I was tasked with solving this problem at work as we are out of coil holders and we bought 140 coils and need to stack them asap. Companies that make the storage holders are on back order for several weeks and no one is willing to share responsibility for this task. Safety first!

Given: 15 steel coils stacked 3 rows high (6 bottom, 5 middle, 4 top). Each weigh 10,000 lbs (4.5 metric tons) max (8500# for most), have an ID of 20" and a length of 60". Some quick calculations will show that the OD will be 32-33in.

Find: All reaction forces at the bottom for the following 2 scenarios: 1) There is only 2 lateral supports at the base on either side of the entire stack. 2) Each coil has its own support laterally.

This question becomes quite complex the more it is looked into. I started with a single top coil and the geometry would make the Normal force act 30 degrees from vertical, and 60 degrees from horizontal so $2 N sin(60)=10000$. So then N=5773.5 lbf going into the second row. This is when it gets confusing and I would like input on the best way to go about it. I was trying to imagine the possibility of a method of joints type of approach. I was going to move on and just look at a free body diagram of each coil independently.

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    $\begingroup$ Can I suggest that you include a diagram, with the relevant forces etc clearly marked. You know, just like the ones you drew at school, before you got the job. :) $\endgroup$ – user108787 Nov 3 '16 at 1:23
  • $\begingroup$ I honestly don't know how to attach images to these threads at the moment. $\endgroup$ – Robert Nov 3 '16 at 2:42
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    $\begingroup$ 2 things .....1. .you could put a link into your post and I will do the rest and 2. Write as much as you think you know about how to solve it and what you have tried yourself. Only do the picture if you think its worth doing. Best of luck with it anyway. $\endgroup$ – user108787 Nov 3 '16 at 3:40
  • $\begingroup$ Is there something holding the bottom row together, because the way you describe it I think it will spread out and the pyramid will collapse. $\endgroup$ – ja72 Nov 3 '16 at 18:55
  • $\begingroup$ You can upload an image by going into 'edit', placing the cursor where you want to insert it, then clicking the 'picture' icon at the top of the edit box. ... What do you mean by "this is when it gets confusing"? Can you be more specific? ... Yes, modelling the stack as a pin-jointed structure, with weights at the joints, should work. Normal reactions are then replaced by compressions in the trusses. Alternatively, drawing a FBD for each coil should also work. $\endgroup$ – sammy gerbil Nov 4 '16 at 1:27

I think that this is a statically indeterminate system, similar to Estimate the reaction force on each leg of a 4-legged table.

Looking at the pyramid as a whole, there are 6 points of contact with the ground and 2 with the side supports. So assuming these reactions are normal only, there are r=8 external reactions. Considering the structure as a truss, there are j=15 joints and m=30 members. Thus r+m=38 which exceeds 2j=30. So the truss is statically indeterminate. (see http://www.ae.msstate.edu/vlsm/truss/.)

The indeterminacy arises because of the ideal nature of the truss - eg members which are perfectly rigid, inextensible. One way to break the indeterminacy is to take account of internal forces and deformations, as with the 4-legged table. In a real truss or pyramid of coils, there will be some deformation in shape, due to the forces between coils, which alters the forces and angles in the structure, and leads to definite values for the external reactions. But even if we know the internal force laws, it becomes a very difficult problem, requiring a computer or spreadsheet.

If you stack the coils as you suggest, perhaps the best you can do is to assume that the total weight is distributed approximately evenly between the 6 coils on the lowest layer. This estimate is not likely to be far from the actual values.


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