text:
User:Eml4500.f08.bottle.barnes/HW6

=Tapered Bar with Constant Modulus of Elasticity=

Let $$\displaystyle E_1 = E_2 = E = 70 GPa $$ and let $$\displaystyle A( \tilde{x}) $$ be linear. Also let $$\displaystyle A_1 = 2400 mm^2 $$, $$\displaystyle A_2 = 600 mm^2 $$ and $$\displaystyle L = 300 mm $$.

Given the following equation

$$\displaystyle {E \over L^{(i)}} {(A_1 + A_2) \over 2} \begin{bmatrix} 1 & -1 \\ -1 & 1  \end{bmatrix} = \textbf{k}^{(i)} $$

$$\displaystyle {(70 GPa) \over (300mm)} {(2400mm^2 + 600 mm^2) \over 2} \begin{bmatrix} 1 & -1 \\ -1 & 1  \end{bmatrix} = \textbf{k}^{(i)} $$

$$\displaystyle \textbf{k}^{(i)} = \begin{bmatrix} 714 & -714 \\ -714 & 714  \end{bmatrix} $$

title: User:Eml4500.f08.bottle.barnes/HW6


text:
User:Eml4500.f08.Lulz.Layton.Eric/Hw6

ElectricPylon Display.m
Command Window Output >> ElectricPylonSystem

d =

0           0            0            0  -0.00019023   0.00017465   -0.0001277 -1.4074e-006 -0.00016006  -0.0002967  -0.00012671   0.00032602  -0.00012798    -0.000128  -0.00012792  -0.00054327   0.00044932   0.00030901   0.00046381  -0.00061057     0.001017   0.00064583    0.0010325  -0.00019427    0.0010502  -0.00097027    0.0017167    0.0010338    0.0017162   0.00023259    0.0016833  -0.00064959     0.001683   -0.0013293    0.0023452   0.00064079    0.0021863   -0.0010297    0.0047122    0.0023902    0.0039576   -0.0024165    0.0063605    0.0057328    0.0063605    0.0037877    0.0063605    0.0029803    0.0063614     0.002384    0.0063657    0.0016463    0.0063711   0.00086194    0.0063182  -0.00024268    0.0062289   -0.0013182    0.0061363   -0.0023672    0.0060413   -0.0033231    0.0058839   -0.0051278    0.0055152    -0.012587    0.0067826    0.0042648     0.006996   -0.0063359     0.007012    0.0034574    0.0074006   -0.0043591    0.0071505    0.0029772    0.0071478    0.0020508    0.0071421    0.0012278     0.007182   0.00032026    0.0072816  -0.00073249    0.0074073   -0.0018633    0.0075643   -0.0033872    0.0090005    0.0029772     0.010455   -0.0033872

reactions =

149.64     -501.65      -149.64       1501.7

results =

1.0781e-005 2.1562e+006       862.48 -6.5356e-006 -1.3071e+006     -522.85 6.7339e-006 1.3468e+006       538.71 6.4556e-006 1.2911e+006       516.45 -3.5152e-006 -7.0304e+005     -281.22 -2.3798e-005 -4.7596e+006     -1903.8 -1.2325e-005 -2.465e+006      -986.01 7.1182e-006 1.4236e+006       569.46 1.9269e-005 3.8538e+006       1541.5 -2.8828e-005 -5.7655e+006     -2306.2 -1.8941e-007      -37882      -15.153 8.778e-009      1755.6      0.70224 1.9328e-005 3.8655e+006       1546.2 1.4698e-005 2.9395e+006       1175.8 -1.8494e-005 -3.6988e+006     -1479.5 -2.883e-005 -5.7661e+006     -2306.4 1.7948e-006 3.5896e+005       143.58 2.621e-005  5.242e+006       2096.8 9.8986e-006 1.9797e+006       791.89 -1.8874e-005 -3.7749e+006       -1510 -3.258e-005 -6.5159e+006     -2606.4 2.7892e-006 5.5784e+005       223.14 3.2194e-006 6.4388e+005       257.55 2.9828e-005 5.9656e+006       2386.3 -4.0852e-006 -8.1704e+005     -326.82 -1.9808e-005 -3.9616e+006     -1584.6 1.16e-005   2.32e+006       928.02 -4.0653e-006 -8.1307e+005     -325.23 -2.8778e-005 -5.7556e+006     -2302.2 -1.4237e-007      -28474       -11.39 -4.9677e-006 -9.9353e+005     -397.41 -7.6072e-008      -15214      -6.0857 2.9608e-005 5.9216e+006       2368.6 2.7907e-007       55813       22.325 -2.4358e-005 -4.8717e+006     -1948.7 6.9573e-006 1.3915e+006       556.59 -1.5116e-008     -3023.3      -1.2093 -2.8794e-005 -5.7588e+006     -2303.5 -2.427e-005 -4.854e+006      -1941.6 6.9522e-006 1.3904e+006       556.17 -9.4154e-007 -1.8831e+005     -75.323 1.129e-005  2.258e+006       903.22 -3.16e-006  -6.32e+005       -252.8 -2.0051e-005 -4.0102e+006     -1604.1 -1.0133e-019 -2.0266e-008 -8.1063e-012 7.3233e-019 1.4647e-007  5.8586e-011 3.2073e-007       64147       25.659 1.3501e-006 2.7001e+005       108.01 1.3575e-006 2.7149e+005        108.6 -1.0389e-005 -2.0778e+006     -831.13 -2.2136e-005 -4.4271e+006     -1770.9 -2.9388e-005 -5.8777e+006     -2351.1 -3.6774e-005 -7.3549e+006     -2941.9 -4.331e-005 -8.662e+006      -3464.8 -4.3478e-005 -8.6957e+006     -3478.3 0           0            0  3.0942e-019  6.1883e-008  2.4753e-011 -2.6985e-019 -5.397e-008 -2.1588e-011 3.0512e-019 6.1024e-008  2.4409e-011 -8.8522e-007 -1.7704e+005     -70.818 1.1034e-006 2.2069e+005       88.276 -9.5965e-007 -1.9193e+005     -76.772 9.9436e-007 1.9887e+005       79.549 1.0732e-005 2.1464e+006       858.56 -1.0415e-005 -2.083e+006      -833.19 1.1448e-005 2.2895e+006       915.81 -1.1321e-005 -2.2643e+006     -905.71 1.0517e-005 2.1035e+006       841.38 -1.0732e-005 -2.1464e+006     -858.56 7.0689e-006 1.4138e+006       565.51 -6.8859e-006 -1.3772e+006     -550.87 7.9177e-006 1.5835e+006       633.42 -1.8464e-005 -3.6928e+006     -1477.1 -4.8696e-007      -97393      -38.957 -1.1845e-007      -23690      -9.4762 1.5604e-007       31208       12.483 4.5239e-005 9.0479e+006       3619.2 4.6964e-019 9.3927e-008  3.7571e-011 4.5158e-005 9.0317e+006       3612.7 -7.9385e-019 -1.5877e-007 -6.3508e-011 4.5256e-005 9.0511e+006       3620.5 -6.5877e-007 -1.3175e+005     -52.702 -1.4822e-006 -2.9645e+005     -118.58 8.9829e-006 1.7966e+006       718.63 2.2438e-005 4.4876e+006         1795 3.3117e-005 6.6233e+006       2649.3 3.8878e-005 7.7755e+006       3110.2 -5.3704e-020 -1.0741e-008 -4.2963e-012 0           0            0 -4.8034e-020 -9.6069e-009 -3.8428e-012 0           0            0

max_compressive_stress =

-8.6957e+006

element_number =

55

max_tensile_stress =

9.0511e+006

element_number =

81

min_eigenvalues =

132.16      2468.4       2940.3

min_eigen_num =

82   85    86

Ethics, Honesty, Innovation, and Imagination
Honesty is one of the most important factors in submitting work. Honesty illustrates strong ethical consideration of other people's work in regards to not plagiarizing. Honesty keeps students true concerning their own work, showing that students can thrive to show their own work. Innovation and imagination are important aspects of any submission of work, but careful consideration must be taken so that no plagiarism occurs. In several homework reports, for example, Team Lulz has striven to improve the innovation of each of our Wikiversity pages.

PVW for Elastic Bar

Continuing with the discussion of the Principle of Virtual Work (PVW) for the dynamics of the elastic bar shown in Figure 1, consider the partial differential equation (PDE) describing the continuous case equation of motion (EOM) for this bar:

 * $$\frac{\partial}{\partial x}[(EA)\frac{\partial u}{\partial x}] + f = m \ddot{u}$$

Deriving the discrete case EOM for the bar from the continuous EOM yields the following:

 * $$- \mathbf{kd} + \mathbf{F} = \mathbf{M} \mathbf{\ddot{d}} => \mathbf{M}\mathbf{\ddot{d}} + \mathbf{kd} = \mathbf{F}$$

Where $$\mathbf{M}\;$$ is the mass matrix. The presence of the mass matrix indicates a multiple degree of freedom system. A similar equation for the single degree of freedom system is given as:

 * $$m\ddot{d} + kd = F$$

The following integral equation was used in the derivation of the discrete EOM from the continous EOM:

 * $$\int^{1}_{0} w(x) \left\{\frac{\partial}{\partial x} [EA \frac{\partial u}{\partial x}] + f - m \ddot{u}\right\} dx = 0$$ for all possible values of $$w(x)\;$$, the weighting function.

Deriving this equation from the continous EOM is a trivial, but to get the continous EOM from this equation is not. In order to show this, let the integral equation be rewritten as follows:

 * $$\int w(x) g(x) dx = 0$$ for all $$w(x)\;$$

Since this equation should hold for all $$w(x)\;$$, selecting $$w(x)\;$$ to be equal to $$g(x)\;$$ is a valid operation. This results in the following:

 * $$\int g^{2} dx = 0$$

From here, it is obvious that $$g(x) = 0\;$$.

 * {| class="collapsible collapsed"

!Integration by Parts Consider two functions $$r(x)\;\;$$ and $$g(x)\;$$. The derivative of the product of these two functions is as follows:
 *
 *

 * $$(rs)^{'} = r^{'}s + rs^{'}\;$$

Where $$r^{'} = \frac{\partial r}{\partial x}$$ and $$s^{'} = \frac{\partial s}{\partial x}$$ Taking the integral of this expression yields the following:

 * $$\int (rs)^{'} =\int r^{'}s + \int rs^{'}\;$$

Noting that the left hand side of this equation is equivalent to $$rs\;$$, with rearranging the following well-known relation can be determined:

 * $$\int r^{'}s = rs - \int rs^{'}\;$$

 * }

Using the integration by parts method on the first term of the continuous EOM, with $$ r(x) = {EA} \frac{\partial u}{\partial x}\;$$ and $$s(x) = w(x)\;$$ yields the following:

 * $$ \int^{L}_{0} w(x) \frac{\partial}{\partial x} \left[(EA) \frac{\partial u}{\partial x}\right] dx = \left[w(EA) \frac{\partial u}{\partial x}\right]^{L}_{0} - \int^{L}_{0} \frac{\partial w}{\partial x}(EA) \frac{\partial u}{\partial x} dx$$

 * $$= \underbrace{w(L)(EA)(L)\frac{\partial u(L,t)}{\partial x}}_{N(L,t)} - \underbrace{w(0)(EA)(0)\frac{\partial u(0,t)}{\partial x}}_{N(0,t)} - \int^{L}_{0} \frac{\partial w}{\partial x}(EA) \frac{\partial u}{\partial x} dx$$

Model Problem
Consider the model problem shown in Figure 2. The boundary conditions for this problem can be found in Homework 5. Additionally, select $$w(x)\;$$ such that $$w(0)=0\;$$; this makes the system "kinematically admissable".

Applying the discrete PVW to a previously studied equation from page 10-1 of the notes yields the following relation:

 * $$\mathbf{W}(\mathbf{K}\begin{Bmatrix} d_{3}\\d_{4}\end{Bmatrix} - \mathbf{F}) = \mathbf{0}$$ for all $$\mathbf{W}$$.

Where $$\mathbf{F}^{T} = \begin{bmatrix}F_1 & F_2 & F_3 & F_4 & F_5 & F_6\end{bmatrix}$$. Since $$\mathbf{W}$$ can be arbitrarily selected, select it such that $$W_1 = W_2 = W_5 = W_6 = 0\;$$. This will eliminate equations involving unknown reactions, eliminating rows 1,2,5, and 6 from the two-bar truss system ans yielding the following relation:

 * $$\mathbf{\bar{K}}\underbrace{\mathbf{\bar{d}}}_{\begin{Bmatrix} d_{3}\\d_{4}\end{Bmatrix}}=\underbrace{\mathbf{\bar{F}}}_{\begin{Bmatrix} F_{3}\\F_{4}\end{Bmatrix}}$$

Note also that $$\mathbf{\bar{W}} \cdot \mathbf{\bar{K}}\mathbf{\bar{d}}=\mathbf{\bar{F}}$$ for all $$\mathbf{\bar{W}}$$, where $$\mathbf{\bar{W}} = \begin{Bmatrix} W_{3}\\W_{4}\end{Bmatrix}$$.

Returning to the continuous case, consider the following unknown reaction:

 * $$N(0,t) = (EA)(0)\frac{\partial u}{\partial x} (0,t)$$

This equation can be written as:

 * $$W(L)F(t) - \int^{L}_{0} \frac{\partial w}{\partial x}(EA) \frac{\partial u}{\partial x} dx + \int^{L}_{0}w(x)[f-m\ddot{u}]dx = 0$$ for all $$w(x)\;$$ such that $$w(0) = 0\;$$.

The weak form of continuous PVW can be expressed as:

 * $$\int^{L}_{0}W(m\ddot{u})dx + \int^{L}_{0} \frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx = W(L)F(t) + \int^{L}_{0} Wfdx$$ for all $$w(x)\;$$ such that $$w(0) = 0\;$$.

