# 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
* $Q_{x}=S_{xx}\varphi _{x}\neq 0\implies$ Non-zero transverse shear.
* $M_{xx}=D_{xx}{\cfrac {d\varphi _{x}}{dx}}=0\implies$ 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:

### 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.