A shear pin is the sacrificial fuse of a mechanical drive: it is sized to shear at a known torque so the rest of the machine survives an overload. This calculator converts a target overload torque into the pin diameter needed, given the pin’s moment arm, the material’s ultimate shear strength, and a safety factor.
How it works
Torque on the shaft produces a shear force on the pin at its moment arm; the pin must just reach its shear strength at the target torque:
shear force = target torque / moment arm
required area = (shear force * safety factor) / ultimate shear strength
(divide area by 2 for double shear)
pin diameter = sqrt(4 * area_per_plane / pi)The result is the minimum diameter; the nearest standard stock size at or above it sets the actual breakaway torque, which the tool reports along with a recommended working torque below the shear point.
Worked example
To protect an auger that should never see more than 800 lb-ft, with the pin on a 1.5-inch radius, in single shear, using mild steel at 35,000 psi and a 1.3 safety factor:
- Shear force = 800 lb-ft × 12 in/ft / 1.5 in = 6,400 lb
- Required area = (6,400 × 1.3) / 35,000 = 0.237 in²
- Minimum diameter = √(4 × 0.237 / π) ≈ 0.549 inches
- Round up to the nearest stock size: a 9/16-inch (0.5625 in) pin
The 9/16-inch pin gives an actual breakaway torque of about 820 lb-ft — close to the target and safely above the working range.
Choosing materials
The material’s ultimate shear strength is roughly 60% of its tensile strength for ductile steels:
| Material | Approx. tensile strength | Approx. shear strength |
|---|---|---|
| Mild steel (1018) | ~60,000 psi | ~35,000 psi |
| Grade-5 bolt stock | ~120,000 psi | ~72,000 psi |
| Grade-8 bolt stock | ~150,000 psi | ~90,000 psi |
| Aluminium 6061-T6 | ~45,000 psi | ~27,000 psi |
Always use the actual material specification, not a generic value, especially when the breakaway torque needs to be precise.
Design tips
- Use a necked-down or grooved pin so failure happens at a predictable cross-section, not somewhere unpredictable along the shaft.
- Single vs. double shear — most hub-and-coupling designs are double shear (the pin is loaded at two planes simultaneously), which halves the required cross-section for the same force. Confirm your geometry before choosing.
- Set working torque conservatively below breakaway — a shear pin that fails every few hours under normal load is too weak; one that never fails even under the fault scenario is too strong. A safety factor of 1.2–1.5 is typical.
- Stock sizes matter — always round up to the next standard drill-rod or dowel diameter, and check the resulting breakaway torque to ensure it still protects the machine.