User:EML4500.f08.JAMAMA/Homework5

 Note: Derivation of 3-D bar element missing. Please don't remove this comment in case you amend your work; just add a comment box to say what you did. Eml4500.f08 11:14, 10 November 2008 (UTC)

Meeting 24
Justification for eliminating rows 1 2 5 6 to reduce the global stiffness matrix for the 2 bar truss system to a 2X2 matrix.

The force displacement relation can be written as follows.

$$\textbf{K}_{6x6}\textbf{d}_{6x1}=\textbf{F}_{6x1}$$

$$\textbf{K}_{6x6}\textbf{d}_{6x1}-\textbf{F}_{6x1}=\textbf{0}$$

The principle of virtual work is shown below is true for all $$\textbf{W}$$.

$$\textbf{W}_{6x1}\bullet \left[ \textbf{K}_{6x6}\textbf{d}_{6x1}-\textbf{F}_{6x1}\right]=\textbf{0}$$

The matrix $$\textbf{W}$$ is a weighting matrix. The weighting matrix can be used to pull out each equation out of the force displacement relation.

Principle of Virtual Work: Now to prove the simplification. From the 2 bar truss example: the boundary conditions: $$d_{1},d_{2},d_{5},d_{6}$$ are all equal to zero. The weighting coefficients must be "kinematically admissible", meaning they can not violate the boundary conditions;

$$W_{1}=W_{2}=W_{5}=W_{6}=0$$

By plugging $$W_{1}=W_{2}=W_{5}=W_{6}=0$$ into $$kd-F=0$$, the row reduction (elimination of rows 1, 2, 5, and 6) is proven because they are fixed, allowing zero displacement.

So, the global stiffness matrix is reduced down to a 6x2 matrix.

$$W\centerdot (\underbrace{kd}_{[6x2]\{2x1\}}-F)$$

$$\begin{bmatrix} K_{31} & K_{41}\\ K_{32} & K_{42}\\ K_{33} & K_{43}\\ K_{34} & K_{44}\\ K_{35} & K_{45}\\ K_{36} & K_{46} \end{bmatrix} \begin{bmatrix} d_{3}\\ d_{4}

\end{bmatrix} - \begin{bmatrix} F_{3}\\ F_{4}

\end{bmatrix} = \begin{bmatrix} 0\\ 0

\end{bmatrix}$$

Then with the weighting matrix being reduced to a 2x1 because there are only two columns to perform the dot product upon. This eliminates rows 1, 2, 5, and 6 in the global stiffness matrix and is reduced to a 2x2.

$$ \begin{bmatrix} W_{3}\\ W_{4}

\end{bmatrix} \bullet \left[ \begin{bmatrix} K_{33} & K_{43}\\ K_{34} & K_{44} \end{bmatrix} \begin{bmatrix} d_{3}\\ d_{4}

\end{bmatrix} - \begin{bmatrix} F_{3}\\ F_{4}

\end{bmatrix} \right] = \begin{bmatrix} 0\\ 0

\end{bmatrix} $$

$$$$

$$$$

This would continue for $$W_{3}=1\text{ and }W_{4}=0,\text{ then for }W_{3}=0\text{ and }W_{4}=1$$

For $$W_{3}=1\text{ and }W_{4}=0$$ (Taken from the matrix $$W^{T}=\left\lfloor \begin{matrix}  0 & 0 & 1 & 0 & 0 & 0  \\ \end{matrix} \right\rfloor $$, providing the equation for $$F_{3}$$:

$$\begin{align} & \left\{ \begin{matrix} W_{3} \\ W_{4} \\ \end{matrix} \right\}=\left\{ \begin{matrix} 1 \\   0  \\ \end{matrix} \right\} \\ & \left\{ \begin{matrix} W_{3} \\ W_{4} \\ \end{matrix} \right\}\centerdot (k_{_{2x2}}d_{_{2x1}}-F_{_{2x1}})=0,\text{ for all matrices }\left\{ \begin{matrix} W_{3} \\ W_{4} \\ \end{matrix} \right\} \\ & \therefore \left\{ \begin{matrix} 1 \\   0  \\ \end{matrix} \right\}\centerdot \left[ \begin{matrix} K_{33} & K_{34} \\ K_{43} & K_{44} \\ \end{matrix} \right]\left\{ \begin{matrix} d_{3} \\ d_{4} \\ \end{matrix} \right\}-\left\{ \begin{matrix} F_{3} \\ F_{4} \\ \end{matrix} \right\}=\left\{ \begin{matrix} 0 \\   0  \\ \end{matrix} \right\} \\ \end{align}$$

