fe1073 lab report c1
TRANSCRIPT
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NANYANG TECHNOLOGICAL UNIVERSITY
First Year Common Engineering Course
FE1072: Laboratory Experiment
Protective Engineering Laboratory (N1.1-B5-02)
FORMAL REPORT
Experiment C1: Equilibrium and Elasticity
Name: YEO SHI JING JACKIE
Matric No.: 083642A03
Group: AL08
Date: 20 FEB 2009
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Abstract
This experiment consisted of two parts. In the first experiment, equilibrium of concurrent
force systems was observed. This experiment used pulleys and hanging masses to setup 2
forces. Equilibrant force was determined from the setup. This force was used to compare with
the resultant force of the 2 forces. In theory, the 2 forces are equal in magnitude and opposite
in direction, so they cancel out each other. However, there may be possible sources of error
which result in equilibrant force do not exactly balance the resultant force. In the later part of
the report, we will discuss the possible sources of error in the measurements and construction,
and possible ways to improve it.
The second experiment was conducted to determine 3 results: strain in a truss member, elastic
modulus of material involved, and truss stiffness. During the experiment, strain increments
on one of the truss member while loading and unloading (increase P and decrease P) were
recorded. From the result, a graph of load P versus strain increment was plotted (See Page
24 of the report for the graph). It was found out that the load P was linearly proportional to
the strain increment . This showed that the truss member had undergone linear elastic
deformation, which meant the truss member obeyed Hooke’s law ( = ). The slope of the
graph was determined, and hence the elastic modulus and truss stiffness. The calculations
were shown on Page 12 of the report.
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Introduction
The purpose of the first experiment was to measure force vectors, force resultants, and
observe equilibrium of concurrent force systems.
Vector is defined both by its direction, the direction of arrow, and by its magnitude, which is
proportional to the length of arrow. An example is shown in Figure 1, a vector ⃗ makes an
angle with the horizontal axis (Direction), with a length λ (Magnitude).
Vector forces acting on the same point of the object are called concurrent forces. The
resultant forces can be determined by adding the vector forces together using the
parallelogram method (as shown in Figure 2).
Another vector shown in Figure 2 is the equilibrant of 1 and 2. is the force needed to
exactly offset the combined effect of 1and 2, which is . has the same magnitude as
, but is in the opposite direction.
Fig. 1 shows a vector ⃗ with length λ and
makes an angle with the horizontal axis.
Fig. 2 shows the resultant
vector force as the
result of 1 + 2. It also
shows, , the equilibrant
force which has the same
magnitude as but is inthe opposite direction.
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The purpose of the second experiment was to measure the deformation (strain) in a truss
member, determine the modulus of elasticity of the material involved and also determine the
stiffness of the truss model.
A truss (as shown in Figure 3(a)) is a structure composed of slender members joined together
at their end points. Each truss member acts as a two-force member. If the force tends to
elongate the member, it is a tensile force (T); whereas if it tends to shorten the member, it is a
compressive force (C).
In a statistically determinate truss, the forces (tension or compression) in all the members can
be calculated by considering equilibrium at joints.
The truss shown in Figure 3(a) has pin-joints at B, D and G. Deformations in DB are
neglected since the member DB is very rigid when compared with GB. There are two
unknowns acting on GB (Tensile force in GB and Compression force along DB) and
we can write two equilibrium equations to solve for and . However, we will be
interested only in . By equilibrium at joint B (Figure 3(b)), we obtain
=
sin =
ℎ/=
ℎ Equation (1)
ℎ
D
B
G
Fig. 3 (a) shows a
truss which
composed of
slender members
joined together at
their end points.
P
B
P
Fig. 3 (b) shows all
forces acting at joint B
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For truss member GB, we define normal strain (Consequences of the load) as deformation
per unit length,
=
where is the total deformation between the ends of the member (See Figure 5).
Note: (a) is the original length before loading.
(b) Strain is always dimensionless.
=
=
ℎ
Given a cross sectional area A of member GB, the stress (force per
unit area) in member GB will be
Note: (a) A is the original area before loading.
