Problem 860 | Deflection by Three-Moment Equation

$L_2 = b$   →   span 3 to 4
 

$M_1 L_1 + 2M_3(L_1 + L_2) + M_4L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 6EI \left( \dfrac{h_1}{L_1} + \dfrac{h_4}{L_2} \right)$

$0 + 2(-\frac{1}{2}w_ob^2)(a + b) + 0 + 0 + \frac{1}{4}w_ob^3 = 6EI \left( 0 + \dfrac{h_4}{b} \right)$

$-w_oab^2 - w_ob^3 + \frac{1}{4}w_ob^3 = \dfrac{6EI}{b}h_4$

$-w_oab^2 - \frac{3}{4}w_ob^3 = \dfrac{6EI}{b}h_4$

$\dfrac{6EI}{b}h_4 = -\frac{1}{4}w_ob^2(4a + 3b)$

$h_4 = -\dfrac{w_ob^3}{6EI}(4a + 3b)$

The negative sign of h₄ indicates that point 4 is below point 3, thus,
$\delta_4 = \dfrac{w_ob^3}{6EI}(4a + 3b) ~ \text{ downward}$           answer

$L_2 = \frac{1}{2}a$   →   span 2 to 3

$h_1 = h_3 = h$
 

$M_1 L_1 + 2M_2(L_1 + L_2) + M_3L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 6EI \left( \dfrac{h_1}{L_1} + \dfrac{h_3}{L_2} \right)$

$0 + 2(-\frac{1}{4}w_ob^2)(\frac{1}{2}a) + \frac{1}{2}a) - \frac{1}{2}w_ob^2(\frac{1}{2}a) + 0 + 0 = 6EI \left( \dfrac{h}{\frac{1}{2}a} + \dfrac{h}{\frac{1}{2}a} \right)$

$-\frac{1}{2}w_oab^2 - \frac{1}{4}w_oab^2 = 6EI \left( \dfrac{4h}{a} \right)$

$-\frac{3}{4}w_oab^2 = \dfrac{24EI}{a}h$

$h = -\dfrac{w_oa^2b^2}{32EI}$
 

The negative sign of h indicates that points 1 and 3 are below point 2, thus,
$\delta_2 = \dfrac{w_oa^2b^2}{32EI} ~ \text{ upward}$           answer