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Anna71 [15]
3 years ago
10

Determine the slopes and deflections at points B and C for the beam shown below by the moment-area method. E=constant=70Gpa I=50

0 (10^6)mm^4
Engineering
1 answer:
inn [45]3 years ago
3 0

Answer:

hello your question is incomplete attached below is the complete question

answer :

Slopes : B = 180 mm , C = 373 mm

Deflection: B = 0.0514 rad ,  C = 0.077 rad

Explanation:

Given data :

I = 500(10^6) mm^4

E = 70 GPa

The M / EI  diagram is attached below

<u><em>Deflection angle at B</em></u>

∅B = ∅BA = [ 150 (6) + 1/2 (300)*6 ] / EI

                 = 1800 / ( 500 * 70 ) = 0.0514 rad

<u><em>slope at B </em></u>

ΔB = ΔBA = [ 150(6)*3 + 1/2 (300)*6*4 ] / EI

                 = 6300 / ( 500 * 70 ) = 0.18 m = 180 mm

<u><em>Deflection angle at C </em></u>

∅C = ∅CA = [ 1800 + 300*3 ] / EI

                 = 2700 / ( 500 * 70 )

                 = 2700 / 35000 = 0.077 rad

<u><em>Slope at C </em></u>

ΔC = [ 150 * 6 * 6 + 1/2 (800)*6*7 + 300(3) *1.5 ]

     = 13050 / 35000 = 373 mm

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4.In a hydroelectric power plant, 100 m3/s of water flows from an elevation of 120 m to a turbine, where electric power is gener
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Answer:

The rate of irreversible loss will be "55.22 MW".

Explanation:

The given values are:

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Now,

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4 0
3 years ago
A boiler is used to heat steam at a brewery to be used in various applications such as heating water to brew the beer and saniti
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Answer:

net boiler heat = 301.94 kW

Explanation:

given data

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to find out

rate of heat does the boiler output

solution

we can say saturated steam is produce at 6 bar from liquid water 18°C

we know at 6 bar from steam table

hg = 2756 kJ/kg

and

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and

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H = 2680.36  × 0.10139

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so net boiler heat = \frac{H}{0.90}

net boiler heat = \frac{271.75}{0.90}

net boiler heat = 301.94 kW

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