Building Structures: Roof Truss Analysis Report

BUILDING STRUCTURES (ARC 2523) PROJECT 1: ROOF TRUSS TRUSS ANALYSIS CALCULATION TUTOR: MS ANN SEE PENG NAME

STUDENT ID

ANG WEI YI

0317885

CHAN YI QIN

0315964

JOYCE WEE YI QIN

0319602

RYAN KERRY JEE JIN YING

0318715

TAN WING HOE

0319333

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STEP 1: Analyse the Reaction Force

Roller Joint has one force acting on Y-axis; Pin Joint has two forces acting on both Y-axis and X-axis.

Diagram above assumes the direction of the force for calculation.

Force Equilibrium: Total Moment = 0 150(1) - 50(3) + 100(1.25) - REy(4) = 0 REy = 31.25kN Total Fx = 0 100 + 100 - REx = 0 REx = 200kN Total Fy = 0 -150 - 150 + 50 + REy + RAy = 0 RAy = 218.75kN

Therefore, forces of each joint: 200kN

218.75kN

31.25kN CASE STUDY 1 | RYAN KERRY JEE JIN YING 0318715 2|Page

STEP 2: Analyse the Internal Forces (i) (ii)

Analyse all the joints Assume all the internal forces are tension

JOINT K: Total Fx = 0 FKJ =0

Total Fy = 0 -150 - FKA = 0 FKA = -150kN = 150kN (Compression)

JOINT A:

tan đ?œƒ =

1.25 1

đ?œƒ = 51.34° FAJx = FAJcos đ?œƒ FAJy = FAJsin đ?œƒ

Total Fy = 0 FKA + Ray + FAJy = 0 -150 + 218.75 + FAJy = 0 FAJsin đ?œƒ = -68.75Kn FAJ = -88.04kN = 88.04kN (Compression) Total Fx = 0 FAB + FAJx = 0 FAB - 88.04 (cos51.34°) = 0 FAB = 55kN (Tension)

JOINT J:

đ?œƒ = 51.34° FAJx = FAJcos đ?œƒ FAJy = FAJsin đ?œƒ

Total Fy = 0 -150 - FJB + FAJy = 0 FJB = -150 + 88.04 (sin51.34°) = -81.25kN = 81.25kN (Compression) Total Fx = 0 FKJ + FJH + FAJx = 0 FJH + 88.04 (cos51.34°) = 0 FJH = -54.998kN = 55kN (Compression)

JOINT B: Total Fy = 0 -FJB + FBHy = 0 FBH (sin51.34°) = 81.25 FBH = 104.05kN (Tension) đ?œƒ = 51.34° FBHx = FBHcos đ?œƒ FBHy = FBHsin đ?œƒ

Total Fx = 0 -FAB + FBC + FBHx = 0 -55 + FBC + 104.05 (cos51.34°) = 0 FBC = -10kN = 10kN (Compression) 3|Page

JOINT H: Total Fx = 0 FJH + FHG - FBHx = 0 55 + FHG - 104.05 (cos51.34°) = 0 FHG = 10kN (Tension) đ?œƒ = 51.34° FBHx = FBHcos đ?œƒ FBHy = FBHsin đ?œƒ

Total Fy = 0 -FHC - FBHy = 0 -FHC - 104.05 (sin51.34°) = 0 FHC = -81.25kN = 81.25kN (Compression)

JOINT C: Total Fy = 0 -FHC + FCGy = 0 -81.25 + FCG (sin51.34°) = 0 FCG = 104.05kN (Tension) đ?œƒ = 51.34° FCGx = FCGcos đ?œƒ FCGy = FCGsin đ?œƒ

Total Fx = 0 FBC + FCD + FCGx = 0 10 + FCD + 104.05 (cos51.34°) = 0 FCD = 75kN (Compression)

JOINT G:

đ?œƒ = 51.34° FCGx = FCGcos đ?œƒ FCGy = FCGsin đ?œƒ

Total Fy = 0 50 - FGD - FCGy = 0 FGD = 50 - 104.05 (sin51.34°) = -31.25kN = 31.25kN (Compression) Total Fx = 0 -FHG + FGF - FCGx = 0 -10 + FGF - 104.05 (cos51.34°) = 0 FGF = 75kN (Tension)

JOINT D:

đ?œƒ = 51.34° FDFx = FDFcos đ?œƒ FDFy = FDFsin đ?œƒ

Total Fy = 0 -FGD + FDFy = 0 -31.25 + FDF (sin51.34°) = 0 FDF = 40.02kN (Tension) Total Fx = 0 FCD + FDE + FDFx = 0 75 + FDE + 40.02 (cos51.34°) = 0 FDE = -100kN = 100kN(Compression) 4|Page

