02
Biot strain and deformation gradient ε = U − I 2 o r ε = C − I 2 o r ε = F T F − I 2 \htmlClass{fea-math-bold-italic}{\boldsymbol{\varepsilon}}=\htmlClass{fea-math-bold-italic}{\boldsymbol{U}}-\htmlClass{fea-math-bold-italic}{\boldsymbol{I}}_{2}\qquad\mathit{or}\qquad\htmlClass{fea-math-bold-italic}{\boldsymbol{\varepsilon}}=\sqrt{\htmlClass{fea-math-bold-italic}{\boldsymbol{C}}}-\htmlClass{fea-math-bold-italic}{\boldsymbol{I}}_{2}\qquad\mathit{or}\qquad\htmlClass{fea-math-bold-italic}{\boldsymbol{\varepsilon}}=\sqrt{\htmlClass{fea-math-bold-italic}{\boldsymbol{F}}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{F}}}-\htmlClass{fea-math-bold-italic}{\boldsymbol{I}}_{2} ε = U − I 2 or ε = C − I 2 or ε = F T F − I 2 In the case of small strains, it can be written:
U = C = F T F ≈ [ 1 + ε x γ x y γ x y 1 + ε y ] \htmlClass{fea-math-bold-italic}{\boldsymbol{U}}=\sqrt{\htmlClass{fea-math-bold-italic}{\boldsymbol{C}}}=\sqrt{\htmlClass{fea-math-bold-italic}{\boldsymbol{F}}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{F}}}\approx\begin{bmatrix}1+\varepsilon_x&\gamma_{xy}\\[4pt]\gamma_{xy}&1+\varepsilon_y\end{bmatrix} U = C = F T F ≈ [ 1 + ε x γ x y γ x y 1 + ε y ] ε x , ε y , γ x y \varepsilon_x,\varepsilon_y,\gamma_{xy} ε x , ε y , γ x y are the engineering strains. The deformation gradient is (see Chapter 10 ):
F = I 2 + [ ∂ u ∂ x ∂ u ∂ y ∂ v ∂ x ∂ v ∂ y ] \htmlClass{fea-math-bold-italic}{\boldsymbol{F}}=\htmlClass{fea-math-bold-italic}{\boldsymbol{I}}_{2}+\begin{bmatrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial u}{\partial y}\\[7pt]\dfrac{\partial v}{\partial x}&\dfrac{\partial v}{\partial y}\end{bmatrix} F = I 2 + ∂ x ∂ u ∂ x ∂ v ∂ y ∂ u ∂ y ∂ v 03
Q4 displacement derivatives For the isoparametric quadrilateral finite element, the displacement derivatives are (see Sections 9.4 and 12.2 ):
with:
b = J − 1 H = [ b 11 b 12 b 13 b 14 b 21 b 22 b 23 b 24 ] \htmlClass{fea-math-bold-italic}{\boldsymbol{b}}=\htmlClass{fea-math-bold-italic}{\boldsymbol{J}}^{-1}\htmlClass{fea-math-bold-italic}{\boldsymbol{H}}=\begin{bmatrix}b_{11}&b_{12}&b_{13}&b_{14}\\[4pt]b_{21}&b_{22}&b_{23}&b_{24}\end{bmatrix} b = J − 1 H = [ b 11 b 21 b 12 b 22 b 13 b 23 b 14 b 24 ] u e l \htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el} u e l contains the nodal displacements of the current finite element:
u e l = ( u 1 v 1 u 2 v 2 u 3 v 3 u 4 v 4 ) T \htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}=\begin{pmatrix}u_1&v_1&u_2&v_2&u_3&v_3&u_4&v_4\end{pmatrix}^{T} u e l = ( u 1 v 1 u 2 v 2 u 3 v 3 u 4 v 4 ) T Therefore:
F = I 2 + [ B u x u e l B u y u e l B v x u e l B v y u e l ] \htmlClass{fea-math-bold-italic}{\boldsymbol{F}}=\htmlClass{fea-math-bold-italic}{\boldsymbol{I}}_{2}+\begin{bmatrix}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_x}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}&\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_y}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}\\[4pt]\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_x}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}&\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_y}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}\end{bmatrix} F = I 2 + [ B u x u e l B v x u e l B u y u e l B v y u e l ] and:
ε = ( I 2 + [ B u x u e l B u y u e l B v x u e l B v y u e l ] ) T ( I 2 + [ B u x u e l B u y u e l B v x u e l B v y u e l ] ) − I 2 \htmlClass{fea-math-bold-italic}{\boldsymbol{\varepsilon}}=\sqrt{\left(\htmlClass{fea-math-bold-italic}{\boldsymbol{I}}_{2}+\begin{bmatrix}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_x}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}&\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_y}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}\\[4pt]\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_x}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}&\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_y}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}\end{bmatrix}\right)^{T}\left(\htmlClass{fea-math-bold-italic}{\boldsymbol{I}}_{2}+\begin{bmatrix}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_x}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}&\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_y}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}\\[4pt]\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_x}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}&\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_y}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}\end{bmatrix}\right)}-\htmlClass{fea-math-bold-italic}{\boldsymbol{I}}_{2} ε = ( I 2 + [ B u x u e l B v x u e l B u y u e l B v y u e l ] ) T ( I 2 + [ B u x u e l B v x u e l B u y u e l B v y u e l ] ) − I 2 04
