Open Access

On the deformation of almost cylindrical elastic beams

International Journal of Mechanical and Materials Engineering201712:8

DOI: 10.1186/s40712-017-0078-2

Received: 7 September 2016

Accepted: 26 February 2017

Published: 9 March 2017

Abstract

Background

The Saint-Venant problem for porous elastic cylinders is of interest both from the technical and mathematical point of view. The intended applications of solution are to mechanics of bone and to some engineering structures.

Method

This work investigates the Saint-Venant problem for almost prismatic bars made of an isotropic porous material. We express the solution in terms of the solutions of two problems concerning the deformation of a right cylinder.

Results and Conclusion

We use the method to study the extension of an almost prismatic conical frustum. It is found that the displacement vector is a polynomial of two degree in the cartesian coordinates. The volume fraction field depends linearly on the axial coordinate. The solution contains terms characterizing the influence of the material porosity and the dependence on the lateral surface.

Keywords

Elastic bars Porous bodies Almost cylindrical beams Elastic materials with voids

Background

A generalization of the classical theory of elasticity is the theory of elastic materials with voids established by Nunziato and Cowin (1979) and Cowin and Nunziato (1983). The intended application of the theory is to elastic bodies with pores which are distributed throughout the material. In the framework of the linear theory of isotropic elastic materials with voids, the deformation of the right cylinders has been the subject of various investigations. Cowin and Nunziato (1983) have studied the pure bending of a cylinder made of a homogeneous material. The problem of extension and bending for nonhomogeneous porous elastic bodies has been investigated by Ciarletta and Iesan (1993), Ieşan and Nappa (1994), Ieşan and Scalia (2007), and Ieşan and Scalia (2009). A treatment of Saint Venant’s problem for homogenous and isotropic porous elastic cylinders has been presented by Dell’Isola and Batra (1997), Ieşan and Quintanilla (1995), Ieşan (2009), and Ieşan (2011).

In the context of the classical elasticity, the mechanical behavior of the noncylindrical elastic bars has been studied in many papers (see, e.g., Dryden (2007), Zupan and Saje (2006), You et al. (2002), and the references therein). These bodies are of interest both from the technical and mathematical point of view. The present paper is concerned with the Saint-Venant problem for almost cylindrical bars made of porous elastic materials. In the classical elastostatics, the deformation of almost cylindrical bars has been studied in various papers (see, e.g., Bors (1973), Chirita (1983), Khatiashvili (1983b), Khatiashvili (1983a), and the references therein). First, we present the basic equations of the linear theory of isotropic porous elastic solids and the formulation of the problem. The next section is devoted to the solution of the Saint-Venant problem for almost cylindrical bars. The problem is reduced to the solving of a problem of Almansi type and to the Saint-Venant problem for a right cylinder. In the next section we use the method to solve the problem of extension of a conical frustum. The solution is expressed in terms of solutions of some problems associated with the deformation of a right circular cylinder.

Formulation of the problem

In this paper, we consider the equilibrium theory of elastic materials with voids. A rectangular cartesian coordinate system O x k (k=1,2,3) is used. We shall employ the usual summation and differentiation conventions: Greek subscripts are understood to range over the integers (1,2) whereas Latin subscripts to the range (1,2,3); summation over repeated subscripts is implied and subscripts preceded by a comma denote partial differentiation with respect to the corresponding cartesian coordinate. We denote by B the interior of a noncilindrical bar of length l, with the ends located at x 3=0 and x 3=l, and with the lateral surface Π defined by
$$ F\left[x_{1}\left(1-{kx}_{3}\right),x_{2}\left(1-{kx}_{3}\right)\right] = 0,\;\; x_{3}\in\, [\!0,l]. $$
(1)

Here, k is a constant small enough for squares and higher powers to be neglected. The bar B is called almost cylindrical. We assume that B is a bounded regular region (Gurtin (1972), Section 5). We denote by Σ 1 the cross section located at x 3=0 and by Σ 2 the cross section located at x 3=l. We call B the boundary of B and denote by n i the components of the outward unit normal of B.

