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Stress-strain curves following ASME VIII div 2 Annex 3-D

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JonOl

Structural
Aug 23, 2021
5
Hi!

I have made a stress-strain curve for 245SMO material (S31254), which has S_y = 310MPa and S_u = 670MPa according to ASME II part D. My material datasheet tells me that the elongation at failure is 35% (engineering strain). Now, using the formulas in AMSE VIII Annex 3-D, I calculate true ultimate tensile stress to be 1042MPa, with a true ultimate strain of 41,8%. This corresponds to an engineering ultimate strain of 52%, way higher than my strain at failure!

If my calculations are correct, can anyone help me understand how this is conservative in any way?

Note: I'm using ASME VIII div 2 2019
 
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Can I get you to show your math, please. And these strains that you are reporting, are these plastic strains or total strains?
 
The value of m2 (= 0,75*(1-R) = 0,413), converted into engineering strain (e^(m2)-1 = 0,511) is much higher than strain at failure. m2 is defined as the true ultimate strain at true ultimate stress. Solving the equations, this is also roughly the strain value I get at true ultimate stress. The value of m2 seems to be way too high, considering the reported failure strain of 35%.

The strains I reported were total strain.
 
I find:
Sy/Su = 0.463 therefore m2=0.402
Therefore, Su,t=1002 MPa. e^(m2)-1 = 0.496.

It would be great to have some working.

The 35% strain is the guaranteed value from a material spec. Perhaps in reality you get a lot more, expecially with the decades of material refinement since that alloy was initially developed.
When you compare the yield stress on the material certificate to the guaranteed value in material specs, there is a 30-50% difference, especially for thinner plate.
 
Sorry, the value of S_u should be 690 MPa which is what I have used in my calculations.
 
According to the definitions, εt at Suts,t should equal m2. It doesn't.
εt equals m2 plus an additional true elastic strain at Suts,t added to it (Strue/E).
Perhaps the definition of m2 should be True Plastic Srain at Suts,t.
 
You're right, though the difference in total and plastic strain at uts is very small.

Regarding your previous post, I should accept the stress-strain curve even if the failure strain is 46% higher than reported in my material datasheet and assume that my material is stronger than reported?

Originally I adjusted the value of m2 down to 0,245 (70% of %EL, which was the necking point in another stainless steel we tested), as I'm not required to follow the code, just using it as a guidance. This gave me both lower strain and stress at the end of the curve, so I assume it is conservative. Looking back at it I see that it gives me lower strain at a given stress, hence it actually wouldn't be conservative considering the local failure criterion. Is this reasoning correct?
 
For SA-516 Gr70 I get an m2 of 0.274 and an Engineering strain of 0.32 at Suts,t.
The material spec quotes a minimum strain in 50mm plate of 21%.

It does seem odd that the material spec requires a certain minimum value, and the Div 2 plastic flow rule allows a considerably higher value in the analysis.
This does seem non-conservative.
 
For the plastic collapse load case this could be an issue if you had highly strained areas causing premature fracture before a cross section had developed its full strength.

But the local failure criteria prevents this from becoming an issue.

 
ASME VIII Div 2 Appendix D-3 uses the MPC (Materials Properties Council) Stress-Strain curve model to determine the strain hardening characteristics of the material used in design calculation. It's emperical, whereas % elongation reported on MTR is physical and actual, usually determined by ASTM A370 (North America). This is the reason for seeing the difference.

GDD
Canada
 
The empirical evidence used to develop the stress-strain curve will be physical evidence, some of it would be obtained by similar tests to the ASTM A370 elongation test (that goes on the material cert).

Still doesn't make a lot of sense.
 
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