text:
User:Eml 4500.f08.delta 6.specht/New HW2

 The following is a revision of the Homework 2 assignment that was submitted for group Delta_6. The following link includes the differences between the two pages that were submitted.

http://en.wikiversity.org/w/index.php?title=User:Eml_4500.f08.delta_6.specht/New_HW2&diff=next&oldid=343668

--Eml 4500.f08.delta 6.specht 15:53, 9 December 2008 (UTC)

Unit Vector Multiplication
The value for m(e) can be calculated in the manner shown below. It is equal to the final solution that is found. It should be noted that the dot product of the i and j unit vectors equals zero, and the dot product of the two j unit vectors is equal to one.

Element Stiffness Matrices
The values that are used to calculate the scalar value k are shown below. The direction cosine matrix is multiplied by the scalar k value, which is the spring constant of each element. The value that is obtained from this operation is the stiffness matrix for each element. The values for the direction cosines of each element are as follows: l(1) = cos θ(1) = cos30° = $$\frac{\sqrt{3}}{2}$$ m(1) = sin θ(1) = sin30° = $$\frac{1}{2}$$ l(2) = cos θ(2) = $$cos(-\frac{\Pi }{4}) = \frac{\sqrt{2}}{2}$$ m(2) = sin θ(2) = $$\sin (-\frac{\Pi }{4}) = -\frac{\sqrt{2}}{2}$$ These values can then be substituted into the matrix of direction cosines in order to calculate the element stiffness matrices.

Verification of Equilibrium of element 1:
Note: The reactions for node 1 were calculated to be

F1(1) = -4.4278

F2(1) = -2.5622

F3(1) = 4.4278

F4(1) = 2.5622

Note: Negative values signify that the direction of the forces in element 1 were assumed to be in the wrong direction.

Redrawing the free body diagram with all positive forces shows more clearly how the resultant forces P1 and P2 cancel each other out.

Now,

F1(1) = 4.4278

F2(1) = 2.5622

F3(1) = 4.4278

F4(1) = 2.5622

Determining Resultants:

P1= [(f1(1))2 + (f2(1))2]1/2= [(4.4278)2+ (2.5622)2]1/2= 5.12

P2= [ (f3(1))2 + (f4(1))2]1/2= [(4.4278)2+ (2.5622)2]1/2= 5.12

Note: P1 = P2, thus element 1 is in equilibrium.

Statics Solution Method

P = 7 L(1) = 4 L(2) = 2 θ(1) = 30° θ(2) = 45°

Observation:

By using trigonometry and the angles given the unknown lengths can be determined. These lengths will be needed when analyzing the summation of the moments of each node.

$$\sin {30^{\circ} } = \frac{L_c}{4} \Rightarrow L_c = 4 \sin {30^{\circ} } = 2$$

$$\cos {30^{\circ} } = \frac{L_x}{4} \Rightarrow L_x = 4 \cos {30^{\circ} } = 3.46$$

$$\sin {45^{\circ} } = \frac{L_b}{2} \Rightarrow L_b = 2 \sin {45^{\circ} } = 1.4$$

$$\cos {45^{\circ} } = \frac{L_y}{2} \Rightarrow L_y = 2 \cos {45^{\circ} } = 1.4$$

Lz = Lx + Ly = 3.46 + 1.4 = 4.86

La = Lc - Lb = 2 - 1.4 = 0.6

Moment equations can now be analyzed since all lengths are known.

(1) M1: -(P)(Lx) + (F4)(Lz) + (F3)(La) = -(7)(3.46) + (F4)(4.86) + (F3)(0.6)

(2) M2: (P)(Ly) + (F1)(La) - (F2)(Lz) = (7)(1.4) + (F1)(0.6) - (F2)(4.86)

(3) M3: (F1)(Lc) - (F2)(Lx) - (F3)(Lb) + (F4)(Ly) = (F1)(2) - (F2)(3.46) - (F3)(1.4) + (F4)(1.4)

Note: There are 4 unknowns with only 3 equations. However, by inspection it is known that the summation of the forces in the x direction are equal to zero, giving us 4 equations with 4 unknowns. (4) $$\sum{F_x} = 0: F_1 - F_3 = 0 $$

All forces can now be determined by using the system of equations. Rearrange the equations solving for a common variable in order to calculate the first unknown force. Next, using the rearrange equations all unknown forces can be calculated.

(1) -24.22 + 4.86 F4 + 0.6 F3 = 0

(2) 9.8 + 0.6 F1 - 4.86 F2 = 0

(3) 2 F1 - 3.46 F2 - 1.4 F3 + 1.4 F4 = 0

(4) F1 = F3

Since F3 is already in terms of F1, F2 and F4 will also solved in terms of F1

(1) $$F_4 = \frac{24.22 - 0.6 F_3}{4.86} = \frac{24.22 - 0.6 F_1}{4.86}$$ (2) $$F_2 = \frac{9.8 + 0.6 F_1}{4.86}$$ (4) F3 = F1

Plugging in these relationships into equation (3), F1 and the rest of the forces can be solved for as follows:

(3) $$2 F_1 - 3.46 (\frac{9.8 + 0.6 F_1}{4.86}) - 1.4 F_1 + 1.4 (\frac{24.22 - 0.6 F_1}{4.86}) = 0 \Rightarrow $$

Combining like terms,

(3) $$2 F_1 - 6.976 - 0.427 F_1 - 1.4 F_1 + 6.976 - 0.173 F_1 = 0 \Rightarrow 0 = 0$$

Thus, F1 = F3 = 0

(2) $$F_2 = \frac{9.8 + 0.6(0)}{4.86} = 2.01$$ (1) $$F_4 = \frac{24.22 + 0.6(0)}{4.86} = 4.98$$

Results

F1 = 0 F2 = 2.01 F3 = 0 F4 = 4.98

Two Bar Truss Example (MATLAB)
It is important to understand how to solve a bar truss system employing a finite element MATLAB code in order to compare the resulting matrices with those obtained from hand calculations.

Take this truss example with the following: Data

MATLAB Code Note: this MATLAB code was obtained from Dr. L. Vu-Quoc website http://clesm.mae.ufl.edu/~vql/courses/fead/2008.fall/codes/twoBarTrussEx.txt

In order to run this MATLAB code the following functions are needed:

Note: this MATLAB code was obtained from the MATLAB.zip Chapter 9 folder given by the Student Companion Site Bhatti: Advanced Topics in Finite Element Analysis of Structures: With Mathematica and MATLAB Computations website http://bcs.wiley.com/he-bcs/Books?action=resource&bcsId=3105&itemId=0471648078&resourceId=7629

Note: this MATLAB code was obtained from the MATLAB.zip Common folder given by the Student Companion Site Bhatti: Advanced Topics in Finite Element Analysis of Structures: With Mathematica and MATLAB Computations website http://bcs.wiley.com/he-bcs/Books?action=resource&bcsId=3105&itemId=0471648078&resourceId=7629

Note: this MATLAB code was obtained from the MATLAB.zip Chapter 9 folder given by the Student Companion Site Bhatti: Advanced Topics in Finite Element Analysis of Structures: With Mathematica and MATLAB Computations website http://bcs.wiley.com/he-bcs/Books?action=resource&bcsId=3105&itemId=0471648078&resourceId=7629

Results

Testing the Steps
It is important to see how the steps to solve truss systems work.

Given

Take this truss example with the following data:

Step 1: Global Picture
Releasing the truss system from its supports and showing the reactions, we have:

There are 6 unknowns for which we need to solve for, therefore we need to analyze each element. The matrix equation is as follows.

 * $$\begin{pmatrix}

f_1\\ ...\\ f_6\end{pmatrix}$$ = $$\begin{bmatrix} & &  \\  & k &  \\ & &  \end{bmatrix}$$ $$\begin{pmatrix} d_1\\ ...\\ d_6\end{pmatrix}$$

which corresponds to

 * $$\mathbf{F}$$=$$\mathbf{K}$$ $$\mathbf{d}$$.

To solve for the reactions, we have to now examine the whole system by breaking it down into elements.

Step 2: Element Picture
Analyzing each element with its own force and dofs yields:

It is important to notice the chosen coordinate axis since they will allow for ease of analyzing each element. The tilda on the coordinate direction names designate a non-horizontal or non-vertical direction.

We now want to creat a matrix equation, just like for the global picture, for the element picture. The matrix equation consists on the basic force-distance relationship, however, this principle is applied at the element level. The notation is altered in that each term is designated to belong to the particular element with an upper script number in parenthesis so that we get, F(e)=k(e) d(e), named as the element force matrix, the element stiffness matrix, and the element distance matrix.

Step 3: Force-Displacement Relationship
To build the element stiffness matrix, we have to follow a "recipe" for calculating each term. This stiffness matrix will end up measuring 4x4.

The recipe is as follows, and in the wise words of Prof. Vu-Quoc, "The notation is beautiful here. It tells you no more, no less."

 * k$$^{(e)}$$ $$\begin{bmatrix}

l^{(e)2} & l^{(e)}m^{(e)} & -l^{(e)2} & -l^{(e)}m^{(e)}\\ l^{(e)}m^{(e)} & m^{(e)2} & -l^{(e)}m^{(e)} & -m^{(e)2}\\ -l^{(e)2} & -l^{(e)}m^{(e)} & l^{(e)2} & l^{(e)}m^{(e)}\\ -l^{(e)}m^{(e)} & -m^{(e)2} & l^{(e)}m^{(e)} & m^{(e)2} \end{bmatrix}$$ = k$$^{(e)}$$

where the axial stiffness k$$^{(e)}$$ of the bar element "e" is

 * $$k^{(e)}=\frac{E^{(e)} A^{(e)}}{L^{(e)}}$$,

where $$E^{(e)}$$ is the element's material Young's Modulus, $$A^{(e)}$$ is the element cross-sectional area, and $$L^{(e)}$$ the element length. Also, $$l^{(e)}$$ and $$m^{(e)}$$ are te director cosines of the rotated, and convenient, coordinate axes with respect to the positive, and horizontal, x-direction (the global axis).

It is very important to make the following observations for the efficiency to construct the stiffness matrix:
 * 1. It is only needed to compute 3 numbers, both of the squares and the product of eachother, and the determination of a negeative or a positive sign.
 * 2. The matrix k$$^{(e)}$$ is symmetric, which means that there is no need to calculate the bottom triangular part once the top has been determined, including the diagonal.

Director Cosines: The director cosines will yield the projections of vectors oriented with an inclination w.r.t. the positive x-axis on the horizontal or vertical directions. These horizontal and vertical directions are more precisely for this course the global x-y coordinates. From this diagram, we can derive our much needed $$l^{(e)}$$ and $$m^{(e)}$$ values. $$l^{(e)}$$ = $$\vec{\tilde{i}}$$ • $$\vec{\hat{i}}$$ = cosθ$$^{(e)}$$ $$m^{(e)}$$ = $$\vec{\tilde{i}}$$ • $$\vec{\hat{j}}$$ = sinθ$$^{(e)}$$ = cos(π/2-θ$$^{(e)})$$ Also, $$\vec{\tilde{i}}$$=cosθ$$^{(e)}$$•$$\vec{\hat{i}}$$+sinθ$$^{(e)}$$•$$\vec{\hat{j}}$$

Global stiffness matrix K.

