Cantilever beam - Revision history

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2022-10-22T19:55:30Z

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2017-10-20T16:17:11Z

‎Exact solution

2017-10-20T16:12:38Z

‎Exact solution

2017-10-20T16:02:05Z

‎Exact solution

2017-10-20T16:01:46Z

‎Exact solution

2017-10-20T15:21:48Z

‎Exact solution

2017-10-18T15:17:22Z

‎Exact solution

2017-10-18T15:16:37Z

‎Exact solution

2017-09-14T14:34:43Z

‎Exact solution

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 {{Box-round|title= Related papers |
 [https://e6.ijs.si/ParallelAndDistributedSystems/publications/31107623.pdf J. Slak, G. Kosec; Refined meshless local strong form solution of Cauchy-Navier equation on an irregular domain, Engineering analysis with boundary elements, vol. 100, 2019]
 }}

← Older revision		Revision as of 19:55, 22 October 2022
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		+	{{Box-round\|title= Related papers \|
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		+	[https://e6.ijs.si/ParallelAndDistributedSystems/publications/31107623.pdf J. Slak, G. Kosec; Refined meshless local strong form solution of Cauchy-Navier equation on an irregular domain, Engineering analysis with boundary elements, vol. 100, 2019]
		+	}}
		+
	Do you want to go back to [[Solid Mechanics]]?		Do you want to go back to [[Solid Mechanics]]?

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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
-$$(\hat{\lambda} + \mu) \nabla(\nabla \cdot \vec{u}) + \nabla^2 \vec{u} = 0, \quad  \hat{\lambda} = \frac{2 \lambda \mu} {\lambda + 2\mu}$$
+$$(\hat{\lambda} + \mu) \nabla(\nabla \cdot \vec{u}) + \nabla^2 \vec{u} = 0, \quad  \hat{\lambda} = \frac{2 \lambda \mu} {\lambda + 2\mu},$$
 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.

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 = Exact solution =
-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> 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 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>
 [[File:cantilever_beam_case.png|800px]]

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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> 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|600px]]
+[[File:cantilever_beam_case.png|800px]]
 The stresses in such a beam are given as:

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 = Exact solution =
-The exact solution to this problem is given by Timoshenko (1951) where it is derived for '''plane stress''' conditions. <ref>Timoshenko, S. and Goodier, J. N. (1951). ''Theory of elasticity'', p. 35 - 39. McGraw-Hill, Inc., New York.</ref> 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:
+The exact solution to this problem is given by Timoshenko (1951) where it is derived for '''plane stress''' conditions. <ref>Timoshenko, S. and Goodier, J. N. (1951). ''Theory of elasticity'', p. 35 - 39. McGraw-Hill, Inc., New York.</ref> 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 $x = L$). The stresses in such a beam are given as:
 \begin{equation}
 \sigma_{xx} = -\frac{Pxy}{I},

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 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_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) &= \frac{P}{6EI}\left(3\nu x y^2 + x^3 - 3L^2 x + 2L^3\right)
+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.

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 \end{align}
 where $E$ is Young's modulus and $\nu$ is the Poisson ratio.
+<!--
 Alternatively we may prefer to see the force applied on the right side at $x = L$, and have the left end at $(x,y) = (0,0)$ fixed. In this case the solution can be found in the following expandable section.
 <div class="toccolours mw-collapsible mw-collapsed">
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 </div>
 </div>
+-->
 A notebook containing the proof of this equations as well as some plots of the solution is available here: [[:File:cantilever_beam.nb]]

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 <syntaxhighlight lang="matlab" line>
 function [sxx, syy, sxy, u, v] = cantilever_beam_analytical(x, y, P, L, D, E, nu)
-% Returns analytical solution to timoshenko cantilever beam.
+% Evaluates closed form solution to timoshenko cantilever beam.
 % P = loading force
 % L = length