axial force generated by 3/8-16 bolt with a given torque 1

[aroundhere]

I have seen the simplified formulas of bolt clamping which is similar to what I am trying to figure, but wanted to know if this is correct. If want to find out how much force is generated on the end of the .375-16 bolt being screwed into a tapped hole when metal hits metal (the end of the .375 screw, not the head), or if I can compress a spring etc... I understand that friction may be a large variable. Either the steel bolt bottoming out on a sleet flat, or if compressing a spring... if I am applying a known torque of 120in-lb.

Do I not need to find the mechanical advantage first of the thread first? (thd dia*pi/pitch) .375*pi/.062= 19 MA
Then divide 120in-lb by thread radius.. 120in-lb/.1875in=640lb then multiply by the mech advantage 640lb*19=12,160lb ? this sounds high. Is this wrong? I understand that friction has not yet been accounted for. Before friction and assuming the head of bolt of screw does not bottom out is what I am curious of. Or, how far off is the above calculation. Please explain if I have missed the boat. If I am close, please comment on how the friction coefficient would then be applied. I have done this before years ago and had our ME check and I believe I was correct. I have forgot much since then. thanks


Dia=.375"
Torque=120 in-lbs
pitch=.062"

 

[desertfox]

Hi

I am a little confused about what your question is but I think what you are asking is :- if you tighten a bolt or screw until the shank hits metal but the head is in fresh air what will the force be exerted on the bottom of the bolt shank as it is torqued to 120 lb -in.
That being the case to me you would exert a force approximately given by the formula F= T/(0.2*d)

So F = 120/(.2*0.375) = 1600lbf

 

[racookpe1978]

That 1600 lbf would only be true if the bottom of the bolt hit the bottom of the drilled hole with 100% contact right?

Or is your 1600 lbf valid as a force (coming from the total of the threads being tightened to failure around the helix of the whole 3/8 dia bolt, but that 1600 lbf being distributed only across the area touching at the bottom of the drilled hole?

(A drilled and tapped hole being assumed to tapped all the way down, and the drilled hole being the usual 118 degree conic end? )

 

[Compositepro]

I get 12,064 lbs. The mechanical advantage is that the bolt will move 1/16" per turn. The 120 lbs is applied to a 1" lever that will move 2pi inches per turn. So F= 120x2xpix16.

 

[Tmoose]

http://www.repairengineering.com/bolt-torque-chart.html

http://www.spaenaur.com/pdf/sectionD/D48.pdf

https://www.kimballmidwest.com/catalog/MarketingText/Torque and Tightening.pdf

page 32 here -

http://www.holo-krome.com/pdf/Consolidated_Tech_Manual.pdf

 

[aroundhere]

I was trying to find out the force created on a certain size thread given a known torque and how the numbers are calculated. I really thought that the mechanical advantage needed to be calculated. .375*pi/.062 is just the wedge or ratio that needs to be determined, correct? Maybe I am wrong

 

[desertfox]

I assumed that the bottom of the screw was hitting a flat base square on and ignoring the threads other than just to obtain the axial force from the torque.

If the threads are being used to raise and lower a load then the equations on this site might be useful.

http://www.roymechx.co.uk/Useful_Tables/Cams_Springs/Power_Screws_1.html

 

[rb1957]

Don't bottom out screws !

at first i thought the question "nuts" but then i thought "aren't you "just" stretching the bolt against the bottom of the hole?"

next thought was "how do you torque engine block studs?" i can't imagine that these bolts are bottomed out; is there just so much friction in the block threads?

Quando Omni Flunkus Moritati

 

[MintJulep]

desertfox has it about right.

Except his coefficient value of 0.2 might be as low as 0.1 or as high as 0.4 depending on material, lubrication, plating, surface finish etc.

"How do you torque engine block studs?" You don't you torque a nut on the end of the stud.

 

[rb1957]

MJ ... no, but you do torque the head of the stud ... no?

Quando Omni Flunkus Moritati

 

[Screwman1]

The .2 that Fox is using is the nut factor for a normal assembled fastener where the head is contacting the bearing surface. around 40-50% of the total friction is underhead friction, so that number will be reduced by the lower friction created by just the tip of the shank contacting the bottom of the hole. How much it will change is pretty hard to say without actually running some tests. My guestimate is that the nut factor for the posters condition will be around .12, which would give a force on the tip of around 2,660 lb.
That is assuming a lot about things like plating, hardness, surface finish etc.
The simple way is to just rig up a little test fixture with a transducer mounted at the end of the tapped hole and run a few tests.

 

[rb1957]

ok, i'm back at "don't bottom out the screw". why would you want to ?

the tapped hole is essentially the nut of a conventional fastener. preload is developed between the head of the bolt and the tapped hole because the end of the bolt isn't restrained and the bolt can stretch.

if you torque a bottomed out screw, very quickly you'll strip the thread.

no?

Quando Omni Flunkus Moritati

 

[aroundhere]

I guess I should have used only the example of the end of the .375 stud compressing a spring, or if I was moving a load. Nothing is bottoming out, Im not torqing bolts, I was just using this as an example as to how the actual numbers are calculated and what they mean. I just wanted to know the compressive force generated at the end of the stud given a known torque. I always thought the MA of the thread came into play.

 

[rb1957]

"when metal hits metal" "the steel bolt bottoming out on a sleet flat" but i'll give you "if compressing a spring".

preload is the force created as the bolt stretches.

in your case, i think you're "just" compressing a spring, F = kx.

Quando Omni Flunkus Moritati

 

[JJPellin]

This is common and neccessary. We call them jack screws and we use them to separate parts that are sealed to one another with sealants or interference fit to one another. The force should be the same whether the bottom of the hole is integral with the threaded hole or is a separate part in the same location.

Johnny Pellin

 

[hydtools]

Here you go:

http://www.engineeringtoolbox.com/screw-jack-d_1308.html

Solve the equation for Q with your givens.

Ted

 

[aroundhere]

Thanks hyd, this is what I was thinking when I started this. Am I wrong to say that this equation can also be used if trying to determine the compressive force at the end of a threaded stud, even if it is metal to metal? You would just have to factor in friction? The reason I earlier said metal hits metal is because I was thinking of a metal to metal seal. I may then have to determine the surface area of the metal to metal... The mechanical advantage is taken into account in this equation. I thought it would also be on a screw.

 

[hydtools]

Yes, that would be the compressive force at the end of the screw. If you use a lubricant, you could neglect friction unless you need to be absolutely accurate.

Ted

 

[Dougt115]

Desertfox supplied the formula for use of a torque wrench at 120 inchpounds. The 0.2 is the generic factor to account for material finish, lubrication, surface contact, incline angle, etc.

If you need the exact force you will need to test and measure. All the other methods listed will just give you a theoretical value.

I used the same torque and screw diameter as first described and after 10 tests the average was a resultant force of 1375 lbs not the 1600lbs. Actual values ranged from 1335 to 1398lbs.

 

[aroundhere]

thanks Dougt...I will see if I can come up with this number. Nothing like actual test numbers. I was after the equation and exactly how the force is obtained more than anything. I am going to buy a good load cell and digital read out. I almost bought one months ago when we were shear testing screws. Again, I appreciate the numbers.