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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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Answer:

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Technician A is mostly correct.

3 0
3 years ago
Give two causes that can result in surface cracking on extruded products.
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Answer:

1. High friction

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Explanation:

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2 years ago
Briefly explain how each of the following influences the tensile modulus of a semicrystalline polymer and why:(a) molecular weig
marin [14]

Answer:

(a) Increases

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It increases with the increase in the molecular weight of the polymer.

(b) Degree of crystallinity:

Tensile strength of the semi-crystalline polymer increases with the increase in the degree of crystallinity of the polymer.

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5 0
3 years ago
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alekssr [168]

Answer:

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4 0
3 years ago
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