User:Eml4500.f08.FEABBQ/HW5

Class Notes: October 20, 2008
Justification of eliminating rows 1,2,3,6 to obtain $$\mathbf{K_{2x2}}$$ in 2-bar truss shown below. refer to pages 10.1 and 23.2

The Force Displacement FD relationship can be expressed as fallows :

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

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

From the principle of virtual work (PVW) :

$$ (\mathbf{W}_{6x1}).(\mathbf{K}_{6x6} \mathbf{d}_{6x1} -\mathbf{F}_{6x1})= \mathbf{0}_{1x1} $$ ,  Eq.(2)

where $$(\mathbf{W}_{6x1}) $$   is Weighting Matrix for all W

Proof that Eq.(1) $$\Leftrightarrow $$ Eq.(2)


 * As seen above from Eq.(1)$$\Rightarrow$$ Eq.(2) is Trivial
 * Now we need to show that Eq.(2)$$\Rightarrow$$ Eq.(1)

$$ (\mathbf{W}).(\mathbf{K} \mathbf{d} -\mathbf{F})= \mathbf{0} $$   for all W


 * Chose 1: select $$\mathbf{W}_{1}=1,\mathbf{W}_{2}=.... =\mathbf{W}_{6}=0$$

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

$$(\mathbf{W}).(\mathbf{K} \mathbf{d} -\mathbf{F})=$$

$$1\cdot \left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}} \right]$$ $$+0\cdot\left[ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}} \right]$$ $$+0\cdot\left[ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}\right]$$

$$+0\cdot\left[ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}\right]$$ $$+0\cdot\left[ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}\right]$$ $$+0\cdot\left[ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}\right]$$

$$\Rightarrow  1\cdot\left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}}\right]$$    (Eq.(1))


 * Chose 2: select $$\mathbf{W}_{1}=0,\mathbf{W}_{2}=1, \mathbf{W}_{2}= .... =\mathbf{W}_{6}=0$$

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

$$(\mathbf{W}).(\mathbf{K} \mathbf{d} -\mathbf{F})=$$

$$0\cdot \left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}} \right]$$ $$+1\cdot\left[ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}} \right]$$ $$+0\cdot\left[ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}\right]$$

$$+0\cdot\left[ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}\right]$$ $$+0\cdot\left[ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}\right]$$ $$+0\cdot\left[ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}\right]$$

$$\Rightarrow  1\cdot\left[ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}}\right]$$    (Eq.(2))


 * Chose 3: select $$\mathbf{W}_{1}=\mathbf{W}_{2}=0,\mathbf{W}_{3}=1, \mathbf{W}_{4}= .... =\mathbf{W}_{6}=0$$

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

$$(\mathbf{W}).(\mathbf{K} \mathbf{d} -\mathbf{F})=$$

$$0\cdot \left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}} \right]$$ $$+0\cdot\left[ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}} \right]$$ $$+1\cdot\left[ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}\right]$$

$$+0\cdot\left[ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}\right]$$ $$+0\cdot\left[ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}\right]$$ $$+0\cdot\left[ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}\right]$$

$$\Rightarrow  1\cdot\left[ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}\right]$$    (Eq.(3))


 * Chose 4: select $$\mathbf{W}_{1}=,... ,=\mathbf{W}_{3}=0,\mathbf{W}_{4}=1, \mathbf{W}_{5}=\mathbf{W}_{6}=0$$

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

$$(\mathbf{W}).(\mathbf{K} \mathbf{d} -\mathbf{F})=$$

$$0\cdot \left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}} \right]$$ $$+0\cdot\left[ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}} \right]$$ $$+0\cdot\left[ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}\right]$$

$$+1\cdot\left[ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}\right]$$ $$+0\cdot\left[ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}\right]$$ $$+0\cdot\left[ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}\right]$$

$$\Rightarrow  1\cdot\left[ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}\right]$$    (Eq.(4))


 * Chose 5: select $$\mathbf{W}_{1}=,... ,=\mathbf{W}_{4}=0,\mathbf{W}_{5}=1, \mathbf{W}_{6}=0$$

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

$$(\mathbf{W}).(\mathbf{K} \mathbf{d} -\mathbf{F})=$$

$$0\cdot \left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}} \right]$$ $$+0\cdot\left[ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}} \right]$$ $$+0\cdot\left[ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}\right]$$

$$+0\cdot\left[ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}\right]$$ $$+1\cdot\left[ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}\right]$$ $$+0\cdot\left[ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}\right]$$

$$\Rightarrow  1\cdot\left[ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}\right]$$    (Eq.(5))


 * Chose 4: select $$\mathbf{W}_{1}=,... ,=\mathbf{W}_{5}=0,\mathbf{W}_{6}=1$$

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

$$(\mathbf{W}).(\mathbf{K} \mathbf{d} -\mathbf{F})=$$

$$0\cdot \left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}} \right]$$ $$+0\cdot\left[ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}} \right]$$ $$+0\cdot\left[ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}\right]$$

$$+0\cdot\left[ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}\right]$$ $$+0\cdot\left[ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}\right]$$ $$+1\cdot\left[ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}\right]$$

$$\Rightarrow  1\cdot\left[ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}\right]$$    (Eq.(6))

