Strain Measurement

UNIVERSITI TENAGA NASIONAL

COLLEGE OF ENGINEERING

DEPARTMENT OF MECHANICAL ENGINEERING

MESB333 – ENGINEERING MEASURAMENT AND LAB

FORMAL REPORT

STRAIN MEASURAMENT

NAME: NAVIN RAJ A/L K SAKARAN

SID: ME091512

GROUP NO. : 4A1

LAB NO. : 1

DATE PERFORMED: 11/06/2015

DATE SUBMITTED: 06/08/2015

TABLE OF CONTENT

No. Tittle Page No.

1 Summary/Abstract

2 Statement of Purpose

3 Theory

4 Equipment

5 Procedure

6 Data, Observation and Results

7 Analysis and Discussion

8 Conclusion

Summary/Abstract

Statement of Purpose

Part 1: The Bending System

- To show how to measure strains in an object that bends and compare the

results with theory.

Part 2: The Torsion System

- To show how to connect and use shear and torque (torsional) strain gauges to

measure strains in an object that twists.

- To show how to compare displayed strains with theory for a torsion beam.

Part 3: The Tension System

- To show how to connect and use strain gauges to measure strains in two

dimensions.

- To show how to connect the displayed tensile strains in two dimensions with

theory and prove Poisson’s Ratio.

Theory

The Wheatstone Bridge

The basis of most strain measurement is the Wheatstone Bridge. It has four identical

resistance (R1, R2, R3, and R4) connected end to end in a diamond shape. An input

voltage (Vi) connects across two opposite connections. The output voltage is

measured at the other two connections.

The output voltage (Vo) depends on the ratio of the resistors, so that

Vo=( R2R1+R2

− R4R3+R4 )Vi

Figure 1

Figure 2

Type of Bridge Connections

1) The quarter bridge connection

When a single strain gauge replaces one of the resistors, the output voltage

Vo is proportional to the strain in the gauge. When all resistors are equal, the

output potential difference is zero. As the stain gauge resistance increase

(tensile strain), the output potential difference becomes more positive. As the

strain gauge resistance decreases (compressive strain), the output potential

difference becomes more negative.

Figure 3

2) Half Bridge 1 (Opposite Arms)

If the resistance of R1 and its opposite resistor (R4) both increase by the

same amount, the output voltage changes twice as much as if only one

resistor changes. This can obtain more output and therefore higher sensitivity

if two identical stain gauges used together. Each gauge is opposite to the

other, so both gauges must measure the same strain, so that they both

change resistance in the same way. So, two opposing gauges must measure

the same type of strain (tensile or compressive) at e same place on the test

structure.

Figure 4

3) Half Bridge 2 (Adjacent Arms)

The changes in the stain gauges resistance will cancel out each other, so they

must measure identical but opposite strains on the same part of the structure

under test. One gauge must measure compressive strain and the other must

measure tensile strain (or the opposite way around). This bridge will also give

twice as much output as Quarter Bridge. When both strains are equal in

magnitude, the output from the bridges is almost linear.

Figure 5

4) Full Bridge Connection

This bridge gives twice as much voltage output and sensitivity than the

standard half bridge (and four times the output of a quarter bridge). As in the

half bridge, each opposite gauge must measure the same part of the

structure. For example –gauges 1 and 4 must measure tensile strain, while

gauges 2 and 3 must measure compressive strain, or the other way around.

When all strains are equal in magnitude, the output from the bridge is linear.

Figure 6

Strain Bridge Equation

To calculate the strain from the dc voltage across the bridge, the Strain Display uses

a standard equation:

ε=4×Vo

GF× Vi× N

Where

ε = Strain

Vo = Voltage measured across the bridge (V)

GF = Gauge Factor

Vi = Fixed Input Voltage applied to the bridge (V)

N = number of active arms (gauges connected)

The output is then multiplied by 106 to give a result in µε (micro strain) (strain x 10-6)

Mass, Weight and Force

Force (N) = Mass (kg) x Acceleration due to gravity (m.s-2)

Acceleration due gravity, g = 9.81 ms-2

Direct Stress, Strain and Young’s Modulus

Stress (σ)

This is the force applied to a material over a known area. It is found by the equation:

σ=ForceArea

= FA

Strain (ε)

This is the changes in length (distortion caused by stress) of a material over its

original length. It is found by the equation:

ε= change∈lengthoriginallength

=∆ ll

Young’s Modulus (E)

This is a ratio of the tensile stress divided by the tensile strain on a material.

