User:Eml4500.f08.group/HW7/Lecture Notes

Lecture Notes

=Rotational Displacement in 2-bar Frames=

Frame element = truss(bar) element + beam element

A truss has axial deformation

A beam has transverse deformation (bends)

Model Frame Problem with 2 Elements:



With the free body diagram of each element with 6 degrees of freedom:



Where $$ \left.\begin{matrix} d_{3}^{(e)}\\ d_{6}^{(e)}

\end{matrix}\right\} $$ rotational dofs and $$ \left.\begin{matrix} f_{3}^{(e)}\\ f_{6}^{(e)}

\end{matrix}\right\} $$ bending moments

Frame Global dofs (9):



2 element stiffness matrices k(e)6x6, where e=1,2

Then the global stiffness matrix K9x9 is the assembly of the two element stiffness matrices shown below.



=Stiffness Matrix with Rotational Displacement Included=

Below is a deformable truss with horizontal, vertical and rotational displacements:



The matrix formulas are as follows:

$$\mathbf{\tilde{k}}^{(e)}_{6x6}\mathbf{\tilde{d}}^{(e)}_{6x6} = \mathbf{\tilde{f}}^{(e)}_{6x6}$$, where

$$\mathbf{\tilde{d}}^{(e)} = \begin{Bmatrix} \tilde{d}^{(e)}_{1}\\ ...\\ \tilde{d}^{(e)}_{6} \end{Bmatrix}$$ and $$\mathbf{\tilde{f}}^{(e)} = \begin{Bmatrix} \tilde{f}^{(e)}_{1}\\ ...\\ \tilde{f}^{(e)}_{6} \end{Bmatrix}$$

Note: $$\tilde{f}_{3}^{(e)} = f_{3}^{(e)}$$ and $$\tilde{f}_{6}^{(e)} = f_{6}^{(e)}$$

Taking the moments about $$\tilde{z}$$ (where $$\tilde{z} = z$$), the displacement matrix is as follows:

.......$$\tilde{d}_{1}$$  .........   $$\tilde{d}_{2}$$   .........  $$\tilde{d}_{3}$$   .........   $$\tilde{d}_{4}$$  .........   $$\tilde{d}_{5}$$    ......... $$\tilde{d}_{6}$$

$$\begin{Bmatrix} \frac{EA}{L} & 0 & 0 & -\frac{EA}{L} & 0 & 0\\ 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}}\\ 0 & \frac{6EI}{L^{2}} & \frac{4EI}{L} & 0 & -\frac{6EI}{L^{2}} & \frac{2EI}{L}\\ \frac{-EA}{L} & 0 & 0 & \frac{EA}{L} & 0 & 0\\ 0 & \frac{12EI}{L^{3}} & -\frac{6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & -\frac{6EI}{L^{2}}\\ 0 & \frac{6EI}{L^{2}} & \frac{2EI}{L} & 0 & -\frac{6EI}{L^{2}} & \frac{4EI}{L} \end{Bmatrix}\begin{matrix} \tilde{d}_{1}\\ \tilde{d}_{2}\\ \tilde{d}_{3}\\ \tilde{d}_{4}\\ \tilde{d}_{5}\\ \tilde{d}_{6} \end{matrix}$$

The dimensional analysis is as follows:

$$[\tilde{d}_{1}] =L=[\tilde{d}_{i}]$$, where i = 1,2,3,4,5

dimensions of rotation = $$[\tilde{d}_{3} = 1 \Rightarrow$$ (no dimension, because the unit is degree)



The length of the arc AB is equal to $$R\theta \Rightarrow \theta = \frac{arcAB}{R}$$

$$[\theta ] = \frac{[arcAB]}{[R]} = \frac{1}{1} = 1$$

To note, the unit is not the same as the dimension. $$\theta$$ of course does not have a unit of 1!

$$\sigma = E\varepsilon \Rightarrow [\sigma ] = [E][\varepsilon ]$$

=Dimensional Analysis=

The validity of the stiffness matrix can be verified by using dimensional analysis. Since this matrix approximated the stiffness of an element as a spring, the resulting dimension of this stiffness should be the same as the dimensions for a spring's spring constant or Force per unit Length.

