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User:EML4500.f08.A-team.VandenBerg/HW2

 Comment: This is HW2 located in the user namespace of Lukas Vandenberg EML4500.f08.A-team.VandenBerg 20:15, 26 September 2008 (UTC)

Axial Stiffness
k(e)d(e)=f(e)

where:

 * k(e) is a 4x4 column matrix, which is the element stiffness matrix for element "e", where e=1,2.

 * d(e) is a 4x1 column matrix, which is the element displacement matrix for element "e".

 * f(e) is a 4x1 column matrix, which is the element force matrix for element "e".

In the book on page 225, the second equation from the bottom states:

$$ \begin{array}{cccc} & \begin{matrix} \quad~ d^{(e)}_1 \quad~ & \quad~ d^{(e)}_2 \quad~ & \quad~ d^{(e)}_3 \quad~ & \quad~ d^{(e)}_4 \quad~ \end{matrix} \\ k^{(e)} & \begin{bmatrix} (\ell^{(e)})^2 & \ell^{(e)}m^{(e)} & -(\ell^{(e)})^2 & -\ell^{(e)}m^{(e)} \\ \ell^{(e)}m^{(e)} & (m^{(e)})^2 & -\ell^{(e)}m^{(e)} & -(m^{(e)})^2 \\ -(\ell^{(e)})^2 & -\ell^{(e)}m^{(e)} & (\ell^{(e)})^2 & \ell^{(e)}m^{(e)} \\ -\ell^{(e)}m^{(e)} & -(m^{(e)})^2 & \ell^{(e)}m^{(e)} & (m^{(e)})^2 \end{bmatrix} & = & k^{(e)} \end{array} $$

Also: k(e) = E(e)A(e)/L(e), which is the actual stiffness of the bar element "e". It is also the equivalent of the spring constant "k".

Director Cosines
l(e) and m(e) are the director cosines of the $$\tilde{x}$$ axis which goes from local node number one to local node number 2 with respect to (w.r.t) the global (x,y) coordinate system.

6-2

Meeting 7; Wednesday, 10 September 2008 7-1

2-Bar Truss System Model
continued....

Element 1 Stiffness:
Θ(1) = 30°

l(1) = cosΘ(1) = cos30° = √3/2

k(1) = E(1)A(1)/L(1) = (3)(1)/(4) = 3/4

$$ k^{(1)} = \begin{bmatrix} k^{(1)}_{11} & k^{(1)}_{12} & k^{(1)}_{13} & k^{(1)}_{14} \\ k^{(1)}_{21} & k^{(1)}_{22} & k^{(1)}_{23} & k^{(1)}_{24} \\ k^{(1)}_{31} & k^{(1)}_{32} & k^{(1)}_{33} & k^{(1)}_{34} \\ k^{(1)}_{41} & k^{(1)}_{42} & k^{(1)}_{43} & k^{(1)}_{44} \end{bmatrix} = \begin{bmatrix} k^{(1)}_{ij} \end{bmatrix} $$

i: row index (=1,2,3,4)

j: column index (=1,2,3,4)

(e): element

4x4 matrix

Observations:

1)Only three numbers need to be computed; other coefficients have the same absolute values just differing by a "+" or "-".

2)Matrix k(1) is symmetric, i.e., kij(1) = kji(1)

k13(1) = k31(1)

just interchanging the row and column indices

k11(1) = k(1) x (l1)2

= (3/4)(√3/2)2 = 9/16

k12(1) = k1(l(1)m(1))

= 3√3/16

k42(1) = -k1(m(1))2

= -3/16

In general kij(1) = kji(1) or k(e)T = k(e) 7-3

 * T = Transpose

The transpose of k(e) is equal to k(e)

When filling in a matrix, the upper triangular portion is the only part that needs to be filled in because the lower triangular portion is symmetrical.

Element 2 Stiffness:
7-4

k(2) = E(2)A(2)/L(2) = (5)(2)/(2) = 5

Θ(2) = -π/4

l(2) = cos(-π/4) = √2/2

m(2) = sin(-π/4) = -√2/2

k(2) = [kij(2)]4x4

k11(2) = k(2)(l(2))2 = (5)(√2/2)2 = 2.5

Observations:

1) The absolute values of all the coefficients kij(e), e=2, (i,j) = 1, ..., 4 are the same.  Comparing one coefficient to another by adding a "+" or a "-".

2) k(2)T = k(2), i.e., k(2) sym.

Element Force Displacement(FD) Relation
7-5

k(e)d(e) = f(e)

with:

k(e) being a 4x4 matrix,

d(e) being a 4x1 matrix,

f(e) being a 4x1 matrix.

and e = 1,2

$$ d^{(e))} = \begin{bmatrix} d^{(e)}_1 \\ d^{(e)}_2 \\ d^{(e)}_3 \\ d^{(e)}_4 \end{bmatrix}

f^{(e))} = \begin{bmatrix} f^{(e)}_1 \\ f^{(e)}_2 \\ f^{(e)}_3 \\ f^{(e)}_4 \end{bmatrix} $$

Global FD Relation ("free-free structure)
Kd = F

where:

K is an nxn matrix,

d is an nx1 matrix, and

F is an nx1 matrix.

there, n = 6 (page 5-5)

Meeting 8, Friday 12, September 2008 8-1

Control from page 7-5:

In compact notation, the equation yields:

[Kij]{dj} = {Fi}

with:

[Kij] being a 6x6 matrix,

{dj} being a 6x1 matrix, and

{Fi} being a 6x1 matrix.

(or more generally noted as nxn)

$$\sum^6_{j = 1} k_{ij} d_j = F_i ~ where ~ i = 1,...,6$$

K nxn = {Kij}nxn = This is the global stiffness matrix

d nx1 = {dj}nx1 = This is the global displacement matrix

F nx1 = {Fi}nx1 = This is the global force matrix 8-2 Recall from page 7-2 the element FD rel....

k4x4(e)d4x1(e) = f4x1(e)

k4x4(e) = [kij(e)] = element stiffness matrix

d4x1(e) = {di(e)} = element displacement matrix

f4x4(e) = {fi(e)} = element force matrix

Element Matrices to Global Matrices
The How to... of going from element to global matrices (i.e. stiffness, displacement, and force)

An assembly process is how it's done!

First things first, the correspondence between the element displacement d.o.f.'s and global d.o.f.'s must be identified. 8-3 Global level: page 5-5

{d1, dSubscript text2 ,....., d6}

Element level: page 4-3, and page 4-2

 * Element 1: {d1(1), d2(1), d3(1), d4(1)}

 * Element 2: {d1(2), d2(2), d3(2), d4(2)}

Global-level DOF Identifying
(Beginning of Assembly Process)

Node 1:

d1 = d1(1)

 * (1) = element 1

 * 1 = first DOF

 * d1 = first Global DOF (LHS)

d2 = d2(1)

Node 2:

d3 = d3(1) = d1(2)

d4 = d4(1) = d2(2)

 * Global Node 2 has 2 displacement components regardless of notation.

Node 3:

d5 = d3(2)

d6 = d4(2)

Conceptual Step of Assembly:
(Topology of K)

The overlap is a result of global node 2 identifications.

Meeting 9, September 15, 2008

Kundefinedk(1)11 = 9/16

Kundefinedk(1)12 = .3247595

K33 = k(1)33 + k(2)11 = 9/16 + 5/2 = 3.0625

K34 = K34 = k(1)34 + k(2)12 = .3247595 + 5/2

The rest of the stiffness matrix can be computed similarly

Elimination of known dofs:

reduce global FD rel.

d1 = d2 = d5 = d6 = 0

EML4500.f08.A-team.kirley 20:20, 26 September 2008 (UTC)

Applying fixed B.C.'s => deleting corresponding collumns i nthe global stiffness matrix K.

By principle of Virtual Work (PVW) => deleting also corresponding rows (here, rows 1,2,3,5,6). Corresponding rows of F are also deleted.

Resulting FD relation

What is K-1 (without calculator) d1 K = (K33 K44) - (K43 K34)

K*K-1 = K-1*K = I

$$K^{-1} \neq \frac 1 {det K} ~ K^{-1}$$

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

| 0.5625  0.32476  ... ...   |          |  0  |    |                             |          |  0  |    |                             |          |  0  |    |                             |4x4       |  0  |4x1

| -4.4348  |       |  f1(1)  | =  |  -2.5622  |  =    |  f2(1)  | | +4.4378  |       |  f3(1)  | | +2.5522  |       |  f4(1)  | f1(1) & f2(1) = reactions f3(1) & f4(1) = internal forces

Observation: Element 1 is in equilibrium summation Fx = f1(1) + f3(1) = 0 summation Fy = f2(1) + f4(1) = 0

p1(1) = [(f1(1))2 + f2(1))2]1/2

Two methods for solving 2-bar truss with Euler cut principle

Method 1:

Method 2:

Back to FE solution: to bring p back into the big picture --> equilibrium of global node n(circled)

Homework Problems for HW 2
9/8/08 P. 1

Write out k(1) and k(2).

First begin with:

Now find the axial stiffness, k:

Next multiply through the stiffness matrix by k to create the k(1) matrix:

The same goes for the k(2) matrix:

and finally:

9/8/08 P. 2

Prove that dotting ĩ with j gives the director cosine equal to sine of the angle theta:

P. 3

Compare the results of statically determining the reactions to the finite element method:

Since this is a statically determinate system, statics can be used to calculate the reaction forces on each of the element. Start by summing the forces on node 2.

Since P is known (P = 7), there are 2 equations with 2 unknowns making this easily solvable. After rearranging the equations and substituting, it can be found that F12 = 5.124 and that F23 = 6.276. The reaction forces can be split up as follows:

These reaction forces are exactly the same as those computed by the MatLab code. Using statics involved less complex mathematics (no matrices), however, rearranging and substituting equations, especially when trig functions are involved can become confusing and tedious for more complex systems. The finite element method reduces everything down to a few matrix multiplications and can easily be put into a computer for computation and saves a great deal of time and headache. The stiffness matrices are the same for either hand or computer calculations, which is not unreasonable to believe since it is merely a scalar being multiplied through a matrix. Displacement matrices were also nearly the same but it can be imagined that if a more complicated system were analyzed that a computer would be a better choice since it will be less apt to make mistakes and can carry out numbers to more decimal places.

