SIAP

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member based on specific cross section and length by using satay sticks and length by using satay sticks and adhesive (super glue).adhesive (super glue). The member will be tested under 3 point bending test or manual deflection test. We, as Group 1, The member will be tested under 3 point bending test or manual deflection test. We, as Group 1, consists of six people, have prepared two specimens with square hollow section based on the consists of six people, have prepared two specimens with square hollow section based on the specific dimension given by using stay commercial satay sticks and super glue. The Course specific dimension given by using stay commercial satay sticks and super glue. The Course Learning Outcome (CLO) of BFC 20903 Mechanics of Materials is to enable us to understand Learning Outcome (CLO) of BFC 20903 Mechanics of Materials is to enable us to understand more and apply the knowledge that we have learned in the class during the progression of this more and apply the knowledge that we have learned in the class during the progression of this project.

project. The The project project aims aims to to expose expose the the strength strength of of the the materials materials used, used, the the influence influence of of eacheach parameter of the member to

parameter of the member to its strength and theory and formula derivatiits strength and theory and formula derivation of the method on of the method chosen.chosen. Moreover, this project trains us to think critically in preparing the flexural member.

Moreover, this project trains us to think critically in preparing the flexural member.

GOAL GOAL

Expose student in applying theoretical knowledge to real life practice by preparing a flexural Expose student in applying theoretical knowledge to real life practice by preparing a flexural member following specification given and doing the

member following specification given and doing the calculations relatedcalculations related

OBJECTIVE OBJECTIVE

1. To determine the yield point, f

1. To determine the yield point, f yyand ultimate load, Pand ultimate load, Pultultof the flexural memberof the flexural member

2. To determine the stress distribution of normal stress and shear stress in the flexural member 2. To determine the stress distribution of normal stress and shear stress in the flexural member 3. To prove

3. To prove



maxmax= -f = -f yy L L33/ 48 EI using specific method given/ 48 EI using specific method given

SCOPE OF STUDY SCOPE OF STUDY

1. The design of the sample (cross section and orientation of the sticks) 1. The design of the sample (cross section and orientation of the sticks)

- Square hollow section with a dimension of 3cm x 4cm x 50cm - Square hollow section with a dimension of 3cm x 4cm x 50cm - The flexural member has two la

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2. No of specimens 2. No of specimens

- Each group is required to p

- Each group is required to prepare two flexural membersrepare two flexural members

3. Type of testing 3. Type of testing

- Manual Deflection Test/ 3 Point Bending Test - Manual Deflection Test/ 3 Point Bending Test

4. Type of analyses 4. Type of analyses

- Double Integration Method - Double Integration Method

5.

5. Others which are Others which are relevance to your relevance to your scope of workscope of work

- Cross section of the flexural member/beam in unit o

- Cross section of the flexural member/beam in unit of meterf meter

Figure 1.1 The cross-section of the flexural beam Figure 1.1 The cross-section of the flexural beam

3cm 3cm 2cm 2cm 4cm 4cm 3cm 3cm

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SIGNIFICANCE OF STUDY SIGNIFICANCE OF STUDY

Mechanics of Materials

Mechanics of Materials is a basic engineering subject that must be understood by anyoneis a basic engineering subject that must be understood by anyone concerned with the strength and physical performance of structures, whether those structures are concerned with the strength and physical performance of structures, whether those structures are man-made or natural.

man-made or natural. The subject matter includes such fundamental concepts as stresses andThe subject matter includes such fundamental concepts as stresses and strains, deformations and displacements, elasticity and inelasticity, strain energy, and strains, deformations and displacements, elasticity and inelasticity, strain energy, and load-carrying capacity. These concepts underlie the design and analysis of a huge variety of mechanical carrying capacity. These concepts underlie the design and analysis of a huge variety of mechanical and structural systems.

and structural systems.

The project is to prepare a flexural member based on specific cross section and length by The project is to prepare a flexural member based on specific cross section and length by using satay sticks and adhesive (super glue). This member will be tested under 3-point bending using satay sticks and adhesive (super glue). This member will be tested under 3-point bending test which loaded until failure. Through this project,

test which loaded until failure. Through this project, students can flourish their creativity skill andstudents can flourish their creativity skill and learn to think critically to solve the

learn to think critically to solve the problem given. Moreover, this project exposes students to real-problem given. Moreover, this project exposes students to real-life engineering, which means they apply the k

life engineering, which means they apply the knowledge from books and class into pnowledge from books and class into practical case.ractical case. These skills are vital when students involve in working environment in

These skills are vital when students involve in working environment in future.future.

