Stress Tensor

The stress vector on a plane passing through a given spatial point $x$ at a given time $t$ depends only on the unit normal vector $n$ to the plane. Thus, let $T$ be the transformation such that \[{{t}_{n}}=Tn.................................(1)\] Let's giving a small tetrahedron which have one normal angle as one of the vertices. This tetrahedron is a small portion from the whole body with $P$ the point on the normal vertrice. The size of the tetrahedron is approching zeros. At that limit, we can think of point $p$ passing incline plane.

 The Newton's second law written for this tetrahedron is \[\sum{F={{t}_{-{{e}_{1}}}}(\Delta {{A}_{1}})}+{{t}_{-{{e}_{2}}}}(\Delta {{A}_{2}})++{{t}_{-{{e}_{3}}}}(\Delta {{A}_{3}})+{{t}_{n}}\Delta {{A}_{n}}=ma.....................(2) \] since assumming the portion very small in size, the mass of the portion which is the product of three lengths will approching zero before the left ide of the equation. therefore, \[\sum{F={{t}_{-{{e}_{1}}}}(\Delta {{A}_{1}})}+{{t}_{-{{e}_{2}}}}(\Delta {{A}_{2}})++{{t}_{-{{e}_{3}}}}(\Delta {{A}_{3}})+{{t}_{n}}\Delta {{A}_{n}}=0.......................(3) \] Let the unit normal vector of the incline plane $ABC$ be \[n={{n}_{1}}{{e}_{1}}+{{n}_{2}}{{e}_{2}}+{{n}_{3}}{{e}_{3}}.....................(4) \] Therefore, the area $\Delta {{A}_{1}}$, $\Delta {{A}_{1}}$, and $\Delta {{A}_{1}}$ are \[\begin{align} & \Delta {{A}_{1}}={{n}_{1}}\Delta {{A}_{n}} \\ & \Delta {{A}_{2}}={{n}_{2}}\Delta {{A}_{n}} .....................(5) \\ & \Delta {{A}_{3}}={{n}_{3}}\Delta {{A}_{n}} \\ \end{align}\] using equation (3) and (5) \[{{t}_{-{{e}_{1}}}{{n}_{1}}}+{{t}_{-{{e}_{2}}}}{{n}_{2}}+{{t}_{-{{e}_{3}}}}{{n}_{3}}+{{t}_{n}}=0.......................(6)\] from the action reaction \[{{t}_{-{{e}_{1}}}}=-{{t}_{{{e}_{1}}}},{{t}_{-{{e}_{2}}}}=-{{t}_{{{e}_{2}}}},{{t}_{-{{e}_{3}}}}=-{{t}_{{{e}_{3}}}}\] therefore, equation (6) is \[{{t}_{{{e}_{1}}}}{{n}_{1}}+{{t}_{{{e}_{2}}}}{{n}_{2}}+{{t}_{{{e}_{3}}}}{{n}_{3}}={{t}_{n}}.......................(7)\] using equation (7) and (4), we can proove equation (1) \[\begin{align} & {{t}_{n}}=Tn=T({{n}_{1}}{{e}_{1}}+{{n}_{2}}{{e}_{2}}+{{n}_{3}}{{e}_{3}})\text{ =}{{n}_{1}}T{{e}_{1}}+{{n}_{2}}T{{e}_{2}}+{{n}_{3}}T{{e}_{3}}={{n}_{1}}{{t}_{1}}+{{n}_{2}}{{t}_{2}}+{{n}_{3}}{{t}_{3}} \\ & \\ & \\ \end{align}\] be a linear transformation. It is called the stress tensor or the Cauchy stress tensor resource: ISBN: 978-0-7506-8560-3

