1 | P a g e

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:

218.75kN 31.25kN

200kN

CASE STUDY 1 | RYAN KERRY JEE JIN YING 0318715

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STEP 2: Analyse the Internal Forces

(i) Analyse all the joints (ii) Assume all the internal forces are tension

JOINT K:

JOINT A:

JOINT J:

JOINT B:

tan 𝜃 =1.25

1

𝜃 = 51.34°

FAJx = FAJcos𝜃

FAJy = FAJsin 𝜃

Total Fx = 0

FAB + FAJx = 0

FAB - 88.04 (cos51.34°) = 0

FAB = 55kN (Tension)

Total Fy = 0

FKA + Ray + FAJy = 0

-150 + 218.75 + FAJy = 0

FAJsin 𝜃 = -68.75Kn

FAJ = -88.04kN

= 88.04kN (Compression)

Total Fy = 0

-150 - FKA = 0

FKA = -150kN

= 150kN (Compression)

Total Fx = 0

FKJ =0

𝜃 = 51.34°

FAJx = FAJcos𝜃

FAJy = FAJsin 𝜃 Total Fx = 0

FKJ + FJH + FAJx = 0

FJH + 88.04 (cos51.34°) = 0

FJH = -54.998kN

= 55kN (Compression)

Total Fy = 0

-150 - FJB + FAJy = 0

FJB = -150 + 88.04 (sin51.34°)

= -81.25kN

= 81.25kN (Compression)

𝜃 = 51.34°

FBHx = FBHcos 𝜃

FBHy = FBHsin𝜃

Total Fy = 0

-FJB + FBHy = 0

FBH (sin51.34°) = 81.25

FBH = 104.05kN (Tension)

Total Fx = 0

-FAB + FBC + FBHx = 0

-55 + FBC + 104.05 (cos51.34°) = 0

FBC = -10kN

= 10kN (Compression)

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JOINT H:

JOINT C:

JOINT G:

JOINT D:

𝜃 = 51.34°

FBHx = FBHcos 𝜃

FBHy = FBHsin𝜃

Total Fx = 0

FJH + FHG - FBHx = 0

55 + FHG - 104.05 (cos51.34°) = 0

FHG = 10kN (Tension)

Total Fy = 0

-FHC - FBHy = 0

-FHC - 104.05 (sin51.34°) = 0

FHC = -81.25kN

= 81.25kN (Compression)

𝜃 = 51.34°

FCGx = FCGcos 𝜃

FCGy = FCGsin 𝜃

Total Fy = 0

-FHC + FCGy = 0

-81.25 + FCG (sin51.34°) = 0

FCG = 104.05kN (Tension)

Total Fx = 0

FBC + FCD + FCGx = 0

10 + FCD + 104.05 (cos51.34°) = 0

FCD = 75kN (Compression)

𝜃 = 51.34°

FCGx = FCGcos 𝜃

FCGy = FCGsin 𝜃

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)

𝜃 = 51.34°

FDFx = FDFcos 𝜃

FDFy = FDFsin𝜃

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:

Diagram:

𝜃 = 51.34°

FDFx = FDFcos 𝜃

FDFy = FDFsin𝜃

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)

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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:

218.75kN 31.25kN

200kN

CASE STUDY 2 | JOYCE WEE YI QIN 0319602

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STEP 2: Analyse the Internal Forces

(i) Analyse all the joints (ii) Assume all the internal forces are tension

JOINT A:

JOINT K:

JOINT J:

JOINT B:

tan 𝜃 =1.25

1

𝜃 = 51.34°

FKBx = FKBcos 𝜃

FKBy = FKBsin𝜃

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)

Total Fy = 0

FKA + 218.75 = 0

FKA = -218.75kN

= 218.75kN (Compression)

Total Fx = 0

FAB = 0

𝜃 = 51.34°

FKBx = FKBcos 𝜃

FKBy = FKBsin𝜃

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)

