Answer and Explanation:
The answer is attached below
A vertical piston-cylinder device contains water and is being heated on top of a range. During the process, 10 kJ of heat is tra
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Answer:
The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa
Explanation:
Given data;
Let,
critical stress required for initiating crack propagation Cc = 112MPa
plain strain fracture toughness = 27.0MPa
surface length of the crack = a
dimensionless parameter = Y.
Half length of the internal crack, a = length of surface crack/2 = 8.8/2 = 4.4mm = 4.4*10-³m
Also for 6.2mm length of surface crack;
Half length of the internal crack = length of surface crack/2 = 6.2/2 = 3.1mm = 3.1*10-³m
The dimensionless parameter
Cc = Kic/(Y*√pia*a)
Y = Kic/(Cc*√pia*a)
Y = 27/(112*√pia*4.4*10-³)
Y = 2.05
Now,
Cc = Kic/(Y*√pia*a)
Cc = 27/(2.05*√pia*3.1*10-³)
Cc = 135.78MPa
The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa
For more understanding, I have provided an attachment to the solution.