Strain Tensor

There are mainly two types of strain tensors; Lagrange Strain Tensor and Infinitesimal Strain Tensor which denote $\nabla u$, the gradient of displacement $u$ Lagrange Strain Tensor is defined by \[\begin{align} & {{E}^{*}}=\frac{1}{2}[\nabla u+{{(\nabla u)}^{T}}+{{(\nabla u)}^{T}}(\nabla u)] \\ & \\ \end{align}\] and Infinitesimal Strain Tensor is defined by \[\begin{align} & {{E}}=\frac{1}{2}[\nabla u+{{(\nabla u)}^{T}}] \\ & \\ \end{align}\] which is the strain tensor assuming for small deformation where its component \[\begin{align} & {{E}_{ij}}=\frac{1}{2}(\frac{\partial {{u}_{i}}}{\partial {{X}_{j}}}+\frac{\partial {{u}_{j}}}{\partial {{X}_{i}}}) \\ & \\ \end{align}\] Therefore, the Lagrange Strain Tensor for a) rectangular coordinates: \[\begin{align} & [E]=\left[ \begin{matrix} \frac{\partial {{u}_{1}}}{\partial {{X}_{1}}} & \frac{1}{2}(\frac{\partial {{u}_{1}}}{\partial {{X}_{2}}}+\frac{\partial {{u}_{2}}}{\partial {{X}_{1}}}) & \frac{1}{2}(\frac{\partial {{u}_{1}}}{\partial {{X}_{3}}}+\frac{\partial {{u}_{3}}}{\partial {{X}_{1}}}) \\ \frac{1}{2}(\frac{\partial {{u}_{2}}}{\partial {{X}_{1}}}+\frac{\partial {{u}_{1}}}{\partial {{X}_{2}}}) & \frac{\partial {{u}_{2}}}{\partial {{X}_{2}}} & \frac{1}{2}(\frac{\partial {{u}_{2}}}{\partial {{X}_{3}}}+\frac{\partial {{u}_{3}}}{\partial {{X}_{2}}}) \\ \frac{1}{2}(\frac{\partial {{u}_{3}}}{\partial {{X}_{1}}}+\frac{\partial {{u}_{1}}}{\partial {{X}_{3}}}) & \frac{1}{2}(\frac{\partial {{u}_{3}}}{\partial {{X}_{2}}}+\frac{\partial {{u}_{2}}}{\partial {{X}_{3}}}) & \frac{\partial {{u}_{3}}}{\partial {{X}_{3}}} \\ \end{matrix} \right] \\ & \\ \end{align}\] b) cylindrical coordinates: \[\begin{align} & [E]=\left[ \begin{matrix} \frac{\partial {{u}_{r}}}{\partial r} & \frac{1}{2}(\frac{1}{r}\frac{\partial {{u}_{r}}}{\partial \theta }-\frac{{{u}_{\theta }}}{r}+\frac{\partial {{u}_{\theta }}}{\partial r}) & \frac{1}{2}(\frac{\partial {{u}_{r}}}{\partial z}+\frac{\partial {{u}_{z}}}{\partial r}) \\ \frac{1}{2}(\frac{1}{r}\frac{\partial {{u}_{r}}}{\partial \theta }-\frac{{{u}_{\theta }}}{r}+\frac{\partial {{u}_{\theta }}}{\partial r}) & \frac{1}{r}\frac{\partial {{u}_{\theta }}}{\partial \theta }+\frac{{{u}_{r}}}{r} & \frac{1}{2}(\frac{\partial {{u}_{\theta }}}{\partial z}+\frac{1}{r}\frac{\partial {{u}_{z}}}{\partial \theta }) \\ \frac{1}{2}(\frac{\partial {{u}_{r}}}{\partial z}+\frac{\partial {{u}_{z}}}{\partial r}) & \frac{1}{2}(\frac{\partial {{u}_{\theta }}}{\partial z}+\frac{1}{r}\frac{\partial {{u}_{z}}}{\partial \theta }) & \frac{\partial {{u}_{z}}}{\partial z} \\ \end{matrix} \right] \\ & \\ \end{align}\] c) Spherical Coordinates: \[\begin{align} & [E]=\left[ \begin{matrix} \frac{\partial {{u}_{r}}}{\partial r} & \frac{1}{2}(\frac{1}{r}\frac{\partial {{u}_{r}}}{\partial \theta }-\frac{{{u}_{\theta }}}{r}+\frac{\partial {{u}_{\theta }}}{\partial r}) & \frac{1}{2}(\frac{1}{r\sin \theta }\frac{\partial {{u}_{r}}}{\partial \phi }-\frac{{{u}_{\phi }}}{r}+\frac{\partial {{u}_{\phi }}}{\partial r}) \\ {{E}_{21}}={{E}_{12}} & \frac{1}{r}\frac{\partial {{u}_{\theta }}}{\partial \theta }+\frac{{{u}_{r}}}{r} & \frac{1}{2}(\frac{1}{r\sin \theta }\frac{\partial {{u}_{\theta }}}{\partial \phi }-\frac{{{u}_{\phi }}\cot \theta }{r}+\frac{1}{r}\frac{\partial {{u}_{\phi }}}{\partial \theta }) \\ {{E}_{31}}={{E}_{13}} & {{E}_{32}}={{E}_{23}} & \frac{1}{r\sin \theta }\frac{\partial {{u}_{\phi }}}{\partial \phi }-\frac{{{u}_{r}}}{r}+\frac{{{u}_{\theta }}\cot \theta }{r} \\ \end{matrix} \right] \\ & \\ & \\ \end{align}\] resource: ISBN: 978-0-7506-8560-3

