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User:Eml4500.f08.group/Homework3/Lecture Notes

=Force-Distance Relationship on the Global System=

Derivation of element fd with respect to the global coordinate system:

eq (1) k (e) d (e) = f (e)

eq (2) $$k^{(e)}\begin{bmatrix} 1 &-1 \\ -1& 1 \end{bmatrix} \begin{Bmatrix} q^{(e)}_{1}\\ q^{(e)}_{2} \end{Bmatrix} = \begin{Bmatrix} P^{(e)}_{1}\\ P^{(e)}_{2} \end{Bmatrix}$$

q(e)i = axial displacement of element e at the local node i. P(e)i = axial force of element e at local node i.

We can derive eq(1) from eq(2) and we can find a relationship between q(e)i and d(e)i as well as the relationship between P(e)i and f(e)i. These relationships can be expressed in the form:

$$q^{(e)}=T^{(e)}d^{(e)}$$

Consider the displacement vector of the local node 1, denoted by d(e)1

$$\vec{d}_{1}^{(e)}=d_{1}^{(e)}\vec{i}+d_{2}^{(e)}\vec{j}$$

$$q_{1}^{(e)}$$ is equal to the axial displacement of node 1, which is the orthogonal projection of the displacement vector $$\vec{d}_{1}^{(e)}$$ of node 1 on the x tilda axis of element e. Therefore:

$$q_{1}^{(e)}=\vec{d}_{1}^{(e)}\cdot \vec{\tilde{i}}$$ $$=(d_{1}^{(e)}\vec{i}+d_{2}^{(e)}\vec{j})\cdot \vec{\tilde{i}}$$ $$=d_{1}^{(e)}(\vec{i}\cdot \vec{\tilde{i}})+d_{2}^{(e)}(\vec{j}\cdot \vec{\tilde{i}})$$ From the definition of l(e) equal to cosθ(e) and m(e) equal to sinθ(e), the above equation reduces to: $$q_{1}^{(e)}=l^{e}d_{1}^{e}+m^{e}d_{2}^{e}$$ In matrix form reduces to: $$q_{1}^{(e)}=\begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix} \begin{Bmatrix} d_{1}^{(e)} \\ d_{2}^{(e)} \end{Bmatrix}$$

HW: Similarly for node 2:

$$\vec{d}_{3}^{(e)}=d_{3}^{(e)}\vec{i}+d_{4}^{(e)}\vec{j}$$

$$q_{2}^{(e)}$$ is equal to the axial displacement of node 2, which is the orthogonal projection of the displacement vector $$\vec{d}_{3}^{(e)}$$ of node 2 on the x tilda axis of element e. Therefore:

$$q_{2}^{(e)}=\vec{d}_{3}^{(e)}\cdot \vec{\tilde{i}}$$ $$=(d_{3}^{(e)}\vec{i}+d_{4}^{(e)}\vec{j})\cdot \vec{\tilde{i}}$$ $$=d_{3}^{(e)}(\vec{i}\cdot \vec{\tilde{i}})+d_{4}^{(e)}(\vec{j}\cdot \vec{\tilde{i}})$$ From the definition of l(e) equal to cosθ(e) and m(e) equal to sinθ(e), the above equation reduces to: $$q_{2}^{(e)}=l^{e}d_{3}^{e}+m^{e}d_{4}^{e}$$ In matrix form reduces to: $$q_{2}^{(e)}=\begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix} \begin{Bmatrix} d_{3}^{(e)} \\ d_{4}^{(e)} \end{Bmatrix}$$

Therefore the relationship between q(e) and d(e) is expressed through the form $$q^{(e)}=T^{(e)}d^{(e)}$$ where $$T=\begin{bmatrix} l^{(e)} & m^{(e)}&0&0\\ 0&0&l^{(e)} & m^{(e)} \end{bmatrix}$$ Leaving the resultant relationship as: $$\begin{Bmatrix} q_{1}^{(e)}\\ q_{2}^{(e)} \end{Bmatrix}=\begin{bmatrix} l^{(e)} & m^{(e)}&0&0\\ 0&0&l^{(e)} & m^{(e)} \end{bmatrix} \begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)} \\ d_{3}^{(e)} \\ d_{4}^{(e)}

\end{Bmatrix} $$

Finding the Stiffness Matrix
If you use the same argument, $$\begin{Bmatrix} p^{(e)}_{1}\\ p^{(e)}_{2} \end{Bmatrix}_{2x1} = \bar{T}^{(e)}_{2x4} \begin{Bmatrix} f^{(e)}_{1}\\ f^{(e)}_{2}\\ f^{(e)}_{3}\\ f^{(e)}_{4} \end{Bmatrix}_{4x1}$$

This yields the symbolic simplification:

P (e) = T (e) f (e)

For each node, the relationship can be expressed as:

q (e)2x1 = T (e)2x4 d (e)4x1

Substituting variables into the element axial Force-Distance Relationship:

$$\hat{k}^{(e)}_{2x2} q^{(e)}_{2x1}= p^{(e)}_{2x1}$$ → $$\hat{k}^{(e)} (T^{(e)}d^{(e)})= (T^{(e)}f^{(e)})$$

The goal is to have k (e) d (e) = f (e). Therefore, we want to move the T (e) from the right hand side to the left hand side by multiplying both sides by T (e)-1 (or the inverse of T). But because T is not a square matrix, it cannot be inverted.

k (e) d (e)= f (e) → $$[T^{(e)T}_{4x2} (transpose) \hat{k}^{(e)}_{2x2}T^{(e)}_{2x4}]_{(4x4)}d^{(e)}_{4x1} = f^{(e)}_{4x1}$$

HW: Verify that $$k^{(e)} = T^{(e)T}\hat{k}^{(e)}T^{(e)}$$ $$\bar{T}^{(e)T}k^{(e)}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}\begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix} = \begin{bmatrix} l^{(e)} & 0\\ m^{(e)} & 0\\ 0 & l^{(e)}\\ 0 & m^{(e)} \end{bmatrix}k^{(e)}\begin{bmatrix} l^{(e)} & m^{(e)} & -l^{(e)} & -m^{(e)}\\ -l^{(e)} & -m^{(e)} & l^{(e)} & m^{(e)} \end{bmatrix} = k^{(e)}\begin{bmatrix} (l^{(e)})^{2} & l^{(e)}m^{(e)} & -(l^{(e)})^{2} & -(l^{(e)}m^{(e)})\\ l^{(e)}m^{(e)} & (m^{(2)})^2 & -(l^{(e)}m^{(e)}) & -(m^{(2)})^2\\ -(l^{(e)})^{2} & -(l^{(e)}m^{(e)}) & (l^{(e)})^{2} & l^{(e)}m^{(e)}\\ -(l^{(e)}m^{(e)}) & -(m^{(2)})^2 & l^{(e)}m^{(e)} & (m^{(2)})^2 \end{bmatrix} = \bar{k}^{(e)}$$

Principle of Virtual Work: K 6x6 d 6x1 = F 6x1 → K' 2x2 d' 2x1 = F' 2x1

We cannot solve for d as d = K -1 F because of the singularity of K -1 where the det[ K ]=0 and therefore we cannot invert K.

In 2D motion, there are three possible rigid body motions: two in translation and one rotation

HW: To find the eigenvalues: $$det\begin{bmatrix} k_{11}^{(1)} -\lambda & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0\\ k_{21}^{(1)} & k_{22}^{(1)} - \lambda & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0\\ k_{31}^{(1)} & k_{32}^{(1)} & (k_{33}^{(1)}  k_{11}^{(2)}) - \lambda & (k_{34}^{(1)}   k_{12}^{(2)}) & k_{13}^{(2)} & k_{14}^{(2)}\\ k_{41}^{(1)} & k_{42}^{(1)} & (k_{43}^{(1)}  k_{21}^{(2)}) & (k_{44}^{(1)}   k_{22}^{(2)}) - \lambda  & k_{23}^{(2)} & k_{24}^{(2)}\\ 0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} - \lambda & k_{34}^{(2)}\\ 0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)} - \lambda \end{bmatrix} = 0$$ This can be reduced to (because of the Principle of Virtual Work) $$det\begin{bmatrix} (k_{33}^{(1)} + k_{11}^{(2)}) - \lambda & (k_{34}^{(1)} + k_{12}^{(2)})\\ (k_{43}^{(1)} + k_{21}^{(2)}) & (k_{44}^{(1)} + k_{22}^{(2)}) - \lambda \end{bmatrix} = 0$$ Solving becomes $$[(k_{33}^{(1)} + k_{11}^{(2)}) - \lambda]*[(k_{44}^{(1)} + k_{22}^{(2)}) - \lambda] - [(k_{34}^{(1)} + k_{12}^{(2)})(k_{43}^{(1)} + k_{21}^{(2)})] = 0$$ But because K43=K34, this can be written as $$\lambda ^{2}- \lambda(k_{33}^{(1)} + k_{11}^{(2)} + k_{44}^{(1)} + k_{22}^{(2)}) + [(k_{33}^{(1)} + k_{11}^{(2)})(k_{44}^{(1)} + k_{22}^{(2)}) - (k_{34}^{(1)} + k_{12}^{(2)})^{2}] = 0 $$ Therefore, the eigenvalues are λ = 0,$$(k_{33}^{(1)} + k_{11}^{(2)})(k_{44}^{(1)} + k_{22}^{(2)}) - (k_{34}^{(1)} + k_{12}^{(2)})^{2}$$, and $$k_{33}^{(1)} + k_{11}^{(2)} + k_{44}^{(1)} + k_{22}^{(2)}$$ There is a zero eigenvalue because there is no rotational component of the system. This means that there is no elastic energy stored in the structure in the rotational mode. → Dynamic eigenvalue as realted to vibration frequency: K v =λ M  v, where K is the stiffness matrix, λ is the eigenvalue, and M is the mass matrix.

Meeting 15 (9/29/08)
Closing the Loop Between The FEM and statistics: virtual displacement

Two Bar Truss System:

From p. 11-2: Free body diagram (FBD) of the 2 force bodies of elements 1 and 2

=> By statics the reactions are unknown and therefore the member forces, $$ p_1^{(1)}, p_2^{(2)}$$, we can compute the axial displacement degrees of freedom or the amount of extension of the bars. This is on p.12-2: $$\displaystyle q_2^{(1)} = \frac{p_1^{(1)}}{k^{(1)}} = \frac{p_2^{(1)}}{k^{(1)}}$$, as shown in figure on p.11-2 $$\displaystyle q_2^{(1)} = A C$$ $$\displaystyle q_1^{(1)} = 0$$ since it is fixed on node 1 \displaystyle q_1^{(2)} = - \frac{p_2^{(2)}}{k^{(2)}} $$\displaystyle q_2^{(2)} = 0$$ since it is fixed on node 3

From the above results how can we obtain the displacement degrees of freedom of node 2?

