University of Florida/Eml4500/f08 Team Homework 3

Homework 3

Derivation of Elemental Free-body Diagram with respect to the Global Coordinate System

${\mathbf{\text{k}}}_{4x4}^{(e)}{\mathbf{\text{d}}}_{4x1}^{(e)}={\mathbf{\text{f}}}_{4x1}^{(e)}$

File:HW-3Picture1.jpg

$k^{(e)}\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]\left\{\begin{array}{c}{q}_1^{(e)}\\ q_2^{(e)}\end{array}\right\}=\left\{\begin{array}{c}{P}_1^{(e)}\\ P_2^{(e)}\end{array}\right\}$

$q_i^{(e)}=$ axial displacement of element 'e' at local node 'i'
$P_i^{(e)}=$ axial force of element 'e' at local node 'i'

We want to find the relationship between ${\mathbf{\text{q}}}^{(e)}$ and ${\mathbf{\text{d}}}^{(e)}$, and ${\mathbf{\text{P}}}^{(e)}$ and ${\mathbf{\text{F}}}^{(e)}$.

These relationships can be expressed in the form: ${\mathbf{\text{q}}}_{2x1}^{(e)}{\mathbf{\text{T}}}_{2x4}^{(e)}={\mathbf{\text{d}}}_{4x1}^{(e)}$

Consider the displacement vector of the local node 1 denoted by ${\overrightarrow{d}}_1^{(e)}$.

File:HW-3Picture2.jpg

${\overrightarrow{d}}_{[1]}^{(e)}=d_1^{(e)}\overrightarrow{i}+d_1^{(e)}\overrightarrow{j}$

$q_1^{(e)}$ = axial displacement of node 1 is the prthagonal projection of the displacement vector ${\overrightarrow{d}}_1^{(e)}$ of node 1 on the $\tilde{x}$ axis of element 'e'.

$q_1^{(e)}={\overrightarrow{d}}_1^{(e)}+\overrightarrow{\tilde{i}}$

$q_1^{(e)}=(d_1^{(e)}\overrightarrow{i}+d_2^{(e)}\overrightarrow{j})+\overrightarrow{\tilde{i}}$

$q_1^{(e)}=d_1^{(e)}(\overrightarrow{i}\cdot \overrightarrow{\tilde{i}})+(d_2^{(e)}(\overrightarrow{j}\cdot \overrightarrow{\tilde{i}})$

$(\overrightarrow{i}\cdot \overrightarrow{\tilde{i}})=\cos {\theta }^{(e)}=l^{(e)}$

$(\overrightarrow{i}\cdot \overrightarrow{\tilde{i}})=\sin {\theta }^{(e)}=m^{(e)}$

$q_1^{(e)}=l^{(e)}{d}_1^{(e)}+m^{(e)}{d}_2^{(e)}$

$q_1^{(e)}={\left[\begin{array}{cc}{l}^{(e)}& m^{(e)}\end{array}\right]}_{1x2}{\left\{\begin{array}{c}{d}_1^{(e)}\\ d_2^{(e)}\end{array}\right\}}_{2x1}$

Here we can see that $q_1^{(e)}$ is a 1x1 scalar.

We do this for node 2 as well. $q_2^{(e)}={\left[\begin{array}{cc}{l}^{(e)}& m^{(e)}\end{array}\right]}_{1x2}{\left\{\begin{array}{c}{d}_3^{(e)}\\ d_4^{(e)}\end{array}\right\}}_{2x1}$

which leads us to:

$\left\{\begin{array}{c}{q}_1^{(e)}\\ q_2^{(e)}\end{array}\right\}=\left[\begin{array}{cccc}{l}^{(e)}& m^{(e)}& 0& 0\\ 0& 0& l^{(e)}& m^{(e)}\end{array}\right]\left\{\begin{array}{c}{d}_1^{(e)}\\ d_2^{(e)}\\ d_3^{(e)}\\ d_4^{(e)}\end{array}\right\}$

where this is the equation we set out to derive, ${\mathbf{\text{q}}}_{2x1}^{(e)}{\mathbf{\text{T}}}_{2x4}^{(e)}={\mathbf{\text{d}}}_{4x1}^{(e)}$

Similarly, (same argument):
$\left\{\begin{array}{c}{P}_1^{(e)}\\ P_2^{(e)}\end{array}\right\}=I^{(e)}\left\{\begin{array}{c}{f}_1^{(e)}\\ f_2^{(e)}\\ f_3^{(e)}\\ f_4^{(e)}\end{array}\right\}$
Where P is a 2x1 matrix, T is a 2x4 matrix, and f is a 4x1 matrix.
This relationship is the same as saying
${\underset{\_}{P}}^{(e)}=I^{(e)}{\underset{\_}{f}}^{(e)}$
Recall the axial Fd relationship:

${\hat{k}}^{(e)}{q}^{(e)}=P^{(e)}$

where k is a 2x2 matrix, q is a 2x1 matrix, and P is a 2x1 matrix.