Furthermore, students learn to solve problems they face while doing this project. This can Furthermore, students learn to solve problems they face while doing this project. This can be considered

be considered as an as an experience for experience for future because future because as a as a civil engineer, civil engineer, we will we will face unpredictableface unpredictable problems

problems throughout throughout the the construction. construction. Besides, Besides, this this project project requires requires cooperation cooperation from from everyevery member in the group.

member in the group. Students learn to communicate and cooperate wiStudents learn to communicate and cooperate with each other in order th each other in order toto accomplish the

accomplish the project given. project given. All soft skills whicAll soft skills which student acquired via this project are all essentialh student acquired via this project are all essential for future working life.

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CHAPTER 2

CHAPTER 2 LITERATURE REVIEWLITERATURE REVIEW

Each bamboo species has its own particular basic and mechanical properties. Most of them Each bamboo species has its own particular basic and mechanical properties. Most of them have hollow stem and some are strong. There are bamboos that grow up to 130 feet tall and 9 have hollow stem and some are strong. There are bamboos that grow up to 130 feet tall and 9 crawls in measurement. However, there are bamboos that develop just 7 inches tall and 0.07 creeps crawls in measurement. However, there are bamboos that develop just 7 inches tall and 0.07 creeps in distance across. Generally, bamboo species that used for development incorporate types of the in distance across. Generally, bamboo species that used for development incorporate types of the variety are

variety are ““GuaduaGuadua””,, “Dendrocalamus” and “Phyllostachys”.“Dendrocalamus” and “Phyllostachys”. The species of The species of “Guadua“Guadua Angustifolia

Angustifolia”” is local to South America and has the best properties for development. is local to South America and has the best properties for development. “Guadua“Guadua Angustifolia

Angustifolia”, also known as "World's m”, also known as "World's most grounded Bamboo" or "Vegetal Steel" is a huge Souost grounded Bamboo" or "Vegetal Steel" is a huge Sou thth American bamboo species with a higher rigidity than steel. Its unprecedented load bearing limit American bamboo species with a higher rigidity than steel. Its unprecedented load bearing limit makes it the most favored bamboo among visionary developers and draftsmen.

makes it the most favored bamboo among visionary developers and draftsmen.

However, as this project is for learning purpose, we use common satay stick, from “Buluh However, as this project is for learning purpose, we use common satay stick, from “Buluh Petung”. We glued the sticks together using super glue and formed them into a beam of which its Petung”. We glued the sticks together using super glue and formed them into a beam of which its size following specifications and dimension given. The diameter for each stick around 2.5mm to size following specifications and dimension given. The diameter for each stick around 2.5mm to 3.0mm. Our group chooses the dimension o

3.0mm. Our group chooses the dimension of 40mm x 40mm rectangular hollow section bef 40mm x 40mm rectangular hollow section beam witham with 50cm length for our project.

50cm length for our project.

For the experiment, we applied the load on the beam to

For the experiment, we applied the load on the beam to get the deformation or deflection. Whenget the deformation or deflection. When the load was applied, the

the load was applied, the beam will deform and change beam will deform and change from its original position. This deformationfrom its original position. This deformation of the beam is called as deflection. We start with applying 2kg load and take deflection reading of the beam is called as deflection. We start with applying 2kg load and take deflection reading from the computer. The load was added by 2kg and the

from the computer. The load was added by 2kg and the same step is repeated until it breaks or fails.same step is repeated until it breaks or fails. The last reading of deflection before it breaks was recorded to get the maximum deflection of the The last reading of deflection before it breaks was recorded to get the maximum deflection of the beam. The deformed shape of beam also known as elastic curve.

beam. The deformed shape of beam also known as elastic curve.

Theoretically, there are a few factors that can affect the magnitude of the deflection such as the Theoretically, there are a few factors that can affect the magnitude of the deflection such as the length (L), cross sectional area (A) and Young mo

length (L), cross sectional area (A) and Young mo dulus (E).dulus (E).

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In order to calculate the deflection of beam, C

In order to calculate the deflection of beam, Castigliano’s Mastigliano’s Methodethod and Macaulay’s (Doubleand Macaulay’s (Double Integration Method) are normally used. In our project, we are required to use Double Integration Integration Method) are normally used. In our project, we are required to use Double Integration Method. Through Double Integration Method, the deflection of the beam and the slope of the beam Method. Through Double Integration Method, the deflection of the beam and the slope of the beam can be obtained.

can be obtained.

