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Ket [755]
3 years ago
11

A structural component in the shape of a flat plate 29.6 mm thick is to be fabricated from a metal alloy for which the yield str

ength and plane strain fracture toughness values are 535 MPa and 38.5 MPa-m1/2, respectively. For this particular geometry, the value of Y is 1.5. Assuming a design stress of 0.5 times the yield strength, calculate the critical length of a surface flaw. mm
Engineering
1 answer:
slega [8]3 years ago
8 0

Answer:

2.93 mm

Explanation:

Plane strain fracture toughness K_{IC}=38.5 Mpa\sqrt{m}

Yield strength, \sigma=535 Mpa

Design stress=0.5*535=267.5 Mpa

Dimensionless parameter Y=1.5

Critical length of surface flaw is given by

a_c=\frac {1}{\pi}\times (\frac {K_{IC}}{Y\sigma})^{2}=\frac {1}{\pi}\times (\frac {38.5}{1.5*267.5})^{2}= 0.00293\approx 2.93 mm

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(a) Precipitation hardening

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(b) Dispersion strengthening

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3 years ago
The air contained in a room loses heat to the surroundings at a rate of 60 kJ/min while work is supplied to the room by computer
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Answer:

The net amount of energy change of the air in the room during a 10-min period is 120 KJ.

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Work supplied to the room(W) = 1.2 KW = 1.2 KJ/s

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So

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4 0
3 years ago
Read 2 more answers
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Answer:

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x_B = 42.8/(248 + 42.8) = 0.1472

∴ \Delta S = - 8.3145(248 \times ln0.83 + 42.8 \times ln 0.1472) =  1066.0279 J/K

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