User:Eml4500.f08.echo.mott/HW5

=HW Report 5= The previous version was submitted.

Axial FD Relation
The axial force-displacement relation for a bar element remains the same whether 2-D or 3-D:

$$k^{(e)}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}\begin{Bmatrix} q_1^{(e)} \\ q_2^{(e)} \end{Bmatrix}=\begin{Bmatrix} P_1^{(e)} \\ P_2^{(e)} \end{Bmatrix}$$

Element dofs & Element Forces in Global x-y-z Coordinates
The transformation matrix in 3-D can be shown by

$$\mathbf{T}= \begin{bmatrix}l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0 \\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)}\end{bmatrix}$$

The axial F-D relation can be written as:

$$\begin{bmatrix}F_{1} \\ F_{2}\end{bmatrix}=\begin{bmatrix}k & -k \\ -k & k\end{bmatrix}\begin{bmatrix}d_{1} \\ d_{2}\end{bmatrix}=\begin{bmatrix}P_{1} \\ P_{2}\end{bmatrix}$$

The conversion from local to global axial coordinates is:

$$\mathbf{d = T {d^{(e)}}_{6x1}}$$

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

Where l, m, and n are the director cosines of each axis.

$$l^{(e)}=\frac{x_{2}-x_{1}}{L}$$

$$m^{(e)}=\frac{y_{2}-y_{1}}{L}$$

$$l^{(e)}=\frac{z_{2}-z_{1}}{L}$$

$$L^2={(x_{2}-x_{1})}^{2} + {(y_{2}-y_{1})}^{2} + {(z_{2}-z_{1})}^{2}$$

Beginning with the element F-D relation and substituting the global transformation matrix,

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

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

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

The global 3-D F-D relation can be written as:

$$ \frac{EA}{L}\begin{bmatrix}l^{2(e)} & m^{(e)}l^{(e)} & n^{(e)}l^{(e)} & -l^{2(e)} & -m^{(e)}l^{(e)} & -n^{(e}l^{(e)} \\ m^{(e)}l^{(e)} & m^{2(e)} & m^{(e)}n^{(e)} & -m^{(e)}l^{(e)} & -m^{2(e)} & -m^{(e)}n^{(e)} \\ n^{(e)}l^{(e)} & m^{(e)}n^{(e)} & n^{2(e)} & -n^{(e)}l^{(e)} & -m^{(e)}n^{(e)} & -n^{2(e)} \\ -l^{2(e)} & -m^{(e)}l^{(e)} & -n^{(e)}l^{(e)} & l^{2(e)} & m^{(e)}l^{(e)} & n^{(e)}l^{(e)} \\ -m^{(e)}l^{(e)} & -m^{2(e)} & -m^{(e)}n^{(e)} & m^{(e)}l^{(e)} & m^{2(e)} & m^{(e)}n^{(e)} \\ -n^{(e)}l^{(e)} & -m^{(e)}n^{(e)} & -n^{2(e)} & n^{(e)}l^{(e)} & m^{(e)}n^{(e)} & n^{2(e)} \\\end{bmatrix}\begin{bmatrix}u_{1} \\ v_{1} \\ w_{1} \\ u_{2} \\ v_{2} \\ w_{2} \\\end{bmatrix}=\begin{bmatrix}F_{1x} \\ F_{1y} \\ F_{1z} \\ F_{2x} \\ F_{2y} \\ F_{2z} \\ \end{bmatrix} $$

Matlab Solutions

 * Please refer to this page.

Monday Oct. 20, 2008
Justification of eliminating rows 1, 2, 5, and 6 to obtain $$\mathbf{\bar{k}}_{2x2}$$ in 2-bar truss: (p 10-1 & 23-2)

FD Relation:

$$\mathbf{K}_{6x6}\mathbf{d}_{6x1}=\mathbf{F}_{6x1} \Rightarrow \mathbf{K}\mathbf{d}-\mathbf{F}=\underline{0}_{6x1} $$ . . . (1)

PVW:

$$\mathbf{W}_{6x1}\cdot (\mathbf{K}\mathbf{d}-\mathbf{F})=0_{1x1}$$ for all $$\mathbf{W}_{6x1}$$. . . (2)

Where the weighting matrix is $$\mathbf{W}_{6x1}$$

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

Proof:

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

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

(2) $$\Rightarrow \mathbf{W}\cdot(\mathbf{Kd}-\mathbf{F})=0$$

for all W

Choice 1: Select W at $$w_1=1, w_2=w_3=w_4=w_5=w_6=0$$

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

$$\mathbf{W}\cdot (\mathbf{kd}-\mathbf{F})=1.\left[\sum_{j=1}^6 k_{1j}d_j-F_1\right]+0\left[\sum_{j=1}^6 k_{2j}d_j-F_2\right]+...+0\left[\sum_{j=1}^6 k_{6j}d_j-F_6\right]=0$$

