User:Eml4500.f08.group/HW5/Lecture Notes

=Lecture Notes=

Lecture 24: 10/20/08
Justification of eliminating rows 1, 2, 5, 6 to obtain the K2x2 in the two bar truss system: FD relation: K6x6d6x6=F6x1 $$Kd-F=0$$ Eq (1) Where 0 is a 6x1 matrix Principle of Virtual Work(PVW): $$W\cdot (Kd-F)=0$$ Eq(2) where W is the the weighting matrix (6x1)

$$Eq(1)\Leftrightarrow Eq(2)$$

Proof:

A) $$(1)\Rightarrow (2)$$ :Trivial

B) Want to show $$(2)\Rightarrow (1)$$

$$(2)\Rightarrow W\cdot (Kd-F)=0$$  for all W

$$W_{1} = 1: \mathbf{W}^{T}_{1x6} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}_{1x6}$$

$$\mathbf{W} * (\mathbf{kd - F}) = 1 * [\sum_{j=1}^{6}{K_{1j}d_{j}} - F_{1}] + 0*[\sum_{j=1}^{6}{K_{2j}d_{j}} - F_{2}] +  0*[\sum_{j=1}^{6}{K_{3j}d_{j}} - F_{3}] +  0*[\sum_{j=1}^{6}{K_{4j}d_{j}} - F_{4}] +  0*[\sum_{j=1}^{6}{K_{5j}d_{j}} - F_{5}] +  0*[\sum_{j=1}^{6}{K_{6j}d_{j}} - F_{6}] = 0 $$

(1) $$\Rightarrow \sum_{j=1}^{6}{K_{1j}d_{j}} = F_{1}$$

$$W_{2} = 1: \mathbf{W}^{T}_{1x6} = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix}_{1x6}$$

$$\mathbf{W} * (\mathbf{kd - F}) = 0*[\sum_{j=1}^{6}{K_{1j}d_{j}} - F_{1}] + 1*[\sum_{j=1}^{6}{K_{2j}d_{j}} - F_{2}] +  ... + 0*[\sum_{j=1}^{6}{K_{6j}d_{j}} - F_{6}] = 0 $$

(2) $$\Rightarrow \sum_{j=1}^{6}{K_{2j}d_{j}} = F_{2}$$


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!HW Weak Form, from team group 0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix}_{1x6}$$
 * $$W_{3} = 1: \mathbf{W}^{T}_{1x6} = \begin{bmatrix}
 * $$W_{3} = 1: \mathbf{W}^{T}_{1x6} = \begin{bmatrix}

$$\mathbf{W} * (\mathbf{kd - F}) = 0*[\sum_{j=1}^{6}{K_{1j}d_{j}} - F_{1}] + ... + 1*[\sum_{j=1}^{6}{K_{3j}d_{j}} - F_{3}] +  ... + 0*[\sum_{j=1}^{6}{K_{6j}d_{j}} - F_{6}] = 0 $$

(3) $$\Rightarrow \sum_{j=1}^{6}{K_{3j}d_{j}} = F_{3}$$

$$W_{4} = 1: \mathbf{W}^{T}_{1x6} = \begin{bmatrix} 0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix}_{1x6}$$

$$\mathbf{W} * (\mathbf{kd - F}) = 0*[\sum_{j=1}^{6}{K_{1j}d_{j}} - F_{1}] + ... + 1*[\sum_{j=1}^{6}{K_{4j}d_{j}} - F_{4}] +  ... + 0*[\sum_{j=1}^{6}{K_{6j}d_{j}} - F_{6}] = 0 $$

(4) $$\Rightarrow \sum_{j=1}^{6}{K_{4j}d_{j}} = F_{4}$$

$$W_{3} = 1: \mathbf{W}^{T}_{1x6} = \begin{bmatrix} 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix}_{1x6}$$

$$\mathbf{W} * (\mathbf{kd - F}) = 0*[\sum_{j=1}^{6}{K_{1j}d_{j}} - F_{1}] + ... + 1*[\sum_{j=1}^{6}{K_{5j}d_{j}} - F_{5}] + 0*[\sum_{j=1}^{6}{K_{6j}d_{j}} - F_{6}] = 0 $$

