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Cantilever truss lab report
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Page 1 of 24
TABLE OF CONTENT
Number Description Page
1.0 Objective of the experiment 2
2.0 Learning Outcome 2
3.0 Theory 2
4.0 Application of Truss 6
5.0 Procedures 10
6.0 Result and Analysis 11
7.0 Discussion 19
8.0 Conclusion 20
9.0 Appendix 21
1.0 OBJECTIVE (Group 7)
(DEPARTMENT OF STRUCTURE AND MATERIAL ENGINEERING)
Page 2 of 24
1.1 The effect of redundant member in a structure is observed
and the method of analyzing type of this structure is
understood.
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
(Group 7) (DEPARTMENT OF STRUCTURE AND MATERIAL
ENGINEERING)
Page 3 of 24
mathematical terms, we have the following necessary condition
for stability:
M +R ≥ 2j
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
(Group 7) (DEPARTMENT OF STRUCTURE AND MATERIAL
ENGINEERING)
Page 4 of 24
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
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.
(Group 7) (DEPARTMENT OF STRUCTURE AND MATERIAL
ENGINEERING)
Page 5 of 24
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 =
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
(Group 7) (DEPARTMENT OF STRUCTURE AND MATERIAL
ENGINEERING)
Page 6 of 24
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
(Group 7) (DEPARTMENT OF STRUCTURE AND MATERIAL
ENGINEERING)
Page 7 of 24
4.0 APPLICATION OF TRUSS
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
(Group 7) (DEPARTMENT OF STRUCTURE AND MATERIAL
ENGINEERING)
Page 8 of 24
Wind Load, plus a factor for your local Snow Load, plus a Safety Factor.
A Structural Engineer can check these figures.
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.
(Group 7) (DEPARTMENT OF STRUCTURE AND MATERIAL
ENGINEERING)
Page 9 of 24
The common types of truss bridge
(Group 7) (DEPARTMENT OF STRUCTURE AND MATERIAL
ENGINEERING)
Page 10 of 24
The common types of roof truss
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.
(Group 7) (DEPARTMENT OF STRUCTURE AND MATERIAL
ENGINEERING)
Page 11 of 24
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.
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.
5.10 Referred to table 4 and entered in the values and carefully
calculated the other terms as required.
5.11 Entered result into Table 3.
6.0 RESULT AND DATA ANALYSIS
Member Strains (με)
Load
(N)
Strain Reading Digital indicator reading
1 2 3 4 5 6 7 8
(Group 7) (DEPARTMENT OF STRUCTURE AND MATERIAL
ENGINEERING)
Page 12 of 24
(mm)0 142 225 -38 -69 109 35 21 22 0.00950 154 218 -50 -89 111 28 32 28 -0.024100 167 213 -58 -58 115 21 45 35 -0.051150 181 209 -67 -67 120 15 58 43 -0.079200 194 204 -76 -76 124 7 72 50 -0.103250 205 200 -84 -84 128 0 83 56 -0.127
Table 1: Strain Reading and Frame Deflection
Member Strains (με)
Load(N)
1 2 3 4 5 6 7 8
0 0 0 0 0 0 0 0 050 12 -7 -12 -20 2 -7 11 6100 25 -12 -20 -51 6 -14 24 13150 39 -16 -29 -60 11 -20 37 21200 52 -21 -38 -69 15 -28 51 28250 63 -25 -46 -77 19 -35 62 34
Table 2: True Strain Reading
Member Experimental Force (N) Theoretical Force (N)1 374.07 2502 -148.44 2503 -273.13 -2504 -457.19 -5005 112.81 06 -207.81 07 368.13 3548 201.88 354
Table 3: Measured and Theoretical in the Redundant Cantilever
Truss
Member Length F n Fnl n2l Pn Pn + f
1 1 250 -0.707 -176.75 0.5 -125.14 -375.14
2 1 -250 -0.707 176.75 0.5 -125.14 124.87
(Group 7) (DEPARTMENT OF STRUCTURE AND MATERIAL
ENGINEERING)
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3 1 -250 0 0 0 0 -250.00
4 1 -500 -0.707 354 0.5 -125.14 -625.14
5 1 0 -0.707 0 0.5 -125.14 -125.14
6 1.414 0 1 0 1.414 177.00 177.00
7 1.414 354 0 0 0 0 354.00
8 1.414 354 1 500.56 1.414 177.00 531.00
Total = 854.6 4.828
Table 4: Table for Calculating the Force in the Redundant
Truss
P =
Data :-
Rod Diameter, D = 6.0 mm
= 0.06 m
Esteel = 2.10 x 10 5 N/mm
EXPERIMENTAL FORCE
Using the Young’s Modulus relationship, we can calculate the
equivalent member force, complete the experimental force in Table 3.
