Stress Concentration Factor

[Cvan]

Hello, in a recent lecture on introductory engineering, our professor made reference to a stress concentration factor that pops up due to inconsistencies or changes in the shape of a material subject to a load.

The introduction he gave referenced a method of stopping cracks by drilling a hole in them, and this point didn't settle well with me. It seems counterintuitive--the only way I was able to reconcile with this idea is in--say a steel rectangular bar, that the hole's added area to the bar creates a larger surface for the crack to attempt to propagate along (the area I mean is (pi*diameter)(depth of material).

Is this an incorrect way of thinking about this situation? Or does it just mean that the stress concentration factor of a crack in a member under load is greater than one with a hole in it--and why?

[Astronuc]

Without knowing the specifics, e.g. geometry, I think the professor is implying that drilling a hole effectively blunts the crack, and the local stress in the material is redistributed over a large surface rather in the vicinity of the crack tip, which is usually very sharp.

[FredGarvin]

Without knowing the specifics, e.g. geometry, I think the professor is implying that drilling a hole effectively blunts the crack, and the local stress in the material is redistributed over a large surface rather in the vicinity of the crack tip, which is usually very sharp.That is exactly the point being made. That is a very common method for stopping/delaying crack propagation in aircraft structures (when in areas allowed) and is known as "stop drilling". The tip of a crack is an extremely complex area. The drill spot opens up that geometry and spreads the loading out over a larger area as Astronuc pointed out. Fatigue in the crack tip zone still is a point of concern though.

[handsomecat]

ok here's the mathematical reply.

If you have an elliptical crack in an infinite plate, the Stress Concentration factor theory says that the stress at the vicinity of the crack is SCF = 1 + 2a/b. So, as the crack goes sharper (a/b increases), the SCF increases. The idea then is to reduce the SCF by making it blunt.