# Increasing Yield Strength with Plastic Deformation

I'm trying to tackle a mechanics problem to be solved using a finite-element analysis, but I'm stuck on one particular bit of theory first.

I've done analysis using a linear-eleastic model, and elasto-plasticity with constant yield stress.

Now I've got to do elasto-plasticity with increasing yield stress. I've been given a table of values to use with this:

plastic strain ￼       0.00  0.01  0.02  0.03  0.04  0.05  0.06
yield stress  (MPa) ￼  200   210   220   225   230   235   240


I don't understand the concept of an 'increasing yield stress', I've only ever come across a material with a single yield point - where it yields...

If anyone can shed some enlightenment that would be great. Happy to provide more info if needed.

• "Increasing yield stress" is not a term I've come across before. By definition, yield stress is the stress at which elastic behaviour ends and plastic behaviour starts. When you say "elasto-plasticity with constant yield", I suspect you mean elastic-perfectly plastic, where once you have hit yield, unlimited further strain happens with no increase in stress. Increasing stress with increasing strain (see graphs in @hazzey's answer) is just "stress" not "yield stress". Feb 23, 2015 at 13:40
• I agree, but this is definitely how it's worded in the question. Thanks for your input. Feb 23, 2015 at 17:48

The yield point of a material is determined by looking at its stress-strain graph.

As you can see from the image, some materials do not exhibit an obvious point where the material starts yielding. When this is the case, the "yield point" is determined by the 0.2% offset rule. This is shown by the dashed line in the graph. As you can see from the graph, even after this yield point, the stress still increases as the strain increases. This is the situation that you have described in your question.

This increase in stress will continue until the material breaks at its tensile strength. The graph below shows a material with an obvious yield point as can be seen from the first peak on the graph. This material also shows an increase in stress as the strain increases even past its yield point.

If one of these materials experiences stress past its yield point, it will experience some elastic shortening, but it will not go back to its original length.

• It's worth pointing out that on the second figure, the graph A (red) is the engineering or apparent stress/strain curve for whatever material, and the graph B (blue) is the true stress/strain curve. The difference is that A is expressed in terms of the original area/length measurement (stress = F/A_0), and B is expressed as an instantaneous measurement of area, which decreases as strain increases in the plastic regime, hence the curve rises. Feb 22, 2015 at 19:37
• How to tie @hazzey's graphs to OP's data: Yield is the end of the initial straight line and at this point plastic strain is zero. This will be 200MPa with the OP's data. stress = Young's modulus * strain in the elastic region (before yield). You can then work out the strain at the yield point. Beyond this point, you use the rest of the data. You can plot a graph of stress against total strain (= strain at yield point + plastic strain). Feb 23, 2015 at 13:43

From a materials science and engineering stand-point, materials which exhibit work-hardening change their structure at lattice and microstructural scales as plastic deformation proceeds in such a way as to resist further deformation. Virtually all metals exhibit work hardening, and thermoplastic polymers may also exhibit work-hardening (or work-softening) depending on temperature, strain-rate, and polymer chain structure and organization.

# Metals

For metals, the general theory holds that plastic deformation occurs due to the motion of crystal lattice dislocations through the material. Dislocations and other lattice defects put the surrounding lattice into various states of stress. As a dislocation moves through the lattice, it may come near another defect, and the stress fields surrounding them will interact. If the stress fields increase as a result of moving the two defects closer together, it takes increasing effort to move the dislocation. In other words, the local lattice yield stress increases with increasing strain.

For constant environmental conditions, the non-dislocation defects have constant concentration, and this only explains the existence of a yield point. However, a number of proposed, and generally accepted, mechanisms exist for building up the concentration of dislocations within the metal as plastic deformation proceeds. The mechanisms include dislocation pileups on inclusions and grain boundaries, and dislocation sources pumping new dislocations into grains with increasing strain. These mechanisms increase the concentration of dislocations in metal grains, making further strain increasingly difficult. The result at large scales is work-hardening: as strain increases, stress needed to cause continued flow increases. Additionally, strain-to-failure decreases.

Metals which undergo work-hardening have increased yield strengths, but only at low homologous temperatures. Homologous temperature is current absolute temperature divided by absolute melting temperature. An increase in work-hardening is associated with an increased level of energy storage in the material, which means an increase in free energy. Work-hardening, therefore, is unstable. Given enough time, or increased temperatures, dislocations will move in such a way as to reduce strain energy, which will reduce the yield strength toward the original, un-strained material strength. Exceedingly pure metals, and metals at sufficiently high homologous temperatures, can recover their yield strength as rapidly as they are deformed. As a result, there are a continuum of heat treatments available to control the properties of metals.

It is worth noting that work-hardening effects are partly anisotropic. I am unsure of the precise microstructural explanations for this, but I imagine it has to do with grains changing shape, different dislocation concentrations along different axes, and dislocation pileup occuring anisotropically. Uniaxial strain will harden a metal to a greater degree in the direction of strain, such as in wire drawing. Biaxial strain will usually harden a metal to a greater degree in the plane of forming than in the thickness direction, as with sheet-forming. It is worth noting that if the strain is tensile, the compressive yield strength of the material in that direction will decrease as the tensile yield strength increases, due to the Bauschinger effect.

The above applies to face-centered cubic and body-centered cubic metals generally. Hexagonal close-packed metals do exhibit work-hardening, but tend to be more brittle, so be careful modeling HCP metals. You may find this article useful in understanding everything going on.

# Polymers

Thermoplastic polymers I know less about, but my understanding is that at a microstructural level the chains are randomly oriented and tangled in amorphous polymers, and in crystalline polymers the chains are laterally organized in crystalline spherulites interspersed by amorphous regions. If the polymer is strained slowly enough, or at high enough temperature, the chains have time to reorganize as they are strained. Generally, the chains will tend to organize in order to maximize the stress on each individual chain, so that the chains align length-wise along the direction of stress. As the fraction of chains aligned with the stress direction increases, the strength of the polymer increases. This process, referred to as polymer chain orientation, is exploited during blow molding of PET soft drink bottles to increase the local in-plane mechanical properties of the bottles. There is also an added benefit of decreasing carbon dioxide permeability in the thickness direction because there is less room between individual chains for gas to pass through. Less free volume in polymers is associated with less ductility and a tendency toward brittle fracture, among other things.

Work-hardening in both thermoplastics and metals increases strength in the direction of stress and decreases further ductility, but there are a number of differences between them. First, the change in strength for thermoplastics is entirely anisotropic, and polymers weaken in directions other than the stress directions, whether uniaxial or biaxial. For metals, due to the different microstructural mechanisms, there is some increase in yield strength in non-stress directions. Second, dislocation density may be reduced via heat treatments without melting the metal, but to remove full polymer orientation virtually requires melting the polymer. Third, the exact chain structure -- i.e. length, side-groups, branching, tacticity, copolymerization -- of a given thermoplastic plays a role in how orientable the chains are, as well as virtually all of its other properties. With metals grain size, concentration of inclusions, and previous working history are the major factors in work hardenability.