Contributions for HW2
Contributors:

Lukas Vandenberg EML4500.f08.A-team.VandenBerg 20:15, 26 September 2008 (UTC)

Brian Morford --EML4500.f08.A-team.morford 20:31, 26 September 2008 (UTC)

Joseph Rieth --EML4500.f08.A-team.rieth 20:13, 19 September 2008 (UTC)

Joseph Melvin --EML4500.f08.A-team.melvin 20:23, 19 September 2008 (UTC)

Brian Kirley --EML4500.f08.A-team.kirley 20:32, 19 September 2008 (UTC)

Daniel Robinson --Eml4500.f08.a-team.robinson 20:32, 19 September 2008 (UTC)

title: User:EML4500.f08.A-team.VandenBerg/HW2


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User:Eml4500.f08.group/Homework4/Contributors

=Contributing Members=

Jeff Wright Eml4500.f08.group.w 05:36, 18 October 2008 (UTC)

Brian Taylor Eml4500.f08.group.taylor 02:49, 19 October 2008 (UTC)

Julia Setlak Eml4500.f08.group.setlak 16:53, 23 October 2008 (UTC)

Tom Cady Eml4500.f08.group.cady 17:54, 23 October 2008 (UTC)

Aaron Clausen Eas4200c.f08.carbon.clausen 12:53, 24 October 2008 (UTC)

Melisa Gaar Eml4500.f08.group.gaar 16:33, 24 October 2008 (UTC)

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User:EML4500.f08.A-team.VandenBerg/HW3

 Comment: This is HW3 located in the user namespace of EML4500.f08.A-team.VandenBerg 19:11, 8 October 2008 (UTC)

HW3
12-1

Meeting 12: Monday, 22 September 2008 EML4500
Topic to be covered in class is the derivation of the element force displacement with respect to the global coordinate system

Derivation of the Element Force Displacement With Respect To the Global Coordinate System
(From page 6-1)

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

12-2 page 4-5

(2) $$k^{(e)} \begin{bmatrix} 1     & -1 \\  -1     & 1 \end{bmatrix} \begin{pmatrix} q_1^{(e)} \\ q_2^{(e)}\end{pmatrix} = \begin{matrix} P_1^{(e)} \\ P_2^{(e)}\end{matrix}$$

where:

$$k^{(e)} \begin{bmatrix} 1     & -1 \\  -1     & 1 \end{bmatrix}$$ = $$\vec{k}^{(e)}$$

qi(e) is equal to the axial displacement of the element (e) at the local node "i". pi(e) is equal to the axial force of the element (e) at the local node "i".

Goal: The goal is to derive equation (1) from equation (2) which was derived previously in meeting 4.

To do this we want to ind the relation between:

$$\underset{2x1}{q}^{(e)}$$ and $$\underset{4x1}{d}^{(e)}$$ $$\underset{2x1}{p}^{(e)}$$ and $$\underset{4x1}{f}^{(e)}$$ 12-3 This relation can also be stated in the form of $$\underset{2x1}{q}^{(e)}$$ = $$\underset{2x4}{T}^{(e)}$$ $$\underset{4x1}{d}^{(e)}$$

Displacement Vector of the Local Node
Considering the displacement vector of the local node "1", also noted in the form of $$d_1^{(e)}$$

where $$\vec{\tilde{i}}$$, $$\vec{i}$$, and $$\vec{j}$$ is the displacement of the local node number.

The displacement vector of element e at local node number is: $$\vec{d}^{(e)}_{1} = d^{(e)}_1 \vec{i} + d_{2}^{(e)} \vec{j}$$

Axial displacement
The orthogonal projection of the displacement vector $$\vec{d}_1^{(e)}$$ of node 1 on the $$\tilde{x}$$ axis of the element e is known as: $$q_1^{(e)}$$ 12-4 $$q_1^{(e)} = d_1^{(e)}$$ • $$\vec{\tilde{i}}$$
 * = ($$d_1^{(e)} \vec{i} + d_2^{(e)} \vec{j}$$) • $$\vec{\tilde{i}}$$
 * = $$d_1^{(e)}$$ ($$\vec{i}$$ • $$\vec{\tilde{i}}$$) + $$d_2^{(e)}$$ ($$\vec{j}$$ • $$\vec{\tilde{i}}$$)

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

($$\vec{j}$$ • $$\vec{\tilde{i}}$$) = sin$$\theta^{(e)}$$

$$q_1^{(e)} = \mathit{l}^{(e)} d_1^{(e)} + m^{(e)} d_2^{(e)} \Rightarrow$$ 1x1 Scalar
 * = $$\llcorner L\mathit{l}^{(e)} m^{(e)} \lrcorner$$ 1x1 matrix $$\begin{Bmatrix}

d_1^{(e)} \\ d_2^{(e)} \end{Bmatrix}$$ 2x1 matrix

12-5

$$\begin{Bmatrix} q_1^{(e)} \\ q_2^{(e)} \end{Bmatrix} = \begin{bmatrix} \mathit{l}^{(e)} & m^{(e)} & 0 & 0    \\ 0 & 0 & \mathit{l}^{(e)} & m^{(e)} \end{bmatrix} \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \\ d_3^{(e)} \\ d_4^{(e)} \end{Bmatrix}$$

$$\underset{2x1}{q}^{(e)}$$ = $$\underset{2x4}{T}^{(e)}$$ $$\underset{4x1}{d}^{(e)}$$

14-1

Meeting 14: Friday, 26 September 2008, EML4500

 * (Meeting 13: Exam 1, Wednesday, 24 September)

Using the same argument as the equation above, you can get:

$$\begin{Bmatrix} p_1^{(e)} \\ p_2^{(e)} \end{Bmatrix}$$ 2x1 = $$T_{2x4}^{(e)}\begin{Bmatrix} f_1^{(e)} \\ f_2^{(e)} \\ f_3^{(e)} \\ f_4^{(e)} \end{Bmatrix}$$ 4x1

Where:

$$\begin{Bmatrix} p_1^{(e)} \\ p_2^{(e)} \end{Bmatrix}$$ 2x1 = $$p^{(e)} = T^{(e)} f^{(e)}$$

Remember the element axial FD rel.: $$\vec{k}$$ * $$\underset{2x1}{q}^{(e)}$$ = $$\underset{2x1}{p}^{(e)}$$

where:

$$\vec{k}$$ 2x2 is a 2x2 vector 14-2

Inverse Matrice
$$\Rightarrow \vec{k}^{(e)} (T^{(e)} d^{(e)}) = (T^{(e)} f^{(e)})$$

Where:

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

$$(T^{(e)} f^{(e)}) = p^{(e)}$$

Goal: We want to have $$k^{(e)} d^{(e)} = f^{(e)}$$ so, we must "move" $$T^{(e)}$$ from the right hand side of the equation to left hand side of the equation by pre-multiplying the equation by $$T^{(e)}$$-1 which is the inverse of $$T^{(e)}$$.

Unfortunately though, $$T^{(e)}$$ is a rectangular matrix, which means it cannot be inverted.

Answer: $$\begin{bmatrix} T^{(e)^{T} } & \vec{k}^{(e)} & T^{(e)} \end{bmatrix} d^{(e)} = f^{(e)} $$

$$\begin{bmatrix} 4x2 & 2x2 & 2x4 \end{bmatrix} 4x1 = 4x1 $$

$$\begin{bmatrix} 4x2 & 2x2 & 2x4 \end{bmatrix} = 4x4 $$

14-3 $$\Rightarrow k^{(e)} d^{(e)} = f^{(e)}$$

Justification for equation (1): $$\mathbf{P}\mathbf{V}\mathbf{W}$$ (this will be covered later)

For the first application of $$\mathbf{P}\mathbf{V}\mathbf{W}$$, see page 10-1 in the notes.

Reduction of Global FD Relationship
$$\underset{6x6}{k}^{(e)}$$ * $$\underset{6x1}{d}^{(e)}$$ = $$\underset{6x1}{f}^{(e)} \Rightarrow \vec{K} \vec{d} = \vec{F}$$

Where:

$$\vec{K}$$ is a 2x2

$$\vec{d}$$ is a 2x2

$$\vec{F}$$ is a 2x1

Rem: Why not solve for the displacement by setting $$d = K^{-1} f$$? 14-4 Because of the singularity of K, the determinant is equal to zero, and thus is not invertible

There are three possible rigid body motions in 2-D for an unconstrained structural system, one rotational motion and two translational motions.

Dynamic Evaluation Point
Kv = λMv

Where:

K = Stiffness matrix M = Mass material eval. (~vibration frequency)

zero eval. corresp. zero stored elastic energy14-5

$$ \Rightarrow $$ rigid body modes

Modes: mode shapes
 * eight vectors

Meeting 15: Monday, 29 September 2008, EML4500
To compute reactions two methods are imployed:

1)Using element force-displacement relations

2)Using global force-displacement relations (described below)

Using Global Free Body Relationships to Compute Reactions
Using force-displacement relations from meeting 10:

(6x2 matrix)x(2x1 mat.)=(6x1 mat.):

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

Only necessary to do computations for rows 1,2,5,6 to get F1,F2,F5,F6 (reactions)

What "the Loop" Is
Clarifying the loop between the F.E.M. and statics:

For two-bar truss system:

FEM --> Compute displacement --> Compute reactions

Statics --> Compute reactions --> Compute displacement

The step in statics from computing reactions to computing displacement is what we define as "closing the loop"

Closing the Loop
FBD's of two-force bodies (elem.'s 1,2):

Figure 8: Axes Forces in Elem. 1

Figure 9: Axes Forces in Elem. 2

By statics, reactions known and therefore the member forces: P1(1), P2(2)

Computing axial displ. dof's (amount of extension of bars):

$$q2^1 = \frac{P1^1}{k^1}=\frac{P2^1}{k^1} $$, $$q1^2 = \frac{-P2^2}{k^2}$$

Fixed node at 1 & 2 means: q1^2 = q2^2 = 0

To back out from above results the displ. dof's of node 2:

Eml4500.f08.a-team.robinson 03:46, 21 October 2008 (UTC)

Infinitesimal displacement

2 unknowns (XD YD)

PQ = (PQ)i^{~} = (X-X_{P})i + (Y-Y_{P})j $$

(PQ)i^{~} = PQ[ \cos{\theta}i + \sin{\theta}j ] $$

Therefore

X - X_{P} = PQ\cdot \cos{\theta} $$

Y - Y_{P} = PQ\cdot \sin{\theta} $$

\frac {Y-Y_{P}}{X-X_{P}} = \tan{\theta} $$

Y-Y{P} = (\tan{\theta})\cdot(X-X_{P}) $$

Equation for a line perpendicular to above line, passing P:

Y-Y_{P} = \tan{(\theta + \frac{\pi}{2})}(X-X{P}) $$

 Three-bar Truss System

Local node numbering by convenience in assembly of K:

EML4500.f08.A-team.kirley 20:55, 8 October 2008 (UTC)

Meeting 17: Friday, 3 October 2008, EML4500
ΣFx = 0, ΣFy = 0, ΣM A = 0 (sum of moments trivial)

Q: How about ΣM B ?

3D explanation

ΣM B  = BA X F

ΣM B  = BA' X F

for A' on linen of action of F

BA = BA + AA'

M B  = (BA + AA') X F

M B  = BA X F + AA' X F

where = AA' X F = 0

Back to 3-Bar Truss:

Node A is in equilibrium : ΣF i  = 0 from i=0 to i=3

Σ i M Bi  = Σ i BA' i  X F i

A' i = any point on the line of action of F i

Σ i M Bi  = Σ i BA' i  X F i

Σ i M Bi  = BA X Σ i F i  = 0 - meaning ΣM B is useless, still only 2 useful equations.

