User:EML4500.F08.RAMROD.B/Homework 2

= Element One Stiffness Analysis =

Taking all of the knowledge we have gained into consideration, we can now begin the analysis of the stiffness matrix for element one. Because we know that the element stiffness matrix is derived with the use of only three variables,

$$\displaystyle \ l^{e}$$     $$\displaystyle \  m^{e}$$     $$\displaystyle \ k^{e}$$

it is imperative to the completion of this problem to compute these values so that the element stiffness matrix for element one can be constructed. And if you have already forgotten, $$\displaystyle \ k^{e}$$ is the axial stiffness of the bar element $$\displaystyle \ e $$ and is given by the following formula,

$$\displaystyle \ k^{e} = \left ( \frac{E^{e} A^{e}}{L^{e}} \right )$$

where $$\displaystyle \ E^{e} $$, $$\displaystyle \ A^{e} $$, and $$\displaystyle \ L^{e} $$ are the modulus of elasticity, cross sectional area, and length of an element $$\displaystyle \ e $$ respectively. If you remember from previous work, the equations for $$\displaystyle \ l^{e}$$ and $$\displaystyle \ m^{e} $$ have already been computed and are as follows;

$$\displaystyle\ l^{e} = \cos{(\theta^{(e)}})$$

$$\displaystyle\ m^{e} = \sin{(\theta^{(e)}})$$

where θ is the angle at which the truss element is tilted. If you recall from the problem statement, the following values for the length, angle, modulus, and cross sectional area were given for element one.

With this information it is possible to begin to calculate the three variables needed to construct the element stiffness matrix.

$$\displaystyle \ l^{1} = cos (30) = \left ( \frac{\sqrt{3}}{2} \right ) $$

$$\displaystyle \ m^{1} = sin (30) = \left ( \frac{1}{2} \right ) $$

$$\displaystyle \ k^{1} = \left ( \frac{(3)(1)}{4} \right ) = \left ( \frac{3}{4} \right ) $$

Having calculated these values, it is now possible to construct the stiffness matrix for element one. Before we defined the element stiffness matrix as the following;

$$\displaystyle \ \mathit{k}^{(e)} = k^{e} \begin{bmatrix} (l^{e})^2 & l^em^e & -(l^e)^2 & -l^em^e\\ l^em^e & (m^e)^2 & -l^em^e & -(m^e)^2\\ -(l^{e})^2 & -l^em^e & (l^e)^2 & l^em^e\\ -l^em^e & -(m^e)^2 & l^em^e & (m^e)^2 \end{bmatrix}$$

To make things a little more condensed and easier to understand, we will multiply the axial stiffness value through and represent the element stiffness matrix as follows;

$$\displaystyle \ k^{(1)} = $$ $$\displaystyle \begin{bmatrix} k_{1,1}^1 & k_{1,2}^1 & k_{1,3}^1 & k_{1,4}^1 \\ \\k_{2,1}^1 & k_{2,2}^1 & k_{2,3}^1 & k_{2,4}^1 \\ \\ k_{3,1}^1 & k_{3,2}^1 & k_{3,3}^1 & k_{3,4}^1 \\ \\ k_{4,1}^1 & k_{4,2}^1 & k_{4,3}^1 & k_{4,4}^1\end{bmatrix}$$,

where   $$\displaystyle \ k_{1,1}^1 = k^{1}* l^{1^{\,\!2}} $$ and    $$\displaystyle \ k_{1,2}^1 = k^{1}*{l^{1}}*{m^{1}} $$ and so on until $$\displaystyle\ k^{e} $$ (axial stiffness) has been multiplied through the director cosines matrix. A short hand form of the same matrix can be constructed in which the entire four by four matrix is represented by a one by one matrix. This form is as follows,

$$\displaystyle \begin{bmatrix} k_{i,j}^{(e)} \end{bmatrix}$$  $$\displaystyle \ 0 \le i \le 4 $$     $$\displaystyle \ 0 \le j \le 4 $$

where $$\displaystyle \ 0 \le i \le 4 $$ and $$\displaystyle \ 0 \le j \le 4 $$. This matrix along with the equivalent matrix above is the element stiffness matrix. All of the required numerical values are at our disposal so it is now time to compute the values that will be placed into the matrix. As stated before,

$$\displaystyle \ k_{1,1}^1 = k^{1}*l^{1^{\,\!2}} $$ and $$\displaystyle \ k_{1,2}^1 = k^{1}*{l^{1}}*{m^{1}} $$

