**Problem 827**

See Figure P-827.

**Solution 827**

$M_1 L_1 + 2M_2(L_1 + L_2) + M_3 L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 0$

$M_1 = 0$

$L_1 = L_2 = 12 ~ \text{ft}$

$\dfrac{6A_1\bar{a}_1}{L_1} = \frac{8}{60}w_o L^3 = \frac{8}{60}(150)(12^3) = 34\,560 ~ \text{lb}\cdot\text{ft}^2$

$\dfrac{6A_2\bar{b}_2}{L_2} = \Sigma \dfrac{Pb}{L}(L^2 - b^2) = \dfrac{400(9)}{12}(12^2 - 9^2) + \dfrac{300(4)}{12}(12^2 - 4^2)$

$\dfrac{6A_2\bar{b}_2}{L_2} = 31\,700 ~ \text{lb}\cdot\text{ft}^2$

Thus,

$0 + 2M_2(12 + 12) + M_3(12) + 34\,560 + 31\,700 = 0$

$48M_2 + 12M_3 = -66\,260$ ← equation (1)

Apply three-moment equation to middle and last spans

$M_2 L_2 + 2M_3(L_2 + L_3) + M_4 L_3 + \dfrac{6A_2\bar{a}_2}{L_2} + \dfrac{6A_3\bar{b}_3}{L_3} = 0$

$L_2 = L_3 = 12 ~ \text{ft}$

$L_3 = 3 ~ \text{ft}$

$M_4 = -4(50)(2) = -400 ~ \text{lb}\cdot\text{ft}$

$\dfrac{6A_2\bar{a}_2}{L_2} = \Sigma \dfrac{Pb}{L}(L^2 - a^2) = \dfrac{400(3)}{12}(12^2 - 3^2) + \dfrac{300(8)}{12}(12^2 - 8^2)$

$\dfrac{6A_2\bar{a}_2}{L_2} = 29\,500 ~ \text{lb}\cdot\text{ft}^2$

$\dfrac{6A_3\bar{b}_3}{L_3} = \frac{1}{4}w_o L^3 = \frac{1}{4}(50)(12^3) = 21\,600 ~ \text{lb}\cdot\text{ft}^2$

Thus,

$M_2(12) + 2M_3(12 + 12) - 400(12) + 29\,500 + 21\,600 = 0$

$12M_2 + 48M_3 = -46\,300$ ← equation (2)

Solving equations (1) and (2) simultaneously

$M_2 = -1215.22 ~ \text{lb}\cdot\text{ft}$ *answer*

$M_3 = -660.78 ~ \text{lb}\cdot\text{ft}$ *answer*