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1.0 TABLE OF CONTENT

NO.

TITLE

PAGES

1.0

TABLE OF CONTENT

1

2.0

SAFETY AND HEALTH

2

3.0

INTRODUCTION

3

4.0

OBJECTIVE

4

5.0

THEORY

5

6.0

APPARATUS/MATERIALS

6

7.0

STEP/PROCEDURE

7

8.0

RESULT

8

9.0

DISCUSSION

31

10.0

CONCLUSION

32

11.0

REFERENCE

32

Page | 1

2.0 SAFETY AND HEALTH

The basic purpose of these safety introductions is to protect students, researchers, technicians and teachers from the many hazards that might be encountered during the use of the various materials and equipment in Structual Laboratory.Please read the information and follow the introductions presented here which are given to safeguard while you in laboratory. 1. No running, eating, drinking or smoking in the lab. 2. Wear Safety jackets while conducting experiments. 3. Know locations of first aid, eye station, safety shower, fire blanket, fire extinguishers and gas masks. 4. Avoid cross contamination from the various microbiological cultures, media and samples taht are handled in this laboratory. 5. Do not overload the test specimen.A Plastically deformed specimen will yield inaccurate results. 6. Do not impact the dial gauge. 7. Do not use hand to touch or press the load cells. 8. Slowly tighten the two screws below the clamping plate until it just touched the load cell bar.Do not over tighten it. 9. After finished every single experiment, loosen all the screws on clamping plate.Push up or lightly vibrate (shake) the test beam.This is to ensure the beam return to its rest position due to the deflection occur when weight applied on the rest beam.Failure to do so may effect the result of your next experiment. 10. Due to the high sensitivity of the measuring device, there might be a slight reading shown on the meters even after tare.These values can be ignored as it shall not be significant.

Page | 2

3.0 INTRODUCTION :

While superposition can be used for many problems in engineering, they are particularly useful for beam deflections. Most beam configuration and loading can be split into simpler beams and loading. For example, the beam at the left with a distributed load and a point moment load can be split into two beams. One with the distributed load and one with the moment load. After the beams are simplified, the deflections of the simplified beams are still needed.

Figure 1 : Some of the Basic Beams

Page | 3

Figure 2 : Superposition for Beam Deflection

The deflection equation of the complex beam is the addition of the two simpler beam equations, or δ(x) = δ1(x) + δ2(x) There are a number of assumptions when using superposition. It is assumed that the beam undergoes linear deflection, all deflections are small, elastic material properties, no shear deflection (i.e. no short thick beams, and normal boundary conditions).

4.0 OBJECTIVE :

Determination of reaction for propped cantilever subjected to point loads and distributed loads.

Comparison between reactions by experimentally and theoretically. (Used Superposition Method)

Page | 4

5.0 THEORY :

A cantilever fixed at one end and simply supported at other end is known as Propped cantilever. A cantilever beam for which one end is fixed and other end is provided support, in order to resist the deflection of the beam, is called a propped cantilever bema. A propped cantilever is a statically indeterminate beam. Such beams are also called as restrained beams, as an end is restrained from rotation. The principle states: If the structural behavior is linearly elastic the forces acting on a structure may be separated or divided in any convenient fashion and the structure analyzed for the separate cases. The final results can then be obtained by adding algebraically the individual results. Many of the analytical techniques used in structural analysis are based on this principle. The following illustration shows how the principle of superposition can be applied to a cantilever beam.

Page | 5

6.0 APPARATUS/MATERIALS

Figure 1 :Propped cantilever beam apparatus

Figure 3: Digital dial test indicator

Figure 2 : Weight

Figure 4 : Micrometer

Figure 5: Knief-edge hanger

Page | 6

7.0

STEP/PROCEDURE

1. Setup the propped cantilever apparatus. 2. Connect the load cell cable to instrument panel, according the number mark on the connector. 3. Remove the load, leave the load hanger with empty load.Switch ON the instrument panel power and record the initial reading of the meters. 4. Apply 10N load at the centre of cantilever beam and measure the Ma, Ra &Rb and filled in table 1.Calculate the net reading. 5. Repets step 4 with different load of 15N & 20N.Measure the Ma, Ra &Rb and fill in table 1. 6. Use the theoretical calculation to calculate the value of Ma, Ra &Rb and filled in table 2. 7. Calculate the percentage of error experiment result as compare with theoretical result.

