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supported beam (again!)
At first I thank and from this position George Mokios and Louie Yaw,
for their help in my previous post.
Trying to figure what is going on, a new query came up.
Suppose we have the rod but now rested on just one table.
[
See here for the figure
https://sites.google.com/site/physicseuler/genike-physike/Doc2.doc?attre...
]
Analysis (if I am right of course!) showed that pure statics can explain the equilibrium of the rod if the supported part of it has length at least 3L/4 where L is the total length.
We know from experience that the rod can be equilibrated even if only half of the rod is rested upon the table (altough unstable).
So let a part of the rod (a little bigger than L/2 in order to avoid the unstable equilibrium) is rested upon the table.
How is it possible to explain its equilibrium?
I consulted my elasticity books (Sadd, Sokolnikof, Barber, Fung)
I took a glance of Hetenei's book on Beams
I search the web,
but I can't find anything relevant.
Thank you, in advance, for your time.
Pure statics
The rigid body (rod) acts on another rigid body (table).
i) The reaction of the table is the force R and moment M applied at any point. There are two unknown values: force R and moment M.
ii) Or the reaction of table is only the force R applied at some unknown point P. In this case there are two unknown values: force R and point P.
Consider ii) way.
It follows from the equilibrium equation of forces that R = Q (Q is the rod weight.)
Using the equilibrium equation of moments about the point P we obtain that Q*c = 0. Thus c = 0.
We note that the reaction R is the resultant of some distributed pressure r(x) which is the true contact pressure. The rod is resting on the table so r(x) is a positive or zero for any x from [0, b]. Thus the resultant R must be applied at point "on the table" (P in [0,b]).
The rod weight Q is applied at the center of rod. So if xQ is negative (i.e. less than half of the rod rests on the table) then c is negative. But this is impossible.
The limit case is xQ = 0, i.e. b = l/2 and xP = 0.
Simpler proof
Two bodies act on the rod. The action of Earth is the weight (force Q). The action of the table is the reaction (force R). if the rod, which is loaded by two forces, is in equilibrium then
1) the magnitude of these forces are equal;
2) the forces act on the same line;
3) the forces are oppositely directed.
We know the magnitude, the direction and the point of application of the rod weight. So we know the reaction of the table.
But the reaction of the table is a contact force. And it can be applied only in the contact zone between the rod and the table.
Thus the equilibrium is possible when more than half or half of the rod rests on the table.