Isotropic Linearly Elastic Solid

Isotropic Linearly Elastic Solid denotes the type of solid material which mechanical properties are the same regardless of directions. We had the linearity with respect to ${{e}_{i}}$ \[{{T}_{ij}}={{C}_{ijkl}}{{E}_{kl}}\] and with respect to ${{{e}'}_{i}}$ \[{{{T}'}_{ij}}={{{C}'}_{ijkl}}{{{E}'}_{kl}}\] For this material, it is obligated that \[{{C}_{ijkl}}={{{C}'}_{ijkl}}\] under all orthogonal transformations of basis (e.g. rotation). For example $I$, the identity matrix, can do this job in any catrtesian coordinate. Also its multiplication with scalar does the job. For isotropic fourth-order tensor, we use the form of identity sencond-order tensor, ${{\delta }_{ij}}$ , multipy to three new isotropic fourth-order tensors \[\begin{align} & {{A}_{ijkl}}={{\delta }_{ij}}{{\delta }_{kl}} \\ & {{B}_{ijkl}}={{\delta }_{ik}}{{\delta }_{jl}} \\ & {{H}_{ijkl}}={{\delta }_{il}}{{\delta }_{jk}} \\ & \\ \end{align}\] Therefore, the isotropic ${{C}_{ijkl}}$ could be defined as \[\begin{align} & {{C}_{ijkl}}=\lambda {{A}_{ijkl}}+\alpha {{B}_{ijkl}}+\beta {{H}_{ijkl}} \\ & \\ \end{align}\] where $\beta$, $\lambda$, and $\alpha$ are constants. Then it becomes \[{{T}_{ij}}={{C}_{ijkl}}{{E}_{kl}}=\lambda {{\delta }_{ij}}{{\delta }_{kl}}{{E}_{kl}}+\alpha {{\delta }_{ik}}{{\delta }_{jl}}{{E}_{kl}} +\beta {{\delta }_{il}}{{\delta }_{jk}}{{E}_{kl}}\] Thus, \[\begin{align} & {{T}_{ij}}=\lambda {{E}_{kk}}{{\delta }_{ij}}+(\alpha +\beta ){{E}_{kl}} \\ & \\ \end{align}\] denoting $(\alpha +\beta )$ by $2\mu $ , we have \[{{T}_{ij}}=\lambda e{{\delta }_{ij}}+2\mu {{E}_{kl}}\] where \[e\equiv {{E}_{kk}}\] denotes the dilation . Therefore, \[T=\lambda eI+2\mu E\] In detail, it says \[\begin{align} & {{T}_{11}}=\lambda ({{E}_{11}}+{{E}_{22}}+{{E}_{33}})+2\mu {{E}_{11}} \\ & {{T}_{22}}=\lambda ({{E}_{11}}+{{E}_{22}}+{{E}_{33}})+2\mu {{E}_{22}} \\ & {{T}_{33}}=\lambda ({{E}_{11}}+{{E}_{22}}+{{E}_{33}})+2\mu {{E}_{33}} \\ & {{T}_{12}}=2\mu {{E}_{12}} \\ & {{T}_{13}}=2\mu {{E}_{13}} \\ & {{T}_{23}}=2\mu {{E}_{23}} \\ & \\ \end{align}\] resource: ISBN: 978-0-7506-8560-3

