User:Eml4500.f08.FEABBQ/HW2/Notes

Class Notes: September 8, 2008


To show that the "vertical" components in the single element pictures from above you must take the moments about node 2 for each element.

Element 1: $$ \sum \mathbf{M_2} = F_1^{x} + (F_1^{y}*L_1) + F_2^{x} + F_2^{y} = 0 + (F_1^{y}*L_1) + 0 + 0 = 0$$

$$ \Rightarrow F_1^{y} = 0 $$

Element 2:

$$ \sum \mathbf{M_2} = F_3^{x} + (F_3^{y}*L_2) + F_2^{x} + F_2^{y} = 0 + (F_3^{y}*L_2) + 0 + 0 = 0$$

$$ \Rightarrow F_3^{y} = 0 $$

3) Global FD

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

$$\underline{k}^{(e)}$$ - Element Stiffness Matrix for Element e

$$\underline{d}^{(e)}$$ - Element Displacement Matrix of Element e

$$\underline{f}^{(e)}$$ - Element Force Matrix of Element e

The equation for $$\underline{k}^{(e)}$$ is,

$$\underline{k}^{(e)} = {k}^{(e)}$$ $$\left[ \begin{matrix} (l^{(e)})^{2} & l^{(e)}m^{(e)} & -(l^{(e)})^{2} & -l^{(e)}m^{(e)}\\ l^{(e)}m^{(e)} & (m^{(e)})^{2} & -l^{(e)}m^{(e)} & -(m^{(e)})^{2} \\ -(l^{(e)})^{2} & -l^{(e)}m^{(e)} & (l^{(e)})^{2} & l^{(e)}m^{(e)} \\ -l^{(e)}m^{(e)} & -(m^{(e)})^{2} & l^{(e)}m^{(e)} & (m^{(e)})^{2} \end{matrix} \right]$$

Axial stiffness of the bar element,

$${k}^{(e)}=\dfrac{E^{e}A^{e}}{(L^{e})}$$ $$ l^{(e)} = cos(\theta) ; m^{(e)} = sin(\theta) $$

$$l^{(e)}, m^{(e)}$$ are found through the director cosines method



$$ l^{(e)} = \mathbf{\tilde{i}} \cdot \mathbf{i} = cos(\theta^{(e)}) $$

$$ m^{(e)} = \mathbf{\tilde{i}} \cdot \mathbf{j} = sin(\theta^{(e)}) $$

$$ \mathbf{\tilde{i}} = cos(\theta^{(e)}) \mathbf{i} + sin(\theta^{(e)}) \mathbf{j} $$

$$ \mathbf{\tilde{i}} \cdot \mathbf{i} = (cos(\theta^{(e)}) \mathbf{i} + sin(\theta^{(e)}) \mathbf{j} ) \cdot \mathbf{i} $$

$$ \mathbf{\tilde{i}} \cdot \mathbf{i} = cos(\theta^{(e)}) \cdot \mathbf{i} \cdot \mathbf{i} + sin(\theta^{(e)}) \cdot \mathbf{i} \cdot \mathbf{j} = cos(\theta^{(e)}) \cdot (1) + sin(\theta^{(e)}) \cdot (0) = cos(\theta^{(e)})$$

$$ \mathbf{\tilde{i}} \cdot \mathbf{j} = (cos(\theta^{(e)}) \mathbf{i} + sin(\theta^{(e)}) \mathbf{j} ) \cdot \mathbf{j} $$

$$ \mathbf{\tilde{i}} \cdot \mathbf{j} = cos(\theta^{(e)}) \cdot \mathbf{i} \cdot \mathbf{j} + sin(\theta^{(e)}) \cdot \mathbf{j} \cdot \mathbf{j} = cos(\theta^{(e)}) \cdot (0) + sin(\theta^{(e)}) \cdot (1) = sin(\theta^{(e)})$$

Director Cosines
The director cosines are the components of the vector $$\mathbf{\tilde{i}}$$ with respect to the global coordinate axes, x and y:

$$\mathbf{\tilde{i}}=\cos \theta^{(e)}\mathbf{i}+\sin \theta^{(e)}\mathbf{j}$$

$$l^{(e)}$$ and $$m^{(e)}$$ are used to denote the magnitude of the i and j components, respectfully:

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

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

2-Bar Truss System, Continued

 * Element 1:
 * $$\theta^{(1)}=30^\circ, \qquad l^{(1)}=\cos 30^\circ = \frac{\sqrt{3}}{2}, \qquad m^{(1)}=\sin 30^\circ = \frac{1}{2}$$


 * $$k^{(1)}=\ \frac{E^{(1})A^{(1)}}{L^{(1)}}\ =\ \frac{(3)(1)}{(4)}\ =\ \frac{3}{4}$$


 * $$\underline{k}^{(1)}\ =\ \begin{bmatrix} k_{11}^{(1)} & \cdots & k_{14}^{(1)} \\ \vdots & \ddots & \vdots \\ k_{41}^{(1)} & \cdots & k_{44}^{(1)}\end{bmatrix}\ =\ \begin{bmatrix} k_{ij}^{(1)}\end{bmatrix}_{[4x4]}$$


 * $$k_{11}^{(1)}\ =\ k^{(1)}(l^{(1)})^2\ =\ \left (\frac{3}{4}\right )\left (\frac{\sqrt{3}}{2}\right )^2 = \frac{9}{16}$$


 * $$k_{12}^{(1)}\ =\ k^{(1)}(l^{(1)}m^{(1)})\ =\ \left (\frac{3}{4}\right )\left (\frac{\sqrt{3}}{2} \cdot \frac{1}{2}\right ) = \frac{3\sqrt{3}}{16}$$


 * $$k_{42}^{(1)}\ =\ -k^{(1)}(m^{(1)})^2\ =\ -\left (\frac{3}{4}\right )\left (\frac{1}{2}\right )^2 = -\frac{3}{16}$$


 * Observations:
 * Looking back to the equation for the element stiffness matrix, k (1), one may notice that only 3 elements need be calculated to determine the matrix. All elements are of one of three magnitudes, with varying signs.
 * Matrix k (1) is symmetric, meaning that kij(1) = kji(1).
 * In general,kij(e)=kji(e), which infers that the transpose of an element stiffness matrix is equal to itself.


 * This quality of the stiffness matrix allows for the following shorthand triangular form of the matrix:


 * $$\underline{k}^{(e)}\ =\ \begin{bmatrix} k_{11}^{(e)} & \cdots & k_{14}^{(e)} \\ & \ddots & \vdots \\ & & k_{44}^{(e)}\end{bmatrix}$$


 * Only the upper triangle of the matrix need be expressed; the empty lower triangle of the matrix is just the reflection of the upper part about the diagonal.


 * Element 2:
 * $$\theta^{(2)}=-45^\circ, \qquad l^{(2)}=\cos -45^\circ = \frac{\sqrt{2}}{2}, \qquad m^{(2)}=\sin -45^\circ = -\frac{\sqrt{2}}{2}$$


 * $$k^{(2)}=\ \frac{E^{(2})A^{(2)}}{L^{(2)}}\ =\ \frac{(5)(2)}{(2)}\ =\ 5$$


 * $$\underline{k}^{(2)}\ =\ \begin{bmatrix} k_{ij}^{(2)}\end{bmatrix}_{[4x4]}$$


 * $$k_{11}^{(2)}\ =\ k^{(2)}(l^{(2)})^2\ =\ 5\left (\frac{\sqrt{2}}{2}\right )^2 = 2.5$$


 * Observations:
 * Since both $$l^{(2)}$$ and $$m^{(2)}$$ have the same magnitude, all elements in k (2) will have the same magnitude just with appropriate sign.
 * Matrix k (2) is symmetric, meaning that its transpose is equal to itself.


