User:EML4500.f08.A-team.VandenBerg/HW3

 Comment: This is HW3 located in the user namespace of EML4500.f08.A-team.VandenBerg 19:11, 8 October 2008 (UTC)

HW3
12-1

Meeting 12: Monday, 22 September 2008 EML4500
Topic to be covered in class is the derivation of the element force displacement with respect to the global coordinate system

Derivation of the Element Force Displacement With Respect To the Global Coordinate System
(From page 6-1)

(1) $$\underset{4x4}{k}^{(e)}$$ * $$\underset{4x1}{d}^{(e)}$$ = $$\underset{4x1}{f}^{(e)}$$



12-2 page 4-5

(2) $$k^{(e)} \begin{bmatrix} 1     & -1 \\  -1     & 1 \end{bmatrix} \begin{pmatrix} q_1^{(e)} \\ q_2^{(e)}\end{pmatrix} = \begin{matrix} P_1^{(e)} \\ P_2^{(e)}\end{matrix}$$

where:

$$k^{(e)} \begin{bmatrix} 1     & -1 \\  -1     & 1 \end{bmatrix}$$ = $$\vec{k}^{(e)}$$

qi(e) is equal to the axial displacement of the element (e) at the local node "i". pi(e) is equal to the axial force of the element (e) at the local node "i".

Goal: The goal is to derive equation (1) from equation (2) which was derived previously in meeting 4.

To do this we want to ind the relation between:

$$\underset{2x1}{q}^{(e)}$$ and $$\underset{4x1}{d}^{(e)}$$ $$\underset{2x1}{p}^{(e)}$$ and $$\underset{4x1}{f}^{(e)}$$ 12-3 This relation can also be stated in the form of $$\underset{2x1}{q}^{(e)}$$ = $$\underset{2x4}{T}^{(e)}$$ $$\underset{4x1}{d}^{(e)}$$

Displacement Vector of the Local Node
Considering the displacement vector of the local node "1", also noted in the form of $$d_1^{(e)}$$



where $$\vec{\tilde{i}}$$, $$\vec{i}$$, and $$\vec{j}$$ is the displacement of the local node number.

The displacement vector of element e at local node number is: $$\vec{d}^{(e)}_{1} = d^{(e)}_1 \vec{i} + d_{2}^{(e)} \vec{j}$$

Axial displacement
The orthogonal projection of the displacement vector $$\vec{d}_1^{(e)}$$ of node 1 on the $$\tilde{x}$$ axis of the element e is known as: $$q_1^{(e)}$$ 12-4 $$q_1^{(e)} = d_1^{(e)}$$ • $$\vec{\tilde{i}}$$
 * = ($$d_1^{(e)} \vec{i} + d_2^{(e)} \vec{j}$$) • $$\vec{\tilde{i}}$$
 * = $$d_1^{(e)}$$ ($$\vec{i}$$ • $$\vec{\tilde{i}}$$) + $$d_2^{(e)}$$ ($$\vec{j}$$ • $$\vec{\tilde{i}}$$)

($$\vec{i}$$ • $$\vec{\tilde{i}}$$) = cos$$\theta^{(e)}$$

($$\vec{j}$$ • $$\vec{\tilde{i}}$$) = sin$$\theta^{(e)}$$

$$q_1^{(e)} = \mathit{l}^{(e)} d_1^{(e)} + m^{(e)} d_2^{(e)} \Rightarrow$$ 1x1 Scalar
 * = $$\llcorner L\mathit{l}^{(e)} m^{(e)} \lrcorner$$ 1x1 matrix $$\begin{Bmatrix}

d_1^{(e)} \\ d_2^{(e)} \end{Bmatrix}$$ 2x1 matrix

12-5

$$\begin{Bmatrix} q_1^{(e)} \\ q_2^{(e)} \end{Bmatrix} = \begin{bmatrix} \mathit{l}^{(e)} & m^{(e)} & 0 & 0    \\ 0 & 0 & \mathit{l}^{(e)} & m^{(e)} \end{bmatrix} \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \\ d_3^{(e)} \\ d_4^{(e)} \end{Bmatrix}$$

$$\underset{2x1}{q}^{(e)}$$ = $$\underset{2x4}{T}^{(e)}$$ $$\underset{4x1}{d}^{(e)}$$

