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	== Finite elements ==		== Finite elements ==
−	[[File:cantilever_beam.png]]	+	[[File:cantilever_beam.png|800px]]

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Revision as of 14:15, 14 November 2016

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On this page we conduct numerical studies of bending of a cantilever loaded at the end, a common numerical benchmark in elastostatics.

Exact solution

The exact solutions to this problem is given in Timoshenko (1951) where it derived for plane stress conditions. Consider a beam of dimensions $L \times D$ having a narrow rectangular cross section. The origin of the coordinate system is placed at $(x,y) = (0,D/2)$. The beam is bent by a force $P$ applied at the end $x = 0$ and the other end of the beam is fixed (at $x = L$). The stresses in such a beam are given as:

$$\begin{equation} \sigma_{xx} = -\frac{Pxy}{I}, \end{equation}$$ $$\begin{equation} \sigma_{yy} = 0, \end{equation}$$ $$\begin{equation}\label{eq:sxy} \sigma_{xy} = -\frac{P}{2I}\left(\left(\frac{D}{2}\right)^2 - y^2 \right), \end{equation}$$ where $I = D^3/12$ is the moment of inertia.

The exact solution in terms of the displacements in $x$- and $y-$ direction is

$$\begin{align}\label{eq:beam_a1} u_x(x,y) &= -\frac{Py}{6EI}\left(3(x^2-L^2) -(2+\nu)y^2 + 6 (1+\nu) \frac{D^2}{4}\right) \\ \label{eq:beam_a2} u_y(x,y) &= \frac{P}{6EI}\left(3\nu x y^2 + x^3 - 3L^2 x + 2L^3\right) \end{align}$$ where $E$ is Young's modulus and $\nu$ is the Poisson ratio.

Numerical solution

For the numerical solution we first choose the following parameters:

• Loading: $P = -1000$ N
• Young's modulus: $E = 3 \times 10^7$ N/m²
• Poisson's ratio: $\nu = 0.3$
• Height of the beam: $D = 12$ m
• Length of the beam: $L = 48$ m

The unloaded beam is discretized with $40 \times 10$ regular nodes. Since the right end of the beam at $x = L$ is fixed, the displacement boundary conditions are prescribed from the known analytical formulae ($\ref{eq:beam_a1}$) and ($\ref{eq:beam_a2}$) :

$$\begin{equation} u_x(L,y) = -\frac{P}{6EI}\left(-(2+\nu)y^2 + 6 (1+\nu) \frac{D^2}{4}\right); \qquad u_y(L,y) = \frac{P}{2EI}(\nu L y^2) \end{equation}$$ The traction boundary at the left end of the beam ($x=0$) is given by ($\ref{eq:sxy}$): $$\begin{equation}\label{eq:trac_a} t_y(L,y) = -\frac{P}{2I}\left(\left(\frac{D}{2}\right)^2 - y^2 \right). \end{equation}$$

An indicator of the accuracy that can be employed is the strain energy error $e$:

$$\begin{equation} e = \left[ \frac{1}{2} \int_\Omega (\b{\varepsilon}^\mathrm{num} - \b{\varepsilon}^\mathrm{exact})\b{C}(\b{\varepsilon}^\mathrm{num} - \b{\varepsilon}^\mathrm{exact}) d\Omega\right]^{1/2} \end{equation}$$ where $\b{\varepsilon}$ is the strain tensor in vector form, $\b{C}$ the reduced stiffness tensor (a matrix) and $\Omega$ the domain of the calculated solution.

Finite elements