 * {| style="text-align:center; border-collapse:collapse;" cellpadding="4"

 * rowspan="6" | K  =
 * rowspan="6" | $$\left.\begin{align} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \end{align} \right($$
 * title="stiffness matrix from element one" style="border-left: 1px solid black; border-top: 1px solid black;" | $$k_{11}^{(1)}$$
 * title="stiffness matrix from element one" style="border-top: 1px solid black;" | $$k_{12}^{(1)}$$
 * title="stiffness matrix from element one" style="border-top: 1px solid black;" | $$k_{13}^{(1)}$$
 * title="stiffness matrix from element one" style="border-top: 1px solid black; border-right: 1px solid black;" | $$k_{11}^{(1)}$$
 * 0
 * 0
 * rowspan="6" | $$\left)\begin{align} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \end{align} \right.$$
 * title="stiffness matrix from element one" style="border-left: 1px solid black;" | $$k_{21}^{(1)}$$
 * title="stiffness matrix from element one" | $$k_{22}^{(1)}$$
 * title="stiffness matrix from element one" | $$k_{23}^{(1)}$$
 * title="stiffness matrix from element one" style="border-right: 1px solid black;" | $$k_{24}^{(1)}$$
 * 0
 * 0
 * title="stiffness matrix from element one" style="border-left: 1px solid black;" | $$k_{31}^{(1)}$$
 * title="stiffness matrix from element one" style="border-right: 1px solid black;" | $$k_{32}^{(1)}$$
 * title="stiffness matrix from element one and element two" style="border-top: 1px solid black;" | $$k_{33}^{(1)} + k_{11}^{(2)}$$
 * title="stiffness matrix from element one and element two" style="border-right: 1px solid black; border-top: 1px solid black;" | $$k_{34}^{(1)} + k_{12}^{(2)}$$
 * title="stiffness matrix from element two" style="border-top: 1px solid black;" | $$k_{13}^{(2)}$$
 * title="stiffness matrix from element two" style="border-right: 1px solid black; border-top: 1px solid black;" | $$k_{14}^{(2)}$$
 * title="stiffness matrix from element one" style="border-left: 1px solid black; border-bottom: 1px solid black;" | $$k_{41}^{(1)}$$
 * title="stiffness matrix from element one" style="border-right: 1px solid black; border-bottom: 1px solid black;" | $$k_{41}^{(1)}$$
 * title="stiffness matrix from element one and element two" style="border-bottom: 1px solid black;" | $$k_{43}^{(1)} + k_{21}^{(2)}$$
 * title="stiffness matrix from element one and element two" style="border-right: 1px solid black; border-bottom: 1px solid black;" | $$k_{44}^{(1)} + k_{22}^{(2)}$$
 * title="stiffness matrix from element two" | $$k_{23}^{(2)}$$
 * title="stiffness matrix from element two" style="border-right: 1px solid black;" | $$k_{24}^{(2)}$$
 * 0
 * style="border-right: 1px solid black;" | 0
 * title="stiffness matrix from element two" | $$k_{31}^{(2)}$$
 * title="stiffness matrix from element two" | $$k_{32}^{(2)}$$
 * title="stiffness matrix from element two" | $$k_{33}^{(2)}$$
 * title="stiffness matrix from element two" style="border-right: 1px solid black;" | $$k_{34}^{(2)}$$
 * 0
 * style="border-right: 1px solid black;" | 0
 * title="stiffness matrix from element two" style="border-bottom: 1px solid black;" | $$k_{41}^{(2)}$$
 * title="stiffness matrix from element two" style="border-bottom: 1px solid black;" | $$k_{42}^{(2)}$$
 * title="stiffness matrix from element two" style="border-bottom: 1px solid black;" | $$k_{43}^{(2)}$$
 * title="stiffness matrix from element two" style="border-right: 1px solid black; border-bottom: 1px solid black;" | $$k_{44}^{(2)}$$
 * }
 * 0
 * style="border-right: 1px solid black;" | 0
 * title="stiffness matrix from element two" | $$k_{31}^{(2)}$$
 * title="stiffness matrix from element two" | $$k_{32}^{(2)}$$
 * title="stiffness matrix from element two" | $$k_{33}^{(2)}$$
 * title="stiffness matrix from element two" style="border-right: 1px solid black;" | $$k_{34}^{(2)}$$
 * 0
 * style="border-right: 1px solid black;" | 0
 * title="stiffness matrix from element two" style="border-bottom: 1px solid black;" | $$k_{41}^{(2)}$$
 * title="stiffness matrix from element two" style="border-bottom: 1px solid black;" | $$k_{42}^{(2)}$$
 * title="stiffness matrix from element two" style="border-bottom: 1px solid black;" | $$k_{43}^{(2)}$$
 * title="stiffness matrix from element two" style="border-right: 1px solid black; border-bottom: 1px solid black;" | $$k_{44}^{(2)}$$
 * }
 * title="stiffness matrix from element two" style="border-bottom: 1px solid black;" | $$k_{43}^{(2)}$$
 * title="stiffness matrix from element two" style="border-right: 1px solid black; border-bottom: 1px solid black;" | $$k_{44}^{(2)}$$
 * }
 * }

The box, within the matrix on the top left, is the stiffness matrix from element one. The box to the lower right is the stiffness matrix from element two. Combined the two become the global stiffness matrix.

Where capital k (K) is the global stiffness coefficient and lowercase k (k) is the element stiffness coefficient. Below is the relationship between the global stiffness coefficients and the element stiffness coefficients for the global stiffness matrix above.

K11 = k11(1) ; K12 = k12(1) ; K13 = k13(1) ; K14 = k14(1) ; K15 = 0 ; K16 = 0

K21 = k21(1) ; K22 = k22(1) ; K23 = k23(1) ; K24 = k24(1) ; K25 = k25(1) ; K26 = k26(1)

K31 = k31(1) ; K32 = k32(1) ; K33 = k33(1) + k11(2) ; K34 = k34(1) + k12(2); K13 = k13(2) ; K14 = k14(2)

K41 = k41(1) ; K42 = k42(1) ; K43 = k43(1) + k21(2) ; K44 = k44(1) + k22(2); K23 = k23(2) ; K24 = k24(2)

K51 = 0 ; K52 = 0 ; K53 = k31(2) ; K54 = k32(2) ; K55 = k33(2) ; K56 = k34(2)

K61 = 0 ; K62 = 0 ; K63 = k41(2) ; K64 = k42(2) ; K65 = k43(2) ; K66 = k44(2) The calculation of the values in the k matrices for element 1 and element 2 are as follows: θ(1) = 30° l(1) = cos θ(1) = cos30° = $$\frac{\sqrt{3}}{2}$$ m(1) = sin θ(1) = sin30° = $$\frac{1}{2}$$ k(1) = E(1)·A(1)/ L(1) = $$\frac{3}{4}$$

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

k11(1) = k(1) · ( l(1) )2 = $$\frac{3}{4}$$ · $$(\frac{\sqrt{3}}{2})^{2}$$ = $$\frac{9}{16}$$

k12(1) = k(1) · l(1) · m(1) = $$\frac{3}{4}$$ · $$\frac{\sqrt{3}}{2}$$· $$\frac{1}{2}$$ = $$\frac{\sqrt{3}}{16}$$

k42(1) = - k(1) · ( m(1) )2 = - $$\frac{3}{4}$$ · $$(\frac{2})^{2}$$ = - $$\frac{3}{16}$$

Observation:

1) Only three numbers need to be computed. Other coefficients have the same absolute value just differ by sign (+ or -) 2) Matrix k(1) is symmetrical i.e., kij(1) = kji(1); just interchange the row and column indices

For example: k13(1) = k31(1)

In general,

kij(e) = kji(e) or kij(1)Т = k(1)

where Т represents the transpose of the matrix.

$$k^{(e)} = \begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)}\\ & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)}\\ & & k_{33}^{(1)} & k_{34}^{(1)}\\ & &  & k_{44}^{(1)} \end{bmatrix}$$

Since the matrix is symmetric about the diagonal, only the upper right triangle part needs to be calculated.

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

Element 2:

k(2) = E(2)·A(2)/ L(2) = 5 θ(2) = $$-\frac{\Pi }{4}$$ l(2) = cos θ(2) = $$cos(-\frac{\Pi }{4}) = \frac{\sqrt{2}}{2}$$ m(2) = sin θ(2) = $$\sin (-\frac{\Pi }{4}) = -\frac{\sqrt{2}}{2}$$

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

Observation:

1) The absolute value of all coefficients kji(e), e=2 (i,j) = 1,...,4 are the same (|l| = |m|) → Only four coefficients need to be computed; all other coefficients add + or -. 2)kij(2)Т = k(2)

$$k^{(2)} = \begin{bmatrix} \frac{5}{2} & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2}\\ -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} & -\frac{5}{2}\\ -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} & -\frac{5}{2}\\ \frac{5}{2} & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2} \end{bmatrix}$$

Compact Notations between Global Diagrams and Element Diagrams
It is necessary to understand the relationships between the global free diagrams and the element free diagrams in order to solve the truss system of matrices going from the element matrices to the global matrices.

Global Free Diagram Relationship
The global free diagram relationship is given by the equation: $$\begin{bmatrix}K_{ij}^{}\end{bmatrix}_{6x6} \left\{d_{ij}\right\}_{6x1} = \left\{F_{i}\right\}_{6x1}$$ where the d and F  elements are composed by 6 x 6 matrices as following:

$$\sum_{j=1}^{6} K_{ij}d_{j}= F_{i}$$ where i goes from 1 through 6 since there are six equations
 * Global Stiffness Matrix:

$$K_{nxn} = \begin{bmatrix}K_{ij}^{}\end{bmatrix}_{nxn}$$
 * Global Displacement Matrix:

$$d_{nx1} = \left\{d_{j}^{}\right\}_{nx1}$$
 * Global Force Matrix:

$$F_{nx1} = \left\{F_{i}^{}\right\}_{nx1}$$

Element Free Diagram Relationship
The element free diagrams relationship is given by the equation: $$k_{4x4}^{(e)} d_{4x4}^{(e)} = f_{4x1}^{(e)}$$ where e indicates the element number. The d and f elements are composed by 4 x 1 matrices as following:

$$d_{4x1}^{(e)} = \left\{d_{j}^{(e)}\right\}_{4x1}$$ $$d_{4x1}^{(e)}=\begin{Bmatrix} d_{1}^{(e)} \\ d_{2}^{(e)} \\ d_{3}^{(e)} \\ d_{4}^{(e)} \end{Bmatrix}$$ $$f_{4x1}^{(e)} = \left\{f_{i}^{(e)}\right\}_{4x1}$$ $$f_{4x1}^{(e)}=\begin{Bmatrix} f_{1}^{(e)} \\ f_{2}^{(e)} \\ f_{3}^{(e)} \\ f_{4}^{(e)} \end{Bmatrix}$$ $$k_{4x4}^{(e)} = \begin{bmatrix}k_{ij}^{(e)}\end{bmatrix}_{4x4}$$ $$k_{4x4}^{(e)}=\mathbf \begin{bmatrix} K_{11}^{(e)} & K_{12}^{(e)} & K_{13}^{(e)} & K_{14}^{(e)}\\K_{21}^{(e)} & K_{22}^{(e)} & K_{23}^{(e)} & K_{24}^{(e)}\\K_{31}^{(e)} & K_{32}^{(e)} & K_{33}^{(e)} & K_{34}^{(e)}\\K_{41}^{(e)} & K_{42}^{(e)} & K_{43}^{(e)} & K_{44}^{(e)}\end{bmatrix}$$
 * Element Displacement Matrix:
 * Element Force Matrix:
 * Element Stiffness Matrix:

Since the matrix k is symmetric in general the upper triangular part of the matrix should transpose to the lower triangular part, i.e. $$k_{ij}^{(e)} = k_{ji}^{(e)}$$

Assembly Process
From the free body general relationships it is possible to generate new associations in order to solve the stiffness, displacement and forces matrices:

 * Global level: $$\mathbf\begin{Bmatrix} d_{1} & d_{2} & d_{3} & d_{4} & d_{5} & d_{6}\end{Bmatrix}$$

 * Element level:
 * Element 1: $$\mathbf\begin{Bmatrix} d_{1}^{(1)} & d_{2}^{(1)} & d_{3}^{(1)} & d_{4}^{(1)} & d_{5}^{(1)} & d_{6}^{(1)}\end{Bmatrix}$$
 * Element 2: $$\mathbf\begin{Bmatrix} d_{1}^{(2)} & d_{2}^{(2)} & d_{3}^{(2)} & d_{4}^{(2)} & d_{5}^{(2)} & d_{6}^{(2)}\end{Bmatrix}$$

 * Nodes:
 * Node 1: $$d_{2} = d_{1}^{(1)}$$
 * Node 2: $$d_{3} = d_{2}^{(1)}$$
 * $$d_{3} = d_{3}^{(1)} = d_{1}^{(2)}$$
 * $$d_{4} = d_{4}^{(1)} = d_{2}^{(2)}$$

Note: These points are from the same node therefore must move in the same direction =>> (same global number = same direction)
 * Node 3: $$d_{5} = d_{3}^{(2)}$$
 * $$d_{6} = d_{4}^{(2)}$$

Step 4: Elimination of known degrees of freedom (dofs)
This Process reduces the global force displacement relationship by eliminating the first and last two columns of the global stiffness matrix.

d1 = d2 = d5 = d6 = 0

The displacements above are zero. Therefore, columns 1, 2, 5 and 6 of the global stiffness matrix go to zero. This reduces the matrix to a 6x2.

