The recently published software aimed at predicting the optimal layouts of trusses of minimal weight, see Sokół (2011a, b), originally based on the truss ground structure filling up a fixed rectangular domain, has been effectively extended to other shapes of feasible domains, including the domains of polygons, allowing also for non-convex shapes. The current version of the program developed makes use of the adaptive ground structure approach with the selective subsets of active bars (Sokół 2011b). This approach is similar in spirit to that proposed by Gilbert and Tyas (2003) but applies a different strategy of adding-removing of the active bars in the fully connected ground structure. The program is capable of solving large-scale problems with the number of potential members exceeding one billion (10\(^{9})\) which provides highly accurate results both for volume and layout of fully stressed truss of minimal weight, designed within a given feasible domain, supported along given parts of the boundary or at fixed supports and subjected to a loading applied at fixed points. In the present paper we shall confine attention to the case of the absolute value of the lower and upper limits for the stress being equal, denoted by \(\sigma _{0}\). Equipped with this new software one can solve numerically a broad class of the optimum design problems of Michell type for the L-shaped domains thus checking the analytical solutions predicted by Lewiński and Rozvany (2008). The noted discrepancies between the numerical predictions and the analytical results mentioned were the impulse for the present authors to look at this class of problems more closely. The result of the current verification is the confirmation of all the layouts predicted in Lewiński and Rozvany (2008) in the case of the point load being directed vertically (or parallel to the vertical edges of the L-shaped domain). However, the formulae for the optimal weight should be corrected. The source of the mistake was neglecting the rigid rotation \(\omega \) at the vicinity of the re-entrant corner. The present paper makes up for this shortcoming, by deriving the new and now fully correct formulae for the virtual (adjoint) displacement fields within the feasible L-shaped domains, leading to analytical formulae for the optimum weight. The new formulae provide the results which compare favorably with the numerical predictions involving up to 10\(^{9}\) truss members.

The cases of the force being non-vertical need a special care. The numerical predictions reveal new, till now never published layouts. Some of them lie within the scope of the formulae derived for the case of the vertical forces, while other are completely different, hence their analytical discussion is planned to be the subject of a future research.

For brevity, Eq.n, Fig.m and Sec.p of Lewiński and Rozvany (2007) (or Lewiński and Rozvany 2008) will be referred to as Eq.II.n, Fig.II.m and Sec.II.p (or Eq.III.n, Fig.III.m and Sec.III.p). The notation used is adopted after the papers referred. The present paper can be viewed as the addendum and extension of the publication by Lewiński and Rozvany (2008).

Optimal topology for an inclined line support and point load

We revisit the Section III.2. Note that Eq.II.78 refers to the circular domain in Fig.III.3; hence the rigid rotation is expressed by: \(\omega (\alpha ,\beta )=-2\beta +d_1 \), which corrects the formula for \(\omega \) given in Sec.III.2 (below Fig.III.3). In fact, the quantity \(\omega (0, 0)\), or the rigid rotation measured at the corner B, see Fig.III.3, is equal to a constant \(d_{1}\) and does not vanish, as assumed in the paper revisited. This shortcoming will change the results concerning the adjoint displacement fields in the domain adjacent to the circular fan.

The Cartesian coordinates of points \((x,y)\) in the domain BPG\(_{1}\), see Fig.III.3, are given by

$$ \begin{array}{rll} x(\alpha, \beta)/a &=& \overline{x}( {\alpha ,\beta} ) \cos ( \beta - \alpha ) - \overline {y} ({\alpha ,\beta} ) \sin ( \beta - \alpha ),\\ y(\alpha, \beta)/a &=& \overline{x}( {\alpha ,\beta} ) \sin ( \beta - \alpha ) + \overline {y} ({\alpha ,\beta} ) \cos ( \beta - \alpha ), \end{array} $$

