Formulation
Let three points: R, N, D are given. In this plane we introduce the Cartesian coordinate system (x, y) of origin R, the x axis being directed along RN, see Fig. 1. The points R, N, D will be nodes of the unknown framework. Only these nodes are subject to the forces: \(\boldsymbol{P} = (P_x, P_y)\), \(\boldsymbol{Q} = (Q_x, Q_y)\), \(\boldsymbol{F} = (F_x, F_y)\). Without loss of generality one of the nodes—the node R—can be assumed to be fixed. Thus the force at R will be viewed as the reaction \(\boldsymbol{P} = (P_x, P_y)\), see Fig. 1.
Our task is to find the lightest framework subjected to the forces \(\boldsymbol{P}\), \(\boldsymbol{F}\), \(\boldsymbol{Q}\) at the nodes R, D, N, respectively, of the areas of cross-sections chosen such that the axial stress in the members is bounded from both sides as follows: \(-\sigma_p \leqslant \sigma \leqslant \sigma_p\).
It will occur that in all members the equality: |σ| = σ
p
is attained, which means that the bars are uniformly stressed.
In the present paper we confine our attention to the sub-class of the problem of three forces given by the layouts consisting of two circular fans: RBA (of radius r
2) and NAC (of radius r
1) and of a fibrous domain ABDC, composed of two families of curvilinear and orthogonal fibres. The force \(\boldsymbol{F}\) at D induces the forces in the bars BD and CD of different signs. Thus the resultant of the components F
x
, F
y
must lie within the dashed lines around D. This sub-class of solutions of the problem of three forces was discovered by Chan (1966). To make the present paper self-contained we shall outline the method of Chan of finding the angular parameters
$$ \gamma_1 = \sphericalangle\mbox{RNA}, \quad \theta_1 = \sphericalangle\mbox{ANC}, \quad \theta_2 = \sphericalangle\mbox{BRA} \label{w2.1} $$
(2.1)
which describe geometry of the whole fibrous domain RBDCNAR.
Geometry of the domain RBDCNAR
The arcs AC and AB have circular shapes, the radii being
$$\begin{array}{rll} r_1 & = & | \mbox{NA} | = | \mbox{NC} |\\ r_2 & = & | \mbox{RA} | = | \mbox{RB} | \label{w2.2} \end{array} $$
(2.2)
and can be expressed by the angles γ
1, γ
2
$$ r_1 = d \cos \gamma_1, \qquad r_2 = d \cos \gamma_2 \label{w2.3} $$
(2.3)
where \(\gamma_1 + \gamma_2 = {\pi \over 2}\); the distance d = |RN| being given. The sides RA and NA of the right angled triangle RAN determine a Cartesian coordinate system (x
0, y
0) of origin at A. The domain ABDC is parameterized by a curvilinear orthogonal system (α, β). The coordinates (α, β) of the vertices of this domain are
$$ \mbox{A}(0,\ 0), \quad \mbox{C}(\theta_1,\ 0), \quad \mbox{B}(0,\ \theta_2), \quad \mbox{D}(\theta_1,\ \theta_2) $$
The analytical construction of the (α, β) net has been for the first time developed by Carathéodory and Schmidt (1923) and then by Chan (1964). The Lamé coefficients at an arbitrary point (λ, μ), referring to the given curvilinear parameterization are expressed by the formulae
$$ \begin{array}{rll} A(\lambda,\ \mu) &=& r_1 G_0 (\lambda,\ \mu) + r_2 G_1 (\mu,\ \lambda) \\ B(\lambda,\ \mu) &=& r_2 G_0 (\lambda,\ \mu) + r_1 G_1 (\lambda,\ \mu) \end{array} \label{w2.4} $$
(2.4)
cf. (6.7) and (6.8) in Graczykowski and Lewiński (2006a); the functions G
n
(λ, μ) are defined by (1) in Lewiński et al. (1994a). Let us define the functions
$$ \begin{array}{rll} \bar{x}(\lambda,\ \mu) &=& r_1F_1(\lambda,\ \mu) + r_2F_2(\mu,\ \lambda) \\ \bar{y}(\lambda,\ \mu) &=& r_1F_2(\lambda,\ \mu) + r_2F_1(\mu,\ \lambda) \end{array} \label{w2.5} $$
(2.5)
where F
n
(λ, μ) are given by (2) in Lewiński et al. (1994a). Let
$$ \phi (\lambda,\ \mu) = \mu - \lambda \label{w2.6} $$
(2.6)
The Cartesian coordinates (x
o, y
o) of a point (λ, μ) are