Comparison Between Continuous and Discrete PVW

 * {| border=1

 * bgcolor="#99ccff"|
 * bgcolor="#99ccff" style="text-align: center;" |Continuous Setting (PVW)
 * bgcolor="#99ccff" style="text-align: center;" |Discrete Setting (PVW)
 * bgcolor="#05fa67"|Inertia
 * bgcolor="#ffffff" style="text-align: center;" | $$\int^{L}_{0}wm\ddot{u}dx$$
 * bgcolor="#ffffff" style="text-align: center;" | $$\mathbf{\bar{W}} \cdot (\mathbf{\bar{M}}\mathbf{\ddot{\bar{d}}})$$
 * bgcolor="#05fa67"|Stiffness
 * bgcolor="#ffffff" style="text-align: center;" | $$\int^{L}_{0} \frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx$$
 * bgcolor="#ffffff" style="text-align: center;" | $$\mathbf{\bar{W}} \cdot (\mathbf{\bar{k}}\mathbf{\bar{d}})$$
 * bgcolor="#05fa67"|Applied Force
 * bgcolor="#ffffff" style="text-align: center;" | $$W(L)F(t) + \int^{L}_{0}wfdx$$
 * bgcolor="#ffffff" style="text-align: center;" | $$\mathbf{\bar{W}} \cdot \mathbf{\bar{F}}$$
 * bgcolor="#05fa67"|Conditions
 * bgcolor="#ffffff" style="text-align: center;" | all $$w(x)\;$$ such that $$w(0)=0\;$$
 * bgcolor="#ffffff" style="text-align: center;" | for all $$\mathbf{\bar{W}}\;$$
 * }
 * bgcolor="#05fa67"|Conditions
 * bgcolor="#ffffff" style="text-align: center;" | all $$w(x)\;$$ such that $$w(0)=0\;$$
 * bgcolor="#ffffff" style="text-align: center;" | for all $$\mathbf{\bar{W}}\;$$
 * }
 * }

Figures 3 and 4 describe the interpolation of displacements along the elastic bar. Figure 5 shows the elastic bar segmented by nodes, whereas Figure 6 shows a close-in view of the ith segment of the bar. Assume the displacement $$u(x)\;$$ for $$x_i \leq x \leq x_i + 1$$, which can also be written as $$(x \in [x_i,x_i+1])$$.

Motivation for Linear Interpolation of u(x)

Consider the deformed truss element seen in Figure 5. The deformed shape is a straight line; i.e., there was an implicit assumption of linear interpolation of displacement between 2 nodes.

Next, consider the case where there are only axial displacements. Expressing $$u(x)\;$$ in terms of $$d_i = u(x_i)\;$$ and $$d_{i+1} = u(x_{i+1})\;$$ as a linaer function of $$\;x$$ (i.e, performing linear interpolation) is as follows:

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

This relationship can be seen graphically in Figure 6. $$N_{i+1}(x)\;$$ can be expressed as $$\frac{x - x_i}{x_{i+1} - x_i}$$.

Lagrange Interpolation
The motivation for the form of $$N_i(x)\;$$ and $$N_{i+1}(x)\;$$ is as follows:

1) $$N_i(x)\;$$ and $$N_{i+1}(x)\;$$ are linear; thus, any linear combination of $$N_i\;$$ and $$N_{i+1}$$ is also linear, and in particular the previously discussed expression for $$u(x)\;$$. This relationship can be further expressed as:

 * $$N_i(x) = \alpha_i + \beta_ix\;$$

 * $$N_{i+1}(x) = \alpha_{i+1} + \beta_{i+1}x\;$$

Where $$\alpha_i,\alpha_{i+1},\beta_i,\;$$ and $$\beta_{i+1}\;$$ are real numbers. A linear combination of these numbers can be written as follows:

 * $$N_id_i + N_{i+1}d_{i+1} = (\alpha_i + \beta_ix)d_i + (\alpha_{i+1} + \beta_{i+1}x)d_{i+1}\;$$

 * $$= (\alpha_i d_i + \alpha_{i+1} d_{i+1}) + (\beta_i d_i + \beta_{i+1} d_{i+1})x\;$$

It is clear that this relation is a function in x.

2) Recall the previously discussed equation for $$u(x_i)\;$$:

 * $$u(x_i) = \underbrace{N_i(x_i)}_{1}d_i + \underbrace{N_{i+1}(x_i)}_{0}d_{i+1}\;$$
 * $$ = d_i\;$$

Determination of Stiffness Matrix for Element i
Applying the interpolation process for w(x) yields the following relation:

 * $$w(x) = N_i(x)W_i + N_{i+1}(x)W_{i+1}\;$$

This equation can be employed to determine the elemental stiffness matrix for the element i. Consider the following integral equation for $$\beta \;$$:

 * $$\beta = \int^{x_{i+1}}_{x_i}\left[N_i^'W_i + N_{i+1}^'W_{i+1}\right](EA)\left[N_i^'d_i + N_{i+1}d_{i+1}\right]dx$$

Where $$N_i^' := \frac{\partial N_i(x)}{\partial x}$$, and similarly for $$N_i^'d_i\;$$. Note also the following two definitions:

 * $$u(x) = \underbrace{\begin{bmatrix}N_i(x) & N_{i+1}(x)\end{bmatrix}}_{\mathbf{N}(x)}\begin{Bmatrix}d_i\\d_{i+1}\end{Bmatrix}$$

 * $$\frac{du(x)}{dx} = \underbrace{\begin{bmatrix}N_i^'(x) & N_{i+1}^'(x)\end{bmatrix}}_{\mathbf{B}(x)}\begin{Bmatrix}d_i\\d_{i+1}\end{Bmatrix}$$

Similarly,

 * $$W(x) = \mathbf{N}(x)\begin{Bmatrix}W_i\\W_{i+1}\end{Bmatrix}$$

 * $$\frac{dW(x)}{dx} = \mathbf{B}(x)\begin{Bmatrix}W_i\\W_{i+1}\end{Bmatrix}$$

Recall the element degrees of freedom as shown in Figure 7. These degrees of freedom can be summarized as:

 * $$\begin{Bmatrix}d_i \\ d_{i+1}\end{Bmatrix} = \begin{Bmatrix}d_1^{(i)} \\ d_2^{(i)}\end{Bmatrix} = \mathbf{d}^{(i)}$$

Similarly for $$\mathbf{W}^{(i)}\;$$,

 * $$\begin{Bmatrix}W_i \\ W_{i+1}\end{Bmatrix} = \begin{Bmatrix}W_1^{(i)} \\ W_2^{(i)}\end{Bmatrix} = \mathbf{W}^{(i)}$$

Substituting the above relations into the integral equation for $$\beta \;$$ yields the following:

 * $$\beta = \int^{x_{i+1}}_{x_i}\left(\mathbf{BW}^{(i)}\right)(EA)\left(\mathbf{Bd}^{(i)}\right)dx = \mathbf{W}^{(i)} \cdot (\mathbf{k}^{(i)}\mathbf{d}^{(i)})$$

Since all of the terms in the integral are constants (1x1 matrices), they can be rearranged as follows:

 * $$\beta = \int^{x_{i+1}}_{x_i}(EA)\left(\mathbf{BW}^{(i)}\right) \cdot \left(\mathbf{Bd}^{(i)}\right)dx = \underbrace{\left(\mathbf{BW}^{(i)}\right)^T}_{\underbrace{\mathbf{W}^{(i)T}\mathbf{B}^T}_{\mathbf{W}^{(i)} \cdot \mathbf{B}^T}}\left(\mathbf{Bd}^{(i)}\right)$$

Finally, $$ \beta \;$$ can be written as:

 * $$\beta = \mathbf{W}^{(i)} \cdot \left(\int \mathbf{B}^T (EA) \mathbf{B}dx\right)\mathbf{d}^{(i)}$$

This leads to the following expression for the element stiffness matrix $$\mathbf{k^{(i)}}\;$$:

 * $$\mathbf{k}^{(i)} = \int^{x_{i+1}}_{x_i} \mathbf{B}^T(x) (EA)(x) \mathbf{B}(x)dx$$

Note also the following definition for $$L^{(i)}$$:

 * $$L^{(i)} = x_{i+1} = x_{i}\;$$

Also, note the transformation of coordinates from $$x\;$$ to $$\tilde{x}\;$$:

 * $$\tilde{x} := x - x_i\;$$

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

This leads to the following version of the element stiffness matrix:

 * $$\mathbf{k}^{(i)} = \int^{\tilde{x}=L^{(i)}}_{\tilde{x}=0} \mathbf{B}^T(\tilde{x}) (EA)(\tilde{x}) \mathbf{B}(\tilde{x})d\tilde{x}$$

Consider the an ith element as shown in Figure 8. The relations for area and Young's modulus as a function of $$\tilde{x}\;$$ can be written as follows:

 * $$A(\tilde{x}) = N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2$$

 * $$E(\tilde{x}) = N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2$$

Combining the above equations with the following equation:

$$k= \underline{k}^{(i)} = \frac{EA}{L^{(i)}}\begin{bmatrix}1 & -1 \\ -1 & 1 \end{bmatrix}$$

it is possible to determine $$\underline{k}^{(i)}$$ as a function of $$\begin{Bmatrix} A_1, & A_2, & E_1, & E_2, & L^{(i)}\end{Bmatrix}$$

$$k^{(i)}(\tilde{x}) = \frac{[N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2][N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2]}{L^{(i)}} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$

Comparison of Element Stiffness Matrix Determination Methods
The shape function $$N_2^{(i)}(\tilde{x})\;$$ can be written as follows:

 * $$N_2^{(i)}(\tilde{x}) = \frac{\tilde{x}}{L^{(i)}}= \left\{\begin{matrix}0\;at\;\tilde{x} = 0 \\ 1\;at\;\tilde{x} = L^{(i)}\end{matrix}\right.$$

9. Average Area]]

Following the book on page 159, setting $$E_1 = E_2 = E\;$$ and letting $$A(\tilde{x})\;$$ be linear, it is possible to obtain $$\mathbf{k}^{(i)}$$ and compare to the equation in the book:

 * $$\frac{E}{L^{(i)}}\frac{(A_1 + A_2)}{2}\begin{bmatrix}1&-1\\-1&1\end{bmatrix} = \mathbf{k}^{(i)}$$

Where the quantity $$\frac{(A_1 + A_2)}{2}\;$$ is the average area. A graphical view of this concept is given in Figure 9. Next, it is useful to compare the general $$\mathbf{k}^{(i)}\;$$ to a stiffness matrix determined using $$.5(A_1+A_2)\;$$ and $$.5(E_1+E_2)\;$$, where $$E_1 \neq E_2\;$$. This yields the following relation for $$\mathbf{k}^{(i)}_{ave}\;$$:

 * $$\frac{(E_1+E_2)(A_1+A_2)}{4L^{(i)}}\begin{bmatrix}1&-1\\-1&1\end{bmatrix} = \mathbf{k}^{(i)}_{ave}$$

The comparison in question is simply the difference $$\mathbf{k}^{(i)} - \mathbf{k}^{(i)}_{ave}$$:

$$\mathbf{k}^{(i)} - \mathbf{k}^{(i)}_{ave}=\frac{(2E-E_1-E_2)(A_1+A_2)}{4L^{(i)}}\begin{bmatrix}1&-1\\-1&1\end{bmatrix}$$

[[Image:Lulz6.9.png|thumb|right|200px|Fig 10. Centroid of Area]

 * {| class="collapsible collapsed"

!Mean Value Theorem Recall the Mean Value Theorem(MVT) and its relation to the centroid:
 *
 *

 * $$\int^{x=a}_{x=b}f(x)dx = f(\bar{x})[b-a]\;$$

for $$\bar{x} \in [a,b]$$. For the centroid of an area, as given in Figure 10, the MVT can be expressed as follows.

 * $$\int_A xdA = \bar{x} \int_A dA = \bar{x}A$$

Applying the MVT to a product of two functions produces a similar result:

 * $$\int^{x=a}_{x=b}f(x)g(x)dx = f(\bar{x})g(\bar{x})[b-a]\;$$

However, it should be noted that $$f(\bar{x})$$ and $$g(\bar{x})$$ are not equivalent to the average values of those functions. This property is shown in equation form below:

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

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

 * }

HW: What if EA is constant?
If EA is constant, prove the below stiffness matrix to be true:

$$\mathbf{k}= \mathbf{k}^{(i)} = \frac{EA}{L^{(i)}}\begin{bmatrix}1 & -1 \\ -1 & 1 \end{bmatrix}$$

We are given this general form of the stiffness matrix:

$$\mathbf{k}^{(i)}=\int^{x_{i+1}}_{x_i}\mathbf{B}^T(x)(EA)(x)\mathbf{B}(x)dx$$

We also can solve $$\mathbf{B}$$ and $$\mathbf{B}^T$$ into terms of $$L^{(i)}=x_{i+1}-x_i$$:

$$\mathbf{B}(x) = \begin{bmatrix} N'_i(x) & N'_{i+}(x)\\ \end{bmatrix} = \begin{bmatrix} \frac{1}{x_i-x_{i+1}} & \frac{1}{x_{i+1}-x_i}\\ \end{bmatrix} = \begin{bmatrix} \frac{1}{-L^{(i)}} & \frac{1}{L^{(i)}}\\ \end{bmatrix}$$

$$\mathbf{B}^T(x) = \begin{bmatrix} \frac{1}{-L^{(i)}}\\ \frac{1}{L^{(i)}}\\ \end{bmatrix}$$

From here it is simple substitution and algebra to solve the proof since (EA)(x) can be removed from the integral as a constant.

$$\mathbf{k}^{(i)}=(EA)\int^{x_{i+1}}_{x_i}\begin{bmatrix} \frac{1}{-L^{(i)}} & \frac{1}{L^{(i)}}\\ \end{bmatrix}\begin{bmatrix} \frac{1}{-L^{(i)}}\\ \frac{1}{L^{(i)}}\\ \end{bmatrix}dx=(EA)\int^{x_{i+1}}_{x_i}\begin{bmatrix} \frac{1}{{L^{(i)}}^2} & -\frac{1}{{L^{(i)}}^2}\\ -\frac{1}{{L^{(i)}}^2} & \frac{1}{{L^{(i)}}^2}\\ \end{bmatrix}dx=(EA)\begin{bmatrix} \frac{x_{i+1}-x_i}{{L^{(i)}}^2} & -\frac{x_{i+1}-x_i}{{L^{(i)}}^2}\\ -\frac{x_{i+1}-x_i}{{L^{(i)}}^2} & \frac{x_{i+1}-x_i}{{L^{(i)}}^2}\\ \end{bmatrix}=(EA)\begin{bmatrix} \frac{L^{(i)}}{{L^{(i)}}^2} & -\frac{L^{(i)}}{{L^{(i)}}^2}\\ -\frac{L^{(i)}}{{L^{(i)}}^2} & \frac{L^{(i)}}{{L^{(i)}}^2}\\ \end{bmatrix}$$

