Ahli Kumpulan 8

1.0 INTRODUCTION

a. A truss is defined as a structure composed of slender element joined together at their end points. b. The joint connection are considered as pinned joint without friction. c. The members commonly used in construction consist of wooden struts, metal bars, angles or channels. In order to determine the force developed in the individual members at a truss, the following assumptions should be make: i. The members are connencted to each other at their ends by friction pins, that is only a force and no moment can be transferred from one member to another. ii. External loads are applied to the truss only at its joints.

1

d. Some consideration can be made in order to idealize the truss : i. The structural members are joined together by pin without friction.

ii. All loadings are applied at the same joints. iii. Each truss member acts as an axial forces member, and therefore the forces acting at the ends of the member must be directed along the axis of the member. e. One of the methods to calculate the force in the member of a truss is using Method of Joint. f. Plane or planar truss composed of members that lie in the same plane and frequently used for bridge and roof support. g. Loads that cause the entire truss to bend are converted into tensile and compressive forces in the members. h. The structures mechanics involves determination of unknown forces on the structures. i. Some of these structures can be completely analyzed by using the equations of equilibrium. ∑Fx = 0 ∑Fy = 0 ∑Má´¢ = 0 j. On the other hand, if there exist extra redundant reaction components, then the structure is said to be statically indeterminate. k. To be in a atate of static equilibrium,a structure must meet the requirements of stability. l. A statically indeterminate structure is a structure that had more unknown forces. m. A truss can be unstable if it is statically determinate or statically indeterminate.

2

The structures are known as statically determinate.

2.0 OBJECTIVE To examine a statically determinate frame and to analyze the frame using simple pin joint theory.

3.0

THEORY A truss is a structure composed of slender member joined together at their end points to form one or more triangles. The joint connections are considered as pinned joint without friction. In order to determine the forces developed in the individual members at a truss, the following assumptions should be make : a. The members are connected to each other at their ends by frictionless pins, that is only a force and no moment can be transferred from one member to another.

b. External loads are applied to the truss only at its joints. One of the methods to calculate the forces in the member of a truss is using Method of Joint.

3

4.0

PROCEDURE 1. The thumbwheel was unscrewed on the “redundant” member. Noted that is effectively no longer part of the structure as the idealized diagram illustrates. 2. The pre-load was applied of 100N downward, the load was re-zero cell and a load was applied of 250N and checked that the frame was stable and secure. 3. The load was returned to zero (leaving the 100N preload), rechecked and the digital indicator was re-zero.

4.

Load was applied in the increment shown in Table 1 recording the strain readings and the digital indicator readings. Table 2 have been completed by subtracting the initial (zero) strain readings.

4

5.0

EQUIPMENT

Cantilever Truss

Digital Display Force

Thumbwheel

Digital Indicator

Vernier Clipper

5

6.0 RESULT

Load (N) 0 50 100 150 200 250

Strain Reading 1 153 162 171 179 188 197 2 213 203 194 186 176 167 3 -23 -32 -41 -50 -59 -68 4 -45 -65 -83 -100 -188 -137 5 119 118 118 117 117 116 6 0 0 0 0 0 0 7 62 76 89 101 113 126 8 83 68 111 123 137 151

Digital Indicator Reading (mm) 0.013 0.019 0.047 0.076 0.116 0.151

Table 1 : Strain Readings and Frame Deflection for Experiment 1

Load (N) 0 50 100 150 200 250

1 0 9 18 26 35 44

2 0 -10 -19 -27 -37 -46

3 0 -9 -18 -27 -36 -45

4 0 -20 -38 -55 -143 -92

5 0 -1 -1 -2 -2 -3

6 0 0 0 0 0 0

7 0 14 27 39 51 64

8 0 -15 28 40 54 68

Table 2 :True Strain Reading for Experiment 1

6

Use : W = 150 N

Member 1 2 3 4 5 6 7 8 Strain reading Experimental Force (N) 159.64 -165.68 -165.68 -337.49 -12.27 0 239.31 245.45 Theoretical Force (N) 150 -150 -150 -300 0 0 212.13 212.13

26 -27 -27 -55 -2 0 39 40

*C = compression, T = Tension Table 3 : Measured and Theoretical Force in the Cantilever Truss

7

7.0 ANALYSIS

Calculation For Experimental Force (N) For Load = 200 N

From the formula:

E = σ/ ε E = Young „s Modulus (Nm-2) ε = Displayed Strain σ = Stress in the member (Nm-2)

