Solution to Problem 656 | Deflections in Simply Supported Beams

$\Sigma M_{R2} = 0$

$4R_1 = 500(3) + 200$

$R_1 = 425 \, \text{ N}$
 

$\Sigma M_{R1} = 0$

$4R_2 + 200 = 500(1)$

$R_2 = 75 \, \text{ N}$
 

 

$EI \, t_{D/A} = (Area_{AD}) \, \bar{X}_D$

$EI \, t_{D/A} = \frac{1}{2}(1)(75)(\frac{2}{3}) + \frac{1}{2}(3)(1275)(2) - \frac{1}{2}(2)(1000)(\frac{5}{3})$

$EI \, t_{D/A} = \frac{6550}{3} \, \text{ N}\cdot\text{m}^3$
 

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

$EI \, t_{C/A} = \frac{1}{2}(3)(1275)(1) - \frac{1}{2}(2)(1000)(\frac{2}{3})$

$EI \, t_{C/A} = \frac{7475}{6} \, \text{ N}\cdot\text{m}^3$
 

$\dfrac{\bar{CE}}{3} = \dfrac{t_{D/A}}{4}$

$\bar{CE} = \dfrac{3}{4}\left( \dfrac{6550}{3EI} \right) = \dfrac{3275}{2EI}$

$EI \, \bar{CE} = \frac{3275}{2} \, \text{ N}\cdot\text{m}^3$
 

$\delta_C = \bar{CE} - t_{C/A}$

$EI \, \delta_C = EI \, \bar{CE} - EI \, t_{C/A}$

$EI \, \delta_C = \frac{3275}{2} - \frac{7475}{6} = \frac{1175}{3}$

$EI \, \delta_C = 391.67 \, \text{ N}\cdot\text{m}^3$           answer