τs = F / (m · π·d²/4)
A clevis pin usually sits in double shear between a fork and a rod; this halves shear stress compared with a single-shear arrangement at the same force and diameter.
A clevis pin usually sits in double shear between a fork and a rod; this halves shear stress compared with a single-shear arrangement at the same force and diameter.
Select a target and calculate.
A clevis pin usually sits in double shear between a fork and a rod; this halves shear stress compared with a single-shear arrangement at the same force and diameter.
5,000 N on a double-shear pin with a 16 mm diameter give τs = 5,000 N / (2 · 201.1 mm²) ≈ 12.4 MPa.
Pure nominal stress from the external force; press-fit preload, local notch stress and the additional bending a clevis pin experiences are not included here and must be checked separately.
This calculator finds the mean shear stress of a transversely loaded pin from transverse force, diameter and shear configuration.
A clevis pin usually sits in double shear between a fork and an in-between rod. Transverse force F then splits across two shear planes; mean shear stress follows from τs = F/(m·A), with A = π·d²/4 the pin cross-section and m the number of shear planes.
Picture a bolt cutter distributing the same load across one or two blades: with two blades instead of one, each blade carries only half the force, halving stress per blade.
τs = F / (m · π·d²/4)
τs = F/(m·π·d²/4)F = τs·m·π·d²/4d = √(4F/(τs·m·π))| Symbol / input | Meaning |
|---|---|
| Shear stress τs | Mean shear stress in the pin cross-section per shear plane. |
| Transverse force F | Joint force acting perpendicular to the pin axis. |
| Pin diameter d | Pin diameter at the shear cross-section. |
| Shear planes m | 1 for single shear, 2 for double-shear (clevis) joints. |
Transverse force F, pin diameter d and shear planes m (1 for single shear, 2 for double shear) set the mean shear stress.
Enter force, diameter and shear configuration to check shear stress. For a sizing question — what diameter an allowable stress needs — use the linked required-pin-diameter calculator.
5,000 N on a double-shear pin with a 16 mm diameter give τs = 5,000 N / (2 · 201.1 mm²) ≈ 12.4 MPa.
A shear stress of 12.4 MPa is far below typical allowable values of roughly 50 to 100 MPa for pin steels; using single shear instead of double shear at otherwise equal values would double the stress to about 24.9 MPa.
Force is given in N, diameter in mm, and the resulting stress in MPa (N/mm²).
A clevis pin is not purely sheared but carries load across distributed bearing zones at the fork and rod, additionally creating a bending moment. For short, thick pins with little distance between the bearing zones, shear stress usually dominates; for slender pins with more distance or clearance, bending stress can markedly exceed shear stress and govern the design. A complete clevis pin check therefore always covers shear, bending and bearing together.
The calculation is used to check and pre-size clevis pins, piston pins and pin joints in machine and vehicle engineering, particularly for fork joints on cylinders, links and levers.
This pure nominal stress considers only the external force. Press-fit preload, local notch stress at transitions and, above all, the additional bending a clevis pin experiences are not included here and must be checked separately with the clevis pin bending calculator.
Common mistake: A common mistake is checking a clevis pin for shear alone and overlooking the often more critical bending stress. Shear plane count is also sometimes set incorrectly, for example in an arrangement with a through-pin lacking a true second shear plane.
No, a clevis pin can also fail in bending, see the section above; bearing pressure must also be checked separately.
Guideline values often fall between roughly 50 and 100 MPa depending on material, safety factor and load type (static, alternating, pulsating).
Dowel pins usually transmit load via an interference (press) fit, while clevis pins typically use a clearance fit to allow joint rotation; the nominal stress formula is the same for both.
Press fits add preload superimposed on external loads; this is not captured by the pure nominal stress formula and is instead covered blanket-style by more conservative allowable stress values.
Yes, the basic principle is identical; the specific application's allowable values should be used for the actual design.