SIAP

CONTENTSCHAPTER 1 INTRODUCTION .................................................................................................... 1 GENERAL .................................................................................................................................. 1 GOAL ......................................................................................................................................... 1 OBJECTIVE ............................................................................................................................... 1 SCOPE OF STUDY.................................................................................................................... 1 SIGNIFICANCE OF STUDY .................................................................................................... 3 CHAPTER 2 LITERATURE REVIEW ......................................................................................... 4 CHAPTER 3 METHODOLOGY ................................................................................................... 8 OVERVIEW OF PROJECT PROGRESS .................................................................................. 8 MATERIALS & APPARATUS USED ...................................................................................... 8 COST IN PREPARING THE FLEXURAL MEMBER/BEAM ................................................ 9 FLOW CHART OF BEAM-MAKING PROGRESS ................................................................. 9 BEAM-MAKING PROCESS ................................................................................................... 10 PROCEDURE OF MANUAL DEFLECTION METHOD (TESTING OF BEAM) ............... 12 CHAPTER 4 ANALYSIS AND DISCUSSION........................................................................... 14 RESULTS ................................................................................................................................. 14 CALCULATION ...................................................................................................................... 16 DISCUSSION ........................................................................................................................... 25 CHAPTER 5 CONCLUSION ....................................................................................................... 27 REFERENCES ......................................................................................................................... 28 APPENDIX ................................................................................Error! Bookmark not defined. CHAPTER 1 INTRODUCTION GENERAL The project of BFC 20903 Mechanics of Materials requires each group to prepare a flexural member based on specific cross section and length by using satay sticks and adhesive (super glue). 