As an example, take stress ($$\sigma$$)which has units of force per unit area. In the case the dimensional analysis would yield:

$$ [\sigma ] = \frac{F}{L^2} $$

We also know that $$ [E] = \frac{F}{L^2},[A] = L^2,[L] = L,[I]=L^4 $$

With this, we can now check and see if all the terms of the element stiffness matrix come out to be Force per unit Length. There are 5 terms that must be checked, these terms are:

$$ \frac{EA}{L},\frac{12EI}{L^3},\frac{6EI}{L^2},\frac{4EI}{L},and \frac{2EI}{L} $$

$$\frac{[E][A]}{[L]}=\frac{F}{L^2}*L^2*\frac{1}{L}=\frac{F}{L}$$

$$\frac{12[E][I]}{[L^3]}=\frac{F}{L^2}*L^4*\frac{1}{L^3}=\frac{F}{L}$$

$$\frac{6[E][I]}{[L^2]}=\frac{F}{L^2}*L^4*\frac{1}{L^2}=F$$

$$\frac{4[E][I]}{[L]}=\frac{F}{L^2}*L^4*\frac{1}{L^2}=F*L$$

$$\frac{2[E][I]}{[L]}=\frac{F}{L^2}*L^4*\frac{1}{L^2}=F*L$$


 * {| class="toccolours collapsible collapsed" width="60%" style="text-align:left"

!MATLAB 2-bar frame with one fixed node, from team Carbon
 * A two bar frame shown below is examine with the left-most node fixed (i.e. no rotational component):
 * A two bar frame shown below is examine with the left-most node fixed (i.e. no rotational component):



This makes the leftmost truss deformation parabolic instead of linear. The MATLAB code is shown below:

close; clear; % model with 3-D beam elements with 6 dofs per node dof = 9;        %  dof per node % 1 : axial disp x,  2 : disp y  , 3 : disp z                    %  4 : torsion x    ,  5 : bend y  , 6 : bend z  % % input the nodal coordinates %  n_node = 3;             % total number of nodes n_elem = 2;            % total number of elements total_dof = 6 * n_node; % total dofs of system Le(1) = 4; % Element 1 beam length Le(2) = 2; % Element 2 beam length position(:, 1) = [     -Le(1) *cosd(30) ; -Le(1) *sind(30); 0]; position(:, 2) = [     0; 0; 0]; position(:, 3) = [      Le(2) *cosd(45) ; -Le(2)*sind(45); 0]; % inital postion of the nodes in the truss system % set up the nodal coordinate arrays x, y, x  x = zeros(1,3); y = zeros(1,3); z = zeros(1,3); for i = 1 : n_node x(i) = position(1,i); y(i) = position(2,i); z(i) = position(3,i); end % set up the element connectivity array node_connect, defined as % follows: % node_connect(local node number, element number) = global node number node_connect(1, 1) = 1;  % element 1 node_connect(2, 1) = 2; node_connect(1, 2) = 2;  % element 2 node_connect(2, 2) = 3; % plot the whole truss system % loop over the elements for i = 1 : n_elem % for element i, do the following: % node_1 = global node number corresponding to the local node 1 node_1 = node_connect(1,i); % node_2 = global node number corresponding to the local node 1 node_2 = node_connect(2,i); % xx : 1x2 array containing x coordinates of node_1 and node_2 xx = [x(node_1),x(node_2)]; % yy : 1x2 array containing y coordinates of node_1 and node_2 yy = [y(node_1),y(node_2)]; % zz : 1x2 array containing z coordinates of node_1 and node_2 zz = [z(node_1),z(node_2)]; % plot the element i in 3-D using command plot3 % use axis command to avoid having the plot window too tight % axis([]) % dashed line plot3(xx,yy,zz,'--') % hold the current figure for the next element hold on  end text(x(1) + .25, y(1)+ .25, z(1),'Node 1'); text(x(2) + .25, y(2)+ .25, z(2),'Node 2'); text(x(3) + .25, y(3)+ .25, z(3),'Node 3'); text((x(2) + x(1))/2, (y(1) + y(2))/2, (z(1) + z(2))/2,'Element 1'); text((x(3) + x(2))/2, (y(3) + y(2))/2, (z(3) + z(2))/2,'Element 2'); view([0 0 1]); x(2) A(1) = 1; A(2) = 2; %   Cross Sectional Area of Elements 1 and 2 E(1) = 3; E(2) = 5; %   Youngs Modulus of Elements 1 and 2 Te(1) = 30; Te(2) = -45; I(1) = (1/8) + 16; I(2) = (2/3) + 8; Ke = zeros(6,6); for i = 1 : n_elem Ke(:,:,i) = [ ((E(i)*A(i))/Le(i)), 0                         , 0                         , -((E(i)*A(i))/Le(i)) , 0                        , 0                          ; 0                , ((12*E(i)*I(i))/(Le(i)^3)) , ((6*E(i)*I(i))/(Le(i)^2))  , 0                   , ((12*E(i)*I(i))/(Le(i)^3)) , ((6*E(i)*I(i))/(Le(i)^2))  ; 0                , ((6*E(i)*I(i))/(Le(i)^2))  , (4*E(i)*I(i))/Le(i)        , 0                   , -((6*E(i)*I(i))/(Le(i)^2)) , (2*E(i)*I(i))/Le(i)        ; -((E(i)*A(i))/Le(i)), 0                        , 0                         , (E(i)*I(i))/Le(i)    , 0                         , 0                         ; 0                 , ((12*E(i)*I(i))/(Le(i)^3)) , -((6*E(i)*I(i))/(Le(i)^2)) , 0                   , ((12*E(i)*I(i))/(Le(i)^3)) , -((6*E(i)*I(i))/(Le(i)^2)) ; 0                 , ((6*E(i)*I(i))/(Le(i)^2))  , (2*E(i)*I(i))/Le(i)        , 0                   , -((6*E(i)*I(i))/(Le(i)^2)) , (4*E(i)*I(i))/Le(i)       ]; %   Element stiffness matrix for Element i     end Kg = zeros(9,9); Kg(4:9,4:9) = Ke(:,:,2); Kg(1:6,1:6) = Ke(1:6,1:6,1); Kg(4:6,4:6) = Ke(4:6,4:6,1) + Ke(1:3,1:3,2) %   Creation of the Global Matrix    % Fg = zeros(9,1); Fg(4,:) = 0; Fg(5,:) = 7; %   Known Forces %Fg(9,:) = 0; %   Set Matrix Size Dg = zeros(9,1); Dg(1,:) = 0; Dg(2,:) = 0; Dg(3,:) = 0; % no rotational component on node 1 Dg(7,:) = 0; Dg(8,:) = 0; %   Known Displacements Dg(4:6,:) = inv(Kg(4:6,4:6)) * Fg(4:6,:) %   d1 = d2 = d7 = d8 = 0, reducing matrix and sloving for d3 and d4   Fe(:,:,1) = Ke(:,:,1) * Dg(1:6,:); Fg(1:3,:) = Fe(1:3,1); %   sloving for f1 and d2   Fe(:,:,2) = Ke(:,:,2) * Dg(4:9,:); Fg(7:9,:) = Fe(4:6,2) %   sloving for f5 and d6   position(:, 1) = [      -Le(1) *cosd(30) ; -Le(1) *sind(30); 0]; position(:, 2) = [      Dg(4)           ;  Dg(5)          ; 0]; position(:, 3) = [      Le(2) *cosd(45) ; -Le(2) *sind(45); 0]; % displaced postion of the nodes in the truss system (only node 2 moves) % set up the nodal coordinate arrays x, y, x  for i = 1 : n_node x(i) = position(1,i); y(i) = position(2,i); z(i) = position(3,i); end % plot the whole truss system % loop over the elements for i = 1 : n_elem % for element i, do the following: % node_1 = global node number corresponding to the local node 1 node_1 = node_connect(1,i); % node_2 = global node number corresponding to the local node 1 node_2 = node_connect(2,i); % xx : 1x2 array containing x coordinates of node_1 and node_2 xx = [x(node_1),x(node_2)]; % yy : 1x2 array containing y coordinates of node_1 and node_2 yy = [y(node_1),y(node_2)]; % zz : 1x2 array containing z coordinates of node_1 and node_2 zz = [z(node_1),z(node_2)]; % plot the element i in 3-D using command plot3 % use axis command to avoid having the plot window too tight %axis([]) % solid line plot3(xx,yy,zz,'-') % hold the current figure for the next element hold on  end text(x(1) + .25, y(1)+ .25, z(1),'Node 1'); text(x(2) + .25, y(2)+ .25, z(2),'Node 2'); text(x(3) + .25, y(3)+ .25, z(3),'Node 3'); text((x(2) + x(1))/2, (y(1) + y(2))/2, (z(1) + z(2))/2,'Element 1'); text((x(3) + x(2))/2, (y(3) + y(2))/2, (z(3) + z(2))/2,'Element 2'); % put the title on the figure title('2-Bar Truss structure w/ 3-D beam elements') % label x axis xlabel('x') % label y axis ylabel('y') % label z axis zlabel('z') % attempt to view the plotted 3-D truss structure from different % viewpoint. % plot viewed from z axis, i.e., projection on xy plane view([0 0 1])     % xy plane view % plot viewed from y axis, i.e., projection on xz plane % view([0 1 0])    % xz plane view % plot viewed from x axis, i.e., projection on yz plane % view([1 0 0])    % yz plane view The results of the code is shown below.