Dividing out $$L^{(i)}$$ from this and reducing the fractions yields the target equation:

$$\mathbf{k}= \mathbf{k}^{(i)} = \frac{EA}{L^{(i)}}\begin{bmatrix}1 & -1 \\ -1 & 1 \end{bmatrix}$$

HW: Find an expression for $$\mathbf{k}^{(i)}(\tilde{x})$$
We are given the general formula for the stifness matrix and want to find a more useful way to express it in the $$\tilde{x}$$ coordinate system:

$$\mathbf{k}^{(i)}=\int^{x_{i+1}}_{x_i}\mathbf{B}^T(x)(EA)(x)\mathbf{B}(x)dx$$

Remembering how to convert the coordinate systems we can convert the general formula to the $$\tilde{x}$$ coordinate system:

$$\tilde{x}:=x-x_i$$

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

$$\mathbf{k}^{(i)}=\int^{\tilde{x}=L^{(i)}}_{\tilde{x}=0}\mathbf{B}^T(\tilde{x})(EA)(\tilde{x})\mathbf{B}(\tilde{x})d\tilde{x}$$

Next we consider the known solutions for A and E from the notes:

$$A(\tilde{x}) = N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2$$

$$E(\tilde{x}) = N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2$$

Substituting these yields:

$$\mathbf{k}^{(i)}=\int^{\tilde{x}=L^{(i)}}_{\tilde{x}=0}\mathbf{B}^T(\tilde{x})[(N_1^{(i)}(\tilde{x})E_1+N_2^{(i)}(\tilde{x})E_2)(N_1^{(i)}(\tilde{x})A_1+N_2^{(i)}(\tilde{x})A_2)]\mathbf{B}(\tilde{x})d\tilde{x}$$

Contributing Team Members
Aaron Fisher - Eml4500.f08.Lulz.fisher 14:26, 21 November 2008 (UTC)

Eric Layton - Eml4500.f08.Lulz.Layton.Eric 21:10, 20 November 2008 (UTC)

Sam Miorelli - Eml4500.f08.lulz.abcd 20:12, 21 November 2008 (UTC)

Benjamin Mitchell - Eml4500.f08.Lulz.mitchell.bm 16:25, 21 November 2008 (UTC)

John Saxon - Eml4500.f08.Lulz.js 07:24, 21 November 2008 (UTC)

Andrew Strack - Eml4500.f08.lulz.strack 16:57, 21 November 2008 (UTC)

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

title: User:Eml4500.f08.Lulz.Layton.Eric/Hw6


text: File:Figure 3 NINE.jpg

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User:Eml4500.f08.gravy.sms/HW6

Modification of 2-bar truss code to accommodate general k(i)
The first modification to the 2-bar truss code was adding the values for E1(i), E2(i), A1(i), and A2(i) to the matrices at the beginning of the code. This was done with a 2x2 matrix for both E and A. The columns represent the elements, and the rows represent the nodes. Therefore, e(1,2) is E1(2). These values were then used to caculate k(i) using the following general formula:

$$k_{2x2}(\tilde{x})^{(i)} = \frac{\begin{bmatrix} 2 \left ( A_1 * E_1 \right ) +  \left ( A_2 * E_1 \right ) + \left ( A_1 * E_2 \right ) + 2\left ( A_2 * E_2 \right ) \end{bmatrix}}{6*L^{(i)}}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$

Then, the values of E1(i), E2(i), A1(i), and A2(i) were fed into the modified versions of PlaneTrussElement.m and PlaneTrussResults.m that are shown below.

The results were then plotted on the same graph as the deformed two bar truss without tapering. The blue dashed line represents the initial conditions, the green line is the non-tapered deformation, and the red line is the deformation of the tapered bars.

title: User:Eml4500.f08.gravy.sms/HW6


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Wikipedia/Suggested essays

Suggested essays in social science

 * Carry out a questionnaire or an interview survey, for example regarding:
 * To what extent is it considered acceptable to use Wikipedia as source in student essays?
 * To what extent does systematic bias exist on Wikipedia?
 * What causes subjective enforcement of the Wikipedia policies?
 * What measures are most efficient in improving the reliability of Wikipedia?
 * Arrange a blind test measurement of the quality of a certain Wikipedia language version in comparison to a commercial encyclopedia
 * Carry out a case study, evaluating the success of some wiki site.
 * Study Wikipedia statistics
 * Write a review paper on a few research articles or some master theses related to wikis and Wikipedia, compare their conclusions and argue for your own opinion on the matter

Suggested technical projects in computer science

 * Install, configure, administrate and build up your own wiki
 * Evaluate some new MediaWiki extension or application
 * Program a MediaWiki robot script
 * Suggest and specify the development of some new MediaWiki application or extension, and develop a first prototype or demonstrator
 * Write a review paper on some technical research articles or a Master of Science theses related to wiki technology.

title: Wikipedia/Suggested essays


text:
Wikipedia service-learning courses/Styles in common in Wikimedia projects

Language

 * Be decent. Avoid insults.

Article names and Headings

 * Article names should not contain special characters like &, |, -, / and +. (aside from a rew exceptions)
 * Only the first letter of articles and headings should be capitalized. (aside from several exceptions)
 * Use spaces and sentences instead OneWordIdentifiers ('Common styles in wikimedia projects' instead of WikimediaStyleCommon)
 * keep Headings short
 * do no repeat parts of within deeper subheadings

Headings format
Introductory paragraph. == Heading == Text. === Subheading === Text. ==== Deeper subheading ==== Text.
 * the same

Content

 * The first use of the title words or words that are akin in sentiment to the title words are marked in bold.
 * When a new term is introduced, the term is italicised.


title: File:Fea.group.cady.6.1.jpg


title: File:Fea.group.cady.6.2.jpg


text: File:Pylon.JPG

title: File:Pylon.JPG


text: File:HW6 p159.JPG

title: File:HW6 p159.JPG


text: Wikipedia service-learning courses/Differences in styles in Wikimedia projects


text:
User:EML4500.f08.A-team.VandenBerg/HW7

 Comment: This is HW7 located in the user namespace of EML4500.f08.A-team.VandenBerg 22:09, 20 November 2008 (UTC)

Meeting 35: Monday, 17 November 2008. EML4500
35-1

Dimensions for Element # 1 of the 2 bar system: $$E_{1}^{(1)}=2, E_{2}^{(1)}=4$$

$$A_{1}^{(1)}=0.5, A_{2}^{(1)}=1.5$$

Dimensions for Element # 2 of the 2 bar system:

$$E_{1}^{(2)}=3, E_{2}^{(2)}=7$$

$$A_{1}^{(2)}=1, A_{2}^{(2)}=3$$

- Compute a solution for the 2 bar truss system with the tapered elements.

- Plot the deformed shape for this model using average E and average A.

- Plot the deformed shape for the previous model again using the average E and average A.

Frame element = truss (bar) element + beam element

- Truss bar element is the axial deformation

- Beam element is the transverse deformation

35-2

Model the frame with 2 elements.

Free Body Diagrams:

Circles on elements represent nodes, not pin connections.

35-3

In general, $$d_{i}^{(e)} \rightarrow f_{i}^{(e)}$$

where $$d_{i}^{(e)}$$ are the generalized displacements,

while $$f_{i}^{(e)}$$ are the generalized forces.

$$ e = 1, 2 $$

$$ i = 1,. . ., 6 $$

$$ \begin{Bmatrix} d_{3}^{(e)} \\ d_{6}^{(e)} \end{Bmatrix} $$ (rotational degrees of freedom) $$ \Rightarrow \begin{Bmatrix} f_{3}^{(e)} \\ f_{6}^{(e)} \end{Bmatrix} $$ (bending moments)

Two dimensional frame global degree of freedoms:

2 element stiffness matrix $$\underline{k}_{6x6}^{(e)}, e = 1 , 2 $$

Global stiffness matrix $$ \underline{K}_{9x9} = \overset{e=2}{\underset{e=1}{A}} \underline{k}_{6x6}^{(e)} $$

35-4

Meeting 36: Wednesday, 19 November 2008. EML4500
36-1

$$ \tilde {\mathbf{k}} _ {6 x 6} ^ {(e)}\ \tilde {\mathbf{d}} _ {6 x 1} ^ {(e)}\ =\ \tilde {\mathbf{f}} _ {6 x 1} ^ {(e)}$$

$$ \tilde {\mathbf{d}} ^ {(e)} = \begin{Bmatrix} d_1^{(e)} \\ \vdots \\ d_6^{(e)} \end{Bmatrix} = \tilde {\mathbf{f}} ^ {(e)} = \begin{Bmatrix} f_1^{(e)} \\ \vdots \\ f_6^{(e)} \end{Bmatrix} $$

Note: $$ \tilde {f} _{3}^ {(e)}\ =\ f_3^{(e)}\, \ \tilde {f} _{6}^ {(e)}\ =\ f_6^{(e)} $$

moments (about $$  \tilde {z}\ =\ z) $$

36-2

$$ \begin{matrix} \tilde {d} _1 & \tilde {d} _2 & \tilde {d} _3 & \tilde {d} _4 & \tilde {d} _5 & \tilde {d} _6 & \cdots\\ EA/L & 0 & 0 & -EA/L & 0 & 0 & \tilde {d}_1\\ \cdots & 12EI/L^3 & 6EI/L^2 & 0 & -12EI/L^3 & 6I/L^2 & \tilde {d}_2\\ \cdots & \cdots & 4EI/L & 0 & -6EI/L^2 & 2EI/L & \tilde {d}_3\\ \cdots & \cdots & \cdots & EA/L & 0 & 0 \tilde {d}_4\\ \cdots & \cdots & \cdots & \cdots & 12EI/L^3 & -6EA/L^2 & \tilde {d}_5\\ SYM & \cdots & \cdots & \cdots & \cdots & -EA/L & \tilde {d}_6 \end{matrix} $$

36-3

Aranuas, David

Dimensional Analysis
$$ \left [ \tilde {d}_1 \right ] = L = \left [ \tilde {d}_i \right ] \qquad i=1, 2, 4, 5$$

where,
 * $$ \left [ \tilde {d}_1 \right ] $$ is a dimension of length

and,
 * $$ \left [ \tilde {d}_i \right ] $$ is displacement

$$ \left [ \tilde {d}_3 \right ] = 1 = \left [ \tilde {d}_6 \right ] $$

where,
 * 1 has no dimensions

and,
 * $$ \left [ \tilde {d}_6 \right ] $$ is the rotational displacement

$$ \left [ \theta \right ] = \frac {[AB]}{[R]} = \frac {L}{L} = 1 $$

$$ \sigma = \Epsilon \varepsilon \rightarrow \left [ \sigma \right ] = \left [ \Epsilon \right ] \left [ \varepsilon \right ] = 1 $$

 * Why?

$$ \overline{AB} = R \theta $$

where,
 * $$ \overline{AB} $$ is the arclength

and,
 * $$ \theta $$ is in radians.

Verifying Dimensions and Element Force Displacement Relations
Using the definition shown above, the following dimensional analysis shows the rotational degrees of freedom are dimensionless. $$\left[\theta \right]=\frac{\left[AB \right]}{\left[R \right]}=\frac{L}{L}=1$$

A dimensional analysis can also be done on other important parameters:

$$ \left[\epsilon\right] = \frac{[du]}{[dx]} = \frac{L}{L} = 1 $$

$$ \left[\sigma\right] = \left[E\right] = \frac{F}{L^2} $$

$$ \left[A\right] = L^2, [I] = L^4 $$

$$ \left[\frac{EA}{L}\right] = [\tilde{k}_{11}] = \frac{\frac{F}{L^2}{L^2}}{L} = \frac{F}{L} $$

$$ \left[\tilde{k}_{11}\tilde{d_1}\right] = [\tilde{k}_{11}][\tilde{d_1}] = F $$

$$ \left[\tilde{k}_{23}\tilde{d_3}\right] = [\tilde{k}_{23}][\tilde{d_3}] $$ where $$ \tilde{d_3} = 1 $$

$$ = \frac{\left[6\right][E][I]}{L^2} $$

$$ = \frac{\left(1\right)\left(\frac{F}{L^2}\right)(L^4)}{L^2} = F $$

Verifying Dimensions of $$ \tilde{k}_{ij} $$ and $$ \tilde{d}_{j} $$ for $$ i $$ and $$ j $$ = 1,...,6:
The matrix $$ \tilde{k_{ij}}\tilde{d_j} $$ is shown below

$$ \begin{bmatrix}\frac{EA}{L} & 0 & 0 & \frac{-EA}{L} & 0 & 0 \\ 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} \\ 0 & \frac{6EI}{L^{2}} & \frac{4EI}{L} & 0 & \frac{-6EI}{L^{2}} & \frac{2EI}{L} \\ \frac{-EA}{L} & 0 & 0 & \frac{EA}{L} & 0 & 0 \\ 0 & \frac{-12EI}{L^{3}} & \frac{-6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & \frac{-6EI}{L^{2}} \\ 0 & \frac{6EI}{L^{2}} & \frac{2EI}{L} & 0 & \frac{-6EI}{L^{2}} & \frac{4EI}{L} \end{bmatrix} $$

The dimension of each term in the $$ \tilde{k_{ij}}\tilde{d_j} $$ matrix was calculated just as the above terms were in the previous notes and the resulting matrix is found to be:

$$ \begin{bmatrix}\frac{F}{L} & 0 & 0 & \frac{-F}{L} & 0 & 0 \\ 0 & \frac{F}{L} & F & 0 & \frac{-F}{L} & F \\ 0 & F & FL & 0 & -F & FL \\ \frac{-F}{L} & 0 & 0 & \frac{F}{L} & 0 & 0 \\ 0 & \frac{-F}{L} & -F & 0 & \frac{F}{L} & -F \\ 0 & F & FL & 0 & -F & FL \end{bmatrix} $$