Page | 7

8.0 RESULT

Inital Ma (N) 7.50

10N

Final Ma (N) 7.28

7.5N/m 20N and 7.5N/m

Net Ma (N) -0.22

Initial Ra ( N) 53.2

Final Ra (N) 58.5

Net Ra (N) 5.3

7.37 7.37

8.05

0.7

Initial Rb (N) 41.1

Final Rb (N) 45.8

55.5 56.1

67.3

11.2

Net Rb (N) 4.7 43.4

44.1

52.6

8.5

Table 1.1:Propped cantilever with point load and distribution load

Calculation Net= Final- Initial

10N

20N and 7.5N/m

Ma = 7.28-7.50 = -0.22 Ra

= 58.5-53.2 = 5.3

Rb

= 45.8-41.1 = 4.7

Ma = 8.05-7.37 = 0.7 Ra = 67.3-56.1 = 11.2 Rb = 52.6-44.1 = 8.5

Page | 8

Theoretical value with 10N load Theoretical value with 7.5N/m Theoretical value with 20N load and 7.5N/m

Ma moment (Nm) 1.86

Reaction Ra (N) 8.14

Reaction Rb (N) 1.86

0.50

3.49

2.14

3.34

17.27

8.36

Table 1.2:Theoretical value of Ma, Ra &Rb

Support beam no. 10N load error%

Ma moment (Nm) -0.112

Reaction Ra (N) -3.489 x 10-3

Reaction Rb (N) 0.0152

7.5N/m load error%

0.1374

1490

0.193

20N load and 7.5N/m error%

-1.9 x 103

-3.5 x 10-3

1.657 x 10-4

Table 1.3: Percentage of error Calculation (experiment result – theoretical result)/theoretical result x 100%

Example calculation = 7.37-0.50 x 100% 0.50 = 0.1374

Page | 9

Question 1 10 N

0.375

0.375

Case 1 : 10 N

Moment eq. Mx–x = 0 Mx + 10 (x-0.375) = 0, Mx= -10 (x-0.375) Mx = EI d2y dx

Page | 10

Slope eq. Mx = EId2y dx EI d2y = -10(x-0.375)

1

dx EI d2y = -10(x-0.375)2 + C1 dx

2

2

EI d2y = -10(x-0.375)3 + C1 x = C2 dx

3

6

Boudary condition At A : x = 0.750 ,

dy = 0,

y=0

dx At B : x = 0.750 ,

dy ≠ 0,

y = YBI @ YB2

dx

Slope eq. EI d2y = -10(x-0.375)2 + C1 dx

2 = -10(0.750 -0.375)2 + C1 2 C1 = 0.70

Page | 11

EI yB1 = -10(x-0.375)3 + C1 x = C2 2 = -10(0.750-0.375)3 + 0.70 (0.750) + C2 2 = -0.26 EI yB1 = -10(x-0.375)3 + C1 x = C2 2 = -10(0-0.375)3 + 0.70 (0) + C2 2 = -0.26 EI

Page | 12

Case 2

YB2 RBY

x ∑Mx – x= 0 Mx –RBY (x) = 0 Mx = RBY (x) Mx = EI d2y = RBY(x)

1

dx EI d2y = RBY(x)2 + C1 dx

2

2

EI d2y = RBY(x)3 + C1 x + C2 dx

3

6

At A : x = 0.750 ,

dy = 0,

y=0

dx

EI d2y = RBY(x)2 + C1 dx

2

2 =(0.750)2 + C1 2 C1 = -0.28 RBY

Page | 13

EI d2y = RBY(x)3 + C1 x + C2 6 = (0.750)3 – 0.28(0.750) + C2 6 = 0.07 RBY – 0.21 +12 C2 = 0.14 RBY

dy ≠ 0,

At A : x = 0,

y = YB2

dx

EI d2y = RBY(x)3 + C1 x + C2 6 = RBY (0)3 – 0.28(0) + 0.14 RBY 6 = 0.14 RBY EI YB1 + YB2 = 0 0.26 + 0.14 EI

=0

E1

RBY = ( 0.26 + EI ) = 0 E1

0.14

= 1.857 kN

Page | 14

10 N

0.375

0.375

∑[email protected]=0 -1.86 ( 0.750 ) +10 (0.375) – Ma = 0 Ma = 1.86 kN

∑ Fy = 0 RAY -10 -1.86 = 0 RAY = 8.14 N

Page | 15

10 N

RAY

0.375

0.375

8.14 N

1.86 N

8.14 N

+ve

RBY

8.14 N

+

=1.86 + 2.65 -

SFD

-1.80 N

= 0.79 N -1.80 N =0.79 – 0.70

( V)

= 0.09 -ve -1.86 N

A1= 8.14 x 0.325 =2.65 A2 = (-1.86)(0.375)

