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User:Eml4500.f08.echo.mott/HW5

=HW Report 5= The previous version was submitted.

Axial FD Relation
The axial force-displacement relation for a bar element remains the same whether 2-D or 3-D:

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

Element dofs & Element Forces in Global x-y-z Coordinates
The transformation matrix in 3-D can be shown by

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

The axial F-D relation can be written as:

$$\begin{bmatrix}F_{1} \\ F_{2}\end{bmatrix}=\begin{bmatrix}k & -k \\ -k & k\end{bmatrix}\begin{bmatrix}d_{1} \\ d_{2}\end{bmatrix}=\begin{bmatrix}P_{1} \\ P_{2}\end{bmatrix}$$

The conversion from local to global axial coordinates is:

$$\mathbf{d = T {d^{(e)}}_{6x1}}$$

$$\begin{bmatrix}d_{1} \\ d_{2}\end{bmatrix}=\begin{bmatrix}l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0 \\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)}\end{bmatrix}\begin{bmatrix}{d_{1}}^{(e)} \\ {d_{2}}^{(e)} \\ {d_{3}}^{(e)} \\ {d_{4}}^{(e)} \\ {d_{5}}^{(e)} \\ {d_{6}}^{(e)}\end{bmatrix}$$

Where l, m, and n are the director cosines of each axis.

$$l^{(e)}=\frac{x_{2}-x_{1}}{L}$$

$$m^{(e)}=\frac{y_{2}-y_{1}}{L}$$

$$l^{(e)}=\frac{z_{2}-z_{1}}{L}$$

$$L^2={(x_{2}-x_{1})}^{2} + {(y_{2}-y_{1})}^{2} + {(z_{2}-z_{1})}^{2}$$

Beginning with the element F-D relation and substituting the global transformation matrix,

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

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

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

The global 3-D F-D relation can be written as:

$$ \frac{EA}{L}\begin{bmatrix}l^{2(e)} & m^{(e)}l^{(e)} & n^{(e)}l^{(e)} & -l^{2(e)} & -m^{(e)}l^{(e)} & -n^{(e}l^{(e)} \\ m^{(e)}l^{(e)} & m^{2(e)} & m^{(e)}n^{(e)} & -m^{(e)}l^{(e)} & -m^{2(e)} & -m^{(e)}n^{(e)} \\ n^{(e)}l^{(e)} & m^{(e)}n^{(e)} & n^{2(e)} & -n^{(e)}l^{(e)} & -m^{(e)}n^{(e)} & -n^{2(e)} \\ -l^{2(e)} & -m^{(e)}l^{(e)} & -n^{(e)}l^{(e)} & l^{2(e)} & m^{(e)}l^{(e)} & n^{(e)}l^{(e)} \\ -m^{(e)}l^{(e)} & -m^{2(e)} & -m^{(e)}n^{(e)} & m^{(e)}l^{(e)} & m^{2(e)} & m^{(e)}n^{(e)} \\ -n^{(e)}l^{(e)} & -m^{(e)}n^{(e)} & -n^{2(e)} & n^{(e)}l^{(e)} & m^{(e)}n^{(e)} & n^{2(e)} \\\end{bmatrix}\begin{bmatrix}u_{1} \\ v_{1} \\ w_{1} \\ u_{2} \\ v_{2} \\ w_{2} \\\end{bmatrix}=\begin{bmatrix}F_{1x} \\ F_{1y} \\ F_{1z} \\ F_{2x} \\ F_{2y} \\ F_{2z} \\ \end{bmatrix} $$

Matlab Solutions

 * Please refer to this page.

Monday Oct. 20, 2008
Justification of eliminating rows 1, 2, 5, and 6 to obtain $$\mathbf{\bar{k}}_{2x2}$$ in 2-bar truss: (p 10-1 & 23-2)

FD Relation:

$$\mathbf{K}_{6x6}\mathbf{d}_{6x1}=\mathbf{F}_{6x1} \Rightarrow \mathbf{K}\mathbf{d}-\mathbf{F}=\underline{0}_{6x1} $$ . . . (1)

PVW:

$$\mathbf{W}_{6x1}\cdot (\mathbf{K}\mathbf{d}-\mathbf{F})=0_{1x1}$$ for all $$\mathbf{W}_{6x1}$$. . . (2)

Where the weighting matrix is $$\mathbf{W}_{6x1}$$

$$Eq (1) \Leftrightarrow Eq (2)$$

Proof:

A) (1) $$\Rightarrow$$ (2): Trivial

B) Want to show (2) $$\Rightarrow$$ (1) :

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

for all W

Choice 1: Select W at $$w_1=1, w_2=w_3=w_4=w_5=w_6=0$$

$$\mathbf{W}_{1x6}^T=\begin{vmatrix} 1 & 0 & 0 & 0 & 0 & 0 \end{vmatrix}$$

$$\mathbf{W}\cdot (\mathbf{kd}-\mathbf{F})=1.\left[\sum_{j=1}^6 k_{1j}d_j-F_1\right]+0\left[\sum_{j=1}^6 k_{2j}d_j-F_2\right]+...+0\left[\sum_{j=1}^6 k_{6j}d_j-F_6\right]=0$$

This leads to:

$$\sum_{j=1}^6 k_{1j}d_j=F_1$$ (1st Equation)

Choice 2: Select W at $$w_1=0, w_2=1, w_3=w_4=w_5=w_6=0$$

$$\mathbf{W}_{1x6}^T=\begin{vmatrix} 0 & 1 & 0 & 0 & 0 & 0 \end{vmatrix}$$

$$\mathbf{W}\cdot (\mathbf{kd}-\mathbf{F})=0\left[\sum_{j=1}^6 k_{1j}d_j-F_1\right]+1\left[\sum_{j=1}^6 k_{2j}d_j-F_2\right]+0\left[\sum_{j=1}^6 k_{3j}d_j-F_3\right]...+0\left[\sum_{j=1}^6 k_{6j}d_j-F_6\right]=0$$

This leads to:

$$\sum_{j=1}^6 k_{2j}d_j=F_2$$ (2nd Equation)

Hence kd=F (or Eq (1))

PVW: Accounting for boundary conditions:

2 bar truss: $$d_1=d_2=d_5=d_6=0$$

Weight coefficient must be "kinematically admissible," i.e. cannot violate the boundary conditions.

Weighting coefficients $$\equiv$$ Virtual Displacements as on p24-4 (Calculus of Variations)

Previously, we wrote:

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

Where k was a 6x2 matrix and d was a 2x1 matrix.

Simplifying results in:

$$=\begin{Bmatrix} w_3 \\ w_4 \end{Bmatrix} \cdot (\mathbf{\bar K}_{2x2}\mathbf{\bar d}_{2x1}-\mathbf{\bar F})=0$$ for all $$\begin{Bmatrix} w_3 \\ w_4 \end{Bmatrix}$$. . . (3)

Specifically:

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

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

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

HW Begin

Show $$(3) \Leftrightarrow \mathbf{\bar K \bar d}=\mathbf{\bar F} (4)$$

Take w3=w4=1

$$\mathbf{1}\cdot\mathbf{\bar K}\mathbf{\bar d}-\mathbf{\bar F}=\mathbf{0}$$

Therefore, $$\mathbf{\bar K}\mathbf{\bar d}=\mathbf{\bar F}$$

Monday Oct. 27, 2008
HW: 6 bar truss, page 226

 * Same E, A for all elements

Bug: for two bar truss used different E's and A's for the elements

$$ \mathbf {d}_{4x1}^{(i)} \harr d(lmm(i,:))$$
 * right hand side is Matlab code

Where: element degree of freedoms in global (x,y) coordinate system for element (i)

$$\mathbf {d}_{6x1}$$ 2-bar truss degree of freedoms in global coordinate (x,y) system.

Bug Explained

In computation of element forces $$\rarr$$ change (e,A) to e(i) and A(i). Correct the code, and rerun the 2-bar truss problem. Interpret the "results".

Back to PVW

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

 * At this point Dr. Vu-Quoc referred to slides 24-1 and 21-2

Course grade $$\alpha_o \cdot$$ Hw grade +$$\sum \alpha_i \cdot Exam_i$$

 * Examine pages 23-3 and 14-3

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

Recall FD relation with axial degree's of freedom $$ \mathbf{q}^{(e)}$$

$$ \mathbf{\hat{k}}_{2x2}^{(e)} \mathbf{q}_{2x1}^{e} = \mathbf{p}_{2x1}^{e}$$

$$\rightarrow \mathbf{\tilde{k}}^{(e)} \mathbf{q}^{(e)} - \mathbf{p}^{(e)}= \mathbf{0}_{2x1}$$. . . (1)

PVW: $$ \mathbf{W}_{2x1} \cdot (\mathbf{\hat{k}}^{(e)} \mathbf{q}^{(e)} - \mathbf{p}^{(e)}) = 0$$. . . (2)

we showed $$ (1) \leftrightarrow (2)$$

Recall: $$ \mathbf{q}_{2x1}^{(e)} = \mathbf{T}_{2x1}^{(e)} \mathbf{d}_{4x1}^{(e)}$$. . . (3)

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

Wednesday Oct. 29, 2008
$$\mathbf{\hat{w}}_{2x1}$$ = virtual displacement corresponding to $$\mathbf{q}_{2x1}^{(e)}$$

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

Replacing Equations 3 and 4 into Equation 2:

$$(\mathbf{T}^{(e)}\mathbf{w})\cdot [\mathbf{\hat{k}}^{(e)}(\mathbf{T}^{(e)}\mathbf{d}^{(e)})-\mathbf{p}^{(e)}]=0$$ for all $$\mathbf{w_{4x1}}$$ (5)

Recall: $$\left ( \mathbf{A}_{pxq} \mathbf{B}_{qxr} \right )^T = \mathbf{B}^T \mathbf{A}^T $$ (6)

Recall: $$\mathbf{a}_{nx1} \cdot \mathbf{b}_{nx1}=\mathbf{a}_{1xn}^T \mathbf{b}_{nx1}$$ (7)

Apply Equation 7 & 6 in 5:

$$(\mathbf{T}^{(e)} \mathbf{w})^T [ \mathbf{\hat{k}}^{(e)}(\mathbf{T}^{(e)}\mathbf{d}^{(e)})-\mathbf{p}^{(e)}]=0_{1x1}$$ for all $$\mathbf{w}_{4x1}$$

Which results in:

$$\mathbf{w}^{T} \mathbf{T}^{(e)T} [ \mathbf{\hat{k}}^{(e)}(\mathbf{T}^{(e)}\mathbf{d}^{(e)})-\mathbf{p}^{(e)}]=0_{1x1}$$ for all $$\mathbf{w}_{4x1}$$

Further simplification results in:

$$\mathbf{w} \cdot \left [ \left (\mathbf{T}^{(e)T} \mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)} \right ) \mathbf{d}^{(e)} - \mathbf{T}^{(e)} \mathbf{p}^{(e)} \right ] = 0 $$ for all $$\mathbf{w}_{4x1}$$

Generalizing:

$$\mathbf{w} \cdot \left [ \mathbf{k}^{(e)} \mathbf{d}^{(e)}-\mathbf{f}^{(e)} \right ] = 0$$ for all w

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

So for discrete case (matrices) (non-continuous)

Now, continuous case (PDEs) Motivational model problem: Elastic bar with varying A(x), E(x), subjected to varying axial (time dependent) load (distributed) + concentrated (time dependent) load + inertial force (dynamic) This problem is illustrated in the figure below.

Friday Oct. 31, 2008
FBD:

$$\sum F_x=0=-N(x,t)+N(x+dx,t)+f(x,t)dx-m(x)\ddot u dx=\frac{\partial N}{\partial x}(x,t)du+h.o.t.+f(x,t)dx-m(x)\ddot u dx $$  eqn(1)

Where h.o.t. = higher order terms

Recall Taylor series expansion:

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

Where the final part of the equation above represents the beginning of the higher order terms.

Eq.(1) yields the Equation of Motion (EoM): $$\frac{\partial N}{\partial x}+f=m\ddot u $$  eqn(2)

$$N_{ }^{ }(x,t)=A(x)\sigma (x,t)$$  eqn(3)

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

and $$\varepsilon (x,t)=\frac{\partial u}{\partial x}(x,t) $$

Where equation 3 is a constitutive relation.