Where,
E = Young’s Modulus (N/m2)
σ = Stress in the member (N/m2)
(Group 7) (DEPARTMENT OF STRUCTURE AND MATERIAL
ENGINEERING)
Page 14 of 24
ε = Displayed strain
And
Where,
F = Force in member (N)
A = Cross section area of the member (m2)
To calculate the experimental force, we use the formula
With,
So,
=
CALCULATION FOR EXPERIMENTAL FORCE
Member 1 ( ε = 63 x 10-6 )
F = 2.10 x 105 N/mm2 x 28.274 mm2 x 63 x 10-6
F =
CALCULATION FOR THEORETICAL FORCE
(Group 7) (DEPARTMENT OF STRUCTURE AND MATERIAL
ENGINEERING)
Page 15 of 24
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
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
(Group 7) (DEPARTMENT OF STRUCTURE AND MATERIAL
ENGINEERING)
Page 16 of 24
Fy = Fy =
FAB = 0 500 + FAC = 0 FAC = -500 kN (C)
JOINT D
Fy = Fy =
FyED = 250 FCD = -FxED
= -250 kN (C)
= = =
(Group 7) (DEPARTMENT OF STRUCTURE AND MATERIAL
ENGINEERING)
Page 17 of 24
FxED = 250 FED = 354 kN
JOINT C
Fy = Fy =
FCE + FyBC = 0 FCD = FAC + FyBC FCE = -FyBC -250 = -500 + FxBC
= -250 kN FxBC = 250 kN
= =
FBC = 354 kN
JOINT B
Fy = Fy =
250 = FBC + FAB FBE + FxBC = 500
(Group 7) (DEPARTMENT OF STRUCTURE AND MATERIAL
ENGINEERING)
Page 18 of 24
= FBE = 500 - FxBC
FxBC = 250 = 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) = 0
HB = 0 N
ΣHX = 0
HA + HB + 1 - 1 = 0
HA = 0 N
ΣFB = 0
-FB + 1 - 1 = 0
FB = 0 N
(Group 7) (DEPARTMENT OF STRUCTURE AND MATERIAL
ENGINEERING)
Page 19 of 24
ALTERNATIVE METHOD
EXAMPLE OF CALCULATION
FOR MEMBER 6
P =
P =
P =
(Group 7) (DEPARTMENT OF STRUCTURE AND MATERIAL
ENGINEERING)
Page 20 of 24
7.0 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.
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.
(Group 7) (DEPARTMENT OF STRUCTURE AND MATERIAL
ENGINEERING)
Page 21 of 24
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.
(Group 7) (DEPARTMENT OF STRUCTURE AND MATERIAL
ENGINEERING)
Page 22 of 24
9.0 APPENDIX
Truss
Digital Dial Gauge
(Group 7) (DEPARTMENT OF STRUCTURE AND MATERIAL
ENGINEERING)
Page 23 of 24
Force Output
Digital Force Reading
Meter
(Group 7) (DEPARTMENT OF STRUCTURE AND MATERIAL
ENGINEERING)
Page 24 of 24
Digital Strain Indicator
Indeterminate truss
(Group 7) (DEPARTMENT OF STRUCTURE AND MATERIAL
ENGINEERING)