Solution to Problem 661 | Deflections in Simply Supported Beams

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

 

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

$EI \, t_{A/B} = \frac{1}{24}PL^3 - \frac{1}{2}Pa(\frac{1}{2}L - a)^2 - \frac{1}{3}P(\frac{1}{2}L - a)^3$

$EI \, t_{A/B} = \frac{1}{24}PL^3 - \frac{1}{2}Pa(\frac{1}{4}L^2 - La + a^2) - \frac{1}{3}P(\frac{1}{8}L^3 - \frac{3}{4}L^2a + \frac{3}{2}La^2 - a^3)$

$EI \, t_{A/B} = \frac{1}{24}PL^3 - \frac{1}{8}PL^2a + \frac{1}{2}PLa^2 - \frac{1}{8}Pa^3 - \frac{1}{24}PL^3 + \frac{1}{4}PL^2a - \frac{1}{2}PLa^2 + \frac{1}{3}Pa^3$

$EI \, t_{A/B} = \frac{1}{8}PL^2a - \frac{1}{6}Pa^3$

$EI \, t_{A/B} = \frac{1}{24}Pa(3L^2 - 4a^2)$           answer
 

When a = ½L
$EI \, t_{A/B} = \frac{1}{24}P(\frac{1}{2}L) [ \, 3L^2 - 4(\frac{1}{2}L)^2 \, ]$

$EI \, t_{A/B} = \frac{1}{48}PL [ \, 3L^2 - L^2 \, ]$

$EI \, t_{A/B} = \frac{1}{48}PL (2L^2)$

$EI \, t_{A/B} = \frac{1}{24}PL^3 \,\, $            → answer

$EI \, t_{A/B} = 2 (\frac{1}{48}PL^3)$   (okay!)