Nonlinear finite elements/Timoshenko beams

Timoshenko Beam

Timoshenko beam.

Displacements

${\begin{aligned}u_{1}&=u_{0}(x)+z\varphi _{x}\\u_{2}&=0\\u_{3}&=w_{0}(x)\end{aligned}}$

Strains

${\begin{aligned}\varepsilon _{xx}&=\varepsilon _{xx}^{0}+z\varepsilon _{xx}^{1}\\\gamma _{xz}&=\gamma _{xz}^{0}\end{aligned}}$

${\begin{aligned}\varepsilon _{xx}^{0}&={\cfrac {du_{0}}{dx}}+{\frac {1}{2}}\left({\cfrac {dw_{0}}{dx}}\right)^{2}\\\varepsilon _{xx}^{1}&={\cfrac {d\varphi _{x}}{dx}}\\\gamma _{xz}^{0}&=\varphi _{x}+{\cfrac {dw_{0}}{dx}}\end{aligned}}$

Principle of Virtual Work

$\delta W_{\text{int}}=\delta W_{\text{ext}}$

where

${\begin{aligned}\delta W_{\text{int}}&=\int _{x_{a}}^{x_{b}}\int _{A}(\sigma _{xx}\delta \varepsilon _{xx}+\sigma _{xz}\delta \gamma _{xz})dA~dx\\&=\int _{x_{a}}^{x_{b}}(N_{xx}\delta \varepsilon _{xx}^{0}+M_{xx}\delta \varepsilon _{xx}^{1}+Q_{x}\delta \gamma _{xz}^{0})~dx\\\delta W_{\text{ext}}&=\int _{x_{a}}^{x_{b}}q\delta w_{0}~dx+\int _{x_{a}}^{x_{b}}f\delta u_{0}~dx\end{aligned}}$

$Q_{x}={K_{s}}\int _{A}\sigma _{xz}~dA$

$K_{s}$ = shear correction factor

Taking Variations

$\varepsilon _{xx}^{0}={\cfrac {du_{0}}{dx}}+{\frac {1}{2}}\left({\cfrac {dw_{0}}{dx}}\right)^{2}$

Take variation

$\delta \varepsilon _{xx}^{0}={\cfrac {d(\delta u_{0})}{dx}}+{\frac {1}{2}}\left(2{\cfrac {dw_{0}}{dx}}\right)\left({\cfrac {d(\delta w_{0})}{dx}}\right)={\cfrac {d(\delta u_{0})}{dx}}+{\cfrac {dw_{0}}{dx}}{\cfrac {d(\delta w_{0})}{dx}}~.$

$\varepsilon _{xx}^{1}={\cfrac {d\varphi _{x}}{dx}}$

Take variation

$\delta \varepsilon _{xx}^{1}={\cfrac {d\delta \varphi _{x}}{dx}}$

$\gamma _{xz}^{0}=\varphi _{x}+{\cfrac {dw_{0}}{dx}}$

Take variation

$\delta \gamma _{xz}^{0}=\delta \varphi _{x}+{\cfrac {d(\delta w_{0})}{dx}}$

Internal Virtual Work

${\begin{aligned}\int _{x_{a}}^{x_{b}}N_{xx}\delta \varepsilon _{xx}^{0}~dx&=\int _{x_{a}}^{x_{b}}N_{xx}\left[{\cfrac {d(\delta u_{0})}{dx}}+{\cfrac {dw_{0}}{dx}}{\cfrac {d(\delta w_{0})}{dx}}\right]~dx\\\int _{x_{a}}^{x_{b}}M_{xx}\delta \varepsilon _{xx}^{1}~dx&=\int _{x_{a}}^{x_{b}}M_{xx}\left[{\cfrac {d\delta \varphi _{x}}{dx}}\right]~dx\\\int _{x_{a}}^{x_{b}}Q_{x}\delta \gamma _{xz}^{0}~dx&=\int _{x_{a}}^{x_{b}}Q_{x}\left[\delta \varphi _{x}+{\cfrac {d(\delta w_{0})}{dx}}\right]~dx\end{aligned}}$

