# University of Florida/Eml4500/f08 Team Homework 3
## Homework 3
### Derivation of Elemental Free-body Diagram with respect to the Global Coordinate System

       ${\textbf {k}}_{4x4}^{(e)}{\textbf {d}}_{4x1}^{(e)}={\textbf {f}}_{4x1}^{(e)}$
${k}^{(e)}{\begin{bmatrix}1&-1\\-1&1\end{bmatrix}}{\begin{Bmatrix}{q}_{1}^{(e)}\\{q}_{2}^{(e)}\end{Bmatrix}}={\begin{Bmatrix}{P}_{1}^{(e)}\\{P}_{2}^{(e)}\end{Bmatrix}}$
${q}_{i}^{(e)}=$ 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 ${\textbf {q}}^{(e)}$ and ${\textbf {d}}^{(e)}$, and ${\textbf {P}}^{(e)}$ and ${\textbf {F}}^{(e)}$.
These relationships can be expressed in the form: ${\textbf {q}}_{2x1}^{(e)}{\textbf {T}}_{2x4}^{(e)}={\textbf {d}}_{4x1}^{(e)}$
Consider the displacement vector of the local node 1 denoted by ${\vec {d}}_{1}^{(e)}$.
${\vec {d}}_{[1]}^{(e)}={d}_{1}^{(e)}{\vec {i}}+{d}_{1}^{(e)}{\vec {j}}$
${q}_{1}^{(e)}$ = axial displacement of node 1 is the prthagonal projection of the displacement vector ${\vec {d}}_{1}^{(e)}$ of node 1 on the ${\tilde {x}}$ axis of element 'e'.
${q}_{1}^{(e)}={\vec {d}}_{1}^{(e)}+{\vec {\tilde {i}}}$
${q}_{1}^{(e)}=({d}_{1}^{(e)}{\vec {i}}+{d}_{2}^{(e)}{\vec {j}})+{\vec {\tilde {i}}}$
${q}_{1}^{(e)}={d}_{1}^{(e)}({\vec {i}}\cdot {\vec {\tilde {i}}})+({d}_{2}^{(e)}({\vec {j}}\cdot {\vec {\tilde {i}}})$
$({\vec {i}}\cdot {\vec {\tilde {i}}})=\cos \theta ^{(e)}=l^{(e)}$
$({\vec {i}}\cdot {\vec {\tilde {i}}})=\sin \theta ^{(e)}=m^{(e)}$
${q}_{1}^{(e)}=l^{(e)}{d}_{1}^{(e)}+m^{(e)}{d}_{2}^{(e)}$
${q}_{1}^{(e)}={\begin{bmatrix}l^{(e)}&m^{(e)}\end{bmatrix}}_{1x2}{\begin{Bmatrix}{d}_{1}^{(e)}\\{d}_{2}^{(e)}\end{Bmatrix}}_{2x1}$
Here we can see that ${q}_{1}^{(e)}$ is a 1x1 scalar.
We do this for node 2 as well. ${q}_{2}^{(e)}={\begin{bmatrix}l^{(e)}&m^{(e)}\end{bmatrix}}_{1x2}{\begin{Bmatrix}{d}_{3}^{(e)}\\{d}_{4}^{(e)}\end{Bmatrix}}_{2x1}$
which leads us to:
${\begin{Bmatrix}{q}_{1}^{(e)}\\{q}_{2}^{(e)}\end{Bmatrix}}={\begin{bmatrix}l^{(e)}&m^{(e)}&0&0\\0&0&l^{(e)}&m^{(e)}\end{bmatrix}}{\begin{Bmatrix}{d}_{1}^{(e)}\\{d}_{2}^{(e)}\\{d}_{3}^{(e)}\\{d}_{4}^{(e)}\end{Bmatrix}}$
where this is the equation we set out to derive, ${\textbf {q}}_{2x1}^{(e)}{\textbf {T}}_{2x4}^{(e)}={\textbf {d}}_{4x1}^{(e)}$
Similarly, (same argument): ${\begin{Bmatrix}P_{1}^{(e)}\\P_{2}^{(e)}\end{Bmatrix}}=I^{(e)}{\begin{Bmatrix}f_{1}^{(e)}\\f_{2}^{(e)}\\f_{3}^{(e)}\\f_{4}^{(e)}\end{Bmatrix}}$
Where P is a 2x1 matrix, T is a 2x4 matrix, and f is a 4x1 matrix.
This relationship is the same as saying ${\underline {P}}^{(e)}=I^{(e)}{\underline {f}}^{(e)}$ 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.
${\hat {k}}^{(e)}\underbrace {({\underline {T}}^{(e)}{\underline {d}}^{(e)})} =\underbrace {({\underline {T}}^{(e)}{\underline {f}}^{(e)})}$
$-\qquad {\underline {q}}^{(e)}\qquad \qquad {\underline {P}}^{(e)}$
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:
$({\underline {T}}^{(e)T}{\underline {\hat {k}}}^{(e)}{\underline {T}}^{(e)}){\underline {d}}^{(e)}={\underline {f}}^{(e)}$
${\underline {k}}^{(e)}{\underline {d}}^{(e)}={\underline {f}}^{(e)}$
${\underline {k}}^{(e)}={\underline {T}}^{(e)T}{\underline {\hat {k}}}^{(e)}{\underline {T}}^{(e)}$
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.
${\underline {K}}_{6x6}d_{6x1}={\underline {F}}_{6x1}\rightarrow {\underline {\bar {K}}}_{2x2}{\underline {\bar {d}}}_{2x1}={\underline {\bar {F}}}_{2x1}$
Remember: Why not solve as follows?
${\underline {d}}={\underline {K}}^{-1}F$
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 ${\begin{matrix}{\frac {1}{detK}}\end{matrix}}$ 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 d$_3$, and d$_4$ in ${\textbf {K}}{\textbf {d}}={\textbf {F}}$ and using boundary conditions to eliminate rows in the global displacement matrix we get:
${\begin{bmatrix}K_{1,3}&K_{1,4}\\K_{2,3}&K_{2,4}\\K_{3,3}&K_{3,4}\\K_{4,3}&K_{4,4}\\K_{5,3}&K_{5,4}\\K_{6,3}&K_{6,4}\end{bmatrix}}_{6x2}{\begin{Bmatrix}d_{3}\\d_{4}\end{Bmatrix}}_{2x1}={\begin{Bmatrix}F_{1}\\F_{2}\\F_{3}\\F_{4}\\F_{5}\\F_{6}\end{Bmatrix}}_{6x1}$
Note: We really only need to do the computations for rows 1,2,5, and 6 to get F$_1$,F$_2$,F$_5$,and F$_6$ because computation from rows 3 and 4 gives the applied load which is already known.