$$\begin{array}{rll} x^o (\lambda,\ \mu) & = & \bar{x} (\lambda,\ \mu) \cos \left( \phi(\lambda,\ \mu) \right) \\ & & - \bar{y} (\lambda,\ \mu) \sin \left( \phi(\lambda,\ \mu) \right) \\ y^o (\lambda,\ \mu) & = & \bar{x} (\lambda,\ \mu) \sin \left( \phi(\lambda,\ \mu) \right) \\ & & + \bar{y} (\lambda,\ \mu) \cos \left( \phi(\lambda,\ \mu) \right) \label{w2.7} \end{array}$$
(2.7)
see (6.16) in Graczykowski and Lewiński (2006a). The coordinates (x, y) of point (λ, μ) are given by
$$ \begin{array}{rll} x(\lambda,\ \mu) & = & \left[ x^o(\lambda,\ \mu) + r_2 \right] \cos \gamma_2 \\ & & - y^o (\lambda,\ \mu) \sin \gamma_2 \\ y(\lambda,\ \mu) & = & \left[ x^o(\lambda,\ \mu) + r_2 \right] \sin \gamma_2 \\ & & + y^o (\lambda,\ \mu) \cos \gamma_2 \label{w2.8} \end{array} $$
(2.8)
and just with using these formulae the net of lines in Fig. 2 was constructed.
By using (2.3) and (7)–(14) in Lewiński et al. (1994a) one can rearrange (2.8) to the form
$$ \begin{array}{rll} \displaystyle {x(\lambda,\ \mu) \over d} &=& \sin \gamma_2\ k_1 (\lambda,\ \mu) - \cos \gamma_2\ k_2(\lambda,\ \mu) \\ \displaystyle {y(\lambda,\ \mu) \over d} &=& -\sin \gamma_2\ h_1 (\lambda,\ \mu) + \cos \gamma_2\ h_2(\lambda,\ \mu) \end{array} \label{w2.9} $$
(2.9)
with
$$ \begin{array}{rll} h_1(\lambda,\ \mu) & = & -\cos \left( \gamma_2 + \phi(\lambda,\ \mu) \right) F_2(\lambda,\ \mu) \\ & & - \sin \left( \gamma_2 + \phi(\lambda,\ \mu) \right) F_1(\lambda,\ \mu) \\ h_2(\lambda,\ \mu) & = & \cos \left( \gamma_2 + \phi(\lambda,\ \mu) \right) F_1(\lambda,\ \mu) \\ & & + \sin \left( \gamma_2 + \phi(\lambda,\ \mu) \right) F_0(\lambda,\ \mu) \label{w2.10} \end{array} $$
(2.10)
and
$$ \begin{array}{rll} k_1(\lambda,\ \mu) & = & -\sin \left( \gamma_2 + \phi(\lambda,\ \mu) \right) F_2(\lambda,\ \mu) \\ & & + \cos \left( \gamma_2 + \phi(\lambda,\ \mu) \right) F_1(\lambda,\ \mu) \\ k_2(\lambda,\ \mu) & = & \sin \left( \gamma_2 + \phi(\lambda,\ \mu) \right) F_1(\lambda,\ \mu) \\ & & - \cos \left( \gamma_2 + \phi(\lambda,\ \mu) \right) F_0(\lambda,\ \mu) \label{w2.11} \end{array} $$
(2.11)
Computation of the reactions at node N
Let x
D = x (θ
1, θ
2), y
D = y(θ
1, θ
2) be coordinates of point D in the coordinate system (x, y) of origin at R.
Let us write down the equilibrium equations involving the reactions Q
x
, Q
y
at node N. The condition of vanishing of the moment of external forces with respect to the hinge R reads
$$ -dQ_y - x_{\rm D} F_y + y_D F_x = 0 \label{w2.12} $$
(2.12)
We need one formula more to find the reactions Q
x
and Q
y
. There are two manners to find such a formula. The first, natural way is very laborious—one should solve the equilibrium problem of the domain ABDC, starting from the equilibrium equations of the node D. These two equilibrium equations result in the values of the axial forces in bars BD and CD. Having found these axial forces (which are constant along the boundary lines DB, DC) one can solve the equilibrium problem of the interior of the domain ABDC, i.e. find the internal forces in the meaning of Hemp (1973) by following the method developed in Graczykowski and Lewiński (2007a). These internal forces are defined by
$$ T_1 = BN_I, \qquad T_2 = AN_{II} \label{w2.13} $$
(2.13)
where N
I
and N
II
are the principal stress resultants within the theory of plane stress; the principal directions (I, II) coincide with the (α, β) trajectories. Having the forces T
1 and T
2 along AC one can find the distribution of T
2 within the fan NAC and then the resultant of these forces at N, being the reaction \(\boldsymbol{Q}\). The component Q
y
thus obtained will coincide with the solution of the algebraic equation (2.12). Additionally we shall be able to compute the component Q
x
.