σ = F/A F = Force in member (N) A = cross section area of the member (m2)

Rod diameter (d) = 6.10 mm E steel = 2.10 x 105 N/mm2 A = π r2 = π (3.05)2 = 29.22 mm2

8

Member 1: σ=Eε E steel = 2.10 x 105 N/mm2 ε = 26 x 10-6 σ = (2.10 x 105 N/mm2) x (26 x 10-6) = 5.46

F = σA σ = 5.46 A = 29.22 mm2 F = 5.46 x 29.22 F = 159.54 N

Member 2 : σ=Eε E steel = 2.10 x 105 N/mm2 ε = -27 x 10-6 σ = (2.10 x 105 N/mm2) x (-27 x 10-6) = -5.67

F = σA σ = -5.67 A = 29.22 mm2 F = -5.67 x 29.22 F = -165.68 N

9

Member 3 : σ=Eε E steel = 2.10 x 105 N/mm2 ε = -27 x 10-6 σ = (2.10 x 105 N/mm2) x (-27 x 10-6) = -5.67

F = σA σ = -5.67 A = 29.22 mm2 F = -5.67x 29.22 F = -165.68 N

Member 4 : σ=Eε E steel = 2.10 x 105 N/mm2 ε = -55 x 10-6 σ = (2.10 x 105 N/mm2) x (-55 x 10-6) = -11.55

F = σA σ = -11.55 A = 29.22 mm2 F = -11.55 x 29.22 F = -337.49 N

10

Member 5 : σ=Eε E steel = 2.10 x 105 N/mm2 ε = -2 x 10-6 σ = (2.10 x 105 N/mm2) x (-2 x 10-6) = -0.42

F = σA σ = -0.42 A = 29.22 mm2 F = -0.42 x 29.22 F = -12.27 N

Member 7 : σ=Eε E steel = 2.10 x 105 N/mm2 ε = 39 x 10-6 σ = (2.10 x 105 N/mm2) x (39x 10-6) = 8.19

F = σA σ = 8.19 A = 29.22 mm2 F = 8.19x 29.22 F = 239.31 N

11

Member 8 : σ=Eε E steel = 2.10 x 105 N/mm2 ε = 40 x 10-6 σ = (2.10 x 105 N/mm2) x (40 x 10-6) = 8.4

F = σA σ = 8.4 A = 29.22 mm2 F = 8.4 x 29.22 F = 245.45.12 N

12

Calculation For Theoretical Force (N) with Using Method Of Joint

RAY RAX A 1 C

5

8

2

7

2.4 m

6

RBX B 4 D 3 E 150N 2.4 m 2.4 m

∑MA

=0

150 ( 4.8 ) - RBX (2.4) = 0 720 – 2.4RBX = 0 - RBX = -720/2.4 RBX = 300.00 N

∑Fx = 0 RAx + RBX = 0 RAx + 300 = 0 RAx = -300.00 N

∑Fy = 0 RAY = 200.00 N

13

150N

MEMBER 3 ∑Fx = 0 - FED – FEC =0

MEMBER 7 ∑Fy = 0 -150 + FEC FEC = 212.13 N =0

- FED – 212.13 (0.707) = 0 FED = -150.00 N

14

FCB=0

MEMBER 1 ∑Fx = 0 –FCA + FCE –FCA + (212.13) FCA = 150.00 N + FCB = 0 + 0= 0

MEMBER 2 ∑Fy = 0 –FCD –FCB – FCE –FCD –212.13 FCD = 150.00 N

FBC =0

6

300N

MEMBER 4

MEMBER 5

∑Fx = 0 300 + FBD = 0 FBD = -300.00 N

∑Fy = 0 FBA = 0 FBA = 0

15

MEMBER 8

∑Fy = 0 -300 +150 -FDA - 150 + FDA FDA = 212.13 N

16

7.0 Discussion 1. Compare the experimental and theoretical result.

From table 3, we get different value from experimental force with theoretical force it is because effect value of strain produced. So, it is produced not accurate value experimental force with theoretical force. During experiment, we have one member (6) to contain force. When calculated theoretical force, two members contain force 0N its member no 5 and no 6.

2. From your result and the theoretical member force, identify which members are in Compression and which member is in tension. Explain your choice. Members no. 1, 7 and 8 is a tension cause have positive value. The value is 150N for member no 1 and 212.13 N for members no. 7 and 8. Furthermore member no. 2, 3 and 4 are compression cause having negative its -300.00 N for member no 4 and -150.00 N for member no 2 and 3.

