# Stress Concentration Factor

Looking through texts such as Shigley's Mechanical Engineering Design, you come across tables of the stress concentration factor (defined as $$K=\sigma_{max}/\sigma_{avg}$$) such as the following:

This plot is sourced from Pilkey's Peterson's Stress Concentration Factors, which, unfortunately, I do not have access to (and even less so now, since interlibrary loans are on hold), so I do not know what underlies this plot. However, the line seems to be related to Kirsch's elastic stress distribution since the value of the stress concentration factor approaches 3 when the ratio r/d approaches zero.

That exact item, though, is what gets me. In the limit of radius going to zero, the bar no longer has a hole in it, and according to Saint-Venant's principle, the stress a a point in the middle of a long bar should be equal to the average stress on that cross section. Hence:

Why does the stress concentration factor go to 3 instead of 1 as r approaches 0?

There is a difference between a very small hole (even of radius zero, theoretically) and a solid bar.

The hole causes the axial stress along the line of its center to be zero, because it is normal to the free surface of the hole. That reduction in the local load-carrying capacity has to cause an increase somewhere else, and that is what causes the stress concentration factor to be greater than 1.

The limiting value of 3 is for any size of circular hole in an infinitely wide plate. For a finite width, the edges of the plate also affect the stress distribution and the stress concentration factor.

• OK - this makes sense. The ratio r/d -> zero is not necessarily the hole radius approaching zero, but the size of the material becoming infinitely large and the hole being infinitesimally small, relative to the bar, but still in existence. Is that correct? – Marius Apr 17 '20 at 15:14

Let's imagine a strap with a width A and let's assume the hole diameter is =A/2.

At initial configuration, the relative stiffness of the two parts of the strap are

$$(A-A/2) = (A/2)=1/2A \ \text{for two blades}=1/4 A\ \text{per blade}$$

As the hole diameter shrinks the relative stiffness of the two sides increases, and the strain contour line becomes flatter. meaning at the interface of the side of the hole and the two blades there is a sharper stress concentration. one could say because the strain has a lot of material as back up to maintain it.

Ultimately at a hole size hypothetically infinitely small but still existing, the width of each blade is $$= 1/2 A$$ and they are very stiff.

It may help to think of the way glazing people cut glass. they score it by a thin shallow line, stress concentration is so high there that by a small shake the glass breaks on the scoreline.

• Are the "blades" you mention the remaining material on each side of the hole? – Marius Apr 17 '20 at 15:13