Solution to Problem 639 | Deflection of Cantilever Beams

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

$t_{A/C} = \dfrac{1}{(1.5 \times 10^6)(60)}[ \, \frac{1}{2}(6)(5400)(6) - \frac{1}{3}(8)(6400)(6) \, ] \, (12^3)$

$t_{A/C} = -0.09984 \, \text{ in}$
 

∴   The free end will move by 0.09984 inch downward.           answer