Question 10.1.2: Beam BDG is part of the frame shown in Figure 1 and is suppo...
DETERMINE LOADS ACTING ON A BEAM
Beam BDG is part of the frame shown in Figure 1 and is supported by pins at B and D. Determine the loads acting on BDG at pins B and D.
Goal Find the loads acting on beam BDG at pins B and D.
Given Information about the geometry and loading of the frame.
Assume The pins at B and D are frictionless, we can treat the system as planar, and that the weights of the members are negligible.
Draw When we isolate BDG (Figure 2) and analyze it for the loads at B and D we find that we have four unknown forces (F_{Bx} , F_{By} , F_{Dx} , F_{Dy} ) and only three equations of planar equilibrium. Thus we need to analyze members that are connected to BD so we can reduce the number of unknowns for member BDG. The first step is to find the loads at supports A. To accomplish this we draw a free-body diagram (Figure 3) of the entire structure. We then disassemble the frame and use the free-body diagram of ABC to determine unknown loads at B (Figure 4). Alternatively we could find the load at support E, which would require us to analyze member CDE to find loads at D.
Formulate Equations and Solve based on the free-body diagram in Figure 3, we set up planar equilibrium equations (5.5) to find the loads at support A.
\sum{F_{x}}=0 (5.5A)
\sum{F_{y}}=0 (5.5B)
\sum{M_{z }}=0 (5.5C)
\sum{M_{z @ E} }\left(\curvearrowleft + \right) =0
– F_{Ay}(16 ft) – 500 lb(2 ft) +500 lb(8 ft) – 400 lb(2 ft) + 1000 lb.ft = 0
F_{Ay}=200 lb
\sum{F_{x} \left(\rightarrow + \right) } =F_{Ax} +500 lb = 0
F_{Ax} = – 500 lb
We next analyze beam BDG to solve for the vertical loads at B and D (Figure 2). based on (5.5C), with the moment center at B:
\sum{M_{z @ B} }\left(\curvearrowleft + \right) = – 500 lb(4 ft) + F_{Dy}(8 ft) -400 lb(14 ft) = 0
F_{Dy}= 950 lb (1)
based on (5.5B):
\sum{F_{y}\left(\uparrow + \right) } =F_{By} + F_{Dy}- 500 lb − 400 lb = 0
Substituting from (1) for F_{Dy},
F_{By}+ 950 lb − 900 lb = 0
F_{By}= −50 lb
The minus sign indicates that F_{By} is acting downward on BDG and therefore upward on ABC.
based on (5.5A):
\sum{F_{x} \left(\rightarrow + \right) } =F_{Bx}-F_{Dx} = 0
F_{Bx}=F_{Dx} (2)
We do not have another equation of planar equilibrium that we can apply to BDG to solve for F_{Bx} and F_{Dx}. However, since F_{Bx} also acts on ABC, we can analyze member ABC to solve for F_{Bx} (Figure 4).
We choose C as the moment center because then there is one unknown in the moment equilibrium equation.
\sum{M_{z @ C} }\left(\curvearrowleft + \right) =0
−200 lb(8 ft) − 500 lb(8 ft) + 500 lb(6 ft) − 50 lb(4 ft) − F_{Bx} (4 ft) = 0
F_{Bx}= −700 lb (3)
Finally, substitute (3) into (2) and solve for F_{Dx}:
−700 lb − F_{Dx} = 0
F_{Dx}= −700 lb
Alternately, we could represent the forces acting on the beam BDG as:
F_{B}= −700 lb i − 50 lb j
F_{D}= 700 lb i + 950 lb j
Comment F_{By} is pulling down on beam BDG to help counteract rotation about pin D that is caused by the 400 lb force at G. (The 500 lb force is also participating in counteracting that rotation).
Check To check our solution we can reconsider the free-body diagram of BDG in Figure 2 and sum the moments about any point other than B. We do not want to sum the moments around B because we used B as the moment center in calculating the results. We want our check to consist of equations that are independent of those used in the analysis.