${\begin{aligned}\delta W_{\text{int}}=\int _{x_{a}}^{x_{b}}&\left\{N_{xx}\left[{\cfrac {d(\delta u_{0})}{dx}}+{\cfrac {dw_{0}}{dx}}{\cfrac {d(\delta w_{0})}{dx}}\right]\right.+\\&M_{xx}\left[{\cfrac {d\delta \varphi _{x}}{dx}}\right]+\left.Q_{x}\left[\delta \varphi _{x}+{\cfrac {d(\delta w_{0})}{dx}}\right]\right\}~dx\end{aligned}}$

Integrate by Parts

Get rid of derivatives of the variations.

${\begin{aligned}\int _{x_{a}}^{x_{b}}N_{xx}&\left[{\cfrac {d(\delta u_{0})}{dx}}+{\cfrac {dw_{0}}{dx}}{\cfrac {d(\delta w_{0})}{dx}}\right]~dx=\left[N_{xx}\delta u_{0}\right]_{x_{a}}^{x_{b}}-\int _{x_{a}}^{x_{b}}{\cfrac {dN_{xx}}{dx}}\delta u_{0}~dx+\\&\qquad \qquad \qquad \left[N_{xx}{\cfrac {dw_{0}}{dx}}\delta w_{0}\right]_{x_{a}}^{x_{b}}-\int _{x_{a}}^{x_{b}}{\cfrac {d}{dx}}\left(N_{xx}{\cfrac {dw_{0}}{dx}}\right)\delta w_{0}~dx\\\\\int _{x_{a}}^{x_{b}}M_{xx}&\left[{\cfrac {d(\delta \varphi _{x})}{dx}}\right]~dx=\left[M_{xx}\delta \varphi _{x}\right]_{x_{a}}^{x_{b}}-\int _{x_{a}}^{x_{b}}{\cfrac {dM_{xx}}{dx}}\delta \varphi _{x}~dx\\\\\int _{x_{a}}^{x_{b}}Q_{x}&\left[\delta \varphi _{x}+{\cfrac {d(\delta w_{0})}{dx}}\right]~dx=\int _{x_{a}}^{x_{b}}Q_{x}\delta \varphi _{x}~dx+\left[Q_{x}\delta w_{0}\right]_{x_{a}}^{x_{b}}-\int _{x_{a}}^{x_{b}}{\cfrac {dQ_{x}}{dx}}\delta w_{0}~dx\end{aligned}}$

Collect terms

${\begin{aligned}&\left[N_{xx}{\delta u_{0}}\right]_{x_{a}}^{x_{b}}-\int _{x_{a}}^{x_{b}}{\cfrac {dN_{xx}}{dx}}{\delta u_{0}}~dx+\\&\left[N_{xx}{\cfrac {dw_{0}}{dx}}\delta w_{0}+Q_{x}\delta w_{0}\right]_{x_{a}}^{x_{b}}-\int _{x_{a}}^{x_{b}}\left[{\cfrac {d}{dx}}\left(N_{xx}{\cfrac {dw_{0}}{dx}}\right)+{\cfrac {dQ_{x}}{dx}}\right]\delta w_{0}~dx+\\&\left[M_{xx}\delta \varphi _{x}\right]_{x_{a}}^{x_{b}}-\int _{x_{a}}^{x_{b}}\left({\cfrac {dM_{xx}}{dx}}-Q_{x}\right)\delta \varphi _{x}~dx\\&=\int _{x_{a}}^{x_{b}}q~\delta w_{0}~dx+\int _{x_{a}}^{x_{b}}f{\delta u_{0}}~dx\end{aligned}}$