The following displays the F matrix calculated two different ways.

$$ \textbf{K}\begin{bmatrix} d_{1}=0\\ d_{2}=0\\ d_{3}\\ d_{4}\\ d_{5}=0\\ d_{6}=0 \end{bmatrix} =\begin{bmatrix} K_{13} & K_{14}\\ K_{23} & K_{24}\\ K_{33} & K_{34}\\ K_{43} & K_{44}\\ K_{53} & K_{54}\\ K_{63} & K_{64} \end{bmatrix} \begin{bmatrix} d_{3}\\ d_{4} \end{bmatrix} = \textbf{F} $$

Here, we delete the corresponding columns in the global stiffness matix, K. Due to the Principle of Virtual Work (PVW), we must also delete the corresponding rows (rows 1, 2, 5, 6).

Below, we have the Resulting Force Displacement relation:

$$ \begin{bmatrix} K_{33} &K_{34} \\ K_{43} & K_{44} \end{bmatrix} \begin{bmatrix} d_{3}\\ d_{4} \end{bmatrix} = \begin{bmatrix} F_{3}\\ F_{4} \end{bmatrix} $$ where the F matrix is found below.

$$ \textbf{F} = \begin{bmatrix} F_{3}\\ F_{4} \end{bmatrix} = \begin{bmatrix} 0\\ P \end{bmatrix} $$ Before we continue, we will find the determinant and inverse of the K matrix. The determinant is simply equal to:

$$ det\; \textbf{K} = K_{33}K_{44}-K_{34}K_{43} $$

Now that we have the determinant, we can calculate the inverse using the formula below.

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

The inverse of the K matrix cannot be found by using the transpose of the matrix, illustrated below.
 * NOTE*

$$ \textbf{K}^{T} = \begin{bmatrix} K_{33} & K_{34}\\ K_{43} & K_{44} \end{bmatrix} , where\; \textbf{K}^{-1}\neq \frac{1}{det\; \textbf{K}} \textbf{K}^{T} $$

With the inverse of the K matrix known, we can now find the unknown forces from the d matrix.

$$ \begin{bmatrix} d_{3}\\ d_{4} \end{bmatrix} = \textbf{K}^{-1}\begin{bmatrix} 0\\ P \end{bmatrix} = \begin{bmatrix} 4.352\\ 6.127 \end{bmatrix} $$

Step 5: Compute Reactions
There are two main methods for solving this 2-bar truss system. Method 1 For the first method, we will use the element force displacement relationship: $$ \textbf{k}^{e}\textbf{d}^{e}=\textbf{f}^{e}, where\; e=1,2 $$

We can now separate the truss into each individual bar, and compute the calculations for each bar alone.

Element 1:

$$where\; \textbf{d}^{(1)}= \begin{bmatrix} 0\\ 0\\ 4.352\\ 6.127 \end{bmatrix} = \begin{bmatrix} d_{1}^{(1)}\\ d_{2}^{(1)}\\ d_{3}^{(1)}\\ d_{4}^{(1)} \end{bmatrix} $$

Element 2:

$$ where\; \textbf{d}^{(2)}= \begin{bmatrix} 4.352\\ 6.127\\ 0\\ 0 \end{bmatrix}= \begin{bmatrix} d_{1}^{(2)}\\ d_{2}^{(2)}\\ d_{3}^{(2)}\\ d_{4}^{(2)} \end{bmatrix} since\; d_{3}=d_{1}^{(2)}, d_{4}=d_{2}^{(2)} $$

Step 6: Computation of Reaction Forces
The displacements for elements 1 and 2 are now known, and can be used in order to solve for the reaction forces at the pins. The force-displacement relationship that is shown below will be used to find the forces at the ends of each element.

The element stiffness matrix and the displacement matrix of element 1 can be reduced based on the values of d(1)1=d(2)2=0. These reduced matrices are shown below.

$$k^{(1)}*d^{(1)}=f^{(1)}= \left[ \begin{matrix} -0.5625&-0.32476\\  -0.32476&-0.1875\\  0.5625&0.32476\\   0.32476&0.1875 \end{matrix}\right]* \left\{ \begin{matrix} 4.352\\  6.1271 \end{matrix} \right\}$$

The forces that are induced in element 1 by the applied force P at node 2 are equivalent to:

 * $$f^{(1)}=\begin{Bmatrix} -4.4378 \\ -2.5622 \\ 4.4378 \\ 2.5622 \end{Bmatrix}=\begin{Bmatrix} f_1^{(1)} \\ f_2^{(1)} \\ f_3^{(1)} \\ f_4^{(1)} \end{Bmatrix}$$

The equations for static equilibrium are listed below. It is evident by the values of the forces in the element 1 force matrix that the first two equations are satisfied. The third equation is satisfied by taking the moment about either one of the nodes of element 1.

$$ \begin{array}{lr} \sum F_x=f_1^{(1)}+f_3^{(1)}=0&(1)\\ \sum F_y=f_2^{(1)}+f_4^{(1)}=0&(2)\\ \sum M_{Any\, Node}=0&(3)\\ \end{array} $$

The two elements of the 2-D Plane Truss problem are shown below. The forces that act along the axis of the element can be found by using the Pythagorean Theorem. It can be noted that the force that acts on one end of each element is equal and opposite to the force acting on the other end of the element. The equation to accomplish this is listed after the diagrams of the elements.

$$P_1^{\left(1\right)}=\left[\left(f_1^{\left(1\right)}\right)^2+\left(f_2^{\left(1\right)}\right)^2\right]^{\frac 12}$$

The forces that are found using the method described can be verified by applying the equations of static equilibrium across each element. The global picture can also be divided in two ways, by using the Euler Cut Principle, in order to perform this operation. This analysis is provided in the homework section at the beginning of this article. The methods that this can be cut are shown below. Each cut is denoted by a dashed circle. The first method includes the global node 2 in the main section of element 1, while the second method has node 2 being analyzed separately.

In order to verify the static equilibrium of the whole system at once, the equilibrium of node 2 can be established. The diagram for this is shown below, while the calculation is performed in the homework section at the beginning of the article. The forces shown are P (applied force), P(1)1 (axial force in element 1), P(2)2 (axial force in element 2).

Team Contributors
Iain Specht--Eml 4500.f08.delta 6.specht 14:24, 22 September 2008 (UTC) Micheal Trent Krueger--Eml4500.f08.delta 6.krueger 01:39, 26 September 2008 (UTC) Nicolas Castrillon --Eml4500.f08.delta 6.castrillon 01:40, 26 September 2008 (UTC) Diana Guzman--Eml4500.f08.delta 6.guzman 05:02, 26 September 2008 (UTC) Sean Lopez--Eml4500.f08.delta 6.lopez 05:25, 26 September 2008 (UTC) Eric Ramirez--Eml4500.f08.delta 6.ramirez 12:42, 26 September 2008 (UTC)


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User:Eml4500.f08.FEABBQ.koby/HW3

Sept 29

Class Notes: September 29, 2008
Using the Global Force-Displacement Relation

Use the global force-displacement relationship matrix from lecture 10:

$$ \left[ \begin{array}{cc} k_{13}^{(1)} & k_{14}^{(1)} \\ k_{23}^{(1)} & k_{24}^{(1)} \\ (k_{33}^{(1)} + k_{11}^{(2)}) & (k_{34}^{(1)} + k_{12}^{(2)}) \\  (k_{43}^{(1)} + k_{21}^{(2)}) & (k_{44}^{(1)} + k_{22}^{(2)})  \\  k_{31}^{(2)} & k_{32}^{(2)} \\ k_{41}^{(2)} & k_{42}^{(2)}  \\ \end{array} \right] $$ $$\left[ \begin{array}{l} d_1 \\ d_2  \\ d_3  \\ d_4 \\ d_5 \\ d_6 \end{array} \right] = \left[ \begin{array}{l} F_1 \\ F_2  \\ F_3  \\ F_4 \\ F_5 \\ F_6 \end{array} \right]$$ (6x2) x (2x1)  =  (6x1)

Only do computations for rows 1,2,5,6 to get F1,F2,F5,F6. ....

Closing the loop between the Finite Element Method and statics: Virtual Displacement

For the two bar truss system:

by statics, reactions are known and therefor the member forces $$P_1^{(1)}$$ and $$P_2^{(1)}$$

Compute the axial displacement of the degrees of freedom (extension of bars):

$$q_2^{(1)} = \frac{P_1^{(1)}}{k^{(1)}} = \frac{P_2^{(1)}}{k^{(1)}}$$

$$ q_1^{(1)} = 0$$, fixed global node 1

$$q_1^{(2)} = \frac{-P_2^{(2)}}{k^{(2)}} = -AB$$

$$q_2^{(2)} = 0 $$, fixed global node 3

Question: How to get the displacement DOF's from the above results for node 2?

title: User:Eml4500.f08.FEABBQ.koby/HW3


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User:Stevenarntson

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User:Eas4200c.f08.aero6.ortega

User:eas4200c.f08.aero6.ortega/HW3

Topics for HW 3 (but I wasn’t in class the day he assigned the NACA code)

Shear flow over curved panels (1.4)(1.5) Derivation of T = 2qA (3.48)(3.49) Prove A = .5bh for odd triangle Description of uniform bar with circular cross-section J= .5 pi a^4 Thin-walled t<<a; ri =a; ro =b; J=.5 pi(b^4 – a^4) =.5 pi (b-a)(b+a)(b^2+a^2) t     2r         2r^2 r = .5(a+b) prove b^2=~ r^2 a^2=~r^2	J = 2 pi t r^3

compare circular cross-section with thin walled fig 1.8a and b Describe warping Roadmap for torsional analysis of wing Problem 1.1 adapted for wing: compute Abar=pi/2(b/2)^2 + ba/2 Twist angle Comparison of sigma and integral symbols (quadrature/ Reimann sum)

title: User:Eas4200c.f08.aero6.ortega


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User:Eas4200c.f08.aero6.ortega/HW3

(1.3)
Problem statement: The dimensions of a steel (300M) I-beam are $$b = 50 \ mm,\!$$ $$t = 5 \ mm,\!$$ and $$h = 200 \ mm,\!$$ as shown in Figure 1. Assume that $$t\!$$ > and $$h\!$$ > are to be fixed for an aluminum (7075-T6) I-beam. Find the width $$b\!$$ > for the aluminum beam such that its bending stiffness $$EI\!$$ > is equal to that of the steel beam. Compare the weight-per-unit length of these two beams. Which is more efficient weightwise? Question quoted from Mechanics of Aircraft Structures by C.T. Sun

Young's Modulus for specified materials:
 * $$E_{steel} = 200 \ GPa\!$$
 * $$E_{aluminum} = 71 \ GPa\!$$

The Moment of Inertia for an I-beam as shown in Figure 1 is found by breaking up the cross-section into three rectangles and using the Parallel Axis Theorem to compensate for the centroids of each shape. The I-beam will be broken up into one $$t$$ by $$h-2t$$ rectangle and two $$b$$ by $$t$$ rectangles. Remembering the general form of the Parallel Axis Theorem and Moment of Inertia calculation, $$I$$ will be as follows:
 * $$I = \frac{t(h-2t)^3}{12} + 2\left(\frac{bt^3}{12} + bt\left(\frac{h-2t}{2}+\frac{t}{2}\right)^2\right)$$ EQN 1

Therefore
 * $$I_{steel} = 7.612 \times 10^{-6} \ m^4\!$$

The problem states that $$E_{steel}\cdot I_{steel} = E_{aluminum}\cdot I_{aluminum}\!$$

Therefore
 * $$I_{aluminum} = \frac{E_{steel} \cdot I_{steel}}{E_{aluminum}} = \frac{(200 \ GPa)(7.612 \times 10^{-6} \ m^4)}{71 \ GPa} = 2.144 \times 10^{-5} \ m^4$$

The next step is to solve EQN 1 for $$b\!$$ >, in terms of the known and fixed values specified in the problem statement, as shown below.
 * $$b = \frac{1}{2}\left(I_{aluminum}-\frac{t(h-2t)^3}{12}\right)\Bigg/\left(\frac{t^3}{12}+t\left(\frac{h-2t}{2}+\frac{t}{2}\right)^2\right) = 0.196 \ m = 196 \ mm$$

Now, to obtain the weight-per-unit length of the beams we need both the density of the materials being used and the cross-sectional areas for the beams.