BMD

= -0.70 N

(M) 0.09 N -0.79 N +ve

Page | 16

Question 2

7.5 N/m

0.750 m Case 1

7.5 N/m YB1

x Moment eq. ∑ Mx–x = 0 Mx = 7.5(x)( x ) = 0 2 Mx = -7.5x 2

Page | 17

Slope eq. Mx = EId2y dx EI d2y = -7.5x dx

1

2

EI d2y = -7.5(x)3 + C1 dx

2

6

EI d2y = -7.5(x)4 + C1 x = C2 dx

3

24

Boudary condition At A : x = 0.750 ,

dy = 0,

y=0

dx At B : x = 0.750 ,

dy ≠ 0,

y = YBI @ YB2

dx

Slope eq. EI d2y = -7.5(x)3 + C1 dx

6 = -7.5(0.750)3 + C1 6 C1 = 0.53 Page | 18

EI yB1 = -7.5(x)4 + C1 x = C2 24 = -7.5(0.750)4 + 0.53 (0.750) + C2 2 = -0.3 EI yB1 = -7.5(x)3 + C1 x = C2 24 = -7.5(0)3 + 0.53 (0) + C2 2 = -0.30 EI

Page | 19

Case 2

YB2

x

∑Mx – x= 0 Mx –RBY (x) = 0 Mx = RBY (x) Mx = EI d2y = RBY(x)

1

dx EI d2y = RBY(x)2 + C1 dx

2

2

EI d2y = RBY(x)3 + C1 x + C2 dx

3

6

At A : x = 0.750 ,

dy = 0,

y=0

dx

EI d2y = RBY(x)2 + C1 dx

2

2 =(0.750)2 + C1 2 C1 = -0.28 RBY

Page | 20

EI d2y = RBY(x)3 + C1 x + C2 6 = (0.750)3 – 0.28(0.750) + C2 6 = 0.07 RBY – 0.21 +C2 C2 = 0.14 RBY

dy ≠ 0,

At A : x = 0,

y = YB2

dx

EI d2y = RBY(x)3 + C1 x + C2 6 = RBY (0)3 – 0.28(0) + 0.14 RBY 6 = 0.14 RBY EI YB1 + YB2 = 0 -0.30 + 0.14 EI

=0

E1

RBY = ( 0.30 + EI ) = 0 E1

0.14

= 2.142 kN

Page | 21

7.5 N/m

RAY

0.750 m

RBY

∑ [email protected] = 0 -MA +7.5(0.75)(0.75 ) -2.14(0.75) =0 2 MA = 0.50

∑ Fy= 0 RAY – 7.5(0.75) -2.14 =0 RAY = 3.49

Page | 22

7.5 N/m

RAY

0.750 m

RBY

3.49

2.14

3.49 + (ve) 0.28

= 3.49 7.5

0.47

= 0.47

(-ve)

-2.14

SFD ( V) (-ve) 0.50 = - 0.50 + 0.8 = 0.30 0.02 BMD (V) (+ve)

0.30

A1= ½ (3.49)(0.47) = 0.82 A2= ½ (-2.14)(0.28) = -0.30 Page | 23

Question 3 20 N 7.5 N/m

0.375

0.375

Case 1 20 N

7.5N/m YB1

x-0.375 Moment eq. Mx–x = 0 Mx + 20 (x - 0.375) + 7.5(x)( x ) = 0 2 Mx = -20(x-0.375)- 7.5x2 2 Mx = EI d2y dx

Page | 24

EI d2y = -20(x-0.375) - 7.5x2 dx

1

2

EI d2y = -20(x-0.375)2 - 7.5x3 + C1 dx

2

6

EI d2y = -20(x-0.375)3 - 7.5x4 + C1 x + C2 dx

6

3

24

Boudary condition At A : x = 0.750 ,

dy = 0,

y=0

dx At B : x = 0.750 ,

dy ≠ 0,

y = YBI @ YB2

dx

Page | 25

Slope eq. EI d2y = -20(x-0.375)2 - 7.5x3- + C1 dx

2

6

= -20(0.750 -0.375)2 - 7.5(0.375)3+ C1 2

6

C1 = 1.93

EI yB1 = -20(x-0.375)3 - 7.5x3 + C1 x + C2 2

6

= -20(0.750-0.375)3 + 7.5(0.375)3+ 1.93(0.750) + C2 2

6

= -1.17 EI yB1 = -20(x-0.375)3 - 7.5x3 + C1 x + C2 26

24

= -20(0 - 0.375)3 + 7.5(0)3+ 1.93(0) + C2 6

24

= -1.17 EI

Page | 26

Case 2 YB2 RBY

x ∑Mx – x= 0 Mx –RBY (x) = 0 Mx = RBY (x) Mx = EI d2y = RBY(x)