K 33 = K(1) 33 + K(2) 11 + K(3) 11

K 34 = K(1) 34 + K(2) 12 + K(3) 12

HW: Fill out the rest!

The filled out matrix would appear as follows:

$${K}= \begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0 & 0 & 0\\ k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0 & 0 & 0\\ k_{31}^{(1)} & k_{32}^{(1)} & k_{33}^{(1)}+k_{11}^{(2)}+k_{11}^{(3)} & k_{34}^{(1)}+k_{12}^{(2)}+k_{12}^{(3)} & k_{13}^{(2)} & k_{14}^{(2)} & k_{13}^{(3)} & k_{14}^{(3)} \\k_{41}^{(1)} & k_{42}^{(1)} & k_{43}^{(1)}+k_{21}^{(2)}+k_{21}^{(3)} & k_{44}^{(1)}+k_{22}^{(2)}+k_{22}^{(3)} & k_{23}^{(2)} & k_{24}^{(2)} & k_{23}^{(3)} & k_{24}^{(3)}\\ 0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)} & 0 & 0 \\0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)} & 0 & 0 \\ 0 & 0 & k_{31}^{(3)} & k_{32}^{(3)} & 0 & 0 & k_{33}^{(3)} & k_{34}^{(3)}\\ 0 & 0 & k_{41}^{(3)} & k_{42}^{(3)} & 0 & 0 & k_{43}^{(3)} & k_{44}^{(3)} \end{bmatrix}$$

Homework 3 Problems
9/26/08:

Show that the k matrix is equal to the T matrix times the transpose of the T matrix times the $$\hat {k}$$

First we show that:

$$\hat {k}^{(e)} = k^{(e)}  \begin{bmatrix} 1 & -1 \\ -1 & 1 \\ \end{bmatrix}$$

Letting the k matrix be made up of k11, k12, etc. and creating it as a 2 by 2 matrix we get:

$$\hat {k}^{(e)} = \begin{bmatrix} k_{11} & -k_{12} \\ -k_{21} & k_{22} \\ \end{bmatrix}$$

and that T is:

$$T^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0     \\ 0 & 0 & l^{(e)} & m^{(e)} \\ \end{bmatrix}$$

The transpose of T becomes:

$$T^{(e)^T} = \begin{bmatrix} l^{(e)} & 0\\ m^{(e)} & 0\\ 0 & l^{(e)}\\ 0 & m^{(e)}\\ \end{bmatrix}$$

Finally, multiplying everything together produces:

$$\begin{bmatrix} k_{11} & k_{12} \\ k_{21} & k_{22} \\ \end{bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0     \\ 0 & 0 & l^{(e)} & m^{(e)} \\ \end{bmatrix}\begin{bmatrix} k_{11} & -k_{12} \\ -k_{21} & k_{22} \\ \end{bmatrix}\begin{bmatrix} l^{(e)} & 0\\ m^{(e)} & 0\\ 0 & l^{(e)}\\ 0 & m^{(e)}\\ \end{bmatrix}$$

Therefore:

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

Find the eigenvalues of the matrix K and comment about the number of zero eigenvalues:

Using MatLab to solve for the eigenvalues we get:

K =

0.5625   0.3248   -0.5625   -0.3248         0         0    0.3248    0.2500   -0.3248   -0.2500         0         0   -0.5625   -0.3248    3.0625   -2.1752   -2.5000    2.5000   -0.3248   -0.2500    2.8248    1.9375    2.5000   -2.5000         0         0   -2.5000    2.5000    2.5000   -2.5000         0         0    2.5000   -2.5000   -2.5000    2.5000

EDU>> eig(K)

ans =

8.4194   2.5880   -0.1730   -0.0218   -0.0000    0.0000

Which shows that two of the eigenvalues are zero.

9/29/08:

Using the global FD relation:

Solve for F1, F2, F5, and F6 and compare results with method 1.

The force matrix f can be solved for by multiplying the global stiffness matrix K by the displacement matrix d, i.e.

$$\begin{bmatrix} k_{11} & k_{12} & k_{13} & k_{14} & k_{15} & k_{16}\\ k_{21} & k_{22} & k_{23} & k_{24} & k_{25} & k_{26}\\ k_{31} & k_{32} & k_{33} & k_{34} & k_{35} & k_{36}\\ k_{41} & k_{42} & k_{43} & k_{44} & k_{45} & k_{46}\\ k_{51} & k_{52} & k_{53} & k_{54} & k_{55} & k_{56}\\ k_{61} & k_{62} & k_{63} & k_{64} & k_{65} & k_{66}\\ \end{bmatrix} \cdot \begin{bmatrix}d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6\end{bmatrix} = \begin{bmatrix}F_1 \\ F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6 \end{bmatrix}$$

Where the K matrix has the same values as the problem above, and the values of the d matrix are such that d1 = d2 = d5 = d6 = 0 and d3 = 4.352 and d4 = 6.127.

After substituting all the numbers in and then simplifying by removing columns 1, 2, 5, and 6 of the K matrix and also rows 1, 2, 5, and 6 of the d matrix, and finally using MatLab to solve for the force matrix f we find that:

k =

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

EDU>> F=k*d

F =

-4.4380  -2.5623    0.0005    6.9998    4.4375   -4.4375

Where F3 and F4 were already known as the applied loads, but where computed for completeness, and the reactions are the other forces shown in the matrix.

In comparison to method 1 (element FD relationship), the numbers are nearly the same, and actually most likely would be had numbers not been truncated by MatLab and the human being who did this problem for simplicity

10/1/08:

Infinitesimal displacement of AC and AB.

These are found by dividing the P matrix by the k matrix

$$AC = \left ( \frac{|P^{(1)}_2|}{k^{(1)}} \right ) = \left ( \frac{5.1243}{3/4} \right ) = 6.8324$$

$$AB = \left ( \frac{|P^{(2)}_1|}{k^{(2)}} \right ) = \left ( \frac{6.276}{5} \right ) = 1.2552$$

Find (XB, YB), (XC, YC), and (XD, YD).

These can be found by multiplying the lengths AB and AC by their respective director cosines as follows:

$$X_B = AB\times\cos{(135)} = 1.2552\cos{(135)} = 0.88756$$

$$Y_B = AB\times\sin{(135)} = 1.2552\sin{(135)} = 0.88756$$

$$X_C = AC\times\cos{(30)} = 6.8324\cos{(30)} = 5.91703$$

$$Y_C = AC\times\sin{(30)} = 6.8324\sin{(30)} = 3.4162$$

Solving for (XD, YD) is not quite as simple however and requires a little bit more work, namely using the following equation for a line perpendicular to another line and passing through P, this equation's derivation should be found in the lecture notes:

$$y - y_P = \tan{(\Theta + \frac{\Pi}{2})} ( x - x_P )$$

$$\overrightarrow{A D} \ = (X_D - x_i)\vec i + (Y_D - y_i)\vec j$$

Since xi and yi drop out (equal to zero) because they are at the origin we get:

$$\overrightarrow{A D} \ = X_D\vec i + Y_D\vec j$$

Therefore:

$$\overrightarrow{A D} \ = d_3\vec i + d_4\vec j$$

Thus XD = d3 and YD = d4 or XD = 4.35 and YD = 6.125

--EML4500.f08.A-team.morford 18:21, 8 October 2008 (UTC)

Contributions for HW3
--EML4500.f08.A-team.rieth 02:39, 8 October 2008 (UTC)

--Eml4500.f08.a-team.robinson 03:33, 8 October 2008 (UTC)

--EML4500.f08.A-team.morford 18:24, 8 October 2008 (UTC)

EML4500.f08.A-team.VandenBerg 19:11, 8 October 2008 (UTC)

EML4500.f08.A-team.kirley 20:55, 8 October 2008 (UTC)

--EML4500.f08.A-team.melvin 21:02, 8 October 2008 (UTC)

title: User:EML4500.f08.A-team.VandenBerg/HW3


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User:Ameetkulkarni

Ameetkulkarni 08:14, 18 October 2008 (UTC)--Ameetkulkarni 08:14, 18 October 2008 (UTC)ameet kulkarni--Ameetkulkarni 08:14, 18 October 2008 (UTC)

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User talk:Vroni

 Hello Vroni, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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User:Greyman

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I've been editing wiki's for more than a few years now, from Wikipedia to a large number of wikis on Wikia. Most of my wiki-time is spent over at Duty & Valour, the Canadian Military History wiki, which I founded.

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User talk:Greyman

 Hello Greyman, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! Jtneill - Talk - c 13:28, 18 October 2008 (UTC)

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User talk:JWSchmidt/JWSchmidt is a whiner

You'll soon be unblocked, so you can stop whining about that. The Jade Knight (d'viser) 18:19, 8 October 2008 (UTC)


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User talk:JWSchmidt/JWSchmidt go fuck yourself

Concerns
Wikiversity is bigger than you, me, or for that matter anyone else. Stop personalizing things. Stop worrying about people's agendas. Stop worrying about your agenda. If you have something in particular that you'd like to learn/teach/research, do the work. If you have a problem with me, cormaggio, or anyone else, bring it up on Community Review. If the only thing you have to offer is your vision, offer it, but don't expect everyone to agree with it.

You seem to have taken the stance that people are either with you or they're against Wikiversity. No matter how firmly you believe that, it's just not true. Stop demonizing, and start talking. Or don't. You've pretty much lost any credibility with me... I'm not alone in feeling that way, and the more you put your agenda above the good of the project, the more other people will feel the same.

You've got a lot of good energy and ideas (or at least you used to). Ditch the hate, embrace the promise. Or just go fuck yourself and go away... at this point I don't care. I really am that disappointed, but I'm starting to get over it now. --SB_Johnny talk 02:52, 18 October 2008 (UTC)

 * I just want to clearly state my opposition to the "go fuck yourself" comment - it is not at all civil, and I very much hope Johnny retracts it. Cormaggio talk 16:23, 19 October 2008 (UTC)

 * If SB_Johnny were not a custodian, he would be blocked for incivility or personal attacks. Anyway, there is something curious, he tells JWSchmidt to "Ditch the hate, embrace the promise." and then he talks in that polite way. This is more like a joke than a post. What if JWSchmidt answered the same way? He would be blocked and everybody would call him troll. But, as a custodian did it, very few complain. And... if he ever replies it will be something like: "oh, dudes, you all know it is saying, I did not mean it. I was just saying it in a free way." I thought those comments were only inside wikia, but in something that tries to be like a university and about education? Oh, maybe I quit before I am told to f*** myself, in a block summary.

 * Stop personalizing things.

 * The other funny aspect is that SB_Johnny suggests that and then he gets into personal insults. If wikiversity is not about education, at least is about humor.