Plugging equation 3 into 2 yields the PDE of Motion:

$$\frac{\partial}{\partial x} \left [ A(x)E(x)\frac{\partial u}{\partial x}\right ] + f(x,t)=m(x)\ddot u$$  eqn(4)

Where $$\ddot u = \frac{d^2 u}{dt^2}$$

In order to proceed further, we need two boundary conditions (2nd order derivative with respect to x)

We also have two initial conditions which are the initial displacement and initial velocity (2nd derive with respect to t)

Applying these to the two possible options shown in the figures to the right:

For scenario 1,

$$u_{ }^{ }(0,t)=u(L,t)=0$$

For scenario 2,

$$u_{ }^{ }(0,t)=0$$ and $$N_{ }^{ }(L,t)=F(t)$$

where $$N_{ }^{ }(L,t)=A(L)\sigma(L,t)$$

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

where $$\varepsilon(L,t)=\frac{\partial u}{\partial x}(L,t)$$

Contributing Team Members
Chris Mott 20:14, 7 November 2008 (UTC)

Bradley LaCroix 20:08, 7 November 2008 (UTC)

Brandon Sell 20:10, 7 November 2008 (UTC)

Charles Marshall 20:15, 7 November 2008 (UTC)

Chris Sell 20:25, 7 November 2008 (UTC)

Jason Bruce 20:28, 7 November 2008 (UTC)

Past Homeworks
[Homework 4]

[Homework 3]

[Homework 2]

[Homework 1]

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User:Eml4500.f08.lulz.abcd/HW5

Prove Kd=F is equivalent to W(kd-F)=0
$$\mathbf{W}\cdot(\mathbf{K}\mathbf{d}-\mathbf{F})=\begin{Bmatrix} w_3\\ w_4\\ \end{Bmatrix}\cdot(\mathbf{\bar{K}}\mathbf{\bar{d}}-\mathbf{\bar{F}})=0 $$ for all $$ \begin{Bmatrix} w_3\\ w_4\\ \end{Bmatrix}$$

Remember and apply our previous derivations for K, d, and F:

k_{33} & k_{34}\\ k_{43} & k_{44}\\ \end{bmatrix}$$ d_3\\ d_4\\ \end{bmatrix}$$ F_3\\ F_4\\ \end{bmatrix}$$
 * $$\mathbf{K}=\begin{bmatrix}
 * $$\mathbf{d}=\begin{bmatrix}
 * $$\mathbf{F}=\begin{bmatrix}

$$\begin{Bmatrix} w_3\\ w_4\\ \end{Bmatrix}\cdot(\begin{bmatrix} k_{33} & k_{34}\\ k_{43} & k_{44}\\ \end{bmatrix}\begin{bmatrix} d_{3}\\ d_{4} \\ \end{bmatrix}-\begin{bmatrix} F_{3}\\ F_{4} \\ \end{bmatrix})=0$$

This matrix can be converted by the dot product and simple matrix math to a single algebraic expression:

$$w_3(k_{33}d_3+k_{34}d_4-F_3)+w_4(k_{43}d_3+k_{44}d_4-F_4)=0$$

This equation will have multiple solutions depending on $$w_3$$ and $$w_4$$. Since by definition the relationships apply for any $$w_3$$ and $$w_4$$ then any arbitrary value can be applied to solve the expression created above. Picking different sets of values will create a system of equations that, when written in matrix form, are identical to $$\mathbf{W}\cdot(\mathbf{K}\mathbf{d}-\mathbf{F})=0$$ and other values will also prove identity.

If we pick 0 for our values of w:

$$w_3=0, w_4=0$$

$$(0)(k_{33}d_3+k_{34}d_4-F_3+(0)(k_{43}d_3+k_{44}d_4-F_4)=0$$

$$0=0$$

If we pick 0 for one value of w and 1 for the other:

$$w_3=0, w_4=1$$

$$(0)(k_{33}d_3+k_{34}d_4-F_3+(1)(k_{43}d_3+k_{44}d_4-F_4)=0$$

$$k_{43}d_3+k_{44}d_4-F_4=0$$

$$F_4=k_{43}d_3 + k_{44}d_4$$

If we pick 1 for our values of w:

$$w_3=1, w_4=1$$

$$(1)(k_{33}d_3+k_{34}d_4-F_3+(1)(k_{43}d_3+k_{44}d_4-F_4)=0$$

$$k_{33}d_3+k_{34}d_4+k_{43}d_3+k_{44}d_4=F_3+F_4$$

$$\begin{bmatrix} k_{33}d_3 + k_{34}d_4\\ k_{43}d_3 + k_{44}d_4\\ \end{bmatrix}=\begin{bmatrix} F_3\\ F_4\\ \end{bmatrix}$$

These three examples each bring us back to our proof that:

\bold{Kd}=\bold{F}

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User talk:Ipatrol

Talk messages
Not sure where you want talk messages, but hopefully here is OK? I've commented out the redirect for the moment. -- Jtneill - Talk - c 00:59, 13 November 2008 (UTC)

Thanks for social psychology edits
Thanks for this. It prompted me to try to put a better indication of the status of those lecture pages, at least for now, with this. Feel free to wander through the unit - there's nothing like fresh eyes. -- Jtneill - Talk - c 00:59, 13 November 2008 (UTC)

Home
This broke Social psychology (psychology)/Banner and Social psychology (psychology). Can you help? -- Jtneill - Talk - c 02:01, 13 November 2008 (UTC)
 * Interestingly, it didn't break Survey research methods and design in psychology. I wonder why. -- Jtneill - Talk - c 02:08, 13 November 2008 (UTC)

No problem
No problem at all - thanks for being bold and trying to improve things (and prompt fixes)! I've known that those lecture pages could be better explained for newcomers but haven't done anything about it - now you've given the impetus for me to improve this; with SPLectureDone2008. I'll add an official welcome aboard below. -- Jtneill - Talk - c 03:20, 13 November 2008 (UTC)

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! Jtneill - Talk - c 03:20, 13 November 2008 (UTC)

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User:Eas4200c.f08.aero.lee/HW5

= Stress - Strain Relation =

To obtain engineering stress results based on an assessment of the engineering strains, we analyze the previously given strain-stress relation equations represented in matrix form as:

\begin{Bmatrix} \epsilon _{xx}\\ \epsilon _{yy}\\ \epsilon _{zz}\\ \gamma _{yz}\\ \gamma _{zx}\\ \gamma _{xy}\\ \end{Bmatrix}=

\begin{bmatrix} \frac{1}{E}& \frac{-\nu }{E} & \frac{-\nu }{E} & 0 & 0 &0 \\ \frac{-\nu }{E}& \frac{1}{E} & \frac{-\nu }{E} & 0 &  0&0 \\ \frac{-\nu }{E}& \frac{-\nu }{E} &   \frac{1}{E}& 0 & 0 &0 \\ 0 & 0 & 0 & \frac{1}{G} & 0 & 0\\ 0 & 0 & 0 & 0 & \frac{1}{G}& 0\\ 0& 0 &0 &0 & 0 & \frac{1}{G} \end{bmatrix} \begin{Bmatrix} \sigma _{xx}\\ \sigma _{yy}\\ \sigma _{zz}\\ \sigma _{yz}\\ \sigma _{zx}\\ \sigma _{xy}\\ \end{Bmatrix} $$

The above matrices can be rearranged to solve directly for stress in terms of strain by inverting the 6x6 coefficient matrix. To do so, consider the 6x6 matrix as a composition of 3x3 matrices, with the upper-right and lower-left quadrants identical to the zero matrix. Such a composition could be represented as:

\begin{Bmatrix} \epsilon _{xx}\\ \epsilon _{yy}\\ \epsilon _{zz}\\ \gamma _{yz}\\ \gamma _{zx}\\ \gamma _{xy}\\ \end{Bmatrix}=

\begin{bmatrix} \bar{A} & \bar{0} \\ \bar{0} & \bar{B} \\ \end{bmatrix} \begin{Bmatrix} \sigma _{xx}\\ \sigma _{yy}\\ \sigma _{zz}\\ \sigma _{yz}\\ \sigma _{zx}\\ \sigma _{xy}\\ \end{Bmatrix} $$

where the matrices $$\bar{A}$$ and $$\bar{B}$$ are given as

\begin{Bmatrix} \bar{A}\\ \end{Bmatrix}=

\begin{bmatrix} \frac{1}{E}& \frac{-\nu }{E} & \frac{-\nu }{E} \\ \frac{-\nu }{E}& \frac{1}{E} & \frac{-\nu }{E} \\ \frac{-\nu }{E}& \frac{-\nu }{E} &  \frac{1}{E} \\ \end{bmatrix}

\begin{Bmatrix} \bar{B}\\ \end{Bmatrix}=

\begin{bmatrix} \frac{1}{G} & 0 & 0\\ 0 & \frac{1}{G}& 0\\ 0 & 0 & \frac{1}{G} \end{bmatrix} $$

Then, the stress-strain relation can be determined by inverting the coefficient matrix.

\begin{Bmatrix} \sigma _{xx}\\ \sigma _{yy}\\ \sigma _{zz}\\ \sigma _{yz}\\ \sigma _{zx}\\ \sigma _{xy}\\ \end{Bmatrix}=

\begin{bmatrix} \bar{A}^{-1} & \bar{0} \\ \bar{0} & \bar{B}^{-1} \\ \end{bmatrix} \begin{Bmatrix} \epsilon _{xx}\\ \epsilon _{yy}\\ \epsilon _{zz}\\ \gamma _{yz}\\ \gamma _{zx}\\ \gamma _{xy}\\ \end{Bmatrix} $$

This solution is possible because of the structure of the coefficient matrix. The inversion of the coefficient matrix is similar to that of a diagonal matrix, whose inverse is solved by inverting each component of the diagonal (assuming nonzero components) while leaving the zero terms off-diagonal ignored. A confirmation of this solution can be found by premultiplying the $$ \bar{A} $$ and $$ \bar{B} $$ matrices by their inverses within the coefficient matrix and obtaining the Identity Matrix, $$ \bar{I} $$.

\begin{bmatrix} \bar{A}^{-1}\bar{A} & \bar{0} \\ \bar{0} & \bar{B}^{-1}\bar{B} \\ \end{bmatrix}=

\begin{Bmatrix} \bar{I}\\ \end{Bmatrix} $$

It is seen by this solution, since $$ \bar{B} $$ is a diagonal matrix, that for strains $$ \epsilon_{xx}, \epsilon_{yy}, \epsilon_{zz}, \gamma_{yz} $$, the corresponding stress components (i.e. $$ \sigma_{xx}, \sigma_{yy}, \sigma_{zz}, \sigma_{yz} $$) are identically zero. Thus the stress-strain relation reduces to

\begin{Bmatrix} \sigma _{zx}\\ \sigma _{xy}\\ \end{Bmatrix}=

\begin{bmatrix} G & 0 \\ 0 & G \\ \end{bmatrix} \begin{Bmatrix} \gamma _{zx}\\ \gamma _{xy}\\ \end{Bmatrix} $$

= Equilibrium Equation for Stresses in a Non-uniform Stress Field =

Consider a 1-D case as a model.

It is apparent from the uniform case that the axial loading is constant along the length of the member. Conversely, in the non-uniform case, the axial loading is non-constant along the length of the member. This results in a non-uniform stress and strain response of the member.

= Bidirectional Bending Recipe = The image below depicts a non-uniform bar under bidirectional bending

Looking at Figure 2, for this case, the bending moments about the y and z axis, respectively, can be expressed as follows:

 * $$M_y = \int_A z\sigma_{xx}\, dA \, \, =\, \int\int z \sigma_{xx} dydz$$

 * $$M_z = \int_A y\sigma_{xx}\, dA \, \, =\, \int\int y \sigma_{xx} dydz$$

where $$\sigma_{xx}$$ is the axial stress due to the bending moments about the y and z axes at the point (x,y).

Likewise, the moment of inertia tensor - $$I_y, I_z, I_{yz}$$, (which can be equivalently denoted as $$ I_{yy}, I_{zz}, I_{xy}$$ or $$I_{22}, I_{33}, I_{23}$$, respectively) - are defined by the expressions

 * $$\qquad I_y=\int_A z^2dA$$

 * $$\qquad I_z=\int_A y^2dA$$

 * $$\qquad I_{yz}= \int_A yzdA$$

Solving these two groups of expressions yields an expression for the axial due to the bending moments about y and z as

 * $$\sigma_{xx}=E\epsilon_{xx}=\frac{I_yM_z-I_{yz}M_y}{I_yI_z-(I_{yz})^2}y+\frac{I_zM_y-I_{yz}M_z}{I_yI_z-(I_{yz})^2}z$$

If one recalls that

 * $$\underline{I} = \begin{bmatrix} I_{11} & I_{12} & I_{13} \\ & I_{22} & I_{23}\\sym.& & I_{33} \end{bmatrix}$$

One can see that denominator of the $$\sigma_{xx}$$ expression is numerically the determinate of the lower right quadrant of the $$\underline{I}$$ matrix. Also, the symmetry of the $$\underline{I}$$ matrix can be seen by comparing the $$ I_{23}$$ and $$I_{32}$$ terms of the matrix.