Much shorter way of computing Q
x
was proposed by Chan (1966). One should make use of the variational equation of equilibrium, cf. (6.1) in Graczykowski and Lewiński (2007b):
$$ \begin{array}{lll} &&{\kern-2pt}\int\!\int \left[ T_1 \left( {\partial u \over \partial \alpha} + v \right) + T_2 \left( {\partial v \over \partial \beta} + u \right) \right] d\alpha d\beta \\ &&{\kern6pt}\qquad\;+\, \mathcal{L}_{\rm NCD} + \mathcal{L}_{\rm RBD} = \mathcal{L}(u,\ v) \label{w2.14} \end{array} $$
(2.14)
where u (α, β), v(α, β) are components of the virtual displacements (referred to the (α, β) parameterization) the integration being taken over the whole structure. The linear form \(\mathcal{L}(u,\ v)\) represents the virtual work of the loading. In the case considered the structure contains two bars (RBD and NCD) of finite cross sections and that is why the left hand side of (2.14) is complemented by \(\mathcal{L}_{\rm NCD} + \mathcal{L}_{\rm RBD}\)—the virtual work of axial forces in these bars on the axial deformations, like in (2.5) of Graczykowski and Lewiński (2006a).
Now we choose the fields u and v in such a manner that the left hand side of (2.14) vanishes
$$ {\partial u \over \partial \alpha} + v = 0, \qquad {\partial v \over \partial \beta} + u = 0 \label{w2.15} $$
(2.15)
Then the axial deformations along the fibres α and β vanish; consequently, the virtual works L
NCD and L
RBD assume zero vales. We shall assume additionally that
$$ \varepsilon_{r\vartheta} = 0 \ \mbox{in}\ \mbox{NCA},\,\, \mbox{RBA} \label{w2.16} $$
(2.16)
The virtual work represented by the right-hand side of (2.14) comprises the virtual work of reactions P
x
and P
y
. To make this virtual work zero it is assumed that
$$ u(\rm R) = 0, \qquad v(\rm R) = 0 \label{w2.17} $$
(2.17)
It will turn out that the family of fields (u, v) satisfying (2.15) and (2.16) is two-parameter—it depends on two independent constants. Thus the (2.14) will imply two algebraic equilibrium equations. These two equations imply (2.12) and, additionally, an equation linking Q
x
, F
x
and F
y
. It was already mentioned that this second equation cannot be easy inferred from the equilibrium equations. Therefore, the kinematic method proposed by Chan (1966) will turn out to be especially effective in the problem discussed.
Let us turn to the details. We shall construct the kinematic fields u, v satisfying the conditions (2.15) within ABDC, the conditions of vanishing of the radial and circumferential strains within the fans RBA and NAC, the continuity conditions as well as the conditions (2.16). We start with construction of these fields in the fan RBA.