 * For steel,
 * $$A_{steel} = 2 \cdot b_{steel} \cdot t+(h-2t) \cdot t = 1450 \ mm^2\!$$
 * $$\rho_{steel} = 7.8 \times 10^{-3} \ \frac{g}{mm^3}$$
 * so the mass-per-unit length for steel is:
 * $$m_{steel}=\rho_{steel} \times A_{steel}\!$$
 * $$m_{steel}=\left(7.8 \times 10^{-3} \ \frac{g}{mm^3}\right) \left(1450 \ mm^2\right)\!$$
 * $$m_{steel}=11.31 \ \frac{g}{mm}\!$$

 * For aluminum,
 * $$A_{aluminum} = 2 \cdot b_{aluminum} \cdot t+(h-2t) \cdot t = 2910 \ mm^2\!$$
 * $$\rho_{aluminum} = 2.78 \times 10^{-3} \ \frac{g}{mm^3}$$
 * so the mass-per-unit length for aluminum is:
 * $$m_{aluminum}=\rho_{aluminum} \times A_{aluminum}\!$$
 * $$m_{steel}=\left(2.78 \times 10^{-3} \ \frac{g}{mm^3}\right) \left(2910 \ mm^2\right)\!$$
 * $$m_{aluminum}=8.09 \ \frac{g}{mm}\!$$

The weight-per-unit length of each beams is then
 * $$W_{steel}=110.84 \ \frac {N}{m}$$
 * $$W_{aluminum}=79.28 \ \frac {N}{m}$$

Therefore, the aluminum beam is more efficient weightwise.

(1.4)
Problem statement: Use AS4/3501-6 carbon/epoxy composite to make the I-beam as stated in Problem 1.3. Compare its weight with that of the aluminum beam.

So, repeating the same process as with problem 1.3, we now use the AS4/3501-6 carbon/epoxy composite. Now we use the Young's modulus of this material:

 * $$E_{composite}=140 \ GPa$$

From problem 1.3, we know the expression for the moment of inertia of this I-beam (EQN1 ). This gave us the moment of inertia of the steel beam:

 * $$I_{steel} = 7.612 \times 10^{-6} \ m^4\!$$

From the problem statement $$E_{steel}\cdot I_{steel} = E_{composite}\cdot I_{composite}\!$$, we find $$I_{composite}\!$$ to be
 * $$I_{composite} = \frac{E_{steel} \cdot I_{steel}}{E_{composite}} = \frac{(200 \ GPa)(7.612 \times 10^{-6} \ m^4)}{140 \ GPa} = 1.087 \times 10^{-5} \ m^4$$

Now, rearranging EQN1 to solve for $$b\!$$ > we obtain the following:

 * $$b_{composite} = \frac{1}{2}\left(I_{composite}-\frac{t(h-2t)^3}{12}\right)\Bigg/\left(\frac{t^3}{12}+t\left(\frac{h-2t}{2}+\frac{t}{2}\right)^2\right) = 0.0843 \ m = 84.3 \ mm$$

This will yield the following cross-sectional area:

 * $$A_{composite} = 2 \cdot b_{composite} \cdot t+(h-2t) \cdot t = 1793 \ mm^2\!$$

Now, knowing $$\rho_{composite} = 1.55 \times 10^{-3} \ \frac{g}{mm^3}$$, we can obtain the mass-per-unit length and our weight-per-unit length

 * $$m_{composite}=\rho_{composite} \times A_{composite}\!$$
 * $$m_{composite}=\left(1.55 \times 10^{-3} \ \frac{g}{mm^3}\right) \left(1793 \ mm^2\right)\!$$
 * $$m_{composite}=2.78 \ \frac{g}{mm}\!$$

The weight-per-unit length of the composite beam is
 * $$W_{composite}=27.2 \ \frac {N}{m}$$

Therefore, the composite beam is more efficient weightwise than both the aluminum and the steel beams.

(1.5)
Problem statement: Derive the relations given by equations [1.4] and [1.5] of the textbook.

If we have constant shear $$\tau\!$$ on a curved panel, we can break down the shear force into two components:

 * $$V_x=\tau\,t\,a \qquad \qquad [1.4] \!$$
 * $$V_y=\tau\,t\,b \qquad \qquad \, [1.5] \!$$

From Figure 2, we can find an expression for the differential resulting force $$d\vec V$$:

 * $$d\vec V=q\,\vec {dl}$$

Where $$q\!$$ is the shear flow and $$\vec {dl}$$ > is the differential length that is being studied.

Knowing $$\vec {dl}$$ > has two components (one in $$x\!$$ and another in $$y\!$$), we are able to work out our answer.

 * $$d\vec V=q\,\left( dl_x \, \hat i + dl_y \, \hat j\right)$$
 * $$=q\,\left( dl\, \cos \theta \, \hat i + dl\, \sin \theta \, \hat j\right)$$
 * $$=q\,\left( dx\, \hat i + dy\, \hat j\right)$$

Now, integrating both sides,

 * $$\int {d \vec V}=q\, \left[ \left( \int_{x_a}^{x_b} {dx} \right)\, \hat i + \left( \int_{y_a}^{y_b} {dy} \right)\, \hat j\right]$$

and we also know $$\int_{x_a}^{x_b} {dx}=a$$ and $$\int_{y_a}^{y_b} {dy}=b$$

therefore,

 * $$\vec V=q\, \left( a\, \hat i + b\, \hat j\right)=qa\, \hat i + qb\, \hat j$$

and knowing $$q = \tau \, t \!$$ we now have

 * $$\vec V=\tau t  a\ \hat i + \tau  t  b\ \hat j$$

So this gives us the two components previously stated,

 * $$V_x=\tau\,t\,a \!$$
 * $$V_y=\tau\,t\,b \!$$

Derivation of $$T = 2q\bar{A}$$
As described in previous assignments, the thin walled cross-section is a special case when considering shear flow. As is common in aircraft structures, thin walled designs require great consideration. Since the wall of a structure forms a closed loop, the sum of the shear forces in the assigned $$x$$ and $$y$$ directions is zero, or $$\sum F_{x} = 0$$ and $$\sum F_{y} = 0$$. However, a shear force produces a resultant torque.

Referring to Figure 3, where the flow of the thin wall denotes a shear flow "q", the torque is described as $$dT = \rho\,qdl$$ where $$\rho$$ is the is the distance from the assigned origin to $$dl$$. Since $$\rho\,dl = 2dA$$, integrating the equation along a closed loop yields equation 3.48.

(3.48)
$$T = \oint \rho\,qdl = \iint_{\bar{A}}2qdA = 2q\bar{A}$$

Where $$\bar{A}$$ is the area enclosed by the shear flow. The torque may also be decomposed into a resultant force $$R$$and a moment arm $$e$$ as described in Mechanics of Aircraft Structures by C.T. Sun.

Thus $$R = qd$$ where $$d$$ is the total straight distance that the shear flow covers. This can be broken down into components in the $$x$$ and $$y$$ directions as illustrated by the figure.

(3.49)
$$F_{x} = qb\!$$

$$F_{y} = qh\!$$

Applying equation 3.48 to this geometry and noting $$\bar{A}$$ is the shaded area, then the location of the moment arm $$e$$ can be obtained.

$$Re = T = 2\bar{A}q$$

-

Area of a Generic Triangle

The area for a rectangle is it's base times it's height:
 * For Figure 5 the area of the rectangle outlined by the purple, red, and blue lines is: $$a \times h \!$$

If you insert a diagonal into the rectangle you will create two triangles with half the area as the rectangle
 * For Figure 5 the area of the triangle outlined by the blue and red lines is: $$\frac{1}{2}a \times h \!$$

Now, to calculate the area of the blue section you can use the above area we calculated for the triangle and subtract the area of the triangle that is outlined by the red lines as follows:
 * Triangle 1:$$\frac{1}{2}a \times h = \frac{ah}{2}\!$$
 * Triangle 2:$$\frac{1}{2}(a-b) \times h = \frac{ah}{2} - \frac{bh}{2}\!$$
 * Subtract Triangle 2 from Triangle 1
 * $$\frac{ah}{2} - \left(\frac{ah}{2} - \frac{bh}{2}\right) = \frac{1}{2}bh\!$$

This proves that the area for the blue section is half of it's base times it's height.

Prove $$J = \frac {1}{2}\,\pi a^{4}$$

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

 * $$dA = 2\pi r \,dr\!$$

 * $$J = \int_{0}^{a}{2\pi r^{3}\,dr}$$

 * $$J = \left[\frac{2\pi r^{4}}{4} \right]_{r=0}^{r=a} $$

which yields...
 * $$J = \frac{\pi a^{4}}{2}$$

Thin-Walled Circular Cross-Section

 * $$r_{i} = a \!$$

 * $$r_{o} = b \!$$

Since t (the thickness) is very small, then

 * $$b^{2} = \bar{r}^{2} $$

 * $$ a^{2} = \bar{r}^{2}$$

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

From the proof above for the area moment of inertia of a circular cross section, it can be noted that a similar derivation can be followed to obtain the area moment of inertia for a thin-walled circular cross section. One would simply change the integration boundaries from 0 to a, to a to b. This will result in the following equation.
 * $$J = \frac{\pi (b^{4} - a^{4})}{2}$$

This equation can be expanded to yield the following
 * $$J = \frac{1}{2}\pi (b-a)(b+a)(b^{2}+a^{2})\!$$

where
 * $$(b-a) = t\,;\ (b+a) = 2\,\bar{r}\,;\ b^{2}+a^{2}= 2\,\bar{r}^{2}$$

Therefore
 * $$J = \pi t\bar{r}^3$$

for a thin walled circular cross section

Comparison of Circular Cross-Sections with Thin Walled Circular Cross-Sections from Fig 1.8 (a) and (b)
To illustrate the effects of optimizing the cross-section of a beam, take the comparison of the torque that the thin walled circular cross-section can withstand, even though it will have the same cross-sectional area as the solid tube. $$T = GJ\Theta$$

Thus for a given material, variations in the torque on the cross-section are directly related to the geometry or polar moment $$J$$.

An approximation of the polar moment for a thin walled cross-section is $$J = 2t\bar{r}^3$$ where $$\bar{r} = (a + b)/2$$. The solid circular cross-section has the polar moment $$J = \frac{1}{2}\pi\,r^{4}$$. Thus the polar moments are as follows:

$$J_{thin wall} = 2*0.1*\pi\,*5.05^{3} = 80.91 cm^{4}$$

$$J_{solid} = \frac{1}{2}\pi\,(1^{4}) = 1.57 cm^{4}$$

Thus the thin walled cross-section containing the same area as the solid circle has a polar moment and torsional stiffness fifty times greater. This is due to the fact that as a beam is put under torsion, the torque is proportional to the deflection θ which is greater at the outer most portion of the cross-section. In other words, the material near the core is underutilized.

Torsion of a Cylinder Without Warping
We begin by describing the torque of a cross-section of the cylinder in order to eventually relate twist angle θ to torque.

$$T = \iint_{A} r\tau\,dA$$ Where $$\tau = G\gamma$$ by Hooke's Law. Continuing, $$\gamma = \frac{rd\alpha}{dx}$$ where $$\frac{d\alpha}{dx} = \Theta$$ by definition. Thus the rate of twist "θ" can now be input to the torque equation as follows: $$T = \iint_{A} rG(r\Theta)dydx$$ Where G and θ are independent of y and z.

Rewriting, $$T = G\Theta(\iint_{A}r^{2}da)$$ where the term in parenthesis is the second polar area moment of inertia J described earlier.

Warping
Warping is the axial displacement along the x-axis. Shafts exhibit warping under pure torque. There are out of plane displacements because of it. The rate of twist is defined as
 * $$\theta = \frac{\alpha }{x}$$

The in plane displacements are defined as
 * $$u_{y} = -\theta xz\!$$ (equation 3.11 in Mechanics of Aircraft Structures)
 * $$u_{z} = \theta xy\!$$ (equation 3.12 in Mechanics of Aircraft Structures)

The warping displacement along the x=axis is defined as
 * $$u_{x} = -\theta \psi\!$$ (equation 3.13 in Mechanics of Aircraft Structures)

Roadmap for torsional analysis of wing

A. Kinematic assumption (3.2)

B. Strain-displacement relationship (3.2)

C. Equilibrium equations for stresses (chapter 2, 3.2)

D. Prandt'l stress function 'φ' (3.2, 3.15)

E. Strain compatibility equation (3.17)

F. Equation for 'φ' (3.19)

G. Boundary condition for 'φ' (3.24)

H. $$T = 2\iint_A \phi\,dA $$(3.25)

 * Where $$T = GJ\Theta\!$$ and $$J = \frac{-4}{\triangledown^{2}\phi}\iint_A \phi\,dA$$ when warping is taken into consideration.

