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

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

Step-by-step 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