In Mathematic, the formula of radius of curvature

In Mathematic, the formula of radius of curvature is y = f(x) is given by:is y = f(x) is given by:

 = [1  (/)2)3/2 / [2/2]

Radius of curvature: Radius of curvature:

  = =   / / 

Slope of elastic curve dy/dx

Slope of elastic curve dy/dx is small and squaring it made it negligible:is small and squaring it made it negligible:

  = = 1 1 / / [2

[2/

/2]2]

= = 1 1 / / 

′′′′

Thus, Thus,

/ = 1/

′′′′



′′′′

_{= = /}

_{/}

If EI constant, it can be written: If EI constant, it can be written:

 

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Figure 2.1 Elastic Curve of a Beam Figure 2.1 Elastic Curve of a Beam x &

x & y y = = Coordinates Coordinates of elastof elastic curve ic curve of beam of beam under loadunder load y

y = = deflection deflection of of beams beams at at any any distance distance of of xx E

E = = Modulus Modulus elasticity elasticity of of beambeam I

I = = Moment Moment of of inertia inertia about about the the neutral neutral axisaxis M

M = = Bending Bending moment moment at at distance distance x x from from end end of of beambeam EI

EI = = Flexural Flexural rigidity rigidity of of beambeam

The first integration y’ yields the slope of elastic curve and integrate twice y” to get the The first integration y’ yields the slope of elastic curve and integrate twice y” to get the deflection of beam at x distance. We must get two constant integration because EI y” = M is of deflection of beam at x distance. We must get two constant integration because EI y” = M is of second order. The two constants are evaluated from known conditions regarding the slope second order. The two constants are evaluated from known conditions regarding the slope deflection at certain point at the beam. For example, if simply supported beam at rigid support deflection at certain point at the beam. For example, if simply supported beam at rigid support where x=0 and x=L, the deflection is 0 (y=0) and it locating the point of maximum deflection. where x=0 and x=L, the deflection is 0 (y=0) and it locating the point of maximum deflection. Thus, slope of elastic curve y’ is equal

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Based on the Double Integration method, we need to prove that

 

_

==









..

Figure 2.2 A Beam that applied with load P at midspan Figure 2.2 A Beam that applied with load P at midspan

 ” =

 ” = 1122  –  <

–  <  

_{  1122 >>}

 ’ =

 ’ = 1144



_{– – 1122 < }

_{ <  1122 >>}



_

_{ }

_

  =

  = 111212



_{– – 1166 < }

_{ <  1122 >>}



_

_{ }

_

_{    }

_

When x = 0, y = 0, When x = 0, y = 0,



_

= = 00

When x = L, y = 0, When x = L, y = 0,

0 0 == 111212



_{– – 1166  << }

_{  1122 >>}



_{ }

_

_

_

0 0 == 111212



_{– – 114848}



_{  }

_

_





= =  111616



Therefore, Therefore,

  =

  = 111212



_{– – 1166 < }

_{ <  1122 >>}



_{  111616}



_

The maximum deflection occur at x = 0.5L (midspan), thus The maximum deflection occur at x = 0.5L (midspan), thus

 



== 111212(0.5)

(0.5)



– – 1166 (0.5  0.5)

 (0.5  0.5)



 111616



(0.5)

 



== 119696



– 0

– 0  113232



 



= =  

_48





Negative sign is just to show the deflection under the underformed neutral axis. Thus, Negative sign is just to show the deflection under the underformed neutral axis. Thus,

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CHAPTER 3

CHAPTER 3 METHODOLOGYMETHODOLOGY

OVERVIEW OF PROJECT PROGRESS OVERVIEW OF PROJECT PROGRESS

MATERIALS & APPARATUS USED MATERIALS & APPARATUS USED

Materials Apparatus

Satay

Satay Sticks Sticks Branch Branch Cutting Cutting ScissorsScissors

Super Glue Super Glue Di Discscusussisionontotoseselelectctththeespspececifificiccrcrosossssesectctioionn anandd lelengngththofofththee beam beam Beam-Making Process Beam-Making Process Laboratory T

Laboratory Testing (Manual esting (Manual Deflection Method)Deflection Method)

Do Report and

Do Report and Prepare for ViPrepare for Video Presentationdeo Presentation

Interview Session with lecturer

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COST IN PREPARING THE FLEXURAL MEMBER/BEAM COST IN PREPARING THE FLEXURAL MEMBER/BEAM

Materials

Materials & & Apparatus Apparatus Quantity Quantity Price Price (RM)(RM) (Each) (Each)

Price(RM) Price(RM)

