Answer:
Yes, fracture will occur
Explanation:
Half length of internal crack will be 4mm/2=2mm=0.002m
To find the dimensionless parameter, we use critical stress crack propagation equation
[tex]\sigma_c=\frac {K}{Y \sqrt {a\pi}}[/tex] and making Y the subject
[tex]Y=\frac {K}{\sigma_c \sqrt {a\pi}}[/tex]
Where Y is the dimensionless parameter, a is half length of crack, K is plane strain fracture toughness, [tex]\sigma_c[/tex] is critical stress required for initiating crack propagation. Substituting the figures given in question we obtain
[tex]Y=\frac {K}{\sigma_c \sqrt {a\pi}}= \frac {40}{300\sqrt {(0.002*\pi)}}=1.682[/tex]
When the maximum internal crack length is 6mm, half the length of internal crack is 6mm/2=3mm=0.003m
[tex]\sigma_c=\frac {K}{Y \sqrt {a\pi}}[/tex] and making K the subject
[tex]K=\sigma_c Y \sqrt {a\pi}[/tex] and substituting 260 MPa for [tex] \sigma_c[/tex] while a is taken as 0.003m and Y is already known
[tex]K=260*1.682*\sqrt {0.003*\pi}=42.455 Mpa[/tex]
Therefore, fracture toughness at critical stress when maximum internal crack is 6mm is 42.455 Mpa and since it’s greater than 40 Mpa, fracture occurs to the material
