CSOL2

Plane or axisymmetric state

Coupled mechanical-flow analysis in large deformations.

Type: 203

The element is defined by 3, 4, 6, 8, 15 or 25 nodes indicated in NODES in the order indicated in the figure.

The flow description can be different from the mechanical description: pressure could be linearly interpolated in a 3 or 4 nodes configurations, while the mechanical DoFs would be parabolically interpolated in a 6 or 8 nodes configuration. In that case, the flow DoF must be fixed for the non-used nodes.


15-node element (12 I.P.)

For the element with 15 nodes, the interpolation functions are the following: \[N_1=\zeta(4\zeta-1)(4\zeta-2)(4\zeta-3)/6 \\ N_2=\xi(4\xi-1)(4\xi-2)(4\xi-3)/6 \\ N_3=\eta(4\eta-1)(4\eta-2)(4\eta-3)/6 \\ N_4=4\zeta\xi(4\zeta-1)(4\xi-1) \\ N_5= 4\xi\eta(4\xi-1)(4\eta-1) \\ N_6= 4\eta\zeta(4\eta-1)(4\zeta-1) \\ N_7=\xi\zeta(4\zeta-1)(4\zeta-2)*8/3 \\ N_8=\zeta\xi(4\xi-1)(4\xi-2)*8/3 \\ N_9=\eta\xi(4\xi-1)(4\xi-2)*8/3 \\ N_{10}=\xi\eta(4\eta-1)(4\eta-2)*8/3 \\ N_{11}=\zeta\eta(4\eta-1)(4\eta-2)*8/3 \\ N_{12}=\eta\zeta(4\zeta-1)(4\zeta-2)*8/3 \\ N_{13}=32\eta\xi\zeta(4\zeta-1) \\ N_{14}=32\eta\xi\zeta(4\xi-1) \\ N_{15}=32\eta\xi\zeta(4\eta-1) \]
25-node element (16 I.P.)

For the element with 25 nodes, the interpolation functions are expressed from the following functions: \[\begin{align*} N_1&=N_1(\xi)N_1(\eta) & N_{14}&=N_1(\xi)N_4(\eta) \\ N_2&=N_2(\xi)N_1(\eta) & N_{15}&=N_1(\xi)N_3(\eta) \\ N_3&=N_3(\xi)N_1(\eta) & N_{16}&=N_1(\xi)N_2(\eta) \\ N_4&=N_4(\xi)N_1(\eta) & N_{17}&=N_2(\xi)N_2(\eta) \\ N_5&=N_5(\xi)N_1(\eta) & N_{18}&=N_3(\xi)N_2(\eta) \\ N_6&=N_5(\xi)N_2(\eta) & N_{19}&=N_4(\xi)N_2(\eta) \\ N_7&=N_5(\xi)N_3(\eta) & N_{20}&=N_4(\xi)N_3(\eta) \\ N_8&=N_5(\xi)N_4(\eta) & N_{21}&=N_4(\xi)N_4(\eta) \\ N_9&=N_5(\xi)N_5(\eta) & N_{22}&=N_3(\xi)N_4(\eta) \\ N_{10}&=N_4(\xi)N_5(\eta) & N_{23}&=N_2(\xi)N_4(\eta) \\ N_{11}&=N_3(\xi)N_5(\eta) & N_{24}&=N_2(\xi)N_3(\eta) \\ N_{12}&=N_2(\xi)N_5(\eta) & N_{25}&=N_3(\xi)N_3(\eta) \\ N_{13}&=N_1(\xi)N_5(\eta) \end{align*}\]

With: \[N_1(X)=1/6X-1/6X^2-2/3X^3+2/3X^4 \\ N_2(X)=-4/3X+8/3X^2+4/3X^3-8/3X^4 \\ N_3(X)=1-5X^2+X^4 \\ N_4(X)=4/3X+8/3X^2-4/3X^3-8/3X^4 \\ N_5(X)=-1/6X-1/6X^2+2/3X^3+2/3X^4 \]
Implemented by: J.D. Barnichon, 1996