Stiffness Matrix:

Kg = 0.7500        0         0   -0.7500         0         0         0         0         0         0    9.0703   18.1406         0    9.0703   18.1406         0         0         0         0   18.1406   48.3750         0  -18.1406   24.1875         0         0         0   -0.7500         0         0   17.0938         0         0   -5.0000         0         0         0    9.0703  -18.1406         0   74.0703   46.8594         0   65.0000   65.0000         0   18.1406   24.1875         0   46.8594  135.0417         0  -65.0000   43.3333         0         0         0   -5.0000         0         0   21.6667         0         0         0         0         0         0   65.0000  -65.0000         0   65.0000  -65.0000         0         0         0         0   65.0000   43.3333         0  -65.0000   86.6667

Displacement Matrix:

Dg = 0        0         0    0.1211   -0.0420         0         0         0         0

Force Matrix:

Fg = 0.3361  -3.2129         0    7.0000         0         0   10.6017    6.0499         0



Contribution by group.w (Jeff Wright)
 * }

Obtain Element Force-Displacement Relationship in Global Coordinates
$$\begin{Bmatrix} \tilde{d_1}\\ \tilde{d_2}\\ \tilde{d_3}\\ \tilde{d_4}\\ \tilde{d_5}\\ \tilde{d_6} \end{Bmatrix}=\begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0 & 0 & 0\\ -m^{(e)}& l^{(e)} & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & l^{(e)} & m^{(e)}\\ 0 & 0 & 0 & 0 & -m^{(e)} & l^{(e)}\\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}*\begin{Bmatrix} d_1\\ d_2\\ d_3\\ d_4\\ d_5\\ d_6 \end{Bmatrix}$$

=PVW for Beams=

$$\displaystyle \int_0^L w(x) [ -\frac{\partial^2}{\partial x^2}((EI)\frac{\partial^2 v}{\partial x^2}) + f_t - m \dot \dot v] dx = 0 $$ for all possible w(x) → Eq. 1

Integration by parts of first term: (p.30-1)

$$\displaystyle \alpha := \int_0^L w(x) \frac{\partial^2}{\partial x^2}[(EI)\frac{\partial^2 v}{\partial x^2}] dx$$

