Difference between revisions of "Linear elasticity"
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== Cantilever beam == | == Cantilever beam == | ||
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| + | The exact solution to this problem is given by Slaughter (2002) where it is derived for '''plane stress''' conditions. <ref>William S. Slaughter (2002). ''The Linearized Theory of Elasticity'', p. 285 - 289. Springer, New York.</ref> | ||
| + | This means we are solving the equation | ||
| + | $$(\hat{\lambda} + \mu) \nabla(\nabla \cdot \vec{u}) + \nabla^2 \vec{u} = 0, \quad \hat{\lambda} = \frac{2 \lambda \mu} {\lambda + 2\mu},$$ | ||
| + | where $\lambda = \frac{E\nu }{(1+\nu )(1-2\nu )}$ and $\mu = \frac {E}{2(1+\nu )}$ are the usual Lame parameters. | ||
| + | Consider a beam of dimensions $L \times D$ having a narrow rectangular cross section. The beam occupies a region of $[0, L] \times [-D/2, D/2]$. The beam is bent by a force $P$ applied at the end $x = 0$ and the other end of the beam is fixed at $x = L$, as illustrated below. | ||
| + | |||
| + | [[File:cantilever_beam_case.png|800px]] | ||
| + | |||
| + | 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(\frac{D^2}{4} - 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) = u(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) = v(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. | ||
Revision as of 14:52, 21 January 2019
On this page we showcase some basic examples from linear elasticity. To read more about the governing equations, refer to the Solid Mechanics page and examples therein, which are considered in more detail.
All examples here will be the solutions of the Cauchy-Navier equation $$ (\lambda + \mu) \nabla(\nabla \cdot \vec{u}) + \mu \nabla^2 \vec{u} = 0. $$
Cantilever beam
The exact solution to this problem is given by Slaughter (2002) where it is derived for plane stress conditions. [1] This means we are solving the equation $$(\hat{\lambda} + \mu) \nabla(\nabla \cdot \vec{u}) + \nabla^2 \vec{u} = 0, \quad \hat{\lambda} = \frac{2 \lambda \mu} {\lambda + 2\mu},$$ where $\lambda = \frac{E\nu }{(1+\nu )(1-2\nu )}$ and $\mu = \frac {E}{2(1+\nu )}$ are the usual Lame parameters. Consider a beam of dimensions $L \times D$ having a narrow rectangular cross section. The beam occupies a region of $[0, L] \times [-D/2, D/2]$. The beam is bent by a force $P$ applied at the end $x = 0$ and the other end of the beam is fixed at $x = L$, as illustrated below.
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(\frac{D^2}{4} - 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) = u(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) = v(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.- ↑ William S. Slaughter (2002). The Linearized Theory of Elasticity, p. 285 - 289. Springer, New York.