 * Element Force-Displacement Relationship:
 * Now that the element stiffness matrices are known (see MATLAB solution at bottom), they can be applied to the element FD relationship:
 * $$\underline{k}_{\ 4x4}^{(e)} \cdot \underline{d}_{\ 4x1}^{(e)}\ =\ \underline{f}_{\ 4x1}^{(e)}$$


 * $$\underline{d}^{(e)}\ =\ \begin{Bmatrix} d_1^{(e)} \\ \vdots \\ d_4^{(e)} \end{Bmatrix} \qquad \qquad \underline{f}^{(e)}\ =\ \begin{Bmatrix} f_1^{(e)} \\ \vdots \\ f_4^{(e)} \end{Bmatrix}$$

Class Notes: September 12, 2008
Expressing the global force-matrix function can be simplified by using the compact notation:

$$\left [ K_{ij} \right ] \left \{ d_j \right \} = \left \{ F_i \right \}$$

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

How do you go from several local element matrices to one global matrix? An assembly plan! First you must identify the correspondence between the various elements and their degree's of freedom:

Class Notes: September 15, 2008
To assemble the global stiffness matrix, $$\underline{\overline{K}}$$, the local stiffness matrices,$$\underline{k}^{(1)}$$ and $$\underline{k}^{(2)}$$, are superimposed as shown in the figure below.



The elements which overlap are summed. The result is the global stiffness matrix shown below.

$$ \underline{\overline{K}} = \left[ \begin{array}{cccccc} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0 \\ k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0 \\ k_{31}^{(1)} & k_{32}^{(1)} & (k_{33}^{(1)} + k_{11}^{(2)}) & (k_{34}^{(1)} + k_{12}^{(2)}) & k_{13}^{(2)} & k_{14}^{(2)} \\ k_{41}^{(1)} & k_{42}^{(1)} & (k_{43}^{(1)} + k_{21}^{(2)}) & (k_{44}^{(1)} + k_{22}^{(2)}) & k_{23}^{(2)} & k_{24}^{(2)} \\ 0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)} \\ 0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)} \\ \end{array} \right] $$

Therefore the global FD relationship, $$\underline{\overline{K}}$$ $$ \underline{\overline{d}} = \underline{\overline{F}}$$, is given by:

$$ \left[ \begin{array}{cccccc} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0 \\ k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0 \\ k_{31}^{(1)} & k_{32}^{(1)} & (k_{33}^{(1)} + k_{11}^{(2)}) & (k_{34}^{(1)} + k_{12}^{(2)}) & k_{13}^{(2)} & k_{14}^{(2)} \\ k_{41}^{(1)} & k_{42}^{(1)} & (k_{43}^{(1)} + k_{21}^{(2)}) & (k_{44}^{(1)} + k_{22}^{(2)}) & k_{23}^{(2)} & k_{24}^{(2)} \\ 0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)} \\ 0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)} \\ \end{array} \right] $$ $$\left[ \begin{array}{l} d_1 \\ d_2 \\ d_3  \\ d_4 \\ d_5 \\ d_6 \end{array} \right] = \left[ \begin{array}{l} F_1 \\ F_2  \\ F_3  \\ F_4 \\ F_5 \\ F_6 \end{array} \right]$$

Class Notes: September 17, 2008
Certain degrees of freedom in our example problem are already known; the boundary conditions set these displacements to equal zero, while the displacements of global node 2 are to be determined. Setting $$ d_1 = d_2 = d_5 = d_6 =0 $$ is the same as deleting columns 1, 2, 5 and 6 in the stiffness matrix. Shown below are columns 3 and 4 and the simplified displacement vector..

$$\underline{K}\left[ \begin{array}{l} d_1=0 \\ d_2=0 \\ d_3=? \\ d_4=? \\ d_5=0 \\ d_6=0 \end{array} \right]= \left[ \begin{array}{ll} K_{13} & K_{14} \\ K_{23} & K_{24} \\ K_{33} & K_{34} \\ K_{43} & K_{44} \\ K_{53} & K_{54} \\ K_{63} & K_{64} \end{array} \right] \left[ \begin{array}{ll} d_3 \\ d_4 \end{array} \right] = \underline{F} $$

Since the stiffness matrix reduces down to a 6-by-2 size and the displacement reduces to a 2-by-1 matrix, the resulting force matrix will have dimensions 6-by-1.

