Transcript Lecture Notes

ME 322: Instrumentation
Lecture 7
February 1, 2016
Strain Gage Introduction,
Demonstration, Wheatstone bridge,
Temperature compensation
Announcements/Reminders
• HW 2 due now (before lecture starts)
• This week in Lab:
– Lab 3 Pressure Transmitter Calibration
– Acquire your own data and use it to calibrate the
pressure transmitter you’re issued
– There are only 4, two-port pressure standards
which can be used by two groups at a time. There
are 16 groups in lab, so please be patient (and
prepared).
– Please bring an electronic copy of the Excel
Workbook you created for homework
Pressure Gage
• Device
– Pressure difference causes diaphragm deformation
– Deformation can be measured using a Strain Gage
Strain Gauge Construction
∆𝐿
•
•
•
•
Thin substrate (0.0002”)
Firmly bounded to surface
Metal or Semiconductor foil on or within substrate
Surface deformation stretches or compresses the foil and
changes its resistance.
– Measure DR = R - RI
– However,
∆𝐿
𝐿
&
∆𝑅
𝑅
are small
• Demo: Gage resistance on an aluminum beam in bending
Stain Gage Applications
• Can measure Strain of a deformed part (or material elastic modulus)
• Can be incorporated in devices to sense
– Applied force or weight
– Acceleration
– Pressure
What is Strain?
L
δ
F
F
• Unit strain 𝜀 =
deformation
initial length
– Microstrain 𝜇𝜀 = 𝜀 × 106
• Strain is caused by stress
–𝜎=
𝐹
𝐴
• For Linear elements 𝜀 =
=
δ
L
𝜇in
in
=
𝜎
E
– E ≡ Elastic or Young's Modulus
in
in
=
cm
cm
=1
𝜇m
m
lbf
in2
= psi
𝑁
m2
• Material property = fn(Temperature, material)
• Describes material stiffness
• In Lab 5 we measure E for aluminum and steel beams
= Pa
To use a Strain Gage, firmly bond it to a
gage
Surface
specimen
• Do this in Lab 4 (next week)
• The gage will experience nearly the same stain
and the object’s (specimen’s) surface
– eGAGE = eSURFACE
– Does not measure internal stains
• The gage deformation will affect the gage
resistance, R
– The initial (un-deformed) resistance is RI
• How to predict DR = R – RI?
Wire Resistance
L, R
A
D
• Depends on length, area and material property
–R =
L
𝜌
A
=
4𝐿𝜌
𝜋D2
– r ≡ Electric Resistivity
• Material property = fn(Temperature, Strain, …)
• Stretching wire changes L, D & ρ by small
amounts, and results in small changes in R
Effect of Small Changes in L, D & ρ on R
• R=
4𝐿𝜌
𝜋D2
• DR = R – RI
• For small changes DR = dR, use the chain rule:
–
–
𝜕𝑅
𝜕𝑅
𝜕𝑅
dR= 𝑑L + 𝑑D + 𝑑𝜌
𝜕𝐿
𝜕𝐷
𝜕𝜌
4𝜌
−2 4L𝜌
𝑑𝑅 =
𝑑L +
𝜋D2
𝜋D3
𝑑D +
• Divide by R to get fractional change
–
•
𝑑𝑅
𝑅
=
𝑑𝐿
𝐿
𝑑𝐷
−2
𝐷
+
4L
𝜋D2
𝑑𝜌
𝑑𝜌
𝜌
• Relates fractional change in 𝑅 to fractional changes in L, D, 𝜌
d( )
How do the
terms relate to strain e ?
()
Evaluate Terms
•
dL
L
= 𝜀a axial strain (assuming 𝜀a = eSpeciman)
• This is what we are tying to measure!
•
dD
D
≡ Transverse strain = 𝜀t = −υ𝜀a
– υ ≡ Gage material Poison’s ratio
• ~ 0 – 0.5
• ~ 0.3 for metals
•
d𝜌
𝜌
= Crystalline realignment effect = Cstrain εa
– Cstrain ≡ Strain Coefficient of Resistivity
• Material property = fn(Temperature, …)
• Can be large, and > or < 0 for semiconductors
Combined Effects
•
•
•
dR
𝑅
dR
R
dR
R
=
dL
L
−
dD
2
D
+
d𝜌
𝜌
= εa − 2 −υ𝜀𝑎 + Cstrain εa
= 1 + 2υ + Cstrain 𝜀𝑎 = 𝑆𝜀𝑎
• S = 1 + 2υ + Cstrain = 𝑆𝑡𝑟𝑎𝑖𝑛 𝐺𝑎𝑔𝑒 𝐹𝑎𝑐𝑡𝑜𝑟
– Dimensionless
– Metal Foils:
S = 1.6 to 4 (typical 2.07)
– Semiconductors: S = -140 to 175
–
dR
R
• S=
= S𝜀𝑎
dR
R
ε
=
dR
dL
R
L
L
Example
D
• Apply 1000 lbf to diameter D = 0.25” steel rod.
