Equilibrium  
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The figure shows the free body diagram of the beam. The reaction forces at C are C subscript x and C subscript y directed horizontally to the right and vertically upwards respectively. A 75 kilo newton load acts on the beam 2.5 meters from point C. The force along the support rod directed away from the point B is F subscript 1 at an angle of 35 degrees with the horizontal.

Begin by considering the equilibrium of beam BC.

Draw the free-body diagram for beam BC. Note that member (1) is a two-force member.

Solve these three equations to find:

∑MC=-(75 kN)(2.5 m)
-(5.0 m) F1 sin(35.0∘)=0
          ∴ F1=-65.38 kN
∑Fx=Cx-F1cos(35.0∘)=0
      ∴Cx=-53.56 kN
∑Fy=Cy-75 kN
-F1sin(35.0∘)=0
      ∴Cy=37.50 kN
The resultant force at bolt C is:
C=(Cx)2+(Cy)2
=65.38 kN