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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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Discuss importance of good communication system​
meriva

Answer:

Effective communication helps decision makers by gathering and providing the information to the right person on right time. Communication performs as a motivator to the employees by notifying the employees about the job task, process of carrying and how it could be done better.

7 0
2 years ago
Select the correct answer. The most frequent maintenance task for a car is: A. Oil changes B. Tire replacements C. Coolant chang
abruzzese [7]

Answer:

  A. Oil changes

Explanation:

It depends on the car and its usage and environment. Usually oil is supposed to be changed every few months, more often if the car is driven a lot. Coolant changes may be indicated as seasons change, so will generally occur less frequently than oil changes.

Tire and brake replacement depend on usage and driving habits. Some owners may never have to replace either one, if they trade their car every year or two. Folks who drive with their foot on the brake pedal may have to replace brakes relatively often.

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5 0
3 years ago
Read 2 more answers
Let A→=(150iˆ+270jˆ) mm , B→=(300iˆ−450jˆ) mm , and C→=(−100iˆ−250jˆ) mm . Find scalars r and s, if possible, such that R→=rA→+s
ioda

Answer: r = 0.8081; s = -0.07071

Explanation:

A = (150i + 270j) mm

B = (300i - 450j) mm

C = (-100i - 250j) mm

R = rA + sB + C = 0i + 0j

R = r(150i + 270j) + s(300i - 450j) + (-100i - 250j) = 0i + 0j

R = (150r + 300s - 100)i + (270r - 450s - 250)j = 0i + 0j

Equating the i and j components;

150r + 300s - 100 = 0

270r - 450s - 250 = 0

150r + 300s = 100

270r - 450s = 250

solving simultaneously,

r = 0.8081 and s = -0.07071

QED!

5 0
3 years ago
A cylindrical metal specimen having an original diameter of 11.34 mm and gauge length of 53.3 mm is pulled in tension until frac
WINSTONCH [101]

Answer:

a) 70.29 %

b) 37%

Explanation:

percent reduction can be found from:

PR = 100*(π(do/2)^2-π(df/2)^2)/π(do/2)^2

     = 100*(π(11.34/2)^2-π(6.21/2)^2)/π(11.34/2)^2

     =70.29 %

percent elongation can be found from:

EL =L_f - Lo/Lo*100

    = (73.17 -53.3/53.3)*100

    = 37%

5 0
3 years ago
Reverse masking forms a soft edge on the panel.
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True I think I’m not sure?
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2 years ago
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