Linearly Elastic Solid

As the name says, linearly elastic solid denote the solid material which there its relation between stress and strain is linear. We give $T$ the stress tensor and $E$ the strain tensor $T$ and $E$ both are second-order tensor (9 members each). Therefore ${{T}_{ij}}$ and ${{E}_{kl}}$ denote the member of the tensor where i,j,k,l are $\in \left\{ 1,2,3 \right\}$ and independent of each other. For linearity relation between $T$ and $E$ ; \[{{T}_{11}}={{C}_{1111}}{{E}_{11}}+{{C}_{1112}}{{E}_{12}}+....................+{{C}_{1133}}{{E}_{33}}\] \[{{T}_{12}}={{C}_{1211}}{{E}_{11}}+{{C}_{1212}}{{E}_{12}}+....................+{{C}_{1233}}{{E}_{33}}\] \[.......................................................\] \[.......................................................\] \[{{T}_{33}}={{C}_{3311}}{{E}_{11}}+{{C}_{3312}}{{E}_{12}}+....................+{{C}_{3333}}{{E}_{33}}\] It involves total of 81 coefficients to do the linear job. This $C$ is the fourth-order tensor where ${{C}_{ijkl}}$ is the member. The above set of equations can be written as ${{T}_{ij}}={{C}_{ijkl}}{{E}_{kl}}$ However, $T$ and $E$ are proof to be symmetric tensors. The fact that $E$ is symmetric tensor, we can combine the sum of two term such as ${{C}_{1112}}{{E}_{12}}+{{C}_{1121}}{{E}_{21}}$ into $({{C}_{1112}}+{{C}_{1121}}){{E}_{12}}$ and make $({{C}_{1112}}+{{C}_{1121}})$ one coefficient. This combination of two term lead to reduction of coefficients from 81 to 54. The fact that $T$ is symmetric reduces the coefficient (by reducing number of equations from above set of equation) to 18. For the simplified set of linear equation, we write \[\left[ \begin{matrix} {{T}_{11}} \\ {{T}_{22}} \\ {{T}_{33}} \\ \begin{matrix} {{T}_{23}} \\ \begin{matrix} {{T}_{31}} \\ {{T}_{12}} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right]=\left[ \begin{matrix} {{C}_{11}} \\ {{C}_{12}} \\ {{C}_{13}} \\ \begin{matrix} {{C}_{14}} \\ \begin{matrix} {{C}_{15}} \\ {{C}_{16}} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix}\begin{matrix} {{C}_{12}} \\ {{C}_{22}} \\ {{C}_{23}} \\ \begin{matrix} {{C}_{24}} \\ \begin{matrix} {{C}_{25}} \\ {{C}_{26}} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix}\begin{matrix} {{C}_{13}} \\ {{C}_{23}} \\ {{C}_{33}} \\ \begin{matrix} {{C}_{34}} \\ \begin{matrix} {{C}_{35}} \\ {{C}_{36}} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix}\begin{matrix} {{C}_{14}} \\ {{C}_{24}} \\ {{C}_{34}} \\ \begin{matrix} {{C}_{44}} \\ \begin{matrix} {{C}_{45}} \\ {{C}_{46}} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix}\begin{matrix} {{C}_{15}} \\ {{C}_{25}} \\ {{C}_{35}} \\ \begin{matrix} {{C}_{45}} \\ \begin{matrix} {{C}_{55}} \\ {{C}_{56}} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix}\begin{matrix} {{C}_{16}} \\ {{C}_{26}} \\ {{C}_{36}} \\ \begin{matrix} {{C}_{46}} \\ \begin{matrix} {{C}_{56}} \\ {{C}_{66}} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right]\left[ \begin{matrix} {{E}_{11}} \\ {{E}_{22}} \\ {{E}_{33}} \\ \begin{matrix} 2{{E}_{23}} \\ \begin{matrix} 2{{E}_{31}} \\ 2{{E}_{12}} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right]\] or \[\left[ \begin{matrix} {{T}_{1}} \\ {{T}_{2}} \\ {{T}_{3}} \\ \begin{matrix} {{T}_{4}} \\ \begin{matrix} {{T}_{5}} \\ {{T}_{6}} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right]=\left[ \begin{matrix} {{C}_{11}} \\ {{C}_{12}} \\ {{C}_{13}} \\ \begin{matrix} {{C}_{14}} \\ \begin{matrix} {{C}_{15}} \\ {{C}_{16}} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix}\begin{matrix} {{C}_{12}} \\ {{C}_{22}} \\ {{C}_{23}} \\ \begin{matrix} {{C}_{24}} \\ \begin{matrix} {{C}_{25}} \\ {{C}_{26}} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix}\begin{matrix} {{C}_{13}} \\ {{C}_{23}} \\ {{C}_{33}} \\ \begin{matrix} {{C}_{34}} \\ \begin{matrix} {{C}_{35}} \\ {{C}_{36}} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix}\begin{matrix} {{C}_{14}} \\ {{C}_{24}} \\ {{C}_{34}} \\ \begin{matrix} {{C}_{44}} \\ \begin{matrix} {{C}_{45}} \\ {{C}_{46}} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix}\begin{matrix} {{C}_{15}} \\ {{C}_{25}} \\ {{C}_{35}} \\ \begin{matrix} {{C}_{45}} \\ \begin{matrix} {{C}_{55}} \\ {{C}_{56}} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix}\begin{matrix} {{C}_{16}} \\ {{C}_{26}} \\ {{C}_{36}} \\ \begin{matrix} {{C}_{46}} \\ \begin{matrix} {{C}_{56}} \\ {{C}_{66}} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right]\left[ \begin{matrix} {{E}_{1}} \\ {{E}_{2}} \\ {{E}_{3}} \\ \begin{matrix} {{E}_{4}} \\ \begin{matrix} {{E}_{5}} \\ {{E}_{6}} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right]\] Where $C$ is symmetric tensor

