Strain Gauge

  • #1


I am studying strain gauges at uni and was wondering if you could clarify for me how they may be used to measure mechanical strains in structures like towers, houses etc..

I know that it is possible to use a mechnical gauge that measures the development of a crack (see wiki)but I can't work out how it may be used in terms of resistance (i.e: an electric gauge)

  • #2
according to my book the gauge is stuck to the object - e.g: a plane wing - a load is applied to the wing, for example, and the resistance is altered.
What I don't get it how this happens

  • #3
The electrical resistance of a conductor depends on its length and cross section area. When you glue the gauge onto a structure, the strain in the gauge is the same as the strain in the structure, so the shape of the conductor in the gauge changes with the strain.

The changes in resistance are small, so you need a something a bit more sensitive than the ohms scale of a multimeter to measure the change, but that is the basic principle.
  • #4
So could I just clarify with an example.
If we assume the strain gauge as a strip of wire on an object, stuck down and then the wire is attached to a device that can accurately measure resistance in a wire.
So the strain gauge is attached to an object (such as a wing). A force is applied to the wing (e.g: by placing a really heavy brick on the wing).
The brick MUST ALSO BE placed over the strain gauge metal wire.

So the metal wire is then deformed and its area / length changes hence so does its resistance

  • #5
No, the strain gauge is deformed by the wing deforming. You can apply the load anywhere. (In fact if you tried to put a heavy weight directly on top of a strain gauge you would probably damage the gauge).

Maybe you are getting the wrong idea thinking of it as a "wire". The piece of the gauge that changes length is small, typically from a few mm long down to less than 1mm. The thickness of the "wire" is of the order of 0.01 to 0.1mm.

Of course the gauge has to to be connected to the measuring equipment by wires, and those wires are often attached to the surface of the component as well, to stop them flapping about or getting tangled up and damaged. If you look casually at a straingauged component, the connecting wires are usually more obvious visually than the gauge at the end of them, which is just a bit of thin film material covered with the glue that fixes it to the component.

If you haven't seen a gauge (but only pictures of them) they look something like the RF security tags used in library books or on expensive items in shops, but very much smaller.

If you were testing something as complex as a plane wing you would have literally hundreds of gauges attached to the wing at different places, to get a picture of the strain pattern all over the wing, but you might only apply the load at one point (for example a force at the wing tip).

For example if you wanted to measure the behaviour of a cantilever I-beam bending, you would put several gauges along the top and bottom flanges to measure the axial strain (if you think about a shear force and bending moment diagram, the stress and strain varies along the length of the beam) and possibly some more gauges on the web of the beam to measure the shear strain in it. There might be 10 or 20 gauges all together. You would then apply a single load at the tip of the beam.
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  • #6
But how does that cause a change in resistance to the gauge.
Or does the wing change shape and hence the wires on the place bend / stretch hence causing a change in resistance in those wires?
  • #8
Ok then, so I read that page (thanks by the way)
Can i just claify then what I read.
A strain gauge is placed (for example) above and below the specimen.
A force is applied to the specimen causing the specimen to bend (deform)
As the gauge is attached to the specimen it bends as well and hence cause either compression or stretching of the gauge and hence modifying the reistsnace

  • #9
Is their a relationship that can be used to measure the stress from the change in resistance
  • #10
Yes. The change in resistance divided by the nominal resistance is equal to the strain multiplied by the gauge factor. If you know the strain then you can calculate the stress as presumably you know the Young's Modulus of the material.
  • #11
How do I know the gauge factor -
I read this up before and saw that GF = (change resistance / orginal resisance) / strain
Therefore if I don't know GF how can I find strain
  • #12
IS the gage factor for an indivudal strain gauge constant
  • #13
btw is my summary from before adequate

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