 * $$I_{23} = I_{yz} = \int_A yz\ dA$$

 * $$I_{32} = I_{zy} = \int_A zy\ dA$$

since

 * $$\int_A yz\ dA = \int_A zy\ dA$$

 * $$\Rightarrow I_{23} = I_{32}$$

The neutral axis is an axis along which the axial stress due to bending moments is zero, i.e.

 * $$\qquad \sigma_{xx}=0$$

For the case when the y or z axis is an axis of symmetry, and hence $$ I_{yz} = 0$$ the above expression for the axial stress can be reduced to

 * $$\sigma_{xx} = M_yy+M_zz = 0$$

an expression for the neutral axis can be taken from this by solving for z in terms of y:

 * $$z=\left (\frac{-M_y}{M_z} \right ) y = \tan\beta\ y$$

where $$\beta$$ is the angle between the neutral and y axes as shown below in Figure 3.

= Equations of Equilibrium in Terms of $$\sigma$$=

The goal is to obtain an equilibrium equation such that $$ \frac{\partial \sigma_{yx}}{\partial y} + \frac{\partial \sigma_{zx}}{\partial z} = 0 $$, or using indicial notataion: $$ \frac{\partial \sigma_{21}}{\partial x_{2}} + \frac{\partial \sigma_{31}}{\partial x_{3}} = 0 $$.

Recall from previous material [Section 2.4, Sun],

 * $$ \frac{\partial \sigma_{xx}}{\partial x} + \frac{\partial \sigma_{yx}}{\partial y} + \frac{\partial \sigma_{zx}}{\partial z}= 0 $$

 * $$ \frac{\partial \sigma_{yy}}{\partial y} + \frac{\partial \sigma_{zy}}{\partial z} + \frac{\partial \sigma_{xy}}{\partial x}= 0 $$

 * $$ \frac{\partial \sigma_{zz}}{\partial z} + \frac{\partial \sigma_{xz}}{\partial x} + \frac{\partial \sigma_{yz}}{\partial y}= 0 $$

It is seen that a cyclic permutation is possible such that

 * $$ \sum_{i=1}^{3}{\frac{\partial \sigma_{ij}}{\partial x_i}} = 0 $$

This notation is particularly powerful. So powerful in fact, that Einstein used it to formulate the Theory of Relativity.

= Using Taylor Series With Infinitesimal Elements = For the purpose of simplicity, we look at the derivation of equation (4) on pg 27-2 for the one-dimensional model first. Looking at the equilibrium of the infinitesimal figure:

$$\sum\ F_x = 0 = -\sigma_{(x)}\ A + \sigma_{(x + dx)}\ A + f_{(x)} dx\,$$

$$ 0 = A [ \sigma_{(x + dx)}\ - \sigma_{(x)}\ ] + f_{(x)} dx\,$$

Next, we apply Taylor series to the term, $$\sigma_{(x + dx)}\ - \sigma_{(x)}$$:

$$\frac{\partial\sigma_{x}}{\partial\ x} dx + higher~order~terms\,$$

Knowing that the general Taylor Series Approximation for the first three series of terms is given as:

$$F_{(x + dx)} = F_{(x)} + \frac{\partial\ F_{(x)}}{\partial\ dx} dx + \frac{1}{2} \frac{\partial ^2 F_{(x)}}{\partial x^2} (dx)^2 + .~.~.~higher~order~terms\,$$

The higher order terms are neglected and yields the final equation of interest shown below:

$$\frac{\partial\sigma}{\partial\ x} + \frac{f_{(x)}}{A} = 0\,$$

 * where $$f_(x)$$ is a $$\frac{force}{length}\,$$ and has dimensional units of $$\frac{N}{m}\,$$

In the equation above, the applied load $$\frac{f_{(x)}}{A}\,$$ is a Body Force and has dimensional units of $$\frac{N}{m^3}\,$$

Now looking at the nonuniform stress field in 3D and without any applied load, we focus on the x-direction in an attempt to view a better representation of the stress componenents without any cluttering.

In Figure 4 above shows an infinitesimal elemental cube where the stress is not completely uniform. Since the infinitesimal cube is in equilibrium, summing the forces of the x-direction is observed to be the sum of stress components of each facet area multiplied by the respective infinitesimal area. Therefore, the sum of the forces in the x-direction is:

$$\sum\ F_x = 0 = dy dz [ -\sigma_{xx}\ (x,y,z) + \sigma_{xx}\ (x+dx,y,z)]~+~ dz dx [ -\sigma_{yx}\ (x,y,z) + \sigma_{yx}\ (x,y+dy,z)]~+~dx dy [ -\sigma_{zx}\ (x,y,z) + \sigma_{zx}\ (x,y,z+dz)]\,$$

Here the terms can be separated and rewritten in terms of their respective Facets as:

Facets with normal X- $$ dy dz [ -\sigma_{xx}\ (x,y,z) + \sigma_{xx}\ (x+dx,y,z)]\,$$

Facets with normal Y- $$ dz dx [ -\sigma_{yx}\ (x,y,z) + \sigma_{yx}\ (x,y+dy,z)]\,$$

Facets with normal Z- $$ dx dy [ -\sigma_{zx}\ (x,y,z) + \sigma_{zx}\ (x,y,z+dz)]\,$$

The next step after is to take the Taylor Series expansion while neglecting the higher order terms.

Combining the terms from all facets, we derive the following equation:

$$0 = (dx dy dz) \left [ \frac{\partial\sigma_{xx}}{\partial\ x} + \frac{\partial\sigma_{yx}}{\partial\ y} + \frac{\partial\sigma_{zx}}{\partial\ z} \right ]\,$$

= Stringers =

In flight, an aircraft wing is subject to bending. This bending moment must be absorbed and transferred by some structure. It is known that the wing skin, because of its relative thinness, is prone to buckling when subjected to compressive loading. The spars, due to a small moment of inertia, are also relatively ineffective at carrying bending loads. Thus, it is desired to add an additional element to the wing structure.

Stringers are placed at the junction of the skin and spars in "nearly" L-shaped brackets. The stringers run the spanwise length of the wing along with the spars. By placing the stringers at the top and bottom of the wing, their moment of inertia about the wing centroid is maximized, allowing for a greater bending moment to be applied to the wing. The figure below provides a generic schematic of stringer placement for one spar - skin joint location.

= NACA Airfoil Matlab Project Continued =

The purpose of this project was to take the previously created Matlab code and add the option to include stringers to the modeled NACA airfoil. Results will show that it is sufficient to consider only the stringers when analyzing the bending moment effects on a wing.

= Contributing team members =

 * Jared Lee--Eas4200c.f08.aero.lee 18:05, 7 November 2008 (UTC)
 * Oliver Oyama--Eas4200c.fo8.aero.oyama 18:55, 7 November 2008 (UTC)
 * Gonzalo Barcia--Eas4200c.f08.aero.barcia 19:25, 7 November 2008 (UTC)
 * William Kurth--Eas4200c.f08.aero.kurth 20:37, 7 November 2008 (UTC)
 * Ray Strods-- Eas4200c.f08.aero.strods 21:01, 7 November 2008 (UTC)

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User:Eml4500.f08.FEABBQ/HW6

 See my comments below. Please don't remove these comment boxes; just add your comment to these comment boxes in case you fix the problems. Eml4500.f08 19:52, 26 November 2008 (UTC)

Class Notes November 03, 2008
Notes from November-03-08

HW: Refine the Mode-Shape Plots
We revisit the 2-bar truss example to refine our mode-shape plot. First, we output the global stiffness matrix, K, and the corresponding eigenvectors and eigenvalues...

Next, we plot each column of the eigenvector matrix as a mode-shape for only the columns which correspond to a zero eigenvector (the first four columns)...

Next, we study the unstable 3 bar truss (Case A in the image below)...

The following global stiffness matrix and eigenvectors and eigenvalues were solved for using Matlab (Note: The matlab code is available upon request, but it is omitted here due to similarity to previous solving and plotting code provided.)

The following are the five mode shapes associated with the five zero-value eigenvectors for the three bar truss (Case A). Note: some magnification has been applied (3x).

END HW

Continuing the derivation:

Initial conditions: At t=0, prescribed

$$\left(x,t=0 \right) =\bar{U}\left(x \right)$$ Known Function Displacement $$\frac{\partial }{\partial t}\left(x,t=0 \right)=\dot{U }\left(x,t=0 \right)=\bar{V}\left(x \right)$$Known Function Velocity

$$\frac{\partial }{\partial x}\left[\left(EA \right)\frac{\partial }{\partial x} \right]+f=m\ddot{U}$$ Eq.(1)
 * PVW (continuous) of the dynamics of elastic bar
 * PDE

Discrete EOM $$\Leftarrow $$ Equation of motion. $$-Kd+F=M\ddot{d}$$

$$\frac{\partial }{\partial x}\left[\left(EA \right)\frac{\partial }{\partial x} \right]\Rightarrow -Kd$$

$$f \Rightarrow F$$

$$m\ddot{U} \Rightarrow M\ddot{d} $$

$$M\ddot{d}+Kd=F$$    Eq.(2)

Goal is to derive Eq.(2) from Eq.(1):
 * MDOF $$\Rightarrow$$ Multiple Degree of Freedom system
 * SDOF $$\Rightarrow$$ Single Degree of Freedom system

Motivation a SDOF spring mass system shown below where : $$M\ddot{d}+Kd=F$$

$$\int_{0}^{x=L}{W\left(x \right)\left\{\frac{\partial }{\partial x}\left[EA\frac{\partial U}{\partial x} \right]+f-m\ddot{U} \right\}}dx=0$$ Eq.(3)    For all posible $$W(x)$$

As seen
 * Eq.(1)$$\Rightarrow$$ Eq.(3) is trivial

 * Eq.(3)$$\Rightarrow$$ Eq.(1) is not trivial

Eq.(3) can be rewritten as $$\int W(x)g(x)dx=0$$

Since Eq.(3) holds for all $$W(x)$$, select $$W(x)=g(x)$$, then Eq.(3) becomes:

$$\int g^{2}dx=0 $$ where $$g^2\geq 0$$  \Rightarrow $$g(x)=0$$

Class Notes November 05, 2008
Continuing with the discussion from November 03, 2008, we revisit the concept of Integration by Parts:

Given $$r(x)$$ and $$s(x)$$, $$(rs)'=r's+rs'$$ where $$r'=\frac{dr}{dx}$$ and $$s'=\frac{ds}{dx}$$. The integral is therefore $$\int{(rs)'}=\int{r's}+\int{rs'}$$. By parts we have $$\int{(r's)}=rs-\int{rs'}$$.

Recall the continuous Principal of Virtual Work. For the first term, think of $$r(x)=(EA)\frac{\partial{u}}{\partial{x}}$$ and $$s(x)=W(x)$$.

Then By Parts: $$\int_{x=0}^{x=L} W(x)\frac{\partial}{\partial{x}}[EA\frac{\partial{u}}{\partial{x}}]dx=[W(EA)\frac{\partial{u}}{\partial{x}}]-\int_0^L \frac{\partial{W}}{\partial{x}}(EA)\frac{\partial{u}}{\partial{x}}dx$$.

This equals $$=W(L)EA(L)\frac{\partial{u}(L,t)}{\partial{x}}-W(0)EA(0)\frac{\partial{u}(0,t)}{\partial{x}}-\int_0^L \frac{dW}{dx}(EA)\frac{\partial{u}}{\partial{x}}dx$$.

Now we can examine the following model problem...

There are two boundary conditions: At $$x=0$$, select $$W(x)$$ at $$ W(0)=0$$, i.e. kindematically admissible. Recall $$ \underline{W}\cdot([]\left\{\begin{array}{c} d_{3} \\ d_{4} \end{array}\right\}-\underline{F})=0$$.

$$\underline{F}^{T}=\left[\begin{array}{cccccc} F_{1} & F_{2} & F_{3} & F_{4} & F_{5} & F_{6} \end{array}\right] $$, where $$ F_{3}$$ & $$F_{4}$$ are known.

Select $$w_{1}=w_{2}=w_{5}=w_{6}=0$$.

