# Nomenclature on Strain

1. Jan 19, 2006

### Cyrus

This kind of had me puzzled for the last week. If you apply a tensile stress on a body, it will elongate, and get a little bit thinner. The amount of stretch will be a function of the stresses. But then I started wondering, "Well, if that’s the case, then it will stretch and get a little thinner. But then, I can say there is a stress still (after it got thinner), and I could repeat my logic using the strain equation based on the stress, over and over again, until this thing becomes thin as a wire." That’s clearly not what’s happening. I think, to be EXTRA precise, the book should say that the strain is a result of a SUDDENLY APPLIED load. You apply a load, it will have a strain. But after that strain, the load is no longer suddenly applied, but steadily applied. So now your strain will be zero, because there is no NEW applied load. In short, strain is a result of a load being suddenly applied or an already applied load changing in magnitude will result in a change in the strain. Because to not say that, would infer that you could continue to use the strain equation to as many times as you like to (because hey, there will always be stresses present right?) , but that’s obviously not correct. (All this assuming the elastic region, of course).

Last edited: Jan 19, 2006
2. Jan 19, 2006

### Homer Simpson

I dont totally get what you are after. When looking at a stress/strain diagram there is a given amount of elongation (strain) per unit of applied stress. If the stress is removed, while in the elastic region, the sample will return to original length. Once into the plastic region, the graph falls apart (usually)

3. Jan 19, 2006

### Homer Simpson

Last edited: Jan 20, 2006
4. Jan 19, 2006

### FredGarvin

Poisson's ratio is a function of strain, not stress.

That would be taking the material well out of the elastic range which is a requirement for Poisson's ratio to hold true. But honestly, it is really difficult to understand what you are getting at here.

The calculation of strain has nothing to do with applied load. As long as there is a deflection of the object under load, there will be a strain. Also, in the elastic range, the part experiencing the load will always be exhibiting a spring like effect and creating an opposing force. As long as you are in the elastic range and have a deflection, you will have to keep applying a force to maintain that displacement. Perhaps I am really reading what you are trying to get at wrong. I am failing to see the distinction you are trying to make by saying that the load has to be suddenly applied. If I have a test specimen and I take 10 days to apply a load or 10 seconds, the result should be pretty much the same. I can think of some things to complicate that last statement, but for the purposes of this thread, I don't think they would apply.

5. Jan 20, 2006

### PerennialII

Strain measures being proportional to gradient of displacement .... yeah, makes sense to me. Thinking about balance equations otherwise 'may' give some understanding about how different systems respond to the "applied"-part.

6. Jan 20, 2006

### FredGarvin

I re-read this thread again this morning and it's still blurry to me. Perennial picked up on it, so I am sure that I am missing something.

7. Jan 22, 2006

### PerennialII

... It think cyrus presented his case in a bit "home-spun" sense, but in principle I think got to the same "plane" as his presentation.
I'm quite not sure whether you're after finite deformations or the system reaching an energy balance, but essentially the system will reach a state of balance (with the constraints of linear-elasticity etc., i.e. limiting everything 'fancy' off) and it will "stop" (in an equilibrium of stress-strain & whatever loading is applied).
The strain results from a change in the configuration, the difference between "suddenly" & "steadily" I'd say is a matter of perspective, and in principle insignificant the deformation process being continuous in the "applied" - scale. What I said above still applies, even if you account for "thinnening" due to stress and use a complex (finite deformation) presentation of strain, there will not be an infinite cycle of thinnening, but a state of equilibrium (I think whether you're thinking about this 'incrementally' or 'continuously' [the response and generation of strains to applied stress] is the source of confusion).

8. Jan 22, 2006

### Cyrus

Yes, I was horrible in my description. I will try to state it again. There are a set of strain equations. They are functions of stress, sigma (x,y,z). Now, if I have a bar, and I apply a stress to it, (it could be a stress that magically 'Just happened' to turn on instantly and uniformly, ie an ideal case:: or :: it could be a stress that starts at zero and increases to some final value).

Ok, so back to the origional point. Lets assume that the stress is just magically 'turned on' and is P/A, initially. The result is that you will have a strain, that is a function of sigma(x,y,z). Now, what happens as it stretches??? It will get longer, and *thinner.* They key word here being *thinner.* Why is this critical? Because P/A will change as it gets thinner, it will be VERY small change, but a change none the less. This would mean that there would have to be a change in strain that results from P/A changing as A gets smaller. In fact, the ratio of P/A would get BIGGER as A(area) gets smaller, which would imply that the strain should become ever LARGER as a result of getting longer and *thinner.* This means that as this thing starts to deform, it would want to always continue to deform because the strain equation, $$\epsilon_{xx} = (1/E) \sigma_{xx}$$ would get larger and larger due to the increasing sigma term. I think the problem lies in trying to use the strain equation to model what is happening in transition, and IT DOES NOT DO THAT. It relates the FINAL strain to the INITIAL stress. What happens inbetween is NOT represnted by this formula, and it would be WRONG to use it as a predictor model for quantifying the strain in transition from applying the load to final equilibrium.

Last edited: Jan 22, 2006
9. Jan 22, 2006

### Q_Goest

Hi Cyrus'. I think I understand what you're getting at, so I'll repeat it back to you and verify. Given a piece of material, such as metal, that is in tension (such as a coupon used for tensile stress testing), you've pointed to the fact that we use the equation for stress which is equal to the force divided by area. Similarly, all stress equations such as beams in bending or pressurized membranes ignore changes in area or changes in inertia due to thinning.

Next, you're noticing that the area will decrease as stress increases (due to strain), and you're asking yourself why we don't take into account this decreased area to calculate the new stress. You also seem to be suggesting that this slightly smaller area (again, due to strain) may somehow compound the stress such that increases in stress which result in reduced area might result in a non-linear increase in stress as opposed to the linear one the equations predict.

