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User:Egm6341.s11.team4.BM/HW3

=Problem 3.1: Taylor Series and Lagrange Approximation Comparison=

As assigned in lecture slide 14-1.

Given
1. $$\displaystyle f(x)=sin(x) $$

Objectives
1. Find the Taylor Series approximation of f(x) about $$\frac{3\pi}{8}$$

2. Find the 4th degree Lagrange approximation of f(x)

3. Find the degree necessary for the error in the Taylor Series to be less than the Lagrange approximation at the point $$ \frac{7\pi}{8}$$.

4. Use the Lagrange Interpolation Estimation Theorem as presented in 11-3 to come up with an upper bound on the Lagrange error and check that the calculated Lagrange error is within this bound.

Solution
All MATLAB code is included in a collapsible box at the end of this problem.

In order to solve this problem, we must first develop our lagrange interpolation polynomial for sin(x) and our Taylor Series approximations of the function as expanded about the point $$x_{0}=\frac{3\pi}{8}$$.

Fortunately, we have already developed the Taylor Series for sin(x) about the point $$x_{0}=\frac{3\pi}{8}$$ previously in homework 2.

Though we will cover the derivation of the Taylor Series approximations here, for a more complete derivation please refer back to homework 2.

First, we will start with the general Taylor Series expansion as presented in the lecture slides

 * $$ P_{n}(x) = f(x_{0}) + \frac{(x-x_{0})}{1!} f^{(1)}(x_{0}) +...+\frac{(x-x_{0})^{n}}{n!}f^{(n)}(x_{0}) \!$$

For the case of sin(x), the equation then becomes

 * $$ f(x) = sin(x_{0}) + \frac{(sin(x)-sin(x_{0}))^{1}}{1!}cos(x_{0}) - \frac{(sin(x)-sin(x_{0}))^{2}}{2!}sin(x_{0}) - ...\!$$

The periodicity of the sine function shows up in our evaluation of the Taylor Series and can be capitalized upon. Since the derivative of sine is the sine function itself every 4 derivatives evaluating the nth derivative is a matter of determining what the remainder is when n is divided by 4 and matching the derivative with the appropriate trigonometric function based on what the remainder is. This process is captured in code below using the sub-function TaylorDer.

The general Taylor Series equation is developed in the code presented below in the section labeled Taylor.

Next, we must develop the Lagrange approximation of the function sin(x). New code had to be developed for this, but the basic premise is to develop the Lagrange basis polynomials and then take a linear combination of them to obtain the Lagrange approximation.

In a Lagrange approximation, we start with a set of data points that are a part of the function that we are attempting to approximate.

 * $$(x_0, y_0),\ldots,(x_j, y_j),\ldots,(x_k, y_k)$$

The Lagrange basis polynomials are then obtained using the following formula.

 * $$\ell_j(x) := \prod_{\begin{smallmatrix}

0\le m\le k\\ m\neq j\end{smallmatrix}} \frac{x-x_m}{x_j-x_m} = \frac{(x-x_0)}{(x_j-x_0)} \cdots \frac{(x-x_{j-1})}{(x_j-x_{j-1})} \frac{(x-x_{j+1})}{(x_j-x_{j+1})} \cdots \frac{(x-x_k)}{(x_j-x_k)}.$$

where k is the degree of the Lagrange approximation. It should also be noted that the notation $$ \ell_j(x) $$ is actually short hand notation for $$ \ell_{j,k}(x) $$. Since the Lagrange basis polynomials are products of k 1st order terms $$ (\frac{x-x_m}{x_j-x_m}) $$, the lagrange basis polynomials are all kth order polynomials.

 * $$ \ell_{j,k}(x) = \mathcal{P}_{k} $$

In our case, we are told to take the 4th degree Lagrange approximation, so k=4. This means our Lagrange basis polynomials will all be 4th order. Now, to obtain our actual Lagrange approximation, we take a linear combination of our Lagrange basis polynomials.

 * $$L(x) := \sum_{j=0}^{k} y_j \ell_j(x)$$

Where $$ y_j $$ is the true function value at our j-th intersection point and is known from our original data set. Since the final Lagrange approximation is simply a linear combination of the basis polynomials, it too will be of the k-th order. The Lagrange approximation intersects with the true function value at k+1 points over the given interval. These intersection points are the original data points from which we developed our Lagrange basis polynomials.

We now can develop the Lagrange approximation and Taylor Series approximations necessary for this problem, but we are also instructed to analyze the Lagrange error using the Lagrange Interpolation Error Theorem as presented in lecture slide 11-3. This is stated below

 * $$ e^{L}_{n}(t)=\frac{q_{n+1}(t)}{(n+1)!}f^{n+1}(\xi)       $$

where

 * $$ q_{n+1}(t):=\prod_{j=0}^{n}=(x-x_{j}) \in \mathcal{P}_{n+1} $$

This equation can be simplified for our problem. Since we are dealing with the function sin(x), a function we know to be cyclic in its derivatives, the term $$ f^{n+1}(\xi) $$ can be bounded. The derivative of the sine function will always be positive or negative sine or cosine. Furthermore, we know the range of these trigonometric functions to be bounded between -1 and 1. This allows to say that

 * $$ |e^{L}_{4}(t)|\leq \frac{|q_{5}(t)|}{5!} $$

Since we want to determine the degree of the Taylor Series about $$ x_{0}=\frac{3\pi}{8} $$ required such that the Taylor Series becomes a better approximation of the function sin(x) than the 4th degree Lagrange approximation when evaluated at the point $$ \frac{7\pi}{8} $$, we expect to see a final relationship of the errors

 * $$ |e^{T}_{4}(\frac{7\pi}{8})| \leq |e^{L}_{4}(\frac{7\pi}{8}) | \leq |\frac{q_{5}(\frac{7\pi}{8})}{5!}| $$

Now that we have a system to develop our Lagrange approximation, our Taylor Series approximations, and an upper bound on our remainder, we have to write code that develops these functions automatically and then determines the Taylor Series degree n at which the Taylor Series error at $$ \frac{7\pi}{8} $$ will be less than the Lagrange error. To do this, we develop the Lagrange approximation and error in its own independent loop. Then, we develop the bound on the remainder in its own independent loop. The Taylor Series approximation and error, meanwhile, is developed in a while loop that develops the next degree of the Taylor Series approximation while the condition that the Taylor Series error is greater than the Lagrange approximation error is met. Once this condition is no longer met the code prints the Taylor Series degree at which the condition was satisfied along with the Taylor Series error at the point $$ \frac{7\pi}{8} $$, the Lagrange approximation error, as well as the upper bound of the Lagrange error. The commented code for this problem is included below.

After running the program below, we obtain the following plot of our function and its approximations.

As can be seen, both our Lagrange and Taylor Series approximations are exceptionally close to the true function over the span 0 to $$ \pi $$ (the three functions are virtually indistinguishable on the above plot). Meanwhile, our code output tells us that the Taylor Series needed to be of degree 6 or higher before the Taylor Series produced a better approximation than the Lagrange Interpolation of degree 4. Additionally, as expected, our Taylor Series error was better than our Lagrange error, which was less than the upper bound in our Lagrange error predicted by the Lagrange Interpolation Error Theorem.

Code Output:

Taylor Series Degree = 6

$$ ( |e^{T}_{4}(\frac{7\pi}{8})|= 0.000905)       <            (|e^{L}_{4}(\frac{7\pi}{8})| = 0.001476)             <              (|\frac{q_{5}(\frac{7\pi}{8})}{5!}|   = 0.008170)   $$

This problem was solved by Brendan Mahon

Given
1.

Function $$ f(x)=\frac{e^{x}-1}{x} $$    on [-1,1]

Discretization Points (uniform) $$ \displaystyle [x_{0},...,x_{4}] $$, n=4

Evaluation Point $$ \displaystyle t=5 $$

Evaluation Point $$ \displaystyle x=0.75 $$

Lagrange Basis Polynomial to Plot $$\displaystyle i=2 \Rightarrow l_{2,4}(\cdot) $$

2.

Function $$ f(x)=\frac{1}{1+4x^{2}} $$    on [-5,5]

Discretization Points (uniform) $$\displaystyle [x_{0},...,x_{8}] $$, n=8

Evaluation Point $$ \displaystyle t=4.5 $$

Evaluation Point $$\displaystyle x=3 $$

Lagrange Basis Polynomial to Plot $$\displaystyle i=3 \Rightarrow l_{3,8}(\cdot) $$

Objectives
Plot 3 figures similar to those on 14-2. Using uniform discretization.

1.

1.1: Find $$ e^{L}_{4}(f,t) $$ and $$ e^{L}_{4}(f,x) $$

1.2: Plot the Lagrange Interpolation and the original function on the same plot.

1.3: Plot the stated Lagrange Basis Polynomial (i=2) over the span of the function, ensuring that it is zero valued at all the basis data points except the i-th data point corresponding that basis polynomial, where has a value of one.

1.4: Plot the remainder component $$ q_{5} $$ to ensure that it is zero valued at all the basis data points. This ensures the Lagrange interpolation is exact when evaluated at the original basis data points.

2.

2.1: Find $$ e^{L}_{8}(f,t) $$ and $$ e^{L}_{8}(f,x) $$

2.2: Plot the Lagrange Interpolation and the original function on the same plot. Additionally, comment on the observed Runge Phenomenon in the Lagrange Interpolation.

2.3: Plot the stated Lagrange Basis Polynomial (i=3) over the span of the function, ensuring that it is zero valued at all the basis data points except the i-th data point corresponding that basis polynomial, where has a value of one.

2.4: Plot the remainder component $$ q_{9} $$ to ensure that it is zero valued at all the basis data points. This ensures the Lagrange interpolation is exact when evaluated at the original basis data points.

Solution
All MATLAB code is included in a collapsible box at the end of this problem.

Both parts 1 and 2 can be solved in the same manner. We will rigorously develop the method used in evaluating part 1 and then show how the same method is used in part 2. MATLAB code, affixed to the end of this solution, was used to develop all plots. The same MATLAB code was used for both parts 1 and 2 with only minor variations to accomodate for the differences in the stated problems.

1.

First, we must obtain the data points for our Lagrage Interpolation. Since we want a uniform discretization, we make a vector x that divides the span over which we're plotting the function into k+1 points. For problem 1, this is 5 points ranging from -1 to 1.

