Reactions at Pin and Roller Supports The beam in Fig. 5.13 has pin and roller supports and is subjected to a 2-kN force. What are the reactions at the supports?

Question

Reactions at Pin and Roller Supports

The beam in Fig. 5.13 has pin and roller supports and is subjected to a 2-kN force. What are the reactions at the supports?

Accepted Answer

Draw the Free-Body Diagram We isolate the beam from its supports and show the loads and the reactions that may be exerted by the pin and roller supports (Fig. a). There are three unknown reactions: two components of force [latex]A_{x}[/latex] and [latex]A_{y}[/latex] at the pin support and a force B at the roller support.
Apply the Equilibrium Equations Summing the moments about point A, the equilibrium equations are

[latex]\sum{F_{x}} = A_{x} - B \sin 30º = 0 [/latex] ,

[latex]\sum{F_{y}} = A_{y} + B \cos 30º - 2 = 0 [/latex],

[latex]\sum{M_{(point A)} } = (5) ( B \cos 30º ) - (3)(2) = 0 [/latex].

Solving these equations, the reactions are [latex] A_{x} = 0.69 kN[/latex] , [latex] A_{y} = 0.80 kN[/latex] , and B = 1.39 kN. The load and reactions are shown in Fig. b. It is good practice to show your answers in this way and confirm that the equilibrium equations are satisfied:

[latex]\sum{F_{x}} = 0.69 - 1.39 \sin 30º = 0 [/latex] ,

[latex]\sum{F_{y}} = 0.80 + 1.39 \cos 30º - 2 = 0 [/latex],

[latex]\sum{M_{(point A)} } = (5) ( 1.39 \cos 30º ) - (3)(2) = 0 [/latex].

Discussion
We drew the arrows indicating the directions of the reactions [latex] A_{x}[/latex] , and [latex] A_{y}[/latex] in the positive x and y axis directions, but we could have drawn them in either direction. In Fig. c we draw the free-body diagram of the beam with the component [latex] A_{y}[/latex] pointed downward. From this free-body diagram we obtain the equilibrium equations

[latex]\sum{F_{x}} = A_{x} - B \sin 30º = 0 [/latex] ,

[latex]\sum{F_{y}} = A_{y} + B \cos 30º - 2 = 0 [/latex],

[latex]\sum{M_{(point A)} } = (5) ( B \cos 30º ) - (3)(2) = 0 [/latex].

The solutions are [latex] A_{x} = 0.69 kN[/latex] , [latex] A_{y} = -0.80 kN[/latex], and B = 1.39 kN. The negative value of [latex] A_{y}[/latex] indicates that the vertical force exerted on the beam by the pin support is in the direction opposite to that of the arrow in Fig. c; that is, the force is 0.80 kN upward. Thus we again obtain the reactions shown in Fig. b.

Question 5.1: Reactions at Pin and Roller Supports The beam in Fig. 5.13 h...

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