A beam is supported by a fixed support A and a cable BD, as shown in Fig. 14.10(a). Determine the tension in the cable by the method of least work.

Question

A beam is supported by a fixed support [latex]A[/latex] and a cable [latex]BD[/latex], as shown in [latex]Fig. 14.10(a)[/latex]. Determine the tension in the cable by the method of least work.

Accepted Answer

We will analyze the structure by considering the tension [latex]T[/latex] in cable [latex]BD[/latex] to be redundant. The magnitude of the redundant will be determined by minimizing the strain energy of the structure with respect to [latex]T[/latex].

Because the structure contains both axially loaded and flexural members, its total strain energy is expressed as the sum of the strain energy due to axial forces and the strain energy due to bending; that is,

TABLE 14.3
Segment	x coordinate		M	F	[latex]\frac{\partial M}{\partial T}[/latex]	[latex]\frac{\partial F}{\partial T}[/latex]
Segment	Origin	Limits (ft)	M	F	[latex]\frac{\partial M}{\partial T}[/latex]	[latex]\frac{\partial F}{\partial T}[/latex]
CB	C	0-4	-15x	0	0	0
BA	C	4-12	-15x + 0.6T(x - 4)	-0.8T	0.6(x - 4)	-0.8
BD	__	__	0	T	0	1

[latex]U=\sum{\frac{F^2L}{2AE}} + \sum{\int{\frac{M^2}{2EI}}}dx[/latex] (1)

According to the principle of least work,

[latex]\frac{\partial U}{\partial T}=\sum{(\frac{\partial F}{\partial T})}\frac{FL}{AE}+\sum{\int{(\frac{\partial M}{\partial T})\frac{M}{EI}}}dx=0[/latex] (2)

The expressions for the bending moments M and the axial forces F in terms of the redundant T and their derivatives with respect to T are tabulated in Table 14.3. By substituting these expressions and derivatives into Eq. (2), we write

[latex]\frac{1}{E}[\frac{(-0.8)(-0.8T)(8)(12)^2}{12}+\frac{1(T)(10)(12)^2}{0.8}[/latex]

[latex]+\frac{(12)^4}{400}\int_{4}^{12}{0.6(x - 4)(-15x + 0.6Tx - 2.4T) dx}]=0[/latex]

[latex]T = 27.612 k[/latex] Ans.

Question 14.7: A beam is supported by a fixed support A and a cable BD, as ...

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