JOINT F:

đ?œƒ = 51.34° FDFx = FDFcos đ?œƒ FDFy = FDFsin đ?œƒ

Total Fy = 0 -FFE - FDFy = 0 FFE = -FDFy = -40.02 (sin51.34°) = -31.25kN = 31.25kN (Compression) Total Fx = 0 -FGF + 100 - FDFx = 0 -75 + 100 - 40.02 (cos51.34°) = 0 (Balance)

Diagram:

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STEP 1: Analyse the Reaction Force

Roller Joint has one force acting on Y-axis; Pin Joint has two forces acting on both Y-axis and X-axis.

Diagram above assumes the direction of the force for calculation.

Force Equilibrium: Total Moment = 0 150(1) - 50(3) + 100(1.25) - REy(4) = 0 REy = 31.25kN Total Fx = 0 100 + 100 - REx = 0 REx = 200kN Total Fy = 0 -150 - 150 + 50 + REy + RAy = 0 RAy = 218.75kN

Therefore, forces of each joint: 200kN

218.75kN

31.25kN CASE STUDY 2 | JOYCE WEE YI QIN 0319602 6|Page

STEP 2: Analyse the Internal Forces (i) (ii)

Analyse all the joints Assume all the internal forces are tension

JOINT A: Total Fx = 0 FAB = 0

Total Fy = 0 FKA + 218.75 = 0 FKA = -218.75kN = 218.75kN (Compression)

JOINT K:

tan đ?œƒ =

1.25 1

đ?œƒ = 51.34° FKBx = FKBcos đ?œƒ FKBy = FKBsin đ?œƒ

Total Fy = 0 -150 + FKA - FKBy = 0 FKBy = -150 + FKA FKB (sin51.34°) = -150 + 218.75 FKB = 88.04kN (Tension) Total Fx = 0 FKJ + FKBx = 0 FKJ = -88.04 (cos51.34°) = -55kN = 55kN (Compression)

JOINT J: Total Fy = 0 -150 - FJB = 0 FJB = -150kN = 150kN (Compression)

Total Fx = 0 FKJ + FJH = 0 FJH = -55kN = 55kN (Compression)

JOINT B:

đ?œƒ = 51.34° FKBx = FKBcos đ?œƒ FKBy = FKBsin đ?œƒ đ?œƒ = 51.34° FBHx = FBHcos đ?œƒ FBHy = FBHsin đ?œƒ

Total Fy = 0 -FJB + FKBy + FBHy = 0 -150 + 88.04 (sin51.34°) + FBH (sin51.34°) = 0 FBH = 104.05kN (Tension) Total Fx = 0 FBC - FKBx + FBHx = 0 FBC - 88.04 (cos51.34°) + 104.05 (cos51.34°) = 0 FBC = -10kN = 10kN (Compression)

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JOINT H:

đ?œƒ = 51.34° FHBx = FHBcos đ?œƒ FHBy = FHBsin đ?œƒ đ?œƒ = 51.34° FHDx = FHDcos đ?œƒ FHDy = FHDsin đ?œƒ

Total Fy = 0 -FHC - FHBy - FHDy = 0 -104.05 (sin51.34°) = FHD (cos51.34°) FHD = -104.05kN = 104.05kN (Compression) Total Fx = 0 FJH + FHG - FHBx - FHDx = 0 55 + FHG - 104.05 (cos51.34°) - 104.05 (cos51.34°) = 0 FHG = 75kN (Tension)

JOINT C: Total Fy = 0 FHC = 0

Total Fx = 0 FBC + FCD = 0 FCD = -10 = 10kN (Compression)

JOINT G: Total Fy = 0 50 + (-FGD) = 0 FGD = 50kN (Tension)

Total Fx = 0 -FHG + FGF = 0 FGF = FHG = 75kN (Tension)

JOINT D:

đ?œƒ = 51.34° FHDx = FHDcos đ?œƒ FHDy = FHDsin đ?œƒ đ?œƒ = 51.34° FDFx = FHDcos đ?œƒ FDFy = FHDsin đ?œƒ

Total Fy = 0 FGD - FHDy + FDFy = 0 50 - 104.05 (sin51.34°) + FDF (sin51.34°) = 0 FDF = 40kN (Tension) Total Fx = 0 FCD + FDE + FHDx + FDFx = 0 10 + FDE + 104.05 (cos51.34°) + 40 (cos51.34°) = 0 FDE = -100kN = 100kN (Compression)