 * $$ [-1, -0.5, 0, 0.5, 1] $$

Now, we must evaluate the function f(x) at these five points to develop our vector y of function values. This is done by simply plugging in the individual x values into our f(x) function. For problem 1, it is worth noting that in Homework 1 Problem 1.1 we found that there is a singularity at 0 for our given function f(x), but that the limit of the function at 0 is valued at 1 when taken from either side. Therefore, when evaluating the function value at 0 instead of plugging the value 0 into our function f(x), we explicitly state that the value is 1. This allows us to proceed with our Lagrange Interpolation as before.

With the five data points required for a 4th degree Lagrange interpolation now obtained, we use the following algorithm to obtain our Lagrange Basis polynomials, as done previously in [ problem 3.1].

 * $$\ell_j(x) := \prod_{\begin{smallmatrix}

0\le m\le k\\ m\neq j\end{smallmatrix}} \frac{x-x_m}{x_j-x_m} = \frac{(x-x_0)}{(x_j-x_0)} \cdots \frac{(x-x_{j-1})}{(x_j-x_{j-1})} \frac{(x-x_{j+1})}{(x_j-x_{j+1})} \cdots \frac{(x-x_k)}{(x_j-x_k)}.$$

Where k is the degree of the Lagrange Interpolation (in our case k=4) and $$ x_j $$ is the jth x value from our data points. These Lagrange Basis Polynomials are obtained in their own for loop and stored as an array (this will be helpful in plotting an individual Langrange Basis over the relevant span later on). They are then used in linear combination to obtain our final Lagrange Interpolation according to the following formula.

 * $$L(x) := \sum_{j=0}^{k} y_j \ell_j(x)$$

Since our individual Lagrange Basis Polynomials are all of degree k, our final Lagrange Interpolation function is also of degree k (k=4 in this problem).

The following plot shows both the true function and the Lagrange Interpolation over the span -1 to 1.

As can be seen, the two functions are a near perfect match over the region. This is because the Lagrange Interpolation weaves in and out of the true function, intersecting at the data points obtained from our uniform discretization. When using the Lagrange Interpolation outside of this region, the error can grow rapidly, since the interpolation no longer weaves about the true function. This can be seen in the following image, a plot of the true function and the Lagrange Interpolation over the region -1 to 5.

The value 5 was selected since we must later on determine the error of the Lagrange interpolation at this point (as can be seen graphically, we should expect the error to be very large since 5 is well outside of the region -1 to 1 from which the interpolation was obtained).

It can be seen from both of these figures that the Lagrange Interpolation is exactly correct at the data points from which the Lagrange Basis Polynomials are obtained. The reason for this can quickly be seen from a plot of an individual Lagrange Basis polynomial over the relevant span.

This is the Lagrange Basis Polynomial corresponding to point i=2. It can be seen that the function is zero valued at every point in our uniform discretization (these points are indicated by the grid lines) except for point $$ x_2 $$, the point to which this Lagrange Basis Polynomial corresponds. At this point, the basis polynomial has a value of 1. This pattern is true for every Lagrange Basis Polynomial. Since the Lagrange Interpolation is a linear combination of the basis polynomials times the function value corresponding them, it should be clear that the value of the Lagrange Interpolation at point $$ x_2 $$ is

 * $$\displaystyle L(x_2)=0+0+1*y_2+0+0 $$
 * $$\displaystyle L(x_2)=y_2 $$

Next, we want to determine the function $$ q_5(t) $$. This is a component of the Lagrange Interpolation Estimation Theorem as presented in Lecture Slide 11-3.

 * $$ e_{n}^{L}(f,t):=f(t)-f_{n}^{L}(t) = \frac{q_{n+1}}{(n+1)!}f^{n+1}(\xi) $$

Where $$ \xi $$ is a point somewhere in the set [-1,1]. Using the following formula to obtain $$ q_{n+1} $$

 * $$ q_{n+1} := \prod_{j=0}^{n}(x-x_j)  \in \mathcal{P}_{n+1}  $$

In our case, n=4. To obtain this function in code, a separate for loop was made to evaluate the equation $$ q_5 $$. After this equation was determined, it was plotted over the relevant span of -1 to 1

As can be seen, the function is zero valued at every one of our data points (indicated by the grid lines). This is to be expected since we have previously shown that the Lagrange Interpolation exactly evaluates the function at the data points from which it is obtained.

Lastly, we wish to evaluate the errors $$ e_{4}^{L}(f,t) $$ and $$ e_{4}^{L}(f,x) $$ at t=5 and x=0.75 respectively.

This is done using the following equation

 * $$ e_{n}^{L}(f,t):=f(t)-f_{n}^{L}(t) $$

Substituting in for t=5, we obtain

 * $$ e_{4}^{L}(\frac{e^{x}-1}{x},5)=\frac{e^{5}-1}{5}-f_{4}^{L}(5)=11.0239 $$

This error is, as expected, very large. We could graphically see from our figure showing the true function and the Lagrange Interpolation that, when the evaluation point is far outside of the region of data points from which we developed our Lagrange Interpolation the error is very large. The Lagrange Interpolation no longer weaves about the true function and the divergence between the two is unbounded.

When determining the error at the point $$x=0.75 $$, however, we will find that the error is very small.

 * $$ e_{4}^{L}(\frac{e^{x}-1}{x},0.75)=\frac{e^{0.75}-1}{0.75}-f_{4}^{L}(0.75)=1.6274e-04 $$

This is also to be expected, since the value 0.75 is within the span from which we obtained the points our Lagrange Interpolation is based. Since the Lagrange Interpolation must intersect with the true function both before and after the point at which we are evaluating the error we expect that Lagrange Interpolation will not stray excessively far from the true function (in part 2, however, we will see that the problem of Runge's Phenomenon means that this assumption does not always hold true).

2.

We now analyze the function

$$ \displaystyle f(x)=\frac{1}{1+4x^{2}} $$ on [-5, 5]

using uniform discretization and an 8th degree Lagrange Interpolation.

First, we must obtain our data points. Using uniform discretization, we obtain the following vector of our x values

$$ [x_0, ..., x_8]=[-5, -3.75, -2.5, -1.25, 0, 1.25, 2.5, 3.75, 5] $$

Now, we must evaluate the function f(x) at these nine points to develop our vector y of function values. This is done by simply plugging in the individual x values into our f(x) function. With these quantities in hand, we proceed with developing our Lagrange Interpolation in an identical manner as was done in part 1 (in fact, the same code was simply copied and pasted with the only adjustment being our value for the degree of the approximation was changed from k=4 to k=8).

After obtaining our Lagrange Interpolation, we plotted the Lagrange Interpolation along with the value of our true function over the span -5 to 5.

As can be seen, our high order Lagrange Interpolation of this function is actually very poor over this region, particularly at the edges. This is due to Runge's Phenomenon, wherein a polynomial interpolation of high order tends to oscillate substantially at the edges of the interval. The reason for this oscillation and increase in error becomes obvious when looking at the Lagrange Interpolation Estimation Theorem.

 * $$ e_{n}^{L}(f,t):=f(t)-f_{n}^{L}(t) = \frac{q_{n+1}}{(n+1)!}f^{n+1}(\xi) $$

Let's isolate the derivative term in this equation. For the equation we previously analyzed in part 1, it can be seen (using Wolfram Alpha that the nth derivative value gets smaller and smaller when evaluated at the interval edge as n approaches infinity. For the function in question, however, the value of increasing derivatives at the interval edges explodes, causing an increasing error in our Lagrange Interpolation Estimation Theorem.  Using Wolfram Alpha, we obtain

 * $$|\frac{\partial ^{4}}{\partial x^{4}}\frac{1}{1+4x^{2}}| \approx 0.00179 $$
 * $$ |\frac{\partial ^{5}}{\partial x^{5}}\frac{1}{1+4x^{2}}| \approx 0.0021  $$
 * $$ |\frac{\partial ^{6}}{\partial x^{6}}\frac{1}{1+4x^{2}}| \approx 0.0029 $$
 * $$ |\frac{\partial ^{7}}{\partial x^{7}}\frac{1}{1+4x^{2}}| \approx 0.0044 $$
 * $$ |\frac{\partial ^{8}}{\partial x^{8}}\frac{1}{1+4x^{2}}| \approx 0.0077 $$

The trend shows that as we use a higher degree Lagrange interpolation, the error at the interval edge actually becomes more severe, explaining the severe oscillations we see in our figure.

Despite these oscillations, our Lagrange Basis polynomials should still be zero valued at every data point except the one corresponding to it, at which point it should have a value of one. Plotting the basis polynomial $$ l_{3,8}(x) $$ we see

The Lagrange Basis Polynomial is, as expected, zero at every evaluation point except the one corresponding to it, where it has a value of one. The grid lines indicate the evaluation points, with $$ x_{3}=-1.25 $$ being the point at which we expect to see the function reach a value of one. This plot matches our expecations.

Next, we evaluate the function $$ q_{9} $$. This function is a significant component of the Lagrange Interpolation Estimation Theorem and, as before, we expect to see it zero valued at every one of evaluation points.

As expected, the function has a value of zero at the grid marks corresponding to our evaluation points. It should be noted, however, that the function oscillates significantly (reaching a value of almost 40,000 near the interval edge). Since the Lagrange Interpolation Estimation Theorem states that the error is proportional to the function value of $$ q_{9}(x) $$, this plot further indicates that our Lagrange Interpolation will have severe errors near the interval edge.

Lastly, we must find our Lagrange errors for t=4.5 and x=3. We will use the same formula used previously

 * $$ e_{n}^{L}(f,t):=f(t)-f_{n}^{L}(t) $$

When evaluated for our function and our point t=4.5

 * $$ e_{8}^{L}(\frac{1}{1+4x^{2}},4.5):=|\frac{1}{1+4*4.5^{2}}-f_{8}^{L}(4.5)|=1.7840 $$

Since the true value is about 0.012, the error in the Lagrange Interpolation at this point very near our edge interval is, as expected, very large due to Runge's Phenomenon. Our next evaluation point is x=3. This point is further away from the edge interval and consequently, since Runge's phenomenon will be less severe away from the edge intervals, we expect the error to be substantially less.