1

dx EI d2y = RBY(x)2 + C1 dx

2

2

EI d2y = RBY(x)3 + C1 x + C2 dx

3

6

At A : x = 0.750 ,

dy = 0,

y=0

dx

EI d2y = RBY(x)2 + C1 dx

2

2 =(0.750)2 + C1 2 C1 = -0.28 RBY

Page | 27

EI d2y = RBY(x)3 + C1 x + C2 6 = (0.750)3 – 0.28(0.750) + C2 6 = 0.07 RBY – 0.21 +C2 C2 = 0.14 RBY

dy ≠ 0,

At B: x = 0,

y = YB2

dx

EI d2y = -20(x-0.375)3 - 7.5x3 + C1 x + C2 26

24

= -20(0 - 0.375)3 + 7.5(0)3+ 0.28(0) + 0.14 RBY 6

24

= 0.14 RBY EI YB1 + YB2 = 0 -1.17 + 0.14 EI

=0

E1

RBY = ( 1.17 + EI ) = 0 E1

0.14

= 8.36 kN

Page | 28

20 N 7.5 N/m

RAY

0.375

0.375

RBY =8.36

∑ MA= 0 -MA + 20(0.375) + 7.5(0.75)(0.75) - 8.36(0.75) MA = 3.34Nm ∑Fy = 0 RAY – 7.5(0.75) + 20 – 8.36 = 0 RAY = 17.27 N

Page | 29

20 N 7.5 N/m

RAY

0.375

0.375

RBY =8.36

y2 = -17.27

17.27 +ve

14.46

0.375 = -7.5 y2 = -7.5(0.375) + 17.27

SFD (V)

=14.46 -5.54

-ve

-8.35

- 3.34

y2- (-5.54) = -7.5 0.375 Y2= -7.5(0.375)-5.54 = 8.35

BMD (M)

= -3.34 + 8.95 = 2.61

A1 = ½ ( 17.27 + 14.46)(0.375) =5.95 A2=½ ( 17.27 + 14.46)(0.375)

+ve

= -2.60

Page | 30

9.0 DISCUSSION: 1. Use superposition method, determine the reaction at fixed end and pinned support. At Fixed End • does not experience any deflection therefore deflection function is zero Slope =>

(gradient horizontal)

Deflection => y = 0 (no deflection) At Free End • no bending moment at the free end • no shearing force acting at the free end 2. The factor that affecting the results of experiments is wind and motion of people that doing the experiment. Thewinds effect the digital dial test indicator reading,we can see in the error table provided.This actually our own fault because we did not switch off the fan while doing the experiment and the wind load is constantly changing,from this mistake we also did not obtain a good result because the theoretical value and the experimental value has a huge different. Besides that,there is also some problem while we doing the determine the length where to hang the knife-edge hanger because we need to use ruler to determine the length, maybe be the length may change while we doing the experiment because there is no lock system or reading number at the cantilever beam. Furthermore, thepeople is doing the experiment also affecting the reading because the people is collide with the tool and the reading also affected. We also determine that the environment of the laboratory is not suitable to do this experiment because the laboratory is too small, because of this the micro-sensitive is easily affected by student when doing the experiment. From this we can say that the error came from both human and nature. Furthermore the other factor that affecting is zero error, due to the high sensitivity if the measuring device, there might be a slight reading shown on the digital meter even before or after tare. These value can be ignore as it shall not be significant, if it a big value it can be +/- (add/subtract)at the final reading. The percentage differences, we can see clearly at data tabulation. Our data is not good, because of some human and environment factor affecting the moment and reaction value. If we do the experiment very carefully with take all the precaution in first place, we can get the small number of percentage or error, or the theoretical and experimental value is almost the same.Now, our percentage of error is until 100%.This data is not good.

Page | 31

10.0 CONCLUSION: 1. The experimental value and the theoretical value has a lot of different or error because as we have mention at discussion, our fix end and pinned end value is disturbed by wind load and human error. Besidesthat, the fix end and pinned end between experiment and theoretical has big different, so from this we can say that the data that we obtain is less impressive and has error. The only pinned end that is similarly with experiment value is RB of theoretical value with 20N load and 7.5N/m that is 11.68N.That RB value is correct and can say that wind do really exert load into the reading because the final load 20N we put when the laboratory is about to close, and the fan is switch off by the laboratory supervisor when we are doing the experiment. 2. The value of RBY at support beam 20N load and 7.5N/m is 11.68N is almost same with the experimental value 11.60N.The rest of the value not same and have percentage of error at small as 27.48N until 100N/mon moment and reaction. At the end of the experiment we have learned how to determine the reaction and moment using super-position method, we also have learned how to do the experiment for propped cantilever subjected to point loads and distributed load correctly. We also have known to use the experiment related tools correctly. At last we also can do comparison between reaction by experimental value and theoretical value correctly. From this we can say that, the objective of the experiment has been archived by our group.

11.0 REFERENCES Book

Mechanics of materials by Dr.B.CPunmia Stability of columns by YI Nagornyi Laboratory Manual Cc505 P (Structural Analysis 1)

Page | 32

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NO.

