Practical Fracture Mechanics Part I

by | 10.08.2024 | Case study

Let me start by pointing out that this is a purely theoretical example because it is difficult to imagine the existence of pressure in a thin-walled object that has a gap running through the entire thickness of the wall.

There is nothing more practical than a good theory!

Suppose we have a pipe or cylindrical pressure vessel that has a slot that is inclined to make it more difficult at a certain angle. To determine if the component can work, the stress intensity factor (SIF) must be determined and compared to the critical value for the material. The pressure is constant and there are no vibrations.

For stresses, the typical formulas apply:

The formula for a crack in a shell undergoing biaxial tension can be obtained from fracture mechanics tables. The coordinate system is attached at the center of the crack.

After substituting the shell stress formulas into the stress intensity factor formulas and using simple trigonometric transformations, we can finally write that:

Below is the calculation for a DN 250 pipe (273 x 10 mm) under pressure of 2 MPa.

Regarding the temptation to obtain K-factors by simulation. In Ansys at the current stage of its development there is a certain limitation, namely the crack must be perpendicular to the surface. In addition, a local coordinate system for the crack must be created, in which the X-axis is directed towards the crack axis. This can be cleverly achieved by selecting "Hit Point Normal" and clicking on the surface of the pipe.

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