text:
User:Eml5526.s11.team7.jb/work51

For Polynomial Basis
The corresponding family $${{\bar{F}}_{P}}=\left\{ {{{\bar{b}}}_{j}} \right\}$$ of polynomial basis function satisfying the Constraint Breaking Solution (CBS) for $$\Gamma_g = \left \{ 0 \right \}$$ is:

$${{F}_{P}}=\left\{ {{\left( x \right)}^{j}},\text{ }j=0,1,2,\ldots \right\}$$

Now that the family of basis functions are determined, we recall that the discrete weak form is defined by

$$ \mathbf{\tilde{d}} \cdot \mathbf{b}( \beta ) = g \qquad \text{and} \qquad \mathbf{\tilde{c}} \cdot \left[ \mathbf{\tilde{M}} \mathbf{\tilde{d}}^{(s)} + \mathbf{\tilde{K}} \mathbf{\tilde{d}} - \mathbf{\tilde{F}} \right] = 0 $$

where the expression $$\displaystyle \mathbf{\tilde{M}} \mathbf{\tilde{d}}^{(s)} = 0 $$ for the static case. Then the discrete weak form is simplified to:

\mathbf{K} \mathbf{d} = \mathbf{F} \qquad \Rightarrow \qquad \mathbf{K} \mathbf{d} = \mathbf{F}_F - \mathbf{K}_{FE} g. $$

The first constant, $$\displaystyle d_0$$, is determined from the formulation of the CBS. It is known that:

 * $$\displaystyle u^h(\beta) = d_0 b_0(\beta)=g$$

 * $$\displaystyle d_0= \frac{g}{b_0(\beta)}=4/1=4$$

Now the matrices used to determine the rest of the coefficients are determined as follows. Note that the number of terms needed in the approximate solution, $$\displaystyle n$$, is determined by the convergence interval given. The first few terms of each matrix computation is shown. A Matlab script was written to iterate these computation for increasing values of $$\displaystyle n$$ until the desired accuracy was reached.

\displaystyle

{F_F} = \left\{ {{F_i},i = 1,2,...,n} \right\} = \left\{ {\tilde f\left( \right),i = 1,2,...,n} \right\}

$$

\displaystyle

\tilde f({b_i}) = {b_i}(\alpha )h + \int\limits_\alpha ^\beta {{b_i}fdx}

$$

\displaystyle

h = {a_2}(\alpha )\frac = (2+3\alpha)\frac = 5*12/5=12

$$

\displaystyle

\tilde f({b_1}) = {b_1}(1)h + \int\limits_0^1 {{b_1}(5x)dx}=1*12+\int\limits_0^1  {{(x)}(5x)dx}=41/3

$$

\displaystyle

\tilde f({b_2}) = {b_2}(1)h + \int\limits_0^1 {{b_2}(5x)dx}=1*12 + \int\limits_0^1  {{(x)^2}(5x)dx}=53/4

$$

$$ \displaystyle

\therefore {F_F} = \left[ {\begin{array}{ccccccccccccccc} {41/3} \\    {53/4} \\   {.} \\   {.} \\  {.} \end{array}} \right]

$$

\displaystyle

{K_{FE}} = \left\{ {{K_{0j}};j = 1,...,n} \right\} = \left\{ {\tilde k\left( {{b_0},{b_j}} \right),j = 1,...,n} \right\}

$$

\displaystyle

\tilde k({b_i},{b_j}) = \int\limits_0^1 {b{'_i}{a_2}b{'_j}dx}

$$

\displaystyle

\tilde k({b_0},{b_1}) = \int\limits_0^1 {b{'_0}{a_2}b{'_1}dx} = \int\limits_0^1 {b{'_0}(2+3x)b{'_1}dx}  = 0 = \tilde k({b_0},{b_2}) = \tilde k({b_0},{b_3}) = 0

$$

\displaystyle

\therefore {K_{EF}} = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   . \\ .\\ . \end{array}} \right]

$$

And：

\displaystyle

\because {K_{FE}} = {K_{EF}}^T = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   . \\ .\\ . \end{array}} \right]

$$

\displaystyle

{K_{FF}} = \left[ {{K_{ij}};i,j = 1,2,...} \right] = \tilde k({b_i},{b_j})

$$

\displaystyle

\tilde k({b_1},{b_1}) = \int\limits_0^1 {b{'_1}{2+3x}b{'_1}dx} = \int\limits_0^1 {1*(2+3x)*1dx}=7/2

$$

\displaystyle

\tilde k({b_1},{b_2}) = \int\limits_0^1 {b{'_1}{2+3x}b{'_2}dx} = \int\limits_0^1 {1*(2+3x)*2(x)dx}=4

$$

\displaystyle

\tilde k({b_2},{b_1}) = \int\limits_0^1 {b{'_2}{2+3x}b{'_1}dx} = \int\limits_0^1 {2(x)*(2+3x)*1dx}=4

$$

\displaystyle

\tilde k({b_2},{b_2}) = \int\limits_0^1 {b{'_2}{2+3x}b{'_2}dx} = \int\limits_0^1 {2(x)*(2+3x)*2(x)dx}=17/3

$$

\displaystyle \therefore {K_{FF}} = \begin{bmatrix} 7/2&4 &.&. \\  4&17/3  & .&.\\ . &.  & .&.\\ . &.  & .&. \end{bmatrix} $$

The matrices are then solved for the coefficients:

\displaystyle

Kd = F $$

\displaystyle

F = {F_F} - K_{FE}g = {F_F} $$

\displaystyle

K_{FF}d_F=F_F $$

\displaystyle

d_F=K_{FF}^{-1} F_F $$

The approximate solution, $$\displaystyle u^h(x)$$ is defined as,

 * $$\displaystyle u^h(x)= d_0b_0+\sum_{j=1}^{n}d_jb_j(x) \qquad \text{where} \quad d_0=4 \ \text{and} \ b_0=1$$.

As stated above a Matlab script was written to iterate $$\displaystyle n$$ until the desired accuracy was reached. At $$\displaystyle n=6$$ the error was $$\displaystyle  O(3.5840*10^{-6})$$. The following plots were generated:

title: User:Eml5526.s11.team7.jb/work51


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

title: File:Fe1.s11.team1.HW5.fig1.1.pdf


title: File:Fe1.s11.team1.HW5.fig1.2.pdf


text:
User:Eml5526.s11.team3.ushnish12/Homework 5

Problem 5.4
i>

Let the LIBF be defined by a function

 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

f(x)=\prod\limits_{0}^{n}{\frac{(x-{{x}_{1}})(x-{{x}_{2}})(x-{{x}_{3}})...(x-{{x}_{n}})}{({{x}_{0}}-{{x}_{1}})({{x}_{0}}-{{x}_{2}})({{x}_{0}}-{{x}_{3}})...({{x}_{0}}-{{x}_{n}})}}

$$
 * }

Now for the given function, at $$\displaystyle x={{x}_{0}}, f(x)= \frac{({{x}_{0}}-{{x}_{1}})({{x}_{0}}-{{x}_{2}})({{x}_{0}}-{{x}_{3}})...({{x}_{0}}-{{x}_{n}})}{({{x}_{0}}-{{x}_{1}})({{x}_{0}}-{{x}_{2}})({{x}_{0}}-{{x}_{3}})...({{x}_{0}}-{{x}_{n}})} =1 $$ Now for the given function, at $$\displaystyle x={{x}_{1}}, f(x)= \frac{({{x}_{1}}-{{x}_{1}})({{x}_{1}}-{{x}_{2}})({{x}_{1}}-{{x}_{3}})...({{x}_{1}}-{{x}_{n}})}{({{x}_{0}}-{{x}_{1}})({{x}_{0}}-{{x}_{2}})({{x}_{0}}-{{x}_{3}})...({{x}_{0}}-{{x}_{n}})}= 0$$

Now for the given function, at $$\displaystyle x={{x}_{2}}, f(x)= \frac{({{x}_{2}}-{{x}_{1}})({{x}_{2}}-{{x}_{2}})({{x}_{2}}-{{x}_{3}})...({{x}_{2}}-{{x}_{n}})}{({{x}_{0}}-{{x}_{1}})({{x}_{0}}-{{x}_{2}})({{x}_{0}}-{{x}_{3}})...({{x}_{0}}-{{x}_{n}})}= 0$$

Now for the given function, at $$\displaystyle x={{x}_{n}}, f(x)= \frac{({{x}_{n}}-{{x}_{1}})({{x}_{n}}-{{x}_{2}})({{x}_{n}}-{{x}_{3}})...({{x}_{n}}-{{x}_{n}})}{({{x}_{0}}-{{x}_{1}})({{x}_{0}}-{{x}_{2}})({{x}_{0}}-{{x}_{3}})...({{x}_{0}}-{{x}_{n}})}= 0$$

So for the basis function, $$\displaystyle {c}_{0} =1 , {c}_{1},{c}_{2},{c}_{n} = 0 $$

Hence the LIBF can be used as a CBS as it satisfies the conditions for the CBS.

Plotting the LIBF used
The LIBF used must satisfy the essential boundary conditions.

The essential boundary conditions to be satisfied are:-

 * {| style="width:100%" border="0"

$$\displaystyle {{b}_{0}}(1)=1 $$ and  $$\displaystyle {{b}_{1...n}}(1)=0 $$ Hence to satisfy the condition, the LIBF basis functions used are:-
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * }

 * {| style="width:100%" border="0"

$$\displaystyle f(x)=\prod\limits_{0}^{n}{\frac{(1-x-{{x}_{1}})(1-x-{{x}_{2}})(1-x-{{x}_{3}})...(1-x-{{x}_{n}})}{({{x}_{0}}-{{x}_{1}})({{x}_{0}}-{{x}_{2}})({{x}_{0}}-{{x}_{3}})...({{x}_{0}}-{{x}_{n}})}} $$
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * }

Using these basis functions, the LIBF are plotted until the convergence of the basis functions with the actual solution takes place where the error is less than $$\displaystyle {{10}^{-6}} $$

The convergence becomes less than $$\displaystyle  {{10}^{-6}} $$ for  $$\displaystyle  n=9 $$ The LIBF are plotted for various values of $$\displaystyle n $$ until convergence.

Plotting the LIBF and Exact Solution and Errors
The exact solution of governing equation is
 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

u(x)=-\frac{118}{27}\ln (\frac{2+3x}{5})+\frac{1}{36}(20x-15{{x}^{2}}+139)

$$
 * }

Using Lagrangian Interpolation basis function
The convergence occurs at $$\displaystyle n=9 $$

Using Matlab to construct above two matrices until satisfying the convergence criterion, we finally have
 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

{{K}_{FF}}=\left( \begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\   -118.406 & 376.137 & -595.975 &  712.071 & -660.485 & 432.555 & -193.210 & 55.498 & -8.185 \\   142.2520 & -595.975 & 1152.701 & -1453.567 & 1351.427 & -919.4265 & 440.528 & -142.253 & 24.315 \\   -178.012 & 712.071 & -1453.569 & 2000.226 & -1954.708 & 1382.496 & -718.900 & 260.149 & -49.751 \\   163.668 & -660.485 & 1351.427 & -1954.709 & 2080.253 & -1583.127 & 880.332 & -349.433 & 72.069\\   -104.8667 & 432.555 & -919.427 & 1382.497 & -1583.123 & 1349.980 & -828.416 & 345.622 & -74.822 \\   44.845 & -193.210 & 440.529 & -718.900 & 880.332 & -828.415 & 593.033 & -276.316 &  58.104 \\    -11.827 &  55.498 & -142.253 & 260.1492 & -349.433 &  345.622 & -276.3164 & 167.423 & -48.861\\   1.519 & -8.185 & 24.315 & -49.752 & 72.069 & -74.822 &  58.104 & -48.861 &  25.609\\ \end{matrix} \right)

$$
 * }
 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

{{F}_{F*1}}=\left( \begin{matrix}  4\\   0.908\\  -0.122\\    1.157\\  -0.400\\   0.694\\ -0.0409\\   0.129\\   12\\ \end{matrix} \right)

$$
 * }

By the equation $$\displaystyle Kd =F $$, we have
 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

{{d}_{F*1}}=\left( \begin{matrix}  4.0000\\    4.3689\\    4.7537\\    5.1595\\    5.5935\\    6.0649\\    6.5867\\    7.1775\\    7.8656\\ \end{matrix} \right)

$$
 * }


title: File:Eml5526.s11.team7.hw5.1poly.jpg


text: File:Eml5526.s11.team7.hw5.1poly 2.jpg

title: File:Eml5526.s11.team7.hw5.1poly 2.jpg


title: File:Eml5526.s11.team7.hw5.2poly.jpg


text: File:Eml5526.s11.team7.hw5.2poly 2.jpg

title: File:Eml5526.s11.team7.hw5.2poly 2.jpg


text:
User:Eml5526.s11.team6.tupsakhare/hwk5

=Problem 5.2 - Solutions using weak form and appropriate families of basis functions=

