User:Eml4500.f08.wiki1.oatley

Section 18: MATLAB Graphics
One of the functions of MATLAB is the ability to produce several different types of curves and plots. *These include the following, followed by their respective commands:

planar plots of curves		                       plot 3-D plots of curves				       plot3 3-D mesh surface plots				       mesh 3-D faceted surface plots				surf

Planar Plots
For example, to plot a cosine function over the interval -5 to 5 on an x-y plot of the elements of x versus the elements of y, the command is as follows:

x = -5 : .01 : 5 y = cos(x) plot(x,y) Another example of a sample function would be to plot the function y=e^(x^3 )over the interval of -3.5 to 3.5 on an x-y plot of the elements of x versus the elements of y, the command is as follows: x = -3.5 : .01 : 3.5 y = exp(x.^3) plot(x,y)

“fplots”
The function “fplot” allows the user to graph a function with great ease and efficiency. To plot the above function, the function must first be defined in an M-file (e.g. “expnormal.m”). The M-file for the above function would read as follows: function y = expnormal(x) y = exp(x.^3) followed by the command fplot(‘expnormal’, [-3.5,3.5])

Parametric Curves
Parametric curves can be created in MATLAB. The commands for an example parametric function are as follows: t = 0: 0.01: 2*pi x = sin(5*t) y = cos(3*t) plot(x,y) MATLAB graphs can be assigned with titles, labeled axes, as well as text placed within the graph. These options with their respective commands are as follows: graph title					title x-axis label					x label y-axis label					y label place text on graph using mouse		       gtext position text on coordinates			text For example, to assign the label “Force (N)” to the x-axis, the appropriate command would be as follows: xlabel(‘Force (N)’)

Axis Reformatting
MATLAB automatically formats the axes of graphs with an auto-scale. The command “axis” following the “plot” command can override this auto-scale. Some additional options with their respective commands are as follows: set axis scale to prescribed limits			axis([xmin, xmax, ymin, ymax]) freeze scale for subsequent graphs			axis(axis) return to auto-scale					axis auto returns vector v showing current scale		       v = axis same scale on both axes				       axis square same scale and tic marks on both axes		       axis equal turn off axis scaling and tic marks			axis off turn on axis scaling and tic marks			axis on

Superimposing Plots on Top of Each Other
In MATLAB it is possible to place multiple plots on a single graph. One example of such a case is listed in the following commands: x = 0: .01: 2*pi y1 = cos(x) y2 = cos(3*x) y3 = cos(5*x) plot(x,y1,x,y2,x,y3) The same plots can be generated using a matrix Y, as follows: x = 0: .01: 2*pi Y = [cos(x)’, cos(3*x)’, cos(5*x)’] plot(x,Y) Another way to superimpose plots on each other is with the command “hold”. “hold on” will freeze the current plot and will superimpose subsequent plots on it. “hold off” will release the specified hold.

Overriding Default Linetypes, Pointtypes, and Colors
One example of overriding defaults for a dotted line and a line with circles placed at each node is as follows: x = 0: .01: 2*pi y1 = cos(x) y2 = sin(3*x) plot(x,y1, ‘:’,x,y2, ‘o’) Options for linetypes includes the following: solid (-) dashed (--) dotted dashdot (-.) Options for marktypes include the following: point (.) plus (+) star(*) circle (o) x-mark (x) Options for colors include the following: yellow (y) magenta (m) cyan (c) red (r) green (g) blue (b) white (w) black (k) For example, to plot a blue dotted line, the commands are as follows: plot(x,y,’b:’)

Graphics Hardcopy
With the command “print”, MATLAB will send a high-resolution copy of the current graphics figure to the computer’s default printer. The command “printopt” specifies the default setting used the by the “print” command. The command “print filename” saves the current graphics figure to the specified filename. 3-D Line Plots MATLAB can also create curves in three dimensions with the command “plot3”. For example, a parametric graph in three dimensions can be created with the following commands: t = .01: .01: 2*pi x = sin(2*t) y = cos(t) z = t.^2 plot3(x,y,z)

3-D Mesh and Surface Plots
The command “mesh” creates a three-dimensional wire mesh surface plot.

The command “surf” creates a three-dimensional faceted surface plot.

For example, the graph of z=e^(y^3+x^2 )over the square [-3,3] x [-3,3] can be created with the following commands: xx = -3: .1 :3 yy = xx [x,y] = meshgrid(xx,yy) z = exp(y.^3 + x.^2) mesh(z)

Color Shading and Color Profile
The command “shading” edits the color shading of different surfaces in MATLAB. The three settings and their respective commands are as follows: faceted (default)		shading faceted interpolated			shading interp flat				shading flat Color profiles in MATLAB can also be edited using the command “colormap”. Possible colormap options include the following: hsv (default) hot cool jet pink copper flag gray bone For example, to set the color profile of a figure to hot, the command is as follows: colormap(hot)

Section 19: Sparse Matrix Computations
MATLAB has the capability of significantly reducing the computation time in regards to solving matrices. MATLAB has two storage modes, full (default) and sparse.