Therefore, proving for $$F_{3}$$, that $$kd=F$$

Meeting 26
Back to the Principle of Virtual Work

Simple Explanation of a Linear Combination:

$$\text{Course Grade }=\alpha _{o}\cdot \text{ Homework Grade }+\sum\limits_{i=1}^{3}{\alpha _{i}\cdot \text{ Exam}_{i}}$$

Deriving $$\underset – {\mathop{k^{(e)}}}\,=\underset – {\mathop{T^{(e)T}}}\,\underset – {\mathop{\hat{k}^{(e)}}}\,\underset – {\mathop{T^{(e)}}}\,$$:

 Recall: The force displacement relationship with axial degrees of freedom, $$q^{(e)}$$



$$\underset – {\mathop{\hat{k}^{(e)}}}\,\underset – {\mathop{q^{(e)}}}\,=\underset – {\mathop{p^{(e)}}}\,$$

 $$\Rightarrow \underset – {\mathop{\hat{k}^{(e)}}}\,\underset – {\mathop{q^{(e)}}}\,-\underset – {\mathop{p^{(e)}}}\,=\underset – {\mathop{0}}\,_{2x1}$$ (Equation 1)

The Definition of the Principle of Virtual Work  $$\underset – {\mathop{W}}\,_{2x1}\underbrace{\text{ }\centerdot \text{  }}_{\text{dot product}}\text{ }\underbrace{\left( \widehat{\underset – {\mathop{k}}\,}^{(e)}\underset – {\mathop{q^{(e)}}}\,-\underset – {\mathop{p^{(e)}}}\, \right)}_{2x1}=0_{_{1x1}}\text{ }\to \text{for all }\underset – {\mathop{W}}\,_{2x1}$$ (Equation 2)

It has now been shown that (Equation 1) $$\Leftrightarrow $$ (Equation 2)

Recalling: $$q^{(e)}=T^{(e)}d^{(e)}$$

And Similarly, for $$q^{(e)}$$ (the axial displacement)

$$\text{ the weighting coefficient, }W,\text{ can be expressed:}$$

 $$\widehat{\underset – {\mathop{W}}\,}_{2x1}=\underset – {\mathop{T}}\,_{2x4}^{(e)}\underset – {\mathop{W}}\,_{4x1}$$

Note: The difference between the weighting coefficients:

$$W$$ is used in conjunction with $$d^{(e)}$$ (the vertical and horizontally described displacements)

and $$\hat{W}$$ is used with $$q^{(e)}$$ (the axially described displacements)

Through the use of: $$q^{(e)}=T^{(e)}d^{(e)}$$ and $$k^{(e)}=T^{(e)T}\hat{k}^{(e)}T^{(e)}$$

Meeting 27
$$\hat{W}$$ = virtual axial displacement

W = Virtual displacement in global coordinate system corresponds to $$d^{(e)}$$ (4X1)

recall last equations:

$$q^{(e)}=T^{(e)}d^{(e)}$$  (3)

$$ \hat{W}=T^{(e)}W $$   (4)

Replace equations (3) and (4) into equation (2). (From page meeting 26, page 2)

$$(T^{(e)}W)\bullet \left[\hat{k}^{(e)}\left(T^{(e)}d^{(e)} \right)-p^{(e)} \right]=0$$ for all $$\hat{W}$$ (4X1) (5)

Recall: $$\left(AB \right)^{T}=B^{T}A^{T}$$ (6)

pxqqxr

Recall:$$a\bullet b=a^{T}b$$ (7)

nx1nx1(1xnnx1)

(1x1 scalar)

Apply (7) and (6) in (5):

$$(T^{(e)}W)^{T}\bullet \left[\hat{k}^{(e)}\left(T^{(e)}d^{(e)} \right)-p^{(e)} \right]=0$$

$$(T^{(e)T}W^{T})\bullet \left[\hat{k}^{(e)}\left(T^{(e)}d^{(e)} \right)-p^{(e)} \right]=0$$

$$\Rightarrow W\bullet \left[\left( T^{(e)}\hat{k}^{(e)}T^{(e)}\right)d^{(e)}-T^{(e)T}p^{(e)} \right]=0$$ for all W (4x1)

$$\Rightarrow W\bullet \left[k^{(e)}d^{(e)}-f^{(e)} \right]= 0 $$ for all W.  $$\Rightarrow k^{(e)}d^{(e)}=f^{(e)}$$

So far, discrete case (non-continuous) matrices. Now, continuous case (Partial Differential Equations) motivational model problem: Elastic bar with varying A(x), E(x), subject to varying axial loads (distributed) + concentrated load + inertia force (dynamic). Axial loads being the dependent.



