text:
User:EML4500.f08.JAMAMA/Homework5

 Note: Derivation of 3-D bar element missing. Please don't remove this comment in case you amend your work; just add a comment box to say what you did. Eml4500.f08 11:14, 10 November 2008 (UTC)

Meeting 24
Justification for eliminating rows 1 2 5 6 to reduce the global stiffness matrix for the 2 bar truss system to a 2X2 matrix.

The force displacement relation can be written as follows.

$$\textbf{K}_{6x6}\textbf{d}_{6x1}=\textbf{F}_{6x1}$$

$$\textbf{K}_{6x6}\textbf{d}_{6x1}-\textbf{F}_{6x1}=\textbf{0}$$

The principle of virtual work is shown below is true for all $$\textbf{W}$$.

$$\textbf{W}_{6x1}\bullet \left[ \textbf{K}_{6x6}\textbf{d}_{6x1}-\textbf{F}_{6x1}\right]=\textbf{0}$$

The matrix $$\textbf{W}$$ is a weighting matrix. The weighting matrix can be used to pull out each equation out of the force displacement relation.

Principle of Virtual Work: Now to prove the simplification. From the 2 bar truss example: the boundary conditions: $$d_{1},d_{2},d_{5},d_{6}$$ are all equal to zero. The weighting coefficients must be "kinematically admissible", meaning they can not violate the boundary conditions;

$$W_{1}=W_{2}=W_{5}=W_{6}=0$$

By plugging $$W_{1}=W_{2}=W_{5}=W_{6}=0$$ into $$kd-F=0$$, the row reduction (elimination of rows 1, 2, 5, and 6) is proven because they are fixed, allowing zero displacement.

So, the global stiffness matrix is reduced down to a 6x2 matrix.

$$W\centerdot (\underbrace{kd}_{[6x2]\{2x1\}}-F)$$

$$\begin{bmatrix} K_{31} & K_{41}\\ K_{32} & K_{42}\\ K_{33} & K_{43}\\ K_{34} & K_{44}\\ K_{35} & K_{45}\\ K_{36} & K_{46} \end{bmatrix} \begin{bmatrix} d_{3}\\ d_{4}

\end{bmatrix} - \begin{bmatrix} F_{3}\\ F_{4}

\end{bmatrix} = \begin{bmatrix} 0\\ 0

\end{bmatrix}$$

Then with the weighting matrix being reduced to a 2x1 because there are only two columns to perform the dot product upon. This eliminates rows 1, 2, 5, and 6 in the global stiffness matrix and is reduced to a 2x2.

$$ \begin{bmatrix} W_{3}\\ W_{4}

\end{bmatrix} \bullet \left[ \begin{bmatrix} K_{33} & K_{43}\\ K_{34} & K_{44} \end{bmatrix} \begin{bmatrix} d_{3}\\ d_{4}

\end{bmatrix} - \begin{bmatrix} F_{3}\\ F_{4}

\end{bmatrix} \right] = \begin{bmatrix} 0\\ 0

\end{bmatrix} $$

$$$$

$$$$

This would continue for $$W_{3}=1\text{ and }W_{4}=0,\text{ then for }W_{3}=0\text{ and }W_{4}=1$$

For $$W_{3}=1\text{ and }W_{4}=0$$ (Taken from the matrix $$W^{T}=\left\lfloor \begin{matrix}  0 & 0 & 1 & 0 & 0 & 0  \\ \end{matrix} \right\rfloor $$, providing the equation for $$F_{3}$$:

$$\begin{align} & \left\{ \begin{matrix} W_{3} \\ W_{4} \\ \end{matrix} \right\}=\left\{ \begin{matrix} 1 \\   0  \\ \end{matrix} \right\} \\ & \left\{ \begin{matrix} W_{3} \\ W_{4} \\ \end{matrix} \right\}\centerdot (k_{_{2x2}}d_{_{2x1}}-F_{_{2x1}})=0,\text{ for all matrices }\left\{ \begin{matrix} W_{3} \\ W_{4} \\ \end{matrix} \right\} \\ & \therefore \left\{ \begin{matrix} 1 \\   0  \\ \end{matrix} \right\}\centerdot \left[ \begin{matrix} K_{33} & K_{34} \\ K_{43} & K_{44} \\ \end{matrix} \right]\left\{ \begin{matrix} d_{3} \\ d_{4} \\ \end{matrix} \right\}-\left\{ \begin{matrix} F_{3} \\ F_{4} \\ \end{matrix} \right\}=\left\{ \begin{matrix} 0 \\   0  \\ \end{matrix} \right\} \\ \end{align}$$

Therefore, proving for $$F_{3}$$, that $$kd=F$$

Meeting 26
Back to the Principle of Virtual Work

Simple Explanation of a Linear Combination:

$$\text{Course Grade }=\alpha _{o}\cdot \text{ Homework Grade }+\sum\limits_{i=1}^{3}{\alpha _{i}\cdot \text{ Exam}_{i}}$$

Deriving $$\underset – {\mathop{k^{(e)}}}\,=\underset – {\mathop{T^{(e)T}}}\,\underset – {\mathop{\hat{k}^{(e)}}}\,\underset – {\mathop{T^{(e)}}}\,$$:

 Recall: The force displacement relationship with axial degrees of freedom, $$q^{(e)}$$

$$\underset – {\mathop{\hat{k}^{(e)}}}\,\underset – {\mathop{q^{(e)}}}\,=\underset – {\mathop{p^{(e)}}}\,$$

 $$\Rightarrow \underset – {\mathop{\hat{k}^{(e)}}}\,\underset – {\mathop{q^{(e)}}}\,-\underset – {\mathop{p^{(e)}}}\,=\underset – {\mathop{0}}\,_{2x1}$$ (Equation 1)

The Definition of the Principle of Virtual Work  $$\underset – {\mathop{W}}\,_{2x1}\underbrace{\text{ }\centerdot \text{  }}_{\text{dot product}}\text{ }\underbrace{\left( \widehat{\underset – {\mathop{k}}\,}^{(e)}\underset – {\mathop{q^{(e)}}}\,-\underset – {\mathop{p^{(e)}}}\, \right)}_{2x1}=0_{_{1x1}}\text{ }\to \text{for all }\underset – {\mathop{W}}\,_{2x1}$$ (Equation 2)

It has now been shown that (Equation 1) $$\Leftrightarrow $$ (Equation 2)

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

And Similarly, for $$q^{(e)}$$ (the axial displacement)

$$\text{ the weighting coefficient, }W,\text{ can be expressed:}$$

 $$\widehat{\underset – {\mathop{W}}\,}_{2x1}=\underset – {\mathop{T}}\,_{2x4}^{(e)}\underset – {\mathop{W}}\,_{4x1}$$

Note: The difference between the weighting coefficients:

$$W$$ is used in conjunction with $$d^{(e)}$$ (the vertical and horizontally described displacements)

and $$\hat{W}$$ is used with $$q^{(e)}$$ (the axially described displacements)

Through the use of: $$q^{(e)}=T^{(e)}d^{(e)}$$ and $$k^{(e)}=T^{(e)T}\hat{k}^{(e)}T^{(e)}$$

Meeting 27
$$\hat{W}$$ = virtual axial displacement

W = Virtual displacement in global coordinate system corresponds to $$d^{(e)}$$ (4X1)

recall last equations:

$$q^{(e)}=T^{(e)}d^{(e)}$$  (3)

$$ \hat{W}=T^{(e)}W $$   (4)

Replace equations (3) and (4) into equation (2). (From page meeting 26, page 2)

$$(T^{(e)}W)\bullet \left[\hat{k}^{(e)}\left(T^{(e)}d^{(e)} \right)-p^{(e)} \right]=0$$ for all $$\hat{W}$$ (4X1) (5)

Recall: $$\left(AB \right)^{T}=B^{T}A^{T}$$ (6)

pxqqxr

Recall:$$a\bullet b=a^{T}b$$ (7)

nx1nx1(1xnnx1)

(1x1 scalar)

Apply (7) and (6) in (5):

$$(T^{(e)}W)^{T}\bullet \left[\hat{k}^{(e)}\left(T^{(e)}d^{(e)} \right)-p^{(e)} \right]=0$$

$$(T^{(e)T}W^{T})\bullet \left[\hat{k}^{(e)}\left(T^{(e)}d^{(e)} \right)-p^{(e)} \right]=0$$

$$\Rightarrow W\bullet \left[\left( T^{(e)}\hat{k}^{(e)}T^{(e)}\right)d^{(e)}-T^{(e)T}p^{(e)} \right]=0$$ for all W (4x1)

$$\Rightarrow W\bullet \left[k^{(e)}d^{(e)}-f^{(e)} \right]= 0 $$ for all W.  $$\Rightarrow k^{(e)}d^{(e)}=f^{(e)}$$

So far, discrete case (non-continuous) matrices. Now, continuous case (Partial Differential Equations) motivational model problem: Elastic bar with varying A(x), E(x), subject to varying axial loads (distributed) + concentrated load + inertia force (dynamic). Axial loads being the dependent.

Meeting 28
Free body diagram of an infinitesimal portion of the bar from lecture 27

$$ \Sigma\ F_x = 0 = -N(x,t) + N(x+dx,t) + f(x,t)dx - m(x)\ddot u$$

$$\Sigma\ F_x = {dN(x,t) \over dx}dx + H.O.T + f(x,t)dx - m(x)\ddot u \qquad \leftrightarrow \qquad Eq(1)$$

Equation (1) can then be written in the form

$${dN \over dx} + f = m \ddot u \qquad \leftrightarrow \qquad Eq(2)$$

By analyzing the diagram of the infinitesmal section the following is determined

$$N(x,t) = A(x) \sigma\ (x,t) \qquad \leftrightarrow \qquad Eq(3)$$Constitutive Relation

where: $$\sigma\ (x,t) = E(x) \varepsilon\ (x,t) $$     and      $$\varepsilon\ (x,t) = {du \over dx}(x,t)$$

By combining equations (2) and (3):

$${d \over dx} \bigg[ A(x)E(x){du \over dx} \bigg] + f(x,t) = m(x) + \ddot u \qquad \leftrightarrow \qquad Eq(4)$$ PDE of Motion

Need 2 boundary conditions (2nd order derivative with respect to x)

$$u(0,t) = 0 \qquad b.c.(1)$$

$$N(L,t) = F(t) \qquad b.c.(2)$$

Need 2 initial conditions (2nd order derivative with respect to t)

$$u(0,t)=0=u(L,t)$$

$$\rightarrow \quad {du(L,t) \over dx}={F(t) \over A(L)E(L)}$$

Matlab Section
{| class="toccolours collapsible collapsed" style="width:100%" ! The corrected 2-Bar truss code

Code Output
k_local = 0.75       -0.75        -0.75         0.75

k = 0.5625     0.32476      -0.5625     -0.32476      0.32476       0.1875     -0.32476      -0.1875      -0.5625     -0.32476       0.5625      0.32476     -0.32476      -0.1875      0.32476       0.1875

k_local = 5   -5    -5     5

k = 2.5        -2.5         -2.5          2.5         -2.5          2.5          2.5         -2.5         -2.5          2.5          2.5         -2.5          2.5         -2.5         -2.5          2.5

K = 0.5625     0.32476      -0.5625     -0.32476            0            0      0.32476       0.1875     -0.32476      -0.1875            0            0      -0.5625     -0.32476       3.0625      -2.1752         -2.5          2.5     -0.32476      -0.1875      -2.1752       2.6875          2.5         -2.5            0            0         -2.5          2.5          2.5         -2.5            0            0          2.5         -2.5         -2.5          2.5

R = 0    0     0     7     0     0

d = 0           0        4.352       6.1271            0            0

reactions = -4.4378     -2.5622       4.4378      -4.4378

results = 1.7081      5.1244       5.1244       0.6276        3.138        6.276

The different output occured because the E and A were not of correct specification. E and A should have been specified with "i" subscripts, such as E(i), A(i), so that with appropriate truss mambers' certain values should be taken.

