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History of Europe to 1648

This learning project is under construction. Building started on November 7, 2008 'Consensus'' required. Please, use the 'discussion' tab at the top of the page or talk to Dilos1 before editing pages of this project. Otherwise, your editing may be reversed.''' Hello! Welcome to the survey of European History to 1648 course page of the Department of European History at Wikiversity. This is an on going course that you can start anytime you like and complete it anytime you like at your own pace. There are no dates set for the completion of assignments. However, you may communicate with the course leader/instructor Dilos1 at any time and send your questions, completed assignments, suggestions for additional material etc to the wseh group. Course Leader/Instructor: Dilos1.

Description
This course is the first part of an introduction to the History of Europe. This first part covers the period from the origins of civilization to the Peace of Westphalia that ended the Thirty Years' War.

Rational
This course will provide students with a general background to the History of Europe until the mid 17th century. Those who will complete the course will be able to move on to either the second half of the History of Europe course or to other more advanced courses relating to the Ancient, Medieval and Early Modern times.

Objectives
As a result of taking this course, the student should be able to demonstrate a general knowledge of the history of Europe and European civilization during ancient, medieval, and early modern times.

Introduction

 * What is History ?

The Emergence of Civilization

 * Ancient Mesopotamia: The Cradle of Civilization
 * Ancient Egypt
 * The Empires of the Near East

The Greek Legacy

 * The Emergence of the Greek Civilization
 * Greek Culture and Colonization
 * The Greek city
 * The Greek Wars
 * Classical Greek Culture
 * Macedonia and Helenistic Times

The Roman Legacy

 * The Rise of Rome and Roman Expansion
 * Social Crisis and Change
 * Roman Family-life, & Culture
 * The End of the Roman Republic
 * The Roman Empire and Social Change
 * The Decline of Rome
 * Rome and Christianity

Medieval Europe

 * The Newcomers in Europe
 * The Western Church
 * The New States
 * The Byzantine Empire
 * Feudalism and Political Order
 * Crisis and Reform in Western Church
 * Heresy and the Church in the Mediterranean world
 * The Crusades
 * The Black Death
 * Trade in Medieval Europe
 * The Hundred Years War and its Aftermath
 * The Papal Crisis
 * Servetus

Renaissance and Reformation

 * Humanism and Art in Italy
 * Scholasticism
 * Religious Problems
 * Martin Luther and Protestantism
 * The Catholics
 * Expansion to new worlds
 * The New Kings
 * The Holy Roman Empire
 * Tudor Origins of the British Empire (1485 - 1603)
 * Philip II of Spain
 * The Thirty Tears War and the Treaty of Westphalia

Course Resources
The reading material includes specific articles from Wikipedia and parts of the textbooks The World History Project, The History of Greece and A Survey of Modern European History from Wikibooks. Also, we will use Wikisource for readings from the sources as well as Wikimedia for maps, visuals, sounds, music etc. Moreover, we will use a limited number of external sources to enchance our study of the History of Europe. In replacement of the electronic textbooks above, any general book on the development of western civilization may be used.

Students interested in history may wish to obtain one or both of the following printed reference books:
 * Sterns, Peter N. Ed. The Encyclopedia of World History. 6th ed. Boston: Houghton Mifflin, 2001 and optional CD-ROM.
 * The Penguin Atlas of World History. Volume 1: From Prehistory to the Eve of the French Revolution Rev. ed. N. York: Penguin, 2004 or any other modern Historical Atlas.

How to Study
Study the Required Reading at your own pace. When you feel that you have absorbed the material for a specific topic, try to answer the respective Review Question at the end of every course section.

How to Test Yourself
When you have revewed the material of a specific topic, you may answer the Test Yourself questions. If you wish, the Course Leader/Instructor of this course may evaluate the outcome of your effort. Please join the wseh group and e-mail your work for evaluation. You may also request an unofficial grade for the work sent.

Additional Information
The resources under this title complement the reading material and may help you understand better some aspects discussed in the relevant course section. However 'Additional Information' material is not required for the completion of this course.

Students who Completed this Course
When you complete this course you may add your login id and/or your name here.
 * 1.

title: History of Europe to 1648


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

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text: User:Eas4200c.f08.aero.E/Matlab 3: Airfoil and Stringers


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User:Eml4500.f08.wiki1.aguilar/TeamHW5

 See my comments below. Eml4500.f08 12:42, 10 November 2008 (UTC)

 Comment: This page is identical to the previous one, except that the failure to parse error has been corrected. LaTeX Formula needed to be modified before working within Wikipedia. Here is a comparison between these two versions. Eml4500.f08.wiki1.aguilar 22:13, 7 November 2008 (UTC)

 Comment: The Two Bar Truss Results Code was not copied from Team Delta_6. There are two sites that the MATLAB files can be downloaded from ( and ). Each site has a different code for the Two Bar Truss Results. Our team and Team Delta_6 happened to use the same website, but we modified the final code so we could obtain the correct results. Eml4500.f08.wiki1.ambrosio 22:01, 10 November 2008 (UTC)

=Team Homework 5= This consists of course notes from 10/20 - 1031.

Two Bar Truss
The following code was used to re-run the two bar truss system.

 Note: This team had to modify the original code to print out the correct array "results" because they used the incomplete code from Team Delta_6; see my comments further below and in "The best of HW5". Eml4500.f08 12:48, 10 November 2008 (UTC)

The following codes were used in the two bar truss code.

 Note: This team used the incomplete code PlaneTrussResults from Team Delta_6 (the last line was missing); see my comments above and in "The best of HW5". Eml4500.f08 12:48, 10 November 2008 (UTC)

When the MATLAB code is ran, the following results are obtained.

Row 1 of the "results" represents element 1, while row 2 represents element 2. Column 1 is the strain, column 2 the stress, and column 3 is the axial force for each element.

Six Bar Truss Example
When the MATLAB code is ran, the following results are obtained.

As with the "results" matrix from the two bar truss code, the row indicates the element while column 1 is the strain, cloumn 2 is the stress, and column 3 is the axial force.

The figure below shows the undeformed truss (dashed blue line) and deformed truss (solid red line) with a weighting factor of 10000.

Six Bar Truss With Different Young's Modulus
When the MATLAB code is ran, the following results are obtained.

As with the "results" matrix from the two bar and six bar truss code, the row indicates the element while column 1 is the strain, cloumn 2 is the stress, and column 3 is the axial force.

The figure below shows the undeformed truss (dashed blue line), deformed truss where all elements have the same Young's Modulus (solid red line), and deformed truss where all elements have a different Young's Modulus (solid green line) with a weighting factor of 10000.

Justification for Eliminating Rows 1,2,5, and 6 to Obtain $$\overline{K}$$ in 2-bar Truss
Force-Displacement Relationship: K*d=F K*d-F=0 $$\rightarrow Equation(1)$$ By the Principle of Virtual Work:W•(Kd-F)=0 for all W $$\rightarrow Equation(2)$$ $$Equation(1)\Leftrightarrow Equation(2)$$ Proof: $$(1) \rightarrow (2)$$: Trivial $$(2) \rightarrow (1)$$: Not Trivial We want to show $$(2) \rightarrow (1)$$. Choice 1: Select W such that W1=1, W2=0, W3=0, W4=0, W5=0, W6=0 W=$$ \begin{Bmatrix} 1\\ 0\\ 0\\ 0\\ 0\\ 0\end{Bmatrix}$$ W•(Kd-F)= 1.) $$[\sum_{j=1}^6 K_{1j}d_j-F_1]+0\cdot [\sum_{j=1}^6 K_{2j}d_j-F_2]+0\cdot [\sum_{j=1}^6 K_{3j}d_j-F_3]+0\cdot [\sum_{j=1}^6 K_{4j}d_j-F_4]$$$$+0\cdot [\sum_{j=1}^6 K_{5j}d_j-F_5]+0\cdot [\sum_{j=1}^6 K_{6j}d_j-F_6]$$ $$\sum_{j=1}^6 K_{1j}d_j=F_1$$   $$\rightarrow (1st Equation)$$ Choice 2: Select W such that W1=0, W2=1, W3=0, W4=0, W5=0, W6=0  W=$$ \begin{Bmatrix} 0\\1\\0\\0\\0\\0\end{Bmatrix}$$  2.) $$0\cdot [\sum_{j=1}^6 K_{1j}d_j-F_1]+[\sum_{j=1}^6 K_{2j}d_j-F_2]+0\cdot [\sum_{j=1}^6 K_{3j}d_j-F_3]+0\cdot [\sum_{j=1}^6 K_{4j}d_j-F_4]$$$$+0\cdot [\sum_{j=1}^6 K_{5j}d_j-F_5]+0\cdot [\sum_{j=1}^6 K_{6j}d_j-F_6]$$ $$\sum_{j=1}^6 K_{2j}d_j=F_2 \rightarrow (2nd Equation)$$ Choice 3: Select W such that W1=0, W2=0, W3=1, W4=0, W5=0, W6=0 W=$$ \begin{Bmatrix} 0\\0\\1\\0\\0\\0\end{Bmatrix}$$ 2.) $$0\cdot [\sum_{j=1}^6 K_{1j}d_j-F_1]+0\cdot[\sum_{j=1}^6 K_{2j}d_j-F_2]+[\sum_{j=1}^6 K_{3j}d_j-F_3]+0\cdot [\sum_{j=1}^6 K_{4j}d_j-F_4]$$$$+0\cdot [\sum_{j=1}^6 K_{5j}d_j-F_5]+0\cdot [\sum_{j=1}^6 K_{6j}d_j-F_6]$$ $$\sum_{j=1}^6 K_{3j}d_j=F_3 \rightarrow (3rd Equation)$$ Choice 4: Select W such that W1=0, W2=0, W3=0, W4=1, W5=0, W6=0 W=$$ \begin{Bmatrix} 0\\0\\0\\1\\0\\0\end{Bmatrix}$$  2.) $$0\cdot [\sum_{j=1}^6 K_{1j}d_j-F_1]+0\cdot[\sum_{j=1}^6 K_{2j}d_j-F_2]+0\cdot[\sum_{j=1}^6 K_{3j}d_j-F_3]+[\sum_{j=1}^6 K_{4j}d_j-F_4]$$$$+0\cdot [\sum_{j=1}^6 K_{5j}d_j-F_5]+0\cdot [\sum_{j=1}^6 K_{6j}d_j-F_6]$$ $$\sum_{j=1}^6 K_{4j}d_j=F_4 \rightarrow (4th Equation)$$ Choice 5: Select W such that W1=0, W2=0, W3=0, W4=0, W5=1, W6=0 W=$$ \begin{Bmatrix} 0\\0\\0\\0\\1\\0\end{Bmatrix}$$ 2.) $$0\cdot [\sum_{j=1}^6 K_{1j}d_j-F_1]+0\cdot[\sum_{j=1}^6 K_{2j}d_j-F_2]+0\cdot[\sum_{j=1}^6 K_{3j}d_j-F_3]+0\cdot [\sum_{j=1}^6 K_{4j}d_j-F_4]$$$$+[\sum_{j=1}^6 K_{5j}d_j-F_5]+0\cdot [\sum_{j=1}^6 K_{6j}d_j-F_6]$$ $$\sum_{j=1}^6 K_{5j}d_j=F_5 \rightarrow (5th Equation)$$ Choice 6: Select W such that W1=0, W2=0, W3=0, W4=0, W5=0, W6=1 W=$$ \begin{Bmatrix} 0\\0\\0\\0\\0\\1\end{Bmatrix}$$  2.) $$0\cdot [\sum_{j=1}^6 K_{1j}d_j-F_1]+0\cdot[\sum_{j=1}^6 K_{2j}d_j-F_2]+0\cdot[\sum_{j=1}^6 K_{3j}d_j-F_3]+0\cdot [\sum_{j=1}^6 K_{4j}d_j-F_4]$$$$+0\cdot [\sum_{j=1}^6 K_{5j}d_j-F_5]+[\sum_{j=1}^6 K_{6j}d_j-F_6]$$ $$\sum_{j=1}^6 K_{6j}d_j=F_6 \rightarrow (6th Equation)$$ $$\begin{bmatrix} K_{11}&\cdots&K_{16}\\ \vdots&\ddots&\vdots\\ K_{61}&\cdots&K_{66}\end{bmatrix}* \begin{Bmatrix} d_1\\d_2\\d_3\\d_4\\d_5\\d_6\end{Bmatrix}= \begin{Bmatrix} F_1\\F_2\\F_3\\F_4\\F_5\\F_6\end{Bmatrix}$$ Hence, Kd=F (or Equation (1) from above)

Principle of Virtual Work, Relating to W
The Principle of Virtual Work accounts for the boundary conditions.

For the two-bar truss system, d1=d2=d5=d6=0 Weighting coefficients must be "kinematically admissible". In other words, the weighting coefficients cannot violate the boundary conditions.

