User:Eml4500.f08.wiki1.aguilar/TeamHW2

EML4500 Fall 2008 Notes
These class notes cover lecture material from Monday, Sept. 8th, 2008 up to Friday, Sept. 19th 2008.

HW1 left off with the description of elements as springs and explained the 6 steps of the FEM process.

In order to continue on with the FEM process and develop the element stiffness matrix, an understanding of director cosines must exits. The director cosines of the $$\tilde{x}$$ axis with respect to the global ($$ \ x $$,$$ \ y $$) coordinates are derived below from the figure below.





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



\ m^{(e)} =  \cos (90^{o} - \theta^{(e)}) $$



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

A further proof is presented below to show the relationship of the director cosines.


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



\begin{array}{lcl} \begin{align} \bar{\tilde{i}} \cdot \bar{i} & = (\cos \theta^{(e)} \bar{i} + \sin \theta^{(e)} \bar{j}) \cdot \bar{i} \\ & = \cos \theta^{(e)} \bar{i} \cdot \bar{i} + \sin \theta^{(e)} \bar{j} \cdot \bar{i} \\ & = \cos \theta^{(e)} (1) + \sin \theta^{(e)} (0)\\ & = \cos \theta^{(e)} \end{align} \\ l^{(e)} =  \cos \theta^{(e)} \end{array} $$



\begin{array}{lcl} \begin{align} \bar{\tilde{i}} \cdot \bar{j} & = (\cos \theta^{(e)} \bar{i} + \sin \theta^{(e)} \bar{j}) \cdot \bar{j} \\ & = \cos \theta^{(e)} \bar{i} \cdot \bar{j} + \sin \theta^{(e)} \bar{j} \cdot \bar{j} \\ & = \cos \theta^{(e)} (0) + \sin \theta^{(e)} (1)\\ & = \sin \theta^{(e)} \end{align} \\ m^{(e)} =  \sin \theta^{(e)} \end{array} $$

In order to describe the global force displacement at the element level, the k(e) matrix must first be determined. The k(e) matrix for a 4x4 matrix is defined below.



\mathbf{k} ^{(e)} = k^{(e)} \begin{bmatrix} (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{bmatrix} $$

Where


 * $$ k^{(e)} =

\frac{E^{(e)} A^{(e)}} {L^{(e)}} $$

And


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

Note: The director cosines are the components of ĩ (unit vector along the x-axis) with respect to basis (î,ĵ) î=cosθ(e)î + sinθ(e)ĵ Model 2-bar truss system (continued from previous lecture) Element 1:

\ \theta^{(1)} = 30^{o} $$



\ l^{(1)} = cos\theta^{(1)}=cos30^{o}=\frac\sqrt{3}{2} $$



\ m^{(1)} = sin\theta^{(1)}= sin30^{o}=\frac{1}{2} $$



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



\mathbf{k} ^{(1)} = \begin{bmatrix} (k_{11}^{(1)}) & (k_{12}^{(1)})    & (k_{13}^{(1)})    & (k_{14}^{(1)})  \\ (k_{21}^{(1)}) & (k_{22}^{(1)})    & (k_{23}^{(1)})    & (k_{24}^{(1)})  \\

(k_{31}^{(1)}) & (k_{32}^{(1)})    & (k_{33}^{(1)})    & (k_{34}^{(1)})  \\ (k_{41}^{(1)}) & (k_{42}^{(1)})    & (k_{43}^{(1)})    & (k_{44}^{(1)}) \end{bmatrix} $$

where

\mathbf{k^{(e)}} = \left[k_{ij}^{(e)}\right] $$ where i = row, j = column, and e = element number. Therefore,

k_{11}^{(1)} = k^{(1)}\left (l^{(1)}\right)^2 = \left (\frac{3}{4} \right)\left(\frac\sqrt{3}{2}\right)=\frac{9}{16}$$
 * $$k_{12}^{(1)} = k^{(1)}\left (l^{(1)}m^{(1)}\right) = \frac{3\sqrt{3}}{16}$$
 * $$k_{42}^{(1)} = -k^{(1)}\left (m^{(1)}\right)^2 = -\frac{3}{16}$$

