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Published in: Continuum Mechanics and Thermodynamics 5/2023

Open Access 23-06-2023 | Original Article

On the thermal stresses in chiral porous elastic beams

Authors: Simona De Cicco, Dorin Ieşan

Published in: Continuum Mechanics and Thermodynamics | Issue 5/2023

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Abstract

This paper is concerned with the strain gradient theory of porous thermoelastic solids. We study the deformation of isotropic chiral cylinders subjected to a temperature field that is linear in the axial coordinate. It is shown that the solution can be reduced to the study of two-dimensional problems. The results are used to investigate the deformation of a circular cylinder subjected to a uniform temperature variation. In contrast to the case of achiral materials, the thermal field in chiral cylinders produces torsional effects.
Notes
Communicated by Andreas Öchsner.

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1 Introduction

The behavior of chiral elastic bodies is a topic of current interest both from a theoretical and an experimental point of view. The chiral elastic solid was used as model for various crystalline materials, carbon nanotubes, bones, honeycomb structures, as well as composites with inclusions. Eringen [1] introduced the theory of microstretch continua as a generalization of Cosserat theory. In this theory, the material particles undergo a uniform microdilatation and a rigid microrotation. The intended applications of the theory are to porous media filled with gas, composite materials reinforced with chopped elastic fibers, bones and other materials with microstructure. In the case of a microstretch continuum, the change of microdeformation tensor, from the constant value in the natural state, has the form \(\varphi \delta _{ij} + \varepsilon _{jik} \psi _{k}\), where \(\varphi \) is the microdilatation function, \(\psi _{k}\) is the microrotation vector, \(\delta _{ij}\) is the Kronecker delta and \(\varepsilon _{jik}\) is the alternating symbol. In the absence of microrotations, the equations of the linear theory of microstretch elastic solids coincide with the equations of the elastic materials with voids introduced by Cowin and Nunziato [2] to study the deformation of porous solids. In the theory of elastic materials with voids, the volume fraction field is a function strictly positive and less than 1. Considering the multiple possibilities of using the solid microstretch model, in this paper we will use the theory of microstretch continua to investigate the deformation of porous bodies. In what follows, the subscripts preceded by a comma denote partial differentiation with respect to the corresponding coordinate. The function \(\varphi \) is dimensionless, and the functions \(\varphi _{,j}\) and \(u_{i,rs}\) have the same dimensions. If the functions \(\varphi \) and \(\varphi _{,j}\) are considered as independent constituent variables, then the second-order partial derivatives of the displacement vector have to be included in the set of independent constituent variables. Thus, the introduction of the microdilatation function requires a strain gradient theory. Toupin [3, 4] and Mindlin [5] have introduced the strain gradient theory of elasticity. Interest in the study of this theory is stimulated by the fact that it is suitable to investigate problems related to the effects of size and nanotechnology. The deformation of chiral elastic materials cannot be described within classical elasticity [6]. In various papers, the authors have studied the behavior of chiral materials by using the linear theory of gradient elasticity (see, e.g., [79] and references therein). The strain gradient elasticity has been used to investigate the behavior of carbon nanotubes (see, e.g., [10, 11]).
This paper is concerned with a theory of thermoelasticity for isotropic microstretch continua, without microrotations, where the second-order displacement gradient is added to the classical set of independent constitutive variables. We have considered this theory since the chirality behavior in strain gradient theory is controlled by a single material parameter, in contrast to the three additional material parameters required in Cosserat theory. We study the equilibrium problem for a cylinder which, in the absence of body forces and lateral loading, is subjected to prescribed surface tractions on bases and to a thermal field that is linear in the axial coordinates. The origin of this problem goes back to the work of Boley and Weiner [12] that is devoted to classical thermoelasticity. The deformation of achiral porous elastic solids has been investigated in various papers (see, e.g., [2, 1318]). In this paper, we focus our attention to the case of homogeneous and isotropic chiral porous elastic materials. Since the cancellous bone is considered as a porous body [19] as well as a chiral material [20], it seems that the linear theory of gradient elasticity of porous solids is adequate to describe the mechanical behavior of bones. This paper is concerned with uncoupled system in the sense that the temperature field can be found by solving the heat flow problem associated with the heat conduction and energy equation. We shall treat the temperature field as having already been determined [21]. The temperature is a prescribed function that is independent of time.
The paper is structured as follows. First, we present the basic equations of chiral porous thermoelastic solids and formulate the problem of deformation of the right cylinders. We have introduced mechanical loads on the ends in order to compare the effects of thermal field with those produced by the resultants of the tractions that act on the bases. Then, we investigate the generalized plane strain problem. In the following section, we establish the solution of the problem when the temperature distribution is independent of the axial coordinate. It is shown that this thermal field produces extension, bending and torsion. The next section deals with the deformation of cylinders subjected to a temperature field that is linear in the axial coordinate. We present a method to reduce the three-dimensional problem to the study of plane problems. The results are used to study the deformation of a circular cylinder subjected to a uniform temperature variation. In contrast to the case of centrosymmetric materials, the thermal field in chiral cylinders produces torsional effects.

2 Preliminaries

In this section, we present the equations of equilibrium in the context of the strain gradient theory of porous thermoelastic materials. Mindlin [5] presented three forms of the linear theory of gradient elasticity. In what follows, we use the strain measures introduced in the first form of the theory. We consider a body that in the undeformed state occupies the bounded region B with Lispchitz boundary \(\partial B.\) The boundary \(\partial B\) consists of the union of a finite number of smooth surfaces, smooth curves (edges) and points (corners). We denote by C the union of the edges. Throughout this paper, a rectangular Cartesian coordinate system \(Ox_{j}, (j=1,2,3)\), is used. We shall employ the usual summation and differentiation conventions.
We assume that B is occupied by a homogeneous and isotropic chiral thermoelastic solid. Let \(u_{i}\) be the components of the displacement vector, and let \(\varphi \) be the microstretch function (microdilatation function). The strain tensors are defined by
$$\begin{aligned} e_{ij} = \frac{1}{2}(u_{i,j} + u_{j,i}),\;\; \kappa _{ijk} = u_{k,ij}. \end{aligned}$$
(1)
In the case of chiral porous thermoelastic bodies, the constitutive equations are given by (see, e.g., [1, 8, 22])
$$\begin{aligned} \tau _{ij}&= \lambda e_{rr}\delta _{ij} + 2\mu e_{ij} + \textrm{d}\varphi \delta _{ij} + f (\varepsilon _{ikm}\kappa _{jkm}+\varepsilon _{jkm}\kappa _{ikm}) - b T \delta _{ij},\nonumber \\ \mu _{ijk}&= \frac{1}{2}\alpha _{1}(\kappa _{rri}\delta _{jk} + 2\kappa _{krr}\delta _{ij} + \kappa _{rrj}\delta _{ik})\nonumber \\&\quad +\alpha _{2}(\kappa _{irr}\delta _{jk} + \kappa _{jrr}\delta _{ik}) +2\alpha _{3}\kappa _{rrk}\delta _{ij} + \beta _{1}\delta _{ij}\varphi _{,k}+ \beta _{2}(\delta _{ik}\varphi _{,j} + \delta _{jk}\varphi _{,i})\nonumber \\&\quad + 2\alpha _{4} \kappa _{ijk} + \alpha _{5}(\kappa _{kji} + \kappa _{kij}) +f(\varepsilon _{iks}e_{js} + \varepsilon _{jks}e_{is}),\nonumber \\ \sigma _{i}&=\beta _{1}\kappa _{rri} + 2\beta _{2}\kappa _{irr} + a_{0}\varphi _{,i}, \;\; g = de_{rr} + \xi \varphi - \beta T, \end{aligned}$$
(2)
where \(\tau _{ij}\) is the stress tensor, \(\mu _{ijk}\) is the dipolar stress tensor, \(\sigma _{i}\) is the microstretch stress vector, g is the intrinsic body force, T is the temperature measured from the constant absolute temperature of reference state, \(\delta _{ij}\) is Kronecker delta, \(\varepsilon _{ijk}\) is the alternating symbol, and \(\lambda , \mu , \alpha _{s}(s=1,2,\ldots ,5)\), \(\beta _{1}, \beta _{2}, \beta ,\) \(b,d,a_{0}, \xi \) and f are constitutive constants. In the case of achiral materials, the coefficient f is equal to zero. In what follows, we assume that the elastic potential is a positive definite quadratic form in the variables \(e_{ij}, \kappa _{ijk},\varphi \) and \(\varphi _{,k}\). The equations of equilibrium, in the absence of body loads, are given by
$$\begin{aligned} \tau _{ji,j} - \mu _{kji,kj}=0, \;\; \sigma _{j,j} - g= 0. \end{aligned}$$
(3)
The equilibrium theory of linear elastic heat conductors has been studied in various books (see., e.g., [13, 22, 23]). We investigate the effects of temperature variation of the deformation of cylinders.
Toupin [3, 4] introduced the functions \(P_{i}, R_{i}\) and \(Q_{i}\) defined by
$$\begin{aligned}&P_{i} = (\tau _{ki} - \mu _{ski,s})n_{k} - D_{j}(n_{r}\mu _{rji}) + n_{s}n_{p}\mu _{spi}(D_{k}n_{k}),\nonumber \\&R_{i} = \mu _{rsi}n_{r}n_{s}, Q_{i} = <\mu _{pji}n_{p}n_{q}>\varepsilon _{jrq}s_{r}\;\; \text {on}\; \partial B, \end{aligned}$$
(4)
where \(n_{j}\) are the components of the outward unit normal of \(\partial B, D_{i}\) are the components of the surface gradient, \(D_{i} = (\delta _{ik} - n_{i}n_{k})\partial /\partial x_{k}, s_{i}\) are the components of the unit vector tangent to C and \(<g>\) denotes the difference of limits of g from both sides of C.
In the case of traction problem, the boundary conditions are [1, 24]
$$\begin{aligned} P_{i} = {\widetilde{P}}_{i}, R_{i} = {\widetilde{R}}_{i}, \sigma _{i}n_{i} = {\widetilde{\sigma }}\;\; \text {on}\; \partial B \backslash C, Q_{i} = {\widetilde{Q}}_{i}\;\; \text {on}\; C, \end{aligned}$$
(5)
where \({\widetilde{P}}_{i}, {\widetilde{R}}_{i}, {\widetilde{\sigma }}\) and \({\widetilde{Q}}_{i}\) are prescribed functions.
We assume that the region B from here on refers to the interior of a right cylinder of length h with the cross section \(\Sigma \) and the lateral boundary \(\Pi \). Let \(\Gamma \) be the boundary of \(\Sigma \). The coordinate system consists of the orthonormal basis \(\{\textbf{e}_{1}, \textbf{e}_{2}, \textbf{e}_{3}\}\) and the origin O. We choose the system \(Ox_{j}\) such that \(x_{3}\)-axis is parallel to the generator of the cylinder and the \(x_{1}Ox_{2}\) plane contains one of the terminal cross sections. We denote by \(\Sigma _{1}\) and \(\Sigma _{2}\), respectively, the cross section located at \(x_{3} = 0\) and \(x_{3}=h\). Let \(\Gamma _{\alpha }\) be the boundary of \(\Sigma _{\alpha }\), \((\alpha =1,2)\). Throughout this paper, the Greek subscripts range over the integers (1, 2).
In what follows, we assume that the thermal field is linear in the axial coordinate,
$$\begin{aligned} T = T_{0}(x_{1}, x_{2}) + x_{3}T_{1}(x_{1},x_{2}), \end{aligned}$$
(6)
where the functions \(T_{0}\) and \(T_{1}\) are prescribed. We suppose that the lateral surface is smooth. This assumption implies that the functions \(Q_{i}\) are equal to zero on \(\Pi \). The cylinder is supposed to be free of lateral loading and subjected to appropriate stress resultants over its ends. On the lateral surface of the cylinder, we have the conditions
$$\begin{aligned} P_{i} = 0, R_{i} = 0, \sigma _{\alpha }n_{\alpha } = 0\;\; \text {on}\; \Pi . \end{aligned}$$
(7)
Let \(\textbf{F} = (F_{1}, F_{2}, F_{3})\) and \(\textbf{M} = (M_{1}, M_{2}, M_{3})\) be prescribed vectors representing the resultant force and resultant moment about O of the tractions acting on \(\Sigma _{1}\). On \(\Sigma _{2}\), there are tractions applied so as to satisfy the equilibrium conditions of the cylinder. We have introduced the mechanical loads \(\textbf{F}\) and \(\textbf{M}\) in order to compare the effects of thermal field with those produced by the resultants \(\textbf{F}\) and \(\textbf{M}\). On the end \(\Sigma _{1}\), we have the boundary conditions
$$\begin{aligned}&\int _{\Sigma _{1}}P_{\alpha }\textrm{d}a + \int _{\Gamma _{1}}Q_{\alpha } \textrm{d}s = F_{\alpha }, \end{aligned}$$
(8)
$$\begin{aligned}&\int _{\Sigma _{1}} P_{3}\textrm{d}a + \int _{\Gamma _{1}}Q_{3}\textrm{d}s = F_{3},\end{aligned}$$
(9)
$$\begin{aligned}&\int _{\Sigma _{1}}(x_{\alpha }P_{3} + R_{\alpha })\textrm{d}a + \int _{\Gamma _{1}}x_{\alpha }Q_{3}\textrm{d}s = \varepsilon _{\beta \alpha 3}M_{\beta },\end{aligned}$$
(10)
$$\begin{aligned}&\int _{\Sigma _{1}} \varepsilon _{\alpha \beta 3}x_{\alpha }P_{\beta }\textrm{d}a + \int _{\Gamma _{1}}\varepsilon _{a\beta 3}x_{\alpha }Q_{\beta } \textrm{d}s = M_{3}. \end{aligned}$$
(11)
From (4), we find that
$$\begin{aligned}&P_{i} = -\tau _{3i} +2\mu _{\alpha 3i,\alpha } + \mu _{33i,3}, R_{i} = \mu _{33i}\;\; \text {on}\; \Sigma _{1},\nonumber \\&Q_{i} = -2\mu _{\alpha 3i}n_{\alpha }\;\;\text {on}\; \Gamma _{1}, \end{aligned}$$
(12)
where \((n_{1}, n_{2},0)\) are the direction cosines of the exterior normal to \(\Pi .\)
The problem consists in finding the functions \(u_{i}\) and \(\varphi \) satisfying the equations (1)–(3) on B, the conditions (7) on \(\Pi \) and the conditions (8)–(11) on the ends, when the temperature T, the constitutive coefficients and the constants \(F_{j}\) and \(M_{j}\) are prescribed.

