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 Unit strain 𝜀= deformation initial length = δ L in in = cm cm =1 Microstrain 𝜇𝜀=𝜀× 10 6 𝜇in in = 𝜇m m Strain is caused by stress 𝜎= 𝐹 𝐴 For Linear elements 𝜀= 𝜎 E E ≡ Elastic or Young's Modulus lbf in 2 =psi 𝑁 m 2 =Pa Material property = fn(Temperature, material) Describes material stiffness In Lab 5 we measure E for aluminum and steel beams

To use a Strain Gage, firmly bond it to a Surface 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? specimen

Wire Resistance Depends on length, area and material property L, R A D Depends on length, area and material property R= L A 𝜌= 4𝐿𝜌 𝜋 D 2 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 𝜋 D 2 DR = R – RI For small changes DR = dR, use the chain rule: dR= 𝜕𝑅 𝜕𝐿 𝑑L+ 𝜕𝑅 𝜕𝐷 𝑑D+ 𝜕𝑅 𝜕𝜌 𝑑𝜌 𝑑𝑅= 4𝜌 𝜋 D 2 𝑑L+ −2 4L𝜌 𝜋 D 3 𝑑D+ 4L 𝜋 D 2 𝑑𝜌 Divide by R to get fractional change 𝑑𝑅 𝑅 = 𝑑𝐿 𝐿 −2 𝑑𝐷 𝐷 + 𝑑𝜌 𝜌 Relates fractional change in 𝑅 to fractional changes in L, D, 𝜌 How do the d( ) ( ) terms relate to strain e ?

Evaluate Terms dL L = 𝜀 a axial strain (assuming 𝜀 a = eSpeciman) This is what we are tying to measure! dD D ≡ Transverse strain = 𝜀 t =−υ 𝜀 a υ ≡ Gage material Poison’s ratio ~ 0 – 0.5 ~ 0.3 for metals d𝜌 𝜌 =Crystalline realignment effect= C strain ε a Cstrain ≡ Strain Coefficient of Resistivity Material property = fn(Temperature, …) Can be large, and > or < 0 for semiconductors

Combined Effects dR 𝑅 = dL L −2 dD D + d𝜌 𝜌 dR R = ε a −2 −υ 𝜀 𝑎 + C strain ε a dR R = 1+2υ+ C strain 𝜀 𝑎 = 𝑆 𝜀 𝑎 S=1+2υ+ C strain =𝑆𝑡𝑟𝑎𝑖𝑛 𝐺𝑎𝑔𝑒 𝐹𝑎𝑐𝑡𝑜𝑟 Dimensionless Metal Foils: S = 1.6 to 4 (typical 2.07) Semiconductors: S = -140 to 175 dR R =S 𝜀 𝑎 S= dR R ε = dR R dL L

Example 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 = 𝐹 𝐸(𝐴) = 4𝐹 𝜋𝐸 𝐷 2 = 4(1000 𝑙𝑏𝑓) 𝜋(30,000,000 𝑙𝑏𝑓 𝑖𝑛 2 ) (0.25𝑖𝑛) 2 =0.000679 𝑖𝑛 𝑖𝑛 =679 𝜇𝑖𝑛 𝑖𝑛 =0.068% ∆𝑅 𝑅 =𝑆 𝜀 =2.07 0.000679 =0.001406 Ω Ω R = 120 + 120(0.001406) = 120.17 W Very small fractional change!

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 =S 𝜀 𝑎 + S T ∆T Can a circuit “automatically” compensate? Desired measurand Gage Temperature Change = undesirable sensitivity

Wheatstone Bridge Circuit Two voltage dividers 𝑉 0 = 𝑉 DC − 𝑉 BC = 𝑅 3 V 𝑠 𝑅 2 + 𝑅 3 − 𝑅 4 𝑉 𝑠 𝑅 1 + 𝑅 4 𝑉 0 𝑉 𝑠 = 𝑅 3 𝑅 1 + 𝑅 4 − 𝑅 4 𝑅 2 + 𝑅 3 𝑅 2 + 𝑅 3 𝑅 1 + 𝑅 4 = 𝑅 1 𝑅 3 − 𝑅 2 𝑅 4 𝑅 2 + 𝑅 3 𝑅 1 + 𝑅 4 R3 Use strain gages for some or all of these resisters If R1R3 ~ R2R4, then the output voltage will be close to 𝑉 0 ≈0

