Solution to Problem 674 | Midspan Deflection

$\dfrac{y}{a} = \dfrac{-P(\frac{1}{2}L + a)}{\frac{1}{2}L + a}$

$y = -Pa$
 

 

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

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

$EI \, t_{B/C} = \frac{1}{24}PL^3 - \frac{1}{24}PL^2a - \frac{1}{12}PL^2(\frac{1}{2}L + a)$

$EI \, t_{B/C} = \frac{1}{24}PL^3 - \frac{1}{24}PL^2a - \frac{1}{24}PL^3 - \frac{1}{12}PL^2a$

$EI \, t_{B/C} = -\frac{1}{8}PL^2a$
 

$2\delta = t_{B/C}$

$\delta = \dfrac{1}{2}\left[ -\dfrac{1}{8EI}PL^2a \right]$

$\delta = -\dfrac{1}{16EI}PL^2a$

$\delta = \dfrac{1}{16EI}PL^2a \,\, \text{ upward}$           answer