Problem 729 | Uniform load over the center part of fixed-ended beam

$\frac{1}{2}(6)(3600) - 6M - \frac{1}{3}(3)(900) = 0$

$10\,800 - 6M - 900 = 0$

$M = 1650 \, \text{ lb}\cdot\text{ft}$           answer
 

$\delta_{max} = t_{A/C}$

$EI ~ \delta_{max} = EI ~ t_{A/C}$

$EI ~ \delta_{max} = (Area_{AC}) \cdot \bar X_A$

$EI ~ \delta_{max} = \frac{1}{2}(6)(3600)[ \, \frac{2}{3}(6) \, ] - 6M(3) - \frac{1}{3}(3)(900)[ \, 3 + \frac{3}{4}(3) \, ]$

$EI ~ \delta_{max} = 10\,800(4) - 18M - 900(5.25)$

$EI ~ \delta_{max} = 10\,800(4) - 18(1650) - 900(5.25)$

$EI ~ \delta_{max} = 8775 \, \text{ lb}\cdot\text{ft}^3$           answer