02
Initial configuration and Jacobian The Jacobian is:
J ( s , t ) = [ ∂ x ∂ s ∂ y ∂ s ∂ x ∂ t ∂ y ∂ t ] = H ( s , t ) [ x 1 y 1 x 2 y 2 x 3 y 3 x 4 y 4 ] \htmlClass{fea-math-bold-italic}{\boldsymbol{J}}(s,t)=\begin{bmatrix}\dfrac{\partial x}{\partial s}&\dfrac{\partial y}{\partial s}\\[4pt]\dfrac{\partial x}{\partial t}&\dfrac{\partial y}{\partial t}\end{bmatrix}=\htmlClass{fea-math-bold-italic}{\boldsymbol{H}}(s,t)\begin{bmatrix}x_1&y_1\\x_2&y_2\\x_3&y_3\\x_4&y_4\end{bmatrix} J ( s , t ) = ∂ s ∂ x ∂ t ∂ x ∂ s ∂ y ∂ t ∂ y = H ( s , t ) x 1 x 2 x 3 x 4 y 1 y 2 y 3 y 4 H ( s , t ) = 1 4 [ 1 + t − ( 1 + t ) − ( 1 − t ) 1 − t 1 + s 1 − s − ( 1 − s ) − ( 1 + s ) ] \htmlClass{fea-math-bold-italic}{\boldsymbol{H}}(s,t)=\frac14\begin{bmatrix}1+t&-(1+t)&-(1-t)&1-t\\1+s&1-s&-(1-s)&-(1+s)\end{bmatrix} H ( s , t ) = 4 1 [ 1 + t 1 + s − ( 1 + t ) 1 − s − ( 1 − t ) − ( 1 − s ) 1 − t − ( 1 + s ) ] x 1 , … , x 4 x_1,\ldots,x_4 x 1 , … , x 4 and y 1 , … , y 4 y_1,\ldots,y_4 y 1 , … , y 4 are the nodal coordinates of the finite element in the initial configuration. The physical area transformation is:
d A = ∣ d e t J ∣ d s d t dA=\left|\mathop{\mathit{det}}\htmlClass{fea-math-bold-italic}{\boldsymbol{J}}\right|\,ds\,dt d A = ∣ det J ∣ d s d t The derivatives of the shape functions with respect to the initial Cartesian coordinates are obtained using the inverse Jacobian; see Section 9.4 :
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}\\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 ] The four displacement-derivative relations are:
The element nodal displacement vector is:
u e l = { u 1 v 1 u 2 v 2 u 3 v 3 u 4 v 4 } \htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}=\begin{Bmatrix}u_1\\v_1\\u_2\\v_2\\u_3\\v_3\\u_4\\v_4\end{Bmatrix} u e l = ⎩ ⎨ ⎧ u 1 v 1 u 2 v 2 u 3 v 3 u 4 v 4 ⎭ ⎬ ⎫ 03
Green–Lagrange strains The two-dimensional Green–Lagrange strain components are:
ε x = ∂ u ∂ x + 1 2 ( ∂ u ∂ x ) 2 + 1 2 ( ∂ v ∂ x ) 2 \varepsilon_x=\frac{\partial u}{\partial x}+\frac12\left(\frac{\partial u}{\partial x}\right)^2+\frac12\left(\frac{\partial v}{\partial x}\right)^2 ε x = ∂ x ∂ u + 2 1 ( ∂ x ∂ u ) 2 + 2 1 ( ∂ x ∂ v ) 2 ε y = ∂ v ∂ y + 1 2 ( ∂ u ∂ y ) 2 + 1 2 ( ∂ v ∂ y ) 2 \varepsilon_y=\frac{\partial v}{\partial y}+\frac12\left(\frac{\partial u}{\partial y}\right)^2+\frac12\left(\frac{\partial v}{\partial y}\right)^2 ε y = ∂ y ∂ v + 2 1 ( ∂ y ∂ u ) 2 + 2 1 ( ∂ y ∂ v ) 2 γ x y = ∂ u ∂ y + ∂ v ∂ x + ∂ u ∂ x ∂ u ∂ y + ∂ v ∂ x ∂ v ∂ y \gamma_{xy}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}+\frac{\partial u}{\partial x}\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}\frac{\partial v}{\partial y} γ x y = ∂ y ∂ u + ∂ x ∂ v + ∂ x ∂ u ∂ y ∂ u + ∂ x ∂ v ∂ y ∂ v After introducing the displacement interpolation, the strains become:
ε x = B u x u e l + 1 2 u e l T G x u e l \varepsilon_x=\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_x}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}+\frac12\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} ε x = B u x u e l + 2 1 u e l T G x u e l ε y = B v y u e l + 1 2 u e l T G y u e l \varepsilon_y=\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_y}\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}+\frac12\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} ε y = B v y u e l + 2 1 u e l T G y u e l γ x y = ( B u y + B v x ) u e l + 1 2 u e l T G x y u e l \gamma_{xy}=(\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_y}+\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_x})\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}+\frac12\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} γ x y = ( B u y + B v x ) u e l + 2 1 u e l T G x y u e l The symmetric 8 × 8 8\times8 8 × 8 matrices are:
G x = B u x T B u x + B v x T B v x \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} 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 \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} 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 \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} 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 04
Strain-displacement matrix The matrix B \htmlClass{fea-math-bold-italic}{\boldsymbol B} B , relating an increment of the Green–Lagrange strain vector to an increment of the element nodal displacement vector, is defined by:
d ε e l = B d u e l d\htmlClass{fea-math-bold-italic}{\boldsymbol{\varepsilon}}_{el}=\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}\,d\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el} d ε e l = B d u e l B = ∂ ε e l ∂ u e l \htmlClass{fea-math-bold-italic}{\boldsymbol B}=\frac{\partial\htmlClass{fea-math-bold-italic}{\boldsymbol{\varepsilon}}_{el}}{\partial\htmlClass{fea-math-bold-italic}{\boldsymbol u}_{el}} B = ∂ u e l ∂ ε e l Component by component:
d ε x = ( B u x + u e l T G x ) d u e l d\varepsilon_x=\left(\htmlClass{fea-math-bold-italic}{\boldsymbol B}_{u_x}+\htmlClass{fea-math-bold-italic}{\boldsymbol u}_{el}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol G}_x\right)d\htmlClass{fea-math-bold-italic}{\boldsymbol u}_{el} d ε x = ( B u x + u e l T G x ) d u e l d ε y = ( B v y + u e l T G y ) d u e l d\varepsilon_y=\left(\htmlClass{fea-math-bold-italic}{\boldsymbol B}_{v_y}+\htmlClass{fea-math-bold-italic}{\boldsymbol u}_{el}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol G}_y\right)d\htmlClass{fea-math-bold-italic}{\boldsymbol u}_{el} d ε y = ( B v y + u e l T G y ) d u e l d γ x y = ( B u y + B v x + u e l T G x y ) d u e l d\gamma_{xy}=\left(\htmlClass{fea-math-bold-italic}{\boldsymbol B}_{u_y}+\htmlClass{fea-math-bold-italic}{\boldsymbol B}_{v_x}+\htmlClass{fea-math-bold-italic}{\boldsymbol u}_{el}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol G}_{xy}\right)d\htmlClass{fea-math-bold-italic}{\boldsymbol u}_{el} d γ x y = ( B u y + B v x + u e l T G x y ) d u e l { d ε x d ε y d γ x y } = [ B u x + u e l T G x B v y + u e l T G y B u y + B v x + u e l T G x y ] ⏟ B d u e l \begin{Bmatrix}d\varepsilon_x\\d\varepsilon_y\\d\gamma_{xy}\end{Bmatrix}=\underbrace{\begin{bmatrix}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_x}+\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{G}}_x\\\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_y}+\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{G}}_y\\\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_y}+\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_x}+\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{G}}_{xy}\end{bmatrix}}_{\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}}d\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el} ⎩ ⎨ ⎧ d ε x d ε y d γ x y ⎭ ⎬ ⎫ = B B u x + u e l T G x B v y + u e l T G y B u y + B v x + u e l T G x y d u e l The matrix is split into a constant part, identical to linear theory, and a nonlinear displacement-dependent part:
B ( u e l ) = B 0 + B L ( u e l ) \htmlClass{fea-math-bold-italic}{\boldsymbol{B}}(\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el})=\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_0+\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_L(\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}) B ( u e l ) = B 0 + B L ( u e l ) [ B u x + u e l T G x B v y + u e l T G y B u y + B v x + u e l T G x y ] ⏟ B = [ B u x B v y B u y + B v x ] ⏟ B 0 + [ u e l T G x u e l T G y u e l T G x y ] ⏟ B L \underbrace{\begin{bmatrix}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_x}+\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{G}}_x\\\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_y}+\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{G}}_y\\\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_y}+\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_x}+\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{G}}_{xy}\end{bmatrix}}_{\htmlClass{fea-math-bold-italic}{\boldsymbol B}}=\underbrace{\begin{bmatrix}\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_x}\\\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_y}\\\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{u_y}+\htmlClass{fea-math-bold-italic}{\boldsymbol{B}}_{v_x}\end{bmatrix}}_{\htmlClass{fea-math-bold-italic}{\boldsymbol B}_0}+\underbrace{\begin{bmatrix}\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}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{G}}_y\\\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}_{el}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{G}}_{xy}\end{bmatrix}}_{\htmlClass{fea-math-bold-italic}{\boldsymbol B}_L} B B u x + u e l T G x B v y + u e l T G y B u y + B v x + u e l T G x y = B 0 B u x B v y B u y + B v x + B L u e l T G x u e l T G y u e l T G x y B 0 \htmlClass{fea-math-bold-italic}{\boldsymbol B}_0 B 0 is the constant part, identical to linear theory, and B L \htmlClass{fea-math-bold-italic}{\boldsymbol B}_L B L is the nonlinear displacement-dependent part.
05
Potential energy and Gauss quadrature The total potential energy is:
Π = U + V = 1 2 A e l = 1 n e l ( ∫ A e l h ε T σ d A ) − u T F \mathit{\Pi}=U+V=\frac12\mathcal{A}_{el=1}^{n_{el}}\left(\int_{A_{el}}h\,\htmlClass{fea-math-bold-italic}{\boldsymbol{\varepsilon}}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{\sigma}}\,dA\right)-\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{F}} Π = U + V = 2 1 A e l = 1 n e l ( ∫ A e l h ε T σ d A ) − u T F Here, V = − u T F V=-\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol F} V = − u T F is the potential of the external loads, u \htmlClass{fea-math-bold-italic}{\boldsymbol u} u is the global nodal displacement vector, and F \htmlClass{fea-math-bold-italic}{\boldsymbol F} F is the global force vector. In natural coordinates:
Π = U + V = 1 2 A e l = 1 n e l ( ∫ − 1 1 ∫ − 1 1 h ( s , t ) ε T ( s , t ) σ ( s , t ) ∣ d e t J ( s , t ) ∣ d s d t ) − u T F \mathit{\Pi}=U+V=\frac12\mathcal{A}_{el=1}^{n_{el}}\left(\int_{-1}^{1}\int_{-1}^{1}h(s,t)\,\htmlClass{fea-math-bold-italic}{\boldsymbol{\varepsilon}}^{T}(s,t)\htmlClass{fea-math-bold-italic}{\boldsymbol{\sigma}}(s,t)\,\left|\mathop{\mathit{det}}\htmlClass{fea-math-bold-italic}{\boldsymbol J}(s,t)\right|\,ds\,dt\right)-\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol F} Π = U + V = 2 1 A e l = 1 n e l ( ∫ − 1 1 ∫ − 1 1 h ( s , t ) ε T ( s , t ) σ ( s , t ) ∣ det J ( s , t ) ∣ d s d t ) − u T F Using 2 × 2 2\times2 2 × 2 Gauss numerical quadrature:
Π = U + V = 1 2 A e l = 1 n e l ( ∑ n G h ( s G , t G ) ε T ( s G , t G ) σ ( s G , t G ) ∣ d e t J ( s G , t G ) ∣ ) − u T F \mathit{\Pi}=U+V=\frac12\mathcal{A}_{el=1}^{n_{el}}\left(\sum_{n_G}h(s_G,t_G)\,\htmlClass{fea-math-bold-italic}{\boldsymbol{\varepsilon}}^{T}(s_G,t_G)\htmlClass{fea-math-bold-italic}{\boldsymbol{\sigma}}(s_G,t_G)\,\left|\mathop{\mathit{det}}\htmlClass{fea-math-bold-italic}{\boldsymbol J}(s_G,t_G)\right|\right)-\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol F} Π = U + V = 2 1 A e l = 1 n e l ( n G ∑ h ( s G , t G ) ε T ( s G , t G ) σ ( s G , t G ) ∣ det J ( s G , t G ) ∣ ) − u T F The integration uses four Gauss points, two in each natural-coordinate direction. For the 2 × 2 2\times2 2 × 2 Gauss rule, all four weights are equal to 1 and are therefore omitted from the following expressions. The four-point sum is:
∑ n G E ( s G , t G ) = E ( 1 3 , 1 3 ) + E ( 1 3 , − 1 3 ) + E ( − 1 3 , 1 3 ) + E ( − 1 3 , − 1 3 ) \sum_{n_G}E(s_G,t_G)=E\left(\frac{1}{\sqrt3},\frac{1}{\sqrt3}\right)+E\left(\frac{1}{\sqrt3},-\frac{1}{\sqrt3}\right)+E\left(-\frac{1}{\sqrt3},\frac{1}{\sqrt3}\right)+E\left(-\frac{1}{\sqrt3},-\frac{1}{\sqrt3}\right) n G ∑ E ( s G , t G ) = E ( 3 1 , 3 1 ) + E ( 3 1 , − 3 1 ) + E ( − 3 1 , 3 1 ) + E ( − 3 1 , − 3 1 ) where the integrand is:
E ( s , t ) = h ( s , t ) ε T ( s , t ) σ ( s , t ) ∣ d e t J ( s , t ) ∣ E(s,t)=h(s,t)\,\htmlClass{fea-math-bold-italic}{\boldsymbol{\varepsilon}}^{T}(s,t)\htmlClass{fea-math-bold-italic}{\boldsymbol{\sigma}}(s,t)\,\left|\mathop{\mathit{det}}\htmlClass{fea-math-bold-italic}{\boldsymbol J}(s,t)\right| E ( s , t ) = h ( s , t ) ε T ( s , t ) σ ( s , t ) ∣ det J ( s , t ) ∣ 06
Virtual work and equilibrium The virtual-work equation is:
δ Π = A e l = 1 n e l ( ∑ n G h δ ε T σ ∣ d e t J ∣ ) − δ u T F = 0 \delta\mathit{\Pi}=\mathcal{A}_{el=1}^{n_{el}}\left(\sum_{n_G}h\,\delta\htmlClass{fea-math-bold-italic}{\boldsymbol{\varepsilon}}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{\sigma}}\,\left|\mathop{\mathit{det}}\htmlClass{fea-math-bold-italic}{\boldsymbol J}\right|\right)-\delta\htmlClass{fea-math-bold-italic}{\boldsymbol{u}}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol F}=0 δ Π = A e l = 1 n e l ( n G ∑ h δ ε T σ ∣ det J ∣ ) − δ u T F = 0 This relation holds for every geometrically admissible virtual displacement field. Therefore, the nonlinear equilibrium system is:
Ψ ( u ) = A e l = 1 n e l ( ∑ n G h B T σ ∣ d e t J ∣ ) − F = 0 \htmlClass{fea-math-bold-italic}{\boldsymbol{\mathit{\Psi}}}(\htmlClass{fea-math-bold-italic}{\boldsymbol u})=\mathcal{A}_{el=1}^{n_{el}}\left(\sum_{n_G}h\,\htmlClass{fea-math-bold-italic}{\boldsymbol B}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{\sigma}}\,\left|\mathop{\mathit{det}}\htmlClass{fea-math-bold-italic}{\boldsymbol J}\right|\right)-\htmlClass{fea-math-bold-italic}{\boldsymbol F}=\htmlClass{fea-math-bold-italic}{\boldsymbol0} Ψ ( u ) = A e l = 1 n e l ( n G ∑ h B T σ ∣ det J ∣ ) − F = 0 Equivalently, using the split of the strain-displacement matrix:
Ψ ( u ) = A e l = 1 n e l ( ∑ n G h [ B 0 T + B L T ( u e l ) ] σ ∣ d e t J ∣ ) − F = 0 \htmlClass{fea-math-bold-italic}{\boldsymbol{\mathit{\Psi}}}(\htmlClass{fea-math-bold-italic}{\boldsymbol u})=\mathcal{A}_{el=1}^{n_{el}}\left(\sum_{n_G}h\left[\htmlClass{fea-math-bold-italic}{\boldsymbol B}_0^{T}+\htmlClass{fea-math-bold-italic}{\boldsymbol B}_L^{T}(\htmlClass{fea-math-bold-italic}{\boldsymbol u}_{el})\right]\htmlClass{fea-math-bold-italic}{\boldsymbol{\sigma}}\,\left|\mathop{\mathit{det}}\htmlClass{fea-math-bold-italic}{\boldsymbol J}\right|\right)-\htmlClass{fea-math-bold-italic}{\boldsymbol F}=\htmlClass{fea-math-bold-italic}{\boldsymbol0} Ψ ( u ) = A e l = 1 n e l ( n G ∑ h [ B 0 T + B L T ( u e l ) ] σ ∣ det J ∣ ) − F = 0 07
Tangent stiffness matrix The first variation of the equilibrium residual is:
δ Ψ = A e l = 1 n e l ( ∑ n G h δ B T σ ∣ d e t J ∣ ) + A e l = 1 n e l ( ∑ n G h B T δ σ ∣ d e t J ∣ ) = K T δ u \delta\htmlClass{fea-math-bold-italic}{\boldsymbol{\mathit{\Psi}}}=\mathcal{A}_{el=1}^{n_{el}}\left(\sum_{n_G}h\,\delta\htmlClass{fea-math-bold-italic}{\boldsymbol B}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{\sigma}}\,\left|\mathop{\mathit{det}}\htmlClass{fea-math-bold-italic}{\boldsymbol J}\right|\right)+\mathcal{A}_{el=1}^{n_{el}}\left(\sum_{n_G}h\,\htmlClass{fea-math-bold-italic}{\boldsymbol B}^{T}\delta\htmlClass{fea-math-bold-italic}{\boldsymbol{\sigma}}\,\left|\mathop{\mathit{det}}\htmlClass{fea-math-bold-italic}{\boldsymbol J}\right|\right)=\htmlClass{fea-math-bold-italic}{\boldsymbol{K}}_T\delta\htmlClass{fea-math-bold-italic}{\boldsymbol u} δ Ψ = A e l = 1 n e l ( n G ∑ h δ B T σ ∣ det J ∣ ) + A e l = 1 n e l ( n G ∑ h B T δ σ ∣ det J ∣ ) = K T δ u Using B = B 0 + B L ( u e l ) \htmlClass{fea-math-bold-italic}{\boldsymbol B}=\htmlClass{fea-math-bold-italic}{\boldsymbol B}_0+\htmlClass{fea-math-bold-italic}{\boldsymbol B}_L(\htmlClass{fea-math-bold-italic}{\boldsymbol u}_{el}) B = B 0 + B L ( u e l ) , the same variation is:
δ Ψ = A e l = 1 n e l ( ∑ n G h δ [ B 0 T + B L T ( u e l ) ] σ ∣ d e t J ∣ ) + A e l = 1 n e l ( ∑ n G h [ B 0 T + B L T ( u e l ) ] δ σ ∣ d e t J ∣ ) = K T δ u \delta\htmlClass{fea-math-bold-italic}{\boldsymbol{\mathit{\Psi}}}=\mathcal{A}_{el=1}^{n_{el}}\left(\sum_{n_G}h\,\delta\left[\htmlClass{fea-math-bold-italic}{\boldsymbol B}_0^{T}+\htmlClass{fea-math-bold-italic}{\boldsymbol B}_L^{T}(\htmlClass{fea-math-bold-italic}{\boldsymbol u}_{el})\right]\htmlClass{fea-math-bold-italic}{\boldsymbol{\sigma}}\,\left|\mathop{\mathit{det}}\htmlClass{fea-math-bold-italic}{\boldsymbol