User:EAS4200C.F08.WIKI.A/My Contribution: 5



We will no continue the equation of equilibrium in terms of $$\sigma$$. The goal is to obtain the following formula.

 $$\displaystyle \frac{\partial \sigma_{yx}}{\partial y} + \frac{\partial \sigma _{zx}}{\partial z} = 0$$

In indicial notation the equation can be represented as follows;

 $$\displaystyle \frac{\partial \sigma_{21}}{\partial x_2} + \frac{\partial \sigma _{31}}{\partial x_3} = 0$$

Now we must recall equation 2.21 from the text which is the above equation but with the term $$\displaystyle \frac{\partial \sigma_{xx}}{\partial x}$$ added in.

 $$\displaystyle \frac{\partial \sigma_{xx}}{\partial x} + \frac{\partial \sigma_{yx}}{\partial y} + \frac{\partial \sigma _{zx}}{\partial z} = 0 $$

SEE EQUATIONS 2.22 AND 2.23 TO COMPARE.

This equation represented in indicial notation is represented as the following  $$\sum_{i=1}^{3}{\frac{\partial \sigma_{ij}}{\partial x_i}} = 0$$ for $$j=1,2,3$$

All three cases ($$j= 1,2,3$$) are shown below.

$$\displaystyle j=1, \frac{\partial \sigma_{11}}{\partial x_1} +  \frac{\partial \sigma_{21}}{\partial x_2} + \frac{\partial \sigma_{31}}{\partial x_3}$$  $$\displaystyle j=2, \frac{\partial \sigma_{12}}{\partial x_1}  + \frac{\partial \sigma_{22}}{\partial x_2}  +  \frac{\partial \sigma_{32}}{\partial x_3}$$  $$\displaystyle j=3, \frac{\partial \sigma_{13}}{\partial x_1}  +  \frac{\partial \sigma_{23}}{\partial x_2}  +  \frac{\partial \sigma_{33}}{\partial x_3}$$

Noting that there are four zer stress components that come from the stress-strain relation, these equations reduce to the following.

$$\displaystyle 0=\sigma_{11}=\sigma_{22}=\sigma_{33}=\sigma_{23}$$  $$\displaystyle j=1, \frac{\partial \sigma_{21}}{\partial x_2} + \frac{\partial \sigma_{31}}{\partial x_3}$$  $$\displaystyle j=2, \frac{\partial \sigma_{12}}{\partial x_1} +  \frac{\partial \sigma_{32}}{\partial x_3}$$  <p style="text-align:center;">$$\displaystyle j=3, \frac{\partial \sigma_{13}}{\partial x_1} + \frac{\partial \sigma_{23}}{\partial x_2}$$

We will now derive the equation of equilibrium in terms of $$\sigma$$. To do so we will call upon the one dimentional model problem that we previously looked at. For conveienence it is depicted below.

[\Image:]

First we must sum the forces in the x direction. doing so results in the following equation.

<p style="text-align:center;">$$\sum{F_x} = 0 = -\sigma_{x}A + \sigma(x+dx)A + f(x)dx = A\left[\sigma(x+dx) - \sigma(x) \right] + f(x)dx$$

Taking the Taylor Series expansion of the part of the equation in the brackets (mutliplied by A) and ignoring the higher order terms yields the following.

<p style="text-align:center;"> $$\displaystyle f(x+dx) = f(x) + \frac{df(x)}{dx}dx + \frac{1}{2}\frac{d^2f}{dx^2}xdx^2....$$    Taylor Series Expansion <br \> <p style="text-align:center;">$$\displaystyle \frac{d\sigma}{dx} + \frac{f(x)}{A} = 0....$$   Result Ignoring Higher Order Terms