$EI \, t_{A/B} = (Area_{AB}) \, \bar{X}_A$
$EI \, t_{A/B} = \frac{1}{3}(\frac{1}{2}L - a)[\,\frac{1}{2}w_o(\frac{1}{2}L - a)^2\,][\,a + \frac{3}{4}(\frac{1}{2}L - a)\,] + \frac{1}{2}(\frac{1}{2}L)(\frac{1}{2}w_oLa)(\frac{1}{3}L) - \frac{1}{3}(\frac{1}{2}L)(\frac{1}{8}w_oL^2)(\frac{3}{8}L)$
$EI \, t_{A/B} = \frac{1}{6}w_oa(\frac{1}{2}L - a)^3 + \frac{1}{8}w_o(\frac{1}{2}L - a)^4 + \frac{1}{24}w_oL^3a - \frac{1}{128}w_oL^4$
$EI \, t_{A/B} = \frac{1}{6}w_oa[\,\frac{1}{2}(L - 2a)\,]^3 + \frac{1}{8}w_o[\,\frac{1}{2}(L - 2a)\,]^4 + \frac{1}{24}w_oL^3a - \frac{1}{128}w_oL^4$
$EI \, t_{A/B} = \frac{1}{48}w_oa(L - 2a)^3 + \frac{1}{128}w_o(L - 2a)^4 + \frac{1}{24}w_oL^3a - \frac{1}{128}w_oL^4$
$EI \, t_{A/B} = \frac{1}{48}w_oa[\,L^3 - 3L^2(2a) + 3L(2a)^2 - (2a)^3\,] + \frac{1}{128}w_o[\,L^4 - 4L^3(2a)+6L^2(2a)^2 - 4L(2a)^3 + (2a)^4\,] + \frac{1}{24}w_oL^3a - \frac{1}{128}w_oL^4$
$EI \, t_{A/B} = \frac{1}{48}w_oL^3a - \frac{1}{8}w_oL^2a^2 + \frac{1}{4}w_oLa^3 - \frac{1}{6}w_oa^4 + \frac{1}{128}w_oL^4 - \frac{1}{16}w_oL^3a + \frac{3}{16}w_oL^2a^2 - \frac{1}{4}w_oLa^3 + \frac{1}{8}w_oa^4 + \frac{1}{24}w_oL^3a - \frac{1}{128}w_oL^4$
$EI \, t_{A/B} = \frac{1}{16}w_oL^2a^2 - \frac{1}{24}w_oa^4$
$EI \, t_{A/B} = \frac{1}{48}w_oa^2(3L^2 - 2a^2)$ answer
Check Problem 653:
wo = 600 N/m; L = 5 m; a = 2 m
$EI \, t_{A/B} = \frac{1}{48}(600)(2^2)[\,3(5^2) - 2(2^2) \, ]$
$EI \, t_{A/B} = 3350 \, \text{ N}\cdot\text{m}^3$ (okay!)
When a = L/2 (the load is over the entire span)
$EI \, t_{A/B} = \frac{1}{48}w_o(\frac{1}{2}L)^2[\,3L^2 - 2(\frac{1}{2}L)^2\,]$
$EI \, t_{A/B} = \frac{1}{192}w_oL^2[\,3L^2 - \frac{1}{2}L^2\,]$
$EI \, t_{A/B} = \frac{1}{192}w_oL^2[\,\frac{5}{2}L^2\,]$
$EI \, t_{A/B} = \frac{5}{384}w_oL^4$
Therefore
$\delta_{max} = \dfrac{5w_oL^4}{384EI}$
