Solution to Problem 613 | Double Integration Method

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Problem 613
If E = 29 × 10⁶ psi, what value of I is required to limit the midspan deflection to 1/360 of the span for the beam in Fig. P-613?

Solution 613

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$\Sigma M_{R2} = 0$

$12R_1 = 2400(6)(5)$

$R_1 = 6000 \, \text{lb}$

$\Sigma M_{R1} = 0$

$12R_2 = 2400(6)(7)$

$R_2 = 8400 \, \text{lb}$

$EI \, y'' = 6000x - \frac{1}{2}(2400) \, \langle \, x - 4 \, \rangle^2 + \frac{1}{2}(2400) \, \langle \, x - 10 \, \rangle^2$

$EI \, y'' = 6000x - 1200 \, \langle \, x - 4 \, \rangle^2 + 1200 \, \langle \, x - 10 \, \rangle^2$

$EI y' = 3000x^2 - 400 \, \langle \, x - 4 \, \rangle^3 + 400 \, \langle \, x - 10 \, \rangle^3 + C_1$

$EI \, y = 1000x^3 - 100\, \langle \, x - 4 \, \rangle^4 + 100\, \langle \, x - 10 \, \rangle^4 + C_1x + C_2$

At x = 0, y = 0, therefore C₂ = 0

At x = 12 ft, y = 0

$0 = 1000(12^3) - 100(12 - 4)^4 + 100(12 - 10)^4 + 12C_1$

$C_1 = -110\,000 \, \text{lb}\cdot\text{ft}$

Therefore
$EI \, y = 1000x^3 - 100\, \langle \, x - 4 \, \rangle^4 + 100\, \langle \, x - 10 \, \rangle^4 - 110\,000x$

E = 29 × 10⁶ psi
L = 12 ft
At midspan, x = 6 ft
y = -1/360 (12) = -1/30 ft = -2/5 in

Thus,
$EI \, y = 1000x^3 - 100\, \langle \, x - 4 \, \rangle^4 + 100\, \langle \, x - 10 \, \rangle^4 - 110\,000x$

$(29 \times 10^6)I (-\frac{2}{5}) = [ \, 1000(6^3) + 100(2^4) - 110\,000(6) \, ](12^3)$

$I = 65.9 \, \text{ in}^4$ answer

Submitted by Romel Verterra on September 19, 2009 - 8:58am

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