(5) $$\Rightarrow \sum_{j=1}^{6}{K_{5j}d_{j}} = F_{5}$$

$$W_{3} = 1: \mathbf{W}^{T}_{1x6} = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}_{1x6}$$

$$\mathbf{W} * (\mathbf{kd - F}) = 0*[\sum_{j=1}^{6}{K_{1j}d_{j}} - F_{1}] + ... + 1*[\sum_{j=1}^{6}{K_{6j}d_{j}} - F_{6}] = 0 $$

(6) $$\Rightarrow \sum_{j=1}^{6}{K_{6j}d_{j}} = F_{6}$$

contribution by carbon.w
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Principle of Virtual Work: Accounting for Boundary Conditions

From the 2 bar truss example: d1=d2=d5=d6=0)

Weighting Coefficients must be "kinematically admissible"(can't violate the boundary conditions)

Weightind coefficients = virtual displacement (W1=W2=W5=W6=0

$$W\cdot (Kd-F)$$ Where $$K=\begin{bmatrix}

k_{33} &k_{34} \\ k_{43}& k_{44} \end{bmatrix}$$ $$d=\begin{Bmatrix} d_{3}\\ d_{4}

\end{Bmatrix}$$ $$F=\begin{Bmatrix} f_{3}\\ f_{4}

\end{Bmatrix}$$


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!HW Force-Distance Relationship from Weak Form, from team group W_{3}\\ W_{4} \end{Bmatrix}*(\mathbf{\bar{K}\bar{d} - \bar{F}}) = 0$$
 * $$\begin{Bmatrix}
 * $$\begin{Bmatrix}

Multiply both sides by $$\mathbf{\bar{W}}^{-1}$$:

$$\begin{Bmatrix} W_{3}\\ W_{4} \end{Bmatrix} * \begin{Bmatrix} W_{3}\\ W_{4} \end{Bmatrix}^{-1}*(\mathbf{\bar{K}\bar{d} - \bar{F}}) = 0 * \begin{Bmatrix} W_{3}\\ W_{4} \end{Bmatrix}^{-1}$$

$$\Rightarrow 1 * (\mathbf{\bar{K}\bar{d} - \bar{F}}) = 0$$

(4) $$\Rightarrow \mathbf{\bar{K}\bar{d} = \bar{F}}$$

contribution by carbon.w
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Lecture 26: 10/27/08
Continued from Lecture 24:

Back to PVW:

Example: $$\displaystyle grade = \alpha_o (HW grade) + \sum_{i=1}^{3} \alpha_i$$

Motivation:


 * 1) Elimination of rows (p 10-1, 23-2)
 * 2) Deriving $$\displaystyle \mathbf{{k^{(e)}_{4x4}} = {T^{(e)T}_{4x2}} \hat{k^{(e)}_{2x2}} {T^{(e)}_{2x4}}}$$ (p 14-3, 23-3)

Recall Force Dispacement (FD) relationship with axial degrees of freedom, $$\displaystyle \mathbf{q^{(e)}}$$

$$\displaystyle => \mathbf{ \hat{k^{(e)}_{2x2}} q^{(e)} = p^{(e)}}$$

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

PVW: $$\displaystyle => \mathbf{ \hat{w} (\hat{k^{(e)}_{2x2}} q^{(e)} - p^{(e)}) = 0_{1x}}$$ for all $$\displaystyle \mathbf{\hat{w_{2x1}}}$$ Eq. (2)

We showed (1) <=> (2) on p 24-1

Recall: $$\displaystyle \mathbf{{q^{(e)}_{2x1}} = T^{(e)}_{2x4} d^{(e)}_{4x1}}$$, Eq. (3), where $$\displaystyle \mathbf$$ is the real displacement

Similarly: $$\displaystyle \mathbf{ \hat{w_{2x1}} = T^{(e)}_{2x4} w_{4x1}}$$, Eq. (4). where $$\displaystyle \mathbf{ \hat{w}}$$ is the virtual displacement

