Stress function and its finite element implementation for elastostatic plain strain orthotropic problems of modified couple stress theory

Abstract

In this work, the stress function of the orthotropic plane strain problems in the modified couple stress theory (MCST) is first proposed and its polynomial analytical solutions are derived. Then, the stress function is used for developing a new 4-node membrane element for the MCST in that the C¹ continuity requirement is satisfied in weak sense using the penalty function method. In the element formulation, the stress function is adopted as the primary parameter for designing element’s stress and couple stress trial functions instead of assuming them directly. Therefore, the deduced stress and couple stress trial functions can satisfy both the equilibrium equations and deformation compatibility equations of the relevant problems a priori. Several numerical tests are examined and the results show that the element has good numerical accuracy and mesh distortion tolerance in simulating the size-dependent behaviors of small-scale orthotropic materials, proving that the usage of analytic trial function in the finite element implementation of the MCST can effectively improve element’s performance.

Appendix

As previously discussed, the stress function for the isotropic plane strain problems of the MCST can also be obtained by the degeneration of Mindlin’s stress functions which are proposed for the TMK-CST [60]. In Mindlin’s development [60], the stresses and couple stresses can be derived from two stress functions $\phi^{\prime}$ and $\psi^{\prime}$:

$$\sigma_{x} = \frac{{\partial^{2} \phi^{\prime}}}{{\partial y^{2} }} - \frac{{\partial^{2} \psi^{\prime}}}{\partial x\partial y},\;\;\;\sigma_{y} = \frac{{\partial^{2} \phi^{\prime}}}{{\partial x^{2} }} + \frac{{\partial^{2} \psi^{\prime}}}{\partial x\partial y}$$

(68)

$$\sigma_{xy} = - \frac{{\partial^{2} \phi^{\prime}}}{\partial x\partial y} - \frac{{\partial^{2} \psi^{\prime}}}{{\partial y^{2} }},\;\;\;\sigma_{yx} = - \frac{{\partial^{2} \phi^{\prime}}}{\partial x\partial y} + \frac{{\partial^{2} \psi^{\prime}}}{{\partial x^{2} }}$$

(69)

$$m_{xz} = \frac{{\partial \psi^{\prime}}}{\partial x},\;\;\;m_{yz} = \frac{{\partial \psi^{\prime}}}{\partial y}$$

(70)

in which $\phi^{\prime}$ and $\psi^{\prime}$ should satisfy

$$\frac{\partial }{\partial x}\left( {l^{2} \nabla^{2} \psi^{\prime} - \psi^{\prime}} \right) = 2\left( {1 - \nu } \right)l^{2} \frac{\partial }{\partial y}\left( {\nabla^{2} \phi^{\prime}} \right)$$

(71)

$$\frac{\partial }{\partial y}\left( {l^{2} \nabla^{2} \psi^{\prime} - \psi^{\prime}} \right) = - 2\left( {1 - \nu } \right)l^{2} \frac{\partial }{\partial x}\left( {\nabla^{2} \phi^{\prime}} \right)$$

(72)

$$\nabla^{4} \phi^{\prime} = 0,\;\;\;l^{2} \nabla^{4} \psi^{\prime} - \nabla^{2} \psi^{\prime} = 0$$

(73)

Considering the difference in the definition of the material length scale parameter l between the MCST and TMK-CST, Eqs. (71), (72) and (73) change into the following expressions for the MCST, respectively:

$$\frac{\partial }{\partial x}\left( {\frac{1}{4}l^{2} \nabla^{2} \psi^{\prime} - \psi^{\prime}} \right) = \frac{1}{2}\left( {1 - \nu } \right)l^{2} \frac{\partial }{\partial y}\left( {\nabla^{2} \phi^{\prime}} \right)$$

(74)

$$\frac{\partial }{\partial y}\left( {\frac{1}{4}l^{2} \nabla^{2} \psi^{\prime} - \psi^{\prime}} \right) = - \frac{1}{2}\left( {1 - \nu } \right)l^{2} \frac{\partial }{\partial x}\left( {\nabla^{2} \phi^{\prime}} \right)$$

(75)

$$\nabla^{4} \phi^{\prime} = 0,\;\;\frac{1}{4}l^{2} \nabla^{4} \psi^{\prime} - \nabla^{2} \psi^{\prime} = 0$$

(76)

It can be seen from Eq. (69) that the stress tensor in the TMK-CST is asymmetric. But in the MCST, the stress tensor is symmetric. By imposing the constraint $\sigma_{xy} = \sigma_{yx}$ to Eq. (69), we can get

$$\nabla^{2} \psi^{\prime} = 0$$

(77)

and accordingly, Eqs. (74), (75) and (76) can be rewritten as

$$\frac{{\partial \psi^{\prime}}}{\partial x} = - \frac{1}{2}\left( {1 - \nu } \right)l^{2} \frac{\partial }{\partial y}\left( {\nabla^{2} \phi^{\prime}} \right)$$

