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Chapter 14

Q. 14.8.2

Fig. 14.8-3(a) shows a propped cantilever supporting a uniformly distributed load w per unit length. If the collapse load factor is to be A, determine the required plastic moment of resistance M_{P}.

 

Step-by-Step

Verified Solution

Let the unknown location of the sagging plastic hinge be defined by x, as in Fig. 14.8-3(b). The work equation is

\qquad \quad \lambda wL\left(\frac{x\phi }{2} \right) =M_{P}\left[\phi +\frac{L\phi }{L-x} \right]

or \qquad \qquad M_{P}=\lambda wL\frac{x(L-x)}{2L-x}

By the unsafe theorem, the value of M_{P} corresponding to an arbitrarily assigned value of x is less than, or at best equal to, the plastic moment required to withstand the load λw. Hence,M_{P} is maximised with respect to x:

\hspace{5em} \frac{dM_P}{dx} =0 \ gives \ x=\underline{0.586L} \\ \hspace{5em} so \ that \qquad M_P=\underline{0.0858 \ wL^2}

COMMENTS
The above solution is based on the unsafe or upper bound theorem. An alternative solution, based on the safe or lower bound theorem is given in Example 14.8-3.

fig14.8-3