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## Q. 10.23

Find $\delta_C, \delta_D, \delta_C$ of the cantilever beam shown in Figure 10.43.

## Verified Solution

Clearly, bending moment at any section at a distance x from end B is given by

$M_x=\left\{\begin{array}{ll} 0 ; & 0 \leq x \leq L / 3 \\ -P_1\left\lgroup x-\frac{L}{3} \right\rgroup ; & L / 3 \leq x \leq 2 L / 3 \\ -P_2\left\lgroup x-\frac{2 L}{3} \right\rgroup -P_1\left\lgroup x-\frac{L}{3} \right\rgroup ; & 2 L / 3 \leq x \leq L \end{array}\right\}$

$\text { where } P_1=P_2=P_1 \text { (let) and } P_{ D }=P_1, P_{ C }=P_2$. Therefore,

$U=\left\lgroup \frac{1}{2 E I} \right\rgroup \int_0^L M_x^2 d x$

or            $U=\left\lgroup \frac{1}{2 E I} \right\rgroup \left[\int_{L / 3}^{2 L / 3} M_x^2 d x+\int_{2 L / 3}^L M_x^2 d x\right]$             (1)

Differentiating under integral sign using Leibneitz’s rule, we get

$\frac{\partial U}{\partial P_1}=\left\lgroup \frac{1}{E I} \right\rgroup \left[\int_{L / 3}^{2 L / 3} M_x \frac{\partial M_x}{\partial P_1} d x+\int_{2 L / 3}^L M_x \frac{\partial M_x}{\partial P_1} d x\right]$

$=\left\lgroup \frac{1}{E I} \right\rgroup\left\{\int_{L / 3}^{2 L / 3} P_1\left\lgroup x-\frac{L}{3} \right\rgroup^2 d x+\int_{2 L / 3}^L\left\lgroup x-\frac{L}{3} \right\rgroup\left[P_1\left\lgroup x-\frac{L}{3} \right\rgroup +P_2\left\lgroup x-\frac{2 L}{3} \right\rgroup\right] d x\right\}$

Therefore,

$\frac{\partial U}{\partial P_1}=\left\lgroup \frac{1}{E I} \right\rgroup\left\{\left[\frac{P_1}{3}\left\lgroup x-\frac{L}{3} \right\rgroup^3\right]_{L / 3}^{2 L / 3}+\left[\frac{P_1}{3}\left\lgroup x-\frac{L}{3} \right\rgroup^3+P_2\left\lgroup \frac{x^3}{3}-\frac{L x^2}{2}+\frac{2 L^2}{9} x \right\rgroup\right]_{2 L / 3}^L\right\}$

$=\frac{1}{E I}\left\lgroup \frac{P_1 L^3}{81}+\frac{8 P_1 L^3}{81}+\frac{P_2 L^3}{18}-\frac{P_1 L^3}{81}-\frac{2 P_2 L^3}{81} \right\rgroup$

$=\frac{1}{E I}\left[\frac{8 P_1 L^3}{81}+\frac{5 P_2 L^3}{(81)(2)}\right]$

By Castigliano’s second theorem

$\delta_{ D }=\underset{\substack{P_1 \rightarrow P \\ P_2 \rightarrow P}}{Lt } \frac{\partial U}{\partial P_1}=\underset{\substack{P_1 \rightarrow P \\ P_2 \rightarrow P}}{Lt } \frac{1}{E I}\left[\frac{16 P_1 L^3}{162}+\frac{5 P_2 L^3}{162}\right]$

or            $\delta_{ D }=\frac{21 P L^3}{162 E I}=\frac{7 P L^3}{54 E I}$

Therefore,

$\delta_{ D }=\frac{7 P L^3}{54 E I}(\downarrow)$

Again from Eq. (1) above

$\frac{\partial U}{\partial P_2}=\left\lgroup \frac{1}{E I} \right\rgroup\left[\int_{L / 3}^{2 L / 3} M_x \frac{\partial M_x}{\partial P_2} d x+\int_{2 L / 3}^L M_x \frac{\partial M_x}{\partial P_2} d x\right]$

or            $\frac{\partial U}{\partial P_2}=\left\lgroup \frac{1}{E I} \right\rgroup\left\{\int_{L / 3}^{2 L / 3}-P_1\left\lgroup x-\frac{L}{3} \right\rgroup \cdot 0 d x+\int_{2 L / 3}^L\left[P_2\left\lgroup x-\frac{2 L}{3} \right\rgroup^2+P_1\left\lgroup x-\frac{L}{3} \right\rgroup\left\lgroup x-\frac{2 L}{3} \right\rgroup\right] d x\right\}$