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 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 more and apply the knowledge that we have learned in the class during the progression of this project. The project aims to expose the strength of the materials used, the influence of each parameter of the member to its strength and theory and formula derivation of the method chosen. Moreover, this project trains us to think critically in preparing the flexural member. GOAL Expose student in applying theoretical knowledge to real life practice by preparing a flexural member following specification given and doing the calculations related OBJECTIVE 1. To determine the yield point, fy and ultimate load, Pult of the flexural member 2. To determine the stress distribution of normal stress and shear stress in the flexural member 3. To prove 𝛿 max = -fy L3/ 48 EI using specific method given SCOPE OF STUDY 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 - The flexural member has two layers in every long side. 1 2. No of specimens - Each group is required to prepare two flexural members 3. Type of testing - Manual Deflection Test/ 3 Point Bending Test 4. Type of analyses - Double Integration Method 5. Others which are relevance to your scope of work - Cross section of the flexural member/beam in unit of meter 3cm 2cm 3cm 4cm Figure 1.1 The cross-section of the flexural beam 2 SIGNIFICANCE OF STUDY Mechanics of Materials is a basic engineering subject that must be understood by anyone concerned with the strength and physical performance of structures, whether those structures are man-made or natural. The subject matter includes such fundamental concepts as stresses 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 and structural systems. 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 test which loaded until failure. Through this project, students can flourish their creativity skill and learn to think critically to solve the problem given. Moreover, this project exposes students to real- life engineering, which means they apply the knowledge from books and class into practical case. These skills are vital when students involve in working environment in future. Furthermore, students learn to solve problems they face while doing this project. This can be considered as an experience for future because as a civil engineer, we will face unpredictable problems throughout the construction. Besides, this project requires cooperation from every member in the group. Students learn to communicate and cooperate with each other in order to accomplish the project given. All soft skills which student acquired via this project are all essential for future working life. 3 CHAPTER 2 LITERATURE REVIEW 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 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 variety are “Guadua”, “Dendrocalamus” and “Phyllostachys”. The species of “Guadua Angustifolia” is local to South America and has the best properties for development. “Guadua Angustifolia”, also known as "World's most grounded Bamboo" or "Vegetal Steel" is a huge South 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. 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 size following specifications and dimension given. The diameter for each stick around 2.5mm to 3.0mm. Our group chooses the dimension of 40mm x 40mm rectangular hollow section beam with 50cm length for our project. For the experiment, we applied the load on the beam to get the deformation or deflection. When the load was applied, the beam will deform and change from its original position. This deformation 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 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 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 length (L), cross sectional area (A) and Young modulus (E). 