Two observations: 1. Only need to compute three values, l2, m2, and l*m. The other coefficients have the same absolute value. They just differ by a (+) or (-). 2. The matrix k(e) is symmetric. i.e. kij(e) = kji(e) or k(e)T = '''k(e)

Element 2:

\ \theta^{(2)}= \frac{-\pi}{4} $$



\ l^{(2)} = cos\theta^{(2)}=cos\frac{-\pi}{4}=\frac\sqrt{2}{2} $$



\ m^{(2)} = sin\theta^{(2)}= sin\frac{-\pi}{4}=-\frac\sqrt{2}{2} $$



\ k^{(2)} = \frac{E^1 A^1}{L^1}=\frac{(5)(2)}{(2)} = 5 $$



\mathbf{k} ^{(2)} = \begin{bmatrix} (k_{11}^{(2)}) & (k_{12}^{(2)})    & (k_{13}^{(2)})    & (k_{14}^{(2)})  \\ (k_{21}^{(2)}) & (k_{22}^{(2)})    & (k_{23}^{(2)})    & (k_{24}^{(2)})  \\

(k_{31}^{(2)}) & (k_{32}^{(2)})    & (k_{33}^{(2)})    & (k_{34}^{(2)})  \\ (k_{41}^{(2)}) & (k_{42}^{(2)})    & (k_{43}^{(2)})    & (k_{44}^{(2)}) \end{bmatrix} $$

Therefore, the three values needed to compute the matrix k(2) are as follows:
 * $$ k_{11}^{(2)} = k^{(2)}\left (l^{(2)}\right)^2 = \left (5 \right)\left(\frac\sqrt{2}{2}\right)^2=2.5$$
 * $$k_{12}^{(2)} = k^{(2)}\left (l^{(2)}m^{(2)}\right) = -2.5$$
 * $$k_{42}^{(2)} = -k^{(2)}\left (m^{(2)}\right)^2 = -2.5$$

The given homework was to compute both k(1) and k(2). They are as follows:



\mathbf{k} ^{(1)} = \begin{bmatrix} (\frac{9}{16})        & (\frac{3\sqrt{3}}{16})    & (-\frac{9}{16})    & (-\frac{3\sqrt{3}}{16})  \\ (\frac{3\sqrt{3}}{16}) & (\frac{3}{16})   & (-\frac{3\sqrt{3}}{16})    & (-\frac{3}{16})  \\

(-\frac{9}{16})       & (-\frac{3\sqrt{3}}{16})    & (\frac{9}{16})    & (\frac{3\sqrt{3}}{16})  \\ (-\frac{3\sqrt{3}}{16})& (-\frac{3}{16})   & (\frac{3\sqrt{3}}{16})    & (\frac{3}{16}) \end{bmatrix} $$

\mathbf{k} ^{(2)} = \begin{bmatrix} 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 \end{bmatrix} $$  Element FD relationship: k(e)d(e)=f(e) k is a 4x4 matrix. d is a 4x1 matrix. $$\mathbf{d} = \begin{Bmatrix} d_1^{(e)} \\ \vdots \\ d_4^{(e)} \end{Bmatrix}$$ f therefore is a 4x1 matrix. $$\mathbf{f} = \begin{Bmatrix} f_1^{(e)} \\ \vdots \\ f_4^{(e)} \end{Bmatrix}$$

Global FD relationship: Kd=F K is a nxn matrix. d is a nx1 matrix. F is a nx1 matrix. Here, n is the number of dof's (degrees of freedom). For this problem, n=6.

$$ \begin{bmatrix} K_{11} & K_{12} & \cdots & K_{16}\\ \vdots & & & \vdots \\ K_{61} & \cdots & \cdots & K_{66} \end{bmatrix} \begin{Bmatrix} d_1 \\ \vdots \\ d_2 \end{Bmatrix} = \begin{Bmatrix} F_1 \\ \vdots \\ F_2 \end{Bmatrix} $$