 * }

{| class="toccolours collapsible collapsed" style="width:100%" ! 2-D 6-bar truss and its deformation (magnified by a factor of 1000) at constant Modulus of Elasticity (E = $$2*10^{11}$$)

Code Output
K = 1.0e+008 * 0.8536   0.3536   -0.5000         0         0         0         0         0   -0.3536   -0.3536    0.3536    0.3536         0         0         0         0         0         0   -0.3536   -0.3536   -0.5000         0    0.8536   -0.3536         0         0         0         0   -0.3536    0.3536         0         0   -0.3536    1.0202         0         0         0   -0.6667    0.3536   -0.3536         0         0         0         0    0.7155   -0.3578         0         0   -0.7155    0.3578         0         0         0         0   -0.3578    0.1789         0         0    0.3578   -0.1789         0         0         0         0         0         0    0.7155    0.3578   -0.7155   -0.3578         0         0         0   -0.6667         0         0    0.3578    0.8456   -0.3578   -0.1789   -0.3536   -0.3536   -0.3536    0.3536   -0.7155    0.3578   -0.7155   -0.3578    2.1382         0   -0.3536   -0.3536    0.3536   -0.3536    0.3578   -0.1789   -0.3578   -0.1789         0    1.0649

d = 1.0e-003 * 0        0    0.2131    0.2500         0         0         0         0   -0.0061    0.0122

d_def = 1.0e-003 * 0        0    0.2131    0.2500         0         0         0         0   -0.0061    0.0122

nodes_def = 0        0    4.0002    0.0002         0    3.0000    4.0000    3.0000    2.0000    2.0000

d_def_aug = 0        0    0.2131    0.2500         0         0         0         0   -0.0061    0.0122

nodes_def_aug = 0        0    4.2131    0.2500         0    3.0000    4.0000    3.0000    1.9939    2.0122

Stress = 1.0e+007 * 1.0655  -0.0927   -0.0977   -1.6665    0.0307   -0.0002

Force_axial = 1.0e+004 * 1.0655  -0.0927   -0.0977   -1.6665    0.0307   -0.0002

6-bar truss image (the deformation was multiplied by a factor of 1000 to show a clearer image)
Blue solid line $$ \Rightarrow $$ deformed truss

Red dashed line$$ \Rightarrow $$ undeformed truss

 * }

{| class="toccolours collapsible collapsed" style="width:100%" ! 2-D 6-bar truss and its deformation (magnified by a factor of 1000) at constant (E = $$2*10^{11}$$) and at a varying specified Modulus of Elasticity

Code Output (Note, the output is only varying the parameter E
K = 1.0e+008 * 0.7639   0.3889   -0.3750         0         0         0         0         0   -0.3889   -0.3889    0.3889    0.3889         0         0         0         0         0         0   -0.3889   -0.3889   -0.3750         0    0.6932   -0.3182         0         0         0         0   -0.3182    0.3182         0         0   -0.3182    0.9849         0         0         0   -0.6667    0.3182   -0.3182         0         0         0         0    0.7155   -0.3578         0         0   -0.7155    0.3578         0         0         0         0   -0.3578    0.1789         0         0    0.3578   -0.1789         0         0         0         0         0         0    0.8944    0.4472   -0.8944   -0.4472         0         0         0   -0.6667         0         0    0.4472    0.8903   -0.4472   -0.2236   -0.3889   -0.3889   -0.3182    0.3182   -0.7155    0.3578   -0.8944   -0.4472    2.3171    0.1602   -0.3889   -0.3889    0.3182   -0.3182    0.3578   -0.1789   -0.4472   -0.2236    0.1602    1.1096

d = 1.0e-003 * 0        0    0.2649    0.2608         0         0         0         0    0.0006   -0.0012

d_def = 1.0e-003 * 0        0    0.2649    0.2608         0         0         0         0    0.0006   -0.0012

nodes_def = 0        0    4.0003    0.0003         0    3.0000    4.0000    3.0000    2.0000    2.0000

d_def_aug = 0        0    0.2649    0.2608         0         0         0         0    0.0006   -0.0012

nodes_def_aug = 0        0    4.2649    0.2608         0    3.0000    4.0000    3.0000    2.0006    1.9988

Stress = 1.0e+007 * 0.9932   0.0096    0.0101   -1.7389   -0.0033   -0.0002

Force_axial = 1.0e+004 * 0.9932   0.0096    0.0101   -1.7389   -0.0033   -0.0002

6-bar truss image (the deformation was multiplied by a factor of 1000 to show a clearler image)
Blue solid line $$ \Rightarrow $$ defored truss of constant E

Red dashed line$$ \Rightarrow $$ undeformed truss

Black solid line$$ \Rightarrow $$ deformed truss of varying E

 * }

Contributing Members
Justin McIntire --Eml4500.f08.jamama.justin 15:46, 6 November 2008 (UTC)

William Mueller --Eml4500.f08.jamama.mueller 04:56, 7 November 2008 (UTC)

Chris Alford -- Eml4500.f08.jamama.chris 18:39, 7 November 2008 (UTC)

Cedric Adam --Eml4500.f08.jamama.adam 18:48, 7 November 2008 (UTC)

Arunas Janulevicius --Eml4500.f08.jamama.jan 19:02, 7 November 2008 (UTC)

Megan Alvarez --EML4500.f08.jamama.megan 20:42, 7 November 2008 (UTC)

title: User:EML4500.f08.JAMAMA/Homework5


text: File:Wikiaholic.svg

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

title: File:Wikiaholic.svg


title: File:6BarPlot1.JPG


text: File:Kmatrix.jpg

title: File:Kmatrix.jpg


text:
User:Eml4500.f08.RAMROD.A/HW5

 This version is very similar to the original but some sections have been rearranged to be placed in the correct order, as there was some confusion as to whose portions went where. Also, a parsing issue was resolved. I, (Eml4500.f08.RAMROD.A 01:11, 14 November 2008 (UTC),) have fixed the HW for final submission and the comparison to the original can be found here: [Comparsion]

Justification of Eliminated rows 1, 2, 5, and 6 to obtain K 2-bar truss:
Force Displacement Relation: K6x6d6x1 = F6x1

Moving F to the other side of the equation gives us Equation (1):

 Equation (1)

$$\mathbf{Kd}-\mathbf{F}=\mathbf{0}_{6x1} $$

Using the Principle of Virtual Work, we can arrive at Equation (2):

 Equation (2)

$$\mathbf{w}_{6x1}\bullet(\mathbf{Kd}-\mathbf{F})=\mathbf{0}_{1x1} $$
 * This is true for all w6x1
 * w is known as the "weighting mat."
 * Note that the dot product is used in this equation.

Equation (1)↔ Equation (2)

Using Principle of Virtual Work
Taking account for the boundary conditions of the 2 bar truss:
 * d1=d2=d5=d6=0

Weighting coefficient must be "kinematically admissible," ie cannot violate the boundary conditions
 * w1=w2=w5=w6=0

Weighting coefficients = virtual displacement

 * (calculus of variations)

From before:

$$\begin{Bmatrix} w_3\\ w_4\end{Bmatrix}\bullet(\mathbf{Kd}-\mathbf{F})=0$$

$$\mathbf{w}\bullet(\mathbf{Kd}-\mathbf{F})$$

$$\begin{Bmatrix} w_3\\ w_4 \end{Bmatrix}\bullet[\begin{bmatrix} K_{13} & K_{14} \\ K_{23} & K_{24} \\ K_{33} & K_{34}\\ K_{43} & K_{44}\\ K_{53} & K_{54}\\ K_{63} & K_{64} \end{bmatrix} \begin{Bmatrix} d_3\\ d_4 \end{Bmatrix} - \begin{Bmatrix} F_1\\ F_2\\ F_3\\ F_4\\ F_5\\ F_6 \end{Bmatrix}] = 0$$
 * This is true for all $$\begin{Bmatrix}

w_3\\ w_4\end{Bmatrix}$$

$$\mathbf{\overline{K}} = \begin{bmatrix} K_{33} & K_{34} \\ K_{43} & K_{44} \end{bmatrix}$$

$$\mathbf{\overline{d}} = \begin{Bmatrix} d_3\\ d_4 \end{Bmatrix}$$

$$\mathbf{\overline{F}} = \begin{Bmatrix} F_3\\ F_4 \end{Bmatrix} $$

Comparing Equations (3) and (4)
Equation (3) : $$(kd-F) = 0$$

Equation (4) : $$w(kd-F) = 0$$

Equation (3) → Equation (4) is trivial

Equation (4) → Equation (3)
 * Since Equation (4) is valid for all w, select w=1

 * $$1(kd-F) = 0$$ Thus proving Equation (3) is true

Principal of Virtual Work Matrix
We are going to try to derive element stiffness matrix by Principle of Virtual Work.

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

Recalling the force displacement with axial degree of freedom $$\underline{q}^{(e)}$$

$$\hat{\underline{K}}^{(e)}_{2x2}\cdot \underline{q}^{(e)}_{2x1} = \underline{P}^{(e)}_{2x1}$$

$$\hat{\underline{K}}^{(e)}_{2x2}\cdot \underline{q}^{(e)}_{2x1} - \underline{P}^{(e)}_{2x1}= \mathbf{0}_{2x1}$$ Equation $$(1)$$

Principle of Virtual Work applied here: $$\hat{\mathbf{W}}\bullet \left(\hat{\underline{K}}^{(e)}_{2x2}\cdot \underline{q}^{(e)}_{2x1} - \underline{P}^{(e)}_{2x1}\right)= \mathbf{0}_{2x1}$$ Equation $$(2)$$

We showed equation $$(1) \Leftrightarrow (2)$$

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

Similarly $$\hat{\underline{W}}^{(e)}_{2x1}= \bar{T}^{(e)}_{2x4}\mathbf{\underline{W}}^{(e)}_{4x1}$$

Virtual Displacement Terms
ŵ2x1 = virtual axial displacement, with corresponding q(e)2x1

w4x1 = virtual axial displacement in global coordinates system, with corresponding d(e)4x1

Equation (5)
$$ (T^{(e)}w)\cdot [\hat{K}^{(e)}(T^{(e)}d^{(e)})-P^{(e)}] = 0 $$   for all w4x1

We want to prove equation (5) using matrix identity equation (6):

Equation 6: $$ (\mathbf{AB})^T=\mathbf{B}^T\mathbf{A}^T $$

Homework Example to verify equation (6)
$$ A=\begin{bmatrix}1 & 2 & 3\\4 & 5 & 6 \end{bmatrix}, B=\begin{bmatrix}7 & 8 & 9\\1 & 2 & 3\\4 & 5 & 6\end{bmatrix} , B^T = \begin{bmatrix}7 & 1 & 4\\ 8 & 2 & 5\\ 9 & 3 & 6\end{bmatrix} , A^T=\begin{bmatrix}1 & 4\\ 2 & 5\\ 3 & 6\end{bmatrix} $$

$$ AB=\begin{bmatrix}1\cdot7+2\cdot1+3\cdot4 & 1\cdot8+2\cdot2+3\cdot5 & 1\cdot9+2\cdot3+3\cdot6 \\4\cdot7+5\cdot1+6\cdot4 & 4\cdot8+5\cdot2+6\cdot5 & 4\cdot9+5\cdot3+6\cdot6 \end{bmatrix} =\begin{bmatrix}21 & 27 & 33\\57 & 72 & 87\end{bmatrix} $$

$$(AB)^T = \begin{bmatrix}21 & 57\\27 & 72\\ 33 & 87\end{bmatrix} $$

$$ B^TA^T = \begin{bmatrix}21 & 57\\27 & 72\\ 33 & 87\end{bmatrix} $$

This proves Equation (6): $$ (\mathbf{AB})^T=\mathbf{B}^T\mathbf{A}^T $$

We can recall that $$ \underline{a}\cdot\underline{b} = \underline{a}^T\underline{b} $$, where a and b are nx1 and aT is 1xn

It follows that from Eq(1):

$$ (T^{(e)}w)^T[\hat{k}^{(e)}(T^{(e)}d^{(e)})-P^{(e)}] = 0 $$, for all w,

and using Eq (6) above, that:

$$ w^TT^{(e)T}[\hat{k}^{(e)}(T^{(e)}d^{(e)})-P^{(e)}] = 0 $$

and that:

$$ w\cdot[(T^{(e)T}\hat{k}^{(e)}T^{(e)})d^{(e)}-(T^{(e)T}P^{(e)})] = 0 $$, for all w4x1

Also:

$$ w\cdot[\hat{k}^{(e)}d^{(e)}-f^{(e)}] = 0 $$, for all w.