14-1

Meeting 14: Friday, 26 September 2008, EML4500

 * (Meeting 13: Exam 1, Wednesday, 24 September)

Using the same argument as the equation above, you can get:

$$\begin{Bmatrix} p_1^{(e)} \\ p_2^{(e)} \end{Bmatrix}$$ 2x1 = $$T_{2x4}^{(e)}\begin{Bmatrix} f_1^{(e)} \\ f_2^{(e)} \\ f_3^{(e)} \\ f_4^{(e)} \end{Bmatrix}$$ 4x1

Where:

$$\begin{Bmatrix} p_1^{(e)} \\ p_2^{(e)} \end{Bmatrix}$$ 2x1 = $$p^{(e)} = T^{(e)} f^{(e)}$$

Remember the element axial FD rel.: $$\vec{k}$$ * $$\underset{2x1}{q}^{(e)}$$ = $$\underset{2x1}{p}^{(e)}$$

where:

$$\vec{k}$$ 2x2 is a 2x2 vector 14-2

Inverse Matrice
$$\Rightarrow \vec{k}^{(e)} (T^{(e)} d^{(e)}) = (T^{(e)} f^{(e)})$$

Where:

$$(T^{(e)} d^{(e)}) = q^{(e)}$$

$$(T^{(e)} f^{(e)}) = p^{(e)}$$

Goal: We want to have $$k^{(e)} d^{(e)} = f^{(e)}$$ so, we must "move" $$T^{(e)}$$ from the right hand side of the equation to left hand side of the equation by pre-multiplying the equation by $$T^{(e)}$$-1 which is the inverse of $$T^{(e)}$$.

Unfortunately though, $$T^{(e)}$$ is a rectangular matrix, which means it cannot be inverted.

Answer: $$\begin{bmatrix} T^{(e)^{T} } & \vec{k}^{(e)} & T^{(e)} \end{bmatrix} d^{(e)} = f^{(e)} $$

$$\begin{bmatrix} 4x2 & 2x2 & 2x4 \end{bmatrix} 4x1 = 4x1 $$

$$\begin{bmatrix} 4x2 & 2x2 & 2x4 \end{bmatrix} = 4x4 $$

14-3 $$\Rightarrow k^{(e)} d^{(e)} = f^{(e)}$$

Justification for equation (1): $$\mathbf{P}\mathbf{V}\mathbf{W}$$ (this will be covered later)

For the first application of $$\mathbf{P}\mathbf{V}\mathbf{W}$$, see page 10-1 in the notes.

Reduction of Global FD Relationship
$$\underset{6x6}{k}^{(e)}$$ * $$\underset{6x1}{d}^{(e)}$$ = $$\underset{6x1}{f}^{(e)} \Rightarrow \vec{K} \vec{d} = \vec{F}$$

Where:

$$\vec{K}$$ is a 2x2

$$\vec{d}$$ is a 2x2

$$\vec{F}$$ is a 2x1

Rem: Why not solve for the displacement by setting $$d = K^{-1} f$$? 14-4 Because of the singularity of K, the determinant is equal to zero, and thus is not invertible

There are three possible rigid body motions in 2-D for an unconstrained structural system, one rotational motion and two translational motions.