E= stressstrain

=σε

Figure 7

Modulus of Rigidity or Shear Modulus (G)

The shear Modulus or Modulus of Rigidity is a measure of the rigidity of the material

when in ‘shear’ when it is twisting. It is a ratio of the shear stress and the shear strain

of the material:

G= Shear StressShear Stress

= F / A∆ x /h

= τγ

Bending of Beams

Second Moment of Area and Stress

The second moment of area for a rectangular cross-section beam is:

I=bd ³12

Figure 8

Bending Moment

For a cantilever beam (supported at one end), the bending moment:

M=F (l−x )

Figure 9

Stress

From the Engineer’s theory of bending, the theoretical stress at any point

along the beam is:

σ=MyI

Strain

The theoretical strain is simply re-arranged equation of Young’s

Modulus:

ε= σE

Torsional Stress and Strain

Polar Moment of Inertia

Second moment of area for circular and solid cross-section beams.

J= π D4

32

The general equation for torque in a beam (bar) is

TJ=Gθ

l

Torque

The twisting force (torque) at the end of the bar is the moment of force on

torque arm:

T = F x Torque Arm Length (m)

Figure 10

Shear Stress

The theoretical shear stress for the solid circular bar is:

τ=TD2J

Shear Strain

The theoretical shear strain for the solid circular bar is

γ= τG

= rθl

Direct Strain

When a force changes the length of an object, the direct strain (ε) is:

ε= change∈lengthoriginallength

So, direct strain is a change in length, but shear strain is caused by a stress in

two dimensions (a change in shape).

Figure 11

The figure 11 shows a force that changes the shape of a rectangle. The force

causes strain in two dimensions to change the diagonal length of the

rectangle (all other dimensions remain the same). The shear strain is the

amount that the diagonal has hanged. For small angles, the approximation is

that α = γ.

From Pythagoras’s theory, the original (unstrained) diagonal length2 = 12 + 12

So the unstrained diagonal length = √2

The strained diagonal length2 = 12 + (1+γ) 2

So the strained diagonal length = √ [12 + (1+γ) 2]

This multiplies out to:

√1+1+2 γ+γ2

For the small strains in this type of application, γ is small (much less than 1),

so γ2 can be ignored and the equation becomes:

√2+2 γ

Which is √2 (1+γ )12 and approximately √2(1+ γ

2)

So, as direct strain = change in length/original length, then the direct strain in

the diagonal is:

ε=√2(1+ γ

2−1)

√2= γ2

So, for this application of the solid, circular cross-section bar, direct strain is

half the shear strain. Or:

ε= γ2

Tensile Stress and Strain, and Poisson’s Ratio

When pulled or pushed by a force, the stress on different specimen is equal to the

force applied for each unit area.

For rectangular specimen:

ε= Fx × z

Figure 12

The strain in the direction of the force is the stress divided by the Young’s Modulus

for the material:

ε= σE

Poisson’s Ratio (υ)

This is the ratio of the “transverse” strain in a material (at right angles to the

applied stress), against longitudinal strain (in the direction of the applied

stress). The French mathematician- Simeon Poisson discovered it when he

noticed that a material’s cross-section decreases as you stretch it.

The equation is:

υ=−εxεy

Figure 13

For most metal, strain in the direction of stress is three times and opposite in

polarity to the strain measured at right angles to the applied stress. So

Poisson’s ratio for metals is usually 0.3.

When the metal is stretched (positive, tensile strain), the transverse strain is

negative (compressive). This also works in reverse, when the metal is

compressed.

Symbol Notation

Equipment

1) The Strain Gauge Trainer

Figure 14

Figure 15

Technical Details

Table 1

Beams and Optional Specimens

Table 2

Procedure


1) Vernier instrument was used to accurately measure the dimensions of the

specimen beam. The measurements then recorded into the result table.

2) The bending system strain gauges connected to the strain display as full

bridge.

3) The knife –edge hanger was carefully sledded onto the beam to the 420mm

position.

4) The equipment was leaved to stabilize for approximately one minute. The

“zero” button was pressed and hold until the display reading becomes 0

(zero).

5) The strain reading was recorded into the table.

6) The small weight hanger then hooked to the knife-edge hanger.

7) 4 x 10g weights was added to the weight hanger to make it to a weight of 50g.

The strain value then recorded into the table.

8) Using a step of 50g, more weight was added to the hanger until a weight of

500g reached. At each step the strain values recorded into the table.


a) To Use Shear and Torque Strain Gauges

1) Blue strain gauge connected to the strain display as a quarter bridge. The

strain Display was adjusted to show the correct gauge factor and the ACT =1.