### Closing the Loop between FEM and Statics: Virtual Displacement

Two-bar truss system:
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?
Note: The displacement of node 2 is in the direction of vector D.

### Infinitesimal displacement (related to virtual displacement)

Global Free-Body Diagram showing Infinitesimal Displacements
${\overline {AC}}={|P_{2}^{(1)}| \over k^{(1)}}={5.1243 \over 3/4}=6.8324$
${\overline {AB}}={|P_{1}^{(1)}| \over k^{(2)}}={6.2762 \over 5}=1.2552$
Find (x,y) coordinates of B and C:
Point B:
$x={\overline {AB}}cos{135^{\circ }}=1.2552*-{{\sqrt {2}} \over 2}=-0.8876$
$y={\overline {AB}}sin{135^{\circ }}=1.2552*{{\sqrt {2}} \over 2}=0.8876$
Point C:
$x={\overline {AC}}cos{30^{\circ }}=6.8324*{{\sqrt {3}} \over 2}=5.917$
$y={\overline {AC}}sin{30^{\circ }}=6.8324*{1 \over 2}=3.4162$
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

PQ Line Segment Diagram
${\overline {PQ}}=(PQ){\tilde {i}}=(PQ)[cos\theta {\vec {i}}+sin\theta {\vec {j}}]=(x-x_{P}){\vec {i}}+(y-y_{P}){\vec {j}}$
$x-x_{P}=(PQ)cos\theta$
$y-y_{P}=(PQ)sin\theta$
${y-y_{P} \over x-x_{P}}=tan\theta$
$y-y_{P}=tan\theta (x-x_{P})$
Equation for line perpendicular to PQ passing through P:
$y-y_{P}=tan(\theta +{\pi \over 2})(x-x_{P})$

### Determination of (X$_D$,Y$_D$)

Line Perpendicular to Point B:
$y-0.8876=tan{225^{\circ }}(x+0.8876)$
$y=x+1.7752$
Line Perpendicular to Point C:
$y-3.4162=tan{120^{\circ }}(x-5.917)$
$y=-1.732x+13.665$
Intersection at Point (X$_D$,Y$_D$):
$y_{D}+1.732(y_{D}-1.7752)=13.665$
$y_{D}=6.127$
$(x_{D}+1.7752)=-1.732x_{D}+13.665$
$x_{D}=4.352$
${\vec {AD}}=(x_{D}-x_{A}){\vec {i}}+(y_{D}-y_{A}){\vec {j}}$
Simplification: $x_{A}=0,y_{A}=0$
${\vec {AD}}=x_{D}{\vec {i}}+y_{D}{\vec {j}}$
${\vec {AD}}=4.352{\vec {i}}+6.127{\vec {j}}$
${\vec {AD}}=d_{3}{\vec {i}}+d_{4}{\vec {j}}$
$d_{3}=4.352$
$d_{4}=6.127$

