Question 9.1.3: The frame in Figure 1 is subjected to the force and moment s...

Goal Find the loads acting on the frame at the supports A and C.
Given Dimensions and support conditions for the frame as well as magnitudes and locations of loads.
Assume The system is planar because the 100-lb force acts in the plane of the frame and the moment acts about an axis perpendicular to the frame.
Draw We draw free-body diagrams for the two members, making sure that forces at B on member AB are equal and opposite to the forces at B on member BC (Figure 2 and Figure 3).1
Formulate Equations and Solve by taking the frame apart and analyzing each member we have six equations with which to find six unknowns (some of which are the loads at A and C).

¹In Figure 3 the two forces F_{Bx}  and F_{By} are drawn in the opposite direction from Figure 2. These are the halves of the internal force pairs at B.

For member AB (Figure 2):

\sum{M_{z @ B} }\left(\curvearrowleft + \right) =- F_{Ay}(6  ft) + (100  lb)(3  ft)= 0
F_{Ay}=100  lb\left\lgroup\frac{3  ft}{6  ft} \right\rgroup \Rightarrow F_{Ay}=50.0  lb                (1)
\sum{F_{x} \left(\rightarrow + \right) } =F_{Ax}- F_{Bx}= 0\Rightarrow F_{Ax}=F_{Bx}                (2)
\sum{F_{y}\left(\uparrow + \right) } =F_{Ay}- F_{By}- 100   lb = 0

Substituting for F_{Ay} from (1) gives

F_{By} =50.0  lb                (3)

For member BC (Figure 3), with geometry indicated in Figure 4:

\sum{M_{z @ C} }\left(\curvearrowleft + \right) =150  lb.ft + F_{By}(2  ft)- F_{Bx} (3.46  ft)=0

We substitute for F_{By}  from (3) to get

F_{Bx}(3.46  ft) = 150  lb.ft + (50.0  lb)(2  ft)\Rightarrow F_{Bx}=72.2  lb           (4)
\sum{F_{x} \left(\rightarrow + \right) } = F_{Bx}- F_{Cx}=0\Rightarrow F_{Cx}=72.2  lb
\sum{F_{y}\left(\uparrow + \right) } =F_{Cy}- F_{By}=0
F_{Cy}= F_{By}\Rightarrow F_{Cy}=50.0  lb

Finally, we substitute (4) into (2) and solve for F_{Ax} :

F_{Ax}= F_{Bx}\Rightarrow F_{Ax}=72.2  lb

The force components F_{Ax}  ,  F_{Ay} and F_{Cx} ,  F_{Cy} are forces from outside the system acting on the system at pins A and C (and where the system is the frame). We combine (F_{Ax}  ,  F_{Ay}) and (F_{Cx} ,  F_{Cy}) into the shear forces acting on the pins:

Pin A shear force = \left\|F_{A}\right\| =\sqrt{F_{Ax}^{2}+F_{Ay}^{2} } \Rightarrow \left\|F_{A}\right\|=87.9  lb
Pin C shear force = \left\|F_{C}\right\| =\sqrt{F_{Cx}^{2}+ F_{Cy}^{2}} \Rightarrow \left\|F_{C}\right\|=87.9  lb

Check As a check, the moment equilibrium equation could be written with the above results for a free body diagram of the entire frame using either A, C, or even B as a moment center. Another check would be to draw the results on each of the member free body diagrams and apply the equilibrium equations, being sure to use a different moment center than was used in the above Formulate Equations and Solve step.
Note: In this solution we generated six linearly independent equations that contained the six unknowns (F_{Ax}, F_{Ay}, F_{Bx}, F_{By}, F_{Cx}, F_{Cy}). The basis for these six equations was the free-body diagrams of members AB (Figure 2) and BC (Figure 3). Alternately, we could have generated six equations based on a free-body diagram of the whole frame (not shown) and either the free-body diagram in Figure 2 or 3, as we do in Example 9.1.4.