𝑃𝐿3 𝑏ℎ3 𝑦𝑚𝑎𝑥 = ; 𝐼 (𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎) = 48𝐸𝐼 12 Based on the formula of maximum deflection, we can indicate that the higher value of Young modulus (E) and Moment of Inertia (I), the lesser the deflection of beam. Meanwhile, the shorter the length of beam, the lower the value of deflection of beam. 4 In order to calculate the deflection of beam, Castigliano’s Method and Macaulay’s (Double 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 can be obtained. In Mathematic, the formula of radius of curvature is y = f(x) is given by: 𝜌 = [1 + (𝑑𝑦/𝑑𝑥)2)3/2 / [𝑑2𝑦/𝑑𝑥2] Radius of curvature: 𝜌 = 𝐸𝐼 / 𝑀 Slope of elastic curve dy/dx is small and squaring it made it negligible: 𝜌 = 1 / [𝑑2𝑦/𝑑𝑥2] = 1 / 𝑦 ′′ Thus, 𝐸𝐼/𝑀 = 1/𝑦 ′′ 𝑦 ′′ = 𝑀/𝐸𝐼 If EI constant, it can be written: 𝐸𝐼 𝑦 ′′ = 𝑀 5 Figure 2.1 Elastic Curve of a Beam x & y = Coordinates of elastic curve of beam under load y = deflection of beams at any distance of x E = Modulus elasticity of beam I = Moment of inertia about the neutral axis M = Bending moment at distance x from end of beam EI = Flexural rigidity of beam 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 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 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 to 0. 6 𝑓𝑦 𝐿3 Based on the Double Integration method, we need to prove that 𝛿𝑚𝑎𝑥 = . 48𝐸𝐼 Figure 2.2 A Beam that applied with load P at midspan 1 1 𝐸𝐼 𝑦” = 𝑃𝑥 – 𝑃 < 𝑥 − 𝐿 > 2 2 1 1 1 𝐸𝐼 𝑦’ = 𝑃𝑥 2 – 𝑃 < 𝑥 − 𝐿 >2 + 𝐶1 4 2 2 1 1 1 𝐸𝐼 𝑦 = 𝑃𝑥 3 – 𝑃 < 𝑥 − 𝐿 >3 + 𝐶1 𝑥 + 𝐶2 12 6 2 When x = 0, y = 0, 𝐶2 = 0 When x = L, y = 0, 1 1 1 0 = 𝑃𝐿3 – 𝑃 < 𝐿 − 𝐿 >3 + 𝐶1 𝐿 12 6 2 1 1 0 = 𝑃𝐿3 – 𝑃𝐿3 + 𝐶1 𝐿 12 48 1 𝐶1 = − 𝑃𝐿2 16 Therefore, 1 1 1 1 𝐸𝐼 𝑦 = 𝑃𝑥 3 – 𝑃 < 𝑥 − 𝐿 >3 − 𝑃𝐿2 𝑥 12 6 2 16 The maximum deflection occur at x = 0.5L (midspan), thus 1 1 1 𝐸𝐼 𝑦𝑚𝑎𝑥 = 𝑃(0.5𝐿)3 – 𝑃 (0.5𝐿 − 0.5𝐿)3 + 𝑃𝐿2 (0.5𝐿) 12 6 16 1 1 𝐸𝐼 𝑦𝑚𝑎𝑥 = 𝑃𝐿3 – 0 + 𝑃𝐿3 96 32 𝑃𝐿3 𝐸𝐼 𝑦𝑚𝑎𝑥 = − 48𝐸𝐼 Negative sign is just to show the deflection under the underformed neutral axis. Thus, 𝑃𝐿3 𝑓𝑦 𝐿3 𝛿 = ; 𝛿𝑚𝑎𝑥 = (𝑝𝑟𝑜𝑣𝑒𝑛) 48𝐸𝐼 48𝐸𝐼 7 CHAPTER 3 METHODOLOGY OVERVIEW OF PROJECT PROGRESS Discussion to select the specific cross section and length of the beam Beam-Making Process Laboratory Testing (Manual Deflection Method) Do Report and Prepare for Video Presentation Interview Session with lecturer MATERIALS & APPARATUS USED Materials Apparatus Satay Sticks Branch Cutting Scissors Super Glue 8 COST IN PREPARING THE FLEXURAL MEMBER/BEAM Materials & Apparatus Quantity Price (RM) Price(RM) (Each) Satay Stick 2 7.50 15.00 Glue 5 2.50 12.50 Branch Cutting Scissors 1 9.00 9.00 Total 36.50 FLOW CHART OF BEAM-MAKING PROGRESS Date: 19 APRIL Date: 22 APRIL Date: 25 APRIL 2017 2017 2017 Time: 4.00-6.00pm Time: 8.00-11.30am Time: 4.00- 6.00pm Place: Library Place: Library Place: Library Work: Trim the Work: Trim the sharp end of satay sharp end of satay Work: Join the sticks sticks satay sticks in row Date: 2 MAY 17 Date: 29 APRIL Date: 26 APRIL 2017 2017 Time: 8.00- 11.00am Time: 7.45-10.00pm Time: 4.00- Place: JRC Center Place: Library 6.00pm Work: Testing of Work: Join the satay Place: Library the beam sticks into hollow Work: Join the beam. satay sticks in row 9 BEAM-MAKING PROCESS 1. First at all, the sharp end of satay sticks is cut out using the branch cutting scissors. 