The linear strain measure is defined by
$$ e_{ij} = \frac{1}{2}\left(u_{i,j} + u_{j,i}\right), $$
(2)
where u i is the components of the displacement vector. Let t ij be the stress tensor and let h i be the equilibrated stress vector. The surface force and the equilibrated surface force at a regular point of B are given by
$$ t_{i} = t_{ji}n_{j},\; h=h_{j}n_{j}, $$
(3)
respectively. Throughout this paper, we assume that the body is homogeneous and isotropic. The constitutive equations are given by Cowin and Nunziato (1983)
$$\begin{array}{*{20}l} & t_{ij} = \lambda e_{rr}\delta_{ij}+2\mu e_{ij}+b \varphi \delta_{ij},\\ &h_{j} = \alpha \varphi_{j}, \;\; g = -b e_{rr} - \xi \varphi, \end{array} $$
(4)

where δ ij is the Kronecker delta, φ is the volume distribution function, g is intrinsic equilibrated body force, and λ,μ,b,α, and ξ are constitutive constants.

The equilibrium equations of porous bodies, in the absence of the body loads, are given by
$$ t_{ji,j} = 0,\;\; h_{j,j} +g = 0. $$
(5)
We assume that the lateral surface is free from tractions. Thus, we have the following conditions
$$ t_{ki}n_{k} = 0,\;\; h_{j}n_{j} = 0\;\; \text{on}\; \Pi. $$
(6)
Let R=(R 1,R 2,R 3) and M=(M 1,M 2,M 3) be prescribed vectors representing the resultant force and resultant moment about O of the tractions acting on Σ 1. On Σ 2, there are tractions applied so as to satisfy the equilibrium conditions of the body. On the end located at x 3=0, we have the conditions
$$ \int_{\Sigma_{1}} t_{3j}da = -R_{j},\;\; \int_{\Sigma_{1}} \varepsilon_{ijk}x_{j}t_{3k}da = -M_{i}, $$
(7)
where ε ijk is the alternating symbol. We note that there is no contribution of the equilibrated surface force in the resultant force and resultant moment (Ciarletta and Iesan 1993). The elastic potential corresponding to the considered continuum is
$$ 2W = \lambda e_{rr}e_{jj} + 2\mu e_{ij}e_{ij} + b \varphi +\xi \varphi^{2} +\alpha \varphi_{,j}\varphi_{,j}. $$
(8)
Throughout this paper, we assume that W is a positive definite quadratic form in the variables e ij ,φ, and φ ,j . Then, following Cowin and Nunziato (1983)
$$ \mu>0,\;\; \alpha>0,\;\; \xi>0,\;\; 3\lambda+2\mu >0, \; (3\lambda+2\mu)\xi >3b^{2}. $$
(9)