Lecture 27: 10/29/08
$$\boldsymbol{\hat {w}_{2x1}} = $$ virtual axial displacement, corresponding to $$\boldsymbol q_{2x1}^{(e)}$$

$$\boldsymbol{\hat {w}_{4x1}} = $$ virtual displacement in the global coordinate system, corresponding to the real displacement in the global coordinate system $$\boldsymbol d_{2x1}^{(e)}$$

Replacing equations (3) and (4) into (2) yields:

$$\boldsymbol {(T^{(e)}w)}[\boldsymbol {\hat k^{(e)}}(\boldsymbol {T^{(e)}d^{(e)})-p^{(e)}]}=0, $$ for all $$\boldsymbol{\hat {w}_{4x1}}$$

Recall that $$\boldsymbol{(A_{pxq}B_{qx1})^T} = \boldsymbol {B^TA^T}$$.

Also, Recall that $$\boldsymbol{a_{nx1}b_{nx1}} = \boldsymbol{a_{1xn}^Tb_{nx1}}$$.

Now, apply equation (7) and (6) into equation (5):

$$\boldsymbol {(T^{(e)}w)^T}[\boldsymbol {\hat k^{(e)}}(\boldsymbol {T^{(e)}d^{(e)})-p^{(e)}]}=0, $$ for all $$\boldsymbol{\hat {w}_{4x1}}$$

$$=>\boldsymbol{w^TT^{(e)}}[\boldsymbol {\hat k^{(e)}}(\boldsymbol {T^{(e)}d^{(e)})-p^{(e)}]}=0$$

$$=>\boldsymbol w [\boldsymbol{( T^{(e)}\hat k^{(e)}T^{(e)})d^{(e)} - (T^{(e)}p^{(e)})}]=0$$, for all vitrual displacements, $$\boldsymbol w_{4x1}$$

$$=>\boldsymbol w [\boldsymbol{k^{(e)}d^{(e)}-f^{(e)}}]=0$$

So far, we have been dealing with the discrete (i.e. not continuous) case for matrices.

Now, let's look at the continuous case (PDE's):

Motivational model problem: An elastic bar with carying cross-section A(x), E(x), subjected to non-uniform (i.e. varying) axial loads (distributed) + a concentric (point) load + inertia forces (dynamic).


 * Note: Axial and concentric loads can be time dependant.



Lecture 28: 10/31/08
To date, our discussions have been limited to using the Finite Element Method for discrete cases, i.e. a finite number of elements. Now we turn our attention to using partial differential equations (PDEs) with the Finite Element Method. As always, we start by drawing a free body diagram. For this example, we will use the structure shown in the figure below.



In order to use PDEs, we must draw the free body diagram of a differential element. This element is shown above, highlighted in red, with the FBD drawn below.



$$\sum{F_y}=0$$ $$\sum{F_x)}=-N(x,t) + N(x+dx,t)+f(x,t)*dx-m(x)*a*dx$$ We use a Taylor series expansion to further evaluate $$\sum{F_x)}$$ Recall Taylor Series Expansion $$f(x+dx)=f(x)+\frac{df(x)}{dx}*dx+\frac{1}{2}\frac{d^2f(x)}{dx^2}*dx^2+ ...$$ where everything from $$\frac{1}{2}\frac{d^2f(x)}{dx^2}*dx^2$$ and above is considered a higher order term (H.O.T.) Applying this expansion to $$\sum{F_x}$$ results in the following equation. $$\sum{F_x}=0=\frac{\partial N(x,t)}{\partial x}*dx + H.O.T.+ f(x,t)*dx -m*a*dx$$ noting that we are neglecing the higher order terms and that the dx variable drops out in the remaining terms, we are left with the following. $$\sum{F_x}=0=\frac{\partial N(x,t)}{\partial x}+ f(x,t) -m*a$$ Equation (1) This equation then becomes $$\frac{\partial N}{\partial x}+f=m*a$$  Equation (2) recall that $$\N(x,t)=A(x)*\sigma (x)$$ where $$\sigma (x)=E(x)*\varepsilon (x,t)$$ and $$\varepsilon (x,t)=\frac{\partial (x,t)}{\partial x}$$ Thus: $$N(x,t)=A(x)*E(x)*\frac{\partial (x,t)}{\partial x}$$ Equation (3) Plugging Equation (3) into Equation (2) yields: $$\frac{\partial }{\partial x}[A(x)*E(x)*\frac{\partial (x,t)}{\partial x}]+f(x,t)=m(x)*a$$ This is called the Partial Differntial Equation of Motion (PDE of motion) In order to solve this equation, we need 4 boundary conditions, 2 position based boundary condiditions and 2 time based boundary conditions. This is because we have 2nd order derivative with respect to x and a 2nd order derivative with respect to time (acceleration) in the PDE of motion.

