2.003SC
Recitation 12 Notes: Modal Analysis
Modal Analysis of Double Pendulum System - Problem Statement
Consider a double pendulum consisting of two masses and one spring as shown in the figure below. Note
that θ1 and θ2 are small-angle displacements.
The system’s equations of motion are,
m1 l1 2
0
0
m2 l2 2
θ¨1
θ¨2
c1 l1 2
0
0
c2 l2 2
+
θ̇1
θ̇2
+
m1 gl1 + kl1 2
−kl1 l2
−kl1 l2
m2 gl2 + kl2 2
θ1
θ2
=
F1 l1
F2 l2
or
M ẍ + Cẋ + Kx = F
For m1 = m2 = 0.5 kg, l1 = 1.1m, l2 = 1.0 m, k = 0.5 N/m, the M and K matrices above are
M=
0.605
0
0
0.5000
K=
6.005 −0.5500
−0.5500 5.4050
and the natural frequencies and natural modes are,
ω1
ω2
=
3.0445
3.3851
rad/sec
U=
1
1
0.7139 −1.6949
Assume that the system damping matrix is of the form
C = αM
Assume that the external forces are
F1 l1
F2 l2
=
0.2cos(ω1 t)
0
[N − m]
1
• Find U T M U , U T KU and U T F (the modal forces in terms of F1 l1 .)
• Check the natural frequencies
• Find α such that the damping ratio for mode 2 is 0.04 .
• Find the steady-state response in terms of the generalized coordinates θ1 and θ2 , due to the contri­
bution of the first mode only.
2
Modal Analysis of Double Pendulum System - Solution
We seek uncoupled equations,
U T M Uq̈ + U T CU q̇ + U T KU q = U T F = Q
U T M U , U T KU and U T F
T
U MU =
T
U KU =
T
U F =
0.8598
0
0
2.0414
7.9699
0
0
23.3918
1 0.7139
1 −1.6949
M1 0
0 M2
K1 0
0 K2
=
=
F1 l1
0
=
F1 l1
F1 l1
Damping
When the system damping matrix is of the form
C = αM
U T CU will be a diagonal matrix.
T
U CU =
αM1
0
0
αM2
=
C1 0
0 C2
We have two independent SDOF equations of motion
M1 q̈1 + C1 q̇1 + K1 q1 = Q1 = F1 l1
M2 q̈2 + C2 q̇2 + K2 q2 = Q2 = F1 l1
So
ζ1 =
ζ2 =
C1
α
=
2ω1 M1
2ω1
C2
α
=
= 0.04
2ω2 M2
2ω2
So
α = 0.04(2ω2 ) = 0.271
3
α
0.271
=
= 0.0445
2(3.0445)
2ω1
ζ1 =
Last bit
So,
θ1
θ2
= U q = [u](1) q1 (t) + [u](2) q2 (t)
q1 (t) = |q1 |cos(ω1 t − φ1 )
|q1 | = |Q1 ||Hq1 /Q1 (ω)|
|q1 | = |Q1 |
1
K1
[(1 −
ω2 2
)
ω12
1
+ (2ζ1 ωω1 )2 ] 2
Q1 1
F1 l1 1
=
K1 2ζ1
K1 2ζ1
|q1 | =
0.22 1
1
· ·
= 0.345
7.9699 2 0.04
|q1 | =
π
2
φ1 =
q1 (t) = 0.345cos(ω1 t −
π
)
2
The contribution of the first mode only is
θ1
θ2
1
0.7139
=
θ1
θ2
=
0.345cos(ω1 t −
0.345
0.2463
cos(ω1 t −
π
)
2
π
)
2
The contribution of the second mode is
θ1
θ2
=
1
−1.6949
q2 (t)
4
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2.003SC / 1.053J Engineering Dynamics
Fall 2011
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