Total Fx = 0

FKJ + FJH = 0

FJH = -55kN

= 55kN (Compression)

Total Fy = 0

-150 - FJB = 0

FJB = -150kN

= 150kN (Compression)

𝜃 = 51.34°

FBHx = FBHcos 𝜃

FBHy = FBHsin𝜃

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JOINT H:

JOINT C:

JOINT G:

JOINT D:

𝜃 = 51.34°

FHBx = FHBcos 𝜃

FHBy = FHBsin𝜃

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)

𝜃 = 51.34°

FHDx = FHDcos 𝜃

FHDy = FHDsin𝜃

Total Fx = 0

FBC + FCD = 0

FCD = -10

= 10kN (Compression)

Total Fy = 0

FHC = 0

Total Fx = 0

-FHG + FGF = 0

FGF = FHG

= 75kN (Tension)

Total Fy = 0

50 + (-FGD) = 0

FGD = 50kN (Tension)

𝜃 = 51.34°

FHDx = FHDcos 𝜃

FHDy = FHDsin𝜃

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)

𝜃 = 51.34°

FDFx = FHDcos 𝜃

FDFy = FHDsin𝜃

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JOINT F:

Diagram:

𝜃 = 51.34°

FDFx = FDFcos 𝜃

FDFy = FDFsin𝜃

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)

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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:

218.75kN 31.25kN

200kN

CASE STUDY 3 | ANG WEI YI 0317885

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STEP 2: Analyse the Internal Forces

(i) Analyse all the joints (ii) Assume all the internal forces are tension

JOINT A:

JOINT K:

JOINT J:

JOINT B:

𝜃 = 51.34°

FKBx = FKBcos 𝜃

FKBy = FKBsin𝜃

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)

Total Fy = 0

FKA + 218.75 = 0

FKA = -218.75kN

= 218.75kN (Compression)

Total Fx = 0

FAB = 0

𝜃 = 51.34°

FJCx = FKCcos𝜃

FJCy = FJCsin𝜃

Total Fx = 0

FKJ + FJH - FJCx = 0

55 + FJH - 104.05 (cos51.34°) = 0

FJH = 10kN (Tension)

Total Fy = 0

-150 + FJB - FJCy = 0

FJC (sin51.34°) = -150 + 68.75

FJC = -104.05kN

= 104.05kN (Tension)

𝜃 = 51.34°

FKBx = FKBcos 𝜃

FKBy = FKBsin𝜃

Total Fy = 0

FJB + FKBy = 0

FJB = -88.04 (sin51.34°)

= -68.75kN

= 68.75kN (Compression)

Total Fx = 0

-FAB + FBC - FKBx = 0

FBC = FKB (cos51.34°)

= 88.04 (cos51.34°)

= 55kN (Tension)

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JOINT H:

JOINT C:

JOINT G:

JOINT D:

Total Fx = 0

-FJH + FHG = 0

FHG = 10kN (Tension)

Total Fy = 0

FHC = 0

𝜃 = 51.34°

FJCx = FJCcos 𝜃

FJCy = FJCsin𝜃

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)

𝜃 = 51.34°

FCGx = FCGcos 𝜃

FCGy = FCGsin 𝜃

𝜃 = 51.34°

FCGx = FCGcos 𝜃

FCGy = FCGsin 𝜃

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)

𝜃 = 51.34°

FDFx = FDFcos 𝜃

FDFy = FDFsin𝜃

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:

Diagram:

𝜃 = 51.34°

FDFx = FDFcos 𝜃

FDFy = FDFsin𝜃

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)

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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:

31.25kN

200kN

218.75kN

CASE STUDY 4 | CHAN YI QIN 0315964

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STEP 2: Analyse the Internal Forces

(i) Analyse all the joints (ii) Assume all the internal forces are tension

JOINT K:

JOINT A:

JOINT B:

JOINT J:

tan 𝜃 =1.25

1

𝜃 = 51.34°

FAJx = FAJcos𝜃

FAJy = FAJsin 𝜃

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)

𝜃 = 51.34°

FAJx = FAJcos𝜃

FAJy = FAJsin 𝜃

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)

𝜃 = 51.34°

FJCx = FJCcos 𝜃

FJCy = FJCsin𝜃

Total Fy = 0

-150 - FKA = 0

FKA = -150kN

= 150kN (Compression)

Total Fx = 0

FKJ =0

Total Fx = 0

-FAB + FBC = 0

FBC = 255kN (Tension)

Total Fy = 0

FJB = 0

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JOINT H:

JOINT C:

JOINT G:

JOINT F:

Total Fx = 0

-FJH + FHG = 0

FHG = FJH

= 10kN (Tension)

Total Fy = 0

FHC = 0

𝜃 = 51.34°

FJCx = FJCcos 𝜃

FJCy = FJCsin𝜃

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)

𝜃 = 51.34°

FCGx = FCGcos 𝜃

FCGy = FCGsin 𝜃

𝜃 = 51.34°

FCGx = FCGcos 𝜃

FCGy = FCGsin 𝜃

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) 𝜃 = 51.34°

FGEx = FGEcos 𝜃

FGEy = FGEsin𝜃

Total Fx = 0

-FGF + 100 = 0

-100 + 100 = 0 (Balance)

Total Fy = 0

FFE = 0

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Diagram:

18 | P a g e

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:

31.25kN

200kN

218.75kN

CASE STUDY 5 | TAN WING HOE 0319333

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STEP 2: Analyse the Internal Forces

(i) Analyse all the joints (ii) Assume all the internal forces are tension

JOINT K:

JOINT A:

JOINT J:

JOINT B:

tan 𝜃 =1.25

1

𝜃 = 51.34°

FAJx = FAJcos𝜃

FAJy = FAJsin 𝜃

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)

𝜃 = 51.34°

FAJx = FAJcos𝜃

FAJy = FAJsin 𝜃

Total Fy = 0

-150 - FJB + FAJy = 0

FJB = -150 + 88.04 (sin51.34°)

= -81.25kN

= 81.25kN (Compression)

Total Fx = 0

FJH + FAJx = 0

FJH = -FAJx

= - 88.04 (cos51.34°)

= -55kN

= 55kN (Compression)

Total Fy = 0

-150 - FKA = 0

FKA = -150kN

= 150kN (Compression)

Total Fx = 0

FKJ =0

FAJx = FAJcos𝜃

FAJy = FAJsin 𝜃

Total Fy = 0

-FJB + FBHy = 0

FBH (sin51.34°) = 81.25

FBH = 104.05kN (Tension)

Total Fx = 0

-FAB + FBC + FBHx = 0

FBC = FAB - FBHx

= 255 - 104.05 (cos51.34°)

= 190kN (Tension)

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JOINT H:

JOINT D:

JOINT G:

JOINT F:

𝜃 = 51.34°

FHBx = FHBcos 𝜃

FHBy = FHBsin𝜃

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) 𝜃 = 51.34°

FHDx = FHDcos 𝜃

FHDy = FHDsin𝜃

FHDx = FHDcos 𝜃

FHDy = FHDsin𝜃

Total Fy = 0

FGD - FHDy = 0

FGD = FHDy

= 104.05 (sin51.34°)

= 81.25kN (Tension)

Total Fx = 0

-FCD + FDE + FHDx = 0

-190 + FDE + 104.05 (cos51.34°) = 0

FDE = 125kN (Tension)

FGEx = FGEcos 𝜃

FGEy = FGEsin𝜃

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)

Total Fx = 0

-FGF + 100 = 0

-100 + 100 = 0 (Balance)

Total Fy = 0

FFE = 0

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Diagram:

22 | P a g e

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.