A pipe made from a low alloy steel is subject to a maximum axial stress of 150 MPa. The pipe has an inner radius Ri = 200 mm and a wall thickness t = 30 mm. The fracture toughness of the steel is 50 MPam.
1) Assuming that a fully circumferential crack exist in the pipe
a) Plot how the stress intensity factor varies as a function of the depth of the crack
b) Calculate the maximum depth of a fully circumferential crack that could exist in the pipe prior to fracture. The stress intensity factor is defined by the following equation:
(1)
where is the applied stress, a is the crack depth and Y is the geometry/loading factor given in Figure 1 (overleaf).
2) Calculate the maximum safe life of this pipe for case a) and b) if stress corrosion cracking occurred.
KISCC = 20 MPam and da/dt = 10-8 mm/s.
a) An initial crack of 5 mm was present at start of life.
b) Repeat calculation for an initial crack depth of 3 mm.
3) Each day, the stress on the pipe is reduced to 90 MPa.
a) Using the stress intensity factor solution given in Fig. 1, calculate the maximum KI, minimum KI, and KI for the pipe assuming an initial crack depth a = 2 mm.
b) Use a spreadsheet to estimate the maximum safe life of this pipe. The fatigue crack growth rate follows the Paris law:
(2)
where da/dN is in m/cycle and KI is in MPam. In an air environment, the parameters C and m are equal to 7.34 x 10-9 and 3.07, respectively.