huangyhg

partially plastic moment for a circular section
hi all,

could anyone please give me the formula for the partially plastic moment for a circular section explaining the terms. i have one here but think it's wrong:-
mpp = yield stress/6 x (2 x (outside dia.**2- elastic dia.**2)**1.5 - outside dia.**2 )
where ** denotes to the power of
dia. is diameter
x is multiplied by
hope you can understand it
replys much appreciated
is this in torsion? ( i hope so).
conceptually all you have to do is split the circle into an inner, elastic part, for which you already know the solution, the outermost edge of which is just yielding, and a tube around that part, which is all yielding. i imagine the expression for the torque developed by a thin part of that tube would be something like
2*pi*r*dr*yield stress*r
a talented engineer will easily be able to sort that out. it's sunday morning, i haven't had enough coffee, and i ain't doing even easy calculus when the sun is shining.
oh all right, i get 1/12 *pi* s* (d2^3-d1^3) for the yielding tube. the elastic centre i leave to you.
cheers
greg locock
are you looking for the plastic section modulus about a diameter? if so, then:
z = (d^3-d^3)/6 where
d is max diameter, and
d is min diameter
(from cisc steel handbook)

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