Calculating Young's Modulus for a Ring Chain of Springs

[galaxion1]

I know that Young's modulus for a spring is Y= K*L/A
where K: is the stiffness of the spring
L: the original length of the spring
A: the cross sectional area
How does this formula change in the case of continuously distributed springs over a ring chain of radius R and a linear density ρ=m/a, where m is the mass of the spring and a is the distance between the neighboring springs? Thanks in advance.

 

[Studiot]

Hello galaxion, Welcome to physics Forums.

You need to describe your system more fully, perhaps a diagram?

Are you not mixing up the formula for springs as a mechanism and the formula for young's modulus in a material?

Springs as a mechanism do not possesses a young's modulus. That is a property of the material from which they are made.

Springs as a system possesses a spring constant, which is a property of the system.

 

[galaxion1]

actually I was talking about a closed chain (a loop) consisting of N identicle particles of mass m each connected by a system of springs which represent the elastic interaction between the particles. At equilibrium the chain is a ring of radius R and the distance between the neighbouring particles is a.
My problem is in the transition from the mechanics of this system of material particles (which is useful in studying classical fields) to continuum mechanics by setting m→ 0 and a→0.. my question was how to find Young's modulus in this case? some books states that Y= Ka but I don't understand why?

 

[Studiot]

Are you studying Madelung theory?

Does page 36 of this look familiar?

http://web.phys.ntu.edu.tw/goan/Courses/M1720/Notes/chapter3.pdf

 

[nasu]

My problem is in the transition from the mechanics of this system of material particles (which is useful in studying classical fields) to continuum mechanics by setting m→ 0 and a→0.. my question was how to find Young's modulus in this case? some books states that Y= Ka but I don't understand why?

It should be Y=K/a, if only from dimensional consideration.

I am not sure what are you trying to do. You want to go from a one dimensional discrete chain to a 1D continuum?

Otherwise, if you want to connect a macroscopic property (Young's modulus) to the microscopic parameters of a balls and springs crystal model (K and a), you can consider a cubic lattice and apply Hooke's law for a unit cell (or for a cubic sample of crystal).

 

[Studiot]

OK so we are talking about lattice theory and the masses on springs of force constant K and spacing a.

We need to also be clear that we are talking about vibrating systems, restoring forces and energies

Let there be a change in separation δa from the equilibrium value.

Then from ordinary mechanics the increase in potential energy ΔU = 0.5K(δa)²

Now to put some values in

If a pair of unit charges increase their separated by 10^-10 (=δa) metres this corresponds to an increase of 1.6 x10^-19 joules (1 electron volt)

This corresponds to a spring constant (K) = 2*1.6 *10^-19 * 10²⁰

That is K ≈ 30 Nm^-1

To relate this to youngs modulus, E


$$E = \frac{{K\delta a/{a^2}}}{{\delta a/a}} = \frac{K}{a}$$

I will leave you to confirm that this leads to average E values for an average value of a at 3x10^-10 metres lattice spacing.

 

[galaxion1]

you are considering 2 dimensional lattice here..right? but actually what I need is to go for 1 dimensional continuum from 1 dimensional discrete chain... my idiot question is about the area in the stress formula if the particles are already point like :shy:
in "Quantum electrodynamics, A.A Sukuruv" book he stated that "Youngs modulus ε=Ka" !

 

[nasu]

you are considering 2 dimensional lattice here..right? but actually what I need is to go for 1 dimensional continuum from 1 dimensional discrete chain... my idiot question is about the area in the stress formula if the particles are already point like :shy:
in "Quantum electrodynamics, A.A Sukuruv" book he stated that "Youngs modulus ε=Ka" !

No, it is 3D lattice actually. And Studiot's example is for the same, I think.
But seems that this is not what you are looking for.

Maybe you need to know how do you define Young's modulus for a 1D continuum (whatever that is). Forget (for a moment) about transitions from chain to continuum. What is Young modulus in 1D? From the above formula is has units of force.

 

[Studiot]

I don't know your book, but I'm not impressed if that is what it says.

Have you done a dimensional analysis on your formula?

The units of the spring constant are Newtons per metre

The units of youngs modulus are Newtons per square metre

How do you get this if you multiply Newtons per metre by metres?

