Force of bolt

[engineerfin]

If I was to use a threaded bolt and nut for a pressing device would the formula to calculate the torque required be t= coefficient x diameter of bolt x force ie 400lbs of force with a 1" dia bolt and .2 coefficient = 80in lbs of torque

 

[israelkk]

Practically, yes. however, you need to do a check to make sure what is the actual coefficient. you can use a load cell or even a known compression spring to test how much torque to produce the needed force.i

 

[desertfox]

Hi engineerfin

It also depends how accurate a bolt load you require because torqueing bolts is subject to a tolerance of +/- about 25%.

 

[rb1957]

i don't clearly understnad the link between "for a pressing device" and bolt preload (from the equation given).

another day in paradise, or is paradise one day closer ?

 

[Screwman1]

It sounds you want to use the bolt as a jack screw and not a compressive fastener. If that is the case, the standard .20 nut factor or K factor will not be accurate because it is including under-head friction and you obviously will not have under-head friction- you will have point friction instead and the radius will be much smaller.
for something like this, about the only way to get an accurate number will be to run some tests and then get the nut factor constant from those tests.

 

[engineerfin]

How come the thread pitch is not a consideration. I would think a fine pitch would have a higher torque. Thats why you see tie rod separators, flaring tools, etc with fine threads. Yes i am using it has a jacking screw, good point on the head friction.

 

[Dougt115]

The .2 is to account for all of the friction. Much of which will be in the threads not on the head. Grease, thread surface finish and head contact can all alter the value of the .2 coefficient. This is why a load test cell is recommended when accurate loading is required.

The equation is accounting for the leverage of the threads relative to the torque. 50 inch lbs on a 1 inch bolt has a leverage of 50 lbs while on a .5 in bolt is has a leverage of 100lbs.

 

[CoryPad]

Thread pitch is part of the consideration for the long-form equations. You are using the short-form equation which grossly oversimplifies the mechanics.

 

[engineerfin]

CoryPad, do you have the long formula? That is what I was originally looking for.

 

[desertfox]

Try pages 2 and 3 of this link

http://www.torqueleader.com/linkservid/EAEA3A24-90F4-30D2-6FA4D3F056E832EE/showMeta/0/

 

[rb1957]

pitch is a consideration if you're looking at bolt extension or "turn of the nut".

another day in paradise, or is paradise one day closer ?

 

[CoryPad]

The link provided by desertfox shows one version of the long form equation, there are others. As stated by rb1957, those equations assume you are clamping a joint using the fasteners. In your case, the bolt is acting like a leadscrew, so use equations like those shown here:

http://en.wikipedia.org/wiki/Leadscrew

 

[tbuelna]

With a jack screw, you need to account for the thread frictions and the end moment frictions at the screw contact face.

 

[CoryPad]

Terry is correct about the contact face, leadscrew equations may miss that. Multi-Jackbolt Tensioning Systems (e.g. Superbolt) use screws in a similar manner to what is described in this thread. The equation they use can be found on page 508 of Handbook of Bolts and Bolted Joints (Google Books shows this page if you do not have the book).

 

[engineerfin]

I cant view page 509 so I dont see all of the variables. Would I be correct in guessing R=length of torque arm on screw P=pitch of threads r=radius of screw ......what would the B(eta) angle be?

 

[CoryPad]

R = mean thread radius
[β] = half thread angle = 30[°]
r = screw end radius
P = pitch

 

[Cockroach]

Just look up Lifting Capacity for threaded fasteners in any decent book on mechanics. Or Google it, many engineering websites spell it out in detail.

Regards,
Cockroach