Page 1 of 23
TABLE OF CONTENT
Number 1.0 2.0 3.0 4.0 5.0 6.0 7.0
Description Objective of the experiment Learning Outcome Theory Application of Truss Procedures Result and Analysis Discussion
Page 2 2 2 6 10 11 19
8.0
Conclusion
20
9.0
Appendix
21
1.0
OBJECTIVE 1.1
The effect of redundant member in a structure is observed and the method of analyzing type of this structure is understood.
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Page 2 of 23
2.0
LEARNING OUTCOME 2.1
Application
of
engineering
knowledge
in
practical
application. 2.2
To enhance technical competency in structure engineering through laboratory application.
3.0
THEORY A truss that is assumed to comprise members that are connected by means of pin joints, and which is supported at both ends by means of hinged joints or rollers, is described as being statically determinate. Newton's Laws apply to the structure as a whole, as well as to each node or joint. In order for any node that may be subject to an external load or force to remain static in space, the following conditions must hold: the sums of all horizontal forces, all vertical forces, as well as all moments acting about the node equal zero. Analysis of these conditions at each node yields the magnitude of the forces in each member of the truss. These may be compression or tension forces. Trusses that are supported at more than two positions are said to be statically indeterminate, and the application of Newton's Laws alone is not sufficient to determine the member forces. In order for a truss with pin-connected members to be stable,
it
must
be
entirely
composed
of
triangles.
In
mathematical terms, we have the following necessary condition for stability: M +R ≥ 2j
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Page 3 of 23 where m
=
total number of truss members
j
=
total number of joints
r
=
number of reactions (equal to 3 generally)
When m = 2j − 3, the truss is said to be statically determinate, because the (m+3) internal member forces and support reactions can then be completely determined by 2j equilibrium equations, once we know the external loads and the geometry of the truss. Given a certain number of joints, this is the minimum number of members, in the sense that if any member is taken out (or fails), then the truss as a whole fails. While the relation (a) is necessary, it is not sufficient for stability, which also depends on the truss geometry, support conditions and the load carrying capacity of the members. Some structures are built with more than this minimum number of truss members. Those structures may survive even when some of the members fail. They are called statically indeterminate structures, because their member forces depend on the relative stiffness of the members, in addition to the equilibrium condition described. In a statically indeterminate truss, static equilibrium alone cannot be used to calculated member force. If we were to try, we would find that there would be too many “unknowns” and we would not be able to complete the calculations. Instead we will use a method known as the flexibility method, which uses an idea know as strain energy. The mathematical approach to the
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Page 4 of 23 flexibility method will be found in the most appropriate text books.
Statically indeterminate can be two types 1.
External Indeterminate • It related with the reaction, it could be determinate if the number of reactions
of the
structure
exceed than
determinate structures by using static equation. 2.
Internal Indeterminate. • It related with the framework construction. Some of framework or trusses should have an adequate number of members for stability indentions. If inadequate members were
detected,
structure
is
classified
as
unstable,
meanwhile, while the redundant number of members were determined, the structures is classified as statically indeterminate.
