Bolt torque vs clamping force
OK I'm officially confused. The torque calculator found here: "http://www.engineersedge.com/cad-forum/posts/280.html", tells me if I apply 44 in/lb to a 1/4 bolt it will apply 880# clamping force. (to the assembly, I think)
44in/lb because, supposedly, that's about what a burly guy can apply w bare hand to a 1.5" diameter knob. PLEASE correct me if that's wrong.
The boss' design team says, "If you have a 1/4-20 bolt, one revolution moves the nut .05 inches. If 44 lbs is applied at a radius of 1 inch (44 inch lbs), the force has to move 2 x pi inches or about 6.28 inches. Therefore the force applied is 44 x 6.28 / .05 = 5580 lbs."
Who is right?
How did it go from 880 to 2.8 tons?
Does that mean, if I have to use a spacer under this knob that will be adjusted frequently, it has to take a 2.8 ton hit ever time it's tightened to 44in/lb.
Shouldn't the calculator referenced above, need to know the thread pitch or why not?
Barnyard engineer jerry