Note:

Use the following method to solve:

$$\displaystyle u_y = R sin\alpha = R \alpha$$ ($$\alpha$$ small)

$$\displaystyle u_x =$$ axial displacement $$\displaystyle = R (1 - cos\alpha = 0$$  (first order)

Meeting 16 (10/1/08)
Closing the loop cont'd:

Look at an infinitesimal displacement (similar to the virtual displacement assosciated with the Principal of Virtual Work):

$$AC = \frac {|P_2^{(1)}|}{k^{(1)}} = \frac {|5.1243|}{0.75} = 6.8324$$

$$AB = \frac {|P_1^{(2)}|}{k^{(2)}}$$

Next, find the $$(x,y)$$ coordinates of points B and C:

$$(x_B, y_B) $$ $$(x_C, y_C) $$

Now there are 2 unknowns left, the x and y coordinates of point D, $$(x_D, y_D)$$. So, we need to find equations for line AB and line BC.

A approach to finding the desired information is given as follows:

$$\vec{PQ} = |\vec{PQ}| \hat{\tilde{i}} = PQ[cos\theta \hat{\tilde{i}} + sin\theta \hat{\tilde{j}}] $$

$$ = (x-x_p) \hat{\tilde{i}} + (y-y_p) \hat{\tilde{j}}$$

$$ => x-x_p = |PQ|cos\theta $$

and

$$ => y-y_p = |PQ|sin\theta $$

$$=> \frac{y-y_p}{x-x_p} = \tan\theta = >y-y_p = \tan\theta(x-x_p)$$

Thus, the equation for a line perpendicular to the above line and passing through point p would be given by:

$$y-y_p = \tan(\theta + \frac{\pi}{2})(x-x_p)$$

$$(x_D, y_D) = $$

Turning now to vector AD:

$$\vec{AD} = (x_D - x_A)\hat{i} + (y_D - y_A)\hat{j}$$

Since point A was chosen to be at the origin, we are left with:

$$\vec{AD} = (x_D)\hat{i} + (y_D)\hat{j}$$

By definition, we have:

$$\vec{AD} = d_3\hat{i} + d_4\hat{j}$$, where $$\vec{AD}$$ is the displacement vector of point A, and $$y_D = d_4$$ and $$x_D = d_3$$ from the finite element method (FEM).

Example: 3-Bar Truss System

$$\theta^{(1)} = 30 E^{(1)} = 2 A^{(1)} = 3 L^{(1)} = 5 $$

$$\theta^{(2)} = -30 E^{(2)} = 4 A^{(2)} = 1 L^{(2)} = 5 $$

$$\theta^{(3)} = 45? E^{(3)} = 3 A^{(3)} = 2 L^{(3)} = 10 $$

$$P = 30$$

Local node numbering should be carried out in such a manner that will make assembling the global stiffness matrix, K, convenient. A convenient local node numbering is demonstrated in the FBD diagram below:

17-1

The problem shown in the figure is a statically indeterminate structure. Though there are 3 equations (ΣFx, ΣFy, ΣMA) and 3 unknowns (R1, R2, and R3) this structure is still statically indeterminate because summing the moments about point A ends up becoming a useless equation. Since each member is a two-force body, it can bee seen through examination that each reaction force acts on a line that runs through point A. Thus there is no moment arm and no moment. The end result is an equation stating 0 = 0. This point can be proved statically as well.

HW: justiy using statics.

ΣMa=0

$$R1sin(\theta 1)*L1cos(\theta 1)+R2sin(\theta 2)*L2cos(\theta 2)-R3sin(\theta 3)*L3cos(\theta 3) +R3cos(\theta 3)*L3sin(\theta 3)-R2cos((\theta 2)*L2sin(\theta 2)-R1cos(\theta 1)*L1sin(\theta 1)=0$$

Note that every tern cancels and the end result is 0 = 0.

17-2

Question: What about summing the moments about any other point? Answer: The result it the same (yields 0 = 0). Proof:

ΣMB=(B*A x F) = (B*A' x F)

B*A' = B*A + A*A'

MB=(B*A+A*A') x (F) = (B*A x F) + (A*A' x F)

(A*A' x F) = 0 because it is along the same line of action

Thus MB=(B*A x F)

17-3
Global Stiffness Matrix $$\begin{bmatrix} K11 & K12 & K13 & K14 & K15 & K16 & K17 & K18\\ K21 & K22 & K23 & K24 & K25 & K26 & K27 & K28\\ K31 & K32 & K33 & K34 & K35 & K36 & K37 & K38\\ K41 & K42 & K43 & K44 & K45 & K46 & K47 & K48\\ K51 & K52 & K53 & K54 & K55 & K56 & K57 & K58\\ K61 & K62 & K63 & K64 & K65 & K66 & K67 & K68\\ K71 & K72 & K73 & K74 & K75 & K76 & K77 & K78\\ K81 & K82 & K83 & K84 & K85 & K86 & K87 & K88 \end{bmatrix}$$

Element Stiffness matrix (same for elements 1, 2, and 3) $$\begin{bmatrix} k11 & k12 & k13 & k14\\ k21 & k22 & k23 & k24\\ k31 & k32 & k33 & k34\\ k41 & k42 & k43 & k44 \end{bmatrix}$$

HW: Define Global Stiffness Matrix in terms of Element Stiffnesses.

K11=k(1)11 K12=k(1)12 K13=k(1)13 K14=k(1)14 K15=0 K16=0 K17=0 K21=k(1)21 K22=k(1)22 K23=k(1)23 K24=k(1)24 K25=0 K26=0 K27=0 K28=0 K31=k(1)31 K32=k(1)32 K33=k(1)33+k(2)11+k(3)11 K34=k(1)34+k(3)12+k(3)12 K35=k(2)13 K36=k(2)14 K37=k(3)13 K38=k(3)14 K41=k(1)41 K42=k(1)42 K43=k(1)43+k(2)21+k(3)21 K44=k(1)44+k(2)22+k(3)22 K45=k(2)23 K46=k(2)24 K47=k(3)23 K48=k(3)24 K51=0 K52=0 K53=k(2)31 K54=k(2)32 K55=k(2)33 K56=k(2)34 K57=0 K58=0 K61=0 K62=0 K63=k(2)41 K64=k(2)42 K65=k(2)43 K66=k(2)44 K67=0 K68=0 K71=0 K72=0 K73=k(3)31 K74=k(3)32 K75=0 K76=0 K77=k(3)33 K78=k(3)34 K81=0 K82=0 K83=k(3)41 K84=k(3)42 K85=0 K86=0 K87=k(3)43 K88=k(3)44


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User:Eml4500.f08.group/Homework3/Matlab

= Matlab Code =

Results

Kg =

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   -2.5000    2.5000   -0.3248   -0.1875   -2.1752    2.6875    2.5000   -2.5000         0         0   -2.5000    2.5000    2.5000   -2.5000         0         0    2.5000   -2.5000   -2.5000    2.5000

Dg =

0        0    4.3520    6.1271         0         0

Fg =

-4.4378  -2.5622         0    7.0000    4.4378   -4.4378

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User:Eml4500.f08.group/Homework3/Matlab


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User:Eml4500.f08.group/Homework3/Contributors

=Contributing Members= Tom Cady Eml4500.f08.group.cady 03:28, 28 September 2008 (UTC)

Brian Taylor Eml4500.f08.group.taylor 14:29, 4 October 2008 (UTC)

Julia Setlak Eml4500.f08.group.setlak 00:41, 7 October 2008 (UTC)

Jeffrey Wright Eml4500.f08.group.w 02:49, 7 October 2008 (UTC)

Aaron Clausen Eas4200c.f08.carbon.clausen 07:49, 8 October 2008 (UTC)

Melisa Gaar Eml4500.f08.group.gaar 19:32, 8 October 2008 (UTC)


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User:BlueNovember

Please allow me to redirect you to my userpage on wikipedia.

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User:Eas4200c.f08.wiki.d/HW3

{| class="toccolours collapsible collapsed" style="width:100%" ! MATLAB Code

NACA 2415 Airfoil in MATLAB command window
EDU>> Airfoil(2,4,15,.5,[0 0 0])

A_bar =

0.0253

EDU>> Airfoil(2,4,15,.5,[0 0.19 0.01])

A_bar =

0.0255

EDU>> Airfoil(2,4,15,.5,[0 0.25 -1])

A_bar =

0.0256

 * }

title: User:Eas4200c.f08.wiki.d/HW3


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title: User talk:Original Spin


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User:Eml4500.f08.lulz.abcd/HW3

Solve for AC and AB
As given in class, AC and AB are dependent on their elemental P's and K's. In HW 2 we solved for both elemental K values and one of the P values: that work is reproduced as appropriate in the proof below

$$AC = \frac{|P^{(1)}_2|}{K^{(1)}} = \frac{\sqrt{\left(f_3^{(1)}\right)^2 + \left(f_4^{(1)}\right)^2}}{\left(\frac{EA}L\right)} = \frac{\sqrt{(4.4378)^2 + (2.5622)^2}} = \frac{5.1243}{0.75} = 6.8324$$

$$AB = \frac{|P^{(2)}_1|}{K^{(2)}} = \frac{\sqrt{\left(f_1^{(2)}\right)^2 + \left(f_2^{(2)}\right)^2}}{\left(\frac{EA}L\right)} = \frac{\sqrt{(-4.438)^2 + (4.438)^2}} = \frac{6.2763}5 = 1.2553$$

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User:Eas4200c.f08.vqcrew.c/Homework 3\my\

MATLAB Question
The task for the MATLAB portion of the HW report involves plotting an arbitrary NACA airfoil, finding the average area of the airfoil, and locating the airfoil's area centroid.

Using the following input (in accordance with the syntax presented in the code header), the code generates Figure ##, a graph of the NACA 2415 airfoil with a cross-hair at the centroid.

 >> abar(1,1,2,4,15,1000,[0;0])

The average area of the airfoil is 0.101 square chord units

The centroid is located at (0.420,0.016) chord units

>>


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User:Eml4500.f08.lulz.strack/hw3

Eigenvalues of the Global Stiffness Matrix
From Homework 2, the MATLAB code for the two bar truss system produced the global stiffness matrix (K) verified by manual calculations. This matrix was used to determine the eigenvalues. After running the M-file "TwoBarTrussSystem" and the command "format", the following output was produced.