Strong Form
The strong form as referred from General one dimensional Model 1.0/Data set 1b on page 2 of Mtg 9:-

Boundary Conditions
The Essential Boundary condition as applied on the boundary $$\Gamma_g = [1]$$, is -- $$ u(1)=4 $$, The Natural Boundary condition as applied on the boundary $$\Gamma_h = [0]$$, is -- $$ -\frac{\partial u(x=0)}{\partial x}=6 $$

Solve for $$ u(x) $$ and $$ u^h(x) $$
To solve the above problem for $$ u(x) $$ and $$ u^h(x) $$ using appropriate basis functions of the following types until convergence of $$u^h(0.5)$$ to $$10^{-6}$$ -- 1. Polynomial basis function 2. Fourier basis function 3. Exponential basis function

Solution
Using weighting function $$ w(x) $$ such as $$ w(1) = 0 $$ and the technique of integration by parts, we have derived the weak form as follows --

Solving the above ordinary differential equation, we get the exact solution of governing equation as follows -

Now, using the technique from Mtg 18 and Mtg 19 we can write the discretized weak form as -

Also we have - as we choose the basis functions such that $$ w(n=0) = 1 $$, using technique from page -1 Mtg 22, we can write - $$ {{d}_{0}}=4, with- : i=1,2,\cdots $$,  $$ j=1,2,\cdots $$

Now we select legitimate families of basis functions as follows -

Fourier basis function
Note - the trial solution need to satisfy the essential boundary condition,$$u(x=1)=4$$ as stated in Eq. 5.2.2 Also, as we choose the basis functions such that $$ w(n=0) = 1 $$, using technique from page -1 Mtg 22, we can write - $$ {{d}_{0}}=4, with- : i=1,2,\cdots $$,  $$ j=1,2,\cdots $$

Now we can write the trial solution as follows -

Now we solve for the remaining coefficients $${{d}_{j}},_ – ^ – j=1,2,\cdots $$, by constructing the stiffness matrix and force matrix as follows:

Note- Here 1st equation is valid when 'i' and 'j' both are even, 2nd expression is valid when 'i' and 'j' both are is odd, 3rd expression is valid for remaining combination of values of 'i' and 'j'

Note- Here 1st expression is valid when 'i' and 'j' both are even, 2nd expression is valid when 'i' and 'j' both are odd

Now we need to have convergence of the solutions to the order of 10^(-6), hence we use a Matlab code demonstrated below to build the above matrices which are obtained then as follows-

By the equation that$$Kd=F$$, we can solve the remaining of $$d$$ as

Hence, the trial solution of the governing equation is

As seen from the Figure 5.2.1 below, it is evident that the approximate solution $$ u^h(x) $$ very closely imitates the exact solution $$ u(x) $$

Also it can be seen from Figure 5.2.2. that the error very rapidly converges to almost zero value when 5 or 6 basis functions are used

Exponential basis function
=Problem 5.7:Solving a strong form with Linear Lagrangian Element Basis Function (LLEBF)=

Given
The given ordinary differential equation is:

Boundary conditions
The Essential Boundary condition as applied on the boundary $$\Gamma_g = [0]$$, is -- $$ u\left(0\right) = 4$$ The Natural Boundary condition as applied on the boundary $$\Gamma_h = [1]$$, is -- $$\frac{du}{dx}\left(1\right) = -6 $$

Descritization
Here we want to use Uniform Descritization such that Element nodes are equidistant with number of elements as $$ nel = 4, 6, 8, ..$$

Find
1. Explain how Linear Lagrangian Element Basis Function (LLEBF) are used as Costraint Breaking Solution. 2. Plot all LLEBF used for $$ nel =4 $$. 3. Solve the system using 3 elements (nel = 4) and increase the number of elements or $$ nel =4,6,8..$$ until the error at x = 0.5 converges to the order of 10^-6 4. Plot for Exact solution $$ u (x)$$ and Approximate solution $$ u^h (x) $$. 5. Convergence - Plot of error vs. number of nodes (n).

LLEBF as Constraint Breaking Solution (CBS)
Consider a basis function $$b_{i}$$ such that we have, $$b_{0}\left(\beta\right) = 1$$, and $$b_{i}\left(\beta\right) = 0; i = 1, 2, ...$$. In such a case as per the discussion in Mtg 20, $$b_{i}$$ becomes a Constraint Breaking Solution (CBS)

LLEBFs (N1 and N2 below) for a 2 node linear element can be used to represent the unknown variable u(x)in a given domain which is written as -

 * {| style="width:100%" border="0"

$$u\left ( x \right )=u_{1}\frac{x-x_{2}}{x_{1} - x_{2}}+u_{2}\frac{x-x_{1}}{x_{2} - x_{1}} =u_{1}N_{1}+u_{2}N_{2} $$
 * $$\displaystyle (5.7.2) $$
 * }
 * }

Here $$u_{1}, u_{2}$$ are the nodal values of the unknown u(x) at the nodes 1 and 2 respectively.

The Lagrange Polynomials follow the property of Partition of Unity i.e. they have the value of 1 at the respective node and 0 at the others nodes. It means the basis function $$ b_{i} $$ carries value as 1 at the i th node and has a zero value at any other nodes in the domain.

Plot all LLEBF used for $$ nel =4 $$
The straight line plots (common for all elements of the 3 used with $$ nel =4 $$) irrespective of the elemental sizes for the 2 LLEBFs given in Eq. 5.7.2, following the Partition of Unity model are shown in the Figure 5.7.1.

Solve the system using 3 elements (nel = 4) and increase the number of elements or $$ nel =4,6,8..$$ until the error at x = 0.5 converges to the order of 10^-6
The weighted residual form with weighting function $$ w (x) $$ is given as:

We develop the corresponding weak form using the technique of integration by parts, with $$ w(0) = 0$$ owing to the fact that the flux is unknown at $$ x = 0 $$.

The expression for the weak form is as below -

To get the element stiffness matrices leading to the global stiffness matrix $$ K $$, we make coordinate change to the coordinate $$\phi$$ with value of -1 at node 1 and 1 at node 2.

The new coordinate $$\phi$$ is given by:

 * {| style="width:100%" border="0"

$$\varphi = \frac{2\left ( x-x_{1} \right )}{x_{2}-x_{1}}-1=\frac{2\left ( x-x_{1} \right )}{l}-1 $$
 * $$\displaystyle (5.7.5) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$ d\varphi =\frac{2dx}{l}$$
 * $$\displaystyle (5.7.6) $$
 * }
 * }

Hence the expression for $$ u(x) $$ in this changed coordinate system becomes-

 * {| style="width:100%" border="0"

$$u\left ( \varphi \right )=\begin{bmatrix} \frac{1-\varphi }{2} & \frac{1+\varphi }{2} \end{bmatrix}\begin{bmatrix} u_{1}\\

u_{2}\end{bmatrix}=\left [ N \right ]\begin{bmatrix} u_{1}\\

u_{2}\end{bmatrix} $$
 * $$\displaystyle (5.7.7) $$
 * }
 * }

Applying same operations on $$ w (x) $$, we can write -

 * {| style="width:100%" border="0"

$$\left [w\left(x\right)= N \right ]\begin{bmatrix} g_{1}\\ g_{2}\end{bmatrix}=\begin{bmatrix} g_{1} & g_{2} \end{bmatrix}\left [N \right ]^{T} $$ Now by using the chain rule of derivation:
 * $$\displaystyle (5.7.8) $$
 * }
 * }

 * {| style="width:100%" border="0"

$$\frac{du}{dx}=\frac{du}{d\varphi }\frac{d\varphi }{dx}=\frac{du}{d\varphi }\frac{2}{l} $$
 * $$\displaystyle (5.7.9) $$
 * }
 * }

 * {| style="width:100%" border="0"

$$\frac{dw}{dx}=\frac{dw}{d\varphi }\frac{d\varphi }{dx}=\frac{dw}{d\varphi }\frac{2}{l} $$ From Eq. (5.7.7) we can take the derivative with respect to $${d\varphi} $$ and then we can write:
 * $$\displaystyle (5.7.10) $$
 * }
 * }

 * {| style="width:100%" border="0"

$$\frac{du}{d\varphi }=\begin{bmatrix} -\frac{1}{2} & \frac{1}{2} \end{bmatrix}\begin{bmatrix} u_{1}\\ u_{2}\end{bmatrix}=\left [ B \right ]\begin{bmatrix} u_{1}\\ u_{2}\end{bmatrix} $$ In similar way for the weighting function $$ w(x) $$, we can say:
 * $$\displaystyle (5.7.11) $$
 * }
 * }

 * {| style="width:100%" border="0"

$$\frac{dw}{d\varphi}=\begin{bmatrix} g_{1} & g_{2} \end{bmatrix}\left [ B \right ]^{T} $$ The stiffness matrix is obtained as below -
 * $$\displaystyle (5.7.12) $$
 * }
 * }

The force matrix is obtained as below -

We assemble the system using a Matlab code given below where integration is performed using the Gauss Quadrature.

Plot for Exact solution $$ u (x)$$ and Approximate solution $$ u^h (x) $$ along x
It can be observed from the Figure 5.7.3 (Using 4 nodes) and Figure 5.7.4 (Using 6 nodes) that the exact solution is nearly linear and can be very well approximated by the LLEBFs considered with least 4 number of nodes and obviously also when the number of nodes is increased to 6. This is as the approximated solution $$ u^h (x) $$ very closely imitates the curve for the exact solution $$ u (x)$$.

Convergence - Plot of error vs. number of nodes (n)
At x = 0.5, to plot the error ( $$ u (x)$$ - $$ u^h (x) $$ ), we use Eq. 5.7.2 with the interpolation functions.

(n = 4), sufficiently low number of elements (around 20 -25) gives satisfactory convergence to an order of to 10^-3. We need around 125-130 elements for convergence to the order of of 10^-6. Figure 7.5.6 depicts the error as a function of the number of elements.

title: User:Eml5526.s11.team6.tupsakhare/hwk5


text:
User:Egm5526.s11.team4.zou/HW5

=Problem 5.5 Continuation of Calculix Practice=

From the lecture slide Mtg23, and this is a continuous effort based on previous work

Problem Statement
Continue to learn Calculix by:

a) Extract information from the “disk” example, including node information: node numbers with coordinates and element information: element number with element nodes.

b) Generate 3 meshes for the disk with triangular element with number of elements increasing.

c) Install ccx module and write a tutorial report.

a) extract information
Firstly run the Calculix command window and type the following command to open .fbd file in “build” mode.

cd disc cgx –b disc.fbd

Now we can only see the geometry, then generate mesh by the following command,

elty all qu4 mesh all plot me all

Then select: viewing->Toggle element edges you can see the shape of your elements (Fig. a-1).

Type,

plot n all

you can see the distribution of the nodes (Fig. a-2)

To see the information for a particular node or element, you may use the “ prnt ” command, such as,

Type,

prnt n 44 prnt e 30

Then you will see (Fig. a-3),

For more information about command “ prnt ” click here.

To acquire all the node and information i.e. nodal coordinates and element component, we type in the command window,

send all abq

By this action a file called “all.msh” is created and contains all nodes number and coordinates, also all element’s composition.

Opening all.msh you’ll see the following,

The node coordinates (Fig. a-4):

The element information(Fig. a-5), since we define element type “ qu4 ” i.e. “4 node shell(displayed S4 here)”, thus every element consists of 4 nodes.

b) Generate 3 meshes for the disk with triangular element with number of elements increasing
To generate new mesh, firs we need to delete the current one:

del me all

Then we create the new mesh with triangular element.

elty all tr3 mesh all plot me all

where the “ tr3 ” stands for triangular element. Alternative choices are “ tr3u ” and “ tr6 ”.

We’ll see the mesh like this (Fig b-1):

After typing “ send all abq ” and reading it, we know the number of elements of the above mesh is 384.

To generate a denser mesh i.e. with more elements, we need to change the division of lines.

Again we delete the current mesh by “ del me all ” command and then type,

plot ld all

This action will yield the following picture (Fig. b-2) telling you the division of the lines.

To get a denser mesh, we need to divide the lines into more parts. Type “ qdiv ”(with Enter) and then “ a ” (no Enter) and then type “ r ” twice (no Enter) to create a selecting window. Use the window to capture lines and change the division of them. Note that division number of 2 digits must type space in advance. Press “ q ” to quit.