Satay

Satay Stick Stick 2 2 7.50 7.50 15.0015.00 Glue

Glue 5 5 2.50 2.50 12.5012.50

Branch

Branch Cutting Cutting Scissors Scissors 1 1 9.00 9.00 9.009.00

Total 36.50

FLOW CHART OF BEAM-MAKING PROGRESS FLOW CHART OF BEAM-MAKING PROGRESS

Date: 19 APRIL Date: 19 APRIL 2017 2017 Time: 4.00-6.00pm Time: 4.00-6.00pm Place: Library Place: Library Work: Trim the Work: Trim the sharp end of satay sharp end of satay

sticks sticks Date: 22 APRIL Date: 22 APRIL 2017 2017 Time: 8.00-11.30am Time: 8.00-11.30am Place: Library Place: Library Work: Trim the Work: Trim the sharp end of satay sharp end of satay

sticks sticks Date: 25 APRIL Date: 25 APRIL 2017 2017 Time: Time: 4.00-6.00pm 6.00pm Place: Library Place: Library Work: Join the Work: Join the satay sticks in row satay sticks in row Date: 26 APRIL Date: 26 APRIL 2017 2017 Time: Time: 4.00-6.00pm 6.00pm Place: Library Place: Library Work: Join the Work: Join the satay sticks in row satay sticks in row Date: 29 APRIL Date: 29 APRIL 2017 2017 Time: 7.45-10.00pm Time: 7.45-10.00pm Place: Library Place: Library Work: Join the satay Work: Join the satay sticks into hollow sticks into hollow

beam. beam. Date: 2 MAY 17 Date: 2 MAY 17 Time: Time: 8.00-11.00am 11.00am Place: JRC Center Place: JRC Center Work: Testing of Work: Testing of the beam the beam

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BEAM-MAKING PROCESS BEAM-MAKING PROCESS

1. First at all, the sharp end

1. First at all, the sharp end of satay sticks is cut out using the bof satay sticks is cut out using the branch cutting scissors.ranch cutting scissors. 2. Next, the satay sticks are joined

2. Next, the satay sticks are joined together into raft-like plate with 4cm width using super together into raft-like plate with 4cm width using super glue.glue. 3. Step

3. Step22 is repeated until there are 4 is repeated until there are 4 pieces of raft-like satay sticks with 50cm long.pieces of raft-like satay sticks with 50cm long.

4. Then, the sata

4. Then, the satay sticks are joined together into raft-like whichy sticks are joined together into raft-like whichless thanless than 3cm by using super glue. 3cm by using super glue. 5. Step

5. Step 44 is repeated until there are 4 pieces of raft-like satay sticks with less than 4cm width and is repeated until there are 4 pieces of raft-like satay sticks with less than 4cm width and have 50cm length.

have 50cm length.

6. After that, the raft-like piece of satay sticks which is less than 4cm is pasted on the piece of 6. After that, the raft-like piece of satay sticks which is less than 4cm is pasted on the piece of satay stick which has 4cm.

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7. Finally, the 4 pieces of satay sticks are joined into a hollow beam. 7. Finally, the 4 pieces of satay sticks are joined into a hollow beam.

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PROCEDURE OF MANUAL DEFLECTION METHOD (TESTING OF BEAM) PROCEDURE OF MANUAL DEFLECTION METHOD (TESTING OF BEAM)

1. The thickness and width of beam is measured. 1. The thickness and width of beam is measured.

2. The position of two lower anvils is adjusted according to the requirement. 2. The position of two lower anvils is adjusted according to the requirement. 3. The beam is put on the two lower anvils.

3. The beam is put on the two lower anvils.

4. The load hanger is hung on the beam. 4. The load hanger is hung on the beam. 5. The reading of dial gauge is set to zero. 5. The reading of dial gauge is set to zero. 6. The load is hung on the mass hanger. 6. The load is hung on the mass hanger.

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7. The reading on dial gauge appears on the data logger and is recorded. 7. The reading on dial gauge appears on the data logger and is recorded.

8. Step 6 and 7 is repeated by increasing the mass of load until the beam fails. 8. Step 6 and 7 is repeated by increasing the mass of load until the beam fails.

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CHAPTER 4

CHAPTER 4 ANALYSIS AND DISCUSSIONANALYSIS AND DISCUSSION RESULTS

RESULTS

Table below shows the result of 3

Table below shows the result of 3 Point Bending/Manual Deflection Testing.Point Bending/Manual Deflection Testing. Graph of Load, P versus deflection,

Graph of Load, P versus deflection,



is plotted. is plotted.

Weight

Weight (kg) (kg) Load Load (N) (N) Deflection Deflection (mm)(mm)

0 0 0 0 00 2 2 19.62 19.62 0.180.18 4 4 39.24 39.24 0.50.5 6 6 58.86 58.86 0.730.73

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Figure P versus Deflection Figure P versus Deflection

0 0 0.1 0.1 0.2 0.2 0.3 0.3 0.4 0.4 0.5 0.5 0.6 0.6 0.7 0.7 0.8 0.8 0 0 1100 2200 3300 4400 5500 6600 7700 D D E E F F L L E E C C T T I I O O N N , , δ δ ( ( m m m m ) ) LOAD,P(N) LOAD,P(N)

Graph of Load,P Versus Deflection,

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CALCULATION CALCULATION

Based on the graph of Load versus Deflection, we can

Based on the graph of Load versus Deflection, we can determine that Pdetermine that Pultultis 58.86 N. The actualis 58.86 N. The actual

total length of the beam is 50cm. However, due to its position during testing (has a 4cm width total length of the beam is 50cm. However, due to its position during testing (has a 4cm width support at each side of the beam), we assume L equals to 46cm, which measures from the middle support at each side of the beam), we assume L equals to 46cm, which measures from the middle of the support to another support’s midspan,

of the support to another support’s midspan, for the calculation purpose. for the calculation purpose.

