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bearhunter [10]
2 years ago
6

I) A sag vertical curve is to be designed to join a 4% grade to a 2% grade. If the design

Engineering
1 answer:
Burka [1]2 years ago
4 0

Answer:

=4/5 because I'm not going to go back in a year meaning that they are you are

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A concentrated load P is applied to the upper end of a 1.47-m-long pipe. The outside diameter of the pipe is D = 112 mm and the
myrzilka [38]

Answer:

Pmax = 38251.73 N

Explanation:

Given info

L = 1.47 m

D = 112 mm ⇒ R = D/2 = 112/2 mm = 56 mm

d = 101 mm  ⇒ r = D/2 = 101/2 mm = 50.5 mm

a) We can apply the following equation in order to get Q (First Moment of Area):

Q = 2*(A₁*y₁-A₂*y₂)

where

A₁ = π*R² = π*(56 mm)² = 3136 π mm²  

y₁ = 4*R/(3*π) = 4*56/(3*π) mm = 224/(3*π) mm

A₂ = π*r² = π*(50.5 mm)² = 2550.25 π mm²

y₂ = 4*r/(3*π) = 4*50.5/(3*π) mm = 202/(3*π) mm

then

Q = 2*(3136 π mm²*224/(3*π) mm-2550.25 π mm²*202/(3*π) mm)

⇒ Q = 62437.833 mm³

b) If  τallow = 83 MPa = 83 N/mm²

P = ?

We can use the equation

τ = V*Q / (t*I)   ⇒  V = τ*t*I / Q

where

t = D - d = 112 mm - 101 mm = 11 mm

I = (π/64)*(D⁴-d⁴) = (π/64)*((112 mm)⁴-(101 mm)⁴) = 2615942.11 mm⁴

Q = 62437.833 mm³

we could also use this equation in order to get Q:

Q = (4/3)*(R³-r³)

⇒  Q = (4/3)*((56 mm)³-(50.5 mm)³) = 62437.833 mm³

then we have

V = (83 N/mm²)*(11 mm)*(2615942.11 mm⁴) / (62437.833 mm³)

⇒ V = 2942.255 N

Finally Pmax = V = 38251.73 N

6 0
3 years ago
Why do engineers play a variety of roles in the engineering process?
katrin2010 [14]

Answer:

b

Explanation:

7 0
2 years ago
Air enters the compressor of a simple gas turbine at 100 kPa, 300 K, with a volumetric flow rate of 5 m3/s. The compressor press
Zepler [3.9K]

Answer:

a) 3581.15067 kw

b) 95.4%

Explanation:

<u>Given data:</u>

compressor efficiency = 85%

compressor pressure ratio = 10

Air enters at:    flow rate of 5m^3/s , pressure = 100kPa, temperature = 300 K

At turbine inlet : pressure = 950 kPa, temperature = 1400k

Turbine efficiency = 88% , exit pressure of turbine = 100 kPa

A) Develop a full accounting of the exergy increase of the air passing through the gas turbine combustor in kW

attached below is a detailed solution to the given question

6 0
2 years ago
technician a uses a current output test to check ac generator output. technician b uses a voltage output test to check output. w
expeople1 [14]

Answer:

both

Explanation:

Both the technician are correct, ac generator output can be tested in both ways. The two ways are  current output test to check ac generator output. and  voltage output test to check output.

7 0
3 years ago
Three tugboats are used to turn a barge in a narrow channel. To avoid producing any net translation of the barge, the forces app
stiks02 [169]

Answer:

a) Fb= 275.77 lb   Fc= 142.75 lb

b) M = -779.97 lb.ft (i.e. 779.97 lb.ft in clockwise direction)

c) Fax = 195 lb

   Fay = 337.75 lb

   Fbx = 195 lb

   Fby = 195 lb

Explanation:

Question: Three tugboats are used to turn a barge in a narrow channel. To avoid producing any net translation of the barge, the forces applied should be couples. The tugboat at point A applies a 390 lb force.

(a) Determine FB and FC so that only couples are applied.

(b) Using your answers to Part (a), determine the resultant couple moment that is produced.

(c) Resolve the forces at A and B into x and y components, and identify the pairs of forces that constitute couples.

Solution:

<u>For this problem Right hand side is positive X direction and Upwards is positive Y direction. Couples and moments will be considered positive in counterclockwise direction.</u>

<u />

a) For no translation condition

∑ F_{x} = 0      &     ∑F_{y} = 0

Hence,

F_{A}cos(30) - F_{B}cos(45) - F_{C} = 0

F_{A}sin(30) - F_{B} sin(45) = 0

and

F_{A} = 390 lb

Inserting the value of F_{A} and solving the remaining equations simultaneously yields (magnitudes),

F_{B} = 275.77 lb\\F_{C} = 142.75 lb

b) Summing up moments

M=45 ( -F_{Ay}-F_{Cy}) +5 (-F_{By})+22(-F_{Ax}-F_{Bx})\\ =45(-390cos(30)-142.75)+5(-275.77cos(45))+22 (-390sin(30)-275.77sin(45))

M = -779.97 lb.ft (i.e. 779.97 lb.ft clockwise)

c)

F_{Ax} = 390 sin(30)  = 195 lb

F_{Ay} = 390 cos(30) = 337.75lb\\

F_{Bx} = 275.77 sin(45) = 195lb\\F_{By} = 275.77 cos(45) = 195 lb

8 0
3 years ago
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