Hi. Sorry just to confirm. I'm using a metal with yield strength of 207Mpa, perfectly plastic. So once I get to yield strength the yield strength doesn't change does it ? Effectively once I get to the yield strength then that's also just about the UTS as because perfectly plastic it will break ?
Thank
If a material (I will assume metallic/ductile) if perfectly plastic, it means that flow/strain/elongation is happening throughout the material. It may or may not be uniform (by 'perfect' do you mean 'uniform'?). This carries on until local instability is reached; the point at which this starts I've always assumed to be a probabilistic problem. When the ultimate tensile strength is reached the yield point is in the distant past and no longer relevant. Long story short, YS can never equal UTS in a ductile alloy.
Of course on the atomic scale nothing is perfect; in fact it is a tangled mess (pun intended).
"Everyone is entitled to their own opinions, but they are not entitled to their own facts."
Well, i bet you are talking about stress-strain curve right?
if it is true stress-stain curve, the strength will keep increasing until rupture.
if it is engineering curve, the strength could decrease after yield, then increase (due to work hardening, reach to the UTS, then it could decrease again (due to necking) until rupture. Note rupture strength can be lower than UTS.
On both curves, the UTS will be larger than YS.
If material is brittle, there is no much strain/deform before rupture, YS is very close to UTS. Or, there is no YS at all if material breaks before experiencing the defined strain.
In stress-strain curve plastic generally just means non-elastic behavior and that permanent deformation is occurring. There are several types of plastic behavior. Most plastics (polymers) are strain strengthening due to molecular alignment increasing with strain. This material will stretch a lot before breaking. Non-strengthening materials like clay or putties neck and break relatively quickly.
So, perfectly plastic and brittle materials both have an UTS close to the yield strength.
There are some very plastic metals that are very ductile and start deforming under very light loads.
The stress strain curves have no straight initial section.
In these materials you have to cheat to report a yield or else you would report a value that is 5% of the UTS.
If there is no work hardening, no necking, and perfect plastic = irreversible deformation, then YS = UTS. but i can hardly imagine the material is ductile with such characteristics.
All it means is that the strength of the material stays constant after yield. In a test machine the strain rate is controlled, so the material just stretches and load decreases because of necking. In the real world where a part is put under load, the part will yield rapidly and basically flow until it breaks. This allows load to transfer to other parts without losing all the strength of the yielding part.
When an assumption of perfect-plasticity is implemented, the material is assumed to have bi-linear stress-strain curve: elastic followed by a zero-slope perfect-plasticity. In this type of material approximation, there is no UTS, per se. This approximation is always performed with a small-displacement assumption, to avoid the true stress-true strain issues mentioned above.
Are you using this in a pressure vessel analysis, perhaps?
I assume you're running a finite element simulation. All the discussion above about material behavior is good stuff, but I think it misses the point. An elastic perfectly plastic material model in an FEA simulation is an idealization, and you need to carefully consider if it is appropriate to use it in your simulation. Maybe it is a suitable estimation of the material's actual behavior, maybe not, or maybe only over a small strain range. It might make sense if you were trying to bound a problem, or maybe find a limit load for plastic collapse, but could be misleading if you were trying to predict an observed failure. Strictly speaking, UTS is meaningless for a perfectly plastic material model because there is no unique stress solution. Failure would be predicted by strain, not stress.
Rick Fischer
Principal Engineer
Argonne National Laboratory