If a closer look is taken at the director cosine matrix, one will notice two important aspects that characterize the values within the matrix. The first is that there are only three numbers need to be calculated in order to completely fill the matrix, two of which are displayed in the two equations above. The second is that this matrix is symmetrical, meaning that if a line was drawn diagonally across the matrix as to cut it into two pieces, the upper portion would be a mirror image of the bottom portion. In general this means

$$\displaystyle \ k_{i,j}^e = k_{j,i}^e $$ or  $$\displaystyle \ K^e = K^{{e}^T} $$

where $$\displaystyle \ T$$ represents the transpose of the element stiffness matrix of an element $$\displaystyle \ e $$.

As was previously stated, the element stiffness matrix is composed of only three values, of which are computed below.

$$\displaystyle \ k_{1,1}^1 = k^{1}*l^{1^{\,\!2}} = \left ( \frac{9}{16} \right )$$ $$\displaystyle \ k_{1,2}^1 = k^{1}*{l^{1}}*{m^{1}} = \left ( \frac{3\sqrt{3}}{16} \right )$$ $$\displaystyle \ k_{2,2}^1 = k^{1}*m^{1^{\,\!2}} = - \left ( \frac {3}{16} \right )$$

It is now possible to construct the stiffness matrix $$\displaystyle \ k $$ for element one.

$$\displaystyle \ k^1 = $$ $$\displaystyle \begin{bmatrix} \left ( \frac{9}{16} \right ) & \left ( \frac{3\sqrt{3}}{16} \right ) & - \left ( \frac{9}{16} \right ) & -\left ( \frac{3\sqrt{3}}{16} \right ) \\ \\ -\left ( \frac{9}{16} \right ) & -\left ( \frac{3\sqrt{3}}{16} \right ) & \left ( \frac{9}{16} \right ) & \left ( \frac{3\sqrt{3}}{16} \right ) \\ \\ -\left ( \frac{3\sqrt{3}}{16} \right ) & \left ( \frac {3}{16} \right ) & \left ( \frac{3\sqrt{3}}{16} \right ) & - \left ( \frac {3}{16} \right ) \\ \\ -\left ( \frac{3\sqrt{3}}{16} \right ) & \left ( \frac {3}{16} \right ) & \left ( \frac{3\sqrt{3}}{16} \right ) & - \left ( \frac {3}{16} \right )\end{bmatrix}$$

=Element Two Stiffness Analysis=

Now that the stiffness matrix for element one is completed, construction of the stiffness matrix for element two can begin. To do so requires us to perform the same method as used for element one. To begin, lets remind ourselves of the parameters for element two given in the problem statement:

With this data we can compute the axial stiffness as well as the director cosine values $$\displaystyle \ l^2, m^2 $$.

$$\displaystyle \ k^{2} = \left ( \frac{E^{2} A^{2}}{L^{2}} \right ) = \left ( \frac{5*2}{2} = 5\right )$$ $$\displaystyle \ l^{1} = cos (-45) = \left ( \frac{\sqrt{2}}{2} \right ) $$ $$\displaystyle \ m^{1} = sin (-45) = -\left ( \frac{\sqrt{2}}{2} \right ) $$

And just as before, the stiffness matrix can be represented as the director cosine matrix multiplied through by the axial stiffness matrix.

$$\displaystyle \ k^2 = $$ $$\displaystyle \begin{bmatrix} k_{1,1}^2 & k_{1,2}^2 & k_{1,3}^2 & k_{1,4}^2 \\ \\k_{2,1}^2 & k_{2,2}^2 & k_{2,3}^2 & k_{2,4}^2 \\ \\ k_{3,1}^2 & k_{3,2}^2 & k_{3,3}^2 & k_{3,4}^2 \\ \\ k_{4,1}^2 & k_{4,2}^2 & k_{4,3}^2 & k_{4,4}^2\end{bmatrix}$$

where

$$\displaystyle \ k_{1,1}^{2} = k^2 * (l^1)^2 $$.