 * $$ e_{8}^{L}(\frac{1}{1+4x^{2}},3):=|\frac{1}{1+4*3^{2}}-f_{8}^{L}(3)|=0.3786 $$

Though this error is, as expected, much smaller than at t=4.5, it is still very large when compared to the true function value of about 0.027. When looking at our plot of both the true function and its Lagrange Interpolation, it can be seen that the value of x=3 lies roughly at an oscillatory peak of our Lagrange Interpolation and, consequently our error is almost at a local maximum at that point. Since the error produced by our Lagrange Interpolation in part 2 was greater than the error produced by our interpolation in part 1, it is clear that one should be aware of Runge's phenomenon before deciding to employ a high order polynomial approximation for function evaluation, especially when evaluating points near the interval edge.

This problem was solved by Brendan Mahon

title: User:Egm6341.s11.team4.BM/HW3


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

title: File:Fe1.s11.team2.HW3.5.fig1a.png


text: File:Fe1.s11.team2.HW3.5.fig1a rev1.png

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University of Florida/Eml5526/s11.team2.reiss.HW3 HW3

Given
where the weak form is A trial solution for u(x) is suggested as with w(x) of the same form. 1) determine the solution to the weak form. 2) verify that the solution satisfies the equilibrium equation of the strong form 3) verify that the natural boundary condition is satisfied

Solution 3.10.1
From the given information the equations for u(x) and w(x) are as follows Using the initial condition $$w(3)=0$$ Using the condition that $$u(3)=0.001$$ The partial derivatives of u(x) and w(x) are Replacing, the weak form becomes After integration the equation becomes Rearranging we get The $$\beta _0$$ and $$\beta_1$$ values must be arbitrary, the solutions to this equation are dependent on the bracketed terms equaling zero. The solution for this, in matrix form, is The solution to the weak form becomes Giving the solution of

Given

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$$  \displaystyle 2\frac{d^2u}{dx^2}+3=0 $$     (7.1)
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Boundary Conditions:
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$$  \displaystyle u(x=1)=0 $$     (7.2)
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$$  \displaystyle -\frac{du}{dx}(x=0)=4 $$     (7.3)
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$$  \displaystyle b_j=(x)^j $$     (7.4) The exact solution to Eq.3.1:
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$$  \displaystyle u(x)=-\frac{3}{4}x^2-4x+\frac{19}{4} $$     (7.5)
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Find
We want to find an approximation to Eq.3.5 in the form of
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$$  \displaystyle u^h(x)=\sum_{i=0}^n a_ib_i $$     (7.6) as well as investigate how the accuracy of our approximation changes with n going from 2 to 4 and then 6.
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Case:n=2
First let's expand Eq.3.6 out to n=2 terms.
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$$  \displaystyle u^h(x)=a_0b_0+a_1b_1+a_2b_2 $$     (7.7) Now consider Eq.7.4, and substitute the appropriate terms into Eq.7.7.
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$$  \displaystyle u^h(x)=a_0(1)+a_1(x)+a_2(x)^2 $$     (7.8) Let's rewrite this equation in terms of a product between 2 vectors.
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$$  \displaystyle u^h(x)=\left[1\quad x\quad (x)^2\right ]\begin{bmatrix} a_0\\a_1\\a_2\end{bmatrix} $$     (7.9) In order for us to solve for the unknown $$a_i's$$ in Eq.7.9, we must have 3 linearly independent equations describing linear combinations of the $$a_i's.$$Two of the equations are easily obtained directly from the given boundary conditions in Eq.7.2 and Eq.7.3. It only makes practical sense that if we want our approximation to closely resemble our exact solution, we must require identical behavior between Eq.7.5 and Eq.7.8 at the boundaries.
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$$  \displaystyle \begin{align} u^h(x=1)=a_0(1)+a_1(1)+a_2(1)^2&=0\\ a_0(1)+a_1(1)+a_2(1)^2&=0 \end{align} $$     (7.10)
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$$  \displaystyle a_0+a_1+a_2=0 $$     (7.11)
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$$  \displaystyle \frac{du}{dx}(x=0)=a_1+a_22(0)=-4 $$     (7.12)
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$$  \displaystyle a_1=-4 $$     (7.13) Now we just need one more linearly independent equation and we can then readily solve for the $$a_i's.$$ To develop another set of equations which are linearly independent of Eq.7.11 and Eq.7.13, we take the inner product between our approximation of u(x), Eq.7.9 and our basis set of functions Eq.7.4. Before getting to the inner product first, let's define our linear differential operator as the following
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$$  \displaystyle P(\cdot)=\frac{d^2}{dx^2}+\frac{3}{2}=0 $$     (7.14) Now, using our operator in Eq.7.14, we can now operate on our approximation in Eq.7.9.
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$$  \displaystyle \begin{align} P(u^h)=\frac{d^2}{dx^2}\left[1\quad x\quad (x)^2\right ]\begin{bmatrix} a_0\\a_1\\a_2\end{bmatrix}+\frac{3}{2}&=0\\ \left[0\quad 0\quad 2\right ]\begin{bmatrix} a_0\\a_1\\a_2\end{bmatrix}+\frac{3}{2}&=0 \end{align} $$     (7.15) Which we can then rewrite Eq.7.15 as
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$$  \displaystyle \left[0\quad 0\quad 2\right ]\begin{bmatrix} a_0\\a_1\\a_2\end{bmatrix}=-\frac{3}{2} $$     (7.16) Therefor
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$$2a_2=-\frac{3}{2}$$ (7.21) Using Eq.7.11 and Eq.7.13, we can form the following matrix equation.
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$$  \displaystyle \begin{bmatrix} 1&1&1\\0&1&0\\0&0&2\end{bmatrix}\begin{bmatrix} a_0\\a_1\\a_2\end{bmatrix}=\begin{bmatrix}0\\-4\\-3/2\end{bmatrix} $$     (7.21) Let's solve for the inverse of the matrix in Eq.7.21 using the Gauss-Jordan Method. Letting the matrix above in Eq.7.21 equal $$\mathbf{K}$$, we can find $$\mathbf{K^{-1}}$$ by
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$$  \displaystyle \begin{align} \mathbf{K}\cdot\mathbf{I}&=\begin{bmatrix} 1&1&1&1&0&0\\0&1&0&0&1&0\\0&0&2&0&0&1\end{bmatrix}\\ &=\begin{bmatrix} 1&0&0&1&-1&-1/2\\0&1&0&0&1&0\\0&0&2&0&0&1\end{bmatrix}\\ &=\begin{bmatrix} 1&0&0&1&-1&-1/2\\0&1&0&0&1&0\\0&0&1&0&0&1/2\end{bmatrix}\\ \end{align} $$     (7.22)
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$$  \displaystyle \therefore \mathbf{K^{-1}}=\begin{bmatrix}1&-1&-1/2\\0&1&0\\0&0&1/2\end{bmatrix} $$     (7.23) Now using Eq.7.23, we can solve Eq.7.21 for our unknown $$a_i's.$$
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$$  \displaystyle \mathbf{K^{-1}}\cdot \mathbf{K}\begin{bmatrix} a_0\\a_1\\a_2\end{bmatrix}=\mathbf{K^{-1}}\begin{bmatrix}0\\-4\\-3/2\end{bmatrix} $$     (7.24)
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$$  \displaystyle \begin{bmatrix} a_0\\a_1\\a_2\end{bmatrix} = \begin{bmatrix}1&-1&-1/2\\0&1&0\\0&0&1/2\end{bmatrix}\begin{bmatrix}0\\-4\\-3/2\end{bmatrix} $$     (7.25)
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$$  \displaystyle \begin{bmatrix} a_0\\a_1\\a_2\end{bmatrix}=\begin{bmatrix}19/4\\-4\\-3/4\end{bmatrix} $$     (7.26)
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Solution

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$$  \displaystyle \therefore u^h(x)=\frac{19}{4}-4(x)-\frac{3}{4}(x)^2 $$     (7.27) Now. let's check the absolute error between our approximate solution above in Eq. 7.26 and our exact solution in Eq.7.5. Comparing $$u^h(x)$$, 7.27, to $$u(x)$$, 7.5, we can see that using this basis gives an exact solution to $$u(x)$$
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$$  \displaystyle u^h(x)= \frac{19}{4}-4(x)-\frac{3}{4}(x)^2 $$     (7.27)
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$$  \displaystyle u(x)= \frac{19}{4}-4(x)-\frac{3}{4}(x)^2 $$     (7.5)
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Case:n=4
We will follow exactly the same methodology as we have above, leaving out the explanations for each step.
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$$  \displaystyle u^h(x)=a_0b_0+a_1b_1+a_2b_2+a_3b_3+a_4b_4 $$     (7.32)
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$$  \displaystyle u^h(x)=a_0(1)+a_1(x)+a_2(x)^2+a_3(x)^3+a_4(x)^4 $$     (7.33)
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$$  \displaystyle u^h(x)=\begin{bmatrix}1 & x & x^2 & x^3 & x^4 \end{bmatrix}\begin{bmatrix}a_0\\a_1\\a_2\\a_3\\a_4 \end{bmatrix} $$     (7.34) Using our given boundary conditions to obtain our first 2 linearly independent equations.
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$$  \displaystyle \begin{align} u^h(x=1)&=a_0(1)+a_1(1)+a_2(1)^2+a_3(1)^3+a_4(1)^4\\ &=a_0+a_1+a_2+a_3+a_4 \end{align} $$     (7.35)
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$$  \displaystyle a_0+a_1+a_2+a_3+a_4=0 $$     (7.36)
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$$  \displaystyle \frac{du^h}{dx}(x=0)=0+a_1+0+0+0 $$     (7.37)
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$$  \displaystyle a_1=-4 $$     (7.38) Using our predefined linear differential operator in Eq.7.14.
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$$  \displaystyle P(u^h)=\frac{d^2}{dx^2}\begin{bmatrix}1 & x & x^2 & x^3 & x^4 \end{bmatrix}\begin{bmatrix}a_0\\a_1\\a_2\\a_3\\a_4 \end{bmatrix} + \frac{3}{2} $$     (7.39) Because this time we have 5 unknown $$a_i's,$$ we will have to choose 3 equations from our inner product operation.
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$$  \displaystyle \left \langle b_i,P(u^h) \right \rangle =\int_0^1b_iP(u^h)=0 $$     (7.40)
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$$  \displaystyle \int_0^1\begin{bmatrix}1\\x\\x^2\\x^3\\x^4\end{bmatrix}\begin{bmatrix}0 & 0 & 2 & 6(x) & 12(x)^2 \end{bmatrix}\,dx \begin{bmatrix}a_0\\a_1\\a_2\\a_3\\a_4\end{bmatrix}=-\frac{3}{2}\int_0^1\begin{bmatrix}1\\x\\x^2\\x^3\\x^4\end{bmatrix}\, dx $$ (7.41)
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$$  \displaystyle \int_0^1\begin{bmatrix} 0 & 0 & 2 & 6(x) & 12(x)^2\\ 0 & 0 & 2(x) & 6(x)^2 & 12(x)^3\\ 0 & 0 & 2(x)^2 & 6(x)^3 & 12(x)^4\\ 0 & 0 & 2(x)^3 & 6(x)^4 & 12(x)^5\\ 0 & 0 & 2(x)^4 & 6(x)^5 & 12(x)^6 \end{bmatrix}\,dx \begin{bmatrix}a_0\\a_1\\a_2\\a_3\\a_4\end{bmatrix}=-\frac{3}{2}\int_0^1\begin{bmatrix}1\\x\\x^2\\x^3\\x^4\end{bmatrix}\, dx $$ (7.42)
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$$  \displaystyle \begin{bmatrix} 0 & 0 & 2 & 3 & 4\\ 0 & 0 & 1 & 2 & 3\\                                       0 & 0 & 2/3 & 3/2 & 12/5\\                                         0 & 0 & 1/2 & 6/5 & 2\\                                         0 & 0 & 2/5 & 1 & 12/7 \end{bmatrix} \begin{bmatrix}a_0\\a_1\\a_2\\a_3\\a_4\end{bmatrix}=-\begin{bmatrix}3/2\\ 3/4 \\ 3/6 \\ 3/8 \\ 3/10 \end{bmatrix} $$     (7.43) Now we can use the bottom three rows of Eq.7.43, along with Eq.7.36 and Eq.7.38 to develop a system of equations
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$$  \displaystyle \begin{bmatrix} 1 & 1 & 1 & 1 & 1\\ 0 & 1 & 0 & 0 & 0\\                                       0 & 0 & 2/3 & 3/2 & 12/5\\                                         0 & 0 & 1/2 & 6/5 & 2\\                                         0 & 0 & 2/5 & 1 & 12/7 \end{bmatrix} \begin{bmatrix}a_0\\a_1\\a_2\\a_3\\a_4\end{bmatrix}=-\begin{bmatrix} 0 \\ 4 \\ 3/6 \\ 3/8 \\ 3/10 \end{bmatrix} $$     (7.44)
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Solving this system using MATLAB, the solution becomes
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$$  \displaystyle \begin{bmatrix}a_0\\a_1\\a_2\\a_3\\a_4\end{bmatrix}=\begin{bmatrix} 19/4 \\ -4 \\ -3/4 \\ 0 \\ 0 \end{bmatrix} $$     (7.45)
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Solution