where \(\overline {x} ( {\alpha ,\beta })\), \(\overline {y}({\alpha ,\beta })\) are given by Eqs.III.3. Thus the position of point P with coordinates (\(\alpha _{p}\), \(\beta _{p}\)) is determined by

$$ a_1 = y \left(\alpha_p, \beta_p\right), \qquad b_p = -x \left(\alpha_p, \beta_p\right), $$

see Figs.III.2, III.3. Note that in Fig.III.3 we have: \(\beta _{p} = \theta \). The incorrect assumption \(\omega (0, 0) = 0\) made in Sec.III.2, is corrected here, thus leading to the following new form of equations III.10:

$$ \begin{array}{rll} u( {\xi ,\eta} ) &=& \overline u_0 ({\xi ,\eta} )+\int\limits_0^{\xi} \left[ {\cos ( {\xi -\alpha} )+\left({\omega_0 +2\eta +2\alpha} \right)}\right.\\ &&\qquad\qquad \qquad\quad\times\left.{\sin ( {\xi -\alpha} )}\right]A( {\alpha ,\eta} )d\alpha,\\ \mathit{v}( {\xi ,\eta} ) &=& \overline {\mathit{v}}_0 ( {\xi ,\eta})+\int\limits_0^{\xi} \left[ {\sin ( {\xi -\alpha} )-\left( {\omega_0 +2\eta +2\alpha} \right)}\right.\\ && \qquad\qquad \qquad\quad\times\left.{\cos ( {\xi -\alpha} )}\right]A( {\alpha ,\eta} )d\alpha, \end{array} $$

with \(\omega _{0}=-\omega (0, 0)\) or \(\omega _{0}=-d_{1}\) and since \(d_1 =-1-\pi \), see the result below Eqs III.6, we have

$$ \omega_0 = 1 + \pi, $$

while \(\overline u_0 ,\overline {\mathit {v}}_0 \) are defined by Eq.III.11. The results III.12, III.13 should be corrected to the form

$$ \begin{array}{rll} u( {\xi ,\eta} ) / a &=& ( {1+k})\left[ {F_1 \left( {\xi ,\eta} \right)-F_3 ( {\xi ,\eta})} \right]\\ &&\quad+2( {\eta -\xi} )G_1 ( {\xi ,\eta})+G_0 ( {\xi ,\eta} )+2F_2 ( {\xi ,\eta}),\\ \mathit{v}( {\xi ,\eta} ) / a&=&( {1+k})\left[ {F_2 ( {\xi ,\eta} )-F_0 ( {\xi ,\eta})} \right]\\ &&\quad-2( {\eta -\xi} )G_0 ( {\xi ,\eta})+G_1 ( {\xi ,\eta} )-2F_1 ( {\xi ,\eta} ), \end{array} $$

where \(k=\pi \). The formula III.15 is replaced by

$$ \begin{array}{rll} u( {\xi ,\xi} )/a &=& ( {1+k})\left[ {F_1 ( {\xi ,\xi} )-F_3 ( {\xi ,\xi})} \right]\\ &&\quad +G_0 ( {\xi ,\xi} )+2F_2 ( {\xi ,\xi}), \end{array} $$

where \(k=\pi \) and the formula III.18 should be corrected accordingly. The rest of Sec.III.2 is correct.

L-shaped domain with a horizontal support

The Sec.III.3 is correct, yet the Table III.1, comprising the results of Sec.III.4 should assume the new form (Table 1).

Table 1 Correction of Table III.1. Properties of the special cases in Section III.4

The volume of an optimal truss is equal to \(V=\overline V V_0 \), \(V_0 =Pa/\sigma _0 \).

The figure caption of Fig.III.8 should read:

  • Case of \(\theta = \pi /3\), \(\overline V =8.849866\), and \(\alpha _{p} = 0.162\) to assure \(a_{1}=a\).

Optimal topology for special cases

The figure caption of Fig.III.10 should read:

  • Case of \(\theta = \pi /2\), \(\bar {{V}}=16.45864\), and \(\alpha _P =\pi /6\).

