Sunday 31 December 2017

homework and exercises - Reaction forces in pyramid stacking of steel coils



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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