3. Observe the reading of member 5. Explain why the reading is almost zero. When located between two supports member 5 always zero because there is no tension or compression applied.

4. Are the strains gauges are an effective transducers for measurement forces in the framework. Not because the reading is too different from theoretical readings. Fear of not reading the same as a real force and will result in failure.

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

17

8.0 CONCLUSION In conclusion, the stability of the truss depending on the circumstances and the number of support staff. Support conditions will determine whether the truss situation: a. Statically determinate truss b. Statically indeterminate truss Truss said stable if it does not collapse or the forces acting on it are balanced. Plane truss is said to be external, static if its support reactions can be determined by using only the equilibrium equations ΣFx = 0 , ΣFy = 0 , ΣM = 0 . If all the forces in members can be determined by static equilibrium equations, is called statically determinate truss . Truss that satisfies equation m = 2 - 3 are trusses in statically indeterminate. There are several methods used to determine the forces in the truss members: a. Method of joints (joint method) b. Methods section (section method) There are also advantages and disadvantages statically indeterminate truss: • Advantages: Member of a small bears that burden Large truss while still in a stable condition. • Pros: Has a tendency to distribute the load Other excessive burden if overloading occurs . • Inadequate: Redundant member will result in Movement in the truss structure . • Inadequate: No cost .

18

9.0 REFERENCES

1. Influence lines for statically determinate structures http://www.engr.mun.ca/~swamidas/ENGI6705-StructuralAnalysis ClassNotes3.ppt 2. Plane Frame Example – http:// www.amazon.com 3. Trusses - http://www.Icsc.edu 4. Definition of indeterminate truss - http://www.wikipedia.org 5. Modul Mekanik Bahan dan Struktur – Penerbit UTM 6. Influence lines for statically determinate structures http://www.engr.mun.ca/~swamidas/ENGI6705-StructuralAnalysis ClassNotes3.ppt 7. Plane Frame Example – http:// www.amazon.com 8. Trusses - http://www.Icsc.edu 9. Definition of indeterminate truss - http://www.wikipedia.org 10. Material of Civil Engineering Module 11. Discussion with group members

19

1.0 INTRODUCTION

a. A truss is defined as a structure composed of slender element joined together at their end points. b. The joint connection are considered as pinned joint without friction. c. The members commonly used in construction consist of wooden struts, metal bars, angles or channels. In order to determine the force developed in the individual members at a truss, the following assumptions should be make: i. The members are connencted to each other at their ends by friction pins, that is only a force and no moment can be transferred from one member to another. ii. External loads are applied to the truss only at its joints.

1

d. Some consideration can be made in order to idealize the truss : i. The structural members are joined together by pin without friction.

ii. All loadings are applied at the same joints. iii. Each truss member acts as an axial forces member, and therefore the forces acting at the ends of the member must be directed along the axis of the member. e. One of the methods to calculate the force in the member of a truss is using Method of Joint. f. Plane or planar truss composed of members that lie in the same plane and frequently used for bridge and roof support. g. Loads that cause the entire truss to bend are converted into tensile and compressive forces in the members. h. The structures mechanics involves determination of unknown forces on the structures. i. Some of these structures can be completely analyzed by using the equations of equilibrium. ∑Fx = 0 ∑Fy = 0 ∑Má´¢ = 0 j. On the other hand, if there exist extra redundant reaction components, then the structure is said to be statically indeterminate. k. To be in a atate of static equilibrium,a structure must meet the requirements of stability. l. A statically indeterminate structure is a structure that had more unknown forces. m. A truss can be unstable if it is statically determinate or statically indeterminate.

2

The structures are known as statically determinate.

2.0 OBJECTIVE To examine a statically determinate frame and to analyze the frame using simple pin joint theory.

3.0

THEORY A truss is a structure composed of slender member joined together at their end points to form one or more triangles. The joint connections are considered as pinned joint without friction. In order to determine the forces developed in the individual members at a truss, the following assumptions should be make : a. The members are connected to each other at their ends by frictionless pins, that is only a force and no moment can be transferred from one member to another.

b. External loads are applied to the truss only at its joints. One of the methods to calculate the forces in the member of a truss is using Method of Joint.