[Kij]6x6{dj}6x1 = {Fi}6x1

Knxn = [Kij]nxn = global stiffness matrix

dnx1 = {dj}nx1 = global displacement matrix

Fnx1 = {Fi}nx1 = global force matrix

k(e)4x4d(e)4x1 = f(e)4x1

k(e)4x4 = [k(e)ij]4x4 = elemental stiffness matrix

d(e)4x1 = [d(e)j]4x1 = elemental displacement matrix

f(e)4x1 = [f(e)i]4x1 = elemental force matrix

To get from  global matrices  to  elemental matrices : Use an  assembly process 

Global level:

{d1, d2, d3, d4, d5, d6}

Elemental level:

Elem. 1: {d1(1), d2(1), d3(1), d4(1)}

Elem. 2: {d1(2), d2(2), d3(2), d4(2)}

Identification of Global-Local (Elemental) Degrees of Freedom (DOF's):

Global Node 1:

d1 = d1(1)

d2 = d2(1)

Global Node 2:

d3 = d3(1) = d1(2)

d4 = d4(1) = d2(2)

Global Node 3:

d5 = d3(2)

d6 = d4(2)





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

K_{12}=k_{12}^{(1)}= {3 \sqrt{3} \over 16}$$

K_{13}=k_{13}^{(1)}= -{9 \over 16}$$

K_{14}=k_{14}^{(1)}= -{3 \sqrt{3} \over 16}$$

K_{21}=k_{21}^{(1)}= {3 \sqrt{3} \over 16}$$

K_{22}=k_{22}^{(1)}= {3 \over 16}$$

K_{23}=k_{23}^{(1)}= -{3 \sqrt{3} \over 16}$$

K_{24}=k_{24}^{(1)}= -{3 \over 16}$$

K_{31}=k_{31}^{(1)}= -{9 \over 16}$$

K_{32}=k_{32}^{(1)}= -{3 \sqrt{3} \over 16}$$

K_{33}=k_{33}^{(1)}+k_{11}^{(2)}= {9 \over 16}+{5 \over 2}=3.0625$$

K_{34}=k_{34}^{(1)}+k_{12}^{(2)}= {3 \sqrt{3} \over 16}+-{5 \over 2}=-2.1752$$

K_{43}=k_{43}^{(1)}+k_{21}^{(2)}= {3 \sqrt{3} \over 16}+-{5 \over 2}=-2.1752$$

K_{44}=k_{44}^{(1)}+k_{22}^{(2)}= {3 \over 16}+{5 \over 2}=2.6875$$ ...

The final global stiffness matrix, K, becomes:

\begin{bmatrix} {9 \over 16} & {3 \sqrt{3} \over 16} & -{9 \over 16} & -{3 \sqrt{3} \over 16} & 0 & 0\\ {3 \sqrt{3} \over 16} & {3 \over 16} & -{3 \sqrt{3} \over 16} & -{3 \over 16} & 0 & 0\\ -{9 \over 16} & -{3 \sqrt{3} \over 16} & 3.0625 & -2.1752 & -{5 \over 2} & {5 \over 2}\\ -{3 \sqrt{3} \over 16} & -{3 \over 16} & -2.1752 & 2.6875 & {5 \over 2} & -{5 \over 2}\\ 0 & 0 & -{5 \over 2} & {5 \over 2} & {5 \over 2} & -{5 \over 2}\\ 0 & 0 & {5 \over 2} & -{5 \over 2} & -{5 \over 2} & {5 \over 2} \end{bmatrix} $$

At this point the global force displacement relationship can now be shown by



\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}

\begin{Bmatrix} 0 \\  0  \\

d_{3} \\ d_{4} \\

0 \\

0 \\

\end{Bmatrix} = \mathbf{F} $$

By applying the boundary conditions we can delete the corresponding first, second, fifth and sixth columns in the global stiffness matrix K. Using the principal of Virtual Work we can also delete the corresponding first, second, fifth and sixth rows in the global stiffness matrix K.