>> K

K =

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   -2.5000    2.5000   -0.3248   -0.1875   -2.1752    2.6875    2.5000   -2.5000         0         0   -2.5000    2.5000    2.5000   -2.5000         0         0    2.5000   -2.5000   -2.5000    2.5000

>> eig(K)

ans =

-0.0000  -0.0000    0.0000    0.0000    1.4705   10.0295 The negative sign in front of the first two zeros implies that all four "zeros" are actually very small values. Due to the precision in MATLAB, it is assumed that the "zeros" are essentially zero.

Under this assumption, the result of four zeros as eigenvalues represents the four degrees of freedom of the two bar truss system that are fixed. Those four degrees of freedom are the x and y directional displacements at global nodes 1 and 3.

Comparison of Global F-d relationships and Elemental F-d relationships
From Homework 2, the elemental F-d relationships produced reactions at global nodes 1 and 3 as shown below.

$$ \begin{bmatrix} F_{1} \\ F_{2} \\ F_{5} \\ F_{6} \end{bmatrix}= \begin{bmatrix} -4.4378 \\ -2.5622 \\ 4.4378 \\ -4.4378 \end{bmatrix} $$

For the global F-d relationships method, the following calculations were performed to produce the reactions at all the nodes. Columns 1, 2, 5, and 6 were removed from the global stiffness matrix, and rows of the same rank were removed from the displacement matrix. This is because the four directional displacements of Nodes 1 and 3 are all zero because the nodes are fixed.

$$ d_1 = d_2 = d_5 = d_6 = 0 $$

$$ \left[ \begin{array}{cc} K_{13} & K_{14} \\ K_{23} & K_{24} \\ K_{33} & K_{34} \\ K_{43} & K_{44} \\ K_{53} & K_{54} \\ K_{63} & K_{64} \end{array} \right]\begin{bmatrix} d_{3} \\ d_{4} \end{bmatrix}= \begin{bmatrix} F_{1} \\ F_{2} \\ F_{3} \\ F_{4} \\ F_{5} \\ F_{6} \end{bmatrix} $$

After running the M-file "TwoBarTrussSystem" and the command "format", the following output was produced. The global stiffness matrix (K) was the same one from the previous section.

>> d

d =

0        0    4.3520    6.1271         0         0

>> K34 = K(1:6,3:4)

K34 =

-0.5625  -0.3248   -0.3248   -0.1875    3.0625   -2.1752   -2.1752    2.6875   -2.5000    2.5000    2.5000   -2.5000

>> d34 = d(3:4)

d34 =

4.3520   6.1271

>> F = K34*d34

F =

-4.4378  -2.5622    0.0000    7.0000    4.4378   -4.4378

The reactions produced using the global F-d relationships method are identical to the results from the elemental F-d relationships method. The additional two values (rows 3 and 4) of the global method verify the load to the system.

title: User:Eml4500.f08.lulz.strack/hw3


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User:Eml4500.f08.echo.mott/HW4

= Homework Report 4=
 * This homework report contains lecture notes from 10/6/08 to 10/17/08 and Matlab solutions requested in class.
 * The previous version of the page was submitted into the HW report table.

Lecture Notes
Connectivity Array

 * Consider the 2-bar truss system described earlier.

 * $$\mathbf {conn}=

\begin{Bmatrix} 1 & 2\\ 2 & 3 \end{Bmatrix}$$

Where
 * Columns represent local nodes (ie. column 1 = local node 1)
 * Rows represent elements (ie. row 1 = element 1)
 * Numbers within matrix represent global node numbers

conn(j,k) = global node number of local node k of element j.

Location Matrix Master Array

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

Where
 * Columns represent the local degree of freedom number (ie. 1 = 1)
 * Rows represent the element numbers (ie. 1= 1)
 * Numbers within matrix represent the global degree of freedom number, or the equivalent number, in k

lmm (i,j)= equivalent number (global degree of freedom number) for the element stiffness coefficient corresponding to the $$ j^{th} $$ local degree of freedom number.

Goal: Want to find $$\mathbf{\tilde{T}}_{4x4}^{(e)}$$ that transforms the set of local elements dof's $$\mathbf d_{4x1}^{(e)}$$ to another set of local element dof's $$\mathbf{\tilde{d}}_{4x1}^{(e)}$$ such that $$\mathbf{\tilde{T}}_{ }^{(e)}$$ is invertible. The tilde symbol refers to displacements and forces in local axial directions.

$$ \mathbf{\tilde{d}}_{4x1}^{(e)}=\mathbf{\tilde{T}}_{4x4}^{(e)}\mathbf{d}_{4x1}^{(e)} $$

$$\tilde{d_1^{(e)}}=q_1^{(e)}= \begin{vmatrix} l^{(e)} & m^{(e)} \end{vmatrix}\begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \end{Bmatrix} \qquad \qquad (1)$$

$$\tilde{d}_{2}^{(e)}=d_1^{(e)}\cdot\vec{\tilde{j}}$$ (component of $$\vec{d}_1^{(e)}$$ along $$\vec{\tilde{j}}$$, i.e. $$\tilde{y}$$ axis)

$$=-sin(\theta^{(e)})d_1^{(e)}+cos(\theta^{(e)})d_2^{(e)}$$

For HW, derive the following: $$\tilde{d}_{2}^{(e)}=\begin{vmatrix} -m_{ }^{(e)} & l_{ }^{(e)} \end{vmatrix} \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \end{Bmatrix} \qquad \qquad (2)$$

Beginning with the transformation equation,

$$ \mathbf{\tilde{d}}_{4x1}^{(e)}=\mathbf{\tilde{T}}_{4x4}^{(e)}\mathbf{d}_{4x1}^{(e)} $$

$$ \begin{bmatrix} \tilde{d}_1^{(e)} \\ \tilde{d}_2^{e)} \\ \tilde{d}_3^{(e)} \\ \tilde{d}_4^{(e)} \end{bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0 \\ -m^{(e)} & l^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & m^{(e)} \\ 0 & 0 & -m^{(e)} & l^{(e)} \end{bmatrix} \begin{bmatrix} d_1^{(e)} \\ d_2^{(e)} \\ d_3^{(e)} \\ d_4^{(e)} \end{bmatrix} $$

$$ \begin{align} \tilde{d}_2^{(e)} &= \begin{bmatrix} -m^{(e)} & l^{(e)} & 0 & 0 \end{bmatrix} \begin{bmatrix} d_1^{(e)} \\ d_2^{(e)} \\ d_3^{(e)} \\ d_4^{(e)} \end{bmatrix} \\ &= \begin{bmatrix} -m^{(e)} & l^{(e)} \end{bmatrix} \begin{bmatrix} d_1^{(e)} \\ d_2^{(e)} \end{bmatrix} \end{align}                    $$

Put (1) & (2) in matrix form: $$\begin{Bmatrix} \tilde{d_1^{(e)}} \\ \tilde{d_2^{(e)}} \end{Bmatrix}=\begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & l^{(e)} \end{bmatrix}\begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \end{Bmatrix}$$

$$\begin{Bmatrix} \tilde{d_1^{(e)}} \\ \tilde{d_2^{(e)}} \\ \tilde{d_3^{(e)}} \\ \tilde{d_4^{(e)}} \end{Bmatrix}=\begin{bmatrix} \mathbf{R}_{2x2}^{(e)} & \mathbf{0}_{2x2} \\ \mathbf{0}_{2x2} & \mathbf{R}_{2x2}^{(e)} \end{bmatrix}\begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \\ d_3^{(e)} \\ d_4^{(e)}\end{Bmatrix}$$

In simpler terms, this can be notated as:

$$\tilde{\mathbf{d}}_{4x1}^{(e)}=\tilde{\mathbf{T}}_{4x4}^{(e)} \tilde{\mathbf{d}}_{4x1}^{(e)}$$

For an example, take the following diagram:

This can be described as:

$$\tilde{\mathbf{f}}_{ }^{(e)}=k_{ }^{(e)}\begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0\end{bmatrix}\tilde{\mathbf{d}}_{ }^{(e)}$$

This can be simplified and summarized as:

$$\tilde{\mathbf{f}}_{4x1}^{(e)}=\tilde{\mathbf{k}}_{4x4}^{(e)}\tilde{\mathbf{d}}_{4x1}^{(e)}$$

Note: Consider the case where $$\tilde{d}_{4}^{(e)} \ne 0$$

and

$$\tilde{d}_1^{(e)}=\tilde{d}_2^{(e)}=\tilde{d}_3^{(e)}=0$$

Starting with the element axial F-D relation,

$$\tilde{\mathbf{f}}_{4x1}^{(e)}=\tilde{\mathbf{k}}_{4x4}^{(e)} \tilde{\mathbf{d}}_{4x1}^{(e)}=\mathbf{0}_{4x1}$$

The zero vector being the 4th column of $$\tilde{\mathbf{k}}^{(e)}$$

Interpretation of the transverse dof's:

$$\tilde{\mathbf{f}}_{4x1}^{(e)}=\tilde{\mathbf{k}}_{4x4}^{(e)} \tilde{\mathbf{d}}_{4x1}^{(e)}=\mathbf{0}_{4x1}$$

We have derived that $$\tilde{\mathbf{d}}^{(e)} = \tilde{\mathbf{T}}^{(e)} {\mathbf{d}}^{(e)}$$.

Similarly we would have the same relation for $$\tilde{\mathbf{f}}$$ and $${\mathbf{f}}$$:

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

We can write the displacement relation differently as

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

By substitution,

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

If $$\tilde{\mathbf{T}}^{(e)}$$ is invertible, then the inverse can be used and the force matrix becomes:

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

$$\tilde{\mathbf{T}}^{(e)}$$ is a block diagonal matrix.

Let's consider a more general block diagonal matrix consisting of s individual matrices :

$$\mathbf {A} = \begin{bmatrix} \mathbf{D}_1 & 0 & 0 \\ 0 & \ \ddots \quad & 0  \\ 0 & 0 & \mathbf{D}_{s} \end{bmatrix}$$

What would be the inverse of this matrix $$\mathbf{A}^{-1}$$?

