Problem of Stress Concentration in Elastostatics of Bodies with Microstructure

Synonyms

Distribution of stress around holes; Maximum stress near discontinuity

Overview

The microcontinuum field theories are extensions of the classical field theory and are intended to study the mechanical behavior of bodies which posses an inner structure. The effects of the inner structure are neglected by the classical theory of elasticity. Thus it is inadequate to describe physical phenomena for which the response of the microstructure becomes relevant.

There are many substances that can be modeled by the microcontinuum theories. For example, polymers with flexible molecules, liquid crystals, suspensions with deformable elements, porous media lattices with base, composites reinforced with chopped elastic fibers, bones. The motion of a microcontinuum is described by the usual displacement field and the deformation of the constituents (microelements). Geometrically each microelement is represented by a point and three deformable vectors attached to it, called directors. We have different theories on the basis of restrictions we impose on the deformation of the directors. A microstretch continuum is characterized by a uniform dilatation and a rigid motion of the directors, whereas a rigid motion without dilatation defines a micropolar continuum. We call micromorphic a continuum with no restrictions on the deformation of the directors.

The literature on nonclassical theories is very extensive. In 1960 Truesdell and Toupin [1] reformulated and developed in modern notation the couple stress theory of the brothers E. and F. Cosserat [2]. Following this work an intense activity began and several hundred papers were published (see e.g., [3–6]). An account of the historical developments as well as references may be found in the monograph [1] and in the book of Kunin [7]. A linear theory of micropolar elasticity was introduced by Eringen [8], who later presented the theory of microstretch elastic solids [9, 10]. A comprehensive and unified treatment of this subject, together with extensive references, can be found in the book by Eringen [11].

In the classical theory of elasticity the problem of stress concentration has been the basic of many studies. The results provide the insight into the possibility of dangerous stress near holes and other irregularities of a body, whereas the nominal stresses are at otherwise safe levels. This problem is of crucial importance in the design of structural members and machine elements.

An early study is due to Kirsh [12], who derived an approximate solution for stress distribution around a circular hole in a rectangular plate under uniform tensile stress. He showed that the maximum tensile stress is locate in two opposite points of the boundary of the hole, and it is three times the uniform stress applied at the ends of the plate. An analogous solution for stress around an elliptical hole was derived by G. V. Kolosov in 1909 [13]. In 1919 N. I. Muskhelishvili [14] pointed out that the Kolosov’s result was a very particular case of the general solution of the first fundamental problem for the infinite plane with an elliptic hole. A variety of cases of plates with circular and elliptical holes are also investigated by Muskhelishvili with the use of complex functions [15]. The method developed by Kolosov and Muskhelishvili made it possible to consider broad classes of problems and constituted a major contribution to stress concentration studies. G. N. Savin [16] showed that for plates weakened by holes commonly occurring in engineering practice, for example square, triangular, rectangular holes, the problem could be handled by a series expansion of the integrand in the Christoffel-Schwartz formula. In his monograph, Savin established various results concerning the stress concentration in an anisotropic medium with an elliptical hole. The cases of noncircular and nonelliptical holes in the anisotropic plates have been solved by the method of perturbation (see, e.g., [17]). More recent studies address the problem of stress states of multiply connected domains. Exact solution has been obtained in the case of an isotropic plate with two circular holes with the use of bipolar coordinates. The special circumstances which arise as the holes draw closer together are also investigated.

The problem of stress distribution around holes has been extended to the linear theory of elastic solids with microstructure. In 1967 Ariman [18] and Kaloni and Ariman [19] established the solution of an isotropic micropolar plate with a circular hole. The stresses and the couple stresses are controlled by Poisson ratio and two new parameters which depend on the material constants of the micropolar medium. It is remarkable that, in a micropolar solid, the stress concentration factor is smaller than that of the classical elastic solid. The case of circular hole has also been investigated by Dyszlewicz [20]. Analogous problem for elliptical hole was solved by Kim and Eringen [21], who calculated the stress concentration factor K and plotted it for different values of parameters dependent on elastic moduli. The problem of stress concentration in the neighborhood of a circular hole in the context of the equilibrium theory of microstretch elastic bodies has been studied in [22]. The behavior of an infinity micropolar elastic body with a spherical cavity has been solved in [23].

Equilibrium Theory

The basic equations of the equilibrium theory of homogeneous and isotropic microstretch elastic materials in the absence of body loads are the equilibrium equations

$$\sigma_{ji,j }=0\quad {m_{ji,j }}+{\epsilon_{irs }}{\sigma_{rs }}=0\quad {\lambda_{i,i }}-s=0$$

(1)

the constitutive equations

$$\begin{array}{ll} {\sigma_{ij }}= \ \lambda {e_{rr }}{\delta_{ij}}+\left( {\mu +\kappa } \right){e_{ij }}+\mu {e_{ji }}+\eta \psi{\delta_{ij }} \\ {m_{ij }}= \ \alpha {\chi_{rr }}{\delta_{ij}}+\beta {\chi_{ji }}+\gamma {\chi_{ij }} \\{\lambda_k}= \ \sigma {\psi_{,k }}\ 3s=\eta {e_{rr }}+b\psi\,\\\end{array}$$

(2)

and the geometrical equations

$$e_{ij }={u_{j,i }}+{\epsilon_{jik }}{\varphi_k}\,\,\,\,\,{\chi_{ij }}={\varphi_{j,i }}$$

(3)

where m _ij is the couple stress tensor, $\lambda_i$ is the microstress vector; s is the scalar microstress function; u, $\varphi$, and $\psi$ are the displacement vector, the microrotation vector, and the microstretch function, respectively; $\lambda$, $\mu$, $\kappa$, $\eta$, $\alpha$, $\beta$, $\gamma$, $\sigma$ and b are constitutive coefficients. We assume that the internal energy is positive definite. This assumption is equivalent with the conditions

$$\begin{array} {ll} {b(3\lambda +2\mu +\kappa )-3{\eta^2}> 0\quad 2\mu+\kappa }\,> 0 \\ \qquad {\kappa\,> 0\,\,\,b> 0\,\quad\,\,3\alpha +\beta+\gamma }\,>0 \\ \quad \,\qquad {\gamma +\beta\,>0\,\,\,\quad\,\,\gamma -\beta }\,>0 \end{array}$$