Methods

In this section, we prove that the problem can be reduced to the study of the deformation of a right cylinder. Let us consider the transformation
$$ y_{1} = x_{1}\left(1-{kx}_{3}\right), \; y_{2} = x_{2}\left(1-{kx}_{3}\right),\;\; y_{3} = x_{3}. $$
(10)
By Eq. (10), the domain B is mapped into the right cylinder D, of length l, with the lateral surface S given by
$$ F\left(y_{1}, y_{2}\right) = 0,\;\; y_{3} \in \,[\!0,l]. $$
(11)
From Eq. (10), we obtain
$$ x_{1} = y_{1}\left(1+{ky}_{3}\right),\;\; x_{2} = y_{2}\left(1+{ky}_{3}\right),\; x_{3} = y_{3}. $$
(12)
If f is a function of point of class C 1, then
$$\begin{array}{*{20}l} &\frac{\partial f}{\partial x_{1}} = \left(1-{ky}_{3}\right)\frac{\partial f}{\partial y_{1}},\;\; \frac{\partial f}{\partial x_{2}} = \left(1-{ky}_{3}\right)\frac{\partial f}{\partial y_{2}},\ \\& \frac{\partial f}{\partial x_{3}} = \frac{\partial f}{\partial y_{3}} - k\left(y_{1}\frac{\partial f}{\partial y_{1}} + y_{2}\frac{\partial f}{\partial y_{2}}\right). \end{array} $$
(13)
Let us introduce the notation
$$f_{;j} = \frac{\partial f}{\partial y_{j}}. $$
We seek the solution of the Saint-Venant problem in the form
$$ u_{i} = u_{i}^{(1)} + k u_{i}^{(2)}, \;\; \varphi = \varphi^{(1)} + k \varphi^{(2)}, $$
(14)
where \(u_{i}^{(\alpha)}\) and φ (α), (α=1,2), are unknown functions. We denote
$$ e_{ij}^{(\alpha)} = \frac{1}{2}(u_{i;j}^{(\alpha)} + u_{j;i}^{(\alpha)}). $$
(15)
It follows from Eqs. (2), (13), and (15) that
$$\begin{array}{*{20}l} e_{ij}& = e_{ij}^{(1)} + k e_{ij}^{(2)} + k E_{ij},\\ \varphi_{,j} &= \varphi_{;j}^{(1)} + k \varphi_{;j}^{(2)} + k\zeta_{j}, \end{array} $$
(16)
where
$$\begin{array}{*{20}l} E_{11} &= -y_{3}e_{11}^{(1)},\; E_{22} = -y_{3}e_{22}^{(1)},\; E_{33} = -y_{\rho}u_{3;\rho}^{(1)},\\ E_{12} &= E_{21} = -y_{3}e_{12}^{(1)},\; E_{13} = E_{31} = -\frac{1}{2}\left(y_{\rho}u_{1;\rho}^{(1)} + y_{3}u_{3;1}^{(1)}\right),\\ E_{23} &= E_{32} =-\frac{1}{2}\left(y_{\rho}u_{2;\rho}^{(1)} + y_{3}u_{3;2}^{(1)}\right),\\ \zeta_{1} &= -y_{3}\varphi_{;1}^{(1)},\; \zeta_{2} = -y_{3}\varphi_{;2}^{(1)},\; \zeta_{3} = -y_{\rho}\varphi_{;\rho}. \end{array} $$
(17)
In view of Eq. (16), from Eq. (4), we obtain
$$\begin{array}{*{20}l} & t_{ij} = t_{ij}^{(1)} + k t_{ij}^{(2)} + k P_{ij},\\ & h_{j} = h_{j}^{(1)} + k h_{j}^{(2)} + k H_{j},\; g = g^{(1)} +k g^{(2)} + k G, \end{array} $$
(18)
where we have used the notations
$$\begin{array}{*{20}l} t_{ij}^{(\rho)} &= \lambda e_{rr}^{(\rho)}\delta_{ij} + 2\mu e_{ij}^{(\rho)} + b\varphi^{(\rho)} \delta_{ij},\\ h_{i}^{(\rho)} &= \alpha \varphi_{;i}^{(\rho)},\; g^{(\rho)} = -b e_{rr}^{(\rho)} - \xi \varphi^{(\rho)},\;\; (\rho=1,2) \end{array} $$
(19)
and
$$\begin{array}{*{20}l} P_{ij} &= \lambda E_{rr}\delta_{ij} + 2\mu E_{ij},\\H_{\alpha} &= -y_{3}\varphi_{3;\alpha}^{(1)},\; H_{3} = y_{\rho}\varphi_{;\rho}^{(1)},\; G = -b E_{rr}. \end{array} $$
(20)
The equilibrium equations, to a second order approximation, reduce to the following systems
$$\begin{array}{*{20}l}& t_{ji;j}^{(1)} = 0,\ h_{j;j}^{(1)} +g^{(1)}= 0, \end{array} $$
(21)
and
$$\begin{array}{*{20}l} &t_{ji;j}^{(2)} + P_{ji;j} - y_{3}t_{\rho i;\rho}^{(1)} - y_{\rho}t_{3i;\rho}^{(1)} = 0,\\& h_{j;j}^{(2)} + g^{(2)} + H_{j;j}-y_{3}h_{\alpha;\alpha}^{(1)} + G = 0, \end{array} $$
(22)

on D.

