Can't Find 2D Elastic Modulus Equation

[Chestermiller]

I just realized that the diagram you drew earlier doesn't include the distance--if any--between the center of the membrane and the center of the contact area...

Yes. That's correct. I assumed that the contact ring on the probe moved vertically downward without causing significant additional stress (over and above the preload stress) in the membrane. That's the same approximation we make when we analyze a preloaded guitar string being loaded transversely.

The rest of the membrane is within the contact patch under the drum stick.

Chet

 

[Chestermiller]

Considering this is a circular membrane, meaning it has but one edge, could we not then use something like the surface tension divided by the circumference? Surface tension, at least, is in N/m...

The term "surface tension" is usually reserved for fluids. The preload force per unit length around the circumference is just equal to the preload stress times the thickness of the membrane. It is uniform over the entire membrane. I indicated in a previous post that, in membrane parlance, this quantity is referred to as the (radial) stress resultant, and is given the symbol N. Unfortunately, the radial force per unit length vector is always pointing toward the center of the drum head (at least in this axisymmetric case), so its direction changes. There are ways of measuring it at the rim by measuring the force over a small arc length at the rim.

 

[Chrono G. Xay]

Here is a sketch of the apparatus:

After a little thought yesterday I realized that, even if still not correct, I could at least somewhat model what you drew earlier, including the distance when the center of the area of contact and the center of the membrane:

 

[Chrono G. Xay]

I understand. I felt nervous asking such a question to begin with, not even remembering the guidelines.

If future questions would fall outside of PF guidelines, please let me know.

I have--what I think is--a conceptual question. Would it not be reasonable to assume that, shape of the implement aside, if we knew the deflection at d=0, that the deflection as one moved away from the center would diminish similar to sqrt( x )?

 

[Chestermiller]

Here is a sketch of the apparatus:

View attachment 91295

After a little thought yesterday I realized that, even if still not correct, I could at least somewhat model what you drew earlier, including the distance when the center of the area of contact and the center of the membrane:

View attachment 91297

I like the experimental setup. To get the preload stress, you would have to determine the tension on each cable, and divide it by the arc length subtended (1/6 the circumference) and by the thickness of the membrane. This will only give you an approximation to the stress. To get a better approximation, you would need to use smaller arc lengths and more cables.

As far as your equations are concerned, I'm not able to follow what you did and I don't want to spend too much time because I'm pretty sure that they are not correct. But, I have figured out how to complete the analysis that you are interested in and will be presenting it to you later. But, in the spirit of how we do things at Physics Forums, I will only present some of the key initial equations, and then work together with you in deriving the final results. How does that sound?

Chet

 

[Chestermiller]

For a spherical indenter (drum stick head), the membrane will be in contact with the indenter up to the separation of the membrane from the indenter (at the contact circle at radial location $r_i$, measured from the vertical axis of the indenter).

If the depth of penetration of the spherical indenter of radius R is large compared to the radius of the contact circle $r_i$, the slope of the indenter at the separation point will be:
$$slope=\frac{r_i}{R}\tag{1}$$
This must match the slope of the membrane immediately above the contact circle:
$$slope=\frac{\delta}{r_i\ln{(r_0/r_i)}}\tag{2}$$
where δ is the depth of penetration of the indenter.

What do you get if you combine these two equations for the slope?

I also previously presented the following equation for the indenter force:
$$F=\frac{2πσ_ph}{\ln(r_o/r_i)}δ\tag{3}$$
We are going to combine the two slope equations to eliminate $r_i$ and obtain the results for the force in terms of $\frac{δR}{r_0^2}$

Chet

 

[Chrono G. Xay]

I just found a free e-Book version of the book called "A Treatise on the Mathematical Theory of Elasticity". Is this what you were referring to?

https://books.google.com/books/about/A_Treatise_on_the_Mathematical_Theory_of.html?id=kZIfAQAAMAAJ

Also, I just found and downloaded a free PDF textbook from MIT Open CourseWare on Calculus, which goes all the way from limits to vector calculus, and includes a number of additional free resources directly relating to the text.

http://ocw.mit.edu/resources/res-18-001-calculus-online-textbook-spring-2005/textbook/

I know I can find even more videos for guidance (and practice problems) on Khan Academy, if needed.

https://www.khanacademy.org

I like the experimental setup[, but] [t]his will only give you an approximation to the stress. To get a better approximation, you would need to use smaller arc lengths and more cables.

