truss report
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FACULTY OF CIVIL AND ENVIRONMENT
ENGINEERING
STRUCTURE AND MATERIAL LABORATORY
BFC21201
FORCE IN A STATICALLY DETERMINATE CANTILEVER TRUSS
GROUP MEMBER :
NAME MATRIC NUMBER
ESTHER MARIE JUIP CF120139
FAEZAH BT MD. ADNAN CF120256
FARADILLAH BINTI ABDUL LAIT CF120250
RABIATUL ADAWIYAH BINTI
ZAINAL ABIDINCF120053
LECTURERS NAME : MR. MOHAMAD HAIRI BIN OSMAN
SECTION : 3
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1.0 OBJECTIVE1.1 To examine a statically determinate frame and to analyze the frame using simple
pin joint theory.
2.0 LEARNING OUTCOME2.1 The application the engineering knowledge in practical application.
2.2 To enhance technical competency in structural engineering through laboratory
application.
2.3 To communicate effectively in group.
2.4 To identify problem, solving and finding out appropriate solution through
laboratory application.
3.0 THEORYA truss is a structure composed of slender member joined together at their
endpoints to form one or more triangles. The joint connections are considered aspinned
joint without friction.
In order to determine the forces developed in the individual members at a truss,the
following assumptions should be make :
1. The members are connected to each other at their ends by friction less pins,that is
only a force and no moment can be transferred from one member toanother.
2. External loads are applied to the truss only at its joints.
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One of the methods to calculate the forces in the member of a truss is using:
Method Of Joints
Suitable to use in calculating all of the member forces for a truss.
This method entails the use of a free body diagram of joints with the
equilibrium equationsFx = 0 and Fy = 0.
Calculation only can be started for joint where the numbers of unknowns are two or less.
4.0 APPARATUS
The strain force meter input Digital indicator meter acting to
acting in the force acting in the rod frame
Apparatus to apply the load Statically determinate cantilever
truss apparatus
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5.0 PROCEDURESi. Unscrew the thumbwheel on the redundant members. Note that it was
effectively no longer part of the structured as the idealized diagram
illustrates.
ii. Applied the pre-load of 100N downward, the load cell was re-zero and
applied it a load of 250N carefully and check that the frame is stabled and
secured.
iii. The load to zero (leaving the 100N preload), recheck and re-zero the
digital indicator. Never applied the loads greater than those specified on
the equipment.
iv. Loads apply in the increment shown in Table 1 recording the strain
readings and the digital indicator readings. Table 2 was completed by
subtracting the initial (zero) strain readings. (be careful with sign)
6.0 RESULTS
Load
(N)
Strain Reading Digital
Indicator
Reading
(mm)1 2 3 4 5 6 7 8
0 131 223 - 13 -37 113 0 31 52 0
50 141 214 -22 -55 113 0 44 66 -0.031
100 149 205 -30 -73 113 0 56 78 -0.059
150 158 196 -40 -92 113 0 69 92 -0.088
200 167 186 -49 -111 113 0 83 106 -0.11
250 174 179 -56 -126 112 0 94 118 -0.132
Table 1: Strain Reading and Frame Deflection for Experiment 1
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Load
(N)1 2 3 4 5 6 7 8
0 0 0 0 0 0 0 0 0
50 10 -9 -9 -18 0 0 13 14
100 18 -18 -17 -36 0 0 25 26
150 27 -27 -27 -55 0 0 38 40
200 36 -37 -36 -74 0 0 52 54
250 43 -44 -43 -89 -1 0 63 66
Table 2: True Strain Reading For Experiment 1
MemberExperiment
force (N)
Theoretical
force (N)
1 239.48 250
2 -244.12 -250
3 -238.57 -250
4 -493.79 -500
5 -5.55 0
6 0 0
7 349.54 353.55
8 366.18 353.55
Table 3: Measured and Theoretical Force in the Cantilever Truss
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Analysis and Calculation
True Strain Reading For Experiment 1(table 2)
Example:
Load = 0, 50, 100, 150, 200, 250
True Strain Reading = BA
B = load at strain needed
A = load at before strain needed
Member 1
Load 50 Load 100