I. Thin walled cross-section with the ad-hoc assumption on shear flow

J. Twist angle θ: Method 1

 * $$\Theta = \frac{1}{2G\bar{A}}\oint\frac{q}{t}ds$$

K. Section 3.6 on multi-celled thin walled cross-section cells

 * $$T = 2\sum_{i=1}^{n_{c}}q_{i}\bar{A}_{i}$$

 * Since $$T_{i} = 2q_{i}\bar{A}_{i}$$ so $$t = \sum_{i=1}^{n_{c}}T_{i}$$

Problem 3.10 adapted for wing: compute $$\bar A=\frac{\pi}{2}\left(\frac{b}{2}\right)^2 + \frac{ba}{2}$$
To calculate the area for Figure 10 split the figure into a triangle and a semi circle. Calculating the areas for these two shapes and then adding them will result in the total area of the figure. Area is calculated as follows:
 * Semicircle Area: $$\frac{\pi \left(\frac{b}{2}\right)^2}{2} = \frac{\pi}{2} m^2$$
 * Triangle Area: $$\frac{1}{2} ba = \left(\frac{1}{2}\right)(2)(4) = 4 m^2$$

Adding the two areas together results in a total area of:
 * $$\bar A = \left(\frac{\pi}{2} + 4\right) m^2$$

Next is the calculation of the twist angle of the airfoil. Assuming this airfoil is rigid leads to the constraint that the twist angle for one section of the airfoil will be equal to the twist angle for another section of the airfoil.
 * In other words: $$\theta = \theta_1 = \theta_2 \!$$

However, warping is experienced out from the plane of the airfoil. An example of this would be an airplane wing in torsion. At any one point there may be a twist angle between the wing tip and wing root; however, any section of the wing (aka. airfoil) will maintain its shape.

The equation for twist angle is given in the road map, part J. This generic equation can be substituted by the Riemann Sum equation as discussed in the section Comparison of Sigma and Integral Symbols (Quadrature/ Riemann Sum). Using this equation we can calculate the twist angle for the airfoil shape in Figure 7 as follows:
 * $$\Theta = \frac{1}{2G\bar{A}} \sum_{j=1}^{n_{c}} \frac{q_{j}l_{j}}{t_{j}} = \frac{1}{2G\bar{A}}\left(\frac{q_{1}l_{1}}{t_{1}} + \frac{q_{2}l_{2}}{t_{2}} + \frac{q_{3}l_{3}}{t_{3}}\right)$$ Factor out q since $$q_1=q_2=q_3\!$$
 * $$\Theta = \frac{q}{2G\bar{A}}\left(\frac{l_{1}}{t_{1}} + \frac{l_{2}}{t_{2}} + \frac{l_{3}}{t_{3}}\right) = \frac{q}{2G\bar{A}}\left(\frac{\frac{\pi}{2}b}{t_1} + \frac{a}{t_2} + \frac{\sqrt{a^2+b^2}}{t_3}\right)$$ Plug in values for b, a , $$\bar A$$, and $$q = \frac{2T}{\bar A}$$
 * $$\Theta = \frac{2T}{2G\left(\frac{\pi}{2}+4\right)^2}\left(\frac{\frac{\pi}{2}(2)}{0.008} + \frac{4}{0.01} + \frac{\sqrt{4^2+2^2}}{0.01}\right) = 39.95\frac{T}{G}$$
 * Where G is the shear modulus and is a material constant and T is the Torque and is a variable.

Determining Internal Area and Centroid of NACA 4-digit Series Airfoil
The NACA 4-digit series airfoil is a family of Airfoils whose shape are governed by the chord length. The four digits define certain characteristics of the wing as shown below:

For a NACA 2412 Airfoil
 * 2 $$\longrightarrow$$ maxmimum camber in percent chord length
 * 4 $$\longrightarrow$$ position of maximum camber in tenths chord length
 * 12 $$\longrightarrow$$ maximum airfoil thickness in percent chord length

More detail on determining the exact shape of a NACA 4-digit series airfoil is available here.

An elegant solution to finding the area of an airfoil is through triangle quadrature. Once an arbitrary point is defined, triangle radiate out of the point and touch the outer surfaces of the airfoil, refer to Figure 11.

The following MATLab code computes the cross-sectional area and centroid of any NACA 4-digit airfoil and plots the shape. Some of the most frequently used are graphed below.

Comparison of Sigma and Integral Symbols (Quadrature/ Riemann Sum)
$$\int$$ is an elongated S, denoting a continuous summation. $$\sum{}$$ denotes a discrete sum.

An example of the summation application would be the twist contributions to a polygon cell of a structure which need not be integrated with calculus.

$$\Theta = \frac{1}{2G\bar{A}} \sum_{j=1}^{n_{c}} \frac{q_{j}l_{j}}{t_{j}}$$

As described by professor Vu-quoc in class, the origin of the summation technique or Reimann Sum dates back to ancient civilizations and the need to estimate land masses for the purpose of property division and agricultural divisions. Any land form may be sub-divided into rectangular sections and then summed, knowing the finer the subsections are, the more accurate the area estimation.

The term for this sub-division system is quadrature. If taken to the third dimension approximation, the term cubiture describes the same principle.

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Contributing Team Members
The following students contributed to this report:

Marlana Behnke Eas4200c.f08.aero6.behnke 20:10, 7 October 2008 (UTC)

Kris Loper Eas4200c.f08.aero6.loper 23:12, 7 October 2008 (UTC)

Matt Inman Eas4200c.f08.aero6.inman 13:30, 8 October 2008 (UTC)

Felipe OrtegaEas4200c.f08.aero6.ortega 16:28, 8 October 2008 (UTC)

Eduardo Villalba Eas4200c.f08.aero6.villalba 17:41, 8 October 2008 (UTC)

Scott Chastain Eas4200c.f08.aero6.chastain 19:58, 8 October 2008 (UTC)

title: User:Eas4200c.f08.aero6.ortega/HW3


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User talk:Jolie

this has all been moved to the account JoliePA
user: JoliePA

--JoliePA 19:02, 13 November 2008 (UTC)

--Jolie 19:03, 13 November 2008 (UTC)

title: User talk:Jolie


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Electrical Engineering Fundamentals/Basics of Electricity

Review

 * Electric current is caused by the flow of electric charge.
 * Two types of materials in the world of electricity are:
 * Conductors - allow the flow of charge.
 * Insulators - stop the flow of charge.
 * Power stations of various sorts provide electricity.
 * Things that "use" electricity are called a "Load" on the circuit.
 * The flow of current is always from positive to negative.
 * Current flow can be either Direct (from positive to negative) or Alternating (Pos/Neg/Pos/Neg) usually 60 cycles per second.
 * Electricity is measured in Volts, Amperes (Amps), and Watts.
 * Electricity can be deadly but it is necessary in modern life.

Electrons & Holes
Electricity comes from the movement of charge. Most commonly, this is the movement of electrons from atom to atom. When an atom gains an electron, it becomes more negatively charged. The atom then pushes away other electrons, similar to the way like poles of a magnet repel one another. However, conductors don't hold onto that newly acquired electron very well. Think of a row of atoms. When one atom in the line becomes positively charged (because it lost its electron), then another electron is drawn to that atom. This same atom gains this other electron from one of the other atoms in the conductor, making that atom a prime candidate for another electron. The process continues like a row of dominoes.

When an electron moves, it leaves a "hole". This is not an actual physical entity but rather a space-holder that becomes useful in much more advanced electronics. In a piece of conductive material that does not have anything connected to it, the process of switching electrons is random and does not cause anything to happen. When there is a voltage difference applied across a chunk of material, electrons will move to the positively charged side of the material, filling holes but also leaving holes behind. This constant process of holes being created and filled causes more electrons to move and creates a current. The current is much like a stream and continues to flow until the voltage difference is zero.

As described, the electrons are traveling toward the positive end. We define "conventional current" as going the opposite way from the flow of electrons. The flow of electrons is negative, because they are negatively charged particles moving toward a positively charged area. We can think of the holes as flowing in the opposite direction, however this is just a mental visualization to help describe behaviors of materials in more advanced electronics. For now, think of electrons flowing from the negative terminal of a voltage source to the positive terminal while the holes that constitute what we think of as "conventional current" flow from the positive terminal of the voltage source to the negative terminal.

Insulators are simple. They are stable atoms of elements not prone to losing their electrons. This means they hardly ever share electrons, except in extreme conditions. Electrical equipment is always designed to operate within a particular temperature range, and this range varies upon the instrument's design. Semi-conductors and super-conductors also exist, but those have characteristics that we don't need to cover at the moment.

What You Need to Know

 * Electricity flows along conductors.
 * Electrons flow from negative terminal to positive terminal.
 * Holes flow from positive terminal to negative terminal. (Conventional Flow)
 * Current, or flow of charge, is stopped by insulators.
 * There are such things as semi-conductors and super-conductors.

The River
Imagine a nice mountain stream. What does it do? It flows downhill until it comes to an area of equilibrium. When the water is at the top of the stream, it has the potential to fall down. If we took away the mountain, the water would fall straight down. We know from physics that the water has a potential energy. The equation for potential energy (PE=mgh, m=mass, g=gravity acceleration, h=height) tells us that the potential energy of this water at the top of the mountain increases with greater mass of water. When thinking of electricity, these ideas of 'potential energy' and 'flow' are similar to what we've discussed about water, but instead, Electrical Engineers deal with a 'potential voltage' applied that causes electrons to 'flow'. In other words, higher amounts of potential energy in water are analogous to higher concentrations of stored charge (therefore higher potential voltage) in a material. The flow of electrons in a conductor can be thought of as the flow of water molecules through a pipe. The electromotive force, EMF, or simply Voltage Potential (V) in the field of Electrical Engineering is analogous to the potential energy which 'motivates' the water to flow downhill to a lower 'potential'.

When the water is flowing, it has a current, and so when electricity flows it also has a current. When the water flows faster, the current is higher, and the faster electrons flow by, the higher electrical current. The electrical term is Amperes which is often shortened to Amps, and notated as A.

In a mountain stream, if you cut a straight channel across a bend, the water will soon only follow the cut. When considering electricity, we call this a 'short circuit'. A short circuit can be very bad if you are part of this new shorter path to relieve the potential. So, always be careful when dealing with electricity! The dangerous part is the current, a static shock between a finger and doorknob has several hundred-thousand volts of potential. A wall socket has 120 V in the US, but the amperage (A) is more than enough to kill.

What You Need to Know

 * Concentrations of charge (electrons or holes) results in a potential voltage (V).
 * The movement of charge is known as current, measured in Amperes (A).
 * Electricity passing through a body can be dangerous, even in small amounts.
 * Be careful with visualizations, electricity can be modeled by several things, but a misconception of one may influence how you learn about electricity.
 * It is important to move beyond these visualizations after some practice with electrical circuits, because electrons have much more strange behavior in special situations that cannot be described by simple analogies.

The Power of Electricity
In the physics classroom, a very common practice problem involves solving for how much power is needed/expended for/during a certain task. The same is true for electricity. Watts are a measure of power, the same as horsepower, but the question never arises as to how much horsepower your sub-woofer is. Usually, the only time horsepower is mentioned when dealing with electricity is when the electrical power is being converted by a motor into mechanical power. Don't worry about equations yet; just know that the power in an electrical system is dependent upon the voltage potential (hereafter simply voltage) and current (ie. flow of charge), amps (A). ..... The power of a light bulb, for example, is given in Watts as an indication of how much current it will use at the rated voltage and how much light it will provide if it is switched on. Light bulbs also are rated in Volts, meaning that they are supposed to be used in a system with that approximate voltage. If the voltage is lower than rated then that light bulb will provide less light if is switched on, and if the voltage is higher then damage to the light bulb can be expected.