All the above proves:

$$ k^{(e)}_{4x4} = T^{(e)T}_{4x2}\hat{k}^{(e)}_{4x2}T^{(e)}_{2x4} $$

Principal of Virtual Work
Equation 2:

$$ \hat{w}\cdot(\hat{k}^{(e)}q^{(e)}-P^{(e)}) = 0 $$, for all w

Equation 3:

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

and finally equation 4:

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

Problems of the Continuous Case (PDE's)
The new problems will contain varying parameters

Example of problem:

Elastic bar with varying cross section A(x), varying modulus E(x), subjective to varying axial load (distributed), a concentrated load and inertia force (dynamic).

note: load will be time dependent

Problem Setup

The following is a FBD of an infinitesimal section area of the elastic bar with varying properties.

The bottom force that is not labeled in the picture can be represented as the following equation;

$$\displaystyle A(x) \partial x \rho(x) \frac{\partial^2 u}{\partial t^2}$$

Within this equation $$ A(x) \partial x \rho(x)$$ is equal to $$\displaystyle m(x)$$ which has the units of $$ \frac{mass}{length}$$. Now we proceed to sum the forces in the x-direction and the results are as follows;

$$\displaystyle \sum{F_x}= -N(x,t) + N(x+dx,t) + f(x,t)dx - m(x)\ddot{u} = 0$$

where $$\ddot{u}$$ is equal to the acceleration as given in Newton's notation. Now we perform a Taylor Series expansion on this equation but we will be ignoring the higher order terms. This is performed below.

$$\displaystyle \sum{F_x}= \frac{\partial N}{\partial x}(x,t) + (Higher Order Terms ) + f(x,t) - m(x)\ddot{u}$$  $$\displaystyle f(x+dx) = f(x) + \frac{df}{dx}(x)dx + \frac{1}{2}\frac{d^2f}{dx^2} +.....$$  $$\displaystyle \frac{\partial N}{\partial x} + f = m\ddot{u}$$

The second equation is part of the expansion while the last equation is known as the equation of motion for the elastic bar which ignores the higher order terms of the expansion. The equation of motion is otherwise abbreviated as $$\displaystyle EOM$$. An expression for the force $$N(x,t)$$ can be obtained and the results are found below.

$$\displaystyle N(x,t) = A(x)\sigma(x) = A(x) E(x) \varepsilon (x,t) = A(x) E(x) \frac{\partial u}{\partial x}(x,t)$$

Putting this expression into the $$EOM$$ yields the partial differential equation of motion (PDE of Motion) for the elastic bar.

$$\displaystyle \frac{\partial }{\partial x}\left[ A(x) E(x) \frac{\partial u}{\partial x}\right] + f(x,t) = m(x) \ddot{u}$$

In order to solve this equation two boundary conditions along with two initial conditions are needed. The two initial conditions are initial displacement and initial velocity. For the boundary conditions, two cases need to be considered. They are depicted below.

For case a, the two boundary conditions are as follows;

$$\displaystyle u(0,t) = u(L,t) = 0$$

$$\displaystyle \frac{\partial u}{\partial x}(L,t) = \frac{F(t)}{A(L)E(L)}$$

For case two the boundary conditions are as follows;

$$\displaystyle u(0,t) = o$$

$$\displaystyle N(L,t) = F(t)$$

The initial conditions are specific to different problems

Debugged 2 Bar Truss System
Below is the code to the debugged two bar truss system. Since the modulus of elasticity and the areas for the given example were not the same the code had to be modified in order for it to run properly. Before the correction the code was running as if the all of the elements had the same modulus of elasticity and the same area. Below is the corrected code and the results from the corrected code. Functions PlaneTrussElemtens.m, NodalSoln.m and PlaneTrussResults.m were used in this code and the discription for these functions can be found in the functions section.

Results from the debugged two bar truss code

Comparison of Results
Once the code was corrected the same reaction results were obtained when comparing it to the statics methods that have been done in class. Below are the results to the statics method.

Finite Element Method:

Using this equation, we can now solve for the f(1) matrix and thus we find the values for the reactions: $$f^{(1)}_{1} = F_{1} = -4.4378$$ $$ f^{(1)}_{2} = F_{2} = -2.5622$$

Using this same procedure applied to element 2, we can solve for the values of the other two unknown reactions giving us: $$f^{(2)}_{3} = F_{5} = 4.4378$$ $$f^{(2)}_{4} = F_{6} = -4.4378$$

Statics

Using the Euler cut method, you find the following components from the resultant P forces.

$$P_x^1 = -4.4378$$ $$P_y^1=-2.5622$$ $$P_x^2=4.4378$$ $$P_y^2=-4.4378$$

Six Bar Truss Code
Below is the code for the six bar truss system found on page 226 in the the textbook. All six of the bars have the same modulus of elasticity (e) and the same area (A). The MATLAB functions of PlaneTrussElement.m, NodalSoln.m, and PlaneTrussResults.m are all found in the code. The descriptions and functions can be found below the results of the six bar truss system.

Results of the above code.

Functions
The functions PlaneTrussElement.m, NodalSoln.m, PlaneTrussResults.m were all called in the code for the 5 truss system. They are all necessary for the completion of the truss problem.

PlaneTrussElement.m

This function take the Young's Modulus (e), the Area of the element (A) and the coordinates of the element ends (coord) and generates the stiffness matrix for a plane truss element. The function first calculates the lengths of the the elements. Once those are calculated the stiffness matrix is calculated.

NodalSoln.m

The function takes the global coefficient matrix (K), the global right hand side vector (R), the list of degrees of freedom with specified values, and the specified values to determine the displacements and reactions at each node. The dof of the the system is first found by using the  command which finds the longest dimension of R. df then finds the difference between the dofs that have known values (a value of zero) and the dof that were found in the previous line. The displacements and the reactions are then calculated.

PlaneTrussResults.m

This function computes the plane truss element results. It takes in the modulus of elasticity (e), the area of the cross-section (A), the coordinates at the element ends (coord), and the displacements at the elemtent ends (disps) and calculates the axial strain (eps), stress(sigma) and force (force). The results are stored in the variable  and sent back to the main program.

Plot of Six Bar Truss System
Below is the code and the plot of the six bar truss system. The plot was made from the results that were found in the above code. The results demonstrate that only node 2 and node 5 had a displacement of (0.2131, 0.2500) and (-0.0061, 0.0122) respectively. The undeformed system is plotted with the dotted lines and the deformed system is plotted with solid lines. The deformed displacements have been scaled up by a factor of 1.2 in order to enhance in the plot.

Six Bar Truss Code for Elements with Varying E
Next, the six bar truss Matlab code used above, was modified to allow for different values of E for each element. This code uses all the same functions as above (i.e. PlaneTrussElement.m, PlaneTrussResults.m and NodalSoln.m). The modified code is shown below. The results from this situation, as well as the constant E case and the undeformed truss system are shown in the plot below. The varying E case is shown by the dotted nodes while the undeformed case is shown dashed. The element and node numbers are suppressed to avoid cluttering the diagram. The code used to create this plot is very similar to that shown above for the six bar truss with constant E, but repeated for the two deformation cases.

Three Bar Space Truss Example
Below is the code for the the 3D three bar truss system found in the textbook on page 230. There are four nodes in the system with each node containing 3 degrees of freedom. The code uses three additional functions for it to be run. SpaceTrussElement.m, NodalSoln.m, and NodalSoln.m were all used for the code. Below this code a description of each function can be found.

Functions
The functions SpaceTrussElement.m, NodalSoln.m, SpaceTrussResults.m were all called in the code for the 5 truss system. They are all necessary for the completion of the truss problem.

SpaceTrussElement.m This function creates the stiffness matrix of a space truss element. It takes in the modulus of elacticity (e), the truss area (A), and the corresponding coordinates for the truss. The function first calculates the length of each element and then calculates the stiffness matrix (k).

NodalSoln.m

The function takes the global coefficient matrix (K), the global right hand side vector (R), the list of degrees of freedom with specified values, and the specified values to determine the displacements and reactions at each node. The dof of the the system is first found by using the  command which finds the longest dimension of R. df then finds the difference between the dofs that have known values (a value of zero) and the dof that were found in the previous line. The displacements and the reactions are then calculated.

SpaceTrussResults.m

This function takes in the modulus of elasticity (e), the area of each truss (A), the coordinate of the nodes and the displacements (disps) at the element ends. It calculates the axial strain (eps), stress(sigma) and force (force). The results are stored in the variable  and sent back to the main program.

3 Bar Space Truss Results
Below are the results from the three bar space truss example.

Code for the lot of the 3 Bar Space Truss
Below is the code for the 3 Bar Space Truss.

Plot of 3 Bar Space Truss
Below is the plot of the 3 bar Space truss with the viewing mode of x = 1m, y = 1m, and z= 1m. The dotted lines represent the undeformed truss system and the solid lines represent the deformed truss system.

]

Below is the same plot but in plane view down the x axis, plane view down the y axis, and plane view down the z axis.

=3-D Truss Relationships=

This process is very similar to that of the 2-D case. Therefore we will be taking the same steps to derive the Force-Displacement relationships. The elemental axial force displacement relationship is identical to the 2-D case and is shown below.

$$\displaystyle (k^e)(q^e)=p^e$$

Since the 3-D case will have three displacements at each node, the element displacement matrix will be a six by one (6x1) as will the element force matrix. These matrices are shown below.

 $$\displaystyle d^e = \begin{Bmatrix}d_1^e\\ d_2^e\\ d_3^e\\ d_4^e\\ d_5^e\\ d_6^e \end{Bmatrix}$$
$$\displaystyle f^e = \begin{Bmatrix}f_1^e\\ f_2^e\\ f_3^e\\ f_4^e\\ f_5^e\\ f_6^e \end{Bmatrix}$$

For the 3-D case the two by six (2x6) transformation matrix $$T^e$$ that relates the axial degrees of freedom to the global degrees of freedom is shown below.

 $$\displaystyle T^e = \begin{bmatrix} l^e & m^e & n^e & 0 & 0 & 0 \\ 0 & 0 & 0 & l^e & m^e & n^e \end{bmatrix}$$

This transformation matrix can be used to relate the axial and global degrees of freedom and forces as shown below.

$$\displaystyle \begin{bmatrix}l^e & m^e & n^e & 0 & 0 & 0 \\ 0 & 0 & 0 & l^e & m^e & n^e\end{bmatrix} \begin{Bmatrix}d_1\\ d_2\\ d_3\\ d_4\\ d_5\\ d_6\end{Bmatrix}=\begin{Bmatrix}q_1\\ q_2\end{Bmatrix}$$

It can also be used to relate the global forces to the local axial forces. This is shown below.

$$\displaystyle \begin{bmatrix}l^e & m^e & n^e & 0 & 0 & 0 \\ 0 & 0 & 0 & l^e & m^e & n^e\end{bmatrix} \begin{Bmatrix}f_1\\ f_2\\ f_3\\ f_4\\ f_5\\ f_6\end{Bmatrix}=\begin{Bmatrix}p_1\\ p_2\end{Bmatrix}$$

Using the Principle of Virtual Work, the global FD relationship is given by:

$$\displaystyle K*(d)=F$$

Also, we can represent the elemental matrix $$k^e$$ in terms of the axial matrix $$\hat{k}$$ and the transformation matrix, $$\displaystyle T$$. This is shown below:

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

Lastly, the relationship between the element forces in global coordinates, $$\displaystyle f^e$$, and the axial forces, $$\displaystyle p^e$$, can be found using the transformation matrix, $$\displaystyle T$$.

$$\displaystyle f^e(T)=p^e$$

Or more detailed:

$$\begin{bmatrix}l^e & m^e & n^e & 0 & 0 & 0 \\ 0 & 0 & 0 & l^e & m^e & n^e\end{bmatrix} \begin{Bmatrix}f_1\\ f_2\\ f_3\\ f_4\\ f_5\\ f_6\end{Bmatrix}=\begin{Bmatrix}p_1\\ p_2\end{Bmatrix}$$

Using the above derived equation for $$\displaystyle k^e$$, we can give the actual values for its elements. Multiplying out the right side of the equation gives us the following:

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

$$k^e = \begin{bmatrix} 1 &-1 & 1\\ -1 & 1 & -1\\ 1&  -1&1 \end{bmatrix}\begin{bmatrix} l^e & m^e & n^e & 0 & 0 & 0\\ 0 & 0 & 0 & l^e & m^e & n^e \end{bmatrix}\begin{bmatrix} l^e & 0\\ m^e & 0\\ n^e & 0\\ 0 & l^e\\ 0 & m^e\\ 0 & n^e \end{bmatrix}$$

This is the elemental stiffness matrix, $$\displaystyle k^e$$ for element e.

