Load distribution on multi lifting points 4

[Truc Do]

Dear all,

Please help me how to calculate load distribution on multi-point truss as illustrated in this photo. How to come up such numbers? Thank you in advance.
Best regards,

Truc Do

https://www.google.com/imgres?imgurl=https://

http://www.hirehop.co.uk/wp-content...hUKEwjFw9z-n_nrAhVry4sBHYLHAPQQMygAegUIARCiAQ

 

[Ron247]

You need to be more specific about what you need. I see the picture, but fail to understand what help you need. Also, your picture is confusing to me. Is the first one 50% of the load at each end?

 

[Truc Do]

Hi Ron247,

Sorry for making you confused. Let me explain for your understanding.
A load of 100%, if there are 02 lifting points at 02 ends, each end carries 50% of the load (it is easy to understand). But when there are 03 lifting points or more, the load distribution are much different (see the photo). I do not understand how to calculate to come up such fihures.
Regards,

Truc Do

 

[Lomarandil]

These are simply the basic solutions for support reactions of any continuous beam supported at regular intervals. You might also find these in AISC Table 3-22, among others.

They are a simplification of a truss lifting problem, which may or may not be appropriate. The simplification relies on the following assumptions:
[ul]
[li]The lifted load has (approximately) uniform weight and stiffness along its length.[/li]
[li]Vertical (or near-vertical) rigging (e.g. multi-crane lifts, or spreader bars used, etc).[/li]
[li]Simple fixed-length rigging without the use of "snatch blocks" or other mechanisms to equalize loads or adjust sling lengths.[/li]
[li]The rigging is assumed to be rigid compared to the bending stiffness of the truss. Depending on sling length and material, this may or may not be true.[/li]
[/ul]

When solving lifting problems, we always allow for some variation from the theoretical solution (e.g. this table). That variation changes with the scale of the load, hazard in case of failure, and how closely the assumptions are met.

Of course, you would still need to evaluate the truss rigging locations chosen for local load effects, size padeyes and rigging accordingly, etc. Depending on the truss, the truss itself may also need global stability analysis to be sure it can be lifted from these support points (most are only designed for their final support conditions, including final bracing).


----
just call me Lo.

 

[r13]

For the 3 load points case:

The deflection at the midspan with two end lift points only: Δ = 5wL⁴/384EI
Deflection due to a concentrated load at the midspan: Δ' = P'L³/48EI
To close the gap, Δ = Δ'
Δ = 5wL⁴/384EI = 5PL³/384EI = P'L³/48EI ---> P' =0.625P (62.5% of P)
Each force at the end lift point, therefore is (100-62.5)*100/2 = 18.75%

The other cases can be calculated using the same approach.

 

[Truc Do]

Thank you retired13.
Regards,

Truc Do

 

[robyengIT]

following the suggestions of retired13 I prepared the attached spreadsheet from where You can see that the procedure is correct only for 4 scheme but slightly different for the last 3 schemes (due to splitting the forces). If anyone can help finding my mistakes. Thanks

 

[Truc Do]

Thank you RobytengIT.
Best regards,

Truc Do

 

[drawoh]

Truc Do,

Do you need to hit those exact percentages, or are you just ensuring the load is distributed fairly evenly?

Read up on Whippletrees.

--
JHG

 

[ldeem]

A lot dependents on the size of the lift. A relatively small and "over" rigged truss is much more forgiving than a large multi-crane lift. As Lomarandil pointed out the configuration of the rigging is critical. You can not just use support reactions from continuous beam tables. Slings are never all the same length so in the case of more than two slings from a single point it is likely the load will not be shared as predicted by statics. For example in a 3 sling lift you would size as if only two slings carried the load.

If you add pad eyes they will likely induce bending moment in the truss chord for the sling horizonal reaction. This can be significant depending on the chord.

 

[robyengIT]

here the correct solution/formulas

 

[BAretired]

Pay attention to Ideem. These tables are not appropriate for lifting beams. To use them in practice would be dangerous because lifted loads are rarely uniform and slings are never uniformly tightened.

BA

 

[robyengIT]

in correlation with my last post