Meeting 28
Free body diagram of an infinitesimal portion of the bar from lecture 27

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

$$\Sigma\ F_x = {dN(x,t) \over dx}dx + H.O.T + f(x,t)dx - m(x)\ddot u \qquad \leftrightarrow \qquad Eq(1)$$

Equation (1) can then be written in the form

$${dN \over dx} + f = m \ddot u \qquad \leftrightarrow \qquad Eq(2)$$

By analyzing the diagram of the infinitesmal section the following is determined

$$N(x,t) = A(x) \sigma\ (x,t) \qquad \leftrightarrow \qquad Eq(3)$$Constitutive Relation

where: $$\sigma\ (x,t) = E(x) \varepsilon\ (x,t) $$     and      $$\varepsilon\ (x,t) = {du \over dx}(x,t)$$

By combining equations (2) and (3):

$${d \over dx} \bigg[ A(x)E(x){du \over dx} \bigg] + f(x,t) = m(x) + \ddot u \qquad \leftrightarrow \qquad Eq(4)$$ PDE of Motion

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

$$u(0,t) = 0 \qquad b.c.(1)$$

$$N(L,t) = F(t) \qquad b.c.(2)$$

Need 2 initial conditions (2nd order derivative with respect to t)

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

$$\rightarrow \quad {du(L,t) \over dx}={F(t) \over A(L)E(L)}$$

Matlab Section
{| class="toccolours collapsible collapsed" style="width:100%" ! The corrected 2-Bar truss code

Code Output
k_local = 0.75       -0.75        -0.75         0.75

k = 0.5625     0.32476      -0.5625     -0.32476      0.32476       0.1875     -0.32476      -0.1875      -0.5625     -0.32476       0.5625      0.32476     -0.32476      -0.1875      0.32476       0.1875

k_local = 5   -5    -5     5

k = 2.5        -2.5         -2.5          2.5         -2.5          2.5          2.5         -2.5         -2.5          2.5          2.5         -2.5          2.5         -2.5         -2.5          2.5

K = 0.5625     0.32476      -0.5625     -0.32476            0            0      0.32476       0.1875     -0.32476      -0.1875            0            0      -0.5625     -0.32476       3.0625      -2.1752         -2.5          2.5     -0.32476      -0.1875      -2.1752       2.6875          2.5         -2.5            0            0         -2.5          2.5          2.5         -2.5            0            0          2.5         -2.5         -2.5          2.5

R = 0    0     0     7     0     0

d = 0           0        4.352       6.1271            0            0

reactions = -4.4378     -2.5622       4.4378      -4.4378

results = 1.7081      5.1244       5.1244       0.6276        3.138        6.276

The different output occured because the E and A were not of correct specification. E and A should have been specified with "i" subscripts, such as E(i), A(i), so that with appropriate truss mambers' certain values should be taken.


 * }

{| class="toccolours collapsible collapsed" style="width:100%" ! 2-D 6-bar truss and its deformation (magnified by a factor of 1000) at constant Modulus of Elasticity (E = $$2*10^{11}$$)

Code Output
K = 1.0e+008 * 0.8536   0.3536   -0.5000         0         0         0         0         0   -0.3536   -0.3536    0.3536    0.3536         0         0         0         0         0         0   -0.3536   -0.3536   -0.5000         0    0.8536   -0.3536         0         0         0         0   -0.3536    0.3536         0         0   -0.3536    1.0202         0         0         0   -0.6667    0.3536   -0.3536         0         0         0         0    0.7155   -0.3578         0         0   -0.7155    0.3578         0         0         0         0   -0.3578    0.1789         0         0    0.3578   -0.1789         0         0         0         0         0         0    0.7155    0.3578   -0.7155   -0.3578         0         0         0   -0.6667         0         0    0.3578    0.8456   -0.3578   -0.1789   -0.3536   -0.3536   -0.3536    0.3536   -0.7155    0.3578   -0.7155   -0.3578    2.1382         0   -0.3536   -0.3536    0.3536   -0.3536    0.3578   -0.1789   -0.3578   -0.1789         0    1.0649