Euler-Lagrange Equations

${\begin{aligned}\int _{x_{a}}^{x_{b}}\left({\cfrac {dN_{xx}}{dx}}+f\right){\delta u_{0}}~dx&=\left[N_{xx}{\delta u_{0}}\right]_{x_{a}}^{x_{b}}\\\int _{x_{a}}^{x_{b}}\left[{\cfrac {d}{dx}}\left(N_{xx}{\cfrac {dw_{0}}{dx}}\right)+{\cfrac {dQ_{x}}{dx}}+q\right]~\delta w_{0}~dx&=\left[N_{xx}{\cfrac {dw_{0}}{dx}}~\delta w_{0}+Q_{x}~\delta w_{0}\right]_{x_{a}}^{x_{b}}\\\int _{x_{a}}^{x_{b}}\left({\cfrac {dM_{xx}}{dx}}-Q_{x}\right)~\delta \varphi _{x}~dx&=\left[M_{xx}~\delta \varphi _{x}\right]_{x_{a}}^{x_{b}}\end{aligned}}$

${\begin{aligned}{\cfrac {dN_{xx}}{dx}}+f&=0\\{\cfrac {d}{dx}}\left(N_{xx}{\cfrac {dw_{0}}{dx}}\right)+{\cfrac {dQ_{x}}{dx}}+q&=0\\{\cfrac {dM_{xx}}{dx}}-Q_{x}&=0\end{aligned}}$

Constitutive Relations

$\sigma _{xx}=E\varepsilon _{xx}~;\qquad \sigma _{xz}=G\gamma _{xz}$

Then,

${\begin{aligned}N_{xx}&=A_{xx}~\varepsilon _{xx}^{0}+B_{xx}~\varepsilon _{xx}^{1}\\\\M_{xx}&=B_{xx}~\varepsilon _{xx}^{0}+D_{xx}~\varepsilon _{xx}^{1}\\\\Q_{x}&=S_{xx}~\gamma _{xz}^{0}\end{aligned}}$

where

$S_{xx}=K_{s}\int _{A}G~dA\qquad \leftarrow \qquad {\text{shear stiffness}}$

Equilibrium Equations

${\begin{aligned}{\cfrac {d}{dx}}\left\{A_{xx}\left[{\cfrac {du_{0}}{dx}}+{\frac {1}{2}}\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]\right\}+f&=0\\{\cfrac {d}{dx}}\left\{S_{xx}\left({\cfrac {dw_{0}}{dx}}+\varphi _{x}\right)+A_{xx}{\cfrac {dw_{0}}{dx}}\left[{\cfrac {du_{0}}{dx}}+{\frac {1}{2}}\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]\right\}+q&=0\\{\cfrac {d}{dx}}\left(D_{xx}{\cfrac {d\varphi _{x}}{dx}}\right)+S_{xx}\left({\cfrac {dw_{0}}{dx}}+\varphi _{x}\right)&=0\end{aligned}}$

Weak Form

${\begin{aligned}\int _{x_{a}}^{x_{b}}A_{xx}{\cfrac {d(\delta u_{0})}{dx}}\left[{\cfrac {du_{0}}{dx}}+{\frac {1}{2}}\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]~dx&=\int _{x_{a}}^{x_{b}}f\delta u_{0}~dx+\left[N_{xx}\delta u_{0}\right]_{x_{a}}^{x_{b}}\\\\\int _{x_{a}}^{x_{b}}A_{xx}{\cfrac {d(\delta w_{0})}{dx}}{\cfrac {dw_{0}}{dx}}\left[{\cfrac {du_{0}}{dx}}+{\frac {1}{2}}\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]~dx&+\int _{x_{a}}^{x_{b}}S_{xx}{\cfrac {d(\delta w_{0})}{dx}}\left({\cfrac {dw_{0}}{dx}}+\varphi _{x}\right)~dx=\\&\int _{x_{a}}^{x_{b}}q\delta u_{0}~dx+\left[\left(N_{xx}{\cfrac {dw_{0}}{dx}}+Q_{x}\right)\delta w_{0}\right]_{x_{a}}^{x_{b}}\\\\-\int _{x_{a}}^{x_{b}}S_{xx}\delta \varphi _{x}\left({\cfrac {dw_{0}}{dx}}+\varphi _{x}\right)~dx&+\int _{x_{a}}^{x_{b}}D_{xx}{\cfrac {d(\delta \varphi _{x})}{dx}}{\cfrac {d\varphi _{x}}{dx}}~dx=\left[M_{xx}\delta \varphi _{x}\right]_{x_{a}}^{x_{b}}\end{aligned}}$

