# Strain Gauge - Aluminum Beam Deflection Help

I have an aluminum beam securely mounted on one side by a block. The
beam is 235 mm long, 3 mm high, and 25 mm wide. The opposite side of
the beam is oscillated up and down.
The center of a strain gauge (with an active grid of 1.57 mm by 1.57
mm) is placed at a point 25 mm away from the block. The vertical
displacement, amplitude A, is measured at a point 100 mm away from the
block.
This can be better seen by the image at

I am estimating the modulus of elasticity of aluminum to be about
10*10^6 psi.
Strain is defined as the change in length over the actual length (in
this case, 100 mm). How can I go about determining the change in
length of the bar?
I appreciate any suggestions or thoughts anyone can give me.
The following is conceptual, or to put it another way, arm-waving.
The usual beam theory for small deflections is premised on the basis of deflecting a straight beam to an arc of a circle. The beam is conceptualized as a set of connected laminae or thin sheets. The thin sheets on the outsides reach their elastic limit first and so control the strength of the ensemble.
If your cantilever is deflected down, then the upper surface stretches and the lower surface contracts. Why shouldn't these changes cancel each other?
It turns out that when you stretch something it gets thinner, and when you compress something it gets thicker. So it stands to reason that the compression face should squeeze less than the top surface stretches. We started by guessing that the neutral axis i.e. the lamina that does not stretch or compress but just bends is situated at the center. But this differential movement puts the neutral axis a little closer to the lower surface - a very little in your case. And so it seems that the top surface overhangs a little more than the lower surface retracts. To go any further, we need to recall that the ratio by which the dimension at right angles to the stress axis changes in comparison to the dimension along the stress axis is called Poisson's ratio. It is about 0.33 for aluminum.
I suspect that if you agreed with this treatment so far, you have probably decided you don't need to know the actual increase in length for i.e one millistrain. A few micrometers at most, as it might be. Is this enough?
Brian Whatcott Altus OK

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