Solution to Problem 643 | Deflection of Cantilever Beams

$EI \, t_{B/A} = (Area_{AB}) \, \bar{X}_B$

$EI \, t_{B/A} = \frac{1}{2}L(PL)(\frac{1}{3}L) - PaL(\frac{1}{2}L) - \frac{1}{2}(L - a)P(L - a)[ \, \frac{1}{3}(L - a) \, ]$

$EI \, t_{B/A} = \frac{1}{6}PL^3 - \frac{1}{2}PL^2a - \frac{1}{6}P(L - a)^3$

$EI \, t_{B/A} = \frac{1}{6}PL^3 - \frac{1}{2}PL^2a - \frac{1}{6}P(L^3 - 3L^2a + 3La^2 - a^3)$

$EI \, t_{B/A} = \frac{1}{6}PL^3 - \frac{1}{2}PL^2a - \frac{1}{6}PL^3 + \frac{1}{2}PL^2a - \frac{1}{2}PLa^2 + \frac{1}{6}Pa^3$

$EI \, t_{B/A} = -\frac{1}{2}PLa^2 + \frac{1}{6}Pa^3$

$EI \, t_{B/A} = -\frac{1}{6}Pa^2(3L - a)$
 

Therefore
$EI \, \delta_{max} = \frac{1}{6}Pa^2(3L - a)$           answer