# Some Strain Gauge Theory Help

1. Mar 9, 2007

### X1088LoD

First, I am not the greatest at LaTex so if I screw this post up I will go back and try to clarify. I will skip some steps, but get the general gist of everything.

Ok, I am following the derivation of the resistive strain gauge equations starting from the basic form of resistance of a wire:
$$R=\rho\frac{l}{A}$$

Question 1: Does this apply to just a long piece of conductive wire, or does it also apply while it is wrapped around in grid form? Does the shape of the grid have any effect?

Ok, examining the rates of change, we get:
$$\frac{dR}{R}=\frac{d\ell}{\ell}-\frac{dA}{A}+\frac{d\rho}{\rho}$$

The first term, $$\frac{d\ell}{\ell}$$ is strain in the longitudinal direction, $$\epsilon_{\ell}$$.

Specifically, the main question I am getting at relies on the change in cross-sectional area, $$\frac{dA}{A}$$, and the change in resistivity $$frac{d\rho}{\rho}$$.

If I look assume the wire is cylindrical with a radius r, then I can determine radial strain, $$\epsilon_{\tau}$$, to be
$$\epsilon_{\tau}=\frac{dr}{r}=-\nu*\frac{d\ell}{\ell}$$
where $$\nu = -\frac{\epsilon_{\tau}}{\epsilon_{\ell}}$$ is Poisson's ratio

The cross sectional area of a cylinder (wire) is given by
$$A=\pi*r^2[\tex] [tex]A_{stretched} = \pi*(r+dr)^2 = A*(1+\epsilon_{\tau})^2$$
$$\frac{dA}{A}=(1+\epsilon_{\tau})^2-1 \approx 2*\epsilon_{\tau} = 2*\nu*\frac{d\ell}{\ell}$$

Ok, so this is the cross-sectional rate of change approached as the radius of a wire. However is this the case for just one leg of the grid? The wire is arranged in a grid, is $$\epsilon_{\tau}$$, the same in each leg of the grid or does it apply for the entire surface? Would I expect the transverse strain to be the same throughout (i kinda think this is the case)? However, would there be some kinda of additive factor that the transverse strain in each direction is added together? The strain gauge I am using has 16 longitudinal legs and 2 segmented transverse legs at the top and bottom. It would kinda seem that there would be some kind of multiplication factor of like (longitudinal path length)/(total path length) or something along those lines.

If not, how is it applied if the grid stretched over an area of width (w) by height (h). Would I follow the same approach starting with:
$$A=w*h$$
and
$$dA = (w+dw)*(h+dh)$$ ?

I appreciate any help anyone can give.

~ Brent Ellis

2. Mar 9, 2007

### FredGarvin

The change in the resistance is for the one wire. When you think of a grid, don't think of a strain gauge grid as a series of boxes. It is one continuous wire that looks like a series of back to back "u" shapes. The gauge is simply a resistor. Each terminal of the gauge is one end of the wire.

Each grid has one orientation that is changing the overall length when a load is applied. In the case of rosettes or stacked gauges, each direction has the same shape, just oriented to each direction.

3. Mar 10, 2007

### X1088LoD

Ok, that makes a ton more sense now.

So in your example image, there appear to be 14 legs of the continuous wire (i'm assuming the solder tabs themselves represent the end of the wire?). In an ideal connection, if I say the transverse strain of the strain gauge, I am essentially saying the transverse strain at any point?

Ok, so, lets say for example, that the image you have attached above is our strain gauge, and it has an active grid of 2 mm by 2 mm. 14 legs make our wire approximately 14*2=28 mm longitudinal, and an additional 2 mm transversely, for a total of approximately $$\ell = 30 mm$$ (is this a safe assumption?) In our definition of longitudinal strain, $$\epsilon_{\ell}=\frac{d\ell}{\ell}$$.

If I am looking at a beam for example, that undergoes a change in length of 0.2 mm. Would the new length be

$$\ell_{new}=14*(2+0.2)+2=32.8$$

Would that be a safe fundamental and approximate way of approaching it.

