FEM: F-bar in total lagrangian shceme

F-bar method is a element technique used in linear elements to alleviate volumetric locking. This method replaces F of each integration point with F-bar.

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Index notation

Displacement interpolation

u_i=N_{iK}d_k

Deformation gradient is calculated as

F_{iA}=\delta_{iA}+N_{iK,A}d_K=\delta_{iA}+B_{iKA}d_K

where d_K is nodal displacement
Total lagrangian shceme without traction and body force is

\delta U=\int P_{iA}\delta F_{iA}dV

and nodal force

R_K=\int P_{iA}B_{iKA}dV

F-bar method: decomposition

\bar{F}_{iA}=(J_0/J)^{1/3}F_{iA}

where J_0 is determinant of the centriod integration point
Replace all F with F-bar in constitutive law

\bar{P}_{iA}=P_{iA}(\bar{F})

Linearization

K_{UU}=\frac{\partial R_K}{\partial d_M}=\int\frac{\partial\bar{P}_{iA}}{\partial\bar{F}_{jB}}\frac{\partial\bar{F}_{jB}}{\partial d_M}dV

where

\begin{align*}
\frac{\partial\bar{F}_{jB}}{\partial d_M}&=\frac{\partial}{\partial d_M}\left(\frac{J_0}{J}\right)^{1/3}F_{jB}+\left(\frac{J_0}{J}\right)^{1/3}\frac{\partial F_{jB}}{\partial d_M}\\
&=\frac13\left(\frac{J_0}{J}\right)^{-2/3}\left[J^{-1}\frac{\partial J_0}{\partial d_M}-\frac{J_0}{J^2}\frac{\partial J}{\partial d_M}\right]F_{jB}+\left(\frac{J_0}{J}\right)^{1/3}B_{jMB}
\end{align*}

and

\frac{\partial J_0}{\partial d_M}=\frac{\partial J_0}{\partial(F_0)_{kC}}\frac{\partial(F_0)_{kC}}{\partial d_M}=J_0(F^{-1}_0)_{Ck}(B_0)_{kMC}

and similarly

\frac{\partial J}{\partial d_M}=JF^{-1}_{Ck}B_{kMC}

now we will have

\frac{\partial\bar{F}_{jB}}{\partial d_M}=\frac13\left(\frac{J_0}{J}\right)^{1/3}[(F^{-1}_0)_{Ck}(B_0)_{kMC}-F^{-1}_{Ck}B_{kMC}]F_{jB}+\left(\frac{J_0}{J}\right)^{1/3}N_{jMB}

Now we will have consistent tangent

K^{UU}_{KM}=\int\bar{C}_{iAjB}B_{iKA}\left\{\frac13[(F^{-1}_0)_{Ck}(B_0)_{kMC}-F^{-1}_{Ck}B_{kMC}]F_{jB}+B_{jMB}\right\}dV

where

\bar{C}_{iAjB}=(J_0/J)^{1/3}\frac{\partial\bar{P}_{iA}}{\partial\bar{F}_{jB}}

ugly huh?

Matrix form

Let’s define an operator to condense 2 indices in one specific order

O[a_{ij}]=\{a\}

Suppose we’re using row-order:

O[a_{ij}]=O\begin{bmatrix}
a_{11}&a_{12}&a_{13}\\
a_{21}&a_{22}&a_{23}\\
a_{31}&a_{32}&a_{33}\\
\end{bmatrix}\rightarrow\begin{Bmatrix}
a_{11}\\a_{12}\\a_{13}\\a_{21}\\a_{22}\\a_{23}\\a_{31}\\a_{32}\\a_{33}\\
\end{Bmatrix}

You can use whatever order you like, just make it consistent
We condense iA,jB at first

\{u\}=[N]\{d\}

and

\{F\}=[B]\{d\}

now

\frac{\partial\{\bar{F}\}}{\partial\{d\}}=\{F\}\{F^{-1}_0\}^T[B_0]-\{F^{-1}\}^T[B]=[\bar{B}]

Note: \{F^{-1}\}=Q[F^{-T}]
So

K_{UU}=\sum_i[\bar{B}]^T[\bar{C}][\bar{B}]^T(\xi_i)w_i

Here Gaussian integration is applied

Box for F-bar in Q4

1. read coord. and disp.
2. calculate F_0, the deforamtion gradient at central intpt. \xi=0
3. Loop over 4 intpt
   1. read \xi_i, calculate F(\xi), [B(\xi)], [N(\xi)]
   2. calculate \bar{F}=(J_0/J)^{1/3}F
   3. calculate stress and tangent P(\bar{F}),[\bar{C}],[\bar{B}]
   4. calculate residual contribution r_i=[N]^T\bar{P}(\xi_i)w_i
   5. calculate stiffness contribution [K]_i=[B]^T[\bar{C}][\bar{B}]w_i
   6. do [K]=[K]+[K]_i and \{r\}=\{r\}+\{r\}_i