From the matrix can you show me how the value stiffness (k) calculated?
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
Related questions
Question
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff330bad6-e473-41b2-adb7-e1c02e00f39b%2Fb0d8d738-eb13-409b-b637-ecc3ad351b16%2F73gn4r33i_processed.jpeg&w=3840&q=75)
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff330bad6-e473-41b2-adb7-e1c02e00f39b%2Fb0d8d738-eb13-409b-b637-ecc3ad351b16%2Fae4o67p_processed.jpeg&w=3840&q=75)
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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