Solution to Problem 646 | Deflection of Cantilever Beams

$M = \frac{1}{2}(5)(60)(2 + \frac{5}{3}) = 550 \, \text{lb}\cdot\text{ft}$

$R = \frac{1}{2}(5)(60) = 150 \, \text{lb}$
 

 

$t_{A/B} = \dfrac{1}{EI}(Area_{AB}) \, \bar{X}_A$

$-0.5 = \dfrac{1}{EI} \Big[ \, \frac{1}{2}(300)(2)(\frac{26}{3}) - 550(2)(9) - \frac{1}{4}(5)(250)(7) \, \Big](12^3)$

$-0.5 = \dfrac{1}{(1.5 \times 10^6)I}(-16394 400)$

$I = 21.8592 \, \text{in}^4$   ←   answer