Free-Body Diagrams for Three Joined Members Determine the forces on the members of the frame in Fig. 6.38.

Question

Free-Body Diagrams for Three Joined Members

Determine the forces on the members of the frame in Fig. 6.38.

Strategy

You can confirm that no information can be obtained from the free-body diagram of the entire frame. To analyze the members, we must deal with an interesting challenge at joint D, where a load acts and three members are connected. We will obtain the free-body diagrams of the members by first isolating member AD. then separating members BD and CD.

Accepted Answer

We first isolate member AD from the rest of the structure, introducing the reactions [latex]D_{x} and D_{y}[/latex] (Fig. a). We then separate members BD and CD, introducing equal and opposite forces [latex]E_{x} and E_{y}[/latex] (Fig. b). In this step we could have placed the 300-N load and the forces [latex]D_{x} and D_{y}[/latex] on either free-body diagram.

Only three unknown forces act on member AD. From the equihbrium equations

[latex] \sum{F_{x} } = A + D_{x} = 0 , [/latex]

[latex] \sum{F_{y} } = D_{y} - 120 = 0 , [/latex]

[latex] \sum{M_{(point D)} } = -(0.3)A + (0.4)(120) = 0 , [/latex]

we obtain A = 160 N, [latex]D_{x} = -160 N, and D_{y} = 120 N.[/latex] Now we consider the free-body diagram of member BD. From the equation

[latex] \sum{M_{(point D)} } = -(0.8)B_{y} + (0.4)(180) = 0 , [/latex]

we obtain [latex]B_{y} = 90 N.[/latex] Now we use the equation

[latex] \sum{F_{y} } = B_{y} - D_{y} + E_{y} - 180 = 90 - 120 + E_{y} - 180 = 0 , [/latex]

obtaining [latex]E_{y} = 210 N.[/latex] Now that we know [latex]E_{y}[/latex] , there are only three unknown forces on the free-body diagram of member CD. From the equilibrium equations

[latex] \sum{F_{x} } = C_{x} - E_{x} = 0 ,[/latex]

[latex] \sum{F_{y} } = C_{y} - E_{y} - 240 = C_{y} - 210 -240 = 0, [/latex]

[latex] \sum{M_{(point C)} } = (0.3)E_{x} - (0.8)E_{y} - (0.4)(240) = (0.3)E_{x} - (0.8)(210) - (0.4)(240) = 0 ,[/latex]

we obtain [latex]C_{x} = 880 N, C_{y} = 450 N, and E_{x} = 880 N.[/latex] Finally, we return to the free-body diagram of member BD and use the equation

[latex] \sum{F_{x} } = B_{x} + E_{x} - D_{x} - 300 = B_{x} + 880 + 160 - 300 = 0[/latex]

to obtain [latex]B_{x} = -740 N[/latex], completing the solution (Fig. c).

Question 6.8: Free-Body Diagrams for Three Joined Members Determine the fo...

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