(78)

$$\frac{{\partial \psi^{\prime}}}{\partial y} = \frac{1}{2}\left( {1 - \nu } \right)l^{2} \frac{\partial }{\partial x}\left( {\nabla^{2} \phi^{\prime}} \right)$$

(79)

$$\nabla^{4} \phi^{\prime} = 0$$

(80)

Thus, the relationships between the stresses, couple stresses and the stress function for the isotropic plane strain problems of the MCST are derived by inserting Eqs. (78) and (79) back into Eqs. (68), (69) and (70):

$$\sigma_{x} = \frac{{\partial^{2} \phi^{\prime}}}{{\partial y^{2} }} + \frac{1}{2}\left( {1 - \nu } \right)l^{2} \frac{{\partial^{2} }}{{\partial y^{2} }}\left( {\nabla^{2} \phi^{\prime}} \right),\;\;\;\sigma_{y} = \frac{{\partial^{2} \phi^{\prime}}}{{\partial x^{2} }} - \frac{1}{2}\left( {1 - \nu } \right)l^{2} \frac{{\partial^{2} }}{{\partial y^{2} }}\left( {\nabla^{2} \phi^{\prime}} \right)$$

(81)

$$\sigma_{xy} = \sigma_{yx} = - \frac{{\partial^{2} \phi^{\prime}}}{\partial x\partial y} - \frac{1}{2}\left( {1 - \nu } \right)l^{2} \frac{{\partial^{2} }}{\partial x\partial y}\left( {\nabla^{2} \phi^{\prime}} \right)$$

(82)

$$m_{xz} = - \frac{1}{2}\left( {1 - \nu } \right)l^{2} \frac{\partial }{\partial y}\left( {\nabla^{2} \phi^{\prime}} \right),\;\;\;m_{yz} = \frac{1}{2}\left( {1 - \nu } \right)l^{2} \frac{\partial }{\partial x}\left( {\nabla^{2} \phi^{\prime}} \right)$$

(83)

in which $\phi^{\prime}$ needs to satisfy Eq. (80).

Finally, it can be easily proved that the above relations are equivalent to those given in Eqs. (15), (30) and (31) in Sect. 2.2, although they have different expressions. For the given deformation state determined by a stress function $\phi^{\prime}$ in accordance with Eqs. (81) to (83), we can find one stress function $\phi$ to describe it in accordance with Eqs. (15), (30) and (31) in Sect. 2.2:

$$\phi = \phi^{\prime} + \frac{1}{2}\left( {1 - \nu } \right)l^{2} \nabla^{2} \phi^{\prime}$$

(84)

More specifically, if $\phi^{\prime}$ is the linear combination of the polynomials that can satisfy Eq. (80) and guarantee the completeness of fourth order as follows:

$$\begin{gathered} \phi^{\prime} = a_{1} x^{2} + a_{2} xy + a_{3} y^{2} + a_{4} x^{3} + a_{5} x^{2} y + a_{6} xy^{2} + a_{7} y^{3} + a_{8} x^{3} y \\ + a_{9} xy^{3} + a_{10} \left( {x^{4} - y^{4} } \right) + a_{11} \left( {6x^{2} y^{2} - x^{4} - y^{4} } \right) \\ \end{gathered}$$

(85)

$\phi$ takes the coming form according to Eq. (84):

$$\begin{gathered} \phi = \left( {a_{1} + 6\eta a_{10} } \right)x^{2} + \left( {a_{2} + 3\eta a_{8} + 3\eta a_{9} } \right)xy + \left( {a_{3} - 6\eta a_{10} } \right)y^{2} + a_{4} x^{3} + a_{5} x^{2} y \\ + a_{6} xy^{2} + a_{7} y^{3} + a_{8} x^{3} y + a_{9} xy^{3} + a_{10} \left( {x^{4} - y^{4} } \right) + a_{11} \left( {6x^{2} y^{2} - x^{4} - y^{4} } \right) \\ \end{gathered}$$

(86)

with

$$\eta = \left( {1 - \nu } \right)l^{2}$$

(87)

Alternatively, Eq. (86) can be expressed as

$$\begin{gathered} \phi = b_{1} x^{2} + b_{2} xy + b_{3} y^{2} + b_{4} x^{3} + b_{5} x^{2} y + b_{6} xy^{2} + b_{7} y^{3} + b_{8} x^{3} y \\ + b_{9} xy^{3} + b_{10} \left( {x^{4} - y^{4} } \right) + b_{11} \left( {6x^{2} y^{2} - x^{4} - y^{4} } \right) \\ \end{gathered}$$

(88)

Note that the constant and linear terms have been ignored in Eq. (86) because they don’t produce stress and couple stress. As could be seen from Eqs. (85) and (88), the two functions $\phi$ and $\phi^{\prime}$ are actually made up of the same polynomials, but with different coefficients.