Contributing Members
Eml4500.f08.RAMROD.A 21:11, 7 November 2008 (UTC)

EML4500.F08.RAMROD.B 19:54, 7 November 2008 (UTC)

EML4500.f08.RAMROD.E 18:05, 6 November 2008 (UTC)

Eml4500.f08.ramrod.D 01:59, 7 November 2008 (UTC)

Eml4500.f08.ramrod.c 19:11, 7 November 2008 (UTC)

Eml4500.f08.RAMROD.F 20:36, 7 November 2008 (UTC)

title: User:Eml4500.f08.RAMROD.A/HW5


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User:Eml4500.f08.bike.viale/homework report 5

=Principal of Virtual Work=

To justify the elimination of rows 1, 2, 5, and 6 in order to obtain K as a 2x2 matrix in a 2-bar truss, the FD relation

$$\mathbf{Kd} = \mathbf{F}$$

Can be rearranged to be:

$$ \mathbf{Kd} - \mathbf{F} = \mathbf{0}$$ (1)

Which is designated to be equation 1.

Utilizing the principle of virtual work, equation 1 becomes:

$$ \mathbf{W}\cdot (\mathbf{Kd} - \mathbf{F}) = \mathbf{0}$$ (2)

Where W is a 6x1 matrix and Kd - F is a 6x1, and 0 is a scalar number. This equation is designated as equation 2. This equation is true for all values of W and W is called the weighting matrix.

Considering this, Equation 2 being true implies Equation one is also true, and vice versa.

The proof going from Equation 1 to 2 is trivial. However, going from two to one is not trivial and to do this we must choose different values for W

1st Choice:

Choose W so that

$$W_{1} = 1, W_{2} = W_{3} = W_{4} = W_{5} = W_{6} = 0$$

Therefore

$$\mathbf{W^T}= \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$

and

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 1[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 0[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 0 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 0 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 0 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 0 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

Thus, the result becomes

$$\sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j= \mathbf{F}_1 $$

Which is designated to be equation (a).

2nd Choice:

Choose W such that W2=1, W1=W3=W4=W5=W6=0 so that:

$$\mathbf{W^T}= \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix}$$

Therefore:

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 0[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 1[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 0 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 0 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 0 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 0 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

Thus, the result becomes

$$\sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j= \mathbf{F}_2 $$

Which is designated to be equation (b)

3rd Choice: C hoose W such that W3=1, W1=W2=W4=W5=W6=0 so that

$$\mathbf{W^T}= \begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix}$$

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 0[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 0[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 1 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 0 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 0 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 0 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

Thus, the result becomes  $$\sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j= \mathbf{F}_3 $$

Which is designated to be equation (c)

Choice 4:

Choose W such that W4=1, W1=W2=W3=W5=W6=0 so that

$$\mathbf{W^T}= \begin{bmatrix} 0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix}$$

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 0[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 0[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 0 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 1 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 0 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 0 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

Thus, the result becomes  $$\sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j= \mathbf{F}_4 $$

This is designated to be equation (d)

Choice 5:

Choose W such that W5=1, W1=W2=W3=W4=W6=0 so that

$$\mathbf{W^T}= \begin{bmatrix} 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix}$$

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 0[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 0[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 0 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 0 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 1 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 0 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

Thus, the result becomes  $$\sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j= \mathbf{F}_5 $$

This is designated to be equation (e)

Choice 6:

Choose W such that W6=1, W1=W2=W3=W4=W5=0 so that

$$\mathbf{W^T}= \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 0[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 0[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 0 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 0 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 0 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 1 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

so that the result becomes  $$\sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j= \mathbf{F}_6 $$

This equation is designated to be equation (f)

Thus, when the equations (a) through (f) are combined, they become $$\mathbf{Kd}= \mathbf{F} $$ which is equation (1) from before.

The weighting coefficients must be kinetically admissible which means they cannot violate the boundary conditions so W1 = W2 =  W5=  W6=0  where the weighting coefficients represent the virtual displacement.

Now,

$$\mathbf{W} \cdot (\mathbf{Kd}-\mathbf{F})$$

Where Kd is a 6x2 matrix minus a 2x1 vector.

Thus

$$= \begin{Bmatrix} W_3 \\ W_4 \end{Bmatrix} (\mathbf{\overline{K}_{2x2} \overline{d}_{2x1}- \overline{F}_{2x1} })$$ (3)

Which is designated as Equation 3.

and is true for all values of $$\begin{Bmatrix} W_3 \\ W_4 \end{Bmatrix} $$

and the K matrix is reduced to $$\mathbf{\overline{K}}=\begin{bmatrix} K_{33} && K_{34} \\ K_{43} && K_{44}\end{bmatrix} $$

as well as:

$$\mathbf{\overline{d}}= \begin{Bmatrix}d_{3} \\ d_{4} \end{Bmatrix} $$

$$\overline{F } = \begin{Bmatrix}F_{3} \\ F_{4} \end{Bmatrix} $$

being the reduced F and d matrices.


text:
User:EML4500.f08.delta 6.ramirez/110308

PVW (Continuous case) Dynamics of Elastic Bar

PDE:

$$ \frac{d}{dx}[ (EA)\frac{\partial u}{\partial x} ] + \mathit{f} = m\ddot{u} $$ (1)

Note: Equation (1) is known as the discrete equation of motion (EOM)

The first term in equation (1) as seen below,

$$ \frac{d}{dx}[ (EA)\frac{\partial u}{\partial x} ] $$

will give rise to

$$ -Kd $$

Likewise, the second term,

$$ \mathit{f} = m\ddot{u} $$

will give rise to

$$ +F = M\ddot{d} $$

where M represents the mass matrix.

Thus,

$$ -Kd + F = M\ddot{d} \Rightarrow M\ddot{d} + Kd = F $$ (2)

Note: Equation (2) is a multi-degree of freedom equation

Goal: Derive (2) from (1)

Review: Associate a single degree of freedom system to a multi-degree of freedom system.

Note: Equation from single degree of freedom system looks very similar to the multi-degree of freedom system

Now deriving (2) from (1) results in equation (3),

$$ \int_{x = 0}^{x = L}{w(x)\left\{\frac{d}{dx} [ (EA) \frac{\partial u}{\partial x}] + f - m\ddot{u}\right\}dx = 0} $$

for all possible w(x), where w(x) represents the weighting function.

Note: (1) → (2) is trivial, but (3) → (1) is not trivial

Further, equation (3) can be rewritten as,

$$ \int w(x) g(x) dx = 0 $$

for all possible w(x)

Since equation (3) holds for all w(x), select w(x) = g(x). Thus, equation (3) becomes,

$$ \int g(x)^2 dx = 0 $$

Since g(x)2 > 0, therefore, g(x) = 0.

title: User:EML4500.f08.delta 6.ramirez/110308


text:
User:Eas4200c.f08.vqcrew.d/hw5

Equation of Equilibrium
The goal of this section is to prove that the equation below must be equal to zero in order for the body to be in equilibrium. This equation is thus called the equilibrium equation. The equilibrium equation in the x direction is:
 * $$\frac{\partial \sigma _{xx}}{\partial x}+ \frac{\partial \sigma _{yx}}{\partial y}+\frac{\partial \sigma _{zx}}{\partial z}=0$$

This equation can be expressed in indicial notation with $$x=1$$, $$y=2$$, and $$z=3$$ and compressed as follows:
 * $$\sum_{i=1}^{3}{\frac{\partial \sigma _{ij}}{\partial x_i}}=0$$ for $$j=1, 2, 3$$

We can also note that in the case of pure torsion, the normal stress component $${\sigma _{xx}}\;$$ is equal to zero, so the equilibrium equation simplifies to:
 * $$\frac{\partial \sigma _{yx}}{\partial y}+\frac{\partial \sigma _{zx}}{\partial z}=0$$

Before moving into the derivation of the equilibrium equation for the 3-D case, The 1-D case shown in Figure ## should first be examined. In order for this stress element to be in equilibrium, the sum of the forces must equal zero.
 * $$\sum{F_x}=0=-\sigma (x)A+\sigma (x+dx)A+f(x)dx$$
 * $$0= A \left[ \sigma (x+dx)-\sigma (x)\right]+f(x)dx$$

Now, recall the Taylor Series expansion:
 * $$f(x+dx)=f(x) + \frac{df(x)}{dx}(dx) + \frac{1}{2}\frac{d^2f(x)}{dx^2}(dx)^2 + ...$$

We can replace the term in square brackets with the first order Taylor Series expansion to simplify the equation as follows:
 * $$\frac{d \sigma }{dx}+\frac{f(x)}{A}=0$$

In this equation, $$f(x)\;$$ is in units of force per length, and the term $$\frac{f(x)}{A}$$ is called the applied load. We will now extend this derivation into the three-dimensional case. We will consider a stress element in a nonuniform stress field, without an applied load, and focus only on the stress components in the x-direction to avoid over-complicating the problem. The derivation will yield the same results if we consider the stresses acting in the y- and z-directions. The 3-D problem is depicted in Figure ##. Note that the first subscript in the stress terms defines the direction normal to the plane of the stress, and the second subscript defines the direction of the stress. The origin of the coordinate system is at the location $$(x,y,x)\;$$.

The sum of the forces is expressed as:

 * $$\sum{F_x}=0=dydz\left[-\sigma _{xx}(x,y,z) + \sigma _{xx}(x+dx,y,z)\right] + dxdz\left[-\sigma _{yx}(x,y,z) + \sigma _{yx}(x,y+dy,z) \right] + dxdy\left[-\sigma _{zx}(x,y,z) + \sigma _{zx}(x,y,z+dz) \right]$$

Simplifying the equation using the Taylor Series expansion, as in the 1-D case, yields the following:
 * $$0=(dxdydz)\left[\frac{\partial \sigma _{xx}}{\partial x} + \frac{\partial \sigma _{yx}}{\partial y} + \frac{\partial \sigma _{zx}}{\partial z}\right]$$

The volume term cancels out, and we are left with the equilibrium equation for stresses in the x direction for the 3-D problem:
 * $$0=\frac{\partial \sigma _{xx}}{\partial x} + \frac{\partial \sigma _{yx}}{\partial y} + \frac{\partial \sigma _{zx}}{\partial z}$$

title: User:Eas4200c.f08.vqcrew.d/hw5


text:
Stargazing

Every to-be observer often wonders, Where do I begin?. Astronomy often is the hobby of looking at very distant object with a telescope... Unfortunately, for so many people, it ends with the skinny unused telescope offered as a kid for a birthday.

Truth is, people rarely know what to look for. Yet, it can be so much simpler if you know the following: Don't start with a telescope, start with your own eyes.

Every dedicated astronomer is first and foremost a stargazer. Stargazing is the act of seeing subtle details, comparing and contrasting what you see from what you know.

What's the Appeal?
A clear sky with no prominent moon, once properly dark adapted, is a beautiful thing. Depending on the time of year, a subtle band of smoky light constitutes the outline of our galaxy and away from bright town and city lights is a showy splendor with bright spots, voids and intricate rivers of light.

Very often, it is similarity and contrast that strike our mind with beauty, and engage our imagination. A beautiful sight can merely be a crescent moon wedged between two bright planets, hovering just above the glow of twilight.

Initial experiences
Seen from the gaze of a beginner, a clear sky can seem almost chaotic and confusing with stars spread throughout the sky, often clearly in groups and elusive patterns. The stars in the sky slowly change hour by hour and season by season. The moon mystically moves throughout the sky; slowly growing in size and brightness then shrinking in size. Many prospective observers remain deeply concerned and intimidated in learning how to get around in the sky and how to find interesting things to observe.

Many people are not aware of the interesting and pleasant observations they might be able to do with little to no equipment. The awareness of the beauty of the universe comes from the famed and well published images of the mighty Hubble telescope; a very powerful telescope however in the boundary of space.

