Solution to Problem 644 | Deflection of Cantilever Beams

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

$R = \frac{1}{2}w_o L$
 

$M = w_o(\frac{1}{2}L)(\frac{3}{4}L)$

$M = \frac{3}{8}w_o L^2$
 

 

$t_{A/B} = \dfrac{1}{EI}(Area_{AB}) \, \bar{X}_A$

$t_{A/B} = \dfrac{1}{EI}[ \, \frac{1}{2}(L)(\frac{1}{2}w_o L^2)(\frac{1}{3}L) - \frac{3}{8}w_o L^2 (L)(\frac{1}{2}L) - \frac{1}{3}(\frac{1}{8}w_o L^2)(\frac{1}{2}L)(\frac{1}{8}L) \, ]$

$t_{A/B} = \dfrac{1}{EI}[ \, \frac{1}{12}w_o L^4 - \frac{3}{16}w_o L^4 - \frac{1}{384}w_o L^4 \, ]$

$t_{A/B} = \dfrac{1}{EI}[ \, -\frac{41}{384}w_o L^4 \, ]$

$t_{A/B} = -\dfrac{41w_o L^4}{384EI}$
 

Therefore
$\delta_{max} = \dfrac{41w_o L^4}{384EI}$           answer