Force in a Statically Determinate Cantilever Truss

7/27/2019 Force in a Statically Determinate Cantilever Truss

1/12

FACULTY OF CIVIL AND ENVIRONMENTAL

ENGINEERING

DEPARTMENT OF STRUCTURE AND MATERIAL

ENGINEERING

LAB MATERIAL

REPORT

Subject Code BFC 21201

Code & Experiment Title 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 24 OCTOBER 2011

Comment by examiner Received


2/12

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 : 24/10/2011

_______________________

Student Signature

Name : MUHAMMAD IKHWAN ZAINUDDIN


Date : 24/10/2011

___________________________

Student Signature

Name : AHMAD FARHAN RAKAWI


Date : 24/10/2011

___________________________

Student Signature

Name : AINUN NAZHIRIN ABD JALIL


Date : 24/10/2011

___________________________

Student Signature

Name : IDAMAZLIZA ISA


Date : 24/10/2011


3/12

1.0 OBJECTIVE

1.1 To examine a statically determinate frame and to analyze the frame using simplepin 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 isonly 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 tocalculate 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 orless


4/12

4.0 EQUIPMENT

Figure 1: Frame Of Truss

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

Figure 4: Digital Strain Display


5/12

5.0 PROCEDURS

1. Unscrew the thumwheel on the redundant member. Note that it is effectively nolonger 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 aload 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 digitalindicator.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 thedigital indicator readings. Complete Table 2 by subtracing the initial (zero) strain

readings. (be careful with your sign)


6/12

6.0 RESULTS

1. Table

Load

(N)

Strain Reading Digital Indicator

Reading (mm)1 2 3 4 5 6 7 8

0 125 200 -42 -86 94 0 5 36 -0.197

50 134 192 -50 -102 96 0 17 48 -0.032

100 144 183 -59 -121 96 0 32 63 -0.066

150 152 174 -68 -139 97 0 45 77 -0.095

200 161 165 -77 -157 97 0 58 91 -0.118

250 169 157 -85 -173 97 0 70 102 -0.138

Table 1: Strain Readings And Frame Deflection For Experiment 1

Table 2: True Strain Reading for Experiment

2. Graphs

I. Choose a member (except member 6), and on the same axis plot a graph ofRecorded Strain against Load (N) and True Strain against Load (N).

Graph Plotted = In The Graph Paper

II. On another graph, do the same for a different member (non member 6).Graph Plotted = In The Graph Paper

III. Plot a separate graph of deflection (mm) against Load (N).Graph Plotted = In The Graph Paper

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.

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

0 0 0 0 0 0 0 0 0

50 9 -8 -8 -16 2 0 12 12

100 19 -17 -17 -35 2 0 27 27

150 27 -26 -26 -53 3 0 40 41

200 36 -35 -35 -71 3 0 53 55

250 44 -43 -43 -87 3 0 65 66


7/12

Using the Youngs Modulus relationship, calculate the equivalent member force.

complete the experimental force in Table 3. (ignore member 6 at this stage)

E = /

Where;

E = Youngs Modulus (Nm-2)

= Stress in the member (Nm-2)

= Displayed strain

and = 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

Use Load 250 N

Member Experimental Force

(N)

Theoretical Force

(N)

1 261.21 250

2 -255.28 -250

3 -255.28 -250

4 -516.49 -500

5 17.81 0

6 0 0

7 385.88 354.2

8 391.82 354.2

Table 3: Measured and Theoretical Force in the Cantilever Truss

Calculate the theoretical force using method of joint and write it down in Table 3 above


8/12

7.0 ANALYSIS DATA

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

From the formula: E =

where;

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

= Displayed Strain

= F

A

F = E A

d = 6mm

A = (6)2 = 28.27 mm2

4

Member 1; F = 2.10 x 105N/mm2 x (44 x 10-6) x 28.27 mm2

= 261.21 N

Member 2; F = 2.10 x 105N/mm2 x (-43 x 10-6 ) x 28.27 mm2

= -255.28 N

Member 3; F = 2.10 x 105N/mm2 x (-43 x 10-6) x 28.27 mm2

= -255.28 N

Member 4; F = 2.10 x 105N/mm2 x (-87x 10-6) x 28.27 mm2

= -516.49 N


= 17.81 N


= 385.88 N

Member 8; F = 2.10 x 105

N/mm2

x (66 x 10-6

) x 28.27 mm2

= 391.82 N


9/12

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

RAY

RAX A 1 C

5 2 7 2.4 m

RBX

B D E 250N

2.4 m 2.4 m

MB = 0

250 (4.8) + RAX (2.4) = 0

1200 + RAX (2.4) = 0

RAX = - 1200

2.4

RAX = - 500 N

FX = 0

RAX + RBX = 0

-500 + RBX = 0

RBX = 500 N

Fy = 0

RAY = 250 N

34

8

AC = DE

AC2 = BA2 + BC2

AC2 = 2.42 + 2.42

AC2 = 11.52

AC = 3.4 m


10/12

Joint Method Calculation

MEMBER 4

Fx = Fx

500 + FBD = 0

FBD = -500.0 N

MEMBER 5

Fy =Fy

FBA = 0 N

MEMBER 7

Fy = Fy

-250 + FEC (2.4/3.4) = 0

FEC (2.4/3.4) = 250

FEC = 354.2 N

MEMBER 3

Fx = Fx

- FED - FEC (2.4/3.4) = 0

- FED354.2 (2.4/3.4) = 0

FED = -250.0 N

MEMBER 2

Fy = Fy

-FCDFCE (2.4/3.4) = 0

-FCD(354.2) (2.4/3.4) = 0

FCD = -250.0 N

MEMBER 1

Fx = Fx

FCA + FCE (2.4/3.4) = 0

FCA + (354.2) (2.4/3.4) = 0

FCA = 250.0 N

MEMBER 8

Fy =Fy

FDC + FDA (2.4/3.4) = 0

(-250) + FDA (2.4/3.4) = 0

FDA = 354.2 N

FED3

E

7

FEC

250

FBA

FBD500N

5

4

FCA

FCDFCE

1

2

7

FDCFDA

FDB FDE

8

2

34


11/12

8.0 DISCUSSION

1. Compare the experimental and theoretical result.

From the experimental results, we obtained that member of cantilever 4 has stated

as the highest force which is -516.49N, and after made some theoreticalcalculations at

cantilever no 4 the value obtained was -100 N which means its not a big different with

experimental result. Besides that, member of cantilever at no 5 was obtained the result

but compare the theoretical result has no result was made. All members show the internal

force equation derived from experimental results and calculation method of the

connection pin.

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

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

We found that the cantilever truss member having the compression at members of

2, 3 and 4. We identified this because sign of the value. All these members have a

negative value. While for the rest members, we found that the cantilever at member of 1,

7 and 8 having a tension. Type of internal force derived from the experimental results

together with the values derived from calculations using the method of connection. Thevalue 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 are almost zero.

From the experimental value we obtained that the reading was 17.81 N and from

the theoretical results we obtained as a 0 N. By the way, these both value are almost

close to 0, this condition happen because there are no force either internal or external

force actually acted on this member and it pinned on both end, this member cannot acting

on any movement either on X or Y axis.


12/12

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

framework.

From the Transducers for Measurement forces gauge reading, we can see the

value obtained is consistent with the load acting on the framework, when the loadincreasing the strain 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, because the value obtained from the experiment only has small difference

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 enoughframe and member, then safety because there is no extra forces that will move the

structure to fall or collapse.

2. Statically determinate structure has 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.

Documents

Force in a Statically Determinate Cantilever Truss