Why is Young's modulus constant below the limit of proportionality?

[Amaterasu21]

TL;DR Summary

Young's modulus = kL/A, so if the spring constant is constant, shouldn't the ratio L/A have to stay the same to keep E constant?

Hi all,

I'm a little confused about something.
Force-extension graphs and stress-strain graphs are always both straight lines up until the limit of proportionality, implying both the spring constant and the Young modulus are constant up until then.

For a force-extension graph, Hooke's Law applies up to the limit of proportionality: F = kx.

However, stress = F/A and strain = x/L, so the Young modulus E = stress/strain = FL/Ax.
Substituting in F = kx we get E = kxL/Ax = kL/A.

So it seems to me that if the spring constant is... well, constant... up until the limit of proportionality, if the Young modulus is constant, the ratio L/A should also be constant, implying they'd both have to increase in the same proportion.
That doesn't sit right with me. It seems that for most materials, the cross-sectional area should decrease as the extension increases. I've heard of auxetic materials which get wider as they get longer, but those aren't the majority of materials that the Young modulus is supposed to apply to, and even so I doubt the length and area are directly proportional there!

So why is it that the Young modulus is constant up to the limit of proportionality?

 

[Kyouran]

It's been some years for me, so I don't have a direct answer, but afaik you have to be careful here as to whether or not you are talking about engineering strain or true strain. In the former case, as is also the case for the spring force, L is considered the initial length, and thus is constant by definition; the same goes for A. In the case of true strain, L and A are indeed a function of time. My recommendation is to think very carefully about what L and A truly represent.

 

[hutchphd]

Caution: this is my poor physicist's understanding.
The Young's modulus is defined for the unstrained material as is the length L. It is really applicable for small values of the strain x/L. The effects you are worried about refer to the effect of the stretch on the "effective" Young's modulus and so the manifestation in the energy will be $(x/L)^ 2$. Most materials are far beyond elastic limit before that is important.

 

[Amaterasu21]

Thanks for the responses!
hutchphd: That's fair enough if you're just going to slap a number named "Young's modulus" on a material and call it a day, but I'm talking about a stress-strain graph and why its slope is constant up to the limit of proportionality. Maybe the term "Young's modulus" is the wrong one to use and I should just say "stress over strain," but still. I can't really see how stress can be proportional to strain and tension be proportional to extension at the same time without the ratio L/A being constant.

Kyouran: That would make sense to me if the graph of engineering stress vs. engineering strain is a straight line, but true stress vs. true strain isn't. But from what I've seen, they both appear to be straight lines up until the limit of proportionality, and that's what's confusing me.

 

[hutchphd]

The engineering stress vs strain is what I am referring to. They cannot both be linear. Check the following:

https://en.wikipedia.org/wiki/Stress–strain_curve#Engineering_stress_and_strainwhere it is shown that true stress is

edit: I guess I should point out explicitly that $$ln(1+x)=x-x^2/2 +x^3/3-...$$ so it all depends upon where the plastic flow starts.

 

[Amaterasu21]

The engineering stress vs strain is what I am referring to. They cannot both be linear. Check the following:

https://en.wikipedia.org/wiki/Stress–strain_curve#Engineering_stress_and_strainwhere it is shown that true stress is

View attachment 277377
edit: I guess I should point out explicitly that $$ln(1+x)=x-x^2/2 +x^3/3-...$$ so it all depends upon where the plastic flow starts.

That makes sense, but it also appears incompatible with this paragraph further down (italics mine):
"The first stage is the linear elastic region. The stress is proportional to the strain, that is, obeys the general Hooke's law, and the slope is Young's modulus."

The two graphs there also seem contradictory.
This one makes sense:

Whereas this one doesn't:

I can't see how what you said can be true and the first part of this graph be true as well.

 

[hutchphd]

The "type" of strain on the graph is not specified. They are not equivalent.

 

[sophiecentaur]

why its slope is constant up to the limit of proportionality.

Sorry but isn't that just a definition of proportionality?

It seems that for most materials, the cross-sectional area should decrease as the extension increases.

But that is something that has bothered me too when stretching wires. If the volume is constant then that must be true. Springs don't usually stretch the material but just bend it so the cross sectional area won't change much. So we have no problem with k.