(b) If force is in Newton (N) and the unit for length is in
millimetre (mm), the stress will be in / or /
or .
Equation (2)
Equation (3)
Fig. 5 shows the totaldeformation between
the ends of the
member GB.
Fig. 4 shows a small portion
of the member GB. It
indicates the variable and
that are needed to
calculate the stress .
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Most engineering structures are designed to function within the linear elastic range, i.e., the
stress is linearly proportional to the strain ,
=
This relation is known as Hooke’s law. Figure 6 shows a graph plotted against . The
coefficient E (Gradient of the graph) is called the modulus of elasticity (or Young’s
modulus). It is a measurement of stiffness of the material involved.
Substitute Equation (2) and Equation (4),
=
=
ℎ
Note: (a) The unit for E is / or .
(b) The strain equals to the measured strain increment ′ multiplied by a conversion factor
due to the instrumentation.
Fig. 6 shows graph plotted against
. The gradient of the graph is called
the modulus of elasticity.
Equation (4)
Equation (5)
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Another quantity which is of interest in this case is the deflection at joint B where force P is
applied. When the deformations in DB can be neglected, B will moved to B’ due to the
extension in the member GB and a rotation of GB about G (See Figure 7). Since the
deflections are small compared with the original length of the members, the circular arcs BB’
and LB’ due to rotation of DB and GL respectively, can be approximated to straight lines. It
follows that the right-angled triangle BLB’, shown in Figure 7, can represent the relationship
between and deflection of B. Therefore,
=
sin =
ℎ/ =
2
ℎ
Define a stiffness coefficient (truss stiffness), k, for the vertical deflection at B, as
=
=
ℎ
2
By plotting the relationship between P and the measured strain for a range of applied P
values, the Young’s modulus E and the stiffness coefficient k can be determined from the
experiment using Equations (5) and (7).
Equation (6)
Equation (7)
BD
G
L
B’
Fig. 7 shows deflection at joint B
where force P is applied.
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Equipments
Equipment was set up as in Experiment No. C1 laboratory manual. Refer to page 3 (For
experiment 1) and 6 (For experiment 2) of the laboratory manual.
Experimental Procedure
Experiment 1:
1) Use pulleys and hanging masses (1 = 55, 2 = 105) to setup the equipment as
shown in Figure 8, so that two known forces, 1 (=1) and 2 (=2), are pulling
the force ring.
2)
Use holding pin to prevent the ring from moving. The holding pin provides a force,
, that is exactly opposite to the resultant of 1 and 2.
1 2
1
2
1
2
Spring Balance
Pulley 1
Pulley 3
Pulley 2
Force Ring
HoldingPin
Figure 8 shows experiment 1 setup.
Experiment Board
Degree ScaleZero-degree line
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3) Adjust the spring balance to determine the magnitude of . As shown in Figure X,
keep the spring balance vertical and use Pulley 3 to direct the force from the spring in
the desired direction. Move the spring balance towards or away from the pulley to
vary the magnitude of the force. Adjust Pulley 3 and the spring balance so that the
holding pin is centred in the force ring.
Note: To minimize the effects of the friction in the pulleys, tap as needed on the
Experiment Board each time you re-position any component. This will help the force
ring come to its true equilibrium position.
4)
Record the values of the hanging masses (1 = 55, 2 = 105); the magnitude of
1, 2, (in Newton); the angles 1, 2and that each vector makes with respect
to the zero-degree line on the degree scale (See Figure 8). The results are shown on
page 21 of the report.
5) Change the hanging masses to 1 = 135, 2 = 205 and repeat step (1) to (4) one
more time.
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Experiment 2:
1)
Before the experiment is carried out, measurements of all necessary dimensions are
made. Use the calipers to measure the breadth and the thickness of the truss member
GB, and compute its cross sectional area. Repeat the procedure 2 more times along
the truss member GB, and find the average cross sectional area of GB.
2) Use a ruler to measure the value of (The length of truss member GB), and ℎ (The
vertical length from joint G to the member DB).
Note: The measurement, , should be measured from the joint point G to the other one, B,
whereas the measurement, ℎ, should be measured from the joint point G to the line that cuts
through both joint point D and B (See Figure 9).