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$$  \displaystyle \therefore u^h(x)=\frac{19}{4}-4(x)-\frac{3}{4}(x)^2 $$     (7.46) We can see from above in Eq.7.46 that our solution did not improve at all by an increase in n from 2 to 4. This should have been intuitive from the start since we showed earlier that using n=2 gives us an equivalent $$u(x)$$
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Graphical Comparison Between Exact and Numerical Approximate Solutions
Egm5526.s11.team-2.langpm 14:38, 14 February 2011 (UTC)

Case:n=6

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$$  \displaystyle \begin{align} u^h(x) &= a_0b_0+a_1b_1+a_2b_2+a_3b_3+a_4b_4+a_5b_5+a_6b_6\\ &=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+a_6x^6 \end{align} $$     (7.47) Using the given boundary conditions in Eq.7.2 and Eq.7.3 we have the following to restrictions on our approximated solution.
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$$  \displaystyle \begin{align} u^h(x=1)=&a_0+a_1+a_2+a_3+a_4+a_5+a_6=0\\ &a_0+a_1+a_2+a_3+a_4+a_5+a_6=0 \end{align} $$     (7.48)
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$$  \displaystyle a_0+a_1+a_2+a_3+a_4+a_5+a_6=0 $$     (7.49)
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$$  \displaystyle \frac{du^h}{dx}(x=0)=0+a_1+0+0+0+0+0 $$     (7.50)
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$$  \displaystyle a_1=-4 $$     (7.51) Using the same linear differential operator as we have for the past 2 cases, Eq.7.14 we have the following.
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$$  \displaystyle P(u^h)=\frac{d^2}{dx^2}\begin{bmatrix} 1 & x & x^2 & x^3 & x^4 & x^5 & x^6\end{bmatrix}\begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ a_3 \\ a_4 \\ a_5 \\ a_6 \end{bmatrix} +\frac{3}{2} $$     (7.52) Forming the inner product.
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$$  \displaystyle \left \langle b_i,P(u^h) \right \rangle = \int_0^1 b_iP(u^h)\,dx = 0 $$     (7.53)
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$$  \displaystyle \int_0^1\begin{bmatrix} 0 & 0 & 2 & 6(x) & 12(x)^2 & 20(x)^3 & 30(x)^4\\ 0 & 0 & 2(x) & 6(x)^2 & 12(x)^3 & 20(x)^4 & 30(x)^5\\ 0 & 0 & 2(x)^2 & 6(x)^3 & 12(x)^4 & 20(x)^5 & 30(x)^6\\ 0 & 0 & 2(x)^3 & 6(x)^4 & 12(x)^5 & 20(x)^6 & 30(x)^7\\ 0 & 0 & 2(x)^4 & 6(x)^5 & 12(x)^6 & 20(x)^7 & 30(x)^8\\ 0 & 0 & 2(x)^5 & 6(x)^6 & 12(x)^7 & 20(x)^8 & 30(x)^9\\ 0 & 0 & 2(x)^6 & 6(x)^7 & 12(x)^8 & 20(x)^9 & 30(x)^{10}\end{bmatrix}\, dx \begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ a_3 \\ a_4 \\ a_5 \\ a_6 \end{bmatrix}=-\frac{3}{2} \int_0^1 \begin{bmatrix} 1 \\ x \\ (x)^2 \\ (x)^3 \\ (x)^4 \\ (x)^5 \\ (x)^6 \end{bmatrix} $$     (7.54)
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$$  \displaystyle \begin{bmatrix} 0 & 0 & 2 & 3 & 4 & 5 & 6\\ 0 & 0 & 1 & 2 & 3 & 4 & 5\\ 0 & 0 & 2/3 &  3/2 & 12/5 &  10/3 &   30/7\\ 0 & 0 &  1/2  & 6/5 &    2 &  20/7 &   15/2\\ 0 & 0 &  2/5 &  1 & 12/7 &  5/2 &     10/3\\ 0 & 0 &  1/3 & 6/7 &  3/2 & 20/9 &  3\\ 0 & 0 & 2/7 & 3/4 & 4/3 &  1/2 & 30/11\end{bmatrix}\begin{bmatrix} a_0\\ a_1\\ a_2\\ a_3\\ a_4\\ a_5\\ a_6\end{bmatrix}=\begin{bmatrix} -3/2\\   -3/4\\    -3/6\\   -3/8\\  -3/10\\   -3/12\\ -3/14\end{bmatrix} $$     (7.55) We take the bottom 5 rows from Eq.7.55 and build a system of equations with Eq.7.49 and Eq.7.51.
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$$  \displaystyle \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1\\ 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 2/3 &  3/2 & 12/5 &  10/3 &   30/7\\ 0 & 0 &  1/2  & 6/5 &    2 &  20/7 &   15/2\\ 0 & 0 &  2/5 &  1 & 12/7 &  5/2 &     10/3\\ 0 & 0 &  1/3 & 6/7 &  3/2 & 20/9 &  3\\ 0 & 0 & 2/7 & 3/4 & 4/3 &  1/2 & 30/11\end{bmatrix}\begin{bmatrix} a_0\\ a_1\\ a_2\\ a_3\\ a_4\\ a_5\\ a_6\end{bmatrix}=\begin{bmatrix} 0\\   -4\\     -3/6\\   -3/8\\  -3/10\\   -3/12\\ -3/14\end{bmatrix} $$     (7.56) As in the $$n=4$$ case, we use a MATLAB to determine the unknown $$ a_i's$$ adn to determine the final form of our approximation to the solution of the differential equation 7.1.
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$$  \displaystyle \begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ a_3 \\ a_4 \\ a_5 \\ a_6 \end{bmatrix}=\begin{bmatrix} 19/4 \\ -4 \\ -3/4 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} $$     (7.57)
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Solution

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$$  \displaystyle \therefore u^h(x)=\frac{19}{4}-4(x)-\frac{3}{4}(x)^2 $$     (7.58) We can see from above in Eq.7.58 that our solution did not improve at all by an increase in n from 2 to 6 as was expected. Comparing this to the results from problem 3.1 we can see that the shift is not necessary for this basis function. It will return an exact solution with or without the shift.
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Talk:Dignity/history is the quest

Seeking an Iconic Image
What image best illustrates the "quest for dignity" during a historic event? I though of the painting of George Washington crossing the Delaware, but believe the painting is not very accurate. Please let me know your suggestions. Thanks! --Lbeaumont 23:34, 12 February 2011 (UTC)

 * Maybe a picture of Nixon and the Watergate Scandal? -- dark lama  23:40, 12 February 2011 (UTC)

title: Talk:Dignity/history is the quest


text:
User:Gregthegeek

Hey. It's Greg.

title: User:Gregthegeek


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User:Eml5526.s11.team04.premchand/HW3

= Problem 3.3 - Direct approach for truss structure =

The following problem is problem 2.3 as taken from Fish and Belytschko's 'A first course in Finite Elements'

Problem Statement
For the truss structure given in Figure 3.1, Nodes A and B are fixed.A force equal to 10N acts in the positive x-direction at node C.Co-ordinates of joints are given in meters.Young's modulus is E=1011Pa and the cross-sectional area for all bars are A=2.10-2m2.

 * 1) Number the elements and nodes.
 * 2) Assemble the global stiffness and force matrix.
 * 3) Partition the system and solve for the nodal displacements.
 * 4) Compute the stresses and reaction forces.

Fig. 3.1 Data for Problem 3

Number the elements and nodes
The nodes are numbered starting with nodes which have prescribed displacements. Numbering in this manner will allow the assembled matrix to be partitioned easily later on.The global node number is given beside the node.While the element number is given in parenthesis beside the elements.

