huangyhg

sf & bm in lifted body
dear all,
perhaps someone could confirm my thinking on how to approach this problem. i have two solid bodies (see below) 'a' and 'b' which are joined together at the interface 'x' via a bolted connection.
lift direction
^
^
|
|
slide bar->|____________|______________|
| |
| |
lift point lift point
|-------------------------------------|
| |
|----------x| |
| body a | body b |
| | |
|----------x| |
| |
|-------------------------------------|
(side elevation)
the bodies are lifted upwards at the lift points shown. the lifting will be by ropes in a "vertical" arrangement (the rope slings will be truly vertical and connected to a sliding bar arrangement - shown). i need to know the shear force and bending moment at the interface x. i have analysed this system as a simply supported beam with two udls (see below) to get the reactions, sf and bm out.
as a beam:
---udl1---------------udl2------------
^ ^
| |
| |
the only thing that worries me is that i don't think the interface x will see the high moments predicted by taking the arrangement as udls. my thinking is that some of the moment will also be taken out by the pivot on the slide bar arrangement. any thoughts?
cheers,
-- drej --
body "a" acts as a simple cantilever, this "system" as you have defined it does not appear to me as a continuous beam. i would also add in the inertia forces from the 'lift' to the weight of "a" resulting in tensile/compressice forces for the resisting moment plus the shear of weight "a" including the additional inertia loads.
first off, it's not clear how the whole system is stable. if this is a crane-type lift, your load will rotate about the point of support until the total cg is under the single lift point at top.
assuming that's not the case, and that the system somehow stays level, you'd just sum moments about the interface, taking body a as a free body diagram, to find the forces at that point.
thanks for the replies. i believe the slider bar arrangement allows the whole unit to stay level by adjusting the bar appropriately.
lift direction
^
^
|
|
slide bar->|__________<_|_>___________|
if this is the case, which is what i've assumed, then i assume from your comments that my simply supported beam model of the unit above is applicable, and that the sum of the moments and forces about the interface (to obtain sf/bm here) would be appropriate?
the term "simply supported" implies that the beam in question is supported at both ends and neither end has any moment resistance. that could be assumed to be the case for body "b" between your 'lift points' but that is not the case for body "a"
drej
based on your two sketches i would say that your original analysis (no load [or very little]) at the slide bar is correct.
the lifting is being done by ropes that can only act in tension. to produce a horizontal reaction at the slide bar you would need to develop a couple through the ropes which would require one of the ropes to carry a compressive force. also there would need to be an additional horizontal reaction in the system to balance the reaction at the slide bar.
i believe that your original analysis is correct. the slide bar will be adjusted so that the pick point is liocated directly above the center of gravity of the lifted mass. this being the case, the moment at interface x is simply the weight of the component outboard of x times the distance from x to the center of gravity of that item. assuming that the center of gravity of that item is at the midpoint between x and the end of the item, then using udl1 will give the same results for the moment.
great! thanks for all your replies.
-- drej --

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