User:Eml4500.f08.wiki1.aguilar/TeamHW3

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): $$ \begin{Bmatrix} P_1^{(e)} \\ P_2^{(e)} \end{Bmatrix} = I^{(e)} \begin{Bmatrix} f_1^{(e)} \\ f_2^{(e)} \\ f_3^{(e)} \\ f_4^{(e)}\end{Bmatrix} $$ b Where P is a 2x1 matrix, I 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 K v = λ M  v  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 $$ \textbf{K}\textbf{d}=\textbf{F} $$ 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 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 $$P^{(1)}_1$$, 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, (XD,YD)

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 (XD,YD)
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 (XD,YD):

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

$$y_D=6.127$$

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

$$x_D=4.352$$

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

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



ΣFx = 0

ΣFy = 0

ΣMA = 0 (Trivial)

→ 2 equations, 3 unknowns → Statically Indeterminate

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



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

(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:

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

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

Ai' = any point on line of action of Fi

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



\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)