We repeat here Fig.III.11 as Fig. 1, with small adjustments.

Fig. 1
figure 1

Case of \(\theta ={6\pi } / {10}\), \(\overline V =19.1096\), and \(\alpha _P =\pi / 6\). The analytical layout

The numerical layout for the problem of Fig. 1 is shown in Fig. 2.

Fig. 2
figure 2

The numerically determined layout for problem of Fig. 1; the volume of the structure equals \(V_{num} = 19.12494V_{0}\), cf. Table 1. The result was obtained using the ground structure with 150 mln potential members of

The layout of Fig.III.12 refers only to the case of the force \(P\) being vertical. Moreover, the line DN is characterized by \(\phi = \theta \) (not \(\phi = 0\), as printed in Fig.III.12).

Extended L-shaped domains with an inclined lower boundary

The formula III.25 holds, where

$$ V_1 =\frac{P}{\sigma_0} u_P \left( {\alpha_P ,\beta_P} \right) $$

and \(u_{P}\) is the virtual displacement along the force \(P\), as in Fig. 3.

Fig. 3
figure 3

Case of the inclined lower boundary of the feasible domain; here \(\mu =\pi /5\)

The displacement \(u_{P}\) is computed by

$$ u_P =u_x \left( {\alpha_P ,\beta_P} \right)\sin \mu +u_y \left( {\alpha _P ,\beta_P} \right)\cos \mu $$


$$ \begin{array}{rll} u_x( {\alpha ,\beta} )&=& u( \alpha, \beta) \cos (\beta-\alpha) - \mathit{v} (\alpha, \beta) \sin (\beta-\alpha), \\ u_y( {\alpha ,\beta} )&=& u( \alpha, \beta) \sin (\beta-\alpha) + \mathit{v} (\alpha, \beta) \cos (\beta-\alpha). \end{array} $$

Substitution of (9) into (8) gives

$$ \begin{array}{rll} u_P ( {\alpha ,\beta} )&=& u( {\alpha ,\beta} )\sin ( {\beta -\alpha +\mu} )\\ &&\quad+v( {\alpha ,\beta} )\cos ( {\beta -\alpha +\mu} ) \end{array} $$

and \(u, \mathit {v}\) are given by (5), where one should put \(k = 2\mu \). These equations for \(u\) and \(\mathit {v}\) can be derived from the integral formulae (3), where one should put \(\omega _0 =1+2\mu \), cf. Eqs.III.24 and II.78.

The cases of the non-vertical point load

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.

Fig. 4
figure 4

Numerical solution of the problem equivalent to Fig.III.15, obtained using ground structure with 565 mln of potential members; \(V_{num} = 63.63408V_{0}\)

Fig. 5
figure 5

Numerical solution of the problem equivalent to Fig.III.16, obtained using ground structure with 30 mln of potential members; \(V_{num} = 19.34832V_{0}\)

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.

Fig. 6
figure 6

Problem of class of Fig. 3 for the case of \(\mu = 95.7106^{\circ }\) and \(\theta = 113.4456^{\circ }\). The analytically found layout of framework

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.

Fig. 7
figure 7

Polar parameterization around point B

We shall use the formulae (III.2) for both the domains

  1. a)

    domain G\(_{0}\)BG\(_{2}\): here \(\beta ^{\ast } \in (0, \alpha _{0})\), where \(\alpha _{0} = 50.309^{\circ }\), cf. Fig.III.9

    $$ \begin{array}{rll} u^{\ast} \left({r,\beta^{\ast}} \right)&=&r, \\ \mathit{v}^{\ast} \left( {r,\beta^{\ast}}\right)&=&-2r\beta^{\ast} +r,\\ \omega^{\ast} \left( {r,\beta^{\ast}}\right)&=&-2\beta^{\ast} +1,\omega_0^{\ast} =-1 \end{array} $$
  2. b)

    domain BG\(_{2}\)D: \(\beta \in (-\mu , \theta )\)

    $$ \begin{array}{rll} u( {r,\beta} )&=&r,\mathit{v}( {r,\beta} )=-2r\beta +d_1 r,\\ \omega ( {r,\beta} )&=&-2\beta +d_1 ,\omega_0 =-d_1 \end{array} $$

