Solution to Problem 637 | Deflection of Cantilever Beams

$R_C = 80(8) = 640 \, \text{ lb}$

$M_C = 80(8)(4) = 2560 \, \text{lb}\cdot\text{ft}$
 

 

$t_{B/C} = \dfrac{1}{EI}(Area_{BC})\bar{X}_B$

$t_{B/C} = \dfrac{1}{EI}[ \, \frac{1}{2}(6)(3840)(2) - 6(2560)(3) - \frac{1}{3}(6)(1440)(1.5) \, ] \, (12^3)$

$t_{B/C} = \dfrac{1}{EI}[ \, 27\,360 \, ] \, (12^3)$

$t_{B/C} = \dfrac{1}{(1.5 \times 10^6)(40)}[ \, 27\,360 \, ] \, (12^3)$

$t_{B/C} = -0.787968 \, \text{ in}$
 

Thus, δ_B = | t_B/C | = 0.787968 in           answer