All values that are needed to complete the stiffness matrix are now known, but if a further observation is made the process of constructing this matrix becomes very simple. If you notice that the absolute value of the director cosine values of $$\displaystyle \ l^2, m^2 $$ are exactly the same, it is only necessary to compute one value of the stiffness matrix. This is possible because the director cosine matrix is a combination of three different processes:

$$\displaystyle \ (l^1)^2, (m^1)^2, m^2 * l^2 $$

Since all of these calculations yield the same values, only one calculation is needed to complete the matrix. Then the value can be placed and the sign of the values can appropriately be assigned as necessary. Below is the once calculation needed in order to carry out this process.

$$\displaystyle \ k_{1,1}^2 = k^2 * (l^1)^2 = \left ( \frac{5}{2} \right )$$

This then makes the element two stiffness matrix to be the following,

$$\displaystyle \ k^1 = $$ $$\displaystyle \begin{bmatrix} \left ( \frac{5{2}}{2} \right ) & -\left ( \frac{5}{2} \right ) & - \left ( \frac{5}{2} \right ) & \left ( \frac{5}{2} \right ) \\ \\ -\left ( \frac{5}{2} \right ) & \left ( \frac{5}{2} \right ) & \left ( \frac{5}{2} \right ) & -\left ( \frac{5}{2} \right ) \\ \\ -\left ( \frac{5}{2} \right ) & \left ( \frac{5}{2} \right ) & \left ( \frac{5}{2} \right ) & -\left ( \frac{5}{2} \right ) \\ \\ \left ( \frac{5}{2} \right ) & -\left ( \frac{5}{2} \right ) & -\left ( \frac{5}{2} \right ) & \left ( \frac{5}{2} \right )\end{bmatrix}$$

Just as the element stiffness matrix for element one, this stiffness matrix is symmetric. In other words,

$$\displaystyle \ k_{i,j}^2 = k_{j,i}^2 $$ or  $$\displaystyle \ K^{(2)} = K^{{2}^T} $$

where $$\displaystyle \ T$$ once again represents the transpose of the element stiffness matrix of an element $$\displaystyle \ 2 $$.

=Element Force-Displacement Relationship=

Now that we have found the stiffness matrices for both elements, we must now find the relationship between the force and displacement. In order to do so we must analysis the following equation:

$$\displaystyle \ k^e * d^e = f^e $$

where $$\displaystyle \ k, d, $$ and $$\displaystyle \ f $$ are the stiffness matrix, displacement matrix , and force displacement of an element $$\displaystyle \ e $$ respectively. The math works as the stiffness matrix is a $$\displaystyle \ 4 X 4 $$ matrix and when multiplied by a $$\displaystyle \ 4 X 1 $$ matrix (element displacment matrix), the product becomes a $$\displaystyle \ 4 X 1 $$ matrix that is the element force matrix.

$$\displaystyle \ k^{(e)} = $$ $$\displaystyle \begin{bmatrix} k_{1,1}^{(e)} & k_{1,2}^{(e)} & k_{1,3}^{(e)} & k_{1,4}^{(e)} \\ \\k_{2,1}^{(e)} & k_{2,2}^{(e)} & k_{2,3}^{(e)} & k_{2,4}^{(e)} \\ \\ k_{3,1}^{(e)} & k_{3,2}^{(e)} & k_{3,3}^{(e)} & k_{3,4}^{(e)} \\ \\ k_{4,1}^{(e)} & k_{4,2}^{(e)} & k_{4,3}^{(e)} & k_{4,4}^{(e)}\end{bmatrix}$$ $$\displaystyle \ d^{(e)} = $$ $$\displaystyle \begin{bmatrix} d_{1}^{(e)} \\ \\ d_{2}^{(e)} \\ \\ d_{3}^{(e)} \\ \\ d_{4}^{(e)}\end{bmatrix}$$ $$\displaystyle \ f^{(e)} = $$ $$\displaystyle \begin{bmatrix} f_{1}^{(e)} \\ \\ f_{2}^{(e)} \\ \\ f_{3}^{(e)} \\ \\ f_{4}^{(e)}\end{bmatrix}$$