Fig. 3.2

Assembling the global stiffness and force matrix
The following matrix is used for solving the individual element matrix

Element 1 is numbered with global nodes 1 and 4. It is positioned at an angle $$\Phi=90^{0}$$ with respect to positive x axis.So cos $$90^{0}$$=0, sin $$90^{0}$$=1, l(1)=l, A(1)E(1)/l(1)= AE/l

Element 2 is numbered with global nodes 2 and 3. It is positioned at an angle $$\Phi=90^{0}$$ with respect to positive x axis.So cos $$90^{0}$$=0, sin $$90^{0}$$=1, l(2)=l, A(2)E(2)/l(2)= AE/l

Element 3 is numbered with global nodes 4 and 3. It is positioned at an angle $$\Phi=0^{0}$$ with respect to positive x axis.So cos $$0^{0}$$=1, sin $$0^{0}$$=0, l(3)=l, A(3)E(3)/l(3)= AE/l

Element 4 is numbered with global nodes 2 and 4. It is positioned at an angle $$\Phi=135^{0}$$ with respect to positive x axis.So cos $$135^{0}$$=$$-\frac{1}{\sqrt{2}}$$, sin $$135^{0}$$=$$\frac{1}{\sqrt{2}}$$, l(4)=l, A(4)E(4)/l(4)= AE/l

The assembled global stiffness matrix is as follows:

The force,displacement and reaction matrices are as follows:
The global system of equations are as follows:

Partitioning the system and solving for nodal displacements
The global system as shown in equation 3.9 is partitioned after four rows and four columns:

Solving using the above equation, the unknown nodal displacements are

Solving for reactions and stresses
Using the above equation the unknown reaction values are

The stresses in the elements are given by the following expression

For element 1:

 * $$ cos90^{\circ} = 0,\quad sin90^{\circ}=1, $$

\mathbf{d}^{2}=l/AE\begin{bmatrix} u_{1x}\\ u_{1y}\\ u_{2x}\\ u_{2y} \end{bmatrix}, \qquad \sigma ^{(2)}=1/A\begin{bmatrix} 0 & -1 &0 &1 \end{bmatrix}\begin{bmatrix} 0\\ 0\\ 38.2885\\ 10 \end{bmatrix} = 10/A = 10/(2*10^{-2})= 500N/m^{2}

$$

For element 2:

 * $$ cos90^{\circ} = 0,\quad sin90^{\circ}=1, $$

\mathbf{d}^{2}=l/AE\begin{bmatrix} u_{2x}\\ u_{2y}\\ u_{3x}\\ u_{3y} \end{bmatrix}, \qquad \sigma ^{(2)}=1/A\begin{bmatrix} 0 & -1 &0 &1 \end{bmatrix}\begin{bmatrix} 0\\ 0\\ 48.2885\\ 0 \end{bmatrix} = 0N/m^{2}

$$

For element 3:

 * $$ cos0^{\circ} = 1,\quad sin0^{\circ}=0, $$

\mathbf{d}^{2}=l/AE\begin{bmatrix} u_{4x}\\ u_{4y}\\ u_{3x}\\ u_{3y} \end{bmatrix}, \qquad \sigma ^{(2)}=1/A\begin{bmatrix} -1& 0 &1 &0 \end{bmatrix}\begin{bmatrix} 38.2885\\ 10\\ 48.2885\\ 0 \end{bmatrix} = 10/A = 10/(2*10^{-2})= 500N/m^{2}

$$

For element 4:

 * $$ cos135^{\circ} = -0.707,\quad sin135^{\circ}=0.707, $$

\mathbf{d}^{2}=l/AE\begin{bmatrix} u_{2x}\\ u_{2y}\\ u_{4x}\\ u_{4y} \end{bmatrix}, \qquad \sigma ^{(2)}=1/A\begin{bmatrix} 0.707 & -0.707 &-0.707 &0.707 \end{bmatrix}\begin{bmatrix} 0\\ 0\\ 38.2885\\ 10 \end{bmatrix} = 10/A = -20/(2*10^{-2})= 1000N/m^{2}

$$

title: User:Eml5526.s11.team04.premchand/HW3


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User talk:Mayur (usurped)~enwikiversity

Hi, i have requested here to claim your user name so as to have global username--Mayurbot 03:36, 13 February 2011 (UTC)

-- Jtneill - Talk - c 04:32, 13 February 2011 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Mayur (usurped). To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Mayur (usurped)~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:38, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:15, 19 April 2015 (UTC)

title: User talk:Mayur (usurped)~enwikiversity


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text:
File talk:DemonWiki.png

License
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title: File talk:DemonWiki.png


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User:EML5526.S11.Team1.HS/HW3

=HW3 Draft=

2.4
Given a three bar structure subjected to the prescribed load at point C equal to $$ 10^3 ~N $$ as shown in Figure 2.19. The Young's modulus is $$ E = 10^{11} ~Pa $$, the cross-sectional area of the bar BC is $$ 2 \times 10^{-2} ~m^2 $$ and that of BD and BF is $$ 10^{-2} ~m^2 $$. Note that point D is free to move in the x-direction. Coordinates of joints are given in meters.

a. Construct the global stiffness matrix and load matrix.

b. Partition the matrices and solve for the unknown displacements at Point B and displacement in the x-direction at point D.

c. Find stresses in the three bars.

d. Find the reactions at nodes C,D,F.

a)
For Element 3, $$ \phi = 45^o,I = 3,J =4 $$

$$ K^{\left(3\right)} = \frac{10^{-2} \times 10^{11}}{\sqrt{2}} \begin{bmatrix} \frac{1}{2} &\frac{1}{2} &-\frac{1}{2} &-\frac{1}{2}\\ \frac{1}{2} &\frac{1}{2} &-\frac{1}{2} &-\frac{1}{2}\\ -\frac{1}{2} &-\frac{1}{2} &\frac{1}{2} &\frac{1}{2}\\ -\frac{1}{2} &-\frac{1}{2} &\frac{1}{2} &\frac{1}{2} \end{bmatrix} $$

For Element 2, $$ \phi = 90^o,I = 2,J = 4 $$

$$ K^{\left(2\right)} = \frac{2 \times 10^{-2} \times 10^{11}}{1} \begin{bmatrix} 0 &0  &0  &0\\  0  &1  &0 &-1\\  0  &0  &0 &0 \\  0  &-1  &0 &1 \end{bmatrix} $$

For Element 1, $$ \phi = 135^o,I = 1,J =4 $$

$$ K^{\left(1\right)} = \frac{10^{-2} \times 10^{11}}{\sqrt{2}} \begin{bmatrix} \frac{1}{2} &-\frac{1}{2} &-\frac{1}{2} &\frac{1}{2}\\ -\frac{1}{2} &\frac{1}{2} &\frac{1}{2} &-\frac{1}{2}\\ -\frac{1}{2} &\frac{1}{2} &\frac{1}{2} &-\frac{1}{2}\\ \frac{1}{2} &-\frac{1}{2} &-\frac{1}{2} &\frac{1}{2} \end{bmatrix} $$

Global Stiffness Matrix

$$ K = 10^{9} \begin{bmatrix} \frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}} &0  &0  &0  &0  &-\frac{1}{2 \sqrt{2}}  &\frac{1}{2 \sqrt{2}} \\ -\frac{1}{2 \sqrt{2}} &\frac{1}{2 \sqrt{2}} &0  &0  &0  &0  &\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}} \\ 0 &0 &0  &0  &0  &0  &0  &0 \\ 0 &0  &0  &2  &0  &0  &0  &-2 \\ 0 &0  &0  &0  &\frac{1}{2 \sqrt{2}}  &\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}} \\ 0 &0 &0  &0  &\frac{1}{2 \sqrt{2}}  &\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}} \\ -\frac{1}{2 \sqrt{2}} &\frac{1}{2 \sqrt{2}} &0  &0  &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}}  &\frac{1}{\sqrt{2}}  &0 \\ \frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}} &0  &-2  &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}}  &0  &\frac{1}{\sqrt{2}}+2 \end{bmatrix} $$

$$ d = \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ u_{3x}\\ 0\\ u_{4x}\\ u_{4y} \end{bmatrix} f = \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 10^3\\ 0 \end{bmatrix} r = \begin{bmatrix} r_{1x}\\ r_{1y}\\ r_{2x}\\ r_{2y}\\ 0\\ r_{3y}\\ 0\\ 0 \end{bmatrix} $$

Therefore

$$ load~matrix = \begin{bmatrix} r_{1x}\\ r_{1y}\\ r_{2x}\\ r_{2y}\\ 0\\ r_{3y}\\ 10^3\\ 0 \end{bmatrix} $$

b)
Global system of equations:

$$ \begin{bmatrix} \frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}} &0  &0 &\mid &0  &0  &-\frac{1}{2 \sqrt{2}}  &\frac{1}{2 \sqrt{2}} \\ -\frac{1}{2 \sqrt{2}} &\frac{1}{2 \sqrt{2}} &0  &0 &\mid  &0  &0  &\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}} \\ 0 &0 &0  &0 &\mid &0  &0  &0  &0 \\ 0 &0 &0  &2 &\mid &0  &0  &0  &-2 \\ - &- &- &- &\mid &- &- &- &- \\ 0 &0 &0  &0 &\mid &\frac{1}{2 \sqrt{2}}  &\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}} \\ 0 &0 &0  &0 &\mid &\frac{1}{2 \sqrt{2}}  &\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}} \\ -\frac{1}{2 \sqrt{2}} &\frac{1}{2 \sqrt{2}} &0  &0 &\mid &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}}  &\frac{1}{\sqrt{2}}  &0 \\ \frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}} &0  &-2 &\mid &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}}  &0  &\frac{1}{\sqrt{2}}+2 \end{bmatrix} \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ --\\ u_{3x}\\ 0\\ u_{4x}\\ u_{4y} \end{bmatrix} = \begin{bmatrix} r_{1x}\\ r_{1y}\\ r_{2x}\\ r_{2y}\\ --\\ 0\\ r_{3y}\\ 10^3\\ 0 \end{bmatrix} $$

Partitions

$$ \bar{d}_{E} = \begin{bmatrix} u_{1x}\\ u_{1y}\\ u_{2x}\\ u_{2y} \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\\ 0 \end{bmatrix},

d_{F}= \begin{bmatrix} u_{3x}\\ 0\\ u_{4x}\\ u_{4y} \end{bmatrix},

r_{E} = \begin{bmatrix} r_{1x}\\ r_{1y}\\ r_{2x}\\ r_{2y} \end{bmatrix} ,

f_{F} = \begin{bmatrix} 0\\ r_{3y}\\ 10^3\\ 0 \end{bmatrix} $$

$$ K_{EF} = \begin{bmatrix} 0 &0 &-\frac{1}{2 \sqrt{2}}  &\frac{1}{2 \sqrt{2}} \\ 0 &0 &\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}} \\ 0 &0 &0  &0 \\ 0 &0  &0  &-2\ \end{bmatrix},