The polar coordinates are denoted by (r, ϑ), the fields u, v being measured along these directions, cf. Fig. 3. Let us recall the formulae defining the strain components in the polar system
$$ \begin{array}{lll} \varepsilon_r &=& \frac{\partial u} {\partial r}, \qquad \varepsilon_\vartheta = \frac{u}{r} +\frac{1}{r} \frac{\partial v}{\partial \vartheta}, \\ 2\varepsilon_{r\vartheta} &=& \frac {1}{r} \frac{\partial u }{\partial \vartheta} + \frac{\partial v }{\partial r} - \frac{v}{r} \label{w2.18} \end{array} $$
(2.18)
The conditions
$$ \varepsilon_r = 0, \qquad \varepsilon_\vartheta =0, \qquad \varepsilon_{r\vartheta} = 0 \label{w2.19} $$
(2.19)
imply the following representation of the displacement fields
$$ \begin{array}{rll} u^{\rm RAB} & = & \bar{C}_1 \cos (\vartheta - \bar{\varphi}), \\ v^{\rm RAB} & = & \bar{\psi} r - \bar{C}_1 \sin (\vartheta - \bar{\varphi}) \label{w2.20} \end{array} $$
(2.20)
These equations follow from (77) of Lewiński and Rozvany (2007), where ϵ = 0 was put. Here \(\bar{C}_1\), \(\bar{\psi}\), \(\bar{\varphi}\) are arbitrary constants. The conditions (2.17) lead to
$$ u^{\rm RAB} = 0, \qquad v^{\rm RAB} = \bar{\psi} r \label{w2.21} $$
(2.21)
Thus, along AB we have
$$ u^{\rm RAB}_{|\rm AB} = 0 \qquad v^{\rm RAB}_{|\rm AB} = \bar{\psi} r_2 \label{w2.22} $$
(2.22)
Let us construct the fields u
NAC, v
NAC within NAC (Fig. 4).
By (2.22) the fields u, v within NC must satisfy the following conditions at A:
$$ u^{\rm NAC}_{\rm A} = \bar{\psi} r_2, \qquad v^{\rm NAC}_{\rm A} = 0 \label{w2.23} $$
(2.23)
The representation (u, v) in NAC is of the form (2.20), or
$$\begin{array}{rll} u^{\rm NAC} (r,\ \vartheta) &=& C_1 \cos (\vartheta - \varphi), \\ v^{\rm NAC} (r,\ \vartheta) &=& \psi r - C_1 \sin (\vartheta - \varphi) \end{array} \label{w2.24} $$
(2.24)
The conditions (2.22) result in
$$ C_1 \cos (\theta_1 - \varphi) = \bar{\psi} r_2 \quad \psi r_1 - C_1 \sin (\theta_1 - \varphi) = 0 \label{w2.25} $$
(2.25)
Let us introduce the constants
$$ C^0_1 = C_1 \sin (\theta_1 - \varphi), \quad C^0_2 = - C_1 \cos(\theta_1 - \varphi) \label{w2.26} $$
(2.26)
Hence
$$ \bar{\psi} r_2 = -C_2^0, \qquad \psi r_1 = C^0_1 \label{w2.27} $$
(2.27)
The representations of (u, v) are
Domain RAB
$$ u = 0, \qquad v = -{C_2^0 \over r_2} r \label{w2.28} $$
(2.28)
Domain NAC
$$ \begin{array}{rll} u &=& C_1^0 \sin \alpha - C_2^0 \cos \alpha, \\ v &=& -C_1^0 \left( \cos \alpha - {r \over r_1} \right) - C_2^0 \sin \alpha \end{array} \label{w2.29} $$
(2.29)
Along the lines AB, AC, the fields (u, v) in ABDC must be compatible with (2.28) and (2.29) or
$$ \begin{array}{rll} u^{\rm ABDC}_{|\rm AB} & = & u^{\rm ARB}_{|\rm AB}, \qquad\;\;\, v^{\rm ABDC}_{|\rm AB} = v^{\rm ARB}_{|\rm AB} \\ u^{\rm ABDC}_{|\rm AC} & = & -v^{\rm NAC}_{|\rm AC}, \qquad v^{\rm ABDC}_{|\rm AC} = u^{\rm NAC}_{|\rm AC} \label{w2.30} \end{array} $$
(2.30)
Consequently, the values of u
ABDC, v
ABDC on the arc AB refer to the line α = 0
$$ u^{\rm ABDC} (0,\ \beta) = 0, \qquad v^{\rm ABDC} (0,\ \beta) = -C^0_2 \label{w2.31} $$
(2.31)
while the values of these fields on the arc AC refer to the case of β = 0
$$ \begin{array}{rll} u^{\rm ABDC} (\alpha,\ 0) &=& C_1^0 (\cos \alpha -1) + C_2^0 \sin \alpha \\ v^{\rm ABDC} (\alpha,\ 0) &=& C_1^0 \sin \alpha - C_2^0 \cos \alpha \end{array} \label{w2.32} $$
(2.32)
Let us pass to the construction of the fields u = u
ABDC, v = v
ABDC.
