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## Q. 7.15

Determine the deflection at point $C$ of the beam shown in Fig. $7.23(\mathrm{a})$ by Castigliano’s second theorem.

## Verified Solution

This beam was previously analyzed by the moment-area, the conjugate-beam, and the virtual work methods in Examples $6.7$, $6.13$, and $7.9$, respectively.

The $12 \mathrm{k}$ external load is already acting at point $C$, where the deflection is to be determined, so we designate this load as the variable $P$, as shown in Fig. $7.23(\mathrm{~b})$. Next, we compute the reactions of the beam in terms of $P$. These are also shown in Fig. 7.23(b). Since the loading is discontinuous at point $B$, the beam is divided into two segments, $A B$ and $B C .$ The $x$ coordinates used for determining the equations for the bending moment in the two segments of the beam are shown in Fig. $7.23$ (b). The equations for $M$ (in terms of $P$ ) obtained for the segments of the beam are tabulated in Table $7.12$, along with the partial derivatives of $M$ with respect to $P$.

 Table 7.12 x Coordinate Segment Origin Limits (ft) M (k-ft) $\frac{\partial M}{\partial P}\left(k-ft/k\right)$ AB A 0-30 $\left\lgroup30-\frac{P}{3} \right\rgroup x-x^{2}$ $-\frac{x}{3}$ CB C 0-10 -Px -x

The deflection at $C$ can now be determined by substituting $P=12 \mathrm{k}$ into the equations for $M$ and $\partial M / \partial P$ and by applying the expression of Castigliano’s second theorem as given by Eq. (7.60):

\begin{aligned} \Delta &=\int_{0}^{L}\left(\frac{\partial M}{\partial P}\right)\left(\frac{M}{E I}\right) d x \\\end{aligned}     (7.60)

\begin{aligned} \Delta_{C} &=\int_{0}^{L}\left(\frac{\partial M}{\partial P}\right)\left(\frac{M}{E I}\right) d x \\ \Delta_{C} &=\frac{1}{E I}\left[\int_{0}^{30}\left(-\frac{x}{3}\right)\left(30 x-\frac{12 x}{3}-x^{2}\right) d x+\int_{0}^{10}(-x)(-12 x) d x\right] \\ &=\frac{1}{E I}\left[\int_{0}^{30}\left(-\frac{x}{3}\right)\left(26 x-x^{2}\right) d x+\int_{0}^{10}(-x)(-12 x) d x\right] \\ &=-\frac{6,500 \mathrm{k}-\mathrm{ft}^{3}}{E I}=-\frac{6,500(12)^{3}}{(29,000)(2,000)}=-0.194 \mathrm{in} . \end{aligned}

The negative answer for $\Delta_{C}$ indicates that point $C$ deflects upward in the direction opposite to that of $P$.

$\Delta_{C}=0.194 \mathrm{in} . \uparrow$