3) Before the weights are loaded, check the strain monitoring equipment is zeroed
(′ = 0). Then load a weight of 10N (which means P = 10N). Wait for the reading to
be stabilised. Record the value of ′ , read off from the strain monitoring machine.
4) Load another 10N weight (which means P = 20N) and record the value of ′ .
5) Repeat step (4) until P = 60N. The results are shown on page 23 of the report.
6) Now unload one 10N weight (which means P = 50N) and record the value of ′ .
Figure 9 shows how to measure
and ℎ for experiment 2 setup.
Line that cuts through both
joint point D and B
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7) Repeat step (6) until P = 0N. The results are shown on page 23 of the report.
8) Compute the values of strain for both loading and unloading.
Strain = strain increment ′ × conversion factor (CF)
Note: The conversion factor (CF) for this experiment = 0.5
Results
Experiment 1:
Appendix A1 shows two tables. Table 1 shows all the recorded values of 1, 2, , 1,
2and for the case where 1 = 55, 2 = 105. Table 2 shows all the recorded values
of 1, 2, , 1, 2and for the case where 1 = 135, 2 = 205. For each case, the
recorded values are used to construct 1, 2, with an appropriate scale (For 1st case:
10cm/Newton, For 2nd
case: 5cm/Newton). are also drawn on the same diagram for each
case using the parallelogram method. These diagrams are shown in Appendix A2.
From the diagrams,
Case 1: Length of = 6.9cm
∴ Magnitude of =6.9
10= 0.69
Case 2: Length of = 6.3cm
∴ Magnitude of =6.3
5= 1.26
From the above calculations, the equilibrant force vector, , does not exactly balance the
resultant vector, , for each case.
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Percentage error: −ℎ
ℎ × 100%
Case 1: Percentage error =0.70−0.69
0.69× 100% = 1.45%
Case 2: Percentage error =1.20−1.26
1.26× 100% = −4.76%
Experiment 2:
Table 3 in Appendix B1 shows the measurements of the breadth and the thickness along the
truss member GB, and the average cross sectional area is found to be 81.72. Also from
the experiment, h is determined to be 427 mm and l is determined to be 450 mm.
Another table, Table 4, shows all the recorded strain increments , while loading and
unloading. The table also includes computed strain , which can be found by
Strain = strain increment ′ × conversion factor (CF)
Note: The conversion factor (CF) for this experiment = 0.5
A graph of load P versus strain increment for the truss model is plotted and it is presented
in Appendix B2 of the report. From the graph, we can see that load P was linearly
proportional (Straight line graph) to the strain increment .
Slope of the straight line graph =
=
60
12×10−6= 5 × 1 06 N
From the slope of the graph, the elastic modulus E of the truss member GB, and the stiffness
k for the vertical deflection at B can be calculated.
Elastic modulus E =
ℎ= 5 × 1 06 ×
450
81.7×427= 64.5 /2
Truss stiffness k =
ℎ2 = 5 × 1 06 × 427
4502 = 10500 /
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Discussion
Experiment 1:
From the diagrams drawn in Appendix A2 and the calculations on Page 11 of the report, the
equilibrant force vector, , does not exactly balance the resultant vector, , for each case.
In next paragraph, we are going to discuss some possible sources of error in the
measurements and constructions that had caused the above result.
All experiments done by students, teachers, and even scientists are not perfect. There are
bound to have errors involved in the experiments. Some possible sources are:
1) There are friction between the contact surface of the string and pulleys. Although
tapping the experiment board minimizes the effect of friction, it does not remove the
effect of friction completely. Hence, the reading obtained from the spring balance is
not what we expected.
2) The portion of string that connects the spring balance is not parallel to the portion of
the string that hangs the 1 or 2 (See Figure 10). Hence, the spring balance may be
slanted at an angle, which results in an inaccurate reading of .
Fig. 10 shows the portion of string
that connects the spring balance
is not parallel to the portion of the
string that hangs the 1 or 2
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3) Even if it is parallel, there may be case where the spring balance is not upright (See
Figure 11), hence results in an inaccurate reading of .
