Respuesta :
Answer:
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[tex]\sigma=[\frac{G_cE}{\pi\ a}]^{\frac{1}{2}}[/tex]
where,
Gc is the critical strain energy releasr
and E is the modulus of elasticity
For case 1, we have
[tex]480\times10^6=[\frac{G_cE}{\pi\ 0.04}]^{\frac{1}{2}}[/tex] .........(1)
and
for case 2
[tex]\sigma=[\frac{G_cE}{\pi\ 0.1}]^{\frac{1}{2}}[/tex] .......(2)
on dividing (2) by (1), we get
[tex]\frac{\sigma}{480\times10^6}=[\frac{0.04}{0.1}]^{\frac{1}{2}}[/tex]
or
σ = 303.57 × 10⁶ Pa
or
σ = 303.57 MPa
