Solution to Problem 666 | Deflections in Simply Supported Beams

$\Sigma M_B = 0$

$aR_A = w_ob(\frac{1}{2}b)$

$R_A = \dfrac{w_ob^2}{2a}$
 

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

$EI \, t_{A/B} = \frac{1}{2}a(\frac{1}{2}w_ob^2)(\frac{2}{3}a)$

$EI \, t_{A/B} = \frac{1}{6}w_oa^2b^2$
 

$EI \, t_{C/B} = (Area_{BC}) \, \bar{X}_C$

$EI \, t_{C/B} = \frac{1}{3}b(\frac{1}{2}w_ob^2)(\frac{3}{4}b)$

$EI \, t_{C/B} = \frac{1}{8}w_ob^4$
 

$\dfrac{y_C}{b} = \dfrac{t_{A/B}}{a}$

$y_C = \dfrac{b}{a}t_{A/B}$

$EI \, y_C = \dfrac{b}{a}EI \, t_{A/B}$

$EI \, y_C = \dfrac{b}{a}(\frac{1}{6}w_oa^2b^2)$

$EI \, y_C = \frac{1}{6}w_oab^3$
 

$\delta_C = y_C + t_{C/B}$

$EI \, \delta_C = EI \, y_C + EI \, t_{C/B}$

$EI \, \delta_C = \frac{1}{6}w_oab^3 + \frac{1}{8}w_ob^4$

$EI \, \delta_C = \frac{1}{24}w_ob^3(4a + 3b)$           answer