A Two-Force Member The L-shaped bar in Fig. 5.39 has a pin support at A and is loaded by a 6-kN force at B. Neglect the weight of the bar. Determine the angle α and the reactions at A. Strategy The bar is a two-force member because it is subjected only to the 6-kN force at B and the force exerted

Question

A Two-Force Member

The L-shaped bar in Fig. 5.39 has a pin support at A and is loaded by a 6-kN force at B. Neglect the weight of the bar. Determine the angle α and the reactions at A.

Strategy
The bar is a two-force member because it is subjected only to the 6-kN force at B and the force exerted by the pin support. (If we could not neglect the weight of the bar, it would not be a two-force member.) We will determine the angle α and the reactions at A in two ways, first by applying the equilibrium equations in the usual way and then by using the fact that the bar is a two-force member.

Accepted Answer

Applying the Equilibrium Equations We draw the free-body diagram of the bar in Fig. a, showing the reactions at the pin support. Summing moments about point A, the equilibrium equations are

[latex]\sum{F_{x} } = A_{x} + 6 \cos α =0 ,[/latex]

[latex]\sum{F_{y} } = A_{y} + 6 \sin α =0 ,[/latex]

[latex]\sum{M_{(point A)} } = (6 \sin α)(0.7) - (6 \cos α)(0.4) =0.[/latex]

From the third equation we see that α = arctan (0.4/0.7). In the range : 0 ≤ α ≤ 360°, this equation has the two solutions α = 29.7° and α = 209.7°. Knowing α, we can determine [latex]A_{x}[/latex] and [latex]A_{y}[/latex] from the first two equilibrium equations. The solutions for the two values of a are

[latex]α = 29.7°, A_{x} = -5.21 kN, A_{y} = -2.98 kN,[/latex]

and

[latex]α = 209.7°, A_{x} = 5.21 kN, A_{y} = 2.98 kN.[/latex]

Treating the Bar as a Two-Force Member We know that the 6-kN force at B and the force exerted by the pin support must be equal in magnitude, opposite indirection, and directed along the line between points A and B. The two possibilities are shown in Figs, b and c. Thus by recognizing that the bar is a two-force member, we immediately know the possible directions of the forces and the magnitude of the reaction at A.

In Fig. b we can see that tan α = 0.4/0.7, so α = 29.7° and the components of the reaction at A are

[latex]A_{x} = - 6 \cos 29.7° = -5.21 kN[/latex]

[latex]A_{y} = - 6 \sin 29.7° = -2.98 kN[/latex]

In Fig. c, α = 180° + 29.7° = 209.7°, and the components of the reaction at A are

[latex]A_{x} = 6 \cos 29.7° = 5.21 kN[/latex]

[latex]A_{y} = 6 \cos 29.7° = 2.98 kN[/latex]

Question 5.11: A Two-Force Member The L-shaped bar in Fig. 5.39 has a pin s...

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