(4)

We denote by B the regular region of three-dimensional space occupied by the body, $\partial B$ the boundary of B, and n the outward unit normal of $\partial B$. Let $\tilde{\bf t}$, $\tilde{\bf m}$, and $\tilde{\lambda}$ be the surface traction, the surface moment, and the microtraction, respectively. The boundary conditions are given by

$$\mathop{\tilde{t}}\nolimits_i={\sigma_{ji }}{n_j}\,\,\,\,\,\mathop{\tilde{m}}\nolimits_i={m_{ji }}{n_j}\,\,\,\,\,\tilde{\lambda}={\lambda_i}{n_i}$$

(5)

Plane Strain

The state of plane strain is characterized by the following restrictions

$$\begin{array}{ll} {u_{\alpha }}= \ {u_{\alpha }}({x_1},{x_2})\qquad{u_3}=0 \\ {\varphi_{\alpha }}= \ 0\,\,\,\,\,{\varphi_3}=\varphi ({x_1},{x_2})\,\,\,\,\,\,\,\psi =\psi ({x_1},{x_2}) \\ \end{array}$$

(6)

It follows from (6) that the strain measures are given by

$$\begin{array}{ll} {e_{{\alpha \beta }}}= \ {u_{{\beta, \alpha }}}+{\epsilon_{{\beta \alpha 3}}}\varphi\,\,\,\,\,{e_{{\alpha 3}}}={e_{{3\alpha }}}={e_{33 }}=0 \\ {\chi_{{\alpha 3}}}= \ {\varphi_{{,\alpha }}}\,\,\,\,\,{\chi_{{\alpha \beta }}}={\chi_{3i }}=0 \end{array}$$

(7)

From the constitutive equations (2), we obtain

$$\begin{array}{ll} {\sigma_{{\alpha \beta }}}&=\lambda {e_{{\rho \rho }}}{\delta_{{\alpha \beta }}}+(\mu +\kappa ){e_{{\alpha \beta }}}+\mu {e_{{\beta \alpha }}}+\eta \psi {\delta_{{\alpha \beta }}} \\ {\sigma_{33 }}&=(\lambda +2\mu +\kappa ){e_{{\rho \rho }}}+\eta \psi \quad{\sigma_{{\alpha 3}}}={\sigma_{{3\alpha }}}=0 \\ {m_{{\alpha 3}}}&=\gamma {\chi_{{\alpha 3}}}\,\quad\,{m_{{3\alpha }}}=\beta {\chi_{{\alpha 3}}}\,\quad {m_{{\alpha \beta }}}={m_{33 }}=0 \\ {\lambda_{\alpha }}&=\sigma \psi {,_{\alpha }}\,\,\,\,{\lambda_3}=0\,\quad 3s=\eta {e_{{\rho \rho }}}+b\psi \end{array}$$

(8)

The equilibrium equations (1) reduce to

$$\sigma_{{\beta \alpha, \beta }}=0\,\,{m_{{\beta 3,\beta }}}+{\epsilon_{{3\alpha \gamma }}}{\sigma_{{\alpha \gamma }}}=0\quad {\lambda_{{\alpha, \alpha }}}-s=0$$

(9)

The functions $\tilde{\bf t}$, $\tilde {\bf m}$, and $\tilde{\lambda}$ must be independent of x ₃ and $\mathop{\tilde{t}}\nolimits_3=0$, $\mathop{\tilde{m}}\nolimits_{\alpha }=0$. The boundary conditions (5) became

$$\sigma_{{\beta \alpha }}{n_{\beta }}=\mathop{\tilde{t}}\nolimits_{\alpha}\,\,\,\,{m_{{\alpha 3}}}{n_{\alpha }}=\mathop{\tilde{m}}\nolimits_3\,\,\,\,{\lambda_{\alpha }}{n_{\alpha }}=\tilde{\lambda}$$

(10)

In what follows we are interested in a plane strain problem with the displacement vector, the microrotation vector and the microstretch function being specified in cylindrical coordinates $(r,\theta, z)$ as follows:

$$\begin{array}{ll} {u_r}= \ u(r,\theta )\,\,\,\,{u_{\theta }}=v(r,\theta )\,\,\,{u_z}=0 \\ {\varphi_r}= \ 0\,\,\,\,{\varphi_{\theta }}=0\,\,\,\,\,\,{\varphi_z}=\varphi (r,\theta ) \\ \psi = \ \psi (r,\theta ) \end{array}$$

(11)

The geometrical equations (7) became

$$\begin{array}{ll} {e_{rr }}= \ \frac{{\partial u}}{{\partial r}}\,\,\,\,\,\,\,{e_{{\theta \theta }}}={r^{-1 }}\left( {\frac{{\partial v}}{{\partial \theta }}+u} \right) \\ {e_{{r\theta }}}= \ \frac{{\partial v}}{{\partial r}}-\varphi\,\,\,\,\,{e_{{\theta r}}}={r^{-1 }}\left( {\frac{{\partial u}}{{\partial \theta }}-v} \right)+\varphi \\ {e_{rz }}= \ {e_{{\theta z}}}={e_{zr }}={e_{{z\theta }}}={e_{zz }}=0 \\ {\chi_{rz }}= \ \frac{{\partial \varphi }}{{\partial r}}\,\,\,\,\,\,{\chi_{{\theta z}}}={r^{-1 }}\frac{{\partial \varphi }}{{\partial \theta }} \\ {\chi_{rr }}= \ {\chi_{{r\theta }}}={\chi_{{\theta r}}}={\chi_{{\theta \theta }}}={\chi_{zz }}={\chi_{zr }}={\chi_{{z\theta }}}=0 \\ \end{array}$$

(12)