Finite element model

Trial Solution

${\begin{aligned}u_{0}(x)&=\sum _{j=1}^{m}u_{j}~\psi _{j}^{(1)}~;\qquad \qquad \delta u_{0}=\psi _{i}^{(1)}\\\\w_{0}(x)&=\sum _{j=1}^{n}w_{j}~\psi _{j}^{(2)}~;\qquad \qquad \delta w_{0}=\psi _{i}^{(2)}\\\\\varphi _{x}(x)&=\sum _{j=1}^{p}s_{j}~\psi _{j}^{(3)}~;\qquad \qquad \delta \varphi _{x}=\psi _{i}^{(3)}\end{aligned}}$

Element Stiffness Matrix

${\begin{bmatrix}\mathbf {K} ^{11}&\vdots &\mathbf {K} ^{12}&\vdots &\mathbf {K} ^{13}\\&\vdots &&\vdots &\\\mathbf {K} ^{21}&\vdots &\mathbf {K} ^{22}&\vdots &\mathbf {K} ^{23}\\&\vdots &&\vdots &\\\mathbf {K} ^{31}&\vdots &\mathbf {K} ^{32}&\vdots &\mathbf {K} ^{33}\\\end{bmatrix}}{\begin{bmatrix}\mathbf {u} \\\\\mathbf {w} \\\\\mathbf {s} \end{bmatrix}}={\begin{bmatrix}\mathbf {F} ^{1}\\\\\mathbf {F} ^{2}\\\\\mathbf {F} ^{3}\end{bmatrix}}$

Choice of Approximate Solutions

Choice 1

$\psi ^{(1)}$ = linear ($m=2$)

$\psi ^{(2)}$ = linear ($n=2$)

$\psi ^{(3)}$ = linear ($p=2$).

Nearly singular stiffness matrix ($6\times 6$).

Choice 2

$\psi ^{(1)}$ = linear ($m=2$)

$\psi ^{(2)}$ = quadratic ($n=3$)

$\psi ^{(3)}$ = linear ($p=2$).

The stiffness matrix is ($7\times 7$). We can statically condense out the interior degree of freedom and get a ($6\times 6$) matrix. The element behaves well.

Choice 3

$\psi ^{(1)}$ = linear ($m=2$)

$\psi ^{(2)}$ = cubic ($n=4$)

$\psi ^{(3)}$ = quadratic ($p=3$)

The stiffness matrix is ($9\times 9$). We can statically condense out the interior degrees of freedom and get a ($6\times 6$) matrix. If the shear and bending stiffnesses are element-wise constant, this element gives exact results.

Shear Locking

Example Case

Linear $u_{0}$, Linear $w_{0}$, Linear $\varphi _{x}$.

${\cfrac {dw_{0}}{dx}}=~~{\text{constant}}.$

But, for thin beams,

${\cfrac {dw_{0}}{dx}}=~~{\text{slope}}~~=-\varphi _{x}~~\leftarrow ~~({\text{linear!}})$

If constant $\varphi _{x}$

${\cfrac {d\varphi _{x}}{dx}}=0$

Also

1. $Q_{x}=S_{xx}\varphi _{x}\neq 0\implies$ Non-zero transverse shear.
2. $M_{xx}=D_{xx}{\cfrac {d\varphi _{x}}{dx}}=0\implies$ Zero bending energy.