4) A spring balance does not retain its accuracy permanently, for no matter how
carefully it is handled, the spring very gradually uncoils even though its limit of
elasticity has not been exceeded. Hence if the spring balance had been use dozens of
times for measurement, we may not obtain a very accurate reading of . Accuracy in
this experiment is important because the force to be measured, , is very small (<
2N).
Further Discussion:
For the given masses 1, 2 and the measured angles 1, 2, calculate the equilibrant
and its direction using the equilibrium conditions. Compare the calculated results with the
measured values.
Case 1:
1 : X-component = 0.54 cos168 = −0.5282
Y-component = 0.54 sin 168 = 0.1123
Fig. 11 shows spring balance is
not upright, hence results in an
inaccurate reading of .
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2 : X-component = 1.03 cos 26 = 0.9258
Y-component = 1.03 sin 26 = 0.4515
: X-component = 0.9258 – 0.5282 = 0.3976N
Y-component = 0.4515 + 0.1123 = 0.5638N
Under equilibrium conditions, | | = | | = 0.690
= 180° − = 180 − tan−1 0.5638
0.3976= 125°
Comparison:
() ()
Calculated 0.69 125
Measured 0.70 126
Case 2:
1 : X-component = 1.32 cos168 = −1.291
Y-component = 1.32 sin 168 = 0.2744
2 : X-component = 2.01 cos 26 = 1.807
Y-component = 2.01 sin 26 = 0.8811
: X-component = 1.807– 1.291 = 0.516N
Y-component = 0.2744 + 0.8811 = 1.156N
Under equilibrium conditions, | | = | | = 1.26
= 180° − = 180 − tan−1 1.156
0.516= 114°
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Comparison:
() ()
Calculated 1.26 114
Measured 1.20 117
Discuss the possible ways to improve the measurement accuracy.
1) Besides tapping the experiment board, we can also lubricate the contact surface of the
pulleys to minimize the effect of friction. This will further minimize error (1) stated in
Page 13 of the report.
2)
Before the experiment starts, place a meter ruler vertically upright in front of the
board (Shown in Figure 12). This is a check to make sure that the spring balance is
upright. Also when you conduct this check, make sure the board and the ruler are
placed on the same flat surface. This will eliminate error (2) & (3) stated in Page 13
&14 of the report.
3) Instead of using a spring balance, use a high tech digital force gauge to obtain a much
accurate reading of . This will minimize error (4) stated in Page 14 of the report.
Me
t
e
r
R
u
l
e
r
Fig. 12 shows a meter
ruler placed vertically
upright in front of the
board to eliminate error
(2) & (3).
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Experiment 2:
Discussion:
What is the significance of observed strains fitting into a straight line in the plot of P vs. ?
• Fitting all the points into a straight line signifies load P is linearly proportional to the
strain increment . This shows that the truss member had undergone linear elastic
deformation, which means the truss member obeys Hooke’s law ( = ).
If member GB is replaced by another member of the same length but different cross sectional
area, will the tensile force be different under the same load P (assuming small
deformation anyway)? What about the deflection at point B?
• From equation (1),
=
sin =
ℎ/=
ℎ
depends on 3 factors, P, l, and h. None of these factors are related to the breadth
and the thickness of member GB. Hence, tensile force will be the same under the
same load with different cross sectional area.
• From equation (6),
=
sin =
ℎ/=
2
ℎ
depends on 3 factors, , l, and h. l, and h are not related to the breadth and the thickness
of member GB. However, is dependant on cross sectional area. Based on equation (5),
=
=
ℎ
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Rearranging equation (5),
=
=
ℎ
Note that E is a constant, and P, l, and h are not related to the breadth and the thickness of
member GB. This shows that is dependent on cross sectional area and hence the deflection
at point B will be different under the same load but with different cross sectional area A of
member GB.
Further Discussion:
Discuss qualitatively how the rigidity of member DB affects the results of the experiment?
