Open Ended - Cantilever Truss

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FACULTY OF CIVIL AND ENVIRONMENTAL

ENGINEERING

DEPARTMENT OF STRUCTURE AND MATERIAL

ENGINEERING

LAB MATERIAL

REPORT

Subject Code BFC 21201

Code & Experiment Title OPEN ENDED - FORCE IN A STATICALLY DETERMINATE

CANTILEVER TRUSS

Course Code 2 BFF/1

Date 03/10/2011Section / Group 2

Name MUHAMAD ASYRAF BIN AB MALIK (DF100108)

Members of Group 1.MUHAMMAD IKHWAN BIN ZAINUDDIN (DF100018)

2.AHMAD FARHAN BIN RAKAWI (DF100142)

3.IDAMAZLIZA BINTI ISA (DF100128)

4.AINUN NAZHIRIN BINTI ABD JALIL (DF100076)

Lecturer/Instructor/Tutor EN MOHAMAD HAIRI BIN OSMAN

Received Date 18 OCTOBER 2011

Comment by examiner Received



STUDENT CODE OF ETHIC

(SCE)

DEPARTMENT OF STRUCTURE AND MATERIAL

ENGINEERING

FACULTY OF CIVIL & ENVIRONMENTAL ENGINEERINGUTHM

We, hereby confess that we have prepared this report on our effort. We also admit not to receive

or give any help during the preparation of this report and pledge that everything mentioned in the

report is true.

___________________________

Student Signature

Name : MUHAMAD ASYRAF AB MALIK

Matric No. : DF100108

Date : 18/10/2011

_______________________

Student Signature

Name : MUHAMMAD IKHWAN ZAINUDDIN


Date : 18/10/2011

___________________________

Student Signature

Name : AHMAD FARHAN RAKAWI


Date : 18/10/2011

___________________________

Student Signature

Name : AINUN NAZHIRIN ABD JALIL


Date : 18/10/2011

___________________________

Student Signature

Name : IDAMAZLIZA ISA


Date : 18/10/2011



1.0 OBJECTIVE

1.1 To examine a statically determinate frame and to analyze the frame using simple

pin joint theory.

2.0 LEARNING OUTCOME

2.1 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 THEORY

A truss is a structure composed of slender member joined together at their end points to

form one or more triangles. The joint connections are considered as 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 frictionless pins, that

is only a force and no moment can be transferred from one member to another

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

calculate the forces in the member of a truss is using Method of Joint.

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

equations ∑Fx = 0 and ∑Fy = 0.

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

less



4.0 EQUIPMENT

Figure 1: Frame Of Truss

Figure 2: Digital Indicator Reading Figure 3: Digital Force Display

Figure 4: Digital Strain Display



5.0 PROCEDURS

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

longer part of the structure as the idealised diagram illustrates.

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.

3. Return the load to zero (leaving the 100N preload), recheck and re-zero the digital

indicator. Never apply loads greater than those specified on the equipment.

4. Apply load in the increment shown in Table 1 recordding the strain readings and the

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

readings. (be careful with your sign)



6.0 RESULTS

Table 1: True Strain Reading for Experiment

2. Graphs

I. Choose a member (except member 6), and on the same axis plot a graph of

Recorded Strain μɛ against Load (N) and True Strain μɛ against Load (N).

Graph Plotted = In The Graph Paper

II. On another graph, do the same for a different member (non member 6).


III. Plot a separate graph of deflection (mm) against Load (N).


IV. Comment on your graph

Both criteria for strain and deflection is inversely proportional showed in the graph, for the strain recorded and the true strain graph is consistence liner but graph

for deflection is not liner graph, it has curve on graph line.

Using the Young’s Modulus relationship, calculate the equivalent member force.

complete the 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 and Esteel = 2.10x105

N/mm2

Load(N) 1 2 3 4 5 6 7 8

123



Use Load 123 N

Member Experimental Force

(N)

Theoretical Force

(N)

1

2

3

4

5

6

7

8

Table 2: Measured and Theoretical Force in the Cantilever Truss

Calculate the theoretical force using method of joint and write it down in Table 2 above

7.0 ANALYSIS DATA

7.1 Calculation For Experimental Force (N), Load = 250 N

From the formula: E = σ

ε

where;