For a matrix A, the command “nnz(A)” would return the number of nonzero elements in matrix A.

To create a sparse matrix, the necessary commands are as follows: m = 5 n = 5 e = ones(n,1) d = -2*e T = spdiags([e,d,e], [-2,0,2], m, n) The sparse analogs of eye, zeros, ones, and randn are as follows: speye sparse spones sprandn The command “sparse” allows for the generation of a sparse matrix from the listing of its nonzero entries. i = [1 2 3 5 5 5] j = [1 2 3 3 2 1] s = [2 3 4 5 6 7] S = sparse(i, j, s, 4, 3), full(S) The command “sparse(i, j, s, m, n) is made up of several components. i and j list the row and column indices and s lists the nonzero entries of S. m and n are the dimensions of the matrix S.
 * }

Section 19: Sparse Matrix Computations
MATLAB has the capability of significantly reducing the computation time in regards to solving matrices. MATLAB has two storage modes, full (default) and sparse.

For a matrix A, the command “nnz(A)” would return the number of nonzero elements in matrix A.

To create a sparse matrix, the necessary commands are as follows:

m = 5 n = 5 e = ones(n,1) d = -2*e T = spdiags([e,d,e], [-2,0,2], m, n)

The sparse analogs of eye, zeros, ones, and randn are as follows:

speye sparse spones sprandn

The command “sparse” allows for the generation of a sparse matrix from the listing of its nonzero entries.

i = [1 2 3 5 5 5] j = [1 2 3 3 2 1] s = [2 3 4 5 6 7] S = sparse(i, j, s, 4, 3), full(S)

The command “sparse(i, j, s, m, n) is made up of several components. i and j list the row and column indices and s lists the nonzero entries of S. m and n are the dimensions of the matrix S.

Meeting 8 Lecture Notes (12 Sep 08)
[Kij]6x6{dj}6x1 = {Fi}6x1

Knxn = [Kij]nxn = global stiffness matrix

dnx1 = {dj}nx1 = global displacement matrix

Fnx1 = {Fi}nx1 = global force matrix

k(e)4x4d(e)4x1 = f(e)4x1

k(e)4x4 = [k(e)ij]4x4 = elemental stiffness matrix

d(e)4x1 = [d(e)j]4x1 = elemental displacement matrix

f(e)4x1 = [f(e)i]4x1 = elemental force matrix

To get from global matrices  to  elemental matrices :
Use an  assembly process 

Global level:
{d1, d2, d3, d4, d5, d6}

Elemental level:
Elem. 1: {d1(1), d2(1), d3(1), d4(1)}

Elem. 2: {d1(2), d2(2), d3(2), d4(2)}

Identification of Global-Local (Elemental) Degrees of Freedom (DOF's):
Global Node 1:

d1 = d1(1)

d2 = d2(1)

Global Node 2:

d3 = d3(1) = d1(2)

d4 = d4(1) = d2(2)

Global Node 3:

d5 = d3(2)

d6 = d4(2)

Meeting 17 Lecture Notes (3 Oct 08)
3-Bar Truss System



ΣFx = 0

ΣFy = 0

ΣMA = 0 (Trivial)

→ 2 equations, 3 unknowns → Statically Indeterminate

Question: How about MB? (3-D Explanation)



$$ \overline{M_B} = \overline{BA} x \overline{F} = \overline{BA'} x \overline{F} $$

(for A' on line of action of $$\overline{F}$$)

$$ \overline{M_B} = (\overline{BA} + \overline{AA'}) x \overline{F} = \overline{BA'} x \overline{F} + \overline{AA'} x \overline{F} $$

Back to 3-bar truss:

Node A is in equilibrium:

$$\sum_{i=0}^3 {\overline{F_i}} = \overline{0}$$

$$\sum_{i} {\overline{M_B,i}} = \sum_{i} {\overline{BA'_i}} x \overline{F_i} $$

Ai' = any point on line of action of Fi

$$\sum_{i} {\overline{M_B,i}} = \sum_{i} {\overline{BA'_i}} x \overline{F_i} = \overline{BA} x \sum_{i=0} {\overline{F_i}} = \overline{0}$$



