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## Q. 14.6.3

(Cambridge University Engineering Tripos: Preliminary Examination for Part I (1975).) Figs. 14.6-7(a) and (b) show respectively the dimensions of a propped uniform cantilever and its cross-section.
(a)  Determine the plastic moment of resistance of the beam section if the yield stress of the material is 250 N/mm2 in tension and in compression.
(b)  If the load W may be applied at any position within the span, determine the minimum value of W that will cause collapse.
(c)  If the load W is removed just after the formation of the collapse plastic hinges, sketch the shapes of the bending moment and shear force diagrams after the removal of the load.

## Verified Solution

(a)The neutral axis divides the cross section into two parts of equal area as in Fig. 14.6-8(a)

$\qquad \begin{matrix}\text{ Therefore }: \ \text{Lever arm } \ & =(10+5)/2=7.5 \ mm \\ \text{Plastic moment } \ M_{p} & =10\times 10\times 250\times 7.5 \\ \quad & = \underline{0.188\ kN\ m } \\ \end{matrix}$

(b) Consider the mechanism in Fig. 14.6-8(b), where the position of the plastic hinge B is defined by the variable x.
The work equation is

$\qquad W(xL\phi )=M_{p} \left( \frac{\phi }{1-x} +\frac{x\phi }{1-x} \right)$

or

$\qquad W = \frac{1+x}{x(1-x)}\frac{M_{p}}{L}$

We do not know the value of x, but we do know that the required value of x is such as to make W a minimum.

$\hspace{5em} \frac{dW}{dx}=0\ gives\ x^{2}+2x-1=0$

Therefore   x = 0.414

and            W=730 N

(c) In the absence of external applied load, the beam is acted on by support reactions only (see Fig. 14.6-8(c)).

Suppose the (unknown) support reactions are M and R. The only possible
shape of the bending moment diagram is a triangle (Fig. 14.6-8(d)); the only possible shape of the shear force diagram is a rectangle (Fig. 14.6-8(e)).