# FUNDAMENTALS AND APPLICATION OF STRAIN GAGES by bjb17276

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FUNDAMENTALS AND APPLICATIONS OF STRAIN GAGES
WHAT IS STRAIN?

L

L


For a member under tensile stress:

L = deformation caused by the tensile stress.
L = original length
 = applied stress (force/area)

By definition:

Strain = L / L = 

Stress =         =
Area

STRESS - STRAIN DIAGRAM OF A MATERIAL

x
STRESS

LINEAR ELASTIC STRAIN LIMIT

STRAIN
FSG 2
ELECTRICAL RESISTANCE OF A WIRE AFFECTED BY STRAIN

l
Recall       R
a

P                                                                           P

A
L                         L

When the wire has a strain of L / L in the longitudinal direction, the
corresponding strains in the lateral directions are:

r      L
                  Where:  (or  ) = Poisson’s ratio
r       L

Then the change in cross-sectional area is:

a = r2 – ( (r - r)2 )= 2rr + (r)2  2rr

L                   L
a = -2r2           , a = r2 = -2a 
L                    L

R due to L:

 R     R     R                                 L     L
R    L    a                                L  2 a  
 l     a                                 a     a      a

R L a            a       L                            R L      L 
         , since     2    ,                                2   
R   L   a             a       L                             R   L      L   

R                  
R  1  2      
= the Gage Factor
L                  L
L                 L
FSG 3
R
R   lies between approximately 1.9 and 2.1
L
L

ELECTRICAL WIRE STRAIN GAGES

1. Wire Gage

2. Etched Foil Gage

3. Semiconductor Gage

 Gage Factors as high as 100

4. Deposited Gage

 Gage is deposited on the tested material
 Gage can be better positioned
 No bonding problem

5. Weldable Strain Gages
 Very convenient – easy to apply
FSG 4
RELATIONSHIP BETWEEN CHANGE IN RESISTANCE AND STRAIN

R
 (GF)  where: GF = Gage Factor and  = strain
R

Example: What is the change in R (resistance) in the Gage?
400 psi

E = 4.0 106 psi
Strain Gage
G.F. = 2.05
R = 120 

Solution:  = 400 / 4 x 106 = 1 x 10-4 R / R = (2.05)(1 x 10-4) = 2.05 x 10-
4

R = (120 ) (2.05 x 10-4) = 0.0246  (not very large)

How can we measure such a small change in resistance?

WHEATSTONE BRIDGE
i1 + i2

R1                      R2

+
VO                                i1        i2
_

R4                      R3

RG

Galvanometer - very
sensitive amp meter
FSG 5

For a balanced bridge, the current through the galvanometer is zero and
the voltage across the galvanometer is also zero.

i1 R1 = i2 R2                                            (1)

i1 R4 = i2 R3                                            (2)

Divide (1) by (2)

R1 R 2                                                                      R2
                             or                          R1 = R4
R4   R3                                                                     R3

For R1 = R2 = R3 = R4 = Ro (initially) and if R1 changes by R
i1 +i2

R0 + R                               R0

+                                           i1     i2
VO
_
iG
R0                       R0

i1 +i2
RG

_        V +

For small R :
Vo R
IG =                        (ignoring the higher order terms)
4R o (R o  R G )

V =
Vo R R G
=
GF          RG
Vo
4R o (R o  R G )                  4         (R o  R G )

                 
GF  1               
=                         VO
4  1 R O           
 R               
     G           
FSG 6
GF
For RG  Ro,      V =            Vo . This is known as a quarter bridge.
4

For a strain gage located at position R1 in a Wheatstone bridge circuit,

GF
V =        Vo
4

In the previous example, if Vo = 20 volts,

V =
2.05
4

1 x10  4 20 V      = 1.025 x 10-3 volts

QUARTER BRIDGE WITH TEMPERATURE COMPENSATION

Active Gage     R1               R2

+
VO
_

R4              R3

Dummy Gage
for Temperature                  V
Compensation

Since the active gage and the dummy gage are identical, the changes in R
in the two gages due to thermal effects are the same.

(R1)thermal = (R4)thermal

R1 R 2
For a balanced bridge:                  
R4 R3
FSG 7
With the thermal effects on both gages, the bridge is still balanced.

R 1 + R 1  R   R
 1  2
R 4 + R 4  R4 R3

The circuit is self-compensating for thermal effects.

