how much can you torque a bolt

Is this significant? " pi * 24 in. / .05 in. ~- 1500 ?????? "" sd

No. I did it again using 0.125" radius and got the same 2 ft-lb torque.

I also did it with an M8*1.25 500MPa bolt and got about the same torque.

There must be something pretty silly we're overlooking.

W/o getting out my hi-school physics book, seems to me the m-advantage of a bolt would be more like the circumference of (THE BOLT / the pitch)--or

3.14*.250/.05~= 15. Then if you exert 200 lbs on the 1-foot lever, you'd get out 1ft*200lbs*15= 3000 ft-lbs to snap it. At least, this 200 ft-lbs sounds more reasonable that 2 ft-lbs, as you said:) HTH, s

My other message seems to have vanished. But that 3000 ft-lbs doesn't mean anything. The tension on the bolt from 200 ft-lbs on the spanner would be about 290,000 lbs, according to the OP's and my way of calculating it. Way too much to break the bolt.

You are calculating it as if the bolt was 2 foot in diameter. If the bolt was that big applying 200 ft-lbs would be able to lift 300,000 lbs. But the bolt is only 1/4" in diameter so 200 ft-lbs will produce about a

100 times less tension on the bolt than what you calculate. But using 1/4" for the bolt diameter, the calculation is still off by a factor of 2 or 3. The cross section of the threaded bolt is less than 1/4" and the effective diameter of the threads for calculating the mechanical advantage should also not be calculated at the tip of the threads. Plus the interface between male and female threads will not be frictionless so the bolt will twist off before it achieves its full rated tension. Probably around 60-80 lbs on a foot long wrench will be enough to snap the the bolt.

-jim

Torque it until the bolt shears off, then back it off 1/4 turn.

HAHA! That's great!

LOL, my buddy's motto for the longest time is a variation:

"Tighten it until it strips, then back it off a 1/4 turn."

He used to tighten lugnuts until they squeaked and that was good enough, which was usually about 50% overtightened. He's since learned about torque wrenches. :)

Ray

I had a reference to a German torque wrench when redoing the injectors on my Nissan. Gudentite.

snipped-for-privacy@gmail.com wrote in news:1193881922.564374.314680 @q3g2000prf.googlegroups.com:

Try here:

In one of my Mech Engineer Design text books, a simplified formula is given. After much derivation the author simplifies it down to this:

T = K*Fi*d where T = torque K = is the lumped together constant that accounts for assumed friction coef, thread pitch, etc. Fi = is the preload desired d = bolt diameter

The diameter is given as the bolt OD or major diameter and the formula doesn't seem to depend on thread pitch (coarse of fine).

I assume by using 60000*pi*d^2/4 you would get a preload of: 2945 lbs Using the K = 0.2 as a constant (the book asserts this to be nearly constant), then you get T = 0.2*2945*0.25/12 = 12.3 ft-lbs

This seems ball park reasonable, don't you think?

How do you calculate that tension? I did it using a 1/4" diameter, and also using an 8mm diameter. Don't be fooled by the OP's equation with a 24" diameter, he didn't seem to actually use that to get the final answer.

What do you mean be "the calculation is still off..."? Do you know what the correct answer should be?

The first and last points would act to reduce the required torque to snap it so they don't matter. I think thread radius won't be very significant.

Even up to 12 ft-lbs like somebody else said, you could do that with your little finger on a 1-foot spanner. That sounds really unbelievable for a bolt holding an engine mount or exhaust manifold.

yes. Just for the heck of it I did this experiment

the 1/4x20 bolt sheared off at around 140-150 in-lb torque.

this is in line with their results

so hey, I guess my 60kpsi estimate was pretty close.

Keep on Torquin',

Viktor Mikhailovich Polesov

Tegger, good stuff there, thanks!

It does sound like a good answer, but the units aren't right. 3000 ft- lbs isn't 3000 lbs. If you try to convert it using the radius of the bolt, you get 19000 lbs tangential load on the bolt, * 15 -> 290,000 lbs tension.

Well you don't need to know the correct answer to know that 200 ft-lbs is too much. Your calculation is essentially correct if you disregard certain realities. If you look at a screw as an incline plane wrapped into a helix, then 20 turns with an incline wrapped around a 12 inch radius is like pushing a 60000lb weight up a 1500 inch incline lifting it one inch with a 200 lb load applied. That is a mechanical advantage of 1500:1. That's what it would be like if the bolt was 24" in diameter and no friction. All your calculation takes into account is the mechanical advantage of an frictionless inclined plane. But if the bolt is a quarter inch in diameter then the mechanical advantage of the lever (wrench) also needs to be considered. The lever mechanical advantage is about 100:1. So in theory you still have the same total mechanical advantage of 1500:1 but now its divided between the lever (100:1) and the incline (15:1). Due to the lever the load on the edge of the bolt from 200 lbs at 12" is

20,000 lbs. Now if 3000 lbs is all the bolt can support in tension it obviously isn't going to support that 20,000 lbs twisting load. There is also the issue of the physical thread size which has the effect of reducing the actual cross-section of the bolt to something less than the 1/4". And also, You need to reduce the calculated length of the incline to account for the actual radius where the load is applied to the thread (circumference of the threads is something less than Pi*1/4)). And none of this accounts for friction which is in reality quite large.

-jim

The maintenance manual I have for my Jeep CJ7 has a basic torque data sheet for the various bolt sizes.

For a 'US thread' size of 1/4" by 20 it lists 6 to 9 ft lb.

I would have to agree with that due to the ease I can twist off a 1/4" bolt with a 1' ratchet wrench.

Mike

86/00 CJ7 Laredo, 33x9.5 BFG Muds, 'glass nose to tail in '00 88 Cherokee 235 BFG AT's - G> Hello,