The constitutive equations (8) take the form

$$\begin{array}{ll} {\sigma_{rr }}= \ (\lambda +2\mu +\kappa ){e_{rr }}+\lambda {e_{{\theta \theta }}}+\eta \psi \\ {\sigma_{{\theta \theta }}}= \ \lambda {e_{rr }}+(\lambda +2\mu +\kappa ){e_{{\theta \theta }}}+\eta \psi \\ {\sigma_{{r\theta }}}= \ (\mu +\kappa ){e_{{r\theta }}}+\mu {e_{{\theta r}}} \\ {\sigma_{{\theta r}}}= \ (\mu +\kappa ){e_{{\theta r}}}+\mu {e_{{r\theta }}} \\ {m_{rz }}= \ \gamma {\chi_{rz }}\quad{m_{{\theta z}}}=\gamma {\chi_{{\theta z}}} \\ {m_{zr }}= \ \beta {\chi_{rz }}\quad{m_{{z\theta }}}=\beta {\chi_{{\theta z}}} \\ {\lambda_r}= \ \sigma \frac{{\partial \psi }}{{\partial r}}\quad{\lambda_{\theta }}={r^{-1 }}\sigma \frac{{\partial \psi }}{{\partial \theta }} \\ 3s= \ \eta \left[ {{r^{-1 }}\frac{\partial }{{\partial r}}(ru)+{r^{-1 }}\frac{{\partial v}}{{\partial \theta }}} \right]+b\psi \\ \end{array}$$

(13)

The equilibrium equations (9) can be rewritten in the form

$$\begin{array} {ll} \frac{{\partial {\sigma_{rr }}}}{{\partial r}}+{r^{-1 }}\frac{{\partial {\sigma_{{\theta r}}}}}{{\partial \theta }}+{r^{-1 }}({\sigma_{rr }}-{\sigma_{{\theta \theta }}})=0 \\ \frac{{\partial {\sigma_{{r\theta }}}}}{{\partial r}}+{r^{-1 }}\frac{{\partial {\sigma_{{\theta \theta }}}}}{{\partial \theta }}+{r^{-1 }}({\sigma_{{r\theta }}}-{\sigma_{{\theta r}}})=0 \\ \frac{{\partial {m_{rz }}}}{{\partial r}}+{r^{-1 }}\frac{{\partial {m_{{\theta z}}}}}{{\partial \theta }}+{r^{-1 }}{m_{rz }}+{\sigma_{{r\theta }}}-{\sigma_{{\theta r}}}=0 \\ {r^{-1 }}\frac{\partial }{{\partial r}}(r{\lambda_r})+{r^{-1 }}\frac{{\partial {\lambda_{\theta }}}}{{\partial \theta }}-s=0 \\ \end{array}$$

(14)

The plane strain problem consists in finding of the functions $u,v,\varphi$, and $\psi$ which satisfy (11)–(14) and the boundary conditions.

Stress Concentration

In this section we study the problem of a cylindrical cavity in an infinite solid. The region B is defined by $\left\{ {({x_1},{x_2},{x_3})\in {R^3}:x_1^2+x_2^2>{a^2}} \right\}$, where $a>0$ is a given constant. We suppose that the body is in equilibrium in the absence of body loads, the surface of the cavity is free of surface loads, and the body is subjected to a field of simple tension at infinity. Let P be the constant tension field at infinity, we assume that the surface moment and surface microtraction vanish at infinity. The body is in a state of plane strain, parallel to the plane $x_1 O{x_2}$ (Fig. 1).

Fig. 1

The boundary conditions on the surface of the cavity can be expressed as

$$t_{rr }=0\,\,\,{t_{{r\theta }}}=0\,\,\,{m_{rz }}=0\,\,\,{h_r}=0$$

(15)

for $r=a$. We have the following conditions at infinity:

$$\begin{array}{ll} {{t_{rr }}} = \ {\frac{1}{2}P(1+\cos 2\theta)} \\ {{t_{{\theta \theta }}}} = \ {\frac{1}{2}P(1-\cos 2\theta )}\\ {{t_{{r\theta }}}} = \ {{t_{{\theta r}}}=-\frac{1}{2}P\sin2\theta } \\ {{m_{rz }}} = \ {{m_{{\theta z}}}=0\quad{h_r}={h_{\theta}}=0} \\ \end{array}$$

(16)

where P is a given constant. We seek the solution of the problem (12)–(14) with the boundary conditions (15), (16) in the form

$$\begin{array}{ll} u = \ {F(r)+U(r)\cos 2\theta \quad v=V(r)\sin2\theta } \\ \varphi = \ {\Phi (r)\sin 2\theta \quad \psi=G(r)+\Psi (r)\cos 2\theta } \\ \end{array}$$

(17)

where $F,U,V,\Phi, G$, and $\Psi$ are functions only on r. It follows from (12), (13), and (17) that

$$\begin{array}{ll} {\sigma_{rr }}= (\lambda +2\mu +\kappa )\frac{dF }{dr }+{r^{-1 }}\lambda F+\eta G \\ \qquad\, + [(\lambda +2\mu +\kappa )\frac{dU }{dr }+{r^{-1 }}\lambda (U+2V) \\ \qquad\, + \eta \Psi ]\cos 2\theta \\ {\sigma_{{\theta \theta }}}= \lambda \frac{dF }{dr }+(\lambda +2\mu +\kappa ){r^{-1 }}F+\eta G \\ \qquad\,\, + \left[ {\lambda \frac{dU }{dr }+(\lambda +2\mu +\kappa ){r^{-1 }}(U+2V)} \right. \\ \qquad\,\, + \left. {\eta \Psi\,} \right]\cos 2\theta \\ {\sigma_{{r\theta }}}= \left[ {(\mu +\kappa )\frac{dV }{dr }-\kappa \Phi -\mu {r^{-1 }}(2U+V)} \right]\sin 2\theta \\ {\sigma_{{\theta r}}}= \left[ {\mu \frac{dV }{dr }+\kappa \Phi -(\mu +\kappa ){r^{-1 }}(2U+V)} \right]\sin 2\theta \\ {m_{rz }}= \gamma \frac{{d\Phi}}{dr}\sin 2\theta\,\,\,\,\,{m_{{\theta z}}}=2\gamma {r^{-1 }}\Phi \\ {h_r}= \sigma \left( {\frac{dG }{dr }+\frac{{d\Psi}}{dr}\cos 2\theta } \right) \\ {h_{\theta }}= -2\sigma {r^{-1 }}\Psi\sin 2\theta \\ 3s = \eta \left( {\frac{dF }{dr }+{r^{-1 }}F} \right)+bG \\ \quad\, = + \left\{ {\eta \left[ {\frac{dU }{dr }+{r^{-1 }}(U+2V)} \right]+b\Psi} \right\}\cos 2\theta \end{array}$$