To help determine these boundary conditions, please consider the two figures below. Since both ends are fixed, the two x boundary conditions can be found by observing: $$\displaystyle u(0,t)=u(L,t)=0$$ Both t boundary conditions can be found by knowing the initial displacement and the initial velocity.

As above, both t based boundary conditions are taken to be the inital velocity and the initial displacement while the x bounday condidtions can be obtained as follows: $$\displaystyle u(0,t)=0$$ $$u(L,t)=\frac{\partial u(L,t)}{\partial x}=\frac{F(t)}{A(L)*E(L)}$$


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!HW Global Force-Distance Relationship from Weak Form, from team group
 * $$\mathbf{W}_{2x1}*(\mathbf{\bar{k}^{(e)}\bar{q}^{(e)} - \bar{p}^{(e)}}) = 0_{1x1}$$
 * $$\mathbf{W}_{2x1}*(\mathbf{\bar{k}^{(e)}\bar{q}^{(e)} - \bar{p}^{(e)}}) = 0_{1x1}$$

Multiply both sides by $$\mathbf{W}^{-1}$$:

$$\mathbf{W}_{2x1} * \mathbf{W}_{2x1}^{-1}*(\mathbf{\bar{k}^{(e)}\bar{q}^{(e)} - \bar{p}^{(e)}}) = 0 * \mathbf{W}_{2x1}^{-1}$$

$$\Rightarrow 1 * (\mathbf{\bar{k}^{(e)}\bar{q}^{(e)} - \bar{p}^{(e)}}) = 0$$

(4) $$\Rightarrow \mathbf{\bar{k}^{(e)}\bar{q}^{(e)} = \bar{p}^{(e)}}$$

contribution by carbon.w
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!HW Ad-hoc Proof of Transpose Property, from team group
 * Prove: $$(\mathbf{AB})^{T} = \mathbf{B}^{T}\mathbf{A}^{T}$$
 * Prove: $$(\mathbf{AB})^{T} = \mathbf{B}^{T}\mathbf{A}^{T}$$

Use the example matrices $$\mathbf{A}_{2x3} = \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix}_{2x3}$$ and $$\mathbf{B}_{3x3} = \begin{bmatrix} 7 & 8 & 9\\ 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix}_{3x3}$$

$$(\mathbf{AB})^{T} = \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix} * \begin{bmatrix} 7 & 8 & 9\\ 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} 21 & 27 & 33\\ 57 & 72 & 87 \end{bmatrix}^{T} = \begin{bmatrix} 87 & 72 & 57\\ 33 & 27 & 21 \end{bmatrix}$$

$$\mathbf{B}^{T}\mathbf{A}^{T} = \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix}^{T} * \begin{bmatrix} 7 & 8 & 9\\ 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix}^{T} = \begin{bmatrix} 6 & 5 & 4\\ 3 & 2 & 1 \end{bmatrix} * \begin{bmatrix} 6 & 5 & 4\\ 3 & 2 & 1\\ 9 & 8 & 7 \end{bmatrix} = \begin{bmatrix} 87 & 72 & 57\\ 33 & 27 & 21 \end{bmatrix}$$

contribution by carbon.w
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!HW Local Force-Distance Relationship from Weak Form, from team group
 * $$\mathbf{W}*(\mathbf{k^{(e)}d^{(e)} - f^{(e)}}) = 0_{1x1}$$
 * $$\mathbf{W}*(\mathbf{k^{(e)}d^{(e)} - f^{(e)}}) = 0_{1x1}$$