TITLE

PAGES

1.0

TABLE OF CONTENT

1

2.0

SAFETY AND HEALTH

2

3.0

INTRODUCTION

3

4.0

OBJECTIVE

4

5.0

THEORY

5

6.0

APPARATUS/MATERIALS

6

7.0

STEP/PROCEDURE

7

8.0

RESULT

8

9.0

DISCUSSION

31

10.0

CONCLUSION

32

11.0

REFERENCE

32

Page | 1

2.0 SAFETY AND HEALTH

The basic purpose of these safety introductions is to protect students, researchers, technicians and teachers from the many hazards that might be encountered during the use of the various materials and equipment in Structual Laboratory.Please read the information and follow the introductions presented here which are given to safeguard while you in laboratory. 1. No running, eating, drinking or smoking in the lab. 2. Wear Safety jackets while conducting experiments. 3. Know locations of first aid, eye station, safety shower, fire blanket, fire extinguishers and gas masks. 4. Avoid cross contamination from the various microbiological cultures, media and samples taht are handled in this laboratory. 5. Do not overload the test specimen.A Plastically deformed specimen will yield inaccurate results. 6. Do not impact the dial gauge. 7. Do not use hand to touch or press the load cells. 8. Slowly tighten the two screws below the clamping plate until it just touched the load cell bar.Do not over tighten it. 9. After finished every single experiment, loosen all the screws on clamping plate.Push up or lightly vibrate (shake) the test beam.This is to ensure the beam return to its rest position due to the deflection occur when weight applied on the rest beam.Failure to do so may effect the result of your next experiment. 10. Due to the high sensitivity of the measuring device, there might be a slight reading shown on the meters even after tare.These values can be ignored as it shall not be significant.

Page | 2

3.0 INTRODUCTION :

While superposition can be used for many problems in engineering, they are particularly useful for beam deflections. Most beam configuration and loading can be split into simpler beams and loading. For example, the beam at the left with a distributed load and a point moment load can be split into two beams. One with the distributed load and one with the moment load. After the beams are simplified, the deflections of the simplified beams are still needed.

Figure 1 : Some of the Basic Beams

Page | 3

Figure 2 : Superposition for Beam Deflection

The deflection equation of the complex beam is the addition of the two simpler beam equations, or δ(x) = δ1(x) + δ2(x) There are a number of assumptions when using superposition. It is assumed that the beam undergoes linear deflection, all deflections are small, elastic material properties, no shear deflection (i.e. no short thick beams, and normal boundary conditions).

4.0 OBJECTIVE :

Determination of reaction for propped cantilever subjected to point loads and distributed loads.

Comparison between reactions by experimentally and theoretically. (Used Superposition Method)

Page | 4

5.0 THEORY :

A cantilever fixed at one end and simply supported at other end is known as Propped cantilever. A cantilever beam for which one end is fixed and other end is provided support, in order to resist the deflection of the beam, is called a propped cantilever bema. A propped cantilever is a statically indeterminate beam. Such beams are also called as restrained beams, as an end is restrained from rotation. The principle states: If the structural behavior is linearly elastic the forces acting on a structure may be separated or divided in any convenient fashion and the structure analyzed for the separate cases. The final results can then be obtained by adding algebraically the individual results. Many of the analytical techniques used in structural analysis are based on this principle. The following illustration shows how the principle of superposition can be applied to a cantilever beam.

Page | 5

6.0 APPARATUS/MATERIALS

Figure 1 :Propped cantilever beam apparatus

Figure 3: Digital dial test indicator

Figure 2 : Weight

Figure 4 : Micrometer

Figure 5: Knief-edge hanger

Page | 6

7.0

STEP/PROCEDURE

1. Setup the propped cantilever apparatus. 2. Connect the load cell cable to instrument panel, according the number mark on the connector. 3. Remove the load, leave the load hanger with empty load.Switch ON the instrument panel power and record the initial reading of the meters. 4. Apply 10N load at the centre of cantilever beam and measure the Ma, Ra &Rb and filled in table 1.Calculate the net reading. 5. Repets step 4 with different load of 15N & 20N.Measure the Ma, Ra &Rb and fill in table 1. 6. Use the theoretical calculation to calculate the value of Ma, Ra &Rb and filled in table 2. 7. Calculate the percentage of error experiment result as compare with theoretical result.

