Solution
The application of the LEFM to fatigue crack growth, through the range of stress intensity factor, varies from simple substitution into the relevant equations (see the theory card), to rather complex. The examples in this section aim to illustrate this variety. This first problem is straightforward and shows a typical application in failure analysis. It should take perhaps 20 minutes to complete.
a) A long pipe has an outer diameter (OD) of 90 mm, an inner diameter (ID) of 70 mm and works at a pressure (p) of 40 MPa. Valve failure downstream in the pipe caused a pressure surge which burst the pipe. Examination of the fracture surface revealed a metallurgical defect at the inner surface of the pipe which was semi-elliptical in shape with a depth of 1.6 mm and a surface length of 4.5 mm. This flaw was orientated perpendicular to the hoop stress in the pipe.
What pressure would have caused this failure?
Note that the formula for hoop stress from thin walled theory is p(ID)/2t, while from thick walled theory it is p[(L2 + 1)/(L2 - 1)] where t is the wall thickness of the pipe and L is the ratio OD/ID. Which formula would you use and why?
The plane strain fracture toughness of the pipe alloy is 25 MPa m½ and the geometry correction factor can be found from the graph below.
Assume that the geometry correction factor has a constant value of 0.7 in this second part of the question and note that a fatigue crack growth rate of 6.25x10-8 mm/cycle corresponds to an applied delta K value of 10 MPa m½. The Paris law exponent m is 4.
Answer: a) A pressure surge of 90.2 MPa would have caused the failure.
b) Nf = 23 723 cycles, therefore the pipe will last.
a) From the OD and ID we can obtain the wall thickness t as 10 mm, i.e. (90 - 70)/2 mm. The hoop stress from thin walled pressure vessel theory is:
This formula is strictly valid only for situations where ID > 10t. This is clearly not true in the present case, and we should really use thick walled pressure vessel theory. In thick walled pressure vessels, the hoop and radial stresses are not constant through the thickness, and the greatest absolute values occur at the inner wall of the vessel.Here:
Thin walled theory gives the average value of the hoop stress through the wall thickness, and it is easy to compare the difference between maximum and average values of hoop stress through the parameter:
The table below shows the difference for values of L from 1.01 to 2.0:
| L | 1.01 | 1.05 | 1.10 | 1.20 | 1.50 | 2.00 |
| S | 1.005 | 1.026 | 1.052 | 1.109 | 1.300 | 1.667 |
We need to add the internal pressure to this value in calculating the stress intensity factor, because K values will arise from the hoop stress and the internal pressure loading of the crack faces. Now:
We can find Y from the graph as a/c = 1.6/2.25 = 0.71 and a/t = 1.6/10 = 0.16. This gives Y approximately as 0.78. Thus at fracture:
Thus a pressure surge of 90.2 MPa would cause fracture of the pipe. If thin walled pressure vessel theory had been used the failure pressure would have been given as 100.5 MPa - a significant difference of 11.4%.
b) To integrate the Paris 'law' to obtain the fatigue life, we need to establish the integration limits in terms of crack size. We are given the initial flaw size as 1.5 mm and we can find the final size causing fracture by substituting into the K equation, the fracture toughness and applied stress arising from the applied load range. The applied load range goes from zero to 40 MPa, i.e. a stress ratio R = 0, hence the peak stress will correspond to the 40 MPa.
The equation above for hoop stress in a thick cylinder indicates that its value at the inner wall is 4.01p, i.e. 160.4 MPa, and to get the total applied stress intensity factor we need to add in the internal pressure (which loads the crack surfaces). This gives a total stress of 200.4 MPa at peak load, hence:
Note that this is, in fact, slightly larger than the wall thickness, so we should limit the final length to 10 mm. The last piece of information we need is the constant C in the Paris law. We can this from the growth rate information given:
We can now separate out the variables in the Paris law and integrate the equation between the crack growth limits.
As 30 years is equivalent to 30 x 365 = 10 950 days (excluding leap years!), the pipe should last the required lifetime.
As many factors can accelerate growth rates, however, it would be prudent to inspect this component at regular intervals. These inspection intervals can be set with reference to the life integration. A curve of a versus N is generated from integration to various lengths. This is an exponential curve, and converting N to time (which can easily be done seeing the frequency is 1 cycle per day) allows inspection intervals to be chosen such that the crack would not become critical in the interval between inspections. In the present case, there is also an initial period of time when inspection is not required.