According to (2.15) these fields satisfy the partial differential equations of the form
$${\partial^2 u \over \partial\alpha \partial\beta} - u = 0, \qquad {\partial^2 v \over \partial\alpha \partial\beta} - v = 0 \label{w2.33} $$
(2.33)
The values of u, v at an arbitrary point (λ, μ) of the domain ABDC are given by the Riemann’s formula, cf. (17) in Lewiński et al. (1994a)
$$ \begin{array}{rll} u(\lambda,\ \mu) & = & u(0,\ 0) G_0 (\lambda,\ \mu) \\ &&+\, \int_0^\lambda G_0 (\lambda -\alpha,\ \mu) {\partial u (\alpha,\ 0) \over \partial \alpha} d\alpha \\ &&+\, \int_0^\mu G_0(\lambda,\ \mu - \beta) {\partial u (0,\ \beta) \over \partial \beta} d\beta \label{w2.34} \end{array} $$
(2.34)
Substitution of (2.31), (2.32) gives
$$ \begin{array}{rll} u(\lambda,\ \mu) &=& -C_1^0 \int_0^\lambda G_0 (\lambda -\alpha,\ \mu) \sin\alpha d\alpha \\ &&+ C_2^0 \int_0^\mu G_0(\lambda -\alpha,\ \mu ) \cos \alpha d\alpha \label{w2.35} \end{array} $$
(2.35)
By using the formulae (16) of Lewiński et al. (1994a) one finds
$$ u(\lambda,\ \mu) = -C_1^0 F_2(\lambda,\ \mu) + C_2^0 F_1(\lambda,\ \mu) \label{w2.36} $$
(2.36)
while applying the differentiation rules (4) of Lewiński et al. (1994a) we arrive at
$$ v(\lambda,\ \mu) = C_1^0 F_1(\lambda,\ \mu) - C_2^0 F_0(\lambda,\ \mu) \label{w2.37} $$
(2.37)
Now we can determine the displacements at the node D
$$ u({\rm D}) = u(\theta_1,\ \theta_2), \qquad v({\rm D}) = v(\theta_1,\ \theta_2) \label{w2.38} $$
(2.38)
The displacements of the point D in the directions x and y are computed by the rules
$$ \begin{array}{rll} w_x({\rm D}) &=& u({\rm D}) \cos(\gamma_2 + \phi_{\rm D}) - v({\rm D}) \sin(\gamma_2 + \phi_D), \\ w_y({\rm D}) &=& u({\rm D}) \sin(\gamma_2 + \phi_{\rm D}) + v({\rm D}) \cos(\gamma_2 + \phi_D) \\ \label{w2.39} \end{array} $$
(2.39)
where
$$ \phi_D = \theta_2 - \theta_1 \label{w2.40} $$
(2.40)
Substitution of (2.36)–(2.38) results in
$$ \begin{array}{rll} w_x({\rm D}) &=& C_1^0 h_1 (\theta_1,\ \theta_2) + C_2^0 h_2 (\theta_1,\ \theta_2), \\ w_y({\rm D}) &=& C_1^0 k_1 (\theta_1,\ \theta_2) + C_2^0 k_2 (\theta_1,\ \theta_2) \end{array} \label{w2.41} $$
(2.41)
where h
1, h
2, k
1, k
2 are given by the (2.10) and (2.11).
We shall find the displacements of the node N in the directions x and y. First we shall compute the displacements in the polar coordinate system within ANC. Substitution of r = 0, α = 0 in (2.29) gives
$$ u^{\rm NAC} ({\rm N}) = - C_2^0, \qquad v^{\rm NAC} ({\rm N}) = - C_1^0 \label{w2.42} $$
(2.42)
Thus, see Fig. 5
$$ \begin{array}{rll} w_x({\rm N}) &=& -u^{\rm NAC} ({\rm N}) \cos\gamma_1 -v^{\rm NAC} ({\rm N}) \sin\gamma_1 \\ w_y({\rm N}) &=& u^{\rm NAC} ({\rm N}) \sin\gamma_1 -v^{\rm NAC} ({\rm N}) \cos\gamma_1 \end{array} \label{w2.43} $$
(2.43)
Substitution of (2.42) gives
$$ \begin{array}{rll} w_x({\rm N}) &=& C_2^0 \cos\gamma_1 + C_1^0 \sin\gamma_1 \\ w_y({\rm N}) &=& -C_2^0 \sin\gamma_1 + C_1^0 \cos\gamma_1 \end{array} \label{w2.44} $$
(2.44)
Now we are prepared to applying the variational equation (2.14). The displacements fields (u, v) assumed are chosen such that the left hand side of (2.14) vanishes, or \(\mathcal{L}(u,\ v) = 0\), where \(\mathcal{L}\) represents the virtual work of all loads. Since the fields u and v vanish at R, the virtual work is done by the other point loads
$$ F_x w_x({\rm D}) + F_y w_y({\rm D}) + Q_x w_x({\rm N}) + Q_y w_y({\rm N}) = 0 \label{w2.45} $$
(2.45)
where w
x
(D), w
y
(D), w
x
(N), w
y
(N) are given by (2.41) and (2.44) depending on two parameters \(C_1^0\) and \(C_2^0\). Since Q
