1 Introduction

To understand the diffusion process of interstitial atoms (later interstitials) near a defect, it is necessary

  • to know the stress field of the defect,

  • to know the stress field of the interstitials, which are placed at several distinct sites (e.g., the so-called octahedral or tetrahedral positions in a cubic lattice) and produce according eigenstrain fields. Here we refer to the recent paper [1].

If we consider, e.g., an edge or screw dislocation, interstitials are densely assembled along the dislocation line, forming something like a cylindrical “cloud,” described by a cylinder with the dislocation line as z axis. As already outlined in the paper [2] (to which this paper is an “addendum”), the cylindrical “cloud” is represented as cylindrical inclusion with the radius R and an eigenstrain state with the components \(\varepsilon _{xx}^\mathrm{{eig}} ,\varepsilon _{yy}^\mathrm{{eig}} ,\varepsilon _{zz}^\mathrm{{eig}} ,\varepsilon _{xy}^\mathrm{{eig}} \). (The coordinates of the cross section are x, y and z is the coordinate in the axial direction.) The paper [2] provides the reader with the stress state inside and outside of the inclusion in analytical form. However, it has become recently evident that also eigenstrain terms \(\varepsilon _{xz}^\mathrm{{eig}} \) and \(\varepsilon _{yz}^\mathrm{{eig}} \) may occur, which are not treated in [2]. The goal of this addendum is to provide the reader also with the according stress state inside and outside of the cylindrical inclusion.

2 Solution concept

The notation is the same as in [2]. According to Eshelby’s seminal work on inclusions, see the references in [2]; the eigenstress state is spatially constant in the cylindrical inclusion. The inside eigenstress state can immediately be calculated in Cartesian coordinates (\(x,y,r=\sqrt{x^{2}+y^{2}}\), z axis of rotation), according to [3, 4], aspect ratio \(\alpha \rightarrow \infty \), Eqs. (4.1–2) and \((\hbox {A}4)_{4}\) there, with the shear modulus \(G=E/2\left( {1+\nu } \right) \), E Young’s modulus, \(\nu \) Poisson’s ratio, as

$$\begin{aligned} 0\le r\le R: \quad \sigma _{xz} =-G\varepsilon _{xz}^\mathrm{{eig}} , \quad \sigma _{yz} =-G\varepsilon _{yz}^\mathrm{{eig}} . \end{aligned}$$
(1)

As boundary condition at the interface (i.e., the radial position \(r=R)\), we use the traction vector \(\mathbf{t}\), which follows with the inside stress tensor \(\varvec{\upsigma }\) (with the only nonzero elements from Eq. (1)) and the normal vector n with the components \(\left( {x/R,y/R,0} \right) \) as \(\varvec{\upsigma }{\cdot } \mathbf{n}{=}{} \mathbf{t}\). The only nonzero component of t is

$$\begin{aligned} r{=}R, \quad t_z {=}\sigma _{xz} x/R+\sigma _{yz} y/R{=}-G\left( {\varepsilon _{xz}^\mathrm{{eig}} x/R+\varepsilon _{yz}^\mathrm{{eig}} y/R} \right) . \end{aligned}$$
(2)

To solve the problem for the stress state outside the inclusion, where no eigenstrain is active, cylindrical coordinates, r and \(\vartheta \), are used. The only displacement is that in z direction, \(w\left( {r,\vartheta } \right) \), which is independent of the z coordinate. As stress components only \(\sigma _{rz} \) and \(\sigma _{\vartheta z} \) exist. Theory of elasticity, see, e.g., [5], Sect. 4.9.1, teaches the equilibrium equation as

$$\begin{aligned} \frac{\partial \sigma _{rz} }{\partial _r }+\frac{1}{r}\sigma _{rz} +\frac{1}{r}\frac{\partial \sigma _{\vartheta z} }{\partial \vartheta }=0, \end{aligned}$$
(3)

and [5], Sect. 5.5.3, Hooke’s law as

$$\begin{aligned} \sigma _{rz} =G\frac{\partial w}{\partial r}, \quad \sigma _{\vartheta z} =\frac{G}{r}\frac{\partial w}{\partial \vartheta }. \end{aligned}$$
(4)

