Difference between revisions of "Cantilever beam"
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Do you want to go back to [[Solid Mechanics]]? | Do you want to go back to [[Solid Mechanics]]? | ||
| − | On this page we conduct numerical studies of the | + | On this page we conduct numerical studies of '''bending of a cantilever loaded at the end''', a common numerical benchmark in elastostatics. |
= Exact solution = | = Exact solution = | ||
| − | Consider a beam | + | 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}\label{eq: | + | \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} | \end{equation} | ||
| − | The exact solution of | + | 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} | \begin{align}\label{eq:beam_a1} | ||
| − | u_x &= -\frac{ | + | 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 &= \frac{P}{6EI}\left(3\nu y^2 | + | u_y(x,y) &= \frac{P}{6EI}\left(3\nu x y^2 + x^3 - 3L^2 x + 2L^3\right) |
\end{align} | \end{align} | ||
| − | where $E$ is | + | where $E$ is Young's modulus and $\nu$ is the Poisson ratio. |
| − | |||
| − | |||
| − | |||
= Numerical solution = | = Numerical solution = | ||
| Line 28: | Line 33: | ||
* 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 | + | 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} | \begin{equation} | ||
| − | u_x( | + | 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 right 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(\frac{D^2}{4}-y^2\right). | ||
\end{equation} | \end{equation} | ||
| − | |||
An indicator of the accuracy that can be employed is the strain energy error $e$: | An indicator of the accuracy that can be employed is the strain energy error $e$: | ||
Revision as of 12:29, 14 November 2016
Do you want to go back to Solid Mechanics?
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 right 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(\frac{D^2}{4}-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.