𝑉 0 𝑉 𝑠 = 𝑅 1 𝑅 3 − 𝑅 2 𝑅 4 𝑅 2 + 𝑅 3 𝑅 1 + 𝑅 4 Initial State R3 - + 𝑉 0 𝑉 𝑠 = 𝑅 1 𝑅 3 − 𝑅 2 𝑅 4 𝑅 2 + 𝑅 3 𝑅 1 + 𝑅 4 - + 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 𝑉 0 Increase 𝑅 1 or 𝑅 3 and/or decrease 𝑅 2 or 𝑅 4

Effect of small resistance changes on VO 𝑉 0 = 𝑅 1 𝑅 3 − 𝑅 2 𝑅 4 𝑅 2 + 𝑅 3 𝑅 1 + 𝑅 4 𝑉 𝑠 Chain Rule: 𝑑 𝑉 0 = 𝜕 𝑉 0 𝜕 𝑅 1 𝑑 𝑅 1 + 𝜕 𝑉 0 𝜕 𝑅 2 𝑑 𝑅 2 + 𝜕 𝑉 0 𝜕 𝑅 3 𝑑 𝑅 3 + 𝜕 𝑉 0 𝜕 𝑅 4 𝑑 𝑅 4 Find all four partial derivatives 𝜕 𝑉 0 𝜕 𝑅 𝑖 and plug in use R1R3= R2R4 and 𝑑 𝑉 0 = 𝑉 0 − 𝑉 0𝐼 = V 0 − 0 = V 0 , Simply ... 𝑉 0 = 𝑉 𝑆 𝑅 2 𝑅 3 𝑅 2 + 𝑅 3 2 𝑑 𝑅 1 𝑅 1 − 𝑑 𝑅 2 𝑅 2 + 𝑑 𝑅 3 𝑅 3 − 𝑑 𝑅 4 𝑅 4 If all four resistances start off roughly the same (satisfies 𝑅 1 𝑅 3 ≈ 𝑅 2 𝑅 4 ) 𝑅 1 = 𝑅 2 = 𝑅 3 = 𝑅 4 and 𝑅 2 𝑅 3 𝑅 2 + 𝑅 3 2 = 𝑅 2 2𝑅 2 = 1 4 𝑉 0 𝑉 𝑆 = 1 4 𝑑 𝑅 1 𝑅 1 − 𝑑 𝑅 2 𝑅 2 + 𝑑 𝑅 3 𝑅 3 − 𝑑 𝑅 4 𝑅 4 For precision resistors, 𝑑 𝑅 𝑖 𝑅 𝑖 =0, but not for stain gages

Incorporate gages into some bridge legs 𝑉 0 𝑉 𝑆 = 1 4 𝑑 𝑅 1 𝑅 1 − 𝑑 𝑅 2 𝑅 2 + 𝑑 𝑅 3 𝑅 3 − 𝑑 𝑅 4 𝑅 4 To increase 𝑉 0 Increase 𝑅 1 and 𝑅 3 and/or decrease 𝑅 2 and 𝑅 4 Install stain gages in some or all legs 𝑑 𝑅 𝑖 𝑅 𝑖 = 𝑆 𝑖 𝜀 𝑖 + 𝑆 𝑇𝑖 ∆ 𝑇 𝑖 =𝑆 𝜀 𝑖 + 𝑆 𝑇 ∆ 𝑇 𝑖 Use gages with the same characteristics Ri=R, Si =S, and STi = ST If identical gages are installed in all four legs, then 𝑑 𝑉 0 𝑉 𝑠 = 1 4 𝑆 𝜀 1 − 𝜀 2 + 𝜀 3 − 𝜀 4 + 𝑆 𝑇 ∆ 𝑇 1 −∆ 𝑇 2 +∆ 𝑇 3 −∆ 𝑇 4 R3

