bolt tension strength

Not as far as I can see. It looks like only a 1.5x or so safety factor and only if the loading is even. I'd certainly want a better safety factor than that even if it isn't a life safety application. Pieces of press falling on my foot from a failure is annoying enough to spend a few more and/or larger bolts on it.

A little more checking puts my calculation at 2.13x. Some 32,000# tensile strength per bolt.

The tensile stress area at the threads for a 5/8-11 is .226 sq.in. and 5/8-18 is .256 sq.in. For grade 5 the max tensile stress is 120,000 psi.

30 ton = 60000 lbs force. For the coarse thread, 4*/.226= .904 sq.in. so .904 * 120000psi= 108484 lbs force, way to close to safety, (108484/60000=1.8 safety factor) and in the event one bolt is stressed to hold all this you have failure. And the 120000 is max, not working strength.

ignator

FYI I'm taking the minimum root/minor diameter for 5/8 NC (0.5589), calculating the area (0.267) and multiplying by the minimum tensile strength spec (120ksi) which comes to the 32,000# per bolt spec.

If you use the 85ksi proof test spec you have even less headroom. If the bolt loading isn't even, the torquing not correct, etc. that would take away from the headroom as well.

Personally I would use larger bolts rather than try to add more bolts and less certainty of even loading. 1"-8 NC would give you over 60,000# cap per bolt, 7/8-9 NC comes in just under, both using the 120ksi ultimate strength. If you go by proof strength then you need 1-1/8" NC.

I was going to say

(tensile strength) * (area) acceptable load = ---------------------------- (normalized testicle size)

And I'm not sure if you should think in terms of bits of the press falling on one's feet -- wouldn't it be more accurate to think in terms of bits of press flying about the shop?

5/8-11, so the minor diameter is something like 0.46", which gives an area of 0.166 sq-in across the threads, 85000 PSI proof strength gives 14000 lb/bolt (why is my number so much lower than yours? -- did you just look at the shank diameter, perhaps?) so you'd need more than four bolts just to barely hold the entire load if you wanted to limit things to the proof strength.

And that's ignoring the fact that if there's torque on the bolt it'll reduce the amount you can pull on it in tension...

Wouldn't Machinery's Handbook (which is beyond arm's reach at the moment, and I'm lazy) have things like the ultimate load for bolts, just for the looking?

Anyway: Karl: LOTS MORE BOLTS!!! Or figure out how to be out of the shop when you test the thing for the first time.

I looked up the minimum root/minor diameter spec for 5/8-11 2A and it's

0.5589". calculated out to the ultimate 120ksi for grade 5 gives the ~32K number, using the proof strength is more like 22K.

Personally, giving the difficulties of placing enough bolts and ensuring even loading, I'd go for four larger bolts sized to handle the full load. By my calculations and using the proof strength I come up with

1-1/8" bolts.

Yes, though I think the greater risk is uneven loading of the four bolts.

...

(snip calcs)

Tim, it looks like there's a serious mistake in the formula above. My understanding of physics is that NTS should be used to multiply the acceptable load, not divide it.

Damn. I think I must have started out by solving for acceptable area, then changed horses in mid stream. Or something.

At any rate -- You're right. It should be:

acceptable load = (tensile strength) * area * NTS

I just went for a sharp-pointed valley, 'cause I don't have the handbook in reach. You have to make sure to look at the actual minimum diameter specified for the bolt -- the nut minimum diameter is bigger.

You're probably right, although that would depend on the design of the machine.