K_{EF}^{T} = \begin{bmatrix} 0 &0 &0  &0 \\ 0 &0  &0  &0 \\ -\frac{1}{2 \sqrt{2}} &\frac{1}{2 \sqrt{2}}  &0  &0 \\ \frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}} &0  &-2 \end{bmatrix},

K_{F} = \begin{bmatrix} \frac{1}{2 \sqrt{2}} &\frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}} \\ \frac{1}{2 \sqrt{2}} &\frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}} \\ -\frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}} &\frac{1}{\sqrt{2}}  &0 \\ -\frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}} &0  &\frac{1}{\sqrt{2}}+2 \end{bmatrix},

K_{E} = \begin{bmatrix} \frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}} &0  &0 \\ -\frac{1}{2 \sqrt{2}} &\frac{1}{2 \sqrt{2}} &0  &0 \\ 0 &0 &0  &0 \\ 0 &0  &0  &2 \end{bmatrix} $$

Unknown displacement at B is solved by

$$ K_{EF}^{T}\bar{d}_{E} + K_{F}d_{F} = f_{F} $$

Subtracting the first term from both sides of the above equation and premultiply by $$ K_F^{-1} $$, we obtain

$$ d_F = K_F^{-1} \left( f_F - K_{EF}^T \bar{d}_E \right) $$

Solving for the displacements we get: $$ d_{F}= 10^{-6} \begin{bmatrix} 1+2\sqrt{2}\\ 0\\ \frac{1}{2}+2\sqrt{2}\\ \frac{1}{2} \end{bmatrix} ~m $$

c)
Element 3 Stress

$$ \sigma = 0 ~Pa $$

Element 2 Stress

$$ \sigma = \frac{E}{l} \begin{bmatrix} 0 &-1 &0  &1 \end{bmatrix} 10^{-6} \begin{bmatrix} \frac{1}{2} + 2\sqrt{2}\\ \frac{1}{2}\\ 0\\ 0 \end{bmatrix} = -5 \times 10^4 ~Pa $$

Element 1 Stress

$$ \sigma = \frac{E}{l} \begin{bmatrix} \frac{\sqrt{2}}{2} &-\frac{\sqrt{2}}{2} &\frac{\sqrt{2}}{2}  &\frac{\sqrt{2}}{2} \end{bmatrix} 10^{-6} \begin{bmatrix} \frac{1}{2} + 2\sqrt{2}\\ \frac{1}{2}\\ 0\\ 0 \end{bmatrix} = \sqrt{2} \times 10^5 ~Pa $$

d)
Reactions can be solved using this equation and the global matrix

$$ r_E = K_E \bar{d}_E + K_{EF} d_F $$

$$ load~matrix = \begin{bmatrix} r_{1x}\\ r_{1y}\\ r_{2x}\\ r_{2y}\\ 0\\ r_{3y}\\ 10^3\\ 0 \end{bmatrix}

= \begin{bmatrix} -10^3\\ 10^3\\ 0\\ -10^3\\ 0\\ 0\\ 10^3\\ 0 \end{bmatrix} ~ N $$

3.3
Consider a trial (candidate) solution of the form $$ u\left ( x \right ) = \alpha_{0} + \alpha_{1}\left ( x-3 \right ) $$ and a weight function of the same form. Obtain a solution to the weak form in Problem 3.1. Check the equilibrium equation in the strong form in Problem 3.1; is it satisfied?

Check the natural boundary condition; is it satisfied?

$$ u\left ( x \right ) = \alpha_{0} + \alpha_{1}\left ( x-3 \right ) $$

$$ w\left ( x \right ) = \beta_{0} + \beta_{1}\left ( x-3 \right ) $$

Using boundary conditions from Problem 3.9

$$ w \left(3 \right) = 0 \therefore  \beta_{0} = 0 $$

$$ u \left(3 \right) = \alpha_{0} = 0.001 $$

Only one unknown parameter and one arbitrary parameter remain

$$ u\left ( x \right ) = 0.001 + \alpha_{1}\left ( x-3 \right ) \frac{du}{dx} = \alpha_{1} $$

$$ w\left( x \right) = \beta_{1} \left(x - 3 \right) \frac{dw}{dx} = \beta_{1} $$

Now substitute into Problem 3.1 weak form

$$ \int_{1}^{3} \frac{\mathrm{d} w }{\mathrm{d} x}AE\frac{\mathrm{d} u}{\mathrm{d} x}dx =  -0.1(wA)_{x=1} + \int_{1}^{3} 2xwdx$$

$$ \int_{1}^{3} \beta_{1}\alpha_{1}EAdx + 0.1(\beta_{1} \left(x-3 \right)A)_{x=1} - \int_{1}^{3} 2\beta_{1}x \left(x-3 \right)dx = 0 $$

Integrating and factoring out $$ \beta_1 $$ we get:

$$ \beta_{1} \left( 2 \alpha_{1} EA - 0.2A + \frac{20}{3} \right) = 0 $$

The $$ \beta_1 $$ goes to zero and $$ \alpha_1 $$ can be solved:

$$ \alpha_{1} = \frac{0.1A - \frac{10}{3}}{EA} $$

Therefore the solution to the weak form is

$$ u \left( x \right) = 0.001 + \frac{0.1A - \frac{10}{3}}{EA} \left( x-3 \right) $$

Checking for equilibrium in the strong form, values are substituted into

$$ \frac{d}{dx} \left( AE \frac{du}{dx} \right) + 2x = 0 $$

This yields $$ 2x = 0 $$ which does not satisfy.

Checking the natural boundary condition, this equation is used

$$ \sigma \left(1 \right) = \left(E \frac{du}{dx} \right)_{x=1} = 0.1 $$

Substituting in $$ \alpha_1 $$ we get

$$ 0.1 - \frac{\frac{10}{3}}{A} = 0.1 $$

Thus, this boundary condition will satisfy only for large values of $$ A $$

title: User:EML5526.S11.Team1.HS/HW3


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text:
Computer Essentials/Archived Homework/PC Term 2, 2010/2011

Desho Sect 2
Specifications

Motherboard: Intel® DH55HC

Processor: Intel® Core™ i7-980X processor Extreme (12MB L2 Cache, 3.33GHz)

Memory: Up to 24GB4 Tri-Channel DDR3 SDRAM at 1333Mhz - 6 DIMMS

Case: Default XPS 9100 Case

Power supply: ENERMAX MODU82+II EMD525AWT-II

Floppy drive: N/A

Hard drive: 4TB Performance RAID 0 (2 x 2TB SATA 3Gb/s 7200 RPM HDDs)

Nistha Sect 2
The best PC computer: Alienware area-51 ALX Desktop

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Sound card: Creative Sound Blaster X-Fi Xtreme Audio

AGP: MATROX G450 MMS 128MB (G450MMS/QPD/B)

Modem: External 56K USB Modem

Paranat Sect 2
CPU CPU InTel Core i7-980x Extreme Edition 3.33GHz RAM RAM 16 GB (4GBx8) DDR3 1600 G.SKILL Monitor LED Samsung XL2370 23″ 1920x1080 5,000,000:1 GRAPHIC CARD Graphic card HIS ATI Radeon HD 5970 2GB DDR5 Bit 256x2 HARDDISK HARDDISK 8TB Segate SATA2 7200RPM (2 Harddisk) Power supply PSU Silverstone 1500w Motherboard M/B ASRock X58 Extreme 3 LGA 1366 Optical drive DVD/Blu-Ray ROW ASUS KeyBoards SteelSeries : 7G Mouse SteelSeries : Xai Laser Case Case Gview GOLF Sound card Creative Sound Blaster X-Fi Elite Pro + 7.1

Atchara Sect 1

 * CHASSIS COLOR ALX Space Black - Anodized Aluminum Chassis PROCESSOR
 * Overclocked Intel® Core™ i7 980x Extreme Six Core Processor (4.0GHz, 12MB Cache)
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My Accessories ALSO INC

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Chanida Sect 1
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Taveechai Karuchit Sect 1
Spec-Sheet

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Processors: Overclocked Intel® CoreTM i7 980x Extreme Six Core Processor (4.0GHz, 12MB Cache)

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Hard Drives: 512GB Raid 0 (2x 256GB Samsung Solid State HDDs)

2nd Hard Drive: 2TB RAID 1 (2x2TB SATA-II, 7,200 RPM, 32MB Cache HDDs)

Floppy Drive: 19-in-1 Media Card Reader

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Sound Card: Creative Sound Blaster X-Fi Titanium

TV Tuners & Remote Controls: Hauppauge WinTV HVR-1250 Hybrid TV Tuner with Remote Control

Optical Drives: Dual Drives: Blu-ray Disc (BD) Burner (Writes to DVD/CD/BD) and DVDRW


text:
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text:
User:Eml5526.s11.t1.ab/HW3

=Problem 2=

1.1 Number the Element and Nodes.
The Spring Systems is shown in the figure.

Superscripts to each of the stiffness values denotes the element number. Furthermore, Node Numbers are also shown in the figure. As shown in the figure, there are 4 nodes.