 * --Davichito 23:27, 27 December 2008 (UTC)

 * It's typical of the double standards that pervades WMF-sponsored projects: Free speech for me, but not for thee.  —Moulton 16:34, 5 January 2009 (UTC)


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User:Eas4200c.f08.vqcrew.c/Homework 4

 Changes: The two MATLAB codes for calculation of the torsional constant J for both a single-celled and multi-celled airfoil have been modified and updated. The results of running the new codes have also been changed in the section.

Link to the pages of differences: Changes

Eas4200c.f08.vqcrew.c 15:49, 30 October 2008 (UTC)

Quadrature of a NACA Airfoil

The area of any shape can be determined by using a collection of many small triangles. The triangles are defined by an arbitrary origin, O, and an infinitesimal length on the shape. Since the infinitesimal length and perpendicular distance from the origin to the surface are known, the infinitesimal area, dA, can be calculated. Figure 1 shows this process for an airfoil.

 * $$ \bar{A}_{airfoil} = \bar{A}_{1} + \bar{A}_{2} = \bar{A}_{1} - \left|\bar{A}_{2}\right|$$

[[Image:vqcrew_triangle.png|thumb|right|200px|Fig. 2 - Determination of the infinitesimal area, dA, using the triangle method. ] In this case, the upper surface yields a positive area and the lower surface yields a negative area due to the location of the origin and the nature of the right-hand rule for vectors. The total area of the airfoil is computed as a sum of all of the small areas for each length along the airfoil contour.  Figure 2 provides a more detailed picture of how each area is calculated.  The example shown is for the lower surface, therefore the cross-product of the vectors  $$\textstyle \vec{r}$$  and $$\textstyle \vec{PQ}$$ is negative.  The negative cross-product leads to a negative area for the lower surface in this configuration.

 * $$\displaystyle d\vec{A} = \frac{1}{2}(\vec{r}\times\vec{PQ})$$

Problems with Trapezoidal Method of Quadrature
The process of dividing the area of a shape in to a series of small trapezoids may be an accurate way to determine the total area for some shapes in some configurations, but it is not a general solution. For example, Figure 3 shows the area of an airfoil being divided into trapezoids horizontally. For this shape in this orientation, the result will not be very accurate due to the errors near the trailing edge. The thin trailing edge is not conducive to being split up in this manner. It is true that if the trapezoids were aligned vertically the result would more accurately represent the true area, but a more elegant and beautiful solution can be reached. Using the triangle method described above, a more accurate result for the area can be found regardless of the location of the origin or orientation of the airfoil.

Single Cell Airfoil

For a single cell airfoil shown in Figure 4 the shear flow is constant.

 * $$\displaystyle q = q_{1} = q_{2} = q_{3}$$

Multi-cell Airfoil
In lecture 19, the professor introduced a specific problem in which he stated he would further generalize after the problem was solved. The image presented in Figure 5 shows a multi-cell airfoil which shows the shear flow through the airfoil as well as the various dimensions of the cells. The values given by the problem statement were as follows:

 * $$a \;= \;30$$ cm
 * $$b \;= \;60$$ cm
 * $$c \;= \;40$$ cm
 * $$t_1 \;= \;0.3$$ cm
 * $$t_2\; = \;0.5$$ cm
 * $$t_{12}\; = \;0.4$$ cm

The overall goal of the problem is to find $$\theta$$ as a function of $$T$$ and to deduce $$J$$ from that.

 * From lecture 17, it was shown that

 * $$T = 2\ \sum_{i=1}^{N_c} q_i\ \bar A$$

which resulted in what will further be defined as equation (1)

 * $$T = T_1 + T_2 = 2\ (q_1\ \bar A_1 + q_2\ \bar A_2)$$

where

 * $$\bar A_1 = a\ c$$
 * $$\bar A_2 = b\ c$$

It can be stated that the two cells have the same rate of twist angles, $$\theta\;$$. That being said, equation (2) can be shown that

 * $$\theta_1 = \theta_2 = ... = \theta_n\;$$

From lecture 17, page 2 of the notes, it can be shown that

 * $$\theta_i = \frac{1}{2\ G\ \bar A_i}\ \oint \frac{q_i}{t_i}\ ds$$

When substituting the values for the two cells, the following is obtained

 * $$\theta_1 = \frac{1}{2\ G\ \bar A_1}\ (\frac{2\ q_1\ a}{t_1} + \frac{q_1\ c}{t_1} + \frac{(q_1 - q_2)\ c}{t_{12}})$$
 * $$\theta_2 = \frac{1}{2\ G\ \bar A_2}\ (\frac{2\ q_2\ a}{t_2} + \frac{q_2\ c}{t_2} + \frac{(q_2 - q_1)\ c}{t_{12}})$$

Which will further be referred to as equation (3) and equation (4), respectively.

Now using equation (3) and equation (4), and using the relation

 * $$T = G\ J\ \theta$$

From this, $$J$$ can be deduced, resulting in

 * $$J = \frac{T}{G\ \theta}$$

Ad-hoc (Engineering) Derivation of Rate of Twist Formula
To derive the twist angle formula
 * $$\theta = \frac{1}{2G\bar{A}}\oint_{C}^{}{\frac{q}{t}ds}$$

we will consider a uniform bar with non-circular cross section subjected to twisting, as shown in Figure 6

To begin, the displacement $$PP'$$ is found to be

 * $$PP'=OPtan \mathbf{\alpha}$$

Then, project $$PP'$$ into the direction perpendicular to $$OP'$$ :
 * $$PP''=PP'cos \mathbf{\alpha}$$

Now the displacement of $$P$$ in the direction tangent to the lateral surface of the beam, $$PP$$, can be expressed in terms of $$OP$$ :
 * $$PP=(OPtan \mathbf{\alpha})cos \mathbf{\alpha} = OPcos \mathbf{\alpha}tan \mathbf{\alpha} = OPtan \mathbf{\alpha}$$

As in Figure 6, $$OP''$$ is defined as the perpendicular distance to the surface, $$\mathbf{\rho}$$. If we assume that the twist angle, $$\mathbf{\alpha}$$, is a small angle, $$PP''$$ simplifies to:
 * $$PP''=\mathbf{\rho}\mathbf{\alpha}$$

The shear strain, $$\mathbf{\gamma}$$, can be defined as the ratio of this displacement distance, $$PP''$$, to a small displacement in the axial (x) direction:
 * $$\gamma = \frac{PP''}{dx}$$

We can then substitute for $$PP''$$, and since we assumed $$\mathbf{\alpha}$$ is very small it can be replaced by it's differential element $$d\mathbf{\alpha}$$. We can also define the rate of twist, $$\mathbf{\theta}$$, as the differential twist angle $$d\mathbf{\alpha}$$ per differential change in the axial direction $$dx$$ :
 * $$\gamma = \frac{\rho d\alpha}{dx} = \rho \theta$$

We can use Hooke's Law to relate the shear stress to the shear strain:
 * $$\mathbf{\tau =G \gamma =G \rho \theta}$$
 * $$\mathbf{\tau(s) =G \rho (s) \theta (x)}$$

If we take the contour integral along C, the equation becomes:
 * $$\oint_{C}^{}{\tau(s)ds} = G \theta(x)\oint_{C}^{}{\rho (s)ds}$$

Next we can substitute in values from previous derivations to arrive at the final form for $$\mathbf{\theta}$$:
 * $$\oint_{C}^{}{\frac{q(s)}{t(s)}ds} = G \theta(x)2\bar{A}$$
 * $$\Rightarrow \theta = \frac{1}{2G\bar{A}}\oint_{C}^{}{\frac{q}{t}ds}$$

What makes this derivation "ad hoc"?

 * 1) The shear strain $$\mathbf{\gamma}$$ must be obtained using the displacement of $$P$$ in the direction tangent to $$C$$ at $$P$$, but $$PP''$$ is not necessarily tangent to C (but is close).
 * 2) The equation for the shear stress, $$\mathbf{\tau} =\frac{q}{t}$$, was obtained from an ad hoc assumption that $$\mathbf{\tau}$$ is uniform across the wall thickness.
 * 3) There is an inconsistency in the assumption on the size of $$\mathbf{\alpha}$$. To get line $$PP'$$, we assumed $$\mathbf{\alpha}$$ was small, but to get $$PP''$$ we assumed a finite $$\mathbf{\alpha}$$ to use the relation $$\mathbf{\rho}=OPcos \mathbf{\alpha}$$. We then reintroduced the small angle approximation for $$\mathbf{\alpha}$$ in the rest of the derivation.

Elastic Theory
For proper understanding of the elastic theory it can help to go back to the Kinematic Assumptions within the road map laid out in previous lectures and stated in the HW 3 report. Recall that in these assumptions that $$\displaystyle \theta$$ is considered to remain constant with respect to 'x' for a uniform bar.

A) Kinematic Assumptions

 * (1) $$u_x (y,z) = \displaystyle \theta \displaystyle \psi (y,z)$$

 * (2) $$u_y (x,z) = -\displaystyle \theta x z$$

 * (3) $$u_z (x,y) = +\displaystyle \theta x y$$

These equations vary slightly from the equations used in the textbook by C.T. Sun, this is due to the use of a different coordinated system. To transform the textbook equations to the uniform notation equations used above it is advised to use cyclic permutation, as seen in Figure 7. The basic relationship can be followed by the figure, but the arrows are used to represent the conversion between the x-axis of C.T. Sun becomes the y-axis in uniform notation and likewise for the other axis.

Other important relations can be generated from use of the Kinematic Assumptions and that is the relationship to certain strains and strain rate, giving the following results:

 * $$\displaystyle \epsilon_{xx} = \displaystyle \epsilon_{yy} = \displaystyle \epsilon_{zz} = \displaystyle \gamma_{yz} = 0$$

Formal Derivation
Now looking at the stains in 3-D. There are a total of six strain components for a 3 dimensional object due to symmetry.

 * $$\mathbf{\displaystyle \epsilon} = \begin{bmatrix}

\epsilon_{xx} & \epsilon_{xy} & \epsilon_{xz} \\ \epsilon_{yx} & \epsilon_{yy} & \epsilon_{yz} \\ \epsilon_{zx} & \epsilon_{zy} & \epsilon_{zz} \end{bmatrix}$$

Now that we have a 3x3 matrix with 9 coefficients, we need to change the notation to that of indicial notation.

 * $$ x \Leftrightarrow 1, y \Leftrightarrow 2 , z \Leftrightarrow 3 $$

Now we can write the matrix in indicial notation.

 * $$\mathbf{\displaystyle \epsilon} = \begin{bmatrix}

\epsilon_{11} & \epsilon_{12} & \epsilon_{13} \\ \epsilon_{21} & \epsilon_{22} & \epsilon_{23} \\ \epsilon_{31} & \epsilon_{32} & \epsilon_{33} \end{bmatrix}$$ $$ = \displaystyle \epsilon_{ij}$$

$$\displaystyle \epsilon_{ij}$$ is known at tensorial notation. For this matrix i,j range from 1 to 3, where i is the row number in the matrix and j in the column designation.

