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kolezko [41]
2 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]2 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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Explanation:

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3 years ago
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Answer:

Explanation:

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0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of  195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

2)  There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

Vt = 195+190+185+180/4

  = 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

2)  There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

  = 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

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The two windings of transformer is c)- Inductively linked

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3 years ago
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