Solution to Problem 653 | Deflections in Simply Supported Beams

By symmetry:
$R_1 = R_2 = 600(2) = 1200 \, \text{N}$
 

 

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

$t_{A/B} = \dfrac{1}{EI} [ \, \frac{1}{2}(2.5)(3000)(\frac{5}{3}) + \frac{1}{3}(0.5)(75)(\frac{19}{8}) - \frac{1}{3}(2.5)(1875)(\frac{15}{8}) \, ]$

$t_{A/B} = \dfrac{3350}{EI}$
 

From the figure
$\delta_{midspan} = t_{A/B}$
 

Thus
$EI \, \delta_{midspan} = 3350 \, \text{ N}\cdot\text{m}^3$           answer