User:Eml4500.f08.gravy.bje/Notes5

3-D Space Trusses
The axial FD relation for 3-D space trusses can be expressed the same as for the 2-D truss element

$$\ \ \frac{EA}{L} \begin{Bmatrix} 1 & -1 \\ -1 & 1 \end{Bmatrix} \begin{Bmatrix} d_{1} \\ d_{2} \end{Bmatrix}  =  \begin{Bmatrix} P_{1} \\ P_{2} \end{Bmatrix} \,$$

where E = elastic modulus of the material, A = area of the cross section of the element, L = length of the element, d1 and d2 are the displacements along the axis of the element, and P1 and P2 are possible applied axial loads at the end.

This can also be expressed as:

$$k_{1} d_{1} = r_{1}\,$$

The degrees of freedom for this system is represented as:

$$\ \ d = \begin{Bmatrix} u_{1} \\ v_{1} \\ w_{1} \\ u_{2} \\ v_{2} \\ w_{2} \end{Bmatrix}_{(6x1)} \,$$

where u, v, and w are your degrees of freedom in the x, y, and z directions respectively

The element forces for this system is represented as:

$$\ \ r = \begin{Bmatrix} F_{x1} \\ F_{y1} \\ F_{z1} \\ F_{x2} \\ F_{y2} \\ F_{z2} \end{Bmatrix}_{(6x1)} \,$$

For 3-D space trusses the transformation matrix is given in terms of three director cosines:

$$l^{(e)} = \frac{x_{2} - x_{1}}{L}$$

$$m^{(e)} = \frac{y_{2} - y_{1}}{L}$$

$$n^{(e)} = \frac{z_{2} - z_{1}}{L}$$

With these director cosines, the transformation between the local and global degrees of freedom can be expressed as:

$$ \begin{Bmatrix}d_{1} \\ d_{2} \end{Bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0 \\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)} \end{bmatrix} \begin{Bmatrix} u_{1} \\ v_{1} \\ w_{1} \\ u_{2} \\ v_{2} \\ w_{2} \end{Bmatrix} $$

This expression is also used to show the relation between axial dofs and element dofs in global coordinates

Similarly, the relationn between axial forces and element forces in global coordinates is:

$$ \begin{Bmatrix}F_{1} \\ F_{2} \end{Bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0 \\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)} \end{bmatrix} \begin{Bmatrix} F_{x1} \\ F_{y1} \\ F_{z1} \\ F_{x2} \\ F_{y2} \\ F_{z2} \end{Bmatrix} $$

The derivation for the transformation matrix is as follewed:

$$\ d_1^{(e)}=l^{(e)}u_1^{(e)} + m^{(e)}v_1^{(e)} + n^{(e)}w_1^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} \end{bmatrix} \begin{Bmatrix} u_1 \\ v_1 \\ w_1 \end{Bmatrix} \,$$

$$ \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \end{Bmatrix}_{(2x1)} = \begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0 \\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)} \end{bmatrix}_{(2x6)} \begin{Bmatrix} u_{1} \\ v_{1} \\ w_{1} \\ u_{2} \\ v_{2} \\ w_{2} \end{Bmatrix}_{(6x1)} $$

where the transformation matrix is equal to:

$$\ \ T = \begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0 \\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)} \end{bmatrix}$$

The global stiffness matrix can be calculated as follows:

$$\ \underline{k}^{(e)} = \underline{T}^{(e)T} \underline{k}^{(e)} \underline{T}^{(e)}$$

Where: $$\ \underline{T}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0 \\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)} \end{bmatrix}$$ and $$\ \underline{k}^{e)} = k^{(e)} * \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$

Since $$k^{(e)}$$ is a scaler, we can move it within the equation without changing the problem. So after moving the scaler and inserting in the matrices, the following equation is found:

$$\ \underline{k}^{(e)} = k^{(e)} \begin{bmatrix} l^{(e)} & 0\\ m^{(e)} & 0 \\ n^{(e)} & 0 \\ 0 & l^{(e)} \\ 0 & m^{(e)} \\ 0 & n^{(e)} \end{bmatrix} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0 \\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)}  \end{bmatrix}$$

