Question

A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 78 MPa (70.98 ksi). If the plate is exposed to a tensile stress of 345 MPa (50040 psi) during use, determine the minimum length of a surface crack that will lead to fracture. Assume a value of 1.04 for Y.

Answer 1

Answer:

minimum length of a surface crack is 15.043 mm

Explanation:

given data

strain fracture toughness K = 78 MPa

tensile stress = 345 MPa

Y = 1.04

to find out

minimum length of a surface crack

solution

we find here length of critical interior flaw from formula that is

α  =  $\frac{1}{\pi} (\frac{K}{\sigma Y})^2$     ....................1

put here value we get

α  =  $\frac{1}{\pi} (\frac{78*\sqrt{10^3} }{345*1.04})^2$

α  = 15.043 mm

so minimum length of a surface crack is 15.043 mm