Edit :

Quantum electrodynamics, A.A Sukuruv

Looking at the title of your book, why are you asking this in the classic physics forum?

There is a forum here for quantum physics and one for modern physics.

 

[Studiot]

you are considering 2 dimensional lattice here..right? but actually what I need is to go for 1 dimensional continuum from 1 dimensional discrete chain... my idiot question is about the area in the stress formula if the particles are already point like

Actually , no it's much simpler than that.

The two points I considered can be in a one two or three D lattice - it doesn't really matter since I only took into account an small displacement δa between two of them which is linear and ignored the effect of the rest of the lattice.

However you original asked about a chain that self intersects to form a closed loop. This can only occur in two or three D, not in one D.

 

[galaxion1]

Looking at the title of your book, why are you asking this in the classic physics forum?

There is a forum here for quantum physics and one for modern physics.

because the subject is in the first chapter "classical field theory" do you feel that my topic is related by any means to modern physics or Quantum mechanics :)

 

[Studiot]

I don't know what your objective is.
The subject you mention is usually covered, in the West, in texts on Solid State Physics.

A good one is the Manchester Physics series book of that title by Hall.

 

[galaxion1]

I don't know what your objective is.
The subject you mention is usually covered, in the West, in texts on Solid State Physics.

A good one is the Manchester Physics series book of that title by Hall.

would you please see page 8 in this link and it all will be clear?

http://www.staff.science.uu.nl/~wit00103/ftip/Ch01.pdf

I am really grateful for your replies and consideration

 

[Studiot]

OK are you referring to sectio 1.2 "The Lagrangian for continuous systems"?

This starts on page 7 and works through page 8.

Since it is about lattice particles in vibratory motion it has terms for both kinetic and potential energy and the partition between them. It is clearly a more sophisticated model than my simplified one.

 

[galaxion1]

OK are you referring to sectio 1.2 "The Lagrangian for continuous systems"?

This starts on page 7 and works through page 8.

Since it is about lattice particles in vibratory motion it has terms for both kinetic and potential energy and the partition between them. It is clearly a more sophisticated model than my simplified one.

anyway Do you have any explanation for Y=Ka, having the units of force as you said?

 

[Studiot]

In order to describe the elastic rod we must take the continuum limit of the
system discussed above. Hence we increase the number of particles to infinity
(n → ∞), keeping the total length l = (n+1)a and the mass per unit length,
μ = m/a, fixed. Furthermore Y = ka must be kept fixed as well. This follows
from Hooke’s law, which tells us that the extension of the rod per unit length
is directly proportional to the force exerted on the rod, with Young’s modulus
being the constant of proportionality

I don't think so, the constant of proportionality is the reciprocal of youngs modulus.

Look here and scroll down to

"General Application to Elastic Materials"

http://en.wikipedia.org/wiki/Hooke's_law


$$\Delta L = \frac{F}{{EA}}L$$


rearrange


$$\frac{{\Delta L}}{L} = \frac{1}{E}\frac{F}{A}$$


Perhaps there's been a publishers' oversight?

 

[nasu]

Assuming that they redefine "Young's modulus" in this way:
"the extension of the rod per unit length is directly proportional to the force exerted on the rod, with Young’s modulus being the constant of proportionality"
or
$$\frac{\Delta l}{l}="Y" F$$
this new Y should have units of inverse force. Still not consistent with Y=ka.
There is something funny going on here.

 

[galaxion1]

I think that they mean that F=Y*ΔL/L instead that is also stated in the folowing pdf in page 10 http://www.phys.ethz.ch/~babis/Teaching/QFTI/qft1.pdf
this is not the problem, I am back again to the question "where is the area?!" how can we just redefine Youngs modulus that way?

 

[Studiot]

Unfortunately science and maths is littered with redefining something, but not bothering to tell anybody and thereby causing general confusion and mayhem.

Don't forget that, even conventionally, the spring constant and youngs modulus are different things.

But a good question anyway.

 

[nasu]

I think that they mean that F=Y*ΔL/L instead that is also stated in the folowing pdf
this is not the problem, I am back again to the question "where is the area?!" how can we just redefine Youngs modulus that way?

Considering that they are talking about a "1D bar", there is no area.
I suppose they redefine Y to fit this specific type of problem. How can they do this? See Studiot's post above.