Solution to Problem 621 | Double Integration Method

$EI \, y'' = [ \, \frac{1}{2}w_o (L + 2a) \, ] (x + \frac{1}{2}L) - \frac{1}{2}w_o (x + \frac{1}{2}L + a)^2$

$EI \, y'' = \frac{1}{2}w_o (L + 2a)x + \frac{1}{4}w_o (L + 2a)L - \frac{1}{2}w_o \, [ \, x^2 + 2x(\frac{1}{2}L + a) + (\frac{1}{2}L + a)^2 \, ]$

$EI \, y'' = \frac{1}{2}w_o (L + 2a)x + \frac{1}{4}w_o (L + 2a)L - \frac{1}{2}w_ox^2 - w_o(\frac{1}{2}L + a)x - \frac{1}{2}w_o(\frac{1}{2}L + a)^2$

$EI \, y'' = \frac{1}{2}w_o (L + 2a)x + \frac{1}{4}w_o (L + 2a)L - \frac{1}{2}w_ox^2 - \frac{1}{2}w_o(L + 2a)x - \frac{1}{8}w_o(L + 2a)^2$

$EI \, y'' = \frac{1}{4}w_o (L + 2a)L - \frac{1}{2}w_ox^2 - \frac{1}{8}w_o(L + 2a)^2$

$EI \, y' = \frac{1}{4}w_o (L + 2a)Lx - \frac{1}{6}w_ox^3 - \frac{1}{8}w_o(L + 2a)^2x + C_1$

$EI \, y = \frac{1}{8}w_o (L + 2a)Lx^2 - \frac{1}{24}w_ox^4 - \frac{1}{16}w_o(L + 2a)^2x^2 + C_1x + C_2$
 

At x = 0, y' = 0, therefore C₁ = 0

At x = ½L, y = 0
$0 = \frac{1}{8}w_o (L + 2a)L(\frac{1}{2}L)^2 - \frac{1}{24}w_o(\frac{1}{2}L)^4 - \frac{1}{16}w_o(L + 2a)^2(\frac{1}{2}L)^2 + C_2$

$0 = \frac{1}{32}w_o (L + 2a)L^3 - \frac{1}{384}w_oL^4 - \frac{1}{64}w_o(L + 2a)^2L^2 + C_2$

$0 = \frac{1}{32}w_oL^4 - \frac{1}{16}w_oL^3a - \frac{1}{384}w_oL^4 - \frac{1}{64}w_o(L^2 + 4La + 4a^2)L^2 + C_2$

$0 = \frac{1}{32}w_oL^4 - \frac{1}{16}w_oL^3a - \frac{1}{384}w_oL^4 - \frac{1}{64}w_oL^4 - \frac{1}{16}w_oL^3a - \frac{1}{16}w_oL^2a^2 + C_2$

$0 = \frac{5}{384}w_oL^4 - \frac{1}{16}w_oL^2a^2 + C_2$

$C_2 = \frac{1}{16}w_oL^2a^2 - \frac{5}{384}w_oL^4$

$C_2 = \frac{1}{384}w_oL^2 (24a^2 - 5L^2)$
 

Therefore,
$EI \, y = \frac{1}{8}w_o (L + 2a)Lx^2 - \frac{1}{24}w_ox^4 - \frac{1}{16}w_o(L + 2a)^2x^2 + \frac{1}{384}w_oL^2 (24a^2 - 5L^2)$
 

At x = 0 (midspan)
$EI \, y = \frac{1}{384}w_oL^2 (24a^2 - 5L^2)$           answer
 

At x = 0 when a = 0
$EI \, y_{max} = \frac{1}{384}w_oL^2 (0 - 5L^2)$

$EI \, y_{max} = -\dfrac{5w_oL^4}{384}$
 

Thus,
$\delta_{max} = \dfrac{5w_oL^4}{384EI}$           answer