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Chapter 7

Q. 7.17

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

Step-by-Step

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}