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)