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Page 5 of 23 Figure 1: Idealized Statically Indetermined cantilever Truss Basically the flexibility method uses the idea that energy stored in the frame would be the same for a given load whether or not the redundant member whether or not. In other word, the external energy = internal energy. In practice, the loads in the frame are calculated in its “released” from (that is, without the redundant member) and then calculated with a unit load in place of the redundant member. The values for both are combined to calculate the force in the redundant member and remaining members. The redundant member load in given by: P= Σ
fnl n2l
The remaining member forces are then given by: Member force = Pn + f Where, P = Redundant member load (N) L = Length of members (as ratio of the shortest) n
= Load in each member due to unit load in place of redundant member (N)
F
= Force in each member when the frame is
“release” (N) Figure 2 shows the force in the frame due to the load of 250 N. You should be able to calculate these values from Experiment: Force in a statically determinate truss
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Page 6 of 23
Figure 2: Force in the “Released” Truss Figure 3 shows the loads in the member due to the unit load being applied to the frame. The redundant member is effectively part of the structure as the idealized in Figure 2
Figure 3: Forces in the Truss due to the load on the redundant members
4.0
APPLICATION OF TRUSS
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Page 7 of 23
Trusses able to allows for the analysis of the structure uses a few assumptions and the application of Newton's laws of motion according to branch of physics known as static. Trusses are assumed to be pin jointed where the straight components meet for purposes of analysis. This assumption means that members of the truss including chords, verticals and diagonals will only act in tension or compression. When rigid joints imposed significant bending loads upon the elements, a more complex of analysis will be required. In the industry of construction, the used of application of truss applied for some construction. There are few products which need to be specifically designed and tailor made for each development. A truss bridge is the one of the example of application of truss. Truss bridge composed of connected elements with typically straight which may be stressed from tension, compression, or sometimes both in response to dynamic loads. Truss bridges are one of the oldest types of modern bridges. The basic types of truss bridges shown in this article have simple designs which could be easily analyzed by nineteenth and early twentieth century engineers. A truss bridge is economical to construct owing to its efficient use of materials. The application of truss also can be apply in the roof construction. Roof trusses are frames made up of timber that is nailed, bolted or pegged together to form structurally interdependent shapes of great strength. Roof trusses have to withstand the weight of the roof timbers and coverings (the ‘Dead Load’), plus a factor for your local Wind Load, plus a factor for your local Snow Load, plus a Safety Factor. A Structural Engineer can check these figures.
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Page 8 of 23
Statically indeterminate truss uses in industry of construction for those structures are built with more than this minimum number of truss members. Those structures may survive even when some of the members fail. It is can be apply for the design of truss or bridge. The basic types of truss bridges shown in this article have simple designs which could be easily analyzed engineers. A truss bridge is economical to construct owing to its efficient use of materials.
Component connections are critical to the structural integrity of a framing system. In buildings with large, clear span wood trusses, the most critical connections are those between the truss and its supports. In addition to gravity-induced forces (a.k.a. bearing loads), these connections must resist shear forces acting perpendicular to the plane of the truss and uplift forces due to wind. Depending upon overall building design, the connections may also be required to transfer bending moment.
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Page 9 of 23
The common types of truss bridge
The common types of roof truss
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Page 10 of 23 5.0
PROCEDURE
5.1 The thumbwheel on the ‘redundant’ member up to the boss was wind and hand–tighten it. Any tools to tighten the thumbwheel are not used. 5.2 The pre-load of 100N downward was applied, re-zero the load cell and carefully zero the digital indicator. 5.3 A load of 250N was carefully applied and checked whether the frame was stable and secure. 5.4 The load to zero (leaving the 100N preload) was returning. Rechecked and re-zero the digital indicator been done. Loads greater than those specified on the equipment never apply. 5.5 A load in the increment shown in table 1 was applied, the strain readings and the digital indicator readings was recorded. 5.6 Subtracted the initial (zero) strain reading (be careful with your signs) and completed table 2.
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Page 11 of 23 5.7 Calculated the equipment member force at 250 N and entered them into table 3. 5.8 A graph of Load vs Deflection was plotted from Table 1 on the same axis as Load vs deflection when the redundant ‘removed’. 5.9 The calculation for redundant truss is made much simpler and easier if the tabular method is used to sum up all of the “Fnl” and “n2l” terms. Referred to table 4 and entered in the values and carefully
5.10
calculated the other terms as required. Entered result into Table 3.