3

4.0

PROCEDURE 1. The thumbwheel was unscrewed on the “redundant” member. Noted that is effectively no longer part of the structure as the idealized diagram illustrates. 2. The pre-load was applied of 100N downward, the load was re-zero cell and a load was applied of 250N and checked that the frame was stable and secure. 3. The load was returned to zero (leaving the 100N preload), rechecked and the digital indicator was re-zero.

4.

Load was applied in the increment shown in Table 1 recording the strain readings and the digital indicator readings. Table 2 have been completed by subtracting the initial (zero) strain readings.

4

5.0

EQUIPMENT

Cantilever Truss

Digital Display Force

Thumbwheel

Digital Indicator

Vernier Clipper

5

6.0 RESULT

Load (N) 0 50 100 150 200 250

Strain Reading 1 153 162 171 179 188 197 2 213 203 194 186 176 167 3 -23 -32 -41 -50 -59 -68 4 -45 -65 -83 -100 -188 -137 5 119 118 118 117 117 116 6 0 0 0 0 0 0 7 62 76 89 101 113 126 8 83 68 111 123 137 151

Digital Indicator Reading (mm) 0.013 0.019 0.047 0.076 0.116 0.151

Table 1 : Strain Readings and Frame Deflection for Experiment 1

Load (N) 0 50 100 150 200 250

1 0 9 18 26 35 44

2 0 -10 -19 -27 -37 -46

3 0 -9 -18 -27 -36 -45

4 0 -20 -38 -55 -143 -92

5 0 -1 -1 -2 -2 -3

6 0 0 0 0 0 0

7 0 14 27 39 51 64

8 0 -15 28 40 54 68

Table 2 :True Strain Reading for Experiment 1

6

Use : W = 150 N

Member 1 2 3 4 5 6 7 8 Strain reading Experimental Force (N) 159.64 -165.68 -165.68 -337.49 -12.27 0 239.31 245.45 Theoretical Force (N) 150 -150 -150 -300 0 0 212.13 212.13

26 -27 -27 -55 -2 0 39 40

*C = compression, T = Tension Table 3 : Measured and Theoretical Force in the Cantilever Truss

7

7.0 ANALYSIS

Calculation For Experimental Force (N) For Load = 200 N

From the formula:

E = σ/ ε E = Young „s Modulus (Nm-2) ε = Displayed Strain σ = Stress in the member (Nm-2)

σ = F/A F = Force in member (N) A = cross section area of the member (m2)

Rod diameter (d) = 6.10 mm E steel = 2.10 x 105 N/mm2 A = π r2 = π (3.05)2 = 29.22 mm2

8

Member 1: σ=Eε E steel = 2.10 x 105 N/mm2 ε = 26 x 10-6 σ = (2.10 x 105 N/mm2) x (26 x 10-6) = 5.46

F = σA σ = 5.46 A = 29.22 mm2 F = 5.46 x 29.22 F = 159.54 N

Member 2 : σ=Eε E steel = 2.10 x 105 N/mm2 ε = -27 x 10-6 σ = (2.10 x 105 N/mm2) x (-27 x 10-6) = -5.67

F = σA σ = -5.67 A = 29.22 mm2 F = -5.67 x 29.22 F = -165.68 N

9

Member 3 : σ=Eε E steel = 2.10 x 105 N/mm2 ε = -27 x 10-6 σ = (2.10 x 105 N/mm2) x (-27 x 10-6) = -5.67

F = σA σ = -5.67 A = 29.22 mm2 F = -5.67x 29.22 F = -165.68 N

Member 4 : σ=Eε E steel = 2.10 x 105 N/mm2 ε = -55 x 10-6 σ = (2.10 x 105 N/mm2) x (-55 x 10-6) = -11.55

F = σA σ = -11.55 A = 29.22 mm2 F = -11.55 x 29.22 F = -337.49 N

10

Member 5 : σ=Eε E steel = 2.10 x 105 N/mm2 ε = -2 x 10-6 σ = (2.10 x 105 N/mm2) x (-2 x 10-6) = -0.42

F = σA σ = -0.42 A = 29.22 mm2 F = -0.42 x 29.22 F = -12.27 N

Member 7 : σ=Eε E steel = 2.10 x 105 N/mm2 ε = 39 x 10-6 σ = (2.10 x 105 N/mm2) x (39x 10-6) = 8.19

F = σA σ = 8.19 A = 29.22 mm2 F = 8.19x 29.22 F = 239.31 N

11

Member 8 : σ=Eε E steel = 2.10 x 105 N/mm2 ε = 40 x 10-6 σ = (2.10 x 105 N/mm2) x (40 x 10-6) = 8.4

