# Nonlinear finite elements/Timoshenko beams

## 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.