3 Two-dimensional problems

In what follows, we will reduce the problem formulated in Sect. 2 to the study of some two-dimensional problems. In this section, we present some results concerning the generalized plane strain of a chiral thermoelastic cylinder. We assume now that a body force \({\mathscr {F}}_{j}\) and a microstretch force density l are prescribed on B. Let us suppose that \({\mathscr {F}}_{j}, l, {\widetilde{P}}_{i}, {\widetilde{R}}_{i}\) and \({\widetilde{\sigma }}\) are independent of the axial coordinate \(x_{3}\). We define the state of generalized plane strain of the cylinder B to be that state in which the displacement vector, the microdilatation function and the temperature are independent of the axial coordinate,
$$\begin{aligned} u_{j}= u_{j}(x_{1}, x_{2}), \varphi = \varphi (x_{1}, x_{2}), T = T(x_{1}, x_{2}), (x_{1}, x_{2})\in \Sigma _{1}. \end{aligned}$$
(13)
From (1), we obtain \(e_{33} = 0, \kappa _{j3i} = 0\) and
$$\begin{aligned} e_{\alpha \beta } = \frac{1}{2}(u_{\alpha ,\beta } + u_{\beta ,\alpha }), 2e_{\alpha 3} = u_{3,\alpha }, \kappa _{\alpha \beta j} = u_{j,\alpha \beta }. \end{aligned}$$
(14)
The constitutive equations (2) reduce to
$$\begin{aligned} \tau _{\alpha \beta }&= \lambda e_{\rho \rho }\delta _{\alpha \beta } + 2\mu e_{\alpha \beta } + \textrm{d}\varphi \delta _{\alpha \beta }+f(\varepsilon _{\alpha \rho 3}\kappa _{\beta \rho 3} + \varepsilon _{\beta \rho 3}\kappa _{\alpha \rho 3}) - b T\delta _{\alpha \beta },\nonumber \\ \tau _{\alpha 3}&= 2\mu e_{\alpha 3} + f\varepsilon _{\rho \beta 3}\kappa _{\alpha \rho \beta },\nonumber \\ \mu _{\alpha \beta \gamma }&= \frac{1}{2}\alpha _{1}(\kappa _{\rho \rho \alpha }\delta _{\beta \gamma } +2\kappa _{\gamma \rho \rho }\delta _{\alpha \beta } + \kappa _{\rho \rho \beta }\delta _{\alpha \gamma }) + \alpha _{2}(\kappa _{\alpha \rho \rho }\delta _{\beta \gamma } + \kappa _{\beta \rho \rho }\delta _{\alpha \gamma })\nonumber \\&\quad + 2\alpha _{3}\kappa _{\rho \rho \gamma }\delta _{\alpha \beta } +2\alpha _{4}\kappa _{\alpha \beta \gamma } + \alpha _{5}(\kappa _{\gamma \beta \alpha } + \kappa _{\gamma \alpha \beta }) + \beta _{1}\delta _{\alpha \beta }\varphi _{,\gamma }\nonumber \\&\quad +2\beta _{2}(\delta _{\alpha \gamma } \varphi _{,\beta } + \delta _{\beta \gamma }\varphi _{,\alpha }) +f(\varepsilon _{\alpha \gamma 3}e_{\beta 3} + \varepsilon _{\beta \gamma 3}e_{\alpha 3}),\nonumber \\ \mu _{\alpha \beta 3}&= 2\alpha _{3}\kappa _{\rho \rho 3}\delta _{\alpha \beta } + 2\alpha _{4}\kappa _{\alpha \beta 3} + f(\varepsilon _{\rho \alpha 3}e_{\beta \rho } + \varepsilon _{\rho \beta 3}e_{\alpha \rho }), \nonumber \\ \sigma _{\alpha }&= \beta _{1}\kappa _{\rho \rho \alpha } + 2\beta _{2}\kappa _{\alpha \rho \rho }+a_{0}\varphi _{,\alpha },\;\; g = de_{\rho \rho } + \xi \varphi -\beta T, \end{aligned}$$
(15)
and
$$\begin{aligned} \tau _{33}&= \lambda e_{\rho \rho }+\textrm{d}\varphi - b T,\nonumber \\ \mu _{3\alpha \beta }&= \frac{1}{2}\alpha _{1}\kappa _{\rho \rho 3}\delta _{\alpha \beta } +\alpha _{5}\kappa _{\beta \alpha 3} + f \varepsilon _{\beta \rho 3}e_{\alpha \rho },\nonumber \\ \mu _{3\alpha 3}&= \frac{1}{2}\alpha _{1}\kappa _{\rho \rho \alpha } + \alpha _{2}\kappa _{\alpha \rho \rho } + f \varepsilon _{\rho \alpha 3}e_{3\rho } + \beta _{2}\varphi _{,\alpha },\nonumber \\ \mu _{33\alpha }&= \alpha _{1}\kappa _{\alpha \rho \rho } + 2\alpha _{3}\kappa _{\rho \rho \alpha } + \beta _{1}\varphi _{,\alpha } + 2f \varepsilon _{3\alpha \rho }e_{3\rho },\nonumber \\ \mu _{333}&= (\alpha _{1}+2\alpha _{3})\kappa _{\rho \rho 3}, \sigma _{3} = \beta _{1}\kappa _{\rho \rho 3}. \end{aligned}$$
(16)
The equations of equilibrium in the presence of body loads can be written as
$$\begin{aligned} \tau _{\alpha i,\alpha } - \mu _{\alpha \beta i, \alpha \beta } + {\mathscr {F}}_{i} = 0, \sigma _{\alpha ,\alpha } - g+l =0\;\;\text {on}\; \Sigma _{1}. \end{aligned}$$
(17)
It follows from (4) that in the plane problems the functions \(P_{i}\) and \(R_{i}\) have the following form on the lateral surface
$$\begin{aligned} P_{i}&= (\tau _{\beta i} - \mu _{\rho \beta i,\rho })n_{\beta } - D_{\rho }(n_{\beta }\mu _{\beta \rho i}) + n_{\beta }n_{\alpha }\mu _{\beta \alpha i}(D_{\rho }n_{\rho }),\nonumber \\ R_{i}&= \mu _{\rho \alpha i}n_{\rho }n_{\alpha }. \end{aligned}$$
(18)
The conditions on the surface \(\Pi \) become
$$\begin{aligned} P_{i} = {\widetilde{P}}_{i}, R_{i} = {\widetilde{R}}_{i}, \sigma _{\alpha }n_{\alpha } = {\widetilde{\sigma }}\;\;\text {on}\; \Gamma _{1}. \end{aligned}$$
(19)
The two-dimensional problem of thermoelasticity consists in finding the functions \(u_{j}\) and \(\varphi \) satisfying the Eqs. (14), (15) and (17) on \(\Sigma _{1}\), and the boundary conditions (19) on \(\Gamma _{1}\). We denote by \(({\mathscr {H}})\) the problem characterized by the Eqs. (14), (15), (17) and the boundary conditions (19). Using the results established by Hlavacek and Hlavacek [25], we can easily deduce the following proposition.
Theorem 1
The necessary and sufficient conditions for the existence of a solution to the problem \(({\mathscr {H}})\) are given by
$$\begin{aligned} \int _{\Sigma _{1}} {\mathscr {F}}_{j}\textrm{d}a + \int _{\Gamma _{1}}{\widetilde{P}}_{j} \textrm{d}s&= 0,\nonumber \\ \int _{\Sigma _{1}} \varepsilon _{3\alpha \beta }x_{\alpha } {\mathscr {F}}_{\beta } \textrm{d}a + \int _{\Gamma _{1}} \varepsilon _{3\alpha \beta }(x_{\alpha }{\widetilde{P}}_{\beta } + n_{\alpha }{\widetilde{R}}_{\beta })\textrm{d}s&= 0. \end{aligned}$$
(20)
In view of (14) and (15), the equations of equilibrium can be written in the form
$$\begin{aligned}&\mu \Delta u_{\alpha } + (\lambda +\mu )u_{\beta ,\beta \alpha } - 2(\alpha _{3}+\alpha _{4})\Delta \Delta u_{\alpha }- 2(\alpha _{1}+\alpha _{2}+\alpha _{5})\Delta u_{\beta ,\beta \alpha } \nonumber \\&\qquad + 2 f \varepsilon _{\alpha \beta 3} \Delta u_{3,\beta } + \textrm{d}\varphi _{,\alpha }- (\beta _{1}+2\beta _{2})\Delta \varphi _{,\alpha } - b T_{,\alpha } + {\mathscr {F}}_{\alpha } = 0,\nonumber \\&\mu \Delta u_{3} - 2(\alpha _{3} + \alpha _{4})\Delta \Delta u_{3} + 2f \varepsilon _{\rho \nu 3}\Delta u_{\nu ,\rho } + {\mathscr {F}}_{3} = 0,\nonumber \\ \qquad&(\beta _{1} + 2\beta _{2})\Delta u_{\rho ,\rho } + a_{0}\Delta \varphi - d u_{\rho ,\rho } - \xi \varphi + \beta T + l =0. \end{aligned}$$
(21)
The functions \(\tau _{33}, \mu _{3\alpha \beta }, \mu _{3\alpha 3}, \mu _{33\alpha },\mu _{333}\) and \(\sigma _{3}\) can be calculated after the determination of displacements and microdilatation.

