User:Eml4500.f08.RAMROD.A/HW2

Statically Indeterminate 3 Truss System
A look at the system with global nodes identifited:



Element Picture
We want to look at the elemental picture with $$\displaystyle {\color{Green} Global }$$ and $$\displaystyle {\color{Blue} Local }$$ coordinate systems:



While the $$\displaystyle {\color{Blue} Local }$$ Coordinates are denoated by $$\displaystyle {\color{Blue} (\tilde{x}, \tilde{y}) } $$.

 * $$\displaystyle \tilde{x} $$. is defined as along the element from local node 1 to 2.

Vector Notation

 * A common occurrence when dealing with forces such as P or the lengths of elements is to use vector notation: $$\displaystyle ({i}, {j})$$ a long with the angle between $$\displaystyle x $$ and $$\displaystyle \tilde{x}$$; denoted as $$\displaystyle \theta $$.



Next Step of Finite Element Method
It is important to go back to the equation described in the recipe from [earlier articles] on Finite Element Method:

K(e) * d(e) = f(e)

Where:


 * K is the element stiffness matrix


 * d is the element displacement matrix


 * f is the element force matrix

It is now possible to determine ke using the aforementioned $$\displaystyle \theta $$ and what we call the direction cosines

Direction Cosines
$$\displaystyle l^e $$ and $$\displaystyle m^e $$.


 * $$\displaystyle l^e = \cos (\theta) $$


 * $$\displaystyle m^e = \cos (90-\theta) = \sin (\theta) $$

Element Stiffness Matrix, k
$$\displaystyle \mathbf{k^e}=\frac{E^e*A^e}{L^e}\left[Direction Cosine Matrix \right] $$

Where: "$\displaystyle L = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} $"
 * E = Young's Modulus of element
 * A = Cross sectional Area of element
 * L = Length of element

When the cosine matrix is added in, we get: $$\displaystyle \mathbf{k^e}=\frac{E^e*A^e}{L^e} \begin{bmatrix} l^2 & (l\cdot m) & -l^2  & -(l\cdot m) \\ \\(l\cdot m) & m^2 & -(l\cdot m) & -m^2 \\ \\-l^2 & -(l\cdot m) & l^2 & (l\cdot m) \\ \\-(l\cdot m) & -m^2 & (l\cdot m) & m^2 \end{bmatrix}$$

= 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.

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

In Compact Notation

 * [kij]6x6{dj}6x1 = {Fi}6x1
 * (more generally, the matrices are nxn)

$$\sum_{j=1}^{6}{ K_{ij}d_j }\! = F_i$$,    i = 1, 2, 3, ...., 6

The three matrices used are shown below: 


 * Knxn = [Kij]nxn = global stiffness matrix


 * dnxn = {dj}nxn = global displacement matrix


 * Fnxn = {Fi}nxn = global force matrix

Recall the element force-displacement relation
 * k(e)4x4d(e)4x1=f(e)4x1


 * kn4x4 = [kij(e)] = element stiffness matrix


 * dnxn = {dj(e)} = element displacement matrix


 * Fnxn = {fi(e)} = elementforce matrix

Combining K matrices
$$\mathbf{K}= \begin{bmatrix} K_{11} & K_{12} & K_{13} & K_{14} & K_{15} & K_{16}\\K_{21} & K_{22} & K_{23} & K_{24} & K_{25} & K_{26}\\K_{31} & K_{32} & K_{33} & K_{34} & K_{35} & K_{36}\\K_{41} & K_{42} & K_{43} & K_{44} & K_{45} & K_{46}\\K_{51} & K_{52} & K_{53} & K_{54} & K_{55} & K_{56}\\K_{61} & K_{62} & K_{63} & K_{64} & K_{65} & K_{66}\end{bmatrix}$$

The appropriate numbers will be inputted into matrix K which is a 6x6 matrix.

$$\mathbf{K}= \begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0\\k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0\\k_{31}^{(1)} & k_{32}^{(1)} & (k_{33}^{(1)}+k_{11}^{(2)}) & (k_{34}^{(1)}+k_{12}^{(2)}) & k_{13}^{(2)} & k_{14}^{(2)}\\k_{41}^{(1)} & k_{42}^{(1)} & (k_{43}^{(1)}+k_{21}^{(2)}) & (k_{44}^{(1)}+k_{22}^{(2)}) & k_{23}^{(2)} & k_{24}^{(2)}\\0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)}\\0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)}\end{bmatrix}$$

