Question 18.6: The propped cantilever AB shown in Fig. 18.14(a) carries a u...

Peak values of bending moment occur at \mathrm{A} and at some point between \mathrm{A} and \mathrm{B} so that plastic hinges will form at \mathrm{A} and at a point \mathrm{C} a distance x, say, from \mathrm{A}; the collapse mechanism is then as shown in Fig. 18.14(b) where the rotations of AC and CB are \theta and \phi respectively. Then, the vertical deflection of \mathrm{C} is given by

\delta=\theta x=\phi(L-x)       (i)

so that

\phi=\theta \frac{x}{L-x}     (ii)

The total load on \mathrm{AC} is w x and its centroid (at x / 2 from \mathrm{A} ) will be displaced a vertical distance \delta / 2. The total load on CB is w(L-x) and its centroid will suffer the same vertical displacement \delta / 2. Then, from the principle of virtual work

w x \frac{\delta}{2}+w(L-x) \frac{\delta}{2}=M_{\mathrm{P}} \theta+M_{\mathrm{P}}(\theta+\phi)

Note that the beam at B is free to rotate so that there is no plastic hinge at B. Substituting for \delta from Eq. (i) and \phi from Eq. (ii) we obtain

w L \frac{\theta x}{2}=M_{\mathrm{P}} \theta+M_{\mathrm{P}}\left(\theta+\theta \frac{x}{L-x}\right)

or

w L \frac{\theta x}{2}=M_{\mathrm{P}} \theta\left(2+\frac{x}{L-x}\right)

Rearranging

w=\frac{2 M_{\mathrm{P}}}{L x}\left(\frac{2 L-x}{L-x}\right)       (iii)

For a minimum value of w,(\mathrm{~d} w / \mathrm{d} x)=0. Then

\frac{\mathrm{d} w}{\mathrm{~d} x}=\frac{2 M_{\mathrm{P}}}{L}\left[\frac{-x(L-x)-(2 L-x)(L-2 x)}{x^{2}(L-x)^{2}}\right]=0

which reduces to

x^{2}-4 L x+2 L^{2}=0

Solving gives

x=0.586 L \quad \text { (the positive root is ignored) }

Then substituting for x in Eq. (iii)

w(\text { at collapse })=\frac{11.66 M_{\mathrm{P}}}{L^{2}}

We can now use the lower bound theorem to check that we have obtained the critical mechanism and thereby the critical load. The internal moment at A at collapse is hogging and equal to M_{\mathrm{P}}. Then, taking moments about \mathrm{A}

R_{\mathrm{B}} L-w \frac{L^{2}}{2}=-M_{\mathrm{P}}

which gives

R_{\mathrm{B}}=\frac{4.83 M_{\mathrm{P}}}{L}

Similarly, taking moments about B gives

R_{\mathrm{A}}=\frac{6.83 M_{\mathrm{P}}}{L}

Summation of R_{\mathrm{A}} and R_{\mathrm{B}} gives 11.66 M_{\mathrm{P}} / L=w L so that vertical equilibrium is satisfied. Further, considering moments of forces to the right of \mathrm{C} about \mathrm{C} we have

M_{\mathrm{C}}=R_{\mathrm{B}}(0.414 L)-w \frac{0.414 L^{2}}{2}

Substituting for R_{\mathrm{B}} and w from the above gives M_{\mathrm{C}}=M_{\mathrm{P}}. The same result is obtained by considering moments about \mathrm{C} of forces to the left of \mathrm{C}. The load therefore satisfies both vertical and moment equilibrium.

The bending moment at any distance x_{1}, say, from B is given by

M=R_{\mathrm{B}} x_{1}-w \frac{x_{1}^{2}}{2}

Then

\frac{\mathrm{d} M}{\mathrm{~d} x_{1}}=R_{\mathrm{B}}-w x_{1}=0

so that a maximum occurs when x_{1}=R_{\mathrm{B}} / w. Substituting for R_{\mathrm{B}}, x_{1} and w in the expression for M gives M=M_{\mathrm{P}} so that the yield criterion is satisfied. We conclude, therefore, that the mechanism of Fig. 18.14(b) is the critical mechanism.