Compute torsional stiffness (k = G·J / L), polar moment of inertia (J), angle of twist for solid or hollow circular shafts. Includes material presets and real‑time visualization.
Torsional stiffness (also called torsional rigidity) quantifies a shaft's resistance to twisting under an applied torque. For circular shafts, the fundamental relation is k = T / θ = G·J / L, where G is the shear modulus of elasticity, J is the polar moment of inertia of the cross‑section, and L is the shaft length. This linear elastic relationship is the backbone of machine design, automotive drivelines, and structural mechanics.
? Polar Moment of Inertia J
Solid circular: J = π·D⁴ / 32 | Hollow circular: J = π·(D⁴ − d⁴) / 32
Torsional stiffness: k = G·J / L (N·m/rad)
Angle of twist: θ = T·L / (G·J) (radians)
In drivetrain shafts, excessive twist leads to angular misalignment, vibration, and fatigue failure. High torsional stiffness improves precision (e.g., robot arms, machine tool spindles). Conversely, in some applications like torsion bars for vehicle suspensions, a controlled compliance is desired. Our calculator helps select diameter, material, and length to meet twist angle limits (typically < 0.25° per meter).
From the torsion formula: τmax = T·r / J, and the angle of twist per unit length: dθ/dx = T/(G·J). Integrating over length L gives θ = T·L/(G·J). By definition, stiffness k = T/θ = G·J/L. This linear relation holds within the elastic range (Hooke’s law in shear). Our calculator uses consistent units: convert mm to meters for J? Internally, we keep J in mm⁴ and G in GPa (N/mm²), L in mm, then k = G·J / L → (N/mm²)·mm⁴ / mm = N·mm/rad, divided by 1000 gives N·m/rad, providing direct industrial relevance.
A rear‑wheel drive vehicle uses a hollow steel shaft (outer dia 76 mm, inner 68 mm, length 1.4 m). Max engine torque = 450 N·m. Using our calculator: G = 79.3 GPa, J = π·(76⁴ − 68⁴)/32 ≈ 5.12×10⁵ mm⁴, stiffness k ≈ 79.3e3 * 5.12e5 / 1400 ≈ 2.90×10⁷ N·mm/rad = 29,000 N·m/rad. Angle of twist at peak torque ≈ 450 / 29,000 = 0.0155 rad ≈ 0.89°, well within design limits. The hollow design saves 30% weight while maintaining adequate stiffness.
Steel (G ≈ 79 GPa) is three times stiffer than aluminum (G ≈ 26 GPa) for identical geometry. For weight‑critical applications, aluminum shafts are common with larger diameters. Composites and titanium offer intermediate values. The calculator includes a custom G input to handle any isotropic linear‑elastic material.
| Material | Shear Modulus G (GPa) | Typical applications |
|---|---|---|
| Structural Steel | 79.3 | Drive shafts, turbine rotors |
| Aluminum 6061-T6 | 26.0 | Lightweight transmission, aerospace |
| Copper Alloy | 44.7 | Electrical rotors, heat exchangers |
| Titanium Grade 5 | 41.0 | High‑performance racing shafts |
| Cast Iron | ~44 | Camshafts, industrial machinery |