Then we return to the familiar $$ \underline{\overline{K}}\underline{\overline{d}}=\underline{\overline{F}} $$, where $$\underline{\overline{d}}=\left\{\begin{array}{c} d_{3} \\ d_{4} \end{array}\right\} $$ and $$\underline{\overline{F}}=\left\{\begin{array}{c} F_{3} \\ F_{4} \end{array}\right\}$$.

Note: $$ \underline{\overline{W}}\cdot(\underline{\overline{K}}\underline{\overline{d}}-\underline{\overline{F}})=0$$ for all $$\underline{\overline{W}}$$, where $$\underline{\overline{W}}=\left\{\begin{array}{c} W_{3} \\ W_{4} \end{array}\right\} $$

Now we have an unkown reaction: $$ N(0,t)=(EA)(0)\frac{\partial{u}(0,t)}{\partial{x}}$$

We use the Principal of Virtual Work (continuous): $$ W(L)F(t)-\int_0^L \frac{dW}{dx}(EA)\frac{\partial{u}}{\partial{x}}dx+\int_0^L W(x)[f-m\ddot{u}]dx=0$$ for all $$W(x)$$ such that $$W(0)=0$$.

This leads to $$ \int_0^L W(m\ddot{u})dx+\int_0^L \frac{dW}{dx}(EA)\frac{\partial{u}}{\partial{x}}dx=W(L)F(t)+\int_0^L Wfdx$$ for all $$W(x)$$ such that $$W(0)=0$$.

Interpolation

Motivation for linear interpolation of $$ u(x) $$

Deformed shape is a straight line. i.e. there was an implicit assumption of linear interpolation of displacement between 2 nodes.

Consider the case where there are only asial displacements (i.e. zero transverse displacement)

Q: Express u(x) in terms of $$ d_{i} = u(x_{i}) $$ and $$ d_{i+1} = u(x_{i+1}) $$ as a linear function in x (i.e. linear interpolation)

$$ u(x) = N_{i}(x)d_{i} + N_{i+1}(x)d_{i+1} $$ where $$ N_{i}(x) $$ and $$ N_{i+1}(x) $$ are linear functions in x.

$$ N_{i+1}(x)= \frac{x-x_{i}}{x_{i+1} - x_{i}} $$

HW: Equation and Picture for $$ N_{i}(x) $$
$$ N_{i}(x)= \frac{x-x_{i+1}}{x_{i+1} - x_{i}} $$

Continuous Principle of Virtual Work to Discrete Principle of Virtual Work (Continued)
Lagrangian Interpolation

What is the motivation for the form of $$N_i(x)\,\!$$ and $$N_{i+1}(x)\,\!$$? 1) $$N_i(x)\,\!$$ and $$N_{i+1}(x)\,\!$$ are both linear, thus any linear combination of $$N_i\,\!$$ and $$N_{i+1}\,\!$$ is also linear.  In particular, the expression for $$u(x)=N_i(x)d_i+N_{i+1}(x)d_{i+1}\,\!$$ must be linear.  Below is a demonstration:

Let: $$N_i(x)=\alpha_i + \beta_ix\,\!$$ -$$\alpha_i, \beta_i\,\!$$ are real numbers

$$N_{i+1}(x)=\alpha_{i+1} + \beta_{i+1}x\,\!$$ -$$\alpha_{i+1}, \beta_{i+1}\,\!$$ are real numbers

Now we can combine $$N_i(x)\,\!$$ and $$N_{i+1}(x)\,\!$$ :

$$u(x)=N_i(x)d_i+N_{i+1}(x)d_{i+1}\,\!$$

$$=(\alpha_i + \beta_ix)d_i+(\alpha_{i+1}, \beta_{i+1})d_{i+1}\,\!$$

$$=(\alpha_i+\alpha_{i+1}+(\beta_i+\beta_{i+1})x\,\!$$ < is a linear function

2) Recall the equation for $$u(x)\,\!$$, an interpolation of x:

$$u(x_i)=\underbrace{N_i(x_i)d_i}_{1}+N_{i+1}(x_i)d_{i+1} = d_i\,\!$$

FEM via PVW:(continued)
Continuing with the discussion of $$u(x_{i})$$ we can find that

$$u(x_{i+1})=N_{i}(x_{i+1})d_{i}+N_{i+1}(x_{i+1})d_{i+1}=d_{i+1}$$ simply because the terms $$N_{i}(x_{i+1})=0$$ and $$N_{i+1}(x_{i+1})=1$$.

Applying the same interpolation for w(x), i.e.,

$$w(x)=N_{i}(x)w_{i}+N_{i+1}(x)w_{i+1}$$

Element stiffness matrix for element i

$$\Beta=\int_{x_{i}}^{x_{i+1}}{[N_{i}'w_{i}+N_{i+1}'w_{i+1}] (EA)[N_{i}'d_{i}+N_{i+1}'d_{i+1}]}dx $$

where the term $$ [N_{i}'w_{i}+N_{i+1}'w_{i+1}] = w'(x) $$ and the term $$ [N_{i}'d_{i}+N_{i+1}'d_{i+1}] = u'(x) $$, along with these terms below.

$$N_{i}'=\frac{d}{dx}N_{i}(x)$$

$$N_{i+1}'=\frac{d}{dx}N_{i+1}(x)$$

Note:

$$ u(x) = \left[ \begin{array}{cc} N_{i}(x) & N_{i+1}(x)\end{array}\right] \left\{\begin{matrix} d_{i}\\ d_{i 1} \end{matrix} \right\} $$

$$\frac{\partial }{\partial x}u(x) = \left[ \begin{array}{cc} N_{i}'(x) & N_{i+1}'(x)\end{array}\right]\left\{\begin{matrix} d_{i}\\ d_{i 1}

\end{matrix} \right\}$$

where the terms $$\left[ \begin{array}{cc} N_{i}(x) & N_{i+1}(x)\end{array}\right]=\underline{N}(x) $$ and $$\left[ \begin{array}{cc} N_{i}'(x) & N_{i+1}'(x)\end{array}\right]=B(x)$$

Similarly:

$$w(x)=\underline{N}(x)\left\{\begin{array}{c} w_{i} \\ w_{i+1} \end{array}\right\} $$

$$\frac{dw(x)}{dx}=\underline{B}(x)\left\{\begin{array}{c} w_{i} \\ w_{i+1} \end{array}\right\} $$

Recall the element dofs

$$\left\{\begin{array}{c} d_{i}\\d_{i+1}\end{array}\right\}=\left\{\begin{array}{c} d^{(i)}_{1}\\d^{(i)}_{2}\end{array}\right\}=\underline{d}^{(i)}$$

$$\left\{\begin{array}{c} w_{i}\\w_{i+1}\end{array}\right\}=\left\{\begin{array}{c} w^{(i)}_{1}\\w^{(i)}_{2}\end{array}\right\}=\underline{w}^{(i)}$$

$$\Beta=\int_{x_{i}}^{x_{i+1}}(\underline{B}\underline{w}^{(i)})(EA)(\underline{B}\underline{d}^{(i)})dx=\underline{w}^{(i)}\cdot(\underline{k}^{(i)}\underline{d}^{(i)})$$

$$\Beta=\int_{x_{i}}^{x_{i+1}}(EA)(\underline{B}\underline{w}^{(i)})\cdot(\underline{B}\underline{d}^{(i)})dx\Rightarrow(\underline{B}\underline{w}^{(i)})^{T}(\underline{B}\underline{d}^{(i)})\Rightarrow\underline{w}^{(i),T}\underline{B}^{T}\Rightarrow\underline{w}^{(i)}\cdot\underline{B}^{T}$$

$$\Beta=\underline{w}^{(i)}\cdot(\int\underline{B}^{T}(EA)\underline{B}dx)d^{(i)}$$

$$\underline{k}^{(i)}=\int_{x_{i}}^{x_{i+1}}\underline{B}^{T}(x)(EA)\underline{B}(x)dx$$

$$\underline{B}(x)=\left[ \begin{array}{cc} -\frac{1}{L^{(i)}} & \frac{1}{L^{(i)}} \end{array} \right] $$

HW: Find an expression for $$\underline{k}^{(i)}$$
The following problem refers to the figure below

where

$$A(\tilde{x})=N^{(i)}_{1}(\tilde{x})A_{1}+N^{(i)}_{2}(\tilde{x})A_{2}$$

and

$$E(\tilde{x})=N^{(i)}_{1}(\tilde{x})E_{1}+N^{(i)}_{2}(\tilde{x})E_{2}$$

Consider the case where $$EA$$ is a constant...

$$ \underline{k}^{(i)}=\frac{EA}{L^{(i)}} \left[ \begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array} \right] $$

The transformation of variables (coordinates) from $$x$$ to $$\tilde{x}$$ is as follows...

$$\tilde{x}:=x-x_{i}$$

$$d\tilde{x}=dx$$

The task is to find an expression for $$\underline{k}^{(i)}$$ using the equation below...

$$\underline{k}^{(i)}=\int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}}\underline{B}^{T}(\tilde{x})(EA)(\tilde{x})\underline{B}(\tilde{x})d\tilde{x}$$

Due to the above equation's similarity to the equation used to find the stiffness matrix for EA=Constant, we assume that the stiffness matrix with varying cross-sectional area and modulus is

$$\underline{k}^{(i)}=\frac{[N^{(i)}_{1}(\tilde{x})A_{1}+N^{(i)}_{2}(\tilde{x})A_{2}][N^{(i)}_{1}(\tilde{x})E_{1}+N^{(i)}_{2}(\tilde{x})E_{2}]}{L^{(i)}} \left[ \begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array} \right] $$

Although this may not be correct...

 Error: You are right; it was not right! See my comments in Team Bike (with annotations). Eml4500.f08 19:52, 26 November 2008 (UTC)

Class Notes November 14, 2008
$$ N_{i}(x)=N_{1}^{(i)}(\tilde{x})=1-\frac{\tilde{x}}{L^{(i)}}= \begin{cases} 1 & \text{ at } \tilde{x}=0 \\ 0 & \text{ at } \tilde{x}=L^{(i)} \end{cases}$$

$$ N_{i+1}(x)=N_{2}^{(i)}(\tilde{x})=\frac{\tilde{x}}{L^{(i)}}= \begin{cases} 0 & \text{ at } \tilde{x}=0 \\ 1 & \text{ at } \tilde{x}=L^{(i)} \end{cases} $$

===HW: Set $$ E_{1}=E_{2}=E_{3}$$, let $$A(\tilde{x})$$ be linear as before, and obtain $$\underline{K}^{(i)}$$ and compare to expression in the book on p.159===

$$A(\tilde{x})=N_{1}^{(i)}(\tilde{x})A_{1}+N_{2}^{(i)}(\tilde{x})A_{2}$$

$$E(\tilde{x})=N_{1}^{(i)}(\tilde{x})E_{1}+N_{2}^{(i)}(\tilde{x})E_{2}$$

Compare general $$\underline{K}^{(i)}$$ from before to $$\underline{K}_{avg}^{(i)}$$ where,

$$ \frac{(E_{1}+E_{2})}{2}\frac{(A_{1}+A_{2})}{2}\frac{1}{L^{(i)}}\begin{bmatrix} 1 &-1 \\ -1 & 1 \end{bmatrix}=\underline{K}_{avg}^{(i)}$$

end HW

HW: Find, $$\underline{K}^{(i)}-\underline{K}_{avg}^{(i)}$$
Solution

end HW

Remark: recall the Mean value Theorem (MVT) and its relations to centroid:

MVT: for $$\bar{x}\epsilon [a,b]$$,     $$a\leq \bar{x}\leq b$$,

$$ \int_{x=a}^{x=b}{f(x)dx}=f(\tilde{x})[b-a] $$

$$\int_{A}^{}{xdA} = \bar{x} \int_{A}^{}{dA} = \bar{x}A$$

$$\int_{x=a}^{x=b}{f(x)g(x)dx} = f(\bar{x})g(\bar{x})[b-a]$$ where  $$ a\leq \bar{x}\leq b$$

But in general,

$$f(\bar{x})\neq \frac{1}{b-a} \int_{a}^{b}{f(x)dx}$$

$$g(\bar{x})\neq \frac{1}{b-a} \int_{a}^{b}{g(x)dx} $$

Matlab HW: Electric Pylon Analysis
The following is the Matlab code which solves the Electric Pylon problem and plots the deformed shape, amongst other things. Please use the comments embedded within the code to clarify the analysis.

The output for the deformed shape overlayed on the undeformed shape is shown below...

