# Thread: Linear Force Generated by a SHCS

1. ## Linear Force Generated by a SHCS

Hi all,

I am trying to find a formula to calculate and check an existing spreadsheet I am working with.

I have a 3/4-16 SHCS, I am putting 214.5 ft/lbs. in the screw and want to know what force the screw is exerting on a fixed object or how the force is multiplied through the fastener., or Linear Force Generated by said input.

I am not interested in the Axial forces, Clamp Load or lifting capacity as that seems to be the only things that seem to come up on the web and here when I looked for it.

Regards,
Jim

2. I'm not sure what you're looking for as I understand that...

Axial forces, Clamp Load or lifting capacity = force the screw is exerting on a fixed object

3. Hi again,I am deforming metal with a Jack screw, so I want to know if I input X force into a screw what is the forces exerted into the fixed object at the other end of the screw/metal being deformed.I have the following formula although not sure the variables:Formula:Torque to RaiseT = ((𝐹∗𝐷&#119898)/2 ∗((𝜋𝜇𝐷𝑚+&#119871/(𝜋𝐷𝑚 −𝜇&#119871)F = ((2∗&#119879)/𝐷𝑚 ∗((𝜋𝐷𝑚 −𝜇&#119871/(𝜋𝜇𝐷𝑚+&#119871)the variables I have are to enter from the spread sheet are the following:Screw Pitch Dia (in): [B1] 0.7094Threads per Inch: [B2] 16Screw Lead (Inch per thread): Formula =1/B2 [B3] 0.0625Coefficient of Friction: [B4] 0.65 (Steel on Steel)Torque Applied (In-Lb): FT-Lb: [B5] 214.5Linear Force Generated (Lb): formula=1/((1/B5)*((B3+(6.2832*B4*B1/2))/(6.2832*B1/2-(B4*B3)))*(B1/2/1)) [B6] 875.6216Not sure I am explaining it correctly, but attempting to validate the above spreadsheet is valid to calculate the force exerted by the screw.Thanks Again,Regards,Jim

4. Agree with Kelly. "Axial force" means "force along the centerline axis", in other words it is the "linear" force I believe you are inquiring about. How is your force different from that?

5. Hi again,

Sorry for the gibberish as not sure what happened as not how I had typed it.

Think I found the inverse formula I need in the machinist handbook just need the inverse formula (any help would be appreciated) or solve for "Q" in the below formula which I guess might be the Axial Load

Force-2024-04-16.jpg

Regards,
Jim

6. Hi Again,

Think this is the formula I need can anyone confirm:

Q = F X ((6.2832r + μp)/(6.2832μr – p)) X (R/r)

Regards,
Jim

7. Originally Posted by jim_moses
Hi Again,

Think this is the formula I need can anyone confirm:

Q = F X ((6.2832r + μp)/(6.2832μr – p)) X (R/r)
Your algebra is correct however do know that Q is still axial force applied by the screw and the variables are those related to 'the motion opposite of load Q"

The unmanipulated equation for F is calculating the estimated force to initially unscrew the force Q applied by the screw.

8. Hi again,

Just to confirm if I put the said screw in a fixture and applied force to the lever (R) the output or Axial force is what I would see if the screw was sitting on a load cell correct?

Regards,
Jim

9. Originally Posted by jim_moses
Hi again,

Just to confirm if I put the said screw in a fixture and applied force to the lever (R) the output or Axial force is what I would see if the screw was sitting on a load cell correct?

Regards,
Jim
As with all analysis it is an nominal estimate with +/- variability.

10. Thanks,

one more question as unsure what the units I should be using as I compared my spreadsheet to your calculator, Estimated Fastener Bolt Clamp Force Torque Calculator and I have put in Ft-lbs, but the wording seems to suggest inch-lbs

either way I would have expected to be closer to Engineers edge calculator so what do I have wrong or am I figuring something different between the 2

see attached:

Regards,
Jim

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