If the direction cosines of the exterior normal to the lateral surface S are (N 1,N 2,0), then, we have
$$ n_{1} = N_{1},\;\; n_{2} = N_{2},\;\; n_{3} = -k y_{\rho}N_{\rho}. $$
(23)
It follows from Eqs. (18) and (23) that the boundary conditions (6) are satisfied if we have
$$ t_{\alpha i}^{(1)}N_{\alpha} = 0,\;\; h_{\alpha}^{(1)}N_{\alpha} = 0\;\; \text{on}\; S $$
(24)
and
$$\begin{array}{*{20}l} t_{\alpha i}^{(2)}N_{\alpha} &= -P_{\alpha i}N_{\alpha} + y_{\rho}N_{\rho}t_{i3}^{(1)},\\ h_{\alpha}^{(2)}N_{\alpha} &= -H_{\rho}N_{\rho} +h_{3}^{(1)}y_{\rho}N_{\rho},\;\; \text{on}\; S. \end{array} $$
(25)
We denote by \(\Sigma _{1}^{*}\) the cross section of D located at y 3=0. In view of Eqs. (10) and (18), the conditions (7) reduce to
$$ \int_{\Sigma_{1}^{*}}t_{3i}^{(1)}da = -R_{i},\;\; \int_{\Sigma_{1}^{*}}\varepsilon_{ijk}y_{j}t_{3k}^{(1)}da = -M_{i}, $$
(26)
and
$$ \int_{\Sigma_{1}^{*}}t_{3i}^{(2)}da = 0,\;\; \int_{\Sigma_{2}^{*}}\varepsilon_{ijk}y_{j}t_{3k}^{(2)}da = 0. $$
(27)

We note that the functions \(u_{i}^{(1)}\) and φ (1) satisfy the equations and the boundary conditions in the Saint-Venant problem characterized by Eqs. (15), (19), and (21) on D and the boundary conditions (24) and (26). The functions \(u_{i}^{(1)}\) and φ (1) can be determined by using the method given by Ciarletta and Iesan (1993). To find the functions \(u_{i}^{(2)}\) and φ (2), we have to solve Eqs. (15), (19), and (22) on D and the boundary conditions (25) and (27). In this problem, the body loads and the surface tractions on the lateral surface S are, in general, different from zero. The functions \(u_{i}^{(2)}\) and φ (2) satisfy a problem of Almansi type, in which the body loads and surface forces depend on the functions \(u_{i}^{(1)}\) and φ (1). A general method to solve the Almansi problem has been presented by Ieşan and Scalia (2009).

Results and discussion

In this section, we use the above method in order to study the extension of a conical frustum made of a porous elastic material. We consider that the domain B is the interior of a circular cone frustum, bounded by plane ends perpendicular to the axis of the cone (Fig. 1). We choose the rectangular cartesian coordinate frame such that x 3-axis can be the axis of the cone. We assume that the ends Σ 1 and Σ 2 are circles of radius r 1 and r 2, respectively, and that r 1<r 2. The lateral surface Π is defined by
$$ x_{1}^{2} + x_{2}^{2} = r_{1}^{2}\left(1+r_{1}^{-1}x_{3}\text{tg}\, \theta\right)^{2},\;\; x_{3}\in \,[\!0,l], $$
(28)
Fig. 1

A conical frustum

where θ is the angle between the generator and the axis of the cone.

