Analytical field for 3D plate test case

Field expression

\[\begin{split}\displaystyle u = \frac{f \cdot \left(1 - 2 \cdot \nu\right) \cdot \left(\nu + 1\right)}{E}\left[\begin{matrix}2 \cdot \nu \cdot \left(y^{2} - z^{2}\right) + x^{2} \cdot \left(\frac{L}{2} - \frac{x}{3}\right)\\- 4 \cdot \nu \cdot x \cdot y\\4 \cdot \nu \cdot x \cdot z\end{matrix}\right]\end{split}\]

Strain

Considering small displacements, the strain is given by:

\[\begin{split}\displaystyle eps = \frac{1}{2}.(u_{i,j}+u_{j,i})=\frac{f \cdot \left(1 - 2 \cdot \nu\right) \cdot \left(\nu + 1\right)}{E}\left[\begin{matrix}x \cdot \left(L - x\right) & 0 & 0\\0 & - 4 \cdot \nu \cdot x & 0\\0 & 0 & 4 \cdot \nu \cdot x\end{matrix}\right]\end{split}\]

Stress

Considering Hook’s law, constrains are given using Lamè coefficients for material properties:

\[\begin{split}\displaystyle sig=2.\mu.eps +\lambda.I.trace(eps) = \left[\begin{matrix}- \frac{f \cdot x \cdot \left(L - x\right) \cdot \left(\lambda + 2 \cdot \mu\right) \cdot \left(\nu + 1\right) \cdot \left(2 \cdot \nu - 1\right)}{E} & 0 & 0\\0 & - \frac{f \cdot x \cdot \left(\nu + 1\right) \cdot \left(2 \cdot \nu - 1\right) \cdot \left(\lambda \cdot \left(L - x\right) - 8 \cdot \mu \cdot \nu\right)}{E} & 0\\0 & 0 & - \frac{f \cdot x \cdot \left(\nu + 1\right) \cdot \left(2 \cdot \nu - 1\right) \cdot \left(\lambda \cdot \left(L - x\right) + 8 \cdot \mu \cdot \nu\right)}{E}\end{matrix}\right]\end{split}\]

Or if Lamè coefficients are replaced by engineering constants:

\[\begin{split}\displaystyle sig=2.\mu.eps +\lambda.I.trace(eps) = \left[\begin{matrix}1.0 \cdot f \cdot x \cdot \left(- L \cdot \nu + L + \nu \cdot x - x\right) & 0 & 0\\0 & f \cdot \nu \cdot x \cdot \left(1.0 \cdot L + 8.0 \cdot \nu - 1.0 \cdot x - 4.0\right) & 0\\0 & 0 & f \cdot \nu \cdot x \cdot \left(1.0 \cdot L - 8.0 \cdot \nu - 1.0 \cdot x + 4.0\right)\end{matrix}\right]\end{split}\]

Loads

Equilibrium equations give the following volume load:

\[\displaystyle fv=-div(sig) = - 1.0 \cdot f \cdot \left(- L + 2 \cdot x\right) \cdot \left(\nu - 1\right)\]

And on the boundary (\(\sigma.n\)):

\[\begin{split}\displaystyle fs (x=L) = \left[\begin{matrix}0\\0.0\\0.0\end{matrix}\right] fs (x=0) = \left[\begin{matrix}0\\0\\0\end{matrix}\right]\end{split}\]

\[\begin{split}\displaystyle fs (y=B) = \left[\begin{matrix}0.0\\f \cdot \nu \cdot x \cdot \left(1.0 \cdot L + 8.0 \cdot \nu - 1.0 \cdot x - 4.0\right)\\0.0\end{matrix}\right] fs (y=0) = \left[\begin{matrix}0\\- f \cdot \nu \cdot x \cdot \left(1.0 \cdot L + 8.0 \cdot \nu - 1.0 \cdot x - 4.0\right)\\0\end{matrix}\right]\end{split}\]

\[\begin{split}\displaystyle fs (z=H) = \left[\begin{matrix}0.0\\0.0\\f \cdot \nu \cdot x \cdot \left(1.0 \cdot L - 8.0 \cdot \nu - 1.0 \cdot x + 4.0\right)\end{matrix}\right] fs (z=0) = \left[\begin{matrix}0\\0\\- f \cdot \nu \cdot x \cdot \left(1.0 \cdot L - 8.0 \cdot \nu - 1.0 \cdot x + 4.0\right)\end{matrix}\right]\end{split}\]