Forced two degree of freedom systems
Today’s Objectives:
Students will be able to:
a) Find the steady state
response of 2DOF forced
systems
b) Use Simulink for MDOF
systems
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Rose-Hulman Institute of Technology
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Finish Example from last time
m, J0
k2
k1
L1
⎡m 0 ⎤ ⎧⎪ &x&G ⎫⎪ ⎡ k1 + k 2
⎢
⎥⎨ ⎬ + ⎢
⎢⎣ 0 J 0 ⎥⎦ ⎪⎩θ&& ⎪⎭ ⎢⎣− L1k1 + L2 k 2
L2
Let’s use the displacement at both
ends as our coordinates
Note:
dPsys x
dt
dLsys0
dt
x2
So
x1
− L1k1 + L2 k 2 ⎤ ⎧⎪ xG ⎫⎪ ⎧⎪0⎫⎪
⎥⎨ ⎬ = ⎨ ⎬
2
2
k1 L1 + k 2 L2 ⎥⎦ ⎪⎩θ ⎪⎭ ⎪⎩0⎪⎭
sinθ ≈ θ =
xG − x1 x2 − xG
=
L1
L2
= ∑ F ⇒ m&x&G = ∑ F
= ∑ M 0 ⇒ J Gθ&& = ∑ M 0
x2 − x1
L1 + L 2
so x G =
L2
L1
x1 +
x2
L1 + L 2
L1 + L 2
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Finish Example from last time
dPsys x
x2
dt
= ∑ F ⇒ m&x&G =
x1
mL2
mL1
&x&1 +
&x&2 = − k1 x − k 2 x2
L1 + L 2
L1 + L 2
mL2
mL1
&x&1 +
&x&2 + k1 x + k 2 x2 = 0
L1 + L 2
L1 + L 2
k2x2
k1x1
dLsysG
dt
= ∑ M 0 ⇒ J 0θ&& = J 0
&x&2 − &x&1
= k1 x1 L1 − k 2 x2 L2
L1 + L2
J0
J0
&x&2 −
&x&1 − k1 x1 L1 + k 2 x2 L2 = 0
L1 + L2
L1 + L2
In matrix form we get
⎡ mL2
⎢ L +L
2
⎢ 1
⎢− J 0
⎢⎣ L1 + L2
mL1 ⎤
L1 + L2 ⎥ ⎧⎪ &x&2 ⎫⎪ ⎡ k1
⎥
+⎢
J 0 ⎥ ⎨⎪ &x& ⎬⎪ ⎢− k L
⎩ 1⎭ ⎣ 1 1
L1 + L2 ⎥⎦
k 2 ⎤ ⎧⎪ x1 ⎫⎪
⎥⎨ ⎬ = 0
k 2 L2 ⎥⎦ ⎪⎩ x2 ⎪⎭
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Rose-Hulman Institute of Technology
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Notes
Finish Example from last time - Matlab
evec1 =
-0.0315 0.0032
0.0059 0.0574
eval1 =
4981.3
0
0 6785.3
freq =
70.5785
82.3732
evec2 =
-1.0000 -0.5187
-0.7780 1.0000
Vibrations
eval2 =
4981.3
0
0 6785.3
freq2 =
70.5785
Rose-Hulman Institute of Technology
82.3732
Mechanical Engineering
Forced vibrations – 2 DOF systems
Consider the 2-DOF system with the following EOM:
⎡m1
⎢
⎢⎣ 0
0 ⎤ ⎧⎪ &x&1 ⎫⎪ ⎡ k11
⎥⎨ ⎬ + ⎢
m2 ⎥⎦ ⎪⎩ &x&2 ⎪⎭ ⎢⎣− k12
− k12 ⎤ ⎧⎪ x1 ⎫⎪ ⎧⎪ F1 ⎫⎪
⎥ ⎨ ⎬ = ⎨ ⎬ sin ωt
k 22 ⎥⎦ ⎪⎩ x2 ⎪⎭ ⎪⎩ 0 ⎪⎭
We know there will be a particular solution (satisfies right hand side) and a
homogeneous solution (satisfied RHS = 0)
{x} = {x}c + {x}p
Let’s look at the particular solution. Since the RHS is a “sin” let’s assume
⎧⎪ x1 ⎫⎪ ⎧⎪ X 1 ⎫⎪
⎨ ⎬ = ⎨ ⎬ sin ωt
⎪⎩ x2 ⎪⎭ ⎪⎩ X 2 ⎪⎭
Substituting into EOM we get
⎡k11 − m1ω 2
⎢
⎢⎣ − k12
⎤ ⎧⎪ X 1 ⎫⎪ ⎧⎪ F1 ⎫⎪
⎥⎨ ⎬ = ⎨ ⎬
2
k 22 − m2ω ⎥⎦ ⎪⎩ X 2 ⎪⎭ ⎪⎩ 0 ⎪⎭
− k12
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Steady-state response (cont.)
