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Forces and reactions when tubing is stacked
- Thread starter bsmet95
- Start date
I imagine that summing the force vectors about a point at the base of the sidewall would give you the resultant force in a horizontal direction. It's just a matter of summing all the forces with the angle that the tubes contact each other and then the horizontal and vertical resultant forces. It appears that it would be symmetrical for each bay, except for the end bay. Depending on the structure and configuration of the retaining walls, (if the forces did not transfer from bay to bay), then the end condition seems to be the only one that would need looking at.
Haven't done that since college, (a very long time ago), but it seems to make sense.
Just a thought. Good luck.
Haven't done that since college, (a very long time ago), but it seems to make sense.
Just a thought. Good luck.
racookpe1978
Nuclear
Wild guess here: But isn't it only the end panel that gets the critical stresses for analysis?
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.
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.)
--
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.
The problem is probably indeterminate, but a good conservative estimate would be to use the fluid analogy.
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.
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.
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