The material of the considered bar is linearly elastic, non-homogeneous and anisotropic. The material properties are smooth functions of the axial coordinate
z. This means that each cross section is made of one material, but the material of one cross section is different from that of its adjoining one. The assumed form of the anisotropy and
z dependence of material properties are given as [
9,
11,
13]:
$$\begin{aligned} \sigma _{x}= & {} g(z)(c_{11}\varepsilon _{x}+c_{12}\varepsilon _{y}+c_{13}\varepsilon _{z}+c_{16}\gamma _{xy}), \end{aligned}$$
(1)
$$\begin{aligned} \sigma _{y}= & {} g(z)(c_{12}\varepsilon _{x}+c_{22}\varepsilon _{y}+c_{23}\varepsilon _{z}+c_{26}\gamma _{xy}), \end{aligned}$$
(2)
$$\begin{aligned} \sigma _{z}= & {} g(z)(c_{13}\varepsilon _{x}+c_{23}\varepsilon _{y}+c_{33}\varepsilon _{z}+c_{36}\gamma _{xy}), \end{aligned}$$
(3)
$$\begin{aligned} \tau _{yz}= & {} g(z)(c_{44}\gamma _{yz}+c_{45}\gamma _{xz}), \end{aligned}$$
(4)
$$\begin{aligned} \tau _{xz}= & {} g(z)(c_{45}\gamma _{yz}+c_{55}\gamma _{xz}), \end{aligned}$$
(5)
$$\begin{aligned} \tau _{xy}= & {} g(z)(c_{16}\varepsilon _{x}+c_{26}\varepsilon _{y}+c_{36}\varepsilon _{z}+c_{66}\gamma _{xy}). \end{aligned}$$
(6)
The non-homogeneity which appears in the axial direction is described by the smooth function
\(g=g(z)\). The stiffness coefficients
\(c_{ij}\) (
\(i, j=1,...,6\)) are constants, their units are [force]/[length]
\(^{{2}}\) and
g is unit-free. In Eq. (
1)
\(\varepsilon _{x}\),
\(\varepsilon _{y}\),
\(\varepsilon _{z}\) are normal strains,
\(\gamma _{yz}\),
\(\gamma _{xz}\),
\(\gamma _{xy }\)are shearing strains,
\(\sigma _{x}\),
\(\sigma _{y},\sigma _{z}\) are normal stresses and
\(\tau _{yz}\),
\(\tau _{xz}\),
\(\tau _{xy}\) are shearing stresses. From Eqs. (
1‐
6) can be read that there exists a plane of symmetry of elasticity. The plane of symmetry of elasticity is the plane
Oxy, thus the non-homogeneous monoclinic system is considered [
7,
11,
13]. From the positivity of strain energy density it follows that [
7,
11] the Hooke’s matrix
$$\begin{aligned} {\mathbf {C}}=\left( {\begin{array}{cccccc} c_{11} &{}\quad {c}_{12}&{}\quad {c}_{13} &{}\quad 0 &{}\quad 0&{}\quad {c}_{16} \\ c_{12}&{}\quad {c}_{22} &{}\quad c_{23}&{}\quad 0&{}\quad 0 &{} \quad c_{26} \\ c_{13}&{}\quad c_{23} &{}\quad c_{33} &{}\quad 0&{}\quad 0&{}\quad {c}_{36} \\ 0&{}\quad 0&{}\quad 0&{}\quad c_{44}&{}\quad {c}_{45} &{}\quad 0 \\ 0&{} \quad 0&{}\quad 0&{}\quad {c}_{45}&{}\quad {c}_{55} &{}\quad 0 \\ c_{16}&{}\quad { c}_{26} &{}\quad c_{36}&{}\quad 0&{}\quad 0 &{}\quad c_{66} \\ \end{array}} \right) \end{aligned}$$
(7)
is a positive definite matrix and
\(g(z) > 0\),
\((0\le z\le L)\). Let the boundary curve
\(\partial A\) of the cross section
A (Fig.
1) be given by the equation
$$\begin{aligned} c_{44}x^{2}-2c_{45}xy+c_{55}y^{2}=k(x,y)\in \partial A, \end{aligned}$$
(8)
where
k is an arbitrary positive real number. Units of
k is [force] and from the positive definitness of matrix
C it follows that
$$\begin{aligned} c_{44}>0,c_{55}>0,\quad { c}_{44}c_{55}-(c_{45})^{2}>0. \end{aligned}$$
(9)
This latter inequalities assures that the points whose
x and
y coordinates satisfy Eq. (
8) are on an ellipse. The centre of the boundary ellipse is the origin of the cross-sectional coordinate system
Oxy. The non-uniform torsion problem for the functionally graded cylinder whose cross section is given by Eq. (
8) is defined by the next mixed type boundary conditions
$$\begin{aligned}&\sigma _{x}n_{x}+\tau _{xy}n_{y}=\tau _{xy}n_{x}+\sigma _{y}n_{y}=\tau _{xz}n_{x}+\tau _{yz}n_{y}=0 \,\,\, {\mathrm{on }}\, \partial A {\times (0,}L{),} \end{aligned}$$
(10)
$$\begin{aligned}&\sigma _{z}\left( x,y,0 \right) =\sigma _{z}\left( x,y,L \right) =0, \quad u\left( x,y,0 \right) =v\left( x,y,0 \right) =0,\nonumber \\&u(x,y,L)=-\varTheta y, \quad v(x,y,L)=\varTheta x. \end{aligned}$$
(11)
In Eq. (
10)
\(n_{x}\,\mathrm{and}\, n_{y}\) are the components of the unit normal vector of boundary curve
\(\partial A\) and in Eqs. (
11)
\(_{{3}}\) and (
11)
\(_{{4}} \varTheta \) is a given constant.