Prepro: CSOL2A.F
Lagamine: CSOL2B.F

For anisotropic case, Biot’s (symmetric) tensor is defined as follows: \[b_{ij}=\delta_{ij}-\frac{C^e_{ijkk}}{3K_S}\] For orthotropic elasticity, it reduces to a diagonal matrix: \[b_{ij}=\delta_{ij}-\frac{C^e_{ijkk}}{3K_S}=\begin{bmatrix} b_{11} & 0 & 0 \\ 0 & b_{22} & 0 \\ 0 & 0 & b_{33} \end{bmatrix}\] The elastic tensor definition for orthotropic elasticity $C^e_{ijkk}$ is available in the mechanical law definition (see ORTHO3D and ORTHOPLA) and reads: \[C^e_{ijkl} = \begin{bmatrix} \frac{1-\nu_{23}\nu_{32}}{E_2E_3det} & \frac{\nu_{21}+\nu_{31}\nu_{23}}{E_2E_3det} & \frac{\nu_{21}\nu_{32}+\nu_{31}}{E_2E_3det} & & & \\ \frac{\nu_{12}+\nu_{13}\nu_{32}}{E_1E_3det} & \frac{1-\nu_{13}\nu_{31}}{E_1E_3det} & \frac{\nu_{32}+\nu_{31}\nu_{12}}{E_1E_3det} & & & \\ \frac{\nu_{13}+\nu_{23}\nu_{12}}{E_1E_2det} & \frac{\nu_{23}+\nu_{21}\nu_{13}}{E_1E_2det} & \frac{1-\nu_{21}\nu_{12}}{E_1E_2det} & & & \\ & & & 2G_{12} & & \\ & & & & 2G_{13} & \\ & & & & & 2G_{23} \\ \end{bmatrix}\] with $det=\dfrac{1-\nu_{31}\nu_{13}-\nu_{21}\nu_{12}-\nu_{32}\nu_{23}-2\nu_{31}\nu_{12}\nu_{23}}{E_1E_2E_3}$
Adopting the micro-homogeneity and micro-isotropy assumptions, in which the bulk modulus of the solid phase Ks is isotropic, the Biot’s coefficient can be expressed as: \[b_{11}=1-\frac{C^e_{1111}+C^e_{1122}+C^e_{1133}}{3K_s}=1-\frac{1-\nu_{23}\nu_{32}+\nu_{21}+\nu_{31}\nu_{23}+\nu_{21}\nu_{32}+\nu_{31}}{E_2E_3\det{3K_s}} \\ b_{22}=1-\frac{C^e_{2211}+C^e_{2222}+C^e_{2233}}{3K_s}=1-\frac{\nu_{12}+\nu_{13}\nu_{32}+1-\nu_{13}\nu_{31}+\nu_{32}+\nu_{31}\nu_{12}}{E_1E_3\det{3K_s}} \\ b_{33}=1-\frac{C^e_{3311}+C^e_{3322}+C^e_{3333}}{3K_s}=1-\frac{\nu_{13}+\nu_{23}\nu_{12}+\nu_{23}+\nu_{21}\nu_{13}+1-\nu_{21}\nu_{12}}{E_1E_2\det{3K_s}} \] For cross anisotropy, let us consider ($e_1$,$e_2$) as the isotropic plane (bedding plane for sedimentary rocks) and $e_3$ the normal to this plane. The subscripts ${\parallel}$ and $\perp$ indicates, respectively, the direction parallel to bedding and perpendicular to bedding. \[{E_1=E_2=E_{\parallel}}\quad , \quad {E_3=E_{\perp}}\] \[\frac{\nu_{\perp\parallel}}{E_{\perp}}=\frac{\nu_{\parallel\perp}}{E_{\parallel}}\] Biot's coefficients become: \[b_{11}=b_{22}=b_{\parallel\parallel}=1-\frac{1+\nu_{\parallel\parallel}+\nu_{\parallel\parallel}\nu_{\perp\parallel}+\nu_{\perp\parallel}}{E_{\parallel}E_{\perp}\det{3K_s}} \\ b_{33}=b_{\perp\perp}=1-\frac{1-\nu_{\parallel\parallel}^2+2\nu_{\parallel\perp}+2\nu_{\parallel\perp}\nu_{\parallel\parallel}}{E_{\parallel}E_{\parallel}\det 3K_s} \\ \det=\frac{1-\nu_{\parallel\parallel}^2-2\nu_{\perp\parallel}\nu_{\parallel\perp}(1+\nu_{\parallel\parallel})}{E_{\parallel}E_{\parallel}E_{\perp}} \] with identical value in the isotropic plane. For isotropy, Biot’s coefficients reduce to $b_{ij}=b\delta_{ij}$ leading to $b_{11}=b_{22}=b_{33}=b$. In that case, use INBIO = 1.

Remark:

• The anisotropy of Biot’s coefficients can only be used with ISOL=9 (see appendix) and ORTHOPLA.
• When variation of the solid density and of the material porosity is taken into account (see flow law), they depend on the mean effective stress $\sigma'=\sigma+b\theta(S_r)p$, which is expressed as a function of a generalized Biot’s coefficient $b=\frac{b_{ii}}{3}$ . The latter is linked to a generalized drained bulk modulus $K_0=\frac{C^e_{iijj}}{9}$ by the relation $b=1-\frac{K_0}{K_s}$.

• Stresses (global axes): $\sigma_x$, $\sigma_y$, $\sigma_{xy}$, $f_x$, $f_y$, $f_{stored}$, 0
• Internal variables: Internal variables of the mechanical law + internal variables of the flow law