Truss Full Report

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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)

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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

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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


ENGINEERING)

http://en.wikipedia.org/wiki/Stiffness

http://en.wikipedia.org/wiki/Statically_indeterminate

http://en.wikipedia.org/wiki/Statically_indeterminate

http://en.wikipedia.org/wiki/Structural_load

http://en.wikipedia.org/wiki/Mechanical_equilibrium

http://en.wikipedia.org/wiki/Stability

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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.


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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


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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


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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


ENGINEERING)

http://www.answers.com/topic/physical-compression

http://www.answers.com/topic/tension-physics

http://www.answers.com/topic/bridge

http://en.wikipedia.org/wiki/Bending

http://en.wikipedia.org/wiki/Statics

http://en.wikipedia.org/wiki/Physics

http://en.wikipedia.org/wiki/Newton's_laws_of_motion

http://en.wikipedia.org/wiki/Truss

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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.


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The common types of truss bridge


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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.


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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


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(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


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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)


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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


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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


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Fy = Fy =

FAB = 0 500 + FAC = 0 FAC = -500 kN (C)

JOINT D

Fy = Fy =

FyED = 250 FCD = -FxED

= -250 kN (C)

= = =


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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


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= 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


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ALTERNATIVE METHOD

EXAMPLE OF CALCULATION

FOR MEMBER 6

P =

P =

P =


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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.


ENGINEERING)

http://en.wikipedia.org/wiki/Stiffness

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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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9.0 APPENDIX

Truss

Digital Dial Gauge


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Force Output

Digital Force Reading

Meter


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Digital Strain Indicator

Indeterminate truss


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