${\hat{k}}^{(e)}\underbrace{({\underset{\_}{T}}^{(e)}{\underset{\_}{d}}^{(e)})}=\underbrace{({\underset{\_}{T}}^{(e)}{\underset{\_}{f}}^{(e)})}$

$-{\underset{\_}{q}}^{(e)}{\underset{\_}{P}}^{(e)}$

Goal: We want to have k^(e)d^(e)=f^(e) so "move" T^e from right side to the left side by pre multiplying equation by T^(e)-1, the inverse of T^(e). Unfortunately, T^(e) is a rectangular matrix of the size 2x4 and cannot be inverted. Only square matricies can be inverted. To solve this issue, transpose T.
Ans:

$({\underset{\_}{T}}^{(e)T}{\underset{\_}{\hat{k}}}^{(e)}{\underset{\_}{T}}^{(e)}){\underset{\_}{d}}^{(e)}={\underset{\_}{f}}^{(e)}$

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

${\underset{\_}{k}}^{(e)}={\underset{\_}{T}}^{(e)T}{\underset{\_}{\hat{k}}}^{(e)}{\underset{\_}{T}}^{(e)}$

where k is on p6.1, k hat is on 12-2, and T is on 12-5.
Justification for (1): PVW see 10-1 for 1st applications of PVW, reduction of global FD relationship.

${\underset{\_}{K}}_{6x6}{d}_{6x1}={\underset{\_}{F}}_{6x1}\to {\underset{\_}{\overline{K}}}_{2x2}{\underset{\_}{\overline{d}}}_{2x1}={\underset{\_}{\overline{F}}}_{2x1}$

Remember: Why not solve as follows?

$\underset{\_}{d}={\underset{\_}{K}}^{-1}F$

Used mathcad to try to solve K^-1 and could not, due to singularity of K., i.e. the determinant of K=0 and thus K is not invertible. Recall that you need to computer $\begin{array}{c}\frac{1}{detK}\end{array}$ to find K^-1.

Why? For an unconstrained structure system, there are 3 possible rigid body mostion in 2-D (2 translational and 1 rotational).

HW: Find the eigenvalues of K and make observations about the number of eigenvalues.
Dyanmic eigenvalue problem Kv = λMv
Where K is the stiffness matrix, M is the mass.
Zero evaluation corresponds to zero stored elastic energy which means rigid body motion.

Using the Global Free-body Diagram Relationship

Using d₃, and d₄ in $\mathbf{\text{K}}\mathbf{\text{d}}=\mathbf{\text{F}}$ and using boundary conditions to eliminate rows in the global displacement matrix we get:

${\left[\begin{array}{cc}{K}_{1,3}& K_{1,4}\\ K_{2,3}& K_{2,4}\\ K_{3,3}& K_{3,4}\\ K_{4,3}& K_{4,4}\\ K_{5,3}& K_{5,4}\\ K_{6,3}& K_{6,4}\end{array}\right]}_{6x2}{\left\{\begin{array}{c}{d}_3\\ d_4\end{array}\right\}}_{2x1}={\left\{\begin{array}{c}{F}_1\\ F_2\\ F_3\\ F_4\\ F_5\\ F_6\end{array}\right\}}_{6x1}$

Note: We really only need to do the computations for rows 1,2,5, and 6 to get F₁,F₂,F₅,and F₆ because computation from rows 3 and 4 gives the applied load which is already known.

MATLAB Coding Showing Computations

K= $\left[\begin{array}{cc}-\frac{9}{16}& -\frac{3\sqrt{3}}{16}\\ \frac{-3\sqrt{3}}{16}& -\frac{3}{16}\\ 3.0625& -2.1752\\ -2.1752& 2.6875\\ -\frac{5}{2}& \frac{5}{2}\\ \frac{5}{2}& -\frac{5}{2}\end{array}\right]$ d= $\left\{\begin{array}{c}4.352\\ 6.1271\end{array}\right\}$ MatLab Code: >> K = [-.5625,(-3*sqrt(3))/16;(-3*sqrt(3))/16,-.1875;3.0625,-2.1752;-2.1752,2.6875;-2.5,2.5;2.5,-2.5]; >> d = [4.352;6.1271]; >> f = K*d f = -4.4378 -2.5622 0.0003 7.0001 4.4377 -4.4377 This gives us:      F= $\left\{\begin{array}{c}-4.4378\\ -2.5622\\ 0.0003\\ 7.0001\\ 4.4377\\ -4.4377\end{array}\right\}$

Closing the Loop between FEM and Statics: Virtual Displacement

Two-bar truss system:

File:HW-3Picture3.jpg

Since our two-bar truss system is statically determinant because we can view element 1 and 2 as two force bodies. By statics, the reactions are known and therefore, so are the member forces $P_1^{(1)}$, and $P_1^{(1)}$.