Further reduction of this system of matrices can be performed by the Principal of Virtual Work, which allows us to delete the same rows from the stiffness matrix as we did the columns (1, 2, 5 and 6) as shown below.

$$\left[ \begin{array}{ll} K_{33} & K_{34} \\ K_{43} & K_{44} \end{array} \right] \left[ \begin{array}{l} d_3 \\ d_4 \end{array} \right] = \left[ \begin{array}{c} F_3=0 \\ F_4=P=7 \end{array} \right] $$

Since the goal now is to determine the displacements $$d_3$$ and $$d_4$$, we can rearrange the equation above to solve for these displacements.

$$\left[ \begin{array}{l} d_3 \\ d_4 \end{array} \right] = \underline{\overline{K}}^{-1} \left[ \begin{array}{c} 0 \\ P \end{array} \right] $$

We now have an inverse matrix. As a review...

$$\underline{\overline{K}}^{(-1)} = \dfrac{1}{det(\underline{\overline{K}})} \left[ \begin{array}{cc} K_{44} & -K_{34} \\ -K_{43} & K_{33} \end{array} \right]; det(\underline{\overline{K}}) = K_{33}K_{44}-K_{34}K_{43}$$

With the values $$K_{34} = K_{43} = -2.175$$, $$K_{33} = 3.0625$$ and $$ K_{44} = 2.6875 $$, we have

$$det(\underline{\overline{K}}) = 3.0625(2.6875)-(-2.175)(-2.175) = 3.4998$$

$$\underline{\overline{K}}^{(-1)} = \left[ \begin{array}{cc} 2.6875/3.4998 & 2.175/3.4998 \\ 2.175/3.4998 & 3.0625/3.4998 \end{array} \right] = \left[ \begin{array}{ll} 0.76790 & 0.62146 \\ 0.62146 & 0.87505 \end{array} \right] $$

$$\left[ \begin{array}{l} d_3 \\ d_4 \end{array} \right] = \left[ \begin{array}{ll} 0.76790 & 0.62146 \\ 0.62146 & 0.87505 \end{array} \right] \left[ \begin{array}{c} 0 \\ 7 \end{array} \right] = \left[ \begin{array}{cc} 4.35022 \\ 6.12535 \end{array} \right]$$

Since finding the inverse stiffness matrix was the prominent calculation, it is worthwhile to check the solution, especially given such a simple check as shown below.

$$\underline{\overline{K}}(\underline{\overline{K}}^{-1})=\underline{\overline{K}}^{-1}(\underline{\overline{K}})=\underline{I}=\left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]$$

The $$\underline{I}$$ above is called the Identity Matrix and results when a matrix is multiplied by it's own inverse.


 * A special note about the transpose of a matrix...

$$\underline{\overline{K}}^{T} = \left[ \begin{array}{cc} K_{33} & K_{43} \\ K_{34} & K_{44} \end{array} \right] $$

The transpose of the matrix $$\underline{\overline{K}}$$ is not the same as the matrix appearing in the $$\underline{\overline{K}}^{-1}$$ equation.

In order to compare to Statics, we compute the reaction forces.

5) Compute Reactions
We use two methods:

5.1) Use the element Force-Displacement relationships; $$\underline{k}^{(e)}\underline{d}^{(e)}=\underline{f}^{(e)}$$, where $$e=1,2$$.

Element 1 (e=1)

$$\underline{k}^{(1)}\underline{d}^{(1)}=\underline{f}^{(1)}$$. The contents of the stiffness matrix were previously calculated, and the displacement matrix is complete from our calculations above for the global displacements, while the force (reaction) matrix is unknown.

$$\underline{d}^{(1)}=\left[ \begin{array}{c} 0\\0\\4.352\\6.1271 \end{array} \right]$$



Element 2 (e=2)

Similarly, for $$\underline{d}^{(2)}$$, the stiffness matrix is already known and the following equation can be used to solve for the reactions of element 2.

$$\underline{k}^{(2)}\underline{d}^{(2)}=\underline{f}^{(2)}$$.

$$\underline{d}^{(2)}=\left[ \begin{array}{c} 4.352\\6.1271\\0\\0 \end{array} \right]$$

