User:Eml4500.f08.Lulz.js/HW4

Array Definitions
Consider the connectivity array "conn" for the two-bar truss system shown in Figure ##:

$$ \mathbf{conn} = \begin{bmatrix} 1 & 2\\ 2 & 3 \end{bmatrix} $$

The connectivity array describes the local nodes of a truss element in terms of the global node numbering convention. The rows correspond to individual truss elements, and the columns to local nodes 1 and 2 of those elements. In other words, conn(e,j) refers to the global node number of local node j for element e.

A related concept is the location master matrix lmm:

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

The location master matrix describes the local degrees of freedom for a truss element in terms of the global degrees of freedom. Like with the connectivity array, the rows correspond to individual truss elements, and the columns refer to local degrees of freedom 1-4 for those elements. In other words, lmm(i,j) refers to the global degree of freedom of local degree of freedom j of element i.

Transformation Matrix
A previous discussion centered on the inability to invert the transformation matrix $$\mathbf{T^{(e)}}$$ used in the axial displacement relationship $$\mathbf{q^{(e)}} = \mathbf{T^{(e)}}\cdot\mathbf{d^{(e)}}$$. Our goal is to convert the 2x4 transformation matrix into a 4x4 matrix $$\mathbf{\tilde{T}^{(e)}}$$ that can be easily inverted.



Figure ## shows a general diagram of axial/transvere displacements $$\tilde{d}^{(e)}_{j}$$ on a truss element e. The relationship describing this case is $$\mathbf{\tilde{d}^{(e)}} = \mathbf{\tilde{T}^{(e)}}\cdot\mathbf{d^{(e)}}$$.

Through the same geometric process used to determine $$\mathbf{q^{(e)}}$$ in a previous homework, it can be shown that the two expressions below are true:

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

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

Combining these two expressions yields the following relationship for the axial/transverse displacements at one local node:

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

This describes a new displacement relationship given by $$\mathbf{\tilde{d}^{(e)}} = \mathbf{R^{(e)}}\cdot\mathbf{d^{(e)}}$$. Extending this to include both local nodes of a truss element yields:

$$\begin{Bmatrix} \tilde{d}^{(e)}_{1}\\\tilde{d}^{(e)}_{2}\\\tilde{d}^{(e)}_{3}\\\tilde{d}^{(e)}_{4}\end{Bmatrix} = \begin{bmatrix} l^{(e)}&m^{(e)}&0&0\\-m^{(e)}&l^{(e)}&0&0\\0&0&l^{(e)}&m^{(e)}\\0&0&-m^{(e)}&l^{(e)}\end{bmatrix}\begin{Bmatrix} d^{(e)}_{1}\\ d^{(e)}_{2}\\d^{(e)}_{3}\\ d^{(e)}_{4} \end{Bmatrix} = \begin{bmatrix} \mathbf{R^{(e)}}&\mathbf{0}\\\mathbf{0}&\mathbf{R^{(e)}}\end{bmatrix}\begin{Bmatrix} d^{(e)}_{1}\\ d^{(e)}_{2}\\d^{(e)}_{3}\\ d^{(e)}_{4} \end{Bmatrix}$$

Figure ## shows the axial/transverse forces for a truss element. The axial/transverse force-displacement relationship is given by $$\mathbf{\tilde{f}^{(e)}} = \mathbf{\tilde{k}^{(e)}}\cdot\mathbf{\tilde{d^{(e)}}}$$. The modified element stiffness matrix $$\mathbf{\tilde{k}^{(e)}}$$is defined below:

$$\mathbf{\tilde{k}^{(e)}} = k^{(e)}\begin{bmatrix} 1&0&-1&0\\0&0&0&0\\-1&0&1&0\\0&0&0&0\end{bmatrix}$$

Consider the case where $$\tilde{d}^{(e)}_{1} = \tilde{d}^{(e)}_{2} = \tilde{d}^{(e)}_{3}= 0$$, but $$\tilde{d}^{(e)}_{4}\neq 0$$. Because the 4th column of matrix $$\mathbf{\tilde{k}^{(e)}}$$ (corresponding to transverse displacement $$\tilde{d}^{(e)}_{4}$$) is a vector of zeros, the following is true:

$$\mathbf{\tilde{f}^{(e)}} = 0$$.

This is the interpretation of transverse degrees of freedom.