y
can be computed by using (2.12) we shall isolate the formula which determines Q
x
. To this end we choose
$$ C_1^0 = C_2^0\tan \gamma_1 \label{w2.46} $$
(2.46)
to make w
y
(N) zero. Substitution (2.46) into (2.41), (2.44) and (2.45) gives the formula we have looked for
$$ \begin{array}{rll} Q_x &=& - \left[ \sin \gamma_1 h_1 (\theta_1,\ \theta_2) + \cos \gamma_1 h_2 (\theta_1,\ \theta_2) \right] F_x \\ &&- \left[ \sin \gamma_1 k_1 (\theta_1,\ \theta_2) + \cos \gamma_1 k_2 (\theta_1,\ \theta_2) \right] F_y \label{w2.47} \end{array} $$
(2.47)
Let us stress once again that this formula was originally found by Chan (1966).
The volume of the optimal framework
The volume of the framework of Fig. 2 will be found by the kinematic method. We shall construct a virtual field of displacements (u, v) satisfying the Michell conditions
$$ \varepsilon_I =1, \qquad \varepsilon_{II} = -1 \label{w2.48} $$
(2.48)
within the whole feasible domain, being the half-plane \(y\geqslant0\). Then we shall determine the volume of the lightest framework by the known rule
$$ V = {1\over \sigma_p} \mathcal{L}(u,\ v) \label{w2.49} $$
(2.49)
where \(\mathcal{L}\) represents the work of the given loads on the given field (u, v). We shall assume that the components (u, v) vanish at R; then \(\mathcal{L}\) becomes the virtual work of the forces Q
x
, Q
y
, F
x
, F
y
.
The construction of the virtual fields (u, v) starts with the triangular domain RAN, parameterized by the Cartesian system (x
0, y
0), see Fig. 6.
Let us assume that the bar RB is in tension. Then the fields (u, v) satisfying (2.48) and vanishing at R are of the form
$$ \begin{array}{rll} u^{\rm RAN} & = & x_0 - Cy_0 + r_2, \\ v^{\rm RAN} & = & -y_0 + Cx_0 + Cr_2 \label{w2.50} \end{array} $$
(2.50)
where C is a constant. Let us consider the domain RBA with the polar parameterization (r, ϑ), see Fig. 7.
According to (77) in Lewiński and Rozvany (2007) we have
$$ u^{\rm RBA} = r, \qquad v^{\rm RBA} = -2r\vartheta + \psi r \label{w2.51} $$
(2.51)
since the bar RB is in tension while point R does not move. The fields (2.50) and (2.51) must be compatible, i.e. the continuity conditions along RA should hold. This links two constants:
$$ C = \psi \label{w2.52} $$
(2.52)
Consider the domain NAC. We assume that the radial fibres are in compression. Thus, according to (77) in Lewiński and Rozvany (2007) we have
$$ \begin{array}{rll} u^{\rm NAC} &=& -r + C_1 \cos(\vartheta - \varphi), \\ v^{\rm NAC} &=& 2r\vartheta + \tilde{\psi} r - C_1 \sin(\vartheta - \varphi) \end{array} \label{w2.53} $$
(2.53)
The compatibility conditions linking the fields (2.50) and (2.53) along NA result in the conditions
$$ \begin{array}{rll} &&\tilde{\psi} = \psi -2\theta_1 \\ &&C_1 \cos (\theta_1 - \varphi) = \psi r_2 + r_1 \\ &&C_1 \sin (\theta_1 - \varphi) = \psi r_1 + r_2 \label{w2.54} \end{array} $$
(2.54)
Substitution of (2.54) into (2.53), along with the formula ϑ = θ
1 − α results in
$$ \begin{array}{rll} u^{\rm NAC} & = & -r + (\psi r_2 + r_1) \cos \alpha + (\psi r_1 + r_2) \sin \alpha \\ v^{\rm NAC} & = & -2r\alpha + \psi r - (\psi r_1 + r_2) \cos \alpha \\ & & +\, (\psi r_2 + r_1) \sin \alpha \label{w2.55} \end{array} $$
(2.55)
The components of displacements of node N (Fig. 8), defined as in domain RAN, are given by
$$ u^{\rm RAN} ({\rm N}) = \psi r_1 + r_2, \qquad v^{\rm RAN} ({\rm N}) = r_1 + \psi r_2 \label{w2.56} $$
(2.56)
where we have made use of: x
0 = 0, y
0 = − r
1, C = ψ.