1.2 Assemble the Global Stiffness and Force Matrix.
For Element Number 1,$$I=1 $$ and $$J=4$$,

$$K^{(1)}=k\begin{bmatrix} 1& -1\\ -1 & 1 \end{bmatrix}\Rightarrow \tilde{K}=\begin{bmatrix} k& 0 & 0 & -k\\ 0 & 0 & 0 & 0\\ 0& 0 & 0 & 0\\ -k& 0 & 0 & k \end{bmatrix}$$

For Element Number 2,$$I=4 $$ and $$J=3$$,

$$K^{(2)}=2k\begin{bmatrix} 1& -1\\ -1 & 1 \end{bmatrix}\Rightarrow \tilde{K}=\begin{bmatrix} 0& 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0& 0 & 2k& -2k\\ 0& 0 & -2k & 2k \end{bmatrix}$$

For Element Number 3,$$I=3 $$ and $$J=2$$,

$$K^{(3)}=k\begin{bmatrix} 1& -1\\ -1 & 1 \end{bmatrix}\Rightarrow \tilde{K}=\begin{bmatrix} 0& 0 & 0 & 0\\ 0 & k & -k & 0\\ 0& -k & k& 0\\ 0& 0 & 0& 0 \end{bmatrix}$$

For Element Number 4,$$I=1 $$ and $$J=3$$,

$$K^{(4)}=3k\begin{bmatrix} 1& -1\\ -1 & 1 \end{bmatrix}\Rightarrow \tilde{K}=\begin{bmatrix} 3k& 0 & -3k & 0\\ 0 & 0& 0& 0\\ -3k& 0 & 3k& 0\\ 0& 0 & 0& 0 \end{bmatrix}$$

Assembled Global Systems Matrix will be given by,
 * {| style="width:100%" border="0"

$$K=\sum_{e=1}^{4}\tilde{K}^{e}=\begin{bmatrix} k+3k & 0 & -3k & -k\\ 0 & k & -k & 0\\ -3k & -k &6k &-2k \\ -k & 0 & -2k & 3k \end{bmatrix}$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style= |
 * }

Where Superscript $$'e'$$ denotes the Element Number.

The Displacement and Global Force Matrices for this system are given by,

1.3 Partition the system and solve for the Nodal Displacements.
Eventually, The Global System of Equations is Given by,

$$\begin{bmatrix} k+3k & 0 & -3k & -k\\ 0 & k & -k & 0\\ -3k & -k &6k &-2k \\ -k & 0 & -2k & 3k \end{bmatrix}\begin{bmatrix} 0\\ 0\\ u_{3}\\ u_{4} \end{bmatrix}=\begin{bmatrix} r_{1}\\ r_{2}\\ 50\\ 0 \end{bmatrix}$$ $$\displaystyle (Eq.2.1) $$

As the first two displacements are prescribed, We can Partition the system after 2 Rows and 2 Columns,

And Can form the Matrix System of Equivalent to,

$$\begin{bmatrix} K_{E} & K_{EF}\\ K_{EF}^{T} & K_{F} \end{bmatrix}\begin{bmatrix} \bar{d}_{E}\\ d_{F} \end{bmatrix}=\begin{bmatrix} r_{E}\\ f_{F} \end{bmatrix}$$

Where,

$$K_{E}=\begin{bmatrix} 4k & 0\\ 0 & K \end{bmatrix}, K_{EF}=\begin{bmatrix} -3k & -k\\ -k& 0 \end{bmatrix},K_{EF} ^T=\begin{bmatrix} -3k & -k\\ -k & 0 \end{bmatrix}, K_{F}=\begin{bmatrix} 6k & -2k\\ -2k & 3K \end{bmatrix}, \bar{d}_{E}=\begin{bmatrix} 0\\ 0\\ \end{bmatrix}, d_{F}=\begin{bmatrix} u_{3}\\ u_{4} \end{bmatrix},r_{E}=\begin{bmatrix} r_{1}\\ r_{2} \end{bmatrix}, f_{F}=\begin{bmatrix} 50\\ 0 \end{bmatrix}$$

Solving the Following Reduced System of Equations,

$$K_{E}\bar{d}_{E}+K_{EF}d_{F}=r_{E}, K_{EF}^T\bar{d}_{E}+K_{F}d_{F}=f_{F}$$

Using $$\displaystyle (Eq.2.1) $$ we can form the Following Simultaneous Equations,

$$ 6k.u_{3}-2k.u_{4}=50$$ $$\displaystyle (Eq.2.2) $$

$$ -2k.u_{3}-3k.u_{4}=0$$ $$\displaystyle (Eq.2.3) $$

Solving These 2 Simultaneous Equations, We get,

1.4 Compute the Reaction Forces.
Now, Utilizing the First row of the math>\displaystyle (Eq.2.1) we can get

And Further, Utilizing the Second row of the math>\displaystyle (Eq.2.1) we can get,

=Problem 3=

Given
===1.1 The truss Structure Shown in the Figure. Nodes A & B are fixed. A Force of 10N Acts in the Positive x-Direction at node C. Young's Modulus is $$E=10^{11}Pa$$ and the Cross Sectional Area for all bars are $$ A= 2.10^{-2}$$===

1.1 Number the Element and Nodes.
The Truss System Along with the Element Numbers and the Node Numbers Written on it, is Shown in the figure.

1.2Assemble the Global Stiffness and Force Matrix.
For Element Number 1,$$I=1 $$ and $$J=3$$, Also Element 1 is inclined at 90 deg to the Positive X-direction,$$\phi^{(1)} =90^{0}$$, $$cos(90)=0,sin(90)=1$$, $$l=1$$,

Using the Following formula for obtaining the Elemental Stiffness Matrix for all the elements,

$$K^{e}=\frac{AE}{l}\begin{bmatrix} (cos\phi ^{e})^{2} & cos\phi ^{e}.sin\phi ^{e} & -(cos\phi ^{e})^{2} & -cos\phi ^{e}.sin\phi ^{e} \\ cos\phi ^{e}.sin\phi ^{e} & (sin\phi ^{e})^{2} & -cos\phi ^{e}.sin\phi ^{e} &-(sin\phi ^{e})^{2} \\ -(cos\phi ^{e})^{2} & -cos\phi ^{e}.sin\phi ^{e} & (cos\phi ^{e})^{2} & cos\phi ^{e}.sin\phi ^{e}\\ -cos\phi ^{e}.sin\phi ^{e} & -(sin\phi ^{e})^{2} & cos\phi ^{e}.sin\phi ^{e} & (sin\phi ^{e})^{2} \end{bmatrix}$$ $$\displaystyle (Eq.3.1) $$

Utilizing above equation for Element 1,

$$K^{(1)}=\frac{AE}{l}\begin{bmatrix} 0 & 0 & 0 & 0\\ 0 &1 &  0&-1 \\ 0 & 0 & 0 & 0\\ 0 & -1 & 0 & 1 \end{bmatrix}$$

For Element Number 2,$$I=2$$ and $$J=4$$, Also Element 2 is inclined at 90 deg to the Positive X-direction,$$\phi^{(2)} =90^{0}$$, $$cos(90)=0,sin(90)=1$$, $$l=1$$, Therefore,

$$K^{(2)}=\frac{AE}{l}\begin{bmatrix} 0 & 0 & 0 & 0\\ 0 &1 &  0&-1 \\ 0 & 0 & 0 & 0\\ 0 & -1 & 0 & 1 \end{bmatrix}$$

For Element Number 3,$$I=2$$ and $$J=3$$, Also Element 3 is inclined at 135 deg to the Positive X-direction,$$\phi^{(3)} =135^{0}$$, $$cos(135)=\frac{-1}{\sqrt{2}},sin(135)=\frac{1}{\sqrt{2}}$$, $$l=\sqrt{2}$$, Therefore,

$$K^{(3)}=\frac{AE}{l}\begin{bmatrix} \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}}\\ -\frac{1}{2\sqrt{2}} &\frac{1}{2\sqrt{2}} &  \frac{1}{2\sqrt{2}}&-\frac{1}{2\sqrt{2}} \\ -\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}}\\ \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} \end{bmatrix}$$

For Element Number 4,$$I=3$$ and $$J=4$$, Also Element 4 is inclined at 90 deg to the Positive X-direction,$$\phi^{(2)} =0^{0}$$, $$cos(0)=1,sin(0)=0$$, $$l=1$$, Therefore,

$$K^{(4)}=\frac{AE}{l}\begin{bmatrix} 1 & 0 & -1 & 0\\ 0 &0 &  0&0\\ -1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}$$

Making the Assembly of all these 4 Elemental Matrices, we get the following Global Stiffness Mtria for the given Truss System,

$$\displaystyle (Eq.3.2) $$ Now, Only force is acting in positive X-Direction at C, and there will be Reactions at A & B.

Therefore, The Force Matrix Can be Assembled as,

$$\displaystyle (Eq.3.3) $$

1.3Partition the system and solve for the Nodal Displacements.
From $$\displaystyle (Eq.3.2) $$ & $$\displaystyle (Eq.3.3)  $$ The Global System os Equation will be,

$$AE \begin{bmatrix} 0 & 0 &0 & 0 & 0 & 0 &  0& 0\\ 0 & 1 & 0 & 0 & 0 & -1 & 0 &0 \\ 0 & 0 & \frac{1}{2\sqrt{2}} & \frac{-1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} &0  &0 \\ 0 & 0 & -\frac{1}{2\sqrt{2}} &1+\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & 0 & -1\\ 0 & 0 & -\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} & 1+\frac{1}{2\sqrt{2}} &-\frac{1}{2\sqrt{2}} & -1 &0 \\ 0 &-1 &\frac{1}{2\sqrt{2}}  & -\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} &1+\frac{1}{2\sqrt{2}}  & 0 & 0\\ 0 & 0 & 0 & 0 & -1 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 & 0 & 0 & 0 & 1 \end{bmatrix}.\begin{bmatrix} 0\\ 0\\ 0\\ 0\\ u_{3x}\\ u_{3y}\\ u_{4x}\\ u_{4y} \end{bmatrix}=\begin{bmatrix} r_{1x}\\ r_{1y}\\ r_{2x}\\ r_{2y}\\ 0\\ 0\\ 10\\ 0 \end{bmatrix}$$

As the Displacements are prescribed untill second node, Partitioning the systems after second row and column and further comparing with the following equation,

$$\begin{bmatrix} K_{E} & K_{EF}\\ K_{EF}^{T} & K_{F} \end{bmatrix}\begin{bmatrix} \bar{d}_{E}\\ d_{F} \end{bmatrix}=\begin{bmatrix} r_{E}\\ f_{F} \end{bmatrix}$$

we get, $$\bar{d_{E}}=\begin{bmatrix} 0\\ 0\\ 0\\ 0 \end{bmatrix},\ d_{F}=\begin{bmatrix} u_{3x}\\ u_{3y}\\ u_{4x}\\ u_{4y} \end{bmatrix},\ r_{E}=\begin{bmatrix} r_{1x}\\ r_{1y}\\ r_{2x}\\ r_{2y} \end{bmatrix},\ f_{F}=\begin{bmatrix} 0\\ 0\\ 10\\ 0 \end{bmatrix},\ K_{F}=\begin{bmatrix} 1+\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & -1 &0 \\ -\frac{1}{2\sqrt{2}} &1+\frac{1}{2\sqrt{2}} &0  & 0\\ -1 & 0 & 1 & 0\\ 0 & 0 & 0 &1 \end{bmatrix},\ K_{EF}=\begin{bmatrix} 0 & 0 & 0 & 0\\ 0 &-1 &0  & 0\\ -\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} &  0& 0\\ \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} &0 &-1 \end{bmatrix}$$

Eventually, we can obtain,

$$d_{F}=K_{F}^{-1}(f_{F}-K_{EF}^T.\bar{d}_{E})$$

But $$\bar{d}_{E}$$ is a Zero Matrix, which will yield, $$d_{F}=K_{F}^{-1}.f_{F}$$ Solving this system in MATLAB, Using the following code, we get,

 Matlab Code:

$$\displaystyle (Eq.3.4) $$

1.4Compute the Stresses & Reactions.
Now the Reactions can be calculated as,

$$r_{E}=K_{E}.\bar{d}_{E}+K_{EF}.d_{F}$$

But again $$\bar{d}_{E}$$ is a Zero Matrix, Which will give, $$r_{E}=K_{EF}.d_{F}$$.