Weighting Coefficients$$\equiv$$Virtual Displacements $$W_1=W_2=W_5=W_6=0$$ W•(Kd-F) $$\begin{Bmatrix}W_3\\W_4\end{Bmatrix}\cdot (\bar{\mathbf{K}} \bar{\mathbf{d}} -\bar{\mathbf{F}})=0$$ for all $$\begin{Bmatrix}W_3\\W_4\end{Bmatrix}$$ (3) $$\bar{\mathbf{K}}=\begin{bmatrix}K_{33}&K_{34}\\K_{43}&K_{44}\end{bmatrix}\qquad \bar{\mathbf{d}}=\begin{Bmatrix}d_3\\d_4\end{Bmatrix}\qquad \bar{\mathbf{F}}=\begin{Bmatrix}F_3\\F_4\end{Bmatrix}$$

$$\begin{Bmatrix}W_3=n\\W_4=n\end{Bmatrix}\cdot (\begin{bmatrix}K_{33}&K_{34}\\K_{43}&K_{44}\end{bmatrix} \begin{Bmatrix}d_3\\d_4\end{Bmatrix} -\begin{Bmatrix}F_3\\F_4\end{Bmatrix})=0$$

Because the $$\begin{bmatrix}K_{33}&K_{34}\\K_{43}&K_{44}\end{bmatrix} \begin{Bmatrix}d_3\\d_4\end{Bmatrix}$$ term is equal to $$\begin{Bmatrix}F_3\\F_4\end{Bmatrix}$$, $$\begin{bmatrix}K_{33}&K_{34}\\K_{43}&K_{44}\end{bmatrix} \begin{Bmatrix}d_3\\d_4\end{Bmatrix} -\begin{Bmatrix}F_3\\F_4\end{Bmatrix}=\begin{Bmatrix}0\\0\end{Bmatrix}$$ $$\begin{bmatrix}K_{33}&K_{34}\\K_{43}&K_{44}\end{bmatrix} \begin{Bmatrix}d_3\\d_4\end{Bmatrix}=\begin{Bmatrix}F_3\\F_4\end{Bmatrix}$$(4)

More Derivations
We want to derive $$\mathbf{k}^{(e)}_{\,4x4}= \mathbf{T}^{(e)T}_{4x2} \,\hat{\mathbf{k}}^{(e)}_{2x2}\, \mathbf{T}^{(e)}_{2x4}$$, an equation that can be found on p14.3 and p23.3 of our class notes.

Recall the force-displacement relationship with axial degrees of freedom, $$q^{(e)}$$.

$$\hat{\mathbf{k}}^{(e)}\, \mathbf{q}^{(e)} = \mathbf{p}^{(e)}$$

Moving some things around we can get

$$\hat{\mathbf{k}}^{(e)}\, \mathbf{q}^{(e)} - \mathbf{p}^{(e)} = \mathbf{0}_{2x1} $$     (1)

Plugging this in for the Principle of Virtual Work:

$$ PVW\Rightarrow \mathbf{\hat{w}}\cdot \underbrace{(\hat{\mathbf{k}}^{(e)}\, \mathbf{q}^{(e)} - \mathbf{p}^{(e)})}_{2x1\mathrm{\;for\;all\;} \mathbf{\hat{w}}_{2x1}} = {0}_{1x1} $$         (2)

We showed that equations (1) and (2) are equivalent on p24.1 of our class notes.

Recall,

Here we can see the similarities between the equation for the real displacement and the virtual displacement.

$${\hat{\mathbf{W}}}_{2x1}$$ is the virtual axial displacement which corresponds to $${\mathbf{q}}^{(e)}_{2x1}$$. On the other hand $${\mathbf{W}}_{4x1}$$ is the virtual displacement in the global coordinate system which corresponds to $${\mathbf{d}}^{(e)}_{4x1}$$. If we combine equations (2), (3) and (4) we get

$$(\mathbf{T^{(e)}}\mathbf{W})[\hat{\mathbf{k^{(e)}}}(\mathbf{T^{(e)}}\mathbf{d^{(e)}})-\mathbf{P^{(e)}}]=0$$ (5)

Which applies for all $${\mathbf{W}}_{4x1}$$. If we recall that

$$\mathbf{(A_{pxq}}\mathbf{B_{qxr})^T}=\mathbf{B^TA^T}$$ (6)

and

$$\mathbf{(a_{nx1}}\mathbf{b_{nx1})}=\mathbf{a_{1xn}^Tb_{nx1}}$$ (7)

we can manipulate (5) using (7). This gives us

$$\mathbf{(T^{(e)}W)^T[\hat{k^{(e)}}(T^{(e)}d^{(e)})-P^{(e)}}]=0 $$

Then by substituting in (6) we get

$$\mathbf{(T^{(e)T}W^T)[\hat{k}^{(e)}(T^{(e)}d^{(e)})-P^{(e)}]}=0$$

which can be simplified to

$$\mathbf{W[k^{(e)}d^{(e)}-f^{(e)}}]=0$$

and finally

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

Up until now we have dealt with the discrete case with matrices. From here forward we will consider the continuous case which involves partial differential equations. The model problem we will use consists of an elastic bar with varying area and modulus. It will be subjected to varying concentrated and axial loads, as well as inertial forces.

Perform literature search for composite materials such that E(x) (varying Young's modulus) (doping, etc.)

→One example includes polyester-50% glass fiber ply. See []

$$\sum{F_x = 0 = -N(x,t) + N(x + dx,t) + f(x,t)dx - m(x)\ddot{u}dx}$$

$$= \frac{\partial N}{\partial x}(x,t)dx + higher order terms + f(x,t)dx - m(x)\ddot{u}dx\; \mathbf{(1)}$$

Neglect higher order terms

Recall Taylor Series Expansion:

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

$$\Rightarrow \frac{\partial N}{\partial x} + f = m\ddot{u}\; \mathbf{(2)}$$ Equation of Motion (EOM)

$$ N(x,t) = A(x)\sigma(x,t)$$

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

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

$$\Rightarrow N(x,t) = A(x)E(x)\frac{\partial u}{\partial x}(x,t)\; \mathbf{(3)}$$ Constitutive Relation

(3) in (2) yields:

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

$$\ddot{u} = \frac{\partial^2 u}{\partial t^2}$$ PDE of Motion

Need 2 boundary conditions (2nd order derivative with respect to x)

Need 2 initial conditions (2nd derivative with respect to t) → (initial displacement, initial velocity)

$$u(0,t) = 0 = u(L,t)$$

1) $$u(0,t) = 0$$

2) $$N(L,t) = F(t)$$

$$N(L,t) = A(L)\sigma(L,t)$$

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

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

Boundary Conditions: $$\Rightarrow \frac{\partial u}{\partial x}(L,t) = \frac{F(t)}{A(L)E(L)}$$

Initial Conditions: (at t=0 prescribed)

$$u(x,t=0) = \bar{u}(x)$$ known function (displacement)

$$\frac{\partial u}{\partial t}(x,t=0) = \dot{u}(x,t=0) = \bar{v}(x)$$ known function (velocity)

Problem Statement
3-D Three Bar Truss Problem p230 in Bhatti Consider the three bar truss problem shown below. The cross sectional are of elements 1 and 2 is 200mm2 and that of element 3 is 600mm2. All elements are made of the same material with E=200Gpa. The applied load is P=20kN. We were also asked to determine if this truss system was statically determinate. If it was, we were to solve it using statics and compare the solution to the FEA solution generated with matlab. Below is the picture, similar to the one given in the book. The nodal coordinates are as follows: The solution given in the book is as follows:

Matlab Code
Below is the code for the 3-D Three Bar Truss System:
 Three Bar Space Truss

Needed m-files for Space Truss Problem
This code uses three m-files. The first mfile, SpaceTrussElement.m, defines the element stiffness matrix. The second m-file, SpaceTrussResults.m, computes the element strain, stress and axial forces. The third m-file, NodalSoln.m, was given in an earlier homework assignment. It computes the nodal solutions.
 SpaceTrussElement.m
 SpaceTrussResults.m
 NodalSoln.m

conn and lmm Arrays of Space Truss Problem
Below are the arrays lmm and conn used to solve the 3-D three-bar truss system: conn= $$ \begin{bmatrix} 1 & 4 \\ 2 & 4 \\ 3 & 4\end{bmatrix} $$ lmm= $$ \begin{bmatrix} 1 & 2 & 3 & 10 & 11 & 12 \\ 4 & 5 & 6 & 10 & 11 & 12 \\ 7 & 8 & 9 & 10 & 11 & 12 \end{bmatrix}$$

Images of Space Truss Problem
Below are the four images of the undeformed shape vs. the deformed shape. The undeformed shape is the blue dotted line and the deformed shape is the red solid line. Their perspective plot is from the point (-2, -2, 3) as suggested.
From point (-2,-2,3)
YZ plane
XZ plane
XY plane

Matlab Code for the Images of Space Truss Problem
Below is the matlab code used to obtain the previous images:
 Image Code

Statics Solution of the Space Truss Problem
This problem, unfortunately for us, is statically determinate. In order to solve this problem, we first note that there will be three reaction forces at node 1, node 2, and node 3. Each one of these reaction forces can be divided into the x, y, and z components. Doing so gives us the following equation: $$ R_1 = R_{1x}\hat i + R_{1y}\hat j + R_{1z}\hat k $$ $$ R_2 = R_{2x}\hat i + R_{2y}\hat j + R_{2z}\hat k $$ $$ R_3 = R_{3x}\hat i + R_{3y}\hat j + R_{3z}\hat k $$ In order to solve for these reactions, we need to find what each reaction components are. Noting that R3x and R3y are both zero, we have the following equations to work off of: $$\mathbf{R_{1x}} = \mathbf{R_1}\cos{\phi_1}\cos{\theta_1}$$ $$\mathbf{R_{1y}} = \mathbf{R_1}\cos{\phi_1}\sin{\theta_1}$$ $$\mathbf{R_{1z}} = \mathbf{R_1}\sin{\phi_1}$$ $$\mathbf{R_{2x}} = \mathbf{R_2}\cos{\phi_2}\cos{\theta_2}$$ $$\mathbf{R_{2y}} = \mathbf{R_2}\cos{\phi_2}\sin{\theta_2}$$ $$\mathbf{R_{2z}} = \mathbf{R_2}\sin{\phi_2}$$ $$\mathbf{R_{3z}} = \mathbf{R_3}$$ Using the pythagorean theorem and trigonometry the following calculations were made in order to find the values for θ and Φ. $$\theta_1 = 63.435^\circ$$ $$\theta_2 = -45^\circ$$ $$\phi_1 = 42.975^\circ$$ $$\phi_2 = 44.482^\circ$$ Using these equations, we set up the equations for the forces in the following steps: $$ \sum F = 0 $$ $$\sum \mathbf{F_x} = \mathbf{R_1}\cos{\phi_1}\cos{\theta_1} + \mathbf{R_2}\cos{\phi_2}\cos{\theta_2} = 0$$ $$\sum \mathbf{F_y} = \mathbf{R_1}\cos{\phi_1}\sin{\theta_1} + \mathbf{R_2}\cos{\phi_2}\sin{\theta_2} - \mathbf{P} = 0$$ $$\sum \mathbf{F_z} = \mathbf{R_1\sin{\phi_1}} + \mathbf{R_2\sin{\phi_2}} + \mathbf{R_3} = 0$$ Plugging in all of the known values, given and calculated, the following results are obtained: R1 = 20374.6588 R2 = 13214.4037 R3 = -23148.1172 These values are just like the ones received using Matlab and the FEA method.

3-Dimensional Derivations
Some of the equations that we have learned over the course of the semester need to be updated for use with 3-dimensional systems or space trusses.

With the local $$\tilde{x}$$ axis along the axis of the element, the local element equations are still the same as the 2-node axial deformation element:

$$\frac{EA}{L}\begin{pmatrix}\;\; 1&-1\\-1&\;\;1\end{pmatrix}\begin{pmatrix}q_1\\q_2\end{pmatrix}=\begin{pmatrix}P_1\\P_2\end{pmatrix}$$

Which gives us our familiar axial displacement relationship: $$\mathbf {\hat {k}}^{(e)} \mathbf q^{(e)} = \mathbf P^{(e)} $$

In the global coordinates, the nodal degrees of freedom and the nodal applied forces for each element are:

$$\mathbf q = \begin{pmatrix}d_1\\d_2\\d_3\\d_4\\d_5\\d_6\end{pmatrix}$$     and       $$\mathbf p = \begin{pmatrix}f_1\\f_2\\f_3\\f_4\\f_5\\f_6\end{pmatrix}$$

In order to derive the transformation matrix, we must do the following:

Consider the displacement vector of the local node 1 denoted by $$\vec{d}_{1}^{(e)}$$.

$$\vec{d}_{[1]}^{(e)}={d}_{1}^{(e)}\vec{i}+{d}_{1}^{(e)}\vec{j}+{d}_{1}^{(e)}\vec{k}$$

$${q}_{1}^{(e)}$$ = axial displacement of node 1 is the orthagonal projection of the displacement vector $$\vec{d}_{1}^{(e)}$$ of node 1 on the $$\tilde{x}$$ axis of element 'e'.

$${q}_{1}^{(e)}=\vec{d}_{1}^{(e)}\cdot\vec{\tilde{i}}$$

$${q}_{1}^{(e)}=({d}_{1}^{(e)}\vec{i}+{d}_{2}^{(e)}\vec{j}+{d}_{3}^{(e)}\vec{k})\cdot\vec{\tilde{i}}$$

$${q}_{1}^{(e)}={d}_{1}^{(e)}(\vec{i}\cdot\vec{\tilde{i}})+{d}_{2}^{(e)}(\vec{j}\cdot\vec{\tilde{i}})+{d}_{3}^{(e)}(\vec{k}\cdot\vec{\tilde{i}})$$

$$L=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$

$$(\vec{i}\cdot\vec{\tilde{i}})=\frac{x_2-x_1}{L}=l^{(e)}$$

$$(\vec{j}\cdot\vec{\tilde{i}})=\frac{y_2-y_1}{L}=m^{(e)}$$

$$(\vec{k}\cdot\vec{\tilde{i}})=\frac{z_2-z_1}{L}=m^{(e)}$$

$${q}_{1}^{(e)}=l^{(e)}{d}_{1}^{(e)}+m^{(e)}{d}_{2}^{(e)}+n^{(e)}{d}_{3}^{(e)}$$

$${q}_{1}^{(e)}= \begin{bmatrix}l^{(e)} & m^{(e)}&n^{(e)} \end{bmatrix}_{1x3}\begin{Bmatrix}{d}_{1}^{(e)}\\{d}_{2}^{(e)}\\{d}_{3}^{(e)}  \end{Bmatrix}_{3x1}$$

Here we can see that $${q}_{1}^{(e)}$$ is a 1x1 scalar.

