Solution to Problem 667 | Deflections in Simply Supported Beams

$\Sigma M_{R2} = 0$]

$9R_1 + 4(60) = \frac{1}{2}(6)(100)(2)$

$R_1 = 40 \, \text{ lb}$
 

$\Sigma M_{R1} = 0$

$9R_2 = \frac{1}{2}(6)(100)(7) + 13(60)$

$R_2 = 320 \, \text{ lb}$
 

 

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

$EI \, t_{A/B} = \frac{1}{2}(9)(360)(6) - \frac{1}{4}(6)(600)(\frac{39}{5})$

$EI \, t_{A/B} = 2700 \, \text{ lb}\cdot\text{ft}^3$
 

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

$EI \, t_{C/B} = -\frac{1}{2}(4)(240)(\frac{8}{3})$

$EI \, t_{C/B} = -1280 \, \text{ lb}\cdot\text{ft}^3$
 

The negative sign indicates that the elastic curve is below the tangent line. It is shown in the figure indicated as t_C/B. See Rules of Sign for Area-Moment Method.
 

$\dfrac{y_C}{4} = \dfrac{t_{A/B}}{9}$

$y_C = \frac{4}{9}t_{A/B}$

$EI \, y_C = \frac{4}{9}EI \, t_{A/B}$

$EI \, y_C = \frac{4}{9}(2700)$

$EI \, y_C = 1200 \, \text{ lb}\cdot\text{ft}^3$
 

Since the absolute value of EI t_C/B is greater than the absolute value of EI y_C, the elastic curve is below the neutral axis (NA) of the beam.
 

Therefore,
$EI \, \delta_C = 1280 - 1200$

$EI \, \delta_C = 80 \, \text{ lb}\cdot\text{ft}^3$   below C (deflection is down)       answer