(18)

If we substitute (18) into equilibrium equations (14), we obtain the following equations

$$\begin{array} {ll} \left( {\lambda +2\mu +\kappa } \right)\frac{d}{dr}\left[ {{r^{-1 }}\frac{d}{dr }(rF)} \right]+\eta \frac{dG }{dr } = 0 \\ \sigma \left( {\frac{{{d^2}G}}{{d{r^2}}}+\frac 1r\frac{{dG}}{dr }-\frac b\sigma G}\right) -\eta r^{-1}\frac{d}{dr }({r F}) =0\\ (\lambda +2\mu +\kappa )\left( {{r^2}\frac{{{d^2}U}}{{d{r^2}}}+r\frac{dU }{dr }} \right) + \\ 2(\lambda +\mu )r\frac{dV }{dr }+\eta {r^2}\frac{{d\Psi}}{dr }-(\lambda +6\mu +5\kappa )U \\ -2(\lambda +3\mu +2\kappa )V+2\kappa r\Phi = 0 \\ (\mu +\kappa )\left( {{r^2}\frac{{{d^2}V}}{{d{r^2}}}+r\frac{dV }{dr }} \right)-2(\lambda +\mu )r\frac{dU }{dr } \\ - \kappa {r^2}\frac{{d\Phi}}{dr }-2(\lambda +3\mu +2\kappa )U \\ - (4\lambda +9\mu +5\kappa )V-2\eta r\Psi = 0 \\ \gamma \left( {{r^2}\frac{{{d^2}\Phi}}{{d{r^2}}}+r\frac{{d\Phi}}{dr }-4\Phi} \right)+\kappa {r^2}\frac{dV }{dr } \\ + \kappa r(2U+V)-2\kappa {r^2}\Phi = 0 \\ \sigma \left( {{r^2}\frac{{{d^2}\Psi}}{{d{r^2}}}+r\frac{{d\Psi}}{dr }-4\Psi -\frac{b}{\sigma }{r^2}\Psi} \right) \\ - \eta {r^2}\frac{dU }{dr }-\eta r(U+2V) = 0 \end{array}$$

(19)

The first equation of (19) implies that

$$r^{-1 }s\frac{d}{dr }(rF)+\frac{\eta }{{\lambda +2\mu +\kappa }}G={C_1}$$

(20)

where C ₁ is an arbitrary constant. In view of (20), the second equation of (19) can be written in the form

$$\frac{{{d^2}G}}{{d{r^2}}}+{r^{-1 }}\frac{dG }{dr }-{\zeta^2}G=\frac{\eta }{\sigma }{C_1}$$

(21)

where

$$\zeta^2=\frac{1}{\sigma}\left( {b-\frac{{{\eta^2}}}{{\lambda +2\mu +\kappa }}} \right)$$

(22)

It follows from (4) that $\zeta^2>0$. The solution of (21) is

$$G={A_1}{K_0}(\zeta r)+A_1^{*}{I_0}(\zeta r)-\frac{\eta }{{\sigma {\zeta^2}}}{C_1}$$

where $I_n$ and $k_n$ are the modified Bessel functions of order n and A ₁ and $A_1^{*}$ are arbitrary constants. Since the function G must be finite at infinity, we have $A_1^{*}=0$. Thus, we get

$$G={A_1}{K_0}(\zeta r)-\frac{\eta }{{\sigma {\zeta^2}}}{C_1}$$

(23)

It follows from (20) and (23) that

$$\begin{array}{ll} F = {\frac{b}{{2\sigma{\zeta^2}}}{C_1}r+{r^{-1 }}{C_2}} {} \,\,\,\, + {\frac{{\eta{A_1}}}{{\zeta (\lambda +2\mu +\kappa )}}{K_1}(\zeta r)} \\ \end{array}$$

(24)

where C ₂ is an arbitrary constant.

We introduce the notations

$$t=\ln r\,\,\,\,\,\,\,D=\frac{d}{dt }$$

Then, (19)₃ and (19)₄ can be written in the form

$$\begin{array} {ll} [{D^2}-(1+4{c_1})]U+2[(1-{c_1})D-(1+{c_1})]V \\ \qquad \qquad =-{e^t}({c_2}D\Psi +2{c_3}\Phi ) \\ \left[ {(1-{c_1})D+(1+{c_1})} \right]U+\left[ {{c_1}{D^2}-(4+{c_1})} \right]V \\ \qquad \qquad={e^t}(2{c_2}\Psi +{c_3}D\Phi ) \\ \end{array}$$

(25)

where

$$\begin{array}{ll} {c_1}=\frac{{\mu +\kappa }}{{\lambda +2\mu +\kappa }}\,\,\,\,\,\,{c_2}=\frac{\eta }{{\lambda +2\mu +\kappa }} \\ {c_3}=\frac{\kappa }{{\lambda +2\mu +\kappa }} \\ \end{array}$$

(26)

The general solution of the homogeneous system (25) which corresponds to a finite stress field at infinity is given by

$$\begin{array}{ll} {{U_0}} = \ {{B_1}{e^{-t }}+{B_2}{e^{-3t }}+{B_3}{e^t}} \\ {{V_0}} = \ {-{c_1}{B_1}{e^{-t }}+{B_2}{e^{-3t }}-{B_3}{e^t}} \\ \end{array}$$

(27)

where B ₁, B ₂ and B ₃ are arbitrary constants. Particular solution of the system (25) can be seen to be

$$\begin{array}{ll} {U^{*}} =-\frac{1}{2}{c_2}({e^t}{S_1}+{e^{-3t }}{S_2})-\frac{1}{{2{c_1}}}{c_3}({e^t}{R_1} \\ \qquad\qquad\qquad\qquad\quad \qquad - {e^{-3t }}{R_2}) \\ {V^{*}} =\frac{1}{2}{c_2}({e^t}{S_1}-{e^{-3t }}{S_2})+\frac{1}{{2{c_1}}}{c_3}({e^t}{R_1} \\ \qquad\qquad\qquad\qquad\quad \qquad - {e^{-3t }}{R_2}) \\ \end{array}$$