    The conditions:

    $$ \begin{array}{rll} u^{\ast} \left({r,\beta^{\ast}} \right) &=& u \left({r,-\beta^{\ast}-\mu} \right) \quad \text{and} \\ \mathit{v}^{\ast} \left({r,\beta^{\ast}} \right) &=& -\mathit{v} \left({r,-\beta^{\ast}-\mu} \right) \end{array} $$

    assure that the formulae (11, 12) describe a continuous displacement field. Note that

    $$ \omega^{\ast} \left( {r,\beta^{\ast}} \right)=-\omega \left({r,-\beta^{\ast} -\mu} \right) $$

    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):

$$ \begin{array}{rll} \overline{u}_0 \left(\xi ^{\ast} ,\eta ^{\ast} \right) &=& a \left[\cos \xi^{\ast} -\sin \xi^{\ast} \left(1-2\eta^{\ast}\right)\right]\\ \overline{\mathit{v}}_0 \left(\xi ^{\ast} ,\eta ^{\ast} \right) &=& a \left[\sin \xi^{\ast} + \cos \xi^{\ast} \left(1-2\eta^{\ast}\right)\right] \end{array} $$

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

$$ u\left( {\xi^{\ast} ,\eta^{\ast}} \right)=au_1 \left( {\xi^{\ast} ,\eta^{\ast}} \right), $$
$$ \begin{array}{rll} u_1 ( {\xi ,\eta} )&=&2( {\eta -\xi} )G_1 ( {\xi,\eta})+G_0 ( {\xi ,\eta} )\\ &&\quad-F_1 ( {\xi ,\eta})+2F_2 ( {\xi ,\eta} )+F_3 ( {\xi ,\eta} ), \end{array} $$


$$ \mathit{v}\left( {\xi^{\ast} ,\eta^{\ast}} \right)=a\mathit{v}_1 \left( {\xi^{\ast} ,\eta^{\ast}} \right), $$
$$ \begin{array}{rll} \mathit{v}_1 ( {\xi ,\eta} )&=&-2( {\eta -\xi} )G_0 ( {\xi,\eta})+G_0 ( {\xi ,\eta} )\\ &&\quad-2F_1 ( {\xi ,\eta})-2F_2 ( {\xi ,\eta} )+G_1 ( {\xi ,\eta} ), \end{array} $$

Let us note that

$$ u_A= au_1 \left( {\alpha_0 ,\alpha_0} \right), \qquad \mathit{v}_A= a\mathit{v}_1 \left( {\alpha_0 ,\alpha_0} \right), $$

where \(\alpha _{0}=\angle \)G\(_{0}\)BG\(_{2}\) \(=\) 50.309\(^{\circ }\). Note that \(\alpha _{0}\) is the root of the equation

$$ F_3 \left( {\alpha_0 ,\alpha_0} \right) - F_1 \left( {\alpha_0 ,\alpha_0} \right) =-1. $$

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

$$ \sigma_0 V_1 =P_3 u_A -P_1 \mathit{v}_A +Pu( P ), $$

where \(u_{A}\), \(\mathit {v}_{A}\) are given by (20) and

$$ u( P )=u( {\theta ,\theta} ), $$

where \(u\) is defined by (6) for \(k = 2\mu \) or

$$ \begin{array}{rll} {u( P )} / a&=&( {1+2\mu})\left( {F_1( {\theta ,\theta} )-F_3 ( {\theta ,\theta} )}\right)\\ &&\quad+G_0 ( {\theta ,\theta} )+2F_2 ({\theta ,\theta}). \end{array} $$

