Solid Shaft Diameter for an Allowable Torsion
The minimum SOLID-shaft diameter that keeps the max surface shear stress within an allowable.
Example
You enter
- Torque T (lb-in) 12000
- Allowable shear stress (psi) 8000
You get
- Minimum solid-shaft diameter 1.969 in
Details, formula, and sources
d = (16 T / (pi tau_allow))^(1/3). A 12,000 lb-in torque at an 8,000 psi allowable needs about a 1.97 in shaft (round up to stock). Sizes for stress only; check the angle of twist (shown with a length and G) against the service limit. Pure torsion, no keyway concentration. A design aid, not the engineer of record.
d = (16 T / (pi tau_allow))^(1/3) (solid shaft), the inverse of tau = 16 T / (pi d^3); J = pi d^4 / 32; theta = T L / (J G).
The standard circular-shaft torsion relations (mechanics of materials), solid shaft, solved for the diameter, by name.
The torsion formulas are published free in any mechanics-of-materials reference. The engineer of record governs the design.
Estimate. AHJ and licensed professional govern.
Field names used by the API: T_lbin, tau_allow_psi, d_in
- Solid circular section d = (16 T / (pi tau_allow))^(1/3); J = pi d^4/32mechanics of materials
- Pure torsion no bending or axial load; elastic and prismatic; check the twist separatelymechanics of materials
- No stress concentration keyways, shoulders, and holes raise the local stress; not includedscope of this tile