How do I Calculate the Force from a Bolt Used as a Jack Bolt

Force gauge....

Sounds like somebody flunked Mechanics 101.

If the surface the bolt heads are thrusting (jacking) against are fixed, then it is based on the pitch of the screws and the torque setting, along with the surface area of the threads, roughly, although if you record the force required to loosen or tighten, you will not get reproducible results, entirely.

There will be differential frictional tearing of the microscopic surfaces that interfere with the results.

If the uppers are not fixed, it depends on weight against the area of the bolt heads.

The moment of friction M (torque) of a cylinder of radius R with pressure P is given by

M=(2/3)μPR

where μ is the coefficient of friction.

So if you know M, R, and μ, solve for P.

https://www.physicsforums.com/threads/friction-of-rotating-object.207053/

If the problem is that a bolt is turned until the head pushes against an immovable machined surface, it will turn relatively freely until the bolt head makes contact with the surface. The further it is turned, the greater the force on the bolt head as it is screwed against the surface, and the more torque required to turn it further.

The bolt is turned until the torque reaches 90 ft lb. Assume most of the torque is due to friction between the bolt head and the surface, and you can approximate the bolt head as circular, of radius R.

The torque required to turn a cylinder against a surface is

T = (2/3)μPR, or P=T / ((2/3)*μ*R)

https://www.physicsforums.com/threads/friction-of-rotating-object.207053/

where P is the pressure (force/area), and μ is the coefficient of friction.

Torque T is 90 ft lb. Given R, we can solve for P.

P=90/(.667*μ*R)

then Force = pressure x area = P*pi*R².

A common formula relating bolt torque to load assumes 90% is lost in friction, so force would be ∼ 0.1 x calculated figure. That might be pessimistic, others say 0.15 - 0.2 x, but not 0.8 - 0.85 x.