Will a 4x4 Steel Tube Fail Under 35-40,000 lbs?

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The discussion centers on the safety and structural integrity of a 4x4 square steel tube under a load of 35-40,000 lbs. Key points include the need for a detailed diagram to assess load distribution and the importance of understanding the tube's dimensions and weld details. Participants emphasize that without professional engineering advice, proceeding with the lift poses significant risks. The conversation highlights the complexity of load calculations, including factors like buckling and load distribution, particularly for different types of loads. Ultimately, the consensus is to avoid the lift without proper engineering validation due to liability concerns.
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Have a lifting yoke with 7/8 inch lifting eyes welded onto the ends of a 3/16 or 1/4 inch wall 4x4 a500 or a513 square steel tube.

Expected UDL is as much as 35-40,000 lbs.

Is this tube going to fail?
 
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Welcome to PF.
You need a diagram.
Look at the triangle of forces.

How long is the tube ?
How long are the risers from the tube ends to the common lifting point ?
 
A500 has a lower UDL before deformation? I want to know the best and worst case scenarios, best steel and thickest wall and worst steel and thinnest wall or please all 4 combos.

I am a stupid labourer being asked to do a sketchy lift and want to know some real material engineering numbers. I could easily refuse already but I want to know some science to either let me go ahead like I want to or refuse .
 
Baluncore said:
Welcome to PF.
You need a diagram.
Look at the triangle of forces.

How long is the tube ?
How long are the risers from the tube ends to the common lifting point ?
I will send a napkin CAD within the hour.
 
Freelander said:
I am a stupid labourer being asked to do a sketchy lift and want to know some real material engineering numbers. I could easily refuse already but I want to know some science to either let me go ahead like I want to or refuse .
PF will not take the risk or liability so you must refuse.
Without a diagram for your sketchy lift it cannot be solved.
Without a diagram we cannot even advise you where to go to solve it.
 
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Sorry is that even remotely a useful diagram?

Also: it is well understood that I am not getting professional advice. I simply lack the google fu or knowledge to even ballpark the safe UDL of such stuff. I do not understand nor can calculate modulus of elasticity stuff or whatever.
 
Two 1", 8.5 ton shackles are insufficient to carry over 40,000 lb load.
You need to consult a qualified rigger.

You do not show the end lug weld detail. Is that a but joint under tension ?
The end lugs must be welded to the ends of the tube, in shear, closing the tube section.

Avoid the liabiliy. Say NO. It is a job for an engineer with insurance.
 
Okay. The weld is the whole way around. It is a butt weld. There's at least 5 passes. It's built up a 1/4 inch. Hard corner on the 4 inch tube was chamfered.

Test lift was already done.

Load safety device on the crane read 35,000. So the actual load on the beam is less: 2,000 lb block, weight of wire rope (3 parts line) and rigging.

I am quickly understanding a few things. Your command of the math is excellent and I need to provide lots of exacting detail and also you are reticent to answer any questions because of the litigious nature of north American society.

Never mind any hypothetical considerations. You may practically apply physics knowledge on a regular basis or not. I am familiar with labour law and construction regulations and will make my own determination.

The question is what is a safe UDL for an A513 3/16" wall 4"×4" square tube 70" in length.
 
  • #10
No, safe is not stupid
'Safe' is stupid. Minimum breaking strength is what I want to know.
 
  • #11
Then I can calculate a design/safety factor. Which I expect to be below 1.
 
  • #12
If such a lift were to happen it might already have been refused because of clearance and wind factors possibly today.
 
  • #13
You show the load is being distributed over the upper surface of the square tube. That upper surface is under compression, with the lower surface under tension.
I expect the failure mode will be by buckling of the upper surface of the tube. The constraint to that buckling is the tube side walls.

So how do you distribute the load on the upper surface?
The loaded beam model employed will depend on the load being lifted.

If the material being supported was a loose bundle of material it might be evenly distributed, but if it was a solid block then the load could be carried at the mid point, or at the two outside edge points, depending on if the load or the square tube deflects more.
 
  • #14
How can I explain. It might be a length of 56 inch wide conveyor belt hanging vertically. A flat one inch wide thing would still be heavier in the middle, hypothetically would it not? In any case a centre point load is the least desirable of the possible situations. Then the question would be what it the minimum breaking strength of that square tube with a centre load. Is that what I am understanding?
 
  • #15
That's not hypothetical. The load would be centred. I mean an evenly thick load is heaviest in the middle no?
 
  • #16
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Like this, no?
 
  • #17
It must be frustrating for you to field questions from stupid people like me. My apologies.
 
  • #18
It is frustrating for us because we want to help. BUT, we cannot because of liability. I am familiar with this subject because I wrote the lifting beam design procedure for a former employer, and it ran about 15 pages. There is much more than just the yield strength of the beam involved in a lifting calculation.

This thread is closed.

And no, you are not stupid. You just need about three years of engineering courses.
 
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