Figure 4.1 Dimension of the Beam in

Figure 4.1 Dimension of the Beam in unit meterunit meter

Figure 4.2 Free Body Diagram of the Beam Figure 4.2 Free Body Diagram of the Beam

58.86N

29.43N

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Figure 4.3 Shear Force Diagram Figure 4.3 Shear Force Diagram

Figure 4.4 Bending Moment Diagram Figure 4.4 Bending Moment Diagram Based on the shear force diagram and bending mo

Based on the shear force diagram and bending moment diagram, maximum shear force andment diagram, maximum shear force and maximum bending moment are determined. V

maximum bending moment are determined. Vmaxmaxis 29.43 N whereas Mis 29.43 N whereas Mmaxmaxis 6.769 Nm.is 6.769 Nm.

-29.43N

29.43N

6.769

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Figure 4.5 Cross Section of Beam Figure 4.5 Cross Section of Beam

Based on Figure 4.5, centroid of the beam is calculated as follow. Based on Figure 4.5, centroid of the beam is calculated as follow.

Part

Part Area Area (m(m44₎₎



_(m)_(m)



_(m)_(m) _A_A



_A_A



1 (0.03)(5x10 1 (0.03)(5x10-3-3)) = 1.5x10 = 1.5x10-4-4 0.04/2 0.04/2 = = 0.02 0.02 (5x10(5x10-3-3)/2)/2 = 0.0025 = 0.0025 0.03x10 0.03x10-6-6 5x105x10-7-7 2 (0.03)(5x10 2 (0.03)(5x10-3-3)) = 1.5x10 = 1.5x10-4-4 (5x10 (5x10-3-3)/2)/2 = 0.0025 = 0.0025 0.03/2 + 5x10 0.03/2 + 5x10-3-3= = 0.02 0.02 3.75x103.75x10-7-7 3x103x10-6-6 3 (0.03)(5x10 3 (0.03)(5x10-3-3)) = 1.5x10 = 1.5x10-4-4 0.04-0.0025 0.04-0.0025 = 0.0375 = 0.0375 0.03/2 + 5x10 0.03/2 + 5x10-3-3= = 0.02 0.02 5.625x105.625x10-6-6 3x103x10-6-6 4 (0.03)(5x10 4 (0.03)(5x10-3-3)) = 1.5x10 = 1.5x10-4-4 0.04/2 0.04/2 = = 0.02 0.02 0.04 0.04 - - (5x10(5x10-3-3)/2)/2 = 0.0375 = 0.0375 0.03x10 0.03x10-6-6 7.5x107.5x10-6-6

ΣΣ

= 6.0x10 = 6.0x10-4-4

ΣΣ

==

ΣΣ

=1.4x10 =1.4x10-5-5 0.02 0.02 0.03 0.03

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Next, moment of inertia is calculated as follow. Next, moment of inertia is calculated as follow.





== ℎℎ

₁₂



For this type of cross section, moment inertia of the

For this type of cross section, moment inertia of the beam is itsbeam is its



_

(outer) minus (outer) minus



_

(inner). Thus, (inner). Thus,





== ℎℎ

₁₂

_{12 ==}



(0.04)(0.04

₁₂



))

 ((0.03

0.03)()(0.03

₁₂

0.03



))

== 1.4

1.455 55  10

 10

−

 



Then, stress distribution of the beam is calculated (bending stress and shear stress) at point of A, Then, stress distribution of the beam is calculated (bending stress and shear stress) at point of A, B, C and D.

B, C and D.