Assumed form of the displacement field of the twisted anisotropic non-homogeneous bar is as follows
$$\begin{aligned} \mathrm{u}=- \varPhi \left( z \right) \mathrm{y},\,\mathrm{v}= \varPhi \left( z \right) \mathrm{x},\quad w=0\,\left( x,y \right) \in A \, \mathrm{and }\, 0 \le z{ \le } L{,} \end{aligned}$$
(12)
where
\(\varPhi =\varPhi (z)\) is an unknown function with the boundary conditions
$$\begin{aligned} \varPhi (0)=0, \quad \varPhi (L)=\varTheta . \end{aligned}$$
(13)
Application of the strain-displacement relationships of the linearized theory of elasticity gives the result [
9,
14]:
$$\begin{aligned} \varepsilon _{x}= & {} \frac{\partial u}{\partial x}=0,\quad { \varepsilon }_{y}=\frac{\partial v}{\partial y}=0,\quad { \varepsilon }_{z}=\frac{\partial w}{\partial z}=0, \quad \gamma _{xy}=\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}=0, \end{aligned}$$
(14)
$$\begin{aligned} \gamma _{xz}= & {} \frac{\partial u}{\partial z}+\frac{\partial w}{\partial x}=-\frac{d\varPhi }{dz}y,\quad \gamma _{yz}=\frac{\partial v}{\partial z}+\frac{\partial w}{\partial x}=\frac{d\varPhi }{dz}x. \end{aligned}$$
(15)
Combination of Eqs. (
1‐
6) with Eqs. (
14) and (
15) yields
$$\begin{aligned} \sigma _{x}= & {} \sigma _{y}=\sigma _{z}=\tau _{xy}=0, \end{aligned}$$
(16)
$$\begin{aligned} \tau _{yz}= & {} g(z)\frac{\mathrm{d}\varPhi }{\mathrm{d}z}(c_{44}x-c_{45}y), \quad \tau _{xz}=g(z)\frac{\mathrm{d}\varPhi }{\mathrm{d}z}(c_{45}x-c_{55}y). \end{aligned}$$
(17)
Starting from Eqs. (
16), (
17) and using the equations of equilibrium [
14]
$$\begin{aligned}&\frac{\partial \sigma _{x}}{\partial x}+\frac{\partial \tau _{xy}}{\partial y}+\frac{\partial \tau _{xz}}{\partial z}=0\,(x,y)\in A \, \mathrm{and }\,0<z{<}L{,} \end{aligned}$$
(18)
$$\begin{aligned}&\frac{\partial \tau _{xy}}{\partial x}+\frac{\partial \sigma _{y}}{\partial y}+\frac{\partial \tau _{yz}}{\partial z}=0(x,y)\in A \, \mathrm{and }\, 0\,<z{<}L{,} \end{aligned}$$
(19)
$$\begin{aligned}&\frac{\partial \tau _{xz}}{\partial x}+\frac{\partial \tau _{yz}}{\partial y}+\frac{\partial \sigma _{z}}{\partial z}=0\,(x,y)\in A\, \mathrm{and}\, 0<z{<}L{.} \end{aligned}$$
(20)
and stress boundary conditions formulated by Eq. (
10) we get
$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}z}\left( g(z)\frac{\mathrm{d}\varPhi }{\mathrm{d}z} \right) =0\,\,0<z<L, \end{aligned}$$
(21)
$$\begin{aligned}&\tau _{xz}n_{x}+\tau _{yz}n_{y}=g(z)\frac{\mathrm{d}\varPhi }{\mathrm{d}z}\left[ (c_{45}x-c_{55}y)\frac{\mathrm{d}y}{\mathrm{d}s}-(c_{44}x-c_{45}y)\frac{\mathrm{d}x}{\mathrm{d}s} \right] =0 \,\, \mathrm{on} \, \partial A. \end{aligned}$$
(22)
Here, we have used
$$\begin{aligned} n_{x}=\frac{\mathrm{d}y}{\mathrm{d}s},\, n_{y}=-\frac{\mathrm{d}x}{\mathrm{d}s}\,\mathrm{on}\, \partial A , \end{aligned}$$
(23)
where
\(x=x(s), y=y(s)\) are the equation of the boundary curve
\(\partial A\) and
s is an arc-length defined on curve
\(\partial A\) (Fig.
1). From Eqs. (
22), we obtain
$$\begin{aligned}&\left[ \left( c_{45}x-c_{55}y \right) \frac{\mathrm{d}y}{\mathrm{d}s}-\left( c_{44}x-c_{45}y \right) \frac{\mathrm{d}x}{\mathrm{d}s} \right] \nonumber \\&\quad = -\frac{\mathrm{d}}{\mathrm{d}s}\left( c_{44}x^{2}-2c_{45}xy+c_{55}y^{2} \right) =-\frac{\mathrm{d}k}{\mathrm{d}s}=0\,\, \mathrm{on}\, \partial A, \end{aligned}$$
(24)
thus all the boundary conditions on the cylindrical surface are satisfied if the contour of elliptical cross section is given by Eq. (
8).