The Young's modulus is defined for the unstrained material as is the length L.

At that point the higher terms in the above series are zero.

Perhaps the apparent contradiction between the graphs is the range where they apply and what they each claim to represent. The upper graph represents the behaviour of an ideal material. The lower graph shows ideal behaviour (corresponding to the bottom cm or so of the upper graph. The rest of the lower graph shows behaviour of a typical substance as it leaves the ideal range. The ideal model doesn't cope with it.

 

[hutchphd]

But that is something that has bothered me too when stretching wires. If the volume is constant then that must be true. Springs don't usually stretch the material but just bend it so the cross sectional area won't change much.

I believe this is more than you want to know about Poisson's Ratio which I think is the appropriate parameter to look at.

 

[Frabjous]

Engineering stress/strain comes from experimental measurements. You need to convert it to true stress/strain in order to map it to material properties. Strain is a function of derivatives of displacements. For small strain, only the linear terms are retained. The non-linear terms can have effects, for example the elastic bending of a thin bar. The moduli are state dependent. For many cases, plasticity comes into play before there is a significant change due to strain. More commonly pressure and temperature dependence are taken into account.

In general, one needs to take into account if one is measuring things relative to the initial state or current state. This is the root of the conversion from “engineering“ to ”true” variables.

 

[sophiecentaur]

I believe this is more than you want to know about Poisson's Ratio which I think is the appropriate parameter to look at.

Yes! I am only familiar with the topic at the level of isotropic solids (not even all metals, I suppose). Measuring the mean diameter of a wire during an experiment seemed to give an answer which agreed with 'the book'. The thing that confused me was that the teacher didn't seem to be aware of my problem and my confusion between coil springs and straight wires. It's reassuring that the same problem still exists for some people, sixty years later.

 

[Amaterasu21]

Sorry but isn't that just a definition of proportionality?

Yes, stress would be proportional to strain, but what I was concerned about is force also being proportional to extension. I've seen both of those relationships labelled as "Hooke's Law," but I struggled to see how both could apply to the same spring at the same time because L/A would have to be constant.

What you said about springs does clear this up for me though! If the spring's cross-sectional area is fairly constant, then so is L/A, and both F = kx and σ = Eε can be true.

Does this mean k is actually not a constant for a solid block of elastic material with a constant volume, then, and F = kx applies only to springs and not to solid blocks of elastic material? Is stress still proportional to strain in solid elastic materials like wires?

 

[hutchphd]

Did you read the Wikipedia article on Poission's Ratio??`

For isotropic materials we can use Lamé's relation[9]

for young's modulus E and bulk modulus K

 

[Frabjous]

Stress and strain are tensor quantities. For example, a wire can be approximated by 1d stress and the constant of proportionality is E and a plate can be approximated by 1d strain and the constant of proportionality is K+4G/3. Force and displacement are integrated quantities and are what are commonly measured. If the shape does not change, equivalences are trivial. Stress and strain are the proper variables, so one should turn your question around and ask when does F=kx work.

 

[JT Smith]

Does this mean k is actually not a constant for a solid block of elastic material with a constant volume, then, and F = kx applies only to springs and not to solid blocks of elastic material? Is stress still proportional to strain in solid elastic materials like wires?

I think you're getting too caught up in what is a first order approximation. As you've surmised it can't really be true that as the material stretches and becomes thinner that Hooke's Law and a constant modulus make sense. But for small extensions it works pretty well. What's the strain when steel yields? It's not much. Compare that to, say, a rubber band.

 

[sophiecentaur]

Does this mean k is actually not a constant for a solid block of elastic material with a constant volume, then, and F = kx applies only to springs and not to solid blocks of elastic material? Is stress still proportional to strain in solid elastic materials like wires?

one should turn your question around and ask when does F=kx work.

This is why coils springs and leaf springs are so often used. k works well enough for those. It would be impossible to achieve a useful value of constant k, using a single straight wire as a spring. The OP seems reluctant to go along with this. Why? (OK - without looking at the Maths in the earlier posts, you end up with an apparent inconsistent theory but that's what you get when relying on intuition.)