Solving the first set of matrices gives:

$$\ \underline{k}^{(e)} = k^{(e)} \begin{bmatrix} l^{(e)} & -l^{(e)} \\ m^{(e)} & -m^{(e)} \\ n^{(e)} & -n^{(e)} \\ -l^{(e)} & l^{(e)}\\ -m^{(e)} & m^{(e)} \\ -n^{(e)} & n^{(e)} \end{bmatrix}  \begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0 \\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)} \end{bmatrix}$$

Solving the second set of matrices gives:

$$\ \underline{k}^{(e)} = k^{(e)}\begin{bmatrix} (l^{(e)})^2 & l^{(e)}m^{(e)} & n^{(e)}l^{(e)} & -(l^{(e)})^2 & -l^{(e)}m^{(e)} & -n^{(e)}l^{(e)}  \\  l^{(e)}m^{(e)} & (m^{(e)})^2 & m^{(e)}n^{(e)} & -l^{(e)}m^{(e)} & -(m^{(e)})^2 & -m^{(e)}n^{(e)} \\ n^{(e)}l^{(e)} & m^{(e)}n^{(e)} & (n^{(e)})^2 & -n^{(e)}l^{(e)} & -m^{(e)}n^{(e)} & -(n^{(e)})^2 \\ -(l^{(e)})^2 & -l^{(e)}m^{(e)} & -n^{(e)}l^{(e)} & (l^{(e)})^2 & l^{(e)}m^{(e)} & n^{(e)}l^{(e)} \\ -l^{(e)}m^{(e)} & -(m^{(e)})^2 & -m^{(e)}n^{(e)} & l^{(e)}m^{(e)} & (m^{(e)})^2 & m^{(e)}n^{(e)} \\ -n^{(e)}l^{(e)} & -m^{(e)}n^{(e)} & -(n^{(e)})^2 & n^{(e)}l^{(e)} & m^{(e)}n^{(e)} & (n^{(e)})^2 \end{bmatrix} $$

Which is the definition of the stiffness matrix, therefore the equation is true.

Combining all of our equations we get our FD relation for the global stiffness matrix, the dofs, and the forces:

$$\ \ \frac{EA}{L} \begin{bmatrix} (l^{(e)})^2 & l^{(e)}m^{(e)} & n^{(e)}l^{(e)} & -(l^{(e)})^2 & -l^{(e)}m^{(e)} & -n^{(e)}l^{(e)} \\  l^{(e)}m^{(e)} & (m^{(e)})^2 & m^{(e)}n^{(e)} & -l^{(e)}m^{(e)} & -(m^{(e)})^2 & -m^{(e)}n^{(e)} \\ n^{(e)}l^{(e)} & m^{(e)}n^{(e)} & (n^{(e)})^2 & -n^{(e)}l^{(e)} & -m^{(e)}n^{(e)} & -(n^{(e)})^2 \\ -(l^{(e)})^2 & -l^{(e)}m^{(e)} & -n^{(e)}l^{(e)} & (l^{(e)})^2 & l^{(e)}m^{(e)} & n^{(e)}l^{(e)} \\ -l^{(e)}m^{(e)} & -(m^{(e)})^2 & -m^{(e)}n^{(e)} & l^{(e)}m^{(e)} & (m^{(e)})^2 & m^{(e)}n^{(e)} \\ -n^{(e)}l^{(e)} & -m^{(e)}n^{(e)} & -(n^{(e)})^2 & n^{(e)}l^{(e)} & m^{(e)}n^{(e)} & (n^{(e)})^2 \end{bmatrix} \begin{bmatrix} u_(e) \\ v_(e) \\ w_(e) \\ u_(e+1) \\ v_(e+1) \\ w_(e+1) \end{bmatrix} = \begin{bmatrix} F_ex \\ F_ey \\ F_ez \\ F_(e+1)x \\ F_(e+1)y \\ F_(e+1)z \end{bmatrix} $$