4 Plane temperature field

In this section, we present the solution of the problem under the following assumptions
$$\begin{aligned} T = T_{0}(x_{1},x_{2}), \;\;\;\textbf{F} = F_{3} \textbf{e}_{3}, \;\;\; \textbf{M} = M_{j}\textbf{e}_{j}. \end{aligned}$$
(22)
It is known that in the classical thermoelasticity, a uniform thermal field produces bending, extension and a plane deformation. We try to solve the problem by combining a solution of Saint-Venant’s type with a solution of the plane problem. We seek the solution in the form
$$\begin{aligned} u_{\alpha }&=-\frac{1}{2}a_{\alpha }x_{3}^{2} + \varepsilon _{3\beta \alpha }a_{4}x_{\beta }x_{3} + \sum _{k=1}^{4}a_{k}u_{\alpha }^{(k)} + w_{\alpha }(x_{1},x_{2}),\nonumber \\ u_{3}&= (a_{1}x_{1} + a_{2}x_{2} + a_{3})x_{3} + \sum _{k=1}^{4}a_{k}u_{3}^{(k)}+w_{3}(x_{1},x_{2}),\nonumber \\ \varphi&= \sum _{k=1}^{4}a_{k}\varphi ^{(k)}+\psi , \end{aligned}$$
(23)
where \(u_{j}^{(k)}, \varphi ^{(k)}, w_{j}\) and \(\psi \) are unknown functions which are independent of \(x_{3}\), and \(a_{k}\) are unknown constants. We denote by \(A^{(k)}\) \((k=1,2,3,4)\), the isothermal \((T=0)\) plane strain problems characterized by the displacement vector \(u_{j}^{(k)}\) and microdilatation function \(\varphi ^{(k)}\). The body loads and the boundary data associated with the problems \(A^{(k)}\) will be precised in what follows. We introduce the notations
$$\begin{aligned} e_{\alpha \beta }^{(k)} = \frac{1}{2}(u_{\alpha ,\beta }^{(k)} + u_{\beta ,\alpha }^{(k)}), 2e_{\alpha 3}^{(k)} = u_{3,\alpha }^{(k)}, \kappa _{\alpha \beta j}^{(k)} = u_{j,\alpha \beta }^{(k)}. \end{aligned}$$
(24)
It follows from (15) that the constitutive equations imply
$$\begin{aligned} \tau _{\alpha \beta }^{(k)}&= \lambda e_{\rho \rho }^{(k)}\delta _{\alpha \beta } + 2\mu e_{a\beta }^{(k)} + \textrm{d}\varphi ^{(k)}\delta _{\alpha \beta } + f(\varepsilon _{\alpha \rho 3}\kappa _{\beta \rho 3}^{(k)} + \varepsilon _{\beta \rho 3}\kappa _{\alpha \rho 3}^{(k)}),\nonumber \\ \tau _{\alpha 3}^{(k)}&= 2\mu e_{\alpha 3}^{(k)} + f \varepsilon _{\rho \beta 3}\kappa _{\alpha \rho \beta }^{(k)},\nonumber \\ \mu _{\alpha \beta \gamma }^{(k)}&= \frac{1}{2}\alpha _{1}(\kappa _{\rho \rho \alpha }^{(k)} \delta _{\beta \gamma } + 2\kappa _{\gamma \rho \rho }^{(k)} \delta _{\alpha \beta }+ \kappa _{\rho \rho \beta }^{(k)}\delta _{\alpha \gamma })\nonumber \\&\quad +\alpha _{2}(\kappa _{\alpha \rho \rho }^{(k)}\delta _{\beta \gamma } + \kappa _{\beta \rho \rho }^{(k)}\delta _{\alpha \gamma }) + 2\alpha _{3}\kappa _{\rho \rho \gamma }^{(k)}\delta _{\alpha \beta }\nonumber \\&\quad +2\alpha _{4}\kappa _{\alpha \beta \gamma }^{(k)} + \alpha _{5}(\kappa _{\gamma \beta \alpha }^{(k)} + \kappa _{\gamma \alpha \beta }^{(k)}) + \beta _{1}\delta _{\alpha \beta }\varphi _{,\gamma }^{(k)}\nonumber \\&\quad +\beta _{2}(\delta _{\alpha \gamma }\varphi _{,\beta }^{(k)} + \delta _{\beta \gamma }\varphi _{,\alpha }^{(k)}) + f(\varepsilon _{\alpha \gamma 3}e_{\beta 3}^{(k)} + \varepsilon _{\beta \gamma 3}e_{\alpha 3}^{(k)}),\nonumber \\ \mu _{\alpha \beta 3}^{(k)}&= 2\alpha _{3}\kappa _{\rho \rho 3}^{(k)}\delta _{\alpha \beta } + 2\alpha _{4}\kappa _{\alpha \beta 3}^{(k)} +f(\varepsilon _{\rho \alpha 3}e_{\beta \rho }^{(k)} + \varepsilon _{\rho \beta 3}e_{\alpha \rho }^{(k)}),\nonumber \\ \sigma _{\alpha }^{(k)}&= \beta _{1}\kappa _{\rho \rho \alpha }^{(k)} + 2\beta _{2}\kappa _{\alpha \rho \rho }^{(k)} + a_{0}\varphi _{,\alpha }^{(k)}, \;\; g^{(k)} = de_{\rho \rho }^{(k)} + \xi \varphi ^{(k)}. \end{aligned}$$
(25)
We shall use the notations
$$\begin{aligned} \tau _{33}^{(k)}&= \lambda e_{\rho \rho }^{(k)} + \textrm{d}\varphi ^{(k)},\;\; \mu _{3\alpha \beta }^{(k)} = \frac{1}{2}\alpha _{1}\kappa _{\rho \rho 3}^{(k)}\delta _{\alpha \beta } + \alpha _{5}\kappa _{\beta \alpha 3}^{(k)} + f\varepsilon _{\beta \rho 3}e_{\alpha \rho }^{(k)},\nonumber \\ \mu _{3\alpha 3}^{(k)}&= \frac{1}{2}\alpha _{1}\kappa _{\rho \rho \alpha }^{(k)} + \alpha _{2}\kappa _{\alpha \rho \rho }^{(k)} + f \varepsilon _{\rho \alpha 3}e_{3\rho }^{(k)} + \beta _{2}\varphi _{,\alpha }^{(k)},\nonumber \\ \mu _{33\alpha }^{(k)}&= \alpha _{1}\kappa _{\alpha \rho \rho }^{(k)} + 2\alpha _{3}\kappa _{\rho \rho \alpha }^{(k)} + \beta _{1}\varphi _{,\alpha }^{(k)} + 2f \varepsilon _{3\alpha \rho }e_{3\rho }^{(k)},\nonumber \\ \mu _{333}^{(k)}&= (\alpha _{1}+2\alpha _{3})\kappa _{\rho \rho 3}^{(k)},\; \sigma _{3}^{(k)} = \beta _{1}\kappa _{\rho \rho 3}^{(k)}. \end{aligned}$$
(26)
We denote by \(({\mathcal {T}})\) the plane thermoelastic problem which corresponds to the temperature \(T_{0}\) and is characterized by the displacement vector \(w_{j}\) and the microdilatation \(\psi .\) Thus, the constitutive equations in the problem \(({\mathcal {T}})\) are
$$\begin{aligned} t_{\alpha \beta }&= \lambda \eta _{\rho \rho }\delta _{\alpha \beta } + 2\mu \eta _{\alpha \beta } + \textrm{d}\psi \delta _{\alpha \beta } + f(\varepsilon _{\alpha \rho 3}\xi _{\beta \rho 3}+\varepsilon _{\beta \rho 3}\xi _{\alpha \rho 3}) - b T_{0} \delta _{\alpha \beta },\nonumber \\ t_{\alpha 3}&= 2\mu \eta _{\alpha 3} + f\varepsilon _{\rho \beta 3}\xi _{\alpha \rho \beta },\nonumber \\ m_{\alpha \beta \gamma }&= \frac{1}{2}\alpha _{1}(\xi _{\rho \rho \alpha }\delta _{\beta \gamma } + 2\xi _{\gamma \rho \rho }\delta _{\alpha \beta } + \xi _{\rho \rho \beta }\delta _{\alpha \gamma }) +\alpha _{2}(\xi _{\alpha \rho \rho }\delta _{\beta \gamma } + \xi _{\beta \rho \rho }\delta _{\alpha \gamma }) \nonumber \\ {}&\quad + 2\alpha _{3}\xi _{\rho \rho \gamma }\delta _{\alpha \beta } + 2\alpha _{4}\xi _{\alpha \beta \gamma } + \alpha _{5}(\xi _{\gamma \beta \alpha } + \xi _{\gamma \alpha \beta }) + \beta _{1}\delta _{\alpha \beta }\psi _{,\gamma },\nonumber \\&\quad +2\beta _{2}(\delta _{\alpha \gamma }\psi _{,\beta } + \delta _{\beta \gamma }\psi _{,\alpha }) + f(\varepsilon _{\alpha \gamma 3}\eta _{\beta 3} + \varepsilon _{\beta \gamma 3}\eta _{\alpha 3}),\nonumber \\ m_{\alpha \beta 3}&= 2\alpha _{3}\xi _{\rho \rho 3}\delta _{\alpha \beta } + 2\alpha _{4}\xi _{\alpha \beta 3} + f(\varepsilon _{\rho \alpha 3}\eta _{\beta \rho } + \varepsilon _{\rho \beta 3}\eta _{\alpha \rho }), \nonumber \\ \pi _{\alpha }&= \beta _{1}\xi _{\rho \rho \alpha } + 2\beta _{2}\xi _{\alpha \rho \rho } + a_{0}\psi _{,\alpha },\;\; \gamma = \textrm{d}\eta _{\rho \rho } +\xi \psi - \beta T_{0}, \end{aligned}$$
(27)
where
$$\begin{aligned} \eta _{\alpha \beta } = \frac{1}{2}(w_{\alpha ,\beta } + w_{\beta ,\alpha }), 2\eta _{\alpha 3} = w_{3,\alpha }, \xi _{\alpha \beta j} = w_{j,\alpha \beta }. \end{aligned}$$
(28)
We shall use the notations
$$\begin{aligned} t_{33}&= \lambda \eta _{\rho \rho } +\textrm{d}\psi - b T_{0},\nonumber \\ m_{3\alpha \beta }&= \frac{1}{2}\alpha _{1}\xi _{\rho \rho 3}\delta _{\alpha \beta } + \alpha _{5}\xi _{\beta \alpha 3} + f \xi _{\beta \rho 3}\eta _{\alpha \rho },\nonumber \\ m_{3\alpha 3}&= \frac{1}{2}\alpha _{1}\xi _{\rho \rho \alpha } + \alpha _{2}\xi _{\alpha \rho \rho } + f \xi _{\rho \alpha 3}\eta _{3\rho } + \beta _{2}\psi _{,\alpha },\nonumber \\ m_{33\alpha }&= \alpha _{1}\xi _{\alpha \rho \rho } + 2\alpha _{3}\xi _{\rho \rho \alpha } + \beta _{1}\psi _{,\alpha } + 2f \varepsilon _{3\alpha \rho }\eta _{3\rho },\nonumber \\ m_{333}&= (\alpha _{1}+2\alpha _{3})\xi _{\rho \rho 3}\xi _{\rho \rho 3}, \pi _{3} = \beta _{1}\xi _{\rho \rho 3}. \end{aligned}$$
(29)
From (1) and (23), we obtain
$$\begin{aligned} e_{\alpha \beta }= & {} \eta _{\alpha \beta } + \sum _{k=1}^{4}a_{k}e_{\alpha \beta }^{(k)}, e_{\alpha 3} = \frac{1}{2}\varepsilon _{3\beta \alpha }x_{\beta }a_{4}\nonumber \\{} & {} +\eta _{\alpha 3} + \sum _{k=1}^{4}a_{k}e_{\alpha 3}^{(k)},e_{33} = a_{1}x_{1}+a_{2}x_{2}+a_{3},\nonumber \\ \kappa _{\alpha \beta \gamma }= & {} \xi _{\alpha \beta \gamma } + \sum _{k=1}^{4}a_{k}\kappa _{\alpha \beta \gamma }^{(k)}, \kappa _{\alpha \beta 3} = \xi _{\alpha \beta 3} + \sum _{k=1}^{4}a_{k}\kappa _{\alpha \beta 3}^{(k)},\nonumber \\ \kappa _{\beta 3\alpha }= & {} \varepsilon _{3\beta \alpha }a_{4},\kappa _{\alpha 33} = -\kappa _{33\alpha } = a_{4},\kappa _{333} = 0. \end{aligned}$$
(30)
In view of (2) and (29), we find that the functions \(\tau _{ij},\mu _{ijk},\sigma _{i}\) and g are given by
$$\begin{aligned} \tau _{\alpha \beta }&= [(a_{1}x_{1}+a_{2}x_{2}+a_{3}) - 2f a_{4}]\delta _{\alpha \beta } + \sum _{k=1}^{4}a_{k}\tau _{\alpha \beta }^{(k)} +t_{\alpha \beta },\nonumber \\ \tau _{\alpha 3}&= \mu a_{4}\varepsilon _{3\beta \alpha }x_{\beta } + 2f \varepsilon _{\alpha \rho 3}a_{\rho } + \sum _{k=1}^{4}a_{k}\tau _{\alpha 3}^{(k)} + t_{\alpha 3},\nonumber \\ \tau _{33}&= (\lambda +2\mu )(a_{1}x_{1}+a_{2}x_{2}+a_{3})+4fa_{4}+t_{33} +\sum _{k=1}^{4}a_{k}\tau _{33}^{(k)},\nonumber \\ \mu _{111}&=2(\alpha _{2}-\alpha _{3})a_{1} + \sum _{k=1}^{4}a_{k}\mu _{111}^{(k)} + m_{111},\nonumber \\ \mu _{222}&= 2(\alpha _{2}-\alpha _{3})a_{2} + \sum _{k=1}^{4}a_{k}\mu _{222}^{(k)}+m_{222},\nonumber \\ \mu _{221}&= (\alpha _{1}-2\alpha _{3})a_{1} - fa_{4}x_{1} + \sum _{k=1}^{4}a_{k}\mu _{221}^{(k)} + m_{221},\nonumber \\ \mu _{112}&= (\alpha _{1} - 2\alpha _{3})a_{2} - f a_{4}x_{2} + \sum _{k=1}^{4}a_{k}\mu _{112}^{(k)}+m_{112},\nonumber \\ \mu _{121}&= \frac{1}{2}(2\alpha _{2}-\alpha _{3})a_{2} + \frac{1}{2}fa_{4}x_{2}+ \sum _{k=1}^{4}a_{k}\mu _{121}^{(k)} + m_{121},\nonumber \\ \mu _{122}&=\frac{1}{2}(2\alpha _{2}-\alpha _{1})a_{1} +\frac{1}{2}fa_{4}x_{1} + \sum _{k=1}^{4}a_{k}\mu _{122}^{(k)} + m_{122},\nonumber \\ \mu _{\alpha 33}&= \frac{1}{2}(2\alpha _{2} - \alpha _{1}+4\alpha _{4})a_{4} - \frac{1}{2}fa_{4}x_{\alpha }+\sum _{k=1}^{4}a_{k}\mu _{\alpha 33}^{(k)} + m_{\alpha 33},\nonumber \\ \mu _{33\alpha }&=(\alpha _{1}-2\alpha _{3} - 2\alpha _{4} + 2\alpha _{5})a_{\alpha }+fa_{4}x_{\alpha }+\sum _{k=1}^{4}a_{k}\mu _{33\alpha }^{(k)} + m_{33\alpha },\nonumber \\ \mu _{\alpha 3\beta }&= \varepsilon _{3\alpha \beta } f(a_{1}x_{1}+a_{2}x_{2}+a_{3}) + \varepsilon _{3\alpha \beta }(2\alpha _{4}-\alpha _{5})a_{4}\nonumber \\&\quad +\sum _{k=1}^{4}a_{k}\mu _{\alpha 3\beta }^{(k)} +m_{\alpha 3\beta },\nonumber \\ \mu _{\alpha \beta 3}&=\sum _{k=1}^{4}a_{k}\mu _{\alpha \beta 3}^{(k)}+m_{\alpha \beta 3},\mu _{333} = \sum _{k=1}^{4}a_{k}\mu _{333}^{(k)} + m_{333},\nonumber \\ \sigma _{\alpha }&= (2\beta _{2} - \beta _{1})a_{\alpha } +\sum _{k=1}^{4}a_{k}\sigma _{\alpha }^{(k)}+\pi _{\alpha },\sigma _{3} = \sum _{k=1}^{4}a_{k}\sigma _{3}^{(k)} +\pi _{3},\nonumber \\ g&= d(a_{1}x_{1}+a_{2}x_{2}+a_{3})+\sum _{k=1}^{4}a_{k}g^{(k)} + \gamma . \end{aligned}$$
