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Using method of joints, find the forces in the truss members (fig.)

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Verification of truss

M=2jr9=2(6)30=0

LHS = RHS … Hence, perfect truss.

FBD

enter image description here

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)

6×29×6+Rb×8=0

Rb=6kNFy=069+Ra+Rb=0Ra=9kN

Solution:

Angle α=tan1(3/2)=56.31Angle β=tan1(3/4)=36.86

At joint A

Fy=0ACsin56.31+Ra=0AC=10.82kNFx=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=7.21kNFx=0BDcos56.31BF=0BF=4kN

At joint F

Fx=0EF+BFFCcos36.86=0FC=2.5kNFy=09+FD+FCsin36.86=0FD=10.5kN

At joint D

Fx=0CD+BDcos56.31=0CD=4kN

enter image description here

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