We denote
$$ k = \frac{1}{r_{1}}\text{tg}\ \theta. $$
(29)
We suppose that k is small enough for squares and high powers to be neglected. Then, Eq. (28) can be expressed as
$$ x_{1}^{2}\left(1-{kx}_{3}\right)^{2} +x_{2}^{2}\left(1-{kx}_{3}\right)^{2} = r_{1}^{2},\quad x_{3}\in\, [\!0,l]. $$
(30)
The surface S is described by
$$ y_{1}^{2}+y_{2}^{2} = r_{1}^{2},\quad y_{3}\in \,[\!0,l]. $$
(31)
Thus, the domain D is a right circular cylinder,
$$D = \left\{\left(y_{1}, y_{2}, y_{3}\right), y_{1}^{2}+y_{2}^{2}<r_{1}^{2}, 0<y_{3}<l\right\}. $$
We assume that R α =0, R 3≠0 and M j =0. In this case, the conditions (26) reduce to
$$ \int_{\Sigma_{1}^{*}}t_{3\alpha}^{(1)}da = 0,\;\; \int_{\Sigma_{1}^{*}}x_{\alpha}t_{33}^{(1)}da = 0,\; \int_{\Sigma_{1}^{*}}\varepsilon_{\alpha\beta3}x_{\alpha}t_{\beta3}^{(1)}da = 0, $$
(32)
and
$$ \int_{\Sigma_{1}^{*}}t_{33}^{(1)}da = -R_{3}. $$
(33)
Let us determine the functions \(u_{i}^{(\alpha)}\) and φ (α), (α=1,2). We seek the functions \(u_{j}^{(1)}\) and φ (1) in the form
$$ u_{\alpha}^{(1)} = a_{3}C_{1}y_{\alpha},\; u_{3}^{(1)} = a_{3}y_{3},\; \varphi^{(1)} = a_{3}C_{2}, $$
(34)
where C α and a 3 are unknown constants. It follows from Eqs. (34), (15), and (19) that
$$\begin{array}{*{20}l} t_{11}^{(1)} &= t_{22}^{(1)} = [\!\lambda\left(1+2C_{1}\right) + 2\mu C_{1} + b C_{2}]a_{3},\;\; t_{12}^{(1)} = 0,\\ t_{33}^{(1)} &=\left(\lambda+2\mu +2\lambda C_{1}+b C_{2}\right)a_{3},\;\; t_{\alpha 3}^{(1)} =0,\\ g^{(1)} &= -\left[\left(1+2C_{1}\right)b +\xi C_{2}\right]a_{3},\;\; h_{j}^{(1)} =0. \end{array} $$
(35)
The boundary conditions (24) are satisfied only if
$$ 2(\lambda+\mu)C_{1} + {bC}_{2} = -\lambda. $$
(36)
The equilibrium Eq. (21) reduces to
$$ 2{bC}_{1} +\xi C_{2} = -b. $$
(37)
It follows from Eqs. (36) and (37) that
$$ C_{1} = \frac{\lambda\xi-b^{2}}{2b^{2} -2\xi (\lambda+\mu)},\;\; C_{2} = \frac{b\mu}{b^{2} - \xi (\lambda+\mu)}. $$
(38)
We note that the conditions (9) imply that b 2ξ(λ+μ) is different from zero. The conditions (32) are identically satisfied. From Eqs. (33) and (35), we find the constant a 3,
$$ a_{3} = \frac{R_{3}\left[b^{2}-\xi(\lambda+\mu)\right]}{A\left[(\lambda+2\mu)(\lambda+\mu)\xi - 3\mu b^{2} - \lambda^{2}\xi\right]}, $$
(39)

where A is the area of the cross section \(\Sigma _{1}^{*}\).

