Turn the load sideways. A bolt in a lap joint is not being pulled along its axis any more; the plates are trying to slide past each other and the bolt is in the way. That one load can break the joint in three different places, each with its own area, and a joint is only as good as the worst of the three. In all of them the load is in newtons and the area in mm², so the stress lands in MPa.

The bolt can be cut. $\tau = \dfrac{V}{A}$, read aloud tau equals V over A. $\tau$, the Greek letter tau, is the average shear stress in the bolt. $V$ is the shear force across it. $A$ is the area being cut: the full shank circle, $\pi d^{2}/4$, counted once for every shear plane. Two lapped plates cut the bolt once. A centre plate between two outer plates cuts it twice, and that doubles the area for the price of one bolt.

The plate can tear. $\sigma = \dfrac{P}{A_{net}}$, sigma equals P over A-net. $P$ is the tension in the plate and $A_{net}$ is the net area, the metal left across the width once the hole is taken out: width minus HOLE diameter, times thickness. The hole is drilled a little larger than the bolt, and it is the hole that removes metal.

The hole can crush. $\sigma_b = \dfrac{P}{d\,t}$, sigma-b equals P over d t. $\sigma_b$ is the bearing stress where the bolt leans on the side of its hole, subscript b for bearing. $d$ is the bolt diameter and $t$ the plate thickness, both in mm. Their product is the projected area: the rectangle you would see looking at the bolt side-on, not the curved surface it actually touches. It is a convention, and every bearing allowable is written against it.

Now the part that decides exam marks. Bearing stress is nearly always the BIGGEST of the three numbers, and it rarely governs, because its allowable is far higher: the plate at the hole is hemmed in by the metal around it and cannot squash away. So never rank the modes by their stress in MPa. Divide each by its OWN allowable, and the one nearest to 1 is the one that fails first. That is the mode that governs.