\begin{bmatrix} K_{11} & K_{12} & K_{13} & K_{14} & K_{15} & K_{16} & K_{17} & K_{18}\\ K_{21} & K_{22} & K_{23} & K_{24} & K_{25} & K_{26} & K_{27} & K_{28}\\ K_{31} & K_{32} & K_{33} & K_{34} & K_{35} & K_{36} & K_{37} & K_{38}\\ K_{41} & K_{42} & K_{43} & K_{44} & K_{45} & K_{46} & K_{47} & K_{48}\\ K_{51} & K_{52} & K_{53} & K_{54} & K_{55} & K_{56} & K_{57} & K_{58}\\ K_{61} & K_{62} & K_{63} & K_{64} & K_{65} & K_{66} & K_{67} & K_{68}\\ K_{71} & K_{72} & K_{73} & K_{74} & K_{75} & K_{76} & K_{77} & K_{78}\\ K_{81} & K_{82} & K_{83} & K_{84} & K_{85} & K_{86} & K_{87} & K_{88} \end{bmatrix}

$$

= \begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0 & 0 & 0\\ k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0 & 0 & 0\\ k_{31}^{(1)} & k_{32}^{(1)} & (k_{33}^{(1)}+k_{11}^{(2)}+k_{11}^{(3)}) & (k_{34}^{(1)}+k_{12}^{(2)}+k_{12}^{(3)}) & k_{13}^{(2)} & k_{14}^{(2)} & k_{15}^{(3)} & k_{16}^{(3)}\\ k_{41}^{(1)} & k_{42}^{(1)} & (k_{43}^{(1)}+k_{21}^{(2)}+k_{21}^{(3)}) & (k_{44}^{(1)}+k_{22}^{(2)}+k_{22}^{(3)}) & k_{23}^{(2)} & k_{24}^{(2)} & k_{25}^{(3)} & k_{26}^{(3)}\\ 0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)} & 0 & 0\\ 0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)} & 0 & 0\\ 0 & 0 & k_{51}^{(3)} & k_{52}^{(3)} & 0 & 0 & k_{55}^{(3)} & k_{56}^{(3)}\\ 0 & 0 & k_{61}^{(3)} & k_{62}^{(3)} & 0 & 0 & k_{65}^{(3)} & k_{66}^{(3)} \end{bmatrix}

$$

Meeting 21 Lecture Notes (13 Oct 08)
Comments on HW 3:
 * closing the loop, infinitesimal displacement
 * eigenvalue problem, zero eigenvalues, rigid body motion, mechanisms

 HW : Plot the eigenvectors corresponding to the zero eigenvalues of the 2-bar truss system (Mtg. 11) and interpret the results: mode shapes

These mode shapes (corresponding to the zero eigenvalues) may come out as a linear combination of the pure mode shapes (pure rigid body motions (3) and 1 pure mechanism).

Eigenvalue problem: $$\mathbf{kv} = \lambda \mathbf{v}$$

Let $$\left\{\mathbf{u_1}, \mathbf{u_2}, \mathbf{u_3}, \mathbf{u_4} \right\}$$ be the pure eigenvectors corresponding to the 4 zero eigenvalues:

$$\mathbf{ku_i} = 0\mathbf{u_i} = \mathbf{0}\left(i = 1,2,3,4 \right)$$

Linear combination of $$\left\{u_i, i = 1,2,3,4 \right\} $$

 = \sum_{i=1}^{4}{\alpha_i\left(\mathbf{ku_i} \right)} = \mathbf{0} = 0 \times \mathbf{W} $$ \sum_{i=1}^{4}{\alpha_i\mathbf{u}_i} \equiv \mathbf{W} $$ (equal by definition)

$$ \alpha_i \Rightarrow $$(real numbers)

W is also an eigenvector corresponding to a zero eigenvalue:

$$ \mathbf{kW} = \mathbf{k}\left(\sum_{i=1}^{4}{\alpha_i\mathbf{u}_i} \right) $$

$$= \sum_{i=1}^{4}{\alpha_i\left(\mathbf{ku_i} \right)} = \mathbf{0} = 0 \times \mathbf{W}$$

 HW : INSERT FIGURE

$$\mathbf{\overline{k}v} = \lambda \mathbf{v}$$

$$\Rightarrow \mathbf{\overline{k}}$$ is stiffness matrix for constrained system

Plot eigenvectors corresponding to zero eigenvalues for case (a).