The output of the code given above displays many things, but most importantly are the following

 * The axial stress in each element; the highest tensile and compressive strengths and their associated elements

Which can be interpretted as  being the highest tensile stress associated with element 81 and   being the highest compressive stress associated with element 55.

 * The plot of the three lowest eigenpairs, which were found to be 132.34 associated with eigenvector column 79, 2471.7 associated with eigenvector column 82 and 2944.2 associated with eigenvector column 83. The plots of the three eigenvectors associated with the three lowest eigenpairs is shown below...

 * The three lowest vibrational periods of the electric pylon were also calculated and the combined output of the section of the code which finds the three lowest eigenpairs and the corresponding three lowest vibrational periods is shown below...

 * As for the electric pylon being statically determinate, we believe it is not so. The two lowest nodes (the nodes set to 0 for boundary conditions) have two elements attached to them.  Cutting the rest of the pylon away and analyzing only those two nodes and the -1000N external force, it is clear that there are too many unknowns because of the added elements connected to the nodes.

 * The general results for the electric pylon code are shown below, placed last due to similarity to the output of previous problems (2-bar truss, 3-bar truss, etc.)... Note: the K matrix is omitted due to its enormity...

Contributing Members:

 * GrantStark 21:05, 21 November 2008 (UTC)
 * J.Jayma 23:25, 17 November 2008 (UTC)
 * Esmail Hamdan 14:55, 21 November 2008 (UTC)
 * Nik Vitt 05:10, 21 November 2008 (UTC)
 * Andy Koby 15:07, 21 November 2008 (UTC)

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Portal talk:Solid mechanics

The division of the Solid Mechanics topic seems very confusing and redundant. From what I understand, the most common divion of the mechanics field seems to be Statics and Dynamics, then beneath Statics is Mechanics of Materials which covers material properties, stress/strain relationships and deflection. It seems like there is a lot of great content in this topic, but it seems confusing when the sub-categories are all visible at a top level. For Example, Elasticity, Plasticity, Fracture, Fatigue could all be sections of Mechanics of Materials. MinosDis 07:53, 8 November 2008 (UTC)

title: Portal talk:Solid mechanics


text:
User talk:Bonzostar

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

 * Take a guided tour and learn to edit.
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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! SB_Johnny talk 10:19, 8 November 2008 (UTC)

title: User talk:Bonzostar


text:
User talk:Dilos1

 Hello Dilos1, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
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 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --The Jade Knight (d'viser) 10:25, 8 November 2008 (UTC)

 * Out of curiosity, what areas in particular did you teach? The Jade Knight (d'viser) 10:25, 8 November 2008 (UTC)
 * I taught the two surveys of western civilization, Europe in the 18th and 19th c., Balkan & East European History, Russian History, Modern Greek history. The last five years though I have researched and written a survey of the history of the ancient world (mainly ancient greece) in Greek. What is your expertise? Dilos1 16:26, 8 November 2008 (UTC)
 * Medieval and Renaissance Western and Northern Europe. In particular, I am an expert on the Normans, and most especially on the formative Norman period (911-1066).  I also have a fair familiarity with Islamic History, US History, Roman History, and 20th century Chinese history.  The Jade Knight (d'viser) 12:13, 9 November 2008 (UTC)
 * Hey Dilos, I am taking a college level European History course and have a mountain of lecture notes. Since you are working on Western Civilizations I and II, should I send the notes relating to those sections to you when I type them? Geoff Plourde 01:25, 15 March 2009 (UTC)

title: User talk:Dilos1


text:
User:Eml4500.f08/HW report table/The best of HW5

 Under construction; not final (but close to being stable after a couple of weeks). The intention here is to document the best features in any HW report for the readers (including you); if you see excellent features in any HW report (including your team's) that I may have missed noticing, don't hesitate to let me know. I don't have time to read all HW reports in detail. In fact, due to time constraint, I only selectively looked at a few important features. On the other hand, I also added some annotations (e.g., related to errors) in the HW reports that I looked through for the benefit of everyone. Eml4500.f08 11:11, 10 November 2008 (UTC)

 * classnotes graded over 100%
 * principle of virtual work (PVW) continued
 * 2-bar truss: elimination of rows to obtain reduced system, formal derivation
 * 2-D bar element (Lecture):
 * axial FD relation: equivalence with PVW, formal derivation
 * general 2-D FD relation: formal derivation by PVW
 * transformation from element global dofs to element axial dofs; similarly for element global force components to element axial force components
 * PVW for axial FD relation; transformation of weighting coefficients from axial coordinates to global coordinates; transformation of axial dofs to global dofs; FD relation in global dofs for 2-D bar element
 * 3-D bar element (HW): derivation of FD relation and element stiffness matrix by PVW, similar to 2-D bar element above. Best presentation: Team ATeam(Sean) even though the details of the matrix multiplication for the final expression for $$\displaystyle \mathbf k ^{(e)}$$ were not provided; see Team Bike, but there was an error in the derivation concerning the use of the PVW. NOTE: Team Delta_6 (with annotations) Team Jamama (with annotations) did not do this part.  Team Bike (with annotations).
 * axial FD relation: equivalence with PVW, formal derivation
 * general 3-D FD relation: formal derivation by PVW
 * transformation from element global dofs to element axial dofs; similarly for element global force components to element axial force components
 * PVW for axial FD relation; transformation of weighting coefficients from axial coordinates to global coordinates; transformation of axial dofs to global dofs; FD relation in global dofs for 3-D bar element
 * continuous system governed by partial differential equations (PDEs)
 * model problem: elastodynamics of a bar with varying cross section and varying Young's modulus, subjected to arbitrary time-dependent loads
 * free-body diagram of infinitesimal element, derivation of PDE of motion
 * boundary conditions: clamped-clamped case, clamped-free case (with distributed load and a concentrated load at the free end)
 * initial conditions: initial displacement, initial velocity

EAS 4200C, The best of HW5
 * matlab problem graded over 100%
 * 2-D truss:
 * two-bar truss: debug, display "results" array, compare to statics solution. Best solution: Team ATeam(Sean), Team Bike, Perfect, nice problem description with figure and table of element properties, clear description of the bug in previous matlab code, presented "results" array in wiki table form, static solution for reactions, comparison with FE solution for element strains, stresses, forces.  Team Wiki1 (with annotations) also did excellent work, except that they used the incomplete matlab function PlaneTrussResults displayed in Team Delta_6 HW2 Fixed (with annotations), where the last line of the code was missing; Team Wiki1 modified the matlab code to make it work due to this missing last line.  NOTE: If you were not careful, and simply copied the incomplete matlab function PlaneTrussResults from the wiki page Team Delta_6 HW2 Fixed (with annotations), you would get only the first column (containing the strains) in the 2x3 array "results", and would unlikely be able to relate these results to the element forces obtained using statics as they did not represent the element forces (but the element strains); see Team Gravy (with annotations)
 * six-bar truss: same Young's modulus E and cross-sectional area A
 * six-bar truss: Different Young's moduli E and same cross-sectional area A
 * 3-D truss with 3 bars: Best solution: Team ATeam(Sean), Team Bike, Team Wiki1 (with annotations)
 * analysis with FE code
 * computation of reactions
 * plots of underformed and deformed shapes
 * static solution; computation of reactions
 * comparison of reactions obtained from FE and from statics


text:
Talk:History of Europe to 1648

I propose we move this to History of Europe (to 1648 AD). The Jade Knight (d'viser) 10:55, 8 November 2008 (UTC)
 * Why? I feel we must be exact with titles of courses. The word Survey emphasizes the introductory content and the dates will make the difference between part one (this one) and part two in the future. What is the rational behind your proposal? Dilos1 16:19, 8 November 2008 (UTC)
 * Well, Jade Night, I though about it myself and you may be right. As we seem to be avoiding details and dates in titles, I'll change it to History of Europe to 1648, which is short, to the point and denotes indirectly the introductory essence of the course. Thank you for the suggestion. By the way, I think I read that you have just been trusted with the role of the custodian. Congratulations! Dilos1 21:17, 8 November 2008 (UTC)
 * Thank you, though I am currently only a probationary Custodian. The Jade Knight (d'viser) 12:49, 9 November 2008 (UTC)

Other courses
Feel free to incorporate other courses or elements within the Department as part of this course; also be aware that forking may occur. The Jade Knight (d'viser) 10:03, 12 November 2008 (UTC)

 * OK. Thanks Dilos1 08:58, 14 November 2008 (UTC)

Depth
Roughly how many hours do you expect it would take a student to finish this course? The Jade Knight (d'viser) 22:05, 18 November 2008 (UTC)

 * Well, it depends on the devotion of the reader. Although the reading links may seem to be too many, actually in most of the cases only a paragraph or two are basically required from each source. If a reader devotes 6 hours per week, I estimate that it will take him about 3-4 months to complete the whole course. The benefit will be that at the end he/she will have a very solid background of the History of Europe in order to move to more specialized courses. I do not want to make this course a quick get in-and-get out thing. I want it to be a high quality learning project. The student can ask for all the help he/she needs and it will be given to him. However, this is really for serious students who would like to attend a survey of European History and learn things. Dilos1 08:42, 19 November 2008 (UTC)
 * Makes perfect sense to me. The Jade Knight (d'viser) 09:22, 20 November 2008 (UTC)

 * You gave me an idea. I'going to add time estimations for each one of the course sections and for the required material because a long list of readings may seem intimidating but the actual required time may not be that much. however, in some sections a student may need to spend more time than in others. Dilos1 09:45, 19 November 2008 (UTC)
 * Great idea. The Jade Knight (d'viser) 09:22, 20 November 2008 (UTC)

Scope
I love what you're doing here, but I just realized that you're covering Egypt and the Middle East, which are not a part of Europe (though they have links to Europe). Any particular reason for this decision? The Jade Knight (d'viser) 09:54, 20 November 2008 (UTC)
 * A valid question. However, history means politics, economy, society and culture. The Greeks who developed the first civilization on technically European soil did not invent everything. On the contrary, they learnt a lot through trade and other contacts from the peoples of the Near East (I would use Middle East only in a modern history context) (e.g. the Persians –Herodotus says so—who in turn had incorporated the whole previous culture going all the way back to the Sumerians) and the Egyptians (Philosopher Thales claims so).

 * A student must know some of the political background of the pre-Greek civilizations in order to appreciate what the Greeks got from them as well as how differently they shaped the European civilization. J.S.Mill, an 18th c. thinker, wrote that if the Greeks had not stopped the Persians at Marathon, the British might still be wondering wild in the woods. OK, it’s an exaggeration but points to the direction that there were mighty empires before the Greeks. Why didn’t they succeed in their expansion westwards, towards Europe? What were the differences (one of my exam questions to come) between the pre-Greek and the Greek systems of government? Dilos1 17:24, 20 November 2008 (UTC)

 * You make several valid points. Being a little bit of a "purist" (you may have noticed), however, I'd prefer less emphasis on this.  Would you mind if I came up with an alternative (not a replacement) section which essentially condenses this entire chapter into a single section, and then presented the two alternatives: yours for those interested in a more in-depth approach to pre-European civilizations, and mine for those who wish only to have a passing familiarity with what came before?  The Jade Knight (d'viser) 06:16, 21 November 2008 (UTC)
 * No problem. It would be nice for the student to have this choice, even for all the sections of this course. Go ahead! Dilos1 06:38, 21 November 2008 (UTC)

Jade Knight, I thought a lot over the weekend and I came to the conclusion that you are right. This introductory part needs simplification and focus on history only. I'll work on it in the days ahead. Thank you for the suggestions. Keep up the good work. Dilos1 07:26, 24 November 2008 (UTC)
 * No need to thank me, you're the one who's been doing all the outstanding work! I've got my fingers in many pies, and I've been more busy elsewhere, but I think the quality of your content here is very high, and while I'm already fairly familiar with this subject matter, I fully intend on being your first student (to shore up any gaps which might exist in my knowledge of the field).  Thanks for doing such a good job—I expect that this course will become the foundation for future work here at the Department of European History.  The Jade Knight (d'viser) 07:32, 24 November 2008 (UTC)

history
FYI: This template adds the page to Category:History, which is supposed to remain clear of specific projects. I'd appreciate it if you could avoid using it in its current incarnation. I believe Geoff Plourde made another template based on it, which allows you to change the category, maybe hist. The Jade Knight (d'viser) 07:52, 22 November 2008 (UTC)

