Reactions at a Hinge Support The bar AC in Fig. 5.33 is 4 ft long and is supported by a hinge at A and the cable BD. The hinge axis is along the z axis. The centerline of the bar lies in the x-y plane, and the cable attachment point B is the midpoint of the bar. Determine the tension in the cable

Question

Reactions at a Hinge Support

The bar AC in Fig. 5.33 is 4 ft long and is supported by a hinge at A and the cable BD. The hinge axis is along the z axis. The centerline of the bar lies in the x-y plane, and the cable attachment point B is the midpoint of the bar. Determine the tension in the cable and the reactions exerted on the bar by the hinge.

Accepted Answer

Draw the Free-Body Diagram We isolate the bar from the hinge support and the cable and show the reactions they exert (Fig. a). The terms [latex] A_{x} , A_{y} ,and A_{z} [/latex] are the components of force exerted by the hinge, and the terms [latex] M_{Ax} and M_{Ay} [/latex] are the couples exerted by the hinge about the x and y axes. (Remember that the hinge cannot exert a couple on the bar about the hinge axis.) The term T is the tension in the cable.

Apply the Equilibrium Equations To write the equilibrium equations, we must first express the cable force in terms of its components. The coordinates of point B are (2 cos 30°, -2 sin 30°, 0) ft, so the position vector from B to D is

[latex] r_{BD} = \left(2 - 2 \cos 30º ight)i + [2 - \left(- 2 \sin 30º ight)] j +\left(- 1 - 0 ight)k = 0.268i +3j - k .[/latex]

We divide this vector by its magnitude to obtain a unit vector [latex] e_{BD}[/latex] that points from point B toward point D:

[latex] e_{BD} = \frac{r_{BD} }{\left|r_{BD} ight| } = 0.084i + 0.945j - 0.315k.[/latex]

Now we can write the cable force as the product of its magnitude and [latex] e_{BD}[/latex]:

[latex] Te_{BD} = T(0.084i + 0.945j - 0.315k).[/latex]

The sums of the forces in each coordinate direction must equal zero:

[latex] \sum{F_{x} } = A_{x} + 0.084T = 0, [/latex]

[latex] \sum{F_{y} } = A_{y} + 0.945T - 100 = 0, [/latex]

[latex] \sum{F_{z} } = A_{z} - 0.315T = 0. [/latex] (5.23)

If we sum moments about A, the resulting equations do not contain the unknown reactions [latex]A_{x} , A_{y}, and A_{z}[/latex]. The position vectors from A to B and from A to C are

[latex]r_{AB} = 2 \cos30°i - 2 \sin30°j.[/latex]

[latex]r_{AC} = 4 \cos30°i - 4 \sin30°j.[/latex]

The sum of the moments about A is

[latex]\sum{M_{(point A)} } = M_{Ax}i + M_{Ay}j + [ r_{AB} imes \left(Te_{BD} ight)] + [r_{AC} imes \left(-100j ight)] = M_{Ax}i + M_{Ay}j +\left|\begin{matrix} i & j & k \ 1.732 & -1 & 0 \ 0.084T &0.945T & -0.315T \end{matrix} ight| +\left|\begin{matrix} i & j & k \ 3.464 & -2 & 0 \ 0 &-100 & 0 \end{matrix} ight| = (M_{Ax} + 0.315T )i + (M_{Ay} + 0.546T )j + (1.72T - 346)k = 0 .[/latex]

From this vector equation we obtain the scalar equations

[latex]\sum{M_{x} } = M_{Ax} + 0.315T = 0 ,[/latex]

[latex]\sum{M_{y} } = M_{Ay} + 0.546T = 0 ,[/latex]

[latex]\sum{M_{z} } = 1.72T - 346 = 0 .[/latex]

Solving these equations, we obtain the reactions

[latex]T= 201 lb, M_{Ax} = -63.4 ft-lb, M_{Ay} = -109.8 ft-lb. [/latex]

Then from Eqs. (5.23) we obtain the forces exerted on the bar by the hinge:

[latex] A_{x} = -17.0 lb, A_{y} = -90.2 lb, A_{z} = 63.4 lb.[/latex]

Question 5.9: Reactions at a Hinge Support The bar AC in Fig. 5.33 is 4 ft...

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