While a photo - however dramatic - will always be a photo, to actually see something with your own eyes is awe-inspiring and boggling to the imagination. With no help from an observation tool, looking at the right place under dark conditions you can see the mighty Andromeda galaxy, a distant and intricate sight so far away that it has taken its image over two million years to reach your very eye.

Thus to begin properly in this hobby, we must explain where things are in the sky, help you develop a plan to learn and use the constellations and to give you interesting sights to begin your observing career. It is important to know when beginning this hobby the five sensible guidelines for beginning astronomy.

Five rules of thumb

 * 1) Don't spend a lot of money on it, at first. The fact is having not observed you don't really know what you find interesting. You may find none of it interesting, or you may find all of it interesting but find yourself encumbered by some limitation. It is better to find some way of sampling many things; then you will know how interesting you find it and what you kind of equipment you need to do what you want.
 * 2) Learn as much as you can. Astronomy is definitely a learning hobby and the dedicated observer takes great joy to learn as much as he can about what he is seeing. In fact, knowledge of what is there will definitely help you learn how to "see" subtle features and give grandeur and scale to your observations.
 * 3) Seeing is not easy. One thing many non-observers do not appreciate is a person's tendency merely looking over (rather than really seeing) things. While 'seeing constellations' in the sky can seem difficult at first, rest assured that the everything you will see in the telescope will be, in its own way, difficult to see.
 * 4) A little social contact can make this hobby enjoyable.  While to some extent stargazing (especially if done seriously) is personal, a little social contact can do wonders. Usually an experienced observer can show someone so much more (and easier) than they can describe it.  Also being up late at night can have its own fatigue and a little social contact can be really welcome.
 * 5) Be comfortable. Dress warmly, don't overdo observing to the point at which you're tired and groggy, make your expectations reasonable. A beginner might observe only long enough to find Cassiopeia, Ursa Major and the North star. Perhaps only 5–10 minutes. But if they were out long enough to appreciate the beauty of the stars, see those constellations and were engaged, interested and warm, they are doing great.

Initial experiences
Having this in mind what is a sensible way of beginning astronomy. For those lucky enough to be able to connect to the amateur astronomical community; by far the best way to begin astronomy is to learn from the experts.

For the many people who aren't able to or would rather begin on there, stargazing will be explained here in details.

 * For a learning course using this procedure to begin astronomy, take the course Amateur astronomy/Initial experiences
 * For a list of ways other people have started this hobby and some of their limitations, have a look at the alternative ways to become an observer

Prepare to observe
So if you can't or don't want to connect with amateur astronomers but want to learn about the basics of astronomy, you should start with what the eye can see, learn the constellations and be aware of interesting astronomical sights.

You will need a dark place where you intend to observe, a chart to show you the constellations, and a plan to get your bearings. It's a good idea to time your initial observation to coincide with some interesting astronomical sight.

Each one will be discussed in detail.

Choose a location

Most people that think about observing celestial objects think of far-away skyscapes in the desert where the man-made lights are eliminated. While this might be useful later on, such a place is rarely easy to reach. Most amateur astronomers live in town; your backyard is a good enough place to start learning.

A few reminders:
 * 1) Keep all bright lights out of sight.  Turn off the porch light, find some kind of shadow to get the neighbors security light out of your eyes.  Use a red flashlight to read charts and make notes - if you don't have one, covering a flashlight with a red cloth will do.
 * 2) The sky should be as clear as possible.  Even broken clouds will prevent you from seeing the patterns you need to see to start learning how to recognize constellations.
 * 3) You need a good horizon.  You need to be able to observe most of the way to the horizon without trees or buildings.  The North horizon is particularly important.
 * 4) Don't go too far.  Being close to your observation spot means you can go there more often, which is a very good thing when learning.

Sometimes the backyards isn't the best place. But often a neighborhood park, river, lake shore, or overlook is great. If your nearest place is particularly dark, you will actually work a little harder. Darker skies means more stars, which will make it tough to see the basic patterns of the bright stars that form the constellations. Take heart! Any difficulty when beginning is greatly rewarded when particularly beautiful cosmic sights is in the backyard or around the corner instead of hours away.

Equipment

If you are going on a trip, one of the most important items would be a map, preferably one showing the major intersections. Similarly in the night sky, you need a map as well. Unfortunately the Earth's wobble through the season creates some definite changes in the sky over time. What happens then is that the sky rotates around a star in the north called Polaris (or the North star). You will either need a chart that rotates (called a planisphere) or you will have to use the correct chart for the year.

It is important that the chart is oriented correctly in the sky. Often this holding the chart above you as you view it. You might become disoriented and confused if you keep the chart on a table. Remember to use a red to view the chart at night. This allows you to read in the dark without spoiling your night vision when looking at the sky.

There are three general options for getting a chart.
 * One of the most flexible is a planisphere; (as shown in the photo). It is resilient to dew, durable and can be set to the proper time and date.  These are quite inexpensive and can be gotten from many commercial astronomical stores for around $5.
 * Another option is to download a chart that shows the approximate position of the stars for about a month, just after sunset. Such a star chart can be obtained free at SkyMaps.com
 * You can also use a web page to show the position of the stars from your location at a particularly time. One example can be found at fourmilab.ch

Binoculars
Main Article: Binoculars

Binoculars are very helpful when it comes to stargazing as they offer a wide field of view and a greater magnification. Many Globular clusters and Nebulae with low magnitudes are also observed with a pair of binoculars.

About the weather
You will soon find that as you decide to regularly observe the sky, weather soon comes into play as the great spoiler of many intended observations.

Of course, clear weather is very variable across the planet, with some locations nearly uninterrupted clear weather (such as deserts) and others nearly uninterrupted dismal precipitation. Regardless of location, a committed observer CAN find a way to pick up this hobby. Below are a few guidelines for dealing with weather.
 * 1) Know your seasons; Many places have better or worse seasons for clear weather.  It is a good idea to seek the most frequently clear weather that your location offers while learning.  You will need repetition to really know the sky.
 * 2) Clear and cloudy weather usually comes in bunches;  thus a clear night is likely to be followed by a clear night.  This highlights the importance of being efficient and limited in any individual observing session.  You might want to do it tomorrow night (and the night after that).
 * 3) Learn when it's clear, catch up on sleep when it's cloudy.  Your health is important!
 * 4) Keep notes! Bad weather might not break until you have forgotten most of what you did last time you observe.  Notes both help you be a better observer and make it easier to resume where you left off.

First time in the field
Now the plan is where this all comes together. You want to get a sense of what is where and how it moves. After you have your supplies that were mentioned earlier, go to your chosen spot to stargaze. It's recommended that you bring a chair with you, or other appropriate furniture. Once you are outside, just look around for about 15 minutes to get your eyes used to the dark. Once your eyes have been adjusted, look to the north.

Why the north? The entire sky revolves around a point in the north (that is, for people in the northern hemisphere). This point is very close to a star called Polaris (also known as the North Star). If you learn the northern constellations, you will see them season after season and hour after hour. You can then more easily learn all the other constellations.

Start with the north: Finding Cassiopeia and Ursa Major

title: Stargazing


text:
User:Eml4500.f08.wiki1.aguilar/hw5

=== Notes 10/27: also show eqn(1) equals eqn(2). Do not do the 6 bar truss coding ===

We want to derive $$\mathbf{k}^{(e)}_{\,4x4}= \mathbf{T}^{(e)T}_{4x2} \,\hat{\mathbf{k}}^{(e)}_{2x2}\, \mathbf{T}^{(e)}_{2x4}$$, an equation that can be found on p14.3 and p23.3 of our class notes.

Recall the force-displacement relationship with axial degrees of freedom, $$q^{(e)}$$.

$$\hat{\mathbf{k}}^{(e)}\, \mathbf{q}^{(e)} = \mathbf{p}^{(e)}$$

Moving some things around we can get

$$\hat{\mathbf{k}}^{(e)}\, \mathbf{q}^{(e)} - \mathbf{p}^{(e)} = \mathbf{0}_{2x1} $$     (1)

Plugging this in for the Principle of Virtual Work:

$$ PVW\Rightarrow \mathbf{\hat{w}}\cdot \underbrace{(\hat{\mathbf{k}}^{(e)}\, \mathbf{q}^{(e)} - \mathbf{p}^{(e)})}_{2x1\mathrm{\;for\;all\;} \mathbf{\hat{w}}_{2x1}} = {0}_{1x1} $$         (2)

We showed that equations (1) and (2) are equivalent on p24.1 of our class notes.

Recall,

Here we can see the similarities between the equation for the real displacement and the virtual displacement.

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User:EricaG~enwikiversity

About me
Hi my name is Erica Grimm I live in Concord NC and it is about 20mins away from Charlotte. I love living in a small town where everyone knows everyone else. I love the Lord and i am very involved in church. I just engaged about a year ago and we are looking to get married when i graduate. I cant wait to get out of school and start teaching. I love children and cant wait to teach them.

title: User:EricaG~enwikiversity


text:
User:Eml4500.f08.delta 6.lopez/HW5

Lecture Notes - Oct. 31

$$ \sum{F_{x}} = 0 = -N(x, t) + N(x + dx, t) + f(x, t)dx - m(x)\ddot{u} $$

$$ = \frac{dN}{dx}(x, t)dx + h.o.t. + f(x, t)dx + m(x)\ddot{u} $$

In the above equation, we can neglect the h.o.t. (higher order terms). If we recall Taylor Series Expansion:

$$ f(x+dx) = f(x) + \frac{df(x)}{dx}dx + \frac{1}{2}\frac{d^{2}f(x)}{dx^{2}} + ... $$

The term with the 1/2 coefficient and everything after that term are know as higher order terms. Equation (1) now turns into:

$$ \frac{dN}{dx} + f = m\ddot{u}\; \; \; \; \; \;\; \; \;\; \; \;Eq. (2) $$

This equation is also know as the equation of motion.

$$ N(x, t) = A(x)\sigma (x, t)\; \; \; \; \; \; \; \; \; \; \; \; Eq. (3) $$

$$ \sigma (x, t) = E(x)\varepsilon (x, t) $$

$$ \varepsilon (x, t) = \frac{du}{dx}(x, t) $$

The equation above is known as the Constitutive Relation. Now, if we substitute Equation (3) into Equation (2):

$$ \frac{d}{dx}[N(x, t) = A(x)\sigma (x, t)] + f(x, t) = m(x)\ddot{u} $$

This is know as the Partial Differential Equation of Motion. We need two boundary conditions, since this is a second order derivative with respect to x. These two conditions are the initial displacements and initial velocities.

where $$ u(0, t) = 0 = u(L, t) $$

$$ N(L, t) = A(L)\sigma (L, t) $$

$$ \sigma (L, t) = E(L)\varepsilon (L, t) $$

$$ \varepsilon (L, t) = \frac{du}{dx}(L, t) $$

$$ \frac{du}{dx}(L, t) = \frac{F(t)}{A(L)E(L)} $$

Homework
If we set w1 = 1, then w2...w6 = 0

$$ \mathbf{w}^{T} = \begin{bmatrix} 1 & 0& 0 &  0& 0 & 0 \end{bmatrix} $$

$$ \mathbf{w}*(\mathbf{k}\mathbf{q}- \mathbf{p}) $$

$$ = (1)[\sum{k_{1j}}q_j- p_1] + (0)[\sum{k_{2j}}q_j- p_2]+ (0)[\sum{k_{3j}}q_j- p_3]+ (0)[\sum{k_{4j}}q_j- p_4]+ (0)[\sum{k_{5j}}q_j- p_5]+ (0)[\sum{k_{6j}}q_j- p_6] $$

$$ =\sum{k_{1j}}q_j- p_1 $$

For all other values of w (w2...w6), the trend remains the same, and the answers can be found below.

w2 = 1, w1,3,4,5,6 = 0 $$ =\sum{k_{2j}}q_j- p_2 $$

w3 = 1, w1,2,4,5,6 = 0 $$ =\sum{k_{3j}}q_j- p_3 $$

w4 = 1, w1,2,3,5,6 = 0 $$ =\sum{k_{4j}}q_j- p_4 $$

w5 = 1, w1,2,3,4,6 = 0 $$ =\sum{k_{5j}}q_j- p_5 $$

w6 = 1, w1,2,3,4,5 = 0 $$ =\sum{k_{6j}}q_j- p_6 $$

title: User:Eml4500.f08.delta 6.lopez/HW5


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User:EML4500.f08.delta 6.ramirez/hw5

Justification of eliminating rows 1,2,5,6 to obtain the K2x2 in 2 bar truss system

Force Displacement Relationship gives,

$$ Kd - F = 0 $$ (1)

Equation 2 is given as,

$$ w(Kd - F) = 0 $$ (2)

for all w.

