CEE575 - Homework 1 Solutions Problem 1: Strain gage design

CEE575 - Homework 1 Solutions
November 19, 2014
Problem 1: Strain gage design
Design a strain gauge of height 50 mm, a width of your choice, and a total
resistance of 250 W. Choose a material, wire length, number of wire turns,
and gage diameter. Draw a digram of your gage and justify your answer. To
help with selection, you can search online for a table of materials and their
corresponding electrical resistivity and conductivity values.
Solution
As an example, choose to make a gage of gold. Resistivity is 2.44 ⇥ 10 8 ⌦ ·
m. If we choose a gage diameter of 0.025mm, we get a specific resistivity
8
⌦·m
250⌦
of 2.44⇥10
= 0.0497⌦/mm. Then 0.0496⌦/mm
gives us a total length of
⇡r 2
5000mm of gage wire, which for a 50mm gage means we have to have 100 50mm
segments/loops.
Problem 2: Quarter bridge vs. half bridge
In class we have introduced the single point Wheatstone Bridge (also known as
a Quarter Bridge) and the Half Bridge. In the ambient (unloaded) case assume
that all resistors and sensors have the same nominal resistance R1 = R2 = R3 =
R4 = R.
a) Show that the half bridge has twice the sensitivity (twice the output) of
the quarter bridge. The sensors are indicated by the red arrows.
b) What are the tradeoffs with using each setup?
1
Figure 1: Quarter bridge and half bridge setups.
Solution:
a) From class, we know that the output of the quarter bridge is

R
1
V0 = V
RX + R 2
Let Rx = R +
R then

R
V = V0
R+ R+R
V0 = V


1
R
= V0
2
2R + R
1
2

2R (2R + R)
1
R
= V0
·
2(2R + R)
2 2R + R
For the half bridge we have:

R
V = V0
R+ R+R

R+ R
R
= V0
R+ R+R
2R + R
which is twice as large of an output as the quarter bridge.
Problem 3: Cantilever load cell
A load cell is constructed by mounting strain gages to the surface of a cantilever
beam (Figure 2). Gages 1 and 2 are mounted to the top of the beam, gages 3
2
and 4 to the bottom. The beam cross section has a moment of inertia I and a
Young’s modulus E. In the unloaded state, all gages have the same resistance
R1 = R2 = R3 = R4 = R, and the gage factor is Gf . A load F is applied
a distance L1 from gage 2 and 4. For all Wheatstone bridge setups, it can be
shown that the relation between bridge output voltage V0 and the input Voltage
can be linearized to take the following form:
✓ ◆
V0
1
R1
R2
R3
R4
=
+
(1)
V
4
R1
R2
R3
R4
This simplification applies to small changes in resistance, which is almost
every problem we will care about in this course.
Using the above information:
a) What is the output voltage V0 in terms of the input voltage V for the
below cantilever setup? As we’ve done in class, your expression should contain
references to material properties E, I, c (distance to beam center, sometimes
also written as half the beam thickness t/2), and Gf .
b) If the distance L1 is increased, how does the bridge output change?
c )What is the advantage of this design? Give an example application.
Figure 2: Force measurement setup.
Solution
a) Strain at any of the locations is given by:
"=
M (x)c
F ·x·c
=
EI
EI
3
then for locations L1 and L2 away from the load, we have for the top of the
beam
"1 =
F L1 c
F L2 c
and "2 =
EI
EI
Since all of the strain gages have the same initial resistance, we get
R1
F (L1 + L2 )c
=
Gf and
R
EI
get
R2
F L2 c
=
Gf
R
EI
The top of the beam is in tension, while the bottom is in compression, so we
R1 =
R3 =
RF Gf c
(L1 + L2 )
EI
and
R2 =
R4 =
RF Gf c
L1
EI
Using the bridge relation in equation 1, we get
✓
◆
✓ ◆
V0
1
R1
R2
R3
R4
1
=
+
=
[2 R1
V
4
R
R
R
R
4R
Substituting for
V0 = V
R,
✓
2
4R
◆
RF Gf c
[(L1 + X)
EI
L] =
2 R2 ]
F Gf c
L2
2EI
b) Since there L1 effectively cancels out of the above equation the output is
not effected at all. It stays the same.
c)The major advantage is that our measurement setup is that the voltage
output is sensitive to the force input, but not the length L1 . This turns our
cantilever setup into a force sensor, or load cell, which should be kind of super
interesting if you think about it.
Problem 4: Photoresistors and getting the most out of measurement setup
It turns out we have not learned about all resistive sensors, yet. A conductor’s
resistance can also be affected by radiation (in this case, more specifically, visible
light). Photoresistors (also often called phototransistors or Cadmium sulphide
photoconductive photocells) are simple resistors that alter resistance depending
on the amount of light they are exposed to. Light is measured in lux, the SI illuminance unit (http://en.wikipedia.org/wiki/Lux). Photoresistors are probably
the most common sensor used in measuring light in every day applications, the
most affordable (only a dollar a pice), and the easiest of all radiation sensors to
implement. A variety of Photoresistors are shown in figure 3.
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The use of this sensor is highly dependent upon the application. Using the
sensor usually involves anticipating which levels of illuminance the sensor will
experience. This often involves exposing the sensor to the brightest environment
it will see and then measuring the resistance at this stage - call it Rb . This is done
for the darkest anticipate illuminance too, by measuring the “dark” resistance
Rd (you can measure this by just covering the sensor with your hand). The
idea is that future readings of resistance will then fluctuate between Rd and Rb
across the lifetime of the sensor deployment.
a) You are using a simple voltage divider circuit (fig 3) to measure the
resistance of the sensor. Given a known ballast resistor R in series (see figure),
and given that you know Rd (dark resistance) and Rb (bright resistance), derive
an expression for V0 (R), the maximum difference in potential that your voltage
divider will output between dark and bright instances.
b) Given that Rd = 3300⌦ , and Rb = 780⌦ , Plot V0 (R) for a series
of values of R, starting at R = 100⌦ and going up to R = 5000⌦ , in 100⌦
increments. What do you notice?
c) Building a voltage divider involves selecting the value of the series resistor
R. We want to select this resistor to maximize the potential difference V0 (R).
This will reduce the need to amplify our output voltage signal, and will make
it easier to discern voltage readings. Derive an expression for the optimal value
of R that will maximize this voltage difference.
d) What is the optimal value of R (in Ohms) given the values of Rd and Rb
above.
Solution
a)
V0 (R) = V (
R
R + Rd
5
R
)
R + Rb
Figure 3: Photoresistors and voltage divider circuit.
b)Assume V=5V
c)
take the derivative, with respect to R
6
V00 (R) =
(R Rd R)
(R + Rd )2
=
(R Rb R)
Rd
=
(R + Rb )2
(R + Rd )2
Rd (R + Rb )2 Rb (R + Rd )2
(R + Rd )2 (R + Rb )2
V0 (R) is maximized when
0=
Rb
(R + Rb )2
V00 (R) is equal to zero. Then
Rd (R + Rb )2 Rb (R + Rd )2
(R + Rd )2 (R + Rb )2
bottom cancels out, then expand
0 = Rd (R + Rb )2
= Rd R2 + 2Rd Rb R + Rd Rb2
Rb (R + Rd )2
Rb R 2
2Rd Rb R
= (Rd
Rb )R2 + (Rb
Rd )(Rd Rb )
(Rd
Rb )R2 = (Rb
Rd )(Rd Rb )
Rb Rd2
then
R 2 = Rd Rb
so
V0 (R) is maximized when
R=
p
Rb Rd
e) Plug in, Roptimal = 1604⌦. We probably can’t buy a rester exactly like
this, so we’ll settle with a 1.6k⌦.
Problem 5: Smart drying machine
You are building a smart clothes dryer. When the clothes lose 50% of their
weight they are ready to be taken out. As an engineer you decided to automate
the process, by creating a smart drying machine. The machine operates by
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detecting the weight of the clothes. A standard load of clothes has a weight of
10 kg when “wet” and 5 kg when fully dried. The dryer has a weight of 50 kg
when empty. The weight sensor is constructed with a strain gauge placed in one
of the 4 “feet” of the dryer. In this position, it detects exactly 1/4 of the weight
of the machine plus the clothes. The feet are cylindrical, but their weight is
negligible. Each foot is essentially a cylinder that has a pre-assembly length