The element force displacement relation in global coordinates from the element force displacement relation in local coordinates yields:

$$ \underline{k}^{(e)}_{6x6}\underline{d}^{(e)}_{6x1} = \underline{f}^{(e)}_{6x1} $$

with $$ \underline{k}^{(e)}_{6x6} = {\underline{\tilde{T}}^{(e)}_{6x6}}^T\underline{\tilde{k}}^{(e)}_{6x6}\underline{\tilde{T}}^{(e)}_{6x6} $$

from $$ \underline{\tilde{k}}^{(e)}_{6x6}\underline{\tilde{d}}^{(e)}_{6x1} = \underline{\tilde{f}}^{(e)}_{6x1} $$

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

The derivation of $$ \tilde{K}^{(e)} $$ from PVW, focusing only on bending effects:

$$ \frac{\partial^{2}}{\partial x^{2}}\left((EI)\frac{\partial^{2}v}{\partial x^{2}}\right) - f_{t}(x) = m(x)\ddot{v} $$

Meeting 38: Monday, 24 November 2008. EML4500
PVW for beams:  $$ \int_{0}^{b} w(\underbrace{x}_{\tilde{x}}) [-\frac{\delta^2}{\delta x^2} ((EI) \frac{\delta^2 v}{\delta x^2}) +ft-m \ddot{v}] \  dx = 0$$ for all possible w(x)

Integration by parts of 1st term

$$ \alpha = \int_{0}^{L} \underbrace{w(x)}_{g(x)} \frac{\delta^2}{\delta x^2}{(EI) \frac{\delta^2 v}{\delta x^2}} \ dx $$

$$ \underbrace{\frac{\delta}{\delta x} (\underbrace{\frac{\delta}{\delta x}(EI) \frac{\delta^2 v}{\delta x^2}}_{r(x)})}_{r'(x)} $$

$$ \alpha = [\underbrace{w \frac{\delta}{\delta x} {(EI) \frac{\delta^2 v}{\delta x^2}}}_{\beta_{1}}]^{2}_{0} - \int_{0}^{L} \underbrace{\frac{dw}{dx}}_{g'(x)} \underbrace{\frac{\delta}{\delta x} {(EI) \frac{\delta^2 v}{\delta x^2}}}_{r(x)} \ dx$$

$$ \alpha = \beta_{1} - [\underbrace{w \frac{\delta}{\delta x} {(EI) \frac{\delta^2 v}{\delta x^2}}}_{\beta_{2}}]^{2}_{0} + \underbrace{\int_{0}^{L} \frac{\delta^2 w}{\delta x^2} (EI) \frac{\delta^2 v}{\delta x^2} \ dx}_{\gamma}$$

Note : symmetric

Eq. (1) becomes: $$ - \beta_{1} + \beta_{2} - \gamma + \int_{0}^{L} wft \ dx - \int_{0}^{L} wm \ddot{v} \ dx = 0 $$ for all possible w(x)

Let's focus on the stiffness term $$ \gamma $$ for now to derive the beam stiffness matrix and to identify the beam shape functions.

$$ v( \tilde{x}) = N_{2} (\tilde{x})\tilde{d}_{2} + N_{3} (\tilde{x})\tilde{d}_{3} + N_{5} (\tilde{x})\tilde{d}_{5} + N_{6} (\tilde{x})\tilde{d}_{6} $$

Recall : $$ u(\tilde{x}) = N_{1} (\tilde{x})\tilde{d}_{1} + N_{4} (\tilde{x})\tilde{d}_{4} $$

p.246:

$$ N_{2} (\tilde{x}) = 1 - \frac{3 \tilde{x}^2}{L^2} + \frac{2 \tilde{x}^3}{L^3} \ \tilde{d}_{2} $$

$$ N_{3} (\tilde{x}) = \tilde{x} - \frac{2 \tilde{x}^2}{L} + \frac{\tilde{x}^3}{L^3} \ \tilde{d}_{3} $$

$$ N_{5} (\tilde{x}) = \frac{3 \tilde{x}^2}{L^2} - \frac{2 \tilde{x}^3}{L^3} \ \tilde{d}_{5} $$

$$ N_{6} (\tilde{x}) = - \frac{\tilde{x}^2}{L} + \frac{\tilde{x}^3}{L^2} $$

Meeting 40: Friday, 5 December 2008. EML4500
Note: Eq. (2) (39-2), dimension analysis:

$$ [u] {=} L $$

(31-3): $$ [N_1] {=} [N_4] {=} 1 $$

$$ Eq. (2) (39-2) \underbrace{[N_1]}_{1} \underbrace{[\tilde{d}_1]}_{L} + \underbrace{[N_4]}_{1} \underbrace{[\tilde{d}_4]}_{L} $$

$$ [V] {=} L $$

$$ \underbrace{[N_2]}_{1} \underbrace{[\tilde{d}_2]}_{L} {=} L $$ (displ. transv.)

$$ \underbrace{[N_2]}_{1} \underbrace{[\tilde{d}_2]}_{L} {=} L $$ (rot.)

$$ {[N_5]}{[\tilde{d}_5]} $$ and $$ {[N_6]}{[\tilde{d}_6]} $$ are to be done for homework

Derivation of the beam shape functions

$$ N_2, N_3, N_5, N_6 $$ (38-4)

p. (38-3) plots

Recall: Governing PDE for beams

p. (37-4), Eq. (1), without $$ f_t $$ (distributed transverse load) and without inertia force $$ m\ddot{v} $$ (static case):

$$ \frac{\delta^2}{\delta x^2} \left \{(EI) \frac{\delta^2v}{\delta x^2}\right \} {=} 0 $$

Further, consider constant EI:

$$ \frac{\delta^4}{\delta x^4} v {=} 0 $$ Integrate 4 times to get 4 constants.

$$ \Rightarrow V(x) {=} C_0 + C_1x^1 + C_2x^2 + C_3x^3 $$

To obtain $$ N_2(x) $$ ($$ \tilde{x} $$ {=} x for simplicity)

V(0) = 1,  V(L) = 0

V'(0)   =   V'(L) = 0

Use above boundary conditions to solve for $$ L_0, ... , L_3 $$

V(0) = 1 = C0

(1) $$ V(L) {=} 1 + C_1(L) + C_2(L)^2 + C_3(L)^3 {=} 0 $$

$$ V^{\prime}(x) {=} C_1 + 2C_2x + 3C_3x^2 $$

$$ V^{\prime}(0) {=} C_1 {=} 0 $$

(2) $$ V^{\prime}(L) {=} 2C_2(L) + 3C_3(L)^2 {=} 0 $$

(2) $$ \rightarrow C_3 {=} - \frac{2}{3} \frac{C_2}{L} $$

(1) $$ \rightarrow 0 {=} 1 + C_2L^2 + (\frac{-2}{3} \frac{C_2}{L})L^3 {=} 1 + C_2L^2[1 - \frac{2}{3}] $$

$$ C_2 {=} - \frac{3}{L^2} $$

$$ C_3 {=} - \frac{2}{3} \frac{1}{L}(- \frac{3}{L^2}) {=} \frac{2}{L^3} $$

Compare with expression for N2 on p. (38-4)

For N3: boundary conditions         $$ \tilde{d}_3 $$ rotation

V(0) = V(L) = 0

V'(0) = 1, V'(L) = 0

For N5: V(0) = 0, V(L) = 1    $$ \tilde{d}_5 $$ displacement

V'(0) = 0, V'(L) = 0

For N6: $$ d_6 $$ rotation

V(0) = V(L) = 0

V'(0) = 0, V'(L) = 1

See p. (39-1) for plots of N5, N6

Derive coefficient in $$ \underline{\tilde{k}} $$ (elem. stiffness matrix)

Coefficient for EA: Done

Coefficient for EI: To Be Done

$$ \tilde{k}_{22} {=} \frac{12EI}{L^3} \int^L_0 \frac{d^2N_2}{dx^2}(EI)\frac{d^2N_2}{dx^2}dx $$

Vibrational Analysis Using FEA
$$\hat{k}_{23} = \frac{6EI}{L^2} = \int_{0}^{L}{\frac{d^2N_2}{dx^2}(EI)\frac{d^2N_3}{dx^2}dx}$$

In general,

$$\hat{k}_{ij} = \int_{0}^{L}{\frac{d^2N_i}{dx^2}(EI)\frac{d^2N_j}{dx^2}dx}$$

With i,j = 2,3,5,6

Elastodynamics (trusses, 2-D frames, 3-D elasticity)

Using modal problem from lecture 31 using discrete Principle of Virtual Works:

$$ \mathbf{\bar{w}} = [\mathbf{\bar{M}}\mathbf{\ddot{\bar{d}}} + \mathbf{\bar{K}}\mathbf{\bar{d}} - \mathbf{\bar{F}}] = 0$$

In the above equation, the boundary conditions are already applied

The following equation is true for all $$\mathbf{\bar{w}}$$:

$$\mathbf{\bar{M}}\mathbf{\ddot{\bar{d}}} + \mathbf{\bar{k}}\mathbf{\bar{d}} = \mathbf{\bar{F}}(t)$$

where $$\mathbf{\bar{d}}(0) = \mathbf{\bar{d}}_0$$

and $$\mathbf{\dot{\bar{d}}}(0) = \mathbf{\bar{V}}_0$$

The above will be referred to as (1)

These are the complete ordinary differential equations (ODEs), second order in time, and initial conditions governing the elastodynamics of the discretized continuous problem with multiple degrees of freedom.

Solving equation (1):

1) Consider the unforced vibrations problem:
$$\mathbf{\bar{M}}_{nxn}\mathbf{\ddot{v}}_{nx1} + \mathbf{\bar{K}}_{nxn}\mathbf{v}_{nx1} = \mathbf{0}_{nx1}$$

Unforced meaning the equation is equal to zero. This will be referred to as equation (2)

Assume: $$\mathbf{v}(t)_{nx1} = (sinwt)\mathbf{\phi _{nx1}}$$

Where the phi matrix is not time dependent

Thus: $$\mathbf{\ddot{v}} = -\omega ^2 sin\omega t\mathbf{\phi }$$

This means equation (2) becomes:

$$-\omega ^2 sin\omega t\mathbf{\bar{M}}\mathbf{\phi } + \omega ^2 sin\omega t\mathbf{\bar{K}}\mathbf{\phi } = \mathbf{0}$$

Therefore:

$$\Rightarrow \mathbf{\bar{k}}\mathbf{\phi } = \omega ^2\mathbf{\bar{M}}\mathbf{\phi }$$

Which is the generalized eigenvalue problem, as it is of the form:

$$\mathbf{A}\mathbf{x} = \lambda \mathbf{B}\mathbf{x}$$

With lambda as the eigenvalue.

A standard eigenvalue problem: $$\mathbf{A}\mathbf{x} = \lambda \mathbf{x}$$ is given when the B matrix is equal to the identity matrix.

This means that

$$\lambda =\omega ^2$$ is an eigenvalue.

$$(\lambda _i,\mathbf{\phi _i})$$ are eigenpairs.

with i = 1 through n

Now the mode i for the animation can be represented as:

$$\mathbf{v}_i(t) = (sinw_it)\mathbf{\phi _i}$$

for i = 1 through n

2) Modal superposition method:
Using orthogonal properties of the eigenpairs:

$$\mathbf{\phi _i}^T_{1xn}\mathbf{\bar{M}}_{nxn}\mathbf{\phi }_{nx1} = \delta _{ij} = \begin{cases} & \text{1 if } i=j \\ & \text{0 if } i\neq j \end{cases}$$

This delta is the Kronecker delta.

Mass orthogonality of the eigenvector.

Now, applying this to Eq (1) and (2) gives:

$$\mathbf{\bar{M}}\mathbf{\phi }_j = \lambda _j \cdot \mathbf{\bar{k}}\mathbf{\phi }_j$$

$$ \mathbf{\phi }_i^T\mathbf{\bar{M}}\mathbf{\phi }_j = \lambda _j\phi _i^T\mathbf{\bar{k}}\mathbf{\phi }_j$$

Therefore:

$$\Rightarrow \phi _i^T\mathbf{\bar{k}}\mathbf{\phi }_j = \frac{1}{\lambda _j}\delta _{ij}$$

Equation (1) can be written as:

$$\mathbf{\bar{M}}(\sum_{j}^{}{\ddot{\zeta }_j\mathbf{\phi }_j}) + \mathbf{\bar{K}}(\sum_{j}^{}{\zeta _j\mathbf{\phi }_j}) = \mathbf{F}$$

In the above equation, the first term in parenthesis is equal to the second derivative of the d matrix, while the second term in parenthesis is equal to the d matrix.

We can also write:

$$\sum_{j}^{}{\ddot{\zeta }_j(\mathbf{\phi }_i^T\mathbf{\bar{M}}\mathbf{\phi }_j} + \sum_{j}^{}{\zeta _j(\mathbf{\phi }_i^T\mathbf{\bar{K}}\mathbf{\phi }_j} = \mathbf{\phi }_i^T\mathbf{F}$$

The first term in parenthesis in this equation is equal to the Kronecker delta, while the second term in parenthesis is equal to an eigenvalue times the Kronecker delta.

Finally,

$$\Rightarrow \ddot{\zeta } + \lambda _i\zeta _i = \mathbf{\phi }_i^T\mathbf{F}$$

with i = 1 through n

MediaWiki VS. WebCT
It is the opinion of this group that neither of these options for completing homework are very appropriate. MediaWiki, while fancy looking and user friendly to viewers, can consume a large portion of time simply in writing the page. This consumption of time takes away from the learning experience in FEA, since FEA is not a web design course. MediaWiki has other problems also, such as having to deal with moderators, administrators, and others who may, intentionally or unintentionally, and with or without malice, change the content of a student's page. WebCT does not have this problem however, but WebCT is not much more viable, it is difficult to work on group projects, and those with older computers or those who devote too much memory to other applications on their computers can find the system slow and prone to crashing. Also, the proposed idea of the student's work lasting far into the future is somewhat ludicrous, since unless that student constantly checks the site, it will most likely become prone to vandals. If a student wishes to have his/her work last into the far future, it would be more logical for them to save their work on their computer, or a flash drive, or even in a filing cabinet. Those student's who would like to be recognized for their work or have their work published for all to see would be advised to contact the appropriate journal/magazine and submit their work to them. It is this groups' conclusion that more time has been spent writing code for MediaWiki than studying the material, which is an inappropriate use of time. Also, receiving/checking grades has become more of a hassle than it should be, and figuring out what was marked wrong and right is practically impossible, considering the only guide the student is given is a selection of other student's homework that is "better" than their own, but it is not known in what way it is "better" or whether or not it is entirely accurate. There lies the possibility that one group's homework may be academically more accurate, but misses the aesthetic quality of another group's work, and is therefor marked with a lower grade.

