User:Eml4500.f08.team.Bengtson/teamHw3

Lecture 14
Goal: Transform from a 4x2 matrix to a 2x2. The transformation relationship

\begin{bmatrix} \boldsymbol{P}_1^{(e)}\\ \boldsymbol{P}_2^{(e)}\\ \end{bmatrix} =\boldsymbol{T}^{(e)} \begin{bmatrix} \boldsymbol{f}_1^{(e)}\\ \boldsymbol{f}_2^{(e)}\\ \boldsymbol{f}_3^{(e)}\\ \boldsymbol{f}_4^{(e)}\\ \end{bmatrix} $$

There is a more general form

$$\mathbf{P}^{(e)}_i=\mathbf{T}^{(e)}\mathbf{f}^{(e)}_i$$, where

$$\mathbf{P}^{(e)}_{i}$$ of element (e) axial force at local node $$i$$

$$\mathbf{T}^{(e)}$$ transforms force components to an axial force

$$\mathbf{f}^{(e)}_{i}$$ is the $$i$$-th force component of element (e)

Goal: to find the relationship between

$$\mathbf{q}^{(e)}_{(2x2)}$$ and $$\mathbf{d}^{(e)}_{(4x2)}$$, (axial displacements of each element node and their respective displacement component)

$$\mathbf{P}^{(e)}_{(2x2)}$$ and $$\mathbf{f}^{(e)}_{(4x2)}$$, (axial forces at each element node and their respective force components)

Let us consider the force transformation relationship: $$\mathbf{P}^{(e)}_{(2x2)}=\mathbf{T}^{(e)}_{(2x4)}\mathbf{f}^{(e)}_{(4x2)}$$

FD relationship is shown by:

$$\mathbf{\hat{k}}^{(e)}_{(2x2)}\mathbf{q}^{(e)}_{(2x2)}=\mathbf{P}_{(2x2)}^{(e)}$$

By introducing a transformation matrix into the FD relationship: $$\mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)}\mathbf{d}^{(e)}=\mathbf{T}^{(e)}\mathbf{f}^{(e)}$$

The ultimate goal is to express the above equation in the form $$\mathbf{k}^{(e)}\mathbf{d}^{(e)}=\mathbf{f}^{(e)}$$. If $$\mathbf{T}^{(e)}$$ were an nxn matrix, then one could simply "move" $$\mathbf{T}^{(e)}$$ from the right hand side to the left hand side by multiplying the entire equation by the inverse $$\mathbf{T^{(e)}}^{-1}$$. Unfortunately, $$\mathbf{T}^{(e)}$$ is a rectangular 2x4 matrix, which means it is non-invertible.

For a more thorough understanding of the Finite Element Method, it is wise to derive the element force displacement with respect to the global coordinate system.

$$k^{(e)}d^{(e)}=f^{(e)}$$ (Equation 1)



The element force displacement matrix can be written as

$$\mathbf{k^{(e)}}\begin{bmatrix} 1 & -1 \\ -1 & 1\\ \end{bmatrix}\begin{pmatrix} q_{1}^{(e)}\\ q_{2}^{(e)}\\ \end{pmatrix}=\begin{pmatrix} P_{1}^{(e)}\\ P_{2}^{(e)}\\ \end{pmatrix} $$

$$q^{(e)}_{i}$$=axial displacementof element e at local node $$i$$

$$P^{(e)}_{i}$$=axial force of element e at local node $$i$$

We want to find equation (1) from equation (2).

We want to find the relationship between:

$$q^{(e)}_{2x1}$$ and $$d^{(e)}_{4x1}$$

$$P^{(e)}_{2x1}$$ and $$f^{(e)}_{4x1}$$

The relationships can be expressed as:

$$q^{(e)}_{2x1}=T^{(e)}_{2x4}d^{(e)}_{4x1}$$

Consider the displacement vector of local node i, denoted by $$d^{(e)}_{i}$$:



$$\vec{d^{(e)}_{[i]}}=d^{(e)}_{1}\vec{i}+d^{(e)}_{2}\vec{j}$$

$$q^{(e)}_{i}$$=axial displacement of node [1] is the orthogonal projection of the displacement vector $$\vec{d_{[1]}^{(e)}}$$onto the $$\tilde{x}$$ axis of element e

Therefore, for node [1], $$q_1^{(e)}$$ can be derived using the following steps.

$$q_{1}^{(e)}=d_{[1]}^{(e)}*\vec{\tilde{i}}$$ $$q_{1}^{(e)}=(d_{1}^{(e)}\vec{i}+d_{2}^{(e)}\vec{j})\vec{\tilde{i}}$$ $$q_{1}^{(e)}=d_{1}^{(e)}(\vec{i}*\vec{\tilde{i}})+d_{2}^{(e)}(\vec{j}*\vec{\tilde{i}})$$ $$q_{1}^{(e)}=l^{(e)}d_{1}^{(e)}+m^{(e)}d_{2}^{(e)}$$ $$q_{1}^{(e)}=\begin{bmatrix} l^{(e)} & m^{(e)}\\ \end{bmatrix}\begin{pmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ \end{pmatrix}$$

The values for $$q_2^{(e)}$$ and $$q_2^{(e)}$$ can now be substituted into the following matrix:

$$\begin{pmatrix} q_{1}^{(e)}\\ q_{2}^{(e)}\\ \end{pmatrix}=\begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)}\\ \end{bmatrix}\begin{pmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ d_{3}^{(e)}\\ d_{4}^{(e)}\\ \end{pmatrix}$$ We cannot solve for the inverse of ($$\mathbf{T}^{(e)}$$ becayse it is not an nxn matrix, we must instead make use of the transpose:

$$\mathbf{T}^{(e)T}_{(4x2)}\hat{k}^{(e)}_{(2x2)}\mathbf{T}^{(e)}_{(2x4)}\mathbf{d}^{(e)}_{(4x1)}=\mathbf{f}^{(e)}_{(4x1)}$$

Note that $$\mathbf{k}^{(e)}=\mathbf{T}^{(e)T}_{(4x2)}\mathbf{\hat{k}}^{(e)}_{(2x2)}\mathbf{T}^{(e)}_{(2x4)}$$

Therefore, $$\mathbf{k}^{(e)}\mathbf{d}^{(e)}=\mathbf{f}^{(e)}$$

Homework: Verify that

$$\mathbf{k}^{(e)}=\mathbf{T}^{(e)T}\mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)}$$