Finally, we have determined that $$\mathbf{\tilde{d}^{(e)}} = \mathbf{\tilde{T}^{(e)}}\cdot\mathbf{d^{(e)}}$$ is valid. A corresponding relationship can be defined for the axial/transverse forces:

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

We also know that $$\mathbf{\tilde{f}^{(e)}} = \mathbf{\tilde{k}^{(e)}}\cdot\mathbf{\tilde{d^{(e)}}}$$ is valid, leading to the following logic:

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

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

Assuming that $$\mathbf{\tilde{T}^{(e)}}$$ is an invertible matrix, the following is true:

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

Inversion of Diagonal Matrices
Consider a general $$n\;x\;n$$ block diagonal matrix $$\mathbf{A}$$:



In order to explore the inverse, first consider an example matrix $$\mathbf{B}$$:



This matrix can alternately be described using the following notation for diagonal matrices:

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

If $$d_{ii} \neq 0$$ for $$i\,=\,1,...,n$$, the inverse of this matrix can be expressed as $$\mathbf{B^{-1}} = Diag[\frac{1}{d_{11}},\frac{1}{d_{22}},...,\frac{1}{d_{nn}}]$$. Applying this procedure to $$\mathbf{A}$$ yields the following:

$$\mathbf{A} = Diag[\mathbf{D_{1}},\mathbf{D_{2}},...,\mathbf{D_{n}}]$$

$$\mathbf{A^{-1}} = Diag[\mathbf{D_{1}^{-1}},\mathbf{D_{2}^{-1}},...,\mathbf{D_{n}^{-1}}]$$

Transformation Matrix Continued
Returning to the modified transformation matrix $$\mathbf{\tilde{T}^{(e)}}$$ and applying the inversion methods discussed above, the matrix $$\mathbf{\tilde{T}^{(e)^{-1}}}$$ can be written as follows:

$$\mathbf{\tilde{T}^{(e)^{-1}}} = Diag[\mathbf{R^{(e)^{-1}}},\mathbf{R^{(e)^{-1}}}]$$, where $$\mathbf{R^{(e)}} = \begin{bmatrix} l^{(e)}&-m^{(e)}\\m^{(e)}&l^{(e)}\end{bmatrix}$$

Additionally, the transpose of matrix $$\mathbf{R^{(e)}}$$ is $$\begin{bmatrix} l^{(e)}&m^{(e)}\\-m^{(e)}&l^{(e)}\end{bmatrix}$$, leading to the following relation:

$$\mathbf{R^{(e)^{T}}}\cdot\mathbf{R^{(e)}} = \begin{bmatrix} l^{(e)}&m^{(e)}\\-m^{(e)}&l^{(e)}\end{bmatrix}\cdot \begin{bmatrix} l^{(e)}&-m^{(e)}\\m^{(e)}&l^{(e)}\end{bmatrix} = \begin{bmatrix} l^{(e)^{2}}+ m^{(e)^{2}}&0\\0&m^{(e)^{2}}+l^{(e)^{2}}\end{bmatrix}$$

Recalling that $$l^{(e)} = \cos{\mathbf{\theta}^{(e)}}$$ and $$m^{(e)} = \sin{\mathbf{\theta}^{(e)}}$$, and the trigonometric identity $$\cos{^{2}\mathbf{\theta}} + \sin{^{2}\mathbf{\theta}} = 1$$, the following is true:

$$\mathbf{R^{(e)^{T}}}\cdot\mathbf{R^{(e)}} = \begin{bmatrix} 1&0\\0&1\end{bmatrix} = \mathbf{I}$$, where $$\mathbf{I}$$ is the 2 x 2 identity matrix.

Since it is a known property that the product of a matrix and it's inverse (e.g. $$\mathbf{M}\cdot\mathbf{M^{-1}} = \mathbf{I}$$), it is apparent that $$\mathbf{R^{(e)^{T}}} = \mathbf{R^{(e)^{-1}}}$$. Thus,

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

The expression for the modified force-displacement relationship derived earlier can now be rewritten as:

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

Not only is the expression inside the brackets easily determined, it also resembles the form of the general force-displacement relationship, and the following can be shown:

$$\mathbf{k^{(e)}} = [\mathbf{\tilde{T}^{(e)^{T}}}\cdot\mathbf{\tilde{k}^{(e)}}\cdot\mathbf{\tilde{T}^{(e)}}]$$

Discussion of Eigenvalues and Eigenvectors
Consider the eigenvalue problem $$\mathbf{kv} = \mathbf{\lambda v}$$. Let $$[\mathbf{u_{1}},\mathbf{u_{2}},\mathbf{u_{3}},\mathbf{u_{4}}]$$ be the eigenvectors for the 4 zero eigenvalues:

$$\mathbf{ku_{i}} = 0\cdot\mathbf{u_{i}} = \mathbf{0}$$, for i = [1,2,3,4].