True Strain Reading = 141131 True Strain Reading = 149 - 131
= 10 = 18
Load 150 Load 200
True Strain Reading = 158131 True Strain Reading = 167 - 131
= 27 = 36
Load 250
True Strain Reading = 174 - 131
= 43
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From the formula:
E = / = F/A
E = Young s Modulus (Nm-2
) F = Force in member (N)
= Displayed Strain A = cross section area of the member (m2)
= Stress in the member (Nm-2
)
= E
F/A = E
F = E x x A
Calculation For Experimental Force (N)
Given ,
Rod diameter (d) = 5.80 mm
E steel = 2.10 x105
N/mm2
A = d2
/4
= (5.80)2
/4
= 26.42 mm3
Member 1
F = E x x A
= (2.10x105) x (43x10
-6) x (26.42)
= 239.48 N
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Member 2
F = E x x A
= (2.10x105) x (-44x10
-6) x (26.42)
= -244.12 N
Member 3
F = E x x A
= (2.10x105) x (-43x10
-6) x (26.42)
= -238.57 N
Member 4
F = E x x A
= (2.10x105) x (-89x10
-6) x (26.42)
= -493.79 N
Member 5
F = E x x A
= (2.10x105) x (-1x10
-6) x (26.42)
= -5.55 N
Member 7
F = E x x A
= (2.10x105) x (63x10
-6) x (26.42)
= 349.54 N
Member 8
F = E x x A
= (2.10x105) x (66x10
-6) x (26.42)
= 366.18 N
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2
34
578
1
B
A
DC
E
Calculation for Theoretical Force (N)
Checking :
Members, m = 7
Reactions, r = 3
m + r = 2j
7 + 3 = 2(5)
10 = 10
.. The structure is statically determinate and stable.
By using equivalent equation :
+ A = 0 + y = 0
-FxB(0.24) + 250(0.48) = 0 FyA250 = 0
FxB = -120/-0.24 FyA = 250 N
FxB = 500 N
FyA
FxA
Fx B
250 N
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FAB
500 N FAB
250 N
-500 N FAE
0
FAC
FACCos
FACSin
+x = 0
FXA + FXB = 0
FXA + 500 = 0
FXA = -500 N
At point B
Fx = 0
500 + FBC = 0
FBC = -500 N (C)
Fy = 0
FAB = 0
At point A
= 45
Fy = 0
250 + FABFACSin = 0
250 + 0 - FACSin45 = 0
FAC = 353.33 N (T)
Fx = 0
-500 + FAE + 353.55cos45 = 0
FAE = 250 N (T)
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FCE
FDEcos
FDEsin FDE
250 N
FCD
-353.55 N
-353.55sin
-353.55cos
At point E
Fx = 0
-250 + FDEcos45 = 0
FDE = 353.55 N (T)
Fy = 0
FCEFDEsin = 0
FCE(-353.55sin45) = 0
FCE = -250 N (C)
At point D
Fx = 0
-353.55cos + FCD
-(-353.55cos45) + FCD = 0
FCD = -250 N (C)
FAB = 0
FAC = 353.33 N (T)
FAE = 250 N (T)
FBC = -500 N (C)
FCD = -250 N (C)
FCE = -250 N (C)
FDE = 353.55 N (T)
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250 N
500 N
-500 N
250 N
250 N
0
353.55 N
-500 N -250 N
250 N
-353.55 N
A
B CD
E
7.0 DATA ANALYSIS
Member 1
0
50
100
150
200
250
0 50 100 150 200 250
TrueandRecordedStrain()
Load (N)
True and Recorded Strain Against Load
Recorded Strain
True Strain
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Member 2
Member 3
0
50
100
150
200
250
0 50 100 150 200 250
TrueandRecordedStrain()
Load (N)
True and Recorded Strain Against Load
True Strain
Recorded Strain
-60
-50
-40
-30
-20
-10
0
0 50 100 150 200 250
Truea
nd
RecordedStrain()
Load (N)
True and Recorded Strain Against Load
Recorded Strain
True Strain
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Member 4
Member 5
-140
-120
-100
-80
-60
-40
-20
0
0 50 100 150 200 250
T
rueand
Recorded
Strain()
Load (N)
True and Recorded Strain Against
Load
Recorded
Strain
True Strain
-20
0
20
40
60
80
100
120
0 50 100 150 200 250True
and
RecordedStrain()
Load (N)
True and Recorded Strain Against
Load
Recorded Strain
True Strain
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Member 7
Member 8
10
20
30
40
50
60
70
80
90
100
0 50 100 150 200 250
Trueand
RecordedStrain()
Load (N)
True and Recorded Strain Against Load
True Strain
Recorded Strain
0
20
40
60
80
100
120
140
0 50 100 150 200 250
Tr
ueand
RecordedStrain()
Load (N)
True and Recorded Strain Against Load
Recorded Strain
True Strain
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Deflection against Load graph
Comment on a graph:
From the graph of True and Recorded Strain against Load above, we obtained that thevalue of recorded and true strain are different. This is because in a statically determinate
cantilever truss is leaving the 100 N pre-load.