Then we have change the division like (Fig b-3),

Now we can regenerate mesh,

mesh all plot me all

Remember to toggle element edge and you’ll see the denser mesh (Fig b-4), this mesh is with 864 elements.

By the same procedure we can divide the line more thoroughly and hence a very dense mesh (Fig b-5), with 1536 elements.

The preparation
For the Windows edition of Calculix, the cgx module and the ccx module are bundled together. So there’s no need to install ccx separately.

To learn the operation of ccx we can go through this tutorial to analyze a simple case: the stress/strain analysis of a disk fixed on the edge with concentrated force acting on the center.

For some reasons – more realistic, more options of element type – it’s better to replace the current 2D disk model with a 3D one. Again here we use “ del me all ” to delete mesh.

Type this command to show the label of surfaces (Fig. c-1):

plot sa all

Now we’re going to use these 4 surfaces to sweep along vertical direction to create a disk with a finite thickness. But in advance we should define them in the same set in order to manipulate them all simultaneously.

Type this command to add a new set: “set1”, with Enter button.

qadd se1

Then press “ a ” and “ r ” twice to make a selecting window (no need to press Enter). When point the window covering all 4 surfaces, press “ s ” for "surface". Finally press “ q ” to quit.

Type:

swep se1 se2 tra 0 0.25 0

This action make the original disk swep along “+y” direction making a 3D disk with 0.25 thickness (Fig. c-2).

Then we mesh de disk by,

elty all he20 mesh all plot me all

where “he20” stands for “20nodes, brick element”, and you can select your element type here.

The mesh is shown below (Fig. c-3).

To add constraint (the disk edge fixed), we need to create a set to contain nodes lying on the edge surface. First type “ plot n all ” and type “ qadd fix ” and like before press “ a ” and then “ r ” twice to select. Press “ n ” if the window is at right position. You can utilize the drop-down menu “orientation->+y direction” to help you locate. Type command “ plus n fix g ” to check whether you have chosen correctly.

When done, the cgx should display Fig. c-4.

If you mistakenly select undesired nodes you can use “ qrem ” command to remove them. The operating procedure of “ qrem ” is like “ qadd ”.

When the constraint job is over, type

send fix abq nam

to restore the selected node number into a file – “fix.nam” - for later use.

Then we can do the same procedure by adding a concentrated force in onto the center of the plate, but no need to restore them (or actually “it” since it’s a single point) into a file. We can read from the command window that it’s node #118. We can just remember this number and announce it in the ccx input file later.

Some steps before we move on to the ccx module:

 * Type “ send all abq ” creating “all.msh” file for ccx module.
 * Type “ send all fbd ” creating a .fbd file restoring geometric entities, also serve as a backup of previous work.
 * Type “ send all frd ” creating a .frd file containing node and element information.

After the all.msh file is generated, open it with WordPad to check if the element type is correct. This is because the cgx and ccx use different acronym for the same element type. And also some element type is compatiable with cgx but not accepted by ccx. This is particularly true in 2D cases. Make sure the ccx can recognize the element type appeared in all.msh. You may check it here.

The final look of the disk (Fig. c-4.5)

Running ccx
The ccx module requires an input file to run the analysis. This kind of input file is .inp format and is also called “input deck”. The input deck is generally consists of two parts: The model definition and some steps. See Fig. c-5.

And the model definition and step definition are all composed of some “keyword” parts, these the keyword with associated data is called a “card”. To help you understand these concepts see Fig. c-6.

The input deck is often written using Sci-TE text editor on Windows OS. Open an Sci-TE and paste such code onto it. And then save it with .inp suffix. This file is identical to what displayed in Fig. c-6

*Heading MODEL: DISC *INCLUDE, INPUT=all.msh *INCLUDE, INPUT=fix.nam *NSET, NSET=LOAD 118 *MATERIAL, Name=steel *ELASTIC 28000000, 0.3 *SOLID SECTION, Elset=Eall, Material=steel *STEP *STATIC *BOUNDARY Nfix,1,2,3 *CLOAD LOAD, 2, 5 *NODE FILE U *EL FILE S, E *END STEP

The heading has no effect on analysis. Let’s go to the model definition part first.

In the model definition we have,

*INCLUDE, INPUT=all.msh *INCLUDE, INPUT=fix.nam

This tells the ccx module the mesh information. The keyword “ INCLUDE ” tells ccx to read additional file which makes this file concise. You can also paste the content of additional file directly here.

*NSET, NSET=LOAD 118

This is the node where the concentrated force acts on and we name it LOAD for later use.

*MATERIAL, Name=steel *ELASTIC 28000000, 0.3 *SOLID SECTION, Elset=Eall, Material=steel

This is the Material description. “28000000” and “0.3” means Young’s Modulus and Poisson’s Ratio respectively. Since this is only a simple tutorial, we only need these parameter by assuming the material linearly elastic.

Then ahead is the step definition. Since we have the static case, only one step is needed for analysis.

*STEP *STATIC

This declare the end of model definition and the begin of step definition, it also tells the ccx module the analysis type is “static”.

*BOUNDARY Nfix,1,2,3 *CLOAD LOAD, 2, 5

This declare the boundary conditions: the edge is fixed and there’s concentrated force. Remember the set of nodes we defined before? Its name’s “fix” just as above, the following digits “123” is the direction of constraint. Keyword “ CLOAD ” stands for concentrated force(load), and “ 2 ” stands for the “y” direction while “ 5 ” stands for the magnitude.

*NODE FILE U *EL FILE S, E *END STEP

This asks the ccx for outputting data into a .frd file. U, S, E stands for data type as displacement, stress and strain respectively.

Then click on the Sci-TE toolbar -> tool -> solve or type in the command window:

ccx disc

Wait until the “Job finished!” appear.

Post Processing
Click on the Sci-TE toolbar -> post process or type in the command window “ cgx –v disc.frd ” to view the result.

The von Mises stress distribution (Fig. c-7),

The XX component of stress distribution (Fig. c-8),

The YY component of strain distribution (Fig. c-9),

As we can we the results generally reflect the fact that the load is concentrated. But the shapes of distribution look weird. This maybe caused my the unsufficiency of nodes and elements - in this 3D analysis we only allocated around 600 nodes and 100 elements. This may also be caused by my inappropriate choice of element type.

After Tutorial
For more information on the ccx module, please refer to the ccx online documentary.

And more examples available.

d) Plot figures showing comparison between exact and numerical solution, convergence vs number of element
From HW5.1/HW5.3 we have known the exact solution is,

The comparison of $$\displaystyle {{u}^{h}}$$ vs $$\displaystyle u$$:

$$\displaystyle \left[ u_{nel}^{h}(0.5)-u(0.5) \right]$$ vs number of element:

We can see that the convergence becomes better when number of elements increase.

=Problem 5.8=

Alternative Matlab code from Fangfang
Fangfang has developed an alternative way of coding and it's with nice outcome too, see part d).

Plots by Fangfang's code
Here's 3 plots about $$\displaystyle {{u}^{h}}$$ vs $$\displaystyle u$$ when $$\displaystyle nel=4,6,8$$ generated by Fangfang's matlab code. We can it's also with nice accuracy.

title: User:Egm5526.s11.team4.zou/HW5


title: File:55-a-1.png


title: File:55-a-2.png


title: File:55-a-3.png


title: File:55-a-4.png


title: File:55-a-5.png


title: File:55-b-1.png


title: File:55-b-2.png


title: File:55-b-3.5.png


title: File:55-b-3.png


title: File:55-b-4.png


title: File:55-b-5.png


title: File:55-c-1.png


title: File:55-c-2.png


title: File:55-c-3.png


title: File:55-c-4.png


title: File:55-c-5.png


title: File:55-c-6.png


title: File:55-c-7.png


title: File:55-c-8.png


title: File:55-c-9.png


text:
User:Eml5526.s11.team6.joglekar/hwk5

Introducing the arbitrary weighted function
Weak form can be found as follows, with weighted function

Now, multiply eq. 5.1.1 by $$ \displaystyle w $$ and integrate between 0 to 1.

Use integration by parts for the first term in eq. 5.1.4

Using eq. (5.1.2) and eq. (5.1.3), we get

Put eq. (5.1.6) in eq. (5.1.5), we get the weak form as follows

a.Using the exponential basis function

For getting appropriate constraint breaking solution, we choose Also, Take $$ \displaystyle \beta=0$$ for the constraint breaking solution. Therefore, From eq. 5.1.2 and eq. 5.1.10, we can say that Therefore, eq. 5.1.10 becomes Similarly, the weighted function can be written as Eq. 5.1.3 gives, Therefore eq. 5.1.13 becomes Hence, from eq. 5.2.12 and 5.2.15 Put this in eq. 5.1.7, we get Using eq. 5.1.15 we can find $$ \displaystyle w(1) $$ and put in eq. 5.1.18, we get By integrating and verifying with| Wolfram Alpha, we get

Following Matlab code is used to find coefficients a_j by changing values of i and j in the above equation After getting the coefficients of numerical solution, we compare numerical solution with analytical solution for different values of n.

Then graph of error that is difference between analytical solution and numerical solution is plotted at x=0.5

As expected, with increase in number of n,numerical solution approaches analytical solution.

From the above graph, error is almost zero for n=6 and above. But to get the error in the order of less than $$10^{-6}$$,we solve it further for higher values of n.

Introducing the arbitrary weighted function
Weak form can be found as follows, with weighted function

Now, multiply eq. 5.1.1 by $$ \displaystyle w $$ and integrate between 0 to 1.

Use integration by parts for the first term in eq. 5.1.4

Using eq. (5.1.2) and eq. (5.1.3), we get

Put eq. (5.1.6) in eq. (5.1.5), we get the weak form as follows

a.Using the exponential basis function

For getting appropriate constraint breaking solution, we choose Also, Take $$ \displaystyle \beta=1$$ for the constraint breaking solution. Therefore, From eq. 5.1.2 and eq. 5.1.10, we can say that Therefore, eq. 5.1.10 becomes Similarly, the weighted function can be written as Eq. 5.1.3 gives, Therefore eq. 5.1.13 becomes Hence, from eq. 5.2.12 and 5.2.15 Put this in eq. 5.1.7, we get Using eq. 5.1.15 we can find $$ \displaystyle w(1) $$ and put in eq. 5.1.18, we get By integrating and verifying with| Wolfram Alpha1 ,| Wolfram Alpha2,| Wolfram Alpha3 we get

Following Matlab code is used to find coefficients a_j by changing values of i and j in the above equation After getting the coefficients of numerical solution, we compare numerical solution with analytical solution for different values of n.

Then graph of error that is difference between analytical solution and numerical solution is plotted at x=0.5

As expected, with increase in number of n,numerical solution approaches analytical solution.

From the above graph, error is almost zero for n=6 and above. But to get the error in the order of less than $$10^{-6}$$,we solve it further for higher values of n.

title: User:Eml5526.s11.team6.joglekar/hwk5


text: File:Social media survey - Laptops.png

title: File:Social media survey - Laptops.png


title: File:55-c-4.5.png


text: File:Social media survey - Proficiency.png


text: File:Social media survey - Usage.png

title: File:Social media survey - Usage.png


text: File:Social media survey - value on unit.png


title: File:Egm6341.team4.BM.pic1.2.jpg


title: File:Egm6341.team4.BM.pic1.3.jpg


title: File:Egm6341.team4.BM.pic1.4.jpg


title: File:Egm6341.team4.BM.pic1.closeup.jpg


title: File:Egm6341.s11.team3.HW5.3.png


title: File:Egm6341.team4.BM.pic2.2.jpg


title: File:Egm6341.team4.BM.pic2.4.jpg


title: File:Egm6341.team4.BM.pic2.7.jpg


title: File:Egm6341.team4.BM.pic3.1.jpg


title: File:Egm6341.team4.BM.pic3.3.jpg


title: File:Egm6341.team4.BM.pic3.7.jpg


title: File:Egm6341.team4.BM.pic4.1.jpg


title: File:Egm6341.team4.BM.pic4.3.jpg


title: File:Egm6341.team4.BM.pic4.7.jpg


title: File:Egm6341.team4.BM.pic5.3.jpg


title: File:Egm6341.team4.BM.pic5.7.jpg


text:
Talk:Introduction to Elasticity/Plate with hole in tension

Shouldn't the symbol S in the first term of the RHS of equation 129 be replaced by the symbol T?? --80.254.146.84 07:58, 22 March 2011 (UTC)


text: File:GanttChart + WBS1.pdf

title: File:GanttChart + WBS1.pdf


text: File:Plan van aanpak.pdf

title: File:Plan van aanpak.pdf


text:
Microfluid Mechanics/Scaling Effects and Governing Equations in Microflows

>back to Chapters of Microfluid Mechanics

Scaling effects in microflows
Microdevices tend to behave differently from the objects we are used to handling in our daily life  because of a simple reason: The influences of forces, which are functions of wetted area, decreases slower than those, which are functions of fluid volume.