Let us look at a simpler diagonal matrix:

$$ \mathbf{B} = \begin{bmatrix} d_{11} & 0 & 0 & 0 \\ 0 & d_{22} & 0 & 0 \\ 0 & 0 & \ddots \quad & 0 \\0 & 0 & 0 & d_{nn} \end{bmatrix} = diag[d_{11}, d_{22} , ... , d_{nn}] $$

Assuming that the values of $$d_{11}, d_{22} , ... , d_{nn}$$ do not equal zero, the inverse of $$\mathbf{B}$$ can be written as:

$$\mathbf{B}^{-1} = diag [ 1/d_{11}, 1/d_{22} , ... , 1/d_{nn} ] $$

For the general block diagonal matrix $$\mathbf{A}$$:

$$\mathbf{A} = diag [ \mathbf{D}_{11}, \mathbf{D}_{22} , ... , \mathbf{D}_{ss} ] $$

and

$$\mathbf{A}^{-1} = diag [ {\mathbf{D}_{11}}^{-1}, {\mathbf{D}_{22}}^{-1} , ... , {\mathbf{D}_{ss}}^{-1} ] $$

Applying this theory to $$ \tilde{\mathbf{T}}$$,

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

From previous notes it is shown that:

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

The identity matrix is found when R is multiplied by its transpose, concluding that the transpose is equivalent to the inverse.

$${\mathbf R}^{(e)^{T}}{\mathbf R}^{(e)} = \begin{bmatrix} l^{(e)} & -m^{(e)} \\ m^{(e)} & l^{(e)} \end{bmatrix} \begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & l^{(e)} \end{bmatrix} = \begin{bmatrix} l^{(e)^{2}} + {m}^{(e){2}}&0 \\0& l^{(e)^{2}} + {m}^{(e)^{2}} \end{bmatrix} = \begin{bmatrix} 1&0 \\0& 1 \end{bmatrix} $$

And so: $${\mathbf R}^{(e)^{-1}}={\mathbf R}^{(e)^{T}}$$

Now $$ \tilde{\mathbf{T}}$$ can be written as:

$$ \begin{align} {\tilde{\mathbf{T}}}^{(e)^{-1}} &= diag [ {\mathbf{R}}^{(e)^{T}} ,{\mathbf{R}}^{(e)^{T}}] \\ &= (diag [ {\mathbf{R}}^{(e)} ,{\mathbf{R}}^{(e)}])^{T}\\ &= {\tilde{\mathbf{T}}}^{(e)^{T}} \\ \end{align} $$

This means that:

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

HW - Verify that $$ {\mathbf{k}}^{(e)} = [{\tilde{\mathbf{T}}}^{(e)^{T}} {\tilde{\mathbf{k}}}^{(e)} {\tilde{\mathbf{T}}}^{(e)}]$$

$$ \begin{align} {\mathbf{k}}^{(e)} &= \begin{bmatrix} l^{(e)} & -m^{(e)} & 0 & 0 \\ m^{(e)} & l^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & -m^{(e)} \\ 0 & 0 & m^{(e)} & l^{(e)} \end{bmatrix} \begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0 \\ -m^{(e)} & l^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & m^{(e)} \\ 0 & 0 & -m^{(e)} & l^{(e)} \end{bmatrix} \\ &= \begin{bmatrix} l^{(e)} & 0 & -l^{(e)} & 0 \\ m^{(e)} & 0 & -m^{(e)} & 0 \\ -l^{(e)} & 0 & l^{(e)} & 0 \\ -m^{(e)} & 0 & m{(e)} & 0 \end{bmatrix} \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0 \\ -m^{(e)} & l^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & m^{(e)} \\ 0 & 0 & -m^{(e)} & l^{(e)} \end{bmatrix} \\ &= \begin{bmatrix} l^{(e)^2} & l^{(e)}m^{(e)} & -l^{(e)^2} & -l^{(e)}m^{(e)} \\ l^{(e)}m^{(e)} & m^{(e)^2} & -l^{(e)}m^{(e)} & -m^{(e)^2} \\ -l^{(e)^2} & -l^{(e)}m^{(e)} & l^{(e)^2} & l^{(e)}m^{(e)} \\ -l^{(e)}m^{(e)} & -m^{(e)^2} & l^{(e)}m^{(e)} & m^{(e)^2}\\ \end{bmatrix} \\ \end{align} $$

Reference [this page]

Eigenvalues problem: $$ \vec{k} \vec{v}=\lambda \vec{v}$$

let $$ \begin{bmatrix} \vec{u}_1 & \vec{u}_2 & \vec{u}_3 & \vec{u}_4 \end{bmatrix}$$ be the pure eigenvalues corresponding to 4 zero eigenvalues

$$ \vec{k}_{6x6} \vec{u}_{6x1}=0 \cdot \vec{u}_i, i=1 , ...4$$

The zero dot product with $$u_i$$ will yield a 6x1 matrix. In this case the zero is a scalar.

linear combination of $$ {\vec{u}_i, i=1, ...4}$$

$$\sum_{i=1}^4 \alpha_i \vec{u}_i=: \vec {W}_{6x1} $$

where $$ \alpha_i $$ represents real numbers.

$$ \vec {W}$$ is also an eigenvector corresponding to a zero eigenvalue:

$$ \vec {k} \vec{W} = \vec {k} (\sum_{i=1}^4 \alpha_i \vec {u}_i)$$

 * $$= \sum_{i=1}^4 \alpha_i ( \vec {k} \vec{u}_i) = \vec{0}_{6x1}$$

 * $$= 0_{1x1} \times \vec{W}_{6x1}$$

HW:

$$\vec{k} \vec{v} = \lambda \vec{v}$$

$$ \vec {k} $$ stiffness matrix for constrained system

Plot eigenvector corresponding to zero eigenvalue for case (a).

Assume $$L_{}^{1} = L_{}^{2} = L_{}^{3} = 1$$, $$E_{}^{1} = E_{}^{2} = E_{}^{3} = 1$$, and $$A_{}^{1} = A_{}^{2} = A_{}^{3} = 1$$

For case A the global Stiffness matrix was found to be the Following:

$$K = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & -1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & -1 & 0 & 0 & 0\\ 0 & -1 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & -1 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & -1\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & -1 & 0 & 1 \end{bmatrix}$$

When run through MATLAB the following results were recieved:

V = 1.0000        0         0         0         0         0         0         0         0   -0.7071         0         0         0         0         0   -0.7071         0         0   -0.7071         0         0   -0.7071         0         0         0   -0.7071         0         0         0         0         0    0.7071         0         0   -0.7071         0         0    0.7071         0         0         0         0         0         0   -0.7071         0   -0.7071         0         0         0         0    1.0000         0         0         0         0         0         0         0         0   -0.7071         0    0.7071         0

D = 0    0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     2     0     0     0     0     0     0     0     0     2     0     0     0     0     0     0     0     0     2

With the diagonal of D being the eigenvectors.

Eigenvector Plot for Case A:

Justification of assembly of element stiffness matrix ($$\mathbf{k}^{(e)}$$) into global stiffness matrix ($$\mathbf{K}$$) where e=1...number of elements

Consider the example of the 2-bar truss from Homework 1. Recall the element force displacement relation: $$\mathbf{k}_{4x4}^{(e)}\mathbf{d}_{4x1}^{(e)}=\mathbf{f}_{4x1}^{(e)}$$

The methods and steps previously examined were:

- The Euler cut principle, method 2 (Equilibrium of global node 2), shown near the end of Homework 2.

- Free body diagrams of element 1 & 2. Element dof's $$\mathbf{d}_{4x1}^{(e)}$$, shown in Homework 1.

- For node 2, identifying the global dof's to element dof's for both element 1 & 2. This is shown in detail in the below illustration. Notice that the forces are equal and opposing.

Using the summation of force and FD relations:

$$\sum F_X=0=-f_3^{(1)}-f_1^{(2)}=0$$

$$\sum F_Y=0=P-f_4^{(1)}-f_2^{(2)}=0$$

Next, use element force displacement relations such that $$\mathbf{k}^{(e)}\mathbf{d}^{(e)}=\mathbf{f}^{(e)}$$

Taking the summation of forces for Node 2 and rearranging:

$$f_3^{(1)}+f_1^{(2)}=0 \qquad (1)$$

$$f_4^{(1)}+f_2^{(2)}=P \qquad (2)$$

Writing the forces in terms of displacements and element stiffness:

$$f_3^{(1)}=\begin{bmatrix} k_{31}^{(1)}d_{1}^{(1)} + k_{32}^{(1)}d_{2}^{(1)} + k_{33}^{(1)}d_{3}^{(1)} + k_{34}^{(1)}d_{4}^{(1)} \end{bmatrix}$$

$$f_1^{(2)}=\begin{bmatrix} k_{11}^{(2)}d_{1}^{(2)} + k_{12}^{(2)}d_{2}^{(2)} + k_{13}^{(2)}d_{3}^{(2)} + k_{14}^{(2)}d_{4}^{(2)} \end{bmatrix}$$

Substituting the previous two equations into Equation 1:

$$\begin{bmatrix} k_{31}^{(1)}d_{1}^{(1)} + k_{32}^{(1)}d_{2}^{(1)} + k_{33}^{(1)}d_{3}^{(1)} + k_{34}^{(1)}d_{4}^{(1)} \end{bmatrix}+\begin{bmatrix} k_{11}^{(2)}d_{1}^{(2)} + k_{12}^{(2)}d_{2}^{(2)} + k_{13}^{(2)}d_{3}^{(2)} + k_{14}^{(2)}d_{4}^{(2)} \end{bmatrix}=0$$

Thus we obtain the 3rd row of the Global Stiffness Matrix. Follow the same link for $$\mathbf{k}_{6x6}\mathbf{d}_{6x1}=\mathbf{F}_{6x1}$$

Homework: Derive details of Equation 2 above & the assembly of row 4 in $$\mathbf{K}\mathbf{d}=\mathbf{F}$$

$$f_4^{(1)}=\begin{bmatrix} k_{41}^{(1)}d_{1}^{(1)} + k_{42}^{(1)}d_{2}^{(1)} + k_{43}^{(1)}d_{3}^{(1)} + k_{44}^{(1)}d_{4}^{(1)} \end{bmatrix}$$

$$f_2^{(2)}=\begin{bmatrix} k_{21}^{(2)}d_{1}^{(2)} + k_{22}^{(2)}d_{2}^{(2)} + k_{23}^{(2)}d_{3}^{(2)} + k_{24}^{(2)}d_{4}^{(2)} \end{bmatrix}$$

Substituting $$f_4^{(1)}$$ and $$f_2^{(2)}$$ into Equation 2:

$$\begin{bmatrix} k_{41}^{(1)}d_{1}^{(1)} + k_{42}^{(1)}d_{2}^{(1)} + k_{43}^{(1)}d_{3}^{(1)} + k_{44}^{(1)}d_{4}^{(1)} \end{bmatrix}+\begin{bmatrix} k_{21}^{(2)}d_{1}^{(2)} + k_{22}^{(2)}d_{2}^{(2)} + k_{23}^{(2)}d_{3}^{(2)} + k_{24}^{(2)}d_{4}^{(2)} \end{bmatrix}=P$$

The assembly of $$\mathbf{K}^{(e)} $$ where e goes from 1 to the number of elements.

Global Stiffness Matrix K:

$$\mathbf{K}_{nxn}=A_{e=1}\mathbf{k}^{(e)}$$

n: Total number of global dof's before eliminating boundary conditions

ned: Number of element dof's.

