Bearing Stress on a Pin or Bolt (σ = P/dt)

A bolted lap joint has three ways to come apart and most people only check two. The plate can tear across the net section beside the hole, which is a tension problem. The bolt can be cut through, which is a shear problem. And the bolt can crush the side of its own hole — elongating it into a slot until the joint slips, the bolt tilts and everything downstream goes out of alignment. That third one is bearing, and it is the failure mode that quietly ruins a connection without ever breaking anything in half.

The arithmetic is deliberately crude: $\sigma_b = P/(dt)$, with $d$ the bolt diameter and $t$ the thickness of the plate being crushed. An M20 bolt carrying 50 kN through a 10 mm plate bears on a projected area of $20 \times 10 = 200$ mm², so $\sigma_b = 50\,000/2.0 \times 10^{-4} = 250$ MPa. In inch units a 3/4 in bolt through a 1/2 in plate at 9000 lbf gives $9000/0.375 = 24\,000$ psi.

Notice what $d \times t$ is: a flat rectangle, the shadow the bolt casts on the plate. The bolt does not touch the hole over that rectangle — it touches over a curved arc, and the true contact pressure peaks well above the nominal figure at the crown. Codes handle this not by refining the geometry but by allowing a much higher stress: AISC permits $2.4F_u$ on the projected area, Eurocode 3 uses $k_1\alpha_b f_u$. So the number this page returns means nothing on its own. It only means something next to a bearing allowable. Held up against a tensile allowable it will look alarming and be wrong, and held up against nothing it is just a number. In a joint of several plates, run the check on the thinnest one, or on the sum of the thicknesses bearing the same direction.