Let $$ \displaystyle p_1(A) $$ and $$ \displaystyle p_2(V)$$ are any two property which scales with wetted area $$ \displaystyle (A)$$ and fluid volume $$ \displaystyle (V) $$. Their change relative to each other with decreasing lengh scale of the system $$ \displaystyle(L)$$ can be written as:

$$ \displaystyle \frac{p_1(A)}{p_2(V)}\propto\frac{L^2}{L^3}\propto\frac{1}{L} $$

Typical order of magnitude is $$ \displaystyle 10^6\,\text{m}^2/\text{m}^3 $$ in microdevices. In other words, surface forces clearly dominates the body forces.

Some of the microflow effects can be best understood when the trends in physical quantities and dimensionless numbers are monitored as a function of decreasing length scales in microflows. These trends are called scaling laws. A scaling law signifies: the law of the variation of physical quantities with the size $$\ L $$ of the system or the object in question. $$\ L $$ is the characteristic dimension in a device, which takes the micron size when the device is miniaturized. For example, a microcanal is anisotropic as its length is generally larger that its width and height. In this case, the quantity $$\ L $$ must be understood to be a scale controling all dimensions of the system; when $$\ L $$ decreases (width, height, etc.), all the dimensions of the system (length, width or height) also decrease while maintaining constant aspect ratios.

Hence, the basic and most relevant physical quantities in microflows scales with $$\ L $$ as follows:

In order to find out the influence of the miniaturization on the dimensionless numbers, the velocity should be treated depending on the constraints. It is easier to assume that in the micro devices we wish to keep the velocity $$ U$$ constant. However, most applications are limited to performance of the pump. Hence, we can assume that we have only the $$ \Delta P $$ supplied by the pump or the micro-devices can sustain a certain internal fluid pressure. If we wish to keep $$ \Delta P $$ constant, already the pressure drop in a laminar pipe flow suggests that the mean velocity scales with $$ L $$

$$ \displaystyle \Delta P=U\frac{32 \mu L_{pipe}}{D_{pipe}^2} $$

therfore the scaling of $$ U $$ reads

$$ \displaystyle U\propto\frac{\Delta P}{32 \mu}L\propto L $$

Hence, the scalings of the non-dimensional numbers with $$\ L $$ for constant $$ U $$ and $$ \Delta P $$ are as follows:

Let us see the other consequences of having dimensions less than a milimeter for constant $$ \Delta P $$:
 *  Reynolds number becomes too small: The inertia does not play a role, i.e. momentum diffuses only through viscosity. Appearance of turbulence is not expected, but at very high $$ \Delta P $$ sufficiently high velocities $$ U $$ can be achieved for the generation turbulent flow.
 *  Knudsen number gets larger: The fluid molecules interact more and more with the walls of the device instead of interacting with the other molecules.
 *  Weber number becomes too small: The capillary forces dominates the inertia of the fluid. In other words, the intermolecular interactions takes the control of the bulk flow.
 *  Froude number get higher: The scaling suggests that influence of gravitation will increase on the flow, however both the inertial forces and gravitation forces ceases down to very low values that no change in the character of the flow is expected.
 *  Capillary number becomes high: The capillary forces dominates over the viscous forces, as they dominate over the inertial forces. Thus, in multiphase flows capillary forces have the most influence on the momentum of the bulk together with the supplied $$ \Delta P $$.
 *  Euler number becomes high: The inertia of the flow remains very low compared to high pressure gradients necesssary to drive the flow in microfluidic devices.
 *  Bond number takes very low values: As in the example of water strider, the gravitational effects is not anymore important in micron scales.
 *  Pecklet number might become low depending on the diffusion constant: The transport of matter is mainly achived by the molecular diffusion processes. Therefore, long mixing times has to expected.

Use scaling laws with care
In the scaling laws given above, positive exponent of $$\ L $$ reveals a reduction of the quantity or the dimensionless number of interest in micro scales, which can be on the order of ten or a hundred micrometers. Note that depending on the flow problem the scaling of velocity has to be properly selected.

We have reasoned here using exponents without discussing in detail the physical iaws in play. This approach has the advantage of being simple and direct, but at times it leads to conclusions that cannot be applied to microsystems, since the range of scales is shifted with respect to their size.

This issue can be highlighted by the problem of mixing in microsystems, and more specifically, the question of knowing whether the agitation of fluids can result in the reduction of mixing time in a microfluidic system.

According to scaling laws given in the table for the physical quantities, a typical hydrodynamic transport time is:
 * $$\ \tau_a \sim \frac{L}{U} \sim L^0 $$

and the molecular diffusion time is expressed as:
 * $$\ \tau_d \sim \frac{L^2}{D} \sim L^2 $$

where $$\ D [\text{m}^2\text{s}^{-1}]$$ is the diffusion coefficient. Hence, we would expect that in microsystems, diffusion phenomena are much faster than hydrodynamic transport phenomena. Given that the effect of molecular diffusion is to mix, we could conclude that in microsystems it is useless to agitate the fluid in order to accelerate the mixing process. However, in practice, such a conclusion would be extreme once actual numerical values are considered. The ratio of these time scales is the dimensionless Péclet number:

$$\ Pe=\frac{\text{Advection velocity}}{\text{Diffusion velocity}}=\frac{\text{Diffusion time}}{\text{Advection time}}=\frac{L^2/D}{L/U}=\frac{U L}{D} $$

To illustrate its significance, we consider the case of fluorescein mixed with water in a device with a characteristic length of 100 µm, flowing at a rate of 30 µm/s. Fluorescein has a diffusion coefficient of 640e-12 [m2s-1] in water. Hence, the Péclet number is $$\ Pe \approx 5 $$. This elevated value is contrary to the previvous analysis based on the length scale exponents. In other words, diffusion phenomena are acting much more slowly than hydrodynamic transport phenomena. We can see that an order of magnitude reduction in size would weaken advective mixing drastically, as $$\ Pe < 1 $$.

Governing equations in microflows
The governing conservation equations can be derived by simplifying the most general form of conservation equations with the help of non-dimensional numbers. The treatments given in this section are limited to an adiabatic flow and Newtonian fluid.

The dimensionless conservation of mass and of momentum equations are (see Fluid_Mechanics_for_MAP_Chapter_7._Dimensional_Analysis):

$$St\frac{\partial \rho^{*}}{\partial t^{*}} + \frac{\partial \rho^{*}U_{i}^{*}}{\partial x_{i}^{*}} = 0$$

$$\rho^{*}\left[St\frac{\partial U_{j}^{*}}{\partial t^{*}} + U_{i}^{*}\frac{\partial U_{j}^{*}}{\partial x^{*}_{i}}\right] = -Eu\frac{\partial P^{*}}{\partial x_{j}^{*}} + \frac{1}{Re}\frac{\partial}{\partial x_{i}^{*}}\left[\mu^{*}\left(\frac{\partial U_{j}^{*}}{\partial x_{i}^{*}} + \frac{\partial U_{i}^{*}}{\partial x_{j}^{*}}\right) - \frac{2}{3}\delta_{ij}\mu^{*} \frac{\partial U_{k}^{*}}{\partial x_{k}^{*}}\right] + \frac{1}{Fr}\rho^{*}g^{*}_{j}$$

Liquid flows
In liquid flows, the fluid can be accepted to be incompressible, which means the reduction of the mass and momentum equations into the following forms:

$$ \frac{\partial U_{i}^{*}}{\partial x_{i}^{*}} = 0 $$

$$\rho^{*}\left[St\frac{\partial U_{j}^{*}}{\partial t^{*}} + U_{i}^{*}\frac{\partial U_{j}^{*}}{\partial x^{*}_{i}}\right] = -Eu\frac{\partial P^{*}}{\partial x_{j}^{*}} + \frac{1}{Re}\frac{\partial}{\partial x_{i}^{*}}\left[\mu^{*}\left(\frac{\partial U_{j}^{*}}{\partial x_{i}^{*}} + \frac{\partial U_{i}^{*}}{\partial x_{j}^{*}}\right)\right] + \frac{1}{Fr}\rho^{*}g^{*}_{j}$$

When we select: $$\Delta P_{c} = \rho_{c}U_{c}^{2} \rightarrow Eu = 1$$

$$\rho^{*}\left[St\frac{\partial U_{j}^{*}}{\partial t^{*}} + U_{i}^{*}\frac{\partial U_{j}^{*}}{\partial x^{*}_{i}}\right] = -\frac{\partial P^{*}}{\partial x_{j}^{*}} + \frac{1}{Re}\frac{\partial}{\partial x_{i}^{*}}\left[\mu^{*}\left(\frac{\partial U_{j}^{*}}{\partial x_{i}^{*}} + \frac{\partial U_{i}^{*}}{\partial x_{j}^{*}}\right) \right] + \frac{1}{Fr}\rho^{*}g^{*}_{j}$$

In one phase flows, this kind of selection of the characteristic pressure is common. Thus, one can see that the pressure gradient can not be neglected.

The Froude number is in general important when gravity starts to play an important role, i.e. when buoyancy effects occurs due to density differences and, especially for multi-phase flows. In liquid microflows, the density difference is in general negligible, Fr becomes very large and the momentum equation reduces to:

$$\rho^{*}\left[St\frac{\partial U_{j}^{*}}{\partial t^{*}} + U_{i}^{*}\frac{\partial U_{j}^{*}}{\partial x^{*}_{i}}\right] = -\frac{\partial P^{*}}{\partial x_{j}^{*}} + \frac{1}{Re}\frac{\partial}{\partial x_{i}^{*}}\left[\mu^{*}\left(\frac{\partial U_{j}^{*}}{\partial x_{i}^{*}} + \frac{\partial U_{i}^{*}}{\partial x_{j}^{*}}\right) \right] $$

Flows can be steady or unsteady. Natural unsteadiness might occur due to, for example flow separation or intermittent laminar to turbulent transition. Flows can be also forced to be unsteady by means of unsteady forcing or flow rates, for example in the case of peristaltic pumps or ultrasound waves generated by means of piezoelectric actuators at the walls of flow chamber. In those cases, the characteristic time scale of the unsteadiness $$\ t_c $$ should be compared to that of the intrinsic time scale of the flow $$\ L_c/U_c $$. For very low $$\ St $$, i.e. when $$\ t_c > L_c/U_c $$, in other words when local acceleration much smaller than spatial acceleration, the flow can be accepted to be steady. Thus the first term at LHS can be neglected, i.e. the momentum equation reads: $$\rho^{*}U_{i}^{*}\frac{\partial U_{j}^{*}}{\partial x^{*}_{i}} = -\frac{\partial P^{*}}{\partial x_{j}^{*}} + \frac{1}{Re}\frac{\partial}{\partial x_{i}^{*}}\left[\mu^{*}\left(\frac{\partial U_{j}^{*}}{\partial x_{i}^{*}} + \frac{\partial U_{i}^{*}}{\partial x_{j}^{*}}\right) \right] $$

This equation shows that the inertial term at the LHS looses its effect with decreasing Reynolds number, whereas the viscous term at the RHS becomes much more important at Reynolds numbers less than unity. Hence for $$\ Re<1 $$, governing equation reduces to Stokes flow equation:

$$0= -\frac{\partial P^{*}}{\partial x_{j}^{*}} + \frac{1}{Re}\frac{\partial}{\partial x_{i}^{*}}\left[\mu^{*}\left(\frac{\partial U_{j}^{*}}{\partial x_{i}^{*}} + \frac{\partial U_{i}^{*}}{\partial x_{j}^{*}}\right) \right] $$

This equation simply tells that the pressure gradient is needed to overcome viscous stresses. There can be flow cases in which other type of forcing might be utilized to drive the flow such as surface tension, electric field or centrifugal forces. Those will be elaborated in the relevant sections.

There are a number of Stokes flow problems which have analytical solution.

