User:Eas4200c.f08.wiki.f/11-3-08

Recall the 1-D case from earlier: Dimensional Analysis
$$\left. [f]=\frac{F}{L} \right \} \;\left [ \frac{f}{A} \right] = \frac{f}{L^3}$$

Notes :


 * The square brackets mean "dimension of"


 * F = Force


 * L = Length

$$[A]=L^2$$

$$[\sigma] = \frac{F}{L^2} \Rightarrow \left [ \frac{d \sigma}{dx} \right] = \frac{[d \sigma]}{[dx]} = \frac{\frac{F}{L^2}}{L} = \frac{F}{L^3}$$


 * F/L3 = Force/Volume or Body Force

$$[dx] = L$$

$$\left. \begin{matrix} \varepsilon = \frac{du}{dx}\\ E = \frac{\Delta L}{L} \end{matrix}  \right \} [\varepsilon] = \frac{[du]}{[dx]} = \frac{L}{L} = 1$$

$$\nu =\frac{\varepsilon_{yy}}{\varepsilon_{xx}} \Rightarrow [\nu] = \frac{[\varepsilon_{yy}]}{[\varepsilon_{xx}]} = 1$$