It is our opinion that neither MediaWiki nor WebCT is an appropriate way to submit homework for this class.

MATLAB PROBLEMS
Here is the Matlab code for the Tapered Two Bar Truss System:

Here is the code for the Two Bar Frame/Truss System:

Here is the code for the Electric Pylon System:

Results are as follows:

Highest Bending Moment was 6.2874 in Element 81 Highest Transverse Shear Stress was 5.4827 in Element 81 The first three eigens were 132.86, 2471.2, and 2942.6 occurring in elements 79, 81, and 82, respectively. The three lowest vibrational periods were 0.54511, 0.12639, and 0.11583

The frame model is not statically determinate since there are far too many unknowns and not enough equations.

The reactions for the frame model are the same as those for the truss model, with the exception of an additional two reactions that relate to the rotations at the nodes.

Below are the M-file functions used to complete the previous Matlab codes:

Contributions
--EML4500.f08.A-team.VandenBerg 22:12, 20 November 2008 (UTC)

--EML4500.f08.A-team.rieth 19:11, 3 December 2008 (UTC)

--EML4500.f08.A-team.melvin 01:26, 9 December 2008 (UTC)

--EML4500.f08.A-team.kirley 02:50, 9 December 2008 (UTC)

--Eml4500.f08.a-team.robinson 01:12, 9 December 2008 (UTC)

--EML4500.f08.A-team.morford 20:11, 9 December 2008 (UTC)

title: User:EML4500.f08.A-team.VandenBerg/HW7


text:
User:Eml4500.f08.Lulz.js/HW6

PVW for Elastic Bar

Continuing with the discussion of the Principle of Virtual Work (PVW) for the dynamics of the elastic bar shown in Figure ##, consider the partial differential equation (PDE) describing the continuous case equation of motion (EOM) for this bar:

 * $$\frac{\partial}{\partial x}[(EA)\frac{\partial u}{\partial x}] + f = m \ddot{u}$$

Deriving the discrete case EOM for the bar from the continuous EOM yields the following:

 * $$- \mathbf{kd} + \mathbf{F} = \mathbf{M} \mathbf{\ddot{d}} => \mathbf{M}\mathbf{\ddot{d}} + \mathbf{kd} = \mathbf{F}$$

Where $$\mathbf{M}\;$$ is the mass matrix. The presence of the mass matrix indicates a multiple degree of freedom system. A similar equation for the single degree of freedom system is given as:

 * $$m\ddot{d} + kd = F$$

The following integral equation was used in the derivation of the discrete EOM from the continous EOM:

 * $$\int^{1}_{0} w(x) \left\{\frac{\partial}{\partial x} [EA \frac{\partial u}{\partial x}] + f - m \ddot{u}\right\} dx = 0$$ for all possible values of $$w(x)\;$$, the weighting function.

Deriving this equation from the continous EOM is a trivial, but to get the continous EOM from this equation is not. In order to show this, let the integral equation be rewritten as follows:

 * $$\int w(x) g(x) dx = 0$$ for all $$w(x)\;$$

Since this equation should hold for all $$w(x)\;$$, selecting $$w(x)\;$$ to be equal to $$g(x)\;$$ is a valid operation. This results in the following:

 * $$\int g^{2} dx = 0$$

From here, it is obvious that $$g(x) = 0\;$$.

 * {| class="collapsible collapsed"

!Integration by Parts Consider two functions $$r(x)\;\;$$ and $$g(x)\;$$. The derivative of the product of these two functions is as follows:
 *
 *

 * $$(rs)^{'} = r^{'}s + rs^{'}\;$$

Where $$r^{'} = \frac{\partial r}{\partial x}$$ and $$s^{'} = \frac{\partial s}{\partial x}$$ Taking the integral of this expression yields the following:

 * $$\int (rs)^{'} =\int r^{'}s + \int rs^{'}\;$$

Noting that the left hand side of this equation is equivalent to $$rs\;$$, with rearranging the following well-known relation can be determined:

 * $$\int r^{'}s = rs - \int rs^{'}\;$$

 * }

Using the integration by parts method on the first term of the continuous EOM, with $$ r(x) = {EA} \frac{\partial u}{\partial x}\;$$ and $$s(x) = w(x)\;$$ yields the following:

 * $$ \int^{L}_{0} w(x) \frac{\partial}{\partial x} \left[(EA) \frac{\partial u}{\partial x}\right] dx = \left[w(EA) \frac{\partial u}{\partial x}\right]^{L}_{0} - \int^{L}_{0} \frac{\partial w}{\partial x}(EA) \frac{\partial u}{\partial x} dx$$

 * $$= \underbrace{w(L)(EA)(L)\frac{\partial u(L,t)}{\partial x}}_{N(L,t)} - \underbrace{w(0)(EA)(0)\frac{\partial u(0,t)}{\partial x}}_{N(0,t)} - \int^{L}_{0} \frac{\partial w}{\partial x}(EA) \frac{\partial u}{\partial x} dx$$

Model Problem
Consider the model problem shown in Figure ##. The boundary conditions for this problem can be found in Homework 5. Additionally, select $$w(x)\;$$ such that $$w(0)=0\;$$; this makes the system "kinematically admissable".

Applying the discrete PVW to a previously studied equation from page 10-1 of the notes yields the following relation:

 * $$\mathbf{W}(\mathbf{K}\begin{Bmatrix} d_{3}\\d_{4}\end{Bmatrix} - \mathbf{F}) = \mathbf{0}$$ for all $$\mathbf{W}$$.

Where $$\mathbf{F}^{T} = \begin{bmatrix}F_1 & F_2 & F_3 & F_4 & F_5 & F_6\end{bmatrix}$$. Since $$\mathbf{W}$$ can be arbitrarily selected, select it such that $$W_1 = W_2 = W_5 = W_6 = 0\;$$. This will eliminate equations involving unknown reactions, eliminating rows 1,2,5, and 6 from the two-bar truss system ans yielding the following relation:

 * $$\mathbf{\bar{K}}\underbrace{\mathbf{\bar{d}}}_{\begin{Bmatrix} d_{3}\\d_{4}\end{Bmatrix}}=\underbrace{\mathbf{\bar{F}}}_{\begin{Bmatrix} F_{3}\\F_{4}\end{Bmatrix}}$$

Note also that $$\mathbf{\bar{W}} \cdot \mathbf{\bar{K}}\mathbf{\bar{d}}=\mathbf{\bar{F}}$$ for all $$\mathbf{\bar{W}}$$, where $$\mathbf{\bar{W}} = \begin{Bmatrix} W_{3}\\W_{4}\end{Bmatrix}$$.

Returning to the continuous case, consider the following unknown reaction:

 * $$N(0,t) = (EA)(0)\frac{\partial u}{\partial x} (0,t)$$

This equation can be written as:

 * $$W(L)F(t) - \int^{L}_{0} \frac{\partial w}{\partial x}(EA) \frac{\partial u}{\partial x} dx + \int^{L}_{0}w(x)[f-m\ddot{u}]dx = 0$$ for all $$w(x)\;$$ such that $$w(0) = 0\;$$.

The weak form of continuous PVW can be expressed as:

 * $$\int^{L}_{0}W(m\ddot{u})dx + \int^{L}_{0} \frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx = W(L)F(t) + \int^{L}_{0} Wfdx$$ for all $$w(x)\;$$ such that $$w(0) = 0\;$$.

Comparison Between Continuous and Discrete PVW

 * {| border=1

 * bgcolor="#99ccff"|
 * bgcolor="#99ccff" style="text-align: center;" |Continuous Setting (PVW)
 * bgcolor="#99ccff" style="text-align: center;" |Discrete Setting (PVW)
 * bgcolor="#05fa67"|Inertia
 * bgcolor="#ffffff" style="text-align: center;" | $$\int^{L}_{0}wm\ddot{u}dx$$
 * bgcolor="#ffffff" style="text-align: center;" | $$\mathbf{\bar{W}} \cdot (\mathbf{\bar{M}}\mathbf{\ddot{\bar{d}}})$$
 * bgcolor="#05fa67"|Stiffness
 * bgcolor="#ffffff" style="text-align: center;" | $$\int^{L}_{0} \frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx$$
 * bgcolor="#ffffff" style="text-align: center;" | $$\mathbf{\bar{W}} \cdot (\mathbf{\bar{k}}\mathbf{\bar{d}})$$
 * bgcolor="#05fa67"|Applied Force
 * bgcolor="#ffffff" style="text-align: center;" | $$W(L)F(t) + \int^{L}_{0}wfdx$$
 * bgcolor="#ffffff" style="text-align: center;" | $$\mathbf{\bar{W}} \cdot \mathbf{\bar{F}}$$
 * bgcolor="#05fa67"|Conditions
 * bgcolor="#ffffff" style="text-align: center;" | all $$w(x)\;$$ such that $$w(0)=0\;$$
 * bgcolor="#ffffff" style="text-align: center;" | for all $$\mathbf{\bar{W}}\;$$
 * }
 * bgcolor="#05fa67"|Conditions
 * bgcolor="#ffffff" style="text-align: center;" | all $$w(x)\;$$ such that $$w(0)=0\;$$
 * bgcolor="#ffffff" style="text-align: center;" | for all $$\mathbf{\bar{W}}\;$$
 * }
 * }

Figures ## and ## describe the interpolation of displacements along the elastic bar. Figure ## shows the elastic bar segmented by nodes, whereas Figure ## shows a close-in view of the ith segment of the bar. Assume the displacement $$u(x)\;$$ for $$x_i \leq x \leq x_i + 1$$, which can also be written as $$(x \in [x_i,x_i+1])$$.

Motivation for Linear Interpolation of u(x)

Consider the deformed truss element seen in Figure ##. The deformed shape is a straight line; i.e., there was an implicit assumption of linear interpolation of displacement between 2 nodes.

Next, consider the case where there are only axial displacements. Expressing $$u(x)\;$$ in terms of $$d_i = u(x_i)\;$$ and $$d_{i+1} = u(x_{i+1})\;$$ as a linaer function of $$\;x$$ (i.e, performing linear interpolation) is as follows:

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

This relationship can be seen graphically in Figure ##. $$N_{i+1}(x)\;$$ can be expressed as $$\frac{x - x_i}{x_{i+1} - x_i}$$.

Lagrange Interpolation
The motivation for the form of $$N_i(x)\;$$ and $$N_{i+1}(x)\;$$ is as follows:

1) $$N_i(x)\;$$ and $$N_{i+1}(x)\;$$ are linear; thus, any linear combination of $$N_i\;$$ and $$N_{i+1}$$ is also linear, and in particular the previously discussed expression for $$u(x)\;$$. This relationship can be further expressed as:

 * $$N_i(x) = \alpha_i + \beta_ix\;$$

 * $$N_{i+1}(x) = \alpha_{i+1} + \beta_{i+1}x\;$$

Where $$\alpha_i,\alpha_{i+1},\beta_i,\;$$ and $$\beta_{i+1}\;$$ are real numbers. A linear combination of these numbers can be written as follows:

 * $$N_id_i + N_{i+1}d_{i+1} = (\alpha_i + \beta_ix)d_i + (\alpha_{i+1} + \beta_{i+1}x)d_{i+1}\;$$

 * $$= (\alpha_i d_i + \alpha_{i+1} d_{i+1}) + (\beta_i d_i + \beta_{i+1} d_{i+1})x\;$$

It is clear that this relation is a function in x.

2) Recall the previously discussed equation for $$u(x_i)\;$$:

 * $$u(x_i) = \underbrace{N_i(x_i)}_{1}d_i + \underbrace{N_{i+1}(x_i)}_{0}d_{i+1}\;$$
 * $$ = d_i\;$$

Determination of Stiffness Matrix for Element i
Applying the interpolation process for w(x) yields the following relation:

 * $$w(x) = N_i(x)W_i + N_{i+1}(x)W_{i+1}\;$$

This equation can be employed to determine the elemental stiffness matrix for the element i. Consider the following integral equation for $$\beta \;$$:

 * $$\beta = \int^{x_{i+1}}_{x_i}\left[N_i^'W_i + N_{i+1}^'W_{i+1}\right](EA)\left[N_i^'d_i + N_{i+1}d_{i+1}\right]dx$$

Where $$N_i^' := \frac{\partial N_i(x)}{\partial x}$$, and similarly for $$N_i^'d_i\;$$. Note also the following two definitions:

 * $$u(x) = \underbrace{\begin{bmatrix}N_i(x) & N_{i+1}(x)\end{bmatrix}}_{\mathbf{N}(x)}\begin{Bmatrix}d_i\\d_{i+1}\end{Bmatrix}$$

 * $$\frac{du(x)}{dx} = \underbrace{\begin{bmatrix}N_i^'(x) & N_{i+1}^'(x)\end{bmatrix}}_{\mathbf{B}(x)}\begin{Bmatrix}d_i\\d_{i+1}\end{Bmatrix}$$

Similarly,

 * $$W(x) = \mathbf{N}(x)\begin{Bmatrix}W_i\\W_{i+1}\end{Bmatrix}$$

 * $$\frac{dW(x)}{dx} = \mathbf{B}(x)\begin{Bmatrix}W_i\\W_{i+1}\end{Bmatrix}$$

Recall the element degrees of freedom as shown in Figure ##. These degrees of freedom can be summarized as:

 * $$\begin{Bmatrix}d_i \\ d_{i+1}\end{Bmatrix} = \begin{Bmatrix}d_1^{(i)} \\ d_2^{(i)}\end{Bmatrix} = \mathbf{d}^{(i)}$$