Result: Zero displacements and rotations $\implies$ Shear Locking!

Recall

${\cfrac {d}{dx}}\left\{A_{xx}\left[{\cfrac {du_{0}}{dx}}+{\frac {1}{2}}\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]\right\}+f=0$

or,

${\cfrac {d}{dx}}\left\{A_{xx}\varepsilon _{xx}^{0}\right\}+f=0$

If $f=0$ and $A_{xx}=$ constant

$A_{xx}{\cfrac {d}{dx}}(\varepsilon _{xx}^{0})=0\qquad \implies \qquad \varepsilon _{xx}^{0}=~{\text{constant}}.$

If there is only bending but no stretching,

$\varepsilon _{xx}^{0}=0={\cfrac {du_{0}}{dx}}+{\frac {1}{2}}\left({\cfrac {dw_{0}}{dx}}\right)^{2}$

Hence,

${\cfrac {du_{0}}{dx}}\approx \left({\cfrac {dw_{0}}{dx}}\right)^{2}$

Also recall:

${\cfrac {d}{dx}}\left\{S_{xx}\left({\cfrac {dw_{0}}{dx}}+\varphi _{x}\right)+A_{xx}{\cfrac {dw_{0}}{dx}}\left[{\cfrac {du_{0}}{dx}}+{\frac {1}{2}}\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]\right\}+q=0$

or,

${\cfrac {d}{dx}}\left\{S_{xx}\gamma _{xz}^{0}+A_{xx}{\cfrac {dw_{0}}{dx}}\varepsilon _{xx}^{0}\right\}+q=0$

If $q=0$ and $S_{xx}=$ constant, and no membrane strains

$S_{xx}{\cfrac {d}{dx}}(\gamma _{xz}^{0})=0\qquad \implies \qquad {\gamma _{xz}=~{\text{constant}}}~={\cfrac {dw_{0}}{dx}}+\varphi _{x}$

Hence,

$\varphi _{x}\approx {\cfrac {dw_{0}}{dx}}$

Shape functions need to satisfy:

${\cfrac {du_{0}}{dx}}\approx \left({\cfrac {dw_{0}}{dx}}\right)^{2}~;\qquad {\text{and}}\qquad \varphi _{x}\approx {\cfrac {dw_{0}}{dx}}$

Example Case 1

Linear $u_{0}$, Linear $w_{0}$, Linear $\varphi _{x}$.

• First condition $\implies$ constant $=$ constant. Passes! No Membrane Locking.
• Second condition $\implies$ linear $=$ constant. Fails! Shear Locking.

Example Case 2

Linear $u_{0}$, Quadratic $w_{0}$, Linear $\varphi _{x}$.

• First condition $\implies$ constant $=$ quadratic. Fails! Membrane Locking.
• Second condition $\implies$ linear $=$ linear. Passes! No Shear Locking.

Example Case 3

Quadratic $u_{0}$, Quadratic $w_{0}$, Linear $\varphi _{x}$.

• First condition $\implies$ linear $=$ quadratic. Fails! Membrane Locking.
• Second condition $\implies$ linear $=$ linear. Passes! No Shear Locking.

Example Case 4

Cubic $u_{0}$, Quadratic $w_{0}$, Linear $\varphi _{x}$.

• First condition $\implies$ quadratic $=$ quadratic. Passes! No Membrane Locking.
• Second condition $\implies$ linear $=$ linear. Passes! No Shear Locking.

Overcoming Shear Locking

Option 1

• Linear $u_{0}$, linear $w_{0}$, linear $\varphi _{x}$.
• Equal interpolation for both $w_{0}$ and $\phi _{x}$.
• Reduced integration for terms containing $\phi _{x}$ - treat as constant.

Option 2

• Cubic $u_{0}$, quadratic $w_{0}$, linear $\varphi _{x}$.
• Stiffness matrix is $9\times 9$.
• Hard to implement.

← Euler Bernoulli beams

Nonlinear finite elements

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