From previous page
⎡k11 − m1ω 2
⎢
⎢⎣ − k12
⎤ ⎧⎪ X 1 ⎫⎪ ⎧⎪ F1 ⎫⎪
⎥⎨ ⎬ = ⎨ ⎬
2
k 22 − m2ω ⎥⎦ ⎪⎩ X 2 ⎪⎭ ⎪⎩ 0 ⎪⎭
− k12
[Z (ω )]
So, we can solve for {X1 X2}T by premultiplying by [Z(ω)]-1
⎧⎪ X 1 ⎫⎪
⎧ F ⎫ adjo int [Z (ω )] ⎧⎪ F1 ⎫⎪
−1 ⎪ 1 ⎪
⎨ ⎬
⎨ ⎬ = [Z (ω )] ⎨ ⎬ =
det [Z (ω )] ⎪⎩ 0 ⎪⎭
⎪⎩ 0 ⎪⎭
⎪⎩ X 2 ⎪⎭
Quick matrix review of the inverse of a 2x2 matrix:
−1
⎡a b ⎤
1 ⎡ d − b⎤
⎢
⎥ =
⎢
⎥
ad
bc
−
⎢⎣ c d ⎥⎦
⎢⎣− c a ⎥⎦
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The determinant of [Z(ω)] is
det ([Z (w)]) = (k11 − m1ω 2 )(k 22 − m2ω 2 ) + k`212
This can be factored and written in terms of the natural
frequencies as
det ([Z (w)]) = m1m2 (ω12 − ω 2 )(ω22 − ω 2 )
(
)(
= m1m2 Ω12 − ω 2 Ω 22 − ω 2
So we get
X1 =
⎧⎪ X 1 ⎫⎪
1
=
⎨ ⎬
⎪⎩ X 2 ⎪⎭ Z (ω )
⎡k 22 − m2ω 2
⎢
⎢⎣
k12
)
⎤ ⎧⎪ F1 ⎫⎪
⎥⎨ ⎬
2
k 22 − m1ω ⎥⎦ ⎪⎩ 0 ⎪⎭
k12
X2 =
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Sketch these
(k − m ω )F
=
m m (Ω − ω )(Ω − ω )
2
X1
1
2
22
2
1
2
2
1
2
2
X2 =
2
X1
k12 F1
m1m2 Ω12 − ω 2 Ω 22 − ω 2
(
)(
)
X2
Ω1
Ω2
ω
Ω1
These are frequency response plots
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Rose-Hulman Institute of Technology
Mechanical Engineering
Ω2
ω
Summary
Homogeneous Solution
1. Find equations of motion:
[M ]{&x&}+ [K ]{x} = {0}
2DOF Steady State Solution
(with harmonic forcing)
1. Find equations of motion:
2. Assume simple harmonic motion
{x} = {X }eiωt
3. This gives
Natural frequency
(− ω 2 [M ] + [K ]){X } = {0}
4. Find natural frequencies
− ω 2 [M ]{x}+ [K ] = {0}
5. Substitute in natural frequencies
and find natural modes Natural
(− ωi2 [M ] + [K ]){X }i = {0}mode
6. Write homogeneous solution
[M ]{&x&}+ [K ]{x} = {F }sin ωt
2. To find the particular solution
assume:
{x} = {X }sin ωt
3. This gives
Known!
(− ω [M ] + [K ]){X } = {F }
2
4. Solve for {X}
Forcing frequency
{X } = (− ω 2 [M ] + [K ])−1{F }
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Magnitude of sinωt
There is a nice website in Australia
http://www.mech.uwa.edu.au/bjs/Vibration/TwoDOF/default.html
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Rose-Hulman Institute of Technology
Mechanical Engineering
What do we do if we have damping?
If we have
[M ]{&x&}+ [C ]{x&}+ [K ]{x} = {F }
(1s 4
[M4]4
+ s[C ] + [K ]){X (s )} = {F (s )}
2444
3
2
If we take the Laplace Transform
⎡
⎢⎣
B (s )⎤⎥⎦
so {X (s )} = [B(s )]−1{F (s )} = [H (s )]{F (s )}
If we let s = jω
[H ( jω )] = frequency response function matrix
We can determine this from
experimental data and use it for
system ID
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Rose-Hulman Institute of Technology
Mechanical Engineering
Using Simulink
Sometimes you have to add small
damping to get rid of transient solution
0.5
0
-0.5
0
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0
50
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0
50
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1
0
-1
0.5
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-0.5
1
0
-1
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Rose-Hulman Institute of Technology
Mechanical Engineering