New Changes
Any particular reason for the removal of some explanatory content? The Jade Knight (d'viser) 07:49, 24 November 2008 (UTC)
 * Simple language for the student as it had become too sophisticated. Dilos1 08:32, 24 November 2008 (UTC)
 * I had removed the suggested books temporarily to check for availability. But it's OK for the moment.Dilos1 08:45, 24 November 2008 (UTC)
 * Both are currently available on Amazon, and the encyclopedia is also currently available on half.com. The Jade Knight (d'viser) 09:10, 24 November 2008 (UTC)
 * OK. Fine. So we can have them on. Dilos1 07:24, 26 November 2008 (UTC)

new editing
im starting to edit the medival history section. I have enough knowledge of history to do such thing (after all its just a wiki) The22 19:58, 18 January 2010 (UTC)

Migration period?
I think we should add the migration period as a whole new part of the course, you cant understand the Mongols without understanding the Huns, you can't understand where Charlamenge came from without knowing who the Franks arrived in France. Or is it under "Newcomers to Europe"? The22 03:04, 31 May 2010 (UTC)

title: Talk:History of Europe to 1648


text:
New antidote

A NEW ANTIDOTE FOR CYANIDE POISONING   By Dr. Mohammad Ali kavianpour Kermanshah medical university Iran May,2008

INTRODUCTION This paper describes a study of the antidotal actions of nickel nitrate against cyanide poisoning.  Hydrocyanic acid acts as a poison because it combines with, and stops the function of, the iron atom in cytochrome oxidase, so blocking the electron transfer through the cytochrome system, and checking the final oxidations involving oxygen uptake of the tricarboxylic acid cycle. It also inhibits many other enzymes.Attempts to find an antidote for cyanide have usually been in the direction of breaking down the cyanide-cytochrome complex, and have looked for substances with a stronger affinity for cyanide ions than has the oxidase.One such method is to convert some of the blood haemoglobin into methaemoglobin,for example,by injection of a nitrite; the methaemoglobin combines with the cyanide ion to form the very stable cyanmethaemoglobin. This has the drawback that it involves the loss of oxygen-carrying power of the blood, so that, for example, to antidote twice the LD50 of cyanide would be equivalent to the loss of about 12% (or in an adult man 550 ml.) of the blood. METHODS

The LD50 of hydrocyanic acid, by various routes of administration, was determined for mice and rabbits. A solution of sodium cyanide was neutralized with acetic acid to about pH 7,the content of hydrocyanic acid was estimated by silver titration, and the subsequent dilution to the required concentration was made with 0.9% saline.At pH 7 the hydrocyanic acid is about 1% ionized. In some experiments pure hydrocyanic acid was used, but this had no advantage and was more difficult to handle. Owing to the volatility of hydrocyanic acid, the solutions were kept in special vessels and, when any doubt existed as to the final concentration, were again titrated at the end of the day’s experiments. It was usually found that after 6 hr the concentration had fallen, often by 5%, and sometimes in hot weather by as much as 12%.  Intravenous injections were made into the tail layer in mice and an ear layer in rabbits; intramuscular injections were into the thigh muscles. Doses are expressed usually in terms of umoles/kg of body weight, and for hydrocyanic acid sometimes also in terms of LD50. Owing to the rapidity of action of hydrocyanic acid when large doses are given, and to the difficulty of giving intravenous injections when the animal is convulsing, it was more usual to give the antidote first, intravenously, and then to follow this in a matter of seconds by the hydrocyanic acid, usually by intramuscular injection, but in some experiments that order was reversed; the antidotal effects were at least as great, and often greater, when that was done, provided the antidote was given soon enough. In order to indicate the molar relationship between the hydrocyanic acid and the antidote in each experiment, the ratio between them was usually expressed (moles of hydrocyanic acid per kg divided by moles of nickel compound per kg) and is called the cyanide/nickel ratio. RESULTS The effectiveness of a nickel nitrate, against hydrocyanic acid was examined mainly on mice and rabbits.for rabbits (4 x LD50), was effective at a molar cyanide/nickel ratio of 4. and for mice (3 x LD50 at a molar ratio of 2), with no apparent side effects. At a molar ratio of 1, it was slightly more effective than on rabbits. Nickel nitrate by mouth was effective against orally administered hydrocyanic acid. The oxygen uptake of the body, reduced by cyanide, improved when nickel antidotes has been successfully administered. Relatively small amounts of nickel salts would be needed for the neutralization of several fatal doses of cyanide. By intravenous injection.The LD50 for mice was found to be about 42 umoles/ kg, and for rabbits 33 umoles/kg. Mice given two to three times the LD50 convulse in a few seconds and die in 50 seconds.By intraperitoneal injection,the results showed a rather large spread, the mean LD50 for rabbits being 52 ,umoles/kg, and for mice 107 umoles/ kg, so that by this route mice were only about half as sensitive as rabbits.This is probably due to a more rapid detoxification by the mouse liver.It will be expected that toxicity would be maximal when the cyanide is given quickly into a vein which drains into the inferior vena cava, so that it is quickly delivered to the central nervous system, less toxic when it enters more slowly, as by intramuscular injection, and least toxic when it enters the circulation slowly, as when given intraperitoneally, subcutaneously or orally. the descending order of toxicity proved to be intravenous intramuscular-intraperitoneal-subcutaneous-oral, The results of inhalation are more vague, but are probably comparable with those by intravenous injection.A method of administration often used, but not employed in the present investigation, is to anaesthetize the animal and to inject the cyanide intravenously at a constant slow rate of about 0.1 mg/kg/min.The results so obtained do not seem to differ greatly from those of the usual procedure for intravenous injection.  The nickel ion is known to be toxic to some micro-organisms, to depress metabolism in tissue slices and, under certain conditions, metallic nickel can be carcinogenic, though this last action has not been demonstrated for the nickel ion. In acute experiments, nickel salts given intravenously cause a fall, followed by a rise and a later fall, of arterial pressure, an increase in breathing, cramps, vomiting and acute diarrhea with intestinal inflammation. These effects have been analyzed and shown to be partly due to central action and partly to peripheral effects. Death is due to respiratory failure when the administration is rapid, to cardiac failure when more slowly given. In more chronic administrations the principal effects are polycythaemia, porphyrinuria, increase in the size of the adrenals, and goitre.Nickel salts have been given orally to human subjects in doses up to 140 mg daily without clearly pronounced harmful effects. Orally administered nickel is very slightly absorbed, and mainly excreted in the faeces; injected intravenously it is mainly, and rapidly, excreted in the urine; according to experiments what remains in the body is to be found mainly in liver, kidneys, pancreas and spleen.In the present experiments, no evidence could be found that nickel was present in any tissues or organs 14 hr after intravenous injection. Post mortem examination showed intense congestion of the mucosa of the intestinal tract.On the rabbit isolated ileum preparation the effect was relaxation, with diminution of the rhythmic contractions on the rectum in situ intravenous injection caused powerful rhythmic contractions, and these probably account for the colic and diarrhea which often follow administration of nickel salts. The toxicity of nickel compounds is related to the ease with which they yield nickel ions, and hence is greatest in salts of nickel. Nickel nitrate yields ions readily; and the compound would be expected to be somewhat toxic.  Exactly what occurs when nickel and cyanide ions are introduced into the body is uncertain; while being at the same time only slightly toxic (LD50=1,000 mg/kg). nickel ions can reach the brain, that is they can penetrate the blood-brain barrier, have no precipitating action on proteins, and, when introduced into the blood stream, are eliminated by the kidney, which is not damaged.The opposition between nickel and cyanide ions can be illustrated by experiments with isolated tissues.  Some experiments were made to find whether nickel ions were able to reverse the inhibition of the cytochrome oxidase system produced by cyanide.using p-phenylenediamine as the oxidation substrate.Only about 10% of the initial oxidase activity was restored, Since the lethal dose of nickel salts is around 110 umoles/kg, the amount given as an antidote has mostly been lower than that; but that dose should be able to antidote 550 umoles of hydrocyanic acid, if the theoretical ratio holds good, and the mutual antidotal action should in theory reduce any risk of nickel poisoning. This would mean that, for a safe dose of a nickel salt, some 10 x LD50 of hydrocyanic acid intraperitoneally for rabbits, and 5 x LD50 intraperitoneally for mice,should be the maximum amounts that could be antidoted.In experiments done the maximum dose antidoted was 10 x LD50, in a goat.The present series, using mice and rabbits, gave results which are satis- factory for rabbits, but for mice fall short of the theoretical predictions mentioned above.  When the molecular ratios of cyanide and nickel are calculated, it is seen that, in rabbits, the ratios are spread over the range 1.5 to 13. then in all cases there should be no hydrocyanic acid left free. In the instances where an excess would be left it would be only a fraction of an LD50, and this should have been nonlethal, but in one instance (cyanide/nickel ratio of 12) was not, which indicates that the case is not so simple as it first seemed.The results show that the nickel salt was more effective as an antidote for rabbits than for mice. For rabbits, provided the molar cyanide/ nickel ratio does not exceed 6, it seems that at least 6 x LD50 can be neutralized, while at a ratio of 12 even twice the LD50 is not antidoted. It should be noted, however, that, in the rabbit series, hydrocyanic acid was usually given intravenously, whether before or after the nickel. With mice that are less sensitive to hydrocyanic acid, and about as sensitive to nickel, as are rabbits, and the results are less regular. It appears from the results that, under the most favorable conditions, only upwards of twice the LD50 can be antidoted, and then only when the cyanide/nickel ratio is below 5. The hydrocyanic acid was given intraperitoneally and the nickel solution intravenously in this species, which might have affected the results. Thiosulphate alone had antidotal action which varied; for mice the value was around 2 x LD50 at the best, but was irregular, and in rabbits it was also around twice the LD50. The results are rather irregular, but show that, for mice, 3 x LD50 can be antidoted when the nickel compound is given with glucose, and when thiosulphate is also given.  Some of the cyanide/nickel ratios were very high, as much as 20, but, as thiosulphate was also given, much of the beneficial effect must have been due to that. For rabbits the results were similar, but were also improved by giving thiosulphate. The most favorable condition for the action of the nickel antidotes would be expected to be if they were mixed with the hydrocyanic acid before administration.It would be expected that, if the resulting compound were not toxic and held the cyanide firmly enough in competition with cytochrome oxidase, the effect of the hydrocyanic acid would be entirely eliminated if enough antidote were present, and that, for instance, nickel salts would be able to fix more molar equivalents of hydrocyanic acid,  A simple experiment shows that the complex formed by nickel salts though not harmless is not very toxic.Solutions of nickel nitrate and sodium cyanide were mixed in such proportions that the mixture contained 260 ,umoles of cyanide and 45 ,umoles of nickel salt per ml. This was injected into an ear vein of a rabbit in a dose of 1 ml./kg. This would be, in terms of hydrocyanic acid, about 8.5-times the LD50. For some minutes after injection there was panting and convulsions, followed by paresis of the hind limbs, and loss of consciousness, as would have been the effects of, say, half a lethal dose of hydrocyanic acid; but after 5 min, apart from a little panting, the animal was apparently normal and made an uninterrupted recovery, with no subsequent diarrhea. We might suppose, since the cyanide/nickel ratio was less than 6 that these transitory symptoms of cyanide poisoning were to be attributed to slight dissociation of the nickelocyanide, but in any case the experiment shows that there are grounds for the belief that the nickel ion can, under the most favorable conditions, antidote the theoretical amount of cyanide. These expectations were only in part borne out by experiment on mice, These results show that when the dose of hydrocyanic acid is low (3 x LD50 or less) and the cyanide/nickel ratio is 6 or less, the expectations from theory hold approximately, but with four or more times the LD50 and cyanide/ nickel ratios above 4 the results are less good. In the experiments the oxygen utilization of rats anaesthetized with a mixture of allobarbitone and urethane was measured by the use of a small-scale respirometer, and the effect of an intraperitoneal dose of hydrocyanic acid, as large as possible without causing arrest of respiration, was observed; nickel derivatives of various kinds were then injected intravenously, and the subsequent oxygen usage measured over a sufficient period. In several experiments, the nickel injection was given too late, but when the animal survived the nickel injection restoration of oxygen usage was prompt and complete and was often followed by increased usage, presumably the expression of oxygen lack.Interaction between nickel salt and hydrocyanic acid given orally when hydrocyanic acid or a cyanide is taken by mouth, one obvious line of treatment would be to give a nickel salt by mouth immediately, or as soon as possible after any appropriate intravenous therapy had been applied. An experiment was carried out by giving hydrocyanic acid by stomach tube to mice, and then following this up after an interval of time by administering by the same route a dilute nickel nitrate solution.  It would seem that nickel nitrate solution, given by mouth within a short time of ingestion of cyanide, might be helpful; it would normally be followed by a wash-out of the stomach, or by chelating any excess of nickel and there would then be little danger from toxic effects due to absorption of nickel into the blood stream  DISCUSSION The consequences of administration of cyanide depend mainly on the rate at which it enters the blood stream, maximum effect resulting when the dose is high and the rate of entry rapid, as for example, when a high concentration of hydrocyanic acid vapor is breathed or when a large dose is injected into a systemic vein, in either of which cases serious symptoms ensue in a matter of seconds and death within minutes. Even if attempts at recovery appear to be successful, permanent damage to the central nervous system may remain if the period of tissue anoxia has been more than a few minutes. As the rate of entry of cyanide into the circulation is slowed, by slower administration , another factor enters into consideration, namely, the rate of destruction of cyanide in the body by conversion into thiocyanate, a change which takes place mainly in the liver by the action of the enzyme rhodanese.  The use of thiosulphate depends on the provision by it of available sulphur for this reaction; the change seems to be somewhat slow, so it is better as a prophylactic than as an antidote. Since cyanide when given orally or intraperitoneally enters initially into the portal circulation, it will go direct to the liver, so that destruction is facilitated and larger doses can therefore be tolerated. If the rate of eventual entry into the systemic circulation is slow enough, as on subcutaneous, oral or respiratory administration.of small doses, the effects are less serious, or may even be absent altogether. The mode of administration therefore has a definite effect on the LD50, and as shown, the LD50 in increasing order is intravenous-intramuscular-intraperitoneal -subcutaneous-oral, with inhalation, which is a common industrial risk, comparable with administration by the intravenous route. Another factor which influences the outcome is whether the antidote precedes or follows the administration of the cyanide and, if the latter, after what interval of time. It would be expected that the maximum antidotal effect would be seen when poison and antidote were mixed before administration,and next to this when the two were injected simultaneously into the circulation, or when antidote preceded the cyanide by a very short time; the minimum effect would be seen when the antidote was given slowly, for example subcutaneously, and at a time when the animal was near to death, when it would be all but useless. In nearly every instance in the present experiments, the antidote was administered intravenously, but intraperitoneal injection has also been found effective. For practical reasons the nickel antidote was usually given to mice intravenously a few seconds before the dose of hydrocyanic acid, intraperitoneally or intramuscularly. The toxic action of nickel ions is the probable reason for this, but this is not great and would largely be offset by the mutual detoxification of the two reactants. The number of mean lethal doses of hydrocyanic acid which can be antidoted by nickel ions is greater for rabbits than for mice, namely five to six for rabbits and two for mice, with maximal cyanide/nickel ratios of 5 and 4 respectively. One reason for this is that mice are less sensitive to hydrocyanic acid than are rabbits, while to nickel ions they are about equal in sensitivity. Doses of nickel nitrate greater than 50 umoles/kg, though lowering the cyanide/nickel ratio, were unsatisfactory for mice. Another reason in most of the present experiments is that in rabbits both substances were given intravenously, and so had more opportunity of reacting. The fact that the cyanide/nickel ratio can be, with rabbits, effective at ratios as high as 5, indicates that the changes involved in the antidotal reaction are much as expected from theoretical considerations; and the fact that, with high doses of hydrocyanic acid the effective cyanide/nickel ratio is usually lower, is probably to be explained on the basis that the difference between the affinities of the cyanide ions for cytochrome and for nickel are not very large.  The results of the present experiments indicate that in that respect there is not much to choose between the two.The more favorable results which have been obtained with the compound by present a disagreement which must be explained; one explanation being that in those experiments cyanide and antidote were given almost simultaneously by intravenous injection, which, gives very favorable conditions for reaction.One fact is against the view that the compound is to be regarded as an ionizable salt of nickel, despite the similarity of their respective toxicities when expressed in terms of molarity;   Ratio at effectivity when no thiosulphate is given is much lower than with nickel. This is also shown for the experiments with previous mixing, and bears out the findings of that the reaction is complete at a molar ratio of about 2,, then the compound is not an ordinary ionizable nickel salt, or it reacts with some blood constituent, or else it was not obtainable in as pure a state. The object of this investigation was to form an opinion as to the best antidote and method of treatment for cyanide poisoning. First, as to the amounts of the various antidotes which would be required for the treatment of a man who had received one LD50 of cyanide, say 50 mg of hydrocyanic acid,  The question is, whether a dose of nickel nitrate of the order of 100 mg would be safe to give intravenously, taking into account the probability that most of its toxicity would be neutralized by the cyanide. The balance of evidence is that it would be a reasonably safe dose under the circumstances. And then to give an intravenous injection of hypertonic glucose.The use of oxygen administration has been advised,although the only explanation of its value which is offered is that it helps to overcome the large oxygen debt which is seen on recovery from the immediate effects of the poison; at all events, it should be tried. So also should the administration of nickel nitrate by mouth, in cases where cyanide has been ingested.