Matlab code: Modify 2-bar truss code

$$ a = b = 1m$$

$$E = 2$$

$$A = 3$$

Eml4500.f08.wiki1.oatley 20:36, 21 October 2008 (UTC)

Meeting 22 Lecture Notes (15 Oct 08)
ADINA Models and Animations (FEA eigen frequency animation)

Justification of assembly of element stiffness matrix $$\left(\mathbf{k^{(e)}}, e = 1,..., nel \right)\Rightarrow nel =$$ number of element

into global stiffness matrix $$\left(\mathbf{K}\right)$$

Consider example of 2-bar truss, (Mtg. 9):

Recall elemental FD relation: $$\mathbf{k}^{(e)}_{4\times 4}\mathbf{d}^{(e)}_{4\times 1} = \mathbf{f}^{(e)}_{4\times 1}$$

Mtg. 11: Euler cut principle, Method 2 (equilibrium of global node 2)

Mtg. 4: Free Body Diagrams of element 1 and element 2 with element degrees of freedom $$\mathbf{d}^{(e)}_{4\times 1}$$

Mtg. 8: For node 2, identify the relationships between the global degrees of freedom and the element degrees of freedom for elements 1 and 2

Equilibrium of node 2:

$$(1) \sum{F_x} = 0 = -f^{(1)}_3 - f^{(2)}_1 = 0$$

$$(2) \sum{F_y} = 0 = P - f^{(1)}_4 - f^{(2)}_2 = 0$$

Next we use the FD relation: $$\mathbf{k}^{(e)}_{4\times 4}\mathbf{d}^{(e)}_{4\times 1} = \mathbf{f}^{(e)}_{4\times 1}$$

Eml4500.f08.wiki1.oatley 20:37, 21 October 2008 (UTC)

Meeting 23 Lecture Notes (17 Oct 08)
Mtg. 22: $$\mathbf{Eq. (1) \ }\Rightarrow f^{(1)}_3 + f^{(2)}_1 = 0  \mathbf{\left( 1\right)}$$

$$\mathbf{Eq. (2) \ }\Rightarrow f^{(1)}_4 + f^{(2)}_2 = P \mathbf{\left( 2\right)}$$

$$\mathbf{(1):}\left[k^{(1)}_{31}d^{(1)}_1 + k^{(1)}_{32}d^{(1)}_2 + ... + k^{(1)}_{34}d^{(1)}_4\right] \equiv f^{(1)}_3$$ $$+ \left[k^{(1)}_{11}d^{(2)}_1 + ... + k^{(2)}_{14}d^{(2)}_4\right] \equiv f^{(2)}_1 = 0$$

 HW:  Fill in for missing forces

Mtg. 8: Rewrite local degrees of freedom to global degrees of freedom:

$$\mathbf{\left(1 \right):} \left[k^{(1)}_{31}d_1 + k^{(1)}_{32}d_2 + k^{(1)}_{33}d_3 + k^{(1)}_{34}d_4\right]$$

$$+ \left[k^{(2)}_{11}d_3 + k^{(2)}_{12}d_4 + k^{(2)}_{13}d_5 + k^{(2)}_{14}d_6\right] = 0$$

$$\Rightarrow $$ Thus obtain the 3rd row of $$\mathbf{k}$$ (Mtg. 9) (See also Mtg. 8)

 HW : Derive the details of Eq. (2) and the assembly of row 4 in $$\mathbf{Kd = F} $$

Assembly of $$\mathbf{k^{(e)}}, e = 1,...,nel $$, into global stiffness matrix $$\mathbf{K}$$:

$$\mathbf{K_{n\times n}= }A^{nel}_{e=1}\mathbf{k^{(e)}_{ned\times ned}}$$

<p style="text-align:center;">A: assembly operator <p style="text-align:center;">n: total number of global dof's before eliminating boundary conditions <p style="text-align:center;">ned: number of element dof's <p style="text-align:center;">(ned<<n)

Principle of Virtual Work (PVW)

Mtg. 10: Elimination of rows corresponding to the boundary conditions to obtain $$\mathbf{\overline{k}_{2\times 2}} $$

Mtg. 12: $$\mathbf{q^{(e)}_{2\times 1}=T^{(e)}_{2\times 4}d^{(e)}_{4\times 1}}$$

Mtg. 14: $$\mathbf{k^{(e)}=T^{(e)^T}\hat{k}^{(e)}T^{(e)}}$$

Deriving FEM for partial differential equations (PDE's):

FD relation for a bar or a spring: $$kd = F$$

<p style="text-align:center;">$$\Rightarrow kd - F = 0\;\;\;\;\; (3)$$

<p style="text-align:center;">$$\Leftrightarrow W(kd - F) = 0\;\;\;\left(for \, all\, W\, \right) \;\;\;\;\;(4)$$

<p style="text-align:center;">W: weighting coefficient <p style="text-align:center;">"weak form" $$\equiv $$ PVW

 Proof :

$$A) \; \left(3 \right)\Rightarrow (4): trivial$$

$$B) \; \left(4 \right)\Rightarrow (3): \mathbf{NOT\; TRIVIAL}$$

Since $$\left(4 \right)$$ is valid for all $$W$$, select $$W=1$$ Then $$\left(4 \right)$$ becomes:

<p style="text-align:center;">$$1\times (kd - F) = 0 \Rightarrow \left(3 \right)$$

Eml4500.f08.wiki1.oatley 20:37, 21 October 2008 (UTC)

Meeting 28 Lecture Notes (31 Oct 08)
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)

Eml4500.f08.wiki1.oatley 17:05, 7 November 2008 (UTC)

Meeting 29 Lecture Notes (3 Nov 08)
 PVW (Continuous) of Dynamics of Elastic Bar:  PDE: <p style="text-align:center;">$$\frac{\partial }{\partial x}\left[(EA)\frac{\partial u}{\partial x} \right]+f=m\ddot{u}\; (1)$$ Discrete EOM (Equation of Motion) $$\Rightarrow \mathbf{-Kd+F=M\ddot{d}}$$ <p style="text-align:center;">$$\Rightarrow \mathbf{M\ddot{d}+Kd=F}\; (2)$$ (for Multiple Degree of Freedom System (MDOF))

SDOF=Single Degree of Freedom:

<p style="text-align:center;">

Derive (2) from (1): <p style="text-align:center;">$$\int_{x=0}^{x=L}{W(x)\left\{\frac{\partial }{\partial x}\left[EA\frac{\partial u}{\partial x} \right]+f-m\ddot{u} \right\}dx}=0\; (3)$$ <p style="text-align:center;">for all possible W(x) (Weighting function)

$$(1)\Rightarrow (3)$$ Trivial $$(3)\Rightarrow (1)$$ NOT Trivial

(3) rewritten as: <p style="text-align:center;">$$\int W(x)g(x)dx=0$$ <p style="text-align:center;">for all W(x)

Since (3) holds for all W(x), select W(x)=g(x), then (3) becomes: <p style="text-align:center;">$$\int g^2dx=0\Rightarrow g(x)=0$$ <p style="text-align:center;">$$(g^2\geq 0)$$

Meeting 30 Lecture Notes (5 Nov 08)
Integration by Parts: r(x), s(x)

<p style="text-align:center;">$$(rs)'=r's+rs'$$

<p style="text-align:center;">$$r'=\frac{dr}{dx}$$ and $$s'=\frac{ds}{dx}$$

<p style="text-align:center;">$$\int (rs)'=\int r's+rs'$$

<p style="text-align:center;">$$\Rightarrow (rs)=\int r's+rs'$$

<p style="text-align:center;">$$\Rightarrow \int r's=rs-\int rs'$$

Recall continuous PVW $$\Rightarrow Eq. (3)$$ from Meeting 29

1st term: $$r(x)=(EA)\frac{\partial u}{\partial x}$$ and $$s(x)= W(x)$$

Through integration by parts:  <p style="text-align:center;">$$\int_{x=0}^{x=L}{W(x)\frac{\partial }{\partial x}\left[(EA)\frac{\partial u}{\partial x} \right]dx}$$

<p style="text-align:center;">$$=\left[ W(EA)\frac{\partial u}{\partial x}\right]-\int_{0}^{L}{\frac{dw}{dx}(EA)\frac{\partial u}{\partial x}dx}$$

<p style="text-align:center;">$$=W(L)(EA)(L)\frac{\partial u}{\partial x}(L,t)-W(0)EA(0)\frac{\partial u}{\partial x}(0,t)-\int_{0}^{L}{\frac{dW}{dx}(EA)\frac{\partial u}{\partial x}dx}$$

Now consider a model problem from Meeting 28:

At x=0, select W(x) such that W(0)=0 and kinematically admissible

<p style="text-align:center;">

Motivation: Discrete PVW applied to Equation from Meeting 10 <p style="text-align:center;">$$\mathbf{W}_{6\times1}\cdot(\mathbf{[K]}_{6\times2}\begin{Bmatrix}d_3\\d_4\end{Bmatrix}_{2\times1}-\mathbf{F}_{6\times1})=0_{1\times1}$$

<p style="text-align:center;">$$\mathbf{F^T=\begin{bmatrix} F_1 & F_2 & F_3 & F_4 & F_5 & F_6 \end{bmatrix} }$$

F3 and F4 are known reactions F1, F2, F5, F6 are unknown reactions

Since W can be selected arbitrarily, select W such that W1 = W2 = W5 = W6 = 0 so to eliminate equations involving unknown reactions →Eliminate rows 1,2,5,6

<p style="text-align:center;">$$\mathbf{\bar{K}_{2x2}\bar{d}_{2x1}=\bar{F}_{2x1}}\; (1)$$