Dynamic Evaluation Point
Kv = λMv

Where:

K = Stiffness matrix M = Mass material eval. (~vibration frequency)

zero eval. corresp. zero stored elastic energy14-5

$$ \Rightarrow $$ rigid body modes

Modes: mode shapes
 * eight vectors

Meeting 15: Monday, 29 September 2008, EML4500
To compute reactions two methods are imployed:

1)Using element force-displacement relations

2)Using global force-displacement relations (described below)

Using Global Free Body Relationships to Compute Reactions
Using force-displacement relations from meeting 10:

(6x2 matrix)x(2x1 mat.)=(6x1 mat.):

$$\begin{bmatrix} K_{13} &  K_{14} \\ K_{23} &  K_{24} \\ K_{33} &  K_{34} \\ K_{43} &  K_{44} \\ K_{53} &  K_{54} \\ K_{63} &  K_{64} \\ \end{bmatrix} \begin{Bmatrix} d_{3}\\ d_{4}\\ \end{Bmatrix} = \begin{Bmatrix} F_{1}\\ F_{2}\\ F_{3}\\ F_{4}\\ F_{5}\\ F_{6}\\ \end{Bmatrix} $$

Only necessary to do computations for rows 1,2,5,6 to get F1,F2,F5,F6 (reactions)

What "the Loop" Is
Clarifying the loop between the F.E.M. and statics:

For two-bar truss system:

FEM --> Compute displacement --> Compute reactions

Statics --> Compute reactions --> Compute displacement

The step in statics from computing reactions to computing displacement is what we define as "closing the loop"

Closing the Loop
FBD's of two-force bodies (elem.'s 1,2):

Figure 8: Axes Forces in Elem. 1

Figure 9: Axes Forces in Elem. 2

By statics, reactions known and therefore the member forces: P1(1), P2(2)

Computing axial displ. dof's (amount of extension of bars):

$$q2^1 = \frac{P1^1}{k^1}=\frac{P2^1}{k^1} $$, $$q1^2 = \frac{-P2^2}{k^2}$$

Fixed node at 1 & 2 means: q1^2 = q2^2 = 0

To back out from above results the displ. dof's of node 2:



Eml4500.f08.a-team.robinson 03:46, 21 October 2008 (UTC)

Infinitesimal displacement


2 unknowns (XD YD)





PQ = (PQ)i^{~} = (X-X_{P})i + (Y-Y_{P})j $$



(PQ)i^{~} = PQ[ \cos{\theta}i + \sin{\theta}j ] $$

Therefore



X - X_{P} = PQ\cdot \cos{\theta} $$



Y - Y_{P} = PQ\cdot \sin{\theta} $$



\frac {Y-Y_{P}}{X-X_{P}} = \tan{\theta} $$



Y-Y{P} = (\tan{\theta})\cdot(X-X_{P}) $$

Equation for a line perpendicular to above line, passing P:



Y-Y_{P} = \tan{(\theta + \frac{\pi}{2})}(X-X{P}) $$

 Three-bar Truss System 



Local node numbering by convenience in assembly of K:



EML4500.f08.A-team.kirley 20:55, 8 October 2008 (UTC)

Meeting 17: Friday, 3 October 2008, EML4500
ΣFx = 0, ΣFy = 0, ΣM A = 0 (sum of moments trivial)

Q: How about ΣM B ?



3D explanation

ΣM B  = BA X F

ΣM B  = BA' X F

for A' on linen of action of F

BA = BA + AA'

M B  = (BA + AA') X F

M B  = BA X F + AA' X F

where = AA' X F = 0

Back to 3-Bar Truss:

Node A is in equilibrium : ΣF i  = 0 from i=0 to i=3

Σ i M Bi  = Σ i BA' i  X F i 

A' i = any point on the line of action of F i 

Σ i M Bi  = Σ i BA' i  X F i 

Σ i M Bi  = BA X Σ i F i  = 0 - meaning ΣM B is useless, still only 2 useful equations.



K 33 = K(1) 33 + K(2) 11 + K(3) 11

K 34 = K(1) 34 + K(2) 12 + K(3) 12

HW: Fill out the rest!