– E = 207 GPa = 30x106 psi
– For a strain gage with: RI = 120 Ω, S = 2.07
• Find final resistance: R = RI + ∆R
• Solution
•
∆𝑅
𝑅
=𝑆 𝜀
–𝜀=
–
(𝜎)
𝐸
=
𝜇𝑖𝑛
679
=
𝑖𝑛
∆𝑅
=𝑆 𝜀
𝑅
𝐹
𝐸(𝐴)
=
4𝐹
𝜋𝐸𝐷2
=
4(1000 𝑙𝑏𝑓)
𝑙𝑏𝑓
𝜋(30,000,000 2 )(0.25𝑖𝑛)2
𝑖𝑛
0.068%
= 2.07 0.000679 = 0.001406
• R = 120 + 120(0.001406) = 120.17 W
– Very small fractional change!
Ω
Ω
=
𝑖𝑛
0.000679
𝑖𝑛
=
Undesired Temperature Sensitivity
• Gauge resistivity 𝜌 is affected by temperature
• Thermal expansion of specimen and gage may be different
– This can stain the gage
• Resistance is determined by measuring the voltage
across the gage while passing a current though it,
which can heat the gage!
• Temperature factor ST:
–
dR
R
= S𝜀𝑎 + ST ∆T
Desired measurand
Gage Temperature Change = undesirable sensitivity
• Can a circuit “automatically” compensate?
Wheatstone Bridge Circuit
• Two voltage dividers
• 𝑉0 = 𝑉DC − 𝑉BC =
V𝑠
𝑉𝑠
𝑅3
− 𝑅4
𝑅2 +𝑅3
R3
•
𝑉0
𝑉𝑠
𝑅1 +𝑅4
=
𝑅3 𝑅1 +𝑅4 −𝑅4 𝑅2 +𝑅3
𝑅2 +𝑅3 𝑅1 +𝑅4
=
𝑅1 𝑅3 −𝑅2 𝑅4
𝑅2 +𝑅3 𝑅1 +𝑅4
• Use strain gages for some or all of these resisters
• If R1R3 ~ R2R4, then the output voltage will be close
to 𝑉0 ≈ 0
Initial State
-
+
-
R3
𝑉0
•
𝑉𝑠
=
𝑅1 𝑅3 −𝑅2 𝑅4
𝑅2 +𝑅3 𝑅1 +𝑅4
+
• Choose initial resistances so that initial VO,I ~ 0
– R1R3~ R2R4 (For example, all could be ~equal)
• Small changes in each Ri will cause a relatively large
fractional change in VO (compared to VO,I ~ 0)
• To increase 𝑉0
– Increase 𝑅1 or 𝑅3 and/or decrease 𝑅2 or 𝑅4
Effect of small resistance changes on VO
• 𝑉0 =
𝑅1 𝑅3 −𝑅2 𝑅4
𝑅2 +𝑅3 𝑅1 +𝑅4
𝑉𝑠
• Chain Rule:
– 𝑑𝑉0 =
R3
𝜕𝑉0
𝑑𝑅1
𝜕𝑅1
– Find all four
𝜕𝑉0
𝜕𝑉
𝜕𝑉
𝑑𝑅2 + 0 𝑑𝑅3 + 0 𝑑𝑅4
𝜕𝑅2
𝜕𝑅3
𝜕𝑅4
𝜕𝑉
partial derivatives 0 and plug in
𝜕𝑅𝑖
+
– use R1R3= R2R4 and 𝑑𝑉0 = 𝑉0 − 𝑉0𝐼 = V0 − 0 = V0 ,
– Simply ...
• 𝑉0 =
𝑉𝑆 𝑅2 𝑅3
𝑅2 +𝑅3 2
𝑑𝑅1
𝑅1
−
𝑑𝑅2
𝑅2
+
𝑑𝑅3
𝑅3
−
𝑑𝑅4
𝑅4
• If all four resistances start off roughly the same (satisfies 𝑅1 𝑅3 ≈ 𝑅2 𝑅4)
– 𝑅1 = 𝑅2 = 𝑅3 = 𝑅4 and
𝑑𝑅2
𝑅2
+
• For precision resistors,
𝑑𝑅𝑖
𝑅𝑖
–
𝑉0
𝑉𝑆
=
1
4
𝑑𝑅1
𝑅1
−
𝑅2 𝑅3
𝑅2 +𝑅3 2
𝑑𝑅3
𝑅3
−
=
𝑅2
2𝑅 2
=
1
4
𝑑𝑅4
𝑅4
= 0, but not for stain gages
Incorporate gages into some bridge legs
•
𝑉0
𝑉𝑆
=
1
4
𝑑𝑅1
𝑅1
−
𝑑𝑅2
𝑅2
+
𝑑𝑅3
𝑅3
−
𝑑𝑅4
𝑅4
• To increase 𝑉0
– Increase 𝑅1 and 𝑅3 and/or decrease 𝑅2 and 𝑅4
• Install stain gages in some or all legs
–
𝑑𝑅𝑖
𝑅𝑖
= 𝑆𝑖 𝜀𝑖 + 𝑆𝑇𝑖 ∆𝑇𝑖 = 𝑆𝜀𝑖 + 𝑆𝑇 ∆𝑇𝑖
– Use gages with the same characteristics
R3
• Ri=R, Si =S, and STi = ST
• If identical gages are installed in all four legs, then
–
𝑑𝑉0
𝑉𝑠
=
1
4
