**Problem 814**

Find the moment at R_{2} of the continuous beam shown in Fig. P-814.

**Solution 814**

$M_1L_1 + 2M_2(L_1 + L_2) + M_3L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 0$

Where

$M_1 = M_3 = 0$

$M_1 = M_3 = 0$

$L_1 = L_2 = 10 \, \text{ ft}$

$\dfrac{6A_1\bar{a}_1}{L_1} = \dfrac{7w_oL^3}{60} + \dfrac{8w_oL^3}{60} = \dfrac{7(200)(10^3)}{60} + \dfrac{8(100)(10^3)}{60}$

$\dfrac{6A_1\bar{a}_1}{L_1} = 36\,666.67 \, \text{ lb}\cdot\text{ft}^2$

$\dfrac{6A_2\bar{b}_2}{L_2} = \dfrac{8w_oL^3}{60} = \dfrac{8(100)(10^3)}{60}$

$\dfrac{6A_2\bar{b}_2}{L_2} = 13\,333.33 \, \text{ lb}\cdot\text{ft}^2$

Thus,

$0 + 2M_2(10 + 10) + 0 + 36\,666.67 + 13\,333.33 = 0$

$M_2 = -1250 \, \text{ lb}\cdot\text{ft}$ *answer*

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