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Question

A two member truss ABC is shown in the figure The force (in kN) transmitted in member AB is

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Solution


Method I:

AB = 1 m
AC = 0.5 m
BC=12+0.52
=1.25=1.118m
Ax+Cx=0
Ay+Cy=10
(from force equilibrium)
MA=0
Cx×0.5=10×1
or Cx=20kN
and Ax=20kN

Mc=0
FAB×0.5=20×0.5
FAB=20kN
Method II:

Free body diagram of point B,

Horizontal Reaction:
FBCcosθ=FAB.....(i)
Vertical Reaction :
FBCsinθ=10....(ii)
Putting the value of FBC from equation (ii) to equation (i)
10sinθ×cosθ=FAB
FAB=10×cotθ=10×10.5=20kN


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