Thus, the functions \(u_{i}^{(1)}\) and φ (1) are determined. Let us study the Almansi problem characterized by Eqs. (15), (19), and (22) on D and the boundary conditions (25) and (27). We note that in the case of extension, we have
$$\begin{array}{*{20}l} t_{\alpha\beta}^{(1)} &= 0,\; t_{33}^{(1)} = \left(\lambda+2\mu +2\lambda C_{1}+b C_{2}\right)a_{3},\\ h_{i}^{(1)} &= 0,\; g^{(1)}=0. \end{array} $$
(40)
From Eqs. (34) and (17), we find that
$$\begin{array}{*{20}l} & E_{11} = E_{22} = -a_{3}C_{1}y_{3},\; E_{\alpha3} = -\frac{1}{2}y_{\alpha}a_{3}C_{1},\\& E_{12} = E_{33} = 0,\; \xi_{i} = 0. \end{array} $$
(41)
In view of Eqs. (20) and (41), we get
$$\begin{array}{*{20}l} P_{11} &= P_{22}= -2(\lambda+\mu)a_{3}C_{1}y_{3},\;\; P_{12} = 0,\\ P_{33} &= -2\lambda C_{1}a_{3}y_{3},\; P_{\alpha} = -\mu C_{1}a_{3}y_{\alpha},\\ H_{j} &= 0,\;\; G=2b C_{1}a_{3}y_{3}. \end{array} $$
(42)
It is a simple matter to see that Eq. (22) reduces to
$$\begin{array}{*{20}l} & t_{j\alpha;j}^{(2)} = 0,\;\; t_{j3;j}^{(2)} -P = 0, \end{array} $$
(43)
$$\begin{array}{*{20}l} &h_{j;j}^{(2)} + g^{(2)} +G = 0, \end{array} $$
(44)
on D, where
$$ P = 2(\lambda+\mu)C_{1}a_{3}. $$
(45)
The direction cosines of the exterior normal to the surface (31) are given by \(N_{\alpha } = y_{\alpha }r_{1}^{-1},\ N_{3} = 0.\) In view of Eqs. (40) and (42), the boundary conditions (25) become
$$\begin{array}{*{20}l} & t_{\rho\alpha}^{(2)}N_{\rho} = P r_{1}^{-1} y_{\alpha},\; t_{\rho 3}^{(2)}N_{\rho} = P_{0},\\& h_{\rho}^{(2)}N_{\rho} = 0\;\; \text{for}\;\; r=r_{1}, \end{array} $$
(46)
where
$$ P_{0} =\ \left[\lambda+2\mu +\left(2\lambda+\mu\right)C_{1}+{bC}_{2}\right]r_{1}a_{3}. $$
(47)
Let us determine the functions \(u_{i}^{(2)}\) and φ (2) which satisfy Eqs. (15), (19), (43), and (44) on D, and the boundary conditions (46) and (27). First, we determine the functions \(u_{i}^{*}\) and φ which satisfy Eqs. (15), (19), (43), and (44) on D and the boundary conditions on the lateral surface (46). We seek these functions in the form
$$\begin{array}{*{20}l} & u_{1}^{*}=A_{1}y_{1} + B_{1}y_{1}y_{3},\;\; u_{2}^{*} = A_{1}y_{2} + B_{1}y_{2}y_{3},\\& u_{3}^{*} = \frac{1}{2}B_{3}y_{3}^{2} + \Psi (y_{1},y_{2}),\; \varphi^{*} = A_{2} +B_{2}y_{3}, \end{array} $$
(48)
on D, where A α and B k are unknown constants, and Ψ is an unknown function. The strain tensor \(e_{ij}^{*}\), corresponding to the displacements \(u_{j}^{*}\), is