Similarly for $$\mathbf{W}^{(i)}\;$$,

 * $$\begin{Bmatrix}W_i \\ W_{i+1}\end{Bmatrix} = \begin{Bmatrix}W_1^{(i)} \\ W_2^{(i)}\end{Bmatrix} = \mathbf{W}^{(i)}$$

Substituting the above relations into the integral equation for $$\beta \;$$ yields the following:

 * $$\beta = \int^{x_{i+1}}_{x_i}\left(\mathbf{BW}^{(i)}\right)(EA)\left(\mathbf{Bd}^{(i)}\right)dx = \mathbf{W}^{(i)} \cdot (\mathbf{k}^{(i)}\mathbf{d}^{(i)})$$

Since all of the terms in the integral are constants (1x1 matrices), they can be rearranged as follows:

 * $$\beta = \int^{x_{i+1}}_{x_i}(EA)\left(\mathbf{BW}^{(i)}\right) \cdot \left(\mathbf{Bd}^{(i)}\right)dx = \underbrace{\left(\mathbf{BW}^{(i)}\right)^T}_{\underbrace{\mathbf{W}^{(i)T}\mathbf{B}^T}_{\mathbf{W}^{(i)} \cdot \mathbf{B}^T}}\left(\mathbf{Bd}^{(i)}\right)$$

Finally, $$ \beta \;$$ can be written as:

 * $$\beta = \mathbf{W}^{(i)} \cdot \left(\int \mathbf{B}^T (EA) \mathbf{B}dx\right)\mathbf{d}^{(i)}$$

This leads to the following expression for the element stiffness matrix $$\mathbf{k^{(i)}}\;$$:

 * $$\mathbf{k}^{(i)} = \int^{x_{i+1}}_{x_i} \mathbf{B}^T(x) (EA)(x) \mathbf{B}(x)dx$$

Note also the following definition for $$L^{(i)}$$:

 * $$L^{(i)} = x_{i+1} = x_{i}\;$$

Also, note the transformation of coordinates from $$x\;$$ to $$\tilde{x}\;$$:

 * $$\tilde{x} := x - x_i\;$$

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

This leads to the following version of the element stiffness matrix:

 * $$\mathbf{k}^{(i)} = \int^{\tilde{x}=L^{(i)}}_{\tilde{x}=0} \mathbf{B}^T(\tilde{x}) (EA)(\tilde{x}) \mathbf{B}(\tilde{x})d\tilde{x}$$

Consider the an ith element as shown in Figure ##. The relations for area and Young's modulus as a function of $$\tilde{x}\;$$ can be written as follows:

 * $$A(\tilde{x}) = N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2$$

 * $$E(\tilde{x}) = N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2$$

Comparison of Element Stiffness Matrix Determination Methods
The shape function $$N_2^{(i)}(\tilde{x})\;$$ can be written as follows:

 * $$N_2^{(i)}(\tilde{x}) = \frac{\tilde{x}}{L^{(i)}}= \left\{\begin{matrix}0\;at\;\tilde{x} = 0 \\ 1\;at\;\tilde{x} = L^{(i)}\end{matrix}\right.$$

Following the book on page 159, setting $$E_1 = E_2 = E\;$$ and letting $$A(\tilde{x})\;$$ be linear, it is possible to obtain $$\mathbf{k}^{(i)}$$ and compare to the equation in the book:

 * $$\frac{E}{L^{(i)}}\frac{(A_1 + A_2)}{2}\begin{bmatrix}1&-1\\-1&1\end{bmatrix} = \mathbf{k}^{(i)}$$

Where the quantity $$\frac{(A_1 + A_2)}{2}\;$$ is the average area. A graphical view of this concept is given in Figure ##. Next, it is useful to compare the general $$\mathbf{k}^{(i)}\;$$ to a stiffness matrix determined using $$.5(A_1+A_2)\;$$ and $$.5(E_1+E_2)\;$$, where $$E_1 \neq E_2\;$$. This yields the following relation for $$\mathbf{k}^{(i)}_{ave}\;$$:

 * $$\frac{(E_1+E_2)(A_1+A_2)}{4}\begin{bmatrix}1&-1\\-1&1\end{bmatrix} = \mathbf{k}^{(i)}_{ave}$$

The comparison in question is simply the difference $$\mathbf{k}^{(i)} - \mathbf{k}^{(i)}_{ave}$$.

[[Image:Lulz6.9.png|thumb|right|200px|Fig #. Centroid of Area]

 * {| class="collapsible collapsed"

!Mean Value Theorem Recall the Mean Value Theorem(MVT) and its relation to the centroid:
 *
 *

 * $$\int^{x=a}_{x=b}f(x)dx = f(\bar{x})[b-a]\;$$

for $$\bar{x} \in [a,b]$$. For the centroid of an area, as given in Figure ##, the MVT can be expressed as follows.

 * $$\int_A xdA = \bar{x} \int_A dA = \bar{x}A$$

Applying the MVT to a product of two functions produces a similar result:

 * $$\int^{x=a}_{x=b}f(x)g(x)dx = f(\bar{x})g(\bar{x})[b-a]\;$$

However, it should be noted that $$f(\bar{x})$$ and $$g(\bar{x})$$ are not equivalent to the average values of those functions. This property is shown in equation form below:

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

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

 * }

title: User:Eml4500.f08.Lulz.js/HW6


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User:Eml4500.f08.gravy.bje/Notes6

Calculation of first term of B(x)
$$ \frac{du(x)}{dx} = \begin{bmatrix} N_i^'(x) & N_{i+1}^'(x) \end{bmatrix} \begin{Bmatrix} d_i \\ d_{i+1} \end{Bmatrix}$$

$$B(x) = \begin{bmatrix} N_i^'(x) & N_{i+1}^'(x) \end{bmatrix}$$

$$ \frac{dW(x)}{dx} = B(x) \begin{Bmatrix} W_i \\ W_{i+1} \end{Bmatrix}$$

$$B(x) = \begin{bmatrix} 1+W & ... \end{bmatrix}$$

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

$$\underline{k}^{(i)}= \begin{Bmatrix} \int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}} AE \frac{1}{L^2} dx & -\int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}} AE \frac{1}{L^2} dx \\ -\int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}} AE \frac{1}{L^2} dx & \int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}} AE \frac{1}{L^2} dx \end{Bmatrix}$$

$$\underline{k}^{(i)}= \frac{A(\tilde{x})E(\tilde{x})}{L} \begin{Bmatrix} \int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}}dx & -\int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}}dx \\ -\int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}}dx & \int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}}dx \end{Bmatrix}$$

$$\underline{k}^{(i)}= \frac{A(\tilde{x})E(\tilde{x})}{L} \begin{Bmatrix} L^{(i)} & -L^{(i)} \\ -L^{(i)} & L^{(i)} \end{Bmatrix}$$

Where:

$$ A(\tilde{x}) = N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2$$

$$ E(\tilde{x}) = N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2$$

Therefore:

$$\underline{k}^{(i)}_{2x2} = \frac{(N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2)(N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2)}{L} \begin{Bmatrix} L^{(i)} & -L^{(i)} \\ -L^{(i)} & L^{(i)} \end{Bmatrix}$$

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Wikipedia service-learning courses/200

Importance of citing
Verifiability is one ot the three fundamental policies in Wikimedia.
 * w:Wikipedia:Verifiability

Overview of citable references
In general, the most reliable sources are peer-reviewed journals and books published in university presses; university-level textbooks; magazines, journals, and books published by respected publishing houses; and mainstream newspapers. As a rule of thumb, the greater the degree of scrutiny involved in checking facts, analyzing legal issues, and scrutinizing the evidence and arguments of a particular work, the more reliable it is.

Where can I find a more detailed guide on reliable sources?
 * w:Wikipedia:Reliable_sources

Where can I ask and discuss if a specific source is reliable?
 * w:Wikipedia:Reliable sources/Noticeboard

 * Concept check
 * Why should I add a page number when citing books?

Overview of copyrights
Can I copy material from the wikipedia?
 * Generally yes with several obligations: w:Wikipedia:Copyrights
 * You do not need to ask the wikimedia foundation.
 * You do not need to ask the authors.
 * You should, however, attribute wikipedia, i.e. refer to the article.
 * When you publish material from the wikipedia you must ensure that its also freely accessible.
 * An exceptions is the material that has been published using the "fair use" principle, which should be indicated in the article or media.

Where can I find a general overview about what is regulated in copyrighs?
 * w:Wikipedia:FAQ/Copyright
 * License your work - Creative Commons

Where can I find more detailed information on copyrights in wikipedia?
 * w:Wikipedia:Copyrights
 * w:Wikipedia:Copyright_FAQ
 * w:Wikipedia:Non-free use rationale guideline

How does the copyright of other wikimedia projects differ from wikipedia?
 * The copyright in other wikimedia projects is essentially the same.
 * Differences
 * so far no reader of this course has noted differences.
 * Wikiversity pages
 * Copyrights
 * Verifiability
 * Wikibooks pages
 * b:Wikibooks:Copyrights
 * b:Wikibooks:Verifiability

 * Concept check
 * When writing wikipedia articles, can you cite other wikipedia articles?
 * Can you cite non-freely available material?
 * Can you include non-freely material in wikipedia articles?
 * Can you include non-freely available material in Wikimedia Commons, and in Wikibooks?
 * Will you get attribution if you write a page in a wikimedia project?

 * See also
 * Address copyright requirements

Wiki format of citing references
How do I include references in content pages?
 * w:Wikipedia:Cite_sources

How should I format the references?
 * generell rule: "must contain enough information for other editors to identify the specific published work you used"
 * w:WP:CITE

Where can I find tools to generate the correct reference formats?
 * w:Wikipedia:Citation_tools

How can I cite a web page and link to its current state?
 * w:Wikipedia:Citing_sources/Further_considerations

 * Concept check
 * Why is it better to cite in a wrong format instead of not citing at all?

 * Assignment
 * 1) Use the tool Google Scholar reference generator to search for the paper Nash 1950 and generate a reference in wiki format
 * 2) Edit your My first page to cite this reference in all of the four commonly used forms of citation.
 * 3) Edit your My first page to cite the online learning course in its current version using the WebCite archive page.

 * See also
 * Learning project Citing Sources

title: Wikipedia service-learning courses/200


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User:Eas4200c.f08.vqcrew.E/Homework6

Traction Force
The traction force is defined as the product of the components of the stress tensor matrix, $$ \begin{bmatrix}\sigma\end{bmatrix} \;$$, and the components of the normal vector, $$  \begin{Bmatrix}n\end{Bmatrix} \;$$.

This relation can be derived using Figures #-#. The normal vector, $$ \begin{Bmatrix}n\end{Bmatrix} \;$$, is a unit vector with magnitude equal to one.

 * $$ \begin{Bmatrix}t\end{Bmatrix}_{3x1} = \begin{bmatrix}\sigma\end{bmatrix}_{3x3} \begin{Bmatrix}n\end{Bmatrix}_{3x1} \;$$, where $$\begin{Vmatrix}n\end{Vmatrix} = 1\;$$

The traction force can then broken into its components in the y and z directions.

 * $$\vec{t} = t_y\vec{j} + t_z\vec{k}\;$$

 * $$\vec{n} = n_y\vec{j} + n_z\vec{k}\;$$

The differential lengths of the section and the components of the normal vector can be found using the surface length, ds, and the angle between the normal vector and the y-axis, $$\theta\;$$.

 * $$dz = ds(cos\theta)\;$$

 * $$dy = ds(sin\theta)\;$$

 * $$n_y = cos\theta\;$$

 * $$n_z = sin\theta\;$$

The sum of the forces in the y and z directions must be zero, so an expression relating the traction force, normal force and stresses can be derived. The depth of the section is set to be equal to one. Below is the development for the component of the traction force in the y-direction. The development of the component in the z-direction is reserved for the homework section.

 * $$\sum F_y = 0 = -\sigma_{yy}(dz)(1) - \sigma_{yz}(dy)(1) + t_y(ds)(1)\;$$

 * $$ 0 = -\sigma_{yy}(ds)(n_y) - \sigma_{yz}(ds)(n_z) + t_y(ds)\;$$

 * $$t_y = \sigma_{yy}(n_y) + \sigma_{yz}(n_z)\;$$

The components of the traction force can be written in matrix form.

 * $$\begin{Bmatrix}t_y\\t_z\end{Bmatrix} = \begin{bmatrix}\sigma_{yy} & \sigma_{yz}\\ \sigma_{zy} & \sigma_{zz}\end{bmatrix}\begin{Bmatrix}n_y\\ n_z \end{Bmatrix}\;$$

The three dimensional case can be expanded from the two dimesional case.

 * $$\begin{Bmatrix}t_1\\t_1\\t_3\end{Bmatrix} = \begin{bmatrix}\sigma_{11} & \sigma_{12} & \sigma_{13}\\ \sigma_{11} & \sigma_{12} & \sigma_{13}\\ \sigma_{11} & \sigma_{12} & \sigma_{13}\end{bmatrix}\begin{Bmatrix}n_1\\ n_2\\n_3 \end{Bmatrix}\;$$

This result then can be compressed into a compacet notation. This brings the derivation back to the origin point with the original definition of the traction force in a general notation.

 * $$t_i = \sum_{j=1}^3\sigma_{ij}n_j$$ where i = 1, 2, 3

 * $$\begin{Bmatrix}t_i\end{Bmatrix}_{3x1} = \begin{bmatrix}\sigma_{ij}\end{bmatrix}_{3x3} \begin{Bmatrix}n_{j}\end{Bmatrix}_{3x1}\;$$

title: User:Eas4200c.f08.vqcrew.E/Homework6


text:
User:Eml4500.f08.bottle.loschak/HW6

HW 6
Equation (1), a partial differential equation (PDE), was given in lecture 29 and used to derive Equation (2).

$$ \frac{\partial }{\partial x}\left[(EA)\frac{\partial u}{\partial x} \right] + f = m\ddot{u} $$             (1)

$$ \mathbf{M\ddot{d}}+\mathbf{kd} = \mathbf{F} $$             (2)

The partial differential equation, Equation (1), represents a summation of forces for infinitesimally thin sections dx over the bounds 0 to length L. Each of the terms in Equation (1) can be represented discretely by a matrix of individual points. Equation (1) is just the continuous expression of the same thing. Therefore, every $$m\ddot{u}$$ between 0 and L added up continuously will have very similar results as the matrix expression $$\mathbf{M\ddot{d}}$$. The same is true for the other terms as well.

