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CSOL2
Plane or axisymmetric state
Description
Coupled mechanical-flow analysis in large deformations.
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
Files
Prepro: CSOL2A.F
Lagamine: CSOL2B.F
Input file
| Title (A5) | |
|---|---|
| TITLE | “CSOL2” in the first 5 columns |
| Control data (4I5) | |
| NELEM | Number of elements |
| ISPSMAS | = 0 → Nothing = 1 → Take into account the specific mass if and only if NTANA<0 |
| INSIG | = 0 → No initial stress = 1 or 2 → Initial stresses |
| INBIO | = 0 → No Biot coefficient = 1 → Isotropic Biot coefficient = 2 → Anisotropic Biot coefficient Only for orthotropic mechanical law ORTHOPLA |
| Specific mass in dynamic analysis - Only if ISPMAS = 1 (1G10.0) | |
| SPEMAS | Specific mass |
| Initial stresses - Only if INSIG > 0 (4G10.0) | |
| If INSIG=1: $\sigma_y=\sigma_{y0}+yd\sigma_{y}$ If INSIG=2: $\sigma_y=min(\sigma_{y0}+yd\sigma_y,0)$ |
|
| SIGY0 | $\sigma_{y0}$ effective stress $\sigma_y$ at the axes origin |
| DSIGY | Effective stress gradient along Y axis |
| AK0X | $k_0$ ratio $\sigma_x/\sigma_y$ |
| AK0Z | $k_0$ ratio $\sigma_z/\sigma_y$ (if AK0Z=0, AK0Z=AK0X) |
| The computation of SIGY0 and DSIGY must take into account the apparent specific mass, defined as \[\rho_a'=[(1-n)\rho_s+nS_w\rho_w]-\rho_w\] where: $\rho_s$ is the solid specific mass - this represents the specific mass of a fictive sample where ther is no porosity, i.e. where the grains occupy the whole volume of the sample $\rho_w$ is the fluid specific mass $n$ is the porosity defined in the flow law related to the element $S_w$ fluid saturation, ∈ [0,1] |
|
| Biot coefficient - Only if INBIO > 0 (3G10.0) | |
| If INBIO = 1 | |
| CBIOT | Biot coefficient |
| If INBIO = 2 | |
| CBIOT1 | Biot coefficient $b_{11}$ |
| CBIOT2 | Biot coefficient $b_{22}$ |
| CBIOT3 | Biot coefficient $b_{33}$ |
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:
\[b_{11}=1-\frac{C^e_{1111}+C^e_{1122}+C^e_{1133}}{3K_s}=1-\frac{1-\nu_{23}\nu_{32}+\nu_{21}+\nu_{31}\nu_{23}+\nu_{21}\nu_{32}+\nu_{31}}{E_2E_3det3K_s} \]