Now that the matrix for $$\displaystyle \epsilon_{ij}$$ is formed, the symmetry of the matrix can now be calculated.

 * $$\displaystyle \epsilon_{ij} = \frac{1}{2} (\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i})$$

The coordinates x,y,z can now be rewritten: $$ x \Leftrightarrow x_1, y \Leftrightarrow x_2 , z \Leftrightarrow x_3 $$

 * $$\epsilon_{11} = \displaystyle \epsilon_{xx} = \frac{1}{2} (\frac{\partial u_1}{\partial x_1} + \frac{\partial u_1}{\partial x_1})$$

 * $$\displaystyle \epsilon_{xx} = \frac{\partial u_1}{\partial x_1} \Rightarrow \frac{\partial u_x}{\partial x}$$

Stress Tensor
The stress tensor can be looked at much like the the strain relations as a 3x3 matrix. Also by symmetry there will be only 6 independent components of $$\displaystyle \sigma$$ in 3-D.

 * $$\displaystyle \sigma = [\sigma]_{3x3}$$

Based of the relations between the stress and strain, several stresses and a shear strain can be found by kinematic assumptions.

 * $$\displaystyle \sigma_{xx} = \displaystyle \sigma_{yy} = \displaystyle \sigma_{zz} = \displaystyle \tau_{yz} = 0$$

Hooke's Law for Three-Dimensional Isotropic Elastic Material
For the purposes of ascertaining strain from given stresses and vice versa, a 6x6 transformation matrix must be developed. But first, individual stress-strain relationships must be determined. From Hooke's Law the normal strains are:

 * $$ \epsilon_{xx} = \frac{\sigma_{xx}}{E} - \frac{\nu \sigma_{yy}}{E} - \frac{\nu \sigma_{zz}}{E} $$
 * $$ \epsilon_{yy} = \frac{\sigma_{yy}}{E} - \frac{\nu \sigma_{xx}}{E} - \frac{\nu \sigma_{zz}}{E} $$
 * $$ \epsilon_{zz} = \frac{\sigma_{zz}}{E} - \frac{\nu \sigma_{yy}}{E} - \frac{\nu \sigma_{xx}}{E} $$

while the shear strains are:

 * $$ \gamma_{xy} = 2\epsilon_{xy} = \frac{\tau_{xy}}{G} $$
 * $$ \gamma_{xz} = 2\epsilon_{xz} = \frac{\tau_{xz}}{G} $$
 * $$ \gamma_{yz} = 2\epsilon_{yz} = \frac{\tau_{yz}}{G} $$

The stresses and strains can be arranged in a column matrix as follows:

 * $$ (\epsilon_{ij})_{6x1} = \begin{bmatrix} \epsilon_{11} \\ \epsilon_{22} \\ \epsilon_{33} \\ \epsilon_{23} \\ \epsilon_{13} \\ \epsilon_{12} \end{bmatrix} $$  and   $$ (\sigma_{ij})_{6x1} = \begin{bmatrix} \sigma_{11} \\ \sigma_{22} \\ \sigma_{33} \\ \sigma_{23} \\ \sigma_{13} \\ \sigma_{12} \end{bmatrix} $$

Combining the normal and shear strains into a transformation matrix yields the relations in the following HW problem solution:

MATLAB Airfoil Code Validation
Part of the first version of the MATLAB problem was to verify that the code was working properly by testing it both against a known area and also for replication of results. In the collapsible table below there is a "validation code" which calls part of the HW3 MATLAB code as a subfunction.

The resulting output from running the validation code is provided below.

The first section of the validation involved a "circular" airfoil. The first task was to determine the area of a circle of radius 1 meter at various numbers of contour segments. A table summarizing the above results is as follows:

 * {| border=1

 * bgcolor="red"|Number of Segments
 * bgcolor="#99ccff" style="text-align: center;" |20
 * bgcolor="#99ccff" style="text-align: center;" |40
 * bgcolor="#99ccff" style="text-align: center;" |60
 * bgcolor="#99ccff" style="text-align: center;" |80
 * bgcolor="#99ccff" style="text-align: center;" |100
 * bgcolor="red"|Calculated Area
 * bgcolor="#99ccff" |3.1045
 * bgcolor="#99ccff" |3.1285
 * bgcolor="#99ccff" |3.1344
 * bgcolor="#99ccff" |3.1369
 * bgcolor="#99ccff" |3.1383
 * }
 * }

The second task was to determine the $$ns_{max}\;$$, the number of segments at which the calculated area is within 1% of the actual area, which in this case is equal to 3.1416 square meters. This was done by calculating the area at various values of n and then comparing them. A plot of the results is given in Figure 9, where $$ns_{max}\;$$ and its corresponding area are denoted by a black square. For reference, the calculated value for $$ns_{max}\;$$ was found to be 23 segments, and the calculated area at this point was 3.1115 square meters, which differs from the actual value by .96%.

The final task for this section was to calculate the circle area at various locations of the origin point $$P_{0}\;$$. The three locations used were the coordinate system origin $$[0,0]\;$$, the circle center $$[1,0]\;$$, and an arbitrary external point $$[1,-4]\;$$. A table summarizing the results is shown below. As expected, the area is identical in all three cases.

 * {| border=1

 * bgcolor="red"|Coordinate of Origin Point
 * bgcolor="#99ccff" style="text-align: center;" |[0,0]
 * bgcolor="#99ccff" style="text-align: center;" |[1,0]
 * bgcolor="#99ccff" style="text-align: center;" |[1,-4]
 * bgcolor="red"|Calculated Area (square meters)
 * bgcolor="#99ccff" |3.1115
 * bgcolor="#99ccff" |3.1115
 * bgcolor="#99ccff" |3.1115
 * }
 * }

The second section of the validation involved a NACA 2415 airfoil of chord length c = 1 meter. The first task was to find $$ns_{max}\;$$ as with the circular airfoil. However, instead of comparing against a known area, the calculated area at a given number of segments is compared with the calculated area using one less segment, such that $$ns_{max}\;$$ is located when the difference between two successive area calculations is less than 1% of the current area calculation. A plot of the results is given in Figure 10. For reference, the calculated value for $$ns_{max}\;$$ was found to be 9 segments, and the calculated area at this point was 0.0244 square meters.

The second task was to show that the area calculation is valid at any choice of the origin point $$P_{0}\;$$. The three locations used were the coordinate system origin $$[0,0]\;$$, the airfoil centroid $$[.210,.008]\;$$ (previously calculated in HW3), and an arbitrary external point $$[.25,-1]\;$$. A table summarizing the results is shown below. Unlike with the circular case, the calculated areas are not identical for each case. However, a quick relative error calculation shows that the calculated values are all within 1% of each other, and such a small difference can be easily explained by the small numerical errors inherent in this kind of analysis.

 * {| border=1

 * bgcolor="red"|Coordinate of Origin Point
 * bgcolor="#99ccff" style="text-align: center;" |[0,0]
 * bgcolor="#99ccff" style="text-align: center;" |[.210,.008]
 * bgcolor="#99ccff" style="text-align: center;" |[.25,-1]
 * bgcolor="red"|Calculated Area (square meters)
 * bgcolor="#99ccff" style="text-align: center;" |.0244
 * bgcolor="#99ccff" style="text-align: center;" |.0246
 * bgcolor="#99ccff" style="text-align: center;" |.0247
 * }
 * }

Description of Airfoil Geometry and Equations

Figure 11 shows the geometry of a generic NACA 4-digit airfoil. The various quantities are defined as follows:

 * $$LE\;$$ - The point corresponding to the leading edge of the airfoil.
 * $$TE\;$$ - The point corresponding to the trailing edge of the airfoil.
 * $$c\;$$ - The chord length, describing the length of the straight line connecting the leading edge to the trailing edge.
 * $$m\;$$ - The maximum camber, corresponding to the first digit of the 4-digit series, and given in percentage of chord.
 * $$p\;$$ - The location of maximum camber, corresponding to the second digit of the 4-digit series, and given in tenths of chord.
 * $$t_{a}\;$$ - The maximum thickness, corresponding to the third and fourth digits of the 4-digit series, and given in percentage of chord.
 * $$y_{u}\;$$ - The contour defining the airfoil's upper surface.
 * $$y_{l}\;$$ - The contour defining the airfoil's lower surface.
 * $$MCL\;$$ - The contour defining the airfoil's mean camber line. Referred to as $$z_{c}\;$$ in the airfoil equations.

A set of general equations used to generate the various contours for an arbitrary NACA 4-digit airfoil can be found here.

Torsional Constant Comparison Between Single- and Multi-celled Airfoils
The new facet to the MATLAB assignment deals with a comparison of single and multi-celled airfoils, in terms of the torsional constant $$J\;$$. The code for the single-celled airfoil is in the table below:

The first portion of the assignment was to determine $$ns_{max}\;$$, as in previous assignments. The next collapsed table contains a "header" that can be applied to the above single celled airfoil analysis code to perform an iterative operation to generate an $$ns_{max}\;$$.The result was a $$ns_{max}\;$$ value of 12 segments and a $$J\;$$ value of $$4.7543x10^{-6} m^{4}\;$$. The second portion involved generating a graph of torsional constant vs. number of segments. The result is shown in Figure 12.

The second portion of the assignment was to determine $$J\;$$ for a NACA2415 airfoil with 3 cells. The MATLAB code used to carry out this task is given in a table below.

The plotted airfoil, with partitions at quarter and three-quarter chord, is shown in Figure 13. The calculated $$J\;$$ for this case, using $$ns = 1000\;$$ is $$5.2500x10^{-6} m^{4}\;$$.

Comparison:

The two calculated values for $$J\;$$ are very close to each other. The difference is likely a result of numerical errors during the generation of the torsional constant in the above MATLAB codes. Accounting for this effect, the overall result echoes the conclusion given in Chapter 3 of Sun: the torsional rigidity of a thin-walled, closed cross-section cannot be appreciably improved by compartmentalization.

