lecture_21

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Modal Mass Matrix
⎛ k11 0⎞ ⎜ ⎟ T X ] [ k ][ X ] = ⎜ O [ ⎟ = [K ] ⎜ ⎟ knn ⎠ ⎝0
For proportional damping:
Modal Stiffness Matrix
⎛ c11 0⎞ ⎜ ⎟ T X ] [ c ][ X ] = ⎜ O [ ⎟ = [C ] ⎜ ⎟ cnn ⎠ ⎝0
[ k ]{ x} = ω 2 [ m]{ x}
Yields “n” eigenvalues and “n” eigenvectors
&& = −ω 2 x x
classic eigenvalue problem
ωi2 for i = 1, n
Computer or calculator solution (MATLAB, Mathematica, etc.) Mathematica,
T T
3 4
ω 2 { X } j [ m ]{ X }i = { X }i [ k ]{ X } j j
T T
Subtract 3 – 4 In general

2 i
−ω2 j
){ X } [m]{ X } = {0}
T j i
ωi ≠ ω j
Also:
{ X } j [ m]{ X }i = {0}
Lecture 21 & 22 General MDOF System
MEEM 3700
5
MDOF: ωi and {X}i from Eigensolution
x [ m]{&&} + [ k ]{ x} = {0}
For Harmonic Motion:
2
( −ω [ m] + [ k ]) {x} = {0}
ωi = λ i
Substitute ωi or λi into original equations for mode shape solution.
Lecture 21 & 22 General MDOF System
MEEM 3700
7
Alternate Method
MDOF: ωi and {X}i from Eigensolution
1
ωi2
for i = 1, n
MEEM 3700
{ X }i
for i = 1, n
8
4
Orthogonality of Mode Shape Vectors
( −ω [ m] + [ k ]) {x} = {0}
2
Using the ith natural frequency and mode shape
MEEM 3700 4
2
MDOF: General Solution to Equation of Motion
1.) Determine natural frequencies (ωi) and mode (ω shapes {X}i from undamped Free Vibration. (eigensolution) eigensolution) 2.) Use orthogonality of mode shapes to transform Equations of Motion into “MODAL SPACE”. SPACE” 3.) Solve “n” single degree of freedom problems. 4.) Use mode shapes to transform back to physical space.
[ k ] − ω 2 [ m] = 0
results in an “n” order polynomial
an ( λ ) + an −1 ( λ )
n n −1
[ k ] − λ [ m] = 0
+ ... + a1 ( λ ) + a0 = 0
roots = λi for i = 1, n
natural frequencies
MEEM 3700 Mechanical Vibrations
Mohan D. Rao Chuck Van Karsen
Mechanical Engineering-Engineering Mechanics Michigan Technological University
Copyright 2003
Lecture 21 & 22 General MDOF Strix
MEEM 3700
12
6
Example
⎡10 0 0 ⎤ [ m] = ⎢ 0 5 0 ⎥ ⎢ ⎥ ⎢ 0 0 20 ⎥ ⎣ ⎦
⎡ 2 −1 0 ⎤ [c ] = ⎢ −1 3 −2⎥ ⎢ ⎥ ⎢ 0 −2 4 ⎥ ⎣ ⎦ 0 ⎤ ⎡ 2000 −1000 ⎢ −1000 3000 −2000 ⎥ [k ] = ⎢ ⎥ ⎢ 0 −2000 4000 ⎥ ⎣ ⎦
{ X }i [ m]{ X }i = M ii
T
Modal Mass Modal Stiffness
{ X }i [ k ]{ X }i = Kii
T
if [ c ] = α [ m ] + β [ k ]
Proportional Damping Modal Damping
{ X }i [c ]{ X }i = Cii
ωi2 { X } j [ m ]{ X }i = { X } j [ k ]{ X }i
T T
3 4
9
ω 2 { X }i [ m ]{ X } j = { X }i [ k ]{ X } j j
T T
Lecture 21 & 22 General MDOF System
MEEM 3700
Orthogonality of Mode Shape Vectors
Because of the symmetry of [k] , [m]
{ X } j [ k ]{ X }i = { X }i [ k ]{ X } j T T { X } j [ m]{ X }i = { X }i [ m]{ X } j
T T
∴ ωi2 { X } j [ m ]{ X }i = { X }i [ k ]{ X } j
Lecture 21 & 22 General MDOF System
MEEM 3700
1
Multiple Degree of Freedom Systems
Topics: 1.) Equations of Motion 2.) Eigenvalue Problem 3.) Orthogonality 4.) Modal Space 5.) Solving MDOF Problems in Simple Fashion
{ X }3 = ⎪−5.16⎪ ⎨ ⎬
⎪ ⎪ ⎩ 1.00 ⎭
13
Lecture 21 & 22 General MDOF System
MEEM 3700
∑F
xi
&& = mxi
i
Rotation:
Lecture 21 & 22 General MDOF System
∑ Mθ
&& = Jθi
3
MEEM 3700
MDOF: Equation of Motion
General Form n-DOF System n& x [ m]{&&} + [c ]{ x} + [ k ]{ x} = {F }
{X}i
for i = 1, n
The natural frequencies are the square roots of the eigenvalues: eigenvalues:
ωi = ωi2
The mode shapes are the corresponding eigenvectors.
ωi2 [ m]{ X }i = [ k ]{ X }i
ω 2 [ m ]{ X } j = [ k ] { X } j j
T
1
Using the jth natural frequency and mode shape 2
T
Pre-multiply 1 by { X } j and 2 by { X }i
( −ω [ m] + [ k ]) {x} = {0}
2
( λ [ k ] − [ m]) { x} = {0}
λ=
1
(λ [k ]
λ [ I ]{ x} = [ D ]{ x}
λi =
Lecture 21 & 22 General MDOF System
−1
[ k ] − [ k ] [ m]) { x} = {0}
ω1 = 9.15
{ X }1 = ⎪1.16⎪ ⎨ ⎬ { X }2
⎧1.00 ⎫
From ωi2 [ m ]{ X } = [ k ]{ X }
ω2 = 14.14
⎪ ⎪ ⎩1.00 ⎭ ⎧ 1.00 ⎫ ⎪ ⎪ = ⎨ 0.00 ⎬ ⎪ ⎪ ⎩ −0.50 ⎭ ⎧ 1.00 ⎫
ω3 = 26.76
Lecture 21 & 22 General MDOF System
MEEM 3700
2
1
MDOF: Equations of Motion
Equations of Motion Using Newton’s Second Law: Newton’