Pin Shear and Bearing Stress Calculator
Checks a pin, rivet or bolt in single or double shear against three separate mechanisms: shear of the pin, bearing crushing of the plate around the hole, and tearing of the net section past the hole. The governing mechanism, the capacity of each, and the joint utilisation are all reported so the weak item is obvious rather than buried.
How to use this calculator
- Enter the pin shank diameter, the number of shear planes, and the shear force carried across the joint.
- Enter the bearing thickness, which is the thickness of the plate the pin actually bears on rather than the total of all plates.
- Enter the plate width at the pin location, and the allowable shear, bearing and tensile stresses from the same source so the three checks are consistent.
- Read the stress in each mechanism, the allowable load for each, and the governing one, then check the edge distance separately because pull out is not covered here.
Frequently asked questions
Is a pin in double shear twice as strong as one in single shear?
For the shear mechanism, yes, because there are two sections resisting the load. The bearing and net section checks do not change, so the joint often stops being governed by shear and starts being governed by the plate. That is why a clevis pin through two outer plates and a central plate is usually limited by crushing of the central plate rather than by the pin.
Why is the allowable bearing stress so much higher than the allowable shear stress?
Bearing is a local crushing stress over a small contact patch, and the material around it constrains the deformed region, so the apparent strength is much higher than a uniaxial test would suggest. Common practice for steel pins allows 1.5 to 2 times the allowable shear. It is not a true stress, because the pressure varies around the hole and peaks at about 1.5 times the average where the load direction crosses the plate edge.
What edge distance do I need from the hole to the side of the plate?
Enough that the plate cannot shear out along two lines running from the hole to the edge. That check is the distance from the hole centre to the edge, times twice the thickness, compared against the allowable shear, and it is a separate mechanism from tearing across the net section. A common rule of thumb is at least 1.5 to 2 times the pin diameter, but the check itself is quick and takes precedence over the rule.
Why does a longer pin or thicker plate not always help?
Shear capacity depends only on the pin cross section, so a longer pin adds nothing to it, and beyond about three diameters of bearing the pin starts to bend rather than shear cleanly, which puts the calculation outside its assumptions. Bearing and net section do benefit from thickness, so on a joint governed by bearing, thickening the plate or increasing the pin diameter both work.