Difference between revisions of "Cantilever beam"
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Consider a beam od dimensions $L \times D$ with the origin of the coordinate system at $(x,y) = (0,D/2)$. The left end of the beam is fixed and the right end of the beam is loaded with a vertical traction: | Consider a beam od dimensions $L \times D$ with the origin of the coordinate system at $(x,y) = (0,D/2)$. The left end of the beam is fixed and the right end of the beam is loaded with a vertical traction: | ||
| − | \begin{equation} | + | \begin{equation}\label{eq:trac_a} |
| − | t_y(y) = \frac{P}{2I}\left(\frac{D^2}{4}-y^2\right). | + | t_y(L,y) = \frac{P}{2I}\left(\frac{D^2}{4}-y^2\right). |
\end{equation} | \end{equation} | ||
The exact solution of this problem is available in (Timoshenko and Goodier, 1977) and can be summarized as follows. The displacements in $x$- and $y-$ direction are given by | The exact solution of this problem is available in (Timoshenko and Goodier, 1977) and can be summarized as follows. The displacements in $x$- and $y-$ direction are given by | ||
| − | \begin{align}\label{eq: | + | \begin{align}\label{eq:beam_a1} |
| − | u_x &= -\frac{P}{6EI}\left(\left(6L-3x\right)x + \left(2+\nu\right)\left(y^2-\frac{D^2}{4}\right)\right) \\ | + | u_x &= -\frac{P}{6EI}\left(\left(6L-3x\right)x + \left(2+\nu\right)\left(y^2-\frac{D^2}{4}\right)\right) \\ \label{eq:beam_a2} |
u_y &= \frac{P}{6EI}\left(3\nu y^2(L-x) + (4+5\nu)\frac{D^2 x}{4} + (3L-x)x^2\right) | u_y &= \frac{P}{6EI}\left(3\nu y^2(L-x) + (4+5\nu)\frac{D^2 x}{4} + (3L-x)x^2\right) | ||
\end{align} | \end{align} | ||
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* Length of the beam: $L = 48$ m | * Length of the beam: $L = 48$ m | ||
| − | The unloaded beam is discretized with $40 \times 10$ regular nodes. Since the left end of the beam at $x = 0$ is fixed, the displacement boundary conditions are prescribed from the known analytical formulae (\ref{eq: | + | The unloaded beam is discretized with $40 \times 10$ regular nodes. Since the left end of the beam at $x = 0$ is fixed, the displacement boundary conditions are prescribed from the known analytical formulae (\ref{eq:beam_a1}) and (\ref{eq:beam_a2}) : |
\begin{equation} | \begin{equation} | ||
u_x(0,y) = -\frac{P}{6EI}(2+\nu)\left(y^2 - \frac{D^2}{4}\right); \qquad u_y(0,y) = \frac{P}{2EI}(\nu L y^2) | u_x(0,y) = -\frac{P}{6EI}(2+\nu)\left(y^2 - \frac{D^2}{4}\right); \qquad u_y(0,y) = \frac{P}{2EI}(\nu L y^2) | ||
\end{equation} | \end{equation} | ||
| + | The traction boundary at the right end of the beam ($x=L$) is given by (\ref{eq:trac_a}). | ||
== Finite elements == | == Finite elements == | ||
Revision as of 14:23, 9 November 2016
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On this page we conduct numerical studies of the cantilever beam, a common numerical benchmark in elastostatics.
Exact solution
Consider a beam od dimensions $L \times D$ with the origin of the coordinate system at $(x,y) = (0,D/2)$. The left end of the beam is fixed and the right end of the beam is loaded with a vertical traction: \begin{equation}\label{eq:trac_a} t_y(L,y) = \frac{P}{2I}\left(\frac{D^2}{4}-y^2\right). \end{equation} The exact solution of this problem is available in (Timoshenko and Goodier, 1977) and can be summarized as follows. The displacements in $x$- and $y-$ direction are given by \begin{align}\label{eq:beam_a1} u_x &= -\frac{P}{6EI}\left(\left(6L-3x\right)x + \left(2+\nu\right)\left(y^2-\frac{D^2}{4}\right)\right) \\ \label{eq:beam_a2} u_y &= \frac{P}{6EI}\left(3\nu y^2(L-x) + (4+5\nu)\frac{D^2 x}{4} + (3L-x)x^2\right) \end{align} where $E$ is Youngs modulus, $\nu$ is the Poisson ratio and $I$ is the moment of inertia for a beam with rectangular cross section and unit thickness: \[ I = \frac{D^3}{12}. \]
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 left end of the beam at $x = 0$ 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(0,y) = -\frac{P}{6EI}(2+\nu)\left(y^2 - \frac{D^2}{4}\right); \qquad u_y(0,y) = \frac{P}{2EI}(\nu L y^2) \end{equation} The traction boundary at the right end of the beam ($x=L$) is given by (\ref{eq:trac_a}).
