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(1 point) A simply supported steel beam shown below Click on the image to enlarge is 64 inches long is designed to carry a lo

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Answer #1

From given data

P=500 lb L=64 in, b= 2.25 in, h= 6 in

E= 30 *10^{6} psi, 490 3 p=4901b/ft3 1231b/in

\rho = 0.2836 lb/ in^{3}

Step1

I= \frac{bh^{3}}{12}

I = \frac{2.25*6^{3}}{12}

I = 40.50 in^{4}

Step 2

For simply supported beam with point load at centre, maximum deflection is

Deflection = \delta ^{_{max}}= \frac{P L^{3}}{48EI}

Deflection = \delta ^{_{max}}= \frac{500* 64^{3}}{48EI}

Deflection = 0.00225 in

Step 3

Volume = L * b * h

Volume = 64 * 2.25 * 6

Volume = 864 in^{3}

Step 4

Weight Density = \frac{Weight}{Volume}

Weight = Density * Volume

Weight = 0.2836* 864

Weight = 245 lb

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