Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the qualitative difference between these two:

As seen on the table Typical yield and ultimate strengths.

I am trying to resolve the meaning of the phrase "contact yield stress" from C. Thorrton 1997 to real world values.

share|cite|improve this question
Could you provide a full citation for "C. Thorrton 1997" please? – deadly Feb 27 '13 at 13:36
@deadly… Thornton CC. Coefficient of Restitution for Collinear Collisions of Elastic-Perfectly Plastic Spheres. J. Appl. Mech.. 1997;64(2):383-386. doi:10.1115/1.2787319. – Mikhail Feb 27 '13 at 20:08
up vote 10 down vote accepted

If you look at a stress-strain diagram, the difference becomes clearer.

The initial slope is where stress is directly proportional to strain (like a spring) and the material behaves like this up to its elastic limit where it reaches its yield strength.

Beyond this the material deforms permanently (like an overstretched spring that won't return to its original shape). The material then becomes strain hardened until you reach the ultimate strength and necking starts to occur and the material becomes weaker again until it breaks apart.

Typical Stress vs. Strain diagram for a ductile material (e.g. Steel).

share|cite|improve this answer
The material does not become weaker after the UTS, but due to necking the engineering strain gets lower because it is defined as $\frac{F}{A_0}$, while the true strain has a higher strain due to a smaller area, caused by the necking. – fibonatic Jul 22 '13 at 11:36
I see what you're saying in that locally (at the neck) the material is under higher stress and so might be considered stronger, but the force required to stretch the overall material (or "sample") becomes less and so in practical terms it is weaker. – deadly Jul 22 '13 at 14:10

Yield stress is the stress at which that the material deforms permanently, ultimate tensile stress is the stress at which it breaks.

There is probably some official ISO/ASME definition of how much it has to deform for it to count as having yielded.

Materials first deform elastically - when you release the stress they return to their original shape, this is what a metal spring does.
Then with more force they deform plastically - when you release the stress they have permanently been stretched into a new shape, this is yield.
Finally they break, this is ultimately tensile stress, or breaking point

share|cite|improve this answer
Can you expound upon the difference between breaking and deformation, for materials such as steel? (i'll give you the check :-) ) – Mikhail Nov 8 '11 at 21:08
The yield stress is often defined as the stress at a 0.2% strain. Probably because this values can be determined with an higher accuracy than the stress at which the deformation will become plastic instead of elastic. I am not 100% sure about the reason for this though. – fibonatic Jul 22 '13 at 11:42
It's more correct to say that the ultimate tensile strength (UTS) is the stress at which a materiel fails (not breaks). Only a brittle material will break at the UTS. – deadly Jul 25 '13 at 9:49

The deformation amount is 0.2% to count as yielded for steel

Hard steels and non-ferrous metals do not have defined yield limit, therefore a stress, corresponding to a definite deformation (0.1% or 0.2%) is commonly used instead of yield limit. This stress is called proof stress or offset yield limit (offset yield strength):

$\sigma t= \frac{F_S}{S_0}$

share|cite|improve this answer

Ultimate Tensile Strength is a common engineering parameter when designing brittle material because there is no yield point. Whereas in designing ductile material, yield stress is a common parameter.

share|cite|improve this answer

Acutaly ultimate stress is maximum stress above which the material will break.

Yield stress is at which it deforms permanently means plastic deformation

share|cite|improve this answer

protected by Kyle Kanos Sep 29 '15 at 11:56

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.