Using the Following MATLAB Code to obtain the Reaction Matrix,

 Matlab Code:

$$\displaystyle (Eq.3.5) $$

Now We can calculate the Stress in each element using the following formula,

$$\sigma ^e=\frac{E^e}{l^e}\begin{bmatrix} -cos\phi ^{e} &-sin\phi ^{e} & cos\phi ^{e} & sin\phi ^{e} \end{bmatrix}.d^e$$ $$\displaystyle (Eq.3.6) $$

Now we can utilize the above equation to calculate the Stresses for all the elements,

For element 1, $$\phi =90^0, l=1$$ &

$$d^1=\begin{bmatrix} u_{1x}\\ u_{1y}\\ u_{3x}\\ u_{3y} \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 38.2843\\ 10 \end{bmatrix}$$ Using the Following MATLAB code,  Matlab Code:

Therefore,

In the similar fashion, For Element 2, Utilizing $$\displaystyle (Eq.3.6) $$ we get,

For Element 3, Utilizing $$\displaystyle (Eq.3.6) $$ we get,

Finally, For Element 4, Utilizing $$\displaystyle (Eq.3.6) $$ we get,

title: User:Eml5526.s11.t1.ab/HW3


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User:Egm6341.s11.team1.arm/HW3

Part 1
From lecture 9, page 9-2, equation 1 :

From lecture 11, page 11-3, equation 3 :

Using the following error definition:

The error at $$t=5 \ $$ and $$x=0.75 \ $$ is

 Matlab code:

Part 2
Similarly,

and the error at $$t = 4.5 \ $$ and $$x = 3 \ $$ is

 Matlab code:

Problem 4a
Information on the history and origin of the bifolium curve is sparse. The earliest found references to the curve originate in papers written by Isaac Newton in the 1690s. The curve was later included in several indexes of quartic curves in the late 1800s. The few sources that could be found that discussed the curve were either mathematics articles or books and did not mention any possible applications of the bifolium curve.

title: User:Egm6341.s11.team1.arm/HW3


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text:
A Translation of the Bible/Exodus

Chapter 1
1 And these are the names of the sons of Israel, who came into Egypt with Jacob; every man came with his household: 2 Reuben, Simeon, Levi and Judah; 3 Issachar, Zebulun and Benjamin; 4 Dan and Naphtali, Gad and Asher. 5 And all the living people that came out of the loins of Jacob were seventy living people; and Joseph was in Egypt already. 6 And Joseph died, and all his brothers, and all that generation. 7 And the children of Israel were fertile, and increased greatly, and multiplied, and grew exceedingly mighty; and the land was filled with them.

8 And there arose a new king over Egypt, who had not known Joseph. 9 And he said to his people: "Behold the people; the children of Israel are too many and too mighty for us. 10 Come, let us deal wisely with them, in case they multiply, and it should happen that, when we get involved in any war, they also join themselves to our enemies, and fight against us, and leave the land." 11 Therefore they set taskmasters over them to afflict them with their burdens. And they built store-cities, Pithom and Raamses, for Pharaoh. 12 But the more they afflicted them, the more they multiplied and the more they spread abroad. And they were terrified because of the children of Israel. 13 And the Egyptians forced the children of Israel into slavery. 14 And they made their lives bitter with hard service, in mortar and in brick, and in all manner of service in the field; in all their service, where they made them work as slaves. 15 And the king of Egypt spoke to the Hebrew midwives, of whom the name of one was Shiphrah, and the name of the other Puah; 16 and he said: "When you act as midwife to the Hebrew women, you shall look at the birth stool; if it is a son, then you shall kill him; but if it is a daughter, then she shall live." 17 But the midwives feared God, and did not do what the king of Egypt commanded them, but saved the male children alive. 18 And the king of Egypt called for the midwives, and said unto them: "Why have you done this thing, and have saved the male children alive?' 19 And the midwives said to Pharaoh: "Because the Hebrew women are not like the Egyptian women; for they are vigorous, and are delivered before the midwife can come to them." 20 And God treated the midwives well; and the people multiplied, and grew very mighty. 21 And it came to pass, because the midwives feared God, that He made them houses. 22 And Pharaoh ordered all his people, saying: "You shall throw every son that is born into the river, and you shall save alive every daughter."

title: A Translation of the Bible/Exodus


text:
User:SB Johnny/4 options

Contributions to the main resolutions page should be at most 3 lines of text, and begin with one of the following template-assisted signals:

 * - Use this if you support the resolution as written. Comments on why you support it will be helpful for other contributors.
 * - Use this if you oppose the resolution in principle. Comments on why you oppose it will be helpful for other contributors.
 * - Use this if you support the goal -- or part of the goal -- that you believe the resolution is aimed at achieving, but think the language needs to be changed or the scope narrowed or expanded. Please be sure to either suggest new language, or state your support for an amendment that someone else has suggested.
 * - Use this to show that you are willing to accept or abandon the resolution for now, in the interests of moving on. This option would normally be used to replace an earlier support or oppose.
 * - Use this to show that you have read the proposed resolution, but either have no preference or haven't made up your mind yet.

Always make sure to sign using ~.

The discussion and resolution approach I've been trying to hash out here is adapted from the consensus discussions used by Quakers, as well in many other consensus based organizations (a number of them are also using Quaker methods, since the Quakers often volunteer to help them structure their decision-making process). It's considerably different in a lot of ways because it's (a) using text, and (b) not in "real time".

The basic outline is:
 * 1) Someone brings up a topic they think need addressing
 * 2) If enough people are interested, the topic is taken up
 * 3) The subject is broken down into the questions needing answers
 * 4) The questions are addressed and discussed, opinions are made clear
 * 5) Resolutions on what to do are floated
 * 6) Resolutions are generally amended to account for everyone's ideas
 * 7) Resolutions are either adopted (unanimously, or mostly-unanimously with some conceding to "the will of the group"), or tabled (if a consensus can't be reached in a reasonable time.
 * 8) Both tabled and adopted resolutions are often brought up at another meeting if circumstances have changed, the course action isn't working out satisfactorily, or if a number of people have changed their mind after further thought.

title: User:SB Johnny/4 options


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User:Egm6341.s11.team4.yang/hw3

=Homework 3.5=

Refer to lecture slide [[media:nm1.s11.mtg16.djvu|mtg-16]] for the problem statement.

Given

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$$ \displaystyle
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{{e}_{n}}(x)=f(x)-{{f}_{n}}(x)

$$     (Eq 1) Where, $$\displaystyle {{f}_{n}}(x)=\sum\limits_{i=0}^{n}{{{l}_{i,n}}(x)f({{x}_{i}})}$$
 *
 * }

Objectives
Prove :

Solutions
Substituting Eq. (2) into Eq. (1) yields to:
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$$ \displaystyle
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{{e}_{n}}(x)=f(x)-{{f}_{n}}(x)=f(x)-\sum\limits_{i=0}^{n}{{{l}_{i,n}}(x)f({{x}_{i}})}

$$     (Eq 3) Where, $$\displaystyle {{l}_{i,n}}(x)=\prod\limits_{j=0,j\ne i}^{n}{\frac{x-{{x}_{i}}}{{{x}_{i}}-{{x}_{j}}}}$$ is a n-order polynomial. Next, take (n+1) –order derivative on both sides of Eq.(4), it becomes:
 *
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$$ \displaystyle
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{{e}^{(n+1)}}(x)={{f}^{(n+1)}}(x)-{{[{{f}_{n}}(x)]}^{(n+1)}}={{f}^{(n+1)}}(x)-{{[\sum\limits_{i=0}^{n}{{{l}_{i,n}}(x)f({{x}_{i}})}]}^{(n+1)}}

$$     (Eq 4) Because $$\displaystyle {{l}_{i,n}}(x)=\prod\limits_{j=0,j\ne i}^{n}{\frac{x-{{x}_{i}}}{{{x}_{i}}-{{x}_{j}}}}$$ is n-order polynomial, $$\displaystyle {{f}_{n}}(x)=\sum\limits_{i=0}^{n}{{{l}_{i,n}}(x)f({{x}_{i}})}$$is also a n-order polynomial, thus taking n+1-order derivative yields:
 *
 * }
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$$ \displaystyle
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{{[{{f}_{n}}(x)]}^{(n+1)}}={{[\sum\limits_{i=0}^{n}{{{l}_{i,n}}(x)f({{x}_{i}})}]}^{(n+1)}}=0

$$     (Eq 5) So,
 *
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$$ \displaystyle
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{{e}^{(n+1)}}(x)={{f}^{(n+1)}}(x)-{{[{{f}_{n}}(x)]}^{(n+1)}}={{f}^{(n+1)}}(x)-0

$$     (Eq 6)
 *
 * }

title: User:Egm6341.s11.team4.yang/hw3


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User:EGM6341.s11.TEAM1.Sujin.HW3

Find
Prove that Eq5.1 is equal to $$\begin{align}e^{(n+1)}(x)=f^{(n+1)}(x)- \cancelto{0}{f_n^{L(n+1)}(x)} \end{align}$$ using Eq5.2 thru Eq 5.4.

Solution
Suppose that $$ f(x) $$ is at leat $$ n+1 $$ times differentiable function. From Eqn.5.3, we can confirm that;

Author

Find
Prove Eq 6.1

Solution
Thus,

Author

title: User:EGM6341.s11.TEAM1.Sujin.HW3


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