Spreader Bar Design

[Leinster]

I am trying to design a spreader beam in which will have 2 possible lifting points at the bottom of the beam (4 nos. lifting lugs). The outside lifting points will be spaced at 4,410mm c/c and the inside ones will be spaced at 2,350mm.

The lifting lugs on the top side of the beam (where the crane hooks will attach) will be spaced at 3,500mm apart from one another. The lifting chains are 6,000mm each and the vertical force is 75 kN on each lug giving a horizontal force of 23 kN.

I need to calculate the maximum bending moment with the force in both positions (not at the same time but 2 lugs loaded and then the other 2 loaded). When the inner lugs (2,350mm c/c) are loaded the moment will obviously be increased due to the axial force (23 kN) caused by the lifting slings on the top side of the beam. How do I factor this into my bending moment calculation?

I hope the above is clear and look forward to your responses.

 

[ishvaaag]

IN each case you need to count point of application of forces and reactions; for the inferior points points at the inferior face of the spreader. So you will have vertical loads, compression loads and moment loads.

So you either introduce a moment or displace the forces to the axis. I post one of the cases, the other would have contrary moments, or displacements.

 

[desertfox]

Hi Leinster

Like the previous poster I have drawn a freebody diagram.
If you can provide a sketch of your beam and loadings we might help further.

desertfox

 

[Leinster]

Many thanks for your replies.

My sketch is exactly as you have done.

So if my x is 200mm my and the 2 closer lugs are loaded my BM calculation would be as follows-Fvert * a) + (-Faxial * H).

Where;
a = Distance between loads and beam ends ((3,500mm-2,350)/2)
Fvert = Vertical Force
Faxial = Axial Force
H = Height of lug

This would give me (-75*0.575) + (-23*.2)= 47.735 kNm

Please advise if this is correct.

 

[desertfox]

Hi Leinster

Please provide a sketch with loads its much easier for us to see your problem.
Depending on where you take moments and whether clockwise moments are negative or positive will determine your moments I can't really be sure your right without seeing the whole of your problem.
I would have thought that the horizontal loads created by the angle of the lifting hook chains would cancel out.

desertfox

 

[BAretired]

Next question: What is the unbraced length of the beam when the outer lugs are used?

BA

 

[rb1957]

Leinster,
your moment calc is not right (IMHO) ...

you've got the axial component Fx*x ok, but not the load, presumably 75/2*2.21/2. more importantly, the axial moment is opposing the load moment ... draw a FDM. This means you should consider the section outbd of the hoist pick-ups. you should also consider the inner pick-ups. i'd draw a moment diagram tip-to-tip for both your load conditions to make sure i had the most critical loads.

also, consider beam column effects ?

 

[BAretired]

rb1957,

Now you're getting us all confused. I believe Leinster has it right for the inside lugs loaded. His sign convention could be argued, but the answer is numerically correct. I would call it positive bending as the tension is on the bottom flange.

For the outside lugs loaded, the moment goes from 0 at the lug to a maximum of 75*0.455 = 34.1 at the support (negative moment). Between supports, the negative moment is reduced by 23*0.2 = 4.6 for a net negative moment of 29.5 kN-m.

I repeat, what is the unbraced length for the latter case? Anyone? I'm sure some will get it wrong.

BA

 

[rb1957]

true enough ... the problem with trying to type and think at the same time (without drawing a FBD). the critical moment is as he posted.

still, what about beam column ? the effect may be small, but it'll lower the allowable.

 

[CTW]

BAretired-

The answer to your question about unbraced length of beam for the outer lugs is 4,410 mm.

 

[BAretired]

Right on, CTW.

BA