ASTM D695 question about how to apply offset yield strength. 2

[KirbyWan]

Howdy all,

I've run a compression test per ASTM D695 and section 3.2.11 indicates the "offset compressive yield strength -- the stress at which the stress-strain curve departs from linearity by a specified percent of deformation." For metals the way this is done (I think) is they take the slope of the elastic region, and shift it to the right by .02 strain. Should I be using .02 strain for plastic as well? It's softer and deforms easier, maybe I should have a higher strain offset. When I look at some TDS for plastic materials it lists "Compressive properties 10% strain" Would that be shifting the slope of the elastic region to the right by .1 strain and then seeing where it hits the stress strain curve or just going up from .1 strain to get the stress? I've included a graph of the data for the material we're characterizing. I've included a picture of the stress-strain curve with the slope of the elastic region shifted by .1 strain.

Thanks,

-Kirby

Kirby Wilkerson

Remember, first define the problem, then solve it.

 

[SWComposites]

Yes, you shift the elastic slope line as shown in your plot.

 

[KirbyWan]

SWComposites,

So offset 'compressive yield strength' and 'compressive properties 10% strain' mean the same thing? In the second one they are just indicating the strain offset used? I guess what's throwing me off is that at .1 strain the stress-strain curve is fully in the plastic region and that sets the strain offset farther away then I would use from the ASTM standards "departs from linearity...". But if that's what everyone does, I guess it's still useful for comparing property data.

Thanks for your response.

-Kirby

Kirby Wilkerson

Remember, first define the problem, then solve it.

 

[SWComposites]

Those two properies are calculated with the same approach - line offset from intial slope. However 10% strain point is not “yield”, more like an ultimate value.

 

[EdStainless]

In most plastics there is no linear region to start with.
And in many plastics you can stress them well into the apparent yield range and they will still rebound nearly all of the way.
We used to report 1%, 5%, and/or 10% for plastics depending on the material and application.

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P.E. Metallurgy, Plymouth Tube

 

[KirbyWan]

EdStainless,

If there was no linear elastic region, did you just get the stress at the 1%, 5% and 10% strain since there was no slope to follow?

For the graph I provided above If I just go up from the .1 strain (actually a strain of .103 to account for the toe of the data) I get a 10% strain of 4302 psi. If I follow the slope of the linear elastic region, I cross that curve at closer to a strain of .162 which gives a stress at 10% offset strain of 4947 psi.

I did a comparison of a plastic I could find a value for Compressive Properties 10% Strain (Starboard ST if you're interested) and the TDS indicated a value of 4790 psi. My test got an average stress at .1 strain of 4701 psi, that's within 2% so I feel like my testing process is reasonably valid, but if I take the strain following the slope of the linear elastic region (it has a linear region) I would be getting closer to 5100 psi or close to 10% high.

Based on this discussion I'm unsure of the correct process. From my testing I'm going to use the stress value at .1 strain, correcting the zero point of the strain for the toe of the data but not following the slope since this seems close to the data I'm getting and is more conservative.

Thanks for your answer.

-Kirby

Kirby Wilkerson

Remember, first define the problem, then solve it.

 

[EdStainless]

We used a slope based on the presumed modulus of the material.

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P.E. Metallurgy, Plymouth Tube