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JOINT F:

đ?œƒ = 51.34° FDFx = FDFcos đ?œƒ FDFy = FDFsin đ?œƒ

Total Fy = 0 -FFE - FDFy = 0 FFE = -FDFy = -40 (sin51.34°) = -31.23kN = 31.23kN (Compression) Total Fx = 0 -FGF + 100 - FDFx = 0 -75 + 100 - 40 (cos51.34°) = 0 (Balance)

Diagram:

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STEP 1: Analyse the Reaction Force

Roller Joint has one force acting on Y-axis; Pin Joint has two forces acting on both Y-axis and X-axis.

Diagram above assumes the direction of the force for calculation.

Force Equilibrium: Total Moment = 0 150(1) - 50(3) + 100(1.25) - REy(4) = 0 REy = 31.25kN Total Fx = 0 100 + 100 - REx = 0 REx = 200kN Total Fy = 0 -150 - 150 + 50 + REy + RAy = 0 RAy = 218.75kN

Therefore, forces of each joint: 200kN

218.75kN

31.25kN CASE STUDY 3 | ANG WEI YI 0317885 10 | P a g e

STEP 2: Analyse the Internal Forces (i) (ii)

Analyse all the joints Assume all the internal forces are tension

JOINT A: Total Fx = 0 FAB = 0

JOINT K:

Total Fy = 0 FKA + 218.75 = 0 FKA = -218.75kN = 218.75kN (Compression) Total Fy = 0 -150 + FKA - FKBy = 0 FKBy = -150 + FKA FKB (sin51.34°) = -150 + 218.75 FKB = 88.04kN (Tension)

đ?œƒ = 51.34° FKBx = FKBcos đ?œƒ FKBy = FKBsin đ?œƒ

Total Fx = 0 FKJ + FKBx = 0 FKJ = -88.04 (cos51.34°) = -55kN = 55kN (Compression)

JOINT J: Total Fx = 0 FKJ + FJH - FJCx = 0 55 + FJH - 104.05 (cos51.34°) = 0 FJH = 10kN (Tension) đ?œƒ = 51.34° FJCx = FKCcos đ?œƒ FJCy = FJCsin đ?œƒ

Total Fy = 0 -150 + FJB - FJCy = 0 FJC (sin51.34°) = -150 + 68.75 FJC = -104.05kN = 104.05kN (Tension)

JOINT B: Total Fy = 0 FJB + FKBy = 0 FJB = -88.04 (sin51.34°) = -68.75kN = 68.75kN (Compression) đ?œƒ = 51.34° FKBx = FKBcos đ?œƒ FKBy = FKBsin đ?œƒ

Total Fx = 0 -FAB + FBC - FKBx = 0 FBC = FKB (cos51.34°) = 88.04 (cos51.34°) = 55kN (Tension)

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JOINT H: Total Fy = 0 FHC = 0

Total Fx = 0 -FJH + FHG = 0 FHG = 10kN (Tension)

JOINT C:

đ?œƒ = 51.34° FJCx = FJCcos đ?œƒ FJCy = FJCsin đ?œƒ đ?œƒ = 51.34° FCGx = FCGcos đ?œƒ FCGy = FCGsin đ?œƒ

Total Fy = 0 FHC - FJCy + FCGy = 0 FCGy = FJCy FCG (sin51.34°) = 104.05 (sin51.34°) FCG = 104.05kN (Tension) Total Fx = 0 -FBC + FCD + FJCx + FCGx = 0 -55 + FCD + 104.05 (cos51.34°) + 104.05 (cos51.34°) = 0 FCD = -75kN = 75kN (Compression)

JOINT G: Total Fy = 0 50 - FGD - FCGy = 0 FGD = 50 - 104.05 (sin51.34°) = -31.25kN = 31.25kN (Compression) đ?œƒ = 51.34° FCGx = FCGcos đ?œƒ FCGy = FCGsin đ?œƒ

Total Fx = 0 -FHG + FGF - FCGx = 0 -10 + FGF - 104.05 (cos51.34°) = 0 FGF = 75kN (Tension)

JOINT D:

đ?œƒ = 51.34° FDFx = FDFcos đ?œƒ FDFy = FDFsin đ?œƒ

Total Fy = 0 -FGD + FDFy = 0 -31.25 + FDF (sin51.34°) = 0 FDF = 40.02kN (Tension) Total Fx = 0 FCD + FDE + FDFx = 0 75 + FDE + 40.02 (cos51.34°) = 0 FDE = -100kN = 100kN(Compression)

