Solution to Problem 647 | Deflection of Cantilever Beams

$R = \frac{1}{2}(\frac{1}{2}L)(w_o) = \frac{1}{4}w_oL$

$M = \frac{1}{2}(\frac{1}{2}L)(w_o)(\frac{5}{6}L) = \frac{5}{24}w_oL^2$
 

 

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

$EI \, t_{A/B} = \frac{1}{2}L(\frac{1}{4}w_oL^2)(\frac{1}{3}L) - L(\frac{5}{24}w_oL^2)(\frac{1}{2}L) - \frac{1}{4}(\frac{1}{2}L)(\frac{1}{24}w_oL^2)(\frac{1}{10}L)$

$EI \, t_{A/B} = \frac{1}{24}w_oL^4 - \frac{5}{48}w_oL^4 - \frac{1}{1920}w_oL^4$

$EI \, t_{A/B} = -\frac{121}{1920}w_oL^4$
 

Therefore
$EI \, \delta_{max} = \frac{121}{1920}w_oL^4$           answer