MATLAB Code Certification
The following students are proficient in MATLAB programming:

Kevin Klauk Eas4200c.f08.vqcrew.E 22:01, 22 October 2008 (UTC)

Adam Edstrand Eas4200c.f08.vqcrew.a 05:09, 24 October 2008 (UTC)

Steven Hepsworth Eas4200c.f08.vqcrew.f 23:59, 23 October 2008 (UTC)

Darin Toscano Eas4200c.f08.vqcrew.d 07:28, 24 October 2008 (UTC)

Keith Javier Stober Eas4200c.f08.vqcrew.b 16:23, 24 October 2008 (UTC)

John Saxon Eas4200c.f08.vqcrew.c 20:41, 24 October 2008 (UTC)

Contributing Team Members
The following students contributed to this report:

Kevin Klauk Eas4200c.f08.vqcrew.E 22:01, 22 October 2008 (UTC)

Adam Edstrand Eas4200c.f08.vqcrew.a 05:09, 24 October 2008 (UTC)

Steven Hepsworth Eas4200c.f08.vqcrew.f 23:59, 23 October 2008 (UTC)

Darin Toscano Eas4200c.f08.vqcrew.d 07:29, 24 October 2008 (UTC)

Keith Javier Stober Eas4200c.f08.vqcrew.b 16:24, 24 October 2008 (UTC)

John Saxon Eas4200c.f08.vqcrew.c 20:42, 24 October 2008 (UTC)

title: User:Eas4200c.f08.vqcrew.c/Homework 4


text:
User:Mu301/Earth-impact

See Earth-impact events

Comments
C--- C C              The flux of near-Earth objects colliding with Earth C C--- C      Abstract: C C      1. Given the kinetic energy of an impacting object, C              calculate how often such an object collides with Earth. C              If 1 or more per year, print total number per year. C              If less than 1 per year, print interval in years. C C      2. Also calculate the range of uncertainty. C C      3. Estimate diameter based on given energy and rough C              assumptions of mean impact velocity and bulk density. C--- C      Equations: C C      N = number of objects colliding per year C C      log (N) = a0 - b0 * log (E) C C              E = kinetic energy in kilotons (TNT) C              a0 = 0.5677  +/- 0.015 C              b0 = 0.90    +/- 0.03 C C              1 kt == 1 kiloton (TNT equivalent) = 4.185E12 Joules C C C      log (N) = c0 - d0 * log (D) C C              D = diameter in meters C              c0 = 1.568  +/- 0.03 C              d0 = 2.70   +/- 0.08 C C              Calculations based on diameter assume: C C              1. bulk density of 3,000 kg / m^3 (2,600 to 3,400) C              2. mean impact velocity of 20.3 km / s (RMS) C C      Both equations are from the paper in Nature, cited below. C C      The second form of the equation (based on diameter) is far less C      accurate due to the added assumptions for density and velocity. C--- C      General Notes: C C      As can be seen from the computed uncertainty range the C      equation is most accurate for input from 100 ton to 25 kt. C      This is the range for which satellites can observe impacts. C C      The equation can be extrapolated to higher and lower energies, C      but the uncertainty range increases. The program allows input C      of E from 0.01 ton to 10 Gt (for which there is a decent fit to C       published estimates which are based on other evidence.) C C      However, the estimated flux for objects with E > 1 Gt does C      deviate significantly from the equation. The most probable C      impact interval falls outside the uncertainty range somewhere C      between 1 - 10 Gt. For instance, table 5 in the Icarus C      paper (cited below) includes the following data: C C              Energy (Mt)       interval (years)      Diameter (m) C                 1,000         63,000  +/-   8,000     277  +/-  7 C                10,000        241,000  +/-  30,000     597  +/- 15 C               100,000        925,000  +/- 121,000    1287  +/- 33 C C      Calculating the range of uncertainty makes use of MIN1 & MAX1 C      due to the critical region near N = 1. C C      Output should be reformatted to give fewer significant digits... C--- C      References: C C      "The flux of small near-Earth objects colliding with the Earth" C      Brown, P.; Spalding, R. E.; ReVelle, D. O.; Tagliaferri, E.; C      Worden, S. P.  Nature, Volume 420, Issue 6913, pp. 294-296. C C      "From Magnitudes to Diameters: The Albedo Distribution of Near C       Earth Objects and the Earth Collision Hazard" C      Morbidelli, A.; Jedicke, R.; Bottke, W. F.; Michel, P.; C      Tedesco, E. F.  Icarus, Volume 158, Issue 2, p. 329-342. C--- C      Related Info: C C      Asteroids with diameter smaller than 50 - 100 meters C      usually detonate in the atmosphere. C C      On June 6, 2002 a bolide detonated above the Mediterranean C      Sea with an estimated energy of 25.77 (+/- 2.16) kt. C C      A Feb. 1, 1994 bolide was perhaps several tens of kilotons. C C      The 1908 Tunguska impact in Siberia is estimated at ~10 Mt. C C       The Chicxulub crater impact is estimated at ~1E8 Mt. C       Such impacts might occur at intervals of tens to hundreds C      of million years. C---

title: User:Mu301/Earth-impact


text: File:RotatedDOF 10-08-08.JPG

title: File:RotatedDOF 10-08-08.JPG


text: File:AxialDOF 10-08-08.JPG

title: File:AxialDOF 10-08-08.JPG


text:
User:EML4500.f08.delta 6.ramirez/10808

Note: 18-2 continued.

Goal: Find $$ \tilde{T}_{4x4}^{(e)} $$ that transforms the set of local element d.o.f.'s $$ d^{(e)} $$ to another set of local element d.o.f.'s $$ \tilde{d}_{4x1}^{(e)} $$ such that $$ \tilde{T}^{(e)} $$ is invertible.

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

$$ (1) \hat{d}_{1}^{(e)} = \begin{bmatrix} l^{(e)}\ m^{(e)}\

\end{bmatrix}

\begin{Bmatrix}d_1^{(e)} \\ d_{2}^{(e)} \end{Bmatrix} $$

Where,

$$ \hat{d}_{1}^{(e)} = q_1^{(e)} $$

$$\hat{d}_{2}^{(e)} = \vec{d}_1^{(e)} \cdot \tilde{j}$$

Note: $$\hat{d}_{2}^{(e)}$$ is the component of $$\vec{d}_1^{(e)}$$ along $$\tilde{j}$$, i.e., $$\tilde{y}$$ axis.

Thus,

$$\tilde{d}_2^{(e)} = -\sin \theta^{(e)}d_1^{(e)} + \cos \theta^{(e)} d_2^{(e)}$$

Equation (2) is as follows:

$$(2) \hat{d}_{2}^{(e)} = \begin{bmatrix} -m^{(e)}\ l^{(e)}\

\end{bmatrix}

\begin{Bmatrix}d_1^{(e)} \\ d_{2}^{(e)} \end{Bmatrix}$$

Next, put (1) & (2) into matrix form,

$$\begin{Bmatrix} \tilde{d}_1^{(e)} \\ \tilde{d}_2^{(e)} \end{Bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & -l^{(e)} \end{bmatrix} \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \end{Bmatrix}$$

Where,

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

Thus,

$$\begin{Bmatrix} \tilde{d}_1^{(e)}\\ \tilde{d}_2^{(e)}\\ \tilde{d}_3^{(e)}\\ \tilde{d}_4^{(e)} \end{Bmatrix} = \begin{bmatrix} R_{2x2}^{(e)} & 0_{2x2}\\ 0_{2x2} & R_{2x2}^{(e)} \end{bmatrix} \begin{Bmatrix} d_1^{(e)}\\ d_2^{(e)}\\ d_3^{(e)}\\ d_4^{(e)} \end{Bmatrix}$$

which represents,

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

Next, rotate the element achieving the following

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

Note: all zero values are zero because transverse force does not stretch the spring.

Next, we can conclude:

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

Note: this is the stopping point of transforming the FD relation back to original (x,y) coordinate system.

title: User:EML4500.f08.delta 6.ramirez/10808


text:
User:Mstewart

–Hi. My name is Michelle Stewart. I am a student at the University of North Carolina Charlotte. My major is elementary education. I am married and have one son. We have 3 dogs, 1 cat and 12 chickens. We live in the country and I absolutely love it. I am currently a preschool teacher in a state funded program for at risk children. I have been teaching preschool for 6 years. I have wanted to teach ever since I can remember.

KWL CHART (template)
Please click on "edit" (here above), copy the code, paste it on your personal user site and insert Your text in the columns.

title: User:Mstewart


text:
User:Lousnews

Greetings!

I am excited to join in on the conversations and idea exchanges within the realm of Wikiversity.

title: User:Lousnews


text:
User:Eml4500.f08.wiki1.ambrosio/hw3

HW3
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.

title: User:Eml4500.f08.wiki1.ambrosio/hw3


text:
User talk:Mstewart

 Hello Mstewart, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --HappyCamper 19:48, 18 October 2008 (UTC)

title: User talk:Mstewart


text: File:Rotatedelement.JPG

title: File:Rotatedelement.JPG


text:
User:Eas4200c.f08.wiki.f/10-6-08

Quadrature Method
Quadratures can be used to find the average area of foils. Below is an image of the quadrature of a foil.

$$\overline{A} = \overline{A}_1 + \overline{A}_2 = \overline{A}_1 - \begin{vmatrix}\overline{A}_2\end{vmatrix}$$
 * $$ \overline{A}_1 > 0 $$
 * $$ \overline{A}_2 > 0 $$

 * $$\overrightarrow{r}x\overrightarrow{PQ}\; goes\; in\; negative\; x\; direction\; $$

Using Trapezoidals to Integrate
There are problems with using this method because of the curvature of the foil. When you split up the area of the foil, not every section will end up looking like a trapezoid. This method should not be used for foils.

Shear Flow in a Single Cell Airfoil
Below is an image of a single cell airfoil.

The shear flow is constant, therefore:

 * $$ q=q_1=q_2=q_3$$

Rate of Twist Angle θ

$$\theta = \frac{1}{2G\overline{A}}\; q\sum_{j=1}^{}{\frac{l_j}{t_j}}$$

$$\theta = \frac{1}{2G\overline{A}}\; q[\frac{\pi (\frac{b}{2})}{t_1} + \frac{a}{t_2} + \frac{\sqrt{a^2 + b^2}}{t_2}]$$

$$\theta = \frac{1}{2G\overline{A}}\; q[\frac{\pi bt_2t_3 + 2at_1t_3 + 2t_1t_2\sqrt{a^2 + b^2}}{t_1t_2t_3}]$$

Shear Stress
Max Shear Stress τmax

$$\tau_{max} = \frac{q}{min(t_1, t_2, t_3)}$$

 * $$if\; \tau_{max} = \tau_{all}$$

 * $$and\; since\; q = \frac{T}{2\overline{A}} \; then$$

$$T_{all} = 2\bar{A}\tau_{all}\;min(t_1, t_2, t_3)$$

title: User:Eas4200c.f08.wiki.f/10-6-08


title: File:Dia1.jpg


text:
User:Davidalanlyon

I am, as the page states, David Alan Lyon, Technical Writer, and participant of the Wikiversity Technical Writing course.

title: User:Davidalanlyon


text:
User talk:Davidalanlyon

Currently working as a Technical Writer, I would be happy to answer any students queries

title: User talk:Davidalanlyon


text:
User:Eml4500.f08.group/Homework4/Matlab Contributors

I, the undersigned, certify that I can read, understand, and write matlab codes, and can thus contribute effectively to my team.

 * Jeff Wright Eml4500.f08.group.w 21:53, 18 October 2008 (UTC)
 * Aaron Clausen Eas4200c.f08.carbon.clausen 19:12, 20 October 2008 (UTC)


text:
User:ErinShipton

Hi, my name is Erin Shipton. I have no clue if this is the right page...

I'm a transfer from UPS in Tacoma (no, not "What can Brown do for you?") and I'm in the DFVP program. I'm particularily interested in screenwriting and cinematography.

title: User:ErinShipton


text:
User:Eas4200c.f08.blue.a/Lecture 20

Problem Statement
Given the following:

1st wall at 1/4*c, 2nd wall at 3/4*c from leading edge.