5.11
6.0
RESULT AND DATA ANALYSIS Member Strains (με)
Loa d (N) 0 50 100 150 200 250
1
2
3
Strain Reading 4 5
142 225 -38 -69 109 154 218 -50 -89 111 167 213 -58 -58 115 181 209 -67 -67 120 194 204 -76 -76 124 205 200 -84 -84 128 Table 1: Strain Reading and
6 35 28 21 15 7 0 Frame
7
8
21 22 32 28 45 35 58 43 72 50 83 56 Deflection
Digital indicator reading (mm) 0.009 -0.024 -0.051 -0.079 -0.103 -0.127
Member Strains (με) Load (N)
1
2
3
4
5
6
7
8
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Page 12 of 23 0 50 100 150 200 250
0 12 25 39 52 63
Member 1 2 3 4 5 6 7 8
0 0 0 -7 -12 -20 -12 -20 -51 -16 -29 -60 -21 -38 -69 -25 -46 -77 Table 2: True Strain
0 0 2 -7 6 -14 11 -20 15 -28 19 -35 Reading
Experimental Force (N) 374.07 -148.44 -273.13 -457.19 112.81 -207.81 368.13 201.88
0 11 24 37 51 62
0 6 13 21 28 34
Theoretical Force (N) 250 250 -250 -500 0 0 354 354
Table 3: Measured and Theoretical in the Redundant Cantilever Truss Member Length 1 1 2 1 3 1 4 1 5 1 6 1.414 7 1.414 8 1.414
F 250 -250 -250 -500 0 0 354 354
n Fnl -0.707 -176.75 -0.707 176.75 0 0 -0.707 354 -0.707 0 1 0 0 0 1 500.56 Total = 854.6
n2l 0.5 0.5 0 0.5 0.5 1.414 0 1.414 4.828
Pn -125.14 -125.14 0 -125.14 -125.14 177.00 0 177.00
Pn + f -375.14 124.87 -250.00 -625.14 -125.14 177.00 354.00 531.00
Table 4: Table for Calculating the Force in the Redundant Truss
P=
Total Fnl Total n 2 l
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Page 13 of 23
Data :Rod Diameter, D = 6.0 mm = 0.06 m = 2.10 x 105 N/mm
Esteel
EXPERIMENTAL FORCE Using the Young’s Modulus relationship, we can calculate the equivalent member force, complete the experimental force in Table 3. E=
σ ε
Where, E = Young’s Modulus (N/m2) σ = Stress in the member (N/m2) ε = Displayed strain
And
σ=
F A
Where, F = Force in member (N) A = Cross section area of the member (m2) To calculate the experimental force, we use the formula F = EA ε
With, A=
πd 2 4
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Page 14 of 23 So, A=
π( 6.0mm ) 4
2
= 28.274mm
2
CALCULATION FOR EXPERIMENTAL FORCE •
Member 1 ( ε = 63 x 10-6 ) F = EAε 5 2 2 -6 F = 2.10 x 10 N/mm x 28.274 mm x 63 x 10
F=
374.07N
CALCULATION FOR THEORETICAL FORCE Using virtual work method, we can calculate the theoretical force of members and calculated the reaction force using the equilibrium equations: ∑M=0
∑ Fx = 0
∑ Fy = 0
Consider moment at point B: ΣMB
=0
-HA(1) + 250(2)
=0
HA = 500 N ΣHX
=0
HA + HB = 0 HB = -500 N
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Page 15 of 23
ΣHY
=0
-VB – 250 = 0 VB
= -250 N
To find out the theoretical force value at each member, we use the joint method. We get the value in Table 3. We ignore for member 6 because it is a redundant member and the truss can be statically determinate trusses after we release a member 6.