Page | 7

8.0 RESULT

Inital Ma (N) 7.50

10N

Final Ma (N) 7.28

7.5N/m 20N and 7.5N/m

Net Ma (N) -0.22

Initial Ra ( N) 53.2

Final Ra (N) 58.5

Net Ra (N) 5.3

7.37 7.37

8.05

0.7

Initial Rb (N) 41.1

Final Rb (N) 45.8

55.5 56.1

67.3

11.2

Net Rb (N) 4.7 43.4

44.1

52.6

8.5

Table 1.1:Propped cantilever with point load and distribution load

Calculation Net= Final- Initial

10N

20N and 7.5N/m

Ma = 7.28-7.50 = -0.22 Ra

= 58.5-53.2 = 5.3

Rb

= 45.8-41.1 = 4.7

Ma = 8.05-7.37 = 0.7 Ra = 67.3-56.1 = 11.2 Rb = 52.6-44.1 = 8.5

Page | 8

Theoretical value with 10N load Theoretical value with 7.5N/m Theoretical value with 20N load and 7.5N/m

Ma moment (Nm) 1.86

Reaction Ra (N) 8.14

Reaction Rb (N) 1.86

0.50

3.49

2.14

3.34

17.27

8.36

Table 1.2:Theoretical value of Ma, Ra &Rb

Support beam no. 10N load error%

Ma moment (Nm) -0.112

Reaction Ra (N) -3.489 x 10-3

Reaction Rb (N) 0.0152

7.5N/m load error%

0.1374

1490

0.193

20N load and 7.5N/m error%

-1.9 x 103

-3.5 x 10-3

1.657 x 10-4

Table 1.3: Percentage of error Calculation (experiment result – theoretical result)/theoretical result x 100%

Example calculation = 7.37-0.50 x 100% 0.50 = 0.1374

Page | 9

Question 1 10 N

0.375

0.375

Case 1 : 10 N

Moment eq. Mx–x = 0 Mx + 10 (x-0.375) = 0, Mx= -10 (x-0.375) Mx = EI d2y dx

Page | 10

Slope eq. Mx = EId2y dx EI d2y = -10(x-0.375)

1

dx EI d2y = -10(x-0.375)2 + C1 dx

2

2

EI d2y = -10(x-0.375)3 + C1 x = C2 dx

3

6

Boudary condition At A : x = 0.750 ,

dy = 0,

y=0

dx At B : x = 0.750 ,

dy ≠ 0,

y = YBI @ YB2

dx

Slope eq. EI d2y = -10(x-0.375)2 + C1 dx

2 = -10(0.750 -0.375)2 + C1 2 C1 = 0.70

Page | 11

EI yB1 = -10(x-0.375)3 + C1 x = C2 2 = -10(0.750-0.375)3 + 0.70 (0.750) + C2 2 = -0.26 EI yB1 = -10(x-0.375)3 + C1 x = C2 2 = -10(0-0.375)3 + 0.70 (0) + C2 2 = -0.26 EI

Page | 12

Case 2

YB2 RBY

x ∑Mx – x= 0 Mx –RBY (x) = 0 Mx = RBY (x) Mx = EI d2y = RBY(x)

1

dx EI d2y = RBY(x)2 + C1 dx

2

2

EI d2y = RBY(x)3 + C1 x + C2 dx

3

6

At A : x = 0.750 ,

dy = 0,

y=0

dx

EI d2y = RBY(x)2 + C1 dx

2

2 =(0.750)2 + C1 2 C1 = -0.28 RBY

Page | 13

EI d2y = RBY(x)3 + C1 x + C2 6 = (0.750)3 – 0.28(0.750) + C2 6 = 0.07 RBY – 0.21 +12 C2 = 0.14 RBY

dy ≠ 0,

At A : x = 0,

y = YB2

dx

EI d2y = RBY(x)3 + C1 x + C2 6 = RBY (0)3 – 0.28(0) + 0.14 RBY 6 = 0.14 RBY EI YB1 + YB2 = 0 0.26 + 0.14 EI

=0

E1

RBY = ( 0.26 + EI ) = 0 E1

0.14

= 1.857 kN

Page | 14

10 N

0.375

0.375

∑[email protected]=0 -1.86 ( 0.750 ) +10 (0.375) – Ma = 0 Ma = 1.86 kN

∑ Fy = 0 RAY -10 -1.86 = 0 RAY = 8.14 N

Page | 15

10 N

RAY

0.375

0.375

8.14 N

1.86 N

8.14 N

+ve

RBY

8.14 N

+

=1.86 + 2.65 -

SFD

-1.80 N

= 0.79 N -1.80 N =0.79 – 0.70

( V)

= 0.09 -ve -1.86 N

A1= 8.14 x 0.325 =2.65 A2 = (-1.86)(0.375)