We do this for node 2 as well. $${q}_{2}^{(e)}= \begin{bmatrix}l^{(e)} & m^{(e)} & n^{(e)}\end{bmatrix}_{1x3}\begin{Bmatrix}{d}_{4}^{(e)}\\{d}_{5}^{(e)}\\{d}_{6}^{(e)}  \end{Bmatrix}_{3x1}$$

which leads us to the transformation between global to local degrees of freedom which are as follows:

$$\begin{Bmatrix} {q}_{1}^{(e)}\\{q}_{2}^{(e)}\end{Bmatrix}= \begin{bmatrix} l^{(e)}& m^{(e)}& n^{(e)}&0&0&0\\0&0&0& l^{(e)}& m^{(e)}& n^{(e)} \end{bmatrix} \begin{Bmatrix}{d}_{1}^{(e)}\\{d}_{2}^{(e)}\\{d}_{3}^{(e)}\\{d}_{4}^{(e)}\\{d}_{5}^{(e)}\\{d}_{6}^{(e)} \end{Bmatrix}\Rightarrow \textbf{q}_{2x1}^{(e)}=\textbf{T}_{2x6}^{(e)}\textbf{d}_{6x1}^{(e)}$$

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

\Rightarrow \textbf{d}_{6x1}^{(e)} = {\textbf{T}_{6x2}^{(e)}}^{T} \textbf{q}_{2x1}^{(e)} $$

The conversion from axial forces and element forces in global coordinates is as follows:

$$\mathbf{f}^{(e)}=\begin{pmatrix}f_{1}^{(e)}\\f_{2}^{(e)}\\f_{3}^{(e)}\\f_{4}^{(e)}\\f_{5}^{(e)}\\f_{6}^{(e)}\end{pmatrix}$$ in global coordinates $$\mathbf{\hat{k}^{(e)}q^{(e)}=p^{(e)}}={EA \over L} \begin{bmatrix}1&-1\\-1&1\end{bmatrix}\begin{pmatrix} q_1^{(e)}\\q_2^{(e)} \end{pmatrix}= \begin{pmatrix}p_1^{(e)}\\p_2^{(e)}\end{pmatrix}$$ in axial coordinates $$\mathbf{q^{(e)}=T^{(e)}d^{(e)}}$$ $$\mathbf{\hat{k}^{(e)}T^{(e)}d^{(e)}=p^{(e)}}$$ $${EA \over L} \begin{bmatrix}1&-1\\-1&1\end{bmatrix}\begin{pmatrix} l^{(e)} & m^{(e)} & n^{(e)}  & 0 & 0 & 0\\ 0 & 0 &0 & l^{(e)}  & m^{(e)} & n^{(e)} \end{pmatrix}\begin{pmatrix} d_1^{(e)}\\ d_2^{(e)}\\ d_3^{(e)}\\ d_4^{(e)}\\ d_5^{(e)}\\ d_6^{(e)} \end{pmatrix}=\begin{pmatrix}p_1^{(e)}\\p_2^{(e)}\end{pmatrix}$$ Multiplying both sides by T(e)T as completed in the 2D case, $$\mathbf{T^{(e)T}\hat{k}^{(e)}T^{(e)}d^{(e)}=T^{(e)T}p^{(e)}}$$ $$\begin{pmatrix} l^{(e)} & 0\\m^{(e)} & 0\\n^{(e)}  & 0\\ 0 &l^{(e)}\\0&m^{(e)}\\ 0&n^{(e)} \end{pmatrix} {EA \over L} \begin{bmatrix}1&-1\\-1&1\end{bmatrix} \begin{pmatrix} l^{(e)} & m^{(e)} & n^{(e)}  & 0 & 0 & 0\\ 0 & 0 &0 & l^{(e)}  & m^{(e)} & n^{(e)} \end{pmatrix} \begin{pmatrix} d_1^{(e)}\\ d_2^{(e)}\\ d_3^{(e)}\\ d_4^{(e)}\\ d_5^{(e)}\\ d_6^{(e)} \end{pmatrix}=\begin{pmatrix} l^{(e)} & 0\\m^{(e)} & 0\\n^{(e)}  & 0\\ 0 &l^{(e)}\\0&m^{(e)}\\ 0&n^{(e)} \end{pmatrix} \begin{pmatrix}p_1^{(e)}\\p_2^{(e)}\end{pmatrix} $$ and with $$\mathbf{k^{(e)}=T^{(e)T}\hat{k}^{(e)}T^{(e)}},\qquad\mathbf{f^{(e)}=T^{(e)T}p^{(e)}}$$ We have $$\mathbf{(T^{(e)T}\hat{k}^{(e)}T^{(e)})d^{(e)}=(T^{(e)T}p^{(e)})}$$ or $$\mathbf{k^{(e)}d^{(e)}=f^{(e)}}$$ $$\begin{pmatrix}f_{1}^{(e)}\\f_{2}^{(e)}\\f_{3}^{(e)}\\f_{4}^{(e)}\\f_{5}^{(e)}\\f_{6}^{(e)}\end{pmatrix}=\begin{pmatrix} l^{(e)} & 0\\m^{(e)} & 0\\n^{(e)}  & 0\\ 0 &l^{(e)}\\0&m^{(e)}\\ 0&n^{(e)} \end{pmatrix}\begin{pmatrix}p_1^{(e)}\\p_2^{(e)}\end{pmatrix}$$

We have already shown that $$\mathbf{k^{(e)}} = \mathbf{T}^{(e)T} \mathbf{{\hat{k}}^{(e)}} \mathbf{{T}^{(e)}} $$ .By using the transformation matrix we get the following.

$$\begin{align} \mathbf{k^{(e)}} & = \mathbf{T}^{(e)T} \mathbf{{\hat{k}}^{(e)}} \mathbf{{T}^{(e)}} \\

\mathbf{{k}^{(e)}} & =

\begin{pmatrix} l^{(e)} & 0 \\ m^{(e)} & 0 \\ n^{(e)} & 0 \\ 0 & l^{(e)} \\ 0 & m^{(e)} \\ 0 & n^{(e)} \end{pmatrix}

{EA \over L} \begin{bmatrix}1&-1\\-1&1\end{bmatrix}

\begin{pmatrix} l^{(e)} & m^{(e)} & n^{(e)}  & 0 & 0 & 0\\ 0 & 0 &0 & l^{(e)} & m^{(e)}  & n^{(e)} \end{pmatrix}

\end{align} $$

$$ \mathbf{k^{(e)}}= \frac{EA}{L} $$ $$ \begin{pmatrix} {l^{(e)}}^2 & m^{(e)}l^{(e)} & n^{(e)}l^{(e)} & -{l^{(e)}}^2 & -m^{(e)}l^{(e)} & -n^{(e)}l^{(e)}\\ m^{(e)}l^{(e)} & {m^{(e)}}^2 & m^{(e)}n^{(e)} & -m^{(e)}l^{(e)} & -{m^{(e)}}^2 & -m^{(e)}n^{(e)}\\ n^{(e)}l^{(e)} & m^{(e)}n^{(e)} & {n^{(e)}}^2 & -n^{(e)}l^{(e)} & -m^{(e)}n^{(e)} & -{n^{(e)}}^2\\ -{l^{(e)}}^2 & -m^{(e)}l^{(e)} & -n^{(e)}l^{(e)} & {l^{(e)}}^2 & m^{(e)}l^{(e)} & n^{(e)}l^{(e)}\\ -m^{(e)}l^{(e)} & -{m^{(e)}}^2 & -m^{(e)}n^{(e)} & m^{(e)}l^{(e)} & {m^{(e)}}^2 & m^{(e)}n^{(e)}\\ -n^{(e)}l^{(e)} & -m^{(e)}n^{(e)} & -{n^{(e)}}^2 & n^{(e)}l^{(e)} & m^{(e)}n^{(e)} & {n^{(e)}}^2

\end{pmatrix} $$ $$ \begin{pmatrix} d_1^{(e)}\\d_2^{(e)}\\d_3^{(e)}\\d_4^{(e)}\\d_5^{(e)}\\d_6^{(e)}\end{pmatrix}= \begin{pmatrix}f_{1}^{(e)}\\f_{2}^{(e)}\\f_{3}^{(e)}\\f_{4}^{(e)}\\f_{5}^{(e)}\\f_{6}^{(e)}\end{pmatrix} $$

=Contributing Team Members=

Eml4500.f08.wiki1.oatley 17:05, 7 November 2008 (UTC) Eml4500.f08.wiki1.ambrosio 15:02, 7 November 2008 (UTC) Eml4500.f08.wiki1.aguilar 19:36, 7 November 2008 (UTC) Eml4500.f08.wiki1.brannon 18:40, 7 November 2008 (UTC) Eml4500.f08.wiki1.schaet 16:55, 7 November 2008 (UTC) Eml4500.f08.wiki1.handy 17:44, 7 November 2008 (UTC)

title: User:Eml4500.f08.wiki1.aguilar/TeamHW5


text: File:3-cell with Stringers II.JPG

title: File:3-cell with Stringers II.JPG


text: File:PDE Model.jpg

title: File:PDE Model.jpg


text: File:CarbonNACAAirfoilBendingMomentAssume.jpg


text: File:CarbonNACAAirfoilBendingMomentEntire.jpg


title: File:Threebartruss3D.JPG


text: File:Threebartruss x.JPG

title: File:Threebartruss x.JPG


text: File:Threebartruss y.JPG

title: File:Threebartruss y.JPG


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title: File:Threebartruss z.JPG


text:
User:Eml4500.f08.team.foskey.ckf/hw5d

Composite Materials - a brief overview

Composite materials are materials that are made up of two or more separate materials, called constituent materials. These composite materials have distinct physical, and sometimes chemical, properties. By combining the materials, the composite material has properties that are different than either of the constituent materials.

Constituent materials are considered either matrix materials or reinforcement materials. All composite materials have at least one constituent material of each kind. Matrix materials act as a medium to hold the reinforcement materials in place, and the reinforcement material is used to provide additional strength for the material.

Generally composite materials are not isotropic, meaning that their physical properties are not equal in all directions. The physical properties depend on the arrangement and orientation of the constituent materials within the composite material.

Composite materials are typically used when weight is a factor. A composite material can be nearly as strong as the reinforcing material, but at a fraction of the weight.

Composite materials include carbon fiber materials, reinforced concrete, and plywood.

For additional information, see:

Composite Materials (wikipedia.com)

Composite Material (encyclopedia.com)

Composite Materials (science.org)

title: User:Eml4500.f08.team.foskey.ckf/hw5d


text:
User:Eml4500.f08.team.foskey.ckf/hw5e

Is the space truss statically indeterminate?

Looking at the three bar space truss problem, the problem is easily solve using the finite element method. However, the question is raised as to whether or not the truss can be solved using statics.

When first looking at the problem, one sees that there will be nine reactions, three for each bar. However, statics only gives a limited number of equations.

The problem is solved statically by using the Euler cut principle.

In doing this, the bars are all cut in half, and the equilibrium of node 4 is solved for. When the bars are cut, their associated axial forces are left. The axial forces must then be resolved into their components for each directions.

To resolve the axial forces into components we first need the lengths of the bars. This is done by using the equation

$$ L= \sqrt{(X_2 - X_1)^2 + (Y_2 -Y_1)^2 + (Z_2 -Z_1)^2}$$

where L is the bar length, X2 is the x coordinate of node 4, Y2 is the y coordinate of node 4, Z2 is the z coordinate of node 4, and X1, Y1, Z1 are the coordinates of the pinned node for a bar.

Using the above formula, the lengths of the bars are found to be L1 =2.9339 L2 = 2.8543 L3 = 2

From there the forces are resolved into components by using a scaling factor. This scaling factor is found by $$ \frac{L_D}{L}$$, where LD is the length in a coordinate direction (x, y, or z) and L is the total length. LD could also be considered the distance between the nodes of a bar in one direction (x, y,or z)

So to find the x component force in the axial force of bar 1, you would use the formula $$ F^1_x = \frac{(X_2-X_1)}{L^1}*F^1$$ where F1 is the axial force in bar 1. This process is repeated for all of the forces in all of the bars, until 9 force components are obtained.

Summing the forces acting on node 4 for the x, y, and z directions will result in the axial forces. These equations are

$$ \Sigma F_x = F^1_x +F^2_x +F^3_x =0$$ $$ \Sigma F_y = F^1_y +F^2_y +F^3_y =-20000$$ $$ \Sigma F_z = F^1_z +F^2_z +F^3_z =0$$

Keep in mind that F1x, F1y, and F1z are all in terms of the axial force F1 (and likewise with F2 and F3) so that there are only three true unknowns in the above equations. Solving those three equations will give the axial forces F1, F2) and F3).

From there, the axial forces are resolved into their components to find the reactions. This is accomplished by using the same scaling factors calculated earlier.

Using this method, the reactions were found to be

The reaction forces calculated by the MATLAB code are:

There is a significant amount of discrepancy between the two methods. This can possibly be attributed to the statics method not taking into account the deformations the bars undergo, the differences in physical properties of the bars. Errors are also possible simply from errors in calculating the values by hand.

To conclude, the three bar space truss can be determined using statics, however there were errors in our calculations, although we believe our methodology to be sound.

title: User:Eml4500.f08.team.foskey.ckf/hw5e


text:
User:Eml4500.f08.team.dwyer.jd/HW5 Oct 27 (Meeting 26)

The Principle of Virtual Work - Part II
$$d^{i}_{4x1}\Rightarrow\mathbf{d(lmm(i,:))}$$

element DOFS in the global (x,y) coordinate system for element (i)

Recall: $$\mathbf{q}^{(i)}=\mathbf{Td}^{(i)}$$

The composition of our course grade is an example of the weighting factor shown below.