Return:Topic:Electrical Engineering

Bgorges 20:50, 6 October 2008 (UTC) William Leismer 22:05, 16 June 2015 (UTC)


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User:Eml4500.f08.delta 6.guzman/Lecture Notes 09/26

Similarly: $$\begin{Bmatrix}P_{1}^{(e)}\\P_{2}^{(e)}\end{Bmatrix}_{2x1}= T^{(e)}\begin{Bmatrix} f_{1}^{(e)} \\ f_{2}^{(e)} \\ f_{3}^{(e)} \\ f_{4}^{(e)} \end{Bmatrix}_{4x4} => P^{(e)}=T^{(e)}*F^{(e)}$$ $$\hat{K}^{(e)}_{2x2}*q^{(e)}_{2x1} = P^{(e)}_{2x1}$$ where $$ q^{(e)}_{2x1}=T^{(e)}*d^{(e)}$$ and $$ P^{(e)}_{2x1}= T^{(e)}*F^{(e)} $$ therefore $$\Rightarrow \hat{K}^{(e)}_{2x2}*T^{(e)}*d^{(e)}= T^{(e)}*F^{(e)}$$

The goal with these matrices is to obtain  $$K^{(e)}*d^{(e)}=F^{(e)}$$  in order to do that we need to move   $$T^{(e)}$$  from the right hand side to the left hand side and this usually can be done by multiplying by the inverse   $$T^{(e)-1}$$  but since   $$T^{(e)}$$  is a rectangular matrix with (2x4) dimensions it can not be inverted therefore what needs to be done is the transpose $$T^{(e)^(T)}$$

$$[T^{(e)^(T)}_{4x2}*\hat{K}^{(e)}_{2x2}*T^{(e)}_{2x4}]$$$$ * d^{(e)}_{4x1}$$ = $$F^{(e)}_{4x1}$$ where $$K^{(e)}_{4x4}=[T^{(e)^(T)}_{4x2}*\hat{K}^{(e)}_{2x2}*T^{(e)}_{2x4}]\Rightarrow $$ $$K^{(e)}*d^{(e)}=F^{(e)}\Rightarrow K^{(e)}= T^{(e)^(T)} * \hat{K}^{(e)} * T^{(e)}$$

By the principle of virtual work (PVW) the reduction of the global force displacement relation is $$\Rightarrow$$

$$K_{6x6}*d_{6x1}=F_{6x1}\Rightarrow K_{2x2}*d_{2x1}=F_{2x1}$$

Then why not solve these matrices at this stage as follows   $$d=K^{-1}*F$$   ? The reason why this can not be done is due to the matrix singularity of  $$ K $$   Since the determinant of matrix   $$ K $$   is equal to zero its inverse will be $$ \frac{1}{det [K]} $$ which will be undefined since the denominator is equal to zero.


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User:Eas4200c.f08.aeris.guan/929lecture

NACA Airfoil and Torsion of Uniform, Non-circular Bars
In homework 3, we are examining the NACA 4 digit airfoil series by using Matlab to do the analysis. Below are some pictures of some examples of the NACA airfoil. The one to the left describes fluid flow around the airfoil cross section (source: rit.edu) and the one to the right is a simple and crude cross sectional representation of the NACA airfoil with a (y,z) coordinate system where the y axis passes through the trailing edge. The y-axis is discretized in "ns" segments. The red circle highlights the infinitesimally small segment that will be analyzed.

Here is the picture of the enlarged segment. As a crucial goal while analyzing the NACA airfoil is to determine the "average" area, we must devise a method to do this. It is possible to use a summation of infinitesimally small trapezoids, however, this method is not very elegant. In turn, we will use a method that utilizes triangles to determine the quadrature of the airfoil. The triangle will be a cross product a vector from an observation point (can be positioned anywhere) to a point on the airfoil and a vector of infinitesimally small length along the airfoil. These two vectors enclose a triangle of infinitesimally small area dA. This is shown in the picture below.  Torsion of uniform, non-circular bars are also examined in homework 3. This is the case in which warping will occur as opposed to the case of uniform, circular bars. Warping is the axial displacement along the x-axis of the bar (along the length of the bar) of a point on the deformed (rotated) cross section. The below pictures describe the bar and the bar's cross section when torsion is applied. Point P is a point on the (y,z) plane which moves to P' as the torque is applied. Beta describes the initial angular displacement measured from the y-axis and alpha is the angular displacement due to the torsion. Alpha is also measured from the y-axis.  Below is the magnified version of point P and its movement and displacement on the (y,z) plane of the cross section. $$u_y$$ and $$u_z$$ are the components of the displacement vector $$\vec {PP'}$$.  Why is the vector PP' perpendicular to OP? Isn't it more intuitive that one would draw an arc with radius |OP| and plot point P' at an angle alpha away from vector OP? The perpendicular nature of PP' to OP is because the angle alpha is very small. The proof is below. Let there be two vectors OP and OP' of equal length (|OP|=|OP'|=R). The displacement differences in the points P and P' are denoted by $$u_z$$and $$u_y$$ where z denotes vertical displacement and y denotes horizontal displacement.

$$\displaystyle Rsin\alpha \cong R\alpha (\alpha small)$$ $$\displaystyle R(1-cos\alpha) \cong 0 (first order)$$ Thus, it is easy to see that since the horizontal displacement of point P' from point P is zero, point P' directly above point P in this case. That makes vectors OP and PP' perpendicular.

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text: Category:Reproductive medicine

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Principles of Marketing

According to the American Marketing Association, "Marketing is the activity, set of institutions, and processes for creating, communicating, delivering, and exchanging offerings that have value for customers, clients, partners, and society at large". Based on this definition, there are four components of marketing: creating, communicating, delivering, and exchanging.

This course introduces basic marketing concepts including strategic planning, consumer behavior, segmentation, supply chains, professional selling, customer satisfaction, and the components included in a marketing plan.

Course Outcomes
Participants will

 * 1) Define and demonstrate an understanding of marketing concepts and strategies
 * 2) Apply necessary marketing skills across the areas of marketing to develop a competitive advantage and build long term customer relationships
 * 3) Describe target market strategies
 * 4) Analyze marketing concepts, specifically product, price, place, and promotion
 * 5) Assess ethical implications related to marketing strategies
 * 6) Create a marketing plan

Lessons

 * 1) /Define Marketing/
 * 2) Strategic Planning
 * 3) Consumer Behavior
 * 4) Segmentation, Targeting, and Positioning (STP)
 * 5) Product Creation and Branding
 * 6) Marketing Channels
 * 7) Supply Chains
 * 8) Marketing Research
 * 9) Integrated Marketing Communications (IMC)
 * 10) Public Relations and Social Media
 * 11) Professional Selling
 * 12) Customer Satisfaction
 * 13) The Marketing Plan

title: Principles of Marketing


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User:Stevenarntson/English Composition

The class is an eclectic survey that touches on largely creative, and some research-oriented writing concerns. It follows a scheme necessitated by the venue of my physical classroom in Seattle, Washington.

assignments and policies

schedule

 * 1) introductions
 * 2) /zines/ and book arts. statement of intent due
 * 3) workshops and research
 * 4) workshops and vispo
 * 5) workshops and metered poetry
 * 6) workshops and /lengths/. midterm assessment due
 * 7) workshops and /collage/
 * 8) workshops and /collaborative forms/
 * 9) workshops and /resources/ for further study. sign up for individual meetings
 * 10) individual meetings
 * 11) final reading. self assessment due on your userpage

title: User:Stevenarntson/English Composition


text:
User:Eml4500.f08.ateam.carr/HW3

Continuation of Three Bar Truss System

Recall the equations used to solve static problems:

\sum{F_x=0} $$

\sum{F_y=0} $$

\sum{M_z=0} $$

Summing the moments about point A is trivial as all members pass through the point. This action will give us no new information about the system. Therefore, there are 2 equations and 4 unknown forces. This problem is statically indeterminate.

Further exploration of Element Coordinate Systems

The three diagrams above describe the element coordinate systems for the Three Bar Truss System.

Moment About Point B
If the sum of the moments are computed about point B, will the sum still be zero?

 * $$\overrightarrow{BA'}=\overrightarrow{BA} + \overrightarrow{AA'}$$

3D- Explanation:

 * $$\sum \overrightarrow{M_B}={\overrightarrow{BA}} \times {\overrightarrow{F}} ={\overrightarrow{BA'}} \times {\overrightarrow{F}}$$

For all A' on line of action of the vector F.

 * $$\sum \overrightarrow{M_B}=({\overrightarrow{BA}} + {\overrightarrow{AA'}}) \times {\overrightarrow{F}}$$
 * $$\sum \overrightarrow{M_B}={\overrightarrow{BA}} \times {\overrightarrow{F}} + {\overrightarrow{AA'}} \times {\overrightarrow{F}} $$

Back to Three Bar Truss
Point A is in equilibrium: $$ \sum_{i=0}^{3}{\overrightarrow {Fi}}=\overrightarrow O $$

Note that the above equation is the same as the two scalar equations summing forces in the X and Y directions.

 * $$ \sum_{i}\overrightarrow {M}_{B,i}=\sum_{i}\overrightarrow {BA'_i} \times \overrightarrow{F_i} $$

 * $$A'_{i}=$$ any point on line of action of $$\overrightarrow{F_i}$$

 * $$ \sum_{i}\overrightarrow {M}_{B,i}=\sum_{i}\overrightarrow {BA} \times \overrightarrow{F_i} = \overrightarrow {BA}\times \sum_{i}\overrightarrow {F_i}$$

Since $$\sum_{i}\overrightarrow {F_i}=0$$, the sum of the moments about point B is zero.

 * $$ \sum_{i}\overrightarrow {M}_{B,i}=\sum_{i}\overrightarrow {BA} \times \overrightarrow{O} $$

The Three Bar Truss problem has 8 DOF (8 x 8 Matrix):

$$k_3$$ is in total a 4 x 4 matrix.

$$k_{33}=k^{(1)}_{33}+k^{(2)}_{11}+k^{(3)}_{11}$$

$$k_{34}=k^{(1)}_{34}+k^{(2)}_{12}+k^{(3)}_{12}$$

$$k^{(1)}_{43} + k^{(2)}_{21} + k^{(3)}_{21}$$

$$k^{(1)}_{44} + k^{(2)}_{22} + k^{(3)}_{22}$$

$$

K_{global}= \left [\begin{array}{cccccccc} k^{(1)}_{11} & k^{(1)}_{12} & k^{(1)}_{13} & k^{(1)}_{14} & 0 & 0 & 0 & 0 \\ k^{(1)}_{21} & k^{(1)}_{22} & k^{(1)}_{23} & k^{(1)}_{24} & 0 & 0 & 0 & 0\\ k^{(1)}_{31} & k^{(1)}_{32} & (k^{(1)}_{33} + k^{(2)}_{11} + k^{(3)}_{11})& (k^{(1)}_{34} + k^{(2)}_{12}+k^{(3)}_{12})& k^{(2)}_{13} & k^{(2)}_{14} & k^{(3)}_{13} & k^{(3)}_{14} \\ k^{(1)}_{41} & k^{(1)}_{42} & (k^{(1)}_{43} + k^{(2)}_{21} + k^{(3)}_{21}) & (k^{(1)}_{44} + k^{(2)}_{22} + k^{(3)}_{22}) & k^{(2)}_{23}) & k^{(2)}_{24} & k^{(3)}_{23} & k^{(3)}_{24} \\ 0 & 0 & k^{(2)}_{31} & k^{(2)}_{32} & k^{(2)}_{33} & k^{(2)}_{34} & 0 & 0 \\ 0 & 0 & k^{(2)}_{41} & k^{(2)}_{42} & k^{(2)}_{43} & k^{(2)}_{44} & 0 & 0 \\ 0 & 0 & k^{(3)}_{31} & k^{(3)}_{32} & 0 & 0 & k^{(3)}_{33} & k^{(3)}_{34} \\ 0 & 0 & k^{(3)}_{41} & k^{(3)}_{42} & 0 & 0 & k^{(3)}_{43} & k^{(3)}_{44}\end{array} \right] $$

title: User:Eml4500.f08.ateam.carr/HW3


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User:Eml4500.f08.delta 6.guzman/Computation of Eigenvalues of Matrix K

The Eigenvalue and Stiffness Matrix K Relationship
$$K= \ \begin{bmatrix} K_{11}^{(1)} & K_{12}^{(1)} & K_{13}^{(1)} & K_{14}^{(1)} & 0 & 0  \\ K_{21}^{(1)}& K_{22}^{(1)} & K_{23}^{(1)} & K_{24}^{(1)} & 0 & 0 \\ K_{31}^{(1)} & K_{32}^{(1)} & (K_{33}^{(1)} + K_{11}^{(2)}) & (K_{34}^{(1)} + K_{12}^{(2)}) & K_{13}^{(2)} & K_{14}^{(2)} \\ K_{41}^{(1)}& K_{42}^{(1)} & (K_{43}^{(1)} + K_{21}^{(2)}) & (K_{44}^{(1)} + K_{22}^{(2)}) & K_{23}^{(2)} & K_{24}^{(2)} \\ 0 & 0 & K_{31}^{(2)} & K_{32}^{(2)} & K_{33}^{(2)} & K_{34}^{(2)} \\ 0 & 0 & K_{41}^{(2)} & K_{42}^{(2)} & K_{43}^{(2)} & K_{44}^{(2)}  \end{bmatrix}_{6X6}\,$$ $$\ \ = \begin{bmatrix}  \frac{9}{16}& \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & 0 & 0 \\  \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & 0 & 0 \\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & (\frac{9}{16} + \frac{5}{2}) & (\frac{3\sqrt{3}}{16} + \frac{5}{2}) & -\frac{5}{2} & -\frac{5}{2} \\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & (\frac{3\sqrt{3}}{16} + \frac{5}{2}) & (\frac{3}{16} + \frac{5}{2}) & -\frac{5}{2} & -\frac{5}{2} \\  0 & 0 & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} \\ 0 & 0 & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2}  \end{bmatrix}_{6X6}\,$$

For an unconstrained structure system there are 3 possible rigid body motions in 2-D (2 in translation plus 1 in rotation) The eigenvalues of the stiffness matrix are really important for design because they are related to the vibrations of the structure. A zero eigenvalue is related to a rigid body movement since it lacks the strain energy according to the equation  $$K*v=\lambda *M*v $$  where $$K$$ is the stiffness matrix, λ is the eigenvalue related to vibrational frequency, and $$M$$ is the mass matrix. A zero eigenvalue means there is no elastic energy therefore there is no energy stored in the structure.


