Solution to Problem 645 | Deflection of Cantilever Beams

$R = \frac{1}{2}(4)(1200)$

$R = 2400 \, \text{N}$
 

$M = \frac{1}{2}(4)(1200)(\frac{8}{3})$

$M = 6400 \, \text{N}\cdot\text{m}$
 

 

$\dfrac{y}{3} = \dfrac{1200}{4}$

$y = 900 \, \text{N/m}$
 

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

$t_{B/A} = \dfrac{1}{EI}[ \, \frac{1}{2}(3)(7200)(1) - 3(6400)(1.5) - \frac{1}{4}(3)(1350)(0.6) \, ](1000^3)$

$t_{B/A} = \dfrac{1}{10000(30 \times 10^6)}[ \, -18607.5 \, ](1000^3)$

$t_{B/A} = -62.025 \, \text{mm}$
 

Therefore:
$\delta_B = 62.025 \, \text{mm}$           answer
 

$\theta_{AB} = \dfrac{1}{EI}(Area_{AB})$

$\theta_{AB} = \dfrac{1}{EI}[ \, \frac{1}{2}(3)(7200) - 3(6400) - \frac{1}{4}(3)(1350) \, ](1000^2)$

$\theta_{AB} = \dfrac{1}{10\,000(30 \times 10^6)}[ \, -9412.5 \, ](1000^2)$

$\theta_{AB} = -0.031375 \, \text{radian}$

$\theta_{AB} = 1.798 \, \text{degree}$           answer