BMD

= -0.70 N

(M) 0.09 N -0.79 N +ve

Page | 16

Question 2

7.5 N/m

0.750 m Case 1

7.5 N/m YB1

x Moment eq. ∑ Mx–x = 0 Mx = 7.5(x)( x ) = 0 2 Mx = -7.5x 2

Page | 17

Slope eq. Mx = EId2y dx EI d2y = -7.5x dx

1

2

EI d2y = -7.5(x)3 + C1 dx

2

6

EI d2y = -7.5(x)4 + C1 x = C2 dx

3

24

Boudary condition At A : x = 0.750 ,

dy = 0,

y=0

dx At B : x = 0.750 ,

dy ≠ 0,

y = YBI @ YB2

dx

Slope eq. EI d2y = -7.5(x)3 + C1 dx

6 = -7.5(0.750)3 + C1 6 C1 = 0.53 Page | 18

EI yB1 = -7.5(x)4 + C1 x = C2 24 = -7.5(0.750)4 + 0.53 (0.750) + C2 2 = -0.3 EI yB1 = -7.5(x)3 + C1 x = C2 24 = -7.5(0)3 + 0.53 (0) + C2 2 = -0.30 EI

Page | 19

Case 2

YB2

x

∑Mx – x= 0 Mx –RBY (x) = 0 Mx = RBY (x) Mx = EI d2y = RBY(x)

1

dx EI d2y = RBY(x)2 + C1 dx

2

2

EI d2y = RBY(x)3 + C1 x + C2 dx

3

6

At A : x = 0.750 ,

dy = 0,

y=0

dx

EI d2y = RBY(x)2 + C1 dx

2

2 =(0.750)2 + C1 2 C1 = -0.28 RBY

Page | 20

EI d2y = RBY(x)3 + C1 x + C2 6 = (0.750)3 – 0.28(0.750) + C2 6 = 0.07 RBY – 0.21 +C2 C2 = 0.14 RBY

dy ≠ 0,

At A : x = 0,

y = YB2

dx

EI d2y = RBY(x)3 + C1 x + C2 6 = RBY (0)3 – 0.28(0) + 0.14 RBY 6 = 0.14 RBY EI YB1 + YB2 = 0 -0.30 + 0.14 EI

=0

E1

RBY = ( 0.30 + EI ) = 0 E1

0.14

= 2.142 kN

Page | 21

7.5 N/m

RAY

0.750 m

RBY

∑ [email protected] = 0 -MA +7.5(0.75)(0.75 ) -2.14(0.75) =0 2 MA = 0.50

∑ Fy= 0 RAY – 7.5(0.75) -2.14 =0 RAY = 3.49

Page | 22

7.5 N/m

RAY

0.750 m

RBY

3.49

2.14

3.49 + (ve) 0.28

= 3.49 7.5

0.47

= 0.47

(-ve)

-2.14

SFD ( V) (-ve) 0.50 = - 0.50 + 0.8 = 0.30 0.02 BMD (V) (+ve)

0.30

A1= ½ (3.49)(0.47) = 0.82 A2= ½ (-2.14)(0.28) = -0.30 Page | 23

Question 3 20 N 7.5 N/m

0.375

0.375

Case 1 20 N

7.5N/m YB1

x-0.375 Moment eq. Mx–x = 0 Mx + 20 (x - 0.375) + 7.5(x)( x ) = 0 2 Mx = -20(x-0.375)- 7.5x2 2 Mx = EI d2y dx

Page | 24

EI d2y = -20(x-0.375) - 7.5x2 dx

1

2

EI d2y = -20(x-0.375)2 - 7.5x3 + C1 dx

2

6

EI d2y = -20(x-0.375)3 - 7.5x4 + C1 x + C2 dx

6

3

24

Boudary condition At A : x = 0.750 ,

dy = 0,

y=0

dx At B : x = 0.750 ,

dy ≠ 0,

y = YBI @ YB2

dx

Page | 25

Slope eq. EI d2y = -20(x-0.375)2 - 7.5x3- + C1 dx

2

6

= -20(0.750 -0.375)2 - 7.5(0.375)3+ C1 2

6

C1 = 1.93

EI yB1 = -20(x-0.375)3 - 7.5x3 + C1 x + C2 2

6

= -20(0.750-0.375)3 + 7.5(0.375)3+ 1.93(0.750) + C2 2

6

= -1.17 EI yB1 = -20(x-0.375)3 - 7.5x3 + C1 x + C2 26

24

= -20(0 - 0.375)3 + 7.5(0)3+ 1.93(0) + C2 6

24

= -1.17 EI

Page | 26

Case 2 YB2 RBY

x ∑Mx – x= 0 Mx –RBY (x) = 0 Mx = RBY (x) Mx = EI d2y = RBY(x)