### 3-bar Truss System

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:
3-bar Truss System Local Node Numbering
3-Bar Truss System
ΣF$_x$ = 0
ΣF$_y$ = 0
ΣM$_A$ = 0 (Trivial)
→ 2 equations, 3 unknowns → Statically Indeterminate
Question: How about M$_B$? (3-D Explanation)
${\overline {M_{B}}}={\overline {BA}}\times {\overline {F}}={\overline {BA'}}\times {\overline {F}}$
(for A' on line of action of \overline{F}
${\overline {M_{B}}}=({\overline {BA}}+{\overline {AA'}})\times {\overline {F}}={\overline {BA'}}\times {\overline {F}}+{\overline {AA'}}\times {\overline {F}}$
Back to 3-bar truss:
Node A is in equilibrium:
$\sum _{i=0}^{3}{\overline {F_{i}}}={\overline {0}}$
$\sum _{i}{\overline {M_{B},i}}=\sum _{i}{\overline {BA'_{i}}}\times {\overline {F_{i}}}$
A$_i$' = any point on line of action of F$_i$
$\sum _{i}{\overline {M_{B},i}}=\sum _{i}{\overline {BA'_{i}}}\times {\overline {F_{i}}}={\overline {BA}}\times \sum _{i=0}{\overline {F_{i}}}={\overline {0}}$
       ${\begin{bmatrix}K_{11}&K_{12}&K_{13}&K_{14}&K_{15}&K_{16}&K_{17}&K_{18}\\K_{21}&K_{22}&K_{23}&K_{24}&K_{25}&K_{26}&K_{27}&K_{28}\\K_{31}&K_{32}&K_{33}&K_{34}&K_{35}&K_{36}&K_{37}&K_{38}\\K_{41}&K_{42}&K_{43}&K_{44}&K_{45}&K_{46}&K_{47}&K_{48}\\K_{51}&K_{52}&K_{53}&K_{54}&K_{55}&K_{56}&K_{57}&K_{58}\\K_{61}&K_{62}&K_{63}&K_{64}&K_{65}&K_{66}&K_{67}&K_{68}\\K_{71}&K_{72}&K_{73}&K_{74}&K_{75}&K_{76}&K_{77}&K_{78}\\K_{81}&K_{82}&K_{83}&K_{84}&K_{85}&K_{86}&K_{87}&K_{88}\end{bmatrix}}$
       $={\begin{bmatrix}k_{11}^{(1)}&k_{12}^{(1)}&k_{13}^{(1)}&k_{14}^{(1)}&0&0&0&0\\k_{21}^{(1)}&k_{22}^{(1)}&k_{23}^{(1)}&k_{24}^{(1)}&0&0&0&0\\k_{31}^{(1)}&k_{32}^{(1)}&(k_{33}^{(1)}+k_{11}^{(2)}+k_{11}^{(3)})&(k_{34}^{(1)}+k_{12}^{(2)}+k_{12}^{(3)})&k_{13}^{(2)}&k_{14}^{(2)}&k_{15}^{(3)}&k_{16}^{(3)}\\k_{41}^{(1)}&k_{42}^{(1)}&(k_{43}^{(1)}+k_{21}^{(2)}+k_{21}^{(3)})&(k_{44}^{(1)}+k_{22}^{(2)}+k_{22}^{(3)})&k_{23}^{(2)}&k_{24}^{(2)}&k_{25}^{(3)}&k_{26}^{(3)}\\0&0&k_{31}^{(2)}&k_{32}^{(2)}&k_{33}^{(2)}&k_{34}^{(2)}&0&0\\0&0&k_{41}^{(2)}&k_{42}^{(2)}&k_{43}^{(2)}&k_{44}^{(2)}&0&0\\0&0&k_{51}^{(3)}&k_{52}^{(3)}&0&0&k_{55}^{(3)}&k_{56}^{(3)}\\0&0&k_{61}^{(3)}&k_{62}^{(3)}&0&0&k_{65}^{(3)}&k_{66}^{(3)}\end{bmatrix}}$

### 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.
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)