JOINT A
Σ Fy ↑ = Σ Fy ↓ FAB = 0
Σ =Σ 500 + FAC = 0 FAC = -500 kN (C)
JOINT D
Σ Fy ↑ = Σ Fy ↓ FyED = 250
Σ =Σ FCD = -FxED
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Page 16 of 23 = -250 kN (C)
=
=
FxED = 250
= FED = 354 kN
JOINT C
Σ Fy ↑ = Σ Fy ↓ FCE + FyBC = 0 FCE = -FyBC
Σ =Σ FCD = FAC + FyBC -250 = -500 +
FxBC = -250 kN =
FxBC = 250 kN
= FBC = 354 kN
JOINT B
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Page 17 of 23
Σ Fy ↑ = Σ Fy ↓ 250 = FBC + FAB = FxBC = 250
FBE
Σ =Σ + FxBC = 500 FBE = 500 - FxBC = 250kN
CALCULATION FOR FORCE DUE TO 1 UNIT LOAD Using the 1 unit load method, we can calculate the forces of each member due to the unit load, 1 N at member 6 and calculate the reaction force using the equation. ∑M=0
∑ Fx = 0
∑ Fy = 0
Consider moment at point A: ΣMA
=
0
HB (1) – 1 (1) + 1 (1) HB ΣHX
=0 =0N
=0 1 1 - 1 = 1 . 414 1.414
HA + HB + 1
0
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Page 18 of 23 HA = 0 N ΣFB
=0 1 1 - 1 1.414 1.414
-FB + 1
FB
=0
=0N
ALTERNATIVE METHOD
EXAMPLE FOR
OF
MEMBER
CALCULATION 6
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Page 19 of 23
7.0
P=
Total Total
P=
854.6 4.828
P=
177.00
Fnl n 2l
N
DISCUSSION
7.1 From table 3, compare your answer to the experimental values. Comment on the accuracy of your result. Refer to table 3, the value in experimental force were differ with the theoretical value. There were in member 1,2,5,6 and 8. It was because parallax, the equipment has not fully function correctly. It is maybe the device were not well maintenance . Secondly, it maybe from environment in the lab. The device were sensitive with vibration and wind. But the member 3,4,7 almost same with theoretical force. 7.2 Compare all of the member forces and the deflection to those from statically determinate frame. Comment on them in terms of economy and safety of the structure. There have positive and negative force with tensile and compression at all member. Some structures are built with more than this minimum number of truss members. Those structures may survive even when some of the members fail or deflection, because their member forces depend on the relative stiffness of the members, in addition to the equilibrium condition described. These can be economy for structure. Failure occurs when the load (L) effect exceeds the ability (R) of the structure, and can be derived by considering the probability density functions of R and L, along with their random variables. The main goal for the safety of the structure is to guarantee an R>L scenario throughout the design life of the structure.
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Page 20 of 23 7.3 What problem could you for seen if you were to use a redundant frame in a “real life’ application. (Hint: look at the zero value for the strain reading once you have included the redundant member by winding up thumbnut). The structure will be failed if the load are exceed the ability. In this experiment, the value and size are not same with ‘real life’ but the application is too same. In my knowledge, the redundant frame always used in bridge construction to stability and the redundant frame are useable for esthetic value sometimes.
8.0
CONCLUSION In this experiment, we use few type of different load from 50N till 250N to evaluate the data from the trusses. The most important of these criteria is the structure’s ability to carry load safely. The limit load for this equipment is 350N. The calculation to evaluate of structural safety can only be done mathematically and the experimental force data that we collected from digital reading than be compared with the theoretical force value that be done manually as we studied in analysis structure module. As the graph load vs. deflection is been plotted, the result was as similar to the linear. Some mistake when reading the value, this is parallax error. And the equipment is not in a good condition. It would be impractical, uneconomical, and unsafe for the structural engineer to evaluate a bridge design by building a full-size prototype. When a structure is built, it must be stiff enough to carry its prescribed loads and fully corrected when reading the value. There will be a small “ralat” in every experiment and it can’t be avoided but any how we should prevent it so that it will not affect the calculation or stiffness of the structure. We suggest making the maintenance for the equipment and exchanging the damage tool. This is because the student can’t get the correct value for those experiments.
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Page 21 of 23
9.0
APPENDIX
Truss
Digital Dial Gauge
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Page 22 of 23
Force Output
Digital Force Reading Meter
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Page 23 of 23
Digital Strain Indicator
Indeterminate truss
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