Choice 1 : Select w so that w1 = 1, w2 = 0, ..., w6 = 0

Thus,

$$ W_{1x6}^T = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} $$

Therefore,

$$ w (Kd - F) = 1 [ \sum_{j = 1}^{6}{k_{1j}d_j - F_1]} + 0 [ \sum_{j = 1}^{6}{k_{2j}d_j - F_2]} +...+ 0 [ \sum_{j = 1}^{6}{k_{6j}d_j - F_6]} = 0 $$

which simplifies to equation (1),

$$ w (Kd - F) = \sum_{j = 1}^{6}{k_{1j}d_j - F_1} = 0 $$

Choice 2 : Select w so that w2 = 1, w1,3,4,5,6 = 0

Thus,

$$ W_{1x6}^T = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix} $$

Therefore,

$$ w (Kd - F) = 0 [ \sum_{j = 1}^{6}{k_{1j}d_j - F_1]} + 1 [ \sum_{j = 1}^{6}{k_{2j}d_j - F_2]} +...+ 0 [ \sum_{j = 1}^{6}{k_{6j}d_j - F_6]} = 0 $$

which simplifies to equation (1),

$$ w (Kd - F) = \sum_{j = 1}^{6}{k_{2j}d_j - F_2} = 0 $$

Choice 3 : Select w so that w3 = 1, w1,2,4,5,6 = 0

Thus,

$$ W_{1x6}^T = \begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix} $$

Therefore,

$$ w (Kd - F) = 0 [ \sum_{j = 1}^{6}{k_{1j}d_j - F_1]} +...+ 1 [ \sum_{j = 1}^{6}{k_{3j}d_j - F_3]} +...+ 0 [ \sum_{j = 1}^{6}{k_{6j}d_j - F_6]} = 0 $$

which simplifies to equation (1),

$$ w (Kd - F) = \sum_{j = 1}^{6}{k_{3j}d_j - F_3} = 0 $$

Choice 4 : Select w so that w4 = 1, w1,2,3,5,6 = 0

Thus,

$$ W_{1x6}^T = \begin{bmatrix} 0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix} $$

Therefore,

$$ w (Kd - F) = 0 [ \sum_{j = 1}^{6}{k_{1j}d_j - F_1]} +...+ 1 [ \sum_{j = 1}^{6}{k_{4j}d_j - F_4]} +...+ 0 [ \sum_{j = 1}^{6}{k_{6j}d_j - F_6]} = 0 $$

which simplifies to equation (1),

$$ w (Kd - F) = \sum_{j = 1}^{6}{k_{4j}d_j - F_4} = 0 $$

Choice 5 : Select w so that w5 = 1, w1,2,3,4,6 = 0

Thus,

$$ W_{1x6}^T = \begin{bmatrix} 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix} $$

Therefore,

$$ w (Kd - F) = 0 [ \sum_{j = 1}^{6}{k_{1j}d_j - F_1]} +...+ 1 [ \sum_{j = 1}^{6}{k_{5j}d_j - F_5]} + 0 [ \sum_{j = 1}^{6}{k_{6j}d_j - F_6]} = 0 $$

which simplifies to equation (1),

$$ w (Kd - F) = \sum_{j = 1}^{6}{k_{5j}d_j - F_5} = 0 $$

Choice 6 : Select w so that w6 = 1, w1,2,3,4,5 = 0

Thus,

$$ W_{1x6}^T = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} $$

Therefore,

$$ w (Kd - F) = 0 [ \sum_{j = 1}^{6}{k_{1j}d_j - F_1]} + 0 [ \sum_{j = 1}^{6}{k_{2j}d_j - F_2]} +...+ 1 [ \sum_{j = 1}^{6}{k_{6j}d_j - F_6]} = 0 $$

which simplifies to equation (1),

$$ w (Kd - F) = \sum_{j = 1}^{6}{k_{6j}d_j - F_6} = 0 $$

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User:Eml4500.f08.bottle.brockmiller/hw5

$$ \begin{bmatrix}1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 1 \cdot \bigg[ \sum_{j=1}^{6}{K_{1j}d_{j}} - F_{1} \bigg]+ 0\cdot \bigg[\sum_{j=1}^{6}{K_{2-6j}d_{j}} - F_{2-6} \bigg]=0 $$ Therefore,

$$ \begin{bmatrix}0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1j}d_{j}} - F_{1} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{2j}d_{j}} - F_{2} \bigg]+ 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{3-6j}d_{j}} - F_{3-6} \bigg]=0 $$ Therefore,

$$ \begin{bmatrix}0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1-2j}d_{j}} - F_{1-2} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{3j}d_{j}} - F_{3} \bigg]+ 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{4-6j}d_{j}} - F_{4-6} \bigg]=0 $$ Therefore,

$$ \begin{bmatrix}0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1-3j}d_{j}} - F_{1-3} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{4j}d_{j}} - F_{4} \bigg]+ 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{5-6j}d_{j}} - F_{5-6} \bigg]=0 $$ Therefore,

$$ \begin{bmatrix}0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1-4j}d_{j}} - F_{1-4} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{5j}d_{j}} - F_{5} \bigg]+ 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{6j}d_{j}} - F_{6} \bigg]=0 $$ Therefore,

$$ \begin{bmatrix}0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1-5j}d_{j}} - F_{1-5} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{6j}d_{j}} - F_{6} \bigg]=0 $$ Therefore,

title: User:Eml4500.f08.bottle.brockmiller/hw5


text:
User talk:Stan Shebs

 Hello Stan Shebs, 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! --mikeu talk 22:12, 6 November 2008 (UTC)

title: User talk:Stan Shebs


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title: User:Eml4500.f08.echo.sell.cm/HW5


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User:Jfrankc08

Hi, my name's Franco R Lucien. I'm originally from Haiti, but I've lived in the states since the age of twelve. Having been practically raised in the US, English has become second nature, and I speak it just as fluently--if not more--than my maternal tongues (French/Haiti Creole). I speak German comfortably enough to survive a short visit in Germany. However, I am committed to mastering the German language, as well as picking up a couple others as the time would allow--such as Portuguese and Dutch.

I take interest mostly in the French Department, and spend sometime there. I have recently became part of the French Mentoring Project, thus have consequently dedicated some time aside for that purpose. I've also introduced the Department of Haitian History, which is currently under construction. There, I hope to share my knowledge of that subject with anyone who is an interested learner. If you are enlightened in the field, your contributions will be most welcomed.

If you need to contact me please email me for faster response. Otherwise leave me a response on my talk page

title: User:Jfrankc08


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User:Eml4500.f08.delta 6.castrillon/HW 5

=Justification=

We can justify the elimination of rows 1,2,5,6 to obtain $$\bar{\mathbf{K}}$$ in a 2-bar truss.

 * We begin with the FD relationship, writing it as $$K$$$$d$$ = $$F$$. Based on the model 2-bar system, we can generalize this method for any kind of application.

 * $$K_{6x6}$$$$d_{6x1}$$ = $$F_{6x1}$$

Therefore,
 * $$K$$$$d$$-$$F$$=$$0_{6x1}$$

 * Principle of Virtual Work:

 * $$w$$$$_{6x1}$$•($$K$$$$d$$-$$F$$)$$_{6x1}$$ = $$0_{1x1}$$; for all $$w$$$$_{6x1}$$

Where w is the weighting matrix, and the equation is equalized to zero as a sacalar (or 1x1 matrix). In other words, the weighting matix is dotted with the FD relationship matrix to obtain a scalar. Also, according to the wise Prof. Vu-Quoc, "$$w$$ was not selected out of frivolity," as it is called the weighting coefficient or matrix.We say that the FD relationship is equivalent to the PVW.

=Proof of FD relationship and PVW equality=

 * 1) FD relationship to PVW is trivial because if FD relationship is true, then $$w$$(0)=0, this being obvious
 * 2) We now want to show how to go from PVW to FD relationship.

 * We have the PVW: $$w$$$$_{6x1}$$•($$K$$$$d$$-$$F$$)$$_{6x1}$$ = $$0_{1x1}$$; for all $$w$$$$_{6x1}$$

 * Choice 1: Select $$w$$ such that $$w_{1}$$=1, $$w_{2}$$=...=$$w_{6}$$=0

 * Therefore,
 * $$w^{T}_{6x1}$$=$$\begin{bmatrix}

1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$
 * $$w$$•($$K$$$$d$$-$$F$$) = 1•$$\left[{\sum_{j=1}^{6}K_{1j}d_{j}-F_1} \right]$$+0•$$\left[{\sum_{j=1}^{6}K_{2j}d_{j}-F_2} \right]$$+...+0•$$\left[{\sum_{j=1}^{6}K_{6j}d_{j}-F_6} \right]$$=0
 * $$\sum_{j=1}^{6}{K_{1j}d_j}=F_1$$ (1$$^{st}$$equation)

 * Choice 2: Select $$w$$ such that $$w_{1}$$=0, $$w_{2}$$=1, $$w_{3}$$=...=$$w_{6}$$=0

 * Therefore,
 * $$w^{T}_{6x1}$$=$$\begin{bmatrix}

0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix}$$
 * $$w$$•($$K$$$$d$$-$$F$$) = 0•$$\left[{\sum_{j=1}^{6}K_{1j}d_{j}-F_1} \right]$$+1•$$\left[{\sum_{j=1}^{6}K_{2j}d_{j}-F_2} \right]$$+0•$$\left[{\sum_{j=1}^{6}K_{3j}d_{j}-F_3} \right]$$+...=0
 * $$\sum_{j=1}^{6}{K_{2j}d_j}=F_2$$ (2$$^{nd}$$equation)

 * Choices 3 through 6 repeat this pattern.

Ultimately, we will have achieved the FD relationship.

=PVW: Furthermore=

According to the boundary conditions (BCs) of the 2-bar truss, $$d_1=d_2=d_5=d_6=0$$.

The weighting cofficient must be kinematically admissible. It cannot violate the BCs: $$w_1=w_2=w_5=w_6=0$$. Therefore, weighting coefficients go hand-in-hand with virtual displacements, as explained by calculus of variations.

For the 2-bar truss at hand, it is evident that $$w$$•($$K$$$$d$$-$$F$$) can be reduced to a rectangular matrix multiplied by a 2x1 matrix with a 6x1 matrix as a result.