The displacements of N in the directions x and y are
$$ \begin{array}{rll} w_x({\rm N}) &=& u^{\rm RAN}({\rm N}) \sin \gamma_1 - v^{\rm RAN} ({\rm N}) \cos \gamma_1 \\ w_y({\rm N}) &=& u^{\rm RAN}({\rm N}) \cos \gamma_1 + v^{\rm RAN} ({\rm N}) \sin \gamma_1 \end{array} \label{w2.57} $$
(2.57)
or
$$ \begin{array}{rll} w_x({\rm N}) &=& {1\over d} \big[ (r_2)^2 - (r_1)^2 \big] \\ \displaystyle w_y({\rm N}) &=& {1\over d} \big[\psi\big((r_1)^2 + (r_2)^2 \big) + 2r_1r_2\big] \end{array} \label{w2.58} $$
(2.58)
We see now that constant ψ represents a small angle of rigid rotation of the whole structure around point R.
Let us pass now to the construction of displacements (u, v) in the domain ABDC, referred to the lines α and β. One should determine the values of these fields along the arcs AB and AC. Due to continuity of displacements on AB we observe that
$$ \begin{array}{rll} u^{\rm ABDC} (\alpha = 0,\ \beta) &=& u^{\rm RBA} (r=r_2,\ \vartheta = \beta) \\ v^{\rm ABDC} (\alpha = 0,\ \beta) &=& v^{\rm RBA} (r=r_2,\ \vartheta = \beta) \end{array} \label{w2.59} $$
(2.59)
or
$$ \begin{array}{rll} u^{\rm ABDC} (0,\ \beta) &=& r_2 \\ v^{\rm ABDC} (0,\ \beta) &=& - 2r_2\beta + \psi r_2 \end{array} \label{w2.60} $$
(2.60)
By continuity conditions along AC we have
$$ \begin{array}{rll} u^{\rm ABDC} (\alpha,\ \beta = 0) &=& -v^{\rm NAC} (r=r_1,\ \alpha) \\ v^{\rm ABDC} (\alpha,\ \beta = 0) &=& u^{\rm NAC} (r=r_1,\ \alpha) \end{array} \label{w2.61} $$
(2.61)
hence
$$ \begin{array}{rll} u^{\rm ABDC} (\alpha,\ 0) & = & 2r_1\alpha - \psi r_1 + (\psi r_1 + r_2) \cos \alpha \\ & & -\, (\psi r_2 + r_1) \sin\alpha \\ v^{\rm ABDC} (\alpha,\ 0) & = & -r_1 + (\psi r_2 + r_1) \cos \alpha \\ & & +\, (\psi r_1 + r_2) \sin\alpha \label{w2.62} \end{array} $$
(2.62)
The quantity ψ is arbitrary, since it represents an angle of rigid body rotation around R. Let us assume this constant such that ψr
1 + r
2 = − (ψr
2 + r
1). Then ψ = − 1.
