UNIVERSITI TUN HUSSEIN ONN MALAYSIA

Centre for Diploma Studies

_________
DEPARTMENT OF CIVIL ENGINEERING
STRUCTURE ANALYSIS LABORATORY
LAPORAN MAKMAL
LABORATORY REPORT

Kod & Nama Kursus

Course Name & Code DAC 21703 ANALISIS STRUKTUR
Tajuk Ujikaji
Title of Experiment
TENSION COEFFICIENT-NO.2
Seksyen 5
Section
Kumpulan
4
Group
Nama Pensyarah/Pengajar
Dr. Nor Farah Atiqah Binti Ahmad
Lecturer/Instructor’s Name
Nama Ketua Kumpulan Muhammad Habil Izuddin Bin No. Matrik : AA210051
Name of Group Leader Norhisham Matric No.

Ahli Kumpulan No. Matrik Penilaian (%)

Group Members Matric No. Assessment (%)
1. Muhammad Habil Izuddin Teori, Objektif & Prosedur 10
AA210051 Theory, Objectives & Procedure
Bin Norhisham
2. Muhammad Arif Bin Data Ujikaji 15
AA210136 Experimental Data
Muhammad Hafiz Saw
3. Zaki Irfan Bin Khairun Analisis 25
AA212778 Analysis
Nidzam
Perbincangan
4. Discussion
25

5. Kesimpulan & Cadangan 15

Conclusion & Recommendation

6. Penyertaan 5
Participation
Tarikh Ujikaji Rujukan
Date of Experiment
27/10/2022
References
5
Tarikh Hantar JUMLAH
Date of Submission
28/12/2022
TOTAL
100
 Lab R eport R ubric (Assessm ent Form ) D AC 21703 STR U C TU R E AN ALISIS Test Title:

C riteria 1 2 3 4 5 SC R WT TS C R (% )

Theory,  Theory and other information  Theory and other information  Theory and other information  Theory and other information  Theory and other information
objecti ve, regarding the laboratory is not regarding the laboratory is regarding the laboratory is partly regarding the laboratory is regarding the laboratory is well 2
pro ce dure presented presented minimally presented presented with some mistakes presented
(10 % )
E xperim ental  Data was not presented  Data lacks precision  Good representation of the  Accurate representation of  Accurate representation of the data
D ata data using tables and/or graphs the data using tables and/or using tables and/or graphs
3
(15 % ) graphs

Analy sis  Trends / patterns are not  Trends / patterns are not  Trends / patterns are logically  Trends / patterns are logically  Trends / patterns are logically
(25 % ) analyzed analyzed analyzed for the most part analyzed analyzed
 Analysis is not relevant  Analysis is inconsistent  Questions are answered in  Questions are answered in  Questions are answered thoroughly
5
complete sentences complete sentences and in complete sentences
 Analysis is general  Analysis is thoughtful  Analysis is insightful

D iscussion  No discussion was included or  A statement of the results is  A statement of the results of  Accurate statement of the  Accurate statement of the results of
(25 % ) shows little effort and reflection incomplete with little reflection the lab indicates whether results results of the lab indicates lab indicates whether results support
on the lab on the lab support the hypothesis whether results support the hypothesis
hypothesis  Possible sources of error and what 5
 Possible sources of error was learned from the lab discussed
identified
C onclu sion &  Conclusion & recommendation  Conclusion & recommendation  Conclusion & recommendation  Conclusion & recommendation  Conclusion & recommendation was
R ecom m en dation was not presented was minimally presented was presented somewhat was presented accurately and presented accurately and relevant to
(15 % ) accurate and relevant to result relevant to result obtain from the result obtain from the laboratory work 3
obtain from the laboratory work laboratory work with minor
mistakes
P articipation  Student was hostile about  Participation was minimal  Did the job but did not appear  Used time pretty well. Stayed  Showed interest, used time very well,
(during participating to be very interested. Focus lost focused on the experiment most guide other students and very focused
experim ent) on several occasion of the time on experiment 1
(5 % )
R eference  Reference was not presented in  Reference was minimally  Reference was partly  Reference was presented in  Reference was presented in proper
(5 % ) the report presented in the report presented in somewhat proper proper format and relevant to format and relevant to the laboratory
format and relevant to the the laboratory work with some work 1
laboratory work minor mistakes

NAME OF LECTURER : SIGNATURE : DATE : TOTAL SCORE : / 100

Note : SCR = SCORE, WT = WEIGHTAGE, TSCR = TOTAL SCORE

 STUDENT CODE OF ETHIC
(SCE)
DEPARTMENT OF CIVIL ENGINEERING
CENTRE FOR DIPLOMA STUDIES

We, hereby confess that we have prepared this report on our own effort. We also admit not to
receive any help from any third party during the preparation of this report and pledge that
everything mentioned in the report is true.