$$\begin{array}{*{20}l} e_{11}^{*} &= e_{22}^{*} = A_{1}+B_{1}y_{3},\; e_{33} = B_{3}y_{3},\\ e_{12} &= 0,\; 2e_{\alpha3} = \Psi_{;\alpha}+B_{1}y_{\alpha}. \end{array} $$
The stress tensor \(t_{ij}^{*}\), the equilibrated stress vector \(h_{j}^{*}\), and the intrinsic equilibrated body force g associated with the deformation described by Eq. (48) are given by
$$\begin{array}{*{20}l} t_{11}^{*}\! &=\! t_{22}^{*} \,=\, 2(\lambda\,+\,\mu)A_{1} \,+\, {bA}_{2} +[2(\lambda\,+\,\mu)B_{1} \,+\, {bB}_{2} +\lambda B_{3}]y_{3},\\ t_{33}^{*} &= 2\lambda A_{1} + {bA}_{2} +[2\lambda B_{1} + {bB}_{2} + (\lambda+2\mu)B_{3}]y_{3},\\ t_{12}^{*} &= 0,\; t_{\alpha3}^{*} = \mu\left(\Psi_{;\alpha} + B_{1}y_{\alpha}\right),\\ h_{\alpha}^{*}\! &=\! 0,\, h_{3}^{*}\! =\! B_{2},\; g^{*}\! =\! -2b A_{1}\,-\,\xi A_{2}\! -\!\left(2{bB}_{1} \,+\, \xi B_{2}\! +\!{bB}_{3}\right)y_{3}. \end{array} $$
(49)
Equation (44) is satisfied if
$$ 2{bA}_{1} + \xi A_{2} = 0 $$
(50)
and
$$ 2{bB}_{1} + \xi B_{2} + b B_{3} = 2b C_{1}a_{3}. $$
(51)
The first two conditions of Eq. (46) reduce to
$$ 2(\lambda+\mu)A_{1} + {bA}_{2} = P, $$
(52)
and
$$ 2(\lambda+\mu)B_{1} + b B_{2} + \lambda B_{3} = 0. $$
(53)
It follows from Eqs. (50) and (52) that
$$ A_{1} = \frac{P\xi}{2\left[\xi(\lambda+\mu)-b^{2}\right]},\;\; A_{2} = -\frac{bP}{\xi(\lambda+\mu)-b^{2}}. $$
(54)
In view of Eq. (49), the equations of equilibrium (43) reduce to
$$ \mu \Delta \Psi = P-2(\lambda+\mu)B_{1} - b B_{2} - (\lambda+2\mu)B_{3}, $$
(55)
on \(\Sigma _{1}^{*}\), where Δ U=U ;α α . The third boundary condition of Eq. (46) becomes
$$ \mu \Psi_{;\alpha}N_{\alpha} = P_{0} - B_{1}\mu y_{\rho}N_{\rho},\;\; \text{for}\; r = r_{1}. $$
(56)
The necessary and sufficient condition to solve the Neumann problem, (55) and (56), is
$$ 2\lambda B_{1} +b B_{2} + (\lambda+2\mu)B_{3} = \Lambda, $$
(57)
where
$$ \Lambda = P-2r_{1}^{-1}P_{0} = 2a_{3}\left(\lambda+2\mu+\lambda C_{1}+{bC}_{2}\right). $$
(58)
The constants B k are determined by the system (50), (53), and (57). We obtain
$$\begin{array}{*{20}l} B_{1}& = \frac{1}{\mathcal{D}}\left[(b^{2}-\xi\lambda)\Lambda - 4\mu b^{2}a_{3}C_{1}\right],\\ B_{2}& = \frac{2}{\mathcal{D}}\left[2{ba}_{3}(\lambda+\mu)(\lambda+2\mu)C_{1} - \mu b \Lambda - 2\lambda^{2}{ba}_{3}C_{1}\right],\\ B_{3}& = \frac{2}{\mathcal{D}}\left\{\Lambda\left[\xi(\lambda+\mu)-b^{2}\right] - 2\mu b^{2}a_{3}C_{1}\right\}, \end{array} $$
(59)
where
$$ \mathcal{D} = 2\mu\left[(3\lambda+2\mu)\xi - 3b^{2}\right]. $$
(60)

