Determining Forces on Members of a Frame The frame in Fig. 6.37 supports a suspended weight W = 40 lb. Determine the forces on members ABCD and CEG.

Question

Determining Forces on Members of a Frame

The frame in Fig. 6.37 supports a suspended weight W = 40 lb. Determine the forces on members ABCD and CEG.

Accepted Answer

Determine the Reactions at tlie Supports We draw the free-body diagram of the entire frame in Fig. a. From the equilibrium equations

[latex]\sum{F_{x} } = A_{x} - D = 0 ,[/latex]

[latex]\sum{F_{y} } = A_{y} - 40 = 0 ,[/latex]

[latex]\sum{M_{(point A) } } = (18)D - (19)(40) = 0 ,[/latex]

we obtain the reactions [latex] A_{x} = 42.2 lb. A_{y} = 40 lb[/latex], and D = 42.2 lb.

Analyze the Members We obtain the free-body diagrams of the members in Fig. b. Notice that BE is a two-force member. The angle α = arctan (6/8) = 36.9°.

The free-body diagram of the pulley has only two unknown forces. From the equilibrium equations

[latex] \sum{F_{x} } = G_{x} - 40 = 0 , [/latex]

[latex] \sum{F_{y} } = G_{y} - 40 = 0 , [/latex]

we obtain [latex] G_{x} = 40 lb and G_{y} = 40 lb.[/latex] There are now only three unknown forces on the free-body diagram of member CEG. From the equilibrium equations

[latex] \sum{F_{x} } = -C_{x} - R \cos α - 40 = 0 , [/latex]

[latex] \sum{F_{y} } = -C_{y} - R \sin α - 40 = 0 , [/latex]

[latex] \sum{M_{(point C)} } = -8R \sin α - (16)(40) =0, [/latex]

we obtain [latex]C_{x} = 66.7 lb, C_{y} = 40 lb[/latex], and R = -133.3 lb, completing the solution (Fig. c).

Question 6.7: Determining Forces on Members of a Frame The frame in Fig. 6...

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