For the frame shown, solve for the reactions and pin forces at A, C, and F. On the frame, draw the moment diagram for each of the members AB, BC, CD, and DF, on the tension side of the members. Solution Va and Vf are vertical reactions. Ha and Hf are horizontal reactions. Va + Vf = 0 & Ha + Hf = -36 M@A = 36 x 6 - Hf x 10 - Vf x 12 = 0...........eqn (1) as internal hinge, M@C = 0 36 x 6- Hf x 22 -Vf x8 = 0..................eqn(2) solving eqn 1 and 2, we get, Vf = 3Hf............put in (1) we get, Hf = 4.69 k Vf = 14.08 k...........put in equilibrium equations we get, Va = -14.08 k & Ha = -40.49 k now, the load is only on member DF therefore all other fixed end moments will be zero except member DF FDF = Wb2a/L2 = 42.84 k.ft FFD = Wba2/L2 = 114.24 k.ft distribution factors for pt D are 0.73 and 0.27 applying moment distribution method, hinges to be released. AB=0, BA=0, BC= 0 CB= 0 CD = -7.81 k.ft DC= 6.67 k.ft DF= -6.67 k.ft FD = 0 diagram to be drawn accordingly.

1. For the frame shown, solve for the reactions and pin forces at A, C, and F. On the frame, draw
the moment diagram for each of the members AB, BC, CD, and DF, on the tension side of the
members.
Solution
Va and Vf are vertical reactions.
Ha and Hf are horizontal reactions.
Va + Vf = 0 & Ha + Hf = -36
M@A = 36 x 6 - Hf x 10 - Vf x 12 = 0...........eqn (1)
as internal hinge,
M@C = 0
36 x 6- Hf x 22 -Vf x8 = 0..................eqn(2)
solving eqn 1 and 2, we get,
Vf = 3Hf............put in (1)
we get, Hf = 4.69 k
Vf = 14.08 k...........put in equilibrium equations
we get,
Va = -14.08 k & Ha = -40.49 k
now, the load is only on member DF therefore all other fixed end moments will be zero except
member DF
FDF = Wb2a/L2 = 42.84 k.ft
FFD = Wba2/L2 = 114.24 k.ft
distribution factors for pt D are 0.73 and 0.27
applying moment distribution method,
hinges to be released.
AB=0, BA=0, BC= 0 CB= 0
CD = -7.81 k.ft DC= 6.67 k.ft DF= -6.67 k.ft FD = 0
diagram to be drawn accordingly