(28)

where

$$\begin{array}{ll} {S_1}(t)=\int^t \Psi (s)ds\,\,\,\,{S_2}(t)=\int ^t {e^{4s }}\Psi (s)ds\, \\ {R_1}(t)=\int ^t \Phi (s)ds\,\,\,\,{R_2}(t)=\int ^t {e^{4s }}\Phi (s)ds\, \\ \end{array}$$

With the help of (27) and (28), we obtain the solution of the system (25)

$$\begin{array}{ll} U = \ {{B_1}{r^{-1 }}+{B_2}{r^{-3 }}+{B_3}r-{\varGamma_1}-{\varGamma_2}} \\ V = \ {-{c_1}{B_1}{r^{-1 }}+{B_2}{r^{-3 }}-{B_3}r+{\varGamma_1}-{\varGamma_2}} \\ \end{array}$$

(29)

where

$$\begin{array}{ll} {\Gamma_1} =\frac{r}{2}\left[ {{c_2}\int ^r{x^{-1 }}\Psi (x)dx+\frac{{{c_3}}}{{{c_1}}}} \int ^r {x^{-1 }}\Phi (x)dx \right] \\ {\Gamma_2} =\frac{{{r^3}}}{2}\left[ {{c_2}\int ^r{x^3}\Psi (x)dx-\frac{{{c_3}}}{{{c_1}}}} \int ^r {x^3}\Phi (x)dx \right] \\ \end{array}$$

If we substitute (29) into (19)₅ and (19)₆, we obtain

$$\begin{array} {ll} {{r^2}\frac{{{d^2}\Phi}}{{d{r^2}}}+r\frac{{d\Phi}}{dr }-(4+{\delta^2}{r^2})\Phi} = {-2{c_4}{B_1}} \\ {{r^2}\frac{{{d^2}\Psi}}{{d{r^2}}}+r\frac{{d\Psi}}{dr }-(4+{\zeta^2}{r^2})\Psi} = {-\frac{{2{c_1}\eta }}{\sigma }{B_1}} \\ \end{array}$$

(30)

where

$$c_4=\frac{\kappa }{\gamma}\,\,\,\,\,\,\,\,{\delta^2}=\frac{{\chi (2\mu +\kappa )}}{{\gamma (\mu +\kappa )}}$$

The solution of (30) which generate finite stresses for $r\to \infty$ are given by

$$\begin{array}{ll} \Phi = {{A_2}{K_2}(\delta r)+\frac{2}{{{\delta^2}}}+{c_4}{B_1}{r^{-2 }}} \\ \Psi = {{A_3}{K_2}(\zeta r)+\frac{2}{{\sigma {\zeta^2}}}{c_1}\eta {B_1}{r^{-2 }}} \\ \end{array}$$

(31)

where A ₂ and A ₃ are arbitrary constants. If we substitute (31) into relations (28), we obtain

$$\begin{array}{ll} U={r^{-1 }} {d_1}{B_1}+{r^{-3}}{B_2}+{B_3}r-\frac{1}{{2\delta {c_1}}}{c_3}{A_2}[{K_3}(\delta r)\\ \qquad\,\,\, - \ {K_1}(\delta r)]+\frac{1}{{2\zeta }}{c_2}{A_3}\left[{{K_3}(\zeta r)+{K_1}(\zeta r)} \right] \\ V={r^{-1}} {c_2}{d_2}{B_1}+{r^{-3 }}{B_2}-{B_3}r-\frac{1}{{2\delta{c_1}}}{c_3}{A_2}[{K_3}(\delta r) \\ \qquad\,\,\, + \ {K_1}(\delta r)]+\frac{1}{{2\zeta }}{c_2}{A_3}\left[ {{K_3}(\zeta r)-{K_1}(\zeta r)} \right] \\ \end{array}$$

(32)

where

$$d_1=1+\frac{{{c_3}{c_4}}}{{{c_1}{\delta^2}}}\quad{d_2}=1+\frac{{{c_2}\eta }}{{\sigma {\zeta^2}}}$$

We introduce the notations

$$\begin{array}{ll} {q_1}={d_1}-2{c_1}{d_2}\quad{q_2}=2{d_1}-{c_1}{d_2} \\ {Q_1}=\frac{1}{{2\mu +\kappa }}\left[ {-(2\mu +\kappa ){d_1}-2\lambda {c_1}{d_2}+\frac{{2{c_1}{\eta^2}}}{{\sigma {\zeta^2}}}} \right] \\ {Q_2}=\frac{1}{{2\mu +\kappa }}\left[ {(\lambda +2\mu +\kappa ){q_1}-\lambda {d_1}+\frac{{2{c_1}{\eta^2}}}{{\sigma {\zeta^2}}}} \right] \\ {Q_3}=\frac{1}{{2\mu +\kappa }}\left[ {(\mu +\kappa ){c_1}{d_2}-\mu {q_2}-\frac{{2\kappa {c_4}}}{{{\delta^2}}}} \right] \\ {Q_4}=\frac{1}{{2\mu +\kappa }}\left[ {\mu {c_1}{d_2}-(\mu +\kappa ){q_2}+\frac{{2\kappa {c_4}}}{{{\delta^2}}}} \right] \\ \chi =(2\lambda +2\mu +\kappa )b-{\eta^2} \\ \end{array}$$

(33)

