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BFC21201 MAKMAL MEKANIK BAHAN (SEKSYEN 6)

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TITLE : OPEN ENDED LABORATORY

( FORCE IN STATICALLY DETERMINATE TRUSS)___________________________________________________________________________

1.0 OBJECTIVE

To examine joint deflection of Statically Determinate Cantilever Truss

2.0 LEARNING OUTCOMES

2.1 Application of theoretical engineering knowledge through practical application.

2.2 To enhance technical competency in structure engineering through laboratory

application.

2.3 Communicate effectively in group.

2.4 To identify problem, solving and finding out appropriate solution through

laboratory application.

3.0 THEORY

In general great stiffness in proportion to their mass. When the deflection of a truss is

significant, it is usually the result of one two causes. For most trusses deflection is essentially

due to the lengthening and shortening of the members caused by the interior forces of tension

and compression.

In relativey simple procedure for the detrmining the deflected shape of a truss due to the

length change of the members is to plot the deformation graphically. His consists simply of

constructing the individual triangles of the truss with the sides equal to the deformed lengths.

The virtual work method can be used to determine the deflection of trusses. We knowfrom the principle of virtual work for trusses that the deflection can be calculated by the equation

with n equal to the virtual force in the member and equal to the change in

length of the member. Therefore, the deflection of a truss due to any condition that causes a

change in length of the members can be calculated. This change in length can be caused by theapplied loads acting on each member, temperature changes, and by fabrication errors.

Axial Deformation:




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From statics we know how to determine member forces in a truss by using either the method of

joints or the method of sections. Once these forces are known we can determine the axialdeformation of each member by using the equation:

The equation for the deflection can be modified with this value for .

where m is equal to the number of members, n is the force in the member due to the virtual load,

N is the force in the member due to the applied load, L is the length, A is the area, and Erepresents Young's Modulus of Elasticity.

Temperature Changes:

The axial deformation of a truss member of length L due to a change in temperature of isgiven by:

where is the coefficient of thermal expansion.

The equation for the deflection is then modified with this value for .

where j is the number of members experiencing temperature change and n is the force in the

member due to the virtual load.

Fabrication Errors:

In the case of fabrication errors, the deformation of each member is known. Therefore, the

original equation for deflection of a truss can be modified.




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where k is the number of members undergoing fabrication errors and n is the force in the

member due to the virtual load and is the change in length of the member due to fabrication

errors.

The total deflection of a truss is made up of the sum of all of these cases.

This equation is now used to find the deflection of a truss. Please refer to an introductory text

book on structural analysis for a complete description of this approach.

APPARATUS

3.1

Cantilever truss

3.2 Digital Dial Test Indicator




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5.O PROCEDURES

4.1 Unscrew the thumwheel on the “redundant” member. Note that it is effectively no

longer part of the structure as the idealised diagram illustrates.

4.2 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.

4.3 Return the load to zero (leaving the 100N preload), re-check and re-zero thedigital indicator. Never apply loads greater than those specified on the

equipment.

Apply load in the increment shown in Table 1 recording the strain readings and

the digital indicator readings. Complete Table 2 by subtracting the initial (zero) strain

readings. (be careful with the sign)




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5.0 DATA

Load

(N) Strain Reading ()

Digital

indicator

reading

(mm)1 2 3 4 5 6 7 8

0 143 220 -48 -78 106 0 7 4 0.000

220 200 200 -86 -136 126 0 63 37 0.107

Table 1 : Strain reading and frame deflection

Table 2: True strain reading

1

5

8 7

34

A

E DC

B

ΣFAX

ΣFEX

2




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Member Experimental force (N) Theoretical Force (N)

1 338.751 220

2 -118.86 220

3 225.834 -220

4 344.694 440

5 118.86 0

6 - -

7 332.808 310.75

8 196.119 310.75

Table 3 : Measured and theoretical force in the cantilever truss

6.0 RESULT AND CALCULATIONS

Calculation of the equivalent member forceusing the Young’s Modulus relationship

. Completethe 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 = 6 x 10-3

m and Esteel = 2.10x105

N/mm2

A = лd2 /4

= л(6 x 10-3)2 / 4

= 2.83 x 10-5 m2




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σ = Eε

F = σA

Calculation for Experiment force :

Use displayed strain at load 220 N

Member 1 : Displayed strain, ε = 57

Stress, σ = Eε

= (2.10 x 105

N/mm2)(57)