(31)
Here we have used the relations (25)–(27).
Let us impose that the equations of equilibrium (3) and the boundary conditions (7) be satisfied by the functions (31). If we require that the coefficients of the constants \(a_{k}\) that appear in the equilibrium equations be equal to zero, then we find that the functions \(\tau _{\alpha j}^{(k)}, \mu _{\alpha \beta i}^{(k)}, \sigma _{\alpha }^{(k)}\) and \(g^{(k)}\) \((k=1,2,3,4)\) satisfy the following equations
$$\begin{aligned} \tau _{\alpha j,\alpha }^{(k)} - \mu _{\alpha \beta j,\alpha \beta }^{(k)} + {\mathscr {F}}_{j}^{(k)} = 0,\;\; \sigma _{\alpha ,\alpha }^{(k)} - g^{(\kappa )} + l^{(k)} = 0\;\; \text {on}\; \Sigma _{1}, \end{aligned}$$
(32)
where we have used the notations
$$\begin{aligned} {\mathscr {F}}_{j}^{(\rho )} = \lambda \delta _{j\rho },\;\; {\mathscr {F}}_{j}^{(3)} = 0,\;\; {\mathscr {F}}_{j}^{(4)}=0,\;\, l^{(\rho )}=-\textrm{d}x_{\rho },\;\; l^{(3)} = -d,\; l^{(4)} = 0. \end{aligned}$$
(33)
The equilibrium equations (3) reduce to
$$\begin{aligned} t_{\alpha i,\alpha } - m_{\alpha \beta i,\alpha } = 0,\;\, \pi _{\alpha ,\alpha } - \gamma = 0\;\;\text {on}\;\; \Sigma _{1}. \end{aligned}$$
(34)
Let us introduce the functions
$$\begin{aligned} P_{i}^{(k)}&= (\tau _{\beta i}^{(k)} - \mu _{\rho \beta i,\rho }^{(k)}) n_{\beta } - D_{\rho }(n_{\beta }\mu _{\beta \rho i}^{(k)})+n_{\beta }n_{\alpha }\mu _{\beta \alpha i}^{(k)}(D_{\rho }n_{\rho }),\nonumber \\ R_{i}^{(k)}&= \mu _{\rho \alpha i}^{(k)} n_{\rho }n_{\alpha },\nonumber \\ P_{i}^{*}&= (t_{\beta i} - m_{\rho \beta i,\rho })n_{\beta } - D_{\rho }(n_{\beta }m_{\beta \rho i})+n_{\beta }n_{\alpha }m_{\beta \alpha i}(D_{\rho }n_{\rho }),\nonumber \\ R_{i}^{*}&= m_{\rho \alpha i}n_{\rho }n_{\alpha },\;\; \text {on}\;\; \Gamma _{1}. \end{aligned}$$
(35)
We require that the boundary conditions (7) be satisfied for any constants \(a_{k}\). Thus, we find that the functions \(\tau _{\alpha j}^{(k)}, \mu _{\alpha \beta i}^{(k)}\) and \(\sigma ^{(k)}\) have to satisfy the following boundary conditions
$$\begin{aligned} P_{i}^{(k)} = {\widetilde{P}}_{i}^{(k)},\;\; R_{i}^{(k)} = {\widetilde{R}}_{i}^{(k)},\;\; \sigma _{\alpha }^{(k)}n_{\alpha } = {\widetilde{\sigma }}^{(k)}\; \text {on}\; \Gamma _{1}, \end{aligned}$$
(36)
where
$$\begin{aligned} {\widetilde{P}}_{1}^{(1)}&= -\lambda x_{1}n_{1} + (\alpha _{1} - 2\alpha _{2})\varepsilon _{3\alpha \nu }(n_{1}n_{2})_{,\nu }n_{\alpha },\nonumber \\ {\widetilde{P}}_{2}^{(1)}&= -\lambda x_{1}n_{2}+ \frac{1}{2}(\alpha _{1} - 2\alpha _{2})\varepsilon _{3\alpha \rho }(n_{1}^{2} - n_{2}^{2})_{,\alpha }n_{\rho }, P_{3}^{(1)} =2f n_{2},\nonumber \\ {\widetilde{R}}_{1}^{(1)}&= 2\alpha _{3} - \alpha _{1}+(\alpha _{1} - 2\alpha _{2})n_{1}^{2}, {\widetilde{R}}_{2}^{(1)}=(\alpha _{1}-2\alpha _{2})n_{1}n_{2},\nonumber \\ {\widetilde{R}}_{3}^{(1)}&= 0, {\widetilde{\sigma }}^{(1)} = (\beta _{1}-2\beta _{2})n_{1},\nonumber \\ {\widetilde{P}}_{1}^{(2)}&= -\lambda x_{2}n_{1} + \frac{1}{2}(\alpha _{1}-2\alpha _{2})\varepsilon _{3\alpha \nu }(n_{1}^{2}-n_{2}^{2})_{,\alpha }n_{\nu },\nonumber \\ {\widetilde{P}}_{2}^{(2)}&= -\lambda x_{2}n_{2} + (\alpha _{1}-2\alpha _{2})\varepsilon _{3\alpha \nu }(n_{1}n_{2})_{,\nu }n_{\alpha }, {\widetilde{P}}_{3}^{(2)} = -2 f n_{1},\nonumber \\ {\widetilde{R}}_{1}^{(2)}&= (\alpha _{1}-2\alpha _{2})n_{1}n_{2}, {\widetilde{R}}_{2}^{(2)} = 2\alpha _{3}-\alpha _{1}+ (\alpha _{1}-2\alpha _{2})n_{2}^{2},\nonumber \\ {\widetilde{R}}_{3}^{(2)}&= 0, {\widetilde{\sigma }}^{(2)} = (\beta _{1}-2\beta _{2})n_{2},\nonumber \\ {\widetilde{P}}_{\alpha }^{(3)}&= -\lambda n_{\alpha }, {\widetilde{P}}_{3}^{(3)} = 0, {\widetilde{R}}_{j}^{(3)} = 0, {\widetilde{\sigma }}^{(3)} =0,\nonumber \\ {\widetilde{P}}_{1}^{(4)}&= \frac{1}{2}f[5n_{1} + D_{1}(x_{2}n_{2})+D_{2}(x_{2}n_{1}-2x_{1}n_{2})-2(x_{2}n_{1}n_{2} - x_{1}n_{2}^{2})(D_{\rho }n_{\rho })],\nonumber \\ {\widetilde{P}}_{2}^{(4)}&= \frac{1}{2}f[5n_{2} + D_{1}(x_{1}n_{2} - 2x_{2}n_{1}) + D_{2}(x_{1}n_{1}) - 2(x_{1}n_{1}n_{2} - x_{2}n_{1}^{2})(D_{\rho }n_{\rho })],\nonumber \\ {\widetilde{P}}_{3}^{(4)}&= \mu \varepsilon _{3\beta \nu }x_{\nu }n_{\beta }, {\widetilde{R}}_{1}^{(4)} = f(x_{1}n_{2}^{2} - x_{2}n_{1}n_{2}),\nonumber \\ {\widetilde{R}}_{2}^{(4)}&= f(x_{2}n_{1}^{2} - x_{1}n_{1}n_{2}), {\widetilde{R}}_{3}^{(4)}=0, {\widetilde{\sigma }}^{(4)}=0. \end{aligned}$$
(37)
We conclude that the problem \(A^{(k)}\) consists in finding the functions \(u_{j}^{(k)}\) and \(\varphi ^{(k)}\) satisfying the Eqs. (24), (25) and (32) on \(\Sigma _{1}\), and the boundary conditions (36). The necessary and sufficient conditions for the existence of the solution are satisfied for each problem \(A^{(k)}\). The solutions of these problems depend only on the cross section \(\Sigma \) and the constitutive constants. The boundary conditions (7) reduce to
$$\begin{aligned} P_{i}^{*} = 0,\;\; R_{i}^{*} = 0, \; \pi _{\alpha }n_{\alpha } = 0\;\; \text {on}\; \Gamma _{1}. \end{aligned}$$
(38)
The problem \(({\mathcal {T}})\) consists in finding the functions \(w_{j}\) and \(\psi \) which satisfy the Eqs. (27), (28) and (34) on \(\Sigma _{1}\), and the boundary conditions (38) on \(\Gamma _{1}\).
Let us determine now the constants \(a_{j}, (j=1,2,3,4)\). By using the divergence theorem and the relations (3), (5) and (12), we obtain
$$\begin{aligned} \int _{\Sigma _{1}} P_{\alpha }\textrm{d}a + \int _{\Gamma _{1}}Q_{\alpha }\textrm{d}s= & {} - \int _{\Gamma _{1}}(x_{\alpha } {\widetilde{P}}_{3} + n_{\alpha }{\widetilde{R}}_{3})\textrm{d}s \nonumber \\ {}{} & {} -\int _{\Sigma _{1}}[x_{\alpha }(\tau _{33,3} - \mu _{333,33})+2\mu _{\alpha 33,3} - \mu _{33\alpha ,3}]\textrm{d}s,\nonumber \\ \int _{\Sigma _{1}}P_{3}\textrm{d}a + \int _{\Gamma _{1}}Q_{3}\textrm{d}s= & {} - \int _{\Sigma _{1}}(\tau _{33} - \mu _{333,33})\textrm{d}a,\nonumber \\ \int _{\Sigma _{1}} (x_{\alpha }P_{3} +R_{\alpha })\textrm{d}a + \int _{\Gamma _{1}}x_{\alpha }Q_{3}\textrm{d}s= & {} -\int _{\Sigma _{1}}[x_{\alpha }(\tau _{33} - \mu _{333,3}) +2\mu _{\alpha 33} - \mu _{33\alpha }]\textrm{d}a,\nonumber \\ \int _{\Sigma _{1}}\varepsilon _{\alpha \beta 3}x_{\alpha }P_{\beta }\textrm{d}s + \int _{\Gamma _{1}}\varepsilon _{\alpha \beta 3}x_{\alpha }Q_{\beta }\textrm{d}s= & {} \int _{\Sigma _{1}}[\varepsilon _{\alpha \beta 3}x_{\alpha } (\mu _{33\beta ,3} - \tau _{3\beta })-2\varepsilon _{\alpha \beta 3}\mu _{\alpha 3\beta }]\textrm{d}a. \end{aligned}$$
(39)
In view of (7) and (31), we find from (39) that
$$\begin{aligned} \int _{\Sigma _{1}}P_{\alpha }\textrm{d}a + \int _{\Gamma _{1}}Q_{\alpha }\textrm{d}s=0. \end{aligned}$$
Thus, the conditions (8), with \(F_{\alpha } = 0\), are identically satisfied. It follows from (31) and (39) that the conditions (9)–(11) reduce to the following system, for the constants \(a_{k}\)
$$\begin{aligned}&\sum _{k=1}^{4}D_{\alpha k}a_{k} = \varepsilon _{3\alpha \beta }(M_{\beta } +M_{\beta }^{*}),\nonumber \\&\sum _{k=1}^{4}D_{3k}a_{k} = -F_{3} - F_{3}^{*},\nonumber \\&\sum _{k=1}^{4}D_{4k}a_{k} = -M_{3} - M_{3}^{*}. \end{aligned}$$
(40)
In (40), we have used the notations
$$\begin{aligned}&D_{\alpha k} = \int _{\Sigma _{1}}(x_{\alpha }S_{33}^{(k)} + 2N_{\alpha 33}^{(k)} - N_{33\alpha }^{(k)})\textrm{d}a,\nonumber \\&D_{3k} = \int _{\Sigma _{1}} S_{33}^{(k)}\textrm{d}a, D_{4k} = \int _{\Sigma _{1}}\varepsilon _{3\alpha \beta }(x_{\alpha }S_{3\beta }^{(k)} +2N_{\alpha 3\beta }^{(k)})\textrm{d}a, \nonumber \\ {}&F_{3}^{*} = \int _{\Sigma _{1}}t_{33}\textrm{d}a, M_{\alpha }^{*} = \varepsilon _{3\alpha \beta }\int _{\Sigma _{1}}(x_{\beta }t_{33} + 2m_{\beta 33} - m_{33\beta })\textrm{d}a,\nonumber \\&M_{3}^{*} =\int _{\Sigma _{1}}\varepsilon _{3\alpha \beta }(x_{\alpha }t_{3\beta } +2m_{\alpha 3\beta })\textrm{d}a, \end{aligned}$$
(41)
where
$$\begin{aligned} S_{33}^{(\rho )}&=(\lambda +2\mu )x_{\rho }+\tau _{33}^{(\rho )}, S_{33}^{(3)}=\lambda +2\mu +\tau _{33}^{(3)},\nonumber \\ S_{33}^{(4)}&= 4f +\tau _{33}^{(4)}, S_{3\alpha }^{(\rho )} =2f \varepsilon _{\alpha \rho 3} + \tau _{\alpha 3}^{(\rho )},\nonumber \\ S_{3\alpha }^{(3)}&= \tau _{\alpha 3}^{(3)}, S_{3\alpha }^{(4)}= \mu \varepsilon _{3\beta \alpha }x_{\beta } + \tau _{\alpha 3}^{(4)},\nonumber \\ N_{\alpha 33}^{(i)}&= \frac{1}{2}(2\alpha _{2} - \alpha _{1} + 4\alpha _{4})\delta _{i\alpha }+\mu _{\alpha 33}^{(i)}, (i=1,2,3),\nonumber \\ N_{\alpha 33}^{(4)}&= -\frac{1}{2}fx_{\alpha }+\mu _{\alpha 33}^{(4)}, N_{33\alpha }^{(i)} = (\alpha _{1} -2\alpha _{3}-2\alpha _{4} + \alpha _{5})\delta _{i\alpha }\nonumber \\ {}&\quad +\mu _{33\alpha }^{(i)},\;\; N_{33\alpha }^{(4)} = fx_{\alpha }+\mu _{33\alpha }^{(4)}, N_{\alpha 3\beta }^{(\rho )} = \varepsilon _{3\alpha \beta }fx_{\rho } + \mu _{\alpha 3\beta }^{(\rho )},\nonumber \\ N_{\alpha 3\beta }^{(3)}&= \varepsilon _{3\alpha \beta }f + \mu _{\alpha 3\beta }^{(3)}, N_{\alpha 3\beta }^{(4)} = \varepsilon _{3\alpha \beta }(2\alpha _{4}-\alpha _{5})+\mu _{\alpha 3\beta }^{(4)}. \end{aligned}$$
(42)
The constants \(D_{mn}\) \((m,n=1,2,3,4)\) can be calculated after the solving of the problems \(A^{(k)}\) \((k=1,2,3,4)\). As in classical elasticity, the positive definiteness of the potential energy implies that [17, 26].
$$\begin{aligned} \text{ det }(D_{mn})\ne 0. \end{aligned}$$
(43)
Thus, the constants \(a_{k}\) are determined by the system (40). By using the reciprocal theorem, we find that
$$\begin{aligned} D_{mn} = D_{nm}. \end{aligned}$$
The functions \(t_{3j}\) and \(m_{j33}\) that appear in \(F_{3}^{*}\) and \(M_{3}^{*}\) depend on the temperature \(T_{0}\). We conclude that a plane temperature field produces axial extension, bending and torsion.