of 1 cm, and a stiffness k of 1,000,000 N/m. A metal film strain gauge with a
nominal resistance of 1 kW and a gage factor Gf of 2.5 is attached to one foot
so that the weight of this machine causes compressive strain in the strain gauge
that is the same magnitude as the strain in the feet. The assembly is shown in
figure 4.
a) If the unloaded length of the foot is 1 cm, what is the length when the
entire machine is assembled and mounted upright on the ground?
b) Assume you placed the gage on the foot of the machine, before placing
the machine upright What is the total resistance of the strain gauge when the
machine is placed upright?
c) After adding a load of clothes, what is the resistance you expect to see on
the strain gage?
d) When the clothes are done drying, what resistance reading would indicate
to you that you should cut power off to the dryer?
e) Assuming you chose to use a simple voltage divider to measure the resistance of the strain gage, use the results you obtained in problem 3 to determine
the optimal size of the required resistor that you would need to use in the circuit. Now go online (digikey.com, sparkfun.com, or any other site) and find a
suitable resistor for this application. Provide the model number and link.
f) The environment can have large temperature variations – leading to resistance variations as large as 4%. Is this a problem for this application? If so,
what can be done in the design of a circuit to reduce or eliminate the effect of
temperature on the sensor output?
8
Figure 4: Industrial-scale drying.
Solutions
a) Divide the total “empty” weight by 4 for each leg.
L=
F
= (12.5kg · 9.8) · 10
k
b)
6
= 1.2x10
4
m
R
L
= Gf
R
L
then
⌦
R = 1000⌦ · 2.5 ·
·
m
✓
1.2x10 4 m
10 2 m
◆
= 30⌦
the total resistance is then
R = 1000 + 30 = 1030⌦
c) The total weight is now 60kg for the machine and clothes. Repeat the
same analysis for parts a and b, above, and you will get
R = 1036.75⌦
d) Same steps as above, but now the total setup weighs 55kg total.
R = 1034.75⌦
e) From problem XX, we know that the optimal resistance is:
p
R = Rwet Rdry = 1035⌦
9
f) It can be a problem. The best thing is to make a whetstone bridge circuit
that has unstrained resistors with the same temperature and same TCR. A
possibility is mounting them on the same legs but rotated 90 degrees so that the
compression of the legs does not produce a strain signal, but they are subjected
to the same temperature variations.’
Problem 6: Measuring flow
In class we learned about thermistors, which change resistance based on a change
in temperature. More specifically, the temperature T at the location of the
thermistor can be obtained by
T = Tamb + Gf R
where Tamb is the ambient temperature (starting temperature) at which the
resistor has a nominal resistance R, and Gf is the thermistor factor. It turns out
we can combine two thermistor to measure flow. Figure 5 shows a sensor which is
called a thermoanemometer. It is composed of three small tubes immersed into a
moving medium. Two tubes contain thermistors R0 and Rs . The detectors are
thermally coupled to the medium and are thermally isolated from the structural
elements and the pipe where the flow is measured. In between the two detectors,
a heating element is positioned. Both detectors are connected to electrical wires
through tiny conductors to minimize thermal loss through conduction 5. The
sensor operates as follows. The first temperature detector R0 measures the
temperature of the flowing medium. A heater warms up the medium in the
middle of the pipe and the elevated temperature is measured by the second
thermistor RS . When the medium flows, heat dissipation increases due to
forced convection. Because the heater is positioned closer to the Rs detector,
that detector will register a higher temperature. The higher the rate of flow, the
higher the heat dissipation and the lower the temperature that will be registered
by the Rs detector. Heat loss can be measured and converted into the flow rate
of the medium. A fundamental relationship for this thermometry setup is based
on a modified version of King’s law, which states
v=
K
⇢
✓
dQ
1
dt Ts T0
◆1.87
(2)
where K is a calibration constant, ⇢ is the density of the liquid or gas flowing
through the pipe, Ts is the surface temperature of thermistor RS , T0 is the
temperature of sensor R0 , and dQ
dt is the thermal loss to the flowing medium
over time. The law of conservation of energy demands that the electric power
in the heating element be equal to thermal loss to flowing medium. The electric
power W across a heater/resistor is given by
W =
10
V2
Rh
Figure 5: Flow anemometerr.
where Rh is the resistance of the heating element, and V is the voltage across
of the heating element.
For this problem, refer to figure 5. R0 , Rs , R1 and R2 all have the same
nominal (ambient) resistance R (e.g. R0 = Rs = R3 = R4 = R). R1 and R2 are
just part of your measurement circuit and are not sensitive to environmental
conditions.
a) Express relation (2) in terms of the voltage V across the heater and its
resistance Rheater.
b) Using the linearized relation (1) from the Wheatstone bridge in problem
1 derive an expression for the temperature difference Ts T0 that only depends
on the input voltage V the measured voltage V0 , the sensor factor Gf , and the
nominal resistance R .
c) Now plug this relation into equation derived in part (a) to obtain a relation
for the flow velocity v based on the measured voltage V0 of your circuit.
d) Plot flow velocity v as a function of the output voltage of your circuit
V0 (from 0 to 300 mV in increments of 0.1 mV), for three fluids: water ⇢ =
1000kg/m3 , propane ⇢ = 500kg/m3 and crude oil ⇢ = 870kg/m3 . Let K =
45000 , V = 5V olts, Rh = 10k⌦, Gf = 0.02, and R = 1k⌦.
11
Solution
a) Just plug in
v=
K
⇢
✓
◆1.87
V2
1
Rh Ts T 0
b) First, express the temperature difference (Ts
terms of the change in the resistance Rs and R0 .
Ts = Tamb + Gf ·
Rs
T0 = Tamb + Gf ·
R0
T0 )between locations in
and
then
Ts
T0 = (Tamb + Gf ·
Rs
Tamb + Gf ·
R 0 ) = Gf · ( R s
R0 )
This circuit can be linearized around a nominal resistance R, just like the circuit
you saw in problem .
✓ ◆
V0
1
R0
RS
R1
R2
=
+
(3)
V
4
R0
RS
R1
R2
V0
1
=
V
4

R0
R
Rs
R
0
0
+
R R
then
( Rs
R0 ) =
4RV0
V
c) plug (b) into (a)
K
v=
⇢
V2
Rh
1
Gf 4RV0
V
d) Plugging in, we get the following image
12
!1.87
13