ALTERNATE CONNECTION FOR DUMMY GAGE (QUARTER BRIDGE)

Dummy Gage
for Temperature
Active Gage    R1                R2    Compensation

+
VO
_

R4               R3

V

R1  R1 R 2  R 2
R1 + R` = R2 + R2                           
R4        R3

SELF - TEMPERATURE - COMPENSATING GAGES

Gages with temperature sensitivity the same as the material it is placed on
- used with specified test materials.

Example: steel or aluminum compensated gages
FSG 8

USE OF TWO ACTIVE GAGES FOR INCREASED SENSITIVITY (more
voltage output for a given strain – higher output to noise ratio)


1        3



Gage 1
R1             R2

+
VO
_
Gage 3
R4             R3

V

R1 = R2 = R3 = R4

For a strain of :
(GF) 
V due to Gage 1 =           Vo
4

(GF) 
V due to Gage 3 =           Vo
4
FSG 9
(GF) 
Total measured V =              Vo    but
2

General Strain Gage Equation:

(GF) 
V =           N Vo
4

Where N = sensitivity factor (or # of active gages) (N = 2 in this
example).

HALF BRIDGE TO MEASURE BENDING STRAINS

Strain Gage 1

Strain Gage 2

1 = strain at Gage 1 (tension)
2 = strain at Gage 2 (compression)

2 = -1

Gage 1                                         Gage 2
R1                   R2
(Tension)                                   (Compression)
+
VO
_

R4                   R3

V

(GF)  1
V due to Gage 1 =                Vo
4
FSG 10
(GF)  2                (GF)  1
V due to Gage 2 =                       Vo       =             Vo
4                       4
(GF)  1
TOTAL V =      Vo                           (N = 2)
2
SHEAR STRAIN MEASUREMENT

X


                 L



x
Shear Strain             
l

MOHR CIRCLE FOR STRAINS


 0 /2


4                               1           
0

-0/2                                      Gage 1
Gage 4

At location of Gage 4,  = 4 = o/2

At location of Gage 1,  = 1 = o/2
FSG 11
BRIDGE CONNECTION FOR SHEAR STRAIN MEASUREMENT

R                 R
Gage 1
+
VO
_
Gage 4
R                 R

V

FOUR ARM (FULL) BRIDGE

Strain Gages 1 & 3
Strain Gages 1 & 3

Strain Gages 2 & 4

Side View                                     Top View

R                 R
Gage 1                                       Gage 2
+
VO
_
Gage 4                                     Gage 3
R                 R

V

V = (GF)  V0 , N = 4
FSG 12

Problems associated with attaching the connecting wires.

For Quarter Bridge

Power                        V
Supply

1. The resistance of the leadwires will add to the resistance of the
gage.

**2. The effects of temperature on the leadwires will affect the strain

Solutions:

1. Use as short as possible leadwires with large cross-sectional area
to minimize resistance.

2. Use dummy gage with lead wires the same as the active gage

Active Gage

Power                         V
Supply
Dummy Gage
FSG 13
The temperature effects on the leadwires will be self-compensating if the
leadwires to both the active gage and dummy gage are identical (same
length and type).

Best Solution - Use a three-wire circuit. No DUMMY gage required

Power                           V
Supply

Thermal effects will be minimized if leadwires are identical. Note the wire to
the voltmeter is not a leadwire and is not part of the circuit.

CALIBRATION OF BRIDGE CIRCUITS (to verify that what you are
reading on the bridge output is correct)

RS
Shunting Resistor

R0                  R0

+
VO
_

R0                  R0

V
FSG 14
When the calibrating resistance, RS is shunted across Ro, the resulting
drop in resistance of that arm is:

2
R oR S            Ro
R  R o                  
Ro  RS           Ro  R S

 R    Ro                  Ro
                                for RS  Ro
 Ro  Ro  RS               RS

1  R       1 Ro
V           Vo       Vo
4  Ro       4 RS

The measured V is similar to that produced by a strain of S

N
V due to a strain, S, =
4
GF  SVo

N                     1 R                       1   Ro
Set
4
GF  S Vo  VS  4 R o Vo      S 
N GF R S
S

When RS is shunted across Ro, the resulting V is the same as that caused
by a strain of S.

1    Ro
Output of VS indicates strain of
N GF  R S
FSG 15

INSTRUMENTATION FOR MEASUREMENT OF DYNAMIC STRAINS

RS

R0                       R0
Active Gage
+
VO
_

R0                   R0

Amplifier   Output to
Oscilloscope

MEASUREMENT OF LARGE STRAINS

R
as high as 10%
R
R
V 
R Vo
              R
Vo

GF  Vo
4 R  2 R           
4 1 
1 R 


4 1 
GF 

    2 R                    2   

END OF TEST MATERIAL NUMBER 1

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