Compute maximum shear stress (τmax), angle of twist (θ), polar moment of inertia (J), and torsional stiffness for circular shafts under pure torque. Interactive cross‑section visualizer included.
The torsion formula, also known as the elastic torsion formula, relates applied torque to shear stress and twist angle in circular shafts. It assumes linear elastic material behavior (Hooke's law), homogeneity, isotropy, and that plane sections remain plane. The maximum shear stress occurs at the outer fiber: τmax = T·ro / J. The angle of twist is given by θ = T·L / (G·J) (radians). These equations are fundamental to mechanical, civil, and aerospace engineering for shaft design, powertrain components, and structural analysis.
Key Assumptions:
Shear Modulus Relationship: The shear modulus G is related to Young's modulus E and Poisson's ratio ν by:
G = E / [2(1 + ν)]
For a circular shaft:
Jsolid = (π/2) · ro4
Jhollow = (π/2) · (ro4 - ri4)
τmax = T · ro / J (maximum shear stress at outer surface)
θ = T·L / (G·J) (twist angle in radians)
Torsional stiffness: kt = G·J / L (N·m/rad)
From the kinematic assumption: shear strain γ = r·(dφ/dx). Using Hooke's law in shear τ = G·γ. The torque equilibrium gives T = ∫ τ·r dA = G·(dφ/dx) ∫ r² dA = G·(dφ/dx)·J. Thus dφ/dx = T/(G·J), integrating over length L yields θ = T·L/(G·J). The maximum shear stress is τmax = G·γmax = G·ro·(dφ/dx) = T·ro/J. This rigorous derivation can be found in classical texts like Mechanics of Materials by Beer & Johnston and Advanced Mechanics of Materials by Boresi.
Detailed Mathematical Derivation:
1. Geometry of Deformation: Consider a circular shaft of radius R. Under torque T, a line on the surface rotates through angle γ. For a small element dx, the angle of twist dφ causes shear strain γ = r·(dφ/dx).
2. Stress-Strain Relation (Hooke's Law): τ = G·γ = G·r·(dφ/dx)
3. Torque Equilibrium: The resultant torque must equal applied torque T:
T = ∫ τ·r dA = ∫ [G·r·(dφ/dx)]·r dA = G·(dφ/dx) ∫ r² dA
4. Polar Moment of Inertia: Define J = ∫ r² dA. For solid circle: J = (π/2)R⁴; for hollow: J = (π/2)(Rₒ⁴ - Rᵢ⁴)
5. Twist Rate: dφ/dx = T/(G·J)
6. Angle of Twist: Integrate over length L: φ = ∫₀ˡ (T/(G·J)) dx = T·L/(G·J) for constant T, G, J
7. Shear Stress Distribution: τ(r) = G·r·(dφ/dx) = (T·r)/J, maximum at r = R: τ_max = T·R/J
This derivation follows Saint-Venant's principle and assumes linear elastic material behavior.
| Material | Shear Modulus G (GPa) | Typical Yield Strength τy (MPa) | Notes |
|---|---|---|---|
| Structural Steel (A36) | 79 – 81 | 150 – 250 | Most common for general shafts |
| Aluminum 6061-T6 | 26 | 55 – 90 | Lightweight, corrosion resistant |
| Stainless Steel 304 | 77 | 170 – 240 | Corrosion resistant, food grade |
| Titanium Ti-6Al-4V | 44 | 350 – 480 | High strength-to-weight ratio |
| Cast Iron (gray) | 40 – 45 | 130 – 200 | Brittle, good damping |
| Copper (C11000) | 44 – 46 | 70 – 100 | High conductivity, moderate strength |
| Brass (C36000) | 37 – 41 | 120 – 200 | Good machinability |
| Nylon 6/6 | 0.9 – 1.2 | 25 – 40 | Polymer, viscoelastic behavior |
Important: Yield strengths are approximate and vary with alloy, temper, and temperature. For critical designs, always consult specific material specifications and apply appropriate safety factors.
An automotive engineer needs to design a hollow transmission shaft transmitting 450 N·m torque, length 1.2 m, using steel (G = 80 GPa). The outer radius is constrained to 45 mm. Using the hollow shaft option with rᵢ = 30 mm, the calculator gives J = 2.57e-6 m⁴, τmax = 70.2 MPa (safe for steel with τallow ≈ 120 MPa using N=2), and twist angle = 2.63 degrees. Compared to a solid shaft of same outer radius, weight reduction is ~44% while maintaining comparable torsional stiffness. This interactive tool enables fast iteration of inner radii to meet stress and weight targets.
A marine propeller shaft transmits 15,000 N·m torque over 4 meters. Using hollow 316 stainless steel (G = 77 GPa, τy = 200 MPa) with outer diameter 120 mm and inner diameter 80 mm. The calculator yields J = 1.64e-5 m⁴, τmax = 54.9 MPa, and θ = 2.12°. With a safety factor of 2.5, the allowable stress is 80 MPa, indicating a safe design. The hollow construction reduces weight by 55% compared to a solid shaft, crucial for marine applications.
For rotating shafts, torque is related to power (P) and angular velocity (ω): T = P / ω, where ω = 2π·N/60 (N in RPM). Example: a motor delivers 100 kW at 1500 RPM → T = (100,000) / (2π·1500/60) = 636.6 N·m. This torque can then be used as input to our calculator to assess shear stress and twist.
Power‑Torque‑Speed Relations:
P (W) = T (N·m) × ω (rad/s)
ω (rad/s) = 2π × N (RPM) / 60
T (N·m) = 60 × P (W) / (2π × N (RPM))
Common conversion: 1 HP = 745.7 W