Students create a tension load cell by bonding two strain gages to a three-inch

Question

Students create a tension load cell by bonding two strain gages to a three-inch length of 1-inch diameter round aluminum rod. Unfortunately the gages selected are mismatched: One is a 350-ohm gage while the other is a 120 ohm gage. Calculate the system load sensitivity (dE_o/dP)in units of milli-volts per pound of load. Data: Nominal unstrained resistance of gage 1 = R1 = 350 Ohm, gage factor F=2.0. Nominal unstrained resistance of gage 2 = R2= 120 Ohm, gage factor F=2.0. Fixed resistor values are both Ro=350 Ohm, circuit input voltage Ei = 10.0 volts. E_aluminium=10*10^6 psi. Neglects any temperature considerations.

Explanation / Answer

We are given following data:

R1 = 350 Ohm and Gauge Factor = 2.0

R2 = 120 Ohm and Gauge factor = 2.0

R0 = 350 Ohm for both fixed resistor

Young's Modulus for Aluminium Ealuminium = 10*106

We will first calculate change in resistance in strain gaugfe for one pound of load.

From change in resistance we can find out unbalanced bridge voltage.

This value of voltage = (dE0) / ( dP )     (Since voltage calculated for one pound of load )

Now Gauge Factor = (dR/R) / (dl/l)      (dR and dl are change in resistance and change length for applied load)   

dR/R = (Gauge Factor) x (dl /l)

dR = (Gauge Factor) x (dl /l) x R

dR = (Gauge Factor) x (dl /l) x R        ......1

Now Young's Modulus E = (Stress ) / (Strain)     (Strain = dl/l)

Now stress = applied load = 1 pound & Young's modulus Ealuminium = 10*106

We can find strain dl/ l from formula for Young's modulus

Therefore Strain (dl/l) = Stress / Ealuminium = 1 / ( 10* 106)

Putting this value of dl/l in equation 1 we get

dR = (Gauge Factor) x (1 / ( 10* 106) x R        ......2

Now we can find change in value of resistance for R1 and R2 from equation 2

dR1 = 2 x (1 / ( 10* 106 ) x 350 = 7*10-5    Ohm

dR2 = 2 x (1 / ( 10* 106 ) x 120 =2.4*10-5    Ohm

These are the change in values of resistance of strain gauge for 1 pound load

Now we can find unbalanced voltage E0 for bridge using Ohm's Law

Therefore E0 = Ei { [ R0 / ( R0 + R2 ) ] -   [ R1 / ( R0 + R1 ) ] } ....... 3

Now we have R0 = 350 & Ei = 10 Volts

Applied strain is tensile therefore change in value of resistance is positive

Hence new values

R1= R1+dR1= 350 + 7*10-5    Ohm   

R2= R2+dR2 = 120 + 2.4*10-5    Ohm

Putting these new values of R1,R2 and give values of R0 & Ei in equation 3 we get

E0 = 244.7 mV    

Now this is the change in bridge voltage for 1 pound of load

(dE0) / (dP) = 244.7 mV / Pound   is the required system load sensitivity

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