Chemistry: 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

They both are white crystal...Explanation:...

A steel with a critical fracture toughness of 150

sanchezp0821

21.07.2021

Explanation:

From the given information:

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

yield strength $\sigma$ = 1500 MPa

surface crack size a_c = ???

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

$K_{IC}= \sigma \sqrt{\pi a_c}$

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

$150= 1500 \sqrt{\pi a_c}$

$\dfrac{150}{ 1500} = \sqrt{\pi a_c}$

$\dfrac{0.1}{1.77} = \sqrt{ a_c}$

a_c = 0.0564²

a_c = 0.0032 m

a_c = 0.32 cm

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