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A steel with a critical fracture toughness of 150 MPa.m1/2 has a yield strength of 1500 MPa. If fracture were to take place at the yield stress, answer the following questions.
Surface crack size at yielding leading to failure is:_____________.
a. 0.497 cm
b. 0.994 cm
c. 0.32 cm

Answer :

ajeigbeibraheem

Answer:

c.

Explanation:

From the given information:

Critical fracture toughness [tex]K_{IC}[/tex] = 150 MPa.[tex]m ^{1/2}[/tex]

yield strength [tex]\sigma[/tex] = 1500 MPa

surface crack size [tex]a_c[/tex] = ???

The formula for the fracture toughness is can be expressed as:

[tex]K_{IC}= \sigma \sqrt{\pi a_c}[/tex]

replacing our values to solve for the surface crack size, we have:

[tex]150= 1500 \sqrt{\pi a_c}[/tex]

[tex]\dfrac{150}{ 1500} = \sqrt{\pi a_c}[/tex]

[tex]\dfrac{0.1}{1.77} = \sqrt{ a_c}[/tex]

[tex]a_c[/tex] = 0.0564²

[tex]a_c[/tex] = 0.0032 m

[tex]a_c[/tex] = 0.32 cm

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