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Chapter 7

Q. 7.15

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

Step-by-Step

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