<p style="text-align:center;">$$\mathbf{\bar{d}_{2x1}}=\begin{Bmatrix} d_3\\d_4\end{Bmatrix}$$ and $$\mathbf{\bar{F}_{2x1}}=\begin{Bmatrix} F_3\\F_4\end{Bmatrix}$$

Note:

<p style="text-align:center;">$$\mathbf{\bar{W}\cdot }(\mathbf{\bar{K}\bar{d}-\bar{F})=0}$$ <p style="text-align:center;">for all $$\mathbf{\bar{W}}$$ <p style="text-align:center;">where $$\mathbf{\bar{W}}=\begin{Bmatrix} W_3\\W_4\end{Bmatrix}$$

Back to Continuous PVW:

Unknown reaction: $$N(0,t)=(EA)(0)\frac{\partial u}{\partial x}(0,t)$$

Continuous PVW gives: <p style="text-align:center;">$$W(L)F(t)-\int_{0}^{L}{\frac{dw}{dx}(EA)\frac{\partial u}{\partial x}dx}+\int_{0}^{L}{W(x)[f-m\ddot{u}]dx=0}$$

<p style="text-align:center;">for all W(x) such that W(0)=0

<p style="text-align:center;">$$\Rightarrow \int_{0}^{L}{W(m\ddot{u})dx+\int_{0}^{L}{\frac{dW}{dx}}(EA)\frac{\partial u}{\partial x}dx}$$

<p style="text-align:center;">$$=W(L)F(t)+\int_{0}^{L}{Wfdx}$$

<p style="text-align:center;">for all W(x) such that W(0) = 0

Meeting 40 Lecture Notes (5 Dec 08)
 Note:  Eq. (2) (p. 39-2, Dimensional Analysis)

[u] = L p. 31-4: [N1] = [N4] = 1

Eq. (2) p. 39-2: $$\left[N_{1} \right]\left[\tilde{d}_{1} \right]+\left[N_{4} \right]\left[\tilde{d}_{4} \right]$$

$$\left[N_{1} \right]=1$$ $$\left[\tilde{d}_{1} \right]=L$$ $$\left[N_{4} \right]=1$$ $$\left[\tilde{d}_{4} \right]=L$$

[v] = L $$\left[N_{2} \right]\left[\tilde{d}_{2} \right]=L$$ $$\left[N_{2} \right]=1$$ $$\left[\tilde{d}_{2} \right]=L$$

$$\left[N_{3} \right]\left[\tilde{d}_{3} \right]=L$$ $$\left[N_{3} \right]=L$$ $$\left[\tilde{d}_{3} \right]=1$$

$$\left[N_{5} \right]\left[\tilde{d}_{5} \right]=L$$ $$\left[N_{5} \right]=1$$ $$\left[\tilde{d}_{5} \right]=L$$

$$\left[N_{6} \right]\left[\tilde{d}_{6} \right]=L$$ $$\left[N_{6} \right]=L$$ $$\left[\tilde{d}_{6} \right]=1$$

Derivation of beam shape functions: $$N_{2}, N_{3}, N_{5}, N_{6}$$ (p. 38-4) p. 38-3 plots Recall: Governing PDE for beams p. 37-4, Eq. (1), without f2 (dist. transverse load) and without inertia force $$m\ddot{v}$$ (static case):

<p style="text-align:center;">$$\frac{\partial }{\partial x^2}\left\{(EI)\frac{\partial^2v }{\partial x^2} \right\}=0$$

Further, consider constant EI : <p style="text-align:center;">$$\frac{\partial^4 }{\partial x^4}v = 0$$

Integrate 4 times 4 constants (c0, c1, c2, c3) <p style="text-align:center;">$$\Rightarrow v(x)=c_{0}+c_{1}x^{1}+c_{2}x^{2}+c_{3}x^{3}$$

To obtain N2(x) ($$\tilde{x}=x$$for simplicity)

<p style="text-align:left;">$$v(0)=1$$ $$v(L)=0$$ $$v'(0)=v'(L)=0$$

Use above boundary conditions to solve for c0, c1, c2, c3

v(0)=1=c0 $$(1) \; \; v(L)=1+c_{1}L+c_{2}L^2+c_{3}L^3=0$$ $$v'(x)=c_{1}+2c_{2x}+3c_{3}x^2$$ v'(0)=c_(1)=0 $$(2)\; \; v'(L)=2c_{2}L+3c_{3}L^2=0$$

$$c_{0}=1$$ $$c_{1}=0$$ $$c_{2}=\frac{-3}{L^2}$$ $$c_{3}=\frac{-2}{3}\frac{1}{L}\frac{-3}{L^2}=\frac{2}{L^3}$$ Compute with expression for N2 on p. 38-4

