User:Eas4200c.f08.gator.edwards/Wk6 Class Notes

Notes for Meeting 29: November 3rd
Eas4200c.f08.gator.reger 05:08, 19 November 2008 (UTC)

Equation of Equilibrium:

Recall 1-D case (from meeting 28): Dimensional Analysis

$$\left[f\right] = \frac{f}{L}$$ $$\left[A\right] = L^2\,$$ Therefore: $$ \left[ \frac{f}{A}\right] = \frac{F}{L^3}$$ Where $$f$$ is 'dimension of', $$F\,$$ is force, $$L\,$$ is length, and $$A\,$$ is area.

Also, $$ \left[dx\right] = L \,$$ And, $$ \left[\sigma\right] = \frac{F}{L^2}\,$$ Therefore: $$ \left[\frac{d \sigma }{dx}\right] = \frac{F}{L^3}\,$$

It can be shown that:

$$\left[ \epsilon \right] = \left[\frac{du}{dx}\right] = \frac{L}{L} = 1  (Non-dimensional)\,$$ And, $$\left[ \mu \right]=\left[\frac{ \epsilon _{yy} }{ \epsilon _{xx}}\right] = 1  (Non-dimensional)\,$$

Torsion Analysis Continued
Case 1: Thin walled cross section (closed)



Outer Surface $$\phi=C_{0}$$ on $$S_{0}$$

Inner Surface $$\phi=C_{1}$$ on $$S_{1}$$

Case 2: Solid cross section



Outer Surface $$\phi=0$$ on $$S_{0}$$

Uniform bar with solid circular cross section

$$\phi (y,z)=C\left(\frac{y^{2}}{a^{2}}+\frac{z^{2}}{b^{2}}-1 \right)$$

Because the cross section is a circle $$a=b$$.



$$\phi=0$$ on $$S_{0}$$ because the cross section is solid.

$$\bigtriangledown^{2} \phi = -2G \theta $$

$$C=-\frac{1}{2}a^{2}G\theta $$

$$T = 2 \int_{A}^{} \phi \partial A = 2 C(\frac{J}{a^{2}} - A)$$

rearrange and substitute the area of a circle for A to get,

$$J = \int_{A}^{} r^{2} \partial A = \frac{1}{2} \pi a^{4}$$

Traction Vector
Relate ds to dz and dy via θ.

$$\partial y = \partial s*sin(\theta)$$

$$\partial z = \partial s*cos(\theta)$$

$$n_{y} = cos(\theta)$$

$$n_{z} = sin(\theta)$$

Now, sum forces in all directions:

$$\sum_{}^{} F_{y} = 0 = \sum_{}^{} \sigma A = \sum_{}^{} \sigma \ell$$

$$ = -\sigma_{yy} \partial z - \sigma_{yz} \partial y + t_{y} \partial s$$

$$ = -\sigma_{yy} \partial s *cos(\theta) - \sigma_{yz} \partial s* sin(\theta) + t_{y} \partial s$$

$$ = -\sigma_{yy}*cos(\theta) - \sigma_{yz}*sin(\theta) + t_{y}$$

$$t_{y} = \sigma_{yy}*n_{y} + \sigma_{yz}*n_{z}$$

$$\sum_{}^{} F_{z} = 0 = \sum_{}^{} \sigma A = \sum_{}^{} \sigma \ell$$

$$ = -\sigma_{zz} \partial y - \sigma_{yz} \partial z + t_{z} \partial s$$

$$ = -\sigma_{zz} \partial s *sin(\theta) - \sigma_{yz} \partial s* cos(\theta) + t_{z} \partial s$$

$$ = -\sigma_{zz}*sin(\theta) - \sigma_{yz}*cos(\theta) + t_{z}$$

$$t_{z} = \sigma_{zz}*n_{z} + \sigma_{yz}*n_{y}$$

The values for ty and tz are pressures and hence have units of pascals or psi. These pressures combine to form the traction vector, which is a surface force.

