I am digging into the field of non-linear static analysis.
As i’m not a specialist for material problems i need some advice from expierienced users regarding the use of plastic / elastic material model.
I’v read some posts, in this forum and on a blogs to the topic: How to derive plastic strain / stress data that can be used as Material model for FEM. PreProMax Post Blog Post
I have followed the procedure of both authors and have come to different conclusions.
When i compare the stress/strain curve from Datsko it seems, that it is still engineering stress/strain because the stress/strain data i derived from the Blog Post (H.N.Hill Equation) is true plastic strain vs true stress (see diagram above).
My questions are:
Do i need to transform the stress/strain from the Datsko approach to true stress/plastic strain that i can use it in a plastic material model in ccx?
Is the approach from the Blog Post (H.N. Hill Equation) + transformation to true plastic strain / true stress suitful for a plastic material model in ccx?
Why not. As long as you get reasonable true stress vs true plastic strain data, you can test it with CalculiX (possibly on a simple, maybe even single element model).
I have calculated true plastic strain from the Datsko approach.
It is showing a quite different charakteristic compared to N.H.Hill equation.
Datsko approach seems to be more conservative than N.H.Hill approach.
Maybe you could find full experimental stress-strain curve for your material in literature. Those approximate approaches may indeed differ significantly.
The first plastic data point should be yield strength vs 0 plastic strain. Abaqus enforces that, CalculiX may not throw an error but it should still be done this way.
Maybe I’m understanding incorrectly Mr. Datsko notes, or I read wrong your results, but I see some unexpected things.
According to Mr. Datsko notes there are some points that could be anticipated on the graph and can help to check if your curve is correct:
1-For a True plastic Strain of 1 (ew=1) —> Syw=Sigma0
Where ew=true plastic strain.
According to Datsko Sigma0=158033 Psi = 1090 MPa
Your last graph is showing you reach aprox 1100MPa>1090Mpa @ 0.09 True Plastic Strain. That’s seems too soon.
2-The instability point or maximum load condition is reached at , True plastic Strain 𝜀# = m
Su is an ultimate Tensile Strength (Engineering value) so you need the Engineering curve to check that against True Plastic Strain.
According to Datsko m=0.0371
3-Your Engineering curve (blue Datsko eng) doesn’t show instability point.
That’s a point in which the enginieering curve starts decreasing. A small amount of additional stress beyond that value results in infinite strain.
It helps a lot to plot clearly which Stress and Strain measure are you considering at every moment. Engineering or True. Plot up to ew=1 and both curves on the same graph.
Ultimate Tensile Strength is the point in which Global Strains become local in the tensile test (Necking begins). Point (3). It is an unstable Point. If your model is force driven, it will collapse.
Maximum elongation is associated to the Strain at the Fracture point (5)
I don’t understand why the blog Post is correlating those two values as if that were a point. Sound weird to me and it is expected both curves end up different. I’m not saying it’s wrong. Maybe it is a conservative hypothesis
Mr.Datsko imposes Instability point is a maximum (Derivative=0) to determine the curve we are searching for.
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I would like to post some comments to clarify the approach to model plasticity.
The Stress - Strain relationship for strain hardening materials first proposed by Hollomon is : The most commonly used constitutive equation is the Hollomon equation, which has the form of a power-law relation between true stress σ and strain ε; σ = K ε^ n , where K is the strength coefficient and n is the strain-hardening exponent (Dieter, 1961, Hollomon, 1945, Kleemola and Nieminen, 1974). (Constitutive equations optimized for determining strengths of metallic alloys - ScienceDirect)
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What my professor Datsko did was rename K to Sigma zero, and n to m. He then worked out the coefficients that need to be obtained from a tension test. His work described in my post allowed the calculation of the strain hardening exponent “n” from the relationships between the Yield Stress (Sy) and the Tensile Strength (Su). Datsko’s correlation Su/Sy = (184.m)^m
Datsko indicated that this relationship is valid for calculated values of m <=0.05.
The wrong interpretation occurs when it is not considered that the true stress-true strain behavior of a metal is the locust of the yield stress, that is, is a material is deformed to a true plastic strain Ew, when it was subjected to a true stress Sw, then its cross sectional area is reduced to Aw. Then if the stress is released and a new tensile stress performed, the initial cross sectional area for the new tensile stress is Aw, and if pulled in tension, its new Yield Stress will be Sw.
Therefore, in the equation Sw=K(Ew)^n represents the true stress-true strain work hardening equation of the material.
Calculix requires that the initial Yield stress will corresponds to a plastic deformation of 0.00, what I do is I assign the calculated Yield Stress at 0.002 to equal to the Initial point of the plasticity data, i.e., assign it the value 0.00. Since the true Yield stress for all the calculated values are used the implicit error is minimal.
That is not correct. You are missing the point:
The true fracture strain is Efracture = Ln (Cross sectional area initial/Cross Sectional Area at fracture). The deformation based on elongation is not valid when the Ultimate load is reached.
The true strain is always defined in a tension test as Ew= Ln (Ao/Ainst)
Constancy of volume results in Ao.Lo = Ainst.Linst or (Ao/Ainst)=(Linst/Lo)
However, after the maximum load is reached, the deformation concentrates in the neck formed and the deformation is not longer uniform
