Problem 870 | Beam Deflection by Three-Moment Equation

$y = 1.2 ~ \text{kN}$
 

$M_B = -\frac{1}{2}(2)(2)(\frac{2}{3} \cdot 2) - \frac{1}{2}(2)(1.2)(\frac{1}{3} \cdot 2)$

$M_B = -\frac{52}{15} ~ \text{kN}\cdot\text{m}$
 

$\begin{align}
\left( \dfrac{6A\bar{a}}{L} \right)_{AB} & = \dfrac{7w_oL^3}{60} + \dfrac{8w_oL^3}{60} \\
& = \dfrac{7(2)(2^3)}{60} + \dfrac{8(1.2)(2^3)}{60} \\
& = \dfrac{236}{75} ~ \text{m}^2
\end{align}$
 

$\begin{align}
\left( \dfrac{6A\bar{b}}{L} \right)_{CB} & = \dfrac{8w_oL^3}{60} \\
& = \dfrac{8(1.2)(3^3)}{60} \\
& = \dfrac{108}{25} ~ \text{m}^2
\end{align}$
 

Another Way to Solve the Three Moment Equation Factors (by Integration)

Apply Three-moment Equation to Span A-B-C
$M_A L_{AB} + 2M_B(L_{AB} + L_{BC}) + M_CL_{BC} + \left( \dfrac{6A\bar{a}}{L} \right)_{AB} + \left( \dfrac{6A\bar{b}}{L} \right)_{CB}$
          $= 6EI \left( \dfrac{h_{BA}}{L_{AB}} + \dfrac{h_{BC}}{L_{BC}} \right)$

$0 + 2(-\frac{52}{15})(2 + 3) + 1.1(3) + \frac{236}{75} + \frac{108}{25} = 6EI\left(\dfrac{\delta_A}{2} + 0 \right)$

$3EI \, \delta_A = -\frac{239}{10}$

$EI \, \delta_A = -\frac{239}{30} ~ \text{kN}\cdot\text{m}^3$       answer