FACULTY OF CIVIL AND ENVIRONMENTAL ENGINEERING DEPARTMENT OF STRUCTURE AND MATERIAL ENGINEERING LAB MATERIAL
REPORT Subject Code Code & Experiment Title Course Code Date Section / Group Name Members of Group
Lecturer/Instructor/Tutor Received Date
Comment by examiner
BFC 21201 FORCE IN A STATICALLY DETERMINATE CANTILEVER TRUSS 2 BFF/1 03/10/2011 2 MUHAMAD ASYRAF BIN AB MALIK (DF100108) 1.MUHAMMAD IKHWAN BIN ZAINUDDIN (DF100018) 2.AHMAD FARHAN BIN RAKAWI (DF100142) 3.IDAMAZLIZA BINTI ISA (DF100128) 4.AINUN NAZHIRIN BINTI ABD JALIL (DF100076) EN MOHAMAD HAIRI BIN OSMAN 24 OCTOBER 2011
Received
STUDENT CODE OF ETHIC (SCE) DEPARTMENT OF STRUCTURE AND MATERIAL ENGINEERING FACULTY OF CIVIL & ENVIRONMENTAL ENGINEERING UTHM
We, hereby confess that we have prepared this report on our effort. We also admit not to receive or give any help during the preparation of this report and pledge that everything mentioned in the report is true. ___________________________ Student Signature Name
___________________________ Student Signature
: MUHAMAD ASYRAF AB MALIK
Name
: IDAMAZLIZA ISA
Matric No. : DF100108
Matric No. : DF100128
Date
Date
: 24/10/2011
___________________________ Student Signature Name
: 24/10/2011
___________________________ Student Signature
: AHMAD FARHAN RAKAWI
Name
: AINUN NAZHIRIN ABD JALIL
Matric No. : DF100142
Matric No. : DF100076
Date
Date
: 24/10/2011
_______________________ Student Signature Name
: MUHAMMAD IKHWAN ZAINUDDIN
Matric No. : DF100018 Date
: 24/10/2011
: 24/10/2011
1.0
OBJECTIVE 1.1
To examine a statically determinate frame and to analyze the frame using simple pin joint theory.
2.0
LEARNING OUTCOME 2.1
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
THEORY A truss is a structure composed of slender member joined together at their end points to form one or more triangles. The joint connections are considered as 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 frictionless pins, that is only a force and no moment can be transferred from one member to another 2. External loads are applied to the truss only at its joints. One of the methods to calculate the forces in the member of a truss is using Method of Joint. 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 equations ∑Fx = 0 and ∑Fy = 0. Calculation only can be started for joint where the numbers of unknowns are two or less
4.0
EQUIPMENT
Figure 1: Frame Of Truss
Figure 2: Digital Indicator Reading
Figure 3: Digital Force Display
Figure 4: Digital Strain Display
5.0
PROCEDURS 1.
Unscrew the thumwheel on the “redundant” member. Note that it is effectively no longer part of the structure as the idealised diagram illustrates.
2.
Apply the pre-load of 100N downward, re-zero the load cell and carefully apply a load of 250N and check that the frame is stable and secure.
3.
Return the load to zero (leaving the 100N preload), recheck and re-zero the digital indicator. Never apply loads greater than those specified on the equipment.
4.
Apply load in the increment shown in Table 1 recordding the strain readings and the digital indicator readings. Complete Table 2 by subtracing the initial (zero) strain readings. (be careful with your sign)
6.0
RESULTS 1. Table Load
Strain Reading
(N)
1
0 50 100 150 200 250
2
3
4
Digital Indicator
5
6
7
Reading (mm)
8
125 200 -42 -86 94 5 36 -0.197 0 134 192 -50 -102 96 17 48 -0.032 0 144 183 -59 -121 96 32 63 -0.066 0 152 174 -68 -139 97 45 77 -0.095 0 161 165 -77 -157 97 58 91 -0.118 0 169 157 -85 -173 97 70 102 -0.138 0 Table 1: Strain Readings And Frame Deflection For Experiment 1
Load(N)
1
0 50 100 150 200 250
0 9 19 27 36 44
2
3
4
5
6
0 0 0 0 0 -8 -8 -16 2 0 -17 -17 -35 2 0 -26 -26 -53 3 0 -35 -35 -71 3 0 -43 -43 -87 3 0 Table 2: True Strain Reading for Experiment
7
8
0 12 27 40 53 65
0 12 27 41 55 66
2. Graphs I. Choose a member (except member 6), and on the same axis plot a graph of Recorded Strain μɛ against Load (N) and True Strain μɛ against Load (N). Graph Plotted = In The Graph Paper II. On another graph, do the same for a different member (non member 6). Graph Plotted = In The Graph Paper III. Plot a separate graph of deflection (mm) against Load (N). Graph Plotted = In The Graph Paper IV. Comment on your graph Both criteria for strain and deflection is inversely proportional showed in the graph, for the strain recorded and the true strain graph is consistence liner but graph for deflection is not liner graph, it has curve on graph line.