It follows from (18), (23), (24), (31), (32) and (33) that

$$\begin{array}{ll} {t_{rr }}= \ \frac{\chi }{{2\sigma{\zeta^2}}}{C_1}-(2\mu +\kappa ){r^{-2 }}{C_2} \\ \quad \,\,\, - \ \frac{{(2\mu+\kappa )}}{\zeta }{c_4}{A_1}{r^{-1 }}{K_1}(\zeta r)+(2\mu +\kappa )\\ \quad\,\,\,\,\,\, \{ {Q_1}{r^{-2 }}{B_1}-3{r^{-4 }}{B_2}+{B_3}\\ \quad \,\,\, + \ \frac{1}{{2\delta {c_1}}}{c_3}{A_2}{r^{-1 }}\left[{{K_1}(\delta r)+3{K_3}(\delta r)} \right] \\ \quad \,\,\, - \frac{1}{{4\zeta}}{c_2}{A_3}[6{r^{-1 }}{K_3}(\zeta r)-\zeta {K_2}(\zeta r) \\ \qquad \ +\zeta {K_0}(\zeta r)]\}\cos 2\theta \\ {t_{{\theta \theta}}}=\frac{\chi }{{2\sigma {\zeta^2}}}{C_1}+(2\mu +\kappa ){r^{-2}}{C_2} \\ \quad \,\,\, \ + (2\mu +\kappa ){c_2}{A_1}\left[ {{K_0}(\zeta r)+\frac{1}{{\zeta r}}{K_1}(\zeta r)} \right] \\ \quad \,\,\, \ + (2\mu +\kappa)\{{Q_2}{r^{-2 }}{B_1}+3{r^{-4 }}{B_2}-{B_3} \\ \quad \,\,\, \ - \frac{1}{{2\delta{c_1}}}{c_3}{A_2}{r^{-1 }}\left[ {{K_1}(\delta r)+3{K_3}(\delta r)}\right] \\ \quad \,\,\, \ + \frac{1}{{4\zeta }}{c_2}{A_3}[3\zeta {K_2}(\zeta r)+\zeta {K_0}(\zeta r) \\ \quad \,\,\, \ - 6{r^{-1 }}{K_3}(\zeta r)]\}\cos 2\theta\\ {t_{{r\theta }}}=(2\mu +\kappa )\{{Q_3}{B_1}{r^{-2 }}-3{r^{-4}}{B_2}-{B_3} \\ \quad \,\,\, \ + \frac{1}{{4\delta {c_1}}}{c_3}{A_2}[6{r^{-1}}{K_3}(\delta r)+\delta {K_0}(\delta r) \\ \quad \,\,\, \ - \delta {K_2}(\delta r)]-\frac{1}{{2\zeta }}{c_2}{A_3}{r^{-1 }}[3{K_3}(\zeta r) \\ \quad \,\,\, \ +{K_1}(\zeta r)]\}\sin 2\theta \\ {t_{{\theta r}}}=(2\mu +\kappa)\{{Q_4}{B_1}{r^{-2 }}-3{r^{-4 }}{B_2}-{B_3} \\ \quad \,\,\, \ + \frac{1}{{4\delta{c_1}}}{c_3}{A_2}[6{r^{-1 }}{K_3}(\delta r)+3\delta {K_2}(\delta r)\\ \quad \,\,\, \ + \delta {K_0}(\delta r)]-\frac{1}{{2\zeta }}{c_2}{A_3}{r^{-1}}[3{K_3}(\zeta r) \\ \quad \,\,\, \ + {K_1}(\zeta r)]\}\sin 2\theta \\ {m_{rz}}=-\{\gamma {A_2}\left[ {\delta {K_1}(\delta r)+2{r^{-1}}{K_2}(\delta r)} \right] \\ \quad \,\,\, \ + 4{B_1}{r^{-3 }}\frac{{\gamma{c_4}}}{{{\delta^2}}}\}\sin 2\theta \\ {m_{{\theta z}}}=\left[{2\gamma {A_2}{r^{-1 }}{K_2}(\delta r)+4{B_1}{r^{-3 }}\frac{{\gamma{c_4}}}{{{\delta^2}}}} \right]\cos 2\theta \\ {\lambda_r}=-\sigma\zeta {A_1}{K_1}(\zeta r)-\{\sigma {A_3}[\zeta {K_1}(\zeta r) \\ \quad \,\,\, \ +2{r^{-1 }}{K_2}(\zeta r)]+4{B_1}{r^{-3 }}\frac{{{c_1}\eta}}{{{\zeta^2}}}]\cos 2\theta \\ {\lambda_{\theta }}=-\left[ {2\sigma{r^{-1 }}{A_3}{K_2}(\zeta r)+4{B_1}{r^{-3 }}\frac{{{c_1}\eta}}{{{\zeta^2}}}} \right]\sin 2\theta \\ \end{array}$$

(34)

On the basis of (34) the conditions at infinity (16) reduce to

$$B_3=\frac{1}{{2(2\mu +\kappa )}}P\,\,\,{C_1}=\frac{{\sigma {\zeta^2}}}{\chi }P$$

(35)

We note that the restrictions (4) imply that $\chi>0$. We introduce the notations

$$\begin{array} {ll} \Lambda (z;p)=6{z^{-1 }}{K_3}(pz)-p{K_2}(pz)+p{K_0}(pz) \\ \Gamma (z;p)=p{K_1}(pz)+\frac{2}{p}{K_2}(pz) \\ \Omega (z)={K_1}(z)+3{K_3}(z) \\ {l_1}=4{a^{-3 }}\gamma {c_4}{\eta^{-2 }}\,\,\,\,{l_2}=4{a^{-3 }}{c_1}\eta {\zeta^{-2 }} \\ J=\gamma \sigma \Gamma (a;\delta )-4b_0^2{a^{-2 }}{K_2}(\zeta a){K_2}(\delta a) \\ {T_1}=\sigma {l_1}\Gamma (a;\zeta )\,\,\,\,{T_2}=\gamma {l_2}\Gamma (a;\delta ) \\ \end{array}$$

The boundary conditions (15) reduce to

$$\begin{array} {ll} {C_2}=\frac{{P{a^2}}}{{2(2\mu +\kappa )}}\,\,\,\,\,{A_1}=0 \\ {A_2}=-\frac{1}{J}{T_1}{B_1}\,\,\,\,{A_3}=-\frac{1}{J}{T_2}{B_1} \\ {H_1}{B_1}-3{a^{-4 }}{B_2}=-\frac{P}{{2(2\mu +\kappa )}} \\ {H_2}{B_1}-3{a^{-4 }}{B_2}=\frac{P}{{2(2\mu +\kappa )}} \\ \end{array}$$

(36)

where

$$\begin{array}{ll} {H_1}={Q_1}{a^{-2 }}-\frac{1}{{2Ja\delta{c_1}}}{c_3}{T_1}\Omega (\delta a) \\ \quad \,\, + \frac{1}{{4\zeta J}}{c_2}{T_2}\Lambda (a;\zeta ) \\ {H_2}={Q_3}{a^{-2}}-\frac{1}{{4\delta {c_1}J}}{c_3}{T_1}\Lambda (a;\delta ) \\ \quad \,\, +\frac{1}{{2\zeta aJ}}{c_2}{T_2}\Omega (\zeta a) \\ \end{array}$$

From (36) we find that

$$\begin{array}{ll} {{B_1}} = {\frac{P}{{(2\mu +\kappa )({H_2}-{H_1})}}} \\ {{B_2}} = {\frac{{P({H_1}+{H_2}){a^4}}}{{6(2\mu +\kappa )({H_2}-{H_1})}}} \\ \end{array}$$

so that all constants $c_{\alpha }$, $A_i$, and $B_i$ are determined. The displacement vector, the microrotation vector and the microstress function are obtained in explicit form.