This leads to:

$$\sum_{j=1}^6 k_{1j}d_j=F_1$$ (1st Equation)

Choice 2: Select W at $$w_1=0, w_2=1, w_3=w_4=w_5=w_6=0$$

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

$$\mathbf{W}\cdot (\mathbf{kd}-\mathbf{F})=0\left[\sum_{j=1}^6 k_{1j}d_j-F_1\right]+1\left[\sum_{j=1}^6 k_{2j}d_j-F_2\right]+0\left[\sum_{j=1}^6 k_{3j}d_j-F_3\right]...+0\left[\sum_{j=1}^6 k_{6j}d_j-F_6\right]=0$$

This leads to:

$$\sum_{j=1}^6 k_{2j}d_j=F_2$$ (2nd Equation)

Hence kd=F (or Eq (1))

PVW: Accounting for boundary conditions:

2 bar truss: $$d_1=d_2=d_5=d_6=0$$

Weight coefficient must be "kinematically admissible," i.e. cannot violate the boundary conditions.

Weighting coefficients $$\equiv$$ Virtual Displacements as on p24-4 (Calculus of Variations)

Previously, we wrote:

$$\mathbf{W} \cdot ( \mathbf{Kd}- \mathbf{F})$$

Where k was a 6x2 matrix and d was a 2x1 matrix.

Simplifying results in:

$$=\begin{Bmatrix} w_3 \\ w_4 \end{Bmatrix} \cdot (\mathbf{\bar K}_{2x2}\mathbf{\bar d}_{2x1}-\mathbf{\bar F})=0$$ for all $$\begin{Bmatrix} w_3 \\ w_4 \end{Bmatrix}$$. . . (3)

Specifically:

$$\mathbf{\bar K}=\begin{bmatrix} K_{33} & K_{34} \\ K_{43} & K_{44} \end{bmatrix}$$

$$\mathbf{\bar d}=\begin{Bmatrix} d_3 \\ d_4 \end{Bmatrix}$$

$$\mathbf{\bar F}=\begin{Bmatrix} F_3 \\ F_4 \end{Bmatrix}$$

HW Begin

Show $$(3) \Leftrightarrow \mathbf{\bar K \bar d}=\mathbf{\bar F} (4)$$

Take w3=w4=1

$$\mathbf{1}\cdot\mathbf{\bar K}\mathbf{\bar d}-\mathbf{\bar F}=\mathbf{0}$$

Therefore, $$\mathbf{\bar K}\mathbf{\bar d}=\mathbf{\bar F}$$

Monday Oct. 27, 2008
HW: 6 bar truss, page 226


 * Same E, A for all elements

Bug: for two bar truss used different E's and A's for the elements

$$ \mathbf {d}_{4x1}^{(i)} \harr d(lmm(i,:))$$
 * right hand side is Matlab code

Where: element degree of freedoms in global (x,y) coordinate system for element (i)

$$\mathbf {d}_{6x1}$$ 2-bar truss degree of freedoms in global coordinate (x,y) system.

Bug Explained

In computation of element forces $$\rarr$$ change (e,A) to e(i) and A(i). Correct the code, and rerun the 2-bar truss problem. Interpret the "results".

Back to PVW

$$ \mathbf{q}_{2x1}^{(i)} = \mathbf{T}_{2x4} \mathbf{d}_{4x1}^{(i)}$$


 * At this point Dr. Vu-Quoc referred to slides 24-1 and 21-2

Course grade $$\alpha_o \cdot$$ Hw grade +$$\sum \alpha_i \cdot Exam_i$$


 * Examine pages 23-3 and 14-3

Deriving $$\mathbf{k}_{4x4}^{(e)}= \mathbf{T}_{4x2}^{(e)T} \mathbf{\hat{k}}_{2x2}^{(e)}\mathbf{I}_{2x4}^{(e)}$$

Recall FD relation with axial degree's of freedom $$ \mathbf{q}^{(e)}$$

$$ \mathbf{\hat{k}}_{2x2}^{(e)} \mathbf{q}_{2x1}^{e} = \mathbf{p}_{2x1}^{e}$$

$$\rightarrow \mathbf{\tilde{k}}^{(e)} \mathbf{q}^{(e)} - \mathbf{p}^{(e)}= \mathbf{0}_{2x1}$$. . . (1)