We note that the relations (9) imply that \(\mathcal {D}\neq \) 0. The last condition from Eq. (46) is identically satisfied.

The solution of the boundary value problem (55) and (56) is given by
$$ \Psi = \frac{1}{4\mu}\left[P-2(\lambda+\mu)B_{1} -b B_{2} - (\lambda+2\mu)B_{3}\right]r^{2}, $$
(61)

where r 2=y α y α .

We seek the functions \(u_{i}^{(2)}\) and φ (2) in the form
$$ u_{i}^{(2)} = u_{i}^{*} +v_{i},\;\; \varphi^{(2)} = \varphi^{*} +\psi. $$
(62)
It is a simple matter to see that the functions v i and ψ must satisfy the equilibrium equations in the absence of the body loads and the conditions on the lateral surface S, in the absence of superficial forces. Let s ij be the stress tensor corresponding to the displacements v i and volume distribution function ψ. It follows from Eqs. (49), (61), and (62) that the conditions (27) reduce to
$$ \int_{\Sigma^{*}}s_{3\alpha}da = 0,\;\; \int_{\Sigma_{1}^{*}}s_{33} da = - R_{3}^{*},\; \int_{\Sigma_{1}^{*}}\varepsilon_{ijk}y_{j}s_{3k}da = 0 $$
(63)
where
$$ R_{3}^{*} = (2\lambda A_{1} + b A_{2})A. $$
(64)
We conclude that v i and ψ satisfy a problem of extension with the axial force \(R_{3}^{*}\). These functions are given by
$$ v_{\alpha} = d_{3}C_{1}y_{\alpha},\; v_{3} = d_{3}y_{3},\; \psi = d_{3}C_{2}, $$
(65)
where C α is defined by Eq. (38) and
$$ d_{3} = \frac{R_{3}^{*}\left[b^{2} - \xi (\lambda+\mu)\right]}{A\left[(\lambda+2\mu)(\lambda+\mu)\xi - 3\mu b^{2} - \lambda^{2}\xi\right]}. $$
(66)
It follows from Eqs. (10), (14), (34), (62), and (65) that the solution of the problem of extension of the porous conical frustum is given by
$$\begin{array}{*{20}l} u_{\alpha} &= x_{\alpha}\left[a_{3}C_{1}+k \left(A_{1}+d_{3}C_{1}\right)+ k(B_{1}-a_{3}C_{1})x_{3}\right],\\ u_{3} &= \left(a_{3} + {kd}_{3}\right)x_{3} + \frac{1}{2}k B_{3}x_{3}^{2} + k Q x_{\alpha}x_{\alpha},\\ \varphi &= \left(a_{3} + k d_{3}\right)C_{2}+ k\left(A_{2} + B_{2}x_{3}\right),\;\; \left(x_{1},x_{2},x_{3}\right)\in B, \end{array} $$
(67)
where we have used the notation
$$Q=\frac{1}{2\mu}\left[(\lambda+\mu)C_{1}a_{3} -\mu B_{3}\right]. $$
The cartesian coordinate frame consists of the orthonormal basis (e 1,e 2,e 3) and the origin O. Let R 3=−F, where F is a positive constant. In this case, the resultant force of the tractions acting on the end located at x 3=l is F e 3 and the point O is fixed. Let x i be the coordinates of the point X 0 in the reference configuration, and let ξ i be the coordinates of the corresponding point X in the deformed configuration. Then, we have ξ i =x i +u i , and from Eq. (67), we obtain
$$\begin{array}{*{20}l} \xi_{\alpha} &= \left(1+T_{1}+T_{2}x_{3}\right)x_{\alpha},\\ \xi_{3} &= \left(1+\Omega_{1}\right)x_{3} + \Omega_{2}x_{3}^{2}+kQ x_{\alpha}x_{\alpha}, \end{array} $$
(68)
where
$$\begin{array}{*{20}l} T_{1} &= a_{3}C_{1} + k\left(d_{3}C_{1} +A_{1}\right),\;\; T_{2} = k\left(B_{1}-a_{3}C_{1}\right),\\ \Omega_{1} &= a_{3}+{kd}_{3}, \;\; \Omega_{2} = \frac{1}{2}{kB}_{3}. \end{array} $$
(69)
The point Y 0 which, prior to deformation, had the coordinates (0,0,l) goes into point Y from the deformed configuration. From Eq. (68), we find that the point Y has the coordinates (0,0,l ) where
$$ l^{*} = \left(1+\Omega_{1} + \Omega_{2}l\right)l. $$
(70)
We note that in the classical elasticity, the value of l for a homogeneous right cylinder is \((1+F/E \mathcal {A})\) l, where E is Young’s modulus and \(\mathcal {A}\) is the area of the cross section. Let us choose magnesium crystal as the hypothetical material for which the values of the constitutive coefficients are (Bachher, 2015)
$$\begin{array}{*{20}l} \lambda &= 2.17 \times 10^{10}N\cdot m^{-1}, \mu=3.278\times 10^{10}N\cdot m^{-1},\\ \alpha &= 3.688\times 10^{-5}N,\;\; b = 1.13849\times 10^{10}N\cdot m^{2},\\ \xi &= 1.475\times 10^{10}N\cdot m^{-2}. \end{array} $$
If R=−e 3 and k=5.10−1, then from Eq. (70), we get
$$l^{*} = (1.01063 + 0.00011l)10^{-6}l. $$
Let us consider a circle \((\mathcal {C})\) of radius ρ, located at the plane x 3=η, where ρ and η are given constants. It follows from Eq. (68) that the image of \((\mathcal {C})\) in the deformed configuration is the circle
$$ \xi_{1}^{2} + \xi_{2}^{2} = \rho^{2}\left(1+T_{1}+T_{2}c\right)^{2}, $$
(71)
located at the plane ξ 3=(1+Ω 1)η+Ω 2 η 2+k Q ρ 2. The relation (71) can be used to describe the deformation of the surface S. Let us assume that r 1=10 mm. We consider the circle of radius r 1 located at the plane x 3=c. Let (Γ) be the image of this circle in the deformed configuration. We denote by R(c) the radius of (Γ). The variation of R(c) with respect to variable c is presented in Fig. 2. The material parameters used are the same as in the previous example.
Fig. 2

Variation of R(c)

We note that in the case of the problem of extension of a right cylinder, the displacements and the volume fraction field depend on the coordinates x i at most linearly.

Conclusions

The results established in this paper can be summarized as follows:
  1. a)

    We have studied the Saint-Venant problem for an almost cylindrical bar made of a porous elastic material. The problem was reduced to the solving of a problem of Saint-Venant type for a right cylinder D and to the problem of Almansi for D.

     
  2. b)

    We have used the above method to investigate the behavior of a conical frustum subjected to extension. In this case, the problem reduces to a problem of Almansi type for a right circular cylindrical. The displacement vector field and the volume fraction field have been determined. The displacement vector is a polynomial of two degree in the cartesian coordinate. The volume fraction field depends linearly on the axial coordinate.

     
  3. c)

    The salient feature of the solution of the problem of extension is that the displacement vector field and the stresses contain new terms characterizing the influence of the material porosity, and their values are therefore modified from the values predicted by the classical elasticity.

     

Declarations

Competing interests

The author declares that he has no competing interests.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License(http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Al. I. Cuza University of Iaşi
(2)
O. Mayer Institute of Mathematics, Romanian Academy

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