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kolezko [41]
3 years ago
7

A beam has a rectangular cross section that is 5 inches wide and 1.5 inches tall. The supports are 60 inches apart and with a 12

5 lb person standing in the middle of the beam and it deflects .15 in.
Find the modulus of elasticity


Find the beam deflection
Engineering
1 answer:
nydimaria [60]3 years ago
6 0

Answer:

The value of Modulus of elasticity E = 85.33 × 10^{6} \frac{lbm}{in^{2} }

Beam deflection is = 0.15 in

Explanation:

Given data

width = 5 in

Length = 60 in

Mass of the person = 125 lb

Load = 125 × 32 = 4000\frac{ft lbm}{s^{2} }

We know that moment of inertia is given as

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

I = \frac{5 (1.5^{3} )}{12}

I = 1.40625 in^{4}

Deflection = 0.15 in

We know that deflection of the beam in this case is given as

Δ = \frac{PL^{3} }{48EI}

0.15 = \frac{4000(60)^{3} }{48 E (1.40625)}

E = 85.33 × 10^{6} \frac{lbm}{in^{2} }

This is the value of Modulus of elasticity.

Beam deflection is = 0.15 in

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Answer / Explanation:

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To do this, we import the data file:

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import java.io.*;

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*/

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4 years ago
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                                  =gcd=(3,131, 1,618)

                                   =gcd(1,618, 1,513)

                                   =gcd(1,513, 105)

                                   =gcd(105, 43)

                                    =gcd(43, 19)

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         Euclid's algorithm is faster by at least 14,142/10 =1400 times

          At most 2*14,142/10 =2800 times.

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