The filled out matrix would appear as follows:

$${K}= \begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0 & 0 & 0\\ k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0 & 0 & 0\\ k_{31}^{(1)} & k_{32}^{(1)} & k_{33}^{(1)}+k_{11}^{(2)}+k_{11}^{(3)} & k_{34}^{(1)}+k_{12}^{(2)}+k_{12}^{(3)} & k_{13}^{(2)} & k_{14}^{(2)} & k_{13}^{(3)} & k_{14}^{(3)} \\k_{41}^{(1)} & k_{42}^{(1)} & k_{43}^{(1)}+k_{21}^{(2)}+k_{21}^{(3)} & k_{44}^{(1)}+k_{22}^{(2)}+k_{22}^{(3)} & k_{23}^{(2)} & k_{24}^{(2)} & k_{23}^{(3)} & k_{24}^{(3)}\\ 0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)} & 0 & 0 \\0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)} & 0 & 0 \\ 0 & 0 & k_{31}^{(3)} & k_{32}^{(3)} & 0 & 0 & k_{33}^{(3)} & k_{34}^{(3)}\\ 0 & 0 & k_{41}^{(3)} & k_{42}^{(3)} & 0 & 0 & k_{43}^{(3)} & k_{44}^{(3)} \end{bmatrix}$$

Homework 3 Problems
9/26/08:

Show that the k matrix is equal to the T matrix times the transpose of the T matrix times the $$\hat {k}$$

First we show that:

$$\hat {k}^{(e)} = k^{(e)}  \begin{bmatrix} 1 & -1 \\ -1 & 1 \\ \end{bmatrix}$$

Letting the k matrix be made up of k11, k12, etc. and creating it as a 2 by 2 matrix we get:

$$\hat {k}^{(e)} = \begin{bmatrix} k_{11} & -k_{12} \\ -k_{21} & k_{22} \\ \end{bmatrix}$$

and that T is:

$$T^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0     \\ 0 & 0 & l^{(e)} & m^{(e)} \\ \end{bmatrix}$$

The transpose of T becomes:

$$T^{(e)^T} = \begin{bmatrix} l^{(e)} & 0\\ m^{(e)} & 0\\ 0 & l^{(e)}\\ 0 & m^{(e)}\\ \end{bmatrix}$$

Finally, multiplying everything together produces:

$$\begin{bmatrix} k_{11} & k_{12} \\ k_{21} & k_{22} \\ \end{bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0     \\ 0 & 0 & l^{(e)} & m^{(e)} \\ \end{bmatrix}\begin{bmatrix} k_{11} & -k_{12} \\ -k_{21} & k_{22} \\ \end{bmatrix}\begin{bmatrix} l^{(e)} & 0\\ m^{(e)} & 0\\ 0 & l^{(e)}\\ 0 & m^{(e)}\\ \end{bmatrix}$$

Therefore:

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

Find the eigenvalues of the matrix K and comment about the number of zero eigenvalues:

Using MatLab to solve for the eigenvalues we get:

K =

0.5625   0.3248   -0.5625   -0.3248         0         0    0.3248    0.2500   -0.3248   -0.2500         0         0   -0.5625   -0.3248    3.0625   -2.1752   -2.5000    2.5000   -0.3248   -0.2500    2.8248    1.9375    2.5000   -2.5000         0         0   -2.5000    2.5000    2.5000   -2.5000         0         0    2.5000   -2.5000   -2.5000    2.5000

EDU>> eig(K)

ans =

8.4194   2.5880   -0.1730   -0.0218   -0.0000    0.0000

Which shows that two of the eigenvalues are zero.

9/29/08:

Using the global FD relation:

Solve for F1, F2, F5, and F6 and compare results with method 1.