1

dx EI d2y = RBY(x)2 + C1 dx

2

2

EI d2y = RBY(x)3 + C1 x + C2 dx

3

6

At A : x = 0.750 ,

dy = 0,

y=0

dx

EI d2y = RBY(x)2 + C1 dx

2

2 =(0.750)2 + C1 2 C1 = -0.28 RBY

Page | 27

EI d2y = RBY(x)3 + C1 x + C2 6 = (0.750)3 – 0.28(0.750) + C2 6 = 0.07 RBY – 0.21 +C2 C2 = 0.14 RBY

dy ≠ 0,

At B: x = 0,

y = YB2

dx

EI d2y = -20(x-0.375)3 - 7.5x3 + C1 x + C2 26

24

= -20(0 - 0.375)3 + 7.5(0)3+ 0.28(0) + 0.14 RBY 6

24

= 0.14 RBY EI YB1 + YB2 = 0 -1.17 + 0.14 EI

=0

E1

RBY = ( 1.17 + EI ) = 0 E1

0.14

= 8.36 kN

Page | 28

20 N 7.5 N/m

RAY

0.375

0.375

RBY =8.36

∑ MA= 0 -MA + 20(0.375) + 7.5(0.75)(0.75) - 8.36(0.75) MA = 3.34Nm ∑Fy = 0 RAY – 7.5(0.75) + 20 – 8.36 = 0 RAY = 17.27 N

Page | 29

20 N 7.5 N/m

RAY

0.375

0.375

RBY =8.36

y2 = -17.27

17.27 +ve

14.46

0.375 = -7.5 y2 = -7.5(0.375) + 17.27

SFD (V)

=14.46 -5.54

-ve

-8.35

- 3.34

y2- (-5.54) = -7.5 0.375 Y2= -7.5(0.375)-5.54 = 8.35

BMD (M)

= -3.34 + 8.95 = 2.61

A1 = ½ ( 17.27 + 14.46)(0.375) =5.95 A2=½ ( 17.27 + 14.46)(0.375)

+ve

= -2.60

Page | 30

9.0 DISCUSSION: 1. Use superposition method, determine the reaction at fixed end and pinned support. At Fixed End • does not experience any deflection therefore deflection function is zero Slope =>

(gradient horizontal)

Deflection => y = 0 (no deflection) At Free End • no bending moment at the free end • no shearing force acting at the free end 2. The factor that affecting the results of experiments is wind and motion of people that doing the experiment. Thewinds effect the digital dial test indicator reading,we can see in the error table provided.This actually our own fault because we did not switch off the fan while doing the experiment and the wind load is constantly changing,from this mistake we also did not obtain a good result because the theoretical value and the experimental value has a huge different. Besides that,there is also some problem while we doing the determine the length where to hang the knife-edge hanger because we need to use ruler to determine the length, maybe be the length may change while we doing the experiment because there is no lock system or reading number at the cantilever beam. Furthermore, thepeople is doing the experiment also affecting the reading because the people is collide with the tool and the reading also affected. We also determine that the environment of the laboratory is not suitable to do this experiment because the laboratory is too small, because of this the micro-sensitive is easily affected by student when doing the experiment. From this we can say that the error came from both human and nature. Furthermore the other factor that affecting is zero error, due to the high sensitivity if the measuring device, there might be a slight reading shown on the digital meter even before or after tare. These value can be ignore as it shall not be significant, if it a big value it can be +/- (add/subtract)at the final reading. The percentage differences, we can see clearly at data tabulation. Our data is not good, because of some human and environment factor affecting the moment and reaction value. If we do the experiment very carefully with take all the precaution in first place, we can get the small number of percentage or error, or the theoretical and experimental value is almost the same.Now, our percentage of error is until 100%.This data is not good.

Page | 31

10.0 CONCLUSION: 1. The experimental value and the theoretical value has a lot of different or error because as we have mention at discussion, our fix end and pinned end value is disturbed by wind load and human error. Besidesthat, the fix end and pinned end between experiment and theoretical has big different, so from this we can say that the data that we obtain is less impressive and has error. The only pinned end that is similarly with experiment value is RB of theoretical value with 20N load and 7.5N/m that is 11.68N.That RB value is correct and can say that wind do really exert load into the reading because the final load 20N we put when the laboratory is about to close, and the fan is switch off by the laboratory supervisor when we are doing the experiment. 2. The value of RBY at support beam 20N load and 7.5N/m is 11.68N is almost same with the experimental value 11.60N.The rest of the value not same and have percentage of error at small as 27.48N until 100N/mon moment and reaction. At the end of the experiment we have learned how to determine the reaction and moment using super-position method, we also have learned how to do the experiment for propped cantilever subjected to point loads and distributed load correctly. We also have known to use the experiment related tools correctly. At last we also can do comparison between reaction by experimental value and theoretical value correctly. From this we can say that, the objective of the experiment has been archived by our group.

11.0 REFERENCES Book

Mechanics of materials by Dr.B.CPunmia Stability of columns by YI Nagornyi Laboratory Manual Cc505 P (Structural Analysis 1)

Page | 32

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