5 Deformation due to a temperature that is linear in the axial coordinate

This section is concerned with the deformation of the beam subjected to the following external data
$$\begin{aligned} T = x_{3}T_{1}(x_{1},x_{2}),\;\; \textbf{F}=F_{\alpha }\textbf{e}_{\alpha },\;\; \textbf{M} = \textbf{0}, \end{aligned}$$
(44)
where \(T_{1}\) and \(F_{\alpha }\) are prescribed.
We try to solve the problem assuming that
$$\begin{aligned} u_{\alpha }&= - \frac{1}{2}c_{\alpha }x_{3}^{2} - \frac{1}{6}b_{\alpha }x_{3}^{3} + \varepsilon _{3\beta \alpha }(c_{4}x_{3} + \frac{1}{2}b_{4}x_{3}^{2})x_{\beta } \nonumber \\ {}&\quad +\sum _{k=1}^{4}(c_{k} + b_{k}x_{3})u_{\alpha }^{(k)} + x_{3}U_{\alpha } + v_{\alpha },\nonumber \\ u_{3}&= (c_{1}x_{1} + c_{2}x_{2}+c_{3})x_{3} + \frac{1}{2}(b_{1}x_{1} + b_{2}x_{2} + b_{3})x_{3}^{2}\nonumber \\&\quad +\sum _{k=1}^{4}(c_{k} + b_{k}x_{3})u_{3}^{(k)} + x_{3}U_{3} + v_{3},\nonumber \\ \varphi&= \sum _{k=1}^{4}(c_{k} + b_{k}x_{3})\varphi ^{(k)} + x_{3}\Phi +\chi , \end{aligned}$$
(45)
where \((u_{j}^{(k)},\varphi ^{(k)})\) is the solution of problem \(A^{(k)} (k=1,2,3,4)\), \(U_{j}, \Phi , v_{j}\) and \(\chi \) are unknown functions of \(x_{1}\) and \(x_{2}\), and \(c_{k}\) \((k=1,2,3,4)\) are unknown constants. Let us consider a plane strain in which the components of the displacement vector are \(v_{j}\) and the microdilatation function is \(\chi \). The strain measures in this problem are defined by
$$\begin{aligned} \gamma _{\alpha \beta } = \frac{1}{2}(v_{\alpha ,\beta } + v_{\beta ,\alpha }), 2\gamma _{\alpha 3} = v_{3,\alpha }, \zeta _{\alpha \beta j} = v_{j,\alpha \beta }. \end{aligned}$$
(46)
We denote by \(E_{\alpha j}\) and \(K_{\alpha \beta j}\) the strain measures in the plane strain problem associated with the displacement vector \(U_{j}\) and microdilatation \(\Phi \),
$$\begin{aligned} E_{\alpha \beta } = \frac{1}{2}(U_{\alpha ,\beta } + U_{\beta ,\alpha }), 2E_{\alpha 3} = U_{3,\alpha }, K_{\alpha \beta j} = U_{j,\alpha \beta }. \end{aligned}$$
(47)
In view of (1) and (45)–(47), we obtain
$$\begin{aligned} e_{\alpha \beta }&= \sum _{k=1}^{4}(c_{k} + b_{k}x_{3})e_{\alpha \beta }^{(k)} + x_{3}E_{\alpha \beta } +\gamma _{\alpha \beta },\nonumber \\ e_{\alpha 3}&= \frac{1}{2}\varepsilon _{3\beta \alpha }(c_{4} + b_{4}x_{3})x_{\beta } + \sum _{k=1}^{4}(c_{k} + b_{k}x_{3}) e_{\alpha 3}^{(k)} + \gamma _{\alpha 3}+x_{3}E_{\alpha 3}\nonumber \\&\quad +\frac{1}{2}\sum _{k=1}^{4}b_{k}u_{\alpha }^{(k)} +\frac{1}{2}U_{\alpha },\nonumber \\ e_{33}&= c_{1}x_{1} + c_{2}x_{2} + c_{3} + (b_{1}x_{1} + b_{2}x_{2}+b_{3}) + \sum _{k=1}^{4}b_{k}u_{3}^{(k)} +U_{3},\nonumber \\ \kappa _{\alpha \beta \gamma }&= \sum _{k=1}^{4}(c_{k} + b_{k}x_{3}) \kappa _{\alpha \beta \gamma }^{(k)} + \zeta _{\alpha \beta \gamma }+x_{3}K_{\alpha \beta \gamma },\nonumber \\ \kappa _{\alpha \beta 3}&= \sum _{k=1}^{4}(c_{k} + b_{k}x_{3})\kappa _{\alpha \beta 3}^{(k)} + \zeta _{\alpha \beta 3} +x_{3}K_{\alpha \beta 3},\nonumber \\ \kappa _{\beta 3\alpha }&= \varepsilon _{3\beta \alpha }(c_{4} + b_{4}x_{3}) + \sum _{k=1}^{4}b_{k}u_{\alpha ,\beta }^{(k)} + U_{\alpha ,\beta },\nonumber \\ \kappa _{\alpha 33}&= c_{\alpha } + b_{\alpha }x_{3} + \frac{1}{2}\sum _{k=1}^{4}b_{k}e_{3\alpha }^{(k)} + \frac{1}{2}E_{\alpha 3},\nonumber \\ \kappa _{33\alpha }&= -c_{k}-b_{\alpha }x_{3} + \varepsilon _{3\beta \alpha }b_{4}x_{\beta },\;\; \kappa _{333} = b_{1}x_{1} + b_{2}x_{2} + b_{3}. \end{aligned}$$
(48)
Let us introduce the notations
$$\begin{aligned} s_{\alpha \beta }&= \lambda \gamma _{\rho \rho }\delta _{\alpha \beta } + 2\mu \gamma _{\alpha \beta } + f(\varepsilon _{\alpha \rho 3}\zeta _{\beta \rho 3} + \varepsilon _{\beta \rho 3}\zeta _{\alpha \rho 3}) + \textrm{d}\chi \delta _{\alpha \beta }, \nonumber \\ s_{\alpha 3}&= 2\mu \gamma _{\alpha 3} +f \varepsilon _{\rho \beta 3}\zeta _{\alpha \rho 3}, s_{33} = \lambda \gamma _{\rho \rho } + \textrm{d}\chi ,\nonumber \\ \nu _{\alpha \beta \gamma }&= \frac{1}{2}\alpha _{1}(\zeta _{\rho \rho \alpha }\delta _{\beta \gamma } + 2\zeta _{\gamma \rho \rho }\delta _{\alpha \beta } + \zeta _{\rho \rho \beta }\delta _{\alpha \gamma })+ \alpha _{2}(\zeta _{\alpha \rho \rho }\delta _{\beta \gamma } + \zeta _{\beta \rho \rho }\delta _{\alpha \gamma }) \nonumber \\&\quad + 2\alpha _{3}\zeta _{\rho \rho \gamma }\delta _{\alpha \beta }+2\alpha _{4}\zeta _{\alpha \beta \gamma } + \alpha _{5}(\zeta _{\gamma \beta \alpha } + \zeta _{\gamma \alpha \beta }) + f(\varepsilon _{\alpha \gamma 3}\gamma _{\beta 3} + \varepsilon _{\beta \gamma 3}\gamma _{\alpha 3})\nonumber \\&\quad +\beta _{1}\delta _{\alpha \beta }\chi _{,\gamma } + 2\beta _{2}(\delta _{\alpha \gamma }\chi _{,\beta } + \delta _{\beta \gamma }\chi _{,\alpha }),\nonumber \\ \nu _{\alpha \beta 3}&= 2\alpha _{3}\zeta _{\rho \rho 3}\delta _{\alpha \beta } + 2\alpha _{4}\zeta _{\alpha \beta 3} + f(\varepsilon _{\rho \alpha 3}\gamma _{\beta \rho } + \varepsilon _{\rho \beta 3}\gamma _{\alpha \rho }),\nonumber \\ \nu _{3\alpha \beta }&= \frac{1}{2}\alpha _{1}\zeta _{\rho \rho 3}\delta _{\alpha \beta } + \alpha _{5}\zeta _{\beta \alpha 3} + f\varepsilon _{\beta \rho 3}\gamma _{\alpha \rho },\nonumber \\ \nu _{3\alpha 3}&= \frac{1}{2}\alpha _{1}\zeta _{\rho \rho \alpha } + \alpha _{2}\zeta _{\alpha \rho \rho } + f \varepsilon _{\rho \alpha 3}\gamma _{3\rho } + \beta _{2}\chi _{,\alpha }, \nonumber \\ \nu _{33\alpha }&= \alpha _{1}\zeta _{\alpha \rho \rho }+2\alpha _{3}\zeta _{\rho \rho \alpha }+2f \varepsilon _{3\alpha \rho }\gamma _{3\rho }+\beta _{1}\chi _{,\alpha },\nonumber \\ \nu _{333}&= (\alpha _{1}+2\alpha _{3})\zeta _{\rho \rho 3}, h_{\alpha } = \beta _{1}\zeta _{\rho \rho \alpha } +2\beta _{2}\zeta _{\alpha \rho \rho } + a_{0}\chi _{,\alpha },\nonumber \\ h_{3}&= \beta _{1}\zeta _{\rho \rho 3}, p= \textrm{d}\gamma _{\rho \rho } + \xi \chi . \end{aligned}$$
(49)
Clearly, \(s_{ij}\) is the stress tensor, \(\nu _{ijk}\) is the dipolar stress tensor, \(\pi _{j}\) is the microstretch stress vector and p is the intrinsic body force in an isothermal plane problem corresponding to the strain measures \(\gamma _{ij}\) and \(\zeta _{ijk}\). We now consider a thermoelastic plane problem associated with the thermal field \(T_{1}\), displacement vector \(U_{j}\) and microdilatation function \(\Phi \). In this problem, we denote the stress tensor, the dipolar stress tensor, the microstretch stress vector and the intrinsic body force by \(T_{ij}, M_{ijk}, H_{\alpha }\) and L, respectively. Thus, we have
$$\begin{aligned} T_{\alpha \beta }&= \lambda E_{\rho \rho }\delta _{\alpha \beta } + 2\mu E_{\alpha \beta }+\textrm{d}\Phi \delta _{\alpha \beta } + f(\varepsilon _{\alpha \rho 3} K_{\beta \rho 3} + \varepsilon _{\beta \rho 3}K_{\alpha \rho 3}) - b T_{1}\delta _{\alpha \beta },\nonumber \\ T_{\alpha 3}&= 2\mu E_{\alpha 3} + f\varepsilon _{\rho \beta 3}K_{\alpha \rho \beta },\nonumber \\ M_{\alpha \beta \gamma }&= \frac{1}{2}\alpha _{1}(K_{\rho \rho \alpha }\delta _{\beta \gamma } + 2K_{\gamma \rho \rho }\delta _{\alpha \beta }+K_{\rho \rho \beta }\delta _{\alpha \gamma })\nonumber \\&\quad + \alpha _{2}(K_{\alpha \rho \rho }\delta _{\beta \gamma } +K_{\beta \rho \rho }\delta _{\alpha \gamma }) + 2\alpha _{3}K_{\rho \rho \gamma }\delta _{\alpha \beta }+ 2\alpha _{4}K_{\alpha \beta \gamma } + \alpha _{5}(K_{\gamma \beta \alpha }+K_{\gamma \alpha \beta })\nonumber \\&\quad +\beta _{1}\delta _{\alpha \beta }\Phi _{,\gamma } +2\beta _{2}(\delta _{\alpha \gamma }\Phi _{,\beta } + \delta _{\beta \gamma }\Phi _{,\alpha }) + f(\varepsilon _{\alpha \gamma 3}E_{\beta 3}+\varepsilon _{\beta \gamma 3}E_{\alpha 3}),\nonumber \\ M_{\alpha \beta 3}&= 2\alpha _{3}K_{\rho \rho 3}\delta _{\alpha \beta } + 2\alpha _{4}K_{\alpha \beta 3}+f(\varepsilon _{\rho \alpha 3}E_{\beta \rho } + \varepsilon _{\rho \beta 3}E_{\alpha 3}),\nonumber \\ H_{\alpha }&= \beta _{1}K_{\rho \rho \alpha } + 2\beta _{2}K_{\alpha \rho \rho } + a_{0}\Phi _{,\alpha }, H_{3} = \beta _{1}K_{\rho \rho 3},\nonumber \\ L&= dE_{\rho \rho } +\xi \Phi - \beta T_{1}, T_{33} = \lambda E_{\rho \rho } + \textrm{d}\Phi - b T_{1},\nonumber \\ M_{3\alpha \beta }&= \frac{1}{2}\alpha _{1}K_{\rho \rho 3}\delta _{\alpha \beta } + \alpha _{5}K_{\beta \alpha 3}+f\varepsilon _{\beta \rho 3}E_{\alpha \rho },\nonumber \\ M_{3\alpha 3}&= \frac{1}{2}\alpha _{1}K_{\rho \rho \alpha } + \alpha _{2}K_{\alpha \rho \rho } + f\varepsilon _{\rho \alpha 3} E_{3\rho }+ \beta _{2}\Phi _{,\alpha }\nonumber \\ M_{33\alpha }&= \alpha _{1}K_{\alpha \rho \rho } + 2\alpha _{3}K_{\rho \rho \alpha }+\beta _{1}\Phi _{,\alpha }+2f \varepsilon _{3\alpha \rho }E_{3\rho },\;\; M_{333} = (\alpha _{1}+2\alpha _{3})K_{\rho \rho 3}. \end{aligned}$$
(50)
From the constitutive equations (2) and the relations (48)–(50), we find that the stress tensor \(\tau _{ij}\) is given by
