(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)).

COMMENTS
(1) With reference to the mechanism in Fig. 14.6-8(b) it is interesting to note that (and the reader should verify this by elastic analysis) the maximum elastic moment at B is 0.174 WL and occurs when x = 0.366 L, as shown in Fig. 14.6-9(a).

However, the reader must not be tempted into thinking that the correct mechanism is that in Fig. 14.6-9(b). As we saw in part (b) of the solution above, the hinge B in the correct mechanism is at 0.414 L from A.

(2) With reference to Part (c) of the question, a common error is to think that
the residual moment diagram has a ‘kink’, at the plastic hinge position.

(2) With reference to Part (c) of the question, a common error is to think that
the residual moment diagram has a ‘kink’, at the plastic hinge position.