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Example 1:Truss analysis using method of joint


Determine the force in each member of the roof truss shown in the photo. The dimensions and loadings are shown in Fig. 1. State whether the members are in tension or compression.



Figure 1





To solve this problem quickly we should start by joint A because it has two unknows only, after that we can check other joints one by one. here this truss is symmetric so no need to calculate forces for both sides. one side will be enough.

For joint A
Fy=0
AG*sin(30)+4=0,    AG=-8 kn(C)
∑Fx=0
AG*cos(30)+AB=0,   AB=6.92Kn(T)

For joint G
Fx=0
-AG*cos(30)+GF*cos(30)+GB*cos*30=0,    8*cos(30)+0.5*GF+0.5*GB=0, GB+GF=-8,  GB=-8-GF
∑Fy=0
-3+GF*sin(30)-GB*sin(30)-AG*sin(30)=0,
-3+GF*sin(30)-GB*sin(30)+8*sin(30)=0,
0.5GF-0.5GB=-1, GF-GB=-2

GF-(-8-GF)=-2, GF=-5(C), GB=-3(C)




For joint B
Fx=0
-AB+BC-GB*cos(30)+BF*cos(60)=0, 
  -(6.92)+BC-(-3)*0.866+0.5*BF=0
BC=-0.5*BF+4.322
∑Fy=0
GB*sin(30)+BF*sin(60)=0,
0.5GB+0.866BF=0,
BF=1.732 kn(T)
BC=-0.5*BF+4.322=3.462kn(T)


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