ASME BTH-1 Major Axis Bending 1

[KB4444]

I am designing a 12.25ft lifting beam that is picked up by lifting lugs on the top flange of the beam. The lifting lugs are welded at the ends, being 12ft (144") apart, which I believe means the unbraced length would be 144"? The load is picked up by lifting lugs on the bottom of the beam spaced uniformly apart.

I am trying to size the beam in bending. Using ASME BTH-1: 3-2.3.1 Major Axis Bending of Compact Sections, I am running into a problem with the unbraced length.

I am finding that I do not have a reasonable beam size that is less then Lp (3-7) or Lr (3-10).

Using a W10x33:

Lp = 1.76 x 1.937" x sqrt(29000ksi / 50 ksi)
= 82"
Lr = sqrt(3.19 x 2.136^2 in^2 x 29000ksi x 1 / 50ksi)
[When the bending moment at any point within an unbraced length is larger than that at both end of this length, Cb shall be taken as unity. Therefore, Cb = 1.0]
= 92"

Lu = 144"?

If I am correct with Lu being the space between supports what is my next step? I wouldn't have Lu < Lr, so I believe I am stuck going to an even larger beam size? The bending stress in this situation wouldn't govern.

 

[KB4444]

That is correct, but I was using ASME BTH-1 for this lifting beam design rather then the CISC handbook. The ASME code is where the unbraced length calculation comes in with bending (i.e. Lr) [3-2.3 Flexural Members].

Using the Beam Selection tables in the handbook I can see how overdesigned the beam size was getting.

 

[BAretired]

Double monorails 12' apart do not work. There can't be a vertical pickup unless the load is correctly placed under the two monorails before the lift.

Also, the ASCE method of analyzing the lift beam does not conform to normal methods of considering unbraced beams.

 

[dik]

top and bottom flanges free to rotate on plan.

With the load below the shear centre, is this possible?

 

[HTURKAK]

With the load below the shear centre, is this possible?

Yes. It is possible but less stringent if compared with loading from top flange . If the load applied below the shear center , that will have stabilizing effect but not restraining.
At the both supports the following conditions exist:
-Beam torsionally not restrained.
-Compression flange laterally not restrained
- Both flanges free to rotation.
I have suggested to use LE = 1.2LLT + 2D assuming the loading from bottom flange . If loaded from top , LE = 1.4LLT + 2D should be used . ( LE would be 1.4*144+2*10= 222 in. Relevant table of BS 5950 :