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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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B. rotor

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A hydraulic jump is induced in an 80 ft wide channel.The water depths on either side of the jump are 1 ft and 10 ft.Please calcu
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Answer:

a) 42.08 ft/sec

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d) 18.225 ft

e) 3.80 ft

Explanation:

Given:

b = 80ft

y1 = 1 ft

y2 = 10ft

a) Let's take the formula:

\frac{y2}{y1} = \frac{1}{5} * \sqrt{1 + 8f^2 - 1}

10*2 = \sqrt{1 + 8f^2 - 1

1 + 8f² = (20+1)²

= 8f² = 440

f² = 55

f = 7.416

For velocity of the faster moving flow, we have :

\frac{V_1}{\sqrt{g*y_1}} = 7.416

V_1 = 7.416 *\sqrt{32.2*1}

V1 = 42.08 ft/sec

b) the flow rate will be calculated as

Q = VA

VA = V1 * b *y1

= 42.08 * 80 * 1

= 3366.66 ft³/sec

c) The Froude number of the sub-critical flow.

V2.A2 = 3366.66

Where A2 = 80ft * 10ft

Solving for V2, we have:

V_2 = \frac{3666.66}{80*10}

= 4.208 ft/sec

Froude number, F2 =

\frac{V_2}{g*y_2} = \frac{4.208}{32.2*10}

F2 = 0.235

d) El = \frac{(y_2 - y_1)^3}{4*y_1*y_2}

El = \frac{(10-1)^3}{4*1*10}

= \frac{9^3}{40}

= 18.225ft

e) for critical depth, we use :

y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3

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User scanf() to get user input.

Print out the sum, product, quotient, and difference between the two numbers.

8 0
3 years ago
Typically each development platform consists of the following components, except:Select one:a.Operating systemb.System softwarec
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Typically each development platform consists of the following components except compilers and assemblers

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700.0 liters of a gas are prepared at 760.0 mmHg and 100.0 °C. The gas is placed into a tank under high pressure. When the tank
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Answer:

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Explanation:

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When the gas is in the thank we have:

V₂ =?

P₂ = 20.0 atm

T₂ = 32.0 °C      

Now, we can find the volume of the gas in the thank by using the Ideal Gas Law:

PV = nRT

V_{2} = \frac{nRT_{2}}{P_{2}}    (1)

Where R is the gas constant

With the initials conditions we can find the number of moles:

n = \frac{P_{1}V_{1}}{RT_{1}}    (2)

By entering equation (2) into (1) we have:

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Therefore, When the gas is placed into a tank the volume of the gas is 11.2 L.

I hope it helps you!                                                                                                                                                                                

5 0
3 years ago
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