2. Next, the satay sticks are joined together into raft-like plate with 4cm width using super glue. 3. Step 2 is repeated until there are 4 pieces of raft-like satay sticks with 50cm long. 4. Then, the satay sticks are joined together into raft-like which less than 3cm by using super glue. 5. Step 4 is repeated until there are 4 pieces of raft-like satay sticks with less than 4cm width and have 50cm length. 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. 10 7. Finally, the 4 pieces of satay sticks are joined into a hollow beam. 11 PROCEDURE OF MANUAL DEFLECTION METHOD (TESTING OF BEAM) 1. The thickness and width of beam is measured. 2. The position of two lower anvils is adjusted according to the requirement. 3. The beam is put on the two lower anvils. 4. The load hanger is hung on the beam. 5. The reading of dial gauge is set to zero. 6. The load is hung on the mass hanger. 12 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. 13 CHAPTER 4 ANALYSIS AND DISCUSSION RESULTS Table below shows the result of 3 Point Bending/Manual Deflection Testing. Graph of Load, P versus deflection, 𝛿 is plotted. Weight (kg) Load (N) Deflection (mm) 0 0 0 2 19.62 0.18 4 39.24 0.5 6 58.86 0.73 14 Graph of Load,P Versus Deflection,δ 0.8 0.7 0.6 DEFLECTION , δ (mm) 0.5 0.4 0.3 0.2 0.1 0 0 10 20 30 40 50 60 70 LOAD,P(N) Figure P versus Deflection 15 CALCULATION Based on the graph of Load versus Deflection, we can determine that Pult is 58.86 N. The actual 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 of the support to another support’s midspan, for the calculation purpose. Figure 4.1 Dimension of the Beam in unit meter 58.86N 29.43N 29.43N Figure 4.2 Free Body Diagram of the Beam + → =0 𝐹𝐴𝑥 = 0 Σ𝐹𝑥 +↺ Σ𝑀𝐴 = 0 - (58.86 x 0.23) + 0.46 𝐹𝐵𝑦 = 0 𝐹𝐵𝑦 = 29.43 N (↑) +↑ Σ𝐹𝑦 = 0 𝐹𝐴𝑦 + 29.43 – 58.86 = 0 𝐹𝐴𝑦 = 29.43 N (↑) 16 29.43N -29.43N Figure 4.3 Shear Force Diagram 6.769 Figure 4.4 Bending Moment Diagram Based on the shear force diagram and bending moment diagram, maximum shear force and maximum bending moment are determined. Vmax is 29.43 N whereas Mmax is 6.769 Nm. 17 0.03 0.02 Figure 4.5 Cross Section of Beam Based on Figure 4.5, centroid of the beam is calculated as follow. Part Area (m4) ̅ (m) 𝒙 ̅ (m) 𝒚 ̅ A𝒙 ̅ A𝒚 1 (0.03)(5x10-3) 0.04/2 = 0.02 (5x10-3)/2 0.03x10-6 5x10-7 = 1.5x10-4 = 0.0025 2 (0.03)(5x10-3) (5x10-3)/2 0.03/2 + 5x10-3 = 0.02 3.75x10-7 3x10-6 = 1.5x10-4 = 0.0025 3 (0.03)(5x10-3) 0.04-0.0025 0.03/2 + 5x10-3 = 0.02 5.625x10-6 3x10-6 = 1.5x10-4 = 0.0375 4 (0.03)(5x10-3) 0.04/2 = 0.02 0.04 - (5x10-3)/2 0.03x10-6 7.5x10-6 = 1.5x10-4 = 0.0375 Σ = 6.0x10-4 Σ = Σ =1.4x10-5 9.435x10-5 ΣA𝑥̅ 9.435x10−5 ΣA𝑦̅ 1.4x10−5 𝑥̅ = = = 0.157𝑚 𝑦̅ = = = 0.023𝑚 Σ𝑥̅ 6.0x10−4 Σ𝑦̅ 6.0x10−4 18 Next, moment of inertia is calculated as follow. 