Dead Load Apply Ratio and Proportion R2 R3 = 5 7.5 7.5 R3 = R 5 2 R 3 = 1.5R 2 ∑ 𝑀𝑅1 = 0 1.5 R2 (1.25) + R2 (5) +0.5 (1.25) +0.92 (1.25) = 0.92(8.75) + 0.5(8.75) + 1.84(7.5) + 1(7.5) + 1.84(6.25) + 1(6.25) + 1.84(5) + 1(5) +1.84(3.75) + 1(3.75) + 1.84(2.5) + 1(2.5) +1.84(1.25) + 1(1.25) R2= 5.24KN R3 = 1.5 (R2) = 1.5(5.24) = 7.86KN ∑ FV = 0 + ↑ R1 + R2 + R3 = (0.92×2) + (0.5×2) + (1.84×7) + (1×7) R1 + 5.24 + 7.86 = 22.72 R1= 22.72 - 5.24 – 7.86 R1=9.62 KN At Joint A At Joint C ∑ 𝑀𝐷 = 0 ∑ FV = 0 + ↑ AB sin(21.8)= 0.92 + 0.5 AB=3.82 KN, T ∑ FV = 0 + ↑ ∑ Fh = 0 + → BC + 1 = 9.62 BC= 8.62 KN, C AC=AB cos(21.8) AC=3.82cos(21.8) AC=3.55 KN, C ∑ Fh = 0 + → AC=3.55 KN AC=3.55 KN, C 9.62(1.25) = 0.92(2.5) + 0.5(2.5) + 1.84(2.5) + 1 (1.25) + BD cos(21.8)(0.5) + BD sin(21.8) (1.25) BD= 5.3 KN, C At Joint E At Joint B ∑ FV = 0 + ↑ ∑ FV = 0 + ↑ 8.62 = 1.84 + 3.82 sin(21.8) + 5.3 sin(21.8) + BE sin(21.8) BE= 9.13 KN, T 9.13 sin(21.8)= DE+ 1 DE= 2.39 KN, C ∑ Fh = 0 + → BE cos(21.8) = 3.82cos(21.8) + 5.3 cos(21.8) BE= 9.13 KN, T ∑ Fh = 0 + → EG + 3.55= 9.13 cos(21.8) EG = 4.92 KN, T ∑ 𝑀𝐺 = 0 9.62(2.5) =0.92(3.75) + 0.5(3.75) + 1.84(2.5) + 1 (2.5) + 1.84(1.25) + 1 (1.25) + DF cos(21.8)(1) + DF sin(21.8)(1.25) DF= 5.8 KN, C At Joint D ∑ FV = 0 + ↑ 2.39 + 5.3 sin(21.8) = 1.84 + 5.8 sin(21.8) + DG sin(21.8) DG= 0.74 KN, T ∑ Fh = 0 + → DG cos(21.8) + 5.3 cos(21.8) = 5.8 cos(21.8) DG= 0.74 KN, T At Joint G ∑ FV = 0 + ↑ 0.74 sin(21.8) + FG = 1 FG= 0.73 KN, T ∑ Fh = 0 + → GI = 0.65cos(21.8) +4.45 GI = 5.61 KN, T ∑ 𝑀𝐼 = 0 9.62(3.75) =0.92(5) + 0.5(5) + 1.84(3.75) + 1(3.75) +1.84(2.5) + 1(2.5) + 1.84(1.25) + 1 (1.25) + FH cos(21.8)(1.5) + FH sin(21.8)(1.25) FH= 4.13 KN, C At Joint F At Joint H ∑ Fh = 0 + → 4.13 cos(21.8)=HJ cos(21.8) HJ=4.13 KN, C ∑ FV = 0 + ↑ 5.8 sin(21.8) + FI sin(21.8) = 1.84 + 0.73 + 4.13 sin(21.8) FI = 3.33 KN, C ∑ Fh = 0 + → FI cos(21.8) + 4.13 cos(21.8) = 5.8 cos(21.8) FI= 3.33 KN, C ∑ 𝑀𝐴 = 0 ∑ FV = 0 + ↑ 2 × 4.13 sin(21.8)= 1.84 + HI HI = 1.23 KN, T 5.24(3.75) + 7.86(1.25) = 1.84(3.75) + 1(3.75) + 1.84(2.5) + 1(2.5) + 1.84(1.25) + 1 (1.25) + IJ cos(21.8)(1.5) + IJ sin(21.8)(3.75) IJ= 2.93 KN, C ∑ 𝑀𝐽 = 0 7.86(1.25) = 1.84(1.25) + 1(1.25) + 1(2.5) + 1.84(2.55) + 0.92(3.75) + 0.5(3.75) + HK (1.5) HK= 3.82 KN, T At Joint J At Joint K ∑ Fh = 0 + → 4.57 cos(21.8) = 2.93cos(21.8) + JL cos(21.8) JL= 1.2 KN, C ∑ Fh = 0 + → KM = 3.82 + 0.44cos(21.8) KM = 4.23 KN, T ∑ FV = 0 + ↑ JK + 1.2 sin(21.8) = 1.84 + 4.57 sin(21.8) + 2.93 sin(21.8) JK = 4.02 KN, T ∑ 𝑀𝐴 = 0 7.86(2.5) = 1.84(2.5) +1(2.5) + 1.84(1.25) + 1(1.25) +KL cos(21.8)(1) + KL sin(21.8) (2.5) KL=0.44 KN, C At Joint L ∑ Fh = 0 + → 1.2 cos(21.8) + 0.44 cos(21.8) = LN cos(21.8) LN= 1.64 KN, C ∑ FV = 0 + ↑ 0.44 sin(21.8) + 1.64 sin(21.8)+ LM = 1.84 + 1.2 sin(21.8) LM= 1.51 KN, C At Joint P At Joint C ∑ FV = 0 + ↑ NP sin(21.8)= 0.92 + 0.5 NP=3.82 KN, T ∑ FV = 0 + ↑ ∑ Fh = 0 + → NO + 1 = 7.86 NO= 6.86 KN, C OP=3.82cos(21.8) OP=3.55 KN, C ∑ Fh = 0 + → MO=3.55 KN MO=3.55 KN, C