You have 3 unknowns q1, q2, and q3

You can now use the following sets of equations and equalities:
 * 1) T=2Σ(qi*Ai)
 * 2) θ1=θ2
 * 3) θ2=θ3

Q: Find J, the Torsional Constant.

Note: You need t, the thickness, but G cancels out in the calculation.

Theory

T=GJθ

T=2qA

Engineering (ad hoc) Derivation

θ= 1/(2GA) Σ (q/t) * ds

This is the formula we shall use in order to find the amount of twist in a uniform bar with non-circular cross section subject to twist (torsion).

Displacement PP' due to α

PP'/OP=tanα =~ α (for small angles)

Project displacement PP' on the direction perpendicular to OP'

PP"=PP'cosα

PP"=OPtanα*cosα

=OP"tanα

Recall:

 * 1) OP=r
 * 2) OP"=ρ

PP"=rcosα*tanα

=ρα

which is the displacement of P in the direction tangent to the lateral surface of the bar.

Computing the strain:

STRAIN:    γ = PP"/dx = ρα/dx = ρθ

with θ=α/dx, which is also known as the rate of twist.

Note: This calculation is based on the assumption that alpha is a very small angle (also denoted as dα).

title: User:Eas4200c.f08.blue.a/Lecture 20


text:
User:EML4500.f08.Ateam.Miller/homework4

Justification of Assembly of Elemental Stiffness Matrices into the Global Stiffness Matrix
The free body diagram at Node 2 yields the following force balance equations.

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

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

Next, the elemental Force-Displacement relation,

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

can be used to rewrite Equation (1) as follows.

$$(1)\rightarrow [k_{31}^{(1)}d_1^{(1)}+k_{32}^{(1)}d_2^{(1)}+k_{33}^{(1)}d_3^{(1)}+k_{34}^{(1)}d_4^{(1)}]+ [k_{11}^{(2)}d_1^{(2)}+k_{12}^{(2)}d_2^{(2)}+k_{13}^{(1)}d_3^{(2)}+k_{14}^{(1)}d_4^{(2)}]$$

Similarly, Equation (2) can be written as follows.

$$(2)\rightarrow [k_{41}^{(1)}d_1^{(1)}+k_{42}^{(1)}d_2^{(1)}+k_{43}^{(1)}d_3^{(1)}+k_{44}^{(1)}d_4^{(1)}]+ [k_{21}^{(2)}d_1^{(2)}+k_{22}^{(2)}d_2^{(2)}+k_{23}^{(1)}d_3^{(2)}+k_{24}^{(1)}d_4^{(2)}]$$

Now, Equations (1) and (2) can be transformed into the global coordinate system as follows.

$$(1)\rightarrow [k_{31}^{(1)}d_1+k_{32}^{(1)}d_2+k_{33}^{(1)}d_3+k_{34}^{(1)}d_4]+ [k_{11}^{(2)}d_3+k_{12}^{(2)}d_4+k_{13}^{(1)}d_5+k_{14}^{(1)}d_6]$$

$$(2)\rightarrow [k_{41}^{(1)}d_1+k_{42}^{(1)}d_2+k_{43}^{(1)}d_3+k_{44}^{(1)}d_4]+ [k_{21}^{(2)}d_3+k_{22}^{(2)}d_4+k_{23}^{(1)}d_5+k_{24}^{(1)}d_6]$$

Now, the elemental stiffness matrices, k(e), can be assembled into the global stiffness matrix, K, through

$$\mathbf{K}=\mathbf{A}\mathbf{k}$$

where K is an n x n matrix, A is nel x 1 matrix and k is an ned x ned matrix. Here, n is the total number of degrees of freedom, nel is the total number of elements, ned is the total number of elemental degrees of freedom, and A is the assembly operator.

As we continue to justify the assembly of the elemental stiffness matrices into the global stiffness matrix, we must use the Principal of Virtual Work, as described previously, to obtain $$\mathbf{\overline{K}}$$ by eliminating the corresponding boundary conditions.

Once the Principal of Vitual Work is applied, the resulting elemental axial force equation and the elemental elemental stiffness equation can be written as follows.

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

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

Now, it is important to derive the Finite Element Method for Partial Differential Equations. Consider the Force-Displacement relation for a bar or spring with

$$kd=F$$

This implies that

$$kd-F=0\rightarrow (3)$$

or equivalently

$$W(kd-F)=0\rightarrow (4)$$

for all W.

Upon investigation, $$(3)\Rightarrow(4)$$ is trivial, but $$(4)\Rightarrow(3)$$ is not trivial.

Since (4) is valid for all W, select W=1. Now, (4) becomes $$1(kd-F)=0\Rightarrow(3)$$.

title: User:EML4500.f08.Ateam.Miller/homework4


text:
Unmanned aerial systems/Readings

Wikibooks

 * Wikibooks:Robotics Design Software

title: Unmanned aerial systems/Readings


text:
Unmanned aerial systems/Specifications

Specifications
🇨🇴 Will include:
 * An on board computer with:
 * Processor
 * Memory
 * Storage
 * Wifi
 * GPS
 * Brushless motor
 * Twin rotors
 * Batteries
 * Requisite wiring
 * Accelerometers

Might include:
 * Cameras
 * With the ability to produce macro images ( w )
 * Microphones
 * Speakers
 * Storage
 * Para-shoot
 * Lights
 * Fuel cells
 * Manipulators
 * Other tools
 * Thin solar cells

Other specifications:
 * Lightweight
 * Aesthetically pleasing
 * Functional
 * Useful
 * Easy to use

title: Unmanned aerial systems/Specifications


text:
User:AdrienneEarl

Hello, my name is Adrienne Earl and I would like to welcome you to my page.

My intention: To create "Chickiversity" on Wikiversity.

Get rad!

title: User:AdrienneEarl


text: User:Maxval

title: User:Maxval


text:
Template talk:Agree

I'm changing to a simple "I agree", which is more intuitive, and leaving the "because" up to the poster. I hope I'm not stepping on anyone's toes with this edit. The Jade Knight (d'viser) 07:13, 19 October 2008 (UTC)

 * The intent in including "because" is to separate it from the support and oppose templates by requiring that a reason be given. Like a fill in the blank type of template "I agree because _______" --dark[[Image:Yin yang.svg|12px]]lama 22:13, 22 October 2008 (UTC)
 * But what if you simply want to announce that you agree with someone else's arguments? The Jade Knight (d'viser) 22:14, 22 October 2008 (UTC)
 * The intention is to also avoid "me too" type of responses as well, and get people to explain exactly what it is a person is agreeing with and why. The point is to encourage better quality discussions. --dark[[Image:Yin yang.svg|12px]]lama 22:18, 22 October 2008 (UTC)
 * But some users specifically don't comment at all on discussions because they simply don't feel they have the time to write out a response. My goal is to encourage more voices to be heard; it is better to have 2 people with lengthy replies and 5 other people saying "me too", than to just have 2 people with lengthy replies.  The Jade Knight (d'viser) 22:20, 22 October 2008 (UTC)
 * Maybe we could have two different templates? -- Jtneill - Talk - c 22:23, 22 October 2008 (UTC)
 * I've thought about that. The Jade Knight (d'viser) 22:24, 22 October 2008 (UTC)
 * I don't consider it helpful to have "me too" responses. I think they can degrade or hinder discussions and compromise since they do not provide any information that can be responded to, could be used to bring about a compromise, aid in finding a solution, bring closure to issues brought up or could be used to determine what consensus is. I think in assents "me too" responses are like yes/no votes or polls which polarize the community since no opinions, thoughts, reasons, etc. are voiced. My creation of this template was an attempt to make it basically impossible to not comment when used. I didn't want this template to be just another simple vote/polling template, but rather an alternative to that approach. --dark[[Image:Yin yang.svg|12px]]lama 22:45, 22 October 2008 (UTC)
 * I think polling is very useful in gauging where consensus lies. Yes, fully responses are better, but I am always glad when people come and chime in "me too"; that means one more perspective, essentially.  The Jade Knight (d'viser) 22:53, 22 October 2008 (UTC)
 * Polls are evil. They don't help people who disagree understand why people who agree agree and people who agree understand why people who disagree disagree when no reasons are given. I made three templates to try to aid in getting people to provide reasons and avoid this problem: this one, and . Removing "because" removes an essential element necessary in carrying out that goal. --dark[[Image:Yin yang.svg|12px]]lama 23:02, 22 October 2008 (UTC)
 * I strongly disagree. I think that people should be encouraged to chime in regardless of whether or not they initially explain their position.  I find polls helpful, not evil.  The Jade Knight (d'viser) 23:11, 22 October 2008 (UTC)
 * I think encouraging people to participate in discussions is way more important than getting people to chiming in with "me too", especially in an environment where we're trying to learn from one another. Getting people to chime in with "me too" votes could be interpreted as an attempt to have people act on your behalf as Meatpuppets — in the sense that they are new to the discussion and may not voted otherwise, or as attempting to canvas people to manipulate the results. --dark[[Image:Yin yang.svg|12px]]lama 00:21, 23 October 2008 (UTC)

I've just parameterized this, with the default using "because" - this allows us to encourage giving reasons, while allowing those who want the text to say something else (such as "I agree" or "Super-duper awesomeness" or anything else) can simply use a paramter. I think this is a compromise all parties should find suitable. – Mike.lifeguard | @en.wb 00:29, 23 October 2008 (UTC)

 * ✅ --mikeu talk 02:26, 23 October 2008 (UTC)
 * ✅. The Jade Knight (d'viser) 03:08, 23 October 2008 (UTC)
 * I'm glad you think so. I'm of the opinion that most templates like this should be as flexible as possible. That's the whole point of parameters, after all! – Mike.lifeguard | @en.wb 04:10, 23 October 2008 (UTC)

title: Template talk:Agree


text:
Talk:Introduction to US History/United We Stand:The History of the United States of America/America Before Columbus