2) The torque arm then screwed into the threaded hole at the end of the torsion

system.

3) The equipment was leaved to stabilize for 1 minute. Then the “zero” button

was pressed and hold until the display reading become 0 (zero).

4) A small weight hanger was added to the end of the torque arm.

5) 49 x 10g weights was added to the weight hanger to make it a total weight of

500g. The strain reading and its polarity (+ or -) was recorded into the table.

6) The weights then removed and the experiment was repeated using red,

yellow and green gauges.

b) Compare Strains

1) Vernier instrument was used to accurately measure the dimensions of the

specimen beam. The measurement data then recorded into the table.

2) The torsion system connected “tensile twist” as opposite using red and green

gauges and the blue and yellow gauges “compressive twist” as opposite to

complete a full bridge.

3) The equipment was leaved to be stabilize for 1 minute, then the “zero” button

was pressed and hold until the display become 0 (zero).

4) The strain reading was recorded into the table.

5) A small weight hanger was added to the end on the torque arm.

6) 24 x 10g weights was added to the weight hanger to make it 250g of total

weight. The strain reading then recorded into the table.

7) More weight then added until the weight reaches 500g. The strain reading

then recorded into the table.

8) The weights was removed and the moment arm then unscrewed.


a) Tensile Strains Only (Red and Yellow Gauges)

1) The Vernier instrument was used to accurately measure the dimensions of the

specimen. Then the measurements was recorded into the table.

2) The red and yellow gauges was connected to the Tension system to the

Strain Display as half bridge (opposite). The ACT was set to 2.

3) The equipment then leaved to stabilize for 1 minute and the “zero” button was

pressed until the display reading becomes 0 (zero).

4) The strain readings recorded into the table.

5) A large weight hanger was fitted to the bottom of the Tension System

specimen. The large weight hanger weighs 500g. A weight of 0.5kg added to

the weight hanger to make it a total load of 1kg.

6) The strain value then recorded into the table.

7) More weights were added with a step of 1kg to the mass hanger until the total

mass reaches 9kg. Strain value at each step were recorded into the table.

b) Compressive Strains Only (Blue and Green Gauges)

1) Procedures for Tensile Strains Only (Red and Yellow Gauges), but the blue

and green gauges used instead of red and yellow gauges.

Data, Observation and Results


Beam Dimension: 20mm x 5mm

Young’s Modulus: 207 x 109 N.m-2

Second Moment of Area: 208.33 mm4

Bridge Connection: Full

Load Position: 420mm

Load

(g)

Force

(N)

Strain

Reading

(µε)

Output

Voltage

(µV)

Bendin

g

Moment

(Nm)

Calculated

Stress (N.m2)

Calculated

Strain (µ)

Error

(%)

0 0 0 0 0 0 0 0

50 0.4905 10 28 0.2060 2.04720 x 106 11.9420 16.26

100 0.981 24 63 0.4120 4.9441 x 106 23 .8845 0.48

150 1.04715 38 101 0.6180 704161 x 106 35.8266 6.07

200 1.962 51 135 0.8240 9.8882 x 106 47.7691 6.76

250 2.4525 65 171 1.0301 12.3614 x 106 59.7169 8.85

300 2.943 78 206 1.2361 14.8334 x 106 71.6589 8.85

350 3.4335 92 243 1.4421 17.3055 x 106 83.6014 10.05

400 3.924 105 277 1.6481 19.7775 x 106 95.5435 9.90

450 4.4145 119 314 1.8541 22.2496 x 106 107.4860 10.71

500 4.905 132 349 2.0601 24.7260 x 106 119.4493 10.51

Table 3

Sample Calculation: (for 50g)

1) Second Moment of Area

I=b d3

12=20mm (5mm) ³

12=208.33mm ⁴

2) Force

F=ma=(0.05 ) (9.81 )=0.4905N

3) Bending Moment

M=Fx=(0.4905 ) (0.42 )=0.2060N . m

4) Calculated Stress

σ=MyI

=(0.2060 ) (0.0025 )(2.0833×10−10)

=2.4720×106 N .m−2

5) Calculated Strain

ε= σE

=2.4720×106

207×109=11.9420×10−6

6) % Error

%=|Theory−Experiment|

Theory×100%=11.9420−10

11.9420×100%=16.26%

0 0.00002 0.00004 0.00006 0.00008 0.0001 0.00012 0.000140

5000000

10000000

15000000

20000000

25000000

30000000

f(x) = 188646012344.185 x

Theoretical Stress Vs Experimental Strain

Experimental Strain, ε

Theo

retic

al S

tres

s (N

/m^2

)

