truss report

7/29/2019 Truss Report

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FACULTY OF CIVIL AND ENVIRONMENT

ENGINEERING

STRUCTURE AND MATERIAL LABORATORY

BFC21201

FORCE IN A STATICALLY DETERMINATE CANTILEVER TRUSS

GROUP MEMBER :

NAME MATRIC NUMBER

ESTHER MARIE JUIP CF120139

FAEZAH BT MD. ADNAN CF120256

FARADILLAH BINTI ABDUL LAIT CF120250

RABIATUL ADAWIYAH BINTI

ZAINAL ABIDINCF120053

LECTURERS NAME : MR. MOHAMAD HAIRI BIN OSMAN

SECTION : 3


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1.0 OBJECTIVE1.1 To examine a statically determinate frame and to analyze the frame using simple

pin joint theory.

2.0 LEARNING OUTCOME2.1 The application the engineering knowledge in practical application.

2.2 To enhance technical competency in structural engineering through laboratory

application.

2.3 To communicate effectively in group.

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

laboratory application.

3.0 THEORYA truss is a structure composed of slender member joined together at their

endpoints to form one or more triangles. The joint connections are considered aspinned

joint without friction.

In order to determine the forces developed in the individual members at a truss,the

following assumptions should be make :

1. The members are connected to each other at their ends by friction less pins,that is

only a force and no moment can be transferred from one member toanother.

2. External loads are applied to the truss only at its joints.


3/20

One of the methods to calculate the forces in the member of a truss is using:

Method Of Joints

Suitable to use in calculating all of the member forces for a truss.

This method entails the use of a free body diagram of joints with the

equilibrium equationsFx = 0 and Fy = 0.

Calculation only can be started for joint where the numbers of unknowns are two or less.

4.0 APPARATUS

The strain force meter input Digital indicator meter acting to

acting in the force acting in the rod frame

Apparatus to apply the load Statically determinate cantilever

truss apparatus


4/20

5.0 PROCEDURESi. Unscrew the thumbwheel on the redundant members. Note that it was

effectively no longer part of the structured as the idealized diagram

illustrates.

ii. Applied the pre-load of 100N downward, the load cell was re-zero and

applied it a load of 250N carefully and check that the frame is stabled and

secured.

iii. The load to zero (leaving the 100N preload), recheck and re-zero the

digital indicator. Never applied the loads greater than those specified on

the equipment.

iv. Loads apply in the increment shown in Table 1 recording the strain

readings and the digital indicator readings. Table 2 was completed by

subtracting the initial (zero) strain readings. (be careful with sign)

6.0 RESULTS

Load

(N)

Strain Reading Digital

Indicator

Reading

(mm)1 2 3 4 5 6 7 8

0 131 223 - 13 -37 113 0 31 52 0

50 141 214 -22 -55 113 0 44 66 -0.031

100 149 205 -30 -73 113 0 56 78 -0.059

150 158 196 -40 -92 113 0 69 92 -0.088

200 167 186 -49 -111 113 0 83 106 -0.11

250 174 179 -56 -126 112 0 94 118 -0.132

Table 1: Strain Reading and Frame Deflection for Experiment 1


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Load

(N)1 2 3 4 5 6 7 8

0 0 0 0 0 0 0 0 0

50 10 -9 -9 -18 0 0 13 14

100 18 -18 -17 -36 0 0 25 26

150 27 -27 -27 -55 0 0 38 40

200 36 -37 -36 -74 0 0 52 54

250 43 -44 -43 -89 -1 0 63 66

Table 2: True Strain Reading For Experiment 1

MemberExperiment

force (N)

Theoretical

force (N)

1 239.48 250

2 -244.12 -250

3 -238.57 -250

4 -493.79 -500

5 -5.55 0

6 0 0

7 349.54 353.55

8 366.18 353.55

Table 3: Measured and Theoretical Force in the Cantilever Truss


6/20

Analysis and Calculation

True Strain Reading For Experiment 1(table 2)

Example:

Load = 0, 50, 100, 150, 200, 250

True Strain Reading = BA

B = load at strain needed

A = load at before strain needed

Member 1

Load 50 Load 100

True Strain Reading = 141131 True Strain Reading = 149 - 131

= 10 = 18

Load 150 Load 200

True Strain Reading = 158131 True Strain Reading = 167 - 131

= 27 = 36

Load 250

True Strain Reading = 174 - 131

= 43


7/20

From the formula:

E = / = F/A

E = Young s Modulus (Nm-2

) F = Force in member (N)

= Displayed Strain A = cross section area of the member (m2)

= Stress in the member (Nm-2

)

= E

F/A = E

F = E x x A

Calculation For Experimental Force (N)

Given ,

Rod diameter (d) = 5.80 mm

E steel = 2.10 x105

N/mm2

A = d2

/4

= (5.80)2

/4

= 26.42 mm3

Member 1

F = E x x A

= (2.10x105) x (43x10

-6) x (26.42)

= 239.48 N


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Member 2

F = E x x A

= (2.10x105) x (-44x10

-6) x (26.42)

= -244.12 N

Member 3

F = E x x A

= (2.10x105) x (-43x10

-6) x (26.42)

= -238.57 N

Member 4

F = E x x A

= (2.10x105) x (-89x10

-6) x (26.42)

= -493.79 N

Member 5

F = E x x A

= (2.10x105) x (-1x10

-6) x (26.42)

= -5.55 N

Member 7

F = E x x A

= (2.10x105) x (63x10

-6) x (26.42)

= 349.54 N

Member 8

F = E x x A

= (2.10x105) x (66x10

-6) x (26.42)

= 366.18 N


9/20

2

34

578

1

B

A

DC

E

Calculation for Theoretical Force (N)

Checking :

Members, m = 7

Reactions, r = 3

m + r = 2j

7 + 3 = 2(5)

10 = 10

.. The structure is statically determinate and stable.

By using equivalent equation :

+ A = 0 + y = 0

-FxB(0.24) + 250(0.48) = 0 FyA250 = 0

FxB = -120/-0.24 FyA = 250 N

FxB = 500 N

FyA

FxA

Fx B

250 N


10/20

FAB

500 N FAB

250 N

-500 N FAE

0

FAC

FACCos

FACSin

+x = 0

FXA + FXB = 0

FXA + 500 = 0

FXA = -500 N

At point B

Fx = 0

500 + FBC = 0

FBC = -500 N (C)

Fy = 0

FAB = 0

At point A

= 45

Fy = 0

250 + FABFACSin = 0

250 + 0 - FACSin45 = 0

FAC = 353.33 N (T)

Fx = 0

-500 + FAE + 353.55cos45 = 0

FAE = 250 N (T)


11/20

FCE

FDEcos

FDEsin FDE

250 N

FCD

-353.55 N

-353.55sin

-353.55cos

At point E

Fx = 0

-250 + FDEcos45 = 0

FDE = 353.55 N (T)

Fy = 0

FCEFDEsin = 0

FCE(-353.55sin45) = 0

FCE = -250 N (C)

At point D

Fx = 0

-353.55cos + FCD

-(-353.55cos45) + FCD = 0

FCD = -250 N (C)

FAB = 0

FAC = 353.33 N (T)

FAE = 250 N (T)

FBC = -500 N (C)

FCD = -250 N (C)

FCE = -250 N (C)

FDE = 353.55 N (T)


12/20

250 N

500 N

-500 N

250 N

250 N

0

353.55 N

-500 N -250 N

250 N

-353.55 N

A

B CD

E

7.0 DATA ANALYSIS

Member 1

0

50

100

150

200

250

0 50 100 150 200 250

TrueandRecordedStrain()

Load (N)

True and Recorded Strain Against Load

Recorded Strain

True Strain


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Member 2

Member 3

0

50

100

150

200

250

0 50 100 150 200 250

TrueandRecordedStrain()

Load (N)


True Strain

Recorded Strain

-60

-50

-40

-30

-20

-10

0

0 50 100 150 200 250

Truea

nd

RecordedStrain()

Load (N)