Gas flows
The gas flows in microdevices suffer from compressibility effects,namely compression and rarefaction. Above a certain level of rarefaction $$\ (Kn > 10) $$, the flow can not be considered as a continuum owing to the large fluctuations of the flow and fluid properties in the volumes of interest. These cases should be analyzed by atomistic models. However, in a many of the microdevices, $$\ Kn $$ is smaller than 10. Hence, continuum models can still be used $$\ (Kn< 0.01) $$, at most with a low order model representing the effect of partial rarefaction occurring next to the walls of flow chamber $$\ (0.01 \leq Kn < 10) $$. Hence for an adiabatic gas flow, it is legitimate to neglect buoyancy forces. If there is no external forcing on the flow and for very low $$\ St $$, the terms with time derivative can also be neglected, which finally results in the following conservation of mass and momentum equations:

$$\frac{\partial \rho^{*}U_{i}^{*}}{\partial x_{i}^{*}} = 0$$

$$\rho^{*}U_{i}^{*}\frac{\partial U_{j}^{*}}{\partial x^{*}_{i}} = -\frac{\partial P^{*}}{\partial x_{j}^{*}} + \frac{1}{Re}\frac{\partial}{\partial x_{i}^{*}}\left[\mu^{*}\left(\frac{\partial U_{j}^{*}}{\partial x_{i}^{*}} + \frac{\partial U_{i}^{*}}{\partial x_{j}^{*}}\right) - \frac{2}{3}\delta_{ij}\mu^{*} \frac{\partial U_{k}^{*}}{\partial x_{k}^{*}}\right] $$

Multi-phase flows
Multi-phase flows comprise flow of matter with different phases and/or immiscible or slowly mixing fluids with different material properties. Examples are bubbly flows, particle laden flows or flows in which locally phase change happens such as boiling or cavitation. In those flows, there is often an interface between the different phase of the fluid or between different fluids. Hence, the surface tension effects, the effects owing to the differences in density (buoyancy) and viscosity has to be considered. When gas and liquid phases exist simultaneously in a flow, incompressibility of the flow can be accepted only under very limited conditions. Moreover, flow of each phase has to be considered separately in the flow.


text: File:FEA HW 5 1 P.jpg

title: File:FEA HW 5 1 P.jpg


text: File:FEA HW 5 1 F.jpg

title: File:FEA HW 5 1 F.jpg


text: File:FEA HW 5 1 E.jpg

title: File:FEA HW 5 1 E.jpg


text: File:FEA HW 5 2 P.jpg

title: File:FEA HW 5 2 P.jpg


text: File:FEA HW 5 2 F.jpg

title: File:FEA HW 5 2 F.jpg


text: File:FEA HW 5 2 E.jpg

title: File:FEA HW 5 2 E.jpg


title: File:Eml5526.s11.team6.hw5.3.1.jpg


title: File:Eml5526.s11.team6.hw5.3.2.jpg


text: File:Solid edge.pdf

title: File:Solid edge.pdf


title: File:5.1.exp.png


text: File:5.1 error.png

title: File:5.1 error.png


title: File:Fe1.s11.team1.HW5.fig1.5.pdf


title: File:Fe1.s11.team1.HW5.fig1.6.pdf


text:
User:Egm6341.s11.team4.KTH/HW5

=Problem 5.3 Cauchy Heavy Tails=

Given
For 1) and 2), Quartile points: $$ Q_1 \ $$, $$ Q_2 \ $$ and $$ Q_3 \ $$.

Cauchy pdf :

Gauss(normal) pdf :

For 3) and 4) $$ x_0 = \mu = 0 \ $$ and $$ \gamma^{C}=1 \ $$ $$ \gamma^{C}= \ $$ half width of $$ C(x_0, \gamma^{C}) \ $$ $$ \gamma^{G}= \ $$ half width of $$ N(\mu, \gamma^{C}) \ $$

Objectives
1) Find $$ \left \{ Q_1, Q_2 \right \} \ $$ for $$ C \left ( x_0, \gamma \right ) \ $$ 2) Find $$ \left \{ Q_1, Q_2 \right \} \ $$ for $$ N \left ( \mu, \sigma \right ) \ $$ 3) Find $$ \sigma\ ^1$$ such that $$ \gamma\ ^G =1 $$, and Plot $$ C \left ( 0,1 \right ) $$ and $$ N \left ( 0, \sigma\ ^1 \right ) $$ 4) Find and plot $$ \left \{ Q_1^c, Q_3^c \right \}  $$ for $$ C \left ( x_0, \gamma\ \right ) $$ and $$ C \left ( 0, 1 \right ) $$ also find and plot $$ \left \{ Q_1^G,Q_3^G  \right \}  $$ for $$ N \left ( \mu\, \sigma\ \right ) $$ and $$ N \left ( 0, \sigma\ ^1 \right ) $$

1) The cumulative distribution function of Cauchy pdf is
Substitue Q1 into x of eq(3.1) we get eq(3.2).

From eq(3.2)

Substitue Q3 into x of eq(3.1) we get eq(3.4).

From eq(3.5)

2) The cumulative distribution function of Gauss pdf is
Substitue Q1 into x of eq(3.8) we get eq(3.8).

Substitue Q3 into x of eq(3.8) we get eq(3.10).

3) Find standard deviation for half width of Gaussian distribution
Gaussian pdf is

The height of Gaussian pdf is

We want to find sigma which will place the half width at 1.

MatLab Code

Plot

===4) Find $$ \left \{ Q_{1}^{C}, Q_{3}^{C} \right \} $$ for $$ C(x_0,\gamma) \ $$ and $$ C(0,1) \ $$ and find $$ \left \{ Q_{1}^{G}, Q_{3}^{G} \right \} $$ for $$ N(\mu,\sigma) \ $$ and $$ N(0,\sigma^1) \ $$.===

From 1)

From 1)

Plot $$ \left \{ Q_{1}^{C}, Q_{3}^{C} \right \} $$ for $$ C(0,1) \ $$ and $$ \left \{ Q_{1}^{G}, Q_{3}^{G} \right \} $$ for $$ N(0,\sigma^1) \ $$

MatLab Code

Plot

title: User:Egm6341.s11.team4.KTH/HW5


title: File:Egm6341.team4.BM.pic6.1.jpg


title: File:Egm6341.team4.BM.pic6.2.jpg


text:
Engineering Experience 4: Design a Small Solar Vehicle/Nl/Team PM7/Design rapport

Materiaalkeuze
Voor de keuze van het materiaal van het frame wordt vooral gekeken naar de specifieke sterkte omdat dit de verhouding is tussen de sterkte en het gewicht. Deze waarde wordt ook in de ruimtevaart vooral gebruikt omdat ze hier zo licht mogelijk materiaal willen gebruiken. Dit is voor de wagen ook goed om sneller te kunnen gaan. De specifieke treksterkte is niet voor alle materialen terug te vinden maar deze worden dan berekend door de treksterkte te delen door de dichtheid. Hier volgen enkele waarden.

Dit waren enkele van de gevonden waarden waaruit blijkt dat balsa het beste materiaal is om te gebruiken. Dit is ook de reden waarom dit het hout is waaruit modelvliegtuigen en dergelijke gemaakt worden.

Het algemeen ontwerp

Het geheel ontwerp bestaat uit het skelet van de wagen, de motor en de overbrenging, de geleidingswieltjes, het zonnepaneel en de wielen.

Het frame
Voor het frame hebben we verschillende ontwerpen bekeken in het programma Salomé en Code Aster (zie ). Deze programma’s kunnen de spanningsverdeling weergeven over het frame onder een opgegeven belasting.

Als eerste ontwerp werd er gedacht aan een Y-structuur zoals op onderstaande figuur. De verticale balk hier op is de balk waar het zonnepaneel op bevestigd zou worden. Bij de simulatie van vervorming bleek dat dit ontwerp het meeste vervormde aan de achteras. Aangezien de vervorming daar zo klein mogelijk moet zijn om wrijving te voorkomen werd er verder gezocht naar andere ontwerpen.

De volgende figuur toont het tweede ontwerp. Zoals hier op te zien is, is de binnenste driehoek overbodig omdat hier niet veel ballast op komt te staan. Daarom werd in het laatste ontwerp hetzelfde model gebruikt maar dan zonder de binnenste driehoek. Er werd beslist dat het frame uiteindelijk zo zal uitgevoerd worden.

De simulaties zijn gedaan met een belasting op het hoogste punt (waar het zonnepaneel op rust). Er werd een belasting van 20N aangebracht, wat ongeveer 4 keer het gewicht van het zonnepaneel is.

Geleidingswieltjes
Het eerste idee van de zelfsturende voorwielen werd al snel verworpen omdat het weinig nut zou hebben. Tijdens de race beweegt de zonnewagen immers niet ver van zijn ideale lijn af.

Vervolgens werden kogellagers als voorwielen voorgesteld. Deze zouden zowel aan de voorkant van de wagen als aan de achterkant geholpen worden door kleine kogellagers die de wagen op de rail moeten houden. Door de kostprijs van de tandwielen, werd er echter gekozen om dit systeem te vervangen door een systeem waar de wagen met de voorkant rust op rail. Hierdoor werden 2 grote en kleine kogellager uitgespaard.

De achter-wielen
De achterwielen zijn bevestigd op een gelagerde as. Voor de achterwielen zijn er verschillende mogelijkheden. Er werd gedacht aan wieltjes van knex, wieltjes maken van CD’s of fenolische wielen (deze moeten dan via het internet besteld worden). De voorkeur ging uit naar het laatste omdat deze vrij weinig wrijving geven.

Uiteindelijk is beslist om de wielen uit plexiglas te snijden. Aangezien deze vrij weinig gewicht moesten dragen, konden ze licht gemaakt worden. Een gekocht wiel zou nog geboord moeten worden om de as in te laten passen. Aangezien het zeer moeilijk zou zijn om dit centrisch te doen, en er anders vrij veel trillingen zouden ontstaan is er voor gekozen dit niet te doen.

Verschillende overbrengingsmethodes
Via de zonnepanelen wordt een DC-motor aangestuurd die op zijn de beurt de wielen aandrijft. Er werden 3 mogelijkheden bedacht voor het overbrengingsmechanisme tussen de DC-motor en de wielen.

Conische as
Als eerste werd er gedacht aan een overbrenging waarbij een touw voor de start opgerold wordt op een conusvormige as die met de wielen verbonden is en bij aandrijving van de as afgerold wordt en op de aandrijvende as weer opgerold wordt. Een voordeel hiervan is dat door op de conusvormige as te wikkelen van de grote diameter naar de kleine men een versnelling zal krijgen. Het nadeel hierbij is dat de wagen niet oneindig zal blijven door gaan. Hij zal stoppen wanneer de draad volledig afgerold is. Wanneer de wagen van de baan gaat, kan het dus zijn dat deze niet meer zal rijden tegen dat hij terug op de baan gezet wordt.

Enkele overbrenging
Als tweede mogelijkheid is er ook een gewone tandwieloverbrenging.

Versnellingskast
Maar omdat het versnellen handig zou zijn werd dit systeem aangepast naar één waarbij men kan schakelen tussen twee snelheden. Hierbij is er wel meer verlies doordat er meerdere tandwielen gebruikt worden, maar kan men toch een tijdswinst maken.

Het principe van deze versnellingsbak, werkt op het veranderen van draairichting van de motor. Hierdoor zal een latje, met tandwielen op, verplaatst worden, zodat een andere overbrenging wordt aangedreven. De draairichting wordt bepaald door een H-brug die aangestuurd wordt door een chip. Deze chip bepaalt het ogenblik dat er geschakeld moet worden. Een drukknop zal de tijd op 0 zetten. Zodra de schakelaar los gelaten wordt, begint de tijd te lopen. De overbrenging heeft dus 2 verschillende paden. In Figuur 9 worden de paden weergegeven. In het groen is het vaste pad aangeduid van de motor tot aan waar het 2e pad gekozen wordt. De dikke as is de as waar de wielen op zitten. De motor is niet zichtbaar op deze figuur aangezien die achter het paneel zit. De voor en zijpanelen zijn weggelaten om de tandwielen zichtbaar te maken.

In Figuur 10 en Figuur 11 worden respectievelijk de aansturing van het blauwe en het rode pad weergegeven. De transparante groene tandwielen liggen in dit perspectief zowel voor de andere tandwielen op het schema, als de bewegende lat. Het kleine transparante tandwiel wordt rechtstreeks aangedreven door de motor. Het groene ondoorzichtige tandwiel ligt in het vlak van het rode en het blauwe tandwiel. Dit tandwiel zal het blauwe of het rode tandwiel aandrijven afhankelijk van de stand, en hangt op de zelfde as als het grote groene tandwiel. In Figuur 10 draait de motor in wijzerzin. Hierdoor kantelt de lat naar rechts, waardoor het groene ondoorzichtige tandwiel het blauwe tandwiel kan aandrijven. Hierdoor wordt het pad vervolledigd en wordt de as met de wielen aangedreven. In Figuur 11 draait de motor in tegenwijzerzin. Hierdoor kantelt de lat naar links en wordt het rode tandwiel aangedreven door het groene ondoorzichtige tandwiel.

Schakelen

Het schakelen zal gebeuren met behulp van een chip. Deze chip zal ofwel na een gegeven tijd schakelen, ofwel de spanning meten en dan bij het optimale vermogen schakelen.

