**Problem 817**

Find the moment under the support R_{2} of the continuous beam shown in Fig. P-817.

**Solution 817**

Where

$L_1 = L_2 = 9 ~ \text{ft}$

$M_1 = M_3 = 0$

$\dfrac{6A_1\bar{a}_1}{L_1} = -\dfrac{M}{L}(3a^2 - L^2) = -\dfrac{300}{9}[ \, 3(3^2) - 9^2 \, ]$

$\dfrac{6A_1\bar{a}_1}{L_1} = 1800 ~ \text{lb}\cdot\text{ft}$

$\dfrac{y}{x - 3} = \dfrac{100}{6}$

$y = \frac{50}{3}(x - 3)$

$P = y \, dx = \frac{50}{3}(x - 3) \, dx$

$\dfrac{6A_2\bar{b}_2}{L_2} = \dfrac{Pb}{L}(L^2 - b^2)$

$\displaystyle \dfrac{6A_2\bar{b}_2}{L_2} = {\int_3^9} \dfrac{[ \, \frac{50}{3}(x - 3) \, dx \, ](9 - x)}{9} [ \, 9^2 - (9 - x)^2 \, ]$

$\displaystyle \dfrac{6A_2\bar{b}_2}{L_2} = \dfrac{50}{27}{\int_3^9} (x - 3)(9 - x)[ \, 81 - (81 - 18x + x^2) \, ] \, dx$

$\displaystyle \dfrac{6A_2\bar{b}_2}{L_2} = \dfrac{50}{27}{\int_3^9} (x - 3)(9 - x)(18x - x^2) \, dx$

$\dfrac{6A_2\bar{b}_2}{L_2} = 4680 ~ \text{lb}\cdot\text{ft}^2$

Another way to solve for $\dfrac{6A_2\bar{b}_2}{L_2}$

$\Sigma M_{V3} = 0$

$9V_2 = \frac{1}{2}(6)(100)[ \, \frac{1}{3}(6) \, ]$

$V_2 = \frac{200}{3} ~ \text{lb}$

$\dfrac{6A_2\bar{b}_2}{L_2} = \frac{6}{9} [ \, \frac{1}{2}(9)(600)(3) - \frac{1}{4}(6)(600)(1.2) \, ]$

$\dfrac{6A_2\bar{b}_2}{L_2} = 4680 ~ \text{lb}\cdot\text{ft}^2$

Thus,

$0 + 2M_2(9 + 9) + 0 + 1800 + 4680 = 0$

$36M_2 = -6480$

$M_2 = -180 ~ \text{lb}\cdot\text{ft}$ *answer*