The optimal trusses equilibrating given forces and reactions are composed of two families of bars, one of which in tension and the second in compression. All the nodes, including the supports should be equilibrated. Thus the layout in figure III.12 is valid only for the case of the vertical force—parallel to the bars AG\(_{2}\) and A\(_{1}\)B. The layouts in Figs III.15, III.16 are incorrect because of the same reason. The numerical solutions to the problems of these figures are shown in Figs. 4 and 5.
The analytical layouts for these two cases (Figs. 4 and 5) will not be constructed, as they lie outside the scope of the layouts discussed here. Instead, let us consider the problem of Fig. 6.
In this problem the reactions \(H_{A}=P_{3}\) and \(P_{1}\) are positive. Thus the bars which meet at the node A will have different signs. The analytical layout shown in Fig. 6 will be now explained and parameterized, see Fig. 7. Our main task is to match two virtual displacement fields within the domains: ABG\(_{2}\) and BG\(_{2}\)D along the radius BG\(_{2}\). Both polar parameterizations (\(r\), \(\beta ^{\ast } )\) and (\(r, \beta \)) are shown in Fig. 7.
We shall use the formulae (III.2) for both the domains
-
a)
domain G\(_{0}\)BG\(_{2}\): here \(\beta ^{\ast } \in (0, \alpha _{0})\), where \(\alpha _{0} = 50.309^{\circ }\), cf. Fig.III.9
$$ \begin{array}{rll} u^{\ast} \left({r,\beta^{\ast}} \right)&=&r, \\ \mathit{v}^{\ast} \left( {r,\beta^{\ast}}\right)&=&-2r\beta^{\ast} +r,\\ \omega^{\ast} \left( {r,\beta^{\ast}}\right)&=&-2\beta^{\ast} +1,\omega_0^{\ast} =-1 \end{array} $$
(11)
-
b)
domain BG\(_{2}\)D: \(\beta \in (-\mu , \theta )\)
$$ \begin{array}{rll} u( {r,\beta} )&=&r,\mathit{v}( {r,\beta} )=-2r\beta +d_1 r,\\ \omega ( {r,\beta} )&=&-2\beta +d_1 ,\omega_0 =-d_1 \end{array} $$
(12)
The conditions:
$$ \begin{array}{rll} u^{\ast} \left({r,\beta^{\ast}} \right) &=& u \left({r,-\beta^{\ast}-\mu} \right) \quad \text{and} \\ \mathit{v}^{\ast} \left({r,\beta^{\ast}} \right) &=& -\mathit{v} \left({r,-\beta^{\ast}-\mu} \right) \end{array} $$
(13)
assure that the formulae (11, 12) describe a continuous displacement field. Note that
$$ \omega^{\ast} \left( {r,\beta^{\ast}} \right)=-\omega \left({r,-\beta^{\ast} -\mu} \right) $$
(14)
because the sign conventions for the rigid rotation are different for both the parameterizations.
The choice \(d_1 =-1-2\mu \) will assure that \(v_{A} = 0\), where \(\mathit {v}_{A}\) is denoted in Fig. 7, while \(u_{A}\) does not vanish. This choice does not satisfy the kinematic conditions of the problem, but the correct formulae for the optimal weight will be achieved by taking into account the virtual work of the reactions.
Our aim is now to find both virtual displacements \((u,\mathit {v})\) at arbitrary point (\(\xi ^{\ast } ,\eta ^{\ast } \)) of the domain G\(_{0}\)AG\(_{2}\). Let us compute the free terms in (3):
$$ \begin{array}{rll} \overline{u}_0 \left(\xi ^{\ast} ,\eta ^{\ast} \right) &=& a \left[\cos \xi^{\ast} -\sin \xi^{\ast} \left(1-2\eta^{\ast}\right)\right]\\ \overline{\mathit{v}}_0 \left(\xi ^{\ast} ,\eta ^{\ast} \right) &=& a \left[\sin \xi^{\ast} + \cos \xi^{\ast} \left(1-2\eta^{\ast}\right)\right] \end{array} $$
(15)
and now compute the displacements: \(u (\xi ^{\ast } , \eta ^{\ast } )\), \(\mathit {v} (\xi ^{\ast } , \eta ^{\ast } )\) by using (3), where we use (15) and put \(\omega _0^{\ast } =-1\). Rather lengthy derivation, with using the integration formulae for Lommel functions lead to
$$ u\left( {\xi^{\ast} ,\eta^{\ast}} \right)=au_1 \left( {\xi^{\ast} ,\eta^{\ast}} \right), $$
(16)
$$ \begin{array}{rll} u_1 ( {\xi ,\eta} )&=&2( {\eta -\xi} )G_1 ( {\xi,\eta})+G_0 ( {\xi ,\eta} )\\ &&\quad-F_1 ( {\xi ,\eta})+2F_2 ( {\xi ,\eta} )+F_3 ( {\xi ,\eta} ), \end{array} $$
(17)
and
$$ \mathit{v}\left( {\xi^{\ast} ,\eta^{\ast}} \right)=a\mathit{v}_1 \left( {\xi^{\ast} ,\eta^{\ast}} \right), $$
(18)
$$ \begin{array}{rll} \mathit{v}_1 ( {\xi ,\eta} )&=&-2( {\eta -\xi} )G_0 ( {\xi,\eta})+G_0 ( {\xi ,\eta} )\\ &&\quad-2F_1 ( {\xi ,\eta})-2F_2 ( {\xi ,\eta} )+G_1 ( {\xi ,\eta} ), \end{array} $$
(19)
Let us note that
$$ u_A= au_1 \left( {\alpha_0 ,\alpha_0} \right), \qquad \mathit{v}_A= a\mathit{v}_1 \left( {\alpha_0 ,\alpha_0} \right), $$
(20)
where \(\alpha _{0}=\angle \)G\(_{0}\)BG\(_{2}\)
\(=\) 50.309\(^{\circ }\). Note that \(\alpha _{0}\) is the root of the equation
$$ F_3 \left( {\alpha_0 ,\alpha_0} \right) - F_1 \left( {\alpha_0 ,\alpha_0} \right) =-1. $$
(21)
Note, also that the displacements (16), (18) can be equivalently found from (5) by specifying \(k=-2\), because this choice corresponds to a correct value of \(\omega _{0}\) for the parameterization in the domain G\(_{1}\)BPE\(_{2}\)G\(_{1}\).