4. Mar 10, 2007

### X1088LoD

Let me also say, I know that just approximating the length of one of the longitudinal legs is generally enough, but i am interested in each one, not just a single one

5. Mar 10, 2007

### FredGarvin

You'd be pretty good to go about it that way. Although that would be a pretty good amount of strain in a specimen. I do have to say that in common practice one doesn't need to worry really about that since all of your calculations are based on gauge data supplied by the manufacturer. They include all of the information such as gauge dimensions, resistivity info and transverse sensitivity to name a few.

6. Mar 10, 2007

### X1088LoD

Indeed, I exaggerated those numbers just to pick a random value, I am trying to get an idea for all these basics for "not so common" practice, so i appreciate the help

7. Mar 11, 2007

### FredGarvin

No problem. You seem to have a good grasp on things.

8. Mar 16, 2007

### X1088LoD

Hey FredGarvin, or anyone else, if you get some time, do you mind looking at this and telling me your thoughts or poking holes where you can:

I am trying to come up with a mathematical approximation of what is happening to the resistance of a strain gauge, as sections of it become either debonded or completely detached.

My assumptions are that the wire is an ideal square cross sectional area, and when the wire is being stretched or compressed, it also keeps and ideal square cross sectional area. Also, the change in resistivity, $$\rho$$, of the wire is minimal due to the properties of the wire. I am trying to fully show how I got my equations, but if you have any questions about where something came from ill explain in full.

Looking at a single leg of the strain gauge the resistance is given by:

$$R=\frac{\rho*\ell}{A}$$

The volume over this length of wire is:

$$V=s^2*\ell$$

where s is a side of the cross sectional area. If the wire experiences an elongation of the surface, the resistance can be written as

$$R_{stretched}=\frac{\rho*(\ell+d\ell)}{(s-ds)^2}$$

where dl is the change in longitudinal length and ds is the change in cross sectional area. The new volume of this stretched length is given by:

$$V_{stretched}=(s-ds)^2*(\ell+d\ell)$$

However, no matter how much the elongation, the volume of the wire will always remain the same, therefore:

$$V-V_{stetched}=0$$

Assuming we know the original length, l, the change in length, dl, and the original length of s, using this property, we can substitute in our two volume equations and determine the value of ds

$$ds=\frac{(\ell+d\ell-\sqrt(\ell^2+\ell*d\ell))*s}{\ell+d\ell}$$

Applying this value to the resistance of a streched wire, we see:

$$R_{stretched}=\frac{\rho*(\ell+d\ell)^2}{s^2*\ell}$$

So, the purpose of bonding a strain gauge to a surface is to fully transfer the elongation characteristics of the surface to the grid of the strain gauge. So in the case that debonding is occuring, the elongation is not fully being transfered to the grid. Let us say that P is the percentage of the grid that is not fully attached to the surface, therefore 100-P is fully attached to the surface.

We can develop an equation of the resistance of one leg of the strain gauge as:

$$\frac{P}{100}*\frac{\rho*(\ell+\eta*d\ell)^2}{s^2*\ell}+\frac{100-P}{100}*\frac{\rho*(\ell+d\ell)^2}{s^2*\ell}$$

where the first term represents a fraction of the grid, P percent, that transfers a percentage, $$\eta$$, of the elongation to the grid, where $$\eta$$ is between 0 and 1. The second term represented the fraction of the grid, 100-P percent, that transfers the full elongation of the grid.

When $$\eta=0$$, the wire is completely detached from the surface, and none of the elongation is detected by the wire
When $$\eta=1$$, the wire is completely attached to the surface, and experiences all of the elongation of the surface

My pitfalls here however are that A) sections of the wire may not be uniformly debonded, so in one point, eta might by 0.2 but at another point it might be 0.5. This assumes an even level of debonding throughout the percentage of the wire. B) the strain gauge has 16 legs, so this assumes that the debonding occurs evenly horizontally, which may not be the case.

I appreciate any advice or criticism that anyone can offer me.

Brent

Last edited: Mar 16, 2007