Engineering strain relations It is easy to show that for small quantities:
[ 1 + ε x γ x y γ x y 1 + ε y ] ≈ [ 1 + 1 2 ε x 1 2 γ x y 1 2 γ x y 1 + 1 2 ε y ] \sqrt{\begin{bmatrix}1+\varepsilon_x&\gamma_{xy}\\[3pt]\gamma_{xy}&1+\varepsilon_y\end{bmatrix}}\approx\begin{bmatrix}1+\dfrac12\varepsilon_x&\dfrac12\gamma_{xy}\\[7pt]\dfrac12\gamma_{xy}&1+\dfrac12\varepsilon_y\end{bmatrix} [ 1 + ε x γ x y γ x y 1 + ε y ] ≈ 1 + 2 1 ε x 2 1 γ x y 2 1 γ x y 1 + 2 1 ε y or:
I 2 + [ ε x γ x y γ x y ε y ] ≈ I 2 + 1 2 [ ε x γ x y γ x y ε y ] \sqrt{\htmlClass{fea-math-bold-italic}{\boldsymbol{I}}_{2}+\begin{bmatrix}\varepsilon_x&\gamma_{xy}\\[3pt]\gamma_{xy}&\varepsilon_y\end{bmatrix}}\approx\htmlClass{fea-math-bold-italic}{\boldsymbol{I}}_{2}+\frac12\begin{bmatrix}\varepsilon_x&\gamma_{xy}\\[3pt]\gamma_{xy}&\varepsilon_y\end{bmatrix} I 2 + [ ε x γ x y γ x y ε y ] ≈ I 2 + 2 1 [ ε x γ x y γ x y ε y ] It results:
ε = [ ε x γ x y γ x y ε y ] ≈ 1 2 ( [ B u x u e l B u y u e l B v x u e l B v y u e l ] + [ B u x u e l B u y u e l B v x u e l B v y u e l ] T + [ B u x u e l B u y u e l B v x u e l B v y u e l ] T [ B u x u e l B u y u e l B v x u e l B v y u e l ] ) \htmlClass{fea-math-bold-italic}{\boldsymbol{\varepsilon}}=\begin{bmatrix}\varepsilon_x&\gamma_{xy}\\[3pt]\gamma_{xy}&\varepsilon_y\end{bmatrix}\approx\frac12\left(\begin{bmatrix}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_x}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}&\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_y}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}\\[3pt]\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_x}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}&\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_y}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}\end{bmatrix}+\begin{bmatrix}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_x}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}&\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_y}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}\\[3pt]\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_x}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}&\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_y}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}\end{bmatrix}^{T}+\begin{bmatrix}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_x}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}&\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_y}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}\\[3pt]\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_x}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}&\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_y}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}\end{bmatrix}^{T}\begin{bmatrix}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_x}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}&\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_y}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}\\[3pt]\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_x}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}&\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_y}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}\end{bmatrix}\right) ε = [ ε x γ x y γ x y ε y ] ≈ 2 1 [ B u x u e l B v x u e l B u y u e l B v y u e l ] + [ B u x u e l B v x u e l B u y u e l B v y u e l ] T + [ B u x u e l B v x u e l B u y u e l B v y u e l ] T [ B u x u e l B v x u e l B u y u e l B v y u e l ] Or in engineering notation:
ε = { ε x ε y γ x y } ≈ { B u x u e l + 1 2 u e l T ( B u x T B u x + B v x T B v x ) u e l B v y u e l + 1 2 u e l T ( B u y T B u y + B v y T B v y ) u e l B u y u e l + B v x u e l + 1 2 u e l T ( B u x T B u y + B u y T B u x + B v x T B v y + B v y T B v x ) u e l } \htmlClass{fea-math-bold-italic}{\boldsymbol{\varepsilon}}=\begin{Bmatrix}\varepsilon_x\\[3pt]\varepsilon_y\\[3pt]\gamma_{xy}\end{Bmatrix}\approx\begin{Bmatrix}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_x}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}+\dfrac12\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}^{T}\left(\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_x}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_x}+\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_x}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_x}\right)\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}\\[7pt]\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_y}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}+\dfrac12\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}^{T}\left(\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_y}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_y}+\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_y}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_y}\right)\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}\\[7pt]\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_y}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}+\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_x}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}+\dfrac12\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}^{T}\left(\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_x}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_y}+\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_y}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_x}+\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_x}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_y}+\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_y}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_x}\right)\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}\end{Bmatrix} ε = ⎩ ⎨ ⎧ ε x ε y γ x y ⎭ ⎬ ⎫ ≈ ⎩ ⎨ ⎧ B u x u e l + 2 1 u e l T ( B u x T B u x + B v x T B v x ) u e l B v y u e l + 2 1 u e l T ( B u y T B u y + B v y T B v y ) u e l B u y u e l + B v x u e l + 2 1 u e l T ( B u x T B u y + B u y T B u x + B v x T B v y + B v y T B v x ) u e l ⎭ ⎬ ⎫ These equations are the same as those obtained in Section 12.2 :
ε = { ε x ε y γ x y } ≈ { B u x u e l + 1 2 u e l T G x u e l B v y u e l + 1 2 u e l T G y u e l B u y u e l + B v x u e l + 1 2 u e l T G x y u e l } \htmlClass{fea-math-bold-italic}{\boldsymbol{\varepsilon}}=\begin{Bmatrix}\varepsilon_x\\[3pt]\varepsilon_y\\[3pt]\gamma_{xy}\end{Bmatrix}\approx\begin{Bmatrix}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_x}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}+\dfrac12\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{G}}_x\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}\\[7pt]\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_y}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}+\dfrac12\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{G}}_y\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}\\[7pt]\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_y}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}+\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_x}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}+\dfrac12\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{G}}_{xy}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}\end{Bmatrix} ε = ⎩ ⎨ ⎧ ε x ε y γ x y ⎭ ⎬ ⎫ ≈ ⎩ ⎨ ⎧ B u x u e l + 2 1 u e l T G x u e l B v y u e l + 2 1 u e l T G y u e l B u y u e l + B v x u e l + 2 1 u e l T G x y u e l ⎭ ⎬ ⎫ where:
{ G x = B u x T B u x + B v x T B v x G y = B u y T B u y + B v y T B v y G x y = B u x T B u y + B u y T B u x + B v x T B v y + B v y T B v x \left\{\begin{aligned}\htmlClass{fea-math-bold-italic}{\boldsymbol{G}}_x&=\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_x}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_x}+\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_x}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_x}\\[4pt]\htmlClass{fea-math-bold-italic}{\boldsymbol{G}}_y&=\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_y}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_y}+\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_y}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_y}\\[4pt]\htmlClass{fea-math-bold-italic}{\boldsymbol{G}}_{xy}&=\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_x}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_y}+\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_y}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_x}+\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_x}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_y}+\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_y}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_x}\end{aligned}\right. ⎩ ⎨ ⎧ G x G y G x y = B u x T B u x + B v x T B v x = B u y T B u y + B v y T B v y = B u x T B u y + B u y T B u x + B v x T B v y + B v y T B v x G x , G y , G x y \htmlClass{fea-math-bold-italic}{\boldsymbol{G}}_x,\htmlClass{fea-math-bold-italic}{\boldsymbol{G}}_y,\htmlClass{fea-math-bold-italic}{\boldsymbol{G}}_{xy} G x , G y , G x y are symmetric 8 × 8 8\times8 8 × 8 matrices.