E = Young Modulus (Nm-2) for steel = 2.10 x 105 N/mm2

ε = Displayed Strain

σ = FA

F = E ε A

d = 6mm

A = π (6)2 = 28.27 mm2

4

Member 1; F = 2.10 x 105 N/mm2 x (44 x 10-6) x 28.274 mm2

= 261.25 N



Member 2; F = 2.10 x 105 N/mm2 x (-43 x 10-6 ) x 28.274 mm2

= -255.31 N

Member 3; F = 2.10 x 105 N/mm2 x (-43 x 10-6) x 28.274 mm2

= -255.31 N

Member 4; F = 2.10 x 105 N/mm2 x (-87 x 10-6) x 28.274 mm2

= -516.57 N


= 17.81 N


= 385.94 N


= 391.88 N



7.2 Calculation For Theoretical Force (N), Load = 250 N

R Ay

R Ax A 1 E

5 2 7 2.4 m

R Bx

B C D 250N

2.4 m 2.4 m

∑MA = 0

250 (4.8) - R Bx = 0

1200 - R Bx = 0

- R Bx = -1200

R Bx = 1200.0 N

34

8



Joint Method Calculation

MEMBER 4

∑Fx = 0

1200 + FBD = 0

FBD = -1200.0 N

MEMBER 5

∑Fy = 0

FBA = 0

FBA = 0 N

MEMBER 3

∑Fx = 0

-FCE (2.4/3.4) – FDE = 0

- FDE - 354 (2.4/3.4) = 0

FDE = -249.9 N

MEMBER 7

∑Fy = 0

-249.9 + FCE (2.4/3.4) = 0

FCE = 354.0 N

MEMBER 2

∑Fy = 0

250 + FCD = 0

FCD = -250.0 N

FDC 3

D

7

FDE

250

4500

5

FBA

FBC

FCE

28

FCD

3C4

FCB

FCA

28



MEMBER 1

∑Fx = 0

1200 – FAD (Cos 45°) – FAC = 0

1200 – 353.6 (Cos 45°) = FAC

FAC = 950.0 N

MEMBER 2

∑Fy = 0

- 250 + FAD (sin 45°) = 0

FAD = 250N

Sin 45°

= 353.6 N

FCA Sin 45

FCA Cos 45

C

4

8

FCB

FCA

FAC Cos 45

FAC Sin 45

FABFAC

FAEA500N



8.0 DISCUSSION

1. Compare the experimental and theoretical result.

Based on the experimental results obtained, no member of a cantilever truss. 4

has the highest internal force of -106.88 N, after performing calculations no.4 truss

members using the connection, the value obtained was -100 N. in addition, a cantilever

truss members who do not have the internal resources of the members of no.5. based on

calculations using the connection, the internal force members also no.5 O N. all member

shows the internal force equation derived from experimental results and calculation

method of the connection pin.

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

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

Cantilever truss member having the internal resources of a compression is

member no. 2.3 and no.4, while the cantilever truss members have the tension internal

force is a member of no.1, 7 and No.8. Type of internal force derived from the

experimental results together with the values derived from calculations using the method

of connection. The value of the internal forces that are compressed as a member no. 2

due to the costs of action are mutually repel members, so members will be trying to fight

out these forces, so there was a compressive force. Tension that occurs at No.7 example

of the burden caused by the attraction of the subject, to fight back, then the internal

forces have to withdraw the action, so there was tension in the member.

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

From the experimental value we obtain the reading is 0 N and fom theoretical value we obtained 0 N, both of this value is almost closest to 0, this condition happen



because there are no force wheither internal or external force actually acting on this

member and it pinned on both end, this member cannot acting on any movement either X

or Y axis.

4. Are the strain gauges are an effective tranducersfor measurement forces in the

framework.

From the Tranducersfor Measurement forces gauge reading, we can see the value

obtained is consistent with the load acting on the framework, when the load increasing the straing gauge value also increasing and it can be accepted because it has small

value.

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

pin joint?.

Yes, beacuse the value obtained from the experiment only has small diffrence

compare to the theoretical value calculated using Euler formula.

9.0 CONCLUSION

1. Statically determinate frame will be more economic and safety because it will use enough

frame and member, then safety because there is no extra force that will move the

structure to fall or collapse.

2. Statically determinate structure have low cost compare to indeterminate structure.

3. We can see that there are large difference value between experimental force and

theoretical force. It is mean that, the accuracy of the result is not exact but for the

compression and tension member, we can conclude that the following tension and

compression is same only the value of the force is different.

Open Ended - Cantilever Truss

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