How about if I used a stronger steel rim for applying force, so that it deforms less under the same stress? (Also, I was too far into drawing the diagram to change it, but the actual drum I was going to attempt to simulate was a 14" snare drum, which typically has either 8 or 10 tuning bolts.) I've read on drumset forums that those who used them felt they were able to get a more even tuning, or had greater ease in the process.

...I have figured out how to complete the analysis that you are interested in and will be presenting it to you later. But, in the spirit of how we do things at Physics Forums, I will only present some of the key initial equations, and then work together with you in deriving the final results. How does that sound?

Only if none of it would get you in any sort of trouble.

 

[Chrono G. Xay]

For a spherical indenter (drum stick head), the membrane will be in contact with the indenter up to the separation of the membrane from the indenter (at the contact circle at radial location $r_i$, measured from the vertical axis of the indenter).

If the depth of penetration of the spherical indenter of radius R is large compared to the radius of the contact circle $r_i$, the slope of the indenter at the separation point will be:
$$slope=\frac{r_i}{R}\tag{1}$$
This must match the slope of the membrane immediately above the contact circle:
$$slope=\frac{\delta}{r_i\ln{(r_0/r_i)}}\tag{2}$$
where δ is the depth of penetration of the indenter.

What do you get if you combine these two equations for the slope?

I also previously presented the following equation for the indenter force:
$$F=\frac{2πσ_ph}{\ln(r_o/r_i)}δ\tag{3}$$
We are going to combine the two slope equations to eliminate $r_i$ and obtain the results for the force in terms of $\frac{δR}{r_0^2}$

Chet

What I'm guessing you mean by "combine the two slope equations" is that they should be set as equal to each other, and then solved for $r_i$ so it can be substituted into $F$ to "eliminate" it.

 

[Chestermiller]

I just found a free e-Book version of the book called "A Treatise on the Mathematical Theory of Elasticity". Is this what you were referring to?

Yes. But I strongly advise starting with a book on Strength of Materials which is typically much less intense mathematically.

How about if I used a stronger steel rim for applying force, so that it deforms less under the same stress? (Also, I was too far into drawing the diagram to change it, but the actual drum I was going to attempt to simulate was a 14" snare drum, which typically has either 8 or 10 tuning bolts.) I've read on drumset forums that those who used them felt they were able to get a more even tuning, or had greater ease in the process.

I thought from your diagram that the rim was not providing any of the support. This is what I was hoping for.

What I'm guessing you mean by "combine the two slope equations" is that they should be set as equal to each other, and then solved for $r_i$ so it can be substituted into $F$ to "eliminate" it.

Basically, yes, but I'll show you some aesthetically nice manipulations of the math to get everything in a more tractable form. Be back a little later.

Chet

 

[Chestermiller]

See if you can manipulate the equations in post #36 into the following form:
$$F=\frac{1}{ξ}\left(\frac{2πσ_phr_0^2}{R}\right)\tag{1}$$
where $ξ=(r_0/r_i)^2$ is the solution to the equation:
$$\frac{\ln{ξ}}{ξ}=\frac{2Rδ}{r_0^2}\tag{2}$$
Note that the right hand side of Eqn. 2 is a "dimensionless displacement" which contains only known geometric quantities. You can make a once-and-for-all graph of Eqn. 2 by choosing a selection of values of ξ, and obtaining the corresponding values of $\frac{2Rδ}{r_0^2}$ by evaluating the left hand side of the equation (at each chosen value of ξ). You then plot a graph. This graph can then be used to evaluate F for any arbitrary situation, given the penetration δ, the indenter radius R, the rim radius r₀, the preload stress σ_p and the membrane thickness h.

Chet

 

[Chrono G. Xay]

Yes. But I strongly advise starting with a book on Strength of Materials which is typically much less intense mathematically.

Ahhh, ok. Strength of Materials first.

I thought from your diagram that the rim was not providing any of the support. This is what I was hoping for.

The rim is there to help make sure the force exerted by each tuning bolt is kept as even as possible over the arc length of the circumference it affects. The stronger it is, be it from a stronger material or thicker walls of the rim, the more even it should be. Now... I could buy a regular rim and drill extra equidistance holes, but as incredibly inexpensive (~$13) as that would be compared to just getting a steel barrel of the needed diameter (~$300), let alone the number spring scales (depending on which range and resolution of force I'll need) and turnbuckles I'll need by default, that's even more money I don't have. Hahaha...