𝑏ℎ3 𝐼𝑥 = 12 For this type of cross section, moment inertia of the beam is its 𝐼𝑥 (outer) minus 𝐼𝑥 (inner). Thus, 𝑏ℎ3 (0.04)(0.043 ) (0.03)(0.033 ) 𝐼𝑥 = = − = 1.455 𝑥 10−7 𝑚4 12 12 12 Then, stress distribution of the beam is calculated (bending stress and shear stress) at point of A, B, C and D. 0.03 0.02 Figure 4.6 Cross Section of Beam 𝑀𝑦 6.769(−0.020) 𝜎𝐴 = 𝜎𝑚𝑎𝑥 (𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛) = = = − 0.930𝑀𝑃𝑎 𝐼 1.455 𝑥 10−7 𝑀𝑦 6.769(−0.015) 𝜎𝐵 = = = −0.698𝑀𝑃𝑎 𝐼 1.455 𝑥 10−7 𝑀𝑦 6.769(0) 𝜎𝑐 = 𝜎𝑁𝐴 = = = 0 𝑀𝑃𝑎 𝐼 1.455 𝑥 10−7 𝑀𝑦 6.769(0.015) 𝜎𝐷 = = = 0.698𝑀𝑃𝑎 𝐼 1.455 𝑥 10−7 19 𝑀𝑦 6.769(0.02) 𝜎𝐸 = = = 0.930 𝑀𝑃𝑎 𝐼 1.455 𝑥 10−7 𝑉𝑄 𝜏𝐴 = = 0𝑀𝑃𝑎 𝐼𝑡 0.005 𝑉𝑄 29.43(0.04𝑥0.005)(0.02 − 2 ) = 0.018𝑀𝑃𝑎 𝜏𝐵1 = = 𝐼𝑡 1.455 𝑥 10−7 (0.04) 0.005 𝑉𝑄 29.43(0.04𝑥0.005)(0.02 − 2 ) 𝜏𝐵2 = = = 0.071𝑀𝑃𝑎 𝐼𝑡 1.455 𝑥 10−7 (2𝑥0.005) 0.005 0.015 𝑉𝑄 29.43[(0.04𝑥0.005) (0.02 − 2 ) + 2(0.005𝑥0.015)( 2 ) 𝜏𝑐 = = = 0.072𝑀𝑃𝑎 𝐼𝑡 1.455 𝑥 10−7 (2𝑥0.005) 0.005 𝑉𝑄 29.43(0.04𝑥0.005)(0.02 − 2 ) = 0.018𝑀𝑃 𝜏𝐷1 = = 𝐼𝑡 1.455 𝑥 10−7 (0.04) 0.005 𝑉𝑄 29.43(0.04𝑥0.005)(0.02 − 2 ) 𝜏𝐷2 = = = 0.071𝑀𝑃 𝐼𝑡 1.455 𝑥 10−7 (2𝑥0.005) 𝑉𝑄 𝜏𝐸 = = 0𝑀𝑃𝑎 𝐼𝑡 - 0.930 -0.698 0.018 0.071 0.072 0.698 0.018 0.071 0.930 Figure 4.7 Bending Stress Distribution Diagram and Shear Stress Distribution Diagram 20 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. Thus, strain, 𝜀 can be calculated. 𝛿 0.5𝑐𝑚 𝜀= = = 0.0109 𝐿 46𝑐𝑚 In order to obtain the elastic modulus of the beam, E, graph of stress, 𝜎 versus strain, 𝜀 is plotted. Stress Vs Strain 8 7 6 5 Stress(MPa) 4 3 2 1 0 0 0.109 Strain The gradient of the graph is the elastic modulus of the beam. Therefore, 𝜎 (0.072 − 0) 𝐸= = = 6.606𝑀𝑃𝑎 𝜀 (0.0109 − 0) Based on the formula given, 𝑓𝑦 𝐿3 (58.86)(0.463 ) 𝛿 𝑚𝑎𝑥 =− = = 1.281𝑥10−3 𝑚 48𝐸𝐼 48 (640.183𝑥106 )(1.455𝑥10−7 ) 21 As the project instruction required us to calculate the deflection of beam using Double Integration Method, the calculation of deflection using this method is shown as follow. Firstly, obtain the bending moment equation using bending moment diagram. 58.86N 29.43N 29.43N 6.769 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 M=mx + c. 6.769 − 0 𝑚1 = = 29.430; 𝑐1 = 0 0.23 − 0 Thus, M1 = (220.726x)Nm. 0 − 6.769 𝑐2 6.769 𝑚2 = = −29.430; = ⟹ 𝑐2 = 13.538 0.46 − 0.23 0.46 0.23 Thus, M2 = (-29.430x + 13.538)Nm. 22 𝐸𝐼𝜃1 = ∫ 𝑀1 = ∫ 29.430𝑥 𝑑𝑥 = 29.430𝑥 2 + 𝐶1 𝐸𝐼𝑦1 = ∫ 𝑀1 = ∫ 29.430𝑥 2 + 𝐶1 𝑑𝑥 = 29.430𝑥 3 + 𝐶1 𝑥 + 𝐶2 𝐸𝐼𝜃2 = ∫ 𝑀2 = ∫(−29.430𝑥 + 13.538) 𝑑𝑥 = −29.430𝑥 2 + 13.538𝑥 + 𝐶3 𝐸𝐼𝑦2 = ∫ 𝑀2 = ∫(−29.430𝑥 2 + 13.538𝑥 + 𝐶3 ) 𝑑𝑥 = −29.430𝑥 3 + 13.538𝑥 2 + 𝐶3 𝑥 + 𝐶4 The boundary condition of the beam is: (i) when x = 0, y = 0 (ii) when x = 0.46, y = 0 When x = 0, y = 0, 𝐸𝐼𝑦1 = 29.430𝑥 3 + 𝐶1 𝑥 + 𝐶2 0 = 𝐶2 ∴ 𝐸𝐼𝑦1 = 29.430𝑥 3 + 𝐶1 𝑥 𝐸𝐼𝑦2 = −29.430𝑥 3 + 13.538𝑥 2 + 𝐶3 𝑥 + 𝐶4 0 = 𝐶4 ∴ 𝐸𝐼𝑦2 = −29.430𝑥 3 + 13.538𝑥 2 + 𝐶3 𝑥 When x = 0.46, y=0, 𝐸𝐼𝑦1 = 29.430𝑥 3 + 𝐶1 0 = 3.581 + 0.46𝐶1 𝐶1 = -6.215 ∴ 𝐸𝐼𝑦1 = 29.430𝑥 3 − 6.215𝑥 23 𝐸𝐼𝑦2 = −29.430𝑥 3 + 13.538𝑥 2 + 𝐶3 𝑥 0 = -2.865 + 2.865 + 0.46𝐶3 𝐶3 = -2.174 ∴ 𝐸𝐼𝑦2 = −29.430𝑥 3 + 13.538𝑥 2 − 2.174𝑥 As the beam is simply supported and load is applied at its midspan, we can indicate that maximum deflection occurs at the midspan. When x = 0.23, 13.538𝑥 3 − 7.785𝑥 13.538(0.23)3 − 7.785(0.23) 𝑦1 = = = −0.017𝑚 𝐸𝐼 (640.183𝑥106 )(1.455𝑥10−7 ) −13.538𝑥 3 + 29.430𝑥 2 − 15.567𝑥 𝑦2 = 𝐸𝐼 −13.538(0.23)3 + 29.430(0.23)2 − 15.567(0.23) = (640.183𝑥106 )(1.455𝑥10−7 ) = −0.023𝑚 24 DISCUSSION 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 deflects downward. Methods Value of Maximum Deflection Manual Deflection Test - 0.73mm 𝑓𝑦 𝐿3 𝛿 𝑚𝑎𝑥 =− - 1.281mm 48𝐸𝐼 Double Integration Method -23.0mm 𝑓𝑦 𝐿3 According to 𝛿 𝑚𝑎𝑥 = − , maximum deflection of the beam should be 1.281mm. 48𝐸𝐼 However, we only obtain 0.73mm of deflection from the laboratory testing. Hence, there are some random error occurs during the testing of the beam such as the gauge does not put completely perpendicular on the top of the beam. The percentage difference between theoretical value and experimental value is: (1.281 − 0.73) 𝑥 100% = 43.01% 1.281 By using Double Integration Method, we obtain 23mm as the maximum deflection of the beam. We can state that this value is the most accurate value as it does not depend on any of the laboratory 𝑓𝑦 𝐿3 testing value. Unlike the double integration method, 𝛿 𝑚𝑎𝑥 = − still need one of the 48𝐸𝐼 laboratory testing value which is 𝑓𝑦 . 𝑓𝑦 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 value of the load, in which the beam fails, as 𝑓𝑦 . 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. For example, some of the sides of beam were not exactly 4cm but were in fact range about 25 ±0.05cm. 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 the beam is affected altogether. On top of that, the pattern which the satay sticks were stick together 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 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 get in the laboratory is quite difficult to be same as the theory calculation result. 26 CHAPTER 5 CONCLUSION 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 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 of 3 point bending test like stress, strain, strain energy and load bearing capacity and how to determine the deflection of beam. 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 stick, making beam model, testing of beam, analysis of deflection (centroid, moment of inertia, bending stress, shear stress, load-deflection graph and deflection equation) until video and report 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 the right person also makes our project goes smooth without any major obstacle. Last of all, we also improve our creativity and problem solving skills during making the report and video. This plays an important role as the report and video need to be clear in explaining every 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 be completed. 27 REFERENCES Books Reference 1. Christopher Jenkins, Sanjeev Khanna (2005). Mechanics of Materials: A Modern Integration of Mechanics and Materials in Structural Design. Elsevier Academic Press. 2. R.C. Hibbeler (2000). Mechanics of Materials.US: Prentice Hall, Inc. 3. Ferdinand P. Beer, E. Russell Johnston, Jr., John T. Dewolf, David F. Mazurek (2012). Mechanics of Materials. US: McGraw Hill Companies, Inc. 4. James M. Gere, Barry J. Goodno (2013). Mechanics of Materials. US: Cengage Learning. Internet Reference 1. https://www.guaduabamboo.com 2. https://www.scribd.com 3. http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/double- integration-method-beam-deflections 28