Solution to Problem 615 | Double Integration Method

$\Sigma M_{R2} = 0$

$10R_1 + 400(3)(1.5) = 1000(4)$

$R_1 = 220 \, \text{lb}$
 

$\Sigma M_{R2} = 0$

$10R_2 = 400(3)(11.5) + 1000(6)$

$R_2 = 1980 \, \text{lb}$
 

 

$EI \, y'' = 220x - 1000 \, \langle \, x - 6 \, \rangle + 1980 \, \langle \, x - 10 \, \rangle - \frac{1}{2} (400) \, \langle x - 10 \, \rangle^2$

$EI \, y'' = 220x - 1000 \, \langle \, x - 6 \, \rangle + 1980 \, \langle \, x - 10 \, \rangle - 200 \, \langle x - 10 \, \rangle^2$

$EI \, y' = 110x^2 - 500 \, \langle \, x - 6 \, \rangle^2 + 990 \, \langle \, x - 10 \, \rangle^2 - \frac{200}{3} \, \langle x - 10 \, \rangle^3 + C_1$

$EI \, y = \frac{110}{3}x^3 - \frac{500}{3} \, \langle \, x - 6 \, \rangle^3 + 330 \, \langle \, x - 10 \, \rangle^3 - \frac{50}{3} \, \langle x - 10 \, \rangle^4 + C_1x + C_2$
 

At x = 0, y = 0, therefore C₂ = 0

At x = 10 ft, y = 0
0 = (110/3)(10³) - (500/3)(4³) + 10C₁
C₁ = -2600 lb·ft²
 

Therefore,
$EI \, y = \frac{110}{3}x^3 - \frac{500}{3} \, \langle \, x - 6 \, \rangle^3 + 330 \, \langle \, x - 10 \, \rangle^3 - \frac{50}{3} \, \langle x - 10 \, \rangle^4 - 2600x$
 

At the right end of the beam, x = 13 ft

$EI \, y = \frac{110}{3}(13^3) - \frac{500}{3}(7^3) + 330(3^3) - \frac{50}{3}(3^4) - 2600(13)$

$EI \, y = -2850 \, \text{lb}\cdot\text{ft}^3$           answer