Note: ned<<n

A: Assembly Operator

Principle of Virtual Work (PVW) - here

 * -Elimination of rows corresponding to boundary conditions to obtain $$\mathbf{\bar k}_{2x2}$$

 * -$$\mathbf{q}_{2x1}^{(e)}=\mathbf{T}_{2x4}^{(e)}\mathbf{d}_{4x1}^{(e)}$$

$$\mathbf{k}_{ }^{(e)}=\mathbf{T}_{ }^{(e)T}\mathbf{\hat k}_{ }^{(e)} \mathbf{T}^{(e)}$$

 * Deriving FEM for Partial Differential Equations (PDE's)

 * FD relation for a bar or spring $$ Kd=F $$ implies:

$$K_{ }^{ }d-F=0 \qquad (3)$$

 * Equivalently, with a weighting coefficient it can be simplified to:

$$w(K_{ }^{ }d-F)=0 \qquad (4)$$  ←for all W

 * Proof that (3) ↔ (4):

A) Going from equation (3)→(4)

If equation (3) = 0, equation (4) must also equal 0 (trivial)

B) Going from equation (4)→(3)

If equation (4) = 0, equation (4) must also equal 0. (not trivial)

Since equation (4) is valid for all w, we select w=1.

Equation (4) becomes $$ (Kd-F)=0 $$

Contributing Team Members
Chris Mott 17:04, 24 October 2008 (UTC)

Brandon Sell 17:06, 24 October 2008 (UTC)

Charles Marshall 18:49, 24 October 2008 (UTC)

Chris Sell 19:10, 24 October 2008 (UTC)

Jason Bruce 19:10, 24 October 2008 (UTC)

Bradley LaCroix 20:42, 24 October 2008 (UTC)

Past Homeworks
[Homework 3]

[Homework 2]

[Homework 1]

title: User:Eml4500.f08.echo.mott/HW4


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German grammar

Greetings! This is just for a rough sketch for now. You are very welcome to contribute.

Here should be focused pages about German grammar and other related material.

Verbs
The following pages are maybe the best examples of close connection between vocabulary and grammar, mentioned at the introduction. One should know what these words mean and how to conjugate them. Here are the pages we have so far:

 * Haben
 * Machen
 * Sein
 * Können
 * more verbs

Note: In the future it would be beneficial to collect all these pages at German verbs.

title: German grammar


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User:Eas4200c.f08.radsam/notes

 * When writing math equations, make sure to put "\," before your closing " " tag, like: : "\, "
 * This will ensure that every equation comes out in LaTex png format! So make sure you have it in there.
 * Also, make sure your sup/superscripts are all-inclusive. I saw stuff like this in our HW2 (just an example):
 * "$$I_y^(1)$$," which looks like this: $$I_y^(1)$$,
 * When all you need to do is put curly brackets around everything you want included in the sub/sup, like this:
 * "$$I_y^{(1)}$$," which looks like this: $$I_y^{(1)}$$
 * So if you want $$I\,$$ to have something more in the subscript, put curlies on the subscript stuff:
 * "$$I_{blah}^{(1)}\,$$," which looks like this: $$I_{blah}^{(1)}\,$$
 * Also also, I know Diego was having this problem, so just to let you guys know, if you want to change the order of equations but seemingly can't because it requires special characters like "\pi," put that character in curly brackets and you can put it anywhere you want:
 * Without curlies you can only put "\pi" after the $$r^2\,$$ because you would be spelling "\pir^2" without them: $$I_y^{(1)} = \frac{\pir^4}{2}$$ which gives you this error:
 * $$I_y^{(1)} = \frac{\pir^4}{2}$$
 * But with curlies you can do whatever you want! "$$I_y^{(1)} = \frac{{\pi}r^4}{2}$$" shows up like $$I_y^{(1)} = \frac{{\pi}r^4}{2}$$
 * Even further, to put spaces in equations use tilde "~" and to take italics out of parts of equations (also lets you put spaces in to, without the use of tildes) use "\mbox{}" ! This is cool:
 * Without mbox: $$blah1blah2blah\,$$ shows up like $$blah1blah2blah\,$$
 * Without mbox but with tildes: $$blah~1~blah~2~blah\,$$ shows up like $$blah~1~blah~2~blah\,$$
 * With mbox: $$\mbox{blah 1 blah 2 blah}\,$$ shows up like $$\mbox{blah 1 blah 2 blah}\,$$

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User talk:Eml4500.f08.FEABBQ/HW4

 * HW; 86.4

 * I'm submitting it. An email says on Friday, So i guess thats when its due. Grant

 * Who is submitting the HW, and when is it Due. it is today or tom. Esmail

 * I suppose the matlab needs to be redone for Oct 6... some jerk took it off. I'll go back through the emails, but if someone could point me to the correct email Dr Vu Quoc sent out awhile ago that might help, thanks. Andy Koby 04:01, 23 October 2008 (UTC)

 * In marston now with Esmail. Jose just left but will be be back after class. call if you want to come. 454-2563 (4:15 PM) --Eml4500.f08.FEABBQ.vitt 20:15, 21 October 2008 (UTC)

 * Alright I finished my notes for the most part I just have to draw two more pictures but we have till Friday. So i'll knock it out Thursday. My section was really small so when we meet today let me know what else I can do to help out. Stark 19:32, 21 October 2008 (UTC)

 * I'll try to do the notes today. What time do you guys want to meet?Stark 18:41, 21 October 2008 (UTC)

 * I agree, if everyone could do their notes tonight or tomorrow morning, it would really help us get the big picture of what to study. If anyone can't do this at least let us know so we can come up with other means of studying that material. Thanks.--Eml4500.f08.FEABBQ.vitt 01:28, 21 October 2008 (UTC)
 * Hey everyone, I think it would be helpful if everybody can finish their part before the exam so we can STUDY. Esmail Hadman 11:54, 20 October 2008 (UTC)

 * I just finished the notes for the 8th. Andy, the code I wrote for Homework 3 requires that the user input the coordinates of the deformed and undeformed nodes. Really it's just a visual aid at this point. But this is all we were asked to do. Maybe later he'll have us do a more complete program.
 * To anyone that's interested, I've already spoken with Andy and Jose about another group study session. Jose and I were thinking that meeting sometime on Tuesday after we all have had a chance to study individually would work best. Apparently Vu Quoc is known to give hard 2nd tests so it wouldn't hurt to do this. If you're interested, let me know when you're available. --Eml4500.f08.FEABBQ.vitt 02:42, 20 October 2008 (UTC)

 * I'm working on modifying the Matlab code from HW3, I had a quick question for whoever did it (Nik or Jose I believe). Where do the numbers for the deformed configuration come in?  Do you have to calculate them by hand and then type them into the code?  Thanks, Andy

 * for some reason Vu-Quoc took 5% off from HW 1 does any of you knowas how to fix this. I dont know what the hill is going on with all those missing links. I dont wanna miss it up. please take a look beore the 5% becomes 55%.......  we still working on HW-1 --- Esmail

 * I guess I'm new El Capitan of the group. Follow me into battle. -Grant

 * I think Vu-Quoc wants that intro page... Jose Jayma 02:31, 13 October 2008 (UTC)

 * I agree with Andy on this, also I will be doing the notes from today the 10thEsmail Hadman 21:33, 10 October 2008 (UTC)

 * HW 4 time!! Unless we have two distinct parts for this hw (like #1), why don't we just keep everything on one page and not have an intro page with the contributing members which makes it confusing to submit?  Just an idea.  Andy K 20:23, 8 October 2008 (UTC)

title: User talk:Eml4500.f08.FEABBQ/HW4


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File:2415airfoil.jpg

Summary
graph of NACA 2415 airfoil

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text: Category:Internet and Project Competence


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User:EML4500.f08.A-team.VandenBerg/Homework 4

HW4
18-1


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University of Florida/Eml4500/f08 Wiki1 Team Collaboration

Current Homework assignment
../f08 Team Homework 3/

Past Homework Submissions
Team Homework 1 Team Homework 2

Team Members
Michael Aguilar

Bryan Ambrosio

Clayton Brannon

Brooks Handy

Benjamin Oatley

Lyndsey Schaet


text:
User:Eas4200c.fo8.aero.lee/HW3

= Curved Panels =

The expression for the finite force applied to a finite unit length of a curved panel is $$d\vec{F}=q\,d\vec{l}$$. However, to analyze the curved panel with respects to an x and y component, we can make use of the fact that $$d\vec{l}=dl_{y}\,\hat{j}+dl_{z}\,\hat{k}$$ to render the equation:

 * $$d\vec{F}=q\left(dl_{y}\,\hat{j}+dl_{z}\,\hat{k} \right)$$

For $$dl_{y}$$ and $$dl_{z}$$ we can implement the polar coordinate transformations:
 * $$dl_{y}=dl\,cos(\theta)$$
 * $$dl_{z}=dl\,sin(\theta)$$

(remembering that we are in the yz plane)

and hence:
 * $$dl\,cos(\theta)=dy$$
 * $$dl\,sin(\theta)=dz$$

Integrate to find the total force
 * $$\vec{F}=\int d\vec{F}=q\int dy\,\hat{j}+dz\,\hat{k}$$
 * $$\vec{F}=q\left(\Delta y\hat{j}+\Delta z\hat{k} \right)=F_{y}\,\hat{j}+F_{z}\,\hat{k}$$

For our application we can solve for the magnitude of the force finally arriving to the equation:
 * $$\parallel\vec{F}\parallel = \sqrt{{F_y}^2+{F_z}^2}=q\sqrt{{\Delta y}^2+{\Delta z}^2}$$

= Closed Thin Walled Cross Sections =

We can relate our resultant force, $$R= \parallel\vec{F}\parallel$$, with the torque, $$T=\,2\,q\,\bar{A}$$ by:
 * $$d\vec{T}={\vec{r}}\times{d\vec{F}}$$

by letting $$\rho$$ be the perpendicular distance from the point of reference to the corresponding finite force, and by remembering that $$d\vec{F}=q\,d\vec{l}$$, we can conclude:
 * $$dT={\rho}\,q\,dl$$
 * $$T=\int dT=q\int \rho\,dl=q\int 2\,dA$$
 * $$T=2\,q\,\bar{A}$$

= Torsion of Uniform, circular Bars =

It is first important to define the rate of twist $$\theta$$ as:
 * $$\theta=\frac{d\alpha}{dx}=rate\,\,of\,\, twist$$

AND
 * $$\gamma=r\frac{d\alpha}{dx}=r\,\theta$$

Also, the second polar area moment of Inertia is defined as:
 * $$J=\int \int _Ar^2\,dA$$

To start, our equation for torque is:
 * $$T=\int \int _Ar\,\tau\,dA$$

By substituting in that $$\tau=G\,\gamma$$ and $$\gamma=r\,\theta$$ we get:
 * $$T=\int \int _A r\,G\,(r\,\theta)=G\,\theta\int \int _A r^2dA$$
 * $$T=G\,\theta\,J$$

= Torsion of Uniform, Non-circular Bars =

Torsion of uniform, non-circular bars typically leads to an occurrence of warping of the cross section. Warping is when a point in the cross section under torsion, subject to deformation, will undergo a axial displacement along the length of bar. A simple example of warping is rolling up a sheet of paper then applying a torque - the rolled sheet will noticeably elongate along its axis.