 * $$w$$•($$K$$$$d$$-$$F$$) =$$\begin{bmatrix}

w_3\\ w_2 \end{bmatrix}$$•($$\bar{K}_{2x2}$$$$\bar{d}_{2x1}$$-$$\bar{F}_{2x1}$$)=0

Where,
 * $$\bar{K}$$=$$\begin{bmatrix}

K_{33} & K_{34}\\ K_{43} & K_{44} \end{bmatrix}$$
 * $$\bar{d}$$=$$\begin{Bmatrix}

d_3\\ d_4 \end{Bmatrix}$$
 * $$\bar{F}$$=$$\begin{Bmatrix}

F_3\\ F_4 \end{Bmatrix}$$


text:
User:Eml4500.f08.gravy.jad/Notes5

$$\ \,$$ $$\  \,$$ $$\  \,$$

Lecture 27

$$\ \hat{\mathbf{w}}_{2x1} = \,$$virtual axial displacement, corresponding $$\ \mathbf{q}^{e}_{2x1} \,$$

$$\ \mathbf{w}_{4x1} = \,$$virtual displacement in global coordinate system, with corresponding $$\ \mathbf{d}^{e}_{4x1} \,$$

Now if you replace Eqs (3) and (4) into Eq (2), the end result is:

$$\ \left ( \mathbf{T}^{(e)} \mathbf{w} \right ) \cdot \left\lbrack \hat{\mathbf{k}}^{(e)} \left ( \mathbf{T}^{(e)} \mathbf{d}^{(e)} \right ) - \mathbf{P}^{(e)} \right\rbrack = 0 \,$$for all $$\ \mathbf{w}_{4x1} \,$$ [This is Eq (5)]

Recall: $$\ \left ( \mathbf{A}_{pxq} \mathbf{B}_{qxr} \right )^{T} = \mathbf{B}^{T} \mathbf{A}^{T} \,$$ [This is Eq 6]

As an Ex:

$$\ \mathbf{A}_{2x3} = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}_{2x3} \,$$ $$\ \mathbf{B}_{3x3} = \begin{bmatrix} 7 & 8 & 9 \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}_{3x3} \,$$

(insert hw5 verify equation 6)

Recall: $$\ \mathbf{a}_{nx1} \cdot \mathbf{b}_{nx1} = \mathbf{a}^{T}_{1xn} \mathbf{b}_{nx1} = \sharp _{1x1} \,$$ where # equals some number, a scalar [Eq 7]

Now, apply (7) and (6) to (5), to reveal: $$\ \left ( \mathbf{T}^{(e)} \mathbf{w} \right )^{T} \left \lbrack \hat{\mathbf{k}}^{(e)} \left ( \mathbf{T}^{(e)} \mathbf{d}^{(e)} \right ) - \mathbf{P}^{(e)} \right \rbrack =0_{1x1} \,$$for all $$\ \mathbf{w}_{4x1} \,$$

$$\ \Rightarrow \left ( \mathbf{w}^{T} {\mathbf{T}^{(e)}}^{T} \right ) \left \lbrack \hat{\mathbf{k}}^{(e)} \left ( \mathbf{T}^{(e)} \mathbf{d}^{(e)} \right ) - \mathbf{P}^{(e)} \right \rbrack =0 \,$$

$$\ \Rightarrow \mathbf{w} \cdot \left \lbrack \left ( {\mathbf{T}^{(e)}}^{T} \hat{\mathbf{k}}^{(e)} \underline{T}^{(e)} \right ) \mathbf{d}^{(e)} - \left ( {\mathbf{T}^{(e)}}^{T} \mathbf{P}^{(e)} \right ) \right \rbrack =0 \,$$for all $$\ \mathbf{w}_{4x1} \,$$

$$\ \Rightarrow \mathbf{w} \cdot \left \lbrack \mathbf{k}^{(e)} \mathbf{d}^{(e)} - \mathbf{f}^{(e)} \right \rbrack =0 \,$$for all $$\ \mathbf{w}_{4x1} \,$$

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

insert hw5 proof

So far, we are dealing with the discrete case [not continuous][matrices], up until now...

NOW, the continuous case: (PDEs)

$$\ \Rightarrow \,$$Motivational Model Problem: Elastic bar with varying A(x), E(x), subjected to varying axial load (distributed) + concentrated load + inirtia force(dynamics) [the distributed and concentrated loads are both time dependent]

HW5 literature search

$$\ \sum F_{x} = 0 = -N(x,t) + N(x + dx, t) + f(x,t)dx - m(x)dx \ddot{u} \,$$

$$\ \sum F_{x} = \frac{dN}{dt} (x,t)dx + H.O.T + f(x,t)dx - m(x)dx \ddot{u} \,$$[Eq 1]

Recall the Taylor series exp: $$\ f(x+dx) = f(x) + \frac{df}{dx}dx+ \frac{1}{2} \frac{d^{2}f}{dx^{2}}(x)dx+... \,$$from the third term onwards are are H.O.Ts = Higher Order Terms

Eq(1) $$\ \Rightarrow \frac{dN}{dx} + f = m \ddot{u} \,$$ Equation of motion (EOM) [Eq 2]

$$\ N(x,t) = A(x) \sigma(x,t) \,$$where $$\ \sigma(x,t) = E(x) \epsilon(x,t) \,$$where $$\ \epsilon(x,t) = \frac{du}{dx}(x,t) \,$$[Eq 3]

Combining (2) and (3) yields: $$\ \frac{d}{dx} \left \lbrack A(x)E(x) \frac{du}{dx} + f(x,t) = m(x) \ddot{u} \right \rbrack \,$$where $$\ \ddot{u} = \frac{d^{2}u}{dt^{2}} \,$$[Eq 4]

Eq 4 describes the PDE of motion.

Now, we need two boundary conditions (2nd order derivative with respect to x), AND two initial conditions (2nd derivative with respect to t) [initial displacement and initial velocity]

$$\ u(0,t) = 0 = u(L,t) \,$$

1)$$\ u(0,t)=0 \,$$, 2)$$\ N(L,t) = F(t) \,$$; $$\ A(L) \sigma (L,t) = A(L)E(L) \epsilon (L,t) \,$$where $$\ \epsilon (L,t) = \frac{du}{dx} (x,t) \,$$

$$\ \Rightarrow \frac{du}{dx} (L,t) = \frac{F(t)}{A(L)E(L)} \,$$@ x=l

Initial conditions: @$$\ t=0, u(x,0) = u(x) \,$$known function displacement, and $$\ \frac{du}{dx}(x,0) = \dot{u}(x,0)= \overline{V}(x) \,$$known function velocity

title: User:Eml4500.f08.gravy.jad/Notes5


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User:Eml4500.f08.delta 6.krueger/HW 2 Bar Truss Sytem

MATLAB Code The code above is the Matlab example of the 2 bar truss system discussed extensively throughout this class. The results are incorrect. Instead of a 2x1 matrix the results array should be a 2x6 matrix. The first column is correct but the absence of the last two columns means that the bug in the code was not corrected.


title: User:Eas4200c.f08.ZYX.reger/Week9,10


text:
User:Eml4500.f08.jamama.jan/Submission5 matlab

{| class="toccolours collapsible collapsed" style="width:100%" ! The corrected 2-Bar truss code

Code Output
k_local = 0.75       -0.75        -0.75         0.75

k = 0.5625     0.32476      -0.5625     -0.32476      0.32476       0.1875     -0.32476      -0.1875      -0.5625     -0.32476       0.5625      0.32476     -0.32476      -0.1875      0.32476       0.1875

k_local = 5   -5    -5     5

k = 2.5        -2.5         -2.5          2.5         -2.5          2.5          2.5         -2.5         -2.5          2.5          2.5         -2.5          2.5         -2.5         -2.5          2.5

K = 0.5625     0.32476      -0.5625     -0.32476            0            0      0.32476       0.1875     -0.32476      -0.1875            0            0      -0.5625     -0.32476       3.0625      -2.1752         -2.5          2.5     -0.32476      -0.1875      -2.1752       2.6875          2.5         -2.5            0            0         -2.5          2.5          2.5         -2.5            0            0          2.5         -2.5         -2.5          2.5

R = 0    0     0     7     0     0

d = 0           0        4.352       6.1271            0            0

reactions = -4.4378     -2.5622       4.4378      -4.4378

results = 1.7081      5.1244       5.1244       0.6276        3.138        6.276

The different output occured because the E and A were not of correct specification. E and A should have been specified with "i" subscripts, such as E(i), A(i), so that with appropriate truss mambers' certain values should be taken.

 * }

{| class="toccolours collapsible collapsed" style="width:100%" ! 2-D 6-bar truss and its deformation (magnified by a factor of 1000) at constant Modulus of Elasticity (E = $$2*10^{11}$$)

Code Output
K = 1.0e+008 * 0.8536   0.3536   -0.5000         0         0         0         0         0   -0.3536   -0.3536    0.3536    0.3536         0         0         0         0         0         0   -0.3536   -0.3536   -0.5000         0    0.8536   -0.3536         0         0         0         0   -0.3536    0.3536         0         0   -0.3536    1.0202         0         0         0   -0.6667    0.3536   -0.3536         0         0         0         0    0.7155   -0.3578         0         0   -0.7155    0.3578         0         0         0         0   -0.3578    0.1789         0         0    0.3578   -0.1789         0         0         0         0         0         0    0.7155    0.3578   -0.7155   -0.3578         0         0         0   -0.6667         0         0    0.3578    0.8456   -0.3578   -0.1789   -0.3536   -0.3536   -0.3536    0.3536   -0.7155    0.3578   -0.7155   -0.3578    2.1382         0   -0.3536   -0.3536    0.3536   -0.3536    0.3578   -0.1789   -0.3578   -0.1789         0    1.0649

d = 1.0e-003 * 0        0    0.2131    0.2500         0         0         0         0   -0.0061    0.0122

d_def = 1.0e-003 * 0        0    0.2131    0.2500         0         0         0         0   -0.0061    0.0122

nodes_def = 0        0    4.0002    0.0002         0    3.0000    4.0000    3.0000    2.0000    2.0000

d_def_aug = 0        0    0.2131    0.2500         0         0         0         0   -0.0061    0.0122

nodes_def_aug = 0        0    4.2131    0.2500         0    3.0000    4.0000    3.0000    1.9939    2.0122

Stress = 1.0e+007 * 1.0655  -0.0927   -0.0977   -1.6665    0.0307   -0.0002

Force_axial = 1.0e+004 * 1.0655  -0.0927   -0.0977   -1.6665    0.0307   -0.0002

6-bar truss image (the deformation was multiplied by a factor of 1000 to show a clearler image)
Blue solid line $$ \Rightarrow $$ defored truss

Red dashed line$$ \Rightarrow $$ undefored truss

 * }

{| class="toccolours collapsible collapsed" style="width:100%" ! 2-D 6-bar truss and its deformation (magnified by a factor of 1000) at constant (E = $$2*10^{11}$$) and at a varying specified Modulus of Elasticity

Code Output (Note, the output is only the varying E parameters)
K = 1.0e+008 * 0.7639   0.3889   -0.3750         0         0         0         0         0   -0.3889   -0.3889    0.3889    0.3889         0         0         0         0         0         0   -0.3889   -0.3889   -0.3750         0    0.6932   -0.3182         0         0         0         0   -0.3182    0.3182         0         0   -0.3182    0.9849         0         0         0   -0.6667    0.3182   -0.3182         0         0         0         0    0.7155   -0.3578         0         0   -0.7155    0.3578         0         0         0         0   -0.3578    0.1789         0         0    0.3578   -0.1789         0         0         0         0         0         0    0.8944    0.4472   -0.8944   -0.4472         0         0         0   -0.6667         0         0    0.4472    0.8903   -0.4472   -0.2236   -0.3889   -0.3889   -0.3182    0.3182   -0.7155    0.3578   -0.8944   -0.4472    2.3171    0.1602   -0.3889   -0.3889    0.3182   -0.3182    0.3578   -0.1789   -0.4472   -0.2236    0.1602    1.1096

d = 1.0e-003 * 0        0    0.2649    0.2608         0         0         0         0    0.0006   -0.0012

d_def = 1.0e-003 * 0        0    0.2649    0.2608         0         0         0         0    0.0006   -0.0012

nodes_def = 0        0    4.0003    0.0003         0    3.0000    4.0000    3.0000    2.0000    2.0000

d_def_aug = 0        0    0.2649    0.2608         0         0         0         0    0.0006   -0.0012

nodes_def_aug = 0        0    4.2649    0.2608         0    3.0000    4.0000    3.0000    2.0006    1.9988

Stress = 1.0e+007 * 0.9932   0.0096    0.0101   -1.7389   -0.0033   -0.0002

Force_axial = 1.0e+004 * 0.9932   0.0096    0.0101   -1.7389   -0.0033   -0.0002

6-bar truss image (the deformation was multiplied by a factor of 1000 to show a clearler image)
Blue solid line $$ \Rightarrow $$ defored truss of constant E