Graph 1


Gauge factor: 2.05

Strain GaugeStrain Reading

(µε)Polarity (+/-) Type of Strain

Blue -23 Negative Compressive

Red 21 Positive Tensile

Yellow -23 Negative Compressive

Green 22 Positive Tensile

Table 4

Gauge Factor: 2.05

Beam Diameter: 10mm

Beam Radius: 5mm

Shear Modulus for the Beam: 79.6 x 109 N.m-2

Bridge Connection: Full

Torque Arm Length: 0.15m

Polar Moment of Inertia: 981.75mm4

Load

(kg)

Force

(N)

Torque

(Nm)

Output

Voltage

(µV)

Strain

reading

(µε)

Calculated

Shear Stress

(N.m-2)

Calculated

Direct Strain

(µε)

0 0 0 0 0 0 0

0.25 2.4525 0.3679 -122 -47 1.8737x106 11.7695

0.5 4.905 0.7358 -242 -94 3.7474x106 23.5389

Table 5

Sample Calculation: (Table 5, for load 0.5kg)

1) Polar Moment of Inertia:

J= π D4

32=

π (0.01 )4

32=981.75×10−12m4

2) Force:

F=ma=0.5×9.81=4.905N

3) Torque :

T=F × armlength=4.905×0.15=0.7358N . m

4) Calculated Shear Stress:

τ=TD2J

=(0.7358 ) (0.01 )2(981.75×10−12)

=3.7474×106N .m−2

5) Calculated Direct Strain:

ε= γ2=

τG2

=

3.7474×106

79.6×109

2=23.5389×10−6


Red and Yellow Gauges

Gauge Factor: 2.15

Specimen dimension: 10mm x 2mm

Specimen cross-section: 20mm2

Young’s Modulus: 105x109 N.m-2

Load

(kg)Force (N)

Displayed

Tensile

Strain (µε)

Calculated

Tensile Stress

(N.m-2)

Calculated

Tensile

Strain (µε)

% Error

0 0 0 0 x 106 0 0

1 9.81 5 0.49 x 106 4.7 6.38

2 19.61 10 0.98 x 106 9.3 7.53

3 29.42 16 1.47 x 106 14.0 14.29

4 39.23 21 1.96 x 106 18.7 12.30

5 49.03 26 2.45 x 106 23.3 11.59

6 58.84 31 2.94 x 106 28.0 10.71

7 68.65 36 3.43 x 106 32.7 10.09

8 78.45 41 3.92 x 106 37.3 9.92

9 88.26 47 4.41 x 106 42.0 11.90

Table 6

Blue and Green Gauges

Load (kg) Force (N) Displayed Strain (µε)

0 0 0

1 9.81 -1

2 19.62 -3

3 29.42 -4

4 39.23 -6

5 49.03 -8

6 58.84 -9

7 68.65 -10

8 78.45 -12

9 88.26 -14

Table 7

Full Bridge

Load (kg) Force (N) Displayed Strain (µε)

0 0 0

1 9.81 -5

2 19.62 -10

3 29.42 -15

4 39.23 -21

5 49.03 -26

6 58.84 -31

7 68.65 -36

8 78.45 -42

9 88.26 -47

Table 8

Sample Calculation: (For Load 4kg)

1) Specimen cross-section area:

A=w ×t= (10 ) (2 )=20mm2

2) Force:

F=ma=(4 ) (9.81 )=39.23N

3) Calculated Tensile Stress:

σ= FA

= 39.23

20×10−6=1.96×106N . m−2

4) Calculated Tensile Strain:

ε= σE

=1.96×106

105×109=18.7×10−6

5) % Error

%=|Theory−ExperimentTheory |×100%=|(18.7×10−6−21×10−6 )

18.7×10−6 |×100% ¿12.30%

0 5 10 15 20 25 30 35 40 45 50

-16

-14

-12

-10

-8

-6

-4

-2

0f(x) = − 0.290647482014389 x

Compressive Strain Vs Tensile Strain

Tensile Strain

Com

pres

sive

Stra

in

Graph 2

Discussion


1) Based on the result we obtain, we get a similar experimental strain compared

to the theoretical strain calculated. The maximum percentage error is 16.26%.

2) Based on the graph that plotted, we obtain a young’s modulus of the beam

which is 200x109 N.m-2. This value is almost similar to the original Young’s

Modulus of the beam which is 207x109 N.m-2.