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JOINT F:

đ?œƒ = 51.34° FDFx = FDFcos đ?œƒ FDFy = FDFsin đ?œƒ

Total Fy = 0 -FFE - FDFy = 0 FFE = -FDFy = -40.02 (sin51.34°) = -31.25kN = 31.25kN (Compression) Total Fx = 0 -FGF + 100 - FDFx = 0 -75 + 100 - 40.02 (cos51.34°) = 0 (Balance)

Diagram:

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STEP 1: Analyse the Reaction Force

Pin Joint has two forces acting on both Y-axis and X-axis; Roller Joint has one force acting on Y-axis only.

Diagram above assumes the direction of the force for calculation.

Force Equilibrium: Total Moment = 0 150(1) - 50(3) + 100(1.25) - REy(4) = 0 REy = 31.25kN Total Fx = 0 100 + 100 - RAx = 0 RAx = 200kN Total Fy = 0 -150 - 150 + 50 + RAy + REy = 0 RAy = 218.75kN

Therefore, forces of each joint: 200kN

31.25kN

218.75kN CASE STUDY 4 | CHAN YI QIN 0315964 14 | P a g e

STEP 2: Analyse the Internal Forces (i) (ii)

Analyse all the joints Assume all the internal forces are tension

JOINT K: Total Fx = 0 FKJ =0

Total Fy = 0 -150 - FKA = 0 FKA = -150kN = 150kN (Compression)

JOINT A:

tan đ?œƒ =

1.25 1

đ?œƒ = 51.34° FAJx = FAJcos đ?œƒ FAJy = FAJsin đ?œƒ

Total Fy = 0 -FKA + 218.75 + FAJy = 0 -150 + 218.75 + FAJ (sin51.34°) = 0 FAJ = -88.04kN = 88.04kN (Compression) Total Fx = 0 -200 + FAB - FAJx = 0 FAB = 200 + FAJx = 200 + 88.04 (cos51.34°) = 255kN (Tension)

JOINT B: Total Fy = 0 FJB = 0

Total Fx = 0 -FAB + FBC = 0 FBC = 255kN (Tension)

JOINT J:

đ?œƒ = 51.34° FAJx = FAJcos đ?œƒ FAJy = FAJsin đ?œƒ đ?œƒ = 51.34° FJCx = FJCcos đ?œƒ FJCy = FJCsin đ?œƒ

Total Fy = 0 -150 + FAJy -FJCy = 0 FJCy = -150 + FAJy FJC (sin51.34°) = -150 + 88.04 (cos51.34°) FJC = -104.05kN = 104.05kN (Compression) Total Fx = 0 FJH + FAJx - FJCx = 0 FJH + 88.04 (cos51.34°) - 104.05 (cos51.34°) = 0 FJH = 10kN (Tension)

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JOINT H: Total Fy = 0 FHC = 0

Total Fx = 0 -FJH + FHG = 0 FHG = FJH = 10kN (Tension)

JOINT C:

đ?œƒ = 51.34° FJCx = FJCcos đ?œƒ FJCy = FJCsin đ?œƒ

đ?œƒ = 51.34° FCGx = FCGcos đ?œƒ FCGy = FCGsin đ?œƒ

Total Fy = 0 FHC - FJCy + FCG (sin51.34°) = 0 -104.05 (sin51.34°) + FCG (sin51.34°) = 0 FCG = 104.05kN (Tension) Total Fx = 0 -FBC + FCD + FJCx + FCGx = 0 -255 + FCD + 104.05 (cos51.34°) + 104.05 (cos51.34°) = 0 FCD = 125Kn (Tension)

JOINT G:

đ?œƒ = 51.34° FCGx = FCGcos đ?œƒ FCGy = FCGsin đ?œƒ đ?œƒ = 51.34° FGEx = FGEcos đ?œƒ FGEy = FGEsin đ?œƒ

Total Fy = 0 50 - FGD - FCGy - FGEy = 0 50 - 104.05 (sin51.34°) - FGE (sin51.34°) = 0 FGE = -40kN = 40kN (Compression) Total Fx = 0 FHG + FGF - FCGx + FGEx = 0 120 + FGF - 104.05 (cos51.34°) + 40 (cos51.34°) = 0 FGF = -80kN = 80kN (Compression)

JOINT F: Total Fy = 0 FFE = 0

Total Fx = 0 -FGF + 100 = 0 -100 + 100 = 0 (Balance)

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Diagram:

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STEP 1: Analyse the Reaction Force

Pin Joint has two forces acting on both Y-axis and X-axis; Roller Joint has one force acting on Y-axis only.

