机械振动学：二自由度振动有阻尼及强迫振动响应-陈 (1)

(0) S x (0) P M 2 x (0) r
•With r(t) known, use the inverse transform to recover the physical solution:
x(t) M
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q(t) M
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Pr(t) Sr(t)
Experiments do not give C. They provide zeta (in modal coordinates) by the half power method. Compute the solution assuming modal damping of:
z1 0.01 and z 2 0.1
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Compute qFra Baidu bibliotekt), Transform back
•To get the proper initial conditions use:
(0) M 2 x (0) q(0) M 2 x(0), and q
1 1
•Use the above to compute q(t) and then:
The C and K matrices have the same form. It follows from the system itself that consisted damping and stiffness elements in a similar manner.
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A Question of matrix decoupling
Compute the modal initial conditions:
0.205 0.263 1.846 1 SL P r0 S x 0 0 . 394 0 . 308 2 . 365 0 0 r
1
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Compute the modal solutions:
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Compute the modal decomposition
L =sqrt(M)
3 0 ~ 1 1 0.667 0.333 L , K L KL 0 2 0 . 333 0 . 500 0.615 0.788 ~ Kv v 1 0.240, v1 , and 2 0.947, v 2 0 . 788 0 . 615 0.615 0.788 P 0 . 788 0 . 615
T
1
1
• In general, these can not be decoupled since K and C can not be diagonalized simultaneously
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A Little Matrix Theory
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More Matrix Stuff and Normal Mode Systems
Then use x(t )=S r(t )
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0.205 0.263 r1 (t ) x(t ) Sr(t ) r (t ) 0.394 0.308 2
x1 (t ) 0.863e 0.004896t sin( 0.49t 1.561) 0.88e 0.096t sin( 0.958t 1.471) x2 (t ) 1.658e 0.004896t sin( 0.49t 1.561) 1.029e 0.096t sin( 0.958t 1.471)
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Matrix form of Equations of Motion:
1 (t ) c1 c2 c2 x 1 (t ) x m1 0 0 m x x ( t ) c c ( t ) 2 2 2 2 2 k1 k2 k 2 x1 (t ) 0 k2 x2 (t ) 0 k2
•
–
Estimate damping due to viscoelasticity using
some approximation method
Model the damping mechanism directly (hard and still
an area of research-good for physicists but engineers
Subject to
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Generic Example:
• If the damping mechanisms are known then • Sum forces to find the equations of motion
Free Body Diagram:
1 c1 x
2 x 1 ) c2 ( x 2 x 1 ) c2 ( x
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Modal Damping (cont)
• Decouple system based on M and K, i.e., use the “undamped” modes • Attribute some zi (zeta) to each mode of the decoupled system (a guess. Not known beforehand. Can be tested with gross data like x):
If the system of equations decouple then the methods of SDOF can be applied
SDOF
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With the modal equation in hand the general solution is given
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The applied force is distributed across the all of the modes except in a special case.
So, first separate solutions in the modal coordinates were found and then the modes were assembled by the use of S. The response in the physical coordinates is therefore a combination of the modal responses just as in the undamped case.
Here the mode shapes and eigenvectors are real valued and form orthonormal sets, even for repeated natural frequencies 1 1 ~ 2 (known because K M KM 2 is symmetric)
x(t ) M
1 2
q(t )
the response in physical coordinates.
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Example
Consider:
9 0 6 2 x x0 0 4 2 2
1 Subject to initial conditions: x0 0 0 0 x 0
Free and Forced Vibration Response of Two Degree of Freedom Systems with Damping
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Systems with Viscous Damping
• Extending the previous sections to include the effects of viscous damping (dashpots)
q(t ) d i e z iit sin di t i vi
i 1
n
where
2 i
M
1
2
KM
T i
1
2
vi v , and di i 1 z
2 i i T i
v q(0) di v q(0) 1 di and i tan T T sin i vi q(0) + z ii vi q(0)
• Can we decouple the system with the same coordinate transformations as before?
Cx Kx 0 M x
2 r 0 I r P M CM Pr diagonal? 2
here
2 i i ri 0 r i 2z ii r z i i t ri (t ) Ai e sin dit i
di i 1 z i2
z i i t r ( t ) e ( Ai sin di t Bi cos di t) Alternately: i
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Viscous Damping in MDOF Systems
• Two basic choices for including damping
–
Modal Damping
• Attribute some amount to each mode based on experience, i.e., an artful guess or
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Proportional Damping
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Proportional Damping (cont)
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Forced Response: the response of an 2 dof system to a forcing term
k1 c1 F1 x1 k2 c2 F2 x2
m1
m2
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Lumped Damping models
• In some cases (FEM, machine modeling), the damping matrix is determined directly from the equations of motion. • Then our analysis must start with:
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Transform Back to Get Physical Solution
•Use modal transform to obtain modal initial conditions and compute Ai and Fi:
r(0) S x(0) P M 2 x(0)
T T 1 1 1 1
Modal Damping by Mode Summation
• • • Can also use mode summation approach Again, modes are from undamped system The higher the frequency, the smaller the effect (because of the exponential term). So just few first modes are enough.
need models that are correct enough).
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Modal Damping Method
Solve the undamped vibration problem as before
(t ) r(t ) 0 (t ) Kx(t ) 0 I M x r
z1 0.01,z 2 0.1, 1 0.49, d 1 0.49, 2 0.963 , d 2 0.958
Yields:
r1 (t ) 4.208e
0.004896 t 0.096t
sin( 0.49t 1.561)
r2 (t ) 3.346e
sin( 0.958t 1.471)