For a metal that has a yield strength of 690 MPa and a plane strain fracture toughness (KIc) of 32 MPa-m1/2, compute the minimum component thickness for which the condition of plane strain is valid.

Answers

Answer 1

Answer:

the minimum component thickness for which the condition of plane strain is valid is 0.005377 m or 5.38 mm

Explanation:

Given the data in the question;

yield strength σ[tex]_y[/tex] = 690 Mpa

plane strain fracture toughness K[tex]_{Ic[/tex] = 32 MPa-[tex]m^{1/2[/tex]

minimum component thickness for which the condition of plane strain is valid = ?

Now, for plane strain conditions, the minimum thickness required is expressed as;

t ≥ 2.5( K[tex]_{Ic[/tex] / σ[tex]_y[/tex] )²

so we substitute our values into the formula

t ≥ 2.5( 32  / 690  )²

t ≥ 2.5( 0.0463768 )²

t ≥ 2.5 × 0.0021508

t ≥ 0.005377 m or 5.38 mm

Therefore,  the minimum component thickness for which the condition of plane strain is valid is 0.005377 m or 5.38 mm


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