Class Notes: September 19, 2008
Revisiting element 1 and the known $$\underline{k}^{(1)}$$ and $$\underline{d}^{(1)}$$ matrices, we can solve for the reactions.

$$\left[ \begin{array}{cccc} 0.5625 & 0.32476 & ... & ...\\ ...\\...\\...\end{array} \right] \left[ \begin{array}{c} 0\\ 0\\ 4.352\\ 6.1271 \end{array} \right] = \left[ \begin{array}{c} \underline{f}^{(1)}_{1}\\ \underline{f}^{(1)}_{2}\\ \underline{f}^{(1)}_{3}\\ \underline{f}^{(1)}_{4} \end{array} \right]$$

Matrix multiplication results in the following values for the reactions of element 1...

$$ \left[ \begin{array}{c} -4.4378\\-2.5622\\4.4378\\2.5622 \end{array} \right] $$ where $$\underline{f}^{(1)}_{1}$$ and $$\underline{f}^{(1)}_{2}$$ are the reactions at the constrained node and $$\underline{f}^{(1)}_{3}$$ and $$\underline{f}^{(1)}_{4}$$ are the internal forces.

Similarly, for the stiffness and displacement matrices of element 2

$$\left[ \begin{array}{cccc} 2.5 & -2.5 & ... & ...\\ ...\\...\\...\end{array} \right] \left[ \begin{array}{c} 4.352\\ 6.1271\\ 0\\ 0 \end{array} \right] = \left[ \begin{array}{c} \underline{f}^{(2)}_{1}\\ \underline{f}^{(2)}_{2}\\ \underline{f}^{(2)}_{3}\\ \underline{f}^{(2)}_{4} \end{array} \right] = \left[ \begin{array}{c} -4.4377\\4.4377\\4.4377\\-4.4377 \end{array} \right]$$

Through the sum of the forces we can see that $$\Sigma F_{x} = \underline{f}^{(1)}_{1}+\underline{f}^{(1)}_{3}=-4.4378+4.4378=0$$ and $$\Sigma F_{y} = \underline{f}^{(2)}_{1}+\underline{f}^{(1)}_{4}=-2.5622+2.5622=0$$. We also note that $$ \Sigma M_{any point}=0$$. These three conditions satisfy the criteria for self-equilibrium.

HW for September 19, 2008
For element 1, if we call the reactions at local node 1, $$ -P^{(1)}_{1} $$, that is

$$ -P^{(1)}_{1}=\underline{f}^{(1)}_{1}+\underline{f}^{(1)}_{2}=-4.4378x-2.5622y$$

and if we also call the internal forces at node 2, $$ P^{(1)}_{2} $$ and $$ P^{(1)}_{2}=\underline{f}^{(3)}_{1}+\underline{f}^{(4)}_{2}=4.4378x+2.5622y$$

then $$ P^{(1)}_{2}=-(-P^{(1)}_{1})=P^{(1)}_{1} $$, further demonstrating equilibrium of element 1.

Similarly, for element 2, if we call local node 2, $$P^{(2)}_{2}$$, that is

$$ P^{(2)}_{2}=\underline{f}^{(2)}_{3}+\underline{f}^{(2)}_{4}=4.4377x-4.4377y$$

then $$ P^{(2)}_{1}= -P^{(2)}_{2} = \underline{f}^{(2)}_{3}+\underline{f}^{(2)}_{4}=-4.4377x+4.4377y$$



5.2) The second method for solving the two-bar truss problem involves using Statics. Here we use the Cut Principal by Euler.

To bring the Finite Element Solution into the picture, we verify the equilibrium of global node 2.

HW for September 19, 2008


For Node 2, $$ \Sigma F_{x}=-4.437+4.437=0$$ for the sum of the forces in the x-direction from both element 1 and element 2. The sum of the forces in the y-direction around global node 2 are also in equilibrium...

$$\Sigma F_{y}=7-4.4377-2.5622=0$$

where the forces are from the external force and the y-components of the forces from element 1 and element 2.

Matlab: Two-Bar Truss Example
The following is a matlab script and output for the two-bar truss example above.