ABSTRACT The effectiveness of a nickel nitrate, against hydrocyanic acid was examined mainly on mice and rabbits.For rabbits (4 xLD50),was effective at a molar cyanide/nickel ratio of 4. For mice (3 x LD50 at a molar ratio of 2), with no apparent side effects. At a molar ratio of 1, it was slightly more effective than on rabbits.Nickel nitrate by mouth was effective against orally administered hydrocyanic acid. The oxygen uptake of the body, reduced by cyanide, improved when nickel antidotes has been successfully administered.Relatively small amounts of nickel salts would be needed for the neutralization of several fatal doses of cyanide.

KEY WORDS  Antidote / Nickel nitrate / Cyanide poisoning

REFFERANCES Ambrose AM, Larson PS, Borzelleca JF, and Hennigar GR, Jr. 1976. Long term toxicologic assessment of nickel in rats and dogs. J. Food Sci. Technol. 13: 181-187. Broder I, Smith JW, Corey P, Holness L. Health status and sulfur dioxide exposure of nickel smelter workers and civic laborers. J. Occup. Med. 31(4):347-353. CARB. 1999. Air toxics emissions data collected in the Air Toxics Hot Spots Program CEIDARS Database as of January 29, 1999. CDTSC. 1985. California Department of Toxic Substances Control. Review: Nickel and compounds. 11/15/85. Cirla AM, Bernabeo F, Ottoboni F, and Ratti R. 1985. Nickel induced occupational asthma: Immunological and clinical aspects. In: Brown SS, and Sunderman FW. (eds.) Progress in Nickel Toxicology. Boston, MA: Blackwell Sci. Pub. Davies JE. 1986. Occupational asthma caused by nickel salts. J. Soc. Occup. Med. 36:29-31. Dunnick JK, Elwell MR, Benson JM, Hobbs CH, Hahn FF, Haly PJ, Cheng YS, and Eidson AF. 1989. Lung toxicity after 13-week inhalation exposure to nickel oxide, nickel subsulfide, or nickel sulfate hexahydrate in F344/N rats and B6C3F1 mice. Fundam. Appl. Toxicol. 12:584- 594. Glaser U, Hochrainer D, Oldiges H, and Takenaka S. 1986. Long-term inhalation studies with NiO and As2O3 aerosols in Wistar rats. Int. Congr. Ser. Excerpta Med. 676: 325-328. Haley PJ, Shopp GM, Benson JM, Cheng YS, Bice DE, Luster MI, Dunnick JK, and Hobbs CH. 1990. The immunotoxicity of three nickel compounds following 13-week inhalation exposure in the mouse. Fundam. Appl. Toxicol. 15:476-487. HSDB. 1995. Hazardous Substances Data Bank. National Library of Medicine, Bethesda, MD (TOMESWP:NOR version) Denver, CO: Micromedex, Inc. (Edition expires 4/30/95). Hueper WC. 1958. Experimental studies in metal cancerigenesis. Arch. Pathol. 65:600-607. Lundborg M, and Camner P. 1984. Lysozyme levels in rabbit lung after inhalation of nickel, cadmium, cobalt, and copper chlorides. Environ. Res. 34:335-342. McConnell LH, Fink JN, Schlueter DP, and Schmidt MG. 1973. Asthma caused by nickel sensitivity. Ann. Int. Med. 78:888-890. Muir DC, Julian J, Jadon N, Roberts R, Roos J, Chan J, Maehle W, Morgan WK. Prevalence of small opacities in chest radiographs of nickel sinter plant workers. Br. J. Ind. Med. 50(5):428- 431. NTP. 1994a. National Toxicology Program. Technical Report on the Toxicology and Carcinogenesis Studies of Nickel Oxide in F344/N Rats and B6C3F1 Mice. NTP TR 451, NIH Publication No. 94-3363. U.S. Department of Health and Human Services. NTP. 1994b. National Toxicology Program. NTP Technical Report on the Toxicology and Carcinogenesis Studies of Nickel Subsulfide in F344/N Rats and B6C3F1 Mice. NTP TR 453, NIH Publication No. 94-3369. U.S. Department of Health and Human Services. NTP. 1994c. National Toxicology Program. NTP Technical Report on the Toxicology and Carcinogenesis Studies of Nickel Sulfate Hexahydrate in F344/N Rats and B6C3F1 Mice. NTP TR 454, NIH Publication No. 94-3370. U.S. Department of Health and Human Services. Novey HS, Habib M, and Wells ID. 1983. Asthma and IgE antibodies induced by chromium and nickel salts. J. Allergy 72(4):407-412. Ottolenghi AD, Haseman JK, Payne WW, Falk HL, and MacFarland HN. 1974. Inhalation studies of nickel sulfide in pulmonary carcinogenesis of rats. J. Natl. Cancer Inst. 54(5):1165- 1172. U.S. EPA. 1985. United States Environmental Protection Agency. Health Assessment Document for Nickel. EPA/600/8-83/012F, pp. 3-3.

title: New antidote


text: User:Killian.levacher

title: User:Killian.levacher


text: User talk:Killian.levacher

title: User talk:Killian.levacher


text: User:Killian.levacher/spanish

title: User:Killian.levacher/spanish


text: File:Unblockrequest.png

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User talk:Mange01

 Hello Mange01, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
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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! --JWSchmidt 17:17, 8 November 2008 (UTC)

Moodle & Quizzes
Hi Mange01, I am also interested in Moodle and Quizzes and would be interested in how you go about moving stuff between MediaWiki and Moodle Quizzes. We're about to start using Moodle in 2009. My Moodle notes are over http://docs.moodle.com/en/User:James_Neill. -- Jtneill - Talk - c 09:19, 18 November 2008 (UTC)

Invitation Wikipedia project
Hi Mange01,

I am more interested in social sciences related topics and joined the Wikipedia service-learning courses, because i liked the idea of having people acquiring certificates for courses they finished. With this tool, a community could arise of responsible people.--Daanschr 17:02, 18 November 2008 (UTC)

Wikipedia
Hi Mange01, thanks for the note. I am interested in developing these materials, though spread rather thinly at the moment, and unsure of how much time/energy I can realistically devote. In general, intrawiki linking is a good idea, in order to deepen/broaden understanding of a particular issue, or to make links with related topics. When doing so, I think it makes sense to briefly explain how the reader can expect to further their knowledge, and to encourage some sort of continuity between pages (or indeed feedback to the page linked from). Overall, there is also Topic:Wikipedia studies, which can act as an organiser page to link to relevant materials. Please continue to edit and leave comments/questions, and I'll stop in from time to time. Cheers, and welcome! Cormaggio talk 11:59, 19 November 2008 (UTC)

hey
a/s/l

Re:Wikipedia/Quizzes
You can combine edits by using the preview function.

title: User talk:Mange01


text: MediaWiki:Categorytree-member-num

title: MediaWiki:Categorytree-member-num


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User:Eas4200c.f08.blue.a/Lecture 29

$$\vec{t}=t_{y}\vec{j}+t_{z}\vec{k}$$

It can be infered that:

$$dz=dscos\theta $$  $$n_{y}=cos\theta $$

$$dy=dssin\theta $$  $$n_{z}=sin\theta $$

Summing the forces in the y directon:

$$\sum{F_{y}}=0=-\sigma _{yy}(dz)-\sigma _{yz}(dy)+t_{y}(ds)$$

and using the above relations:

$$t_{y}=\sigma _{yy}n_{y}+\sigma _{yz}n_{z}$$

Similarly,

$$\sum{F_{z}}=0=-\sigma _{zz}(dy)-\sigma _{yz}(dz)+t_{z}(ds)$$

and again, substituting, we find:

$$t_{z}=\sigma _{zz}n_{z}+\sigma _{yz}n_{y}$$

In matrix form this becomes:

$$\begin{Bmatrix} t_{y}\\ t_{z} \end{Bmatrix}=

\begin{bmatrix} \sigma _{yy} & \sigma _{yz}\\ \sigma _{zy}& \sigma _{zz} \end{bmatrix}

\begin{Bmatrix} n_{y}\\ n_{z} \end{Bmatrix}$$

Note: $$\sigma _{zy}= \sigma _{yz}$$ (because $$\sum{M_{x}}=\sum{M_{y}}=\sum{M_{z}}=0$$), but in order to keep with the unified notation, and to be compatible with the indicial notation, the difference is retained.