We then compute axial displacement degrees of freedom. This is the amount of extension within the bars.

$q_2^{(1)}=\frac{P_1^{(1)}}{k^{(1)}}=\frac{P_2^{(1)}}{k^{(1)}}=AC$

$q_1^{(1)}=0$    (Node 1 is fixed)

$q_2^{(1)}=\frac{{-P}_2^{(2)}}{k^{(2)}}=AB$

$q_2^{(2)}=0$    (Node 2 is fixed)

How do we back out the displacement DOFs of node 2 from above results?

File:HW-3Picture4.jpg
Note: The displacement of node 2 is in the direction of vector D.

Statics reason that previous method cannot be done with arcs of a compass
File:HW-3Picture5.jpg $U_y=R\sin \alpha \cong R\alpha$  (if alpha is small) $U_x=R(1-\cos \alpha )\cong 0$  (first order) File:HW-3Picture6.jpg

Infinitesimal displacement (related to virtual displacement)

File:Infinitesimaldisp1.JPG

Global Free-Body Diagram showing Infinitesimal Displacements

$\overline{AC}=\frac{|P_2^{(1)}|}{k^{(1)}}=\frac{5.1243}{3/4}=6.8324$

$\overline{AB}=\frac{|P_1^{(1)}|}{k^{(2)}}=\frac{6.2762}{5}=1.2552$

Find (x,y) coordinates of B and C:

Point B:

$x=\overline{AB}cos135^{\circ }=1.2552*-\frac{\sqrt{2}}{2}=-0.8876$

$y=\overline{AB}sin135^{\circ }=1.2552*\frac{\sqrt{2}}{2}=0.8876$

Point C:

$x=\overline{AC}cos30^{\circ }=6.8324*\frac{\sqrt{3}}{2}=5.917$

$y=\overline{AC}sin30^{\circ }=6.8324*\frac{1}{2}=3.4162$

We now have two unknowns, (X_D,Y_D)

We need the equations for line $\overline{AB}$ and $\overline{BC}$

Method for the Determination of the Slope between Two Arbitrary Points

File:Pqpicture.JPG

PQ Line Segment Diagram

$\overline{PQ}=(PQ)\tilde{i}=(PQ)[cos\theta \overrightarrow{i}+sin\theta \overrightarrow{j}]=(x-x_P)\overrightarrow{i}+(y-y_P)\overrightarrow{j}$

$x-x_P=(PQ)cos\theta$

$y-y_P=(PQ)sin\theta$

$\frac{y-y_P}{x-x_P}=tan\theta$

$y-y_P=tan\theta (x-x_P)$

Equation for line perpendicular to PQ passing through P:

$y-y_P=tan(\theta +\frac{\pi }{2})(x-x_P)$

Determination of (X_D,Y_D)

Line Perpendicular to Point B:

$y-0.8876=tan225^{\circ }(x+0.8876)$

$y=x+1.7752$

Line Perpendicular to Point C:

$y-3.4162=tan120^{\circ }(x-5.917)$

$y=-1.732x+13.665$

Intersection at Point (X_D,Y_D):

$y_D+1.732(y_D-1.7752)=13.665$

$y_D=6.127$

$(x_D+1.7752)=-1.732x_D+13.665$

$x_D=4.352$

$\overrightarrow{AD}=(x_D-x_A)\overrightarrow{i}+(y_D-y_A)\overrightarrow{j}$

Simplification: $x_A=0,y_A=0$

$\overrightarrow{AD}=x_D\overrightarrow{i}+y_D\overrightarrow{j}$

$\overrightarrow{AD}=4.352\overrightarrow{i}+6.127\overrightarrow{j}$

$\overrightarrow{AD}=d_3\overrightarrow{i}+d_4\overrightarrow{j}$

$d_3=4.352$

$d_4=6.127$

3-bar Truss System

File:3bartruss.JPG

3-bar Truss System

$E^{(1)}=2,A^{(1)}=3,L^{(1)}=5$

$E^{(2)}=4,A^{(2)}=1,L^{(2)}=5$

$E^{(3)}=3,A^{(3)}=2,L^{(3)}=10$

${\theta }^{(1)}=30^{\circ },{\theta }^{(2)}=-30^{\circ },{\theta }^{(3)}=45^{\circ }$