Figure 4.6 Cross Section of Beam Figure 4.6 Cross Section of Beam





== 



((

)) == 



_{ ==}

6.769(0.020)

_{1.455  10}

_−

==  0.9

 0.930

30







== 



_{ ==}

6.769(0.015)

_{1.455  10}

_−

= 0.698





== 



== 



_{ ==}

_{1.455  10}

6.769(0)

_−

= = 0 

0 





== 



_{ ==}

6.769(0.015)

_{1.455  10}

_−

= = 0.6

0.698 98



0.02 0.02 0.03 0.03

(21)





== 



_{ ==}

_{1.455  10}

6.769(0.02)

_−

== 0.9

0.930 30 







== 



_{ =0}

=0





== 



_{ ==}

29.43(0.040.005)(0.02 0.005

29.43(0.040.005)(0.02

_{1.455  10}

_−

_(0.04)

0.005 22 ))

=0.018





== 



_{ ==}

29.43(0.040.005)(0.02 0.005

29.43(0.040.005)(0.02

_{1.455  10}

_−

_(20.005)

0.005 22 ))

=0.071





== 



_{ ==}

29.43[

29.43[((0.040.005

0.040.005))0.02

_{1.455  10}

0.02 0.005

0.005

_−

22 2(0.0050.015)(

_(20.005)

2(0.0050.015)(0.015

0.015 22 ))

=0.072





== 



_{ ==}

29.43(0.040.005)(0.02 0.005

29.43(0.040.005)(0.02

_{1.455  10}

_−

_(0.04)

0.005 22 ))

=0.018





== 



_{ ==}

29.43(0.040.005)(0.02 0.005

29.43(0.040.005)(0.02

_{1.455  10}

_−

_(20.005)

0.005 22 ))

=0.071





== 



_{ =0}

=0

- 0.930 - 0.930 -0.698 -0.698 0.072 0.072 0.018 0.071 0.018 0.071

(22)

After the testing of beam is being conducted, we obtain that the maximum bending stress After the testing of beam is being conducted, we obtain that the maximum bending stress of the beam is 0.072 MPa. The beam has an elongation of 0.5cm after the load is being removed. of the beam is 0.072 MPa. The beam has an elongation of 0.5cm after the load is being removed. Thus, strain,

Thus, strain,



can be calculated. can be calculated.

 ==  == 0.5

0.5

46

46 =0.0109

=0.0109

In order to obtain the elastic modulus of

In order to obtain the elastic modulus of the beam, E, graph of stress,the beam, E, graph of stress,



versus strain, versus strain,



isis plotted.

plotted.

The gradient of the graph

The gradient of the graph is the elastic modulus of the beam. Theis the elastic modulus of the beam. Therefore,refore,

 ==  ==

((0.01090

((0.0720

0.01090)) =6.606

0.0720))

=6.606

Based on the formula given, Based on the formula given,





==   







48

48 ==

48 48 ((640.18310

640.18310

((58.86

58.86)()(0.46



0.46 )()(1.45510

1.45510



))

−

)) =1.28110

=1.28110

−



0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 0 0 00..110099 S S t t r r e e s s s s ( ( M M P P a a ) ) Strain Strain

Stress Vs Strain

(23)

As the project instruction required us to calculate the d

As the project instruction required us to calculate the d eflection of beam using Doubleeflection of beam using Double Integration Method, the calculation of deflection using this method

Integration Method, the calculation of deflection using this method is shown as follow.is shown as follow. Firstly, obtain the bending moment equation using bending moment diagram. Firstly, obtain the bending moment equation using bending moment diagram.

Figure 4.8 Free Body Diagram and Bending Moment Diagram Figure 4.8 Free Body Diagram and Bending Moment Diagram

As the bending moment diagram is in linear form, we can assume that y=mx + c is same As the bending moment diagram is in linear form, we can assume that y=mx + c is same as M=mx + c. as M=mx + c.





== 6.7690

6.7690

_0.230

_{0.230 =29.430; }

=29.430; 



== 00

58.86N

29.43N

6.769

(24)





= = ∫∫



= ∫29.430 =29.430



  







= = ∫∫



= ∫29.430



  





 == 2299.4.4330

0



  



   







= = ∫∫



= = ∫(

∫(2

29.9.43

430

0 13

13.5.538 38) 

)  == 2

29.9.43

430

0



13.538 







= = ∫∫



= ∫(29.430



1133..553388   



) )  == 2

29.9.443300



13.538







  



The boundary condition of the beam is: The boundary condition of the beam is: (i) when x = 0, y = 0 (i) when x = 0, y = 0 (ii) when x = 0.46, y = 0 (ii) when x = 0.46, y = 0 When x = 0, y = 0, When x = 0, y = 0,





=29.430



  



   



0 = 0 =



_

∴



=29.430



  









=29.430



13.538



 



  



0 = 0 =



_

∴ ∴ 



=29.430



13.538



 





When x = 0.46, y=0, When x = 0.46, y=0,





=29.430



  



0 = 3.581 + 0.46 0 = 3.581 + 0.46



_





= -6.215= -6.215

∴



=29.430



 6.215

(25)





=29.430



13.538



 





0 = -2.865 + 2.865 + 0.46 0 = -2.865 + 2.865 + 0.46



_





= -2.174 = -2.174

∴ ∴ 



=29.430



13.538



2.174

As the beam is simply supported and load is applied at its midspan, we can indicate that As the beam is simply supported and load is applied at its midspan, we can indicate that maximum deflection occurs at the midspan.

