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Bolt Sizing for Push/Pull Blocks

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dmbypsi

Aerospace
Sep 15, 2017
4
US
Hi,

We currently have a project to where we are moving a heavy (10 ton) steel base around into position. We are doing this with the use of what we call "push/pull" blocks. They are simply a 2" thick block with a threaded hole. We are sending a bolt through this threaded hole which will push our base into position. My question is, what calculation will I use in determining if my bolt (currently 3/4") will be able to push this 10 ton base around. By around I mean +/- 1/4". The base is steel on steel so I know we will need a coefficient of friction in there somewhere.

Any help would be appreciated. See photo for an example of what we are doing.

Thanks,
Jamie
 
 http://files.engineering.com/getfile.aspx?folder=9a51c725-005a-4b76-af3c-ef2096bf10fb&file=PushPull.JPG
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My experience with piping systems dictates that these devices are commonly called "jacking bolts" or "screw jacks"

I have seen these in use specifically for separating large piping flanges and aligning pumps. motors and large compressors.

One calculation method is here :
Your design looks like both the bolt and pad are made to be removable. The vertical bolts on the pad are loaded in shear and it doesn't look robust enough.

I believe that, IMHO, a fully welded design would be more suitable ...



Please share with us your strength calculations for these bolts....

MJCronin
Sr. Process Engineer
 
"... we are moving a heavy (10 ton) steel base around into position..." does that mean rotation in lieu of translation ( linear motion)?
 
Chicopee,
We will have no rotation. Only X/Y linear motion of the base.
 
Hi dmbypsi

Multiply a coeff of friction with the 10 tons and that's you're axial load the screw will have to move approximately, then select a screw size that can withstand the axial load you've calculated so you don't strip the thread.
Select a reasonable margin of safety on the bolt and you should be okay.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
 
Steel on steel is pretty slippery, as long as there are no mechanical "stops" in the way, like Weld spatter on the floor, Floor plates with mismatched un-beveled edges presented as to prevent motion.

If you can apply a couple of hundred foot pounds to a (edit - grade 5) 3/4" coarse thread bolt I'd expect it exert about 20,000 lbs of force.

When you are done moving the steel base around, what happens next?
Is rotating machinery involved?
 
Tmoose,

Yes, we have gearboxes and motors on these frames once the base is in place. However, these jack screws will no longer be holding these frames in place as they will be mounted to grouted-in floor plates once they are in position. There will be no final forces on the jack screws.

 
desertfox,

How would I choose this bolt? Is it just tensile strength of the bolt?
 
Hi dmbypsi

Check out this link and look at the proof loads for given bolt sizes and grades


So I figure if you have a friction coefficient of .25 then the axial bolt load in pushing the 10 tons would be 2.5 tons which is roughly 25kN so a grade 8.8 M20 bolt would be more than adequate

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
 
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