1. Reference problem#

1.1. Geometry#

It is a semi-elliptical longitudinal crack opening into a narrow internal cavity plunged into a straight pipe. Symmetry planes are used in order to model only a quarter of the pipe.

_images/100002010000032800000224DC7A72A08238C33C.png

Figure 1.1-2: Defect geometry

The tube has a length of 2,000 mm, a thickness of 60 mm, and an internal radius of 270 mm.

The defect is 7.5 mm deep and 22.5 mm long.

1.2. Material properties#

A rational traction curve of the Ramberg-Osgood type is used.

\(\frac{ϵ}{{ϵ}_{0}}=\frac{\sigma }{{\sigma }_{0}}+\alpha \mathrm{.}{(\frac{\sigma }{{\sigma }_{0}})}^{n}\), with \({ϵ}_{0}=\frac{{\sigma }_{0}}{E}\)

Where:

\(n=6\)

\(\alpha =1\)

\({\sigma }_{0}=163\mathit{MPa}\)

\(E=174700\mathit{MPa}\)

\(\nu \mathrm{=}0.3\)

The law of material behavior is elasto-plastic (” VMIS_ISOT_TRAC “)

Below is its traction curve.

_images/10000201000008EB00000669AB1710074C1F0D82.png

Figure 1.2-1: Ramberg-Osgood traction curve n=6, a=1 provided by [bib1].

1.3. Boundary conditions and loads#

Symmetry with respect to the 2 main planes:

\({U}_{Y}=0.\) in the \(Y=0\) plan.

\({U}_{Z}=0.\) in the \(Z=0.\) plane out of the crack

Rigid mode on the \(X\) axis.

Average \({U}_{X}\) in nodes NODE_CENTRE_FISS_T and NODE_CENTRE_FISS_TM is zero.

The loading conditions are applied in two steps:

  1. A pressure load \(P\) is gradually applied to the inner pipe wall.

A tensile load \(\text{EFFET\_FOND}\) at the end of the pipe is also applied to model the background effect.

\(\text{EFFET\_FOND}=P\frac{{R}_{\text{int}}^{2}}{{R}_{\text{ext}}^{2}-{R}_{\text{int}}^{2}}\)

Until maximum pressure \(P=36.97\mathit{MPa}\)

  1. By staying with a maximum pressure applied to the pipe, a next bending moment \(M\) (\(-Y\)) applying pressure to open the semi-elliptical defect is applied progressively until \(M=-5989.460\mathit{kNm}\).

    It is applied to the end of the tube via a linear distribution following \(x\) of axial stresses \({\sigma }_{\mathit{zz}}\)

\({\sigma }_{\mathit{zz}}=\frac{Mx}{I}\) in the \(Z=2000\mathit{mm}\) plan

where \(I\) is the squared moment of bending. \(I=\pi \frac{({R}_{\text{ext}}^{4}-{R}_{\text{int}}^{4})}{4}\)

Below is a figure with the pipe loading path given by [bib1].

_images/100002010000031000000208AD583E5CF16C276A.png

Figure 1.3-1: load path provided by [bib1].