Multiply both sides by $$\mathbf{W}^{-1}$$:

$$\mathbf{W} * \mathbf{W}^{-1}*(\mathbf{k^{(e)}d^{(e)} - f^{(e)}}) = 0 * \mathbf{W}^{-1}$$

$$\Rightarrow 1 * (\mathbf{k^{(e)}d^{(e)} - f^{(e)}}) = 0$$

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

contribution by carbon.w
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Lecture 29: 11/3/08

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!HW Literature on Composite Materials, from team group
 * In the article "The effect of interphase on residual thermal stresses. 1. Single fiber composite materials" from the journal "Mechanics of Composite Materials", the authors Kushnevsky, Wacher, Chate, and Bledzki treat the composite material as a region with variable Young's Modulus. They make this conclusion as "a direct consequence of the changes in the microstructure of the matrix near the fiber surface".
 * In the article "The effect of interphase on residual thermal stresses. 1. Single fiber composite materials" from the journal "Mechanics of Composite Materials", the authors Kushnevsky, Wacher, Chate, and Bledzki treat the composite material as a region with variable Young's Modulus. They make this conclusion as "a direct consequence of the changes in the microstructure of the matrix near the fiber surface".

Source: http://www.springerlink.com/content/v3735152h75775mu/

contribution by carbon.w
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!Contribution History of Calculus Notations, from team group
 * Calculus has two primary systems of notations developed by Leibniz and Newton (however Leibniz's notation is the most popular notation used today). Newton used dots over the variables as such.
 * Calculus has two primary systems of notations developed by Leibniz and Newton (however Leibniz's notation is the most popular notation used today). Newton used dots over the variables as such.

Newton first derivative of x:       $$\dot{x}$$

While Leibniz used a "d" in the numerator and denominator to symbolize "derivative". Leibniz also developed the integral symbol commonly used today.

Leibniz first derivative of x with respect to time:     $$\frac{dx}{dt}$$

contribution by carbon.w =E-mail Homework=
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$$\begin{Bmatrix} f^{(e)}_1\\ f^{(e)}_2\\ f^{(e)}_3\\ f^{(e)}_4 \end{Bmatrix}=$$$$*\begin{Bmatrix} d^{(e)}_1\\ d^{(e)}_2\\ d^{(e)}_3\\ d^{(e)}_4 \end{Bmatrix}$$ $$l_s=cos(\theta)$$ $$m_s=sin(\theta)$$

transformation matrix T (2x6) to transform the 6 element dofs in global xyz coordinates to the 2 axial dofs
$$\underline{k}^{(e)} = k^{(e)} \begin{bmatrix} (l^{(e)})^2   & l^{(e)}m^{(e)} & l^{(e)}n^{(e)} &-(l^{(e)})^2    &-l^{(e)}m^{(e)} &-l^{(e)}n^{(e)} \\ l^{(e)}m^{(e)} & (m^{(e)})^2   & m^{(e)}n^{(e)} &-l^{(e)}m^{(e)} &-(m^{(e)})^2    &-m^{(e)}n^{(e)} \\ l^{(e)}n^{(e)} & m^{(e)}n^{(e)} & (n^{(e)})^2   &-l^{(e)}n^{(e)} &-m^{(e)}n^{(e)} &-(n^{(e)})^2    \\ -(l^{(e)})^2   &-l^{(e)}m^{(e)} &-l^{(e)}n^{(e)} & (l^{(e)})^2    & l^{(e)}m^{(e)} & l^{(e)}n^{(e)} \\ -l^{(e)}m^{(e)} &-(m^{(e)})^2   &-m^{(e)}n^{(e)} & l^{(e)}m^{(e)} & (m^{(e)})^2    & m^{(e)}n^{(e)} \\ -l^{(e)}n^{(e)} &-m^{(e)}n^{(e)} &-(n^{(e)})^2   & l^{(e)}n^{(e)} & m^{(e)}n^{(e)} & (n^{(e)})^2    \\ \end{bmatrix} $$