Where: $$\displaystyle s(x) = w(x)$$ $$\displaystyle r'(x) = \frac{\partial^2}{\partial x^2}[(EI)\frac{\partial^2 v}{\partial x^2}]$$ $$\displaystyle r(x) = \frac{\partial}{\partial x}[(EI)\frac{\partial^2 v}{\partial x^2}]$$

$$\displaystyle \alpha = \beta_1 - \int_0^L \frac{dw}{dx} \frac{\partial}{\partial x}[(EI)\frac{\partial^2 v}{\partial x^2}] dx$$

where: $$\displaystyle \beta_1 = [w \frac{\partial}{\partial x} [(EI) \frac{\partial^2 v}{\partial x^2}] ]_0^L $$

Repeat intergration by parts

Therefore $$\displaystyle \alpha = -(\beta_1 - \beta_2 + \gamma)$$

Where: $$\displaystyle \beta_2 = [\frac{dw}{dx} (EI) \frac{\partial^2 v}{\partial x^2}]_0^L $$ $$\displaystyle \gamma = \int_0^L \frac{d^2 w}{dx^2}(EI)\frac{\partial^2 v}{\partial x^2} dx$$

Equation 1 becomes: $$\displaystyle -\beta_1 + \beta_2 - \gamma) + \int_0^L w f_t dx - \int_0^L w m \dot \dot v dx = 0$$ for all possible w(x)

Lets focus on the stiffness term $$\displaystyle \gamma$$ for now to derive the beam stiffness matrix and to identify the shape functions of a beam.

$$ \displaystyle V(\tilde x) = N_2(x) \tilde d_2 + N_3(x) \tilde d_3 + N_5(x) \tilde d_5 + N_6(x) \tilde d_6$$ → Eq. 2

Recall:

$$\displaystyle u(\tilde x) = N_1(x) \tilde d_1 + N_4(x) \tilde d_4$$

pg. 246 in book: $$\displaystyle N_2(\tilde x) = 1 -\frac{3 \tilde x^2}{L^2} + \frac{2 \tilde x^3}{L^3}$$

$$\displaystyle N_3(\tilde x) = \tilde x -\frac{2 \tilde x^2}{L} + \frac{\tilde x^3}{L^2}$$

$$\displaystyle N_5(\tilde x) = \frac{3 \tilde x^2}{L^2} - \frac{2 \tilde x^3}{L^3}$$

$$\displaystyle N_6(\tilde x) = -\frac{\tilde x^2}{L} + \frac{\tilde x^3}{L^2}$$

=Deformation Analyzed Graphically=



From Lecture 37-3: $$\boldsymbol{\tilde d^{(e)}_{6x1} = \tilde T^{(e)}_{6x6}d^{(e)}_{6x1}}$$

Compute $$u(\tilde x), v(\tilde x)$$ using Equations (1) and (2)from Lecture 38-3.



$$\boldsymbol u(\tilde x) = u(\tilde x)\hat{\tilde i}+v(\tilde x)\hat{\tilde j} = u_x(\tilde x)\hat i + u_y(\tilde x)\hat j$$

$$\begin{Bmatrix} u_x(\tilde x) \\ u_y(\tilde x)\end{Bmatrix}= R^T\begin{Bmatrix} u(\tilde x) \\ v(\tilde x)\end{Bmatrix} $$

$$\begin{Bmatrix} u(\tilde x) \\ v(\tilde x)\end{Bmatrix}=\begin{bmatrix} N_1 & 0 & 0 & N_4 & 0 & 0 \\ 0 & N_2 & n_3 & 0 & N_5 & N_6 \end{bmatrix} \begin{Bmatrix} \tilde d_1^{(e)} \\\tilde d_2^{(e)} \\\tilde d_3^{(e)} \\\tilde d_4^{(e)} \\\tilde d_5^{(e)} \\ \tilde d_6^{(e)}\end{Bmatrix}$$

where, $$\begin{bmatrix} N_1 & 0 & 0 & N_4 & 0 & 0 \\ 0 & N_2 & n_3 & 0 & N_5 & N_6 \end{bmatrix} = \mathbb N$$

$$\begin{Bmatrix} u_x(\tilde x) \\ u_y(\tilde x)\end{Bmatrix}=\boldsymbol R^T\mathbb N \boldsymbol T^{(e)}\boldsymbol d^{(e)}$$