3) There is some errors that occurs during the experiment which makes the

theoretical value and the experimental value to be different. Those are:-

a) There is external air flow which causes the beam to not in stationary

position during the reading was taken.

b) The load not placed exactly at 420mm on the beam due to parallax error.

c) The small weight on the hanger may be not evenly distributed due to non-

linear arrangement of the weight.

4) Some precaution can be taken to prevent errors from occurring:-

a) Make sure there is no air flow around the experiment area so that the

reading will be accurate.

b) Try to avoid parallax error during placing the weight hanger at the 420mm

on the beam.


1) Strain readings for gauges are same value just have different polarity/ gauges

blue and yellow have the same strain reading with negative polarity, while red

and green have the same strain reading with positive polarity. Positive polarity

indicates tensile strain while negative strain indicates compressive strain.

2) Comparing the displayed and theoretical direct strain value shows us that the

displayed strain readings is 4 times more than the calculated theoretical

value.

The percentage difference calculated below:

Load (kg)Displayed Strain

(µε)

Theoretical Direct

Strain (µε)% Error

0 0 0 0

0.25 -47 11.7695 499.34

0.5 -94 23.5389 499.34

Table 9

3) Errors and precautions from this experiment:

a) Errors

- The reading of the strain may be not accurate due to high sensitivity of the

sensor that records even a slight change.

- The hook of the load is not stable during the reading was taken because of air

flow in the surrounding.

b) Precautions

- Make sure the hook of the load does not swing during the reading is taken.

- Balance the load properly so that the load is properly distributed along the

hook.


1) The theoretical strain is different with the displayed reading that obtained from

the experiment. This difference is with a small margin only with the highest

percentage error of 14.29%.

2) Based on the graph that plotted, we can see that the compressive strain

decreases as the tensile strain increases. From the graph the obtained

gradient is -0.2906.

3) To get more accurate graph, use small intervals of load for each reading. For

example instead of adding 1kg of load, add less load such as 500g or less. By

doing his the graph will be more reliable and we can determine the property of

the test specimen more accurately.

4) The strain formula is

ε=4×Vo

GF× Vi× N

Where

ε = Strain

Vo = Voltage measured across the bridge (V)

GF = Gauge Factor

Vi = Fixed Input Voltage applied to the bridge (V)

N = number of active arms (gauges connected)

From the equation we can found that that when the number of gauge is

inversely proportional to the strain value.

ε α1N

This means when the number of gauges is maximum (full bridge) N=4, this

will give the exact value of strain reading.

Conclusion


The experiment was conducted to measure strains in an object that bends

and to compare the results with theory values.

Based on the experiment it found that the experimental values is quite similar

to the theoretical values. Based on the graph plotted, it been found that the Young’s

Modulus of the beam is 200x109 N.m-2 which is slightly different than the actual

Young’s Modulus of the beam which is 207x109 N.m-2.

There is some errors in the experiment that can be eliminated by taking some

precaution measures into account. Overall the objective of this experiment was

achieved.


The experiment was conducted to show how to connect and use shear and

torque (torsional) strain gauges to measure strains in an object that twist and to

show how to compare displayed strains with theory for a torsion beam.

From the experiment it been able to identify which gauges reads the same

strain value and which gauges reads compressive or tensile strain. Blue and yellow

gauges measures compressive strain while red and green gauges reads tensile

strain.

It been also identify that how to compare the displayed strain value that

obtained from the experiment with the calculated direct strain value which is 4 times

less than the displayed strain value.

Thus, the objective of this experiment was achieved.


The experiment was conducted to show how to connect and use strain

gauges to measure strains in two dimensional and to show how to compare the

displayed tensile strains in two dimensions with theory and prove Poisson’s ratio.

From the experiment in been found that the compressive strain value can be

obtained when the blue and green gauges are connected as half bridge and the

tensile strain can be obtained when the red and yellow gauges connected as half

bridge.

Plotting a graph of compressive strain vs tensile strain able to identify the

relationship between these two different strains. A Poisson’s ratio of 0.2906 was

obtained from the gradient of the graph plotted. This value is almost near to the

usual value of Poisson’s ratio in metals which is 0.30.

It’s also been found that connecting the gauges in full bridge setup enables

more accurate result to be obtained based on the strain equation which is

ε=4×Vo

GF× Vi× N

Thus, the objective of this experiment was achieved.

Strain Measurement

Documents

voltage output

strain gauge resistance

compressive strain

wheatstone bridge

strain measurement

single strain gauge

output voltage changes

opposite strains