A linear combination of {$$\mathbf{u_{i}}$$} can be defined as $$\sum_{i=1}^{4} \alpha_{i}\mathbf{u_{i}} = \mathbf{W}$$, where $$\alpha_{i}\;$$ refers to real number constants.

The vector $$\mathbf{W}$$ is also an eigenvector corresponding a zero eigenvalue, as shown below:

$$\mathbf{kW} = \mathbf{k}(\sum_{i=1}^{4} \alpha_{i}\mathbf{u_{i}}) = \sum_{i=1}^{4} \alpha_{i}\mathbf{ku_{i}} = \mathbf{0}$$

Justification of Assembly of Element Stiffness Matrices into Global Stiffness Matrix
ERIC YOUR STUFF GOES HERE



Figure ## shows the generic two-bar truss system, with global node 2 isolated for analysis. Two equations are immediately evident from the diagram:

$$(1)\;\Sigma F_{x} = -f_{3}^{(1)} - f_{1}^{(2)} = 0$$

$$(2)\;\Sigma F_{y} = \mathbf{P} - f_{4}^{(1)} - f_{2}^{(2)} = 0$$

Equation 2 can also be written as $$\mathbf{P} = f_{4}^{(1)} + f_{2}^{(2)}$$ Using the general force-displacement relationship $$\mathbf{k^{(e)}}\cdot\mathbf{d^{(e)}}= \mathbf{f^{(e)}}$$, Equation 1 can be rewritten as follows:

$$0 = [k_{31}^{(1)}d_{1}^{(1)} + k_{32}^{(1)}d_{2}^{(1)} + k_{32}^{(1)}d_{3}^{(1)} + k_{34}^{(1)}d_{4}^{(1)}] + [k_{11}^{(2)}d_{1}^{(2)} + k_{12}^{(2)}d_{2}^{(2)} + k_{13}^{(2)}d_{3}^{(2)} + k_{14}^{(2)}d_{4}^{(2)}]$$

Replacing the local displacement degrees of freedom with the global degrees of freedom produces the following:

$$0 = [k_{31}^{(1)}d_{1} + k_{32}^{(1)}d_{2} + k_{32}^{(1)}d_{3} + k_{34}^{(1)}d_{4}] + [k_{11}^{(2)}d_{3} + k_{12}^{(2)}d_{4} + k_{13}^{(2)}d_{5} + k_{14}^{(2)}d_{6}]$$

Comparison of the above with the third row of $$\mathbf{K}$$ as described in Meeting 8 shows that the two are identical.

An expression for the assembly of $$ned\;x\;ned$$ element stiffness matrices $$\mathbf{k^{(e)}}$$ for $$e\;=\; 1,...,nel$$, where $$nel$$ refers to the number of elements, into the $$n\;x\;n$$ global stiffness matrix $$\mathbf{K}$$ can be expressed as follows:

$$ \mathbf{K} = \begin{matrix} nel\\A\\e=1\end{matrix}\mathbf{k^{(e)}}$$

$$n\;$$ is the number of global degrees of freedom before the application of boundary conditions. $$ned\;$$ is the number of element degrees of freedom. $$ned\;$$ must be less than $$n\;$$. $$A\;$$ is a symbol representing the operation of assembly.

Principle of Virtual Work (PVW)
PVW was previously used as the validation for elimination of rows in the global stiffness matrix that corresponded to zero-valued boundary conditions to produce $$\mathbf{\bar{k}}$$. It was also used to derive $$\mathbf{q^{(e)}} = \mathbf{T^{(e)}} \cdot \mathbf{d^{(e)}}$$, the axial displacement relationship, and also $$\mathbf{k^{(e)}} = \mathbf{T^{(e)^{T}}} \cdot \mathbf{\hat{k}^{(e)}} \cdot \mathbf{T^{(e)}}$$, the alternate expression for the element stiffness matrix based on the transformation matrix.

Deriving FEM for PDEs
The force-displacement relationship for a bar of spring is given by $$\mathbf{kd}=\mathbf{F}$$. This relationship implies the following:

$$(3)\;\mathbf{kd} - \mathbf{F} = 0$$

An equivalent expression to Equation 3 is also implied:

$$(4)\;\mathbf{W}(\mathbf{kd} - \mathbf{F}) = 0$$ for all $$\mathbf{W}$$.

Proof:

$$(3)\leftrightarrow(4)$$ -  trivial $$(4)\leftrightarrow(3)$$ -  NOT TRIVIAL

Since $$(4)\;$$ is valid for all $$\mathbf{W}$$ selecting $$\mathbf{W}\,=\,1$$ is an acceptable operation.

$$(4)\;$$ then becomes $$\mathbf{kd} - \mathbf{F} = 0$$, which is the same as $$(3)\;$$.