PVW: $$ \mathbf{W}_{2x1} \cdot (\mathbf{\hat{k}}^{(e)} \mathbf{q}^{(e)} - \mathbf{p}^{(e)}) = 0$$. . . (2)

we showed $$ (1) \leftrightarrow (2)$$

Recall: $$ \mathbf{q}_{2x1}^{(e)} = \mathbf{T}_{2x1}^{(e)} \mathbf{d}_{4x1}^{(e)}$$. . . (3)

Similarly $$\mathbf{\hat{W}}_{2x1} = \mathbf{T}_{2x4}^{(e)} \mathbf {W}_{4x1}$$. . . (4)

Wednesday Oct. 29, 2008
$$\mathbf{\hat{w}}_{2x1}$$ = virtual displacement corresponding to $$\mathbf{q}_{2x1}^{(e)}$$

$$\mathbf{w}_{4x1}$$ = virtual displacement in global coordinate system corresponding to $$\mathbf{d}_{4x1}^{(e)}$$

Replacing Equations 3 and 4 into Equation 2:

$$(\mathbf{T}^{(e)}\mathbf{w})\cdot [\mathbf{\hat{k}}^{(e)}(\mathbf{T}^{(e)}\mathbf{d}^{(e)})-\mathbf{p}^{(e)}]=0$$ for all $$\mathbf{w_{4x1}}$$ (5)

Recall: $$\left ( \mathbf{A}_{pxq} \mathbf{B}_{qxr} \right )^T = \mathbf{B}^T \mathbf{A}^T $$ (6)

Recall: $$\mathbf{a}_{nx1} \cdot \mathbf{b}_{nx1}=\mathbf{a}_{1xn}^T \mathbf{b}_{nx1}$$ (7)

Apply Equation 7 & 6 in 5:

$$(\mathbf{T}^{(e)} \mathbf{w})^T [ \mathbf{\hat{k}}^{(e)}(\mathbf{T}^{(e)}\mathbf{d}^{(e)})-\mathbf{p}^{(e)}]=0_{1x1}$$ for all $$\mathbf{w}_{4x1}$$

Which results in:

$$\mathbf{w}^{T} \mathbf{T}^{(e)T} [ \mathbf{\hat{k}}^{(e)}(\mathbf{T}^{(e)}\mathbf{d}^{(e)})-\mathbf{p}^{(e)}]=0_{1x1}$$ for all $$\mathbf{w}_{4x1}$$

Further simplification results in:

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

Generalizing:

$$\mathbf{w} \cdot \left [ \mathbf{k}^{(e)} \mathbf{d}^{(e)}-\mathbf{f}^{(e)} \right ] = 0$$ for all w

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

So for discrete case (matrices) (non-continuous)

Now, continuous case (PDEs) Motivational model problem: Elastic bar with varying A(x), E(x), subjected to varying axial (time dependent) load (distributed) + concentrated (time dependent) load + inertial force (dynamic) This problem is illustrated in the figure below.



Friday Oct. 31, 2008
FBD:



$$\sum F_x=0=-N(x,t)+N(x+dx,t)+f(x,t)dx-m(x)\ddot u dx=\frac{\partial N}{\partial x}(x,t)du+h.o.t.+f(x,t)dx-m(x)\ddot u dx $$  eqn(1)

Where h.o.t. = higher order terms

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 the final part of the equation above represents the beginning of the higher order terms.

Eq.(1) yields the Equation of Motion (EoM): $$\frac{\partial N}{\partial x}+f=m\ddot u $$  eqn(2)

$$N_{ }^{ }(x,t)=A(x)\sigma (x,t)$$  eqn(3)

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

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

Where equation 3 is a constitutive relation.

Plugging equation 3 into 2 yields the PDE of Motion:

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



Where $$\ddot u = \frac{d^2 u}{dt^2}$$

In order to proceed further, we need two boundary conditions (2nd order derivative with respect to x)

We also have two initial conditions which are the initial displacement and initial velocity (2nd derive with respect to t)

Applying these to the two possible options shown in the figures to the right:



For scenario 1,

$$u_{ }^{ }(0,t)=u(L,t)=0$$

For scenario 2,

$$u_{ }^{ }(0,t)=0$$ and $$N_{ }^{ }(L,t)=F(t)$$

where $$N_{ }^{ }(L,t)=A(L)\sigma(L,t)$$

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

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

Contributing Team Members
Chris Mott 20:14, 7 November 2008 (UTC)

Bradley LaCroix 20:08, 7 November 2008 (UTC)

Brandon Sell 20:10, 7 November 2008 (UTC)

Charles Marshall 20:15, 7 November 2008 (UTC)

Chris Sell 20:25, 7 November 2008 (UTC)

Jason Bruce 20:28, 7 November 2008 (UTC)

Past Homeworks
[Homework 4]

[Homework 3]

[Homework 2]

[Homework 1]