 

[Frabjous]

Let’s put some numbers in. The yield strength of steel is on the order of 1 GPa. The Young’s modulus is on the order of 200 GPa. Plastic flow therefore starts at strains of roughly 0.5%. This gives an engineering strain of 5.01%. Let’s increase it by an order of magnitude to 5%. This gives an engineering strain of 5.1%, a two percent difference.

 

[sophiecentaur]

Compare that to, say, a rubber band.

How to inject a bit of cognitive dissonance into a lesson. Good sport - just when they think they've found a Law to remember.

 

[JT Smith]

This is why coils springs and leaf springs are so often used. k works well enough for those. It would be impossible to achieve a useful value of constant k, using a single straight wire as a spring.

But that's because of the limit elasticity not because steel isn't Hookian within that limit. Steel has a fairly constant k, until it doesn't.

I measured a guitar string one time, hanging weights and using a micrometer to determine the stretch. It looks pretty linear to me:

 

[sophiecentaur]

But that's because of the limit elasticity not because steel isn't Hookian within that limit. Steel has a fairly constant k, until it doesn't.

I think you have missed the whole point of this argument. When you take a coil spring, the effective CSA of the steel changes very little over a huge range of 'strain' because the square section steel wire is only twisting and not stretching. It is only when the spring is stretched so that the pitch of the coils is, say 45° that the CSA of the wire starts to change appreciably. That's a gross oversimplification of course. By that stage the outer parts of the cross section have distorted appreciably so the stress is actually increasing (due to geometry) nonlinearly with spring length - but not a lot till then.
Once that starts to happen, the spring will be inelastically deformed and the spring gets stretched permanently. Once the spring becomes nearly a straight wire, you are in an entirely different domain (beyond Hooke's experiment and k)
What you have asserted is that "Steel has a fairly constant k, until it doesn't." when you actually mean "A spring has a fairly constant k etc. etc." You have ignored the Geometry.

Something you should bear in mind is that 'the spring' experiment usually involves hanging weights on it and measuring the length (GCSE Physics). BUT studying the behaviour of a wire involves stretching it by a certain length and measuring the Force. That's the other way round. If you tried just loading a steel wire, it would suddenly give up and you would have no idea about the stress / strain relationship after the weight falls to the ground. The apparatus for a proper stress / strain measurement is too expensive to provide a whole class with the kit for each pair.

What actual increase in length did your guitar string exhibit over that experiment? (A few mm in 60cm?) What increase in length does a coil spring exhibit over the straight line portion of the Force / Extension curve?

 

[JT Smith]

I think you have missed the whole point of this argument. When you take a coil spring, the effective CSA of the steel changes very little over a huge range of 'strain' because the square section steel wire is only twisting and not stretching. It is only when the spring is stretched so that the pitch of the coils is, say 45° that the CSA of the wire starts to change appreciably. That's a gross oversimplification of course. By that stage the outer parts of the cross section have distorted appreciably so the stress is actually increasing (due to geometry) nonlinearly with spring length - but not a lot till then.
Once that starts to happen, the spring will be inelastically deformed and the spring gets stretched permanently. Once the spring becomes nearly a straight wire, you are in an entirely different domain (beyond Hooke's experiment and k)
What you have asserted is that "Steel has a fairly constant k, until it doesn't." when you actually mean "A spring has a fairly constant k etc. etc." You have ignored the Geometry.

I understand the difference between a coiled spring and a straight wire. Perhaps I misunderstood your statement, the one where you said that a straight wire wouldn't have a constant K. That's obviously not true so you must have meant something else. Maybe you just meant that a wire doesn't stretch as far as a spring made out of the same material?

Something you should bear in mind is that 'the spring' experiment usually involves hanging weights on it and measuring the length (GCSE Physics). BUT studying the behaviour of a wire involves stretching it by a certain length and measuring the Force. That's the other way round.

I am aware of that but I'm not sure I understand the difference. In either case you correlate force and extension. Does it really matter whether you stretch and measure force or apply force and measure stretch? And if so, why?

What actual increase in length did your guitar string exhibit over that experiment? (A few mm in 60cm?) What increase in length does a coil spring exhibit over the straight line portion of the Force / Extension curve?

