indeterminate-truss-lab-report.pdf

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Indeterminate Truss Lab Report
Structural engineering lab (Universiti Teknologi MARA)
StuDocu is not sponsored or endorsed by any college or university
Indeterminate Truss Lab Report
Structural engineering lab (Universiti Teknologi MARA)
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3. TABLE OF CONTENT

CONTENT PAGE
INTRODUCTION 1
OBJECTIVE 1
PROBLEM STATEMENT 1
APPARATUS/ MATERIAL 1
PROCEDURES 2
DATA & CALCULATION 3-16
ANALYSIS/ DISCUSSION 17
CONCLUSION 18
PRECAUTION 18
REFERRENCES 18










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4. 1.0 INTRODUCTION

In statics, a structure is statically indeterminate when the static equilibrium equations are
insufficient for determining the internal forces and reactions on that structure. In order to
analyse the indeterminate structures, considerations in the material properties and
compatibility in deformations are taken to solve statically indeterminate. A statically
indeterminate truss can be determined using the formula as below:

DOI = (m + r) – (2j + c), where DOI > 0

Where m = member, r = reaction, j = joint, c = internal hinge

2.0 OBJECTIVE

To compare the member forces of indeterminate truss determined in theoretical structural
analysis with respect to experimental result.

3.0 PROBLEM STATEMENT

A truss is a structure that is made of straight, slender bars that are joined together to form
a pattern of triangles. Trusses are usually designed to transmit forces over relatively long
spans. Common examples of trusses are bridge trusses and roof trusses.

4.0 APPARATUS/ MATERIAL

1. Portal truss apparatus
2. Ruler
3. Computer




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5. 5.0 PROCEDURES
1. The computer and data acquisition module is switched. For stability of the reading,
the data acquisition must be switched on 10 minutes before taking readings.
2. A truss configuration is selected. (See attachment for possible configuration)
3. The truss is assembled according to the selected configuration using the members
available.
4. One end of the plane truss is placed on the roller support and the other on the pin
support (Please ensure that the center of the joint is located on the knife edge and at
the center of the roller).
5. Ensure that the pinned support is properly secured to the frame.
6. Attached the screw jack to the joint to be loaded.
7. The screw jack is loosened so that the truss is free from applied load.
8. The wire is connected from the load cell to the data acquisition module, each load cell
occupying one channel of the module.
9. The Win view CP Plus software is run.
10. The ‘setting’ option to set the modules and channels to be acquired is selected.
11. When the setting is complete, return to the sub menu. Click the start button and
choose the overwrite file option.
12. Some figures will be displayed in the boxes of the chosen channels. These figures are
the voltage outputs from the load cells. They are in millivolt units. For this apparatus
1 millivolt is equivalent to 100 N forced.
13. Allowed approximately 20 seconds of readings to be captured. After 20 seconds press
the stop button (watch the graph for the time span). This will be the initial readings for
each load cell.
14. The ‘start’ button is clicked on the sub menu. The append option is chosen.
15. The screw jack handle is turned to apply loads in the downward direction and the
readings of the screw jack are observed. When the desired load is reached, stop
turning the screw jack.
16. Allow approximately 20 seconds of readings to be captured. After 20 seconds, the
stop button is pressed.
17. Step 14 to 16 is repeated for a few more load increments.
18. At the end of the experiment, the data is exported to the excel spreadsheet and the
necessary analysis is carried out.
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6. 6.0 DATA & CALCULATION


Degree of indeterminacy, DOI = ( m+ r) – (2j + c)
= [ 8 + 3 ] - [ 2(5) + 0 ]
= 1 statically indeterminate structure.

External redundant = reaction - 3
= 3 – 3
= 0 redundant in external

Internal = DOI – external redundant
= 1 – 0
= 1 redundant in internal (Internally Indeterminate Truss)


3
Data Set
4
Reading
Load
(N)
Member
Forces 1
(N)
Member
Forces 2
(N)
Member
Forces 3
(N)
Member
Forces 4
(N)
Member
Forces 5
(N)
Member
Forces 6
(N)
Member
Forces 7
(N)
1 13.662 95.040 35.937 -23.463 -31.185 105.430 26.136 21.087
2 25.542 130.980 53.163 -35.937 -31.185 139.890 -12.474 39.204
3 41.283 166.320 75.438 -49.005 -39.501 168.100 -48.114 45.441
4 54.351 187.110 93.852 -60.885 -56.133 179.390 25.542 38.610
5 69.795 211.170 116.130 -75.438 -77.814 191.860 8.019 28.809
6 84.942 234.330 137.210 -90.882 -100.090 203.440 -49.599 19.305
7 94.743 249.480 151.470 -102.470 -115.830 211.170 25.542 13.068
8 106.330 267.000 167.810 -115.830 -134.840 219.780 -40.986 5.643
9 123.85 294.030 193.350 -132.760 -160.080 232.850 12.474 -6.237
10 139.890 319.570 215.920 -145.230 -178.500 246.510 12.177 -15.147
0.5
0.5
0.5
All dimension in m
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7. Using Flexibility Method (Superposition Method)

