User:Eml4500.f08.wiki1.aguilar/part6

Some of the equations that we have learned over the course of the semester need to be updated for use with 3-dimensional systems or space trusses.

With the local $$\tilde{x}$$ axis along the axis of the element, the local element equations are still the same as the 2-node axial deformation element:

$$\frac{EA}{L}\begin{pmatrix}\;\; 1&-1\\-1&\;\;1\end{pmatrix}\begin{pmatrix}q_1\\q_2\end{pmatrix}=\begin{pmatrix}P_1\\P_2\end{pmatrix}$$

Which gives us our familiar axial displacement relationship: $$\mathbf {\hat {k}}^{(e)} \mathbf q^{(e)} = \mathbf P^{(e)} $$



In the global coordinates, the nodal degrees of freedom and the nodal applied forces for each element are:

$$\mathbf q = \begin{pmatrix}d_1\\d_2\\d_3\\d_4\\d_5\\d_6\end{pmatrix}$$     and       $$\mathbf p = \begin{pmatrix}f_1\\f_2\\f_3\\f_4\\f_5\\f_6\end{pmatrix}$$

In order to derive the transformation matrix, we must do the following:

Consider the displacement vector of the local node 1 denoted by $$\vec{d}_{1}^{(e)}$$.



$$\vec{d}_{[1]}^{(e)}={d}_{1}^{(e)}\vec{i}+{d}_{1}^{(e)}\vec{j}+{d}_{1}^{(e)}\vec{k}$$

$${q}_{1}^{(e)}$$ = axial displacement of node 1 is the orthagonal projection of the displacement vector $$\vec{d}_{1}^{(e)}$$ of node 1 on the $$\tilde{x}$$ axis of element 'e'.

$${q}_{1}^{(e)}=\vec{d}_{1}^{(e)}\cdot\vec{\tilde{i}}$$

$${q}_{1}^{(e)}=({d}_{1}^{(e)}\vec{i}+{d}_{2}^{(e)}\vec{j}+{d}_{3}^{(e)}\vec{k})\cdot\vec{\tilde{i}}$$

$${q}_{1}^{(e)}={d}_{1}^{(e)}(\vec{i}\cdot\vec{\tilde{i}})+{d}_{2}^{(e)}(\vec{j}\cdot\vec{\tilde{i}})+{d}_{3}^{(e)}(\vec{k}\cdot\vec{\tilde{i}})$$

$$L=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$

$$(\vec{i}\cdot\vec{\tilde{i}})=\frac{x_2-x_1}{L}=l^{(e)}$$

$$(\vec{j}\cdot\vec{\tilde{i}})=\frac{y_2-y_1}{L}=m^{(e)}$$

$$(\vec{k}\cdot\vec{\tilde{i}})=\frac{z_2-z_1}{L}=m^{(e)}$$

$${q}_{1}^{(e)}=l^{(e)}{d}_{1}^{(e)}+m^{(e)}{d}_{2}^{(e)}+n^{(e)}{d}_{3}^{(e)}$$

$${q}_{1}^{(e)}= \begin{bmatrix}l^{(e)} & m^{(e)}&n^{(e)} \end{bmatrix}_{1x3}\begin{Bmatrix}{d}_{1}^{(e)}\\{d}_{2}^{(e)}\\{d}_{3}^{(e)}  \end{Bmatrix}_{3x1}$$

Here we can see that $${q}_{1}^{(e)}$$ is a 1x1 scalar.

We do this for node 2 as well. $${q}_{2}^{(e)}= \begin{bmatrix}l^{(e)} & m^{(e)} & n^{(e)}\end{bmatrix}_{1x3}\begin{Bmatrix}{d}_{4}^{(e)}\\{d}_{5}^{(e)}\\{d}_{6}^{(e)}  \end{Bmatrix}_{3x1}$$

which leads us to the transformation between global to local degrees of freedom which are as follows:

$$\begin{Bmatrix} {q}_{1}^{(e)}\\{q}_{2}^{(e)}\end{Bmatrix}= \begin{bmatrix} l^{(e)}& m^{(e)}& n^{(e)}&0&0&0\\0&0&0& l^{(e)}& m^{(e)}& n^{(e)} \end{bmatrix} \begin{Bmatrix}{d}_{1}^{(e)}\\{d}_{2}^{(e)}\\{d}_{3}^{(e)}\\{d}_{4}^{(e)}\\{d}_{5}^{(e)}\\{d}_{6}^{(e)} \end{Bmatrix}\Rightarrow \textbf{q}_{2x1}^{(e)}=\textbf{T}_{2x6}^{(e)}\textbf{d}_{6x1}^{(e)}$$

$$\begin{Bmatrix}{d}_{1}^{(e)}\\{d}_{2}^{(e)}\\{d}_{3}^{(e)}\\{d}_{4}^{(e)}\\{d}_{5}^{(e)}\\{d}_{6}^{(e)} \end{Bmatrix}= \begin{bmatrix} l^{(e)}&0\\m^{(e)}&0\\ n^{(e)}&0\\0&l^{(e)}\\0&m^{(e)}\\0& n^{(e)}\end{bmatrix}\begin{Bmatrix}{q}_{1}^{(e)}\\{q}_{2}^{(e)}\end{Bmatrix}

\Rightarrow \textbf{d}_{6x1}^{(e)} = {\textbf{T}_{6x2}^{(e)}}^{T} \textbf{q}_{2x1}^{(e)} $$