 For N3:  (Boundary Conditions) v(0)=v(L)=0 v'(0)=1 v'(L)=0

 For N5:  ($$\tilde{d_{5}}$$ rotation) v(0)=0 v(L)=1 v'(0)=0 v'(L)=0

 For N6:  ($$\tilde{d_{6}}$$ rotation) v(0)=0 v(L)=0 v'(0)=0 v'(L)=1

See p. 39-1 for plots of N5, N6

From p. 36-2: $$\tilde{k}_{22}=\frac{12EI}{L^3}$$

From p. 38-2: $$\tilde{k}_{22}=\int_{0}^{L}{\frac{d^2N_2}{dx^2}(EI)}\frac{d^2N_2}{dx^2}dx$$

Meeting 41 Lecture Notes (8 Dec 08)
In general:

<p style="text-align:center;">$$\tilde{k}_{ij}=\int_{0}^{L}{\frac{d^2N_i}{dx^2}(EI)}\frac{d^2N_j}{dx^2}dx$$ (i,j)=2,3,5,6

Therefore, knowing that the stiffness matrix $$\tilde{K}_{6x6}$$ is symmetric about the diagonal:

From p. 36-2: $$\tilde{k}_{23}=\tilde{k}_{32}=\frac{6EI}{L^2}$$

From p. 38-2: $$\tilde{k}_{23}=\int_{0}^{L}{\frac{d^2N_2}{dx^2}(EI)}\frac{d^2N_3}{dx^2}dx=\tilde{k}_{32}=\int_{0}^{L}{\frac{d^2N_3}{dx^2}(EI)}\frac{d^2N_2}{dx^2}dx$$

Similarly: $$\tilde{k}_{25}=\tilde{k}_{52}=\frac{-12EI}{L^3}$$

$$\tilde{k}_{25}=\int_{0}^{L}{\frac{d^2N_2}{dx^2}(EI)}\frac{d^2N_5}{dx^2}dx=\tilde{k}_{52}=\int_{0}^{L}{\frac{d^2N_5}{dx^2}(EI)}\frac{d^2N_2}{dx^2}dx$$

$$\tilde{k}_{26}=\tilde{k}_{62}=\frac{6EI}{L^2}$$

$$\tilde{k}_{26}=\int_{0}^{L}{\frac{d^2N_2}{dx^2}(EI)}\frac{d^2N_6}{dx^2}dx=\tilde{k}_{62}=\int_{0}^{L}{\frac{d^2N_6}{dx^2}(EI)}\frac{d^2N_2}{dx^2}dx$$

$$\tilde{k}_{33}=\frac{4EI}{L}$$

$$\tilde{k}_{33}=\int_{0}^{L}{\frac{d^2N_3}{dx^2}(EI)}\frac{d^2N_3}{dx^2}dx$$

$$\tilde{k}_{35}=\tilde{k}_{53}=\frac{-6EI}{L^2}$$

$$\tilde{k}_{35}=\int_{0}^{L}{\frac{d^2N_3}{dx^2}(EI)}\frac{d^2N_5}{dx^2}dx=\tilde{k}_{53}=\int_{0}^{L}{\frac{d^2N_5}{dx^2}(EI)}\frac{d^2N_3}{dx^2}dx$$

$$\tilde{k}_{36}=\tilde{k}_{63}=\frac{2EI}{L}$$

$$\tilde{k}_{36}=\int_{0}^{L}{\frac{d^2N_3}{dx^2}(EI)}\frac{d^2N_6}{dx^2}dx=\tilde{k}_{63}=\int_{0}^{L}{\frac{d^2N_6}{dx^2}(EI)}\frac{d^2N_3}{dx^2}dx$$

$$\tilde{k}_{55}=\frac{12EI}{L^3}$$

$$\tilde{k}_{55}=\int_{0}^{L}{\frac{d^2N_5}{dx^2}(EI)}\frac{d^2N_5}{dx^2}dx$$

$$\tilde{k}_{56}=\tilde{k}_{65}=\frac{-6EI}{L^2}$$

$$\tilde{k}_{56}=\int_{0}^{L}{\frac{d^2N_5}{dx^2}(EI)}\frac{d^2N_6}{dx^2}dx=\tilde{k}_{65}=\int_{0}^{L}{\frac{d^2N_6}{dx^2}(EI)}\frac{d^2N_5}{dx^2}dx$$

$$\tilde{k}_{66}=\frac{4EI}{L}$$

$$\tilde{k}_{66}=\int_{0}^{L}{\frac{d^2N_6}{dx^2}(EI)}\frac{d^2N_6}{dx^2}dx$$

Elastodynamics (used with trusses, frames, 2-D, 3-D, elasticity)

p. 31-1: Model problem

Discrete PVW: (Boundary Conditions already applied) <p style="text-align:center;">$$\mathbf{\bar{w}\bullet \left[\bar{M}\ddot{\bar{d}}+\bar{K}\bar{d}-\bar{F} \right]}=0$$ <p style="text-align:center;">for all $$\mathbf{\bar{w}}$$ where $$\mathbf{\bar{K}}$$ is the reduced stiffness matrix

