Problem 816 | 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
$L_1 = L_2 = L$

$M_1 = M_2 = M_3 = M = -w_ox(\frac{1}{2}x) = -\frac{1}{2}w_ox^2$

$\dfrac{6A_1\bar{a}_1}{L_1} = \dfrac{6A_2\bar{b}_2}{L_2} = \frac{1}{4}w_oL^3$
 

Thus,
$ML + 2M(L + L) + ML + \frac{1}{4}w_oL^3 + \frac{1}{4}w_oL^3 = 0$

$6ML = -\frac{1}{2}w_oL^3$

$6(-\frac{1}{2}w_ox^2)L = -\frac{1}{2}w_oL^3$

$3w_ox^2L = \frac{1}{2}w_oL^3$

$3x^2 = \frac{1}{2}L^2$

$x^2 = \frac{1}{6}L^2$

$x = \frac{1}{\sqrt{6}}L$           answer