User:Eas4200c.f08.blue.a/Lecture 29

$$\vec{t}=t_{y}\vec{j}+t_{z}\vec{k}$$

It can be infered that:

$$dz=dscos\theta $$  $$n_{y}=cos\theta $$

$$dy=dssin\theta $$  $$n_{z}=sin\theta $$

Summing the forces in the y directon:

$$\sum{F_{y}}=0=-\sigma _{yy}(dz)-\sigma _{yz}(dy)+t_{y}(ds)$$

and using the above relations:

$$t_{y}=\sigma _{yy}n_{y}+\sigma _{yz}n_{z}$$

Similarly,

$$\sum{F_{z}}=0=-\sigma _{zz}(dy)-\sigma _{yz}(dz)+t_{z}(ds)$$

and again, substituting, we find:

$$t_{z}=\sigma _{zz}n_{z}+\sigma _{yz}n_{y}$$

In matrix form this becomes:

$$\begin{Bmatrix} t_{y}\\ t_{z} \end{Bmatrix}=

\begin{bmatrix} \sigma _{yy} & \sigma _{yz}\\ \sigma _{zy}& \sigma _{zz} \end{bmatrix}

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

Note: $$\sigma _{zy}= \sigma _{yz}$$ (because $$\sum{M_{x}}=\sum{M_{y}}=\sum{M_{z}}=0$$), but in order to keep with the unified notation, and to be compatible with the indicial notation, the difference is retained.

The generalized form is as follows:

$$\begin{Bmatrix} t_{1}\\ t_{2}\\ t_{3} \end{Bmatrix}=

\begin{bmatrix} \sigma _{11} & \sigma _{12}&\sigma _{13}\\ \sigma _{21}& \sigma _{22}&\sigma _{23}\\ \sigma_{31} &\sigma _{32}&\sigma _{33} \end{bmatrix}

\begin{Bmatrix} n_{1}\\ n_{2}\\ n_{3} \end{Bmatrix}$$

ShorthandSymbollically it would written as

$$\begin{Bmatrix} t_{i} \end{Bmatrix}_{3x1}

=\begin{bmatrix} \sigma _{ij} \end{bmatrix}_{3x3}

\begin{Bmatrix} n_{j} \end{Bmatrix}_{3x1}$$