Figure 4 below shows a typical depiction of displacement in a cross section view of a non-circular bar in torsion.

As can be seen above, the displacement vector, $$\overrightarrow{P P'}$$ is perpindicular to the original position vector, $$\overrightarrow{O P}$$. This method of describing the displacement vector is chosen because with the assumption that $$\alpha$$ is small the y and z components of the displacement vector can easily be calculated. Figure 4a below shows the y and z components of displacement for the case $$\beta = 0$$.

By applying basic trigonometry to the figure above, $$u_y$$ and $$u_z$$ can be solved for in terms of $$\alpha$$:
 * $$ u_y = R(1-\cos{\alpha}) \qquad u_z = R\sin{\alpha}$$

If the assumption that $$\alpha\approx0$$ is made, then $$\cos{\alpha}\approx1$$ and $$\sin{\alpha}\approx0$$. These equations then become:
 * $$u_y\approx0 \qquad u_z\approx{R\alpha}$$

For the general case as shown in figure 4, R becomes the initial distance OP and the y-component of the displacement of point P can be expressed as
 * $$ u_{y} = -(PP')sin\beta = -(OP\alpha)sin\beta = -\alpha(OP\sin\beta) = -\alpha z_{p} $$

Similarly the z-component of the displacement of point P is

 * $$ u_{z} = (PP')cos\beta = (OP\alpha)cos\beta = \alpha(OP\cos\beta)=\alpha y_{p} $$

Recalling the definition of $$ \theta $$

 * $$ u_{y} = -\theta xz_{p} $$

 * $$ u_{z} = \theta xy_{p} $$

Generalizing for all points:

 * $$ u_{y} = -\theta xz $$ (Eq. 1)

 * $$ u_{z} = \theta xy $$ (Eq. 2)

For warping displacement in the axial direction (i.e. x-direction) it can be shown that

 * $$ u_{x} = \theta \psi (y,z) $$ (Eq. 3)

= Road Map for Analysis of Wing Torsion of a Multicell Section =

 * 1) Kinematic Assumptions (Equations (1), (2) and (3) above) [Sun, Sec. 3.2]
 * 2) Strain Displacement Relationship [Sun, Sec. 3.2]
 * 3) Equilibrium Equation for Stresses [Sun, Ch. 2, Sec. 3.2]
 * 4) Prandtl Stress Function $$ \phi $$ [Sun, Sec. 3.2, Eqn. 3.15]
 * 5) Strain Compatibility Equation [Sun, Sec. 3.2, Eqn. 3.17]
 * 6) Equation for $$ \phi $$ [Sun, Sec. 3.2, Eqn. 3.19]
 * 7) Boundary Conditions for $$ \phi $$ [Sun, Sec. 3.2, Eqn. 3.24]
 * 8) Calculation of Torque

 * $$ T = 2\int\int_{A}^{}{\phi dA} = GJ\theta $$ [Sun, Sec. 3.2, Eqn. 3.25]

 * $$ J = \frac{-4}{\bigtriangledown^2 \phi } \int \int_{A}^{}{\phi dA} $$

 * 9. Formal Derivation of Torque and Shear Flow Relationship

 * $$ T = 2q\bar{A} $$ [Sun, Sec. 3.5, Eqn. 3.48]

 * 10. Calculation of Twist Angle $$ \theta $$

 * $$ \theta = \frac{1}{2G\bar{A}}\oint_{}^{}{\frac{q}{t}ds} $$ [Sun, Sec. 3.5, Eqn. 3.56]

 * 11. Solution of Multicell Thin-walled Sections  [Sun, Sec. 3.6]

= Use of "RoadMap" to Solve a Single Cell Section Airfoil =

We can simplify the analysis of a complex, multi-cell airfoil by looking at a semi-circle attached to a triangle.

Average Area ($$\bar{A}$$)
To find $$\bar{A}$$ we simply look at the area of each shape sepratly.
 * $$\bar{A}=\bar{A}_{Semi-Circle}+\bar{A}_{Triangle}$$
 * $$\bar{A}=\frac{1}{2}\left[\pi\left(\frac{b}{2} \right)^2\right]+\frac{1}{2}\,b\,a$$
 * $$\bar{A}=5.57$$

Shear flow ($$q$$)
Remembering our derived equation for Torque, $$T=2\,q\,\bar{A}$$, we can solve for the shear flow:
 * $$q=\frac{T}{2\,\bar{A}}=\frac{T}{11.14}$$

Twist angle ($$\theta$$)
Our twist angle equation is:
 * $$\theta=\frac{1}{2\,G\,\bar{A}}\oint{\frac{q}{t}ds}$$

However when looking at each segment of the wall, we have 3 separate sections with constant thickness, so the integral portion becomes a sum:
 * $$\theta=\frac{1}{2\,G\,\bar{A}}\sum_{i=1}^{3}{\frac{q_i\,l_i}{t_i}}$$
 * $$\theta=\frac{q}{11.14\,G}\left (\frac{2\,\pi}{2\,t_1}+\frac{4}{t_2}+\frac{4.47}{t_3} \right)$$
 * $$\theta=10\frac{T}{G}$$

= Proof of Triangle Area Formula =

The area of a triangle is given as:

 * $$ A = \frac{1}{2}bh $$

To prove this equation, the area of triangle CDE is subtracted from the area of triangle BDE. Notice that area of a right triangle is exactly half the area of a square whose sides are equal to the legs of the triangle. Thus:

 * $$ A_{BDE} = \frac{1}{2}(BD)(DE) = \frac{1}{2}(BD)h $$

and

 * $$ A_{CDE} = \frac{1}{2}(CD)(DE) = \frac{1}{2}(CD)h $$

Subtracting the areas to solve for the triangle of interest:

 * $$ A_{BCE} = A_{BDE} - A_{CDE} = \frac{1}{2}(BC-CD)h = \frac{1}{2}bh $$

= Second Polar Area Moment of Inertia for a Circle = The definition of the Second Polar Area Moment of Inertia is given as:

 * $$ J = \int r^2 dA $$

For a circle of radius a, the second polar moment of inertia can be solved via a double integral as

 * $$ J = \int r^2 dA = \int_{0}^{2\pi}\int_{0}^{a}{r^2}rdrd\theta = \frac{1}{4}a^4\left[\int_{0}^{2\pi}d\theta \right] = \left[\frac{1}{4}a^4 \right]\left[2\pi \right] = \frac{\pi }{2}a^4 $$

= Second Polar Moment of Inertia of a Thin-walled Cylinder =

For a thin-walled cylinder of inner radius r_{i} = a </math and outer radius $$ r_{o} = b $$ the average radius can be given as:

 * $$ \bar{r} = \frac{b+a}{2} $$

The square of the average radius is:

 * $$\bar{r}^2 = \left( \frac{b+a}{2} \right)^2 = \frac{1}{4}\left(b^2 + a^2 + 2ab \right)$$

If the assumption is made that the wall is thin such that thickness, $$ t = b-a $$, is much less than a or b, then

 * $$ \bar{r}^2 = \frac{1}{4}\left(2b^2 + 2a^2 \right) = \frac{1}{2}\left(b^2 + a^2 \right) $$

Then,the second polar area moment of inertia for the thin-walled section can be given as

 * $$ J = \int r^2 dA = \int_{0}^{2\pi}\int_{a}^{b}{r^2}rdrd\theta = \frac{1}{4}(b^4 - a^4)\left[\int_{0}^{2\pi}d\theta \right] = \left[\frac{1}{4}(b^4 - a^4) \right]\left[2\pi \right] = \frac{\pi }{2}(b^4 - a^4) $$

Expanding the last term provides

 * $$ J = \frac{\pi }{2}(b-a)(b+a)(b^2 - a^2) $$

Recalling the definitions for $$ t $$, $$ \bar{r} $$ and $$ \bar{r}^2 $$ from above

 * $$ J = \frac{\pi }{2}t(2\bar{r})(2\bar{r}^2) = 2\pi t\bar{r}^3 $$

= Comparison of a solid, circular cross-section to a hollow, thin-walled, circular cross-section =

Given

A solid, circular cross-section has radius $$ r_{o}^{(a)} = 1 cm $$. A hollow, thin-walled, circular cross-section has inner radius $$ r_{i}^{(b)} = 5 cm $$ and thickness $$ t = 0.1 cm $$.

Find
a) compute $$ A^{(a)} $$ and $$ A^{(b)} $$

b) compute $$ J^{(a)} $$ and $$ J^{(b)} $$

c) compare $$ \frac{J^{(a)}} {J^{(b)}} $$

d) find radius $$ r_{i}^{(c)} $$ of thin-walled section with thickness $$ t = 0.02r_{i}^{(c)} $$ such that $$ J^{(c)} = J^{(a)} $$ and compare $$ \frac{A^{(a)}} {A^{(c)}} $$

a)

 * $$ A^{(a)} = \pi (r_{o}^{(a)})^2 = \pi(1 cm^2) = \pi cm^2 $$

 * $$ A^{(b)} = \pi ((r_{i}^{(b)}+t)^2 - (r_{i}^{(b)})^2) = \pi((5.1 cm)^2 -(5cm)^2) = 1.01\pi cm^2 $$

b)

 * $$ J^{(a)} = \frac{\pi }{2}(r_{o}^{(a)})^4 = \frac{\pi }{2}(1cm)^4 = \frac{\pi }{2}cm^4 $$

 * $$ J^{(b)} = \frac{\pi }{2}((r_{i}^{(b)} t) - r_{i}^{(b)})^4 = \frac{\pi }{2}((5.1cm)^4 - (5cm)^4 = 25.76\pi cm^4 $$

c)

 * $$ \frac{J^{(a)}}{J^{(b)}} = \frac{\frac{\pi}{2}cm^4}{25.76\pi cm^4} = 0.0194 $$

d)
To determine the proper $$ r_{i}^{(c)} $$, set the second polar area moment of inertia equations for a thin-walled section and a solid section equal to one another and solve for the unkonwn:

 * $$ \frac{\pi }{2}\left[\left(1.02r_{i}^{(c)} \right)^4 - \left(r_{i}^{(c)}\right)^4 \right] = \frac{\pi}{2}\left(r^{(a)}\right)^4 = \frac{\pi}{2} cm^4 $$

thus,

 * $$ r_{i}^{(c)} = 12.13 cm $$

To compare areas, calculate the area $$ A^{(c)} $$ and compare to $$ A^{(a)} $$ determined above.