It is worth noting that the equations containing the matrices G x \htmlClass{fea-math-bold-italic}{\boldsymbol{G}}_x G x , G y \htmlClass{fea-math-bold-italic}{\boldsymbol{G}}_y G y , and G x y \htmlClass{fea-math-bold-italic}{\boldsymbol{G}}_{xy} G x y are identical to those obtained in Chapter 12 for the Green–Lagrange strain formulation.
05
Improved stiff subprogram Only a few changes must be made in the stiff subprogram given in Section 17.2 . The lines shown in red were added or modified in the new stiff subprogram, as illustrated below:
Figure 1. MATLAB modifications introduced in stiff.Recall that the following equation (see the blue line in Figure 1) is exact:
f e l = h 1 ∑ n G B T σ d e t ( J ) \htmlClass{fea-math-bold-italic}{\boldsymbol{f}}_{el}=h_1\sum_{n_G}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{\sigma}}\mathop{\mathit{det}}\left(\htmlClass{fea-math-bold-italic}{\boldsymbol{J}}\right) f e l = h 1 n G ∑ B T σ det ( J ) h 1 h_1 h 1 is the current finite-element thickness, which accounts for the thickness modification due to transverse contraction (see Section 15.1 ):
h ⟶ h 1 = [ 1 − ν 1 − ν ( ε x + ε y ) ] h h\longrightarrow h_1=\left[1-\frac{\nu}{1-\nu}\left(\varepsilon_x+\varepsilon_y\right)\right]h h ⟶ h 1 = [ 1 − 1 − ν ν ( ε x + ε y ) ] h The strain–displacement matrix B \htmlClass{fea-math-bold-italic}{\boldsymbol{B}} B is the same matrix derived in Section 17.2 and is computed, for each element and Gauss point, by the same B_matrix subroutine. In the program, B_matrix is called before the computation of the element internal force vector f e l \htmlClass{fea-math-bold-italic}{\boldsymbol{f}}_{el} f e l and tangent stiffness matrix k e l \htmlClass{fea-math-bold-italic}{\boldsymbol{k}}_{el} k e l . The improvement introduced here concerns the tangent stiffness matrix, not the definition or computation of B \htmlClass{fea-math-bold-italic}{\boldsymbol{B}} B .
Compared with the expression used in Section 17.2 , the tangent stiffness matrix becomes:
k e l = h 1 ∑ n G B T D B d e t ( J ) ⟶ k e l = h 1 ∑ n G ( B T D B + σ x G x + σ y G y + τ x y G x y ) d e t ( J ) \htmlClass{fea-math-bold-italic}{\boldsymbol{k}}_{el}=h_1\sum_{n_G}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{D}}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}\mathop{\mathit{det}}\left(\htmlClass{fea-math-bold-italic}{\boldsymbol{J}}\right)\quad\longrightarrow\quad\htmlClass{fea-math-bold-italic}{\boldsymbol{k}}_{el}=h_1\sum_{n_G}\left(\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{D}}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}+\sigma_x\htmlClass{fea-math-bold-italic}{\boldsymbol{G}}_x+\sigma_y\htmlClass{fea-math-bold-italic}{\boldsymbol{G}}_y+\tau_{xy}\htmlClass{fea-math-bold-italic}{\boldsymbol{G}}_{xy}\right)\mathop{\mathit{det}}\left(\htmlClass{fea-math-bold-italic}{\boldsymbol{J}}\right) k e l = h 1 n G ∑ B T D B det ( J ) ⟶ k e l = h 1 n G ∑ ( B T D B + σ x G x + σ y G y + τ x y G x y ) det ( J ) Therefore, the additional geometric terms significantly improve convergence.