$$k_{11}^{(1)}=\frac{9}{16}=0.5625$$

$$k_{12}^{(1)}=\frac{3\sqrt{3}}{16}=0.3248$$

$$k_{13}^{(1)}=-k_{11}^{(1)}=-0.5625$$

$$k_{14}^{(1)}=-k_{12}^{(1)}=-0.3248$$

$$k_{22}^{(1)}=\frac{3}{16}=0.1875$$

$$k_{24}^{(1)}=-k_{22}^{(1)}=-0.1875$$

$$k_{13}^{(2)}=-k_{11}^{(2)}=\frac{-5}{2}=-2.5$$

$$k_{14}^{(2)}=-k_{12}^{2}=2.5$$

$$k_{24}^{(2)}=\frac{-5}{2}=-2.5$$

$$k_{33}^{(2)}=\frac{5}{2}=2.5$$

$$k_{34}^{(2)}=\frac{-5}{2}=-2.5$$

$$k_{44}^{(2)}=\frac{5}{2}=2.5$$

$$K_{15}^{}=K_{16}=K_{25}=K_{26}=K_{51}=K_{52}=K_{61}=K_{62}=0$$

$$\mathbf{}= \begin{bmatrix} 0.5625 & 0.32475 & -0.5625 & -0.32475 & 0 & 0\\0.32475 & 0.1875 & -0.3247 & -0.1875 & 0 & 0\\-0.5625 & -0.32475 & (0.5625+2.5) & (0.32475-2.5) & -2.5 & 2.5\\-0.32475 & -0.1875 & (0.32475-2.5) & (0.1875+2.5) & 2.5 & -2.5\\0 & 0 & -2.5 & 2.5 & 2.5 & -2.5\\0 & 0 & 2.5 & -2.5 & -2.5 & 2.5\end{bmatrix}$$

$$\mathbf{K}=\begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3248 & 0 & 0\\0.3248 & 0.1875 & -0.3248 & -0.1875 & 0 & 0\\-0.5625 & -0.3248 & 3.0625 & -2.1752 & -2.5 & 2.5\\-0.3248 & -0.1875 & -2.1752 & 2.6875 & 2.5 & -2.5\\0 & 0 & -2.5 & 2.5 & 2.5 & -2.5\\0 & 0 & 2.5 & -2.5 & -2.5 & 2.5\end{bmatrix}$$

Elimination of known degrees of freedom in order to reduce the global Force-Displacement relationship
For this problem, the reduction is due to the fact that global nodes 1 and 3 are fixed and therefore have zero displacement, or:

$$d_{1} = d_{2} = d_{5} = d_{6} = 0$$

Setting these displacements to zero is the equivalent of eliminating the corresponding columns (1,2,5 & 6) of the global stiffness matrix, K. This elimination is shown below.

$$K = \begin{bmatrix} K_{13} & K_{14} \\ K_{23} & K_{24}\\ K_{33} & K_{34}\\ K_{43} & K_{44}\\ K_{53} & K_{54}\\ K_{63} & K_{64} \end{bmatrix}$$

In addition to removing these columns because of the fixed boundary conditions, rows 1, 2, 5, & 6 can also be eliminated due to the principle of virtual work (PVW). After deletion of these rows, the resulting K becomes:

$$K = \begin{bmatrix} K_{33} & K_{34}\\ K_{43} & K_{44}\\ \end{bmatrix}$$

And, the resulting global Force-Displacement relationship is:

$$\begin{bmatrix} K_{33} & K_{34}\\ K_{43} & K_{44}\\ \end{bmatrix} * \begin{Bmatrix} d_{3}\\ d_{4} \end{Bmatrix} = \begin{Bmatrix} F_{3}\\ F_{4} \end{Bmatrix}$$

We have already solved for each element in K previously such that

$$K = \begin{bmatrix} 3.0625 & -2.175\\ -2.175 & 2.6875\\ \end{bmatrix}$$

Looking at the problem drawing, we can see that F3 is zero and that F4 is equal to -7. Plugging these values and the values of K into the above global F-D relationship gives the relationship:

$$ \begin{bmatrix} 3.0625 & -2.175\\ -2.175 & 2.6875\\ \end{bmatrix} * \begin{Bmatrix} d_{3}\\ d_{4} \end{Bmatrix} = \begin{Bmatrix} 0\\ -7 \end{Bmatrix}$$

Now, to solve for the unknown displacements, we can multiply both sides by K-1, where:

$$K^{-1} = \frac{1}{det(K)}\begin{bmatrix} K_{33} & -K_{34}\\ -K_{43} & K_{44} \end{bmatrix} = \begin{bmatrix} 0.875 & 0.621\\ 0.621 & 0.768 \end{bmatrix}$$

It is important to not that K-1 is not the same as the transpose of K(KT).