The generalized form is as follows:

$$\begin{Bmatrix} t_{1}\\ t_{2}\\ t_{3} \end{Bmatrix}=

\begin{bmatrix} \sigma _{11} & \sigma _{12}&\sigma _{13}\\ \sigma _{21}& \sigma _{22}&\sigma _{23}\\ \sigma_{31} &\sigma _{32}&\sigma _{33} \end{bmatrix}

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

ShorthandSymbollically it would written as

$$\begin{Bmatrix} t_{i} \end{Bmatrix}_{3x1}

=\begin{bmatrix} \sigma _{ij} \end{bmatrix}_{3x3}

\begin{Bmatrix} n_{j} \end{Bmatrix}_{3x1}$$

title: User:Eas4200c.f08.blue.a/Lecture 29


text:
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w

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text: User:Kamnet

title: User:Kamnet


text:
User:EML4500.f08.FEABBQ.Hamdan/HW-6

Notes from November-03-08

 * Redo Matlab Rectangular Trusses

Continuing the derivation:

Initial conditions: At t=0, prescribed

$$\left(x,t=0 \right) =\bar{U}\left(x \right)$$ Known Function Displacement $$\frac{\partial }{\partial t}\left(x,t=0 \right)=\dot{U }\left(x,t=0 \right)=\bar{V}\left(x \right)$$Known Function Velocity

$$\frac{\partial }{\partial x}\left[\left(EA \right)\frac{\partial }{\partial x} \right]+f=m\ddot{U}$$ Eq.(1)
 * PVW (continuous) of the dynamics of elastic bar
 * PDE

Discrete EOM $$\Leftarrow $$ Equation of motion. $$-Kd+F=M\ddot{d}$$

$$\frac{\partial }{\partial x}\left[\left(EA \right)\frac{\partial }{\partial x} \right]\Rightarrow -Kd$$

$$f \Rightarrow F$$

$$m\ddot{U} \Rightarrow M\ddot{d} $$

$$M\ddot{d}+Kd=F$$    Eq.(2)

Goal is to derive Eq.(2) from Eq.(1):
 * MDOF $$\Rightarrow$$ Multiple Degree of Freedom system
 * SDOF $$\Rightarrow$$ Single Degree of Freedom system

Motivation a SDOF spring mass system shown below where : $$M\ddot{d}+Kd=F$$

$$\int_{0}^{x=L}{W\left(x \right)\left\{\frac{\partial }{\partial x}\left[EA\frac{\partial U}{\partial x} \right]+f-m\ddot{U} \right\}}dx=0$$ Eq.(3)    For all posible $$W(x)$$

As seen
 * Eq.(1)$$\Rightarrow$$ Eq.(3) is trivial

 * Eq.(3)$$\Rightarrow$$ Eq.(1) is not trivial

Eq.(3) can be rewritten as $$\int W(x)g(x)dx=0$$

Since Eq.(3) holds for all $$W(x)$$, select $$W(x)=g(x)$$, then Eq.(3) becomes:

$$\int g^{2}dx=0 $$ where $$g^2\geq 0$$  \Rightarrow $$g(x)=0$$

title: User:EML4500.f08.FEABBQ.Hamdan/HW-6


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title: File:Spring 11-03-08.jpg


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title: File:ElasticNode.jpg


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text:
File:Fisk self.JPG

Summary
Diagram of aspects of self adapted from Fiske, S. T. (2004). The self: social to the core. In Social beings: A core motives approach to social psychology. (Ch 5, pp. 169-214). Hoboken, NJ: John Wiley & Sons.

title: File:Fisk self.JPG


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User:IQBAL88

HY I AM IQBAL.I MA STUSENT OF IUBAT,BANGLADESH AT UNDERGRADUATE LAVEL.I AM IANTERESTED TO STUDT IN AUSTRALIA/UK/NORWAY/HUNGRAY FOR MY HIGHER EDUCATION Insert non-formatted text here underline text

title: User:IQBAL88


text:
User talk:IQBAL88

 Hello IQBAL88, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! Jtneill - Talk - c 09:39, 9 November 2008 (UTC)

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title: User talk:IQBAL88


text:
User:WikimediaNotifier/translation

User:WikimediaNotifier/notifications
 * Here you can add the title of a translation page for User:WikimediaNotifier/notifications

title: User:WikimediaNotifier/translation


text:
Brezhoneg Unan/Lesson 5

{| cellpadding="10" cellspacing="5" style="width: 99%; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="background-color: #eed153; border: 1px solid #777777; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" colspan="2" |

Pronunciation rules
'''Foreword : This lesson does not treat of exceptions. Please refer to the next lesson on pronunciation for a full explanation including exceptions.'''

Pouezh mouezh / Accentuation
Accentuation is strong in Breton.

Accentuation may vary :
 * according to the length of the word
 * according to the group the word belongs to (word alone or name + adjective, etc.).

 * }

{| cellpadding="10" cellspacing="5" style="width: 99%; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="background-color: #EED153; border: 1px solid #777777; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" colspan="1" |

Polysyllabic words
Accentuation is on the penultimate syllable.

 * }

{| cellpadding="10" cellspacing="5" style="width: 99%; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="background-color: #EED153; border: 1px solid #777777; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" colspan="1" |

Monosyllabic words alone
Monosyllabic words alone are accentuated.

Name with a short adjective
In the case of a monosyllabic name with a short adjective (no more than 2 syllables), the adjective is accentuated.

Name with a long adjective
In the case of a monosyllabic name with a long adjective (3 syllables or more), the name is accentuated.

 * }

{| cellpadding="10" cellspacing="5" style="width: 99%; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="background-color: #EED153; border: 1px solid #777777; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" colspan="1" |

Exceptions
The exceptions are not treated in this lesson. Please refer to the next lesson on pronunciation for a description of exceptions.

 * }

title: Brezhoneg Unan/Lesson 5


text:
User:Eas4200c.f08/HW report table/The best of HW5

 Under construction; not final (but close to being stable after a couple of weeks). The intention here is to document the best features in any HW report for the readers (including you); if you see excellent features in any HW report (including your team's) that I may have missed noticing, don't hesitate to let me know. I don't have time to read all HW reports in detail. In fact, due to time constraint, I only selectively looked at a few important features. On the other hand, I also added some annotations (e.g., related to errors) in the HW reports that I looked through for the benefit of everyone. Eml4500.f08 11:11, 10 November 2008 (UTC)

 * classnotes graded over 100%
 * torsional analysis (cont'd)
 * strain-stress relation for isotropic linear 3-D elasticity in Voigt notation
 * stress-strain relation:
 * inverse of block-diagonal matrix
 * inversion of strain-stress relation
 * deduce the 4 zero stress components from the 4 zero strain components due to kinematic relations using stress-strain relation
 * equilibrium equation for stresses
 * uniform stress field vs non-uniform stress field: 1-D case as example
 * derivation of 1-D equation of equilibrium with distributed axial load
 * 3-D equation of equilibrium in indicial notation
 * derivation for x direction
 * deduce equation of equilibrium in terms shear stresses for torsional analysis
 * bidirectional bending
 * definition of bending moments in terms of normal stress and cross-sectional coordinates
 * components of moment of inertia tensor for cross section
 * normal bending stress in terms of bending moments and moments of inertia
 * neutral axis, definition, formula

 * matlab problem graded over 100%: develop matlab codes to analyze the bending behavior of the NACA 2415 airfoil reinforced with stringers. Best overall solution: Team VQCrew, complete description of the problem with figures, complete matlab codes, figures describing the results, narrative on the results, comparison of the results of Part I and Part II in table form; but it is not clear why Team VQCrew chose not to use their rear stringers with L brackets. NOTE: Team ZYX (with annotations) did not do this problem.  Team Aero_(Eelman) (with annotations), there are a lot of problems to be fixed in this wiki page.  Team Aeris (with annotations).
 * description of the problem
 * Part I: neglect bending resistance from skin and spar webs
 * location of centroid of stringers
 * components of moments of inertia tensor
 * slope of neutral axis
 * highest normal bending stress and its location
 * ultimate bending moments, assuming stringers made of 300 M steel
 * a figure and a narrative explaining the above results
 * Results: $$\displaystyle I_{22}, I_{33}, I_{23}, \beta$$ in $$\displaystyle m^4, m^4, m^4, rad \ {\rm or} \ ^\circ$$; $$\displaystyle M_y^{ult}, M_z^{ult}$$ in $$\displaystyle Nm$$
 * Team VQCrew $$\displaystyle 6.6875E-007, 3.4375E-005, 9.3322E-007, 0.0345 \,rad = 1.219^\circ$$; Moments $$\displaystyle 28355, -11342$$.

 * Team Radsam $$\displaystyle 6.3403E-007, 3.6111E-005, -2.8043E-007, 0.7694$$. Later corrected in Team Radsam Fixed to $$\displaystyle 6.8320E-007, 8.1250E-005, 4.7132E-007, 1.3690^\circ$$ (Results in close agreement with Team VQCrew for $$\displaystyle \beta$$, except for $$\displaystyle I_{22}$$); Moments $$\displaystyle -41013, 16405$$ (Results NOT in agreement with Team VQCrew, likely incorrect)
 * Team Carbon $$\displaystyle 6.3913E-007, 8.3333E-006, -7.6055E-008, ???$$, $$\displaystyle \beta$$ not given ($$\displaystyle I_{22}$$ agreed with Team VQCrew and Team Radsam, but not other moments of inertia); Moments $$\displaystyle -7176, 2870$$ (Too small, likely incorrect).
 * Team Gator $$\displaystyle 8.7409E-007, 1.6664E-005, -6.4413E-007, -3.3609$$ (Results did not agree with VQCrew, Radsam; likely incorrect); $$\displaystyle -8087, -3235$$ (Results likely incorrect).

 * Part II: with bending resistance from skin and spar webs, and with detailed geometry of stringers
 * location of centroid of stringers
 * components of moments of inertia tensor
 * slope of neutral axis
 * highest normal bending stress and its location
 * now neglect bending contribution from the skin and spar webs
 * highest normal bending stress in any stringer
 * ultimate bending moments, assuming stringers made of 300 M steel
 * a figure and a narrative explaining the above results and a comparison with the results from Part I
 * Results
 * Team VQCrew $$\displaystyle 6.2884E-007, 1.4533E-004, 4.8673E-006, 0.0347$$ (Agreed with Part I of Team VQCrew); Moments $$\displaystyle 20970, -8388$$ (consistent with results in Part I; likely correct).
 * Team Radsam $$\displaystyle 7.4915E-005, 5.3120E-003, -9.6928E-003, 56.6522$$ !!! Later corrected in Team Radsam Fixed to $$\displaystyle 6.3373E-007, 1.4810E-004, 5.3883E-006, 2.2781$$; That's better, but still did not agree with Team VQCrew for $$\displaystyle \beta$$; Moments $$\displaystyle -24662, 9865$$ (closer to results of Team VQCrew, but signs seem incorrect)
 * Team Carbon $$\displaystyle 1.0893E-004, 1.7658E-004, -3.4076E-007, ???$$ ($$\displaystyle \beta$$ not given; results did not agree with Team VQCrew, Team Radsam; like not correct); Moments $$\displaystyle -490920, 196370$$ (too large, likely incorrect).
 * Team Gator $$\displaystyle 3.3178E-007, 1.4315E-004, 6.6650E-006, 2.6622$$ (Results did not agree with VQCrew; likely incorrect; $$\displaystyle \beta$$ switch from negative to positive); Moments $$\displaystyle 8630, 3452 $$ (magnitude did not agree with Team VQCrew; signs changed compared to Part I; results likely incorrect).

EML4500, The best of HW5


text:
Talk:Solid state physics

funny that the reference are linked to Mceuen lab at Cornell. I don't think those are good for general reference for ssp

title: Talk:Solid state physics