I agree with Sergey and Muhammad. With the Finite Element Method, plastic strains are computed at integration points by solving constitutive equations of elastoplasticity. If the values calculated at integration points are positive or zero, the fact that you find negative values at nodes comes from the extrapolation technique.

1) In FEA all derivatives, including strain, are usually computed at integration points and can be extrapolated to nodes. 2) Plastic strain in FEA is usually computed by subtracting from total strain elastic one.

So, I think that the problem may reside in computational errors associated with 1) and/or 2)

If its a small (negative) value then it can occur dur to extrapolation.

I agree with Sergey and Muhammad. With the Finite Element Method, plastic strains are computed at integration points by solving constitutive equations of elastoplasticity. If the values calculated at integration points are positive or zero, the fact that you find negative values at nodes comes from the extrapolation technique.

Hi David,

Do you know how Abaqus does the extrapolation? Does it use the same shape functions as for approximating the displacements? If so, which elements did you use? For linear elements, the shape functions are non-negative, and if the integration point plastic strains are positive, you cannot get negative nodal values.

Bests,

Phu

If you are using gauss integration, integration points are placed in the interior of the element. If a gauss point (A) has a positive value, and another (B) has a zero value, the extrapolated value at the node near the (B) gauss point will be less than zero, while the one near the (A) gauss point will be positive and larger than the value in the gauss point.

Anyway you should not warry about the values (of variables such as plastic strain) at nodes, they are extrapolated just for visualization purposes.

By definition of equivalent plastic strain, it cant be negative so dont worry at all about negative values on nodes. I see these all the time but the values are very small so i dont bother. If you plot element results you will not see the negative values.

1) Extrapolation from integration points to a node involves adjacent to the node elements; 

2) Equivalent plastic strains (NE and LE) can be negative, there is no restriction for their sign

Dear Sergey. Can you explain NE and LE? Generally the strain hardening (von mises strain) or work hardening criterion are used to calculate eqivalent plastic strain. With these criterion one cant get negative equivant plastic strain.

Btw what is the the physical explanation of negative eq. Plastic strain?

Dear Muhammad,

1) NE is Nominal strain, while LE is Logarithmic strain. NE is preferrable, as it is more consistent with true strain at large deformations, both LE and NE are referred to total strain.

2) See this simple sketch for uniaxial loading-unloading of the isotropic elastic-plastic material (epsilon_t is total strain, epsilon_e is elastic strain, and epsilon_pl is plastic strain). All strains finally become negative, while at the first two stages of active loading they were positive.

Dear Sergey. I completely agree that plastic strain can be negative but we are talking about EQUIVALENT plastic strain which cant be negative. Equivalent plastic strain is a scalar quantity which represent the plastic strain. Similar to equivalent stress (which can also not be negative). Even if the normal or shear components of the stress or strain tensor are negative, the equivalent is always positive.

Hi David,

The plastic strain at Gauss points of an element can be positive while nodal plastic strain of the element is negative.

It is because the nodal value depends on the plastic strain values from the adjacent elements that share the same node.

The nodal plastic strain of a FEA model is calculated as follows:

 (1) Start the calculation from node #1.

 (2) Determine the elements that share the node in question.

 (3) For each adjacent element found in (1), determine the closest Gauss point to the node in question.

 (4) Calculate the nodal plastic strain by averaging plastic strain values at Gauss points found in (2).

 (5) Repeat (2) to (4) for the other nodes.

Check if the plastic strain at Gauss points from adjacent elements is negative.

Besides the above-mentioned explanations, your question certainly regards the process involved in interpolating from nodes to Gauss points. With most finite elements, interpolation is of Lagrange type because interpolation is based on uniformly spaced nodes. If you use linear elements (T3 or Q4 in 2D, T6 or H8 in 3D, etc.), interpolation generally causes no surprise, as it involves uniform increase or uniform decrease; while with quadratic or higher degree interpolants, you are almost always at risk of overshoots and meandering behavior.

Dear David,

Most probably, it is due to

- Interpolation of strain from nodal displacements to integration points. This is usually less problematic as it is an "interpolation"

- "extrapolation" and "averaging" of eq.plastic strain to nodal positions. In most displacement based FE code, strains are calculated at integration points. Then, there is no forced continuity condition on strain and stresses. Mostly for post-processing, those results are extrapolated to element nodes. As nodes are shared by neighbor elements, those results are also "averaged" (each element will give a different extrapolation result). 

The things above may lead negative cumulative plastic strains (or other non-physical results). If you don`t do any other calculation on nodal strains and if their magnitude is rather small , you can ignore them. Notice that surface elements are more prone to this problem as they don`t have neighbors in normal direction ! 

If the magnitude is not so small, this may indicate that you should refine your mesh in that region to decrease this effect. You may try to use quadratic elements (considering other penalties with them). But, in some FE software this won`t change anything because for visualisation they use linear interpolation and averaging. So the plot may not look good but if you probe nodal values, they should be improved.

Even if the negative magnitude is small, If you are performing something on nodal results you should eliminate this ! Otherwise your routine may crash or give non-sense results. For example, in many cases you need to take square root of something, or log of something. Negative number (does not matter how small it is) would cause a problem.