# Plane Trusses Finite Elements

1. Oct 15, 2012

### bugatti79

Folks,

The element equations for a uniform bar element with constant EA according to the attachment is given as

$\displaystyle \frac{E_a A_e}{h_e}\begin{bmatrix} 1 &0 &-1 &0 \\0 &0 &0 &0 \\-1 &0 &1 &0 \\ 0 &0 &0 &0 \end{bmatrix}\begin{Bmatrix} u^e_1\\v^e_1 \\u^e_2 \\ v^e_2 \end{Bmatrix}=\begin{Bmatrix} F^e_1\\0\\F^e_2 \\0 \end{Bmatrix}$

I am just wondering, can this not be also written as

$\displaystyle \frac{E_a A_e}{h_e}\begin{bmatrix} 1 &-1 &0 &0 \\0 &0 &0 &0 \\-1 &1 &0 &0 \\ 0 &0 &0 &0 \end{bmatrix}\begin{Bmatrix} u^e_1\\u^e_2 \\v^e_1 \\ v^e_2 \end{Bmatrix}=\begin{Bmatrix} F^e_1\\0\\F^e_2 \\0 \end{Bmatrix}$...?

#### Attached Files:

• ###### IMAG0125.jpg
File size:
15.2 KB
Views:
51
2. Oct 15, 2012

### AlephZero

You could do that, but things will get very confusing in the long run if the vector of displacements is in a different order to the vector of forces. For example the work done by the external forces will not be just $F^T x$ and the strain energy in the structure will not be $(1/2)x^TKx$. You would need to keep track of the different order of the two vectors.

You could use a different consistent ordering of the variables, like

$\displaystyle \frac{E_a A_e}{h_e}\begin{bmatrix} 1 &-1 &0 &0 \\ -1 &1 &0 &0 \\ 0 &0 &0 &0 \\ 0 &0 &0 &0 \end{bmatrix}\begin{Bmatrix} u^e_1\\u^e_2 \\v^e_1 \\ v^e_2 \end{Bmatrix}=\begin{Bmatrix} F^e_1\\F^e_2 \\0\\0 \end{Bmatrix}$

3. Oct 16, 2012

### bugatti79

OK thanks. Further on I am just wondering how he determined the transformations between the 2 sets of coordinates systems global $(x,y)$ and local $(x_e,y_e)$ based on the attachment.

He writes

$x_e=x \cos \theta_e+y \sin \theta_e$, $y_e=-x \sin \theta_e+y \cos \theta_e$

$x=x_e \cos \theta_e- y_e \sin \theta_e$, $y=x_e \sin \theta_e+y_e \cos \theta_e$

I would have written $x_e=x \cos \theta_e-y \sin \theta_e$, $y_e=x \sin \theta_e+y \cos \theta_e$ for $(x_e,y_e)$

Then based on my above i write it in matrix form and get the inverse to arrive at

$x=x_e \cos \theta_e+y_e \sin \theta_e$, $y=-x_e \sin \theta_e+y_e \cos \theta_e$
(On the side, I would like to know how I would obtain $(x,y)$ graphically)

Last edited: Oct 16, 2012
4. Oct 16, 2012

### bugatti79

It is ok, I have it sorted. The book is correct. Cheers