$$\begin{aligned} \tau _{\alpha \beta }&=s_{\alpha \beta } + x_{3}T_{a\beta }+\{\lambda [c_{1}x_{1}+c_{2}x_{2} + c_{3} + (b_{1}x_{1} + b_{2}x_{2}+b_{3})x_{3}]\nonumber \\&\quad -2f (c_{4} + b_{4}x_{3})\} \delta _{\alpha \beta } + \sum _{k=1}^{4}(c_{k}+b_{k}x_{3})\tau _{\alpha \beta }^{(k)} + G_{\alpha \beta },\nonumber \\ \tau _{\alpha 3}&= s_{\alpha 3} + x_{3}T_{\alpha 3} + 2f \varepsilon _{\alpha \rho 3}(c_{\rho } + b_{\rho }x_{3}) + \mu \varepsilon _{3\beta \alpha }(c_{4}+b_{4}x_{3}) x_{\beta } \nonumber \\ {}&\quad + \sum _{k=1}^{4}(c_{k} + b_{k}x_{3}) \tau _{\alpha 3}^{(k)} + G_{\alpha 3},\nonumber \\ \tau _{33}&= s_{33}+x_{3}T_{33} + (\lambda +2\mu )[c_{1}x_{1} + c_{2}x_{2} + c_{3}+ (b_{1}x_{1} + b_{2}x_{2}+b_{3})x_{3}]\nonumber \\&\quad +4f (c_{4} +b_{4}x_{3}) +\sum _{k=1}^{4}(c_{k}+b_{k}x_{3}) \tau _{33}^{(k)} + G_{33}, \end{aligned}$$
(51)
where
$$\begin{aligned} G_{a\beta }&= f[\varepsilon _{3\rho \alpha }U_{\rho ,\beta } + \varepsilon _{3\rho \beta }U_{\rho ,\alpha } + \sum _{k=1}^{4}b_{k} (\varepsilon _{3\rho \alpha }u_{\rho ,\beta }^{(k)} + \varepsilon _{3\rho \beta }u_{\rho ,\alpha }^{(k)})],\nonumber \\ G_{\alpha 3}&= \mu (U_{\alpha } + \sum _{k=1}^{4}b_{k}u_{\alpha }^{(k)}) + f\varepsilon _{\alpha \rho 3}(U_{3,\rho } + \sum _{k=1}^{4}b_{k}u_{3,\rho }^{(k)}) -fb_{4}x_{\alpha },\nonumber \\ G_{33}&= (\lambda +2\mu )(U_{3} + \sum _{k=1}^{4}b_{k}u_{3}^{(k)}) + 2f \varepsilon _{3\alpha \beta }(U_{\alpha ,\beta }+\sum _{k=1}^{4}b_{k}u_{\alpha ,\beta }^{(k)}). \end{aligned}$$
(52)
The functions \(\mu _{ijk}\), \(\sigma _{j}\) and g can be expressed as
$$\begin{aligned} \mu _{111}&= \nu _{111} + x_{3}M_{111} + 2(\alpha _{2} - \alpha _{3})(c_{1}+b_{1}x_{3}) + \sum _{k=1}^{4}(c_{k} + b_{k}x_{3}) \mu _{111}^{(k)}+J_{111},\nonumber \\ \mu _{222}&= \nu _{222}+x_{3}M_{222} + 2(\alpha _{2}-\alpha _{3})(c_{2}+b_{2}x_{3})+\sum _{k=1}^{4}(c_{k} + b_{k}x_{3})\mu _{222}^{(k)} + J_{222},\nonumber \\ \mu _{221}&= \nu _{221} + x_{3}M_{221}+(\alpha _{1}-2\alpha _{3})(c_{1}+b_{1}x_{3}) - f(c_{4} +b_{4}x_{3})x_{1}\nonumber \\&\quad +\sum _{k=1}^{4}(c_{k} + b_{k}x_{3})\mu _{221}^{(k)} + J_{221},\nonumber \\ \mu _{112}&= \nu _{112} + x_{3}M_{112} + (\alpha _{1}-2\alpha _{3})(c_{2} +b_{2}x_{3}) - f(c_{4} +b_{4}x_{3})x_{2}\nonumber \\&\quad +\sum _{k=1}^{4}(c_{k} + b_{k}x_{3})\mu _{112}^{(k)} + J_{112}, \nonumber \\\mu _{121}&= \nu _{121} + x_{3}M_{121} + \frac{1}{2}(2\alpha _{2} -\alpha _{1})(c_{2}+b_{2}x_{3})+\frac{1}{2} f(c_{4} + b_{4}x_{3})x_{2} \nonumber \\&\quad + \sum _{k=1}^{4}(c_{k} + b_{k}x_{3})\mu _{121}^{(k)} + J_{121},\nonumber \\ \mu _{122}&= \nu _{122} + x_{3}M_{122} + \frac{1}{2}(2\alpha _{2} -\alpha _{1})(c_{1}+b_{1}x_{3}) +\frac{1}{2}f(c_{4} + b_{4}x_{3})x_{1}\nonumber \\ {}&\quad +\sum _{k=1}^{4}(c_{k} + b_{k}x_{3})\mu _{122}^{(k)} + J_{122},\nonumber \\\mu _{\rho 33}&= \nu _{\rho 33}+x_{3}M_{\rho 33} + \frac{1}{2}(2\alpha _{2}-\alpha _{1}+4\alpha _{4})(c_{\rho } + b_{\rho }x_{3})\nonumber \\&\quad -\frac{1}{2}f (c_{4} +b_{4}x_{3})x_{\rho } + \sum _{k=1}^{4}(c_{k} + b_{k}x_{3})\mu _{\rho 33}^{(k)} + J_{\rho 33},\nonumber \\ \mu _{33\rho }&= \nu _{33\rho } + x_{3}M_{33\rho } + (\alpha _{1}-2\alpha _{3} - 2\alpha _{4} + 2\alpha _{5})(c_{\rho }+b_{\rho }x_{3})\nonumber \\ {}&\quad + f(c_{4} + b_{4}x_{3})x_{\rho } + \sum _{k=1}^{4}(c_{k} + b_{k}x_{3})\mu _{33\rho }^{(k)} + J_{33\rho }, \nonumber \\ \mu _{\alpha 3\beta }&= \nu _{\alpha 3\beta } + x_{3}M_{\alpha 3\beta } + (2\alpha _{4} - \alpha _{5})(c_{4} + b_{4}x_{3})\varepsilon _{\alpha \beta 3}\nonumber \\&\quad +f[c_{1}x_{1} + c_{2}x_{2} + c_{3} + (b_{1}x_{1} + b_{2}x_{2} + b_{3})x_{3}]\varepsilon _{\alpha \beta 3} \nonumber \\ {}&\quad +\sum _{k=1}^{4}(c_{k} + b_{k}x_{3})\mu _{\alpha 3\beta }^{(k)} + J_{\alpha 3\beta },\nonumber \\ \mu _{\alpha \beta 3}&= \nu _{\alpha \beta 3} + x_{3}M_{\alpha \beta 3} + \sum _{k=1}^{4}(c_{k} + b_{k}x_{3})\mu _{\alpha \beta 3}^{(k)}+J_{\alpha \beta 3}, \nonumber \\\mu _{333}&= \nu _{333} + x_{3}M_{333} + (\alpha _{1}+2\alpha _{3})\sum _{k=1}^{4}(c_{k} + b_{k}x_{3})\kappa _{\alpha \alpha 3}^{(k)} + J_{333}, \nonumber \\\sigma _{\alpha }&= h_{\alpha } + x_{3}H_{\alpha } + \sum _{k=1}^{4}(c_{k} + b_{k}x_{3})\sigma _{\alpha }^{(k)} + (2\beta _{2} - \beta _{1})(c_{\alpha } +b_{\alpha }x_{3}) \nonumber \\ {}&\quad +\beta _{1}\varepsilon _{3\beta \alpha }b_{4}x_{\beta } + \beta _{2}(\sum _{k=1}^{4}b_{k} e_{\alpha 3}^{(k)} + E_{\alpha 3}),\nonumber \\ \sigma _{3}&= h_{3}+x_{3}H_{3} + \sum _{k=1}^{4}(c_{k} + b_{k}x_{3})\sigma _{3}^{(k)} + (b_{1}+2\beta _{2})(b_{1}x_{1} + b_{2}x_{2} +b_{3})\nonumber \\&\quad + 2\beta _{2}(\sum _{k=1}^{4}b_{k} u_{\rho ,\rho }^{(k)} + U_{\rho ,\rho }) + a_{0}(\sum _{k=1}^{4}b_{k} \varphi ^{(k)} + \Phi ),\nonumber \\ g&= p+x_{3}L + \sum _{k=1}^{4}(c_{k} + b_{k}x_{3}) g^{(k)} + d[c_{1}x_{1} + c_{2}x_{2} + c_{3} \nonumber \\ {}&\quad + (b_{1}x_{1} + b_{2}x_{2}+b_{3})x_{3} + \sum _{k=1}^{4}b_{k}u_{3}^{(k)} + U]. \end{aligned}$$
(53)
In these relations, we have used the following notations
$$\begin{aligned} J_{111}&= - (\alpha _{1}+2\alpha _{3})b_{4}x_{2} + (\alpha _{1}+2\alpha _{2})(U_{3,1} + \sum _{k=1}^{4}b_{k}u_{3,1}^{(k)}), \nonumber \\ J_{222}&= (\alpha _{1}+2\alpha _{3})b_{4}x_{1} + (\alpha _{1} + 2\alpha _{2})(U_{3,2} + \sum _{k=1}^{4} b_{k}u_{3,2}^{(k)}), \nonumber \\ J_{221}&= -2\alpha _{3}b_{4}x_{2} + \alpha _{1}(U_{3,1} + \sum _{k=1}^{4}b_{k}u_{3,1}^{(k)}) - f(U_{2} + \sum _{k=1}^{4}b_{k}u_{2}^{(k)}), \nonumber \\J_{112}&= \alpha _{1}(U_{3,2} + \sum _{k=1}^{4} b_{k}u_{3,2}^{(k)}) + 2\alpha _{3}(b_{4} x_{1}+\sum _{k=1}^{4}b_{k}u_{1}^{(k)}) \nonumber \\ {}&\quad + f(U_{1} + \sum _{k=1}^{4} b_{k}u_{1}^{(k)}),\nonumber \\ J_{121}&= \frac{1}{2}\alpha _{1}b_{4}x_{1} + \alpha _{2}(U_{3,2} + \sum _{k=1}^{4} b_{k}u_{3,2}^{(k)}) -\frac{1}{2}f (U_{1} + \sum _{k=1}^{4}b_{k}u_{1}^{(k)}),\nonumber \\ J_{122}&= -\frac{1}{2}\alpha _{1}b_{4}x_{2} + \alpha _{2}(U_{3,1} + \sum _{k=1}^{4}b_{k}u_{3,1}^{(k)}) + \frac{1}{2}f (U_{2} + \sum _{k=1}^{4} b_{k}u_{2}^{(k)}), \nonumber \\ J_{\rho 33}&= -\frac{1}{2}(\alpha _{1} + 2\alpha _{5})\varepsilon _{3\rho \beta }b_{4}x_{\beta } + (\alpha _{2} + 2\alpha _{4} + \alpha _{5}) (U_{3,\rho }\nonumber \\&\quad + \sum _{k=1}^{4}b_{k}u_{3,\rho }^{(k)}) + \frac{1}{2}f \varepsilon _{3\rho \beta }(U_{\beta }+\sum _{k=1}^{4} b_{k}u_{\beta }^{(k)}),\nonumber \\ J_{33\rho }&= - 2(\alpha _{3}+\alpha _{4})\varepsilon _{3\rho \beta } b_{4}x_{\beta } + (\alpha _{1}+2\alpha _{5}) (U_{3,\rho }+\sum _{k=1}^{4} b_{k}u_{3,\rho }^{(k)}) \nonumber \\ {}&\quad + f \varepsilon _{3\rho \beta }(U_{\beta } + \sum _{k=1}^{4} b_{k}u_{\beta }^{(k)}), \nonumber \\J_{\alpha 3\beta }&= 2\alpha _{4}(U_{\beta ,\alpha }+\sum _{k=1}^{4} b_{k}u_{\beta ,\alpha }^{(k)}) + \alpha _{5}(U_{\alpha ,\beta } +\sum _{k=1}^{4} b_{k}u_{\alpha ,\beta }^{(k)}) \nonumber \\ {}&\quad + \delta _{\alpha \beta }[\frac{1}{2}(\alpha _{1}+2\alpha _{2})(b_{1}x_{1} + b_{2}x_{2}+b_{3}) + \alpha _{2}(U_{\rho ,\rho } + \sum _{k=1}^{4} b_{k}u_{\rho ,\rho }^{(k)})] \nonumber \\ {}&\quad + f(U_{3} + \sum _{k=1}^{4} b_{k}u_{3}^{(k)})\varepsilon _{a\beta 3},\nonumber \\ J_{\alpha \beta 3}&= \delta _{\alpha \beta }[(\alpha _{1}+2\alpha _{3})(b_{1}x_{1} + b_{2}x_{2} + b_{3}) +\alpha _{1}(U_{\rho ,\rho } + \sum _{k=1}^{4}b_{k}u_{\rho ,\rho }^{(k)})\nonumber \\&\quad + \beta _{1}(\Phi + \sum _{k=1}^{4}b_{k}\varphi ^{(k)})] + 2\alpha _{5}(E_{\alpha \beta } + \sum _{k=1}^{4}b_{k}e_{\alpha \beta }^{(k)}),\nonumber \\ J_{333}&= 2(\alpha _{1} + \alpha _{2}+\alpha _{3} + \alpha _{4}+\alpha _{5})(b_{1}x_{1} + b_{2}x_{2}+b_{3}) + (\alpha _{1} + 2\alpha _{2})U_{\rho ,\rho } \nonumber \\&\quad +(\beta _{1}+2\beta _{2})(\Phi +\sum _{k=1}^{4} b_{k}\varphi ^{(k)}). \end{aligned}$$
(54)
If we take into account (32), (33), (51) and (53), then the equilibrium equations (3) reduce to the equations
$$\begin{aligned} s_{\beta j, \beta } - \nu _{\alpha \beta j, \alpha \beta } + \Psi _{j} = 0, \;\; h_{\alpha ,\alpha } - p +\mathcal {H} = 0\;\; \text {on}\; \Sigma _{1}, \end{aligned}$$
(55)
and
$$\begin{aligned} T_{\alpha j,\alpha } - M_{\alpha \beta j, \alpha \beta } = 0,\;\; H_{\alpha ,\alpha }-L = 0\;\; \text {on}\; \Sigma _{1}. \end{aligned}$$
(56)
In equations (55) and (56), we have used the notations
$$\begin{aligned} \Psi _{\alpha }= & {} G_{\beta \alpha ,\beta } +T_{\alpha 3} - 2M_{3\rho \alpha ,\rho } - \sum _{k=1}^{4}(2\mu _{3\rho \alpha ,\rho }^{(k)} - \tau _{\alpha 3}^{(k)})b_{k}\nonumber \\{} & {} +4f \varepsilon _{3\alpha \beta }b_{\beta } + \mu \varepsilon _{3\rho \alpha }b_{4}x_{\rho } - J_{\rho \eta \alpha ,\rho \eta }, \nonumber \\\Psi _{3}= & {} G_{\alpha 3,\alpha } + T_{33} - 2M_{\alpha 33,\alpha } - J_{\alpha \beta 3,\alpha \beta }\nonumber \\{} & {} +(\lambda +2\mu )(b_{1}x_{1} + b_{2}x_{2} + b_{3}) + 6f b_{4} - \sum _{k=1}^{4} (2\mu _{\alpha 33,\alpha }^{(k)} - \tau _{33}^{(k)})b_{k},\nonumber \\ \mathcal {H}= & {} H_{3}+\beta _{2}(E_{\alpha 3,\alpha } + \sum _{k=1}^{4} b_{k}e_{\alpha 3,\alpha }^{(k)}) -d (U_{3} + \sum _{k=1}^{4}b_{k}u_{3}^{(k)}) + \sum _{k=1}^{4} b_{k}\sigma _{3}^{(k)}. \end{aligned}$$
(57)
Following (4) and (18), we define the functions