J}\right|\right)+\mathcal{A}_{el=1}^{n_{el}}\left(\sum_{n_G}h\left[\htmlClass{fea-math-bold-italic}{\boldsymbol B}_0^{T}+\htmlClass{fea-math-bold-italic}{\boldsymbol B}_L^{T}(\htmlClass{fea-math-bold-italic}{\boldsymbol u}_{el})\right]\delta\htmlClass{fea-math-bold-italic}{\boldsymbol{\sigma}}\,\left|\mathop{\mathit{det}}\htmlClass{fea-math-bold-italic}{\boldsymbol J}\right|\right)=\htmlClass{fea-math-bold-italic}{\boldsymbol{K}}_T\delta\htmlClass{fea-math-bold-italic}{\boldsymbol u} δ Ψ = A e l = 1 n e l ( n G ∑ h δ [ B 0 T + B L T ( u e l ) ] σ ∣ det J ∣ ) + A e l = 1 n e l ( n G ∑ h [ B 0 T + B L T ( u e l ) ] δ σ ∣ det J ∣ ) = K T δ u Because B 0 \htmlClass{fea-math-bold-italic}{\boldsymbol B}_0 B 0 is constant:
δ B = δ ( B 0 + B L ) = δ B L \delta\htmlClass{fea-math-bold-italic}{\boldsymbol B}=\delta(\htmlClass{fea-math-bold-italic}{\boldsymbol B}_0+\htmlClass{fea-math-bold-italic}{\boldsymbol B}_L)=\delta\htmlClass{fea-math-bold-italic}{\boldsymbol B}_L δ B = δ ( B 0 + B L ) = δ B L Taking into account that 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 symmetric 8 × 8 8\times8 8 × 8 matrices, the first contribution becomes:
δ B T σ = δ B L T σ = δ ( [ ( u e l T G x ) T ( u e l T G y ) T ( u e l T G x y ) T ] ) σ \delta\htmlClass{fea-math-bold-italic}{\boldsymbol B}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{\sigma}}=\delta\htmlClass{fea-math-bold-italic}{\boldsymbol B}_L^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{\sigma}}=\delta\left(\begin{bmatrix}\left(\htmlClass{fea-math-bold-italic}{\boldsymbol u}_{el}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol G}_x\right)^{T}\\\left(\htmlClass{fea-math-bold-italic}{\boldsymbol u}_{el}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol G}_y\right)^{T}\\\left(\htmlClass{fea-math-bold-italic}{\boldsymbol u}_{el}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol G}_{xy}\right)^{T}\end{bmatrix}\right)\htmlClass{fea-math-bold-italic}{\boldsymbol{\sigma}} δ B T σ = δ B L T σ = δ ( u e l T G x ) T ( u e l T G y ) T ( u e l T G x y ) T σ δ B T σ = ( σ x G x + σ y G y + τ x y G x y ) δ u e l \delta\htmlClass{fea-math-bold-italic}{\boldsymbol B}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{\sigma}}=\left(\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)\delta\htmlClass{fea-math-bold-italic}{\boldsymbol u}_{el} δ B T σ = ( σ x G x + σ y G y + τ x y G x y ) δ u e l The stress variation is:
δ σ = D δ ε = D B δ u e l \delta\htmlClass{fea-math-bold-italic}{\boldsymbol{\sigma}}=\htmlClass{fea-math-bold-italic}{\boldsymbol D}\,\delta\htmlClass{fea-math-bold-italic}{\boldsymbol{\varepsilon}}=\htmlClass{fea-math-bold-italic}{\boldsymbol D}\htmlClass{fea-math-bold-italic}{\boldsymbol B}\,\delta\htmlClass{fea-math-bold-italic}{\boldsymbol u}_{el} δ σ = D δ ε = D B δ u e l Consequently, the global tangent stiffness matrix is:
K T = A e l = 1 n e l ( ∑ n G h [ σ x G x + σ y G y + τ x y G x y ] ∣ d e t J ∣ ) + A e l = 1 n e l ( ∑ n G h B T D B ∣ d e t J ∣ ) \htmlClass{fea-math-bold-italic}{\boldsymbol{K}}_T=\mathcal{A}_{el=1}^{n_{el}}\left(\sum_{n_G}h\left[\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]\left|\mathop{\mathit{det}}\htmlClass{fea-math-bold-italic}{\boldsymbol J}\right|\right)+\mathcal{A}_{el=1}^{n_{el}}\left(\sum_{n_G}h\,\htmlClass{fea-math-bold-italic}{\boldsymbol B}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol D}\htmlClass{fea-math-bold-italic}{\boldsymbol B}\,\left|\mathop{\mathit{det}}\htmlClass{fea-math-bold-italic}{\boldsymbol J}\right|\right) K T = A e l = 1 n e l ( n G ∑ h [ σ x G x + σ y G y + τ x y G x y ] ∣ det J ∣ ) + A e l = 1 n e l ( n G ∑ h B T D B ∣ det J ∣ ) The tangent stiffness matrix contains two contributions: the material stiffness and the geometric (stress-dependent) stiffness. For one finite element:
k e l = ∑ n G h ( σ x G x + σ y G y + τ x y G x y + B T D B ) ∣ d e t J ∣ \htmlClass{fea-math-bold-italic}{\boldsymbol k}_{el}=\sum_{n_G}h\left(\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}+\htmlClass{fea-math-bold-italic}{\boldsymbol B}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol D}\htmlClass{fea-math-bold-italic}{\boldsymbol B}\right)\left|\mathop{\mathit{det}}\htmlClass{fea-math-bold-italic}{\boldsymbol J}\right| k e l = n G ∑ h ( σ x G x + σ y G y + τ x y G x y + B T D B ) ∣ det J ∣ The equilibrium system can also be written in terms of the element internal force vectors:
Ψ ( u ) = A e l = 1 n e l ( f e l ) − F = 0 \htmlClass{fea-math-bold-italic}{\boldsymbol{\mathit{\Psi}}}(\htmlClass{fea-math-bold-italic}{\boldsymbol u})=\mathcal{A}_{el=1}^{n_{el}}\left(\htmlClass{fea-math-bold-italic}{\boldsymbol f}_{el}\right)-\htmlClass{fea-math-bold-italic}{\boldsymbol F}=\htmlClass{fea-math-bold-italic}{\boldsymbol0} Ψ ( u ) = A e l = 1 n e l ( f e l ) − F = 0 f e l = ∑ n G h B T σ ∣ d e t J ∣ \htmlClass{fea-math-bold-italic}{\boldsymbol f}_{el}=\sum_{n_G}h\,\htmlClass{fea-math-bold-italic}{\boldsymbol B}^{T}\htmlClass{fea-math-bold-italic}{\boldsymbol{\sigma}}\,\left|\mathop{\mathit{det}}\htmlClass{fea-math-bold-italic}{\boldsymbol J}\right| f e l = n G ∑ h B T σ ∣ det J ∣ The element tangent stiffness matrix is the derivative of the element internal force vector:
k e l = ∂ f e l ∂ u e l \htmlClass{fea-math-bold-italic}{\boldsymbol k}_{el}=\frac{\partial\htmlClass{fea-math-bold-italic}{\boldsymbol f}_{el}}{\partial\htmlClass{fea-math-bold-italic}{\boldsymbol u}_{el}} k e l = ∂ u e l ∂ f e l 08
Newton–Raphson solution At Newton–Raphson iteration i i i :
Δ u ( i ) = − ( K T ( i ) ) − 1 Ψ ( i ) \mathit{\Delta}\htmlClass{fea-math-bold-italic}{\boldsymbol u}^{(i)}=-\left(\htmlClass{fea-math-bold-italic}{\boldsymbol K}_T^{(i)}\right)^{-1}\htmlClass{fea-math-bold-italic}{\boldsymbol{\mathit{\Psi}}}^{(i)} Δ u ( i ) = − ( K T ( i ) ) − 1 Ψ ( i ) u ( i + 1 ) = u ( i ) + Δ u ( i ) \htmlClass{fea-math-bold-italic}{\boldsymbol u}^{(i+1)}=\htmlClass{fea-math-bold-italic}{\boldsymbol u}^{(i)}+\mathit{\Delta}\htmlClass{fea-math-bold-italic}{\boldsymbol u}^{(i)} u ( i + 1 ) = u ( i ) + Δ u ( i ) Ψ ( i ) = Ψ ( u i ) \htmlClass{fea-math-bold-italic}{\boldsymbol{\mathit{\Psi}}}^{(i)}=\htmlClass{fea-math-bold-italic}{\boldsymbol{\mathit{\Psi}}}\left(\htmlClass{fea-math-bold-italic}{\boldsymbol u}_i\right) Ψ ( i ) = Ψ ( u i ) The element internal-force vectors f e l \htmlClass{fea-math-bold-italic}{\boldsymbol f}_{el} f e l and tangent stiffness matrices k e l \htmlClass{fea-math-bold-italic}{\boldsymbol k}_{el} k e l are assembled to obtain Ψ ( i ) \htmlClass{fea-math-bold-italic}{\boldsymbol{\mathit{\Psi}}}^{(i)} Ψ ( i ) and K T ( i ) \htmlClass{fea-math-bold-italic}{\boldsymbol K}_T^{(i)} K T ( i ) , respectively. A convenient convergence measure is:
