Changing weight at beam ends for different angles

[jrg_1411]

Hi, first time poster, sorry if this is in the wrong area or has already been asked. I'm looking at a situation like this: except that instead of being lifted via hydraulics, a crane will be attached to one end. The other end will be attached to a concrete slab via a flange, so that it does not move, but eventually allows all the weight to be transferred to the slab and the cylinder will stand vertically

I'm looking for the formula to determine the weights on each crane at different angle. Obviously when the beam is horizontal, the weights will be evenly distributed, and at vertical, all the weight will be on the bottom end/concrete slab flange. But what about in between? At 45 degrees will the bottom take 60% of the load and the crane take 40%?

 

[nvn]

jrg_1411: I assume the crane cable will always be vertical, at the free end of your beam. The force on each end of your beam (V1 and V2) will always be vertical.

Throughout the lift, the value of force V1 and V2 will always remain the same as their value when the beam was horizontal. No change.

The only exception is when the beam has rotated upward beyond approximately 88 deg, past the tipping point. When you pass the tipping point, then the beam will "fall" to the vertical (90 deg) position, and the crane cable tensile force (V2) can be reduced.

 

[jrg_1411]

Thanks nvn. What if the bottom end is atop another pole and transferring the weight onto it?
Is it only past 88 degrees that the weight transfers? Do you have a formula based on angle that I can use?

Cheers.

 

[nvn]

jrg_1411: Yes, even if the beam bottom end is sitting on top of another pole, the answer is still the same as given in post 2.

Compute the force at each end of your beam (V1 and V2) when your beam is horizontal. Afterwards, V1 and V2 remain constant for all angles throughout the lift, from 0 to approximately 88 deg. Constant means V1 and V2 are not a function of beam angle.

 

[sci-phy]

I have attached the solved problem as pdf. Hope it is what u wanted.

We have to take the beam to be 2D. This is true when diameter is not negligible. We have to do this because we are taking the beam to pivot about it's edge.

I assumed the crane rope to be always vertical and in tension (i.e.) it does not slack at any time. So, when the beam is vertical, the entire load will be taken up by the crane rope. When the rope is released, the entire load will be taken up by the ground. I have also given the angle at which the beam automatically falls to stable/vertical position when no tension rope is provided. This comes out to be between 88- 90 degrees for long beams.

cheers!