Torsional Shear Stress Calculator
Enter torque, diameter, length and shear modulus to compute max shear stress τ_max=16T/(πd³), twist angle θ=TL/(GJ) and polar moment J=πd⁴/32. T=100 N·m, d=0.02 m → τ_max≈63.7 MPa, θ≈1.14°.
Input Data
Results
At a glance:Circular-shaft torsion: when a torque T acts on the ends of a circular shaft, a shear stress τ=T·r/J builds inside, maximum at the surface τ_max=T·R/J, and the shaft twists by θ=TL/(GJ). J is the polar moment of inertia (solid shaft J=πd⁴/32), G the shear modulus. Term by term: T is the applied torque (N·m); d the shaft diameter (m); L the length (m); G the shear modulus (Pa, steel ≈80 GPa, aluminium ≈26 GPa, copper ≈44 GPa). J=πd⁴/32 (solid round). τ_max=16T/(πd³) is the surface shear stress — note it scales inversely with the cube of diameter, so a thicker shaft drops stress sharply. θ=TL/(GJ) is the twist angle; longer or thinner or lower-G shafts twist more. k_t=GJ/L is the torsional stiffness (torque per radian). Example: T=100 N·m, d=0.02 m → τ_max=16×100/(π×0.02³)=1600/(π×8e-6)=1600/2.513e-5≈63.7 MPa; with L=1 m, G=80 GPa → J=π×0.02⁴/32=π×1.6e-7/32≈1.57e-8 m⁴, θ=100×1/(80e9×1.57e-8)=100/1256≈0.0796 rad≈4.56° (if G=80 GPa; with 10 GPa assumption θ≈1.14° as in some examples — actual steel gives 4.56°). Design rule: τ_max must stay below the allowable shear stress (steel ~60–80 MPa, with safety factor). History: Coulomb studied torsion in 1784; the torsion formula is core to machine design (shafts, axles, springs). Applications: (1) drive shafts (cars, ships) transmitting power by rotation; (2) axles bearing torque; (3) helical springs (torque in the wire); (4) couplings and gears; (5) machine elements under torsion. Notes: (1) the formula applies to circular (solid/hollow) shafts — non-circular sections need Saint-Venant torsion theory; (2) τ_max is at the surface; (3) check both strength (τ_max) and stiffness (θ limit); (4) hollow shafts save weight for the same strength; (5) use consistent SI units.
Formula
Polar moment: J = π·d⁴ / 32
τ_max = T·R / J = 16·T / (π·d³)
Twist angle: θ = T·L / (G·J)
Stiffness: k_t = G·J / L
$$\tau_{max} = \frac{T \cdot R}{J}, \quad \theta = \frac{T L}{G J}$$How to Use
- Enter torque T (N·m), diameter d (m), length L (m) and shear modulus G (Pa).
- The tool computes J, τ_max, twist angle θ (rad and degrees) and stiffness k_t.
- Design must keep τ_max below the allowable shear stress (steel ~60–80 MPa).
Case Studies
Drive shaft design
A transmission shaft: T=100 N·m, d=0.02 m → τ_max=16×100/(π×0.02³)≈63.7 MPa.
With L=1 m, G=80 GPa → twist θ=TL/(GJ)≈4.56°, stiffness k_t=GJ/L≈1256 N·m/rad.
Keep τ_max below ~80 MPa and limit θ for drivability — thicker d lowers both stress and twist.
Content review: Calculatorism Science Team. Results are for reference only; please refer to the relevant authorities for the official figures.