BTW, I'd recommend citing as much as possible; if you have specific sources you're taking things from, cite them. That way the text will be both effective and authoritative. The Jade Knight (d'viser) 07:19, 19 October 2008 (UTC)
 * (✅), I will go back and cite many of the facts that I build my narrative from after I finish the first chapter. From an acedemic point of view. some people may not be pleased with heavy citing of wikipedia.  However, wikipedia has been a great resource for me.  This US introductery textbook is by nature simple and general, thus wikipedia allows me a chance to understand underlying ideas without having to gather them from all over the internet.
 * Acedemically, people would surely prefer big researches at a university library (with definitive books, acedemic studies ,etc). I do not have access to that. This is where you have to accept that a chemist ,with no expertise in history and limited resources (including time), is not going to equal the panel of history experts that normally would author a definitive textbook.--Jolie 16:19, 20 October 2008 (UTC)
 * Citing Wikipedia, while certainly not ideal, is better than citing nothing at all. BTW, if you cite Wikipedia, it is important that you either link to EXACTLY which version of the page you are citing, or that you give the EXACT time you referenced the Wikipedia article.  The Jade Knight (d'viser) 21:36, 20 October 2008 (UTC)
 * The cites can be fixed at another time. I am sure we can find sources on the Wikipedia article. Geo.plrd 06:21, 21 October 2008 (UTC)


text:
Talk:Introduction to US History/Foreign Exploration

New question: Are these intended to be "wiki"-lectures, editable by all, or are these intended to specifically be your notes exclusively? The Jade Knight (d'viser) 07:26, 19 October 2008 (UTC)
 * They are editable by all, Jolie rewrote my last set. Right now they are coming out as notes. Geo.plrd 06:29, 21 October 2008 (UTC)


text:
User:Eml4500.f08.jamama.adam/Homework4

Meeting 19
The objective is to transform the set of local degrees of freedom $$\tilde{d}^{(e)}$$ to a different set degrees of freedom $$\tilde{d}^{(e)}_{4X1}$$ using an invertible matrix $$\tilde{T}^{(e)}{_{4X4}}$$.

$$\tilde{d}^{(e)}_{4X1} = \tilde{T}^{(e)}_{4X4}d^{(e)}_{4X1}$$

Lets define equation (1) bellow:

$$\tilde{d}^{(e)}_{1} = [l^{(e)}m^{(e)}]\begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ \end{Bmatrix}$$

$$\tilde{d}_{2}^{(e)} = \overrightarrow{d}_{1}^{(e)}\cdot \overrightarrow{j}$$ (comp. of $$ \overrightarrow{d}_{1}^{(e)}$$ along $$\vec\tilde{j}$$ i.e. y-axis)
 * $$= sin(\theta^{(e)}) + cos(\theta^{(e)}) d_{2}^{(e)}$$

And lets define equation (2) bellow:

$$\tilde{d}_{2}^{(e)} = [-m^{(e)}l^{(e)}]\begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ \end{Bmatrix}$$

By putting (1) and (2) in matrix form:

$$\begin{Bmatrix} \tilde{d}_{1}^{(e)}\\ \tilde{d}_{2}^{(e)}\\ \end{Bmatrix} = \underbrace{\begin{bmatrix} l^{(e)}&m^{(e)}\\ -m^{(e)}&l^{(e)}\\ \end{bmatrix}}_{R^{(e)}}\begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ \end{Bmatrix}$$

$$\underbrace{\begin{Bmatrix} \tilde{d}_{1}^{(e)}\\ \tilde{d}_{2}^{(e)}\\ \tilde{d}_{3}^{(e)}\\ \tilde{d}_{4}^{(e)}\\ \end{Bmatrix}}_{\tilde{d}_{4X1}^{(e)}} = \underbrace{\begin{bmatrix} R_{2X2}^{(e)}&0_{2X2}\\ 0_{2X2}&R_{2X2}^{(e)}\\ \end{bmatrix}}_{\tilde{T}_{4X4}^{(e)}}\underbrace{\begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ d_{3}^{(e)}\\ d_{4}^{(e)}\\ \end{Bmatrix}}_{d_{4X1}^{(e)}}$$

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

$$\tilde{f}^{(e)}_{4X1} = \tilde{k}^{(e)}_{4X4}\tilde{d}^{(e)}_{4X1}$$

title: User:Eml4500.f08.jamama.adam/Homework4


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Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

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Computer Architecture Lab/WS2008/Group2

Project page of Group 2.

 * Instruction Set Comparison.

title: Computer Architecture Lab/WS2008/Group2


text:
Computer Architecture Lab/WS2008/Group2/ISA comparison

The three chosen processors are Zilog Z80, Motorola 68000, ARM7DI. You first have a quick overlook of each processor in separate and a comparison at the end.

ARM7DI
The ARM is an enhanced Reduced Instruction Set Computer (RISC) and consists of such typical architecture features like a large uniform register file, a load/store architecture where data operations are performed on register contents only and not directly on memory contents. Moreover it consists of simple addressing modes where all load store addresses are determined from register contents and instruction fields only. To simplify the instruction decoding process, all ARM instructions have a fixed length. In addition to these basic RISC features, ARM CPUs also provide control over the ALU as well as over the shifter in most of the data processing instructions. Moreover auto-increment and auto-decrement addressing modes provide an optimization to program loops. Another enhancement in contrast to the basic RISC architecture is the ability to load and store multiple instructions to increase the data throughput, as well as conditional execution of many instructions to get a better execution throughput.

These general features of the ARM CPU series are shown more concrete on the ARM7D processor in the following section.

Register Layout
The processor consist of 37 registers, where 31 are 32 bit wide general purpose registers and the remaining 6 are status registers. 16 general purpose registers and 2 status registers are available to the programmer.

Instruction Set
Concerning the instructions it must be said that all ARM7DI instructions are conditionally executed depending on the values of the N, Z, C and V flags in the CPSR (Computed Program Status Register). The following list shows the meaning of these flags:

 * zero (Z)
 * carry / borrow / extend (C)
 * overflow (V)
 * negative / less than (N)

Nevertheless there is a condition which overules the above written. If the always (AL) condition is specified, the instruction will be executed irrespective of the set flags.

Two distinctive marks of the ARM7DI instruction set are that, it can also handle undefined instructions which are not available at this specific type of ARM cpu, but might be at another. This works in a way that the cpu offers the instruction to any available coprocessor and if the coprocessor is available but busy, the cpu waits until the coprocessor is able to deal with the offer again. If on the other hand there is no such coprocessor available which can deal with the instruction the cpu rises a trap which can be captured by the programmer who might want to emulate the missing instruction. The other distincitive mark is a branch feature called branch-link which does the following: If the link bit is set the PC(Programm Counter) is written to the register R14, where it is adjusted in way that it contains the correct return address of the function call.

The table shows a summary of the most relevant data processing instructions of the ARM7DI. The instructions TST,TEQ,CMP,CMN are compare operations which only set status bits in the CPSR. The logical operations like AND, EOR, TST, TEQ, ORR, MOV, BIC and MVN are performed on every bit to produce the result.

Motorola 68000
The 68000 is a 32bit CISC microprocessor. One important aspect of the design of the 68000 is its forward compatibility. Introduced in 1979 the 68000 is still in use, e.g. in the TI-89 and TI-92 series and Voyage 200 calculators of Texas Instruments.

Registers and Address Bus
The address bus is implemented by 24 bit which allows to address up to 16 MB of physical memory. But the address storage and computation use 32 bits ignoring the high order byte. This means that the Motorola 68000 is a 32-bit microprocessor. The reason for this design is that Motorola wanted to force forward compatibility. The endianness of the 68000 family is big-endian. The CPU consists of 8 general purpose data registers and 8 address registers. One advantage of this amount of registers is that it is small enough to guarantee a small interrupt response time on the one hand, and it is also large enough to do calculations relatively fast on the other hand. Here the list of available registers:

 * 8 data registers (32 bit)
 * 8 address registers (32 bit)
 * 1 status register (16 bit)

The last of the address registers is also used as stack pointer.

Instruction Set
The Motorola 68000 is a complex instruction set computer (CISC). This means that one instruction can execute several low-level operations. The instruction set is called to be orthogonal. It means that instructions are divided into operation and address modes. Further nearly all address modes are available for almost all instructions. The instruction set consists of 56 instructions. The comparison, arithmetic and logic operations set the following bit flags in the status register:

 * zero (Z)
 * carry (C)
 * overflow (V)
 * extend (X)
 * negative (N)

The extend-flag is set to the value of the carry-flag, but only for arithmetic operations, otherwise it is not affected. These instructions can be categorized into the following groups:

 * Arithmetic
 * Binary Coded Decimal Arithmetic
 * Logic
 * Shift
 * Bit manipulation in memory
 * Multiprocessing control
 * Flowcontrol
 * Branch, Decrement-and-branch

Very interesting here are the instructions for multiprocessing control, which can perform indivisible bus operation, as well as handling semaphores for processors sharing a single memory and even more.

Zilog Z80
The Zilog Z80 is a very successful 8 bit CISC-microprocessor which is in use since the mid 70s. Its design is based on the Intel 8080, so that programs for the 8080 CPU would also run on the Z80. One of the most famous examples for usage of the microprocessor is the Nintendo Gameboy (it uses an adopted version of the Z80).

Register Layout
There are six pairs of 8-bit general purpose registers, of which each pair consists of the register itself and a corresponding flag register. Each pair can either be used as 8-bit or 16-bit register. Two 8-bit accumulators serve for storing results of 8-bit arithmetic and logic operations. Both accumulator registers have a corresponding 8-bit flag register. Furthermore, there are 6 special purpose registers: two index registers (each 16-bit), a stack pointer (16-bit), a program counter (16-bit), an interrupt page address register (8-bit) and the memory refresh register (8-bit).

Instruction Set
The Z80 has a set of 252 different instructions which are combined to 158 different instruction types. These instruction types are further categorized into the following groups:

 * Load and Exchange
 * Block Transfer and Search
 * Arithmetic and Logical
 * Rotate and Shift
 * Bit Manipulation (Set, Reset, Test)
 * Jump, Call and Return
 * Input/Output
 * Basic CPU Control

The load and exchange instructions serve to move data between and among registers and external memory. The block transfer and search instructions are rather complex instructions that would not be found in a RISC architecture. For example there exists a single instruction to move a whole block of memory between two locations. The groups of arithmetic and logical, and rotate and shift operations are quite commonly implemented. The group of bit manipulation instructions allows to test, set and reset any bit of the accumulator, the general purpose registers, or any external memory location with a single instruction. The jump, call and return instructions are used for program control flow. The rather mighty set of input/output instructions allows for data transfer between external memory, internal registers and I/O devices. The last group, the basic CPU control instructions, serve to control different modes of the CPU like certain interrupt handling.

As a consequence of being based on the Intel 8080, the Zilog Z80 has a von Neumann architecture, stores words in little endian format, and comprises the whole instruction set of the 8080 CPU. Also it originally did not implement pipelining.

Comparison
Obviously the ARM processor is the only RISC processor. Nevertheless it can interact with a coprocessor, but has no instructions for synchronization like the Motorola 68000. The ARM tries to execute instructions with an operation code it does not understand by sending them to a coprocessor. Clearly the set of synchronization instruction of the Motorola is more complex than the instructions ARM executes. Also the Z80, has very powerful instructions like search. When concerning registers both CISC processors have a lot less registers than the ARM, even if you consider the duplicate registers in the Z80. Whereas the 68000 only uses 8 registers to react fast in case of an interrupt, the Z80 has them for compatibility to the 8080 family of Intel, even if they are duplicated.

Each processor complies with the requirements of a set of applications, even if it might was designed with other uses in mind. The ARM for embedded systems, the Motorola 68000 now for the TI200, but has been used in a lot of other applications like printers. And the Z80 was used in embedded systems as well as in desktop pcs.