maximum deflection occurs at the midspan. When x = 0.23, When x = 0.23,





== 13.538

13.538



_

  7.7

7.785 85

== 13.538(0.23)

13.538(0.23)

_{((640.18310}

_640.18310



_

_{)()(1.45510}

  7.78

_1.45510

7.785(0

5(0.23

.23))

_−

_{)) =0.017}

=0.017





== 13.538

13.538



29.430

_



15.567

== 13.538

13.538((0.23

0.23))

_{((640.18310}

_640.18310



29.430

29.430((0.23

_

_{)()(1.45510}

_1.45510

0.23))



15.567((0.23

15.567

_−

₎₎

0.23))

= 0.023

(26)

DISCUSSION DISCUSSION

Below are the results of the maximum deflection obtained using different methods. The Below are the results of the maximum deflection obtained using different methods. The negative value of the maximum deflection is negligible as it is only to show that the beam

negative value of the maximum deflection is negligible as it is only to show that the beam deflectsdeflects downward.

downward.

Methods

Methods Value Value of of Maximum Maximum DeflectionDeflection

Manual

Manual Deflection Deflection Test Test - - 0.73mm0.73mm





==   







48

- 1.281mm- 1.281mm

Double

Double Integration Integration Method Method -23.0mm-23.0mm

According to



_

== 









, maximum , maximum deflection of the deflection of the beam should beam should be 1.281mm.be 1.281mm.

However, we only obtain 0.

However, we only obtain 0.73mm of deflection from the laboratory testing. Hence, there are 73mm of deflection from the laboratory testing. Hence, there are somesome random error occurs during the testing of the beam such as the gauge does not put completely random error occurs during the testing of the beam such as the gauge does not put completely perpendicular

perpendicular on on the the top top of of the the beam. beam. The The percentage percentage difference difference between between theoretical theoretical value value andand experimental value is:

experimental value is:

(1.2810.73)

1.281 1.281  10

 100%

0% == 443.3.01

01%%

By using Double Integration Method, we obtain 23

By using Double Integration Method, we obtain 23 mm as the maximum deflection of the beam.mm as the maximum deflection of the beam. We can state that this value is the most accurate value as it does not depend o

We can state that this value is the most accurate value as it does not depend o n any of the laboratoryn any of the laboratory testing value. Unlike the double integration method,

testing value. Unlike the double integration method,





== 









still need one still need one of of thethe

laboratory testing value which is

 

_

..

 

_

is the yield point of the beam. We assume that the beam is the yield point of the beam. We assume that the beam may not reach its yield point but it eventually fail due to certain reason. Hence, we only take the may not reach its yield point but it eventually fail due to certain reason. Hence, we only take the value of the load, in whic

value of the load, in which the beam fails, ash the beam fails, as

 

_

..

Besides, the manual deflection test did not get the value of deflection same as double Besides, the manual deflection test did not get the value of deflection same as double integration method because there may be some minor technical inaccuracies in making the beam. integration method because there may be some minor technical inaccuracies in making the beam. For example, some of the sides of beam were not exactly 4cm but were in fact range about For example, some of the sides of beam were not exactly 4cm but were in fact range about

(27)

±±

0.05cm. Moreover, the surface of the beam is also affected as we made the beam on the0.05cm. Moreover, the surface of the beam is also affected as we made the beam on the cardboard, so some cardboard skins are stick on the surface of the beam. Hence, the strength of cardboard, so some cardboard skins are stick on the surface of the beam. Hence, the strength of the beam is affected altogether. On to

the beam is affected altogether. On top of that, the pattern which p of that, the pattern which the satay sticks were stick togetherthe satay sticks were stick together is also one of the factors that influence the result such as horizontal layer or vertical layer and is also one of the factors that influence the result such as horizontal layer or vertical layer and single layer or double layers. As we can know, the double integration method calculates the single layer or double layers. As we can know, the double integration method calculates the deflection of beam as a whole without any glue joint. However, the beam we made are stick deflection of beam as a whole without any glue joint. However, the beam we made are stick together with several joints using super glue. Therefore, the result we

together with several joints using super glue. Therefore, the result we get in the laboratory is quiteget in the laboratory is quite difficult to be same as the theory calculation

(28)

CHAPTER 5

CHAPTER 5 CONCLUSIONCONCLUSION

As a conclusion, this Mechanic of Materials project had taught us a lot of lessons including As a conclusion, this Mechanic of Materials project had taught us a lot of lessons including theoretical stuff and also integrity experience like cooperation between each other. All of these are theoretical stuff and also integrity experience like cooperation between each other. All of these are important for students when they go into the working field. Although the result we get from the important for students when they go into the working field. Although the result we get from the manual deflection test is not same as the theoretical result, we are still able to learn the concept manual deflection test is not same as the theoretical result, we are still able to learn the concept ofof 3 point bending test like stress, strain, strain energy and load bearing capacity and how

3 point bending test like stress, strain, strain energy and load bearing capacity and how to determineto determine the deflection of beam.

the deflection of beam.

