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Gnesinka [82]
3 years ago
15

Calculate the maximum internal crack length allowable for a 7075-T651 aluminum alloy component that is loaded to a stress one-ha

lf of its yield strength. Assume a yield strength 495 MPa, a plane strain fracture toughness of 24 MPa and that Y = 1.39.
Engineering
1 answer:
aleksley [76]3 years ago
7 0

Answer: 0.0033 m = 3.3 mm (0.13 in.)

Explanation:

since this problem requires us to calculate the maximum internal crack length allowable for the 7075-T651

aluminum alloy is given that it is loaded to a stress level equal to one-half of its yield strength.

For this alloy, KIc 24 MPa \sqrt{m}(22 ksi \sqrt{in}); also, σ= σ_y/2 = (495 MPa)/2 = 248 MPa (36,000 psi).

Now solving for 2ac gives us;

2a_c=2/π (K_IC/γσ)^2=2/π [(24MPa √m)/(1.35)(248MPa) ]^2=0.0033m=3.3mm (0.13 in.)

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The true statement about the dot plot is 1 has 4 and 0 dots.

Explanation:

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You are testing a new jet engine in a test cell at sea level conditions. You measure the mass flow through the engine and find i
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Answer:

43248 newtons.

Explanation:

Force = mass x accelerations and units of force are newtons which are given in the question.

here mass = 125 of air and 2.2 of fuel, total = 125+2.2=127.5kg/s  and the velocity of the exhaust is 340m/s.

force = 340m/s * 127.5kg/s = 43248 newtons technically this is wrong (observe units) but i will expalin how i have taken acceleration as a velocity here and mass/unit time as simply mass.

see force is mass times acceleration or deceleration, here our velocity is not changing therefore it is constant 340m/s but if it were to change and become 0 in one second then there would be -340m/s^2 (note the units ) of deceleration and there would be force associated with it and that force is what i have calculated here. similarly there would be mass in flow rate of mass per second, which is also in that one second of time.

let's calculate error.

error = (actual-calculated)/actual. = (43248-60000)/43248= -38.734% less is ofcourse greater than 2%.

So the load cell is not reading correct to within 2% and it should read 43248newtons.

5 0
3 years ago
Read 2 more answers
A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. If the gas
ch4aika [34]

Answer:

gauge pressure is 133 kPa

Explanation:

given data

initial temperature T1 = 27°C = 300 K

gauge pressure = 300 kPa = 300 × 10³ Pa

atmospheric pressure = 1 atm

final temperature T2 = 77°C = 350 K

to find out

final pressure

solution

we know that gauge pressure is = absolute pressure - atmospheric pressure so

P (gauge ) = 300 × 10³ Pa - 1 × 10^{5} Pa

P (gauge ) = 2 × 10^{5} Pa

so from idea gas equation

\frac{P1*V1}{T1} = \frac{P2*V2}{T2}   ................1

so {P2} = \frac{P1*T2}{T1}

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P2 = 2.33 × 10^{5} Pa

so gauge pressure = absolute pressure - atmospheric pressure

gauge pressure = 2.33 × 10^{5}  - 1.0 × 10^{5}

gauge pressure = 1.33 × 10^{5} Pa

so gauge pressure is 133 kPa

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