d = 1.0e-003 * 0        0    0.2131    0.2500         0         0         0         0   -0.0061    0.0122

d_def = 1.0e-003 * 0        0    0.2131    0.2500         0         0         0         0   -0.0061    0.0122

nodes_def = 0        0    4.0002    0.0002         0    3.0000    4.0000    3.0000    2.0000    2.0000

d_def_aug = 0        0    0.2131    0.2500         0         0         0         0   -0.0061    0.0122

nodes_def_aug = 0        0    4.2131    0.2500         0    3.0000    4.0000    3.0000    1.9939    2.0122

Stress = 1.0e+007 * 1.0655  -0.0927   -0.0977   -1.6665    0.0307   -0.0002

Force_axial = 1.0e+004 * 1.0655  -0.0927   -0.0977   -1.6665    0.0307   -0.0002

6-bar truss image (the deformation was multiplied by a factor of 1000 to show a clearer image)
Blue solid line $$ \Rightarrow $$ deformed truss

Red dashed line$$ \Rightarrow $$ undeformed truss




 * }

{| class="toccolours collapsible collapsed" style="width:100%" ! 2-D 6-bar truss and its deformation (magnified by a factor of 1000) at constant (E = $$2*10^{11}$$) and at a varying specified Modulus of Elasticity

Code Output (Note, the output is only varying the parameter E
K = 1.0e+008 * 0.7639   0.3889   -0.3750         0         0         0         0         0   -0.3889   -0.3889    0.3889    0.3889         0         0         0         0         0         0   -0.3889   -0.3889   -0.3750         0    0.6932   -0.3182         0         0         0         0   -0.3182    0.3182         0         0   -0.3182    0.9849         0         0         0   -0.6667    0.3182   -0.3182         0         0         0         0    0.7155   -0.3578         0         0   -0.7155    0.3578         0         0         0         0   -0.3578    0.1789         0         0    0.3578   -0.1789         0         0         0         0         0         0    0.8944    0.4472   -0.8944   -0.4472         0         0         0   -0.6667         0         0    0.4472    0.8903   -0.4472   -0.2236   -0.3889   -0.3889   -0.3182    0.3182   -0.7155    0.3578   -0.8944   -0.4472    2.3171    0.1602   -0.3889   -0.3889    0.3182   -0.3182    0.3578   -0.1789   -0.4472   -0.2236    0.1602    1.1096

d = 1.0e-003 * 0        0    0.2649    0.2608         0         0         0         0    0.0006   -0.0012

d_def = 1.0e-003 * 0        0    0.2649    0.2608         0         0         0         0    0.0006   -0.0012

nodes_def = 0        0    4.0003    0.0003         0    3.0000    4.0000    3.0000    2.0000    2.0000

d_def_aug = 0        0    0.2649    0.2608         0         0         0         0    0.0006   -0.0012

nodes_def_aug = 0        0    4.2649    0.2608         0    3.0000    4.0000    3.0000    2.0006    1.9988

Stress = 1.0e+007 * 0.9932   0.0096    0.0101   -1.7389   -0.0033   -0.0002

Force_axial = 1.0e+004 * 0.9932   0.0096    0.0101   -1.7389   -0.0033   -0.0002

6-bar truss image (the deformation was multiplied by a factor of 1000 to show a clearler image)
Blue solid line $$ \Rightarrow $$ defored truss of constant E

Red dashed line$$ \Rightarrow $$ undeformed truss

Black solid line$$ \Rightarrow $$ deformed truss of varying E




 * }

Contributing Members
Justin McIntire --Eml4500.f08.jamama.justin 15:46, 6 November 2008 (UTC)

William Mueller --Eml4500.f08.jamama.mueller 04:56, 7 November 2008 (UTC)

Chris Alford -- Eml4500.f08.jamama.chris 18:39, 7 November 2008 (UTC)

Cedric Adam --Eml4500.f08.jamama.adam 18:48, 7 November 2008 (UTC)

Arunas Janulevicius --Eml4500.f08.jamama.jan 19:02, 7 November 2008 (UTC)

Megan Alvarez --EML4500.f08.jamama.megan 20:42, 7 November 2008 (UTC)