Brochure

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 MPam.

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 MPam 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 MPam. In an air environment, the parameters C and m are equal to 7.34 x 10-9 and 3.07, respectively.