According to the True and Recorded Strain against Load, it shows the line graph are
obtaining from the result. For member 1, 7 and 8 we obtained the linear graph, the graph true
strain and recorded proportional with load. This is because the value of true strain and recorded
can increases when the load increases. The recorded strain are larger than true strain values.
From the graph, it shows the true and recorded strain for the member 2, 3 and 4 aredecrease when the load is increase. For the member 5, the graph shows it does not have an
increasing or decreasing with a large value. The values are same with different load.
From the deflection graph, we obtained that the graph are in linear. When the load is
increase, the deflection decrease.
-0.14
-0.12
-0.1
-0.08
-0.06
-0.04
-0.02
0
0 50 100 150 200 250
Deflection
(mm)
Load (N)
Deflection Against Load
Deflection
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8.0 DISCUSSION AND CONCLUSION
1. Compare the experimental and theoretical result:
MemberExperiment
force (N)
Theoretical
force (N)
1 239.48 250
2 -244.12 -250
3 -238.57 -250
4 -493.79 -500
5 -5.55 0
6 0 0
7 349.54 353.55
8 366.18 353.55
From the table above, we obtained the differences value between the value of
experimental force and theoretical force. This may because of several factors. Such as:
One of the factor is machine error.
The applied pre-load of 100N downward maybe less or more than 100N.
The imperfection of members used for the particular experiment also affects
the result of the experiment.
The load cell was not accurately point to zero value that gave affect to the
value of force.
After the apply load of 250N, the frame not checking.
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2. From your result and the theoretical member force, identify which members are
in compression and which members are in tension. Explain your choice.
Base on the figure 1, we can conclude that members are in compression is member
3,4 and 7. The members are in tension is member 1, 2 and 8. This is because the value of
force is negative are considered as compression force and force with positive values are
considered as tensional forces. The negative value shows that the force is pointing inwards
and the positive values denote the force pointing outward.
Axial force member :
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FAB
500 N FAB
3. Observe the reading of members 5. Explain why the readings is almost zero.
At point BFx = 0
500 + FBC = 0
FBC = -500 N (C)
Fy = 0
FAB = 0
From the results on member 5, we noticed that the reading force is almost zero. We
know that, member 5 is attached by a pin joint and a roller joint at both ends. On the pin
joint, two forces acting towards it on the horizontal axis and vertical axis. On the roller joint,
there is only one force acting towards it on the horizontal axis. Therefore, the reading force
of member 5 is almost zero due to these three forces.
4. Are the strains gauges are an effective transducers for measurement forces in the
framework.
Yes, the strains gauges are effective transducers for measurement of forces in the
framework. We can get the effective transducers from the data we obtain.
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5. Does the framework comply with pin joint theory even though the joint are not truly
pin joint?
Based on the experiment, we know that the values between experimental force and
theoretical force are totally different. This result indicates that the framework comply with
the pin joint theory even though the joint are not truly pined joint.
One of the objective of this experiment is to examine a statically determinate frame
and to analyze the frame using simple pin joint theory. As on conclusion, we can summaries
that our experiment is achieved the objective of experiment.
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