Habil
Student Signature (Group Representative)

Name : Muhammad Habil Izuddin Bin Norhisham

…………………………………………
AA210051
Matric No. : …………………………………………
28/12/2022
Date : …………………………………………
 Centre for Diploma Studies Page 4
Department of Civil Engineering Edition 1
Tension Coefficient – No. 2
Effective Date 16/10/2022
Tension Coefficient Member
Amendment Date

1.0 OBJECTIVE

1.1 To find theoretical forces calculated by tension coefficient method

2.0 LEARNING OUTCOME

2.1 To 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 THEORY

The tension coefficient for member of a frame is defined as the pull or tension in that member divided by its
T
length; t  . Where, t is tension coefficient of the member, T is the pull in member and L is length of
L
member. If a member of a truss system is situated not in a 2D plane, then the truss is defined as a space frame
truss. In other words, space truss has component in 3 axis; x,y, and z. Consider a member with
node A (xA, yA) and B (xB, yB);

TAB
Assume the force in the member in TAB is tension (+ve), the definition of tension coefficient, t is; t AB 
L AB

At A, the horizontal component TAB is;

t AB L AB cos 
TAB cos    t AB L AB  x B  x A 
L AB
 t AB  x B  x A 
Using the same method, the vertical component at A is  t AB  y B  y A 
 Centre for Diploma Studies Page 5
Department of Civil Engineering Edition 1
Tension Coefficient – No. 2
Effective Date 16/10/2022
Tension Coefficient Member
Amendment Date

At B, the horizontal component TAB is =  t AB x A  x B 

Using the same method, the vertical component at B is  t AB  y A  y B 
Using static equation and the coordinate value, tension at each joint can be solve.

TAB  t AB L AB  xB  x A 2   y B  y A 2
Space frames usually utilize a multidirectional span and are often used to accomplish long spans with few
supports. They derive their strength from inherent rigidity of triangular frame; flexing loads (bending
moments) are transmitted as tension and compression loads along the length of each strut.
 Centre for Diploma Studies Page 6
Department of Civil Engineering Edition 1
Tension Coefficient – No. 2
Effective Date 16/10/2022
Tension Coefficient Member
Amendment Date

4.0 PROCEDURES

1. Adjust the length of jib by turning male threaded rods clockwise or anticlockwise depending on
either to increase or decrease the length.
2. Adjust the length of tie rods using S.S chain and turnbuckle.
3. Fix the post length using sliding square bar and measure it from pivot joint.
4. Note down the initial readings of tie rods and jib unloaded.
5. Put weight on hanger and note down the readings of force on jib and tie rods along with their
respective lengths.
6. Repeat the procedure with different post heights. Different chain lengths and with different loads.
7. Recod all the experimental forces and change in lengths along with coordinates
 Centre for Diploma Studies Page 7
Department of Civil Engineering Edition 1
Tension Coefficient – No. 2
Effective Date 16/10/2022
Tension Coefficient Member
Amendment Date

5.0 RESULTS

DATA:

Initial Length
Tie A : 530 mm
Tie B : 530 mm
Jib : 530 mm

Final Length (mm) Initial Force (kg) Final Force (kg) Observed Force (kg)
Weight Tie Tie Jib Tie Jib Tie Jib
(kg) Jib
A B A B compression A B compression A B compression

10 580 560 500 1.0 0.5 1.6 11.5 7.7 7.0 10.2 7.2 5.4

1 kg = 10 N

ANALYSIS:

1. Draw the Free Body Diagram of the members

2. Identify the coordinates of each point member.
3. Analyze theoretical technique using tension coefficient method.
4. Present the obtained result in appropriate table.
 Centre for Diploma Studies Page 8
Department of Civil Engineering Edition 1
Tension Coefficient – No. 2
Effective Date 16/10/2022
Tension Coefficient Member
Amendment Date
 Centre for Diploma Studies Page 9
Department of Civil Engineering Edition 1
Tension Coefficient – No. 2
Effective Date 16/10/2022
Tension Coefficient Member
Amendment Date
 Centre for Diploma Studies Page 10
Department of Civil Engineering Edition 1
Tension Coefficient – No. 2
Effective Date 16/10/2022
Tension Coefficient Member
Amendment Date

6.0 DISCUSSIONS

1. Discuss your obtained data and results regarding truss member (tension and compression).

Initial forces for ties A and B are 1.0 kg and 0.5 kg, respectively. Jib's initial forces are 1.6kg, however.
Finally, the forces for A and B are 11.5 kg and 7.7 kg, respectively, while the force for the jib is 7 kg.
Determine if the truss member is in tension or compression using the formula (final - initial forces). The
results show that A and B are in tension, while the jib is in compression.

2. Make your summary analysis based on the result obtained.

Using the tension coefficient approach, the data from the experiment are required to compute the force and
identify whether each member is under tension or compression. We can find the coordinates for each
member point by sketching the FBD. An equation of equilibrium can be used to express each joint in terms
of the tension coefficient and joint coordinates. The answer of this equation was then translated into forces.