𝑆 𝜀1 − 𝜀2 + 𝜀3 − 𝜀4 + 𝑆𝑇 ∆𝑇1 − ∆𝑇2 + ∆𝑇3 − ∆𝑇4
Quarter Bridge
+
-
R3
-
• In general
–
𝑉0
𝑉𝑆
=
1
4
𝑑𝑅1
𝑅1
−
𝑑𝑅2
𝑅2
+
𝑑𝑅3
𝑅3
−
𝑑𝑅4
𝑅4
• For quarter bridge with a gage only at 3
– 𝑑𝑅1 = 𝑑𝑅2 = 𝑑𝑅4 = 0
•
–
𝑑𝑅3
𝑅3
= 𝑆𝜀3 + 𝑆𝑇 ∆𝑇3
𝑉0
𝑉𝑆
1
4
𝑆𝜀3 + 𝑆𝑇 ∆𝑇3
=
undesired sensitivity
+
+
Half Bridge
•
𝑉0
𝑉𝑆
=
1
4
𝑑𝑅1
𝑅1
−
𝑑𝑅2
𝑅2
+
𝑑𝑅3
𝑅3
−
𝑑𝑅4
𝑅4
• For this half bridge
-
-
R3
+
– 𝑑𝑅1 = 𝑑𝑅4 = 0
–
•
𝑉0
𝑉𝑆
𝑑𝑅2
𝑅2
=
= 𝑆𝜀2 +
1
4
𝑑𝑅3
𝑆𝑇 ∆𝑇2 ;
𝑅3
= 𝑆𝜀3 + 𝑆𝑇 ∆𝑇3
− 𝑆𝜀2 + 𝑆𝑇 ∆𝑇2 + 𝑆𝜀3 + 𝑆𝑇 ∆𝑇3
– If Gage 3 is placed on a deformed specimen (𝜀3 , ∆𝑇3 ) and Gage 2
is placed on an identical but un-deformed specimen, then
– ∆𝑇2 = ∆𝑇3 , 𝜀2 = 0
–
𝑉0
𝑉𝑆
=
1
4
𝑆𝜀3 (Automatic Temperature Compensation)
Beam in Bending: Half Bridge
ε3
ε2 = -ε3
• Place gage 2 on the side opposite of gage 3, so ε2 = -ε3
•
𝑉0
𝑉𝑆
–
=
𝑉0
𝑉𝑆
1
4
𝑆 𝜀3 − 𝜀2 + 𝑆𝑇 ∆𝑇3 − ∆𝑇2
1
2
ε2 = -ε3
=
2
𝑆𝜀3
4
= 𝑆𝜀3
– Twice the output amplitude of a quarter bridge, and with
temperature compensation
Full Bridge
-
•
𝑉0
𝑉𝑆
=
1
4
𝑑𝑅1
𝑅1
−
𝑑𝑅2
𝑅2
-
+
+
𝑑𝑅3
𝑅3
−
R3
𝑑𝑅4
𝑅4
• All four legs are identical stain gages
–
𝑑𝑉0
𝑉𝑠
=
1
4
𝑆 𝜀1 − 𝜀2 + 𝜀3 − 𝜀4 + 𝑆𝑇 ∆𝑇1 − ∆𝑇2 + ∆𝑇3 − ∆𝑇4
• In bridge, opposite legs (1, 3) and (2, 4) reinforce
– Adjacent legs (1, 2) and (3, 4) oppose
+
Beam in Bending: Full Bridge
3
2
•
•
𝑉0
𝑉𝑠
𝑉0
𝑉𝑠
=
=
1
+
-
-
R3
4
+
1
4
𝑆 𝜀1 − 𝜀2 + 𝜀3 − 𝜀4 + 𝑆𝑇 ∆𝑇1 − ∆𝑇2 + ∆𝑇3 − ∆𝑇4
1
4
𝑆 4𝜀3 + 𝑆𝑇 0
= e3
= -e3
= -e3
= DT3
= 𝑆𝜀3
• V0 is 4 times larger than quarter bridge
– And has temperature compensation.
= DT3
= DT3
Tension
2
3
R3
4
1
ε4=-υ ε3
ε2=-υ ε3
ε1=ε3
General Guidelines for HWs
• Use your Course ID numbers (which you can find in MyNevada), not your
name or student ID number, when submitting your HWs
• Units and significant digits always!
• No hand-drawn plots! Starting from HW3, whenever you are asked to plot
your data, plot them using computer software, i.e. Excel, Matlab, Mathcad
etc.
– Include labels for both axes with the units. If necessary, include legends too.
Make it look good!
• Show you work! Do not skip the steps and just write your final answer.
Whenever applicable, list your assumptions, write out your formulas and
work through to your final answer. If you use a Table (or graph) on your
solutions, give reference to that table in your book, i.e from Table 6.3, z=1.28.
• Be clear with your solutions and work neatly! If the grader needs to spend
more than 3 minutes to figure out what you write, you may not get even
partial credit.