Problem 818 | Continuous Beam by Three-Moment Equation

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

 

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

$M_1 = M_2 = M_3 = 0$

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

$\dfrac{6A_2\bar{b}_2}{L_2} = 4680 ~ \text{lb}\cdot\text{ft}^2$   ←   see Solution 817 for details.
 

Thus,
$0 + 0 + 0 + 6M + 4680 = 0$

$M = -780 ~ \text{lb}\cdot\text{ft}$

$M = 780 ~ \text{lb}\cdot\text{ft counterclockwise}$           answer