Everything explained in this section also applies to Hencky strains because the strains are small:
E H = l n U ≈ U − I 2 = E B \htmlClass{fea-math-bold-italic}{\boldsymbol{E}}_{H}=\mathop{\mathit{ln}}\htmlClass{fea-math-bold-italic}{\boldsymbol{U}}\approx\htmlClass{fea-math-bold-italic}{\boldsymbol{U}}-\htmlClass{fea-math-bold-italic}{\boldsymbol{I}}_{2}=\htmlClass{fea-math-bold-italic}{\boldsymbol{E}}_{B} E H = ln U ≈ U − I 2 = E B 06
Strain update and reference-configuration correction During the iterations corresponding to load step i + 1 i+1 i + 1 , the strains are updated:
e i + 1 ≈ e i + Δ e \htmlClass{fea-math-bold-italic}{\boldsymbol{e}}_{i+1}\approx\htmlClass{fea-math-bold-italic}{\boldsymbol{e}}_{i}+\mathit{\Delta}\htmlClass{fea-math-bold-italic}{\boldsymbol{e}} e i + 1 ≈ e i + Δ e Neither the Hencky strain tensor nor the Biot strain tensor is generally additive under finite successive deformations. In the present small-strain UL formulation, the update e i + 1 ≈ e i + Δ e \htmlClass{fea-math-bold-italic}{\boldsymbol{e}}_{i+1}\approx\htmlClass{fea-math-bold-italic}{\boldsymbol{e}}_{i}+\mathit{\Delta}\htmlClass{fea-math-bold-italic}{\boldsymbol{e}} e i + 1 ≈ e i + Δ e is therefore only an incremental approximation.
A formulation that avoids this approximate additive accumulation by updating the deformation gradient multiplicatively is presented in Section 17.4 : UL formulation with multiplicative update of the deformation gradient .
The terms e i \htmlClass{fea-math-bold-italic}{\boldsymbol{e}}_{i} e i and Δ e \mathit{\Delta}\htmlClass{fea-math-bold-italic}{\boldsymbol{e}} Δ e are calculated for the configuration at load step i i i , while e i + 1 \htmlClass{fea-math-bold-italic}{\boldsymbol{e}}_{i+1} e i + 1 must be calculated for the slightly different configuration corresponding to load step i + 1 i+1 i + 1 . This error decreases as the number of load steps increases. Another method for correcting this error is described in Section 15.2 .
To account for the fact that e i \htmlClass{fea-math-bold-italic}{\boldsymbol{e}}_{i} e i and Δ e \mathit{\Delta}\htmlClass{fea-math-bold-italic}{\boldsymbol{e}} Δ e are evaluated with respect to slightly different configurations, the strain tensor is transferred to the current configuration before the stresses are evaluated. The stiff subroutine therefore includes the following correction:
e i + 1 = F − T ( e i + Δ e ) F − 1 \htmlClass{fea-math-bold-italic}{\boldsymbol{e}}_{i+1}=\htmlClass{fea-math-bold-italic}{\boldsymbol{F}}^{-T}\left(\htmlClass{fea-math-bold-italic}{\boldsymbol{e}}_{i}+\mathit{\Delta}\htmlClass{fea-math-bold-italic}{\boldsymbol{e}}\right)\htmlClass{fea-math-bold-italic}{\boldsymbol{F}}^{-1} e i + 1 = F − T ( e i + Δ e ) F − 1 The deformation gradient F \htmlClass{fea-math-bold-italic}{\boldsymbol{F}} F is already calculated to obtain the right stretch tensor U \htmlClass{fea-math-bold-italic}{\boldsymbol{U}} U ; see the green line in Figure 1. In the final stiff subprogram, the red lines shown below are included:
Figure 2. MATLAB strain transformation included in stiff.As in Section 17.2 , the same stiff subroutine can be used with either Hencky or Biot strains by setting istrain=1 or istrain=2.
