Question

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

Answer 1

σ = 303.57 MPa

Step-by-step 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
`\sigma=[(G_cE)/(\pi\ a)]^{(1)/(2)}`

where,

Gc is the critical strain energy releasr

and E is the modulus of elasticity

For case 1, we have

`480*10^6=[(G_cE)/(\pi\ 0.04)]^{(1)/(2)}` .........(1)

and

for case 2

`\sigma=[(G_cE)/(\pi\ 0.1)]^{(1)/(2)}` .......(2)

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

`(\sigma)/(480*10^6)=[(0.04)/(0.1)]^{(1)/(2)}`

or

σ = 303.57 × 10⁶ Pa

or

σ = 303.57 MPa