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Using method of joints, find the forces in the truss members (fig.)
1 Answer
written 8.7 years ago by |
Verification of truss
M=2j−r9=2(6)–30=0
LHS = RHS … Hence, perfect truss.
FBD
To apply method of joints at points A, E, B, F and D. It is essential to find reactions at A and B
∑M=0 (about point A)
−6×2–9×6+Rb×8=0
Rb=6kN∑Fy=0−6−9+Ra+Rb=0Ra=9kN
Solution:
Angle α=tan−1(3/2)=56.31Angle β=tan−1(3/4)=36.86
At joint A
∑Fy=0ACsin56.31+Ra=0AC=−10.82kN∑Fx=0ACcos56.31+AE=0AE=6kN
At joint E
By special Case,
EC=6kN and EF=AE=6kN
At joint B,
∑Fy=0BDsin56.31+Rb=0BD=−7.21kN∑Fx=0−BDcos56.31−BF=0BF=4kN
At joint F
∑Fx=0−EF+BF–FCcos36.86=0FC=−2.5kN∑Fy=0−9+FD+FCsin36.86=0FD=10.5kN
At joint D
∑Fx=0−CD+BDcos56.31=0CD=−4kN