 GTotal = $$\alpha _{o} \cdot (HW Grade)+\sum_{i=1}^{3}{\alpha _{i}}\cdot (Exam_{i})$$

Our overall goal is to derive $$\mathbf{k}_{4x4}^{(e)} = \mathbf{T}_{4x2}^{(e)T} \hat\mathbf{k}_{2x2}^{(e)} \mathbf{T}_{2x4}^{(e)}$$ via the Principle of Virtual Work Recall: $$\hat\mathbf{k}_{2x2}^{(e)} \mathbf{q}_{2x2}^{(e)} = \mathbf{p}_{2x1}^{(e)}$$

$$ \Rightarrow \hat\mathbf{k}^{(e)}_{2x2} \mathbf{q}^{(e)}_{2x1} - \mathbf{p}^{(e)}_{2x1} = \mathbf{0}_{2x1} \qquad \rightarrow \qquad (1) $$ By the Principle of Virtual Work (PVW): $$\mathbf{w}_{2x1} * ( \hat\mathbf{k}^{(e)} \mathbf{q}^{(e)} - \mathbf{p}^{(e)} )_{2x1} = 0_{1x1} \quad $$ for all $$\; \mathbf{w}_{2x1}\qquad \rightarrow \qquad (2) $$ $$\therefore \;$$ We showed that $$ (1) \longleftrightarrow (2)$$ Recall:

$$\mathbf{q}{(e)} = \mathbf{T}^{(e)} \mathbf{d}^{(e)} \qquad \rightarrow \qquad (3)$$ Similarly,
 * $$\mathbf{q}^{(e)} \;$$ is the real displacement matrix defined in the axial coordinate system
 * $$\mathbf{d}^{(e)} \;$$ is the real displacement matrix defined in the global coordinate system

$$ \; \hat\mathbf{w}_{2x1} = \mathbf{T}_{2x4}^{(e)} \mathbf{w}_{4x1}^{(e)} \qquad \rightarrow \qquad (4)$$
 * $$\hat\mathbf{w}_{2x1} \;$$ is the virtual displacement matrix defined in the axial coordinate system
 * $$\mathbf{w}_{4x1} \;$$ is the virtual displacement matrix defined in the global coordinate system


text:
User:Eas4200c.f08.wiki.c/homework report 5

=Matlab Code=

The Matlab Code used in the following analysis is provided below.

'Part 1' is commented out when doing the analysis for 'Part 2' and vice versa.

=Stringer Analysis=

The above shows the diagram of the NACA2415 airoil with only the stringers considered in the analysis and ignoring the contributions of the skin. The important results are given below.

I_22 =

3.5173e-005

I_23 =

9.5347e-007

I_33 =

6.6954e-007

beta =

1.5010

maxsigmaxx_ylocation =

0.4999

maxsigmaxx_zlocation =

-7.8576e-004

maxsigma_xx =

4.1620e+008

M_zult =

2.1625e+005

M_yult =

-8.6498e+004

=Full NACA2415 Airfoil Analysis=

The diagram above shows the centroid location and neutral axis of the airfoil when the skin is considered along with the stringers. The important results are as follows:

I_22combined =

3.5628e-005

I_23combined =

7.9850e-006

I_33combined =

1.2106e-004

beta =

0.1771

maxsigmaxx_ylocation =

0.4999

maxsigmaxx_zlocation =

-7.8576e-004

maxsigma_xx =

4.1620e+008

A close up of the difference in the centroids of the stringers and the total airfoil is provided below for easier analysis.

As seen, the location and magnitude of the max normal stress barely changes, which is expected due to the centroid barely changing with the addition of the skin to the analysis.

The highest bending stress is in a stringer occurs in stringer H due to its distance from the neutral axis. It's value is

maxsigma_xx1 =

3.3823e+008

This causes the ultimate bending moments experienced by the stringers to be:

M_zult =

3.1349e+003

M_yult =

-1.2539e+003


title: File:HW5-1.GIF


text:
User:Eml4500.f08.wiki1.schaet/hw5

Problem Statement
3-D Three Bar Truss Problem p230 in Bhatti Consider the three bar truss problem shown below. The cross sectional are of elements 1 and 2 is 200mm2 and that of element 3 is 600mm2. All elements are made of the same material with E=200Gpa. The applied load is P=20kN. We were also asked to determine if this truss system was statically determinate. If it was, we were to solve it using statics and compare the solution to the FEA solution generated with matlab. Below is the picture, similar to the one given in the book. The nodal coordinates are as follows: The solution given in the book is as follows:

Matlab Code
Below is the code for the 3-D Three Bar Truss System:  Three Bar Space Truss

Needed m-files for Space Truss Problem
This code uses three m-files. The first mfile, SpaceTrussElement.m, defines the element stiffness matrix. The second m-file, SpaceTrussResults.m, computes the element strain, stress and axial forces. The third m-file, NodalSoln.m, was given in an earlier homework assignment. It computes the nodal solutions.  SpaceTrussElement.m  SpaceTrussResults.m  NodalSoln.m

conn and lmm Arrays of Space Truss Problem
Below are the arrays lmm and conn used to solve the 3-D three-bar truss system: conn= $$ \begin{bmatrix} 1 & 4 \\ 2 & 4 \\ 3 & 4\end{bmatrix} $$ lmm= $$ \begin{bmatrix} 1 & 2 & 3 & 10 & 11 & 12 \\ 4 & 5 & 6 & 10 & 11 & 12 \\ 7 & 8 & 9 & 10 & 11 & 12 \end{bmatrix}$$

Images of Space Truss Problem
Below are the four images of the undeformed shape vs. the deformed shape. The undeformed shape is the blue dotted line and the deformed shape is the red solid line. Their perspective plot is from the point (-2, -2, 3) as suggested. From point (-2,-2,3) YZ plane XZ plane XY plane

Matlab Code for the Images of Space Truss Problem
Below is the matlab code used to obtain the previous images:  Image Code

Statics Solution of the Space Truss Problem
This problem, unfortunately for us, is statically determinate. In order to solve this problem, we first note that there will be three reaction forces at node 1, node 2, and node 3. Each one of these reaction forces can be divided into the x, y, and z components. Doing so gives us the following equation: $$ R_1 = R_{1x}\hat i + R_{1y}\hat j + R_{1z}\hat k $$ $$ R_2 = R_{2x}\hat i + R_{2y}\hat j + R_{2z}\hat k $$ $$ R_3 = R_{3x}\hat i + R_{3y}\hat j + R_{3z}\hat k $$ In order to solve for these reactions, we need to find what each reaction components are. Noting that R3x and R3y are both zero, we have the following equations to work off of: $$\mathbf{R_{1x}} = \mathbf{R_1}\cos{\phi_1}\cos{\theta_1}$$ $$\mathbf{R_{1y}} = \mathbf{R_1}\cos{\phi_1}\sin{\theta_1}$$ $$\mathbf{R_{1z}} = \mathbf{R_1}\sin{\phi_1}$$ $$\mathbf{R_{2x}} = \mathbf{R_2}\cos{\phi_2}\cos{\theta_2}$$ $$\mathbf{R_{2y}} = \mathbf{R_2}\cos{\phi_2}\sin{\theta_2}$$ $$\mathbf{R_{2z}} = \mathbf{R_2}\sin{\phi_2}$$ $$\mathbf{R_{3z}} = \mathbf{R_3}$$ Using the pythagorean theorem and trigonometry the following calculations were made in order to find the values for θ and Φ. $$\theta_1 = 63.435^\circ$$ $$\theta_2 = -45^\circ$$ $$\phi_1 = 42.975^\circ$$ $$\phi_2 = 44.482^\circ$$ Using these equations, we set up the equations for the forces in the following steps: $$ \sum F = 0 $$ $$\sum \mathbf{F_x} = \mathbf{R_1}\cos{\phi_1}\cos{\theta_1} + \mathbf{R_2}\cos{\phi_2}\cos{\theta_2} = 0$$ $$\sum \mathbf{F_y} = \mathbf{R_1}\cos{\phi_1}\sin{\theta_1} + \mathbf{R_2}\cos{\phi_2}\sin{\theta_2} - \mathbf{P} = 0$$ $$\sum \mathbf{F_z} = \mathbf{R_1\sin{\phi_1}} + \mathbf{R_2\sin{\phi_2}} + \mathbf{R_3} = 0$$ Plugging in all of the known values, given and calculated, the following results are obtained: R1 = 20374.6588 R2 = 13214.4037 R3 = -23148.1172 These values are just like the ones received using Matlab and the FEA method.

title: User:Eml4500.f08.wiki1.schaet/hw5


title: File:HW5-2.GIF


text: File:Hw 5 three bar truss.jpg

title: File:Hw 5 three bar truss.jpg


title: File:GroupFEA6bartruss.jpg


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title: File:HW5-3.GIF


text: File:Airfoil sketch.JPG

title: File:Airfoil sketch.JPG


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text:
User:Eml4500.f08.team.dwyer.jd/HW5 Oct 31 (Meeting 28)

Introduction to Finite Element Analysis using Differential Equations
Looking back at meeting 27, we take: $$\sum{F_x}=0= -N(x,t) +N(x +dx,t) + f(x,t)dx - m(x) \ddot{u}dx$$

$$= \frac{\partial N}{\partial x}(x,t)dx + h.o.t. + f(x,t)dx + m(x)\ddot{u}dx\;\;(1)$$

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

Expanding

$$-N(x,t)+N(x+dx,t)$$ to $$-N(x,t)+N(x,t)+N(dx,t)=N(dx,t)$$

Equation 1

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

Where H.O.T. stands for higher order terms which can be neglected by recalling the Taylor series expansion:

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

Therefore Equation 1 can be re-written as Equation 2- The Equation of Motion for an Elastic Bar

$$\frac{\partial N}{\partial x}+f=m\ddot{u}$$

$$N(x,t)=A(x)\sigma(x)\;\;(3)$$ is the constructive relation where,

$$\sigma(x) = E(x)\epsilon(x,t)$$ where,

$$\epsilon(x,t)=\frac{\partial U}{\partial x}(x,t)$$

PDE of motion

(3) in (2) yields: $$\frac{\partial}{\partial x}[A(x)E(x)\frac{\partial u}{\partial x}] + f(x,t) = m(x) \ddot{u}\;\;(4)$$ where, $$\ddot{u} = \frac{{\partial^2}u}{\partial t^2}$$

Single Degree of Freedom Example

A single degree of freedom (SDOF) situation is shown above. The equation of motion is: $$m\ddot{d}+kd=F$$

 * 2 initial conditions are needed since it is a 2nd order derivative.

Boundary Conditions

 * Need 2 b.c.'s ( 2nd order derivation with respect to x)
 * Need 2 initial conditions (2nd derivative with respect to t) - initial displacement, initial velocity

2 initial conditions(initial displacement, initial velocity)

$$u(o,t)=0=u(L,t)$$

(2nd derivative with respect to t)

1) $$u(o,t) = 0$$

$$N(L,t) = F$$

$$N(L,t)=A(L)\sigma(L,t)$$ where $$(\sigma(L,t) = E(L)\epsilon(L,t)$$ where $$\epsilon(x,t)=\frac{\partial U}{\partial x}(x,t)$$


text:
User:Eml4500.f08.wiki1.aguilar/part6

Some of the equations that we have learned over the course of the semester need to be updated for use with 3-dimensional systems or space trusses.

With the local $$\tilde{x}$$ axis along the axis of the element, the local element equations are still the same as the 2-node axial deformation element:

$$\frac{EA}{L}\begin{pmatrix}\;\; 1&-1\\-1&\;\;1\end{pmatrix}\begin{pmatrix}q_1\\q_2\end{pmatrix}=\begin{pmatrix}P_1\\P_2\end{pmatrix}$$

Which gives us our familiar axial displacement relationship: $$\mathbf {\hat {k}}^{(e)} \mathbf q^{(e)} = \mathbf P^{(e)} $$

In the global coordinates, the nodal degrees of freedom and the nodal applied forces for each element are:

$$\mathbf q = \begin{pmatrix}d_1\\d_2\\d_3\\d_4\\d_5\\d_6\end{pmatrix}$$     and       $$\mathbf p = \begin{pmatrix}f_1\\f_2\\f_3\\f_4\\f_5\\f_6\end{pmatrix}$$

In order to derive the transformation matrix, we must do the following:

Consider the displacement vector of the local node 1 denoted by $$\vec{d}_{1}^{(e)}$$.

$$\vec{d}_{[1]}^{(e)}={d}_{1}^{(e)}\vec{i}+{d}_{1}^{(e)}\vec{j}+{d}_{1}^{(e)}\vec{k}$$

$${q}_{1}^{(e)}$$ = axial displacement of node 1 is the orthagonal projection of the displacement vector $$\vec{d}_{1}^{(e)}$$ of node 1 on the $$\tilde{x}$$ axis of element 'e'.

$${q}_{1}^{(e)}=\vec{d}_{1}^{(e)}\cdot\vec{\tilde{i}}$$

$${q}_{1}^{(e)}=({d}_{1}^{(e)}\vec{i}+{d}_{2}^{(e)}\vec{j}+{d}_{3}^{(e)}\vec{k})\cdot\vec{\tilde{i}}$$

$${q}_{1}^{(e)}={d}_{1}^{(e)}(\vec{i}\cdot\vec{\tilde{i}})+{d}_{2}^{(e)}(\vec{j}\cdot\vec{\tilde{i}})+{d}_{3}^{(e)}(\vec{k}\cdot\vec{\tilde{i}})$$

$$L=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$

$$(\vec{i}\cdot\vec{\tilde{i}})=\frac{x_2-x_1}{L}=l^{(e)}$$

$$(\vec{j}\cdot\vec{\tilde{i}})=\frac{y_2-y_1}{L}=m^{(e)}$$

$$(\vec{k}\cdot\vec{\tilde{i}})=\frac{z_2-z_1}{L}=m^{(e)}$$

$${q}_{1}^{(e)}=l^{(e)}{d}_{1}^{(e)}+m^{(e)}{d}_{2}^{(e)}+n^{(e)}{d}_{3}^{(e)}$$

$${q}_{1}^{(e)}= \begin{bmatrix}l^{(e)} & m^{(e)}&n^{(e)} \end{bmatrix}_{1x3}\begin{Bmatrix}{d}_{1}^{(e)}\\{d}_{2}^{(e)}\\{d}_{3}^{(e)}  \end{Bmatrix}_{3x1}$$

Here we can see that $${q}_{1}^{(e)}$$ is a 1x1 scalar.