07
Numerical example Example. Consider a cantilever beam of length L = 400 m m L=400\ \mathit{mm} L = 400 mm , with a rectangular cross-section of thickness 5 m m 5\ \mathit{mm} 5 mm and height 20 m m 20\ \mathit{mm} 20 mm . The material properties are E = 1000 M P a E=1000\ \mathit{MPa} E = 1000 MPa and ν = 0.3 \nu=0.3 ν = 0.3 . The free end is loaded by a force F = 100 N F=100\ \mathit{N} F = 100 N . The mesh contains 60 × 8 = 480 60\times8=480 60 × 8 = 480 4-node isoparametric elements and 549 nodes. The load is applied in 20 steps.
The table below shows that the reference-configuration correction significantly improves the results, even for a small number of load steps.
UL + Hencky strain tensor without reference-configuration correction UL + Hencky strain tensor with reference-configuration correction ANSYS 20 load steps v m a x = − 287.12 m m u m a x = 2.25 m m u m i n = − 157.99 m m \begin{aligned}v_{\mathop{\mathit{max}}}&=-287.12\ \mathit{mm}\\u_{\mathop{\mathit{max}}}&=2.25\ \mathit{mm}\\u_{\mathop{\mathit{min}}}&=-157.99\ \mathit{mm}\end{aligned} v max u max u min = − 287.12 mm = 2.25 mm = − 157.99 mm 20 load steps v m a x = − 286.61 m m u m a x = 2.44 m m u m i n = − 155.76 m m \begin{aligned}v_{\mathop{\mathit{max}}}&=-286.61\ \mathit{mm}\\u_{\mathop{\mathit{max}}}&=2.44\ \mathit{mm}\\u_{\mathop{\mathit{min}}}&=-155.76\ \mathit{mm}\end{aligned} v max u max u min = − 286.61 mm = 2.44 mm = − 155.76 mm v m a x = − 286.32 m m u m a x = 2.35 m m u m i n = − 155.68 m m \begin{aligned}v_{\mathop{\mathit{max}}}&=-286.32\ \mathit{mm}\\u_{\mathop{\mathit{max}}}&=2.35\ \mathit{mm}\\u_{\mathop{\mathit{min}}}&=-155.68\ \mathit{mm}\end{aligned} v max u max u min = − 286.32 mm = 2.35 mm = − 155.68 mm 5 load steps v m a x = − 288.94 m m u m a x = 2.23 m m u m i n = − 162.32 m m \begin{aligned}v_{\mathop{\mathit{max}}}&=-288.94\ \mathit{mm}\\u_{\mathop{\mathit{max}}}&=2.23\ \mathit{mm}\\u_{\mathop{\mathit{min}}}&=-162.32\ \mathit{mm}\end{aligned} v max u max u min = − 288.94 mm = 2.23 mm = − 162.32 mm 5 load steps v m a x = − 286.49 m m u m a x = 2.47 m m u m i n = − 155.39 m m \begin{aligned}v_{\mathop{\mathit{max}}}&=-286.49\ \mathit{mm}\\u_{\mathop{\mathit{max}}}&=2.47\ \mathit{mm}\\u_{\mathop{\mathit{min}}}&=-155.39\ \mathit{mm}\end{aligned} v max u max u min = − 286.49 mm = 2.47 mm = − 155.39 mm
Figure 3. Vertical displacement obtained with the reference-configuration correction and five load steps.08
Programs and download The modified MATLAB programs can be downloaded below.
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