Here is the most misunderstood fact in machine design. Hang a working load on a properly preloaded bolted joint, and the bolt does NOT pick up that load. It picks up a small fraction of it. The rest simply un-squeezes the metal that was clamped.

The reason is that the bolt and the clamped members are two springs in parallel, and a spring's share of a movement follows its stiffness. $C = \dfrac{k_b}{k_b + k_m}$ — read aloud C equals k-b over k-b plus k-m. $C$ is the joint stiffness ratio, a bare number with no units. The subscripts are the whole convention, so fix them now: b is the bolt and m is the members. $k_b$ is the bolt's stretch stiffness and $k_m$ the stiffness of the material squeezed under its head, both in MN/m. A metal joint lands $C$ somewhere around 0.1 to 0.3, so an external load of 100 kN raises the bolt's tension by perhaps 20 kN — and the other 80 kN comes off the members' compression.

That gift lasts exactly as long as the joint stays shut. Keep unloading the members and eventually there is nothing left to unload: the faces part, and from that instant the bolt takes every further newton alone. $P_0 = \dfrac{F_i}{1 - C}$ — P-nought equals F-i over one minus C. $P_0$ is the external load at separation in kilonewtons; $F_i$ is the preload put in at assembly, subscript i for installed. Read the denominator as what it is: $(1 - C)$ is the MEMBERS' share, and it is the members' compression that is being spent, so it is their share that divides.

Two things follow, and they run against instinct. Separation comes at MORE than the preload, not at it — a joint at 40 kN of clamp with $C = 0.20$ does not open until 50 kN. And a joint that cycles through separation dies fast, because the bolt's stress jumps discontinuously every time the faces part. When a joint fatigues, the answer is very often more preload, not a bigger bolt.