The structure shown in Fig. (a) is obtained by fixing point B of the mechanism in Sample Problem 10.2. For θ1 = 30º, determine the pin reactions Bx and By.

Question

The structure shown in Fig. (a) is obtained by fixing point [latex]B[/latex] of the mechanism in Sample Problem 10.2. For [latex]θ_{1}=30^\circ[/latex], determine the pin reactions [latex]B_{x}[/latex] and [latex]B_{y}[/latex].

Accepted Answer

Using [latex]θ_{1}=30^\circ[/latex], we found from geometry in Sample Problem 10.2 that [latex]θ_{2}=50.66^\circ[/latex].

Because the system in Fig. (a) is a structure rather than a mechanism, it has no [latex]DOF[/latex]. To determine the reactions at [latex]B[/latex] by the method of virtual work, we must turn [latex]B_{x}[/latex] and [latex]B_{y}[/latex] into active forces by removing the support at [latex]B[/latex].

Figure (b) shows the system with both [latex]B_{x}[/latex] and [latex]B_{y}[/latex] as active forces. In this case, angles [latex]θ_{1}[/latex] and [latex]θ_{2}[/latex] are independent position coordinates; consequently, the system possesses two [latex]DOF[/latex]. The principle of virtual work states that at equilibrium

[latex]δU=W\:δy_{1}+W\:δy_{2}+B_{x}\:δx_{B}+B_{y}\:δy_{B}=0[/latex] (a)

The virtual displacements in Eq. (a) can be related to [latex]δθ_{1}[/latex] and [latex]δθ_{2}[/latex] as follows:

[latex]y_{1}=\frac{L}{2}\cos θ_{1}[/latex]

[latex]δy_{1}=\frac{dy_{1}}{dθ_{1}}δθ_{1}=−\frac{L}{2}\sin θ_{1}\:δθ_{1}[/latex] (b)

[latex]y_{2}=L\cos θ_{1}+\frac{L}{2}\cos θ_{2}[/latex]

[latex]δy_{2}=\frac{∂y_{2}}{∂θ_{1}}δθ_{1}+\frac{∂y_{2}}{∂θ_{2}}δθ_{2}[/latex]

[latex]δy_{2}=−L\sin θ_{1}\:δθ_{1}−\frac{L}{2}\sin θ_{2}\:δθ_{2}[/latex] (c)

[latex]x_{B}=L\sin θ_{1}+L\sin θ_{2}[/latex]

[latex]δx_{B}=\frac{∂x_{B}}{∂θ_{1}}δθ_{1}+\frac{∂x_{B}}{∂θ_{2}}δθ_{2}[/latex]

[latex]δx_{B}=L\cos θ_{1}\:δθ_{1}+L\cos θ_{2}\:δθ_{2}[/latex] (d)

[latex]y_{B}=L\cos θ_{1}+L\cos θ_{2}[/latex]

[latex]δy_{B}=\frac{∂y_{B}}{∂θ_{1}}δθ_{1}+\frac{∂y_{B}}{∂θ_{2}}δθ_{2}[/latex]

[latex]δy_{B}=−L\sin θ_{1}\:δθ_{1}−L\sin θ_{2}\:δθ_{2}[/latex] (e)

Substituting Eqs. (b)–(e) into Eq. (a) and regrouping terms, we obtain

[latex]δU=\left(−\frac{W}{2}\sin θ_{1}−W\sin θ_{1}+B_{x}\cos θ_{1}−B_{y}\sin θ_{1}\right)L\:δθ_{1}+\left(−\frac{W}{2}\sin θ_{2}+B_{x}\cos θ_{2}−B_{y}\sin θ_{2}\right)L\:δθ_{2}=0[/latex] (f)

Because [latex]δθ_{1}[/latex] and [latex]δθ_{2}[/latex] are independent, Eq. (f) will be satisfied only if each of the terms in parentheses is zero; that is,

[latex]−\frac{3W}{2}\sin θ_{1}+B_{x}\cos θ_{1}−B_{y}\sin θ_{1}=0[/latex] (g)

and

[latex]−\frac{W}{2}\sin θ_{2}+B_{x}\cos θ_{2}−B_{y}\sin θ_{2}=0[/latex] (h)

Substituting [latex]θ_{1}=30^\circ[/latex] and [latex]θ_{2}=50.66^\circ[/latex], and solving Eqs. (g) and (h) simultaneously, we obtain

[latex]B_{x}=1.096\:W[/latex] and [latex]B_{y}=0.398\:W[/latex]

Question 10.3: The structure shown in Fig. (a) is obtained by fixing point ...

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