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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 strength 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

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Answer:

2.93 mm

Step-by-step explanation:

Plane strain fracture toughness
K_(IC)=38.5 Mpa√(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}* (\frac {K_(IC)}{Y\sigma})^(2)=\frac {1}{\pi}* (\frac {38.5}{1.5*267.5})^(2)= 0.00293\approx 2.93 mm

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