This project took us about 2 weeks to complete it, starting from deciding the dimension and This project took us about 2 weeks to complete it, starting from deciding the dimension and method of calculation to report and video making. We go from doing research about the bamboo method of calculation to report and video making. We go from doing research about the bamboo stick, making beam model, testing of beam, analysis of deflection (centroid, moment of inertia, stick, making beam model, testing of beam, analysis of deflection (centroid, moment of inertia, bending stress, shear stress, load-deflection

bending stress, shear stress, load-deflection graph and degraph and deflection equation) until video flection equation) until video and reportand report making. Throughout the whole period, we have learnt and experienced the importance of good making. Throughout the whole period, we have learnt and experienced the importance of good cooperation between each other and time management. Other than that, the delegation of task to cooperation between each other and time management. Other than that, the delegation of task to the right person also makes our project goes

the right person also makes our project goes smooth without any major obstacle.smooth without any major obstacle. Last of all, we also improve our creativity and

Last of all, we also improve our creativity and problem solving skills during making the reportproblem solving skills during making the report and video. This plays an important role as the report and video need to be clear in ex

and video. This plays an important role as the report and video need to be clear in ex plaining everyplaining every details of the project so that the readers and audiences can understand them easily. Without the details of the project so that the readers and audiences can understand them easily. Without the guidance from the lecturer, the project would face a lot more challenges and take a longer time to guidance from the lecturer, the project would face a lot more challenges and take a longer time to be completed.

(29)

REFERENCES REFERENCES Books Reference Books Reference

1.

1. Christopher Jenkins, Christopher Jenkins, Sanjeev Khanna Sanjeev Khanna (2005). Mechanics of Materials: A Mod(2005). Mechanics of Materials: A Modern Integration ofern Integration of Mechanics and Materials in Structural Design. Elsevier Academic Press.

Mechanics and Materials in Structural Design. Elsevier Academic Press. 2. R.C. Hibbeler (2000). Mechanics of Materials.US: Prentice Hall, 2. R.C. Hibbeler (2000). Mechanics of Materials.US: Prentice Hall, Inc.Inc.

3. Ferdinand P. Beer, E. Russell Johnston, Jr., John T. Dewolf, David F. Mazurek (2012). 3. Ferdinand P. Beer, E. Russell Johnston, Jr., John T. Dewolf, David F. Mazurek (2012). Mechanics of Materials. US: McGraw Hill Companies, Inc.

Mechanics of Materials. US: McGraw Hill Companies, Inc.

4. James M. Gere, Barry J. Goodno (2013). Mechanics of Materials. US: Cengage Learning. 4. James M. Gere, Barry J. Goodno (2013). Mechanics of Materials. US: Cengage Learning.

Internet Reference Internet Reference 1. 1. https://www.guaduabamboo.comhttps://www.guaduabamboo.com 2. 2. https://www.scribd.comhttps://www.scribd.com 3. 3. http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/double- http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/double-integration-method-beam-deflections integration-method-beam-deflections

SIAP

CONTENTS

CONTENTS



 = [1  (/)2)3/2 / [2/2]

 = [1  (/)2)3/2 / [2/2]

  = =   / / 

  = = 1 1 / / [2

[2/

/2]2]

= = 1 1 / / 

′′′′

/ = 1/

/ = 1/

′′′′



′′′′

= = /

/

 

 





==











 ” =

 ” = 1122  –  <

–  <  

  1122 >>

 ’ =

 ’ = 1144



– – 1122 < 

 <  1122 >>





 



  =

  = 111212



– – 1166 < 

 <  1122 >>





 



    







= = 00

0 0 == 111212



– – 1166  << 

  1122 >>



 







0 0 == 111212



– – 114848



  









= =  111616



  =

  = 111212



– – 1166 < 

_{= = /}

_{/}

_

_

_{  1122 >>}

_{– – 1122 < }

_{ <  1122 >>}

_

_{ }

_

_{– – 1166 < }

_{ <  1122 >>}

_

_{ }

_

_{    }

_

_

_{– – 1166  << }

_{  1122 >>}

_{ }

_

_

_

_{– – 114848}

_{  }

_

_

_{– – 1166 < }

_{ <  1122 >>}

_{  111616}

_

_48

_48

₁₂

₁₂

_

_

₁₂

_{12 ==}