We do this for node 2 as well. $${q}_{2}^{(e)}= \begin{bmatrix}l^{(e)} & m^{(e)} & n^{(e)}\end{bmatrix}_{1x3}\begin{Bmatrix}{d}_{4}^{(e)}\\{d}_{5}^{(e)}\\{d}_{6}^{(e)}  \end{Bmatrix}_{3x1}$$

which leads us to the transformation between global to local degrees of freedom which are as follows:

$$\begin{Bmatrix} {q}_{1}^{(e)}\\{q}_{2}^{(e)}\end{Bmatrix}= \begin{bmatrix} l^{(e)}& m^{(e)}& n^{(e)}&0&0&0\\0&0&0& l^{(e)}& m^{(e)}& n^{(e)} \end{bmatrix} \begin{Bmatrix}{d}_{1}^{(e)}\\{d}_{2}^{(e)}\\{d}_{3}^{(e)}\\{d}_{4}^{(e)}\\{d}_{5}^{(e)}\\{d}_{6}^{(e)} \end{Bmatrix}\Rightarrow \textbf{q}_{2x1}^{(e)}=\textbf{T}_{2x6}^{(e)}\textbf{d}_{6x1}^{(e)}$$

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

\Rightarrow \textbf{d}_{6x1}^{(e)} = {\textbf{T}_{6x2}^{(e)}}^{T} \textbf{q}_{2x1}^{(e)} $$

title: User:Eml4500.f08.wiki1.aguilar/part6


title: File:HW5Picture2.gif


title: File:HW5Picture2.jpg


title: File:TeckHw5.PNG


text: File:Cutspacetruss.png

title: File:Cutspacetruss.png


title: File:Tlc.fig.5.1.jpg


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text:
User:Eml4500.f08.echo.mott/HW5matlab

Matlab Solutions

 * Use matlab to plot the underformed shape (dotted lines) and deformed shape (solid lines) with a magnifying factor.

 * 1-Plane view down the x axis

 * 2-Plane view down the y axis

 * 3-Plane view down the z axis

 * 4-Perspective view from an observer point

 * Run the 6-bar truss matlab code, and plot the underformed (dotted lines) and deformed (solid lines) shapes of the truss. Put the global node numbers and element numbers on the deformed shape in the figure.

 * Next, modify the 6-bar truss code for the case where the Young's modulus E is not the same for all elements (but the cross sectional area remains the same for all elements); use the following values:

$$E ^{(1)} = 150 GPa$$

$$E ^{(2)} = 180 GPa$$

$$E ^{(3)} = 200 GPa$$

$$E ^{(4)} = 200 GPa$$

$$E ^{(5)} = 220 GPa$$

$$E ^{(6)} = 250 GPa$$

 * In another figure, plot the deformed shape of the truss for both cases:

 * Case 1. Original 6-bar truss example:

 * $$E ^{(i)} = 200 GPa $$ for all i = 1, ..., 6

 * $$A ^{(i)} = 0.001 m^2 $$ for all i = 1, ..., 6

 * Case 2. As shown above:

 * $$E ^{(1)} = 150 GPa$$

 * $$E ^{(2)} = 180 GPa$$

 * $$E ^{(3)} = 200 GPa$$

 * $$E ^{(4)} = 200 GPa$$

 * $$ E ^{(5)} = 220 GPa$$

 * $$E ^{(6)} = 250 GPa$$

 * $$A ^{(i)} = 0.001 m^2$$ for all i = 1, ..., 6

 * Using the same magnification factor, with different line styles; plot also the underformed shape in dotted line, but do not put the global node numbers and element numbers to avoid cluttering up this figure; this information was given already in the previous figure.

title: User:Eml4500.f08.echo.mott/HW5matlab


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User:Eml4500.f08.team.foskey.ckf/teamHW6

Team "TEAM"

HW6 Wiki Report

Lecture Notes

Nov 3 - James Dwyer

Nov 5 - Joseph Figuerrez (done)

Nov 7 - Jin Guan (done)

Nov 10 - Jin Guan (done)

Nov 12 - Tim Bengtson

Nov 14 - Jin Guan (done)

Determining if the electric pylon is statically indeterminate - Joseph Figuerrez (done)

MATLAB electric pylon -Cory

eigenvalues of the pylon - Cory (might need help though)

HW problems:

Nov 3 - redo rectangle truss problem from hw 4 (redraw the eigenvector mode diagrams) see []

Nov 5 - Provide the step before getting the equation Kd=F

Nov 7 - provide function for N_i(x)

Nov 10 - evaluate u(x_i+1)=d_i+1

Nov 12 - fill in the B matrix's first column $$B(x)=\begin{bmatrix} HW & \frac{1}{L_{(i)}}\\ \end{bmatrix} $$, verify $$k^{(i)}=\frac{EA}{L^{(i)}} \begin{bmatrix} 1 &-1\\ -1 &1\\ \end{bmatrix} $$ by evaluating (working out the multiplication of matrices and integration)$$k_{2x2}^{(i)}=\int_{x_i}^{x_{i+1}}{B(x)^T_{2x1}EA(x)_{1x1}B(x)_{1x2}dx}$$ the details of these can be found in the book on p. 153-154 verify $$k^{(i)}=\frac{EA}{L^{(i)}} \begin{bmatrix} 1 &-1\\ -1 &1\\ \end{bmatrix} $$ by evaluating (working out the multiplication of matrices and integration) $$k_{2x2}^{(i)}=\int_{0}^{\tilde{x}}B(\tilde{x})^T_{2x1}EA(\tilde{x})_{1x1}{B(\tilde{x})_{1x2}d \tilde{x}}$$ the details of these can be found in the book on p. 153-154 find $$k^{(i)}$$ for $$A(\tilde{x})=N_1^{(i)}(\tilde{x})A_1+N_12^{(i)}(\tilde{x})A_2$$ and $$E(\tilde{x})=N_1^{(i)}(\tilde{x})E_1+N_12^{(i)}(\tilde{x})E_2$$ Nov 14 -


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

6-Bar Truss System Simulation
The following describes the reactions and displacements of a 6-bar truss system shown on pg 226 of the textbook.

Same cross-sectional areas and Moduli
Code and results of the reaction forces and displacements:

The following displays the code for the visual representation of the truss system and its deformation.

Different cross-sectional areas and Moduli
Code and results of the reaction forces and displacements of the six-bar truss system, each element with a different cross-sectional area and Modulus value combination, chosen without much thought:

>

The following displays the code for the visual representation of the truss system and its deformation.


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User:Eml4500.f08.team.Bengtson/teamHW5

Meeting 28: Friday, 31st Oct 08

This is the full body diagram of a composite material with force acting on it(shows the different reactions a displacement)

The equation of force in the x direction for the above diagram is

$$ \sum{F_{x}}=0=-N(x,t) +N(x+dx,t)+ f(x,t)dx-M(x)\ddot{u}dx$$

$$ =\frac{\partial N}{\partial x}(x,t)dx + HOT+f(x,t)dx-M\ddot{u}dx $$
 * HOT = Higher order term

The HOT is neglected.

The dx can be factored out of all the terms leaving

$$ =\frac{\partial N}{\partial x}(x,t) + HOT+f(x,t)-M\ddot{u} $$

Review of Taylor Series

This is to show why the HOT are $$ f(x+dx)=f(x) + \frac{\partial F(x)}{\partial x}dx + \frac{1}{2}\frac{\partial^2 f(x)}{\partial x^2}dx^2 +..... $$

After the first two terms, the terms are considered HOT and as such in this class be neglected.

eq(1)$$\Rightarrow \frac{\partial N}{\partial x}+f=m\ddot{u}$$ (2)

The above is the equation of motion.

$$ N(x,t) = A(x) \sigma (x,t) $$(3)

$$ \sigma (x,t)= E(x),\epsilon (x,t) $$

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

We substitute (3) into (2)

This yields a new equation representative of the equation of motion $$ \frac{\partial }{\partial x}[A(x) E(x) \frac{\partial u}{\partial x}]+ F(x,t) = m(x)\ddot{u} $$

Please Note:

$$ \ddot{u}=\frac{\partial^2 u}{\partial x^2}$$

It is found that in order to solve this equation we need two boundary conditions and two initial conditions.

These can be two initial conditions, such as the initial displacement or initial velocity.

$$ u(0,t)=0=u(L,t)$$

This is the initial conditions of a bar that is fixed at both ends.

Shown as:

2nd condition

$$ N(L,t) = F(t)$$

$$ N(L,t)=A(L) \sigma (L,t) (3) $$

$$ \sigma (L,t)= E(L),\epsilon (L,t) $$

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

$$ \frac{\partial u}{\partial x}(L,t) = \frac{F(t)}{A(L)(E(L)} $$

The velocity of the bar at length L at time t of the bar below.

title: User:Eml4500.f08.team.Bengtson/teamHW5


text:
User:Eas4200c.f08.nine.s

Homework 1

Homework 2

Homework 3

Homework 4

Homework 5

Homework 6

Group nine - Homework 5

Stress-Strain Relation
$\left\{\epsilon_{ij}\right\}_{6x1} = \begin{bmatrix}A_{3x3}&O_{3x3}\\ O_{3x3}&B_{3x3}\end{bmatrix}_{6x6}\left\{\sigma_{ij}\right\}_{6x1}$  where the {6x6} determinant is denoted by  $C$

$\sigma-\epsilon$  relation:

$\left\{\epsilon_{ij}\right\}_{6x1} = \begin{bmatrix}A_{3x3}^{-1}&O_{3x3}\\ O_{3x3}&B_{3x3}^{-1}\end{bmatrix}_{6x6}\left\{\sigma_{ij}\right\}_{6x1}$  where the {6x6} determinant is denoted by  $C^{-1}$

Verification of the Identity Matrix: $C_{6x6}^{-1} C_{6x6} = I_{6x6}$

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

$$\left\{\sigma_{ij} \right\} = \begin{bmatrix} \bar{A}^{-1} & 0 \\ 0 & \bar{B}^{-1} \end{bmatrix}\begin{Bmatrix} 0\\ 0\\ 0\\ 0\\ \epsilon_{31}\\ \epsilon_{12} \end{Bmatrix}$$

Stress Tensors

four zero stress components using this equation:

$$\sum_{i=1,j=1}^3\frac{\delta \sigma_{ij}}{\delta x_i}=0$$

gives:

$$\frac{\delta \sigma_{11}}{\delta x_{1}}+\frac{\delta \sigma_{12}}{\delta x_{2}}+\frac{\delta \sigma_{13}}{\delta x_{3}}=0$$

$$\frac{\delta \sigma_{21}}{\delta x_{1}}+\frac{\delta \sigma_{22}}{\delta x_{2}}+\frac{\delta \sigma_{23}}{\delta x_{3}}=0$$

$$\frac{\delta \sigma_{31}}{\delta x_{1}}+\frac{\delta \sigma_{32}}{\delta x_{2}}+\frac{\delta \sigma_{33}}{\delta x_{3}}=0$$

Bidirectional Bending

$$M_{y}=\int_{aA}^{}{z\sigma _{xx}dA}$$

$$M_{z}=\int_{aA}^{}{y\sigma _{xx}dA}$$

Moment of Inertia Tensors:

$$I_{yy}=\int_{A}^{}{z^{2}dA}$$

$$I_{zz}=\int_{A}^{}{y^{2}dA}$$

$$I_{yz}=\int_{A}^{}{zydA}$$

Hooke's Law

$$\sigma _{xx}=E\varepsilon _{xx}$$

$$\sigma=\frac{IyMz-IyzMy}{IyIz-(Iyz)^{2}}(y)+\frac{IzMy-IyzMz}{IyIz-(Iyz)^{2}}(z)$$

$$I=\begin{bmatrix}I_{11}&I_{12}&I_{13}\\|&I_{22}&I_{23}\\symm&-&I_{33}\end{bmatrix}$$

Nonuniform Stress Field
Nonuniform Stress Field in 3-D. $$\sum Fx=0=-\sigma(x)A+\sigma(x+dx)A+f(x)dx$$

0=A[$\sigma(x+dx)-\sigma(x)$]+f(x)dx

$$\frac{d\sigma}{dx}+\frac{f(x)}{A}=0$$ were f(x) is the force per unit length and A is the applied load.

$$\sum{Fx}=0=dydz[-\sigma_{xx}(x,y,z)+\sigma_{xx}(x+dx,y,z)]+dzdx[-\sigma_{yx}(x,y,z)+\sigma_{yx}(x,y+dy,z)]

+dxdy[-\sigma_{zx}(x,y,z)+\sigma_{zx}(x,y,z+dz)]$$

$$0=(dxdydz)\left[\frac{\delta\sigma_{xx}}{\delta x}+\frac{\delta\sigma_{yx}}{\delta x}+\frac{\delta\sigma_{zx}}{\delta z}\right]$$

Similarly for the forces in the y direction:

$$\sum Fy =0= dy dz [-\sigma_{yx}(x,y,z)+\sigma_{yx}(x+dx),y,z)]+dzdx[-\sigma_{yy}(x,y,z)+\sigma_{yy}(x,y+dy,z)]+dxdy[\sigma_{yz}(x,y,z)+\sigma_{yz}(x,y,z+dz)]$$

which becomes:

$$0=(dx dy dz)\left[\frac{\delta \sigma_{yx}}{\delta_x}+\frac{\delta \sigma_{yy}}{\delta_y}+\frac{\delta \sigma_{yz}}{\delta_z}\right]$$

Z direction:

$$\sum Fy =0= dy dz [-\sigma_{zx}(x,y,z)+\sigma_{zx}(x+dx),y,z)]+dzdx[-\sigma_{zy}(x,y,z)+\sigma_{zy}(x,y+dy,z)]+dxdy[\sigma_{zz}(x,y,z)+\sigma_{zz}(x,y,z+dz)]$$

which becomes:

$$0=(dx dy dz) \left[\frac{\delta \sigma_{zx}}{\delta_x}+\frac{\delta \sigma_{zy}}{\delta_y}+\frac{\delta \sigma_{zz}}{\delta_z}\right]$$

