Solution to Problem 612 | Double Integration Method

$\Sigma M_{R2} = 0$

$6R_1 = 600(3)(3.5)$

$R_1 = 1050 \, \text{N}$
 

$\Sigma M_{R1} = 0$

$6R_2 = 600(3)(2.5)$

$R_2 = 750 \, \text{N}$
 

$EI \, y'' = 1050x - \frac{1}{2}(600)\langle \, x - 1 \, \rangle^2 + \frac{1}{2}(600)\langle \, x - 4 \, \rangle^2$

$EI \, y'' = 1050x - 300\langle \, x - 1 \, \rangle^2 + 300\langle \, x - 4 \, \rangle^2$

$EI \, y' = 525x^2 - 100\langle \, x - 1 \, \rangle^3 + 100\langle \, x - 4 \, \rangle^3 + C_1$

$EI \, y = 175x^3 - 25\langle \, x - 1 \, \rangle^4 + 25\langle \, x - 4 \, \rangle^4 + C_1x + C_2$
 

At x = 0, y = 0, therefore C₂ = 0
 

At x = 6 m, y = 0
$0 = 175(6^3) - 25(6 - 1)^4 + 25(6 - 4)^4 + 6C_1$

$C_1 = -3762.5 \, \text{N}\cdot\text{m}^2$
 

Therefore,
$EI \, y = 175x^3 - 25\langle \, x - 1 \, \rangle^4 + 25\langle \, x - 4 \, \rangle^4 - 3762.5x$
 

At midspan, x = 3 m
$EI \, y_{midspan} = 175(3^3) - 25(3 - 1)^4 - 3762.5(3)$

$EI \, y_{midspan} = -6962.5 \, \text{N}\cdot\text{m}^3$
 

Thus,
$EI \, \delta_{midspan} = 6962.5 \, \text{N}\cdot\text{m}^3$           answer