Therefore, the equation to solve for the unknown displacements are:

$$\begin{bmatrix} 0.875 & 0.621\\ 0.621 & 0.768 \end{bmatrix}*\begin{Bmatrix} 0\\ -7 \end{Bmatrix} = \begin{Bmatrix} d_{3}\\ d_{4} \end{Bmatrix}$$

Solving for these displacements gives

$$d_{3} = 4.352 $$ $$d_{4} = 6.1271$$

Finding Reactions
The final step for this problem is to solve for the unknown reactions at global nodes 1 and 3. For this we use the element F-D relationships:

$$K^{(1)}* d^{(1)} = F^{(1)}$$ $$K^{(2)}* d^{(2)} = F^{(2)}$$

In the above equation, K(e), d(e), and  half of f(e) are known for each element, e. For this situation, the F-D relationships with all known values substituted is:

$$\begin{Bmatrix} \frac{9}{16} & \frac{3\sqrt{3}}{16}\\ \frac{3\sqrt{3}}{16}\ & \frac{3}{16} \end{Bmatrix} \begin{Bmatrix} 4.352\\ 6.1271 \end{Bmatrix}= \begin{Bmatrix} f^{(1)}_1\\ f^{(1)}_2\end{Bmatrix}$$

$$\begin{Bmatrix} \frac{5}{2} & \frac{-5}{2}\\ \frac{-5}{2} & \frac{5}{2} \end{Bmatrix} \begin{Bmatrix} 4.352\\ 6.1271 \end{Bmatrix}= \begin{Bmatrix} f^{(2)}_3\\ f^{(2)}_4 \end{Bmatrix}$$

Using this equation, we can now solve for the f(1) matrix and thus we find the values for the reactions: $$f^{(1)}_{1} = F_{1} = -4.4378$$ $$ f^{(1)}_{2} = F_{2} = -2.5622$$

Using this same procedure applied to element 2, we can solve for the values of the other two unknown reactions giving us: $$f^{(2)}_{3} = F_{5} = 4.4378$$ $$f^{(2)}_{4} = F_{6} = -4.4378$$

Explaining Resulting Forces
Element 1:

K(1) · d(1) = F(1)

$$= \begin{bmatrix} 0.5625 & 0.032476 & -0.5625 & -0.032476 \\ 0.032476 &

0.1875 & -0.032476 & -0.1875 \\ -0.5625 & -0.032476 & 0.5625 & 0.032476 \\ -0.032476 & -0.1875 & 0.032476 & 0.1875 \end{bmatrix} \begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{Bmatrix}$$

You may simplify the matrix because of the boundary conditions e.g. the first two numbers in the displacement matrix are zero.

$$= \begin{bmatrix} -0.5625 & -0.032476 \\ -0.032476 & -0.1875 \\ 0.5625 & 0.032476 \\ 0.032476 & 0.1875 \end{bmatrix} \begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix}$$

$$=\begin{Bmatrix} -4.4378 \\ -2.5622 \\ +4.4378 \\ +2.5622 \end{Bmatrix} = \begin{Bmatrix} F_1^1 \\ F_2^1 \\ F_3^1 \\ F_4^1 \end{Bmatrix}$$

Reactions forces are actually $$F_1^1$$ & $$F_2^1$$

Internal forces are actually $$F_3^1$$ & $$F_4^1$$

Element 1:

$$P_1^1 = \sqrt((F_1^1)^2 + (F_2^1)^2)$$

$$P_1^1 = \sqrt((-4.4378)^2 + (-2.5622)^2$$

$$ P_1^1 = 5.69064                     $$



Element 2:

$$P_1^2 = \sqrt((F_1^2)^2 + (F_2^2)^2)$$

$$P_1^2 = \sqrt((-4.4378)^2 + (-2.5622)^2$$

$$P_1^2 = 5.69064  $$



Method Two: Statics
Solving the two bar truss system using Euler Cut Principle.



Back to FE Solution: How to bring P back into the big picture?