Primary structure, No Redundant Structure, N1


Lets take P = 13.662N
+↑Σ��=�
13.662+VA=0
VA= -13.662N
+ ��=�
-13.662+0.5HB=0
-13.662+0.5HB=0
HB=27.324N
+ →Σ��=�
HA+HB= 0
HA+27.324=0
HA= -27.324




4
0.5m
0.5m 0.5m
0.5m
0.5m
0.5m
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8. Using method of joint to calculate the member forces.
5
At point A



+↑Σ��=�
-13.662 + FAB = 0
FAB = 13.662N

+ →Σ��=�
27.324 + FAE = 0
FAE = -27.324N

At point D




+↑Σ��=�
13.662 + FDC sin 45 = 0
FDC = -19.32N

+ →Σ��=�
-(-19.32 cos 45)- FDE = 0
FDE = 13.66N


At point B



+↑Σ��=�
-13.662 – FBE sin 45= 0
FBE = -19.32N

+ →Σ��=�
-27.324 + FBC + FBE cos 45 = 0
-27.324 + FBC - 19.32 cos45 = 0
FBC = 40.985N

Checking at point E


+↑Σ��=�
13.662 +(- 19.32 sin 45) = 0
0 = 0
+ →Σ��=�
-13.661 + 27.324 + 13.66 = 0
0 = 0




HA=27.324

VA=-13.662
FAB
FAE

-13.662
FDC sin ᶿ45
FDC cos ᶿ 45
FDE
FDC
FBE sin ᶿ45
FBE cos ᶿ 45
-27.324
FBA
FBE
FBC
FEA
FED=13.66
FEB sin ᶿ45
FEB cos ᶿ 45
FEC=-13.662
FEB = -19.32
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9. Example calculation for P = 13.662N
MEMBERS LENGTH �
� �
� ∆
�=
�
��
��
��
∆
�=
�
�
�
�
��

N =�
�+
�
��
� (�)
FAE 0.5 -27.324 -0.707 9.659 0.250 -16.007
FAB 0.5 13.662 0 0 0 13.662
FDC 0.5 -19.321 0 0 0 -19.321
FDE 0.5 13.662 0 0 0 13.662
FBE 0.707 -19.321 1 -13.660 0.707 -35.328
FBC 0.5 40.986 -0.707 -14.489 0.250 52.303
FCE 0.5 13.662 -0.707 -4.830 0.250 24.979
∑ =
−23.319
��
∑ =
1.457
��


Compatibility Equation:
∆
1= ∆
10+ ∆
11
Assume no support settlement, ∆
10 = 0
0 =
−23.319
��
+
1.457
��
(�
1
)
�
1= −16.005N






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10. Example calculation for P = 25.542N
MEMBERS LENGTH �
� �
� ∆
�=
�
��
��
��
∆
�=
�
�
�
�
��

N =�
�+
�
��
� (�)
FAE 0.5 -51.084 -0.707 18.058 0.250 -29.926
FAB 0.5 25.542 0 0.000 0.000 25.542
FDC 0.5 -36.122 0 0.000 0.000 -36.122
FDE 0.5 25.542 0 0.000 0.000 25.542
FBE 0.707 -36.122 1 -25.538 0.707 -66.049
FBC 0.5 76.626 -0.707 -27.087 0.250 97.784
FCE 0.5 25.542 -0.707 -9.029 0.250 46.700
∑=
−43.597
��
∑=
1.457
��


Compatibility Equation:
∆
1= ∆
10+ ∆
11
Assume no support settlement, ∆
10 = 0
0 =
−43.597
��
+
1.457
��
(�
1
)
�
1= −29.922N






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11. Example calculation for P = 41.283N
MEMBERS LENGTH �
� �
� ∆
�=
�
��
��
��
∆
�=
�
�
�
�
��

N =�
�+
�
��
� (�)
FAE 0.5 -82.566 -0.707 29.187 0.250 -48.368
FAB 0.5 41.283 0.000 0.000 0.000 41.283
FDC 0.5 -58.384 0.000 0.000 0.000 -58.384
FDE 0.5 41.283 0.000 0.000 0.000 41.283
FBE 0.707 -58.384 1.000 -41.277 0.707 -106.754
FBC 0.5 123.849 -0.707 -43.781 0.250 158.047
FCE 0.5 41.283 -0.707 -14.594 0.250 75.481
∑ =
−70.464
��
∑ =
1.457
��


Compatibility Equation:
∆
1= ∆
10+ ∆
11
Assume no support settlement, ∆
10 = 0
0 =
−70.464
��
+
1.457
��
(�
1
)
�
1= -48.362N






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12. Example calculation for P =54.351 N
MEMBERS LENGTH �
� �
� ∆
�=
�
��
��
��
∆
�=
�
�
�
�
��

N =�
�+
�
��
� (�)
FAE 0.5 -108.702 -0.707 38.426 0.250 -100.977
FAB 0.5 54.351 0 0.000 0 54.351
FDC 0.5 -76.864 0 0.000 0 -76.864
FDE 0.5 54.351 0 0.000 0 54.351
FBE 0.707 -76.864 1 -54.343 0.707 -87.790
FBC 0.5 163.053 -0.707 -57.639 0.250 170.778
FCE 0.5 -163.053 -0.707 57.639 0.250 -155.328
∑=
−15.917
��
∑=
1.457
��


Compatibility Equation:
∆
1= ∆
10+ ∆
11
Assume no support settlement, ∆
10 = 0
0 =
−15.917
��
+
1.457
��
(�
1
)
�
1= -10.925N






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13. Example calculation for P = 69.795N
MEMBERS LENGTH �
� �
� ∆
�=
�
��
��
��
∆
�=
�
�
�
�
��

N =�
�+
�
��
� (�)
FAE 0.5 -139.59 -0.707 49.345 0.250 -81.774
FAB 0.5 69.795 0 0 0 69.795
FDC 0.5 -98.706 0 0 0 -98.706
FDE 0.5 69.795 0 0 0 69.795
FBE 0.707 -98.706 1 -69.795 0.707 -180.483
FBC 0.5 209.385 -0.707 -74.018 0.250 267.201
FCE 0.5 69.795 -0.707 -24.673 0.250 127.611
∑=
−119.13
��
∑=
1.457
��


Compatibility Equation:
∆
1= ∆
10+ ∆
11
Assume no support settlement, ∆
10 = 0
0 =
−119.13
��
+
1.457
��
(�
1
)
�
1= −81.764�






10
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14. Example calculation for P = 84.942N
MEMBERS LENGTH �
� �
� ∆
�=
�
��
��
��
∆
�=
�
�
�
�
��

N =�
�+
�
��
� (�)
FAE 0.5 -108.702 -0.707 38.426 0.250 -63.679
FAB 0.5 54.351 0.000 0.000 0.000 54.351
FDC 0.5 -76.8647 0.000 0.000 0.000 -76.865
FDE 0.5 54.351 0.000 0.000 0.000 54.351
FBE 0.707 -76.8647 1.000 -54.343 0.707 -140.546
FBC 0.5 163.053 -0.707 -57.639 0.250 208.075
FCE 0.5 54.351 -0.707 -19.213 0.250 99.373
∑=
−92.769
��
∑=
1.457
��


Compatibility Equation:
∆
1= ∆
10+ ∆
11
Assume no support settlement, ∆
10 = 0
0 =
−92.769
��
+
1.457
��
(�
1
)
�
1= 63.671�






11
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15. Example calculation for P =94.743 N
MEMBERS LENGTH �
� �
� ∆
�=
�
��
��
��
∆
�=
�
�
�
�
��

N =�
�+
�
��
� (�)
FAE 0.5 -189.486 -0.707 66.983 0.250 -122.486
FAB 0.5 84.942 0 0 0.000 84.942
FDC 0.5 -120.127 0 0 0.000 -120.127
FDE 0.5 84.942 0 0 0.000 84.942
FBE 0.707 -120.127 1 -84.930 0.707 -214.894
FBC 0.5 254.826 -0.707 -90.081 0.250 321.826
FCE 0.5 84.942 -0.707 -24.673 0.250 151.942
∑ =
132.701
��
∑ =
1.457
��


Compatibility Equation:
∆
1= ∆
10+ ∆
11
Assume no support settlement, ∆
10 = 0
0 =
132.70
��
+
1.457
��
(�
1
)
�
1= -91.092N






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16. Example calculation for P =106.33 N
MEMBERS LENGTH �
� �
� ∆
�=
�
��
��
��
∆
�=
�
�
�
�
��

N =�
�+
�
��
� (�)
FAE 0.5 -212.66 -0.707 75.175 0.250 -124.579
FAB 0.5 106.33 0.000 0.000 0 106.330
FDC 0.5 -150.375 0.000 0.000 0 -150.375
FDE 0.5 106.33 0.000 0.000 0 106.330
FBE 0.707 -150.375 1.000 -106.315 0.707 -274.959
FBC 0.5 318.99 -0.707 -112.763 0.250 407.071
FCE 0.5 106.33 -0.707 -37.588 0.250 194.411
∑=
−181.49
��
∑=
1.457
��


Compatibility Equation:
∆
1= ∆
10+ ∆
11
Assume no support settlement, ∆
10 = 0
0 =
−181.490
��
+
1.457
��
(�
1
)
�
1= -124.584






13
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17. Example calculation for P =123.85 N
MEMBERS LENGTH �
� �
� ∆
�=
�
��
��
��
∆
�=
�
�
�
�
��

N =�
�+
�
��
� (�)
FAE 0.5 -247.7 -0.707 87.562 0.250 -145.106
FAB 0.5 123.85 0.000 0.000 0.000 123.850
FDC 0.5 -175.152 0.000 0.000 0.000 -175.152
FDE 0.5 123.85 0.000 0.000 0.000 123.850
FBE 0.707 -175.152 1.000 -123.832 0.707 -320.263
FBC 0.5 371.55 -0.707 -131.343 0.250 474.143
FCE 0.5 123.85 -0.707 -43.781 0.250 226.443
∑=
−211.394
��
∑=
1.457
��


Compatibility Equation:
∆
1= ∆
10+ ∆
11
Assume no support settlement, ∆
10 = 0
0 =
−211.394
��
+
1.457
��
(�
1
)
�
1= -145.111






14
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18. Example calculation for P =139.89 N
MEMBERS LENGTH �
� �
� ∆
�=
�
��
��
��
∆
�=
�
�
�
�
��

N =�
�+
�
��
� (�)
FAE 0.5 -279.78 -0.707 98.902 0.250 -299.662
FAB 0.5 139.89 0.000 0.000 0.000 139.890
FDC 0.5 -197.836 0.000 0.000 0.000 -197.836
FDE 0.5 139.86 0.000 0.000 0.000 139.860
FBE 0.707 197.836 1.000 139.870 0.707 225.958
FBC 0.5 419.67 -0.707 -148.353 0.250 399.788
FCE 0.5 139.89 -0.707 -49.451 0.250 120.008
∑=
40.968
��
∑=
1.457
��


Compatibility Equation:
∆
1= ∆
10+ ∆
11
Assume no support settlement, ∆
10 = 0
0 =
40.968
��
+
1.457
��
(�
1
)
�
1= 28.122






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19. Example of percentage error calculation.
���������� ����� =
�ℎ��������� ����� − ������������
�ℎ��������� �����

=_________%
Percentage Error Comparison
P = 13.662N
Member Theoretical Experimental Percentage Error
1 -16.007 95.040 6.93
2 13.662 35.937 -1.63
3 -19.321 -23.463 2.21
4 13.662 -31.185 3.28
5 -35.328 105.430 3.98
6 52.303 26.136 0.50
7 24.979 21.087 0.15



















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20. 7.0 ANALYSIS/ DISCUSSION

When conducting indeterminate truss laboratory in order to analyze this laboratory
activity of indeterminate truss structure, the consideration on compatibility in deformation
and redundant factor are taken to solve statically indeterminate. By using superposition
method to solve this indeterminate structure, the theoretical structure analysis determined
with respect to experimental result obtained from both results of experimental and theoretical
analysis cases that obtained, the graph of forces in member versus applied load and the forces
in the member of truss. In comparison of actual and theoretical cases, the percentages error
by calculation the slope of each graph of each member determined.
Based on data analysis for this experiment, there is an error occurred. There is slightly
different of the results obtained for actual value and the theoretical value. From that, a
comparison can be made. The errors may occur due to several factors that causes:

a) Improperly installed or inadequate bracing
b) Overloading the trusses before permanent bracing or sheathing has been installed
c) Improper or inadequate connections to the supporting structure or broken connections
d) Improper or unauthorized field changes made to trusses
e) Installing damaged, broken or improperly repaired trusses
Percentage Error = Theoretical Slope – Experimental Slope x 100

Theoretical Slope:
���������� ����� =
�ℎ��������� ����� − ������������
�ℎ��������� �����







17

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21. 8.0 CONCLUSION

The objectives of this experiment have been achieved when able to compare the
member forces of indeterminate truss determined in theoretical structural analysis with
respect to experimental result. There is a different between the results of experimental and
theoretical results which is caused by few errors. The truss is an important element in
current structure trends. Trusses are able to allow for analysis of the structure uses a few
assumptions and the application of Newton’s Law of Motion according static. In the
industry of construction, the used of application for truss applied to some construction site
such as truss bridge and roof construction.


9.0 PRECAUTION

1. Make sure when turning the handle is to the correct direction (anti-clockwise) when
applying the load to the truss.
2. For stability of the reading, the data acquisition module must be switched on 10
minutes before taking readings.




10.0 REFERENCES

➢ MOHD SAMSUDIN ABDUL HAMID premiered October 31 2020
https://www.youtube.com/watch?v=OLn2mZ642a0.
➢ Open-Ended Manual For Structural Engineering Laboratory (UITM Penang),
March 2013.








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