Quarter Bridge In general For quarter bridge with a gage only at 3 + - - + In general 𝑉 0 𝑉 𝑆 = 1 4 𝑑 𝑅 1 𝑅 1 − 𝑑 𝑅 2 𝑅 2 + 𝑑 𝑅 3 𝑅 3 − 𝑑 𝑅 4 𝑅 4 For quarter bridge with a gage only at 3 𝑑 𝑅 1 =𝑑 𝑅 2 =𝑑 𝑅 4 =0 𝑑 𝑅 3 𝑅 3 =𝑆 𝜀 3 + 𝑆 𝑇 ∆ 𝑇 3 undesired sensitivity 𝑉 0 𝑉 𝑆 = 1 4 𝑆 𝜀 3 + 𝑆 𝑇 ∆ 𝑇 3

Half Bridge - + For this half bridge 𝑉 0 𝑉 𝑆 = 1 4 𝑑 𝑅 1 𝑅 1 − 𝑑 𝑅 2 𝑅 2 + 𝑑 𝑅 3 𝑅 3 − 𝑑 𝑅 4 𝑅 4 For this half bridge 𝑑 𝑅 1 =𝑑 𝑅 4 =0 𝑑 𝑅 2 𝑅 2 =𝑆 𝜀 2 + 𝑆 𝑇 ∆ 𝑇 2 ; 𝑑 𝑅 3 𝑅 3 =𝑆 𝜀 3 + 𝑆 𝑇 ∆ 𝑇 3 𝑉 0 𝑉 𝑆 = 1 4 − 𝑆 𝜀 2 + 𝑆 𝑇 ∆ 𝑇 2 + 𝑆 𝜀 3 + 𝑆 𝑇 ∆ 𝑇 3 If Gage 3 is placed on a deformed specimen ( 𝜀 3 , ∆ 𝑇 3 ) and Gage 2 is placed on an identical but un-deformed specimen, then ∆ 𝑇 2 =∆ 𝑇 3 , 𝜀 2 =0 𝑉 0 𝑉 𝑆 = 1 4 𝑆 𝜀 3 (Automatic Temperature Compensation)

Beam in Bending: Half Bridge ε3 ε2 = -ε3 Place gage 2 on the side opposite of gage 3, so ε2 = -ε3 𝑉 0 𝑉 𝑆 = 1 4 𝑆 𝜀 3 − 𝜀 2 + 𝑆 𝑇 ∆ 𝑇 3 −∆ 𝑇 2 = 2 4 𝑆 𝜀 3 𝑉 0 𝑉 𝑆 = 1 2 𝑆 𝜀 3 Twice the output amplitude of a quarter bridge, and with temperature compensation ε2 = -ε3

Full Bridge R3 + - 𝑉 0 𝑉 𝑆 = 1 4 𝑑 𝑅 1 𝑅 1 − 𝑑 𝑅 2 𝑅 2 + 𝑑 𝑅 3 𝑅 3 − 𝑑 𝑅 4 𝑅 4 All four legs are identical stain gages 𝑑 𝑉 0 𝑉 𝑠 = 1 4 𝑆 𝜀 1 − 𝜀 2 + 𝜀 3 − 𝜀 4 + 𝑆 𝑇 ∆ 𝑇 1 −∆ 𝑇 2 +∆ 𝑇 3 −∆ 𝑇 4 In bridge, opposite legs (1, 3) and (2, 4) reinforce Adjacent legs (1, 2) and (3, 4) oppose

Beam in Bending: Full Bridge 3 1 + - - + 2 4 𝑉 0 𝑉 𝑠 = 1 4 𝑆 𝜀 1 − 𝜀 2 + 𝜀 3 − 𝜀 4 + 𝑆 𝑇 ∆ 𝑇 1 −∆ 𝑇 2 +∆ 𝑇 3 −∆ 𝑇 4 𝑉 0 𝑉 𝑠 = 1 4 𝑆 4 𝜀 3 + 𝑆 𝑇 0 =𝑆 𝜀 3 V0 is 4 times larger than quarter bridge And has temperature compensation. = e3 = -e3 = -e3 = DT3 = DT3 = DT3

Tension R3 2 3 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.