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Wikiversity talk:Colloquium/What should we do about JWSchmidt?

What is the perma link that this was forked from so edit histories can be seen more clearly? Emesee 00:57, 7 October 2008 (UTC)


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User:Stevenarntson/English Composition/Research

Please visit the following research pages:

 * Why It's A Good Idea to Evaluate Web Resources by Susan E. Beck
 * Seattle Literacy Resources
 * Online Research
 * Ten Things You Should Know about the Seattle Public Library


text:
Enterprise Resource Planning/Project Apprenticeship

Programme
This is a module part of the Open Source ERP/Executive Masters course conducted by a private university.

Subject
ERP Project Apprenticeship - Worldwide Exposure "There is a need as well as an opportunity to jump on a growing bandwagon of online resources of both guidance and mentoring as well as business leads and employment. In a robust and dynamic community project such as ADempiere, a newbie can grow into it easily."

Objectives
The objectives of this subject are:
 * To work hands-on real global projects on a remote basis as apprentices
 * To network with diverse community contexts and cultures
 * To be accepted member of the Project Site and Resources

Learning Outcomes
On completion of this subject the students shall be able to:
 * Start working with community and commercial projects worldwide
 * Assist in the many tasks leading up to executing on their own
 * Use the project tools to communicate and manage project milestones
 * Brand themselves as contributor material

Sypnosis
This subject is the hands-on practical approach part to the real challenge of ERP Projects both locally and out there. Open Source allows a great opportunity as well as new dimension to manage and deliver services via the web and community. The length and breadth of true ERP is very daunting to beginners and having an apprentice program in a real community project allows at least a handy reference on where to look for things and start off.

Subject Details
Here the student can select different tracks of focus:
 * Documentation works which involves alot of room to contribute such as improving existing or reorganising them.
 * Bug testing and fixing those project ones particularly those in priority 9.
 * Community champion to assist other newbies in chatroom and forums.
 * Overall managing and promoting the project.

Project Track Focus
(20hrs)
 * Selecting the track and mentor
 * Select the project and tasks
 * Getting on-board knowing the client
 * Establishing working mode and scope
 * Documenting progress

Role and Scope
(20hrs)
 * Debugger
 * Tester
 * Documenter
 * Project Manager
 * Saviour
 * Anti-Patterns

Debugging
(20 hrs)
 * Bugs Selection Strategy
 * Establishing A New Bug
 * Commiting the solution
 * Tracker Documentation

Tester
(20 hrs)
 * Selecting Areas of Concern
 * Documenting the Test and Results
 * Documenting the Subject Matter

Documentation
(20 hrs)
 * Making Static Content
 * Making Audio-Visual Content
 * Editing Resources
 * Adding Sample Data and Help Screen

Netiquette
(20 hrs)
 * Web Culture
 * Global Cultural Norms
 * Etiquette
 * Discussion Handling
 * Flames and Poison

Reference Materials

 * Posionous Behaviour On Open Source Projects


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User:Eml4500.f08.ateam.rivero/w3

Homework Report 3: Monday, September 29
An alternative way of solving for the reactions is by using the Global Force Displacement relation. Since one knows the values for the displacement column matrix ( d : 6x1) and the global stiffness matrix ( K : 6x6). One can solve for the force column matrix ( F : 6x1).

\begin{bmatrix} &  &   &   &   &  & \\         &   &   &   &   &  & \\         &   &   &   K   &  & \\ &  &   &   &   &  & \\         &   &   &   &   &  & \\         &   &   &   &   &  & \\     \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} $$

Example Calculation:

\begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3248 & 0.00 & 0.00\\   0.3248 & 0.1875 & -0.3248 & -0.1875 & 0.00 & 0.00\\    -0.5625 & -0.3248 & 3.0625 & -2.1752 & 2.5 & -2.5\\    -0.3248 & -0.1875 & -1.9375 & 2.6875 & -2.5 & 2.5\\    0.00 & 0.00 & 2.5 & -2.5 & 2.5 & -2.5\\    0.00 & 0.00 & -2.5 & 2.5 & -2.5 & 2.5\\ \end{bmatrix} \cdot \begin{bmatrix} 0\\      0\\      4.352\\      6.127\\       0\\       0\\     \end{bmatrix} =      \begin{bmatrix} F_1\\ F_2\\ F_3\\ F_4\\ F_5\\ F_6\\ \end{bmatrix} $$

The form above can be reduced to the form below :

\begin{bmatrix} -0.5625 & -0.3248\\    -0.3248 & -0.1875\\       3.0625& -2.1752\\     -1.9375 &  2.6875\\      -2.5   &   2.5  \\      -2.5   &   2.5  \\ \end{bmatrix} \cdot \begin{bmatrix} 4.352\\     6.127\\     \end{bmatrix} =      \begin{bmatrix} F_1\\ F_2\\ F_3\\ F_4\\ F_5\\ F_6\\ \end{bmatrix} $$

How to solve in MATLAB:

>> K = [-0.5625,-0.32476;-0.32476,-0.1875;3.0625,-2.1752;-2.1752,2.6875;-2.5,2.5;2.5,-2.5]

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

>> d = [4.352;6.1271]

d =

4.3520   6.1271

>> F = K*d

F =

-4.4378  -2.5622    0.0003    7.0001    4.4377   -4.4377

Note: F3 and F4 return the given input values for P. Therefore, to make computations easier there is no      need to compute rows 3, 4 since row 3 and 4 will return the values for F3 and F4.

Results Comparison:

Method 1: Using the element force displacement relationship to calculate the force reactions.
 * F 1 = -4.4378
 * F 2 = -2.5622
 * F 5 = 4.4378
 * F 6 = -4.4378

Method 2: Using the global force displacement relationship to calculate the force reactions.
 * F 1 = -4.4378
 * F 2 = -2.5622
 * F 5 = 4.4377
 * F 6 = -4.4377

Method 3: Using statics to calculate the force reactions.
 * F 1 = -5.124cos(30o)= -4.4375
 * F 2 = -5.124sin(30o)= -2.5622
 * F 5 = 6.276cos(-45o)= 4.4378
 * F 6 = 6.276sin(-45o)= -4.4378

The following image to the left is a visual representation of closing the loop between the relationship of the finite element method and statics, which will be further described below.To be able to close the loop in the diagram above we need to know how to go from the computed force reaction to an unknown displacement. By the statics method described above, we know the reactions of each element. Therefore, we also know the axial force being applied to that member. If we know both the axial load Pe and the element stiffness ke. Then we can calculate the axial displacement of the element. See calculations below for further explanation.

Recall Equations for Axial Displacement :

$$ q_{2}^{(1)} = \frac{|P_{1}^{(1)}|}{k^{(1)}} =\frac{5.1243}{3/4} = 6.8324 $$

$$ q_{1}^{(2)} = \frac{|P_{2}^{(2)}|}{k^{(2)}} =\frac{6.276}{5} = 1.2552 $$

The image on the left is a geometric representation of the deformed shape of a two bar truss. Length AB is the geometric representation of the axial displacement of element 2. Length AC is the geometric representation of the axial displacement of element 1. The intersection of the perpendicular projection of line segment AB and the perpendicular projection of line segment AC, would be the new location of node 2 after the initial load is applied. If you connect a line from node 1 and node 3 to the new location of node 2 this would give you what is known as the deformed shape of the truss after being loaded.

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User:Eml4500.f08.delta 6.ramirez/globalstiffmat

The following variables are known: E(1) = 2 A(1) = 3 L(1) = 5 θ(1) = 30° E(2) = 4 A(2) = 1 L(2) = 5 θ(2) = -30° E(3) = 3 A(3) = 2 L(3) = 10 θ(3) = -135° The following variables can be determined:

$$k_{(1)} = \frac{E^{(1)}A^{(1)}}{L^{(1)}} = \frac{6}{5} = 1.2$$ $$k_{(2)} = \frac{E^{(2)}A^{(2)}}{L^{(2)}} = \frac{4}{5} = 0.8$$ $$k_{(3)} = \frac{E^{(3)}A^{(3)}}{L^{(3)}} = \frac{6}{10} = 0.6$$

$$l^{(1)} = cos\theta ^{(1)} = 0.866$$ $$m^{(1)} = sin\theta ^{(1)} = 0.5$$ $$l^{(2)} = cos\theta ^{(2)} = 0.866$$ $$m^{(2)} = sin\theta ^{(2)} = -0.5$$ $$l^{(3)} = cos\theta ^{(3)} = -0.707$$ $$m^{(3)} = sin\theta ^{(3)} = -0.707$$ Note: l(3) = m(3) The following are the element stiffness matrices, which are needed to compute the global stiffness matrix.

$$k^{(e)} = k_{(e)}\begin{bmatrix} l^{(e)2} & l^{(e)}m^{(e)} & -l^{(e)2} & -l^{(e)}m^{(e)}\\ l^{(e)}m^{(e)} & m^{(e)2} & -l^{(e)}m^{(e)} & -m^{(e)2} \\ -l^{(e)2} & -l^{(e)}m^{(e)} & l^{(e)2} & l^{(e)}m^{(e)}\\ -l^{(e)}m^{(e)} & -m^{(e)2} & l^{(e)}m^{(e)} & m^{(e)2} \end{bmatrix}$$ Thus,

$$k^{(1)} = \begin{bmatrix} 0.9 & 0.52 & -0.9 & -0.52\\ 0.52 & 0.3 & -0.52 & -0.3 \\ -0.9 & -0.52 & 0.9 & 0.52\\ -0.52 & -0.3 & 0.52 & 0.3 \end{bmatrix}$$

$$k^{(2)} = \begin{bmatrix} 0.6 & -0.346 & -0.6 & 0.346\\ -0.346 & 0.2 & 0.346 & -0.2 \\ -0.6 & 0.346 & 0.6 & -0.346\\ 0.346 & -0.2 & -0.346 & 0.2 \end{bmatrix}$$

$$k^{(3)} = \begin{bmatrix} 0.3 & 0.3 & -0.3 & -0.3\\ 0.3 & 0.3 & -0.3 & -0.3\\ -0.3 & -0.3 & 0.3 & 0.3\\ -0.3 & -0.3 & 0.3 & 0.3 \end{bmatrix}$$

Now the global stiffness matrix can be computed.

$$K = \begin{bmatrix} 0.9 & 0.52 & -0.9 & -0.52 & 0 & 0 & 0 & 0\\ 0.52 & 0.3 & -0.52 & -0.3 & 0 & 0 & 0 & 0\\ -0.9 & -0.52 & 1.8 & 0.474 & -0.6 & 0.346 & -0.3 & -0.3\\ -0.52 & -0.3 & 0.474 & 0.8 & 0.346 & -0.2 & -0.3 & -0.3\\ 0 & 0 & -0.6 & 0.346 & 0.6 & -0.346 & 0 & 0\\ 0 & 0 & 0.346 & -0.2 & -0.346 & 0.2 & 0 & 0\\ 0 & 0 & -0.3 & -0.3 & 0 & 0 & 0.3 & 0.3\\ 0 & 0 & -0.3 & -0.3 & 0 & 0 & 0.3 & 0.3 \end{bmatrix}$$