$$\begin{aligned}&\Pi _{j}^{(1)} = (s_{\beta j} - \nu _{\rho \beta j,\rho })n_{\beta } -D_{\beta }(n_{\rho }\nu _{\rho \beta j}) + (D_{\alpha }n_{\alpha })n_{\rho }n_{\eta }\nu _{\rho \eta j}, \nonumber \\ {}&\Lambda _{k}^{(1)} = \nu _{\rho \beta j}n_{\rho }n_{\beta }, \Lambda _{j}^{(2)} = M_{\rho \beta j}n_{\rho }n_{\beta },\nonumber \\&\Pi _{j}^{(2)} = (T_{\beta j} - M_{\rho \beta j,\rho })n_{\beta } - D_{\beta }(n_{\rho }M_{\rho \beta j}) + (D_{\nu }n_{\nu }) n_{\rho }n_{\eta }M_{\rho \eta j}. \end{aligned}$$
(58)
With the help of relations (36), (37), (51), (53) and (58), we see that the conditions on the lateral surface (7) reduce to
$$\begin{aligned} \Pi _{j}^{(1)} = B_{j}, \Lambda _{j}^{(1)} = C_{j}, h_{\alpha }n_{\alpha } = h\;\; \text {on}\; \Gamma _{1}, \end{aligned}$$
(59)
and
$$\begin{aligned} \Pi _{j}^{(2)} =0, \Lambda _{j}^{(2)} = 0, H_{\alpha }n_{\alpha } = 0\; \text {on}\; \Gamma _{1}, \end{aligned}$$
(60)
where
$$\begin{aligned} B_{\alpha }&= 2n_{\rho }(M_{3\rho \alpha } + \sum _{k=1}^{4} b_{k}\mu _{3\rho \alpha }^{(k)}) - 2\varepsilon _{\alpha \rho 3}n_{\rho }[f(b_{1}x_{1} + b_{2}x_{2} + b_{3}) \nonumber \\ {}&\quad +(2\alpha _{4} - \alpha _{5})b_{4}] - B_{\alpha }^{*}, \nonumber \\ B_{3}&= 2n_{\rho }[M_{\rho 33} + \frac{1}{2}(2\alpha _{2} - \alpha _{1}+4\alpha _{4})b_{\rho } - \frac{1}{2}f b_{4}x_{\rho }] \nonumber \\ {}&\quad +\sum _{k=1}^{4} b_{k}\mu _{\rho 33}^{(k)} - B_{3}^{*},\; C_{j} = -J_{\alpha \beta j}n_{\alpha }n_{\beta },\nonumber \\ h&= [\beta _{1}\varepsilon _{3\alpha \beta } b_{4}x_{\beta } - \beta _{2}(E_{\alpha 3} +\sum _{k=1}^{4} b_{k}e_{\alpha 3}^{(k)})]n_{\alpha }, \nonumber \\ B_{j}^{*}&= (G_{\beta j} - J_{\rho \beta j,\rho })n_{\beta } - D_{\nu }(n_{\rho }J_{\rho \nu j}) + (D_{\alpha }n_{\alpha })n_{\rho }n_{\nu }J_{\rho \nu j}. \end{aligned}$$
(61)
We denote by \(({\mathcal {P}}_{1})\) the isothermal plane problem which consist in finding the functions \(v_{j}\) and \(\chi \) that satisfy the geometrical equations (46), the constitutive equations (49), the equilibrium equations (55) and the boundary conditions (59). Let us denote by \(({\mathcal {P}}_{2})\) the thermoelastic plane problem associated with the temperature \(T_{1}\) and characterized by the geometrical equations (47), the constitutive equations (50), the equilibrium equations (56) and the boundary conditions (60). Clearly, the necessary and sufficient conditions to solve the problem \(({\mathcal {P}}_{2})\) are satisfied for any thermal field. The necessary and sufficient conditions to solve the problem \(({\mathcal {P}}_{1})\) are
$$\begin{aligned}&\int _{\Sigma _{1}}\Psi _{j}\textrm{d}a +\int _{\Gamma _{1}}B_{j}\textrm{d}s = 0,\nonumber \\&\int _{\Sigma _{1}} \varepsilon _{3\alpha \beta }x_{\alpha }\Psi _{\beta }\textrm{d}a + \int _{\Gamma _{1}}\varepsilon _{3\alpha \beta } (x_{\alpha }B_{\beta } + n_{\alpha }C_{\beta })\textrm{d}s=0. \end{aligned}$$
(62)
By using the divergence theorem, we find that
$$\begin{aligned}&\int _{\Sigma _{1}} (G_{\beta j,\beta } - J_{\rho \eta j, \rho \eta })\textrm{d}a + \int _{\Gamma _{1}} B_{j}^{*}\textrm{d}s = 0,\nonumber \\&\int _{\Sigma _{1}}\varepsilon _{3\alpha \beta }x_{\alpha }(G_{\rho \beta ,\rho } - J_{\nu \rho \beta , \nu \rho })\textrm{d}a + \int _{\Gamma _{1}} \varepsilon _{3\alpha \beta }(x_{\alpha }B^{*}_{\beta } + n_{\alpha }C_{\beta })\textrm{d}s=0. \end{aligned}$$
(63)
It follows from (51), (57), (61) and (63) that
$$\begin{aligned} \int _{\Sigma _{1}}\Psi _{\alpha }\textrm{d}a + \int _{\Gamma _{1}} B_{\alpha }\textrm{d}s = \int _{\Sigma _{1}} \tau _{\alpha 3,3}\textrm{d}a. \end{aligned}$$
(64)
By using the equilibrium equations, we find
$$\begin{aligned} \tau _{\alpha 3}&= \tau _{\alpha 3} + x_{\alpha }(\tau _{j3,j} -\mu _{rs3,rs}) = [x_{\alpha }(\tau _{\beta 3} - \mu _{\beta \nu 3,\nu })]_{,\beta }\nonumber \\&\quad + x_{\alpha }(\tau _{33,3} - 2\mu _{3\beta 3,3\beta } - \mu _{333,33}) + \mu _{\alpha \nu 3,\nu }. \end{aligned}$$
(65)
The condition \(P_{3} = {\widetilde{P}}_{3}\) on the lateral boundary can be expressed in the form
$$\begin{aligned} (\tau _{\beta 3} - \mu _{\beta \nu 3,\nu }) n_{\beta } = {\widetilde{P}}_{3} + [2\mu _{3\beta 3,3} - (\mu _{\alpha \beta 3}n_{\alpha }n_{\gamma } - \mu _{\rho \gamma 3}n_{\rho }n_{\beta })_{,\gamma }]n_{\beta }, \end{aligned}$$
(66)
so that the relation (65) implies that
$$\begin{aligned} \int _{\Sigma _{1}}\tau _{\alpha 3}\textrm{d}a&= \int _{\Gamma _{1}}[\mu _{\alpha \nu 3,\nu } + x_{\alpha }(\tau _{33,3} - 2\mu _{3\beta 3,3\beta } - \mu _{333,33})] \textrm{d}a\nonumber \\&\quad +\int _{\Gamma _{1}} x_{\alpha } \{{\widetilde{P}}_{3} + (\mu _{\rho \gamma 3}n_{\rho }n_{\beta } - \mu _{\alpha \beta 3}n_{\alpha }n_{\gamma })_{,\gamma }n_{\beta } + 2\mu _{3\beta 3,3}n_{\beta }\}\textrm{d}s. \end{aligned}$$
(67)
Let us note the identity
$$\begin{aligned} \int _{\Sigma _{1}}x_{\alpha }(\mu _{\rho \gamma 3}n_{\rho }n_{\beta } - \mu _{\alpha \beta 3}n_{\alpha }n_{\gamma })_{,\gamma } n_{\beta } \textrm{d}s = \int _{\Gamma _{1}}n_{\alpha } {\widetilde{R}}_{3}\textrm{d}s - \int _{\Sigma _{1}} \mu _{\alpha \nu 3,\nu }\textrm{d}a. \end{aligned}$$
(68)
From (65) and (68), we get
$$\begin{aligned} \int _{\Sigma _{1}}\tau _{\alpha 3}\textrm{d}a = \int _{\Sigma _{1}} [x_{\alpha }(\tau _{33,3} - \mu _{333,33}) + 2\mu _{\alpha 33,3}]\textrm{d}a + \int _{\Gamma _{1}} (x_{\alpha }{\widetilde{P}}_{3} +n_{\alpha }{\widetilde{R}}_{3})\textrm{d}s. \end{aligned}$$
(69)
In view of (7), (51), (53) and (69), we find that
$$\begin{aligned} \int _{\Sigma _{1}} \tau _{\alpha 3,\alpha }\textrm{d}a = 0, \end{aligned}$$
so that the first two conditions from (61) are satisfied. With the help of (40), (57) and (60), we obtain
$$\begin{aligned}&\int _{\Sigma _{1}} \Psi _{3}\textrm{d}a + \int _{\Gamma _{1}}B_{3}\textrm{d}s = \int _{\Sigma _{1}}T_{33}\textrm{d}a + \sum _{k=1}^{4}D_{3k}b_{k},\\&\int _{\Sigma _{1}}\varepsilon _{3\alpha \beta }x_{\alpha }\Psi _{\beta }\textrm{d}a + \int _{\Gamma _{1}} \varepsilon _{3\alpha \beta } (x_{\alpha }B_{\beta } + n_{\alpha }C_{\beta })\textrm{d}s\\ {}&= \int _{\Sigma _{1}}\varepsilon _{3\alpha \beta }x_{\alpha }T_{\beta 3}\textrm{d}a + \sum _{k=1}^{4}D_{4k}b_{k}. \end{aligned}$$
The last two conditions from (62) reduce to
$$\begin{aligned} \sum _{k=1}^{4}D_{3k}b_{k} = -\int _{\Sigma _{1}} T_{33}\textrm{d}a, \sum _{k=1}^{4}D_{4k}b_{k} = -\int _{\Sigma _{1}}\varepsilon _{3\alpha \beta }x_{\alpha }T_{\beta 3}\textrm{d}a. \end{aligned}$$
(70)
Let us impose the conditions (8). In view of (12), we get
$$\begin{aligned} \int _{\Sigma _{1}} P_{\alpha }\textrm{d}a + \int _{\Gamma _{1}}Q_{s}\textrm{d}s = -\int _{\Sigma _{1}} (\tau _{3\alpha } - \mu _{33\alpha ,3})\textrm{d}a. \end{aligned}$$
By using (67), we obtain
$$\begin{aligned} \int _{\Sigma _{1}}P_{\alpha }\textrm{d}a + \int _{\Gamma _{1}}Q_{\alpha }\textrm{d}s&= -\int _{\Gamma _{1}}(x_{\alpha }{\widetilde{P}}_{3} + n_{\alpha }{\widetilde{R}}_{3})\textrm{d}s - \int _{\Sigma _{1}} [x_{\alpha }(\tau _{33,3} - \mu _{333,33})\nonumber \\ {}&\quad +2\mu _{\alpha 33,3} - \mu _{33\alpha ,3}]\textrm{d}a. \end{aligned}$$
(71)
With the help of (7), (40), (51), (53) and (71), we see that the conditions (8) reduce to
$$\begin{aligned} \sum _{k=1}^{4}D_{\alpha k}b_{k} = -F_{\alpha } - \int _{\Sigma _{1}}(x_{\alpha }T_{33} +2M_{\alpha 33} - M_{33\alpha })\textrm{d}a. \end{aligned}$$
(72)
The system (70), (72) uniquely determines the constants \(b_{k} (k=1,2,3,4)\).
If we take into account the relations (38), (40), (51) and (53), we find that the conditions (9)–(11) take the form
$$\begin{aligned}&\sum _{k=1}^{4}D_{\alpha k}c_{k} = -\varepsilon _{3\alpha \beta }(M_{\beta } + M_{\beta }^{0}),\nonumber \\&\sum _{k=1}^{4} D_{3k}c_{k} = -F_{3} - F_{3}^{0}, \sum _{k=1}^{4} D_{4k}c_{k} = -M_{3} - M_{3}^{0}, \end{aligned}$$
(73)
where we have used the notations
$$\begin{aligned} F_{3}^{0}&= \int _{\Sigma _{1}} [s_{33} + G_{33} - M_{333} - (\alpha _{1}+2\alpha _{3})\sum _{k=1}^{4}b_{k}\kappa _{333}^{(k)}]\textrm{d}a, \nonumber \\M_{\alpha }^{0}&= \varepsilon _{3\alpha \beta }\int _{\Sigma _{1}}\{ x_{\beta }[s_{33} + G_{33} - M_{333} - (\alpha _{1}+2\alpha _{3})\sum _{k=1}^{4}b_{k}\kappa _{333}^{(k)}] \nonumber \\ {}&\quad + 2\nu _{\beta 33} +2J_{\beta 33} -\nu _{33\beta } - J_{33\beta }\}\textrm{d}a,\nonumber \\ M_{3}^{0}&= \int _{\Sigma _{1}}\{\varepsilon _{\alpha \beta 3}[s_{\beta 3} + G_{\beta 3} - M_{33\beta } - (\alpha _{1}-2\alpha _{3} -2\alpha _{4}+2\alpha _{5}) b_{\beta }\nonumber \\&\quad - f b_{4}x_{\beta } - \sum _{k=1}^{4}b_{k}\mu _{33\beta }^{(k)}] + 2\varepsilon _{\alpha \beta 3}(\nu _{\alpha 3\beta } + J_{\alpha 3\beta })\}\textrm{d}a. \end{aligned}$$
(74)
The constant \(c_{k}\) are determined by the system (73). First, we have to find the solutions of the plane problems \(A^{(k)}\) \((k=1,2,3,4)\). Then, we can determine the constants \(D_{mn}\) from (41) and the solution of the problem \(({\mathcal {P}}_{2})\). Next, we determine the constants \(b_{k}\) from the system (70), (72). Now, we can solve the problem \(({\mathcal {P}}_{1})\) and to find from (73) the constants \(c_{k}\) \((k=1,2,3,4)\). The above results can be used to study the case when the temperature field is a polynomial in the axial coordinate.