Dimensional Analysis
$$[F] = \frac{F}{L} = \frac{force}{lenght} $$

$$[ \frac{F}{A}] = \frac{F}{L^3}$$

$$[A] = L^2 $$

$$[\sigma] = \frac{f}{L^2} \rightarrow [\frac{d \sigma}{dx}] = \frac{[F/L^2]}{[L]} = MLT^{-2}L^{-3} = ML^{-2}T^{-2} $$

$$[dx] = L $$

$$ [EI] = \frac{F}{L^3} \times L^4 = FL^2 $$

$$ [D] = [E][h^3]= \frac {F}{L^2} \times L^3 = FL $$

$$ \epsilon = \frac{du}{dx} = \frac{\Delta L}{L} \rightarrow [\epsilon] = \frac {[du]}{[dx]} = \frac {L}{L} = 1 $$

$$ \nu = \frac{\epsilon_{yy}}{\epsilon_{xx}} \rightarrow [\nu] = \frac{[\epsilon_{yy}]}{[\epsilon_{xx}]} = 1 $$

Sample Run of Code (NACA Plot)
NACA Airfoil calculation program Enter first digit of airfoil: 2 Enter second digit: 4 Enter the third and fourth digits: 15 Enter Py: 0 Enter Pz: 0 Enter number of segments: 60

The average area is: 0.103 The minumum number segments required to have the average area accurate within 1 percent is: 24.000

Figure 1 shows the cross-section of the NACA airfoil and the centroid line

Contributing Team Members
The following students contributed to this report:

Felix Izquierdo Eas4200c.f08.nine.F 18:34, 6 November 2008 (UTC) Ricardo Albuquerque Eas4200c.f08.nine.R 18:39, 6 November 2008 (UTC) Dave Phillips Eas4200C.f08.nine.d 18:46, 6 November 2008 (UTC) Stephen Featherman Eas4200c.f08.nine.S 18:49, 6 November 2008 (UTC)

title: User:Eas4200c.f08.nine.s


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User:Eml4500.f08.delta 6.guzman/Composites with Moduli that Change with Length

'''Composite materials

'''

Refer to engineered materials composed of usually two constituent materials with considerably different mechanical properties.

'''Types '''

In order to have the composite at least one fraction of each type is required.

Matrix: by maintaining a given position the matrix materials surround and support the reinforcement materials.

Reinforcement: by providing mechanical and electrical properties reinforcement materials enhance the matrix materials

Examples

Examples of composite materials can be dated to primitive times. One of the most elementary would be the composite for bricks which were made out of mud and straw. More modern examples can be encountered from spacecraft to roadways applications such as asphalt concrete.

Some examples of composite materials where the modulus of elasticity is a function of length are:

 * Hardmetal: is a composite of metal and carbide. The modulus of metal is significantly different than that of carbide in order to increase the composite’s strength.
 * Bone: is a composite of hydroxyapatite with collagen fibers. The modulus of bone is altered by the collagen fibers embedded inside of the dydroxyapatite to lend it the force necessary to support the body movements.
 * Concrete walls (reinforced): the modulus of concrete is changed by using reinforced steel in order to reinforce the concrete in walls.
 * Composite panel: the modulus of two aluminum sheets is changed by inserting in between a non-aluminum material.

For additional information please visit:

 * http://www.mech.utah.edu/~rusmeeha/labNotes/composites.html

 * http://www.societyofrobots.com/materials_carbonfiber.shtml

 * http://www.science.org.au/nova/059/059key.htm

 * http://www.matter.org.uk/matscicdrom/manual/co.html#_Toc360435310

For more in depth research about composite materials please visit the University of Florida engineering databases http://libguides.uflib.ufl.edu/content.php?pid=7641&sid=48706 Other resources are available by engineering subjects guides http://www.uflib.ufl.edu/msl/subjects/engin.html or by consulting the University of Florida engineering subject specialists:
 * Denise Bennett
 * Amy Buhler
 * Vernon Kisling
 * Carrie Newsom


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User:Eas4200c.f08.gator.edwards/Wk5 Matlab

This MATLAB program calculates the maximum bending moments, maximum normal bending stresses, and ultimate moments for an airfoil and stringers. Part 1 uses only the four sets of stringers while Part 2 uses the stringers and the skin. The figure shows the centroid of the stringers, the total centroid, and the neutral axis, along with the airfoil outline and spars.

The results show around a 6% increase in ultimate moment between Parts 1 and 2. This is a fairly small amount so the approximation of using only the stringers is good for many applications. The maximum normal bending stress is two orders of magnitude higher in Part 2, indicating an error in the calculations


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User:Eml4500.f08.echo.mott/hw6

=HW Report 6=

Electric Pylon
% Cross sectional area of elements A = [400]; % mm^2

% Elastic Modulus e = [200]*10^6; % kPa

% Density rho = [7.8]*1000; % g/mm^3

% Coordinates of the nodes coord = [1.5 0 0; 4.53 0 0; 1.88 1.18 0; 3.03 1.18 0; 4.17 1.18 0; 2.2 2.181 0; 3.03 2.181 0; 3.85 2.181 0; 2.53 3.181 0; 3.53 3.181 0; 2.34 3.681 0;3.03 3.681 0; 3.71 3.681 0; 2.16 4.181 0; 2.62 4.181 0; 3.44 4.181 0; 3.89 4.181 0; 2.36 4.5 0; 3.7 4.5 0; 1.8 5.172 0; 4.25 5.172 0; 0 6 0; 1.06 6 0; 1.5 6 0; 1.82 6 0; 2.22 6 0; 2.71 6 0; 3.34 6 0; 3.84 6 0; 4.23 6 0; 4.55 6 0; 5 6 0; 6.05 6 0; .8 6.23 0; 5.25 6.23 0; 1.24 6.355 0;  4.82 6.355 0; 1.5 6.43 0; 2 6.43 0; 2.48 6.43 0; 3.03 6.43 0; 3.58 6.43 0; 4.05 6.43 0; 4.55 6.43 0; 1.5 7.43 0; 4.55 7.43 0]*1000; %mm

% Connectivity Array conn = [ 1 3; 1 4; 4 2; 2 5; 3 4; 4 5; 3 6; 3 7; 5 7; 5 8; 6 7; 7 8; 6 9; 7 9; 7 10; 8 10; 9 10; 9 11; 9 12; 10 12; 10 13; 11 12; 12 13; 11 14; 11 15; 12 15; 12 16; 13 16; 13 17; 14 15; 15 16; 16 17; 14 20; 14 18; 15 18; 16 19; 17 19; 17 21; 18 20; 19 21; 20 24; 20 26; 21 29; 21 31; 22 23; 23 24; 24 25; 25 26; 26 27; 27 28; 28 29; 29 30; 30 31; 31 32; 32 33; 22 34; 23 34;  23 36; 24 36; 24 38; 25 38; 25 39; 26 39; 26 40; 27 40; 27 41; 28 41; 28 42; 29 42; 29 43; 30 43; 30 44; 31 44; 31 37; 32 37;  32 35; 33 35; 34 36; 35 37; 36 38; 37 44; 38 39; 39 40; 40 41; 41 42; 42 43; 43 44; 38 45; 39 45; 43 46; 44 46];

% Location Master Matrix lmm = [1 2 5 6; 1 2 7 8; 7 8 3 4; 3 4 9 10; 5 6 7 8; 7 8 9 10; 5 6 11 12; 5 6 13 14; 9 10 13 14; 9 10 15 16; 11 12 13 14; 13 14 15 16; 11 12 17 18; 13 14 17 18; 13 14 19 20; 15 16 19 20; 17 18 19 20; 17 18 21 22; 17 18 23 24; 19 20 23 24; 19 20 25 26;  21 22 23 24;   23 24 25 26; 21 22 27 28; 21 22 29 30; 23 24 29 30; 23 24 31 32; 25 26 31 32; 25 26 33 34; 27 28 29 30; 29 30 31 32; 31 32 33 34; 27  28 39 40; 27 28 35 36; 29 30 35 36; 31 32 37 38; 33 34 37 38; 33 34 41 42; 35 36 39 40; 37 38 41 42; 39 40 47 48; 39 40 51 52; 41 42  57 58; 41 42 61 62; 43 44 45 46; 45 46 47 48; 47 48 49 50; 49 50 51 52; 51 52 53 54; 53 54 55 56;  55 56 57 58; 57 58 59 60; 59 60 61  62; 61 62 63 64; 63 64 65 66; 43 44 67 68; 45 46 67 68; 45 46 71 72; 47 48 71 72; 47 48 75 76;  49 50 75 76; 49 50 77 78; 51 52 77 78;      51 52 79 80; 53 54 79 80; 53 54 81 82; 55 56 81 82; 55 56 83 84; 57 58 83 84; 57 58 85  86; 59 60 85 86; 59 60 87 88; 61 62 87 88; 61  62 73 74; 63 64 73 74; 63 64 69 70; 65 66 69 70; 67 68 71 72; 69 70 73 74; 71 72 75   76; 73 74 87 88; 75 76 77 78; 77 78 79 80; 79 80  81 82; 81 82 83 84; 83 84 85 86; 85 86 87 88; 75 76 89 90; 77 78 89 90; 85 86 91   92; 87 88 91 92];

% Generating the stiffness matrix K

K = zeros(92); M = zeros(92);

for i = 1:91 l = lmm(i, :); c = conn(i, :); k = PlaneTrussElement(e, A, coord(c,:)); m = LumpMassMatrix(rho, A, coord(c,:)); K(l, l) = K(l, l) + k; M(l, l) = M(l, l) + m;

end

% Force matrix

R = zeros(92, 1); R(66,1) = -1000; [d, rf] = NodalSoln(K, R, [1,2,3,4], zeros(4,1)); truss_reax = [];

for i = 1:91 truss_reax = [truss_reax; PlaneTrussResults(e, A, coord(conn(i,:),:), d(lmm(i,:)))]; [Deps(i), sigma(i,1), force(i,1)] = PlaneTrussResults(e, A, coord(conn(i,:),:), d(lmm(i,:))); end d = d*1000;

% Plot the "superman" structure % model with 3-D beam elements

dof = 2;         %  dof per node:  1= axial disp x,  2= disp y

% obtain the coordinates of all nodes

n_node = 46;            % number of nodes n_elem = 91;            % number of elements total_dof = 2 * n_node; % total dof of system

position(:, 1) = [ 1.5; 0; 0]; position(:, 2) = [ 4.53; 0; 0];

position(:, 3) = [ 1.88; 1.18; 0]; position(:, 4) = [ 3.03; 1.18; 0]; position(:, 5) = [ 4.17; 1.18; 0];

position(:, 6) = [ 2.2; 2.181; 0]; position(:, 7) = [ 3.03; 2.181; 0]; position(:, 8) = [ 3.85; 2.181; 0];

position(:, 9) = [ 2.53; 3.181; 0]; position(:, 10) = [ 3.53; 3.181; 0];

position(:, 11) = [ 2.34; 3.681; 0]; position(:, 12) = [ 3.03; 3.681; 0]; position(:, 13) = [ 3.71; 3.681; 0];

position(:, 14) = [ 2.16; 4.181; 0]; position(:, 15) = [ 2.62; 4.181; 0]; position(:, 16) = [ 3.44; 4.181; 0]; position(:, 17) = [ 3.89; 4.181; 0];

position(:, 18) = [ 2.36; 4.5; 0]; position(:, 19) = [ 3.7; 4.5; 0];

position(:, 20) = [ 1.8; 5.172; 0]; position(:, 21) = [ 4.25; 5.172; 0];

position(:, 22) = [ 0; 6; 0]; position(:, 23) = [ 1.06; 6; 0]; position(:, 24) = [ 1.5; 6; 0]; position(:, 25) = [ 1.82; 6; 0]; position(:, 26) = [ 2.22; 6; 0]; position(:, 27) = [ 2.71; 6; 0]; position(:, 28) = [ 3.34; 6; 0]; position(:, 29) = [ 3.84; 6; 0]; position(:, 30) = [ 4.23; 6; 0]; position(:, 31) = [ 4.55; 6; 0]; position(:, 32) = [ 5; 6; 0]; position(:, 33) = [ 6.05; 6; 0];

position(:, 34) = [ .8; 6.23; 0]; position(:, 35) = [ 5.25; 6.23; 0];

position(:, 36) = [ 1.24; 6.355; 0]; position(:, 37) = [ 4.82; 6.355; 0];

position(:, 38) = [ 1.5; 6.43; 0]; position(:, 39) = [ 2; 6.43; 0]; position(:, 40) = [ 2.48; 6.43; 0]; position(:, 41) = [ 3.03; 6.43; 0]; position(:, 42) = [ 3.58; 6.43; 0]; position(:, 43) = [ 4.05; 6.43; 0]; position(:, 44) = [ 4.55; 6.43; 0];

position(:, 45) = [ 1.5; 7.43; 0]; position(:, 46) = [ 4.55; 7.43; 0]; % Deformed Truss Position dposition(:, 1) = [ 1.5; 0; 0]; dposition(:, 2) = [ 4.53; 0; 0];

dposition(:, 3) = [ 1.88 + d(5); 1.18 + d(6); 0]; dposition(:, 4) = [ 3.03 + d(7); 1.18 + d(8); 0]; dposition(:, 5) = [ 4.17 + d(9); 1.18 + d(10); 0];

dposition(:, 6) = [ 2.2 + d(11); 2.181 + d(12); 0]; dposition(:, 7) = [ 3.03 + d(13); 2.181 + d(14); 0]; dposition(:, 8) = [ 3.85 + d(15); 2.181 + d(16); 0];

dposition(:, 9) = [ 2.53 + d(17); 3.181 + d(18); 0]; dposition(:, 10) = [ 3.53 + d(19); 3.181 + d(20); 0];

