Torque (Pitch/Bolt) to Axial Force 1

[KevinH673]

I have an assembly in which I must translate torsional force (a torque from screwing in a piece), that pushes against a silicon ring, which pushes against a metal piece. I'm interested to see if the piece will deform or fracture.

I do not have the specs for the spring constant for the silicon ring. I'm just going to do calculations assuming the piece that is being torqued is directly pushing on the metal piece. I'm not entirely sure on how to go about calcuating this.

I know of the torque to axial force equation that is:
F=.2*D*T
However, this does not account for pitch....

I've included a hand calculation my boss did. I'm not sure if it is correct or not...

I will include the hand calculations with an illustration, as well as two pictures of the assembly modelled in CAD if it helps.

The torque is 15 in-lbs. The thread is a 7/8 - 32 pitch.

 

[KevinH673]

Here is one picture

 

[KevinH673]

Here is another picture.

 

[CoryPad]

Could you explain what you mean by "silicon ring"? Your hand drawn image makes it look like it is an elastomeric seal (O ring), so is it SILICONE (a polymer)? Or, do you mean elemental silicon, the semiconductor used for integrated circuits?

The hand calculation is way off. The torque (15 in-lb) can be applied by your fingers, yet the thread diameter is almost an inch in diameter, so there is no way it could generate > 1000 lb of force.

You can find equations to calculate torque-tension relationship in MIL-HDBK-60 Threaded Fasteners - Tightening to Proper Tension, available free of cost here:

http://assist.daps.dla.mil/quicksearch/

 

[rb1957]

a simple google found this ...

http://euler9.tripod.com/fasteners/preload.html

which derives the 0.2 factor, and includes the effect of pitch.

i suggest that in your case, because you're compression a Si gasket, the your "joint" stiffness is much reduced ... maybe the derivation shows you how to account to that; maybe you need to determine your joint stiffness by test (controlled clamping of the joint).

 

[KevinH673]

Thanks. The hand calculations were not my own. I thought the value was too high, and when applying FEA to the existing piece, it fractured, when in actuality it assembles fine. I am redesigning the piece, which is why I have interest in properly calculating it's deformation.

 

[rb1957]

http://www.roymech.co.uk/Useful_Tables/Screws/Preloading.html

another site ... the pic is why i thought to post it

 

[KevinH673]

CoryPad, I did mean silicone, as in an O-Ring. I feel it can be neglected though, and the force just translated directly to the part the stress will be on without accounting for compression for now.


If I use the T= k*d*F, I get a value of about 83 lbf. This is much lower than my bosses estimated 1500 lbf...

 

[desertfox]

Hi KevinH673

Using the 15lbf-in torque on a 7/8" diameter screw I get a load of 2.625lbf with the formula you have provided.
Normally to create a seal with an 'o' ring you compress the thickness or cross section of the 'o' ring by a %, usually off the top of my head about 15-30% of the section size.
One full turn of your screw will compress the ring by 0.031".
Your design as it stands does not appear to have any great control on how much the 'o' ring gets compressed, given the error on using torque setting your 'o' ring could get severly over or undercompressed.
In your last post you appear to be working out torque with your formula but your answers are in lbf.

regards

desertfox

 

[KevinH673]

desertfox, I am indeed solving for the "F" value in lbf, I just wrote the equation out as it appears on the website.

I am getting the 85.71 lbf from

F= T/(K*D) = 15 in-lb/(.2*.875) = 85.71 lbf

Thanks for the help, I will look into the O-Ring compression.

 

[desertfox]

Hi KevinH

I found why I got a different figure from you, in your original post you put:-

know of the torque to axial force equation that is:
F=.2*D*T

I should have checked and didn't I just used it.
So I agree with your later figures now.

regards

desertfox