$P=30$

Convenient Local Node Numbering:

File:3bartrusslocalnodes.JPG

3-bar Truss System Local Node Numbering

Template:Center top 3-Bar Truss System Template:Center bottom

File:Hw3-1.JPG

ΣF_x = 0
ΣF_y = 0
ΣM_A = 0 (Trivial)

→ 2 equations, 3 unknowns → Statically Indeterminate

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

File:Hw3-2.JPG

$\overline{M_B}=\overline{BA}\times \bar{F}=\overline{B{A}^{\prime }}\times \bar{F}$

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

$\overline{M_B}=(\overline{BA}+\overline{A{A}^{\prime }})\times \bar{F}=\overline{B{A}^{\prime }}\times \bar{F}+\overline{A{A}^{\prime }}\times \bar{F}$

Back to 3-bar truss:

Node A is in equilibrium:

$\sum _{i=0}^3\overline{F_i}=\bar{0}$

$\sum _i\overline{M_B,i}=\sum _i\overline{BA{\text{'}}_i}\times \overline{F_i}$

A_i' = any point on line of action of F_i

$\sum _i\overline{M_B,i}=\sum _i\overline{BA{\text{'}}_i}\times \overline{F_i}=\overline{BA}\times \sum _{i=0}\overline{F_i}=\bar{0}$

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

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

MatLab Homework

Part of this assignment required the team to develop a MATLAB code that plots the deformed and un-deformed configurations of the two-bar truss system solved in class. The code is based on two models that were created by Dr. Vu-Quoc and X.G. Tan.

MATLAB Code

%HW 3 %Assignment: %Write a matlab code to plot the initial underformed %configuration of the two-bar truss system solved in class %and then superpose the deformed configuration on the same figure. function truss_plot dof = 2; %number of dofs per node n_node = 3; %number of nodes n_elem = 2; %number of elements total_dof = 2*n_node; %total number of dofs %******************************************************************* % UNDEFORMED BEAM %******************************************************************* %coordinates of undeformed position position(:,1) = [0;0]; position(:,2) = [2*sqrt(3);2]; position(:,3) = [2*sqrt(3)+sqrt(2);2-sqrt(2)]; %set up the nodal coordinate arrays for undeformed beams for i = 1 : n_node x(i) = position(1,i); y(i) = position(2,i); end %define elements %element 1 node_connect(1, 1) = 1; node_connect(2, 1) = 2; %element 2 node_connect(1, 2) = 2; node_connect(2, 2) = 3; %******************************************************************* % DEFORMED BEAM %******************************************************************* %the deformed beam variables are denoted with a '2' %coordinates of deformed position %(NOTE: node 1 and node 2 remain undeformed) position2(:,1) = [0;0]; position2(:,2) = [2*sqrt(3)+4.352;2+6.1271]; position2(:,3) = [2*sqrt(3)+sqrt(2);2-sqrt(2)]; %set up the nodal coordinate arrays for deformed beams for i = 1 : n_node x2(i) = position2(1,i); y2(i) = position2(2,i); end %define elements %element 1 node_connect2(1, 1) = 1; node_connect2(2, 1) = 2; %element 2 node_connect2(1, 2) = 2; node_connect2(2, 2) = 3; %******************************************************************* % CONNECT BEAMS AND PRINT %******************************************************************* for i = 1 : n_elem; %undeformed beam node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [x(node_1),x(node_2)]; yy = [y(node_1),y(node_2)]; %deformed beam node_1_deformed = node_connect2(1,i); node_2_deformed = node_connect2(2,i); xx_deformed = [x2(node_1_deformed),x2(node_2_deformed)]; yy_deformed = [y2(node_1_deformed),y2(node_2_deformed)]; %plot hold on plot(xx,yy,'--b') plot(xx_deformed,yy_deformed,'-r') end end

File:Plot for truss.GIF

The blue dashed line represents the un-deformed truss, while the solid red line shows the truss after it has been deformed.

Authors

Eml4500.f08.wiki1.oatley 20:02, 8 October 2008 (UTC)

Eml4500.f08.wiki1.brannon 01:31, 8 October 2008 (UTC)

Eml4500.f08.wiki1.ambrosio 21:27, 7 October 2008 (UTC)

Eml4500.f08.wiki1.aguilar 20:45, 8 October 2008 (UTC)

Eml4500.f08.wiki1.schaet 19:42, 8 October 2008 (UTC)

Eml4500.f08.wiki1.handy 20:25, 8 October 2008 (UTC)