Two Force Member AB

Three Force Member BC

Begin with free-body diagrams isolating each piece of the frame as shown in the diagrams at the left.

The free-body diagram of the quarter-circular arch shows that it is a two-force member. Thus, the reaction forces at pins A and B must be equal, opposite, and collinear. From geometry, it is known that F_A and F_B act at a 45° incline and

     F_A = - F_B

The free-body diagram of the beam shows that it is a three-force member. Thus, the forces acting on it must be concurrent at point P. Note that the force acting on the beam at point B is equal and opposite the force acting on point B of the arch.

Using geometry, you know that the angle θ of the reaction force at C is a function of the distance x:

     θ = tan^-1 [x / (24 - x ]

Notice that BC is a three force member. Thus, the forces are concurrent, and must all act through point P. This helps solve the problem without using the moment equalibrium equation.

Applying the equilibrium equations to the beam BC gives

     ΣF_x = F_B cos45 - F_C cosθ = 0

     ΣF_y = F_B sin45 - 200 + F_C sinθ = 0

Solving the equations simultaneously for F_B and F_C,

     F_B = 200 cosθ / [sin45 cosθ + cos45 sinθ ]

     F_C = 200 cos45 / [sin45 cosθ + cos45 sinθ ]
          = 200 / [cosθ + sinθ ]

Recall, the problem specifically requires the force magnitude at B and C to be equal. Thus, the value of θ can be obtained when F_B is equal to F_C,

     F_B = F_C ==> cosθ/cos45 = 1

         θ = 45^o

Substitute θ = 45° into the equation for θ to determine the value of x for which F_B is equal to F_C,

     45^o = tan^-1 [x / (24 - x ]

       x = 12 ft

Substitute θ = 45° into F_B or F_C to obtain the value of F_B and F_C,

     F_B = F_C = 200 / (cos45 + sin45)

          = 141.4 lb

Since the arch is a two-force member,

     F_A = F_B = 141.4 lb