Solution (Problem 6)
This question should take about 15 minutes to complete. It illustrates the difference between design for 'failure' using the traditional solid mechanics approach of yield criteria, and the fracture mechanics criterion of energetically favourable fast fracture.
A cylindrical pressure vessel, with a diameter of 6.1 m and a
wall thickness of 25.4 mm, underwent catastrophic fracture when the internal pressure reached 17.5 MPa.
The steel of the pressure vessel had E = 210 GPa, a yield strength of 2450 MPa a value of GC = 131 kJ/m2.
a) Show that failure would not have been expected if the von Mises yield criterion had served for design purposes:
b) Based on Griffith's analysis determine the size of crack that might have caused this failure, stating assumptions that you have made.
Answer: 2.2 mm
There are a couple of assumptions to be made here. The first relates to fabrication and the orientation of the fatal defect. It is likely that this vessel is made from welded plates with welds running perpendicular to both hoop and longitudinal stresses. As the hoop stress is the maximum principal
stress, one should assume that the defect is perpendicular to this stress direction. Secondly, although Poisson's ratio is not given here, 25.4 mm plate is quite thick and the stress state is likely to be close to plane strain. Hence one should take a value of Poisson's ratio as 0.3 and assume plane strain conditions apply.
a) We need to find all three principal stresses to substitute into von Mises criterion; using thin walled pressure vessel theory we get:
and substituting these values into the von Mises yield criterion gives:
hence meeting the criterion, thus failure would NOT have been expected on the basis of yield.
b) To determine the likely critical crack size that caused failure, all we need to do is substitute into the Griffith formula:
Hence:
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