From the above derivations we can conclude that: $$ \mathbf{K}_{6x6} \mathbf{d}_{6x1} = \mathbf{F}_{6x1} $$

PVW: Accounting for boundary conditions (b.c's)
fro the same 2-bar trusses above. $$\mathbf{d}_{1}=\mathbf{d}_{2}=\mathbf{d}_{5}=\mathbf{d}_{6}=0$$

The weighting coefficient must be "kinematically admissible"-cannot violate the (b.c's) $$\mathbf{W}_{1}=\mathbf{W}_{2}=\mathbf{W}_{5}=\mathbf{W}_{6}=0$$ Weighting coefficient$$\equiv$$ Virtual displacement, (calculus of variations)

we can conclude the derivations as fallows:

$$(\mathbf{W}).(\mathbf{K} \mathbf{d} -\mathbf{F})=0$$

$$ (W)_{6x1}{K}_{6x6}\left[ \begin{array}{l} d_1=0 \\ d_2=0 \\ d_3=? \\ d_4=? \\ d_5=0 \\ d_6=0 \end{array} \right]_{6x1}-F_{6x1}= (W)_{6x1}\left[ \begin{array}{ll} K_{13} & K_{14} \\ K_{23} & K_{24} \\ K_{33} & K_{34} \\ K_{43} & K_{44} \\ K_{53} & K_{54} \\ K_{63} & K_{64} \end{array} \right]_{6x2} \left[ \begin{array}{ll} d_3 \\ d_4 \end{array} \right]_{2x1}-F_{2x1} = 0 $$

$$\begin{Bmatrix} W_3\\W_4

\end{Bmatrix}\left(K_{2x2}\cdot d_{2x1}-F_{2x1} \right)=0 $$

$$K =\begin{bmatrix} k_{33} &k_{34} \\ k_{43}& k_{44} \end{bmatrix} $$

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

$$F=\begin{Bmatrix} F_3\\F_4

\end{Bmatrix}$$

Notes Mon. 27 Oct. 08
Deriving $$ \mathbf{k_{4x4}^{(e)}} = \mathbf{T_{4x2}^{(e)T} \hat{k}_{2x2}^{(e)} T_{4x4}^{(e)}}$$,

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

$$ \mathbf{\hat{k}_{2x2}^{(e)} q_{2x1}^{(e)} = p_{2x1}^{(e)} } $$ → $$ \mathbf{\hat{k}^{(e)} q^{(e)} - p^{(e)} = 0_{2x1}} $$ [1]

PVW that is equivalent to [1]

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

NOTE: $$ (\mathbf{\hat{k}^{(e)} q^{(e)} - p^{(e)}}) $$ is 2x1 matrix

Recall: $$\mathbf{q_{2x1}^{(e)} = T_{2x4}^{(e)} d_{4x1}^{(e)}}$$

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

Notes Wed. 29 Oct 08
$$\mathbf{\hat{W}}_{2x1}$$ = Virtual axial displacement, corresponding to $$\mathbf{q^{(e)}_{1x1}}$$

$$\mathbf{W}_{4x1}$$ = Virtual displacement in the global coordinate system.

Replace equations (3) and (4) into equation (2)

$$\left( \mathbf{T^{(e)}w} \right)\bullet \left[ \mathbf{\hat{k}^{(e)}\left(T^{(e)}d^{(e)} \right)-p^{(e)}} \right]=0$$ for all $$\mathbf{W}_{4x1}$$  (5)

Recall $$\mathbf{\left(AB \right)^{T}=B^TA^T}$$  (6)

Recall: $$\mathbf{a_{nx1} \cdot b_{nx1}=a^T_{1xn}b_{nx1}=S_{1x1}}$$ (7)

Apply (6) and (7) into equation (5):

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

$$\Rightarrow \mathbf{ w^TT^{(e)T}} \left[ \mathbf{\hat{k}^{(e)}\left(T^{(e)}d^{(e)} \right)-p^{(e)}} \right]=0_{1x1}$$

$$\Rightarrow \mathbf{ w^T}\cdot \left[ \mathbf{ \underbrace{ \left( T^{(e)T}\hat{k}^{(e)}T^{(e)} \right) }_{k^{(e)}} }d^{(e)} -\underbrace{ T^{(e)T}p^{(e)}}_{f^{(e)}} \right]=0_{1x1}$$

$$\Rightarrow \mathbf{ w^T}\cdot \left[ \mathbf{{k}^{(e)}d^{(e)} -f^{(e)}} \right]=0_{1x1}$$

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

The Continuous Case

So far we have only been analyzing discrete cases (with matrices). These cases are not continuous. The next phase of the class will look at the continuous case (with Partial Differential Equations, PDEs). There are many motivational problems such as an elastic bar with varying A(x), E(x), varying axial load (distributed and concentrating), and inertia forces (dynamics).