 

[Chrono G. Xay]

I was just reading back over Hooke's Law out of a WikiBooks entry called 'Strength of Materials',

https://en.m.wikibooks.org/wiki/Strength_of_Materials/Introductory_Concepts

and only just remembered reading in the past that Mylar is a class of product called BoPET, or Biaxially-Oriented Polyethylene Terephthalate...

https://en.m.wikipedia.org/wiki/BoPET

Then again, maybe the type of Mylar used for drumheads is at least *more* axially isotropic than others... or rather, wouldn't it HAVE to be? or else you would be, technically, distorting the drum rims whenever the drumhead had tension on it, since the x and y axes would stretch at different rates...

For my own sanity I'm going to assume that.

 

[Chrono G. Xay]

Whenever I try to solve for $r_i$ I keep getting stuck with it either 1) both inside and outside a logarithm, as when I started, or 2) when solved for $r_0$, it is both a coefficient and an exponent of an exponential function.

 

[Chrono G. Xay]

I tried solving for the logarithm and $r_i$ separately, but I still don't feel as if I've gotten any warmer,

because by substituting those expressions in it seems I wind up losing $δ$ ...

Regarding your expression $\frac{ln(ξ)}{ξ}$ , I know it is equal to $\frac{ξ'}{ξ}$ , but it's just coincidence that I ran across that little fact.

 

[Chrono G. Xay]

For a spherical indenter (drum stick head), the membrane will be in contact with the indenter up to the separation of the membrane from the indenter (at the contact circle at radial location $r_i$, measured from the vertical axis of the indenter).

If the depth of penetration of the spherical indenter of radius R is large compared to the radius of the contact circle $r_i$, the slope of the indenter at the separation point will be:
$$slope=\frac{r_i}{R}\tag{1}$$
This must match the slope of the membrane immediately above the contact circle:
$$slope=\frac{\delta}{r_i\ln{(r_0/r_i)}}\tag{2}$$
where δ is the depth of penetration of the indenter.

What do you get if you combine these two equations for the slope?

I also previously presented the following equation for the indenter force:
$$F=\frac{2πσ_ph}{\ln(r_o/r_i)}δ\tag{3}$$
We are going to combine the two slope equations to eliminate $r_i$ and obtain the results for the force in terms of $\frac{δR}{r_0^2}$

Chet

The more I work on this the more I wonder if you had accidentally defined $r_i$ twice, as both a radial location and as a radius of its own, because I'm getting frustrated with the problem, and feel as though I can't do anything.

 

[Chrono G. Xay]

I have tried showing your posts to my Pre-Calculus professor as a 'last ditch' effort (explaining the forum rules beforehand), and after mulling it over even they don't know how to solve for $r_i$. They did make the concession, though, that its method could be rooted in physics. Way back when they were getting their bachelor's degree, they started in computer science before switching to math.

 

[Chestermiller]

I have tried showing your posts to my Pre-Calculus professor as a 'last ditch' effort (explaining the forum rules beforehand), and after mulling it over even they don't know how to solve for $r_i$. They did make the concession, though, that its method could be rooted in physics. Way back when they were getting their bachelor's degree, they started in computer science before switching to math.

I'm sorry. I lost track of your posts in this thread. I will get back with you a little later and show you what I mean.

Chet

 

[Chestermiller]

This is all algebraic. I'm surprised that your professor was not able to help you. From the two slope equations, you obtained:
$$\frac{δR}{r_i^2}=\ln(r_o/r_i)\tag{1}$$
If I multiply both sides of this equation by $\left(\frac{r_i}{r_o}\right)^2$, I obtain:
$$\frac{δR}{r_o^2}=\left(\frac{r_i}{r_o}\right)^2\ln(r_o/r_i)=\frac{\ln(r_o/r_i)}{(r_o/r_i)^2}\tag{2}$$
But,$$\ln(r_o/r_i)=\frac{1}{2}\ln(r_o/r_i)^2\tag{3}$$
So, we have:
$$\frac{2δR}{r_o^2}=\frac{\ln(r_o/r_i)^2}{(r_o/r_i)^2}\tag{4}$$
OK so far?

Chet