= 11.97 x 106N/m

2

Force, F = σA

= (11.97 x 106

N/m2)( 2.83 x 10-5 m

2)

= 338.751 N

Member 2 : Displayed strain, ε = -20

Stress, σ = (2.10 x 105

N/mm2)(-20)

= -4.20 x 106 N/m2

Force, F = (-4.20 x 106

N/m2)( 2.83 x 10

-5 m2)

= -118.86 N




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Member 3 : Displayed strain, ε = 38

Stress, σ = (2.10 x 105

N/mm2)( 38)

= 7.98 x 106

N/m2

Force, F = σA

= (7.98 x 106 N/m2)( 2.83 x 10-5 m2)

= 225.834 N

*Calculation of experiment force for others member same with above

Calculation for Theoretical force :

Checking :

m = 2j – R (Joint = 5, Member = 7, Reaction = 3)

7 = 2(5) – 3

7 = 7 The structure is statically determinate

∑ Fy = 0

RD – F = 0

RD – 220 = 0

RD = 220 N




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∑MD = 0

220 (480) – HA(240) = 0

HA = 440 N

∑FX = 0

HA – HD = 0

440 – HD = 0

HD = 440 N

Joint A ∑FX = 0

440 – FAB = 0

FAB = 440 N

∑ Fy = 0

FAD = 0 N

FAD

AFAB 440N




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Joint D

∑Fy= 0

220 – 0 – FDB = 0

FDB = 220

FDB = 310.75 N

∑FX = 0

FDE - 440+ (310.75) = 0

FDE = 220 N

Joint C

∑Fy = 0

-220 + FCE = 0

FCE = 220

FCE = 310.75N

∑FX = 0

– FCB – (310.75) = 0

FCB = – 220 N

Joint E ∑FX = 0

– FED + FEC = 0

– 220 + (310.75) = 0

0 = 0

220N

D FDE 440N

FAD = 0N FDB

FCE

CFCB

220N

EFED= 220 N

FEC= 310.75NFEB




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∑Fy = 0

FEB – FEC = 0

FEB =

(310.75)

FEB = 220 N

Joint B ∑Fy = 0

– FBE + FBD = 0

– 220 + (310.75N) = 0

0 = 0

∑FX = 0

FBA – FBC – FBD = 0

440 – 220 – (310.75) = 0

0 = 0

FBE = 220N

FBD = 310.75N

FBC =220N FBA=440 NB




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220 N

440 N - 220 N

220 N310.75 N

220 N

310.75 N

HA=440

HD=440 N

RD =220 N

A

E

CB

D

0 N




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∑ Fy = 0

RD – F = 0

RD – 1 = 0

RD = 1 N

∑MD = 0

1 (480) – HA(240) = 0

HA = 2 N

∑FX = 0

HA – HD = 0

2 – HD = 0

HD =2 N

Joint A ∑FX = 0

2 – FAB = 0

FAB

=2 N

∑ Fy = 0

FAD = 0 N

FAD

AFAB 2N




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Joint D

∑Fy= 0

1 – 0 – FDB = 0

FDB = 1

FDB = 1.41 N

∑FX = 0

FDE - 2+ (1.41) = 0

FDE = 1 N

Joint C

∑Fy = 0

-1+ FCE = 0

FCE =1

FCE = 1.41N

∑FX = 0

– FCB – (1.41) = 0

FCB = – 1 N

Joint E ∑FX = 0

– FED + FEC = 0

– 220 + (310.75) = 0

0 = 0

1N

D FDE 2N

FAD = 0N FDB

FCE

CFCB

1N

EFED=1 N

FEC=1.41NFEB




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∑Fy = 0

FEB – FEC = 0

FEB =

(1.41)

FEB = 1 N

Joint B ∑Fy = 0

– FBE + FBD = 0

– 1 + (1.41N) = 0

0 = 0

∑FX = 0

FBA – FBC – FBD = 0

2 – 1 – (1.41) = 0

0 = 0

FBE = 1N

FBD = 1.41N

FBC =220N FBA=2 NB




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1 N

2 N - 1 N

1 N1.41 N

1N

1.41 N

HA=2N

HD=2N

RD =1N

A

E

CB

D

0

1N




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∂ = ΣPOPIL

Experimental

MEMBER PO (N) PI (N) L (M) POPIL

1 338.751 1 0.24 81.30

2 -118.86 1 0.24 -28.32

3 225.834 -1 0.24 -54.20

4 344.694 2 0.24 165.45

5 118.86 0 0.24 0

7 332.808 1.41 0.34 159.55

8 196.119 1.41 0.34 94.02

Σ=417.8

∂ = ( )()417.8 = 70.30

Theoretical

MEMBER PO PI L POPIL

1 220 1 0.24 52.8

2 220 1 0.24 52.8

3 -220 -1 0.24 52.8

4 440 2 0.24 211.2

5 0 0 0.24 0

7 310.75 1.41 0.34 148.97

8 310.75 1.41 0.34 148.97

Σ = 667.54

∂ = ( )()