Literature Search
Composite Materials at E(x) varying Young's modulus

Although many links on the subject can be found, a very thorough article is provided at http://www.iop.org/EJ/article/0022-3727/3/5/319/jdv3i5p778.pdf?request-id=7d001542-606d-4bda-9cd9-6cd05e4c4042, which discusses two-layer composites and their varying elastic moduli. In this experiment, a composite of steel-brass and steel-araldite shows an elastic modulus vs. material thickness/transition graph wherein the steel has a typically higher elastic modulus with a greater thickness showing a trend of decreasing elastic modulus towards the transition to brass/araldite.

Another good read is the article found at http://www.mech.utah.edu/~rusmeeha/labNotes/composites.html, which describes, amongst other things, the nature of composite materials to have different moduli in compression and tension.

Continuing with the discussion of Partial Differential Equations (continuous case), we have the following model image and equation of forces...



$$ \Sigma{F}_{x}=0=-N(x,t)+N(x+dx,t)+f(x,t)dx-m(x)\ddot{u}$$

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

We can call the equation above Eq(1), where h.o.t. are the higher order terms, which can be neglected...

Recall Taylor Series

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

where the second order derivative and beyond are the higher order terms.

We now have, from Eq(1) $$ \frac{\partial N}{\partial x}+f=m\ddot{u}$$ and we call this Eq(2), the equation of motion.

Further expanding on the $$N(x,t)$$ term, we have

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

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

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

We call these relations Eq(3).

Putting Eq(3) into Eq(2) yields

$$\frac{\partial}{\partial x}(A(x)E(x)\frac{\partial u}{\partial x})+f(x,t)=m(x)\ddot{u}$$

which we call Eq(4), the Partial Differential Equation of Motion.

We need two boundary conditions (2nd order derivative with respect to x), two initial conditions (2nd derivative with respect to t) and the initial displacement, initial velocity...



For the image above,$$ u(0,t)=0=u(L,t)$$



For the image above

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

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

where $$ N(L,t)$$ is the Area x Stress, $$ A(L)\sigma(L,t)$$

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

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

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

=Rederivation of 3-D Matrix Relations= Axial F-D Relation

Now that a 3 dimensional system is being considered, there is now one more additional degree of freedom at each node. However, in any element there can still only be two forces and those forces always be in the direction of the element. This means that the relation of P to Q remains unchanged:

$$\begin{bmatrix} P^{(e)}_1\\ P^{e)}_2 \end{bmatrix} = k^{(e)}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix} \begin{bmatrix} q^{(e)}_1\\ q^{e)}_2 \end{bmatrix}$$

Element DOF's and Element Forces in global XYZ coordinates (6x1 matrices)

$$ \mathbf{d^{(e)}}=\begin{bmatrix} d^{(e)}_1 \\ d^{(e)}_2 \\ d^{(e)}_3 \\ d^{(e)}_4 \\ d^{(e)}_5 \\ d^{(e)}_6 \end{bmatrix} $$

$$ \mathbf{f^{(e)}}=\begin{bmatrix} f^{(e)}_1 \\ f^{(e)}_2 \\ f^{(e)}_3 \\ f^{(e)}_4 \\ f^{(e)}_5 \\ f^{(e)}_6 \end{bmatrix} $$

=Matlab Matters=

2-Bar Truss Code Correction
The original 2-Bar Truss Matlab Code given had a small but significant error in the use of one of the functions, PlaneTrussResults.m. Since the code assumed a constant cross-sectional area and Modulus of Elasticity, the code looked as follows...

It should have read...

The following is the revised code with the results following...

6-Bar Truss Example From Textbook
Below is the Matlab code which is used to solve the 6-bar truss example given on page 226 of the text.


 * Note that this script requires the following three functions:

PlaneTrussElement.m

NodalSoln.m

PlaneTrussResults.m

Below is a figure showing the deformed and undeformed truss for the book's example.



The following Matlab code was appended to the given code above to facilitate the plot above:

6-Bar Truss with Varying Moduli
The following is the 6-Bar Truss code modified to handle various elasticities.



Three-Bar Truss in 3-space
The following 3 functions were used in solving this problem, and were supplied by Bhatti, 2005.

SpaceTrussResults.m
The following -.m file enters the parameters such as the original positions of the nodes, the cross sectional areas of the trusses, the material modulus of elasticity, as well as the applied load. This -.m file also calls the required functions and solves the three-bar truss example on page 230.

ThreeBarSpaceTrussEx.m
The following code was appended to the given code to server the function of creating plots from different perspectives to illustrate the results

Results
to be completed in the future....work in progress

=Contributing members:=
 * Esmail Hadman 14:02, 1 November 2008 (UTC)
 * Andy Koby 15:27, 3 November 2008 (UTC)
 * Stark 19:48, 3 November 2008 (UTC)
 * J.Jayma 22:23, 6 November 2008 (UTC)
 * Nik Vitt16:04, 5 November 2008 (UTC)
 * William Kurth 21:04, 7 November 2008 (UTC)