6.0 RESULTS
1. Table
Load (N)
Strain reading 1
2
3
4
5
6
7
8
0
101
192
-50
-112
63
0
-5
18
50
109
185
-57
-128
64
0
6
29
100
119
175
-67
-148
64
0
20
44
150
128
165
-76
-167
64
0
35
60
200
138
156
-85
-185
64
0
48
74
250
144
149
-92
-200
63
0
59
85
Table 1: Strain readings and frame deflection for experiment 1 Load
1
2
3
4
5
6
7
8
0
0
0
0
0
0
0
0
0
50
8
-7
-7
-15
1
0
11
11
100
18
-17
-17
-35
1
0
25
26
150
27
-27
-26
-54
1
0
40
42
200
37
-36
-35
-72
1
0
53
56
250
43
-43
-42
-87
0
0
64
67
(N)
Table 1: True strain readings for experiment 1
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 = I n T he Graph Paper
II. On another graph, do the same for a different member (non member 6). Graph Plotted = I n T he Graph Paper
III. Plot a separate graph of deflection (mm) against Load (N). Graph Plotted = I n T he 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 equiva lent 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) 2
A = cross section area of the member (m )
Rod diameter = 6 mm and Esteel = 2.10x10 5 N/mm2 Use L oad 250 N
Member
Experimental force (N)
Theoretical force (N)
1
255.28
250
2
-255.28
-250
3
249.34
-250
4
-516.49
-500
5
0
0
6
0
0
7
379.95
354.2
8
397.75
354.2
Table 3: Measured and Theoretical Force in the Cantilever Truss
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
28.27 mm2 5
2
-6
Member 1; F = 2.10 x 10 N/mm x (43 x 10 ) x 28.27 mm
2
= 255.28 N 5
-6
2
Member 2; F = 2.10 x 10 N/mm2 x (-43 x 10 ) x 28.27 mm
= -255.28 N 5
2
-6
2
5
2
-6
2
5
2
Member 3; F = 2.10 x 10 N/mm x (-42 x 10 ) x 28.27 mm
= -249.34 N
Member 4; F = 2.10 x 10 N/mm x (-87 x 10 ) x 28.27 mm
= -516.49 N -6
2
Member 5; F = 2.10 x 10 N/mm x (0 x 10 ) x 28.27 mm
=0N
5
2
-6
2
5
2
-6
2
Member 7; F = 2.10 x 10 N/mm x (64 x 10 ) x 28.27 mm
= 379.95 N
Member 8; F = 2.10 x 10 N/mm x (67 x 10 ) x 28.27 mm
= 397.76 N
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 -500 N which means it’s not a big different wit h experimental result. Besides that, member of cantilever at no 5 no result obtained which are equal to 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 and 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 the re is no extra forces that will move the structure to fall or collapse. 2. Statically determinate structure has low cost compa re 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 o nly the value of the force is different.