The value of $t_{{\theta \theta }}$ at the periphery of the cavity is given by

$$t_{{\theta \theta }}=P(1-\frac{2}{1+f}\cos 2\theta )$$

(37)

where

$$\begin{array}{ll} f = \ {{N^{-1 }}[8J\delta {c_1}a\zeta({H_2}-{H_1})-N]} \\ N = \ {4J\delta {c_1}\zeta {a^{-1}}{Q_2}+4J\delta {c_1}a\zeta {H_1}} \\ {} \quad + \ {2{c_3}{T_1}\zeta\Omega (\delta a)-{c_1}{c_2}a\delta {T_2}[3\zeta {K_2}(\zeta a)} \\{} \quad + \ {\zeta {K_0}(\zeta a)-6{a^{-1 }}{K_3}(\zeta a)]} \\\end{array}$$

The maximum value of $t_{{\theta \theta }}$ occurs at $\theta =\pm \frac{\pi }{2}$, leading to the stress concentration factor

$$\begin{array}{ll} K = {\frac{{t_{{\theta \theta}}^m}}{P}=\frac{3+f }{1+f }} \\ \end{array}$$

Micropolar Elasticity

The motion of a micropolar elastic solid is described by a displacement vector field $\bf u$ and a microrotation vector field $\varphi$. The stretch is absent. The basic equations are obtained by (1)–(3) with $\psi =0$ and $\eta =0$

$$\sigma_{ji,j }=0\,\,\,\,{m_{ji,j }}+{\epsilon_{irs }}{\sigma_{rs }}=0$$

(38)

$$\begin{array}{rr} {{\sigma_{ij }}} = {\lambda {e_{rr }}{\delta_{ij }}+(\mu +\kappa ){e_{ij }}+\mu {e_{ji }}} \\ {{m_{ij }}} = {\alpha {\chi_{rr }}{\delta_{ij }}+\beta {\chi_{ji }}+{\gamma_{ij }}} \end{array}$$

(39)

$$e_{ij }={u_{i,j }}+{\epsilon_{ijk }}+{\varphi_k}\,\,\,\,{\chi_{ij }}={\varphi_{j,i }}$$

(40)

The boundary conditions are given by

$$\sigma_{ji }{n_j}=\mathop{\tilde{t}}\nolimits_i\quad{m_{ji }}{n_j}=\mathop{\tilde{m}}\nolimits_i$$

(41)

The restrictions on the constitutive coefficients became

$$\begin{array}{ll} {3\lambda +2\mu +\kappa },>{0\,\,\,\,2\mu+\kappa\,> 0\,\,\,\,\kappa\,> 0} \\ {3\alpha +\beta +\gamma }\,> {0\,\,\,\,\beta +\gamma\,>0\,\,\,\,\,\gamma -\beta\,>0} \end{array}$$

(42)

In the case of plane strain (38) reduce to

$$\sigma_{{\alpha \beta, \alpha }}=0\,\,\,\,\,{m_{{\alpha 3,\alpha }}}+{\epsilon_{{3\rho \nu }}}{\sigma_{{\rho \nu }}}=0$$

(43)

The general solution of (43) is given by stress and couple stress functions S and T

$$\sigma_{{\alpha \beta }}={\epsilon_{{3\alpha \rho }}}{\epsilon_{{3\beta \nu }}}S{,_{{\rho \nu }}}-{\epsilon_{{3\alpha \rho }}}T{,_{{\rho \beta }}}\,\,\,\,{m_{{\alpha 3}}}=T{,_{\alpha }}$$

(44)

The constitutive equations (39) can be solved for e _ij and $\chi_{ij }$. In the case of plane strain, we obtain

$$\begin{array}{ll} {e_{{\alpha \beta }}}= \ \frac{1}{{2\mu +\kappa }}\left( {-\frac{\lambda }{{2\lambda +2\mu +\kappa }}{\sigma_{{\rho \rho }}}} \right. \\ \quad + \ \left. {\frac{{\mu +\kappa }}{\kappa }{\sigma_{{\alpha \beta }}}-\frac{\mu }{\kappa }{\sigma_{{\beta \alpha }}}} \right) \\ {\chi_{{\alpha 3}}}= \ \frac{1}{\gamma }{m_{{\alpha 3}}} \\ \end{array}$$

(45)

The functions $e_{{\alpha \beta }}$ and $\chi_{{\alpha 3}}$ cannot be arbitrary; they should obey the compatibility conditions

$$\begin{array}{ll} {{e_{11,2 }}-{e_{21,1 }}+{\chi_{13 }}} = 0 \\ {{e_{22,1 }}-{e_{12,2 }}-{\chi_{23 }}} = 0 \\ {{\chi_{13,2 }}-{\chi_{23,1 }}} = 0 \\ \end{array}$$

(46)

From (44)–(46) follows

$$\begin{array}{ll} {(T-{c^2}\Delta T){,_1}} = {-2{b^2}(1-{\nu^{*}})\Delta S{,_2}} \\ {(T-{c^2}\Delta T){,_2}} = {2{b^2}(1-{\nu^{*}})\Delta S{,_1}} \\ \end{array}$$

(47)

where

$$\begin{array}{ll} {c^2}=\frac{{\gamma (\mu +\kappa )}}{{\kappa (2\mu +\kappa )}}\,\,\,\,\,{b^2}=\frac{\gamma }{{(2\mu +\kappa )}} \\ {\nu^{*}}=\frac{\lambda }{{2\lambda +2\mu +\kappa }} \\ \end{array}$$

and $\varDelta$ is the Laplacian.

