User:Eml4500.f08.ramrod.D/HW4

October 10th, 2008
Note: Consider case where $$\tilde{d}_{4}^{(e)} \ne 0$$

$$\tilde{d}_{1}^{(e)}=\tilde{d}_{2}^{(e)}=\tilde{d}_{3}^{(e)}= 0$$

$$\tilde{\mathbf{f}}_{4x1}^{(e)}=\tilde{\mathbf{k}}_{4x4}^{(e)}\tilde{\mathbf{d}}_{4x1}^{(e)}=\mathbf{0}_{4x1}$$

The $$\mathbf{0}_{4x1}$$ is the 4th column of $$\mathbf{k}^{(e)}$$

Interpretation of the the transverse degrees of freedom

$$\tilde{\mathbf{d}}^{(e)}=\tilde{\mathbf{T}}^{(e)}{d}^{(e)}$$

Similarly: $$\tilde{\mathbf{f}}^{(e)}=\tilde{\mathbf{T}}^{(e)}{f}^{(e)}$$

Also:

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

Therefore:

$$\tilde{\mathbf{k}}^{(e)}\tilde{\mathbf{T}}^{(e)}\mathbf{d}^{(e)}=\tilde{\mathbf{T}}^{(e)}\mathbf{f}^{(e)}$$

If $$\tilde{\mathbf{T}}^{(e)}$$ is invertible, then:

$$ [\tilde{\mathbf{T}}^{(e)-1}{\tilde{\mathbf{k}}^{(e)}}\tilde{\mathbf{T}}^{(e)}]\mathbf{d}^{(e)}=\mathbf{f}^{(e)}$$

$$\tilde{\mathbf{T}}^{(e)} $$ block diagonal matrix: $$\begin{bmatrix} \mathbf{R}_{2x2}^{(e)} & \mathbf{0}_{2x2} \\ \mathbf{0}_{2x2} & \mathbf{R}_{2x2}^{(e)} \end{bmatrix}$$

Consider a general block diagonal matrix: $$ \mathbf{A}=\begin{bmatrix} \mathbf{D}_{1}& \mathbf{0} \\ \mathbf{0} & \mathbf{D}_{s} \end{bmatrix}$$

Question: What is $$\mathbf{A}^{-1}$$?

Here is a simple example

$$ \mathbf{B}=\begin{bmatrix} \mathbf{d}_{11}& \mathbf{0}& \mathbf{0} \\ \mathbf{0} & \mathbf{d}_{22} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} & \mathbf{d}_{nn} \end{bmatrix}$$

$$\mathbf{B}=Diag[d_{11}, d_{22}, ..., d_{nn}]$$ (This is the shortened notation for $$ \mathbf{B} $$)

$$ \mathbf{B}^{-1}=Diag[1/d_{11}, 1/d_{22}, ..., 1/d_{nn}]$$

Assumming $$d_{ii}\ne 0$$ for $$ i=1, ..., n$$

For a block diagonal matrix A:

$$\mathbf{A}=Diag[d_{1} ..., d_{s}]$$

$$\mathbf{A}^{-1}=Diag[d_{1}^{-1} ..., d_{s}^{-1}]$$

$$\tilde{\mathbf{T}}^{(e)-1}=Diag[\mathbf{T}^{(e)-1} ..., R^{(e)-1}]$$

We know:

$$\mathbf{R}^{(e)T}=\begin{bmatrix} l^{(e)}& -m^{(e)} \\ m^{(e)} & -l^{(e)} \end{bmatrix}$$

Also

$$\mathbf{R}_{2x2}^{(e)}\mathbf{R}_{2x2}^{(e)T}=\begin{bmatrix} 1& 0 \\ 0 & 1 \end{bmatrix}_{2x2}=\mathbf{I}_{2x2}$$ (This is the Identity matrix)

In conclusion: $$\mathbf{R}^{(e)-1}=\mathbf{R}^{(e)T}$$

So $$\tilde{\mathbf{T}}^{(e)-1}=(Diag[\mathbf{R}^{(e)} ,\mathbf{R}^{(e)}])^{T}$$ where $$Diag[\mathbf{R}^{(e)} ,\mathbf{R}^{(e)}]=\tilde{\mathbf{T}}^{(e)}$$

This shows that: $$\tilde{\mathbf{T}}^{(e)-1}=\tilde{\mathbf{T}}^{(e)T}$$

5 Truss MATLAB code
The 5 bar truss system in example 1.4 in the book can be solved by the below MATLAB code. analysis and explanation of the MATLAB code is shown below. This code and the the functions were downloaded from the books accompanying website:

http://bcs.wiley.com/he-bcs/Books?action=index&itemId=0471648086&bcsId=2256

Functions
The functions PlaneTrussElement.m, NodalSoln.m, PlaneTrussResults.m were all called in the code for the 5 truss system. They are all necessary for the completion of the truss problem.

PlaneTrussElement.m

This function take the Young's Modulus (e), the Area of the element (A) and the coordinates of the element ends (coord) and generates the stiffness matrix for a plane truss element. The function first calculates the lengths of the the elements. Once those are calculated the stiffness matrix is calculated.

NodalSoln.m

The function takes the global coefficient matrix (K), the global right hand side vector (R), the list of degrees of freedom with specified values, and the specified values to determine the displacements and reactions at each node. The dof of the the system is first found by using the  command which finds the longest dimension of R. df then finds the difference between the dofs that have known values (a value of zero) and the dof that were found in the previous line. The displacements and the reactions are then calculated.

PlaneTrussResults.m

This function computes the plane truss element results. It takes in the modulus of elasticity (e), the area of the cross-section (A), the coordinates at the element ends (coord), and the displacements at the elemtent ends (disps) and calculates the axial strain (eps), stress(sigma) and force (force). The results are stored in the variable  and sent back to the main program.

Results from the 5 Truss MATLAB example
The results from the 5 Truss System MATLAB code is shown below. As stated in the PlaneTrussResults.m function the  columns go in the order of axial strain, axial stress, and last axial force.

Code for the plot of the undeformed and deformed five truss system
Below is the code for the plot of the five bar system. Using the results that were obtained from the above MATLAB solution the deformed plot was created. The dotted lines are the undeformed system and the solid lines are the deformed system. The the displacements were multiplied by a factor of 500 so that they would appear on the plot.

Three Bar Truss System
A program was also written to solve the three bar truss system. The code follows the model code that was given to us in HW 2. The functions PlaneTrussElement.m, NodalSoln.m, and PlaneTrussResults.m were once again used in the code to obtain the results. Descriptions of these functions can be found in the section for the Five bar truss system.

Below are the results from the above code.

Plot of undeformed and deformed three bar truss system
Below is the code and the plot of the three bar truss system. The plot was made from the results that were found in the above code. The results demonstrate that only node two had a displacement of (-15.167, 48.231). The undeformed system is plotted with the dotted lines and the deformed system is plotted with solid lines. The deformed displacements have been scaled down by a tenth in order to enhance in the plot.