Het vermogen kan gemeten worden door de spanning te meten die uit het zonnepaneel komt. Met een spanningsverdeler wordt voorkomen dat er een te hoge spanning op de chip komt, en dat er teveel stroom verloren gaat via de meting. Aangezien de stroom-spannings karacteristiek van de zonnecellen bekend is, kan het vermogen hier uit afgeleid worden.

Het omwisselen van de draairichting gebeurt via een H-brug die door de chip gestuurd wordt.

Benodigde tandwielen berekenen via genetisch algoritme
Voor een enkele overbrenging is het vinden van de benodigde tandwielen triviaal. Voor een overbrenging met een versnellingsmechanisme is dit echter niet meer zo vanzelfsprekend. Er zijn immers 2 keer exact 3 overbrengingen die gebruikt moeten worden. De eerste 2 overbrengingen zijn voor beide versnelling de zelfde. De laatste overbrenging bepaalt echter welke overbrengingsverhouding bekomen zal worden en is dus verschillend voor de 2 versnellingen. Aangezien de eerste 2 overbrengingen echter gemeenschappelijk zijn, is er dus weinig keuzevrijheid.

Indien er voor elk aantal tanden, tandwielen ter beschikking zouden zijn en de maximum grootte niet van belang zou zijn, zou de oplossing voor de hand liggen. Dan kan men immers de eerste 2 overbrengingen neutraal houden en de overbrengingsverhouding in de laatste (niet gemeenschappelijke) overbrenging toepassen. In de praktijk is dit echter niet realistisch. De grootte van het maximum tandwiel is beperkt door de grootte van het wiel. Gebaseerd op de catalogus van SDP/SI, bestaat er ook niet voor elk aantal tanden een tandwiel. Een mogelijkheid is het zelf maken van tandwielen met de laser-cutter van het FabLab en Solid Edge. Vermoedelijk zou dit echter geen optimale resultaten leveren. Daarom is er een genetisch algoritme geschreven in Maple. Dit algoritme probeert aan de hand van de bestaande tandwielen een combinatie van tandwielen te zoeken die voldoet aan de juiste verhouding. Een andere optie zou zijn om zoals bij de overbrengingssimulatie in Maple, brute-force alle mogelijkheden af te gaan. Aangezien dit voor 30 soorten tandwielen echter een oplossingsruimte van 656 100 000 000 mogelijkheden oplevert, bekomt het genetisch algoritme sneller een resultaat en zal het indien er geen exacte oplossing is, een goede benadering als resultaat geven. Ter vergelijking, met het genetisch algoritme is er al na 1000 stappen een oplossing die de ideale overbrenging voldoende benadert om geen merkbaar verschil op te leveren.

Bij de uiteindelijke uitwerking kwamen er echter nog 3 andere voorwaarden bij.

De moeilijkste om rekening mee te houden was de kostprijs. Er zijn immers 10 tandwielen nodig. De prijzen per tandwiel gaan van 7 dollar tot 36,5 dollar voor een enkel tandwiel. Als hier niet op gelet wordt, zal het budget van 200 euro dus gemakkelijk overschreden kunnen worden. Om aan deze voorwaarde te voldoen, werd de prijs mee in de fitheid functie gestoken van het algoritme. Hoe duurder, hoe minder fit.

Vervolgens moeten de laatste tandwielen in de reeks ook voldoende groot zijn. De as met de wielen op heeft een diameter van 10mm. De tandwielen moeten dus groter zijn dan 10mm en nog plaats over hebben om de tanden op een stevige manier te bevatten. Ook hier voor werd een voorwaarde in de fitheidsfunctie gestoken.

Tenslotte kon niet elke tandwielcombinatie in de praktijk in een tandwielkast gestoken worden. De assen en de tandwielen hadden soms de neiging door elkaar te lopen. Een regel vinden voor dit probleem was echter minder gemakkelijk. Na een aantal combinaties te proberen, werd er een regel gevonden die combinaties gaf waarmee meestal een praktisch realiseerbaar resultaat bekomen kan worden.

Voor de overbrenging van 15,6 en 6,8 worden de volgende tandwielen gebruikt (aantal tanden) :
 * Gemeenschappelijk : 12-25 en 12-36.
 * Voor de overbrenging van 15,6 : 32-80
 * Voor de overbrenging van 6,8 : 60-65

Voor meer detail over het algoritme en de broncode, kan je op de volgende pagina terecht.

Constructie
De constructie bleek niet gemakkelijk te zijn.

Eerst was er gekozen voor carbon assen voor de tandwielen op te monteren omdat deze vrij stijf zijn. Deze bleken echter teveel wrijving te hebben met het plexiglazen frame van de versnellingskast. De assen werden dan vervangen door messing assen. Deze hadden veel minder wrijving, maar nog steeds te veel. Vervolgens hebben we de contactpunten eerst met potlood gekleurd (grafiet) en tenslotte met machine olie. Ook dit bleek tijdens de test nog altijd vrij veel wrijving te hebben. Wat uiteindelijk in de race ook nog bevestigd is geweest. De 2e versnelling had namelijk teveel wrijving, waardoor de wagen stil viel. Er is dan ook besloten om enkel op de laagste versnelling te racen.

Met meer tijd hadden we de wrijving nog naar beneden kunnen halen. Door bijvoorbeeld de assen in aluminium te draaien, zouden de tandwielen beter op de assen gemonteerd kunnen worden. De binnendiameter van de tandwielen was namelijk niet bij elk tandwiel het zelfde. De assen zijn in dit geval met plakband dikker gemaakt tot de tandwielen er op pasten. Met gedraaide tandwielen zou dit perfect kunnen aangesloten worden.

Verder zou een beter inschatting van de wrijving ook 2 andere overbrengingscombinaties hebben opgeleverd.

Experimenten met de smering en eventueel kogellagers zouden de wrijvingscoëfficiënt tenslotte ook nog naar beneden kunnen halen.

Het zonnepaneel
De energie die men uit een zonnepaneel kan krijgen is afhankelijk van de lichtsterkte per oppervlakte-eenheid. Deze lichtsterkte is gericht. Zo komt het dat de energie zal verschillen afhankelijk van de richting die het zonnepaneel aanneemt. Dit kan men afleiden uit de volgende figuur. Over een doorsnede van 1 meter van de lichtstraal zal de lichtsterkte gelijk zijn maar onder een hoek van 90 graden zal er dus op een zelfde oppervlak twee keer meer licht kunnen invallen dan onder een hoek van 30 graden. Zo zal men onder een hoek van 90 graden dus maximale energie bekomen.

Het zonnepaneel wordt bovenop het skelet bevestigd met behulp van een kogelgewricht. Dit maakt het mogelijk om het zonnepaneel te richten om meer zonlicht op te vangen.

Constructie zonnewagen
Zoals eerder vernoemd, zat ver weg de meeste tijd in de constructie van de versnellingskast. Aangezien er 10 tandwielen in een vrij kleine plaats moesten gemonteerd worden, heeft dit aardig wat tijd in beslag genomen.

Het grootste probleem zat bij de montage van de as op de motor. Aangezien wij een lange as nodig hadden voor ons systeem konden we niet onmiddellijk een tandwiel op deze as plaatsen. Een extra overbrenging zou extra verliezen met zich mee brengen.

Er is achtereenvolgens getracht om de as te lijmen met een lijmpistool, te solderen, te lijmen met secondelijm en de as toe te knijpen. Dit laatste tenslotte leverde een bruikbaar, doch nog altijd niet betrouwbaar resultaat op.

Resultaat
In de poule waar we in zaten hebben we 2 keer gewonnen, waardoor we de poule dus gewonnen hebben. De eerst volgende race hebben we echter verloren.

De eerste race is er getracht om te vertrekken met het versnellingssysteem. Toen de wagen stil viel in 2e versnelling zijn we overgeschakeld naar het rechtstreeks aansturen van de motor met het zonnepaneel in 1e versnelling. De tegenstander in deze race had eveneens problemen, waardoor we de race toch nog gewonnen hebben.

De volgende races hebben we op 1e versnelling gereden.

Algemene bronnen
a technical guide to building fighting robots. (n.d.). Retrieved from http://robots.freehostia.com/Materials/MaterialsBody.html

matbase a leap forward in material data. (n.d.). Retrieved from http://www.matbase.com/material/wood/class4-5-10-years/balsa/properties

natuurinformatie/darwinjaar 2009. (n.d.). Retrieved from http://www.natuurinformatie.nl/nnm.dossiers/natuurdatabase.nl/i002046.html

northwest caster and equipment. (n.d.). Retrieved from http://www.nwcaster.com/dept.aspx?d_id=25123

toledo. (n.d.). Retrieved from https://cygnus.cc.kuleuven.be/webapps/portal/frameset.jsp?tab=null&url=/webapps/blackboard/execute/courseMain?course_id=_374226_1

working with balsawood. (n.d.). Retrieved from http://pldaniels.com/flying/balsa/balsa-properties.html

world lingo translation localization globalization. (n.d.). Retrieved from http://www.worldlingo.com/ma/enwiki/nl/Specific_strength

Eindige elementen analyse
De eindige elementen analyse is gedaan onder CAE Linux (Computer Aided Linux). Dit is een Ubuntu gebaseerde Linux distributie, waar stadaard allerhande ingenieurspakketen op geïnstalleerd staan. De gebruikte pakketen zijn hier specifiek Salomé en Code Aster. Salomé is een overkoepelend pakket, wat een workflow creëert voor de eindige elementen analyse. Code Aster is de toepassing die de eigenlijke analyse doet.

CAElinux : http://www.caelinux.com/CMS/

Salome : http://www.salome-platform.org/

Code Aster : http://www.code-aster.org/V2/spip.php?rubrique2


title: File:5.2.exp.png


text: File:5.2.exp error.png

title: File:5.2.exp error.png


text:
Preschool Language and Skills/5

 * Created: 2011 03 22 | Percent completed: [[Image:50%.png]]

Quick Assessment Results
The items below are not necessarily taught in order.

The fastest way to get a student to catch up with peers is table work. With appropriate pairing of the work environment, teachers, and shaping of skills, students can learn at a fast and fun pace without behavior problems.
 * 1) Student sits at table ready to work
 * 1) *A score of 1 means that there will need a lot of pairing to take place before attempting to do table work. Use the NET approach (Natural Environment Teaching) to do pairing and initial teaching.
 * 2) *A score of 2 means that the student has started to be able to work at the table but really there needs to be more pairing and probably more initial skill building in manding before attempting table work.
 * 3) *A score of 3 means that either there isn't enough pairing between demands, the student is very young, or the student has very low skills and will need some NET time in addition to table work.
 * 4) *A score of 4 means the student is on track.
 * 5) Play appropriately with toys
 * 6) *Playing appropriately with toys is an important skill for imagination, safety, and social skills.
 * 7) *A score of 1 means the student will need to have toys paired with other favorite toys or activities.
 * 8) *A score of 2 means the student has started to play with toys appropriately but will need to slowly develop more skills with other toys.
 * 9) *A score of 3 means the student is well on his or her way to learning to play with toys appropriately.
 * 10) *A score of 4 means the student is competent at playing with toys appropriately.
 * 11) Motor Imitation
 * 12) *It would seem that imitating other people/students is a strange skill to teach. Much research indicates that imitation leads to a great deal of learning.  But what happens if the student is not aware of others thereby is unable to imitate?  You teach the skill slowly, discretely, with patience and time.
 * 13) *A score of 1 means the student likely does not find the teacher motivating. A whole lot of pairing will need to take place.
 * 14) *A score of 2 means the student is starting to learn motor imitation but has a long way to go. More pairing.
 * 15) *A score of 3 means the student is well on his or her way to learning imitation skills.
 * 16) *A score of 4 means the student is competent at imitation skills and the teacher is likely well paired.
 * 17) Matching
 * 18) *A score of 1 means the student will need to have matching objects paired along with the teacher. Keep session on this skill short and shape the matching skill with prompting.  Don't forget to fade the prompts and reinforce attempts/correct matching without prompts differentially (more reward for independently matching without prompt)
 * 19) *A score of 2 means student is starting to learn matching skills but has a long way to go. More pairing of the objects to match and teacher.
 * 20) *A score of 3 means the student is well on his or her way to learning matching of objects, pictures, shapes or colors.
 * 21) *A score of 4 means the student is competent at matching skills and the teacher is likely well paired.
 * 22) Vocal
 * 23) *A score of 1 means the student is non-vocal and will need intensive speech intervention.
 * 24) *A score of 2 means the student has some sound capability but will likely need intensive speech intervention.
 * 25) *A score of 3 means the student is on his or her way to learning vocal sounds.
 * 26) *A score of 4 means the student is vocal and can likely imitate any sound.
 * 27) Tact (Identify/label)
 * 28) *A score of 1 means
 * 29) *A score of 2 means
 * 30) *A score of 3 means
 * 31) *A score of 4 means
 * 32) Follow Directions (get shoes, point to item, touch item)
 * 33) *A score of 1 means
 * 34) *A score of 2 means
 * 35) *A score of 3 means
 * 36) *A score of 4 means
 * 37) Mand (Request)
 * 38) *A score of 1 means
 * 39) *A score of 2 means
 * 40) *A score of 3 means
 * 41) *A score of 4 means
 * 42) Receptive Feature, Function, Category (with item present)
 * 43) *A score of 1 means
 * 44) *A score of 2 means
 * 45) *A score of 3 means
 * 46) *A score of 4 means
 * 47) Intraverbals (Fill in the blank)
 * 48) *A score of 1 means
 * 49) *A score of 2 means
 * 50) *A score of 3 means
 * 51) *A score of 4 means

title: Preschool Language and Skills/5


title: File:Fe1.s11.team1.HW5.fig6.1.pdf


title: File:Fe1.s11.team1.HW5.fig6.2.pdf


title: File:Fe1.s11.team1.HW5.fig6.3.pdf


title: File:Fe1.s11.team1.HW5.fig6.4.pdf


text:
User:Egm6341.s11.team4.yang/hw5

=HW 5.4 problem: Drive equation of motion and plot system=

Refer to lecture slide [[media:nm1.s11.mtg29.djvu|mtg-29]] for the problem statement.