 * $$ A^{(c)} = \pi\left[\left(r_{o}^{(c)}\right)^{2} - \left(r_{i}^{(c)}\right)^{2}\right] = 5.945\pi cm^2 $$

Comparing the two areas,

 * $$ \frac{A^{(a)}}{A^{(c)}} = 0.168 $$

= Analysis of Multi-Cell Sections =

It is important to understand how to analyze a multi-cell section because of their widespread use in common structures such as wings. We can apply what we already know about a single cells to our analysis of multi-cells by simply adding or integrating all of the single cell sections.

If we look at the torque generated by the $$i^{th}$$ single cell, we get the formula:
 * $$T_i=2\,q_i\,\bar{A}_i$$

To apply this equation to a multi-cell structure, we simply add all of the cells. Given that $$n_c$$ is the number of cells:
 * $$T=\sum_{i=1}^{n_c}{T_i}=2\sum_{i=1}^{n_c}{q_i\,\bar{A}_i}$$

As for the rate of twist, $$\theta$$, since the structure is a rigid body, $$\theta={\theta}_1=...={\theta}_{n_c}$$ such that the rate of twist attributed to the $$i^{th}$$ cell is:
 * $${\theta}_i=\frac{1}{2\,G_i\,\bar{A}_i}\oint{\frac{q_i}{t_i}ds}$$

= NACA Airfoil Matlab Project =

The purpose of this project is to form the basis for future structural analysis of thin walled bodies, with emphasis on airfoils. The code should be able to plot the body, determine the average area (Ā), and find the location of the centroid. These values are found through splitting the perimeter of the bodies into small discrete segments and then using properties of the cross product to determine the area from a point in space and a quadrature to determine the centroid.

= Contributing Members = The following Team Aero members contributed to this report.

Jared Lee --Eas4200c.f08.aero.lee 03:06, 1 October 2008 (UTC)

Ray Strods --Eas4200c.f08.aero.strods 03:53, 7 October 2008 (UTC)

Oliver Oyama --Eas4200c.fo8.aero.oyama 17:14, 7 October 2008 (UTC)

William Kurth --Eas4200c.f08.aero.kurth 11:53, 8 October 2008 (UTC)

Gonzalo Barcia --Eas4200c.f08.aero.barcia 12:30, 8 October 2008 (UTC)

title: User:Eas4200c.fo8.aero.lee/HW3


text:
University of Florida/Eml4500/f08 Team Homework 3

Derivation of Elemental Free-body Diagram with respect to the Global Coordinate System

 * $$ \textbf{k}_{4x4}^{(e)}\textbf{d}_{4x1}^{(e)}=\textbf{f}_{4x1}^{(e)}

$$

$$ {k}^{(e)} \begin{bmatrix} 1&-1\\ -1&1 \end{bmatrix} \begin{Bmatrix} {q}_{1}^{(e)}\\ {q}_{2}^{(e)} \end{Bmatrix} = \begin{Bmatrix} {P}_{1}^{(e)}\\ {P}_{2}^{(e)} \end{Bmatrix} $$

$${q}_{i}^{(e)}=$$ axial displacement of element 'e' at local node 'i' $${P}_{i}^{(e)}=$$ axial force of element 'e' at local node 'i'

We want to find the relationship between $$\textbf{q}^{(e)}$$ and $$\textbf{d}^{(e)}$$, and $$\textbf{P}^{(e)}$$ and $$\textbf{F}^{(e)}$$.

These relationships can be expressed in the form: $$ \textbf{q}_{2x1}^{(e)}\textbf{T}_{2x4}^{(e)}=\textbf{d}_{4x1}^{(e)} $$

Consider the displacement vector of the local node 1 denoted by $$\vec{d}_{1}^{(e)}$$.

$$\vec{d}_{[1]}^{(e)}={d}_{1}^{(e)}\vec{i}+{d}_{1}^{(e)}\vec{j}$$

$${q}_{1}^{(e)}$$ = axial displacement of node 1 is the prthagonal projection of the displacement vector $$\vec{d}_{1}^{(e)}$$ of node 1 on the $$\tilde{x}$$ axis of element 'e'.

$${q}_{1}^{(e)}=\vec{d}_{1}^{(e)}+\vec{\tilde{i}}$$

$${q}_{1}^{(e)}=({d}_{1}^{(e)}\vec{i}+{d}_{2}^{(e)}\vec{j})+\vec{\tilde{i}}$$

$${q}_{1}^{(e)}={d}_{1}^{(e)}(\vec{i}\cdot\vec{\tilde{i}})+({d}_{2}^{(e)}(\vec{j}\cdot\vec{\tilde{i}})$$

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

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

$${q}_{1}^{(e)}=l^{(e)}{d}_{1}^{(e)}+m^{(e)}{d}_{2}^{(e)}$$

$${q}_{1}^{(e)}= \begin{bmatrix}l^{(e)} & m^{(e)}\end{bmatrix}_{1x2}\begin{Bmatrix}{d}_{1}^{(e)}\\{d}_{2}^{(e)}  \end{Bmatrix}_{2x1}$$

Here we can see that $${q}_{1}^{(e)}$$ is a 1x1 scalar.

We do this for node 2 as well. $${q}_{2}^{(e)}= \begin{bmatrix}l^{(e)} & m^{(e)}\end{bmatrix}_{1x2}\begin{Bmatrix}{d}_{3}^{(e)}\\{d}_{4}^{(e)}  \end{Bmatrix}_{2x1}$$

which leads us to:

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

where this is the equation we set out to derive, $$ \textbf{q}_{2x1}^{(e)}\textbf{T}_{2x4}^{(e)}=\textbf{d}_{4x1}^{(e)} $$

Similarly, (same argument): $$ \begin{Bmatrix} P_1^{(e)} \\ P_2^{(e)} \end{Bmatrix} = I^{(e)} \begin{Bmatrix} f_1^{(e)} \\ f_2^{(e)} \\ f_3^{(e)} \\ f_4^{(e)}\end{Bmatrix} $$ Where P is a 2x1 matrix, T is a 2x4 matrix, and f is a 4x1 matrix. This relationship is the same as saying $$ \underline{P}^{(e)}= I^{(e)} \underline{f}^{(e)} $$ Recall the axial Fd relationship:

$$ \hat k^{(e)} q^{(e)} = P^{(e)} $$

where k is a 2x2 matrix, q is a 2x1 matrix, and P is a 2x1 matrix.

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

$$ - \qquad \underline{q}^{(e)} \qquad \qquad \underline{P}^{(e)} $$

Goal: We want to have k (e) d (e)= f (e) so "move" T e from right side to the left side by pre multiplying equation by T (e)-1, the inverse of T (e). Unfortunately, T(e) is a rectangular matrix of the size 2x4 and cannot be inverted. Only square matricies can be inverted. To solve this issue, transpose T. Ans:

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

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

$$ \underline{k}^{(e)} = \underline{T}^{(e)T} \underline{\hat k}^{(e)} \underline{T}^{(e)} $$

where k is on p6.1, k hat is on 12-2, and T is on 12-5. Justification for (1): PVW see 10-1 for 1st applications of PVW, reduction of global FD relationship.

$$ \underline {K}_{6x6} d_{6x1} = \underline{F}_{6x1} \rightarrow \underline{\bar K}_{2x2} \underline{\bar d}_{2x1} = \underline {\bar F}_{2x1} $$

Remember: Why not solve as follows?

$$\underline {d} = \underline{K}^{-1} F$$

Used mathcad to try to solve K-1 and could not, due to singularity of K ., i.e. the determinant of K =0 and thus K is not invertible. Recall that you need to computer $$\begin{matrix} \frac{1}{detK} \end{matrix}$$ to find K -1. Why? For an unconstrained structure system, there are 3 possible rigid body mostion in 2-D (2 translational and 1 rotational).

HW: Find the eigenvalues of K and make observations about the number of eigenvalues. Dyanmic eigenvalue problem K v = λ M  v  Where K is the stiffness matrix, M is the mass. Zero evaluation corresponds to zero stored elastic energy which means rigid body motion.

Using the Global Free-body Diagram Relationship
Using d3, and d4 in $$ \textbf{K}\textbf{d}=\textbf{F} $$ and using boundary conditions to eliminate rows in the global displacement matrix we get:

$$ \begin{bmatrix} K_{1,3} & K_{1,4} \\ K_{2,3} & K_{2,4} \\ K_{3,3} & K_{3,4} \\ K_{4,3} & K_{4,4} \\ K_{5,3} & K_{5,4} \\ K_{6,3} & K_{6,4} \end{bmatrix}_{6x2} \begin{Bmatrix}d_3\\d_4\end{Bmatrix}_{2x1}=\begin{Bmatrix}F_1\\F_2\\F_3\\F_4\\F_5\\F_6\end{Bmatrix}_{6x1}$$

Note: We really only need to do the computations for rows 1,2,5, and 6 to get F1,F2,F5,and F6 because computation from rows 3 and 4 gives the applied load which is already known.

Closing the Loop between FEM and Statics: Virtual Displacement
Two-bar truss system:

Since our two-bar truss system is statically determinant because we can view element 1 and 2 as two force bodies. By statics, the reactions are known and therefore, so are the member forces $$P^{(1)}_1$$, and $$P^{(1)}_1$$.

We then compute axial displacement degrees of freedom. This is the amount of extension within the bars.

$${q}_{2}^{(1)} = \frac{{P}_{1}^{(1)}}{{k}^{(1)}}= \frac{{P}_{2}^{(1)}}{{k}^{(1)}}=AC$$

$${q}_{1}^{(1)} = 0$$    (Node 1 is fixed)

$${q}_{2}^{(1)} = \frac{{-P}_{2}^{(2)}}{{k}^{(2)}}=AB$$

$${q}_{2}^{(2)} = 0$$    (Node 2 is fixed)

How do we back out the displacement DOFs of node 2 from above results?

Note: The displacement of node 2 is in the direction of vector D.

