Calculation
Mb = F/4 · (t + b); Wb = π·d³/32; σb = Mb/Wb
A clevis pin is never in pure shear; the distributed support at fork and rod additionally creates a bending moment that can govern for slender pins.
Mb = F/4·(t+b); σb = Mb/Wb
Clevis pin bending stress
A clevis pin is never in pure shear; the distributed support at fork and rod additionally creates a bending moment that can govern for slender pins.
MINTSI
02
Result
Select a target and calculate.
Understand the inputs
- Bending stress σb — Maximum bending stress in the pin cross-section between fork and rod.
- Joint force F — Transverse force transmitted through the pin.
- Fork-leg thickness t — Thickness of each fork leg (assuming both legs equal).
- Rod-eye width b — Width of the center rod eye between the fork legs.
- Pin diameter d — Diameter of the clevis pin.
Example
5,000 N, a 10 mm fork-leg thickness, 20 mm rod-eye width and 16 mm pin diameter give Mb = 37.5 N·m and σb ≈ 93.3 MPa.
Assumptions and limits
Idealized model with clearance fit at both fork and rod and uniformly distributed bearing load per component (per the transversely loaded pin-joint figure); press fits, joint clearance and real load distribution shift the magnitude and location of the maximum bending moment.
Understand Clevis pin bending stress
This calculator finds the maximum bending stress of a double-shear clevis pin from joint force, fork-leg thickness, rod-eye width and pin diameter.
What does this quantity describe?
A clevis pin is never in pure shear: the distributed support at the fork legs and rod eye additionally creates a bending moment. With clearance fit at both components, the maximum bending moment is Mb = F/4·(t+b), with section modulus Wb = π·d³/32 and bending stress σb = Mb/Wb.
Picture the pin as a short beam supported at both ends by the fork legs and loaded in the middle by the rod eye — similar to a simple beam on two supports with a central point load, except here the load itself is applied over some width rather than at a single point.
Formula and variables
Mb = F/4 · (t + b); Wb = π·d³/32; σb = Mb/Wb
Mb = F/4·(t+b)Wb = π·d³/32σb = Mb/Wb
| Symbol / input | Meaning |
|---|---|
| Bending stress σb | Maximum bending stress in the pin cross-section between fork and rod. |
| Joint force F | Transverse force transmitted through the pin. |
| Fork-leg thickness t | Thickness of each fork leg (assuming both legs equal). |
| Rod-eye width b | Width of the center rod eye between the fork legs. |
| Pin diameter d | Diameter of the clevis pin. |
Choose the inputs correctly
Joint force F, fork-leg thickness t, rod-eye width b and pin diameter d set the maximum bending stress.
How to use the calculator
Enter force, fork-leg thickness, rod-eye width and pin diameter to check bending stress. Compare the result against the pin material's allowable bending stress, and additionally against the shear and bearing checks.
Worked example
5,000 N, a 10 mm fork-leg thickness, 20 mm rod-eye width and 16 mm pin diameter give Mb = 37.5 N·m and σb ≈ 93.3 MPa.
Understand the result and units
A bending stress of 93.3 MPa at this pin diameter already represents a noticeable share of the allowable bending stress for many steels — markedly higher than the associated shear stress of about 12.4 MPa for the same geometry, showing that bending, not shear, governs the design here.
Force is given in N, lengths in mm, and the resulting stress in MPa (N/mm²).
Why a clevis pin is never purely sheared
Ideal shear loading would require load at fork and rod to act exactly in an infinitely thin plane each. In reality, load distributes over the finite thickness of the fork legs and the width of the rod eye, so the pin behaves like a short beam supported at both ends, developing bending stress in addition to shear stress. The greater the distance between load-introduction points relative to pin diameter, the more bending stress dominates over shear stress — a slender, long pin is therefore more bending-critical than a short, thick one.
Typical applications
The calculation is used to design and check fork joints on hydraulic cylinders, steering linkages, rocker arms and similar structures where a pin transmits force perpendicular to its axis between a fork and a rod.
Assumptions, limits and common mistakes
The model assumes clearance fit at both components and a bearing load evenly distributed across fork legs and rod eye (idealized per the transversely loaded pin-joint figure). With a press fit at fork or rod, both the magnitude and location of the maximum bending moment shift; real joint clearance can additionally alter load introduction.
Common mistake: A common mistake is checking a clevis pin only for shear instead of also for bending, even though bending often governs for slender pins with a larger leg or eye spacing. It is also easy to overlook that different fits (press fit instead of clearance fit) change the idealized bending moment.
Frequently asked questions
Why does bending stress arise in the pin at all?
Because load acts distributed over the fork-leg and rod-eye width rather than in a single plane, making the pin behave like a short bending beam; see the section above.
When does bending dominate over shear?
For slender pins with a large distance between bearing zones relative to diameter; short, thick pins tend to be shear-critical instead.
Does the bending moment change with a press fit?
Yes, a press fit at fork or rod instead of a clearance fit shifts the idealized bending moment to different values, as given in the source literature for each fit combination.
Is the bending check alone enough for a complete verification?
No, shear stress and bearing pressure at fork and rod must also be checked.
How do I reduce bending stress for a given force?
Use a larger pin diameter (acts with the cube on section modulus), reduce fork-leg thickness or rod-eye width, or place fork and rod closer together.