calculting torque required

Hi experts!

I am struggling to find a formula which satisfies a problem I would like to have numbers for. I am working on a way to apply pressure to a transducer via a large bolt pressing on a plate.

Torque required on the bolt = how much pressure applied by the end of the bolt to a flat plate?

Assume fixture with threads to allow bolt to react when in contact with the plate.

Thoughts?

Thanks
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Commented:
If I understand your problem correctly, this should be fairly straightforward.

The force axially along the bolt is equal to the torque you apply to the wrench x length of wrench / radius of bolt.

This is actually the force that you are applying tangentially along the threads of the bolt.  Assuming that there is no friction between the bolt and whatever it threads into, this same force will be applied axially along the bolt to create the pressure between the plates.
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Author Commented:
Just to clarify, I have looked at the following which seems to satisfy if two plates are being clamped together?

Torque = coefficient of friction x bolt diameter x bolt tension

which gives me torque of 18,896 in/lbs required for 20,000 lbs clamping force...

Can bolt tension be considered the same as the axial force of the bolt pressing on the plate?.
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Author Commented:
Thanks CompProbSolv,

Do these numbers look correct?

Torque      1000      in/lbs
Arm Length      2      ft
Diameter of bolt      0.5      in

Force =      96000      lbs

Thanks
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Commented:
I would think that is correct.

What is interesting is that the pitch of the bolt doesn't factor into this.  It DOES factor into how fast you have to turn the bolt for a specific torque.
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Author Commented:
Perfect, thanks, I will be doing the real-world testing and will note the numbers...
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Commented:
Do these numbers look correct?

Torque      1000      in/lbs
Arm Length      2      ft
Diameter of bolt      0.5      in

Force =      96000      lbs
Assuming you mean
Torque      1000      in*lbs
then
Force =      96000      lbs
might look correct if the bolt was threaded at around 30.5 per inch, which seems a lot for a 0.5 in diameter bolt, especially one meant to hold that much force.
Friction might increase the torque to tighten, and decrease the torque to hold.

Or do you mean 1000 lbs of force at the end of a 2 ft arm, which would be 24000 in*lbs of torque, in which case  96000      lbs of force on the plate could be achieved with a more reasonable thread pitch
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Author Commented:
Hi ozo,

Yes, 1000 lbs torque is applied at the end of the 24" arm in the calculation.

I do not need to apply such heavy presure on a plate. these were just nice numbers to work with :)

I will be trying this in the real world later next week, so I will have numbers for both dry and lubricated thread setups.

Changing the threads depending on durability is one of the things to consider depending on results, but I am starting with M30 x 1.5.
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Commented:
1000 in*lbs torque means 41.6 lbs force applied at the end of the 24" arm
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Commented:
Ozo is correct both in units (it is in*lbs and not in/lbs; I overlooked that error originally) and in the calculation of torque.  The force applied multiplied by the arm length is the torque.
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