two-degree-of-freedom system with translation & rotation unit 45 vibrationdata
TRANSCRIPT
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Two-degree-of-freedom System with Translation & Rotation
Unit 45 Vibrationdata
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Introduction Vibrationdata
• Rotational degree-of-freedom dynamic problem requires mass moment of inertia
• Measure, calculate, or estimate MOI
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Measure Inertia via Bifilar Pendulum Vibrationdata
R2d,L4
dgmJ
2n
2
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Two-Plane Spin Balance Vibrationdata
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Inertia & Radius of Gyration Vibrationdata
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Rectangular Prism Vibrationdata
Z
Y
X
a c
b
12
2c2bm
zI12
2c2am
yI12
2b2am
xI
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Inertia & Radius of Gyration Vibrationdata
2rmJ
The mass moment of inertia J
m is the mass
r is the radius of gyration -
The distance from an axis at which the mass of a body may be assumed to be concentrated and at which the moment of inertia will be equal to the moment of inertia of the actual mass about the axis, equal to the square root of the quotient of the moment of inertia and the mass.
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Avionics Component Vibrationdata
What is mass moment of inertia?
• Seldom if ever have measurements
• Could treat as solid block with constant mass density
• Or just estimate radius of gyration
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Two DOF Example Vibrationdata
m, J
k 1 k 1
L1
k 2
L2
x
m mass
J mass moment of inertia about the C.G.
k spring stiffness
L length from spring attachment to C.G.
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Free Body Diagram Vibrationdata
x
Reference
k2 (x+L2)k1 (x-L1)
L1
L2
xmF
JM
Derive two equations of motion
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Equations of Motion Vibrationdata
0
0x
L kL kL kL k
L kL kkkx
J0
0m2
222
112211
221121
2211 L kL k
The vibration modes have translation and rotation which will be coupled if
Assemble equation into matrix form.
0
JL kL kL kL k
L kL kmkkdet
2222
2112211
22112
21
The natural frequencies are found via the eigenvalue problem:
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Sample Values Vibrationdata
Variable Value Value
m 3200 lbm -
L1 4.5 ft 54 in
L2 5.5 ft 66 in
k1 2400 lbf / ft 200 lbf/in
k2 2600 lbf / ft 217 lbf/in
R 4.0 ft 48 in
From Thomson, Theory of Vibration with Applications
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Vibrationdata
vibrationdata > Structural Dynamics > Two-DOF System Natural Frequencies
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Vibrationdata
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Natural Frequency Results Vibrationdata
mass matrix 8.29 0 0 1.91e+04
stiffness matrix 417 1482 1482 1.504e+06
Natural Frequencies No. f(Hz)1. 1.1234 2. 1.4166 Modes Shapes (column format)
0.3445 0.0444 -0.0009 0.0072
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Vibrationdata
C.G.
Combined rotation and translation
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Vibrationdata
C.G.
Combined rotation and translation
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Apply Base Excitation Vibrationdata
• The corresponding base excitation problem is complicated because the base motion directly couples with each of the two different degree-of-freedom types
• Thus use an “enforced motion” approach
• Must add a base mass for this method
• Motion will then be enforced on the base degree-of-freedom
• The mass value of the base is arbitrary
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Three Dof System Vibrationdata
m2, J
k 1 k 1
L1
k 2
L2
x2
m1x1
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Free Body Diagram Vibrationdata
k2 (x2+L2 - x1)k1 (x2-L1 - x1)
L1
L2
k1 (x2-L1 - x1)k2 (x2+L2 - x1)
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Derive Equation of Motion Vibrationdata
11 xmF
22 xmF
JM
0
0
0
x
x
L kL kL kL kLk-Lk
L kL kkkkk-
Lk-Lkkk-kk
x
x
J00
0m0
00m
2
1
222
21122112211
22112121
22112121
2
1
2
1
Homogeneous for now
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Avionics Component Vibrationdata
Partition the matrices and vectors as follows
0
0
u
u
KK
KK
u
u
MM
MM
f
d
fffd
dfdd
f
d
fffd
dfdd
Subscripts
d is drivenf is free
f
d
u
uwhere the displacement vector is
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Transformation Matrix Vibrationdata
ff1
ddIT
0I
fd1
ff1 KKT
Transformation matrix
where
I is the identify matrix
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Decoupling via Transformation Matrix Vibrationdata
0
0
u
u
KK
KK
u
u
MM
MM
w
d
fffd
dfddT
w
d
fffd
dfddT
Substitute transformed displacement and pre-multiply mass and stiffness matrices by T as follows
w
d
f
d
u
u
u
u
Let
Intermediate goal is to decouple the partition-wise stiffness matrix.
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Apply Transformation Matrix Vibrationdata
0
0
u
u
K̂0
0K̂
u
u
m̂m̂
m̂m̂
w
d
ww
dd
w
d
wwwd
dwdd
ww
dd
ff1fffd
ffT
1df1ffT
1dffdT
1ddT
K̂0
0K̂
KTKK
KTKTKTKKTKK
wwwd
dwdd
ff1fffd
ffT
1df1ffT
1dffdT
1ddT
m̂m̂
m̂m̂
MTMM
MTMTMTMMTMM
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Avionics Component Vibrationdata
dwdwwwwww um̂uK̂um̂
w
d
f
du
u
u
u
The equation of motion is thus
The final displacement are found via
Non-homogenous term due to enforce acceleration
du
The equation of motion can solved either in the frequency or time domain.
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Reference Vibrationdata
• Too many equations for a Webinar!
• Further details given in:
T. Irvine, Spring-Mass System Subjected to Enforced Motion, Vibrationdata, 2015
• Similar to Craig-Bampton modeling
• Damping can be applied later as modal damping
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Sample Avionics Box with C.G. Offset Vibrationdata
Variable Value
m 5 lbm
L1 3 in
L2 5 in
k1 2000 lbf/in
k2 2000 lbf/in
R 2 in
Q= 10 for both modes
Calculate natural frequencies, modes, shapes and frequency response functions for base excitation
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Vibrationdata
vibrationdata > Structural Dynamics > Two-DOF System Base Excitation
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Avionics Component Vibrationdata
Next Enter Uniform Damping Q=10
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Natural Frequency Results Vibrationdata
mass matrix 0.01295 0 0 0.05181
stiffness matrix 4000 4000 4000 6.8e+04
Natural Frequencies No. f(Hz)1. 85.052 2. 183.93 Modes Shapes (column format)
8.6887 1.3065 -0.6532 4.3443
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Vibrationdata
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Transmissibility Vibrationdata
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Translational Acceleration Transmissibility Vibrationdata
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Rotational Acceleration Transmissibility Vibrationdata
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Relative Displacement Transmissibility Vibrationdata
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Angular Displacement Transmissibility Vibrationdata
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Vibrationdata
Import: SRS 2000G Acceleration & NAVMAT PSD Specification
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Vibrationdata
Apply the Navmat PSD as the base input.
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Angular Displacement Transmissibility Vibrationdata
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Vibrationdata
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Vibrationdata
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Angular Displacement Transmissibility Vibrationdata
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Angular Displacement Transmissibility Vibrationdata
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Angular Displacement Transmissibility Vibrationdata
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Angular Displacement Transmissibility Vibrationdata