Question

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.

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$_y$ = 690 Mpa

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

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$_{Ic$ / σ$_y$ )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