non-prismatic shafts with abrupt cross-section changes.

The tool supports solid circular, hollow circular, and thin-walled tubular cross-sections. Material presets provide standard shear modulus G values from engineering references. Note that real-world shafts experience stress concentrations at fillets, keyways, and splines that this idealized model does not capture. Apply appropriate stress concentration factors K_t separately. Pro tip: for design, most codes limit twist to 0.25 - 1.0 °/m depending on application.

Formulas

The angle of twist for a prismatic shaft under uniform torque along its length is derived from the equilibrium and compatibility conditions of linear elasticity:

θ = T ⋅ LG ⋅ J

Where T = applied torque (N⋅m), L = shaft length (m), G = shear modulus of elasticity (Pa), J = polar moment of inertia (m⁴). The result θ is in radians. Convert to degrees by multiplying by 180π.

The polar moment of inertia depends on cross-section geometry:

Solid circle: J = π d⁴32

Hollow circle: J = π(D⁴ − d⁴)32

Thin-walled tube: J ≈ 2π r_m³ t

Where d = diameter (solid) or inner diameter (hollow), D = outer diameter, r_m = mean radius, t = wall thickness.

Maximum shear stress at the outer surface:

τ_max = T ⋅ cJ

Where c = distance from the centroid to the outermost fiber (outer radius for circular sections).

Reference Data

Material	Shear Modulus G (GPa)	Shear Yield Strength τ_y (MPa)	Density (kg/m³)	Typical Use
Structural Steel (A36)	79.3	145	7850	General shafts, beams
Stainless Steel (304)	77.2	170	8000	Corrosion-resistant shafts
Alloy Steel (4140)	80.0	415	7850	High-strength drive shafts
Aluminum 6061-T6	26.0	207	2700	Lightweight shafts
Aluminum 2024-T4	27.6	283	2780	Aerospace components
Copper (C11000)	44.7	69	8940	Electrical conductors
Brass (C36000)	37.0	124	8500	Fittings, valves
Bronze (C93200)	38.0	110	8800	Bearings, bushings
Titanium (Ti-6Al-4V)	44.0	550	4430	Aerospace, medical
Cast Iron (Gray)	41.0	170	7200	Machine frames (brittle)
Nickel Alloy (Inconel 718)	77.2	650	8190	High-temp turbine shafts
Magnesium (AZ31B)	17.0	130	1770	Ultra-light structures
Monel 400	65.0	240	8800	Marine shafts
Beryllium Copper (C17200)	50.0	690	8250	Springs, non-sparking tools
Tungsten	161.0	360	19250	High-temp, radiation
AISI 1045 Steel	80.0	310	7870	Medium-carbon shafts
Polycarbonate	0.83	41	1200	Prototype shafts
Nylon 6/6	0.76	40	1140	Low-load bushings

Frequently Asked Questions

Shear modulus G decreases with increasing temperature for most metals. For structural steel, G drops approximately 5% at 200°C and roughly 30% at 500°C compared to room temperature values. Since θ is inversely proportional to G, a 30% reduction in G increases θ by approximately 43%. For high-temperature applications, use the modulus value at operating temperature from material datasheets, not room-temperature defaults.

The thin-walled approximation J ≈ 2πr_m³t assumes t < 0.1 ⋅ r_m. When the wall thickness exceeds roughly 10% of the mean radius, the radial stress gradient across the wall becomes significant. Error relative to the exact hollow-circle formula exceeds 5% at t/r_m ≈ 0.15. For thick-walled tubes, always use the hollow circle formula.

Most mechanical engineering design standards recommend limiting twist to 0.25°/m to 1.0°/m depending on application. Power transmission shafts for precision machinery typically use 0.25°/m. General-purpose industrial shafts allow up to 1.0°/m. ASME standards for marine propeller shafts specify a maximum total twist that depends on shaft length and rotational speed. Always check the applicable code for your specific application.

This calculator computes nominal stress τ_nom assuming a smooth, uniform cross-section. Real shafts with keyways, splines, or stepped diameters experience local stress amplification. The actual peak stress is τ_max = K_t ⋅ τ_nom, where K_t ranges from 1.3 to 3.0 for common features. A standard sled-runner keyway in a shaft has K_t ≈ 1.6. Apply these factors manually to the shear stress output for fatigue or yield checks.

No. The formula θ = TL / (GJ) strictly applies to circular and thin-walled closed cross-sections where warping is zero or negligible. Rectangular, I-beam, and open sections experience warping during torsion, requiring Saint-Venant's torsion constant C (not the polar moment J) and additional warping stiffness terms. For a rectangular bar with sides a and b, the torsion constant is C ≈ βab³ where β depends on the aspect ratio.

If τ_max ≥ τ_y, the material enters plastic deformation. The linear-elastic formula no longer applies. The shaft develops a plastic zone growing inward from the outer surface. Beyond the fully plastic torque (approximately 1.33 times the yield torque for a solid circle), the entire cross-section yields. The calculator flags a warning when the computed stress exceeds the yield value for the selected material. Redesign by increasing diameter, selecting a higher-strength material, or reducing the applied torque.