# University of Florida/Eml4500/f08 Team Homework 3

## Homework 3

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

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

${\displaystyle {k}^{(e)}{\begin{bmatrix}1&-1\\-1&1\end{bmatrix}}{\begin{Bmatrix}{q}_{1}^{(e)}\\{q}_{2}^{(e)}\end{Bmatrix}}={\begin{Bmatrix}{P}_{1}^{(e)}\\{P}_{2}^{(e)}\end{Bmatrix}}}$

${\displaystyle {q}_{i}^{(e)}=}$  axial displacement of element 'e' at local node 'i'
${\displaystyle {P}_{i}^{(e)}=}$  axial force of element 'e' at local node 'i'

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

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

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

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

${\displaystyle {q}_{1}^{(e)}}$  = axial displacement of node 1 is the prthagonal projection of the displacement vector ${\displaystyle {\vec {d}}_{1}^{(e)}}$  of node 1 on the ${\displaystyle {\tilde {x}}}$  axis of element 'e'.

${\displaystyle {q}_{1}^{(e)}={\vec {d}}_{1}^{(e)}+{\vec {\tilde {i}}}}$

${\displaystyle {q}_{1}^{(e)}=({d}_{1}^{(e)}{\vec {i}}+{d}_{2}^{(e)}{\vec {j}})+{\vec {\tilde {i}}}}$

${\displaystyle {q}_{1}^{(e)}={d}_{1}^{(e)}({\vec {i}}\cdot {\vec {\tilde {i}}})+({d}_{2}^{(e)}({\vec {j}}\cdot {\vec {\tilde {i}}})}$

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

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

${\displaystyle {q}_{1}^{(e)}=l^{(e)}{d}_{1}^{(e)}+m^{(e)}{d}_{2}^{(e)}}$

${\displaystyle {q}_{1}^{(e)}={\begin{bmatrix}l^{(e)}&m^{(e)}\end{bmatrix}}_{1x2}{\begin{Bmatrix}{d}_{1}^{(e)}\\{d}_{2}^{(e)}\end{Bmatrix}}_{2x1}}$

Here we can see that ${\displaystyle {q}_{1}^{(e)}}$  is a 1x1 scalar.

We do this for node 2 as well. ${\displaystyle {q}_{2}^{(e)}={\begin{bmatrix}l^{(e)}&m^{(e)}\end{bmatrix}}_{1x2}{\begin{Bmatrix}{d}_{3}^{(e)}\\{d}_{4}^{(e)}\end{Bmatrix}}_{2x1}}$

${\displaystyle {\begin{Bmatrix}{q}_{1}^{(e)}\\{q}_{2}^{(e)}\end{Bmatrix}}={\begin{bmatrix}l^{(e)}&m^{(e)}&0&0\\0&0&l^{(e)}&m^{(e)}\end{bmatrix}}{\begin{Bmatrix}{d}_{1}^{(e)}\\{d}_{2}^{(e)}\\{d}_{3}^{(e)}\\{d}_{4}^{(e)}\end{Bmatrix}}}$

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

Similarly, (same argument):
${\displaystyle {\begin{Bmatrix}P_{1}^{(e)}\\P_{2}^{(e)}\end{Bmatrix}}=I^{(e)}{\begin{Bmatrix}f_{1}^{(e)}\\f_{2}^{(e)}\\f_{3}^{(e)}\\f_{4}^{(e)}\end{Bmatrix}}}$
Where P is a 2x1 matrix, T is a 2x4 matrix, and f is a 4x1 matrix.
This relationship is the same as saying
${\displaystyle {\underline {P}}^{(e)}=I^{(e)}{\underline {f}}^{(e)}}$
Recall the axial Fd relationship:

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

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

${\displaystyle {\hat {k}}^{(e)}\underbrace {({\underline {T}}^{(e)}{\underline {d}}^{(e)})} =\underbrace {({\underline {T}}^{(e)}{\underline {f}}^{(e)})} }$

${\displaystyle -\qquad {\underline {q}}^{(e)}\qquad \qquad {\underline {P}}^{(e)}}$

Goal: We want to have k(e)d(e)=f(e) so "move" Te 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:

${\displaystyle ({\underline {T}}^{(e)T}{\underline {\hat {k}}}^{(e)}{\underline {T}}^{(e)}){\underline {d}}^{(e)}={\underline {f}}^{(e)}}$

${\displaystyle {\underline {k}}^{(e)}{\underline {d}}^{(e)}={\underline {f}}^{(e)}}$

${\displaystyle {\underline {k}}^{(e)}={\underline {T}}^{(e)T}{\underline {\hat {k}}}^{(e)}{\underline {T}}^{(e)}}$

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.

