Solution to Problem 606 | Double Integration Method

From the figure below
$EI \, y'' = \frac{1}{2}w_o Lx - w_o x ( \frac{1}{2} x)$

$EI \, y'' = \frac{1}{2}w_o Lx - \frac{1}{2}w_o x^2$
 

 

$EI \, y' = \frac{1}{4}w_o Lx^2 - \frac{1}{6}w_o x^3 + C_1$

$EI \, y = \frac{1}{12}w_o Lx^3 - \frac{1}{24}w_o x^4 + C_1x + C_2$
 

At x = 0, y = 0, therefore C₂ = 0
 

At x = L, y = 0
$0 = \frac{1}{12}w_o L^4 - \frac{1}{24}w_o L^4 + C_1L$

$C_1 = -\frac{1}{24}w_o L^3$
 

Therefore,
$EI \, y = \frac{1}{12}w_o Lx^3 - \frac{1}{24}w_o x^4 - \frac{1}{24}w_o L^3 x$
 

Maximum deflection will occur at x = ½ L (midspan)

$EI \, y_{max} = \frac{1}{12}w_o L (\frac{1}{2}L)^3 - \frac{1}{24}w_o (\frac{1}{2}L)^4 - \frac{1}{24}w_o L^3 (\frac{1}{2}L)$

$EI \, y_{max} = \frac{1}{96}w_o L^4 - \frac{1}{384}w_o L^4 - \frac{1}{48}w_o L^4$

$EI \, y_{max} = -\frac{5}{384}w_o L^4$
 

$\delta_{max} = \dfrac{5w_o L^4}{384EI}$           answer
 

Taking W = w_oL:
$\delta_{max} = \dfrac{5(w_o L)(L^3)}{384EI}$

$\delta_{max} = \dfrac{5W L^3}{384EI}$           answer