The resulting 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} $$

Because our only force, P, is in the X direction and equal to seven



\mathbf{\bar{F}} = \begin{Bmatrix} 0\\ 7 \end{Bmatrix} $$

Now we can solve for the displacements. However, first we must find the inverse of K. In order to do this without a calculator we use the determinate method.



det\mathbf{\bar{K}} = K_{33}K_{44}-K_{34}K_{43} $$



\mathbf{K^{-1}} = \frac{1}{det\mathbf{\bar{K}}}\begin{bmatrix} K_{44} & -K_{34}\\ -K_{43} & K_{33} \end{bmatrix} $$

In order to verify that we do indeed have the inverse matrix we can check



\mathbf{\bar{K}}\mathbf{\bar{K}^{-1}}=\mathbf{\bar{K}^{-1}}\mathbf{\bar{K}}=\mathbf{\bar{I}}=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} $$

Now that we have the inverse K matrix we can plug it into



\mathbf{\bar{d}}=\mathbf{\bar{K}^{-1}}\mathbf{\bar{F}} $$

The result is then



\mathbf{\bar{d}}=\begin{Bmatrix} d_{3}\\ d_{4} \end{Bmatrix}= \begin{Bmatrix} 4.352\\ 6.1271 \end{Bmatrix} $$

Now that we have solved for the displacements, we can solve for the reactions using the element force displacement relationship



\mathbf{k^{e}}\mathbf{d^{e}}=\mathbf{f^{e}} $$

where



\mathbf{d^{1}}=\begin{Bmatrix} 0\\ 0\\ 4.352\\ 6.1271 \end{Bmatrix} $$

and



\mathbf{d^{2}}=\begin{Bmatrix} 4.352\\ 6.1271\\ 0\\ 0 \end{Bmatrix} $$

k1,k2,d1, and d2 are known, so we can solve for f1 and f2

Element 1: k(e)d(e)=f(e)

\mathbf{k} ^{(1)} = \begin{bmatrix} (\frac{9}{16})        & (\frac{3\sqrt{3}}{16})    & (-\frac{9}{16})    & (-\frac{3\sqrt{3}}{16})  \\ (\frac{3\sqrt{3}}{16}) & (\frac{3}{16})   & (-\frac{3\sqrt{3}}{16})    & (-\frac{3}{16})  \\

(-\frac{9}{16})       & (-\frac{3\sqrt{3}}{16})    & (\frac{9}{16})    & (\frac{3\sqrt{3}}{16})  \\ (-\frac{3\sqrt{3}}{16})& (-\frac{3}{16})   & (\frac{3\sqrt{3}}{16})    & (\frac{3}{16}) \end{bmatrix} \begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{Bmatrix} = \begin{bmatrix} (-\frac{9}{16})   & (-\frac{3\sqrt{3}}{16})  \\ (-\frac{3\sqrt{3}}{16})   & (-\frac{3}{16})  \\ (\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} -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} $$ where f1 and f2 are the reaction forces and f3 and f4 are the internal forces for element 1. Element 1 is in equilibrium so ΣFx=f1(1) + f3(1) = 0     (1) ΣFy=f2(1) + f4(1) = 0     (2)

Method 2: Statics for a 2-bar truss Use Euler Cut Method Back to FE solution. The question arises, how do you bring P back into the picture. Find the equilibrium at node 2. This is part of the homework.

P^{(1)} = [(f_1^{(1)})^2 + (f_2^{(1)})^2]^{\frac{1}{2}}$$

P^{(2)} = [(f_1^{(2)})^2 + (f_2^{(2)})^2]^{\frac{1}{2}}$$ Therefore,

P = [((f_1^{(1)})+ (f_1^{(2)}))^2 + ((f_2^{(1)}) + (f_2^{(2)}))^2]^{\frac{1}{2}} $$ Which equals our value for P. Therefore, node 2 is in equilibrium.

Authors
Eml4500.f08.wiki1.aguilar 20:23, 26 September 2008 (UTC) Eml4500.f08.wiki1.oatley 19:51, 26 September 2008 (UTC) Eml4500.f08.wiki1.schaet 20:45, 26 September 2008 (UTC) Eml4500.f08.wiki1.brannon 20:44, 26 September 2008 (UTC) Eml4500.f08.wiki1.handy 19:58, 26 September 2008 (UTC) Eml4500.f08.wiki1.ambrosio 14:36, 29 September 2008 (UTC)