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User:Stevenarntson/Ten Things You Should Know About the Seattle Public Library

access

 * Anyone who lives, works, or goes to school in Seattle can have a free SPL card. Fill out the form, present your ID, and you can check out materials immediately.
 * You can return materials checked out from SPL to any SPL branch--if you check out a book at the Central branch, you can return it to Columbia City. But don't get SPL confused with our other local library systems--KCLS or Sno-Isle.
 * The library is currently threatened with budget cuts that may limit its hours, and cause some closures.

collections
SPL’s collection includes books, magazines, newspapers, CDs, DVDs, comics, maps, sheet music, plays, genealogy research resources, and many other things.

services
The Central branch has 2 pianos in private practice rooms that you can sign up to play, free wireless Internet throughout the building (and plugs for laptops at many of the reading desks), hundreds of Internet-connected computers you can access with your library card number, and many librarians wandering around who want to help you.

reviews
You can write reviews of books you've read/recordings you've listened to to post to the library catalog. I wrote one for this book.

spl.org
Here, you can verify what you have checked out, place holds, and even renew items.

events
The Central branch has an auditorium that hosts diverse events from readings to concerts to screenings of films. Check the website for up-to-date information, or sign up here for an email newsletter.

taxes and classes
Trying to figure out your taxes in mid-January/April? Go to the central library. Volunteer CPAs are sitting at tables waiting for you—they will work with you for free. SPL also has many classes and book clubs that are offered throughout the year.

parking
There is underground parking, I hear, although I have not used it; I ride my bike. Did you know that the bicycle is the most efficient form of transportation in the world? It is even more efficient than walking. You can buy a bike at the downtown Goodwill for about $20 (see the transportation resources page).

databases
Of immense value to research are the many databases you have to choose from, such as
 * employment/career resources
 * newspapers & magazines from around the world
 * &c.

the building
Some people find the inside of the Central library a bit confusing; I did too, at first. Take a few minutes and explore it—for me, I knew where everything was after two or three visits. Natural light, excellent views, high ceilings, and places to sit abound. You can sit in the library all day and not be arrested for loitering! Isn’t it strange that being able to sit down seems so subversive?
 * There have been several different libraries on this spot, and around town.
 * Herbert Muschamp's review in the New York Times.
 * Paul Goldberger's review in the New Yorker.
 * In fact, there's a whole book about the building.


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User:SaroopChand

This is Saroop Chand's home page.

--SaroopChand

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User:Stevenarntson/Five Places In Seattle To Get A Semi-Functional Bike For Under Twenty Dollars

 * 1) The Goodwill Outlet 1765 6th Ave. S. Mon-Sat: 9 to 6 Sun: 11 to 5
 * 2) Seattle Goodwill 1400 S. Lane St. (Deerborn, off Ranier Ave S) Mon-Sat: 9 to 9 Sun: 11 to 6
 * 3) Recycled Cycles free pile 1007 NE Boat Street (near the Fremont neighborhood)
 * 4) The Seattle Salvation Army 1010 4th Ave S
 * 5) The Bikery will open soon in the Central District

Need Help Fixing Up the Bike You Bought?

 * The Bike Shack @ the Wayward Cafe 55th and Roosevelt in the University District Hours vary - check the website
 * Bike Works 3709 S. Ferdinand Tues-Fri: Noon-6:00 Sat: 11:00-6:00 Sun: 11:00pm-5:00pm
 * 2020 Cycle bike shop 2020 Union St Mon-Sat 11-7 Sun 11-4


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User talk:Stevenarntson

= Bio =
 * I was born in 1973 in Washington State, and studied piano, music composition, and creative writing at Fairhaven College in Bellingham. I earned an MFA in fiction writing from the Iowa Writers’ Workshop. I perform frequently as a concertina soloist and with The Toy Boats, a quartet. More information is available at stevenarntson.info. I also write an assessment of my teaching every quarter here.

=FAQ=
 * Q: "Where did my old group space go?" A: It went here.
 * Q: "How do I find my old userspace?" A: Look at the Student Hub Archives
 * Q: "Why is there a dotted blue box around what I wrote?" A: It means you have spaces at the beginning of a line. Delete them, and the blue box will disappear.

= Discussion Archives = winter 2010 archive fall 2009 archive summer 2009 archive spring 2008 archive winter 2008 archive = Question/Comment? Post It Below: =

Throttle
There is a feature with Wiki that prevents messages from being sent too fast so that such actions can be regulated. Normally, only "bots" or wikiscripts have that ability. Normally, the "rollback permission" would allow you to get around it, I think, but I don't have the ability to grant it. Ottava Rima (talk) 15:12, 3 April 2010 (UTC)

bot help?
if you wish to post to a large number of students, you may ask a bot-operator for help. we have such people locally and on as well. Regards, Hillgentleman | //\\ |Talk 15:36, 3 April 2010 (UTC)

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User:Eml4500.f08.ateam.didomenico/hw 3

% Matlab Script %********************************************************************* % filename: twobar.m % % PURPOSE: %   Example of Two Bar Truss deformation % % AUTHOR: Daniel DiDomenico % % Modified on: % Created on : Mon, 06 Oct 2008, 11:32:47 EST % % DEPENDENCIES: %  call: % % REMARKS: % %*********************************************************************  clear; close; dof = 2; n_node = 4; n_elem = 4; total_dof = 2 * n_node; position(:, 1) = [ 0; 0]; position(:, 2) = [ 3.4641; 2]; position(:, 3) = [ 4.87831; 0.585786]; position(:, 4) = [ 7.8161; 8.1271]; for i = 1 : n_node; x(i) = position(1,i); y(i) = position(2,i); end node_connect(1, 1) = 1; node_connect(2, 1) = 2; node_connect(1, 2) = 2; node_connect(2, 2) = 3; node_connect(1, 3) = 1; node_connect(2, 3) = 4; node_connect(1, 4) = 4; node_connect(2, 4) = 3; for i = 1 : 2 node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [x(node_1),x(node_2)]; yy = [y(node_1),y(node_2)]; axis([0 10 0 10]) plot(xx,yy,'--') hold on   end for i = 3 : 4 node_3 = node_connect(1,i); node_4 = node_connect(2,i); xx = [x(node_3),x(node_4)]; yy = [y(node_3),y(node_4)]; axis([-2 10 -2 10]) plot(xx,yy,'-r') hold on   end text(x(node_connect(1,1)),y(node_connect(1,1)),'Global Node 1','HorizontalAlignment', 'center') text(x(node_connect(2,1)),y(node_connect(2,1)),'Global Node 2','HorizontalAlignment', 'center') text(x(node_connect(2,2)),y(node_connect(2,2)),'Global Node 3','HorizontalAlignment', 'center') text(x(node_connect(2,1))/2,y(node_connect(2,1))/2,'Element 1','HorizontalAlignment', 'center') text(x(node_connect(1,2))/2 + x(node_connect(2,2))/2,y(node_connect(1,2))/2+y(node_connect(2,2))/2,'Element 2','HorizontalAlignment', 'center') text(x(node_connect(2,3)),y(node_connect(1,4)),'Global Node 3','HorizontalAlignment', 'center') text(x(node_connect(2,3))/2,y(node_connect(2,3))/2,'Element 1','HorizontalAlignment', 'center') text(x(node_connect(2,3))/2 + x(node_connect(2,4))/2,y(node_connect(2,3))/2+y(node_connect(2,4))/2,'Element 2','HorizontalAlignment', 'center') title('Two Bar Truss Deformation') xlabel('x') ylabel('y')

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Advanced Classical Mechanics/Hamilton's Equations

Hamilton's Equations
Hamilton's equations are an alternative to the Euler-Lagrange equations to find the equations of motion of a system. They are applied to the Hamiltonian function of the system. Hamilton's equations are

$$\dot{q_i} = \frac{\partial H}{\partial p_i}$$

$$\dot{p_i} = -\frac{\partial H}{\partial q_i}$$

$$\frac{\partial L}{\partial t} = -\frac{\partial H}{\partial t}$$

Where $$H$$ is the Hamiltonian, $$q_i$$ are the generalized coordinates, $$p_i$$ are the generalized momenta, and a dot represents the total time derivative. The Hamiltonian can be found by performing a Legendre transformation on the Lagrangian of the system:

$$H = \dot{q_i}p_i - L(q,\dot{q},t)$$

where the Einstein summation notation is used, a dot represents the total time derivative, qi are the generalized coordinates, pi are the generalized momenta.

These momenta are found by differentiating the Lagrangian with respect to the generalized velocities $$\dot{q_i}$$. Mathematically

$$p_i = \frac{\partial L}{\partial \dot{q_i}}$$.

In some cases, $$H = T + V$$ is the total energy of the system. There are two conditions for this. First, the equations defining the generalized coordinates q don't depend on time explicitly. Second, the forces involved in the system are derivable from a scalar potential. The forces must be conservative (such as gravity).

References:


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User:Eas4200c.f08.wiki.c/homework report 3

=Meeting 14=

The analysis of a hollow circular cross section with thin walls (ie. t << a), is conducted as follows:

$$r_{i} = a$$ (inner radius)

$$r_{o} = b$$ (outer radius)

The equation for the polar moment is done by superimposing 'empty' circle on top of one solid circle, and the resulting equation is as follows:

$$J = \frac{1}{2}\pi (b^{4} - a^{4})$$

This can be expanded as:

$$J = \frac{1}{2}\pi (b - a)(b + a)(b^{2} + a^{2}$$

If an average radius of the cross section is then defined as:

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

And it is noted that, since t is very small,

$$ b \cong \bar{r}$$

and

$$ a \cong \bar{r}$$

Therefore,

$$ a^{2} \cong \bar{r}^{2}$$

$$ a^{2} \cong \bar{r}^{2}$$

and noting that $$b - a = t$$

J can then be expressed as:

$$J = 2\pi t\bar{r}^{3}$$

J can then be expressed in terms of the average radius by noting that:

$$J = (2\pi ^{-\frac{1}{2}}t)(\pi \bar{r}^{2})^{3/2}$$

With $$\pi \bar{r}^{2} = \bar{A}$$ it's easily seen that J is proportional to $$\bar{A}^{3/2}$$, with $$(2\pi ^{-\frac{1}{2}}t)$$ being a proportionality factor.

=Meeting 15=

Matlab Project Discription
For HW 3, the team was asked to analyze a NACA 4-digit airfoil series. The discription of the problem can be seen here. A diagram showing the shape of the airfoil in this series can be found on this website. The group is asked to break up the airfoil into a number of discrete sections (ns) in order to analyze the geometry of the airfoil.

In order to complete this problem the student must first understand how to discretize an object into triangular segments in order to find the area of the object. A diagram of how this is done is provided below.

Using this, the following equations can be used in order to find the area of the shaded region.

$$d\vec{T} = \vec{r} \times d\vec{F} = q\vec{r} \times \vec{PQ}$$

Here, $$d\vec{F} = \vec{PQ}$$ and $$q\vec{r} \times \vec{PQ} = (2dA)\vec{i}$$

With

$$||\vec{PQ}|| = d\mathit{l} $$

Now that the basic foundation of the analysis has been discussed, the execution of the code will be done later in this report.

Torsion
Torsion of uniform, non circular bars leads to warping of the cross section, which is defined as an axial displacement along the x-axis (ie, along the bars length due to the conventions used in class) of a point on the deformed, or rotated, cross section. A schematic representation is shown below.

$$u_{y}$$ = the y-component of the displacement vector $$\vec{PP^{'}}$$

and

$$u_{z}$$ = the z-component of the displacement vector $$\vec{PP^{'}}$$

It should be noted (and easily seen) from the figure that $$\vec{PP^{'}}$$ is not on a circle centered at the origin with radius $$\vec{OP^{'}}$$ and $$\vec{OP^{'}}$$ in fact has a larger magnitude than $$\vec{OP}$$

This is due to the line $$\vec{PP^{'}}$$ having a vertical and horizontal displacement from $$\vec{PP^{'}}$$

However, since the angles associated with twist in torsion are likely to be relatively small, one can use small angle approximations and assume that the magnitude of $$\vec{OP} = \vec{OP^{'}} = R$$

$$u_{z} = Rsin\alpha \cong R\alpha$$

$$u_{y} = R(1 - cos\alpha )$$

With $$\alpha = 0$$, this makes $$u_{y} = 0$$ and $$u_{z} = \alpha R$$