The force matrix f can be solved for by multiplying the global stiffness matrix K by the displacement matrix d, i.e.

$$\begin{bmatrix} k_{11} & k_{12} & k_{13} & k_{14} & k_{15} & k_{16}\\ k_{21} & k_{22} & k_{23} & k_{24} & k_{25} & k_{26}\\ k_{31} & k_{32} & k_{33} & k_{34} & k_{35} & k_{36}\\ k_{41} & k_{42} & k_{43} & k_{44} & k_{45} & k_{46}\\ k_{51} & k_{52} & k_{53} & k_{54} & k_{55} & k_{56}\\ k_{61} & k_{62} & k_{63} & k_{64} & k_{65} & k_{66}\\ \end{bmatrix} \cdot \begin{bmatrix}d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6\end{bmatrix} = \begin{bmatrix}F_1 \\ F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6 \end{bmatrix}$$

Where the K matrix has the same values as the problem above, and the values of the d matrix are such that d1 = d2 = d5 = d6 = 0 and d3 = 4.352 and d4 = 6.127.

After substituting all the numbers in and then simplifying by removing columns 1, 2, 5, and 6 of the K matrix and also rows 1, 2, 5, and 6 of the d matrix, and finally using MatLab to solve for the force matrix f we find that:

k =

-0.5625  -0.3248   -0.3248   -0.1875    3.0625   -2.1752   -2.1752    2.6875   -2.5000    2.5000    2.5000   -2.5000

EDU>> F=k*d

F =

-4.4380  -2.5623    0.0005    6.9998    4.4375   -4.4375

Where F3 and F4 were already known as the applied loads, but where computed for completeness, and the reactions are the other forces shown in the matrix.

In comparison to method 1 (element FD relationship), the numbers are nearly the same, and actually most likely would be had numbers not been truncated by MatLab and the human being who did this problem for simplicity

10/1/08:

Infinitesimal displacement of AC and AB.

These are found by dividing the P matrix by the k matrix

$$AC = \left ( \frac{|P^{(1)}_2|}{k^{(1)}} \right ) = \left ( \frac{5.1243}{3/4} \right ) = 6.8324$$

$$AB = \left ( \frac{|P^{(2)}_1|}{k^{(2)}} \right ) = \left ( \frac{6.276}{5} \right ) = 1.2552$$

Find (XB, YB), (XC, YC), and (XD, YD).

These can be found by multiplying the lengths AB and AC by their respective director cosines as follows:

$$X_B = AB\times\cos{(135)} = 1.2552\cos{(135)} = 0.88756$$

$$Y_B = AB\times\sin{(135)} = 1.2552\sin{(135)} = 0.88756$$

$$X_C = AC\times\cos{(30)} = 6.8324\cos{(30)} = 5.91703$$

$$Y_C = AC\times\sin{(30)} = 6.8324\sin{(30)} = 3.4162$$

Solving for (XD, YD) is not quite as simple however and requires a little bit more work, namely using the following equation for a line perpendicular to another line and passing through P, this equation's derivation should be found in the lecture notes:

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

$$\overrightarrow{A D} \ = (X_D - x_i)\vec i + (Y_D - y_i)\vec j$$

Since xi and yi drop out (equal to zero) because they are at the origin we get:

$$\overrightarrow{A D} \ = X_D\vec i + Y_D\vec j$$

Therefore:

$$\overrightarrow{A D} \ = d_3\vec i + d_4\vec j$$

Thus XD = d3 and YD = d4 or XD = 4.35 and YD = 6.125

--EML4500.f08.A-team.morford 18:21, 8 October 2008 (UTC)

Contributions for HW3
--EML4500.f08.A-team.rieth 02:39, 8 October 2008 (UTC)

--Eml4500.f08.a-team.robinson 03:33, 8 October 2008 (UTC)

--EML4500.f08.A-team.morford 18:24, 8 October 2008 (UTC)

EML4500.f08.A-team.VandenBerg 19:11, 8 October 2008 (UTC)

EML4500.f08.A-team.kirley 20:55, 8 October 2008 (UTC)

--EML4500.f08.A-team.melvin 21:02, 8 October 2008 (UTC)