Let us note that \(\mathit {v}_{A} = 0\). Indeed, by (19) we compute

$$ \begin{array}{rll} {\mathit{v}_A} / a&=&G_0 \left( {\alpha_0 ,\alpha_0} \right)-2F_2\left( {\alpha_0 ,\alpha_0} \right)\\ &&\quad-2F_1 \left( {\alpha_0 ,\alpha_0}\right)+G_1 \left( {\alpha_0 ,\alpha_0} \right). \end{array} $$

but \(G_0 =F_0 +F_2 \), \(G_1 =F_1 +F_3 \) (cf. (7) in Lewiński et al. 1994) hence

$$ {\mathit{v}_A} / a=F_0 \left( {\alpha_0 ,\alpha_0} \right)-F_2 \left( {\alpha_0 ,\alpha_0} \right)+F_3 \left( {\alpha_0 ,\alpha_0}\right)-F_1 \left( {\alpha_0 ,\alpha_0} \right). $$

Recalling the identity (cf. (8) in Lewiński et al. 1994)

$$ F_0 ( {\beta ,\alpha} )-F_2 ( {\beta ,\alpha} )=\cos( {\beta -\alpha} ) $$

and taking into account (21) we conclude that \(\mathit {v}_{A} = 0\).

The equilibrium equations for the structure of Fig. 6 read

$$ \begin{array}{rll} P_1 &=& P[ L /a + \cos \mu], \\ P_2 &=& P[ L/a + \sin \mu + \cos \mu], \\ P_3 &=& -P\cos \mu, \end{array} $$

and \(L=| {\text {G}_{\text {1}} \text {P}} |\) or, see Eq.III.26

$$ L=a\left( {F_1 ( {\theta ,\theta} )-F_3 ({\theta ,\theta})} \right). $$

Let us compute \(u_{A}/a\) by (23), (17), (21):

$$ u_A=a\left[ {G_0 \left( {\alpha_0 ,\alpha_0} \right)+ 2 F_2 \left( {\alpha_0 ,\alpha_0}\right)}-1 \right]. $$

We compute \(V\) by (22), (28), (29), taking into account that \(v_{A} = 0\):

$$\begin{array}{rll} V/V_0&=&-\cos \mu \left(u_A/a\right)+ u(P)/a \\ &&+\, L/a + \sin \mu +\cos \mu. \end{array} $$

Let us define the function

$$ f( \alpha )=G_0 ( {\alpha ,\alpha} )+2F_2 ({\alpha ,\alpha} ). $$

We rearrange (31) to the form

$$ \begin{array}{rll} V /{V_0} &=&2( {1+\mu} )( {L / a} )+\sin \mu \\ &&\quad+\cos \mu \left[{2-f\left( {\alpha_0} \right)} \right]+f( {\theta}). \end{array} $$

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:

$$ a( {1-\cos \mu} )+L\sin \mu =5a. $$

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.

Fig. 8
figure 8

The numerically found layout of problem of Fig. 6 for the data: \(\mu = 95.7106^{\circ }\), \(\theta = 113.4456^{\circ }\) and ground structure with 474 mln of potential bars; \(V_{num} = 42.91234V_{0}\)

Final remarks

The present paper provides new analytical and numerical solutions to the Michell problem, of transmitting a given point load to given supporting nodes, corresponding to the feasible domain being L-shaped, thus being concave, with one reentrant corner. The solutions displayed are characterized by the following features:

  • the optimal structures do not make use of the whole feasible domain; some empty domains, around the boundary as well as within the feasible domain appear, see Fig. 5;

  • the solutions turn out to be highly sensitive to the load direction, see Fig. 1; the domain between the vertical bars will be (partly) filled up if the load deviates slightly from the vertical direction, see Fig. 6.

All analytical solutions presented in this paper have been confirmed with four-digits precision by the numerical solutions corresponding to highly dense ground structures.