F = σA σ = 8.4 A = 29.22 mm2 F = 8.4 x 29.22 F = 245.45.12 N

12

Calculation For Theoretical Force (N) with Using Method Of Joint

RAY RAX A 1 C

5

8

2

7

2.4 m

6

RBX B 4 D 3 E 150N 2.4 m 2.4 m

∑MA

=0

150 ( 4.8 ) - RBX (2.4) = 0 720 – 2.4RBX = 0 - RBX = -720/2.4 RBX = 300.00 N

∑Fx = 0 RAx + RBX = 0 RAx + 300 = 0 RAx = -300.00 N

∑Fy = 0 RAY = 200.00 N

13

150N

MEMBER 3 ∑Fx = 0 - FED – FEC =0

MEMBER 7 ∑Fy = 0 -150 + FEC FEC = 212.13 N =0

- FED – 212.13 (0.707) = 0 FED = -150.00 N

14

FCB=0

MEMBER 1 ∑Fx = 0 –FCA + FCE –FCA + (212.13) FCA = 150.00 N + FCB = 0 + 0= 0

MEMBER 2 ∑Fy = 0 –FCD –FCB – FCE –FCD –212.13 FCD = 150.00 N

FBC =0

6

300N

MEMBER 4

MEMBER 5

∑Fx = 0 300 + FBD = 0 FBD = -300.00 N

∑Fy = 0 FBA = 0 FBA = 0

15

MEMBER 8

∑Fy = 0 -300 +150 -FDA - 150 + FDA FDA = 212.13 N

16

7.0 Discussion 1. Compare the experimental and theoretical result.

From table 3, we get different value from experimental force with theoretical force it is because effect value of strain produced. So, it is produced not accurate value experimental force with theoretical force. During experiment, we have one member (6) to contain force. When calculated theoretical force, two members contain force 0N its member no 5 and no 6.

2. From your result and the theoretical member force, identify which members are in Compression and which member is in tension. Explain your choice. Members no. 1, 7 and 8 is a tension cause have positive value. The value is 150N for member no 1 and 212.13 N for members no. 7 and 8. Furthermore member no. 2, 3 and 4 are compression cause having negative its -300.00 N for member no 4 and -150.00 N for member no 2 and 3.

3. Observe the reading of member 5. Explain why the reading is almost zero. When located between two supports member 5 always zero because there is no tension or compression applied.

4. Are the strains gauges are an effective transducers for measurement forces in the framework. Not because the reading is too different from theoretical readings. Fear of not reading the same as a real force and will result in failure.

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

17

8.0 CONCLUSION In conclusion, the stability of the truss depending on the circumstances and the number of support staff. Support conditions will determine whether the truss situation: a. Statically determinate truss b. Statically indeterminate truss Truss said stable if it does not collapse or the forces acting on it are balanced. Plane truss is said to be external, static if its support reactions can be determined by using only the equilibrium equations ΣFx = 0 , ΣFy = 0 , ΣM = 0 . If all the forces in members can be determined by static equilibrium equations, is called statically determinate truss . Truss that satisfies equation m = 2 - 3 are trusses in statically indeterminate. There are several methods used to determine the forces in the truss members: a. Method of joints (joint method) b. Methods section (section method) There are also advantages and disadvantages statically indeterminate truss: • Advantages: Member of a small bears that burden Large truss while still in a stable condition. • Pros: Has a tendency to distribute the load Other excessive burden if overloading occurs . • Inadequate: Redundant member will result in Movement in the truss structure . • Inadequate: No cost .

18

9.0 REFERENCES

1. Influence lines for statically determinate structures http://www.engr.mun.ca/~swamidas/ENGI6705-StructuralAnalysis ClassNotes3.ppt 2. Plane Frame Example – http:// www.amazon.com 3. Trusses - http://www.Icsc.edu 4. Definition of indeterminate truss - http://www.wikipedia.org 5. Modul Mekanik Bahan dan Struktur – Penerbit UTM 6. Influence lines for statically determinate structures http://www.engr.mun.ca/~swamidas/ENGI6705-StructuralAnalysis ClassNotes3.ppt 7. Plane Frame Example – http:// www.amazon.com 8. Trusses - http://www.Icsc.edu 9. Definition of indeterminate truss - http://www.wikipedia.org 10. Material of Civil Engineering Module 11. Discussion with group members

19