Calculation
p = F/(d·l)
For the rod eye, bearing length is its own thickness; for the fork, it is the sum of both fork-leg thicknesses; the thinner side usually governs.
p = F/(d·l)
Transversely loaded pin bearing pressure
For the rod eye, bearing length is its own thickness; for the fork, it is the sum of both fork-leg thicknesses; the thinner side usually governs.
MINTSI
02
Result
Select a target and calculate.
Understand the inputs
- Bearing pressure p — Nominal pressure between pin surface and component bore.
- Joint force F — Transverse force transmitted through the pin.
- Pin diameter d — Pin diameter at the eye considered.
- Effective bearing length l — Rod eye: its own thickness; fork: sum of both leg thicknesses (2·t).
Example
5,000 N, a 16 mm pin diameter and a 20 mm effective bearing length give p = 5,000 N / (16 mm · 20 mm) ≈ 15.6 MPa.
Assumptions and limits
Pressure evenly distributed over the projected area; real pressure distribution is uneven, especially with joint clearance or edge loading from pin deflection.
Understand Transversely loaded pin bearing pressure
This calculator finds the nominal bearing pressure between pin and component from joint force, pin diameter and effective bearing length.
What does this quantity describe?
Bearing pressure of a transversely loaded pin is referenced to the projected rectangular area of diameter times bearing length: p = F/(d·l). For the rod eye, l is its own thickness; for the fork, it is the sum of both fork-leg thicknesses (2·t).
This is comparable to a pipe resting on a support: what matters for support pressure is not the curved contact surface but the projected area of diameter times support width.
Formula and variables
p = F/(d·l)
p = F/(d·l)F = p·d·ll = F/(p·d)
| Symbol / input | Meaning |
|---|---|
| Bearing pressure p | Nominal pressure between pin surface and component bore. |
| Joint force F | Transverse force transmitted through the pin. |
| Pin diameter d | Pin diameter at the eye considered. |
| Effective bearing length l | Rod eye: its own thickness; fork: sum of both leg thicknesses (2·t). |
Choose the inputs correctly
Joint force F, pin diameter d and effective bearing length l (rod-eye thickness or sum of fork-leg thicknesses) set bearing pressure.
How to use the calculator
Enter force, diameter and bearing length to check bearing pressure. Evaluate fork and rod eyes separately by using the matching bearing length for each, comparing both against their respective allowable value.
Worked example
5,000 N, a 16 mm pin diameter and a 20 mm effective bearing length give p = 5,000 N / (16 mm · 20 mm) ≈ 15.6 MPa.
Understand the result and units
A bearing pressure of 15.6 MPa is already noticeable for a lubricated sliding fit in a joint, where allowable pressures are often only 5 to 15 MPa — markedly lower than a fixed dowel-pin joint's over 100 MPa.
Force is given in N, diameter and bearing length in mm, and the resulting pressure in MPa (N/mm²).
Sliding fit versus fixed fit: very different allowable pressures
Moving joints with a lubricated sliding fit have markedly lower allowable bearing pressures than fixed dowel-pin joints, since repeated relative motion can cause wear, galling and lubricant film failure if pressure is set too high. Typical guideline values for lubricated steel-on-steel sliding fits are around 15 MPa, for bronze bushings around 10 MPa — well below the over 100 MPa that can be allowable for a fixed joint with no relative motion. When designing a clevis pin, always check whether a sliding or fixed joint is present before choosing an allowable value.
Typical applications
The calculation is used to check both the fork and rod eyes of a clevis pin separately against their respective allowable bearing pressure, particularly for moving joints with sliding fits, where markedly lower limits apply than for fixed joints.
Assumptions, limits and common mistakes
The model assumes pressure evenly distributed over the projected area. Real pressure distribution is uneven, especially with joint clearance or when the pin deflects under load, concentrating load at the edges of the bearing area (edge loading).
Common mistake: A common mistake is assuming the same allowable bearing pressure for fork and rod, even though moving sliding fits in joints have markedly lower allowable values than fixed fits. It is also easy to use only one leg's thickness for the fork instead of the sum of both leg thicknesses.
Frequently asked questions
Why is the projected area used instead of the cylindrical surface?
Because load transfer at the bore wall behaves approximately like pressure on the rectangular area seen from the load direction.
What bearing length do I use for the fork?
The sum of both fork-leg thicknesses (2·t), since both legs jointly support the rod's load.
Why are allowable pressures so much lower for joints?
Because repeated relative motion at excessive pressure can cause wear and galling; see the section above.
What happens with edge loading from pin deflection?
Load concentrates at the edges of the bearing area instead of distributing evenly, so actual peak pressure can exceed the nominal value computed here.
Do I need to check fork and rod separately?
Yes, both have different bearing lengths and often different materials with different allowable pressures.