Objective
1. drive eq of motion in terms of d, c, k, m, u 2. let $$\underline{x}=\{\begin{matrix} d \\ {\dot{d}} \\ \end{matrix}\}$$. Find ( $$\underline{F}$$, $$\underline{G}$$ ). 3. find C in terms of k, m, such that systen is critically damped. 4. let k=1, m=1/2, x0=$${{[0.8,-0.4]}^{T}}$$ a. For u=0, plot $$\underline$$ for $$c=0.5*{{C}_{cr}},{{C}_{cr}},1.5*{{C}_{cr}}$$ b. For u=0.5 gaussian noise and c=$$1.5*{{C}_{cr}}$$, plot $$\underline$$ c). For u=0.5 Cauchy noise and c=$$1.5*{{C}_{cr}}$$, plot $$\underline$$

solution
1. According the balance of force on the mass of M, we obtain that:
 * {| style="width:100%" border="0"

$$\begin{align} & {{F}_{c}}=c\centerdot \dot{d} \\ & {{F}_{k}}=k\centerdot d \\ \end{align}$$ $$u-{{F}_{c}}-{{F}_{k}}=m\centerdot a=m\centerdot \ddot{d}$$ So, the equation of motion can be represented as: $$u-c\centerdot \dot{d}-k\centerdot d=m\centerdot \ddot{d}$$ 2. From above, we can drive the following equation:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

 * style="width:95%" |
 * style="width:95%" |

$$\ddot{d}=-\frac{k}{m}\centerdot d-\frac{c}{m}\centerdot \dot{d}+\frac{u}{m}$$

And it is obvious that :
 *
 * }
 * {| style="width:100%" border="0"

 * style="width:95%" |
 * style="width:95%" |

$$\dot{d}=0\centerdot d+1\centerdot \dot{d}+0$$

Then, we can form the matrix equation using the above two equations:
 *
 * }
 * {| style="width:100%" border="0"

 * style="width:95%" |
 * style="width:95%" |

$$\left[ \begin{matrix} {\dot{d}} \\ {\ddot{d}} \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 1 \\   -\frac{k}{m} & -\frac{c}{m}  \\ \end{matrix} \right]\centerdot \left[ \begin{matrix} d \\ {\dot{d}} \\ \end{matrix} \right]+\frac{1}{m}\centerdot \left[ \begin{matrix} 0 \\   u  \\ \end{matrix} \right]$$

Discretization of model with $$\underline=({{\underline{x}}_{k+1}}-{{\underline{x}}_{k}})/h$$, then:
 *
 * }
 * {| style="width:100%" border="0"

 * style="width:95%" |
 * style="width:95%" |

$${{\underline{x}}_{k+1}}=[\underline{I}+h\centerdot \left[ \begin{matrix} 0 & 1 \\   -\frac{k}{m} & -\frac{c}{m}  \\ \end{matrix} \right]]\centerdot {{\underline{x}}_{k}}+\frac{h}{m}\centerdot \left[ \begin{matrix} 0 \\   u  \\ \end{matrix} \right]$$

Simplying the above equation,
 *
 * }
 * {| style="width:100%" border="0"

$${{\underline{x}}_{k+1}}=\left[ \begin{matrix} 1 & h \\ -\frac{kh}{m} & 1-\frac{hc}{m} \\ \end{matrix} \right]\centerdot {{\underline{x}}_{k}}+\frac{h}{m}\centerdot \left[ \begin{matrix} 0 \\   u  \\ \end{matrix} \right]$$ So,
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$\begin{align} & \underline{F}=\left[ \begin{matrix} 1 & h \\ -\frac{kh}{m} & 1-\frac{hc}{m} \\ \end{matrix} \right] \\ & \underline{G}=\frac{h}{m} \\ \end{align}$$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

3. The equation of motion is represented as:
 * {| style="width:100%" border="0"

 * style="width:95%" |
 * style="width:95%" |

$$\ddot{d}=-\frac{k}{m}\centerdot d-\frac{c}{m}\centerdot \dot{d}+\frac{u}{m}$$

 *
 * }
 * Then, the natural frequency and damping ration are: $$\omega_0 = \sqrt{\frac{k}{m}}$$ and $$\zeta = \frac{c}{2 m \omega_0} \, .$$
 * Critically damped(ζ=1): The system returns to equilibrium as quickly as possible without oscillating.

 * Then, $${{C}_{cr}}=2\centerdot \sqrt{km}$$

4. a). For u=0, plot $$\underline$$ for $$c=0.5*{{C}_{cr}},{{C}_{cr}},1.5*{{C}_{cr}}$$

b). For u=0.5 gaussian noise and c=$$1.5*{{C}_{cr}}$$, plot $$\underline$$

c). For u=0.5 Cauchy noise and c=$$1.5*{{C}_{cr}}$$, plot $$\underline$$

This problem was solved by shengfeng yang

title: User:Egm6341.s11.team4.yang/hw5


text:
Primary English 1

<<Back to English Language Welcome to the Primary English 1 course.

Introduction
The Primary English 1 course is aimed at 5-12 year old children who can read and write English at a very basic level.

Units
Here is a list of units you will look at:
 * 1) /Tenses/
 * 2) /Should, Would and Could/
 * 3) /Question Words/
 * 4) /Writing Stories/
 * 5) /Pronouns/
 * 6) /Story Elements/

Teachers

 * Atcovi (Talk - Contribs) 23:31, 28 October 2015 (UTC)

To Do
If anyone can help create this course, it would be greatly appreciated. This course should be a basis for young English-speakers, as currently there is not an available course for the very young. Here is a list of objectives that need to be completed for Primary English 1, if you are willing to help:

 * Completion of units
 * Review, e.g. is the material suitable for its purpose?
 * Completion of PDF/teacher's version

title: Primary English 1


text:
User:Eml5526.s11.team7.jb/57work

Problem Description
Solve the general 1-D model for an elastic bar with data set 1b (G1DM1.0/D1b) using the Lagrangian Interpolation Basis Function until convergence of $$\displaystyle u^h (0.5) \rightarrow O(10^{-6}) $$.
 * 1. Explain how the basis functions satisfy the constraint breaking solution (CBS).
 * 2.Plot these basis functions.
 * 3.Use matlab to integrate.
 * 4.Make Plots of u^h vs u and u^h(.5)-u(.5) vs m

Solution
The Lagrangian Interpolation Basis Function (LIBF) is given by the following equation and generates a polynomial of degree $$\displaystyle n-1$$:

 * $$L_{i,n}\left (x\right )=\prod_{k=1}^{n}\frac{x-x_{k}}{x_{i}-x_{k}},k\neq i$$

For a linear function the number of nodes is 2 ($$\displaystyle n=2$$). For uniform discretization with $$\displaystyle nel=4$$ the domain $$\displaystyle \Omega$$ will be subdivided into 4 equal length elements each having two nodes. The LIBFs for the first element $$\displaystyle n=1$$ over the elemental domain $$\displaystyle \omega^{(1)}=[0,.25]$$ can be written as follows:

 * $$L_{1,2}^{(1)}\left (x\right)=b_{1}^{(1)}=\frac{(x-x_2)}{x_1-x_2}=\frac{(x-.25)}{0-.25}=-4(x-.25)$$

 * $$L_{2,2}^{(1)}\left (x\right)=b_{2}^{(1)}=\frac{(x-x_1)}{x_2-x_1}=\frac{(x-0)}{.25-0}=4(x)$$

Recall the constraint breaking solution (CBS) states the basis functions must satisfy $$\displaystyle \left\{ b_i(x), i=0,1,2,...\right\}$$ such that $$\displaystyle b_0(\beta) \neq 0$$ and $$\displaystyle b_i(\beta) = 0$$ for $$\displaystyle i=1,2,...,n$$.

It can be seen that these basis functions for the first element satisfy the CBS at each node. For the first node, the basis function has a value of one at that node and a value of zero at every other node (there is only one other node for the linear LIBF). In other words the basis function is equal to the kronicker delta at each node.

 * $$b_{1}^{(1)}(\beta^{(1)}_1)=L_{1,2}^{(1)}\left (0\right)=b_{1}^{(1)}(0)=-4(0-.25)=1$$

 * $$L_{1,2}^{(1)}\left (.25\right)=b_{1}^{(1)}(.25)=-4(.25-.25)=0$$

Similarly for the second node, the basis function has a value of one at that second node and a value of zero at the first node.

A Matlab script was written to plot the LIBFs at each node for each element. The Matlab script is given followed by the plots. The first plot shows each of the LIBFs plotted across the entire domain. The second plot is more informative as the plot is divided into the elemental domains and lines corresponding to the values zero and one are present. The legend states the node number, $$n$$, and element number, $$e$$, of each LIBF.

The global matrices are formed by summing the matrices pertaining to each element:

 * $$\displaystyle \mathbf{K}_{ij} = \sum_{e=1}^{nel}\mathbf{K}^{e}_{ij}$$

The element matrices are formed as in problems 5.1 and 5.2 as follows:

 * $$\displaystyle \mathbf{F}^{e}_{F}=b_{i}^{(1)}(\alpha_{i}^{e} )h + \int_{\omega^{e}}^{.} {{b_{i}^{(1)}}fdx}$$

\displaystyle

h = {a_2}(\alpha )\frac = (2+3\alpha)\frac = 5*12/5=12

$$

 * $$\displaystyle \mathbf{F}^{(1)}_{1}=b_{1}^{(1)}(0)h + \int_{0}^{.25} {{b_{1}^{(1)}}fdx}=1*12+\int_{0}^{.25} -4(x-.25)(5x)=12.0520833$$

 * $$\displaystyle \mathbf{F}^{(1)}_{2}=b_{2}^{(1)}(.25)h + \int_{0}^{.25} {{b_{2}^{(1)}}fdx}=1*12+\int_{0}^{.25} 4x(5x)=12.104167$$

$$ \displaystyle

\therefore {F_F} = \left[ {\begin{array}{ccccccccccccccc} {12.0520833} \\    {12.104167} \end{array}} \right]

$$

 * $$\displaystyle \mathbf{K}^{e}_{ij}=\int_{\omega^{e}}^{.} b_{i}^{e'}(x^e)a_2(x^e)b_{j}^{e'}(x^e)dx$$

\displaystyle

\mathbf{K^{(1)}_{FF}} = \left\{ {\mathbf{K^{(1)}_{ij}};i,j = 1,...,n} \right\}

$$

\displaystyle

\mathbf{K^{(1)}_{11}}= \int\limits_0^{.25} b_{1}^{(1)'}a_2b_{1}^{(1)'}=\int\limits_0^{.25}-4*(2+3x)*-4=32(.25)+24(.25)^2=9.5

$$

\displaystyle

\mathbf{K^{(1)}_{12}}= \int\limits_0^{.25} b_{1}^{(1)'}a_2b_{2}^{(1)'}=\int\limits_0^{.25}-4*(2+3x)*4=-32(.25)-24(.25)^2=-9.5

$$

\displaystyle

\mathbf{K^{(1)}_{21}}=\mathbf{K^{(1)}_{12}} =-9.5

$$

\displaystyle

\mathbf{K^{(1)}_{22}}= \int\limits_0^{.25} b_{1}^{(1)'}a_2b_{1}^{(1)'}=\int\limits_0^{.25}4*(2+3x)*4=32(.25)+24(.25)^2=9.5

$$

\displaystyle \therefore \mathbf{K^{(1)}_{FF}} = \begin{bmatrix} 9.5&-9.5\\ -9.5&9.5 \end{bmatrix} $$

\mathbf{d_{F}^{(1)}} = \mathbf{K}_{FF}^{(1)-1} \mathbf{F_{F}^{(1)}} $$

title: User:Eml5526.s11.team7.jb/57work


text:
Primary English 1/Tenses

<<Back to Primary English 1

1 What are TENSES?
Tenses show when something happens. For example:

Yesterday, Maria lost her book.

