I have a silo 43 feet tall and 13 feet in diameter. It is a cylinder and it weighs 28000 lbs. It will be bolted to a concrete slab. If the concrete is 20 x 20 and weighs 60000 lbs, how do I calculate what wind speed it would take to knock it over?
you will locate the traditional acceleration by skill of utilising very final velocity - preliminary velocity/ (exchange in time). the priority you're handling is complicated, however counting on how the spring is released. If a tension is utilized to a minimum of one end and the spring isn't bounded, that's perplexing to p.c. no remember if element of the ability will bypass to compressing the spring and yet another element to the action of the spring by using area, or if the spring won't compress in any respect and the all the ability will bypass to the action of the spring by using area. Intuitively, however imagining a spring being pushed on a floor might let us know reckoning on the spring tension consistent how lots the spring will compress as a consequence of the tension of friction. when you consider that's already compressed, i might think of which you would be able to cope with it the comparable as an merchandise which comprise a ball. the traditional tension ought to then be calculated by skill of utilising F=ma and utilising regular acceleration as calculated by skill of very final and preliminary velocity.
You have to know the wind pressure, then: F= P*A, F:Wind force, P:Pressure, A:Area facing wind= 43*13 then, you should consider 2 failing scenarios (1 of them should happen 1st, hence should be the determining factor): 1. Failure of concrete (which I believe will happen 1st, because concrete is very weak in opposing tension loads) 2. Failure of bolts (it can be easily solved by cheaply using thicker bolts) Suggested solution for 1: bury the silo's legs inside the concrete (so forces are more of compressive) NB am not very sure, but I guess 20*20 is larger than needed (or would better be distributed into smaller bases).
