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.(adsbygoogle = window.adsbygoogle || []).push({});

Ok, I am following the derivation of the resistive strain gauge equations starting from the basic form of resistance of a wire:

[tex]R=\rho\frac{l}{A}[/tex]

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:

[tex]\frac{dR}{R}=\frac{d\ell}{\ell}-\frac{dA}{A}+\frac{d\rho}{\rho}[/tex]

The first term, [tex]\frac{d\ell}{\ell}[/tex] is strain in the longitudinal direction, [tex]\epsilon_{\ell}[/tex].

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

If I look assume the wire is cylindrical with a radius r, then I can determine radial strain, [tex]\epsilon_{\tau}[/tex], to be

[tex]\epsilon_{\tau}=\frac{dr}{r}=-\nu*\frac{d\ell}{\ell}[/tex]

where [tex]\nu = -\frac{\epsilon_{\tau}}{\epsilon_{\ell}}[/tex] is Poisson's ratio

The cross sectional area of a cylinder (wire) is given by

[tex]A=\pi*r^2[\tex]

[tex]A_{stretched} = \pi*(r+dr)^2 = A*(1+\epsilon_{\tau})^2[/tex]

[tex]\frac{dA}{A}=(1+\epsilon_{\tau})^2-1 \approx 2*\epsilon_{\tau} = 2*\nu*\frac{d\ell}{\ell}[/tex]

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 [tex]\epsilon_{\tau}[/tex], 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:

[tex]A=w*h[/tex]

and

[tex]dA = (w+dw)*(h+dh)[/tex] ?

I appreciate any help anyone can give.

~ Brent Ellis

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# Some Strain Gauge Theory Help

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