From the graph I posted you can see I stretched it about 1%. The string was around 90cm long. Using the same steel I imagine you could design a coiled spring that stretched 100% or more and be linear over most of its range. It wouldn't sound as good though.

 

[hutchphd]

For isotropic media apparently any two of the "moduli" are sufficient to characterize. In particular as mentioned earlier the ratio of extension (Young's) modulus E to bulk (compressive) modulus K tells the tale. But even for isotropic stuff it depends upon the details of the molecular folding. This is fascinating ( from wikipedia article on Bulk modulus):
https://en.wikipedia.org/wiki/Bulk_... modulus ( or,relative decrease of the volume.

 

[sophiecentaur]

I understand the difference between a coiled spring and a straight wire. Perhaps I misunderstood your statement, the one where you said that a straight wire wouldn't have a constant K. That's obviously not true so you must have meant something else. Maybe you just meant that a wire doesn't stretch as far as a spring made out of the same material?

I am aware of that but I'm not sure I understand the difference. In either case you correlate force and extension. Does it really matter whether you stretch and measure force or apply force and measure stretch? And if so, why?

From the graph I posted you can see I stretched it about 1%. The string was around 90cm long. Using the same steel I imagine you could design a coiled spring that stretched 100% or more and be linear over most of its range. It wouldn't sound as good though.

I can see we are not disagreeing basically. All I have been pointing out is that different experiments examine different ranges of conditions. A Guitar string is made of appropriate material and has dimensions to allow it to resonate at the appropriate frequency with the 90cm length. Its "k" is constant enough over the range of conditions needed for the tuning (It has to). I can't see any reason why a short fat steel bar would not also have a constant k range if it's isotropic. Probably the 1% range of extension? But a fat bar will not remain parallel sided (hence the importance of using tensors in those cases).
Any graph that's continuous and differentiable can have a near enough linear region. Aren't we just dealing with the particular application of the material?
Question: When you bend a guitar note, the increase in frequency would be due to length and linear density changes but would the stretch take it beyond the constant k region and affect the frequency significantly? Does bending take light strings into the plastic region? i.e. how far would the constant k region extend? How often do Power Players need to re-tune during a performance (if they can actually hear what they're doing)?

 

[JT Smith]

I can see we are not disagreeing basically. All I have been pointing out is that different experiments examine different ranges of conditions. A Guitar string is made of appropriate material and has dimensions to allow it to resonate at the appropriate frequency with the 90cm length. Its "k" is constant enough over the range of conditions needed for the tuning (It has to). I can't see any reason why a short fat steel bar would not also have a constant k range if it's isotropic. Probably the 1% range of extension? But a fat bar will not remain parallel sided (hence the importance of using tensors in those cases).
Any graph that's continuous and differentiable can have a near enough linear region. Aren't we just dealing with the particular application of the material?
Question: When you bend a guitar note, the increase in frequency would be due to length and linear density changes but would the stretch take it beyond the constant k region and affect the frequency significantly? Does bending take light strings into the plastic region? i.e. how far would the constant k region extend? How often do Power Players need to re-tune during a performance (if they can actually hear what they're doing)?

For sure you wouldn't suspend an automobile or construct a bridge using straight wires.

I don't believe that a constant K is necessary for a guitar string. Does the string frequency depend on that? I don't think so. As for tuning, why would it matter if the tension changes nonlinearly as you twist the tuning knobs, as long as it changes in a consistent way?

Normal bending (a full step) doesn't increase the tension all that much. Really big bends I don't know. But my suspicion is that the difference between entering the plastic region and the string breaking is very, very small.

People need to re-tune sometimes after big bends or whammy bar dives because of the poor design of the tuning systems of the vast majority of guitars. There's way too much slop and friction, particularly in the tuning machines. This is why alternate systems have been developed (e.g. Floyd Rose) that lock the strings so whammy bar enthusiasts aren't continually retuning.

If the strings were non-Hookian one thing that would change is the feel of bends. The bending force does involve K. But since that force is already slightly nonlinear, and discontinuous (since one often ends up bending adjacent strings as well), it would probably just mean the guitarist would get used to that response, whatever it is.