Solution to Problem 614 | Double Integration Method

$\Sigma M_{R2} = 0$

$8R_1 + 240(2) = 100(4)(6)$

$R_1 = 240 \, \text{lb}$
 

$\Sigma M_{R1} = 0$
$8R_2 = 240(10) + 100(4)(2)$

$R_2 = 400 \, \text{lb}$
 

 

$EI \, y'' = 240x - \frac{1}{2} (100)x^2 + \frac{1}{2} (100) \, \langle \, x - 4 \, \rangle^2 + 400 \, \langle \, x - 8 \, \rangle$

$EI \, y'' = 240x - 50x^2 + 50 \, \langle \, x - 4 \, \rangle^2 + 400 \, \langle \, x - 8 \, \rangle$

$EI \, y' = 120x^2 - \frac{50}{3}x^3 + \frac{50}{3} \, \langle \, x - 4 \, \rangle^3 + 200 \, \langle \, x - 8 \, \rangle^2 + C_1$

$EI \, y = 40x^3 - \frac{25}{6}x^4 + \frac{25}{6} \, \langle \, x - 4 \, \rangle^4 + \frac{200}{3} \, \langle \, x - 8 \, \rangle^3 + C_1x + C_2$
 

At x = 0, y = 0, therefore C₂ = 0

At x = 8 ft, y = 0
0 = 40(8³) - (25/6)(8⁴) + (25/6)(4⁴) + 8C₁
C₁ = -560 lb·ft²
 

Thus,
$EI \, y' = 120x^2 - \frac{50}{3}x^3 + \frac{50}{3} \, \langle \, x - 4 \, \rangle^3 + 200 \, \langle \, x - 8 \, \rangle^2 - 560$
 

At the right support, x = 8 ft

$EI \, y' = 120(8^2) - \frac{50}{3}(8^3) + \frac{50}{3}(4^3) - 560$

$EI \, y' = -\frac{1040}{3} \, \text{lb}\cdot\text{ft}^2$

$y' = -\dfrac{1040}{3EI} \, \text{lb}\cdot\text{ft}^2$           answer