Recorded Strain

True Strain


14/20

Member 4

Member 5

-140

-120

-100

-80

-60

-40

-20

0

0 50 100 150 200 250

T

rueand

Recorded

Strain()

Load (N)

True and Recorded Strain Against

Load

Recorded

Strain

True Strain

-20

0

20

40

60

80

100

120

0 50 100 150 200 250True

and

RecordedStrain()

Load (N)

True and Recorded Strain Against

Load

Recorded Strain

True Strain


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Member 7

Member 8

10

20

30

40

50

60

70

80

90

100

0 50 100 150 200 250

Trueand

RecordedStrain()

Load (N)


True Strain

Recorded Strain

0

20

40

60

80

100

120

140

0 50 100 150 200 250

Tr

ueand

RecordedStrain()

Load (N)


Recorded Strain

True Strain


16/20

Deflection against Load graph

Comment on a graph:

From the graph of True and Recorded Strain against Load above, we obtained that thevalue of recorded and true strain are different. This is because in a statically determinate

cantilever truss is leaving the 100 N pre-load.

According to the True and Recorded Strain against Load, it shows the line graph are

obtaining from the result. For member 1, 7 and 8 we obtained the linear graph, the graph true

strain and recorded proportional with load. This is because the value of true strain and recorded

can increases when the load increases. The recorded strain are larger than true strain values.

From the graph, it shows the true and recorded strain for the member 2, 3 and 4 aredecrease when the load is increase. For the member 5, the graph shows it does not have an

increasing or decreasing with a large value. The values are same with different load.

From the deflection graph, we obtained that the graph are in linear. When the load is

increase, the deflection decrease.

-0.14

-0.12

-0.1

-0.08

-0.06

-0.04

-0.02

0

0 50 100 150 200 250

Deflection

(mm)

Load (N)

Deflection Against Load

Deflection


17/20

8.0 DISCUSSION AND CONCLUSION

1. Compare the experimental and theoretical result:

MemberExperiment

force (N)

Theoretical

force (N)

1 239.48 250

2 -244.12 -250

3 -238.57 -250

4 -493.79 -500

5 -5.55 0

6 0 0

7 349.54 353.55

8 366.18 353.55

From the table above, we obtained the differences value between the value of

experimental force and theoretical force. This may because of several factors. Such as:

One of the factor is machine error.

The applied pre-load of 100N downward maybe less or more than 100N.

The imperfection of members used for the particular experiment also affects

the result of the experiment.

The load cell was not accurately point to zero value that gave affect to the

value of force.

After the apply load of 250N, the frame not checking.


18/20

2. From your result and the theoretical member force, identify which members are

in compression and which members are in tension. Explain your choice.

Base on the figure 1, we can conclude that members are in compression is member

3,4 and 7. The members are in tension is member 1, 2 and 8. This is because the value of

force is negative are considered as compression force and force with positive values are

considered as tensional forces. The negative value shows that the force is pointing inwards

and the positive values denote the force pointing outward.

Axial force member :


19/20

FAB

500 N FAB

3. Observe the reading of members 5. Explain why the readings is almost zero.

At point BFx = 0

500 + FBC = 0

FBC = -500 N (C)

Fy = 0

FAB = 0

From the results on member 5, we noticed that the reading force is almost zero. We

know that, member 5 is attached by a pin joint and a roller joint at both ends. On the pin

joint, two forces acting towards it on the horizontal axis and vertical axis. On the roller joint,

there is only one force acting towards it on the horizontal axis. Therefore, the reading force

of member 5 is almost zero due to these three forces.

4. Are the strains gauges are an effective transducers for measurement forces in the

framework.

Yes, the strains gauges are effective transducers for measurement of forces in the

framework. We can get the effective transducers from the data we obtain.


20/20

5. Does the framework comply with pin joint theory even though the joint are not truly

pin joint?

Based on the experiment, we know that the values between experimental force and

theoretical force are totally different. This result indicates that the framework comply with

the pin joint theory even though the joint are not truly pined joint.

One of the objective of this experiment is to examine a statically determinate frame

and to analyze the frame using simple pin joint theory. As on conclusion, we can summaries

that our experiment is achieved the objective of experiment.

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