Solution to Problem 655 | Deflections in Simply Supported Beams

$\Sigma M_{R2} = 0$

$8R_1 = 200(5) + 400(1)$

$R_1 = 175 \, \text{ lb}$
 

$\Sigma M_{R1} = 0$

$8R_2 = 200(3) + 400(7)$

$R_2 = 425 \, \text{ lb}$
 

 

$\dfrac{y_{C1}}{7} = \dfrac{1400}{8}$

$y_{C1} = 1225 \, \text{ lb}$
 

$\dfrac{y_{C2}}{4} = \dfrac{-1000}{5}$

$y_{C2} = -800 \, \text{ lb}$
 

$\dfrac{y_B}{3} = \dfrac{1400}{8}$

$y_B = 525 \, \text{ lb}$
 

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

$EI \, t_{D/A} = \frac{1}{2}(8)(1400)(\frac{8}{3}) - \frac{1}{2}(5)(1000)(\frac{5}{3}) - \frac{1}{2}(1)(400)(\frac{1}{3})$

$EI \, t_{D/A} = 10 \, 700 \, \text{ lb}\cdot\text{ft}^3$
 

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

$EI \, t_{C/A} = \frac{1}{2}(7)(y_{C1})(\frac{7}{3}) - \frac{1}{2}(4)(y_{C2})(\frac{4}{3})$

$EI \, t_{C/A} = \frac{1}{2}(7)(1225)(\frac{7}{3}) - \frac{1}{2}(4)(800)(\frac{4}{3})$

$EI \, t_{C/A} = \frac{47 \, 225}{6} \, \text{ lb}\cdot\text{ft}^3$
 

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

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

$EI \, t_{C/A} = \frac{1}{2}(3)(525)(1)$

$EI \, t_{C/A} = \frac{1575}{2} \, \text{ lb}\cdot\text{ft}^3$
 

By ratio and proportion:
$\dfrac{\bar{BE}}{3} = \dfrac{\bar{CF}}{7} = \dfrac{t_{D/A}}{8}$

$\bar{BE} = \frac{3}{8}t_{D/A} = \frac{3}{8}(10 \, 700) = \frac{8025}{2}$

$\bar{CF} = \frac{7}{8}t_{D/A} = \frac{7}{8}(10 \, 700) = \frac{18\,725}{2}$
 

Deflections:
$\delta_B = \bar{BE} - t_{B/A}$

$EI \, \delta_B = EI \, \bar{BE} - EI \, t_{B/A} = \frac{8025}{2} - \frac{1575}{2}$

$EI \, \delta_B = 3225 \, \text{ lb}\cdot\text{ft}^3 \,\,$            → answer
 

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

$EI \, \delta_C = EI \, \bar{CF} - EI \, t_{C/A} = \frac{18\,725}{2} - \frac{47\,225}{6}$

$EI \, \delta_C = \frac{4475}{3} = 1491.67 \, \text{ lb}\cdot\text{ft}^3$           answer