# Thread: Bolt force Equation - Why no need for Pitch

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May 2018
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## Bolt force Equation - Why no need for Pitch

Hi all,

I was trying to calculate the clamping load from a vise based on the leadscrew TPI. All of the calculators I can find online use the equation P = (KxD)/T where:

• T Target tighten torque (the result of this formula is in inch pounds, dividing by 12 yields foot pounds)
• K Coefficient of friction (nut factor), always an estimation in this formula
• D Bolts nominal diameter in inches
• P Bolt's desired tensile load in pounds
-Techshop

So, why doesn't the pitch of the screw come into play? Surely a 3/4"-5TPI Acme thread takes way more torque than a 3/4"-20 screw. Yet the formula does not account for this, and so I do not really trust it. Advice? Thanks.

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The pitch (lead, to be more precise) is buried in the variable K. Shigley's Mechanical Engineering Design, 1989, p. 346, says, "The interesting fact...is that K~0.20 for µ=µC=0.15 [the "typical" coefficient of friction] no matter what size bolts are employed and no matter whether the threads are coarse or fine." The main reason is that the wedging action caused by the 60° thread flank included angle dominates over the effect of the lead angle. Thus, the formula would not work for square-threaded screws.

At the end of the day, the formula is only an approximation, so you should only trust it as an approximation. The torque-tension relationship is fairly loosey-goosey. A better-but-still-imperfect method of bolt tightening is to tighten to some given torque less than "handbook" full torque and then tighten another 60°, i.e., turn one by more point on the hex head.

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Originally Posted by rklopp
The pitch (lead, to be more precise) is buried in the variable K. Shigley's Mechanical Engineering Design, 1989, p. 346, says, "The interesting fact...is that K~0.20 for µ=µC=0.15 [the "typical" coefficient of friction] no matter what size bolts are employed and no matter whether the threads are coarse or fine." The main reason is that the wedging action caused by the 60° thread flank included angle dominates over the effect of the lead angle. Thus, the formula would not work for square-threaded screws.

At the end of the day, the formula is only an approximation, so you should only trust it as an approximation. The torque-tension relationship is fairly loosey-goosey. A better-but-still-imperfect method of bolt tightening is to tighten to some given torque less than "handbook" full torque and then tighten another 60°, i.e., turn one by more point on the hex head.
I see. Thanks. So are there any good formulas for power transmission/square thread?

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Because of this disclaimer:

"This relationship is based on the assumption that regular series nuts and bolts with rolled threads are used, acting on surfaces with industry standard thread pitch and flank angle."

Also that K is variable. Ops, I see that's been covered.

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Originally Posted by The Metal Butcher
I see. Thanks. So are there any good formulas for power transmission/square thread?
To quote the wascally wabbit... "mhhyeehh, could be."

https://www.engineersedge.com/mechan...sign_13982.htm

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Thank you so much sir. I did some searches but wasn't as successful as you.

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I think You are asking for what the force is between the jaws when you turn the handle ?
That is a simple leverage calculation and is a function of Pitch,PD, and lever arm (handle length) and thread can be Buttress, square,Acme , of V Thread. Friction occurs is all but for a lubricated thread, I would assume a 15 % loss.
You want the Circumference of the Pitch Diameter versus the Pitch for the leverage
Say the screw is .950 in diameter at Pitch Center and the OD is 1 " ( Just for grins here) and the pitch is 8 TPI or .125"
So the Circumference is .95 times Pi ( 3.14 ) which is 3"
Divide the C by the Pitch ( 3 /.125 ) and you get 24
So if your Handle is one foot in length and you put one pound of force on the end of the handle ( one FP) , the thread exerts 24 pounds of force (minus friction of 15 % approx)
So (PD x Pi) /P x FP = F x .85

Rich

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Interesting forumula Rich.

Using a 1 1/2-4 thread (1.375? Pitch) and the amesweb calculator above, 400 ft-lbs gives a 24000lbs pull. Using your calculations, I get a 8000lbs force, which seems more reasonable. These are quite different though but from what I've seen from Royce (the guy that built the big 1400lb vise) yours seems much more accurate. I guess I should take some time to make sense of the formulas on the amesweb site.

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Originally Posted by Rich Carlstedt
I would assume a 15 % loss.
Maybe for a very well lubricated system, ball nut, ball thrust, etc. I bet the real friction loss is greater for a typical machine vise. If it was really only 15% the vise might not even stay tight. Might unwind as soon as you let go of the handle. Think about it. This is one of the objections to using ball lead screws on manual machines. Cutting forces can backdrive the screws.

Friction loss for bolted joints is more like 80%. This is why thread pitch does not appear in the formula. The difference between coarse and fine thread is a very small percentage, like 1 or 2%. Lost in the noise.

http://www.boltscience.com/

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As an approximation, presumably it is only useful over a small range covering typical bolts (what it's for). A pitch of 100mm on a 25mm dia round object would not be within that range.

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