dposition(:, 11) = [ 2.34 + d(21); 3.681 + d(22); 0]; dposition(:, 12) = [ 3.03 + d(23); 3.681 + d(24); 0]; dposition(:, 13) = [ 3.71 + d(25); 3.681 + d(26); 0];

dposition(:, 14) = [ 2.16 + d(27); 4.181 + d(28); 0]; dposition(:, 15) = [ 2.62 + d(29); 4.181 + d(30); 0]; dposition(:, 16) = [ 3.44 + d(31); 4.181 + d(32); 0]; dposition(:, 17) = [ 3.89 + d(33); 4.181 + d(34); 0];

dposition(:, 18) = [ 2.36 + d(35); 4.5 + d(36); 0]; dposition(:, 19) = [ 3.7 + d(37); 4.5 + d(38); 0];

dposition(:, 20) = [ 1.8 + d(39); 5.172 + d(40); 0]; dposition(:, 21) = [ 4.25 + d(41); 5.172 + d(42); 0];

dposition(:, 22) = [ 0 + d(43); 6 + d(44); 0]; dposition(:, 23) = [ 1.06 + d(45); 6 + d(46); 0]; dposition(:, 24) = [ 1.5 + d(47); 6 + d(48); 0]; dposition(:, 25) = [ 1.82 + d(49); 6 + d(50); 0]; dposition(:, 26) = [ 2.22 + d(51); 6 + d(52); 0]; dposition(:, 27) = [ 2.71 + d(53); 6 + d(54); 0]; dposition(:, 28) = [ 3.34 + d(55); 6 + d(56); 0]; dposition(:, 29) = [ 3.84 + d(57); 6 + d(58); 0]; dposition(:, 30) = [ 4.23 + d(59); 6 + d(60); 0]; dposition(:, 31) = [ 4.55 + d(61); 6 + d(62); 0]; dposition(:, 32) = [ 5 + d(63); 6 + d(64); 0]; dposition(:, 33) = [ 6.05 + d(65); 6 + d(66); 0];

dposition(:, 34) = [ .8 + d(67); 6.23 + d(68); 0]; dposition(:, 35) = [ 5.25 + d(69); 6.23 + d(70); 0];

dposition(:, 36) = [ 1.24 + d(71); 6.355 + d(72); 0]; dposition(:, 37) = [ 4.82 + d(73); 6.355 + d(74); 0];

dposition(:, 38) = [ 1.5 + d(75); 6.43 + d(76); 0]; dposition(:, 39) = [ 2 + d(77); 6.43 + d(78); 0]; dposition(:, 40) = [ 2.48 + d(79); 6.43 + d(80); 0]; dposition(:, 41) = [ 3.03 + d(81); 6.43 + d(82); 0]; dposition(:, 42) = [ 3.58 + d(83); 6.43 + d(84); 0]; dposition(:, 43) = [ 4.05 + d(85); 6.43 + d(86); 0]; dposition(:, 44) = [ 4.55 + d(87); 6.43 + d(88); 0];

dposition(:, 45) = [ 1.5 + d(89); 7.43 + d(90); 0]; dposition(:, 46) = [ 4.55 + d(91); 7.43 + d(92); 0];

% print the node coord. for i = 1 : n_node; x(i) = position(1,i)*8.3; y(i) = position(2,i)*8.3; z(i) = position(3,i)*8.3; end

for i = 1 : n_node; dx(i) = dposition(1,i)*8.3; dy(i) = dposition(2,i)*8.3; dz(i) = dposition(3,i)*8.3; end

node_connect(1, 1) = 1;  % element 1 node_connect(2, 1) = 3;

node_connect(1, 2) = 1;  % element 2 node_connect(2, 2) = 4;

node_connect(1, 3) = 3;  % element 5 node_connect(2, 3) = 4;

node_connect(1, 4) = 4;  % element 3 node_connect(2, 4) = 2;

node_connect(1, 5) = 4;  % element 6 node_connect(2, 5) = 5;

node_connect(1, 6) = 2;  % element 4 node_connect(2, 6) = 5;

node_connect(1, 7) = 3;  % element 8 node_connect(2, 7) = 7;

node_connect(1, 8) = 5;  % element 9 node_connect(2, 8) = 7;

node_connect(1, 9) = 3;  % element 7 node_connect(2, 9) = 6;

node_connect(1, 10) = 5;  % element 10 node_connect(2, 10) = 8;

node_connect(1, 11) = 6;  % element 11 node_connect(2, 11) = 7;

node_connect(1, 12) = 7;  % element 12 node_connect(2, 12) = 8;

node_connect(1, 13) = 6;  % element 13 node_connect(2, 13) = 9;

node_connect(1, 14) = 7;  % element 14 node_connect(2, 14) = 9;

node_connect(1, 15) = 7;  % element 15 node_connect(2, 15) = 10;

node_connect(1, 16) = 8;  % element 16 node_connect(2, 16) = 10;

node_connect(1, 17) = 9;  % element 17 node_connect(2, 17) = 10;

node_connect(1, 18) = 9;  % element 18 node_connect(2, 18) = 11;

node_connect(1, 19) = 9;  % element 19 node_connect(2, 19) = 12;

node_connect(1, 20) = 10;  % element 20 node_connect(2, 20) = 12;

node_connect(1, 21) = 10;  % element 21 node_connect(2, 21) = 13;

node_connect(1, 22) = 11;  % element 22 node_connect(2, 22) = 12;

node_connect(1, 23) = 12;  % element 23 node_connect(2, 23) = 13;

node_connect(1, 24) = 11;  % element 24 node_connect(2, 24) = 14;

node_connect(1, 25) = 11;  % element 25 node_connect(2, 25) = 15;

node_connect(1, 26) = 12;  % element 26 node_connect(2, 26) = 15;

node_connect(1, 27) = 12;  % element 27 node_connect(2, 27) = 16;

node_connect(1, 28) = 13;  % element 28 node_connect(2, 28) = 16;

node_connect(1, 29) = 13;  % element 29 node_connect(2, 29) = 17;

node_connect(1, 30) = 14;  % element 30 node_connect(2, 30) = 15;

node_connect(1, 31) = 15;  % element 31 node_connect(2, 31) = 16;

node_connect(1, 32) = 16;  % element 32 node_connect(2, 32) = 17;

node_connect(1, 33) = 14;  % element 33 node_connect(2, 33) = 20;

node_connect(1, 34) = 14;  % element 34 node_connect(2, 34) = 18;

node_connect(1, 35) = 15;  % element 35 node_connect(2, 35) = 18;

node_connect(1, 36) = 16;  % element 36 node_connect(2, 36) = 19;

node_connect(1, 37) = 17;  % element 37 node_connect(2, 37) = 19;

node_connect(1, 38) = 17;  % element 38 node_connect(2, 38) = 21;

node_connect(1, 39) = 18;  % element 39 node_connect(2, 39) = 20;

node_connect(1, 40) = 19;  % element 40 node_connect(2, 40) = 21;

node_connect(1, 41) = 20;  % element 41 node_connect(2, 41) = 24;

node_connect(1, 42) = 20;  % element 42 node_connect(2, 42) = 26;

node_connect(1, 43) = 21;  % element 43 node_connect(2, 43) = 29;

node_connect(1, 44) = 21;  % element 44 node_connect(2, 44) = 31;

node_connect(1, 45) = 22;  % element 45 node_connect(2, 45) = 23;

node_connect(1, 46) = 23;  % element 46 node_connect(2, 46) = 24;

node_connect(1, 47) = 24;  % element 47 node_connect(2, 47) = 25;

node_connect(1, 48) = 25;  % element 48 node_connect(2, 48) = 26;

node_connect(1, 49) = 26;  % element 49 node_connect(2, 49) = 27;

node_connect(1, 50) = 27;  % element 50 node_connect(2, 50) = 28;

node_connect(1, 51) = 28;  % element 51 node_connect(2, 51) = 29;

node_connect(1, 52) = 29;  % element 52 node_connect(2, 52) = 30;

node_connect(1, 53) = 30;  % element 53 node_connect(2, 53) = 31;

node_connect(1, 54) = 31;  % element 54 node_connect(2, 54) = 32;

node_connect(1, 55) = 32;  % element 55 node_connect(2, 55) = 33;

node_connect(1, 56) = 22;  % element 56 node_connect(2, 56) = 34;

node_connect(1, 57) = 23;  % element 57 node_connect(2, 57) = 34;

node_connect(1, 58) = 23;  % element 58 node_connect(2, 58) = 36;

node_connect(1, 59) = 24;  % element 59 node_connect(2, 59) = 36;

node_connect(1, 60) = 24;  % element 60 node_connect(2, 60) = 38;

node_connect(1, 61) = 25;  % element 61 node_connect(2, 61) = 38;

node_connect(1, 62) = 25;  % element 62 node_connect(2, 62) = 39;

node_connect(1, 63) = 26;  % element 63 node_connect(2, 63) = 39;

node_connect(1, 64) = 26;  % element 64 node_connect(2, 64) = 40;

node_connect(1, 65) = 27;  % element 65 node_connect(2, 65) = 40;

node_connect(1, 66) = 27;  % element 66 node_connect(2, 66) = 41;

node_connect(1, 67) = 28;  % element 67 node_connect(2, 67) = 41;

node_connect(1, 68) = 28;  % element 68 node_connect(2, 68) = 42;

node_connect(1, 69) = 29;  % element 69 node_connect(2, 69) = 42;

node_connect(1, 70) = 29;  % element 70 node_connect(2, 70) = 43;

node_connect(1, 71) = 30;  % element 71 node_connect(2, 71) = 43;

node_connect(1, 72) = 30;  % element 72 node_connect(2, 72) = 44;

node_connect(1, 73) = 31;  % element 73 node_connect(2, 73) = 44;

node_connect(1, 74) = 31;  % element 74 node_connect(2, 74) = 37;

node_connect(1, 75) = 32;  % element 75 node_connect(2, 75) = 37;

node_connect(1, 76) = 32;  % element 76 node_connect(2, 76) = 35;

node_connect(1, 77) = 33;  % element 77 node_connect(2, 77) = 35;

node_connect(1, 78) = 34;  % element 78 node_connect(2, 78) = 36;

node_connect(1, 79) = 35;  % element 79 node_connect(2, 79) = 37;

node_connect(1, 80) = 36;  % element 80 node_connect(2, 80) = 38;

node_connect(1, 81) = 37;  % element 81 node_connect(2, 81) = 44;

node_connect(1, 82) = 38;  % element 82 node_connect(2, 82) = 39;

node_connect(1, 83) = 39;  % element 83 node_connect(2, 83) = 40;

node_connect(1, 84) = 40;  % element 84 node_connect(2, 84) = 41;

node_connect(1, 85) = 41;  % element 85 node_connect(2, 85) = 42;

node_connect(1, 86) = 42;  % element 86 node_connect(2, 86) = 43;

node_connect(1, 87) = 43;  % element 87 node_connect(2, 87) = 44;

node_connect(1, 88) = 38;  % element 88 node_connect(2, 88) = 45;

node_connect(1, 89) = 39;  % element 89 node_connect(2, 89) = 45;

node_connect(1, 90) = 43;  % element 90 node_connect(2, 90) = 46;

node_connect(1, 91) = 44;  % element 91 node_connect(2, 91) = 46;

% connect all beam elements by connectivity array for i = 1 : n_elem; node_1 = node_connect(1,i); node_2 = node_connect(2,i); %dnode_1 = dnode_connect(1,i); %dnode_2 = dnode_connect(2,i); xx = [x(node_1),x(node_2)]; dxx = [dx(node_1),dx(node_2)]; yy = [y(node_1),y(node_2)]; dyy = [dy(node_1),dy(node_2)]; zz = [z(node_1),z(node_2)]; dzz = [dz(node_1),dz(node_2)]; plot3(xx,yy,zz,'r--') hold on     plot3(dxx,dyy,zz,'-') hold on  end

title('Electric Pylon') xlabel('x') ylabel('y') zlabel('z')

view([0 0 1])     % xy plane view % view([0 1 0])    % xz plane view %view([1 0 0])    % yz plane view

% produce postscript file for later printing % print -dps electric_pylon.ps  %print -

Results

[Max_ax_Stress, Loc] = max(Deps)

% Finding the Maximum Tensile stress and where it is located Max_ax_Stress =

4.5256e-008

Loc =

81 % This means that the maximum tensile stress is located in the 81st dof which correlates to element number 66.

[Min_ax_Stress, Loc] = min(Deps)

% Finding the Maximum Compressive Stress and where it is located.

Min_ax_Stress =

-4.3478e-008

Loc =

55

% This means that the maximum compressive stress is located in the 81st dof which correlates to element number 67.

In order to run this coding, the following three codes used in previous homeworks were used, i.e. PlaneTrussElements, PlaneTrussResults, and NodalSln, and they were all taken from the website for the text Fundamental Finite Element Analysis and Applications by M. Asghar Bhatti.

In addition to these codes, the following m.file had to be created in order to assemble the lump mass matrix for the Electric Pylon system:

This code created the following mass matrix for the pylon:

Monday, November 3rd
Initial conditions can be found when $$t = 0$$.

Notation:

1. Prescribe $$u(x, t=0) = \bar{u} (x)$$

 * Where $$\bar{u} (x)$$ is a known function of displacement

2. Also let $$\frac{\partial u}{\partial t} (x, t=0) = \dot{u} (x, t=0) = \bar{v} (x)$$

 * Where $$\bar{v}(x)$$ is a known function of velocity

 * Now proceed with the Finite Element Method to solve Partial Differential Equations.

First, look at the principle of virtual work (continuous) for the dynamics of an elastic bar:

 * $$ {(1)} \qquad { \frac{\partial}{\partial x} ( E A \frac{\partial u}{\partial x}) + f = m  \ddot u } $$

The discrete equation of motion for the elastic bar can be written as

 * $$ {(2)} \qquad {\underline{M} \underline{\ddot d} + \underline{K}\ \underline{d} = \underline{F}} $$

By inspection and comparison of equations (1) & (2) it is clear that

 * $${\frac{\partial}{\partial x} [ E A \frac{\partial u}{\partial x}]} \equiv {-\underline{K}\ \underline{d}}$$

 * $${f} \equiv {\underline{F}}$$

 * $$ {m} \equiv {\underline{M}\ \underline}$$

Equations (1) & (2) are applicable to a multiple degree of freedom system.

