somebody double check me. I'm using 4 each 5/8 bolts grade 5 in pure
tension to hold 30 ton force. good to go?

Karl

Karl Townsend wrote:

somebody double check me. I'm using 4 each 5/8 bolts grade 5 in pure
tension to hold 30 ton force. good to go?

Karl

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.

"Pete C." wrote:

Karl Townsend wrote:

somebody double check me. I'm using 4 each 5/8 bolts grade 5 in pure
tension to hold 30 ton force. good to go?

Karl

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.

On Monday, April 30, 2012 10:08:21 AM UTC-5, Karl Townsend wrote:
somebody double check me. I'm using 4 each 5/8 bolts grade 5 in pure
tension to hold 30 ton force. good to go?

Karl

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

"Pete C." wrote:

"Pete C." wrote:

Karl Townsend wrote:

somebody double check me. I'm using 4 each 5/8 bolts grade 5 in pure
tension to hold 30 ton force. good to go?

Karl

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.

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.

On Mon, 30 Apr 2012 10:11:54 -0600, Pete C. wrote:

Karl Townsend wrote:

somebody double check me. I'm using 4 each 5/8 bolts grade 5 in pure
tension to hold 30 ton force. good to go?

Karl

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.

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.

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

Tim Wescott wrote:

On Mon, 30 Apr 2012 10:11:54 -0600, Pete C. wrote:

Karl Townsend wrote:

somebody double check me. I'm using 4 each 5/8 bolts grade 5 in pure
tension to hold 30 ton force. good to go?

Karl

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.

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?)

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.

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.

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.


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...

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

thanks guys, I'm glad i asked. I'll move up to 3/4

Karl

On Mon, 30 Apr 2012 12:11:43 -0500, Tim Wescott wrote:
On Mon, 30 Apr 2012 10:11:54 -0600, Pete C. wrote:
Karl Townsend wrote:
somebody double check me. I'm using 4 each 5/8 bolts grade 5 in pure
tension to hold 30 ton force. good to go?
....
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.

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?
(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.

--
jiw

On Mon, 30 Apr 2012 17:56:57 +0000, James Waldby wrote:

On Mon, 30 Apr 2012 12:11:43 -0500, Tim Wescott wrote:
On Mon, 30 Apr 2012 10:11:54 -0600, Pete C. wrote:
Karl Townsend wrote:
somebody double check me. I'm using 4 each 5/8 bolts grade 5 in pure
tension to hold 30 ton force. good to go?
...
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.

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?
(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

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

On Mon, 30 Apr 2012 11:20:22 -0600, Pete C. wrote:

Tim Wescott wrote:

On Mon, 30 Apr 2012 10:11:54 -0600, Pete C. wrote:

Karl Townsend wrote:

somebody double check me. I'm using 4 each 5/8 bolts grade 5 in pure
tension to hold 30 ton force. good to go?

Karl

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.

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?)

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.

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.

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.

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.


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...

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

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

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com