Forces and reactions when tubing is stacked
Forces and reactions when tubing is stacked
(OP)
I have an application in which tubing will be stacked in a rack with sides, (9) pieces on the bottom, (8) pieces above that, then (9), (8), etc. See the attached.
How can I calculate forces on the sides of the rack?
How can I calculate forces on the sides of the rack?





RE: Forces and reactions when tubing is stacked
-handleman, CSWP (The new, easy test)
RE: Forces and reactions when tubing is stacked
Haven't done that since college, (a very long time ago), but it seems to make sense.
Just a thought. Good luck.
RE: Forces and reactions when tubing is stacked
The interior panel loads are balanced by the opposing forces from the other side - and if one panel area is empty and its neighbor is filled, then the empty panel must act like an "end panel." Therefore, every intermediate panel division must be fully capable of holding an end=panel's load.
The load against the end panel is going to be transmitted at a 30 degree from the horizontal because that "angled row" of pipes goes up at 30 degrees from the bottom. Assume every pipe is going to roll on the one below it as if it had no friction. Then, at 30 degrees from vertical, the maximum side force is going to be developed. Add up all the side forces at each 30 degree "slant"
Rows 1 through row 6 have 7 pipes in the "slant row"
Row 7 has 5 pipes x wt/pipe x cos 30.
Row 8 has 3 pipes x wt/pipe x cos 30
Row 9 has 1 pipe x wt/pipe x cos 30.
(Cos 30 x wt/pipe because some of the weight of each pipe is carried down and some sideways at the 30 degree point where it "touches" the support pipe below.)
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I strongly suspect, but really don't know, that a "real" quarry or "chute and solids" engineer who works with gravel and stone transportation in vertical conical-bottom tanks would know how to approximate this.
RE: Forces and reactions when tubing is stacked
Get the effective density of the tubes and simply multiply the stack height by that density to get the unit force at the bottom. This will yield a conservatively high value since the cylinders will not transmit the forces ideally.