Why is plastic deformation a non-linearity if I am using Bilinear hardening to model it in FEA?

Attached you may find a picture for isotropic bilinear hardening up until the ultimate tensile strength. The first line is the elastic region while the second one is plastic. I couldn't understand that why is plasticity even a non-linearity in FEA? I mean if I just consider the graph below, then a point in a FEA model will just move over this graph only. If it switches from first line to the other, then there is just a change in the elastic modulus. If I am conducting a geometrically linear analysis, then how could inputting a material plasticity model, like the one showed below, make the analysis still be non-linear? • Err its clearly nonlinear on account of it having two line segments. Everything that is not a single line in its whole domain is not linear. You need a whole different class of solvers to solve nonlinear equations because you now need to keep track of the actual path that was taken for the solution whereas for a linear solution is guaranteed to be a simple solution. You can just disregard the path and jump directly to the solution. Jul 6 '21 at 19:30
• "LInear" means "if you double the loads, you get exactly twice the displacements". It should be obvious this is not true for your stress-strain curve. Also, in a structure with this type of material behavior and redundant load paths, there may not be a unique solution for every possible set of applied loads. Jul 7 '21 at 9:39
• Understood. Makes sense. Thanks, to both of you. Jul 7 '21 at 12:59
• I am curious about where you get this diagram, can you please provide the source? Thanks.
– r13
Jul 7 '21 at 17:08

In the plastic range, the stress-strain relationship is non-linear as shown in the graph below. With consideration of geometric changes after yield, the true stress-strain curve (dotted line in the graph) shall be used instead of the normal curve, and non-linearity needs to be considered in the analysis. Note: Due to the shrinking of section area and the ignored effect of developed elongation to further elongation, true stress and strain are different from engineering stress and strain.

$$\delta_t = \delta(1 + \epsilon)$$, and

$$\epsilon_t = ln(1 + \epsilon)$$

• The question has nothing to do with plastic behaviour. The stress-strain curve in the OP's example is completely reversible. A real material showing a similar effect, but with the graph curving the other way, is rubber. Jul 7 '21 at 9:37