Introduction & Analytical Solutions

A simple study is considered to verify that the micromorphic upscaling workflow is functioning properly. The goal is to conduct a linear elastic simulation in a DNS code, homogenize the DNS results via the Micromorphic Filter, calibrate the simplified “two parameter” micromorphic linear elasticity (i.e. Saint Venant-Kirchhoff) model which should recover the input material properties of the DNS, and run a macroscale simulation in Tardigrade-MOOSE.

A uni-axial stress state is considered for a cylindrical geometry with a height, \(h_0\), and diameter, \(d_0\), of 5 mm. A coordinate system is assumed in which the x-, y-, and z-directions correspond to unit vectors identified with subscripts 1, 2, and 3, respectively. The axis of the cylinder is aligned with the z-direction. A compressive displacement, \(\Delta h\), of -0.05 mm is applied to the top surface of the cylinder corresponding to a nominal strain, \(\varepsilon_{33}\), of -1%. The bottom surface is restricted from motion in the z-direction. For this study, nominal elastic properties for the FK-800 binder are chosen with an elastic modulus, \(E^*\), of 165.0 MPa and a Poisson’s ratio, \(\nu^*\), of 0.39 [6, 7, 8].

Solution to small strain linear elasticity for uni-axial stress

The solution to infinitesimal (small) strain linear elasticity for uni-axial stress is as follows. All stresses are zero except for the \(33\) component which is homogeneous through the cylinder.

\[ \begin{align}\begin{aligned}\sigma_{11} &= \sigma_{22} = \sigma_{12} = \sigma_{13} = \sigma_{23} = 0\\\varepsilon_{33} &= \frac{\Delta_h}{h_0} = \frac{-0.05 mm}{5.0 mm} = -1.0\%\\\sigma_{33} &= E^* \varepsilon_{33} = 165.0 MPa \cdot -1.0\% = -1.650 MPa\end{aligned}\end{align} \]

The cylinder will expand laterally due to Poisson’s effect corresponding to the lateral strain (\(\varepsilon_L = \varepsilon_{11} = \varepsilon_{22}\)). The total force acting on the cylinder, \(F_3\) may then be calculated according to the resulting stress and final area, \(A_f\).

\[ \begin{align}\begin{aligned}\varepsilon_L &= - \nu^* \cdot \varepsilon_{33} = 0.0039\\F_3 &= \sigma_{33} A_f = \frac{\pi \sigma_{33}}{4} \left[ d_0 \left( 1 + \varepsilon_{L} \right) \right]^2 = -32.651 N\end{aligned}\end{align} \]

Solution to finite deformation linear elasticity for uni-axial stress

The Micromorphic Filter provides homogenized stresses and deformations in the finite strain regime. Therefore, it is possible that homogenized DNS results may differ from the input DNS results which may be run in either the small or finite strain regimes. Here, the applied nominal strain of -1% corresponds to a stretch, \(\alpha_3\), of 0.99. The lateral stretches, (\(\alpha_L = \alpha_1 = \alpha_2\)), will be equal and unknown. The applied uniform deformation gradient is therefore:

\[\begin{split}\left[\mathbf{F}\right] = \begin{bmatrix} \alpha_L & 0 & 0 \\ 0 & \alpha_L & 0 \\ 0 & 0 & \alpha_3 \\ \end{bmatrix}\end{split}\]

The Green-Lagrange strain is:

\[\begin{split}\mathbf{E} &= \frac{1}{2} \left(\mathbf{F}^T \mathbf{F} - \mathbf{I} \right) &= \frac{1}{2} \begin{bmatrix} \left(\alpha_L^2 - 1\right) & 0 & 0 \\ 0 & \left(\alpha_L^2 - 1\right) & 0 \\ 0 & 0 & \left(\alpha_3^2 - 1\right) \\ \end{bmatrix}\end{split}\]

The second Piola-Kirchhoff stress for the Saint Venant-Kirchhoff model [5] may then be calculated. First, the elastic Modulus, \(E^*\), and Poisson ratio, \(\nu^*\), need to be converted to the Lame'e constant, \(\lambda^*\), and shear modulus, \(\mu^*\), according to:

(12)\[ \begin{align}\begin{aligned} \lambda^* &= \frac{E^* \nu^*}{\left(1 + \nu^* \right) \left(1 - 2\nu^*\right)} ~ 210.4 MPa\\ \mu^* &= \frac{E^*}{2\left(1 + \nu^* \right)} ~ 59.35 MPa\end{aligned}\end{align} \]

The second Piola-Kirchhoff stress is thus:

\[\mathbf{S} = \lambda^* tr\left(\mathbf{E}\right) \mathbf{I} + 2\mu^*\mathbf{E}\]

where \(tr(\mathbf{E})\) is the sum of the diagonal components of the Green-Lagrange strain:

\[tr\left(\mathbf{E}\right) = E_{kk} = E_{11} + E_{22} + E_{33} = \left(\alpha_L^2 -1\right) + \frac{1}{2}\left(\alpha_3^2 -1\right)\]

The stress may be pushed forward to the current configuration to calculate the Cauchy stress, which is calculated by the Micromorphic Filter, as:

\[\mathbf{\sigma} = \frac{1}{J} \mathbf{F S F^T}\]

The total applied force is then calculated based on the lateral stretch, \(\alpha_L\), as:

\[F_3 = \sigma_{33} A_f = \frac{\pi \sigma_{33}}{4} \left(d_0 \alpha_L\right)^2\]

To solve this system of equations, one must first determine the unknown lateral stretch, \(\alpha_L\), which may be determined by recognizing that the lateral normal stresses will be zero.

(13)\[ \begin{align}\begin{aligned} S_{11} &= S_{22} = 0\\ S_{11} &= \lambda^* \left[\left(\alpha_L^2 -1\right) + \frac{1}{2}\left(\alpha_3^2 - 1 \right)\right] + \mu^* \left(\alpha_L^2 - 1\right) = 0\\ &\rightarrow \alpha_L = \sqrt{\frac{\left(3 - \alpha_3^2\right)\frac{\lambda^*}{2} + \mu^*}{\lambda^* + \mu^*}}\end{aligned}\end{align} \]

Upon substitution of \(\alpha_3 = 0.99\), the lateral stretch is found to be \(\alpha_L = 1.003873\).

Finally, the stresses of interest and total force will be equal to:

(14)\[ \begin{align}\begin{aligned} S_{33} &= -1.64175 MPa\\ \sigma_{33} &= -1.61282 MPa\\ F_3 &= -31.91 N\end{aligned}\end{align} \]