By cross differentiation (47) gives

$$\begin{array}{ll} {{\Delta^2}S} = 0 \\ {\Delta ({c^2}\Delta T-T)} = 0 \\ \end{array}$$

(48)

Thus, the solution of a plain strain problem requires the solution of (47) and (48).

Stress Concentration Around a Circular Hole

We consider a plate with a circular hole of radius a subjected to a uniform tension P at $x=\infty$. The boundary conditions in polar coordinates read

$$\sigma_{rr }={\sigma_{{r\theta }}}={m_{rz }}=0\,\,\,\,\mathrm{ on}\,\,\,\,r=a$$

(49)

$$\begin{array}{rr} {\sigma_{rr }}=\frac{P}{2}(1+\cos 2\theta )\,\,\,\,{\sigma_{{r\theta }}}=-\frac{P}{2}\sin 2\theta \\ {m_{rz }}=0\,\,\,\,\,\mathrm{ on}\,\,\,\,\,r=\infty \\ \end{array}$$

(50)

The solution of the problem of stress concentration for a micropolar plate can be obtained putting $\psi =0$ and $\eta =0$ in the previous solution of the microstretch plate. In the following an alternative method is considered. A solution of (48) is given by

$$\begin{array} {ll} S=\frac{P}{4}{r^2}(1-\cos 2\theta )+{A_1}\ln r+({A_2}{r^{-2 }}+{A_3})\\ \qquad\cos 2\theta \\ T=\left[ {{A_4}{r^{-2 }}+{A_5}{K_2}(r/c)} \right]\sin 2\theta \\ \end{array}$$

(51)

The equations of compatibility (46) are satisfied if

$$A_4=8(1-{\nu^{*}}){c^2}{A_3}$$

(52)

The stress and couple stress fields are given by

$$\begin{array}{ll} {\sigma_{rr }} =\frac{P}{2}(1+\cos 2\theta)+{A_1}{r^{-2 }}-(6{A_2}{r^{-4 }}+4{A_3}{r^{-2 }} \\ \quad \,\,\, - 6{A_4}{r^{-4}})\cos 2\theta +\frac{{2{A_5}}}{cr }[3c{r^{-1 }}{K_0}(r/c) \\ \quad \,\,\, +(1+6{c^2}{r^{-2 }}){K_1}(r/c)]\cos 2\theta \\ {\sigma_{{\theta\theta }}} =\frac{P}{2}(1-\cos 2\theta )-{A_1}{r^{-2 }}+(6{A_2}{r^{-4}} \\ \quad \,\,\, + 6{A_4}{r^{-4 }})\cos 2\theta -\frac{{2{A_5}}}{cr }[3c{r^{-1}}{K_0}(r/c) \\ \quad \,\,\, + (1+6{c^2}{r^{-2 }}){K_1}(r/c)]\cos 2\theta \\{\sigma_{{r\theta }}} =-\left( {\frac{P}{2}+6{A_2}{r^{-4}}+2{A_3}{r^{-2 }}-6{A_4}{r^{-4 }}} \right)\sin 2\theta \\ \quad \,\,\, +\frac{{{A_5}}}{cr}\left[ 6 \right.c{r^{-1}}{K_0}(r/c)+(1+12{c^2}{r^{-2 }}) \\ \,\,\,\,\,\,\,\left.{{K_1}(r/c)} \right]\sin 2\theta \\ {\sigma_{{\theta r}}} =-\left({\frac{P}{2}+6{A_2}{r^{-4 }}+2{A_3}{r^{-2 }}-6{A_4}{r^{-4 }}}\right)\sin 2\theta \\ \quad \,\,\, + \frac{{{A_5}}}{cr}\left[ {(1+6{c^2}{r^{-2}}{K_0}(r/c)+(3c{r^{-1 }}} \right. \\ \quad \,\,\, + \left. {12{c^3}{r^{-3}}){K_1}(r/c)} \right]\sin 2\theta \\ {m_{rz }}= -2{A_4}{r^{-3 }}\sin2\theta -\frac{{{A_5}}}{c}\left[ {2c{r^{-1 }}{K_0}(r/c)} \right. \\ \quad \,\,\, + \left. {(1+4{c^2}{r^{-2 }}){K_1}(r/c)} \right]\sin 2\theta \\{m_{{\theta z}}}= \left\{ {2{A_4}{r^{-3 }}+2{A_5}{r^{-1 }}\left[{{K_0}(r/c)} \right.} \right. \\ \quad \,\,\, + \left. {\frac{2c }{r}{K_1}\left.{(r/c)} \right\}} \right]\cos 2\theta \\ \end{array}$$

(53)

From the boundary conditions (50) follows

$$\begin{array} {ll} {A_1}=-\frac{P}{2}{a^2}\,\,\,\,\,\,\,{A_2}=\frac{{P{a^4}(1-{f_1})}}{{4(1+{f_1})}} \\ {A_3}=-\frac{{P{a^2}}}{{2(1+{f_1})}}\,\,\,\,\,\,{A_4}=\frac{{4(1-{\nu^{*}}){a^2}{b^2}P}}{{1+{f_1}}} \\ {A_5}=-\frac{{Pac{f_1}}}{{(1+{f_1}){K_1}(a/c)}} \\ \end{array}$$

where

$$f_1=8(1-{\nu^{*}})\frac{{{b^2}}}{{{c^2}}}\mathop{{\left[ {4+\frac{{{a^2}}}{{{c^2}}}+\frac{2a }{c}\frac{{{K_0}(a/c)}}{{{K_1}(a/c)}}} \right]}}\nolimits^{-1 }$$

The value of $\sigma_{{\theta \theta }}$ for $r=a$ is given by

$$\sigma_{{\theta \theta }}=P\left( {1+\frac{{2\cos 2\theta }}{{1+{f_1}}}} \right)$$

The maximum value of this occurs at $\theta =\pm \frac{\pi }{2}$. Thus the stress concentration factor yields

$$K=\frac{{3+{f_1}}}{{1+{f_1}}}$$