Diagram above assumes the direction of the force for calculation.

Force Equilibrium: Total Moment = 0 150(1) - 50(3) + 100(1.25) - REy(4) = 0 REy = 31.25kN Total Fx = 0 100 + 100 - RAx = 0 RAx = 200kN Total Fy = 0 -150 - 150 + 50 + RAy + REy = 0 RAy = 218.75kN

Therefore, forces of each joint: 200kN

31.25kN

218.75kN CASE STUDY 5 | TAN WING HOE 0319333 18 | P a g e

STEP 2: Analyse the Internal Forces (i) (ii)

Analyse all the joints Assume all the internal forces are tension

JOINT K: Total Fx = 0 FKJ =0

Total Fy = 0 -150 - FKA = 0 FKA = -150kN = 150kN (Compression)

JOINT A:

tan đ?œƒ =

1.25 1

đ?œƒ = 51.34° FAJx = FAJcos đ?œƒ FAJy = FAJsin đ?œƒ JOINT J:

Total Fy = 0 -FKA + 218.75 + FAJy = 0 -150 + 218.75 + FAJ (sin51.34°) = 0 FAJ = -88.04kN = 88.04kN (Compression) Total Fx = 0 -200 + FAB - FAJx = 0 FAB = 200 + FAJx = 200 + 88.04 (cos51.34°) = 255kN (Tension) Total Fy = 0 -150 - FJB + FAJy = 0 FJB = -150 + 88.04 (sin51.34°) = -81.25kN = 81.25kN (Compression)

đ?œƒ = 51.34° FAJx = FAJcos đ?œƒ FAJy = FAJsin đ?œƒ

Total Fx = 0 FJH + FAJx = 0 FJH = -FAJx = - 88.04 (cos51.34°) = -55kN = 55kN (Compression)

JOINT B: Total Fy = 0 -FJB + FBHy = 0 FBH (sin51.34°) = 81.25 FBH = 104.05kN (Tension) FAJx = FAJcos đ?œƒ FAJy = FAJsin đ?œƒ

Total Fx = 0 -FAB + FBC + FBHx = 0 FBC = FAB - FBHx = 255 - 104.05 (cos51.34°) = 190kN (Tension)

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JOINT H:

đ?œƒ = 51.34° FHBx = FHBcos đ?œƒ FHBy = FHBsin đ?œƒ đ?œƒ = 51.34° FHDx = FHDcos đ?œƒ FHDy = FHDsin đ?œƒ

Total Fy = 0 -FHC - FHBy - FHDy = 0 FHDy = -FHBy FHD (sin51.34°) = -104.05 (sin51.34°) FHD = -104.05 = 104.05 (Compression) Total Fx = 0 FJH + FHG - FHBx + FHDx = 0 55 + FHG - 104.05 (cos51.34°) - 104.05 (cos51.34°) = 0 FHG = 75kN (Tension)

JOINT D: Total Fy = 0 FGD - FHDy = 0 FGD = FHDy = 104.05 (sin51.34°) = 81.25kN (Tension) FHDx = FHDcos đ?œƒ FHDy = FHDsin đ?œƒ

Total Fx = 0 -FCD + FDE + FHDx = 0 -190 + FDE + 104.05 (cos51.34°) = 0 FDE = 125kN (Tension)

JOINT G:

FGEx = FGEcos đ?œƒ FGEy = FGEsin đ?œƒ

Total Fy = 0 50 - FGD - FGEy = 0 FGEy = 50 - FGD FGE (sin51.34°) = 50 - 81.25 FGE = -40kN = 40kN (Compression) Total Fx = 0 -FHG + FGF - FGEx = 0 -75 + FGF - 40 (cos51.34°) = 0 FGF = 100kN (Tension)

JOINT F: Total Fy = 0 FFE = 0

Total Fx = 0 -FGF + 100 = 0 -100 + 100 = 0 (Balance)

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Diagram:

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Conclusion

Case study 1 is chosen to be the most efficient truss among all 5 trusses, it has the most effective truss arrangement for the load system.

Reasons: 1.

It has only one zero force acting on the horizontal truss KJ, hence, with that, the other trusses are still withstanding their forces to uphold the load.

2. Forces are distributed evenly among all members in the truss. Therefore, case study 1 is more stable as compared to other case studies having more than one zero forces, whereas some members have to withstand extremely heavy load which will weaken the entire load system.

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