Infinitesimal displacement (related to virtual displacement)
Global Free-Body Diagram showing Infinitesimal Displacements

$$ \overline{AC}= {|P_2^{(1)}| \over k^{(1)}}={5.1243 \over 3/4}=6.8324 $$

$$ \overline{AB}= {|P_1^{(1)}| \over k^{(2)}}={6.2762 \over 5}=1.2552 $$

Find (x,y) coordinates of B and C:

Point B:

$$ x= \overline{AB} cos{135^\circ}=1.2552*-{\sqrt 2 \over 2}=-0.8876 $$

$$ y= \overline{AB} sin{135^\circ}=1.2552*{\sqrt 2 \over 2}=0.8876 $$

Point C:

$$ x= \overline{AC} cos{30^\circ}=6.8324*{\sqrt 3 \over 2}=5.917 $$

$$ y= \overline{AC} sin{30^\circ}=6.8324*{1 \over 2}=3.4162 $$

We now have two unknowns, (XD,YD)

We need the equations for line $$\overline{AB}$$ and $$\overline{BC}$$

Method for the Determination of the Slope between Two Arbitrary Points
PQ Line Segment Diagram

$$ \overline{PQ}=(PQ) \tilde{i}=(PQ)[cos \theta \vec i +sin \theta \vec j]=(x-x_P)\vec i +(y-y_P)\vec j $$

$$ x-x_P=(PQ)cos \theta$$

$$ y-y_P=(PQ)sin \theta$$

$$ {y-y_P \over x-x_P}=tan \theta$$

$$ y-y_P=tan \theta(x-x_P)$$

Equation for line perpendicular to PQ passing through P:

$$ y-y_P=tan( \theta + {\pi \over 2})(x-x_P)$$

Determination of (XD,YD)
Line Perpendicular to Point B:

$$y-0.8876=tan{225^\circ}(x+0.8876)$$

$$y=x+1.7752$$

Line Perpendicular to Point C:

$$y-3.4162=tan{120^\circ}(x-5.917)$$

$$y=-1.732x + 13.665$$

Intersection at Point (XD,YD):

$$y_D+1.732(y_D-1.7752)=13.665$$

$$y_D=6.127$$

$$(x_D+1.7752)=-1.732x_D + 13.665$$

$$x_D=4.352$$

$$\vec {AD} = (x_D-x_A) \vec i + (y_D-y_A) \vec j$$

Simplification: $$x_A=0, y_A=0$$

$$\vec {AD} = x_D \vec i + y_D \vec j$$

$$\vec {AD} = 4.352 \vec i + 6.127 \vec j$$

$$ \vec {AD} = d_3 \vec i + d_4 \vec j$$

$$ d_3=4.352$$

$$d_4=6.127$$

3-bar Truss System
3-bar Truss System

$$ E^{(1)}=2, A^{(1)}=3, L^{(1)}=5$$

$$E^{(2)}=4, A^{(2)}=1, L^{(2)}=5$$

$$E^{(3)}=3, A^{(3)}=2, L^{(3)}=10$$

$$\theta^{(1)}=30^\circ, \theta^{(2)}=-30^\circ , \theta^{(3)}=45^\circ$$

$$P=30$$

Convenient Local Node Numbering:

3-bar Truss System Local Node Numbering

3-Bar Truss System

ΣFx = 0 ΣFy = 0 ΣMA = 0 (Trivial)

→ 2 equations, 3 unknowns → Statically Indeterminate

Question: How about MB? (3-D Explanation)

$$ \overline{M_B} = \overline{BA}\times\overline{F} = \overline{BA'} \times\overline{F} $$

(for A' on line of action of \overline{F}

$$ \overline{M_B} = (\overline{BA} + \overline{AA'}) \times\overline{F} = \overline{BA'} \times\overline{F} + \overline{AA'} \times\overline{F} $$

Back to 3-bar truss:

Node A is in equilibrium:

$$\sum_{i=0}^3 {\overline{F_i}} = \overline{0}$$

$$\sum_{i} {\overline{M_B,i}} = \sum_{i} {\overline{BA'_i}} \times \overline{F_i} $$

Ai' = any point on line of action of Fi

$$\sum_{i} {\overline{M_B,i}} = \sum_{i} {\overline{BA'_i}} \times \overline{F_i} = \overline{BA} \times \sum_{i=0} {\overline{F_i}} = \overline{0}$$

\begin{bmatrix} K_{11} & K_{12} & K_{13} & K_{14} & K_{15} & K_{16} & K_{17} & K_{18}\\ K_{21} & K_{22} & K_{23} & K_{24} & K_{25} & K_{26} & K_{27} & K_{28}\\ K_{31} & K_{32} & K_{33} & K_{34} & K_{35} & K_{36} & K_{37} & K_{38}\\ K_{41} & K_{42} & K_{43} & K_{44} & K_{45} & K_{46} & K_{47} & K_{48}\\ K_{51} & K_{52} & K_{53} & K_{54} & K_{55} & K_{56} & K_{57} & K_{58}\\ K_{61} & K_{62} & K_{63} & K_{64} & K_{65} & K_{66} & K_{67} & K_{68}\\ K_{71} & K_{72} & K_{73} & K_{74} & K_{75} & K_{76} & K_{77} & K_{78}\\ K_{81} & K_{82} & K_{83} & K_{84} & K_{85} & K_{86} & K_{87} & K_{88} \end{bmatrix}

$$

= \begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0 & 0 & 0\\ k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0 & 0 & 0\\ k_{31}^{(1)} & k_{32}^{(1)} & (k_{33}^{(1)}+k_{11}^{(2)}+k_{11}^{(3)}) & (k_{34}^{(1)}+k_{12}^{(2)}+k_{12}^{(3)}) & k_{13}^{(2)} & k_{14}^{(2)} & k_{15}^{(3)} & k_{16}^{(3)}\\ k_{41}^{(1)} & k_{42}^{(1)} & (k_{43}^{(1)}+k_{21}^{(2)}+k_{21}^{(3)}) & (k_{44}^{(1)}+k_{22}^{(2)}+k_{22}^{(3)}) & k_{23}^{(2)} & k_{24}^{(2)} & k_{25}^{(3)} & k_{26}^{(3)}\\ 0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)} & 0 & 0\\ 0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)} & 0 & 0\\ 0 & 0 & k_{51}^{(3)} & k_{52}^{(3)} & 0 & 0 & k_{55}^{(3)} & k_{56}^{(3)}\\ 0 & 0 & k_{61}^{(3)} & k_{62}^{(3)} & 0 & 0 & k_{65}^{(3)} & k_{66}^{(3)} \end{bmatrix}

$$

MatLab Homework
Part of this assignment required the team to develop a MATLAB code that plots the deformed and un-deformed configurations of the two-bar truss system solved in class. The code is based on two models that were created by Dr. Vu-Quoc and X.G. Tan.

The blue dashed line represents the un-deformed truss, while the solid red line shows the truss after it has been deformed.

Authors
Eml4500.f08.wiki1.oatley 20:02, 8 October 2008 (UTC)

Eml4500.f08.wiki1.brannon 01:31, 8 October 2008 (UTC)

Eml4500.f08.wiki1.ambrosio 21:27, 7 October 2008 (UTC)

Eml4500.f08.wiki1.aguilar 20:45, 8 October 2008 (UTC)

Eml4500.f08.wiki1.schaet 19:42, 8 October 2008 (UTC)

Eml4500.f08.wiki1.handy 20:25, 8 October 2008 (UTC)


text:
Relosophy/Introduction

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Philosophy into Religion
While generally considered distinct, the lines between philosophy and religion are nebulous at best. Many historical examples exist of individuals who wrote philosophy (and from a philosophical perspective), but drew conclusions that could only be described as relating to religion. Nietzsche and Descartes provide two such examples:
 * Nietzsche - Nietzsche, writing philosophically, declared that "Gott ist tot" (God is dead) based on the grounds that God has made it impossible for us to know Him. While this explanation of Nietzsche's arguments is overly simplistic, Nietzsche clearly used philosophical grounds to make an assertion into religion—and for Nietzsche, the natural result of his conclusion was the rejection of religion for nihilism.
 * Descartes, writing several centuries before Nietzsche, was a philosopher and mathematician who developed several advanced theories which are still used today. One of Descartes' most significant contributions to Science has been the beginnings of the modern version of the Scientific Method (Descartes approached science and philosophy as a skeptic).  Descartes referred to this as "la méthode" (the Method).  In a philosophical work which he wrote called Méditations, Descartes explores the possibility of using the Method philosophically to explore whether or not God exists.  Ultimately, Descartes reaches the conclusion that, rationally, God must exist on a philosophical/scientific basis.  Descartes' application of Science to make a statement about Religion clearly crosses relosophical borders.

Religion as Philosophy
Religions often fail to distinguish between religion and philosophy themselves, or contain clearly reflective and philosophical elements. Buddhism is an excellent example of this:
 * Buddhism: Three of The Four Noble Truths of Buddhism, while considered central tenets of the religious practice of Buddhism, are essentially philosophical, and the fourth arguably so.  An individual could reach the same understandings of the Four Noble Truths without ever touching on Buddhist theology or feeling any "religious" experience.  However, these tenets are critical to the foundation of Buddhism.
 * Many other religions contain truths which are philosophical in nature, though they may have a less central position within the theology of those religions.

Religion into Philosophy
Not only can philosophy make forays into religion, but religions often shape or lead to new philosophical thought and discourse, often within the context of a religious background. Centuries of philosophy accompany Christianity and Islam, for example, and the relosophical background of most philosophers tints their perceptions and philosophies, even if that background is Agnostic or Atheistic in nature.
 * Islamic Philosophy: After the introduction and spread of Islam in the Middle East, a vast "empire" of culture and learning arose, giving rise to a number of new philosophies and philosophers working within the religious context provided by the Qur'an.  One significant example of this is Al-Ghazali's assertion that when fire and cotton are placed in contact, God, and not the fire, burns the cotton directly—Al-Ghazali, working from a religious viewpoint, then used logic to develop new philosophies within that context.  Historically, such philosophies sometimes have become entirely separated from the religious context which may have spawned them, and become independent concepts in their own right.  Even when they are not, it is clear that religion spawns thoughts that are clearly philosophical, and which are approached in a logical and rational manner.
 * Christian philosophy: Christian philosophy is similar in some regards, though it arose in different contexts.  Examples of the crossing of relosophical borders include the writings of Origen, who reconciled Platonism with Christianity, Martin Luther who refused to believe that a Church which violated its own doctrines could rationally contain God's truth, and Søren Kierkegaard, who wrote on a variety of philosophical topics which frequently touched the realms of theology.
 * A less "orthodox" example is Maoist Philosophy—while not religious on its face, Maoism, by encouraging the establishment of a personality cult, shares many significant elements with other religions, and became the driving force for the development of Communist philosophy in China. In some other historical contexts, similar personality cults have actually marked the commencement of new religious movements.

The influence of religion on philosophy has been long and convoluted, but, in practice, one finds that it is very difficult to separate the two entirely, and some of the most influential philosophers in history (such as Descartes) have merged the two seamlessly, at least in their own minds.

title: Relosophy/Introduction