Let us introduce the auxiliary field within ABDC
$$ u^0 (\alpha,\ \beta) = u (\alpha,\ \beta) - 2\alpha A(\alpha,\ \beta) \label{w2.63} $$
(2.63)
see (66) in Lewiński et al. (1994a). We compute the values of the Lamé coefficient A along AB and AC
$$ \begin{array}{rll} A(0,\ \beta) &=& r_1 G_0(0,\ \beta) + r_2 G_1 (\beta,\ 0) = r_1 + \beta r_2\\ A(\alpha,\ 0) &=& r_1 G_0(\alpha,\ 0) + r_2 G_1 (0,\ \alpha) = r_1 \end{array} \label{w2.64} $$
(2.64)
as well as the field u = u
ABDC
$$ \begin{array}{rll} u(0,\ \beta) &=& r_2 \\ u(\alpha,\ 0) &=& 2 r_1\alpha + r_1 + (r_2 - r_1) (\cos \alpha + \sin \alpha) \end{array} \label{w2.65} $$
(2.65)
The values of the field u
0 on AB and AC are
$$ \begin{array}{rll} u^0 (0,\ \beta) &=& r_2 \\ u^0 (\alpha,\ 0) &=& r_1 + (r_2 - r_1) (\cos \alpha + \sin \alpha) \end{array} \label{w2.66} $$
(2.66)
The field u
0 satisfies (2.33). Its solution has the form (2.34). We compute the derivatives
$$ \begin{array}{rll} {\partial u^0 (\alpha,\ 0) \over \partial \alpha} &=& (r_2 - r_1) (\cos \alpha - \sin \alpha) \\ {\partial u^0 (0,\ \beta) \over \partial \beta} &=& 0, \qquad u^0(0,\ 0) = r_2 \end{array} \label{w2.67} $$
(2.67)
Substitution of these results into (2.34) gives
$$ {\kern-5pt} \begin{array}{rll} u^0(\lambda,\ \mu) &\!\!\!=\!\!\!&\! r_2 G_0(\lambda,\ \mu)\!\! \\ &&\!+\, (r_2 - r_1) \int_0^\lambda G_0 (\lambda - \alpha,\ \mu) \cos \alpha d\alpha \!\!\\ &&\!-\,(r_2 - r_1) \int_0^\lambda G_0 (\lambda - \alpha,\ \mu) \sin \alpha d\alpha \end{array} \label{w2.68} $$
(2.68)
By using the formulae (16) of Lewiński et al. (1994a) one finds
$$ \begin{array}{rll} u^0 (\lambda,\ \mu) & = & r_2 G_0(\lambda,\ \mu) + (r_2 - r_1) \\ && \times\left[ F_1(\lambda,\ \mu) - F_2(\lambda,\ \mu) \right] \label{w2.69} \end{array} $$
(2.69)
and then we find u(λ, μ) by (2.63) and (2.4); the result reads
$$ \begin{array}{rll} u(\lambda,\ \mu) & = & 2r_1\lambda G_0(\lambda,\ \mu) \\ && +\, r_2 \left[ G_0(\lambda,\ \mu) + 2\mu G_1(\lambda,\ \mu) \right] \\ && +\, (r_2 - r_1) \left[ F_1(\lambda,\ \mu) - F_2(\lambda,\ \mu) \right] \label{w2.70} \end{array} $$
(2.70)
The field v is found from the equation
$$ v(\lambda.\ \mu) = A(\lambda,\ \mu) - {\partial u (\lambda,\ \mu) \over \partial \lambda} \label{w2.71} $$
(2.71)
see (65) in Lewiński et al. (1994a). By using the differentiation rules (4) of the latter paper one finds
$$ \begin{array}{rll} v(\lambda,\ \mu) & = & -2r_1\mu G_1(\lambda,\ \mu) \\ & & - r_2 (1+2\mu) G_0(\lambda,\ \mu) \\ & & + (r_2 - r_1) \left[ F_1(\lambda,\ \mu) + F_2(\lambda,\ \mu) \right] \label{w2.72} \end{array} $$
(2.72)
The displacements w
x
(D), w
y
(D) of point D of coordinates (θ
1, θ
2) are determined by (2.39), where
$$ u({\rm D}) = u(\theta_1,\ \theta_2), \qquad v({\rm D}) = v(\theta_1,\ \theta_2) \label{w2.73} $$
(2.73)
The displacements w
x
(N), w
y
(N) of point N are given by (2.58), where ψ = − 1. Having found the displacements of the nodes N and D as well as the values of the forces Q
x
and Q
y
given by (2.12) and (2.47), one can compute the volume of the whole framework by using (2.49), or
$$ \begin{array}{lll} V & = & {1\over \sigma_p} \left[ F_x w_x({\rm D}) + F_y w_y({\rm D}) + Q_x w_x({\rm N}) \right. \\ & & {\kern15pt} \left. +\, Q_y w_y({\rm N}) \right] \label{w2.74} \end{array} $$
(2.74)
Let us emphasize once again that the result above is not valid in general; it refers to the case of such inclination of the force \(\boldsymbol{F} =(F_x,\ F_y)\) that the bar RBD is in tension and the bar NCD is in compression. The resultant \(\boldsymbol{F}\) must lie between the lines l
1 and l
2 shown in Fig. 2.