=Displacements in Deformed Trusses=

$$[u]=L$$

$$[N_{1}]=[N_{4}]=1$$ and [d]=L

$$[N_{1}][\tilde{d_{1}}] + [N_{4}][\tilde{d_{4}}] = L$$

[v]=L

$$[N_{2}][\tilde{d_{2}}] = L$$

$$[N_{3}][\tilde{d_{3}}] = L$$

$$[N_{5}][\tilde{d_{5}}] = L$$

$$[N_{6}][\tilde{d_{6}}] = L$$

Derivation of beam shape functions

N2, N3, N5, N6

Recall the governing partial differential equation for beams, but without $$f$$ (distributed frames load) and without the inertia force $$m\ddot{v}$$

$$\frac{\partial ^{2}}{\partial x^{2}}\left\{(EI)\frac{\partial ^{2}V}{\partial x^{2}} \right\}=0$$

Then when (EI) is considered constant the equation reduces to $$\frac{\partial ^{4}}{\partial x^{4}}V=0$$ Integrate 4 times and you get:

$$V(x)=C_{o}+C_{1}x +C_{2}x^{2}+C_{3}x^{3}$$

The boundary conditions are: $$V(0)=1, V(L)=0, V^{'}(0)=V^{'}(L)=0$$

Coefficient Analysis
$$ v(x) = C_0+C_1x^1+C_2x^2+C_3x^3 $$ $$ v'(x) = C_1+2C_2x+3C_3x^2 $$

N_2 Coefficients
Boundary Conditions

$$ v(0) = 1,v(L) = 0,v'(0) = 0, v'(L) = 0 $$

$$ v(0) = 1 = C_0 $$ $$ v(L) = 1 + C_1L+C_2L^2 + C_3L^3 = 0 $$ $$ v'(L) = 2C_2L + 3C_3L^2$$

$$ C_2L^2 + C_3L^3 = -1 $$ $$ 2C_2l + C_3L^2 = 0 $$

This yields:

$$ C_2 = \frac{-3}{L^2},C_3 = \frac{2}{L^3} $$

Therefore:

$$ v(x) = 1 - \frac{-3}{L^2}x^2 + \frac{2}{L^3}x^3 $$

HW: Find the Coefficients for Shape Functions N_3, N_5, and N_6

N_3 Coefficients
Boundary Conditions

$$ v(0) = 0,v(L) = 0,v'(0) = 1, v'(L) = 0 $$

Using $$ v(0) = 0 $$, $$ 0 = C_0 + 0 + 0 + 0 $$ $$ C_0 = 0 $$

Using $$ v'(0) = 1 $$, $$ 1 = C_1 + 0 + 0 $$ $$ C_1 = 1 $$

Using $$ v(L) = 0 $$, $$ 0 = C_0 + C_1L + C_2L^2 + C_3L^3 $$ $$ 0 = L + C_2L^2 + C_3L^3 $$ $$ C_2L^2 + C_3L^2 = -L $$

Using $$ v'(L) = 0 $$, $$ 0 = C_1 + 2C_2L + 3C_3L^2 $$ $$ 0 = 1 + 2C_2L + 2C_3L^2 $$ $$ C_22L + C_33L^2 = -1 $$

Solving the 2 above equations simultaneously yields the last 2 coefficients. $$ C_2 = \frac{-2}{L}, C_3 = \frac{1}{L^2} $$

N_5 Coefficients
Boundary Conditions

$$ v(0) = 0,v(L) = 1,v'(0) = 1, v'(L) = 0 $$

Using $$ v(0) = 0 $$, $$ 0 = C_0 + 0 + 0 + 0 $$ $$ C_0 = 0 $$

Using $$ v'(0) = 1 $$, $$ 1 = C_1 + 0 + 0 $$ $$ C_1 = 1 $$

Using $$ v(L) = 1 $$, $$ 1 = C_0 + C_1L + C_2L^2 + C_3L^3 $$ $$ 1 = L + C_2L^2 + C_3L^3 $$ $$ C_2L^2 + C_3L^2 = 1-L $$