ϵ = Δ u T Δ u n \epsilon=\sqrt{\frac{\mathit{\Delta}\htmlClass{fea-math-bold-italic}{\boldsymbol u}^{T}\mathit{\Delta}\htmlClass{fea-math-bold-italic}{\boldsymbol u}}{n}} ϵ = n Δ u T Δ u where n n n is the number of equations.
Important remark #1 The Total Lagrangian formulation presented here is not based on an incremental strain approximation. The equilibrium equations are written directly with respect to the initial configuration. Therefore, the accuracy of the formulation itself does not depend on the number of load steps. For problems with moderate nonlinearity, a single load step may be sufficient, provided that the Newton–Raphson iterations converge. Additional load steps may nevertheless be useful to improve convergence in more strongly nonlinear problems.
09
Example and numerical results The cantilever beam has length L = 400 m m L=400\ \mathit{mm} L = 400 mm and a rectangular cross-section with 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 . A vertical force F = 100 N F=100\ \mathit{N} F = 100 N is applied at the free end.
The mesh has 120 quadrilateral elements along the beam and 16 elements through its height: 1,920 isoparametric 4-node quadrilateral finite elements and 2,057 nodes.
Figure 1. Horizontal displacement u u u .Figure 2. Vertical displacement v v v .Figure 3. Green–Lagrange axial strain ε x \varepsilon_x ε x .Figure 4. Second Piola–Kirchhoff normal stress σ x \sigma_x σ x .Figure 5. Von Mises stress σ V M \sigma_{\mathit{VM}} σ VM .The MATLAB and ANSYS results are in very good agreement.
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Stress-map interpretation Important remark #2 In the Total Lagrangian formulation, strains and stresses are referred to the initial undeformed configuration. Therefore, the mathematically consistent stress map is the one represented on the initial geometry.
For a more intuitive visualization, the same second Piola–Kirchhoff stresses may also be represented on the deformed geometry. In this representation, the material directions rotate together with the finite elements; therefore, the x x x -stress direction becomes approximately tangent to the deformed beam axis.
Cauchy stresses are defined in the current configuration. When strains are small, their values referred to the corresponding local, rotated material directions are very close to the second Piola–Kirchhoff stresses. The distinction becomes important when comparing stress components expressed in different reference frames.
This topic will be discussed in a later chapter.
Figure 6. Second Piola–Kirchhoff normal stress represented on the initial configuration and, for intuitive visualization, on the deformed geometry. The original explanatory arrows and labels are preserved.PREVIOUS SECTION ← 12.1. CST Finite Element NEXT SECTION 12.3. Isoparametric 8-Node Quadrilateral Finite Element →