From the matrix can you show me how the value stiffness (k) calculated?

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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From the matrix can you show me how the value stiffness (k) calculated? 

6 for [k
ma -Ka+ K2X =0
mi
m, o
Ki+k2
- Ka
Ma
- K2
Ka
mamin By Eigenvawes *
Figen veeture.
Soivihg the
个
Ka (K2-Xx)
I= Edentiy mounix =
LCF Dynamic mamux [m'K]
Inl
ma
mi
m] =
> ml= adj m
mima
%3D
mi
1)
m2
K+K2
- K2
Kitka
K2
mi
IC]= [m] K] =
%3D
-K2
K2
K2
ma
m2
Then
-K2
2-Kitkz
mi
mi
K2
K2
m2
K2
K2
K2
> (a - Kitka
mi
M2
M2
2.
K2
2(kitk2),
Ke (Kitkz)
mim2
ma
mi
K2
m2
K2
Ka
ma
KitK2
mi
ml
Iwot
IBy soiving tue egn
Solving
we find the
K = m,2,
m1
う
Ka = m, 12
m2
Transcribed Image Text:6 for [k ma -Ka+ K2X =0 mi m, o Ki+k2 - Ka Ma - K2 Ka mamin By Eigenvawes * Figen veeture. Soivihg the 个 Ka (K2-Xx) I= Edentiy mounix = LCF Dynamic mamux [m'K] Inl ma mi m] = > ml= adj m mima %3D mi 1) m2 K+K2 - K2 Kitka K2 mi IC]= [m] K] = %3D -K2 K2 K2 ma m2 Then -K2 2-Kitkz mi mi K2 K2 m2 K2 K2 K2 > (a - Kitka mi M2 M2 2. K2 2(kitk2), Ke (Kitkz) mim2 ma mi K2 m2 K2 Ka ma KitK2 mi ml Iwot IBy soiving tue egn Solving we find the K = m,2, m1 う Ka = m, 12 m2
(4.31 x 10
Advanced Research in Natural Fibers, Vol. 2 No. 1 (2020) p. 15-20
First iteration.
Given (X)trielH
Therefore [K]|M] {X} = {Xnew}
Isolator pad
2.66 X 10-
The Largest (X) = 2.66 x 10-4
Nomalized (X)new
holator pad
Figure
FigureY
1.
Fig. 1- Stackable washing and dryer machine and
its two DOFS mass spring system
[2.66 x 10-41
2.66 x 10-412.66 × 10-4)
- H
Second iteration.
Given (X)triat =I
Therefore (K]|M] {X} = {Xnew}
%3D
15.97 >
The Largest {Xew) = 5.97 x 10-4
Nomalized (X}new
Fig. 2- Free body diagram of stackable washing and
dryer machine
[4.31 x 10-4
5.97 x 10-415.97 x 10-4]
After the free body diagram (refer Fig. 2) show all the related
direction and force applied, an equation of motion can be
derived as follows.
M,X, – k2x2 + (k2 + k,)x, = 0
M2X2 - k2x1 + k2x2 =0
5. Based on the above iteration, after multiply [K]-'|M] {X}
- {Xnew), the (Xnew} can be normalized by using the largest
Xpew
[k, +k2
-k2
k2
6. Then, check the converge and use for a new trial {Xaew}.
7. The calculation of {Xnew} is repeated until the largest Xmew
become constant where is the 7" iteration.
Then, the lowest natural frequency for both Neoprene and
Sorbothone materials using matrix iteration method.
1. Identify matrix [K] & [M]
8. Lastly, the lowest natural frequency is calculated by using
the formula given:
|M]= [93 01
lo 58l
1
[K]-[ 700000
l-350000 350000
-3500001
Xnormalized
2. To find the lowest natural frequency, turn [K] to [K]
5.11 x 10-4= 44.24 rad/s
[K]-1 = 2.86 x 10-6 2.86 x 10-6
12.86 x 10-6 5.71 x 10-6
2.2 Validation coding
The Mathlab coding used to find the lowest natural
frequency is shown as follows:
3. Define the initial trial vector as:
{X} -
n-input ('Enter dimension of the matrix, n:
A - zeros (n, n) ;
x - zeros (1,n);
y - zeros (1, n) :
tol - input ('Enter the tolerance, tol: '):
m - input ('Enter maximum number of iterations,
');
4. Multiply [K]-1|M] {X} = {Xnew}
[K]-'|M]-|2.66 x 10- 1.66 x 10-*
12.66 x 10- 3.31 x 10-)
m:
17
Published by FAZ Publishing
http://www.fazpublishing.com/arnf
Transcribed Image Text:(4.31 x 10 Advanced Research in Natural Fibers, Vol. 2 No. 1 (2020) p. 15-20 First iteration. Given (X)trielH Therefore [K]|M] {X} = {Xnew} Isolator pad 2.66 X 10- The Largest (X) = 2.66 x 10-4 Nomalized (X)new holator pad Figure FigureY 1. Fig. 1- Stackable washing and dryer machine and its two DOFS mass spring system [2.66 x 10-41 2.66 x 10-412.66 × 10-4) - H Second iteration. Given (X)triat =I Therefore (K]|M] {X} = {Xnew} %3D 15.97 > The Largest {Xew) = 5.97 x 10-4 Nomalized (X}new Fig. 2- Free body diagram of stackable washing and dryer machine [4.31 x 10-4 5.97 x 10-415.97 x 10-4] After the free body diagram (refer Fig. 2) show all the related direction and force applied, an equation of motion can be derived as follows. M,X, – k2x2 + (k2 + k,)x, = 0 M2X2 - k2x1 + k2x2 =0 5. Based on the above iteration, after multiply [K]-'|M] {X} - {Xnew), the (Xnew} can be normalized by using the largest Xpew [k, +k2 -k2 k2 6. Then, check the converge and use for a new trial {Xaew}. 7. The calculation of {Xnew} is repeated until the largest Xmew become constant where is the 7" iteration. Then, the lowest natural frequency for both Neoprene and Sorbothone materials using matrix iteration method. 1. Identify matrix [K] & [M] 8. Lastly, the lowest natural frequency is calculated by using the formula given: |M]= [93 01 lo 58l 1 [K]-[ 700000 l-350000 350000 -3500001 Xnormalized 2. To find the lowest natural frequency, turn [K] to [K] 5.11 x 10-4= 44.24 rad/s [K]-1 = 2.86 x 10-6 2.86 x 10-6 12.86 x 10-6 5.71 x 10-6 2.2 Validation coding The Mathlab coding used to find the lowest natural frequency is shown as follows: 3. Define the initial trial vector as: {X} - n-input ('Enter dimension of the matrix, n: A - zeros (n, n) ; x - zeros (1,n); y - zeros (1, n) : tol - input ('Enter the tolerance, tol: '): m - input ('Enter maximum number of iterations, '); 4. Multiply [K]-1|M] {X} = {Xnew} [K]-'|M]-|2.66 x 10- 1.66 x 10-* 12.66 x 10- 3.31 x 10-) m: 17 Published by FAZ Publishing http://www.fazpublishing.com/arnf
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