${\displaystyle {\underline {K}}_{6x6}d_{6x1}={\underline {F}}_{6x1}\rightarrow {\underline {\bar {K}}}_{2x2}{\underline {\bar {d}}}_{2x1}={\underline {\bar {F}}}_{2x1}}$

Remember: Why not solve as follows?

${\displaystyle {\underline {d}}={\underline {K}}^{-1}F}$

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 ${\displaystyle {\begin{matrix}{\frac {1}{detK}}\end{matrix}}}$  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 d3, and d4 in ${\displaystyle {\textbf {K}}{\textbf {d}}={\textbf {F}}}$  and using boundary conditions to eliminate rows in the global displacement matrix we get:

${\displaystyle {\begin{bmatrix}K_{1,3}&K_{1,4}\\K_{2,3}&K_{2,4}\\K_{3,3}&K_{3,4}\\K_{4,3}&K_{4,4}\\K_{5,3}&K_{5,4}\\K_{6,3}&K_{6,4}\end{bmatrix}}_{6x2}{\begin{Bmatrix}d_{3}\\d_{4}\end{Bmatrix}}_{2x1}={\begin{Bmatrix}F_{1}\\F_{2}\\F_{3}\\F_{4}\\F_{5}\\F_{6}\end{Bmatrix}}_{6x1}}$

Note: We really only need to do the computations for rows 1,2,5, and 6 to get F1,F2,F5,and F6 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 ${\displaystyle P_{1}^{(1)}}$ , and ${\displaystyle P_{1}^{(1)}}$ .

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

${\displaystyle {q}_{2}^{(1)}={\frac {{P}_{1}^{(1)}}{{k}^{(1)}}}={\frac {{P}_{2}^{(1)}}{{k}^{(1)}}}=AC}$

${\displaystyle {q}_{1}^{(1)}=0}$     (Node 1 is fixed)

${\displaystyle {q}_{2}^{(1)}={\frac {{-P}_{2}^{(2)}}{{k}^{(2)}}}=AB}$

${\displaystyle {q}_{2}^{(2)}=0}$     (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

${\displaystyle {\overline {AC}}={|P_{2}^{(1)}| \over k^{(1)}}={5.1243 \over 3/4}=6.8324}$

${\displaystyle {\overline {AB}}={|P_{1}^{(1)}| \over k^{(2)}}={6.2762 \over 5}=1.2552}$

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

Point B:

${\displaystyle x={\overline {AB}}cos{135^{\circ }}=1.2552*-{{\sqrt {2}} \over 2}=-0.8876}$

${\displaystyle y={\overline {AB}}sin{135^{\circ }}=1.2552*{{\sqrt {2}} \over 2}=0.8876}$

Point C:

${\displaystyle x={\overline {AC}}cos{30^{\circ }}=6.8324*{{\sqrt {3}} \over 2}=5.917}$

${\displaystyle y={\overline {AC}}sin{30^{\circ }}=6.8324*{1 \over 2}=3.4162}$

We now have two unknowns, (XD,YD)

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

### Method for the Determination of the Slope between Two Arbitrary Points

PQ Line Segment Diagram

${\displaystyle {\overline {PQ}}=(PQ){\tilde {i}}=(PQ)[cos\theta {\vec {i}}+sin\theta {\vec {j}}]=(x-x_{P}){\vec {i}}+(y-y_{P}){\vec {j}}}$

${\displaystyle x-x_{P}=(PQ)cos\theta }$

${\displaystyle y-y_{P}=(PQ)sin\theta }$

${\displaystyle {y-y_{P} \over x-x_{P}}=tan\theta }$

${\displaystyle y-y_{P}=tan\theta (x-x_{P})}$

Equation for line perpendicular to PQ passing through P:

${\displaystyle y-y_{P}=tan(\theta +{\pi \over 2})(x-x_{P})}$

### Determination of (XD,YD)

Line Perpendicular to Point B:

${\displaystyle y-0.8876=tan{225^{\circ }}(x+0.8876)}$

${\displaystyle y=x+1.7752}$

Line Perpendicular to Point C:

${\displaystyle y-3.4162=tan{120^{\circ }}(x-5.917)}$

${\displaystyle y=-1.732x+13.665}$

Intersection at Point (XD,YD):