In order to derive (2) from (1), look at a single dof system - a spring/mass combination:

 * [[Image:SpringMassSystem.jpg|270px]]

For the system shown above, $$ m \ddot d + k d = F$$

And,

 * $$(3) \qquad \int_{0}^{L} W(x) \Big\{ \frac{\partial}{\partial x} [ E A \frac{\partial u}{\partial x}] +f - m  \ddot u  \Big\} dx = 0 \qquad for \ all \ W(x)$$

Going from equation (1) to (3) is trivial but going from (3) to (1) is not.

(3) Can be re-written as, $$ \int_{0}^{L} W(x)\ G(x) dx = 0 \qquad for \ all \ W(x)$$

Since (3) is valid for all values of W, then let $$W(x) = G(x)$$

Now (3) becomes:

$$ \int_{0}^{L} G^2(x) dx = 0 $$

$$ G^2 \ge 0 $$ and so $$ G(x) = 0 $$

Wednesday, November 5th
Integration by Parts:

$$r^{ }_{ }(x), s(x)$$

$$(rs)' = r's^{ }_{ } + rs'$$

$$r'=\frac{dr}{dx^{ }_{ }}$$, $$s'=\frac{ds}{dx^{ }_{ }}$$

$$\int (rs)' = \int r's + \int rs'$$, $$\int (rs)' = rs$$

so, $$ rs = \int r's + \int rs'$$

rearranging: $$ \int r's = rs - \int rs'$$

Recalling the Continuous PVW method: 1st term: $$r(x)=(EA)\frac{\partial u}{\partial x}$$,  $$ s(x)=w(x)^{ }_{ }$$

Using Integration by parts:

$$s^{ }_{ }=w(x)$$, $$r^{ }_{ }=(EA)\frac{\partial u}{\partial x}$$

$$\begin{align} \int_{x=0}^{x=L}w(x)\frac{\partial}{\partial x} \Bigg[(EA)\frac{\partial u}{\partial x}\Bigg]dx &=\Bigg[w(EA)\frac{\partial u}{\partial x}\Bigg]^{x=L}_{x=0} -\int^{L}_{0}\frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx \\ &=w(L)(EA)(L)\frac{\partial u}{\partial x}(L,t)-w(0)(EA)(0)\frac{\partial u}{\partial x}(0,t) -\int^{L}_{0}\frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx \\ \end{align} $$

Model Problem:

Set two boundary conditions:

At x=0, we choose w(x) such that w(0)=0 (i.e. kinematically admissible)

We now have the unknown reaction $$N(0,t)=(EA)(0)\frac{\partial u}{\partial x}(0,t)$$

Using continuous PVW, we have $$w(L)$$ and $$F(t)$$:

$$-\int_{0}^{L}\frac{dw}{dx}(EA)\frac{\partial u}{\partial x}dx+\int_{0}^{L}w(x)[f-m\ddot{u}]dx=0$$ for all $$w^{}_{}(x)$$ such that $$w^{}_{}(x)=0$$.

This leads to the following "weak form" for continuous PVW: $$\int_{0}^{L}w(m\ddot{u})dx+\int_{0}^{L}\frac{dw}{dx}(EA)\frac{\partial u}{\partial x}dx=w(L)F(t)+\int_{0}^{L}wfdx$$ for all $$w^{}_{}(x)$$ such that $$w^{}_{}(0)=0$$

Motivation: Discrete PVW applied to Equation on pg. 10-1

$$\mathbf{w}_{6x1}\cdot([\ ]_{6x2} \begin{Bmatrix} d_3 \\  d_4 \end{Bmatrix} -\mathbf{F}_{6x1})=\mathbf{0}_{1x1}$$

$$\mathbf{F}^T= \begin{bmatrix} F_1 & F_2 & F_3 & F_4 & F_5 & F_6 \end{bmatrix}$$

$$F_3$$ and $$F_4$$ are known forces, but $$F_1$$, $$F_2$$, $$F_5$$, and $$F_6$$ are unknown reactions.

Since w can be selected arbitrarily, we select w such that $$w_1=w_2=w_5=w_6=0$$.

To eliminate equations involving unknown reactions, we eliminate rows 1,2,5, and 6 to obtain the following equation:

$$\mathbf{\bar{K}}_{2x2}\mathbf{\bar{d}}_{2x1}=\mathbf{\bar{F}}_{2x1}$$ or $$\mathbf{\bar{K}}_{2x2}\begin{Bmatrix}d_3\\d_4\end{Bmatrix}=\begin{Bmatrix}F_3\\F_4\end{Bmatrix}$$

It should be noted that $$\mathbf{\bar{w}}\cdot(\mathbf{\bar{K}}\mathbf{\bar{d}}-\mathbf{\bar{F}})=0$$ for all $$\mathbf{\bar{w}}$$ where $$\mathbf{\bar{w}}=\begin{Bmatrix}w_3\\w_4\end{Bmatrix}$$

Friday, November 7th

Assume displacement u(x) for $$x_i^{ } \le x \le x_{i+1}$$ (i.e. $$x \epsilon \left [ x_i, x_{i+1} \right ]$$)

Motivation for linear interpolation of the u(x) 2-bar truss is shown in the figure below.

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

Consider the case where there are only axial displacements (i.e. zero transverse displacement) Express u(x) in terms of di=u(xi) & di+1=u(xi+1) as a linear function in x (i.e., linear interpolation)

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

where Ni(x) & Ni+1(x) are linear functions in x

HW

Ni(x)=? (in book)

END HW

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

Monday, November 10th
 Continuous PVW to Discrete PVW  (continued)

Lagrangian interpolation

Motivation for form of $$N_i(x)$$ and $$N_{i+1}$$:

1. $$N_i(x)$$ and $$N_{i+1}(x)$$ are linear (straight lines), thus any linear combination of $$N_i$$ and $$N_{i+1}$$ is also linear, and in particular, the expression for $$u(x)$$.
 * $$N_i^{}(x)=\alpha_i+\beta_i x$$, where $$\alpha _i$$ and $$\beta_i$$ are numbers
 * $$N_{i+1}^{}=\alpha_{i+1}+\beta_{i+1} x$$, where $$\alpha_{i+1}$$ and $$\beta_{i+1}$$ are numbers

The linear combination of $$N_i$$ and $$N_{i+1}$$, $$ N_i^{} d_i + N_{i+1}d_{i+1} = (\alpha_i + \beta_i x)d_i + (\alpha_{i+1} + \beta_{i+1}x)d_{i+1} = (\alpha_i d_i + \alpha_{i+1}d_{i+1}) + (\beta_i d_i + \beta_{i+1} d_{i+1})x $$ is clearly a linear function of $$x$$

2. Recall the equation for $$u(x)$$ (interpolation of $$u(x)$$):
 * $$u(x_i)=N_i^{}(x_i)d_i+N_{i+1}^{}(x_{i})d_{i+1}$$ where $$N_i^{}(x_i)=1$$ and $$N_{i+1}^{}(x_{i})=0$$, so $$u^{}_{}(x_i)=d_i$$

Wednesday, November 12th
FEM via PVW (continued)

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

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

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

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

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

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

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

Element Stiffness Matrix for element i

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

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

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

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

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

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

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

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

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

Recall the elemental degrees of freedom:

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

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

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

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

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

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

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

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

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

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

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

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

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

$$d \tilde x = dx$$

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

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

HW6

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

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

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

Friday, November 14th
HW6

Using the figures from 31-4: $$N_1^{(i)}(\tilde{x})=HW6$$

$$N_2^{(i)}(\tilde{x})=\frac{\tilde{x}}{L^{(i)}}=\begin{cases} \mbox{0 at }\tilde{x}=0\\ \mbox{1 at }\tilde{x}=L^{(i)} \end{cases}$$

Shape functions $$N_1^{(i)}, N_2^{(i)}$$ (basis)

HW6: On page 159 of the book, set $$E_1=E_2=E$$

Let $$A(\tilde{x})$$ be linear as on p33-5. Obtain $$\mathbf{k}^{(1)}$$ from the previous problem and compare to the expression given in book.

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

Next, compare the general $$\mathbf{k}^{(i)}$$ on p33-5 to the stiffness matrix obtained by using $$\frac{(A_1+A_2)}{2}$$ and $$\frac{(E_1+E_2)}{2}$$

Note $$E_1 \ne E_2$$

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

Find $$\mathbf{k}^{(i)}-\mathbf{k}_{average}^{(i)}$$

Remember: Recall the mean value theorem (MVT) & its relation to the centroid:

$$\int_{x=a}^{x=b} f(x)\,dx=f(\bar x)[b-a]$$ for $$\bar x \epsilon [a,b]$$

$$ \mathbf{a} \le \bar x \le \mathbf{b}$$

$$\int_A x \,dA = \bar x \int_A dA = \bar x A$$

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

However, in general: $$f(\bar x) \ne \frac{1}{b-a}\int_a^b f(x)\,dx$$ Where the right hand side of the equation is the average value of f

Similarly: $$g(\bar x) \ne \frac{1}{b-a}\int_a^b g(x)\,dx$$ Where the right hand side of the equation is the average value of g

Modify the 2-bar truss code to accomodate the general $$\mathbf{k}^{(i)}$$ as on p33-5

Contributing Team Members
Chris Mott 19:51, 21 November 2008 (UTC)

Bradley LaCroix 20:16, 21 November 2008 (UTC)

Brandon Sell 20:25, 21 November 2008 (UTC)

Chris Sell 20:40, 21 November 2008 (UTC)

Jason Bruce

Charles Marshall 21:12, 21 November 2008 (UTC)

title: User:Eml4500.f08.echo.mott/hw6


text: File:Hw5 fea.jpg

title: File:Hw5 fea.jpg


text:
Talk:Gravity

Exercise
 On July 20, Neil Armstrong and Buzz Aldrin became the first humans to land on the Moon. While Armstrong and Aldrin did important research on the surface of the moon, the third astronaut, Michael Collins orbited in the command module above. Determine the height (read: distance from the surface of the moon) at which Collins orbited in the command module by exploiting the following data. The moon has been approximated to weigh 7,3477×1022kg, and the radius to 1 737,10 km. The mass of the command module was 30,320 kg. Collins stayed in lunar orbit for 59 h 30 m. The command module did exactly 30 laps around the moon in this time. Assume that the orbit was circular and that the command module was very light compared to the moon.

[Solution]

The command module which stays in lunar orbit is constantly accelerating towards the moon, since a force acts on it. However, since the module has a large velocity component which is right to the gravitational force component (see schematic), it manages to overcome the acceleration towards the moon, so that the distance to the surface remains constant. It is then constantly falling without ever reaching the surface. Imagine throwing a rock. Eventually it will hit the ground since a gravitational force acts on it. But if you were to throw the rock fast enough the earth would "curve away" from the stone as it moves forward, and thus the rock would never hit the ground. The rock would then be in orbit.

Let $$m_1$$ be the command module's mass, and $$m_2$$ the moon's mass.

From Newton's Law of Universal Gravitation we get

$$F_G=G\frac{m_1m_2}{r^2}\,$$

Since the command module is in an orbit around the moon, we can also express the gravitational force which acts on the command module as

$$F_G=m_1a_n$$

By substituting this into Newton's Law of universal Gravitation we get

$$m_1a_n=G\frac{m_1m_2}{r^2}\,$$

By dividing the equation with $$m_1$$ we get.

$$a_n=G\frac{m_2}{r^2}$$

By rewriting the centripetal acceleration $$a_n$$ we get

$$\frac{v^2}{r}=G\frac{m_2}{r^2}\,$$

By solving the equation on $$v^2$$ we get

$$v^2=G\frac{m_2}{r}$$

We can express the velocity of the command module in an other way. From the relation of velocity and angular velocity we get

$$v=r\omega$$

And from the relation of orbital period and angular velocity we get

$$T=\frac{2\pi}{\omega}$$

(The orbital period $$T$$ is equal to the time it takes for the Command Module to to one lap around the moon.) By combining both expressions and solving for $$v$$ we get

$$v=\frac{2\pi r}{T}$$

By substituting this expression into the expression we have from Newton's Law of Universal Gravitation we get

 * $$\mathbf{\left(\frac{2\pi r}{T}\right)}^2=G\frac{m_2}{r}$$

By solving for $$r$$ we get

$$r=\sqrt[3]{\frac{Gm_2T^2}{(2\pi)^2}}$$

Notice that the height at which the Command Module orbits the moon doesn't depend on the modules mass. As seen in the schematic, the value $$r$$ in the equation stands for the distance between the center of the moon and the center of the Command Module. We want to know the distance "$$h$$" to the surface, which naturally is

$$h=r-radius_{moon}$$

Notice that we are making a small mistake by stating the above. We are actually calculating the distance from the center of the command module to the surface, not the distance from the side closest to the surface. This error is, however, so small that it makes no difference.

This gives

$$h=\sqrt[3]{\frac{Gm_2T^2}{(2\pi)^2}}-radius_{moon}$$

Inserting values gives

$$h=\sqrt[3]{\frac{(6,67428 \times 10^{-11} \ \mbox{m}^3 \ \mbox{kg}^{-1} \ \mbox{s}^{-2}) (7,3477 \times 10^{22}kg) \mathbf{\left(\frac{59 \cdot 60^2 \text{s}+30\cdot 60 \text{s}}{30}\right)}^2}{(2\pi)^2}} -1 737,10 \times 10^{3}m$$

$$\approx{113} \text{ km}$$

--Ofey 20:23, 7 November 2008 (UTC)

title: Talk:Gravity


text: File:SixBarVarE.jpg

title: File:SixBarVarE.jpg