Combining Eqs. (3) and (4) one finds with \(w=Cf\left( r \right) g\left( \vartheta \right) \), C is an integration constant, \(f\left( r \right) \) has to be derived, \(g\left( \vartheta \right) =\cos \vartheta \) or \(g\left( \vartheta \right) =\sin \vartheta \), \({g}''\left( \vartheta \right) =-g\left( \vartheta \right) \), the following differential equation

$$\begin{aligned} {f}''+\frac{1}{r}{f}'-\frac{1}{r^{2}}f=0 \end{aligned}$$
(5)

with the solution

$$\begin{aligned} r\ge R: \quad w=\frac{C}{r}g\left( \vartheta \right) . \end{aligned}$$
(6)

Remark: We denote now first derivatives by \(^\prime \) and second derivatives by\(''\).

Note that two eigenstrain cases are superposed, see Eq. (1). Therefore, two superposed displacements exist as \(w_x \) (subscript x according to \(\varepsilon _{xz}^\mathrm{{eig}} )\) with \(C=C_x \), \(g\left( \vartheta \right) =g_x \left( \vartheta \right) \) and \(w_y \) (subscript y according to \(\varepsilon _{yz}^\mathrm{{eig}} )\) with \(C_y ,g_y \left( \vartheta \right) \). The total outside component \(\sigma _{rz} \), Eq. (4)\(_{1}\), reads with Eq. (6) for both cases as

$$\begin{aligned} r\ge R: \quad \sigma _{rz} =-\frac{G}{r^{2}}\left( {C_x g_x \left( \vartheta \right) +C_y g_y \left( \vartheta \right) } \right) . \end{aligned}$$
(7)

Since a jump of the traction vector t must not exist at the interface \(r=R\), \(t_z =\sigma _{rz} \), comparison between Eqs. (2) with \(\cos \vartheta =x/r\), \(\sin \vartheta =y/r\) and (7) yields

$$\begin{aligned} r\ge R: \quad w_x =\varepsilon _{xz}^\mathrm{{eig}} \frac{R^{2}}{r}\cos \vartheta , \quad w_y =\varepsilon _{yz}^\mathrm{{eig}} \frac{R^{2}}{r}\sin \vartheta . \end{aligned}$$
(8)

The stress state follows with Eq. (4) as

$$\begin{aligned} r\ge R: \quad \sigma _{rz} =-G\frac{R^{2}}{r^{2}}\left( {\varepsilon _{xz}^\mathrm{{eig}} \cos \vartheta +\varepsilon _{yz}^\mathrm{{eig}} \sin \vartheta } \right) ,\quad \sigma _{\vartheta z} =-G\frac{R^{2}}{r^{2}}\left( {\varepsilon _{xz}^\mathrm{{eig}} \sin \vartheta -\varepsilon _{yz}^\mathrm{{eig}} \cos \vartheta } \right) . \end{aligned}$$
(9)

It may be of interest to mention that a decay of the stress state proportional to \({R^{2}}/{r^{2}}\) in outside can also be seen in [6], Sect. 4.1.

A final note: If a plane strain configuration (it is, e.g., a disk of infinite extension in the x, y plane with no constraints in the x, y direction and a finite thickness h) is assumed, with the z axis being the axis of the cylindrical inclusion, \(0\le z\le h\), then the displacement field \(w=w_x +w_y \),

$$\begin{aligned} 0\le x\le R: \quad w=\left( {\varepsilon _{xz}^\mathrm{{eig}} \cos \vartheta +\varepsilon _{yz}^\mathrm{{eig}} \sin \vartheta } \right) r, \quad r\ge R \hbox { see Eq}.~(9) \end{aligned}$$
(10)

must be suppressed at \(z=h\). However, the average value \(\tilde{w}\) of \(w\left( {r,\vartheta } \right) \) is zero, due to the integration with respect to \(\vartheta \). Therefore, according to St. Venant’s principle, only a local stress state, i.e., \(\sigma _{zz} \) and corresponding values of \(\sigma _{xx} ,\sigma _{yy} \), will develop due to \(w\equiv 0\) at \(z=h\), which, however, will disappear in certain distance from the surface at \(z=h\). Therefore, we omit, for the sake of simplicity, any estimation of the stress component \(\sigma _{zz}\).