Question

A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 50.0 MPa-m1/2 . If, during service use, the plate is exposed to a tensile stress of 200 MPa (29,000 psi), determine the minimum length of a surface crack that will lead to fracture. Assume a value of 1.0 for Y.

Answer 1

The answer is "
`0.0199044586 \ m\\\\`".

Step-by-step explanation:

The crucial stress essential for activating the spreading of the crack is
`\sigma`, the hardness of the strain break is K, as well as the area long of a break is a, for dimensionless Y. Its equation of the length of its surface of the fracture is 50.1 MPa
`√(m)` on K, 200MPa on
`\sigma`, and 1 on Y.

`a = (1)/(\pi) ((K)/(Y \ \sigma) )^2 \\\\`

`= (1)/(\pi) ( (50.0)/(1 * 200) )^2 \\\\= (1)/(3.14) ( (1)/(4))^2 \\\\= (1)/(3.14) ( (1)/(16)) \\\\=(1)/(50.24)\\\\=0.0199044586 \ m\\\\`