Matrix Notation of t The components of the traction vector can be displayed in matrix notation as such:

$$\begin{pmatrix} {t_{y}}\\ {t_{z}} \end{pmatrix} = \begin{pmatrix} {\sigma_{yy}}&{\sigma_{yz}}\\ {\sigma_{zy}}&{\sigma_{zz}} \end{pmatrix}

\begin{pmatrix} {n_{y}}\\ {n_{z}} \end{pmatrix}$$

For the three dimensional consideration with axes 1, 2, and 3:

$$\begin{pmatrix} {t_{1}}\\ {t_{2}}\\ {t_{3}} \end{pmatrix} = \begin{pmatrix} {\sigma_{11}}&{\sigma_{12}}&{\sigma_{13}}\\ {\sigma_{21}}&{\sigma_{22}}&{\sigma_{23}}\\ {\sigma_{31}}&{\sigma_{32}}&{\sigma_{33}} \end{pmatrix} \begin{pmatrix} {n_{1}}\\ {n_{2}}\\ {n_{3}} \end{pmatrix}$$

or in a more condensed form,

$$\begin{pmatrix} {t_{i}} \end{pmatrix}_{3x1} = \begin{pmatrix} {\sigma_{ij}} \end{pmatrix}_{3x3} \begin{pmatrix} {n_{j}} \end{pmatrix}_{3x1}$$

Notes from meeting 34, November 14, 2008
Unsymmetric thin walled cross sections



General equation for Unsymmetric cross sections $$\int_{A_{s}}^{} \frac{\partial \sigma_{xx}}{\partial x} \partial A = -q_{s} = -q(s)$$

Generalized nonsymmetric case:

$$\sigma_{xx} = (k_{y} M_{z} - k_{yz} M_{y}) y + (k_{z} M_{y} - k_{yz} M_{z}) z$$

$$k_{y} = \frac{I_{y}}{det I}$$ $$k_{z} = \frac{I_{z}}{det I}$$ $$k_{yz} = \frac{I_{yz}}{det I}$$

$$I = \begin{pmatrix} {I_{11}}&{I_{12}}&{I_{13}}\\ {I_{21}}&{I_{22}}&{I_{23}}\\ {I_{31}}&{I_{32}}&{I_{33}} \end{pmatrix} $$

$$\sigma_{xx} = \begin{pmatrix} {y}&{z} \end{pmatrix} \begin{pmatrix} {k_{y}}&{-k_{yz}}\\ {-k_{yz}}&{k_{z}} \end{pmatrix} \begin{pmatrix} {M_{z}}\\ {M_{y}} \end{pmatrix}$$

$$\sigma_{xx} = \begin{pmatrix} {y}&{z} \end{pmatrix} \begin{pmatrix} {k_{y} M_{z} - k_{yz} M_{y}}\\ {-k_{yz} M_{z} - k_{z} M_{y}} \end{pmatrix} = (k_{y} M_{z} - k_{yz} M_{y}) y + (-k_{yz} M_{z} + k_{z} M_{y}) z$$

Consider $$M_{z} = 0$$

$$I_{yz} = 0 \Rightarrow det\ I = I_{y} - I_{z}$$

$$k_{y} = \frac{1}{I_{z}},\qquad k_{z} = \frac{1}{I_{y}},\qquad k_{yz} = 0$$

$$q(s) = -(k_{y} V_{y} - k_{yz} V_{z}) Q_{z} - (k_{z} V_{z} - k_{yz} V_{y}) Q_{y}$$

$$Q_{z} = \int_{A_{s}}^{} y \partial A$$ $$Q_{y} = \int_{A_{s}}^{} z \partial A$$

For the case of a cross section that is symmetric about the y-axis:

$$\sigma_{xx} = \frac{M_{y} z}{I_{y}}$$

$$q(s) = \frac{-V_{z} Q_{y}}{I_{y}}$$

$$Q_{y} = \int_{A_{s}}^{} z \partial A = z_{c} A_{s}$$

Example 5.2 From Textbook
(The beginning of Example 5.2) Symmetric w.r.t. the y-axis:

$$A_3=A_2$$

$$A_4=A_1$$

Non-Symmetric:

$$A_1=A$$

$$A_2=2A$$

$$A_3=3A$$

$$A_4=4A$$

A Note on Matrix Brackets
Column matrices use curly brackets, { and }

Row matrices use "l" brackets

Larger matrices use the straight brackets, [ and ]

Contributing Members
Eas4200c.f08.spars.prey 07:27, 17 November 2008 (UTC)

William Mueller Eas4200c.f08.gator.mueller 18:33, 17 November 2008 (UTC)

Jacob Papp Eas4200c.f08.gator.papp 23:55, 18 November 2008 (UTC)

Robert Reger Eas4200c.f08.gator.reger 05:09, 19 November 2008 (UTC)

Eas4200c.f08.spars.prey 20:53, 19 November 2008 (UTC)

Justin Dickinson User:Eas4200c.f08.gator.dickinson 05:34, 19 November 2008 (UTC)

Greg Edwards Eas4200c.f08.gator.edwards 18:28, 21 November 2008 (UTC)