V1/2, so Vo = V1/2-V1/2 = 0. When the gauge is strained, the voltage
drop across the gauge is VsR1(1+x)/[R1+R1(1+x)], where x is the
fractional change in resistance due to strain, and Vo =
VsR1(1+x)/[R1+R1(1+x)]-Vs/2 = Vsx/2(2+x)
Since the maximum strain in the gauge should not be more than
0.01, the maximum fractional change in the resistance of the gauge =
Ge =0.021 (the gauge factor of my strain gauge is 2.1), since this is
only 1% of 2, we can ignore the x in the term 2+x, which,
substituting x with Ge, gives Vo ? VsGe/4. The problem with a single
gauge bridge is that it doesn’t compensate for changes in resistance
due to temperature changes, which not only change the resistance in
the wire directly, but also make the object to which the gauge is
attached expand or contract, which increases or decreases the strain
in the gauge, and therefore its resistance. This problem can be
solved by including a dummy gauge which is attached to another
object of the same material in close proximity to the one with the
active gauge on, but which isn’t being bent, so that the difference
between the resistances of the two is the fractional change in
resistance due to strain. The gauges should be set up as in the single
gauge bridge, but with the dummy gauge between A and C. If y =
the fractional change in resistance due to temperature, then the
resistance of the dummy gauge (R’g) = R1(1+y) and Rg =
R1(1+x)(1+y), and Vo = VsRg/(Rg+R’g)-Vs/2 ?
VsGe/4. This method also has drawbacks, though: that an unstrained
specimen of the original material has to be provided and that the
dummy gauge is not necessarily at the same temperature as the
active one. These problems can be solved by mounting the dummy
gauge on the same member as the active one and at right angles to
the direction of strain, so that the gauges are unlikely to have a
measurable difference in temperature. The dummy gauge will be
strained at right angles to its active axis, which will make it slightly
shorter along its active axis, as explained above, which means that
the resistance will decrease by an amount proportional to Poisson’s
ratio (v). If Rg = R1(1+x)(1+y) as before, and R’g = R1(1-vx)(1+y),
then Vo = VsR1(1+x)(1+y)/[R1(1+x)(1+y)+ R1(1-vx)(1+y)]- Vs/2 ?
Vs(1+v)Ge/4.
This method can be used to give a measurement of strain in a
member under tensile stress, but I’m planning to use a cantilever, the
end of which I will be putting weights on. In this situation I will be
able to increase my output readings by putting active gauges on both
sides of the cantilever, because as the cantilever bends the gauge on
one side will be put under tensile stress and the gauge on the other
side will be put under equal compressive stress2. This means that the
difference between the two will be twice the difference between a
single gauge and a fixed resistor, in other words, Vo ? VsGe/2, and if
dummy gauges are included Vo ? Vs(1+v)Ge/2 (in both cases the
active gauges are between B/C and D, and the resistors/dummy
gauges occupy the other two arms of the gauge.
According to Hooke’s law, until the elastic limit is reached, which
it won’t be in this experiment, the weight on the end of the cantilever
will be proportional to the amount it bends, which will be
proportional to the amount of strain that the gauges are placed
under, which is proportional to their resistance and so the output
voltage, therefore the output voltage should be proportional to the
weight on the end of the cantilever.
Because the Wheatstone bridge won’t be balanced, or give a
reading of zero, at the start of my experiment, I put an 11 ohm
potentiometer in the bridge at B. Adjusting the resistance of this
should allow me to balance the bridge before I start
For my experiment I attached an active and dummy gauge on
either side of the cantilever, for which I chose a long, flat piece of
metal that was not so thick as to make the strain too small to
measure. I attached all of the gauges on a line perpendicular to the
direction of stress, because the stress is not constant along the
length of the cantilever.
To calibrate my sensor I attached a bucket to the end of the
cantilever, zeroed the output, then put in known weights, measured
using a set of scales that I trust, before checking the output voltage
and recording my results. Once I have calibrated my sensor I will
test it by arriving at a weight from the output voltage of an unknown
weight, then comparing it with the weight found with the scales.
Analysis
As I was using Excel to tabulate and plot the data, I was able to
add an accurate line of best fit using a feature in the program. This
gave me an equation for the line of best fit of y=0.0026x, where y is
the reading in mV and x is the weight in N. The error margin in the
voltage readings is mainly due to the inaccuracy of the multimeter
that I used: its most accurate setting could only measure millivolts,
therefore the voltage is accurate to 0.1mV, there was also a small
error in the resistance of the gauge, but this was only 0.015%/?C3,
and as I estimate that the temperature would not have changed more
than a degree in the time that I took to do the experiment, this is only
a error of ?0.5?V in the highest voltage reading of 3.4mV. For this
reason, I have disregarded this error, and put ?0.1mV error bars on
the graph. Any error in the weight could only have caused by errors
in reading the scales, which would ideally have been digital but in
this case were not. To determine the error in the readings I had to
make an estimate based on my observations of the same weight
being measured several times, which gave me an error of ?10N.
When I weighed my teacher on my equipment I got an output
voltage of 3.4mV, which, when taking the error of ?0.1mV into
account gives a weight of between 1269N and 1346N. Using the
scales, his weight was found to be 1350 N, which is within 10N of
the upper weight found from the voltage.
Analysis
Overall I was happy with my procedure, although I encountered
several problems with the equipment that I used. The piece of metal
that I used as a cantilever was very dirty when I found it, and
required a lot of time spent cleaning it before I was satisfied that the
strain gauges would attach sufficiently reliably to give accurate
results. Even so, the gauges would not stick at all the proper way
round, with the plastic backing stuck directly to the cantilever, so I
had to glue them on with the foil on the metal, and hope that the
superglue would insulate the gauge from the cantilever, which it
seems as though it did.
The choice of what to use as a cantilever is a very important one,
as if it is too rigid it will not bend much and so not give a big enough
reading, but if it is too flexible it will bend too much and the gauges
limit will be exceeded. I think that the metal I used as the cantilever
was too thick, and gave too small a reading, but I was restricted in
my choice by the materials available to me
I think that my technique was a suitable one, as it gave accurate
results without causing major problems or being too complicated:
the most complicated part of the project was assembling the bridge
itself. It is also an adaptable one, with different cantilevers being
used for different weight ranges, and different arrangments of the
bridge available.
Bibliography
Strain gauges supplied by RS (http://rswww.com)
1 Instrumentation units 1 and 2 by the instrumentation course team
(The Open University Press, Walton Hall, Milton Keynes, MK7
6AA)
2 Advanced Design and Tecnology by Eddie Norman, Joyce
Cubitt, Syd Urry, Mike Whittaker (Addison Wesley Longman
Limited, Edinburgh Gate, Harlow, Essex CM20 2JE)
3 RS Electronic Catalogue data sheets (http://rswww.com)
