User:Eml4500.f08.delta 6.castrillon/HW 3

=Determination of Element FD (wrt global coordinates)=

We have the FD for springs:

k $$^{(e)}$$ d $$^{(e)}$$ = f $$^{(e)}$$   (1) Where each represents a 4x4, 4x1, and 4x1 matrix, respectively.

This is represented by the following diagram.



Also, we can treat this element as a spring element turned 45° with the following analysis.



The reaction forces have been taken along the direction of the element so that we can analyze the stretching or shrinkage of the bar element as a force is applied on it. The p vectors translate to be the force on the node and the q vectors are the displacement of the node as the bar stretches or shrinks. These pertain to the FD relationship mentioned above as:

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

In long-hand form:

$$k^{(e)} \begin{bmatrix} 1 & -1 \\ -1 & 1\\ \end{bmatrix} \begin{pmatrix} q^{(e)}_{1}\\ q^{(e)}_{2} \end{pmatrix} = \begin{pmatrix} p^{(e)}_{1}\\ p^{(e)}_{2} \end{pmatrix}$$

Where $$q^{(e)}_i$$ is the axial displacement of element e at local node i, and $$p^{(e)}_i$$ is the axial force of element e at local node i. Also, these matrices are sized as 2x2, 2x1, and 2x1, respectively.

=Derivation of eqn (1) from eqn (2)=

We want to find the relationship between $$q^{(e)}_{2x1}$$ and $$d^{(e)}_{4x1}$$, and $$p^{(e)}_{2x1}$$ and $$f^{(e)}_{4x1}$$. It can be expressed as follows:


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

Consider the displacement of node 1, denoted as \vec{d}^{(e)}.




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

We say that $$q^{(e)}_{1}$$, the axial displacement of local node 1, is the orthogonal projection of the displacement. It is defined as:

$$q^{(e)}_{1}$$ = $$\vec{d}^{(e)} \vec{\tilde{i}}$$
 * =($$d^{e}_{1} \vec{i}$$ + $$d^{e}_{2} \vec{j}$$)•$$\vec{\tilde{i}}$$
 * =$$d^{e}_{1}$$ ($$\vec{i}$$•$$\vec{\tilde{i}}$$) + $$d^{e}_{2}$$ ($$\vec{j}$$•$$\vec{\tilde{i}}$$)

where
 * $$\vec{i}$$•$$\vec{\tilde{i}}$$ = cosθ^{(e)} = l $$^{(e)}$$
 * $$\vec{j}$$•$$\vec{\tilde{i}}$$ = sinθ^{(e)} = m $$^{(e)}$$

Which leads to the construction of the matrix equation that relates $$d$$ and $$q$$:

$$q^{(e)}_{1}$$ = l$$^{(e)}$$ $$d^{e}_{1}$$ + m$$^{(e)}$$ $$d^{e}_{2}$$
 * = $$\begin{bmatrix}

l^{(e)} & m^{(e)}\\ \end{bmatrix}_{1x2}$$ $$\begin{pmatrix} d^{e}_{1}\\ d^{e}_{2}\\ \end{pmatrix}_{2x1}$$

Similarily for node 2:

$$q^{(e)}_{2}$$ = l$$^{(e)}$$ $$d^{e}_{2}$$ + m$$^{(e)}$$ $$d^{e}_{2}$$
 * = $$\begin{bmatrix}

l^{(e)} & m^{(e)}\\ \end{bmatrix}_{1x2}$$ $$\begin{pmatrix} d^{e}_{3}\\ d^{e}_{4}\\ \end{pmatrix}_{2x1}$$

These two matrix equations combine to yield:


 * $$\begin{pmatrix}

q^{e}_{1}\\ q^{e}_{2}\\ \end{pmatrix}_{2x1}$$ = $$\begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)}\\ \end{bmatrix}_{2x4} \begin{pmatrix} d^{e}_{1}\\ d^{e}_{2}\\ d^{e}_{3}\\ d^{e}_{4}\\ \end{pmatrix}_{4x1}$$

and in short-hand form:


 * $$q^{(e)}$$ = $$T^{(e)}$$ $$d^{(e)}$$