How to write Math equations in Blogger

I am long gone from blogger because I could not find the way to write math in Blogger. I have done some google and combine all the clues and finally it 's work. This Is how I accomplish...

 1. Go to your blog's template option. You will see this page look. Click "Edit HTML"

2. Press "Ctrl+f" for the word "</head>". Copy the script ...

<script type="text/javascript"     src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX- AMSMML_HTMLorMML"></script>  <script type="text/x-mathjax-config">    MathJax.Hub.Config({  jax: ["input/TeX","output/HTML-CSS"],  extensions: ["tex2jax.js"],  tex2jax: {   inlineMath: [ ['$','$'], ['\\(','\\)'] ],   processEscapes: true  }    });  

</script>

...before the line "</head>". Then Click "Save Template"

 before

  after

This is the setup for your blogger page to accept Tex. However, ordinary people usually don't know how to write Tex. If you don't

3. Download and install MathType Software (https://www.dessci.com/en/products/mathtype/) or any software that does the job. I did downloaded the trail version.

4. Go to your Microsoft Word (I hope you have one on your window PC). Your will see the new tab "MathType" appear. Click that tab.

5. Click "Inline" . A new window will popup

6. Take your time to use this window to write equation by clicking around. It's not difficult. After you finish your equation, just close the window. Your eqaution will appear on Word document.

7. Then press "Toggle TeX". The equation will turn into TeX. So you just copy TeX and past onto your bolg post. Note that you should change your post to HTML mode before publish it

8. That's it. You should now have your equation on your blog like this...

$\frac{2}{5}\sqrt{{{a}^{2}}+{{b}^{2}}}\frac{n!}{r!\left( n-r \right)!}$ 


<I acknowledge the information from this
https://www.youtube.com/watch?v=zuXeU3l042Y
and modify by using the script from Xinyu Wang  from the youtube's comment>