${\displaystyle y_{D}+1.732(y_{D}-1.7752)=13.665}$

${\displaystyle y_{D}=6.127}$

${\displaystyle (x_{D}+1.7752)=-1.732x_{D}+13.665}$

${\displaystyle x_{D}=4.352}$

${\displaystyle {\vec {AD}}=(x_{D}-x_{A}){\vec {i}}+(y_{D}-y_{A}){\vec {j}}}$

Simplification: ${\displaystyle x_{A}=0,y_{A}=0}$

${\displaystyle {\vec {AD}}=x_{D}{\vec {i}}+y_{D}{\vec {j}}}$

${\displaystyle {\vec {AD}}=4.352{\vec {i}}+6.127{\vec {j}}}$

${\displaystyle {\vec {AD}}=d_{3}{\vec {i}}+d_{4}{\vec {j}}}$

${\displaystyle d_{3}=4.352}$

${\displaystyle d_{4}=6.127}$

### 3-bar Truss System

3-bar Truss System

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

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

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

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

${\displaystyle P=30}$

Convenient Local Node Numbering:

3-bar Truss System Local Node Numbering

3-Bar Truss System

ΣFx = 0
ΣFy = 0
ΣMA = 0 (Trivial)

→ 2 equations, 3 unknowns → Statically Indeterminate

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

${\displaystyle {\overline {M_{B}}}={\overline {BA}}\times {\overline {F}}={\overline {BA'}}\times {\overline {F}}}$

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

${\displaystyle {\overline {M_{B}}}=({\overline {BA}}+{\overline {AA'}})\times {\overline {F}}={\overline {BA'}}\times {\overline {F}}+{\overline {AA'}}\times {\overline {F}}}$

Back to 3-bar truss:

Node A is in equilibrium:

${\displaystyle \sum _{i=0}^{3}{\overline {F_{i}}}={\overline {0}}}$

${\displaystyle \sum _{i}{\overline {M_{B},i}}=\sum _{i}{\overline {BA'_{i}}}\times {\overline {F_{i}}}}$

Ai' = any point on line of action of Fi

${\displaystyle \sum _{i}{\overline {M_{B},i}}=\sum _{i}{\overline {BA'_{i}}}\times {\overline {F_{i}}}={\overline {BA}}\times \sum _{i=0}{\overline {F_{i}}}={\overline {0}}}$

${\displaystyle {\begin{bmatrix}K_{11}&K_{12}&K_{13}&K_{14}&K_{15}&K_{16}&K_{17}&K_{18}\\K_{21}&K_{22}&K_{23}&K_{24}&K_{25}&K_{26}&K_{27}&K_{28}\\K_{31}&K_{32}&K_{33}&K_{34}&K_{35}&K_{36}&K_{37}&K_{38}\\K_{41}&K_{42}&K_{43}&K_{44}&K_{45}&K_{46}&K_{47}&K_{48}\\K_{51}&K_{52}&K_{53}&K_{54}&K_{55}&K_{56}&K_{57}&K_{58}\\K_{61}&K_{62}&K_{63}&K_{64}&K_{65}&K_{66}&K_{67}&K_{68}\\K_{71}&K_{72}&K_{73}&K_{74}&K_{75}&K_{76}&K_{77}&K_{78}\\K_{81}&K_{82}&K_{83}&K_{84}&K_{85}&K_{86}&K_{87}&K_{88}\end{bmatrix}}}$
${\displaystyle ={\begin{bmatrix}k_{11}^{(1)}&k_{12}^{(1)}&k_{13}^{(1)}&k_{14}^{(1)}&0&0&0&0\\k_{21}^{(1)}&k_{22}^{(1)}&k_{23}^{(1)}&k_{24}^{(1)}&0&0&0&0\\k_{31}^{(1)}&k_{32}^{(1)}&(k_{33}^{(1)}+k_{11}^{(2)}+k_{11}^{(3)})&(k_{34}^{(1)}+k_{12}^{(2)}+k_{12}^{(3)})&k_{13}^{(2)}&k_{14}^{(2)}&k_{15}^{(3)}&k_{16}^{(3)}\\k_{41}^{(1)}&k_{42}^{(1)}&(k_{43}^{(1)}+k_{21}^{(2)}+k_{21}^{(3)})&(k_{44}^{(1)}+k_{22}^{(2)}+k_{22}^{(3)})&k_{23}^{(2)}&k_{24}^{(2)}&k_{25}^{(3)}&k_{26}^{(3)}\\0&0&k_{31}^{(2)}&k_{32}^{(2)}&k_{33}^{(2)}&k_{34}^{(2)}&0&0\\0&0&k_{41}^{(2)}&k_{42}^{(2)}&k_{43}^{(2)}&k_{44}^{(2)}&0&0\\0&0&k_{51}^{(3)}&k_{52}^{(3)}&0&0&k_{55}^{(3)}&k_{56}^{(3)}\\0&0&k_{61}^{(3)}&k_{62}^{(3)}&0&0&k_{65}^{(3)}&k_{66}^{(3)}\end{bmatrix}}}$

### 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)