It is a thoughtfull observation, and you're correct in that we ignore this decrease in area when doing stress analysis. It is simply ignored completely. Although Poison's ratio provides the constant needed to determine the thinning of the test specimen, we simply ignore it and base our calculations on the original area. Did I state your question correctly? Is that the basis for your post?

We use the original area (unstressed area) as the basis for our analysis because in general, we are working with materials who's change in area is very small. If we calculated the new area, it wouldn't make any significant difference. The stress is not significantly affected by the area change. Note that this is true as long as the stress remains below the yield strength of the material. Once we reach the plastic range (assuming we have a ductile material) all bets are off. Stress analysis is only valid within the range at which we have no significant yielding. This is the primary reason it is notoriously difficult to predict material failure, the equations for stress and strain provided by textbooks, don't apply once we enter the plastic region where deformation becomes permanent.

10. Jan 22, 2006

### Cyrus

More or less you are close. What Im saying is that you should see a run off effect, (if you were to JUST use the equations of strain, thinking that they explain what goes on during the transition from the initial length to the final length after total deformation). Because the specimin would get thinner and increase the stress (even if minute) which would cause runoff that would just get worse and always increase with time. I know that it is ignored in the calculations, but im forcing you by saying DONT ignore it. Look at what happens when you do not. You get alot of contradictions. And this is only ressurected if we place the condition of cause and effect. Cause (an applied initial stress), effect (strain based on the INTIAL Stress state). You can NOT use the stress-strain equations during the actual kinematics of the deformation, (Ie it would be wrong to use those equations during the period of transition to the newly increased lengths) to 'have a feel' for what is going on, so to speak.

Last edited: Jan 22, 2006
11. Jan 23, 2006

### PerennialII

Yeah, sounds 'still' reasonable. If you're interested in getting to know how the change in "A" is incorporated in continuum mech, you can take a peek what is said about 'finite deformations', or with respect to a uniaxial axisymmetric tensile test, the Bridgman model.
Yeah, although would say 'in principle'. Say for typical metals in the elastic-plastic regime 'classical' models don't deviate significantly from those having 'failure terms' in them (i.e. models able to predict material failure) with strains of some tens of percents (or in a tensile test can use initiation of necking as an approximate point for where classical models get totally lost).

12. Jan 23, 2006

### FredGarvin

I do see what you are saying now. I only have problems with the way you are starting your reasoning and it may be just the way I was taught (i.e. not written in stone). You state:
I was always taught that you start with deformations, i.e. strains and then stress is calculated from those strains. Therefore stress is a function of strain, not the other way around. One does not apply a stress. One applies a force, that causes deformation that results in a stress. I'm probably nit picking here and again, it may be the way I was taught.

This rang a bell for me so I had to dig up my advanced book because I remember a small discussion about Hooke's Law in it.

Looking at the standard 3D cube element (and leaving out the shear terms because I am lazy):

$$\epsilon_x = \frac{1}{E_x}\sigma_x - \frac{\nu_{yx}}{E_y}\sigma_y - \frac{\nu_{zx}}{E_z}\sigma_z + \alpha_x T$$

$$\epsilon_y = -\frac{\nu_{xy}}{E_x}\sigma_x - \frac{1}{E_y}\sigma_y - \frac{\nu_{zy}}{E_z}\sigma_z + \alpha_y T$$

$$\epsilon_z = -\frac{\nu_{xz}}{E_x}\sigma_x - \frac{\nu_{yz}}{E_y}\sigma_y - \frac{1}{E_z}\sigma_z + \alpha_z T$$

The general form of Hooke's law is not the simple P/A. It also reminds us that Hooke's Law is not really a law, but an approximation valid for small strains and certain materials (Ref: Advanced Mechanics of Materials, Cook & Young).

This is exactly what Q was referring to.

13. Jan 23, 2006

### Q_Goest

What we're talking about here is the affect of thinning due to stress/strain and it's affect on strength, right?

If you do the calculations, you'll find there's no "run off (on?) effect". Even if you assume the affect is real, which is highly debatable, the affect is exceedingly small and the error created is generally much less than the error incurred due to tolerances in manufacture of the actual part. For example, a 1" diameter steel rod under a 100,000 psi load will only shrink on the diameter about 0.001" which is generally MUCH less than any tolerance in manufacture, which could be 0.015" or more! In addition, the actual yield strength of the material can vary significantly, so the actual dimension becomes even less important yet.

No one wants to ignore real errors, but no one calculates this affect because it isn't significant and I'm not even sure it is real. I doubt you could even measure any reduction in strength due to thinning in a lab environment with precision machined specimens.

14. Jan 31, 2006

### rdt2

This is a deeper matter than it first appears. Unless unloading occurs, it doesn't matter whether you call the deformation 'elastic' or 'plastic' so it's not about elasticity or plasticity. Nor is it about whether you use 'nominal' stress (=F/Ao) or 'true' stress (=F/A), either of which is perfectly acceptable when accompanied by the appropriate (i.e energy conjugate) strain measure. Nor can it matter whether you regard stress as a function of strain or vice versa. All of these are choices are made on the basis of mathematical convenience. The answer may be that, for any given force, the reduction in area (which not all of us can ignore and which depends on Poisson's Ratio), is small enough for the increase in stress to cause an even smaller reduction in area... and so on, the system settling down to a stable state.

For normal materials, Poisson's Ratio can't exceed 0.5 (otherwise tension would cause an decrease in volume) but you might imagine some exotic material in which Poisson's Ratio was large enough for the area to reduce dramatically (perhaps after a phase change) and for the material to be unstable under a moderate applied force.