6 Application

In this section, we use the solution (23) to investigate the thermoelastic deformation of a chiral circular cylinder subjected to a uniform temperature. We consider a homogeneous cylinder that occupies the domain \(B = \{(x_{1}, x_{2}, x_{3}): x_{1}^{2} + x_{2}^{2}<a^{2}, 0<x_{3}<h\}\), \((a>0)\). We assume that the mechanical loads are absent. In this case, we have
$$\begin{aligned} T = T^{*},\;\; F_{j} = 0,\;\;\, M_{j} = 0, \end{aligned}$$
(75)
where \(T^{*}\) is a prescribed constant. We seek the solution of the problem \(({\mathcal {T}})\) in the form
$$\begin{aligned} w_{\alpha } = E_{1}x_{\alpha },\;\, w_{3} = 0,\;\, \psi = E_{2}, \end{aligned}$$
(76)
where \(E_{1}\) and \(E_{2}\) are arbitrary constants. From (28), (27) and (76), we obtain
$$\begin{aligned}&\eta _{\alpha \beta } = E_{1}\delta _{\alpha \beta }, \eta _{\alpha 3} =0, \xi _{ijk} = 0, \nonumber \\&t_{\alpha \beta } = [2(\lambda +\mu )E_{1} + d E_{2} - b T^{*}]\delta _{\alpha \beta }, t_{\alpha 3} = 0, m_{\alpha \beta j} = 0, \nonumber \\ {}&\pi _{\alpha } = 0, \gamma = 2d E_{1}+\xi E_{2} - \beta T^{*}, m_{3\alpha \beta }= f E_{1}\varepsilon _{\beta \alpha 3}, m_{3\alpha 3} = 0,\nonumber \\ {}&m_{33j} = 0, P_{\alpha }^{*} = [2(\lambda +\mu )E_{1} + dE_{2}- bT^{*}]n_{\alpha }, P_{3}^{*} = 0,R_{i}^{*} = 0. \end{aligned}$$
(77)
The equilibrium equations (34) and the boundary conditions (38) are satisfied if the constants \(E_{1}\) and \(E_{2}\) are given by
$$\begin{aligned} E_{1} = (\beta d - b \xi )T^{*}/(2D), E_{2} = [db - \beta (\lambda +\mu )]T^{*}/D, \end{aligned}$$
(78)
where
$$\begin{aligned} D = d^{2}-\xi (\lambda +\mu ). \end{aligned}$$
(79)
The positive definiteness of the elastic potential implies that \(D\ne 0\), Thus, the solution of the problem \(({\mathcal {T}})\) is given by (76). It follows from (40) and (77) that
$$\begin{aligned} F_{3}^{*} = \pi a^{2}(2\lambda E_{1} + d E_{2}- bT^{*}), M_{j}^{*} = 0. \end{aligned}$$
(80)
In a similar way, we can study the problems \(A^{(3)}\) and \(A^{(4)}\). The solution of the problem \(A^{(3)}\) is
$$\begin{aligned} u_{\alpha }^{(3)} = S_{1}x_{\alpha }, u_{3}^{(3)} = 0, \varphi ^{(3)} = S_{2}, \end{aligned}$$
(81)
where
$$\begin{aligned} S_{1} = (\lambda \xi -d^{2})/(2D), \;\; S_{2} = \textrm{d}\mu / D, \end{aligned}$$
and D is given by (79). The problem \(A^{(4)}\) has the following solution
$$\begin{aligned} u_{\alpha }^{(4)} = K_{1}x_{\alpha }, u_{3}^{(4)} = 0, \varphi ^{(3)} = K_{2}, \end{aligned}$$
(82)
where
$$\begin{aligned} K_{1} = -f \xi / D,\;\; K_{2} = 2bf/D. \end{aligned}$$
It follows from (81), (82) and (41) that
$$\begin{aligned} \begin{aligned}&D_{\alpha 3} = 0, D_{\alpha 4} = 0, D_{33} = \pi \mu a^{2}[3d^{2} - (3\lambda +2\mu )\xi ]/D, \\ {}&D_{44} = \mu \frac{\pi a^{4}}{2} + 4(2a_{4} - \alpha _{5}-K_{1}f)\pi a^{2},\\ {}&D_{34} = D_{43} = 2\pi a^{2}f [3d^{2} - (3\lambda +2\mu )\xi ]/D. \end{aligned} \end{aligned}$$
(83)
In view of symmetry of \(D_{mn}\), we obtain \(D_{3\alpha } = 0\) and \(D_{4\alpha } = 0.\) Thus, with the help of (82) and (83) we find that the system (40) reduces to
$$\begin{aligned} D_{\alpha \beta }a_{\beta } = 0, D_{33}a_{3} + D_{34}a_{4} = -F_{3}^{*}, D_{43}a_{3} + D_{44}a_{4} = 0. \end{aligned}$$
(84)
It follows from (42) that this system uniquely determines the constants \(a_{k}\). We find that
$$\begin{aligned} a_{1} = a_{2} = 0, a_{3} = - D_{44}F_{3}^{*}q, a_{4} = D_{34}F_{3}^{*}q, \end{aligned}$$
(85)
where
$$\begin{aligned} q^{-1} = D_{33}D_{44} - D_{34}^{2}. \end{aligned}$$
Let us note that we have solved the problem without using the solutions of the problems \(A^{(1)}\) and \(A^{(2)}\).
In view of (23), (76), (81), (82) and (85), we see that the solution of the problem is given by
$$\begin{aligned}&u_{\alpha } = \varepsilon _{3\beta \alpha }a_{4}x_{\beta }x_{3} + a_{3}S_{1}x_{\alpha } + a_{4}K_{1}x_{\alpha } + w_{\alpha },\nonumber \\&u_{3} = a_{3}x_{3}, \varphi = a_{3}S_{2} + a_{4}K_{2}+\psi . \end{aligned}$$
(86)
From (86), we conclude that the thermal field produces torsion, extension and a variation of microdilatation. In the case of an achiral material, we have \(f=0\) and (83) implies that \(D_{34} = 0.\) Then, from (85) we obtain \(a_{4} = 0\). In this case, the thermal field does not produce torsion.

7 Conclusions

The results presented in this paper can be summarized as follows:
a.
In the context of the strain gradient theory of porous thermoelastic solids, we study the deformation of isotropic chiral cylinders subjected to a temperature field that is linear in the axial coordinate.
 
b.
We introduce the problem of generalized plane deformation of a chiral cylinder, and we express the equilibrium equations in terms of displacements and microdilatation function.
 
c.
We establish the solution of the problem when the temperature distribution is independent of the axial coordinate. It is shown that the thermal field produces extension, bending and torsion.
 
d.
We study the thermoelastic deformation of cylinders subjected to a thermal field which is linear in the axial coordinate. We present a method to reduce the three-dimensional problem to the study of plane problems.
 
e.
We use the results to study the deformation of a circular cylinder subjected to a uniform temperature variation. In contrast to the case of achiral materials, the thermal field in chiral cylinders produces torsional effects.
 

Funding

Open access funding provided by Universitá degli Studi di Napoli Federico II within the CRUI-CARE Agreement.
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Metadata
Title
On the thermal stresses in chiral porous elastic beams
Authors
Simona De Cicco
Dorin Ieşan
Publication date
23-06-2023
Publisher
Springer Berlin Heidelberg
Published in
Continuum Mechanics and Thermodynamics / Issue 5/2023
Print ISSN: 0935-1175
Electronic ISSN: 1432-0959
DOI
https://doi.org/10.1007/s00161-023-01236-6

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