Truss Analysis Calculation

1. 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

2. 2 | P a g e
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

3. 3 | P a g e
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 = FAJcos 𝜃
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 = FAJcos 𝜃
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 = FBHsin 𝜃
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)

4. 4 | P a g e
JOINT H:
JOINT C:
JOINT G:
JOINT D:
𝜃 = 51.34°
FBHx = FBHcos 𝜃
FBHy = FBHsin 𝜃
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 = FDFsin 𝜃
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)

5. 5 | P a g e
JOINT F:
Diagram:
𝜃 = 51.34°
FDFx = FDFcos 𝜃
FDFy = FDFsin 𝜃
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)

6. 6 | P a g e
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

7. 7 | P a g e
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 = FKBsin 𝜃
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 = FKBsin 𝜃
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 = FBHsin 𝜃

8. 8 | P a g e
JOINT H:
JOINT C:
JOINT G:
JOINT D:
𝜃 = 51.34°
FHBx = FHBcos 𝜃
FHBy = FHBsin 𝜃
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 = FHDsin 𝜃
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 = FHDsin 𝜃
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 = FHDsin 𝜃

9. 9 | P a g e
JOINT F:
Diagram:
𝜃 = 51.34°
FDFx = FDFcos 𝜃
FDFy = FDFsin 𝜃
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)

10. 10 | P a g e
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

11. 11 | P a g e
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 = FKBsin 𝜃
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 = FKCcos 𝜃
FJCy = FJCsin 𝜃
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 = FKBsin 𝜃
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)

12. 12 | P a g e
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 = FJCsin 𝜃
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 = FDFsin 𝜃
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)

13. 13 | P a g e
JOINT F:
Diagram:
𝜃 = 51.34°
FDFx = FDFcos 𝜃
FDFy = FDFsin 𝜃
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)

14. 14 | 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 4 | CHAN YI QIN 0315964

15. 15 | P a g e
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 = FAJcos 𝜃
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 = FAJcos 𝜃
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 = FJCsin 𝜃
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

16. 16 | P a g e
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 = FJCsin 𝜃
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 = FGEsin 𝜃
Total Fx = 0
-FGF + 100 = 0
-100 + 100 = 0 (Balance)
Total Fy = 0
FFE = 0

17. 17 | P a g e
Diagram:

18. 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

19. 19 | P a g e
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 = FAJcos 𝜃
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 = FAJcos 𝜃
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 = FAJcos 𝜃
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)

20. 20 | P a g e
JOINT H:
JOINT D:
JOINT G:
JOINT F:
𝜃 = 51.34°
FHBx = FHBcos 𝜃
FHBy = FHBsin 𝜃
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 = FHDsin 𝜃
FHDx = FHDcos 𝜃
FHDy = FHDsin 𝜃
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 = FGEsin 𝜃
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

21. 21 | P a g e
Diagram:

22. 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.