Knowing that Equations (1) and (2) are the continuous and discrete versions of the same system we can derive the continuous Equation (3) and use the discrete case to verify that our process is correct.

First, we know that $$\int w(x)g(x)dx = 0$$ for all $$w(x)$$. In the case that $$w(x) = g(x)$$, our above equation becomes $$\int g^{2}(x)dx$$. As long as $$g(x) \geq 0$$ then $$g(x) = 0$$.

Therefore, since Equation (1) can also be rearranged to equal zero, Equation (3) is true.

$$ 0 = \int_{x = 0}^{x = L}{W(x) \left\{ \frac{\partial }{\partial x}\left[ EA\frac{\partial u}{\partial x}\right] + f - m\ddot{u} \right\} dx} $$             (3)

To verify this process, we recall a proof for discrete matrices from lecture 24 that was used to justify weighting matrices.

For the case where

$$\textbf{W}_{6x1}\cdot (\textbf{Kd}-\textbf{F})_{6x1}=0_{1x1}$$

We use the following proof from Team Bottle HW 5.

Proof: Select W at $$ \textbf{W}^{T}_{1x6}=\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}_{1x6}\ $$ plugging W back into the Principle of Virtual Work equation gives: $$ \begin{bmatrix}1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 1 \cdot \bigg[ \sum_{j=1}^{6}{K_{1j}d_{j}} - F_{1} \bigg]+ 0\cdot \bigg[\sum_{j=1}^{6}{K_{2-6j}d_{j}} - F_{2-6} \bigg]=0 $$ Therefore, $$ \textbf{W}\cdot (\textbf{Kd}-\textbf{F})=\sum_{j=1}^{6}{K_{1j}d_{j}=F_{1}}\ $$ Following through as before for the rest of the matrices gives these results: $$ \textbf{W}^{T}_{1x6}=\begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix}_{1x6}\ $$ $$ \begin{bmatrix}0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1j}d_{j}} - F_{1} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{2j}d_{j}} - F_{2} \bigg]+ 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{3-6j}d_{j}} - F_{3-6} \bigg]=0 $$ Therefore, $$ \textbf{W}\cdot (\textbf{Kd}-\textbf{F})=\sum_{j=1}^{6}{K_{2j}d_{j}=F_{2}}\ $$ $$ \textbf{W}^{T}_{1x6}=\begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix}_{1x6}\ $$ $$ \begin{bmatrix}0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1-2j}d_{j}} - F_{1-2} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{3j}d_{j}} - F_{3} \bigg]+ 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{4-6j}d_{j}} - F_{4-6} \bigg]=0 $$ Therefore, $$ \textbf{W}\cdot (\textbf{Kd}-\textbf{F})=\sum_{j=1}^{6}{K_{3j}d_{j}=F_{3}}\ $$ $$ \textbf{W}^{T}_{1x6}=\begin{bmatrix} 0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix}_{1x6}\ $$ $$ \begin{bmatrix}0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1-3j}d_{j}} - F_{1-3} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{4j}d_{j}} - F_{4} \bigg]+ 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{5-6j}d_{j}} - F_{5-6} \bigg]=0 $$ Therefore, $$ \textbf{W}\cdot (\textbf{Kd}-\textbf{F})=\sum_{j=1}^{6}{K_{4j}d_{j}=F_{4}}\ $$ $$ \textbf{W}^{T}_{1x6}=\begin{bmatrix} 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix}_{1x6}\ $$ $$ \begin{bmatrix}0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1-4j}d_{j}} - F_{1-4} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{5j}d_{j}} - F_{5} \bigg]+ 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{6j}d_{j}} - F_{6} \bigg]=0 $$ Therefore, $$ \textbf{W}\cdot (\textbf{Kd}-\textbf{F})=\sum_{j=1}^{6}{K_{5j}d_{j}=F_{5}}\ $$ $$ \textbf{W}^{T}_{1x6}=\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}_{1x6}\ $$ $$ \begin{bmatrix}0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1-5j}d_{j}} - F_{1-5} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{6j}d_{j}} - F_{6} \bigg]=0 $$ Therefore, $$ \textbf{W}\cdot (\textbf{Kd}-\textbf{F})=\sum_{j=1}^{6}{K_{6j}d_{j}=F_{6}}\ $$ Hence, adding up all of these results gives us $$ \textbf{Kd}=\textbf{F} $$.

HW from slide 33-5:

Given the following equation, find an expression for $$ \mathbf{k^{(i)}} $$.

$$ \mathbf{k}^{(i)}_{2x2}=\int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}}\mathbf{B^{T}}(\tilde{x})(EA)(\tilde{x})\mathbf{B}(\tilde{x})d\tilde{x}$$

It is already known that

$$ \mathbf{B}(\tilde{x}) = \begin{bmatrix}N_{i}(\tilde{x}) & N_{i+1}(\tilde{x})\end{bmatrix} $$

$$ \mathbf{B^{T}}(\tilde{x}) = \begin{bmatrix}N_{i}(\tilde{x})\\ N_{i+1}(\tilde{x})\end{bmatrix} $$

Also, from lecture 33, it is known that

$$A(\tilde{x})=N^{(i)}_1(\tilde{x})A_1+N^{(i)}_2(\tilde{x})A_2$$ $$E(\tilde{x})=N^{(i)}_1(\tilde{x})E_1+N^{(i)}_2(\tilde{x})E_2$$

Since Ni is assigned to Node 1 and Ni+1 is assigned to Node 2, the following two equations are true.

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

Plugging these expressions back into our original $$ \mathbf{k^{(i)}} $$ equation results in

$$ \mathbf{k^{(i)}} = \begin{Bmatrix} N_{1}^{(i)}(L)\\ N_{2}^{(i)}(L) \end{Bmatrix} \begin{bmatrix}N_{1}^{(i)}(L) & N_{2}^{(i)}(L)\end{bmatrix} E(L)A(L) - \begin{Bmatrix} N_{1}^{(i)}(0)\\ N_{2}^{(i)}(0) \end{Bmatrix} \begin{bmatrix}N_{1}^{(i)}(0) & N_{2}^{(i)}(0)\end{bmatrix} E(0)A(0) $$ Multiplying the 2x1 and 1x2 matrices together, the equation above becomes $$ \mathbf{k^{(i)}} = \left[   (N_{1}^{(i)}(L))^{2} + (N_{2}^{(i)}(L))^{2}   \right] E(L)A(L) - \left[   (N_{1}^{(i)}(0))^{2} + (N_{2}^{(i)}(0))^{2}   \right] E(0)A(0) $$

According to the definition of N at the point x = 0, N1 = 1, N2 = 0 At the point x = L, N1 = 0, N2 = 1.

Therefore,

$$ \mathbf{k^{(i)}} = \left[   (0)^{2} + (1)^{2}   \right] \left[(0)E_{1} + (1)E_{2} \right] \left[ (0)A_{1} + (1)A_{2}  \right] - \left[   (1)^{2} + (0)^{2}   \right] \left[(1)E_{1} + (0)E_{2} \right] \left[ (1)A_{1} + (0)A_{2}  \right] $$

$$ \mathbf{k^{(i)}}_{2x2} = \left[ E_{2}A_{2} - E_{1}A_{1} \right] \begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix} $$

title: User:Eml4500.f08.bottle.loschak/HW6


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Organic chemistry/Stereochemistry

Organic chemistry involves infinitely varied structures arising form a series of limited elements. Because of this, just giving the empirical formula of a compound is insufficient. For example, the empirical formula C2H6O, a molecule of only 9 atoms, can refer to the ether dimethyl ether or ethanol, depending on whether the oxygen is in the middle of or at the end of the carbon chain. The term for different compounds arising from the same empirical formula is isomers.

There are multiple types of isomers, and dimethyl ether and ethanol are examples of structural isomers. They have the same atoms, but their bond structure is different, leading to different physical and chemical properties. For example, ethanol is liquid at room temperature, whereas diethyl ether is gaseous. Another term for structural isomerism is constitutional isomerism.

If the compound has the same bonds, but their arrangement in space differs, they are considered stereoisomers. This notably occurs in cyclic hexane compounds. If two bromines are substituted for two hydrogen in cyclcohexane then it has either a cis or trans configuration. If both bromines lie in the same plane, either both above cyclohexane or both below, they are considered the cis isomer. If one is up and the other down they are trans isomer. cis-1,3 dibromocyclohexane trans-1,3 dibromocyclohexane

This can have an effect on the stability and chemical and physical properties of the element. These difference often occur because of steric strain, when two large groups are forced close together. The different properties are much less pronounced than in constitutional isomers.

Another form of isomerism is enantiomerism, or chirality. Two molecules are enantiomers have the same bonds to the same atoms, but do not have superimposable mirror images. A good example of chirality is your hands. Your left hand is exactly the same as your right hand, but you cannot align them on top of each other when they are both facing the same direction. This often arises when a carbon has four different groups bound to it. Chiral molecules have the same physical properties and exactly the same chemical properties, except when reacting with other chiral molecules. Thus, chiral molecules have potentially drastic differences in physiology. Enantiomers also have the notable difference of rotating plane polarized light the same number of degrees but in opposite directions. This leads to the designation of enantiomers based on the direction they rotate the light as either positive (+) or negative (-), However, since this is relative to which side is arbitrarily chosen to be positive, the absolute configuration is determined as either sinister (S) or rectus (R) based on the Cahn-Ingold-Prelog priority system.

title: Organic chemistry/Stereochemistry


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User:Unionhawk/monobook.js

/*** BEGIN WIKIBREAK ENFORCER ***/ $(function {	/*** Start editing here ***/	// When you want to end your break?	// no leading zeroes. (example: 7 - correct, 07 - incorrect)	var date = { year: 2008, month: 12, day: 1 };	var time = { hours: 0, minutes: 0, seconds: 0 };	/*** Stop editing here ***/	var currentDate = new Date;	var enforcedBreakEnd = new Date( date.year,date.month-1,date.day,time.hours,time.minutes,time.seconds);	if (currentDate <= enforcedBreakEnd) {		alert("Enforced wikibreak until "+enforcedBreakEnd.toLocaleString + "\n(now is "+currentDate.toLocaleString+")\n\nBye!");		location = "http://"+location.host+"/w/index.php?title="			+ "Special:Userlogout&returnto=Main_Page";	} }); /*** END WIKIBREAK ENFORCER ***/

title: User:Unionhawk/monobook.js


text:
User:Eml4500.f08.delta 6.krueger/HW N1^i(x)tilda

From the diagram above it can be seen that $$\tilde{x}$$ is equal to $$x-x_{i}$$.Therefore the formula for $$N_{i}(\tilde{x})$$ from the above diagram is: $$N_{i}(\tilde{x})={x-x_{i} \over x_{i}-x_{i-1}}$$ It is also know that: $$N_{i}(\tilde{x})=N_{i}(x)=1-{x\over L^{(i)}}$$ Now let x=0 and x=$$L^{(i)}$$ and it can be seen that when: x=0 $$\Rightarrow$$ $$N_{i}(\tilde{x})=1$$ and when x=$$L^{(i)}$$ $$\Rightarrow$$ $$N_{i}(\tilde{x})=0$$


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title: User:Eas4200c.f08.wiki.d/HW5


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User:Eas4200c.f08.wiki.d/HW6

= Airfoil =

Above is a graph of $$\sigma_{xx}$$ at all y positions in the middle wing section on the top skin. Due to the graph being concave up, it can be seen that the load is compressive on the top skin of the wing section.

Above is a graph of $$\sigma_{xx}$$ at all y positions in the middle wing section on the bottom skin. Due to the graph being concave down, it can be seen that the load is tensile on the top skin of the wing section.

The average compressive stress was found to be 144.2 MPa.

= Plate Buckling =

Simply-Supported Boundary Conditions
A simply-supported rectangular plate under an in-plane distributed compressive load is to be evaluated:

Buckling mode shapes can be determined by the following equation: $$\displaystyle \psi (x,y) = c_{mn}\sin\left(\frac{m \pi x}{a}\right)\sin\left(\frac{n \pi y}{b}\right)\, \ {\rm for} \ m,n = 1, 2, 3, \ldots$$ 1

Perspective views of the buckling shape of a simply-supported rectangular plate for various cases of $$\displaystyle\ m$$ and $$\displaystyle\ n$$

Now considering the factor with $$\displaystyle\ x$$ as the variable in the expression for $$\displaystyle\ \psi_{mn}(x,y)$$, the period, $$\displaystyle\ T$$, is calculated as follows:

$$ \sin ( \frac{m \pi (x + T)}{a} ) = \sin ( \frac{m \pi x}{a} )$$

$$\frac{m\pi (x+T)}{a}=\frac{m\pi (x)}{a}+2\pi n$$ where $$\displaystyle\ n=1,2,3,...$$

Evaluating at $$\displaystyle\ n=1$$

$$\frac{m\pi (x+T)}{a}=\frac{m\pi (x)}{a}+2\pi$$

$$ x+T = x + \frac{2\pi a}{m\pi}$$

$$ T = \frac{2a}{m}$$

The values of $$\displaystyle\ T$$ along with plots of the function are calculated and displayed below for values of $$\displaystyle\ m=1,...,5$$ to show that $$\displaystyle\ m$$ is indeed the number of half wave-lengths generated.

The critical buckling load $$\displaystyle\ (P_x)_{cr}$$ is calculated as follows:

$$\displaystyle (P_x)_{cr}=k_c\frac{\pi ^2 D}{b}$$ 1

where $$\displaystyle k_c(m, a/b):=\left(\frac{m b}{a}+\frac{a}{m b}\right)^2$$

and $$\displaystyle D:=\frac{E h^3}{12 (1 - \nu^2)}$$

A graph of $$\displaystyle\ k_c$$ versus aspect ratio, $$\displaystyle\ a/b$$, for various values of $$\displaystyle\ m$$ is shown below

Going one step further, the critical buckling stress can be calculated by

$$\displaystyle (\sigma_{xx})_{cr}=\frac{(P_x)_{cr}}{bh}=k_c\frac{\pi ^2 D}{b^2 h}$$ 1

Clamped Boundary Conditions
Next a rectangular plate with clamped boundary conditions is to be examined

The critcal buckling stress under an in-plane compressive load can be calculated as follows:

$$\displaystyle\ (\sigma_{xx})_{cr}=K\frac{E}{1-\nu ^{2}}(\frac{t}{b})^2$$

where $$\displaystyle K$$ is a function different from $$\displaystyle k_c$$, and $$\displaystyle t \equiv h$$ the plate thickness.

