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