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harkovskaia [24]
3 years ago
10

An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40MPa. It has been dete

rmined that fracture results at a stress of 300 MPa when the maximum (or critical) internal crack length is 4.0 mm. For this same component and alloy, will fracture occur at a stress level of 260 MPa when the maximum internal crack length is 6.0 mm? Why or why not?
Engineering
1 answer:
Nataly [62]3 years ago
6 0

Answer:

Yes, fracture will occur

Explanation:

Half length of internal crack will be 4mm/2=2mm=0.002m

To find the dimensionless parameter, we use critical stress crack propagation equation

\sigma_c=\frac {K}{Y \sqrt {a\pi}} and making Y the subject

Y=\frac {K}{\sigma_c \sqrt {a\pi}}

Where Y is the dimensionless parameter, a is half length of crack, K is plane strain fracture toughness, \sigma_c  is critical stress required for initiating crack propagation. Substituting the figures given in question we obtain

Y=\frac {K}{\sigma_c \sqrt {a\pi}}= \frac {40}{300\sqrt {(0.002*\pi)}}=1.682

When the maximum internal crack length is 6mm, half the length of internal crack is 6mm/2=3mm=0.003m

\sigma_c=\frac {K}{Y \sqrt {a\pi}}  and making K the subject

K=\sigma_c Y \sqrt {a\pi}  and substituting 260 MPa for \sigma_c  while a is taken as 0.003m and Y is already known

K=260*1.682*\sqrt {0.003*\pi}=42.455 Mpa

Therefore, fracture toughness at critical stress when maximum internal crack is 6mm is 42.455 Mpa and since it’s greater than 40 Mpa, fracture occurs to the material

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In an orthogonal cutting operation, the tool has a rake angle = 12°. The chip thickness before the cut = 0.32 mm and the cut yie
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Answer:

The shear plane angle and shear strain are 28.21° and 2.155 respectively.

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(a)

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Step3

Shears train is calculated as follows:

\gamma=cot\phi+tan(\phi-\alpha)

\gamma=cot28.21^{\circ}+tan(28.21^{\circ}-12^{\circ})\gamma = 2.155.

Thus, the shear strain rate is 2.155.

6 0
3 years ago
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