Solution to Problem 673 | Midspan Deflection

$EI \, t_{A/M} = (Area_{AM}) \, \bar{X}_A$

$EI \, t_{A/M} = \frac{1}{2}(\frac{1}{2}L)(\frac{1}{2}PL) [ \, \frac{2}{3}(\frac{1}{2}L) \, ] - \frac{1}{2}(\frac{1}{2}L - b)P(\frac{1}{2}L - b) [ \, b + \frac{2}{3}(\frac{1}{2}L - b) \, ]$

$EI \, t_{A/M} = \frac{1}{24}PL^3 - \frac{1}{2}Pb(\frac{1}{2}L - b)^2 - \frac{1}{3}P(\frac{1}{2}L - b)^3$

$EI \, t_{A/M} = \frac{1}{24}PL^3 - \frac{1}{2}Pb \left( \dfrac{L - 2b}{2} \right)^2 - \frac{1}{3}P \left( \dfrac{L - 2b}{2} \right)^3$

$EI \, t_{A/M} = \frac{1}{24}PL^3 - \frac{1}{8}Pb(L - 2b)^2 - \frac{1}{24}P (L - 2b)^3$

$EI \, t_{A/M} = \frac{1}{24}PL^3 - \frac{1}{8}Pb(L^2 - 4Lb + 4b^2) - \frac{1}{24}P(L^3 - 6L^2b + 12Lb^2 - 8b^3)$

$EI \, t_{A/M} = \frac{1}{24}PL^3 - \frac{1}{8}PL^2b + \frac{1}{2}PLb^2 - \frac{1}{2}Pb^3 - \frac{1}{24}PL^3 + \frac{1}{4}PL^2b - \frac{1}{2}PLb^2 + \frac{1}{3}Pb^3$

$EI \, t_{A/M} = \frac{1}{8}PL^2b - \frac{1}{6}Pb^3$

$EI \, t_{A/M} = \frac{1}{24}Pb(3L^2 - 4b^2)$

$t_{A/M} = \dfrac{Pb}{24EI}(3L^2 - 4b^2)$
 

$\delta_{midspan} = \frac{1}{2} t_{A/M}$

$\delta_{midspan} = \dfrac{Pb}{48EI}(3L^2 - 4b^2)$       (okay!)