Red dashed line$$ \Rightarrow $$ undefored truss

Black solid line$$ \Rightarrow $$ defored truss of varying E

 * }


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User talk:Poseurtech

Hello,

I've just deleted your userpage since our wikis are not for promotion. You are welcome to contribute constructively, however linking as you have done on many wikis is unwelcome. – Mike.lifeguard | @en.wb 04:01, 7 November 2008 (UTC)

title: User talk:Poseurtech


text: File:3D Stress Field.JPG

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title: File:Spacetruss.JPG


text: File:Element DOF's-Forces in global xyz coord for 6x1 matrixes.jpg


text: File:Stress.GIF

title: File:Stress.GIF


text: File:Axial FD Relation.jpg

title: File:Axial FD Relation.jpg


text:
User:EML4500.f08.Ateam.Miller/homework5

Expression for element forces in terms of transformation matrix and element axial forces
From the previous derivations, it was determined that the element axial force was related to the element forces in global coordinates by

$$p^{(e)}_{2x1}=T^{(e)}_{2x6}f^{(e)}_{6x1}$$

Therefore, the expression for the element forces in global coordinates in terms of the transformation matrix and the element axial forces is

$$f^{(e)}_{6x1}=T^{(e)T}_{6x2}p^{(e)}_{2x1}$$

title: User:EML4500.f08.Ateam.Miller/homework5


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text: File:VITELLO xz.jpg

title: File:VITELLO xz.jpg


text: File:VITELLO yx.jpg

title: File:VITELLO yx.jpg


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text: File:VITELLO p3p3p3.jpg

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text:
Talk:Debates

* (cur) (last) 06:54, 2 November 2008 Moonriddengirl (Talk | contribs) (641 bytes) (Proposing article for deletion per WP:PROD. using TW) (undo) * (cur) (last) 08:41, 28 October 2008 Sarnowski (Talk | contribs) (197 bytes) (←Created page with 'The debate was held on October, 3rd in 2007. The participants were collegues professor Richard Dawkins and professor John Lennox. The topic was God's existence usin...')
 * (cur) (last) 17:17, 6 November 2008 Dennis Brown (Talk | contribs) (650 bytes) (agree) (undo)

title: Talk:Debates


text:
User talk:Mrak~enwikiversity

Welcome
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User talk:Collinito

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User talk:Nanakaly

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User talk:Arbranique

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User talk:Carol+2

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User talk:Bolazz

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User:Thomas Bliem

See http://www.mi.uni-koeln.de/~tbliem/.

title: User:Thomas Bliem


text:
User:Eas4200c.f08.vqcrew.c/Homework 5/mememe

MATLAB Airfoil Problem Continued
The MATLAB airfoil problem, continued for HW 5, consisted of two parts. Both parts dealt with analysis of the effects of bidirectional bending on an arbitrary 4-digit NACA airfoil. Part I refers to an idealized case where only the axial effects of the stringers are considered. Part 2 considers the "full-blown" case - the bending effects of the airfoil skin, spar webs at 1/4 and 3/4 chord, and the stringers are considered. Below is a list of given and researched values used during the problem:

 * $$ A_B = 2 x 10^{-4}\;m^{2}\;$$ - Cross sectional area of the stringer at point B
 * $$ A_E = 2 x 10^{-4}\;m^{2}\;$$ - Cross sectional area of the stringer at point E
 * $$ A_F = 1 x 10^{-4}\;m^{2}\;$$ - Cross sectional area of the stringer at point F
 * $$ A_H = 1 x 10^{-4}\;m^{2}\;$$ - Cross sectional area of the stringer at point H
 * $$ t_s = .002\;m\;$$ - Airfoil skin thickness
 * $$ t_w = .003\;m\;$$ - Spar web thickness
 * $$ t_{str} = .005\;m\;$$ - Stringer flange thickness
 * $$ M_y = -1250\;N\cdot m\;$$ - y-direction bending moment (compressive)
 * $$ M_z = 500\;N\cdot m\;$$ - z-direction bending moment (tensile)
 * $$ \sigma_{u} = 1860\;N/m\;$$ - Ultimate tensile stress for steel 300M

Part I: Ideal Case

Figure ## shows a generic airfoil with spars located at 1/4 and 3/4 chord. The points B, E, H, and F, representing the intersection of the spars with the airfoil contour, are labeled. For the purposes of Part I, these 4 points represent the location of the idealized stringers. The stringer cross-section is not considered, and each stringer is represented as a "point area" during the calculations. Below is the full MATLAB code for this part followed by a sample output run.

 * {| class="collapsible collapsed"

!MATLAB Code Part I - Bending Effects, Only Considering Stringers
 *
 *
 * }

 * {| class="collapsible collapsed"

!MATLAB Code Part I Output >> vqcrew_4str(.5,2,4,15,1000);
 *
 *
 * }

Figuree ## is the output plot from the MATLAB code for Part I. Several features are notable. First, the solid black circles located at the spar-airfoil intersections represent each stringer. The blue dashed lines represent the airfoil skin and spar webs. The red dash-dotted line running across the airfoil represents the neutral axis, and the black crosshair is the centroid of the stringers. The solid red circle represents the location of maximum normal stress on the airfoil.

A notable result from the output are the ultimate bending moments for the supplied ultimate stress. The calculated moments are of opposite sign to the original given moments. This is clearly explained by the result of the maximum bending stress analysis for the original loading condition. The determined stress was negative, indicating a compressive stress. The provided ultimate stress was given as a tensile stress, and since the neutral axis and thus the point of maximum normal stress do not change for this portion of the problem, the signs on the moments must switch to produce a tensile stress at that point. This also explains why much smaller magnitude moments produce a stress in excess of the ultimate stress by an order of magnitude - the ultimate stress provided is for tensile stresses, not compressive.

Part II: Full-Blown Case

For part II, the entirety of the airfoil was considered, including the cross-section of the stringers. In order to generate this cross-section, the assumption was made that the stringers were manufactured by stamping from a metal sheet of consistent thickness $$t_{str}\;$$. The following equation was used to determine the length of one flange of each stringer:

 * $$L_{brack} = .5*(A_{str}/2)/t_{str}\;$$

The stringers were assumed to be flush with the inner airfoil contour and the spar web wall. One stringer would consist of two "L-shaped" brackets on either side of the spar. Since the airfoil is assumed to have curvature, the brackets will not be identical, and the assumed manufacturing process means the angle between the two flanges will not be 90 degrees.

Figures ## and ## shows a generic depiction at both the 1/4 chord spar and the 3/4 chord spar. The diagram of the 1/4 chord spar is more representative of the general stringer shape. The stringers for the 3/4 chord spar are presented differently, however. The result of the above equation for the two stringers at the 3/4 spar was a bracket length of .005 m, which is identical to the provided stringer thickness. If one goes by the geometry provided in Figure ##, this essentially provides a quadrilateral cross-section, which is what is drawn in Figure ## and also used in the code for Part II, which is provided below. After the code is a sample output run.

 * {| class="collapsible collapsed"

!MATLAB Code Part II - Full-Blown NACA Airfoil
 *
 *
 * }

 * {| class="collapsible collapsed"

!MATLAB Code Part II Output
 *
 *
 * }

Figure ## depicts the full output plot of the Part II code. The airfoil skin and spar webs were plotted as solid blue lines. The stringer cross-sections are plotted as small black circles, with each portion of the stringer wall represented with 5 points for ease of viewing. Zoomed in plots of the 1/4 spar and 3/4 spar are provided in Figures ## and ## respectively. The neutral axis is again plotted as a red dash-dotted line.

There are two centroids plotted on this airfoil; a zoomed in view of them is available in Figure ##. The green crosshair represents the centroid of the stringers when the stringer cross-sections are considered. The blue crosshair is the crosshair calculated in Part I. Also plotted is a red solid circle again representing the location of maximum bending stress on the airfoil, and a green solid circle represents the location of maximum bending stress in a stringer. Comparison and conclusions about these results follows in the next section.

Comparison and Conclusions
Below is a table showing selected data from the output of both cases. The stringer centroids, locations of maximum bending stresses, and neutral axis angle for both cases are more or less identical, which is not surprising given both are based off of the same airfoil with the same input data. The most apparent differences are in the data for ultimate bending moments and maximum normal bending stress. The higher maximum stress at lower supplied moment indicates the full-blown airfoil is slightly less efficient at carrying bending loads than the idealized airfoil. This is probably due primarily to the stringers being much, much better at carrying bending loads than either the spars or the airfoil skin. The only other notable point from the analysis is the stringer which carries the most stress in the full-blown case, which was found to be stringer B. This is not surprising, as this stringer is closest to the position of maximum thickness, which in combination with the positive slope of the neutral axis means it would be the farthest stringer from the neutral axis, and thus experience the highest stress.

 * {| border=1

 * bgcolor="red"|
 * bgcolor="#99ccff" style="text-align: center;" |Ideal Case
 * bgcolor="#99ccff" style="text-align: center;" |Full-Blown Case
 * bgcolor="red"|Stringer centroid (meters,meters)
 * bgcolor="#99ccff" style="text-align: center;" | (0.2083 0.0080)
 * bgcolor="#99ccff" style="text-align: center;" | (0.2088 0.0080)
 * bgcolor="red"|Moment of Inertia Tensor Components (m^4,m^4,m^4)
 * bgcolor="#99ccff" style="text-align: center;" | (6.6875e-007,3.4375e-005,9.3322e-007)
 * bgcolor="#99ccff" style="text-align: center;" | (6.2884e-007,1.4533e-004,4.8673e-006)
 * bgcolor="red"|Neutral Axis Slope (radians)
 * bgcolor="#99ccff" style="text-align: center;" | 0 .0345
 * bgcolor="#99ccff" style="text-align: center;" | 0.0347
 * bgcolor="red"|Maximum normal bending stress (Pascals)
 * bgcolor="#99ccff" style="text-align: center;" | -8.1995e+007 Pascals
 * bgcolor="#99ccff" style="text-align: center;" | -1.1641e+008 Pascals
 * bgcolor="red"|Location of maximum normal bending stress (meters,meters)
 * bgcolor="#99ccff" style="text-align: center;" | (0.1436 0.0467)
 * bgcolor="#99ccff" style="text-align: center;" | (0.1434 0.0478)
 * bgcolor="red"|Ultimate Bending Moments (N-m, N-m)
 * bgcolor="#99ccff" style="text-align: center;" | (28355,-11342)
 * bgcolor="#99ccff" style="text-align: center;" | (20970,-8388)
 * }
 * bgcolor="#99ccff" style="text-align: center;" | (0.1434 0.0478)
 * bgcolor="red"|Ultimate Bending Moments (N-m, N-m)
 * bgcolor="#99ccff" style="text-align: center;" | (28355,-11342)
 * bgcolor="#99ccff" style="text-align: center;" | (20970,-8388)
 * }
 * }
 * }


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User:Eml4500.f08.gravy.sms/HW5

FEM Analysis 3-Bar Space Truss Using MATLAB
The following Matlab code was taken from the 3-bar space truss system found in the textbook on page 230. The code uses three other functions: SpaceTrussElement.m, NodalSoln.m, and SpaceTrussResults.m. These functions were obtained from the online resources located at the book's website here.

Results for 3-bar Space Truss
Note that the conn and lmm arrays are also shown.

Plots for 3-bar Space Truss
This is a 3-D plot with a view from (3,3,3):

This is a 2-D plot down the X-axis:

This is a 2-D plot down the Y-axis:

This is a 2-D plot down the Z-axis:

title: User:Eml4500.f08.gravy.sms/HW5


text:
User:Thuvack/Mind-Logs/2008 November

November-07-2008: While I was away
I have just completed the 9 weeks of primary site work & Related inspections for a major shopping centre that My employing company was involved in. To be honest I quite enjoyed my self out there doing field work. I had ample opportunities to explore the barrier between Theory and Practical application of Electrical Reticulation. The insight gained by working with veterans of the field has been both inspiring and eye opening.

While I was away, I've been popping in every now and again to check on my watch list and other areas of interest around this virtual campus. Needless to say that a lot has happened. Some unbelievable, and some down right unforgivable.

Yet through all that I dare say this community emerged out of it ( obviously not in one piece ), and we continue to work towards a common goal despite any differences we may have. To me conflict is not so important as the methods employed to resolve it. I dare say we have learned our lesson when it comes to conflict resolution and following protocol in achieving it. We obviously need to improve on policy of conflict resolution.

What i yearn for is a way to get back those skilled editors we have lost as a result of the recent events. I don’t think it is at all possible to replace a skilled editor, and we have to face the truth that they were very good at what they did.

I have also started preparing my self for activities I would to get involved or start here in 2009. I am also planning to attend the Wikimania 2009, depending on funding I will get to supplement what i will have saved up that point. For now though, I want to concentrate on those courses I started during the course of the year.

Happy wikiediting -- Thuvack | talk 12:50, 7 November 2008 (UTC)

title: User:Thuvack/Mind-Logs/2008 November


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