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## Q. 7.17

Determine the rotation of joint $C$ of the frame shown in Fig. $7.25(\mathrm{a})$ by Castigliano’s second theorem.

## Verified Solution

This frame was previously analyzed by the virtual work method in Example $7.11$.

No external couple is acting at joint $C$, where the rotation is desired, so we apply a fictitious couple $\bar{M}(=0)$ at $C$, as shown in Fig. $7.25(\mathrm{~b})$. The $x$ coordinates used for determining the bending moment equations for the three segments of the frame are also shown in Fig. $7.25(\mathrm{~b})$, and the equations for $M$ in terms of $\bar{M}$ and $\partial M / \partial \bar{M}$ obtained for the three segments are tabulated in Table $7.13$. The rotation of joint $C$ of the frame can now be determined by setting $\bar{M}=0$ in the equations for $M$ and $\partial M / \partial \bar{M}$ and by applying the expression of Castigliano’s second theorem as given by Eq. (7.65):

\begin{aligned} \theta &=\sum \int\left(\frac{\partial M}{\partial \bar{M}}\right) \frac{M}{E I} d x \\ \end{aligned}      (7.65)

\begin{aligned} \theta_{C} &=\sum \int\left(\frac{\partial M}{\partial \bar{M}}\right) \frac{M}{E I} d x \\ &=\int_{0}^{30}\left(\frac{x}{30}\right)\left(38.5 x-1.5 \frac{x^{2}}{2}\right) d x \\ &=\frac{6,487.5 \mathrm{k}-\mathrm{ft}^{2}}{E I}=\frac{6,487.5(12)^{2}}{(29,000)(2,500)}=0.0129 \mathrm{rad} \\ \end{aligned}

\begin{aligned} \theta_{C} &=0.0129 \mathrm{rad} \end{aligned}

 Table 7.13 x Coordinate Segment Origin Limits (ft) M (k-ft) $\frac{\partial M}{\partial \bar{M} }\left(k-ft/k-ft\right)$ AB A 0-12 40x 0 CB C 0-12 480 0 DC D 0-30 $\left\lgroup38.5+\frac{\bar{M} }{30} \right\rgroup x-1.5\frac{x^{2}}{2}$ $\frac{x}{30}$