User:Eas4200c.f08.aero.e/Week 9

=Monday= Revisiting the road map,

Kinematic assumption (Compare to pg. 70)

4 - Stain components (Compare to pg. 70)

4 zero stress components (Compare to pg. 70)

$$ \epsilon - \sigma $$ relationship (2 forms: tensorial or engineering)

Rewrite $$ \epsilon - \sigma $$ relation

$$ \varepsilon_{ij} = \begin{bmatrix} {\underline A}_{3x3} & {0}_{3x3}\\ {0}_{3x3} & {\underline B}_{3x3}\\ \end{bmatrix} \sigma_{ij} $$

Let $$ \begin{bmatrix} {\underline A}_{3x3} & {0}_{3x3}\\ {0}_{3x3} & {\underline B}_{3x3}\\ \end{bmatrix} = \underline C $$

An advantage of using this form is that $$\underline C$$ can be easily inverted.

$$ \sigma_{ij} = \begin{bmatrix} {\underline A}^{-1} & \underline 0\\ \underline 0 & {\underline B}^{-1}\\ \end{bmatrix} \varepsilon_{ij} $$

$$ \begin{bmatrix} {\underline A}^{-1} & \underline 0\\ \underline 0 & {\underline B}^{-1}\\ \end{bmatrix} = \underline {C}^{-1} $$

When the two matrices are dotted into each other we get the identity matrix.

$$\underline C \underline {C}^{-1} = \underline I$$

Goal: prove 0 stress components.

$$\sigma _{xx}, \sigma _{yy} , \sigma _{zz}, \tau_{yz} = 0$$

$$\epsilon _{xx}, \varepsilon_{yy} , \varepsilon_{zz}, \gamma_{yz} = 0$$

This implies,

$$ \sigma_{ij} = \begin{bmatrix} {\underline A}^{-1} & \underline 0\\ \underline 0 & {\underline B}^{-1}\\ \end{bmatrix} \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ \varepsilon_{31}\\ \varepsilon_{12}\\ \end{bmatrix} $$

(Note, using numerical notation as introduced in MIT OCW assigned reading)

$$\sigma _{11}, \sigma _{22} , \sigma _{33} = 0$$

$$\sigma _{23} = 2G\varepsilon_{23} = \frac{1}{2}\gamma_{23} = \frac{1}{2}\gamma_{yz}$$

$$\sigma _{31} = 2G\varepsilon_{31}$$

Roadmap step C:

Implicit reading: Section 2.4 in book (Non-Uniform stress field)

Consider 1-D case as a model:



=Wednesday=

 Exam 2

=Friday=

 Homecoming