The weight of the optimal structure is given by \(V=V_1 +\Delta V\), where \(\sigma _{0}V_{1}\) is equal to the virtual work of the forces \(P_{1}\), \(P_{3}\), \(P\) (see Fig. 6) on the virtual displacement field and \(\Delta V\) is the weight of the bar BA\(_{1}\): \(\Delta V=\frac {P_2} {P}V_0 \). Thus
$$ \sigma_0 V_1 =P_3 u_A -P_1 \mathit{v}_A +Pu( P ), $$
(22)
where \(u_{A}\), \(\mathit {v}_{A}\) are given by (20) and
$$ u( P )=u( {\theta ,\theta} ), $$
(23)
where \(u\) is defined by (6) for \(k = 2\mu \) or
$$ \begin{array}{rll} {u( P )} / a&=&( {1+2\mu})\left( {F_1( {\theta ,\theta} )-F_3 ( {\theta ,\theta} )}\right)\\ &&\quad+G_0 ( {\theta ,\theta} )+2F_2 ({\theta ,\theta}). \end{array} $$
(24)
Let us note that \(\mathit {v}_{A} = 0\). Indeed, by (19) we compute
$$ \begin{array}{rll} {\mathit{v}_A} / a&=&G_0 \left( {\alpha_0 ,\alpha_0} \right)-2F_2\left( {\alpha_0 ,\alpha_0} \right)\\ &&\quad-2F_1 \left( {\alpha_0 ,\alpha_0}\right)+G_1 \left( {\alpha_0 ,\alpha_0} \right). \end{array} $$
(25)
but \(G_0 =F_0 +F_2 \), \(G_1 =F_1 +F_3 \) (cf. (7) in Lewiński et al. 1994) hence
$$ {\mathit{v}_A} / a=F_0 \left( {\alpha_0 ,\alpha_0} \right)-F_2 \left( {\alpha_0 ,\alpha_0} \right)+F_3 \left( {\alpha_0 ,\alpha_0}\right)-F_1 \left( {\alpha_0 ,\alpha_0} \right). $$
(26)
Recalling the identity (cf. (8) in Lewiński et al. 1994)
$$ F_0 ( {\beta ,\alpha} )-F_2 ( {\beta ,\alpha} )=\cos( {\beta -\alpha} ) $$
(27)
and taking into account (21) we conclude that \(\mathit {v}_{A} = 0\).
The equilibrium equations for the structure of Fig. 6 read
$$ \begin{array}{rll} P_1 &=& P[ L /a + \cos \mu], \\ P_2 &=& P[ L/a + \sin \mu + \cos \mu], \\ P_3 &=& -P\cos \mu, \end{array} $$
(28)
and \(L=| {\text {G}_{\text {1}} \text {P}} |\) or, see Eq.III.26
$$ L=a\left( {F_1 ( {\theta ,\theta} )-F_3 ({\theta ,\theta})} \right). $$
(29)
Let us compute \(u_{A}/a\) by (23), (17), (21):
$$ u_A=a\left[ {G_0 \left( {\alpha_0 ,\alpha_0} \right)+ 2 F_2 \left( {\alpha_0 ,\alpha_0}\right)}-1 \right]. $$
(30)
We compute \(V\) by (22), (28), (29), taking into account that \(v_{A} = 0\):
$$\begin{array}{rll} V/V_0&=&-\cos \mu \left(u_A/a\right)+ u(P)/a \\ &&+\, L/a + \sin \mu +\cos \mu. \end{array} $$
(31)
Let us define the function
$$ f( \alpha )=G_0 ( {\alpha ,\alpha} )+2F_2 ({\alpha ,\alpha} ). $$
(32)
We rearrange (31) to the form
$$ \begin{array}{rll} V /{V_0} &=&2( {1+\mu} )( {L / a} )+\sin \mu \\ &&\quad+\cos \mu \left[{2-f\left( {\alpha_0} \right)} \right]+f( {\theta}). \end{array} $$
(33)
The above equation is applicable if the net of bars is composed of one family of bars in tension and the second family of bars in compression, mutually orthogonal.
Consider now the problem of Fig. 6 for the data:
$$\mu = 95.7106^{\circ}, l= 5a, $$
where \(l\) is the horizontal coordinate of point P, see Fig. 6. From the equation:
$$ a( {1-\cos \mu} )+L\sin \mu =5a. $$
(34)
we find \(L = 3.91995a\), and by solving (29) we find \(\theta = 113.4456^{\circ }\). We remember that \(\alpha _{0} = 50.309^{\circ }\). The volume of the optimal truss is computed by (33), hence \(V/V_0 = 42.85857\). The analytical layout is shown in Fig. 6, while the numerical layout is given in Fig. 8. The numerically found value of the volume equals: \(V/V_{0} = 42.91234\), which compares favorably with the analytical result.