effect of local stress concentrations on the strength 2

[Trajano]

Good morning/evening, Sirs

I am thinking of the effect of local stress concentrations on the strength of the metal parts. In order to clarify my ideas, I have thought of a simple test:

Let's imagine we have a test specimen A (made of aluminium, or steel) with a rectangular section of width "a" and thickness "t". The length is "l".
In the center of the length there are two notches, depth "n", therefore the minimum net section is b = (a-2n)*t.

Let's have a second test specimen B (same material), with a rectangular secion of width b = a-2n and thickness "t". The length is also "l". No notches this time.
Therefore the minimum net section in both specimes is the same

1. If we did a tension testing, what would happen? Both specimens would fail at the same load level, or the stress concentration around the notches would induce an earlier failure in Specime A?
2. Do you think the relative effect of the notches depends on the ratio b/a?

What do you think?

 

[rb1957]

i think that specimen A (with notches) will fail at a slightly smaller load than B (the narrow specimen).

at some small load the notched specimen will have already yielded some material near the notches (due to Kt); at the same load the narrow specimen is all below yield.

another day in paradise, or is paradise one day closer ?

 

[Dougt115]

This is a simple notched bar analysis. Every engineering design book has the chart for the notch/width to Kt relationship.

For any notched bar Kt>1.

Meaning the notched bar will always fail at a lower load if the bars are otherwise perfect.

 

[tbuelna]

Take a look at Peterson's Stress Concentration Factors.

As others noted, the notched specimen described will fail at a lower tension load than the un-notched specimen. The impact of the stress concentration(s) depend greatly on factors like the notch geometry, ratio of a/b, etc. How the tension load is applied also matters. Is this like a fatigue test where the max load applied produces stress below yield, and the load is cycled from zero to max repeatedly until failure occurs? Or is it like a simple tension test where the load is applied once, increasing gradually until failure occurs?

While the OP did not ask about this, in practice it is common to offset the effect of elevated tensile stress at stress concentration locations by mechanically working the surface material (using a process like shot peen or rolling) to impart a residual compressive stress. This approach is especially effective for applications where fatigue is a concern, such as the shank/head fillet of bolts or the root fillets of gear teeth.

 

[Trajano]

Thank you for all your answers. I am prone to think that things would happen as rb1957 describes. "A (notched) will fail at a slightly smaller load than B".

Dougt115 and tbuelna: I do not think it is as simple as you describe it. If we just applied the Kt factor from Peterson to the nominal stress, we would reach the (wrong) conclussion that the notched specimenn fails much earlier than the unocthed one, because these are ELASTIC srtress concentration factors. When we are close to the failure, the metal is in the plastic range. The material around the notch will start yielding before the rest of the specimen and therefore the load will be redistributed to the adjacent areas. If you want to regard the effect in terms of Kt, we could say that the Kt (stress) is reduced due to the pasticity.
And here comes my question: The load redsitribution due to the palsticity, is "perfect", i.e. will it result in a perfectly uniform stress throughout the section or still there will be a higher stress around the notch, that will induce an early failure?

PS: @tbuelna, I am talking about a (quasi-)static strength test, not fatigue

 

[RPstress]

This Hawker-Siddeley data is relevant. Max reduction in strength is 0.8 times for 2014-T3. Usually it's above 0.9. I don't have any hard data for steel or titanium.

 

[Dougt115]

Think of it in terms of the energy required to deform the specimen or energy stored per volume.

The weakest part of a notched specimen is at the notch. Once a load is applied the stresses will be maximum at the notch and so will the deformations. Now think of the deformation as the distribution of energy to deform the specimen.
The notch cross sectional area is small and the area of deformation is small too relative to the specimen as a whole. Therefore most of the energy will focus in the notch and a smaller amount to the adjacent areas.

If a specimen is uniform then the stresses will be uniform and the deformation will be uniform. In terms of energy to deform it will be distributed over the entirety, a greater volume.

The notch focuses the area that can store the energy of deformation to a smaller area. The greater the notch the smaller the area and the smaller the load required to break it.

 

[3DDave]

If there is plastic deformation it will start at the notches, while the center of the part will remain elastic because the stress is not uniform. By the time the full width is yielding, the edges will have greater strain than the center. For a given stretch of the entire bar the notched bar will always have more damage done near the notches than a uniform section will have.

 

[Trajano]

RPstress, thank you for the document you left. It answers my question.
Thank you to everybody else for your contributions.