Strain Measurement
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Transcript of 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
Part 1: The Bending System
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.
Part 2: The Torsion System
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.
Part 3: The Tension System
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
Part 1: The Bending System
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
Part 2: The Torsion System
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
Part 3: The Tension System
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
Part 1: The Bending System
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.
Part 2: The Torsion System
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.
Part 3: The Tension System
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
Part 1: The Bending System
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.
Part 2: The Torsion System
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.
Part 3: The Tension System
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.