Using $$ v'(L) = 0 $$, $$ 0 = C_1 + 2C_2L + 3C_3L^2 $$ $$ 0 = 1 + 2C_2L + 2C_3L^2 $$ $$ C_22L + C_33L^2 = -1 $$

Solving the 2 above equations simultaneously yields the last 2 coefficients. $$ C_2 = \frac{-2L+3}{L^3}, C_3 = \frac{L-2}{L^3} $$

N_6 Coefficients
Boundary Conditions

$$ v(0) = 0,v(L) = 0,v'(0) = 0, v'(L) = 1 $$

Using $$ v(0) = 0 $$, $$ 0 = C_0 + 0 + 0 + 0 $$ $$ C_0 = 0 $$

Using $$ v'(0) = 0 $$, $$ 0 = C_1 + 0 + 0 $$ $$ C_1 = 0 $$

Using $$ v(L) = 0 $$, $$ 0 = C_0 + C_1L + C_2L^2 + C_3L^3 $$ $$ 0 = L + C_2L^2 + C_3L^3 $$ $$ C_2L^2 + C_3L^2 = 0 $$

Using $$ v'(L) = 1 $$, $$ 1 = C_1 + 2C_2L + 3C_3L^2 $$ $$ 1 = 2C_2L + 2C_3L^2 $$ $$ C_22L + C_33L^2 = 1 $$

Solving the 2 above equations simultaneously yields the last 2 coefficients. $$ C_2 = \frac{-1}{L}, C_3 = \frac{1}{L^2} $$

Derive Stiffness Coefficient $$\displaystyle \tilde k$$
$$\displaystyle \tilde k_{23} = \frac{6EI}{L^3} = \int_0^L \frac{d^2 N_2}{dx^2} (EI) \frac{d^2 N_3}{dx^2} dx$$

In general:

$$\displaystyle \tilde k_{ij} = \int_0^L \frac{d^2 N_i}{dx^2} (EI) \frac{d^2 N_j}{dx^2} dx$$ for $$\displaystyle i,j = 2,3,5,6$$

=Elastodynamics=

Discrete PVW:

$$\displaystyle w * [M \dot \dot d + Kd - F] = 0$$ for all w

=Wikiversity Vs. E-Learning=

wikiversity pros and cons
Pros:


 * Can collaborate with team members even if you can not meet with them in person.


 * Many people can be adding to a page simultaneously.


 * The wiki network was never down, so could always be accessed.


 * The class can see the best team's homework and notes which each student can then use when studying for the test.

Cons:


 * Have to use basic code to type notes.


 * Have to use outside sources to make decent formulas.


 * The pages can be changed by other people possibly resulting in wrong or incomplete sections of notes.


 * Cheating is easy to do with no restricted access to each teams pages.


 * studying other teams notes can lead to misleading information.

E-Learning pros and cons
Pros:


 * Everyone in the group was previously familiar with the system and how it works.


 * A chat feature where you can communicate with any students also in the class without actually changing any posted notes.


 * Cheating is very difficult because homework is submitted directly to the teacher and other students do not have access to it.

Cons:


 * The organization of group pages would not be as efficient.


 * Usually only instructors have access to adding or changing folders and links on a class page.


 * It is a tool meant for easy communication between a teacher and students, and between students via discussion topics, not as a tool for students to type up homework and class notes for everyone to see.


 * The web page is sometimes down or slow due to the number of people trying to access it at once.

=Useful software for typing notes and homework on Wikiversity=

For writing wiki articles the members of our group just typed directly into the text area, using basic html code. We did not use any of the mentioned software in class such as WikED, AutoWikiBrowser, MWiki-Browser so we cannot provide a comparison between them.

Equations
For the Equations most of us found it easiest to use the latex equation editor, without it many of the equations would have been very difficult to type. Although sometimes the latex site was down and then our team members used Microsoft Word Equation Editor and saved the equation as a picture to put on the wiki page.

Pictures and Figures
Our team members preferred to use paint or Microsoft Powerpoint to create our drawings. We used which ever programs we were already familiar with to create the figures in our notes. No one used Inkscape so a comparison cannot be made.