The buckling shape is now calculated using

$$\displaystyle\psi (x,y)=\frac{c}{4}\left(1 - \cos \frac{2 \pi x}{a}\right)\left(1 - \cos \frac{2 \pi y}{b}\right)$$

= Conclusions =

The figure below shows the critical buckling stress as a function of aspect ratio for the top panel (under compressive load from prescribed conditions) of the NACA 2415 Airfoil under both simply-supported and clamped boundary conditions. Also shown in green is the average compressive stress in the panel. Since stringers support the panel, it is neither exactly simply-supported or clamped; therefore, the actual critical stress must fall within the envelope prescribed by the simply-supported and clamped cases. It can be seen that the average compressive stress of the panel does encounter the critical stress for the clamped case but not the simply-supported one; consequently, depending on how close the panel relates to the two cases, buckling is a possibility if the panel behaves closely to the clamped boundary condition.

title: User:Eas4200c.f08.wiki.d/HW6


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User:Eml4500.f08.echo.sell.cm/HW6

Wednesday, November 12th
I was working on this date during class one day to kill time. I'll let someone else take over from here though. I'll go ahead and start working on this section and let someone else work on the homework portions. Bradley LaCroix 21:20, 20 November 2008 (UTC)

FEM via PVW (continued)

HW: Show that $$u^{ }_{}(x_{i+1})=d_{i+1}$$

$$u(x_{i+1})=N_i^{}(x_{i+1})d_i+N_{i+1}(x_i)d_{i+1}$$

In this case, $$N_i^{}(x_{i+1})=0$$ and $$N_{i+1}^{}(x_{i+1})=1$$

Substituting these values into the equation, we get $$u(x_{i+1}^{})=(0)\cdot d_i+(1)\cdot d_{i+1}$$

Or $$u(x_{i+1}^{})=d_{i+1}$$

END HW

From page 31-3: Interpolate for u(x) Apply same interpolation for u(x), i.e.,

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

Element Stiffness Matrix for element i

$$\beta=\int_{x_i}^{x_{i+1}} \left [ N_i^'w_i + N_{i+1}^'w_{i+1} \right ] (EA) \left [ N_i^'d_i+N_{i+1}^'d_{i+1} \right ]\, dx$$

Where $$w'(x)=N_i^'w_i + N_{i+1}^'w_{i+1}$$

and $$u'(x)=N_i^'d_i+N_{i+1}^'d_{i+1}$$

and $$N_i^':=\frac{dN_i(x)}{dx}$$

Likewise for $$N_{i+1}^'$$

Note $$u(x)=\begin{bmatrix} N_i^'(x) & N_{i+1}(x) \end{bmatrix}_{1x2} \begin{Bmatrix} d_i \\ d_{i+1} \end{Bmatrix}_{2x1}$$

$$\frac{du(x)}{dx}=\begin{bmatrix} N_i^'(x) & N_{i+1}^'(x) \end{bmatrix}_{1x2} \begin{Bmatrix} d_i \\ d_{i+1}\end{Bmatrix}_{2x1}$$

Similarly: $$w(x)=\mathbf{N}(x) \begin{Bmatrix} w_i \\ w_{i+1} \end{Bmatrix}$$

$$\frac{dW(x)}{dx}=B(x) \begin{Bmatrix} w_i \\ w_{i+1} \end{Bmatrix}$$

Recall the elemental degrees of freedom:

$$\begin{Bmatrix} d_i \\ d_{i+1} \end{Bmatrix}=\begin{Bmatrix} d_1^{(i)} \\ d_2^{(i)} \end{Bmatrix} = \mathbf{d}^{(i)}$$

$$\begin{Bmatrix} w_i \\ w_{i+1} \end{Bmatrix}=\begin{Bmatrix} w_1^{(i)} \\ w_2^{(i)} \end{Bmatrix} = \mathbf{w}^{(i)}$$

$$\beta=\int_{x_i}^{x_{i+1}}(\mathbf{Bw}^{(i)})_{1x1}(EA)_{1x1}(\mathbf{Bd}^{(i)})_{1x1}\,dx=\mathbf{w}^{(i)}\cdot (\mathbf{k}^{(i)} \mathbf{d}^{(i)})$$

$$\beta=\int_{x_i}^{x_{i+1}}(EA)(\mathbf{Bw}^{(i)}) \cdot (\mathbf{Bd}^{(i)})\, dx$$

$$(\mathbf{Bw}^{(i)})^T (\mathbf{Bd}^{(i)})$$

Where $$(\mathbf{Bw}^{(i)})^T=\mathbf{w}^{(i)T} \mathbf{B}^T=\mathbf{w}^{(i)} \cdot \mathbf{B}^T$$

$$\beta=\mathbf{w}^{(i)}\cdot \int \mathbf{B}^T (EA) \mathbf{B}$$

$$\mathbf{k}_{2x2}^{(i)}=\int_{x_i}^{x_{i+1}} \mathbf{B}^T(x)_{2x1}(EA)_{1x1}\mathbf{B}(x)_{1x2}\,dx$$

HW $$B(x)=\begin{bmatrix} \mathbf{HW} & \frac{1}{L^{(i)}} \end{bmatrix}$$

L(i)=xi+1-xi (length of element i)

HW6: Consider EA=const $$\mathbf{k}^{(i)}=\frac{EA}{L^{(i)}}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$

Transfer of variable coordinates from x to $$\tilde x$$

$$\tilde x := x-x_i$$

$$d \tilde x = dx$$

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

HW6: Find expression for $$\mathbf{k}^{(i)}$$ using the above forumula

HW6

$$A(\tilde{x})=N_1^{(i)}(\tilde{x})A_1+N_2^{(i)}(\tilde{x})A_2$$

$$E(\tilde{x})=N_1^{(i)}(\tilde{x})E_1+N_2^{(i)}(\tilde{x})E_2$$

Find $$\mathbf{k}^{(i)}$$

title: User:Eml4500.f08.echo.sell.cm/HW6


text:
Mergers and Acquisitions/Introduction/Session 1

Merge for "Economies of Scale" Spin-Off for "Shareholder Value"


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User:Eml4500.f08.team.figuerrez/hw6b

Determining if the Electric Pylon is Statically Indeterminate
The electric pylon structure is declared to have 46 global nodes and 91 elements. Each element contains two nodes supporting four unknown forces. There are three equations, $$\Sigma{\boldsymbol{F_{x}=0}}$$, $$\Sigma{\boldsymbol{F_{y}=0}}$$ and $$\Sigma{\boldsymbol{M=0}}$$. You can't entirely know the vector sum of the components to compute the shear stress across the face of the cross section of the nodes. The solution set is infinite because it has more unknowns than equations. The electric pylon structure is therefore statically indeterminate.

title: User:Eml4500.f08.team.figuerrez/hw6b


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User:Eml4500.f08.delta 6.castrillon/Square 3-bar truss MATLAB simulation

The Three-Bar Square Truss
The following truss was assigned in homework 4, and an analysis of its behavior is performed.

The elements of this truss have the following properties:

 * $$A^{(1)}=A^{(2)}=A^{(3)}=1$$
 * $$E^{(1)}=E^{(2)}=E^{(3)}=1$$
 * $$L^{(1)}=L^{(2)}=L^{(3)}=1$$

The following is the MATLAB code for determining the global stiffness matrix, $$\mathbf{K}$$, and the dof values.

It is evident that the Finite Element Method is not enough to solve the behavior of this truss. It is evident that this truss is very inestable, and so there is an undetermined number of ways this truss will behave under any load. It is necessary that the engineer add another bar element for support of the structure.

The Eigenvalues and Eigenvectors
The following command yielded the eigenvalues (D) and eigenvectors (V) for the global stiffness matrix assembled above.

We can see that there are five zero-value eigenvalues. Below are the plots of these five eigenvalues of the three-bar square truss and the generic code for obtaining the plots below.

First Zero Eigenvalue

Second Zero Eigenvalue

Third Zero Eigenvalue

Fourth Zero Eigenvalue

Fifth Zero Eigenvalue

The Four-Bar Square Truss
The following truss is a more rigid version of the above three-bar square truss. Another bar element has been placed diagonally from the global node 1 to the global node 3.

This new truss system has slightly different properties:

 * $$A^{(1)}=A^{(2)}=A^{(3)}=3$$
 * $$E^{(1)}=E^{(2)}=E^{(3)}=2$$
 * $$L^{(1)}=L^{(2)}=L^{(3)}=1$$ and $$L^{4}=\sqrt{2}$$

The source code for assembling the global stiffness matrix, and determining the dof values is below:

It is now evident that by the addition of the fourth bar element the truss system is now stable.

The Eigenvalues and Eigenvectors
The eigenvalues (D) and eigenvectors (V) were determined with the code below. Also, the actual results are displayed.

We can see that there are four zero eigenvalues. The plots of these eigenvalues can be found below, as well as the code for the fourth zero eigenvalue, as an example.

First Zero Eigenvalue

Second Zero Eigenvalue

Third Zero Eigenvalue

Fourth Zero Eigenvalue


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Talk:President of the United States/George Washington

How is this intended to be different from an encyclopedia article? The Jade Knight (d'viser) 06:24, 21 November 2008 (UTC)

 * Instead of providing a lot of information about the man, as an encyclopedia article would do, these lessons are intended to focus on the term of presidency and place them in context of US and World events. At present, I'm simply trying to flesh out the pages enough to provide summary information. The overall objective is to provide an introductory course to US History that will use the Presidents as a mnuemonic device, versus mundane memorization of lists and dates. Long term, the personalities and policies of the individual's should be added to show how they shaped history. Sri-ganesh 18:22, 21 November 2008 (UTC)

 * Sounds like a good idea. Try to make a particular effort while you're at it to focus on participation, instead of simply information—without participatory elements, this may as well be a Wikibook.  The Jade Knight (d'viser) 07:40, 22 November 2008 (UTC)


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The US Presidents/John Adams

For an in depth look at the 2nd US President, follow this link: John Adams.

Elected: 1st Term in 1796

Presidency: 1797 to 1801

Major US Laws
The following is a list of major laws in US History signed by John Adams during his presidency:
 * Alien and Sedition Acts, multiple acts designed to protect the United States from alien citizens of foreign powers and to prevent seditious attacks from weakening the government.

Interesting Facts
John Adams was the first president to:
 * Live in the White House.
 * Only serve one term.
 * Serve as president in the 19th century.

title: The US Presidents/John Adams


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User:Eas4200c.f08.vqcrew.d/hw6

Boundary Conditions for Phi
For solid cross sections, the constant value of $$\phi \;$$ is arbitrary, therefore it can be set to zero. Thus, we have the boundary condition $$\phi =0 \;$$ on the lateral surface of a solid bar. This boundary condition is depicted in Figure #, where $$\phi \;$$ is equal to zero on $$S_0$$. In the case of a thin-walled cross section, $$\phi \;$$ has a different constant value on each surface. This is depicted in Figure #, where $$\phi =C_0\;$$ on $$S_0$$ and $$\phi =C_1\;$$ on $$S_1$$.

For a solid bar with circular cross section, as shown in Figure #, the Prandtl stress function is assumed to be
 * $$\phi = C\left(\frac{y^2}{a^2} + \frac{z^2}{a^2}-1 \right)$$
 * {| class="collapsible collapsed"

!Proof that this equation satisfies the boundary condition We can prove that this satisfies the surface boundary condition of $$\phi =0\;$$ as follows:
 *
 *
 * $$\phi = C\left(\frac{y^2}{a^2} + \frac{z^2}{a^2}-1 \right) = C\left(\frac{y^2+z^2}{a^2}-1 \right)$$

Substituting in the Pythagorean theorem, $$a^2=y^2+z^2$$, we get:
 * $$\phi = C\left(\frac{a^2}{a^2}-1 \right)=C(1-1)=0$$

To find C, we can plug this function into the compatibility equation:
 * }
 * $$\frac{\partial^2 \phi }{\partial y^2} + \frac{\partial^2 \phi }{\partial z^2}=-2G \theta$$

which results in:
 * $$\frac{2C}{a^2} + \frac{2C}{a^2} = -2G \theta \Rightarrow C=-\frac{1}{2}a^2G\theta$$

This can also be expressed in terms of torque $$T$$ and polar moment of inertia $$J$$:
 * $$C=-\frac{1}{2}a^2TJ$$

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User talk:Sri-ganesh

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

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! SB_Johnny talk 10:54, 21 November 2008 (UTC)

US Presidents
Sounds like a timely project! A few thoughts though: the pages for each president should be subpages (e.g., The US Presidents/John Adams rather than John Adams)... presumably there could at some point even be a separate learning resource just about John Adams. I'll move those around for you (your watchlist will show the new names).

On a side note, I started Digital media and democracy a few days ago after hearing a good bit of NPR discussions on how the Obama campaign used the new forms of media, and it got me thinking to how advances in communication have changed things over the years (telegraph, phone, fax, etc.) for campaigns. Any thoughts? --SB_Johnny talk 10:54, 21 November 2008 (UTC)

 * Fixed the links too... if you want to link to wikipedia, you can just use the prefix "w:" in the link, so cheese turns up as cheese :-). --SB_Johnny talk 11:04, 21 November 2008 (UTC)

 * Looks like you found the move button... I was thinking that "George Washington" would be better too, but didn't want to step on your toes :-). --SB_Johnny talk 19:17, 21 November 2008 (UTC)

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User:Mu301/ProAm

 * Pro-Am Collaboration at Sky & Telescope magazine

title: User:Mu301/ProAm


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User:Eas4200c.f08.gator.edwards/Wk6 Matlab

 Airfoil Program

 Plate Buckling


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