667.54 = 112.32




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Member TYPE Theoretical

Force (N)

AREA

(10-5 m2)

LENGTH

(M)

LENGTH CHANGE

(M)

TRUE10

-6

FOR PLOT10

-3

1 COMPRESSION 220 2.83 0.24 8.88 x 10-6 2.22

2 COMPRESSION 220 2.83 0.24 8.88 x 10- 2.22

3 TENSION 220 2.83 0.24 8.88 x 10- 2.22

4 COMPRESSION 440 2.83

0.24 1.78 x 10-5 4.45

5 0 2.83

0.24 0 0

7 COMPRESSION 310.75 2.83 0.34 1.78 x 10-5

4.45

8 COMPRESSION 310.75 2.83 0.24 1.78 x 10-5

4.45

Member TYPE Experimental

force (N)

AREA

(10-5

m2)

LENG

TH

(M)

LENGTH CHANGE

(M)

TRUE FOR PLOT10

-3

1 COMPRESSION 338.751 2.83 0.24 1.37 x 10-5 3.43

2 TENSION 118.86 2.83 0.24 4.80 x 10- 1.20

3 COMPRESSION 225.834 2.83 0.24 9.12 x 10- 2.28

4 COMPRESSION 344.694 2.83

0.24 1.39 x 10-5 3.48

5 COMPRESSION 118.86 2.83 0.24 4.80 x 10-

1.20

7 COMPRESSION 332.808 2.83 0.34 1.90 x 10-5

4.75

8 COMPRESSION 196.119 2.83 0.24 1.12 x 10-5

2.80




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7.0 DATA ANALYSIS

Graph 1

-100

-50

0

50

100

150

200

250

1 2 3 4 5 7 8

P o P i L ( N ² M )

MEMBER

PoPiL (N²M) VERSUS MEMBER

EXPERIMENTAL

THEORETICAL




Page | 20

Graph 2

DISCUSSION

The tool used is an error and causes an error reading. This is probably because alt is used

loosely there any place that affect the reading. In addition, the device used can not be shaken or

there is any movement of the tool. Other issues discussed camp, the reading is taken too quickly

without waiting for a static recitation.

From grap that have been obtained, the first graph shows comparable diaantara PoPiL

Theoretical and expremental between the member. Can be seen that, comparable between the

kedua2dua not reading too much. In graph 1, shows the readings of the member 1 and member of

three readings were the same for Theoretical while reading from 1 to 3 experimental experience

menurunan. Theoretical and experimental readings then began to rise in the member-3 to be in

the same reading on the to-5 member. As a member of the 7 Theoretical and experimental

readings and the readings theorical increased in the same reading on the member 8 and the

experimental decrease.

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

1 2 3 4 5 7 8

L E G T H C H A N G E ( M )

MEMBER

DIFFERENT LENGTH CHANGE BETWEEN

THEORITICAL AND EXPERIMENT

THEORITICAL

EXPERIMENTAL




Page | 21

To graph the length-2 showed different change Between Theoretical and experiment in

the reading and Theoretical experimtal. Theoretical readings can dilahat member to the

horizontal so that the reading of experimental-3 and decreased significantly from member to

member -1 to-2 and upgrading of the member to the member-2 to-4. Theoretical readings to rise

in member-4. Further reading went down from member to member on-4 to-5 and increased again

from member to member-5-7 as well as the experimental readings. In the latest data, while the

experimental data decreased Theoretical horizontal.

From what you see there are changes that affect the reading of whether the data is

dierolehi in Theoretical or experimental. Decimal point is also playing a role in getting the data

in Theoretical data obtained.

CONCLUSION

We have learned a great deal about how the truss deflection depends on change of length. Can be

seen in the experiment was carried out differences between the two readings are not too far away

and it can use to determine the deflection of the truss.

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