Ans: Using the Euler cut method to isolate node two, equilibrium can be established at global node 2. Statics can be applied to the system by using summation of the forces in the x and y directions. This provides two equations with two unknowns that can be solved by using a system of equations.



$$\sum{F_x}= P_1cos30 - P_2cos45=0$$

$$\sum{F_y}=7- P_1cos30 -P_2cos45=0$$

$$\sum{F_x}= P_1 \frac{\sqrt{3}}{2} - P_2\frac{\sqrt{2}}{2}=0$$

$$\sum{F_y}=7- \frac{P_1}{2}-P_2\frac{\sqrt{2}}{2}=0$$

$$P_1=5.122436\;$$

$$P_2 = 6.27603\;$$

Two Bar Truss System MATLAB Code
This section goes over the finite element MATLAB code for a two bar truss system. All of the resulting matrices are below the code. The MATLAB example code can also be found on Dr. Vu-Quoc's website:

http://clesm.mae.ufl.edu/~vql/courses/fead/2008.fall/codes/twoBarTrussEx.txt

% Two bar truss example clear all; e = [3 5]; A = [1 2]; P = 7; L=[4 2]; alpha = pi/3; beta = pi/4; nodes = [0, 0; L(1)*cos(pi/2-alpha), L(1)*sin(pi/2-alpha); L(1)*cos(pi/2-alpha)+L(2)*sin(beta),L(1)*sin(pi/2-alpha)-L(2)*cos(beta)]; dof=2*length(nodes); conn=[1,2; 2,3]; lmm = [1, 2, 3, 4; 3, 4, 5, 6]; elems=size(lmm,1); K=zeros(dof); R = zeros(dof,1); debc = [1, 2, 5, 6]; ebcVals = zeros(length(debc),1); %load vector R = zeros(dof,1); R(4) = P; % Assemble global stiffness matrix K=zeros(dof); for i=1:elems lm=lmm(i,:); con=conn(i,:); k_local=e(i)*A(i)/L(i)*[1 -1; -1 1] k=PlaneTrussElement(e(i), A(i), nodes(con,:)) K(lm, lm) = K(lm, lm) + k; end K R % Nodal solution and reactions [d, reactions] = NodalSoln(K, R, debc, ebcVals) results=[]; for i=1:elems results = [results; PlaneTrussResults(e, A, ...            nodes(conn(i,:),:), d(lmm(i,:)))]; end format short g results k_local = 0.75       -0.75         -0.75         0.75 k = 0.5625     0.32476      -0.5625     -0.32476       0.32476       0.1875     -0.32476      -0.1875       -0.5625     -0.32476       0.5625      0.32476      -0.32476      -0.1875      0.32476       0.1875 k_local = 5   -5     -5     5 k = 2.5        -2.5         -2.5          2.5          -2.5          2.5          2.5         -2.5          -2.5          2.5          2.5         -2.5           2.5         -2.5         -2.5          2.5 K = 0.5625     0.32476      -0.5625     -0.32476            0            0       0.32476       0.1875     -0.32476      -0.1875            0            0       -0.5625     -0.32476       3.0625      -2.1752         -2.5          2.5      -0.32476      -0.1875      -2.1752       2.6875          2.5         -2.5             0            0         -2.5          2.5          2.5         -2.5             0            0          2.5         -2.5         -2.5          2.5 R = 0     0      0      7      0      0 d = 0            0         4.352        6.1271             0             0 reactions = -4.4378      -2.5622        4.4378       -4.4378 results = 1.7081      5.1244       8.5406       5.1244       17.081        0.6276       1.8828        3.138       1.8828        6.276

Comparison of Methods
The Matlab code produced the exact same results for the reaction forces located at global nodes 1 and 3, as did the Finite Element Method learned in class:

Matlab reactions:

reactions = -4.4378      -2.5622        4.4378       -4.4378

Finite Element Method:

Using this equation, we can now solve for the f(1) matrix and thus we find the values for the reactions: $$f^{(1)}_{1} = F_{1} = -4.4378$$ $$ f^{(1)}_{2} = F_{2} = -2.5622$$

Using this same procedure applied to element 2, we can solve for the values of the other two unknown reactions giving us: $$f^{(2)}_{3} = F_{5} = 4.4378$$ $$f^{(2)}_{4} = F_{6} = -4.4378$$

Statics

Using the Euler cut method, you find the following components from the resultant P forces.

$$P_x^1 = -4.4378$$ $$P_y^1=-2.5622$$ $$P_x^2=4.4378$$ $$P_y^2=-4.4378$$

Contributing Team Members
Eml4500.f08.ramrod.D 19:19, 25 September 2008 (UTC)

Eml4500.f08.ramrod.c 04:24, 26 September 2008 (UTC)

Eml4500.f08.RAMROD.A 13:30, 26 September 2008 (UTC)

EML4500.f08.RAMROD.E 16:56, 26 September 2008 (UTC)

Eml4500.f08.RAMROD.F 19:38, 26 September 2008 (UTC)

EML4500.F08.RAMROD.B 19:38, 26 September 2008 (UTC)