Complete Ordinary Differential Equations (ODE's) (2nd order in time) and initial conditions governing the elastodynamics of discretized continuity problem (MDOF) $$(1)\Rightarrow \mathbf{\bar{M}\ddot{\bar{d}}+\bar{K}\bar{d}=\bar{F}(t)}$$ $$\; \; \; \; \mathbf{\bar{d}(0)=\bar{d_{0}}}$$ $$\; \; \; \; \mathbf{\dot{\bar{d}}(0)=\bar{v}_{0}}$$

'''Solving Eq. (1): 1)Consider unforced vibration problem:''' $$\mathbf{\bar{M}_{nxn}\ddot{v}_{nx1}+\bar{K}_{nxn}v_{nx1}=0_{nx1}}$$ Assume $$\mathbf{v(t)_{nx1}=(sin\omega t)\phi _{nx1}}$$ $$\mathbf{\ddot{v}=-\omega ^2sin\omega t\phi}$$ $$\mathbf{-\omega ^2sin\omega t\bar{M}\phi+sin\omega t\bar{K}\phi=0}$$ Generalized eigenvalue problem $$\Rightarrow \mathbf{\bar{K}\phi =\omega ^2\bar{M}\phi}\; \; \; (1)$$ General form: $$\mathbf{Ax=\lambda Bx}$$, where $$\lambda$$ is an eigenvalue Standard eigenvalue problem: $$\mathbf{Ax=\lambda x}$$,where $$\mathbf{(B=I)}$$ The identity matrix $$\mathbf{I}=\begin{bmatrix}1 &  & 0\\ & ... & \\ 0 &  & 1\end{bmatrix}$$ $$\lambda =\omega ^2$$ $$(\lambda _i,\mathbf{\phi _i)}$$ are eigenpairs and i = 1,...,n Mode i $$\Rightarrow \mathbf{v_i(t)=(sin\omega _it)\phi _i}$$ where i = 1,...,n

2) Model superposition method

Orthogonal properties of eigenpairs:

<p style="text-align:center;">$$\mathbf{\phi _{i}^T\bar{M}\phi _j=\delta _{ij}=\begin{cases}1 & \text{ if } i=j \\0 & \text{ if } i\neq j\end{cases}}$$ where $$\delta _{ij}$$ is the Kronecker delta

Mass Orthogonality of eigenvectors: Eq. (2) p. 41-2: Eq. (1) p. 41-3:

<p style="text-align:center;">$$\mathbf{\bar{M}\phi _j=\lambda _j^{-1}\bar{K}\phi _j}$$

<p style="text-align:center;">$$\mathbf{\phi _i^T\bar{M}\phi _j=\lambda _j^{-1}\phi _i^T\bar{K}\phi _j}$$ where $$\mathbf{\phi _i^T\bar{M}\phi _j=\delta _{ij}}$$

<p style="text-align:center;">$$\Rightarrow \mathbf{\phi _i^T\bar{K}\phi _j=\lambda _j\delta _{ij}}$$

Eq. (1) p. 41-2: $$\mathbf{\bar{d}(t)_{nx1}=\sum_{i=1}^{n}{\xi _i(t)_{1x1}}\phi _{nx1}}$$

<p style="text-align:center;">$$\mathbf{M\left(\sum_{j}^{}{}\ddot{\xi _j}\phi _j\right)+\bar{K}\left(\sum_{j}^{}{}\xi _j\phi _j \right)=F}$$ where $$\sum_{j}^{}{}\ddot{\xi _j}\phi _j=\mathbf{\ddot{\bar{d}}}$$ and $$\sum_{j}^{}{}\xi _j\phi _j=\mathbf{\bar{d}}$$

<p style="text-align:center;">$$\mathbf{\sum_{j}^{}{}\ddot{\xi _j}\left(\phi _i^T\bar{M\phi _j}\right)+\sum_{j}^{}{}\xi _j\left(\phi _i^T\bar{K}\phi _j \right)=\phi _i^TF}$$ <p style="text-align:center;">where $$\phi _i^T\bar{K}\phi _j=\delta _{ij}$$ and $$\phi _i^T\bar{K}\phi _j=\lambda _j\delta _{ij}$$

<p style="text-align:center;">$$\Rightarrow \mathbf{\ddot{\xi _i+\lambda _i\xi _i=\phi _i^TF}}$$ where i = 1,...,n