• The rigidity of member DB indicates whether there is deformation in DB. In this
experiment, DB is rigid with respect to GB. DB is non-deformable – that is, for ideal
rigid body, the relative locations of all particles of which the object is composed
remain constant. Hence, there will not be any deflection shown (in DB with respect to
GB) at joint B which will make our calculation much simple. However, if the DB
used in the experiment is not rigid, then the deformations in DB cannot be neglected
and hence deflections will be significant and we will have another unknown .
Hence, Equation (6) (on Page 7 of the report) is not valid because circular arcs BB’
and LB’ due to rotation of DB and GL respectively, cannot be approximated to
straight lines. Therefore, making our calculation much more difficult.
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Why is it preferable to measure the strain along GB rather than the deflection with a dial
gauge mounted on the truss at B?
•
A dial gauge is able to detect smallest dimensional variations (Up to millimetres).
However, the dial gauge is not preferable. Due to the moving parts (Levers), friction
is generated within the gauge, hence reducing the accuracy of the reading obtained.
The accuracy of the reading is important because the deflection of B is very small in
this experiment.
• It can also generate error due to parallax with the dial gauge as pointer moves over a
fixed scale. However we do not have this problem if we use strain monitoring
equipment (digital) to measure strain along GB.
• The dial gauge is also sensitive to small vibration because the mechanisms in these
gauges have more inertia. It takes a longer time to get an accurate reading because the
weights tend to sway a little in the air while we were loading and unloading (increase
P and decrease P respectively) during the experiment, and hence the reading fluctuates
before it is stabilized.
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Conclusion
Experiment 1:
In theory, the equilibrant vector force exactly offsets the combined effect of 1and 2,
which is the resultant force . However, in this experiment we observed that does not
exactly balance out . This may due to possible sources of error in our measurements and
constructions (This had been discussed in Page 13 &14 of the report).
Experiment 2:
From the graph, it was found out that the load P was linearly proportional to the strain
increment . This showed that the truss member had undergone linear elastic deformation,
which meant the truss member obeyed Hooke’s law ( = ). The slope of the graph was
determined,
Slope of the straight line graph = = 60
12×10−6 = 5 × 1 06 N
And hence, the elastic modulus
Elastic modulus E =
ℎ= 5 × 1 06 ×
450
81.7×427= 64.5 /2
and truss stiffness.
Truss stiffness k =
ℎ
2= 5 × 106 ×
427
4502= 10500 /
References
1) Laboratory Manual Experiment No. C1.
2) R.A. Serway & J.W. Jewett, 2008, "Physics for Scientists and Engineers with Modern
Physics", 7th Edition, Thomson Brooks/Cole Publishing.
3) William D. Callister, Jr., 2007, “Materials Science and Engineering, An
Introduction”, 7th Edition, John Wiley & Sons, Inc.
4) A/P Tan Ming Jen, 2009, FE1005 Materials Science Lecture Notes.
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Appendices
Table 1: (Case 1) 1 = 55, 2 = 105
1 (g)
2 (g)
1 (degree)
2 (degree)
1 = 1 (N)
2 = 2 (N)
(N)
(degree)
55 105 168 26 0.54 1.03 0.70 126
Table 2: (Case 2) 1 = 135, 2 = 205
1 (g)
2 (g)
1 (degree)
2 (degree)
1 = 1
(N)
2 = 2
(N) (N)
(degree)
135 205 168 26 1.32 2.01 1.20 117
Appendix A1
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Appendices
Appendix A2
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Appendices
Table 3: Cross sectional area of truss member GB
1 2 3
Breadth (mm) 12.75 12.75 12.70
Thickness (mm) 6.40 6.45 6.40
Area (2) 81.6 82.2 81.3
Average Area, A =81.6+82.2+81.3
3= 81.72
Other measurements: h = 427mm l =
P
(N)
450mm
Table 4:
Strain Increment
Loading Unloading
, (× 10−6) (× 10−6) , (× 10−6) (× 10−6)
0 0 0 0 0
10 4 2 4 2
20 8 4 8 4
30 12 6 12 6
40 16 8 16 8
50 20 10 20 10
60 24 12 24 12
Appendix B1
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Appendices
Appendix B2