bending stress upon lifting

[leecoop]

I have a structural member x" long of varying cross-sections. The total weight is 1970 lbf. In the middle of the member, there are 4 SAE8 grade bolts. I will be lifting one end up using a crane. I am concerned that the bolts will not be able to support the bending stresses. Any suggestions?

Thanks

 

[JAE]

What are the bolts connected to? Are they embedded anchor bolts in concrete? You say you are lifting one end of the member - I assume then that the crane would be prying up the member from one end, with the bolts in the middle?

 

[leecoop]

The member is an assembly laying on a shop floor. The bolts are parallel to the floor holding together two sections of the assembly. The crane will lift the member up by one end until it is upright. Imagine picking up a pen on a desk by one end.

 

[rb1957]

so these bolts are carrying the bending load (due to the weight of the beast) in tension and compression (well really the compression is carried by the flange around the bolt, but ...) so the maximum moment is the hoist load at the start of the lift (2*1970*x/2 = 1970x) ... pedantically you could say the further 1/2 (closer to the lifting point) is relieving this (2*1970*x/2-(1970/2)*(x/4) = 7/8*(1970x) ... but i'd keep this in my back pocket).

you know how far apart your bolts are y", so the bolt load is 1970x/y.

 

[leecoop]

There are 4 of them in a 6 1/2" bolt circle. In finding my bending stress, I am not sure what to use for the moment of inertia since there are various cross sections (hollow square beam and then hollow cylinder).
Thanks

 

[rb1957]

let me 1st correct my post, the lifting tension is about 1/2 the weight (not double, sigh).

i don't think you need MoI as the moment is being reacted by a couple, rather than a stress distribution.

the moment at the joint is about 1970/2*x/2, and the couple is BM/(6.5/sqrt(2)) = (1970x/4)/4.5; but this is reacted by two bolts ...
bolt load is 1970x/4/9 = 55x

but i don't know what your bolt is (nor waht a grade 8 bolt means) ...

 

[leecoop]

rb1957,

Thank you for being so helpful. The bolts are 3/4" diameter with ultimate tensile strength of 150 ksi, yield of 130 ksi.

What is "BM"?

Thanks

 

[rb1957]

Bending Moment

your bolts should be good for 44,000 lbs,

which means x < 800" should be ok,
but then the structure surrounding the bolts needs to be able to take this load too.

btw, how well connected are the various sections, i'm guessing they're welded. also, i'm fixating abit on the bolts, what does the section near the bolts look like ? i suspect that your sections are ok, and that you're joints are the biggest problem. in either case, you'll have a problem if the effective material (beit welds or steel) is lumped about the neutral axis (so that it can't react bending moments very well.

 

[DTGT2002]

You may be able to help yourself out with your choice of rigging. If you can have a pick point at or near the midpoint or critical point as well as one at the end, you can use some rolling rigging - think sheave - to allow the load to be shared between points as the assembly is raised. This would be easy with unlimited vertical access, difficult with a ceiling height close to the assembly height.

Will the crane and rigging be performed in house? If you'll be using a local crane service, talk to them for rigging ideas.

Alternately, if you have material available, could you rig a temporary section to the assembly for erection purposes?

Good luck,

Daniel Toon

 

[Ron]

The bolts are more likely in direct tension rather than bending. The force couple through the bolts needs to be computed, with two bolts in tension, two in compression.

I would expect that with 3/4" dia. grade 8 bolts, you don't have a problem.

 

[JAE]

...or worse, I guess you could have the bolts in a diamond type arrangement where one bolt is on top, one on the bottom, and two are opposite horizontally in the middle...then one bolt in tension and one in compression.

Wouldn't that be a worse case? Larger depth but only one bolt?

 

[rb1957]

the diamond arrangement gives 40% higher loads, sqrt(2)- 1, but then we're just guessing, and the OP knows.

 

[leecoop]

rb1957,

Can you explain the formula below?

BM/(6.5/sqrt(2)) = (1970x/4)/4.5

I agree that the diamond shape is less preferable.

Thanks

 

[rb1957]

the bending moment at the joint due to lifting is (1970/2)*(x/2) = 1970x/4.

This moment is reacted by a couple applied to the bolts, on a PCD of 6.5". I've assumed that the bolts are pitched on lines at 45deg to the loading line (say vertical) so that the bolt group is acting as two pairs of bolts (instead of as a diamond pattern). The distance between these two bolt pairs is D/sqrt(2) = 6.5"/1.414 = 4.5" (near enough).

So ... 1970x/4 = (2P)*4.5, P being the load in one bolt.

So the load in the bolt is (1970x/4)/9 = 55x

 

[leecoop]

rb1957,

Thanks.

 

[Kwan]

i picture a beam laying on the floor that has a splice in the middle. The splice is perhaps a lap splice with 4 bolts loaded in shear (arranged in a square pattern centered on the neutral axis of the beam)

So just as you lift, consider it a pin/pin beam with the 1970 acting in the middle. Hence, BM is 1970x/2 in the middle. Now the bolt pattern must carry this moment and the shear (1970/2=985). Assume the bolts are centered on the neutral axis and are all the same size, then each bolt gets an even shear load (985/2) acting up and the moment puts a load acting perpendicular to a line running from center of bolt pattern to bolt. This load due to moment is BM/[(radius of bolt pattern)4]. Now vector sum the worst case location and this is your max bolt load.

Be careful with your failure mode as RB said. The bolt may be plenty strong and the beam material might fail. Check the bearing stress due to the bolt load. Also, worry about the beam if your cross section changes at the joint (maybe the flanges run out and thus decrease MOI).

Please forgive me if I'm not seeing the whole picture.