At the moment, Maria likes her book a lot.

Maria will look for her book tomorrow.

2 The PAST Tense
The first sentence happened yesterday. When we talk about things that have already happened, we use the PAST tense. All of these sentences are also in the past tense:

I learned my spellings last week.

I did do my homework!

I was walking to school, when the bell rang.

There are lots of ways to make the past tense. These can be using "did" or "has".

3 The PRESENT Tense
The PRESENT tense shows things that are happening now, or things that you usually do. All of these sentences are in the present tense:

I am running to school.

I like eating chocolate.

I usually wake up at ten o'clock!

There are two easy ways of making the present tense. The first way is to say you do something, for example "I eat", "you jump" and "she has fun". The second way is to say you are doing something. You can do this by adding -ing to the end of an action. For example "I am eating", "you are jumping" and "she is having fun".

title: Primary English 1/Tenses


text: File:Cauchy-noise.jpg

title: File:Cauchy-noise.jpg


title: File:Fe1.s11.tm7.HW5.figPB5.6v3.png


text: File:Gaussian-noise.jpg

title: File:Gaussian-noise.jpg


text: File:Nm1 s11 team1 HW5 p3 fig1.png

title: File:Nm1 s11 team1 HW5 p3 fig1.png


text: File:Nm1 s11 team1 HW5 p3 fig2.png

title: File:Nm1 s11 team1 HW5 p3 fig2.png


text:
User:Eml5526.s11.team6.kurth/HW5

=Problem 5.8: Solving Data Set G1DM1.0/D1 Using Linear Lagrangian Element Basis Functions and Uniform Discretization=

Boundary Conditions
The Essential Boundary condition as applied on the boundary $$\Gamma_g = [1]$$, is

The Natural Boundary condition as applied on the boundary $$\Gamma_h = [0]$$, is

Descritization
Here we want to use Uniform Descritization such that Element nodes are equidistant with number of elements as $$ nel = 4, 6, 8, ..$$

Find
1. Explain how Linear Lagrangian Element Basis Function (LLEBF) are used as Costraint Breaking Solution. 2. Plot all LLEBF used for $$ nel =4 $$. 3. Use matlab to numerically integrate and find the solution for $$n_{el}$$ = 2,4,6,8... until the error evaluated at x=0.5 converges to the order of $$10^{-6}$$ 4. Plot for Exact solution $$ u (x)$$ and Approximate solution $$ u^h (x) $$. 5. Convergence - Plot of error vs. number of nodes (n).

Linear Lagrangian Element Basis Functions as a Constraint Breaking Solution
Linear Lagrangian element basis functions assume the value of the Kronecker Delta at nodes, i.e,

For this case, where $$\displaystyle \Gamma_g = 1$$, define the terminating node, $$\displaystyle x_n = 1$$ so that all except the last of the LLEBF will be equal to zero when evaluated at this point while simultaneously the final basis function, $$ b_n$$ will be equal to unity, i.e.

This breaks the constraints of the weighting coefficients $$\displaystyle \{c_i,\;\;i = 0,2,...,n-1\}$$ and as such any arbitrary set may be chosen, as simply selecting $$\displaystyle c_n = 0$$ will satisfy the homogeneous essential boundary condition.

Solution of the System
In the element sense the basis functions are composed of linear Lagrange equations of the form

Now creating the element shape function matrix for this linear 2-node element:

Now, substituting $$\displaystyle l^e = x_2^e-x_1^e\;$$ into $$ \displaystyle (Eq.5.8.8)$$

The element shape function derivative matrix for these elements is

Now constructing the element stiffness matrix

The element force matrix is

Finally the global stiffness and force matrices are assembled via equations (5.13) and (5.14) (F&B p. 96)

where $$\displaystyle L^e$$ is the element dependent scatter operator.

Plot for Exact solution $$ u (x)$$ and Approximate solution $$ u^h (x) $$ along x
It can be observed from the Figure 5.8.2 (Using 4 nodes) and Figure 5.8.3 (Using 6 nodes) that the exact solution is nearly linear and can be very well approximated by the LLEBFs considered with least 4 number of nodes and obviously also when the number of nodes is increased to 6. This is as the approximated solution $$ u^h (x) $$ very closely imitates the curve for the exact solution $$ u (x)$$.

Convergence - Plot of error vs. number of nodes (n)
Even with (n = 4), sufficiently low number of elements, the system gives considerable convergence to an order of to 10^-3. We need more elements for it to further converge to the order of of 10^-6. Figure 5.7.4 depicts the error as a function of the number of elements.

title: User:Eml5526.s11.team6.kurth/HW5


text: File:Nm1 s11 team1 HW5 p3 fig cdf.png

title: File:Nm1 s11 team1 HW5 p3 fig cdf.png


text:
User talk:Wallaby11

 Hello Wallaby11, 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. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse 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.
 * Explore Wikiversity with the links to your left.

 * 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.
 * Follow Wikiversity on twitter (http://twitter.com/Wikiversity) and identi.ca (http://identi.ca/group/wikiversity).

You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --mikeu talk 20:41, 22 March 2011 (UTC)

title: User talk:Wallaby11


title: File:Hw5.4-model.jpg


text: Template:OERcenter

title: Template:OERcenter


text:
Instructional design/Adult learning

Choosing Methods for Adult Learners
When you are asked to give a presentation for a group of adults, knowing what you want to teach the group is just one piece of the puzzle. Adults learn differently than children and knowing a little bit about how adults like to learn can help you put together instruction that will be appealing and effective. Walking away at the end feeling like you have reached your audience and made an impact is a good feeling. Choosing Methods for Adult Learners is a wiki module designed to help you put together effective and appealing instruction for adult learners.

Lesson Goal
This lesson is designed to help subject matter experts responsible for creating classroom instruction use methods that address the needs of adult learners. The resulting instruction will create a relevant, practical, engaging, supportive, and socially interactive learning experience for adult learners.

Key Learning Objectives
At the completion of this Wikiversity module you will:
 * 1) Choose instructional methods that address the needs of adult learners.
 * 2) Identify four characteristics of adult learners.
 * 3) Identify how learning methods address the needs of adult learners.
 * 4) Acknowledge the benefit of using methods effective for engaging adult learners.

Meet Tammy
Tammy is an account manager for a consulting firm in Indiana. Like many of her colleagues, Tammy sits in a cubicle day-in and day-out. In her free time Tammy enjoys doing yoga as a stress reliever. In a recent staff meeting Tammy’s supervisor asked her to present some desktop yoga exercises to her unit to promote stress management in the office. Tammy is passionate about yoga and is eager to share the benefits she enjoys from it. However, Tammy has never taught a class before and is not sure where to begin.

If you are like Tammy, you can benefit from this module on Choosing Methods for Adult Learners.

Click Adult Learner Characteristics to begin.

title: Instructional design/Adult learning


text:
Instructional design/Adult learning/Characteristics of Adult Learners

Characteristics of Adult Learners
Adult Learning Theory is rooted in the idea that adults learn differently than children. Adults are independent and self-directed, they have a wealth of life experiences to draw on, and are motivated to learn when they can see the connection to how their life is affected by the instruction. The following chart identifies five characteristics of adult learners and an example of what each one means when it comes to designing instruction. Use this chart as a guide when creating instruction for adult learners.

Taking Adult Learner Characteristics into Account
The following are two scenarios. In the first scenario the presenter does not take adult learner characteristics into account. In the second scenario the presenter has taken into acount adult learner characteristics. Compare the outcomes of each scenario.

The case of Jimmy
Jimmy works in the Human Resources area of his company. He has been asked to give a one hour presentation to new employees about how to complete payroll and travel documents. Jimmy puts together a powerpoint presentation which he lectures through for 45 minutes. Jimmy asks the participants to hold their questions until the end of the presentation. At first the group seems attentive, but then Jimmy notices that some of participants begin to zone out and are reading the documents instead of listening. For the last 15 minutes of the presentation Jimmy hands out the payroll and travel documents and asks participants to complete them. Jimmy intends to circulate the room while the documents are completed, but there are so many questions after the presentation that the forms do not get completed. The participants seem agitated and comment that they don’t feel prepared to complete the forms, nor do they feel all of their questions were answered.

The case of Steve
Steve works in the Human Resources area of his company. He has been asked to give a one hour presentation to new employees about how to complete payroll and travel documents. Steve knows that getting paid is important to people so he is sure they will be ready to learn. Steve understands that adult learners are oriented to learning such that they want to apply their new knowledge immediately. Steve decides to have participants complete the payroll and travel documents as he is presenting the information rather than having everyone wait until the end of his presentation. Steve also knows that adult learners have an independent and self-directed self-concept so he decides to let the participants choose the order in which they would like to go through the documents. Steve presents the information in small chunks and allows participants to ask questions after each chunk so that if he needs to switch directions he can be flexible and allow the participants to explore the materials. At the end of the session everyone has successfully completed their payroll and travel documents.

Self Quiz
Read the examples below and identify which adult learner characteristic it describes. If you score less than four out of five, review the information above and retake the quiz.

{Charles is looking for a course in leadership development because he knows that a supervisory position will be opening up in his area and he wants to show he is capable of the job. What adult learner characteristic prompts Charles to attend this training? - Motivation to learn - Orientation to learn - Self concept + Readiness to learn
 * type=""}

{Robert’s company is switching to a new piece of software for tracking inventory. Robert is very adept with technology having designed several websites on the side for small businesses. In the training for the new software Robert breezes through the materials and tries to answer questions and help the participants seated near him. The instructor comes over and asks Robert to stop, stating that they need to run the training without his help. Robert gives the instructors poor reviews. What adult learner characteristic has the instructor not taken into account? + Experience - Motivation to learn - Self concept - Orientation to learn
 * type=""}

{Jackie signed up for a community education class in Cooking Basics. From the course description it looked like participants would have opportunities to put together meals that fit their lifestyle. After the first session Jackie was frustrated that the instructor had chosen the meals everyone would prepare and did not allow her to have any input. Jackie decides to stop attending. What adult learner characteristic has the instructor not taken into account? - Readiness to learn + Self concept - Motivation to learn - Experience
 * type=""}

{Angela is taking a course on creating bird houses. After reviewing safety information the instructor immediately has participants pick up their tools and start construction. Angela is excited that they get to actually construct a bird house and not just learn about them. What adult learner characteristic has the instructor tapped into? - Experience - Motivation to learn - Self concept + Orientation to learn
 * type=""}

{Maureen started a new job at a domestic violence shelter and is attending training on the basics of the cycle of domestic violence. During the training participants are given many opportunities to discuss case scenarios and how they would approach them with a client. Maureen has a chance to express her views and feels supported and validated by the instructor and other participants. What adult learner characteristic has the instructor taken into account? - Readiness to learn - Self concept + Motivation to learn - Experience
 * type=""}

Click Instructional Methods to continue.


text:
Instructional design/Adult learning/Instructional Methods

Common Instructional Methods
When working with adult learners, there are instructional strategies that work well, and others that don't. For example, a drill-and-response method where instructor provides terms and figures for the learner to memorize might not be very effective. As the previous slide mentioned, the adult learner's readiness and orientation to learn bring a desire to immediately apply new learning to a context that the learner finds important (such as at work). So, memorizing terms and phrases apart from providing contexts or appropriate suggestions for application would likely seem frustrating to the learner.

How do we increase the likelihood that the learner finds relevance in the instruction (and, as a result, retains and uses the information)? The following table outlines seven common instructional strategies, along with how individual adult learning characteristics may be addressed by the strategy. Finally, the last paragraph provides examples of how the individual strategies might be utilized when delivering subject specific content.

For the last two instructional methods, the adult learning characteristic and example are blank. Please take a moment to determine which characteristic(s) the method addresses, and provide an example of how it might play out in an instructional setting.

More Information
To read more about Instructional Methods:

Olson, Dorsey, & Reigeluth - as found on page 22 of Instructional -Design Theories and Models

Click Choosing Methods to continue.