Using the Young’s Modulus relationship, calculate the equivalent member force. complete the experimental force in Table 3. (ignore member 6 at this stage) E = σ/ε Where; E = Young’s Modulus (Nm-2) σ = Stress in the member (Nm-2) ε = Displayed strain and σ = F/A where, F = Force in member (N) A = cross section area of the member (m2) Rod diameter = 6 mm and Esteel = 2.10x105 N/mm2
Use Load 250 N Member
Experimental Force
Theoretical Force
(N)
(N)
1
261.21
250
2
-255.28
-250
3
-255.28
-250
4
-516.49
-500
5
17.81
0
6
0
0
7
385.88
354.2
8
391.82
354.2
Table 3: Measured and Theoretical Force in the Cantilever Truss Calculate the theoretical force using method of joint and write it down in Table 3 above
7.0
ANALYSIS DATA 7.1
Calculation For Experimental Force (N), Load = 250 N From the formula: E = σ ε where; E = Young Modulus (Nm-2) for steel = 2.10 x 105 N/mm2 ε = Displayed Strain σ=F A F=EεA d = 6mm A = π (6)2 = 28.27 mm2 4 Member 1;
F = 2.10 x 105 N/mm2 x (44 x 10-6) x 28.27 mm2 = 261.21 N
Member 2;
F = 2.10 x 105 N/mm2 x (-43 x 10-6 ) x 28.27 mm2 = -255.28 N
Member 3;
F = 2.10 x 105 N/mm2 x (-43 x 10-6) x 28.27 mm2 = -255.28 N
Member 4;
F = 2.10 x 105 N/mm2 x (-87 x 10-6) x 28.27 mm2 = -516.49 N
Member 5;
F = 2.10 x 105 N/mm2 x (3 x 10-6) x 28.27 mm2 = 17.81 N
Member 7;
F = 2.10 x 105 N/mm2 x (65 x 10-6) x 28.27 mm2 = 385.88 N
Member 8;
F = 2.10 x 105 N/mm2 x (66 x 10-6) x 28.27 mm2 = 391.82 N
7.2
Calculation For Theoretical Force (N), Load = 250 N
RAY RAX
A
1
5
C
2
8
7
2.4 m
RBX 4
B
2.4 m
∑MB
250 (4.8) + RAX (2.4) = 0 1200 + RAX (2.4) = 0 RAX = - 1200 2.4 RAX = - 500 N
∑FX = 0 RAX + RBX = 0 -500 + RBX = 0 RBX = 500 N
RAY = 250 N
3
E
250N
2.4 m AC = DE
=0
∑Fy = 0
D
AC2 = BA2 + BC2 AC2 = 2.42 + 2.42 AC2 = 11.52 AC = 3.4 m
Joint Method Calculation MEMBER 4 ∑Fx = ∑Fx 500 + FBD = 0
MEMBER 5
FBA
∑Fy =∑Fy FBA = 0 N
5
FBD = -500.0 N 500N MEMBER 7 ∑Fy = ∑Fy
MEMBER 3
FEC
∑Fx = ∑Fx
-250 + FEC (2.4/3.4) = 0
- FED - FEC (2.4/3.4) = 0
FEC (2.4/3.4) = 250
- FED – 354.2 (2.4/3.4) = 0
FEC = 354.2 N
FED = -250.0 N
FBD
4
7
FED
E
3 250
MEMBER 2 ∑Fy = ∑Fy
MEMBER 1 ∑Fx = ∑Fx
FCA
-FCD – FCE (2.4/3.4) = 0
–FCA + FCE (2.4/3.4) = 0
-FCD – (354.2) (2.4/3.4) = 0
–FCA + (354.2) (2.4/3.4) = 0
FCD = -250.0 N
FCA = 250.0 N
1 7 2
FCD
FCE
MEMBER 8 ∑Fy =∑Fy
FDC
FDA
FDC + FDA (2.4/3.4) = 0 (-250) + FDA (2.4/3.4) = 0
2 8
FDA = 354.2 N FDB
4
3
FDE
8.0
DISCUSSION 1. Compare the experimental and theoretical result. From the experimental results, we obtained that member of cantilever 4 has stated as the highest force which is -516.49N, and after made some theoretical calculations at cantilever no 4 the value obtained was -100 N which means it’s not a big different with experimental result. Besides that, member of cantilever at no 5 was obtained the result but compare the theoretical result has no result was made. All members show the internal force equation derived from experimental results and calculation method of the connection pin.
2. From your result and the theoretical member force, identify which members are in compression and which members are in tension. Explain your choice. We found that the cantilever truss member having the compression at members of 2, 3 and 4. We identified this because sign of the value. All these members have a negative value. While for the rest members, we found that the cantilever at member of 1, 7 and 8 having a tension. Type of internal force derived from the experimental results together with the values derived from calculations using the method of connection. The value of the internal forces that are compressed as a member no. 2 due to the costs of action are mutually repel members, so members will be trying to fight out these forces, so there was a compressive force. Tension that occurs at No.7 example of the burden caused by the attraction of the subject, to fight back, then the internal forces have to withdraw the action, so there was tension in the member. 3. Observe the reading of member 5. Explain why the readings are almost zero. From the experimental value we obtained that the reading was 17.81 N and from the theoretical results we obtained as a 0 N. By the way, these both value are almost close to 0, this condition happen because there are no force either internal or external force actually acted on this member and it pinned on both end, this member cannot acting on any movement either on X or Y axis.
4. Are the strain gauges are effective transducers for measurement forces in the framework. From the Transducers for Measurement forces gauge reading, we can see the value obtained is consistent with the load acting on the framework, when the load increasing the strain gauge value also increasing and it can be accepted because it has small value.
5. Does the framework comply with pin joint theory even though the joint are not truly pin joint? Yes, because the value obtained from the experiment only has small difference compare to the theoretical value calculated using Euler formula.
9.0
CONCLUSION 1. Statically determinate frame will be more economic and safety because it will use enough frame and member, then safety because there is no extra forces that will move the structure to fall or collapse. 2. Statically determinate structure has low cost compare to indeterminate structure. 3. We can see that there are large difference value between experimental force and theoretical force. It is mean that, the accuracy of the result is not exact but for the compression and tension member, we can conclude that the following tension and compression is same only the value of the force is different.
