User:Eml4500.f08.bottle.butler/HW 4

Before we continue with our analysis of the 3-bar truss system we will make a quick comment on modifying a 2-bar truss program in MATLAB to solve a 3-bar truss system.

First, we define the connection array as "conn". Consider the 2-bar truss system:

$$ conn = \begin{bmatrix} 1 & 2\\ 2& 3 \end{bmatrix}$$ where column 1 corresponds to local node $$ 1 \! $$ and column 2 corresponds to local node $$ 2 \! $$. Similarly, row 1 represents element $$ 1 \! $$ and row 2 represents element $$ 2 \! $$. This array can then be represented by conn(e,j) = global node # of local node $$ j \! $$ of element $$ e \! $$

We denote the location master array by "lmm". Therefore

$$ lmm = \begin{bmatrix} 1 & 2 & 3 & 4\\ 3& 4 & 5 & 6 \end{bmatrix} $$

where row 1 represents element $$ 1 \! $$ and row 2 represents element $$ 2 \! $$, each column represents a local degree of freedom number with each number in the matrix representing the global degree of freedom number (or equation number) in $$ K \! $$.

This can be rewritten as limm(i,j)= equation number (global degree of freedom) for the element stiffness matrix coefficient corresponding to the $$jth \! $$ local degree of freedom number

Returning to our analysis of FEA, our goal is to find $$ \tilde{T}^{(e)}_{4X4} $$ that transforms the set of local element dof's $$ d^{(e)}_{4X1} $$ to another set of local element $$ \tilde{d}^{(e)}_{4X1}$$ such that $$ \tilde{T}^{(e)} $$ is invertible.

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From previous lectures we know that:

$$ \underbrace {\tilde{d}_{1}^{(e)}}_{q_{1}^{(e)}} = \begin{vmatrix} l^{(e)} & m^{(e)} \end{vmatrix}\begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)} \end{Bmatrix} $$

and $$ \tilde{d}_{2}^{(e)}= \vec{d}^{(e)}_{1}\circ \vec{\tilde{j}} $$ $$= -\sin \theta ^{(e)}d_{1}^{(e)}+\cos \theta ^{(e)}d_{2}^{(e)}$$

therefore:

$$ \tilde{d}_{2}^{(e)} = \begin{vmatrix} -m^{(e)} & l^{(e)} \end{vmatrix}\begin{Bmatrix} d^{(e)}_{1}\\ d_{2}^{(e)} \end{Bmatrix} $$

If we combine the two we get the following equation in matrix form: $$ \begin{Bmatrix} \tilde{d}_{1}^{(e)}\\ \tilde{d}_{2}^{(e)} \end{Bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)} & l^{(e)} \end{bmatrix}\begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)} \end{Bmatrix} $$

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

This can be expanded to : $$\underbrace{\begin{Bmatrix} \tilde{d}^{(e)}_{1}\\ \tilde{d}^{(e)}_{2}\\ \tilde{d}^{(e)}_{3}\\ \tilde{d}^{(e)}_{4} \end{Bmatrix}}_{\tilde{d}^{(e)}_{4X1}}= \underbrace{\begin{bmatrix} R^{(e)} & 0\\ 0 & R^{(e)} \end{bmatrix}}_{\tilde{T}^{(e)}_{4X4}}\underbrace{\begin{Bmatrix} d^{(e)}_{1}\\ d^{(e)}_{2}\\ d^{(e)}_{3}\\ d^{(e)}_{4} \end{Bmatrix}}_{d^{(e)}_{4X1}}$$

Examining the element below

we find $$ \tilde{f}^{(e)}= k^{(e)}\begin{bmatrix} 1 &0 & -1 & 0\\ 0&  0& 0 & 0\\  -1&  0&  1& 0\\  0& 0 & 0 &0  \end{bmatrix}\tilde{d}^{(e)}$$

where $$ \tilde{f}^{(e)}_{4X1}=\tilde{k}^{(e)}_{4X4}\tilde{d}^{(e)}_{4X1} $$

In previous meetings we found the eigenvalues associated with a 2-bar truss system. Four eigenvalues were found, three of which were due to rigid body motion and the forth resulting from mechanisms. To understand the concept of a mechanism, we imagine that the support on a 2-bar truss system is cut and the structure can now move freely. The two bars can now move in a "scissor" motion about the pin joint. This is possible because there is no stored energy in the pin.

For the eigenvalue problem we have: $$ Kv=\lambda v \! $$, where $$ Kv $$ is the mass matrix and $$ \lambda v $$ is the identity matrix.

Let $$ \begin{Bmatrix} (u_{1})_{6X1} & (u_{2})_{6X1} & (u_{3})_{6X1} & (u_{4})_{6X1} \end{Bmatrix} $$ be pure eigenvalues corresponding to the four eigenvalues.

$$ K_{6X6}(u_{i})_{6X1}= \underbrace {0 u_{i}}_{0_{6X1}}, i = 1, ... ,4 $$

We can write the linear combination of $$ \begin{Bmatrix} u_{i}, i = 1, ... , 4 \end{Bmatrix}$$ as $$ \Sigma (\alpha _{i})_{1X1}(u_{i})_{6X1} = W_{6X1} $$

where $$\alpha _{i} = real numbers $$

Physically this means that the linear combination of movement in the  x \! and $$ y \! $$ directions and rotation.

$$ W \! $$ is also an eigenvalue that corresponds to a zero eigenvalue $$ KW=K(\sum_{i = 1}^{4} \alpha_{i}u_{i} ) $$ $$= \sum_{i = 1}^{4}{\alpha _{i}\underbrace{(Ku_{i})}_{0_{6X1}}} = 0 $$ $$ = 0_{1x1} W_{6X1} \!$$ We will now begin the justification of the assembly of the elemental stiffness matrix $$ k^{(e)} \!$$ into the global stiffness matrix.

Consider the example of the 2-bar truss and recall the following: 1. The elemental FD relationship can be written as $$ k^{(e)}d^{(2)}=f^{(e)} \!$$ 2. We used the Euler cut principle to find the equivalent of global node $$ 2 \! $$ 3. We drew the FBD's of element $$ 1 \! $$ and $$ 2 \! $$ to find the elemental degrees of freedom $$ d^{(e)}_{4X1} $$ 4. We identified the global degrees of freedom to elemental degrees of freedom for node $$ 2 \! $$  for element  $$ 1 \! $$ and $$ 2 \! $$

We can draw the following FBD to represent the equilibrium for node $$ 2 \! $$

We can now see that the addition of factors in the global stiffness matrix is because of the equilibrium position of node $$ 2 \! $$. The equations of equilibrium are $$ \sum{F_{x}}= 0 = -f^{(1)}_{3}-f^{(2)}_{1}$$ $$ \sum{F_{y}}= 0 = P-f^{(1)}_{4}-f^{(2)}_{2} $$