A.2 Transformation matrices

Consider the two right-handed coordinate systems of Fig. A.3, defined by orthogonal unit vectors $\mathbf{i, j, k,}$ and $\mathbf{i^{\ast}, j^{\ast}, k^{\ast}}$. Let xyz be global coordinates and $x^{\ast}y^{\ast}z^{\ast}$ a local coordinate system.

The vector $\mathbf{{\stackrel{\rm\rightarrow}{F}}}$ in Fig. A.3 can be written as the sum of two vectors along the coordinate axes $\mathbf{i, k}$ with magnitude $F_{x}$, $F_{z}$ and along the coordinate axes $\mathbf{i^{\ast}, k^{\ast}}$ with magnitude $F^{\ast}_{x}$, $F^{\ast}_{z}$.

Figure A.3: Coordinate transformation

$${\stackrel{\rm\rightarrow}{\mathbf{F}}} = F_{x}\cdot {\stackrel{\... ...thbf{i}}} \\ \cdot {\stackrel{\rm\rightarrow}{\mathbf{k}}} \end{array} \right.$$ (A.15)

To find the components of the vector ${\stackrel{\rm\rightarrow}{\mathbf{F}}}$ of Eq. (A.15), we multiply this equation by $\stackrel{\rm\rightarrow}{\mathbf{i^{\ast}}}$ and $\stackrel{\rm\rightarrow}{\mathbf{k^{\ast}}}$. The scalar products are:

$$\begin{array}{ccccccc} {\stackrel{\rm\rightarrow}{\mathbf{F}}}\cd... ...w}{\mathbf{k}}}\cdot {\stackrel{\rm\rightarrow}{\mathbf{k}^{\ast}}} \end{array}$$ (A.16)

where the scalar product of the two orthogonal vectors is zero.

To find the inverse transformation, we multiply Eq. (A.15) by $\stackrel{\rm\rightarrow}{\mathbf{i}}$ and $\stackrel{\rm\rightarrow}{\mathbf{k}}$. The scalar products are:

$$\begin{array}{ccccccc} {\stackrel{\rm\rightarrow}{\mathbf{F}}}\cd... ...w}{\mathbf{k}^{\ast}}}\cdot {\stackrel{\rm\rightarrow}{\mathbf{k}}} \end{array}$$ (A.17)

The scalar product of the two unit vectors is related to the cosine of the angle between these vectors (Fig. A.3).

$$\begin{array}{ccccccccc} {\stackrel{\rm\rightarrow}{\mathbf{i}}}\... ...\stackrel{\rm\rightarrow}{\mathbf{k}}} & = & \cos\alpha_{zx^{\ast}} \end{array}$$ (A.18)

In Fig. A.3, we show the direction cosines of the vector $\mathbf{\stackrel{\rm\rightarrow}{F}}$:

$$\begin{array}{ccccccc} \cos\alpha_{xx^{\ast}} & = & \cos\alpha , ... ...ha , \quad & \quad & \cos\alpha_{xz^{\ast}} & = & - \cos\beta \quad \end{array}$$ (A.19)

Be careful using cosine and sine angles associated with coordinate system transformation: $\cos\alpha_{xx^{\ast}}$ = $\cos\alpha$ and $\cos\alpha_{zx^{\ast}}$ = $\cos\beta$ ( = $\cos \left(90\,^{\circ} + \alpha\right)$ = $- \sin\alpha$).

Figure A.4: Direction cosines of an element

The length l and the direction cosines of an element can be calculated using coordinates $x_{A}$, $z_{A}$ of the node at the beginning and coordinates $x_{L}$, $z_{L}$ of the node at the end of the element (Fig. A.4):

$$l = \sqrt{\left(z_{L} - z_{A}\right)^{2} + \left(x_{L} - x_{A}\right)^{2}}$$ (A.20)

$$\cos\alpha = \frac{x_{L} - x_{A}}{l}$$ (A.21)

$$\cos\beta = \frac{z_{L} - z_{A}}{l}$$ (A.22)

We now consider the transformation of the vector ${\stackrel{\rm\rightarrow}{\mathbf{F}}}$ components $F_{x}$, $F_{z}$ of Eq. (A.15) from the global xy coordinate system to the components $F^{\ast}_{x}$, $F^{\ast}_{z}$ in the local $x^{\ast}y^{\ast}$ coordinate system.

$$\left[\begin{array}{c} F^{\ast}_{x} \\ F^{\ast}_{z} \end{array} ... ... \end{array} \right] \left[\begin{array}{c} F_{x} \\ F_{z} \end{array} \right]$$ (A.23)

Taking into account that $\cos\beta$ = $\cos \left(90\,^{\circ} + \alpha\right)$ = $- \sin\alpha$, we can write the above equation as

$$\left[\begin{array}{c} F^{\ast}_{x} \\ F^{\ast}_{z} \end{array} ... ... \end{array} \right] \left[\begin{array}{c} F_{x} \\ F_{z} \end{array} \right]$$ (A.24)

The inverse transformation of the vector ${\stackrel{\rm\rightarrow}{\mathbf{F}}}$ components $F^{\ast}_{x}$, $F^{\ast}_{z}$ of Eq. (A.15) from the local $x^{\ast}y^{\ast}$ coordinate system to the components $F_{x}$, $F_{z}$ of the global xy coordinate system:

$$\left[\begin{array}{c} F_{x} \\ F_{z} \end{array} \right] = \lef... ...right] \left[\begin{array}{c} F_{x}^{\ast} \\ F_{z}^{\ast} \end{array} \right]$$ (A.25)

Taking into account that $\cos\beta$ = $\cos \left(90\,^{\circ} + \alpha\right)$ = $- \sin\alpha$, we can write Eq. (A.25) in the form

$$\left[\begin{array}{c} F_{x} \\ F_{z} \end{array} \right] = \lef... ...right] \left[\begin{array}{c} F^{\ast}_{x} \\ F^{\ast}_{z} \end{array} \right]$$ (A.26)

Comparing the transformation matrix (A.23) with that of (A.25), we can see that they are transposed (rows and columns reversed). The multiplication of the matrix (A.23) by a transpose of itself of Eq. (A.25) gives the identity matrix

$$\left[\begin{array}{cc} \cos\alpha & \cos\beta \\ - \cos\beta & ... ...nd{array} \right] = \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]$$ (A.27)

We have thus proved that the matrices are orthogonal - an orthogonal matrix has the property that its transpose equals the inverse.