The fracture stress of a large sheet of steel with a central crack of 40 mm is 480 mpa. then, the fracture stress of similar sheet with crack of 100 mm is

σ = 303.57 MPa

Explanation:

Given data:

Crack length for case 1 = 40 mm = 0.04 m

Fracture stress for the case 1 = 480 MPa = 480 × 10⁶ Pa

Crack length for the case 2 = 100 mm = 0.1 m

Now,

using the Griffith equation, we have

where,

Gc is the critical strain energy releasr

and E is the modulus of elasticity

For case 1, we have

 .........(1)

and

for case 2

 .......(2)

on dividing (2) by (1), we get

or

σ = 303.57 × 10⁶ Pa

or

σ = 303.57 MPa

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