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Forces from elevator to tower structure 1

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mekafime

Mechanical
Aug 14, 2015
88
Hi engineers,
I am analyzing the forces of an elevator in the tower that supports it, could you please help me to see if any forces are missing?

elevador_1_rieysv.png


The tower has a rack in the center of the horizontal profiles and the elevator has the gear, in addition to wheels in the corners, due to the eccentricity of the elevator load I obtain a couple of forces, which I divide between the pulleys of the upper part 500 kg traction and in the lower part a load of 500 kg on each compression pulley.

elevador_2_iffbq5.png


However, I have the doubt of how I divide the vertical load due to the weight between the 4 pulleys and the rack (which loads the horizontal profiles)
Very attentive to your comments, thank you.
 
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Hi mekahfime

Well you have accounted for the offset correctly and I would divide the vertical force equally between the top and bottom pulleys that’s assuming I am correctly interpreting your post, the way I see it, you are transferring a vertical offset load to another reference point i.e. the pulleys and therefore it needs a direct vertical force and a moment force at the pulleys.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
 
Hi desertfox,

I think the system is in balance?

elevador_3_tgdyxr.png
 
Hi again

Yes it’s in balance however we are making the assumption that the load is equally shared based on everything being symmetrical, in truth the system is statically indeterminate and you need other techniques to solve the actual loads on the parts but as a starter for ten it’s not bad.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
 
Thanks desertfox,

Now the issue is, I need to introduce those loads in sap2000 without drawing the elevator, that is, only the tower that supports it and the loads. Do you have an idea of ​​what charges to enter?
I was thinking of entering the torques but as I represent the reaction forces of the structure, or perhaps I should draw the ascessor to obtain those reactions.
elevador_4_jnpopd.png
 
Hi mekafime

I am not familiar with SAP2000, however I can see you
havke four corner vertical reactions equal just less than half the vertical reaction and the remainder being on the central gear but I can’t see how you have come tO that conclusion, to me if you are assuming symmetry then the vertical load would be shared equally between the four corners and the central portion but I don’t know what information you have to come to that conclusion. Further you are assuming that the four corners take all the 2000kg due to them offset but doesn’t the central gear also take some of the offset load?

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
 
Hello,
it is for the tributary area the pinion covers more reaction than the corners
 
Hi mekafime

Okay but how have you got to that conclusion?

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
 
with a free-body diagram, my concern is now to calculate the braking force of the elevator, will you have knowledge of it?
 
Hi mekafime

I can’t see how you got these forces because the free body diagrams above are statically indeterminate as mentioned previously. To solve and get the correct forces you need other techniques like compatibility,superposition etc. The braking force would be determined from the elevator mass plus the mass of people in it and the velocity of the lift and how quickly you wish to stop it, designing a lift is not something we can do on an engineering forum it’s a very involved and complex problem, have you any standards that you can refer to?

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
 
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