$=\left\lgroup \frac{1}{E I} \right\rgroup\left[\frac{P_2\left\lgroup x-\frac{2 L}{3} \right\rgroup ^3}{3}+P_1\left\lgroup \frac{x^3}{3}-\frac{L x^2}{2}+\frac{2 L^2 x}{9} \right\rgroup\right]_{2 L / 3}^L$

$=\left\lgroup =\frac{1}{E I} \right\rgroup\left[\frac{P_2 L^3}{81}+\frac{P_1 L^3}{18}-\frac{2 P_2 L^3}{81}\right]$

$=\left\lgroup \frac{1}{E I} \right\rgroup \left[\frac{P_1 L^3}{18}-\frac{P_2 L^3}{81}\right]=\frac{L^3}{162 E I}\left(9 P_1-2 P_2\right)$

Therefore,

$\delta_{ C }=\underset{\substack{P_1 \rightarrow P \\ P_2 \rightarrow P}}{Lt }\left\lgroup \frac{\partial U}{\partial P_2} \right\rgroup$

or          $\delta_{ C }=\frac{7 P L^3}{162 E I}(\downarrow)$

To find $\delta_{ B }$, we note there is no generalised force acting on the structure and hence we put a dummy load Q at B and calculate the strain energy of the beam (Figure 10.44). The bending moment, $M_x$ of the beam at any distance x from free end of the beam is

$M_x=\left\{\begin{array}{lrl} -Q x ; & & 0 \leq x \leq \frac{L}{3} \\ -P\left\lgroup x-\frac{L}{3} \right\rgroup-Q x ; & & \frac{L}{3} \leq x \leq \frac{2 L}{3} \\ -P\left\lgroup x-\frac{2 L}{3} \right\rgroup -P\left\lgroup x-\frac{L}{3} \right\rgroup -Q x ; & \frac{2 L}{3} & \leq x \leq L \end{array}\right\}$

Therefore,

$U=\left\lgroup \frac{1}{2 E I} \right\rgroup \int_0^L M_x^2 d x$

$=\left\lgroup\frac{1}{2 E I} \right\rgroup \left[\int_0^{L / 3} M_x^2 d x+\int_{L / 3}^{2 L / 3} M_x^2 d x+\int_{2 L / 3}^L M_x^2 d x\right]$

Differentiating with respect to Q, we get

$\frac{\partial U}{\partial Q}=\left\lgroup\frac{1}{E I} \right\rgroup\left[\int_0^{L / 3} M_x\left\lgroup\frac{\partial M_x}{\partial Q} \right\rgroup d x+\int_{L / 3}^{2 L / 3} M_x\left\lgroup\frac{\partial M_x}{\partial Q} \right\rgroup d x+\int_{2 L / 3}^L M_x\left\lgroup\frac{\partial M_x}{\partial Q} \right\rgroup d x\right]$

$=\left\lgroup\frac{1}{E I} \right\rgroup \left\{\int_0^{L / 3} Q x^2 d x+\int_{L / 3}^{2 L / 3}\left[P\left\lgroup x^2-\frac{L x}{3} \right\rgroup+Q x^2\right] d x+\int_{2 L / 3}^L\left[P\left(2 x^2-x L\right)+Q x^2\right] d x\right\}$

Now, by Castigliano’s second theorem

$\delta_{ B }=\underset{Q \rightarrow 0}{Lt}\left\lgroup \frac{\partial U}{\partial Q} \right\rgroup$

$=\frac{1}{E I}\left[\int_{L / 3}^{2 L / 3} P\left\lgroup x^2-\frac{L x}{3} \right\rgroup d x+\int_{2 L / 3}^L P\left(2 x^2-x L\right) d x\right]$

Thus,

$\delta_{ B }=\left\lgroup \frac{1}{E I} \right\rgroup\left\{P\left[\frac{x^3}{3}-\frac{L x^2}{6}\right]_{L / 3}^{2 L / 3}+\left[P\left\lgroup \frac{2 x^3}{3}-\frac{x^2 L}{2} \right\rgroup\right]_{2 L / 3}^L\right\}$

$=\left\lgroup \frac{1}{E I} \right\rgroup \left[P\left\lgroup \frac{8 L^3}{81}-\frac{2 L^3}{27}-\frac{L^3}{81}+\frac{L^3}{54} \right\rgroup +P\left\lgroup \frac{2 L^3}{3}-\frac{L^3}{2}-\frac{16 L^3}{81}+\frac{2 L^3}{9} \right\rgroup \right]$

$=\left\lgroup \frac{1}{E I} \right\rgroup \left\lgroup \frac{5 P L^3}{162}+\frac{31 P L^3}{162} \right\rgroup =\frac{36 P L^3}{162 E I}=\frac{2 P L^3}{9 E I}$

Thus,          $\delta_{ B }=\frac{2 P L^3}{9 E I}(\downarrow)$