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Two Degree of Freedom Systems
Dr./ Ahmed NagibDecember 15, 2015
Two Degree of Freedom Systems
Real systems can not be modelled as one degree of freedom system, and are modelled by using multiple degree of freedom systems.We will extend the previous chapters for two degree of freedom system.
Examples Two Degree of Freedom SystemsSystems that require two independent coordinates to describe their motion are called two degree of freedom systems. Some examples of systems having two degrees of freedom are shown.
Examples Two Degree of Freedom Systems
Examples Two Degree of Freedom Systems
Two Degree of Freedom – Undamped SystemExample 1
Two Degree of Freedom – Undamped SystemExample 1
Two Degree of Freedom – Undamped SystemExample 1
Two Degree of Freedom – Undamped SystemExample 1
Two Degree of Freedom – Undamped SystemExample 1
Two Degree of Freedom – Undamped SystemExample 1
Two Degree of Freedom – Undamped SystemExample 1
Two Degree of Freedom – Undamped SystemExample 1
Two Degree of Freedom – Undamped SystemExample 1
Two Degree of Freedom – Undamped SystemExample 1
Two Degree of Freedom – Undamped SystemExample 1
Two Degree of Freedom – Undamped SystemExample 2
Two Degree of Freedom – Undamped SystemExample 2
Two Degree of Freedom – Undamped SystemExample 2
Two Degree of Freedom – Undamped SystemExample 2
Two Degree of Freedom – Undamped SystemExample 2
Two Degree of Freedom – Undamped SystemExample 2
Two Degree of Freedom – Undamped SystemExample 2
Two Degree of Freedom – Undamped SystemExample 2
Two Degree of Freedom – Undamped SystemExample 2
Two Degree of Freedom – Undamped SystemExample 2
Two Degree of Freedom – Undamped SystemExample 2
Two Degree of Freedom – Undamped SystemExample 2
Two Degree of Freedom – Undamped SystemExample 2
Two Degree of Freedom – Undamped SystemExample 2
Two Degree of Freedom – Undamped SystemExample 2
Two Degree of Freedom – Undamped SystemExample 2
Two Degree of Freedom – Forced Vibration
Two Degree of Freedom – Forced Vibration
Two Degree of Freedom – Forced Vibration
Two Degree of Freedom – Forced Vibration ExampleFind The Steady state response of the following System
Two Degree of Freedom – Forced VibrationExample
Two Degree of Freedom – Dynamic Absorber
Two Degree of Freedom – Dynamic AbsorberA machine or system may experience excessive vibration if it is acted upon by a force whose excitation frequency nearly coincides with a natural frequency of the machine or system. In such cases, the vibration can be reduced by using a vibration neutralizer or dynamic vibration absorber, which is simply another spring-mass system. The dynamic vibration absorber is designed such that the natural frequencies of the resulting system are away from the excitation frequency. The analysis of the dynamic vibration absorber will be considered by idealizing the machine as a single degree of freedom system.
Two Degree of Freedom – Dynamic Absorber
Two Degree of Freedom – Dynamic Absorber
Two Degree of Freedom – Dynamic Absorber
Two Degree of Freedom – Dynamic Absorber
We are primarily interested in reducing the amplitude of the machine (X1) In order to make the amplitude of zero, the numerator of Eq. (X1) should be set equal to zero.
This gives
If the machine, before the addition of the dynamic vibration absorber, operates near its resonance, .Thus if the absorber is designed such that
where
Two Degree of Freedom – Dynamic Absorber
The equations of X1 and X2 can be rewritten as
Two Degree of Freedom – Dynamic Absorber
The equations of X1 and X2 can be rewritten as
Two Degree of Freedom – Dynamic Absorber
As seen before, X1 = 0 at ω =ω2 =ωa. At this frequency :
𝑋" = −𝑘&𝑘"𝛿() = −
𝑘&𝑘"×𝐹,𝑘&= −
𝐹,𝑘"
This shows that the force exerted by the auxiliary spring is opposite to the impressed force (k2 X2 = - Fo) and neutralizes it, thus reducing X1 to zero. The size of the dynamic vibration absorber can be found from
k2 X2 = m2 ω2 X2 = - Fo
Thus the values of k2 and m2 depend on the allowable value of X2
Two Degree of Freedom – Dynamic Absorber
The two peaks correspond to the two natural frequencies of the composite system. The values of the two natural frequencies can be found by equating the denominator of the following equation to zero
, which leads to:
1 +𝑘"𝑘&−
𝜔𝜔(
"1 −
𝜔𝜔1
"−𝑘"𝑘&= 0 𝑜𝑟,
𝜔6 − 𝜔(" +𝜔1" +𝑘"𝑚&
𝜔" + 𝜔(". 𝜔1" = 0
Two Degree of Freedom – Dynamic Absorber
The roots of the previous equation is given by
which can be seen to be functions of and
Two Degree of Freedom – Dynamic Absorber
The roots of the previous equation is given by
which can be seen to be functions of and
Two Degree of Freedom – Dynamic Absorber
Two Degree of Freedom – Dynamic AbsorberNotes:
1. It can be seen, that ωn1 is less than andωn2 is greater than the operating speed(ω) of the machine. Thus the machinemust pass through ωn1 during start-upand stopping. This results in largeamplitudes.
2. Since the dynamic absorber is tuned toone excitation frequency (ω), the steady-state amplitude of the machine is zeroonly at that frequency. If the machineoperates at other frequencies or if theforce acting on the machine has severalfrequencies, then the amplitude ofvibration of the machine may becomelarge.
Two Degree of Freedom – Dynamic AbsorberNotes:
3. It can be seen from equations (31) and(32) that the amplitude of the absorber’smass (X2) is always much greater than thatof the main mass (X1). Thus the designshould be able to accommodate the largeamplitudes of the absorber mass.4. Since the amplitudes of m2 are expected tobe large, the absorber spring (k2) needs to bedesigned from a fatigue point of view.
Two Degree of Freedom – Dynamic AbsorberNotes:
5. The variation of ωn1/ωa and ωn2/ωa asfunctions of the mass ratio m2/m1 are shownfor three different values of the frequencyratio ωa/ωs. It can be seen that the differencebetween ωn1 and ωn2 increases withincreasing the ratio of m2/m1.
Two Degree of Freedom – Dynamic Absorber
• If ω hits ω:& or ω:& resonance occurs• Using ;<
;=><1, defines useful operating
range of absorber• In this range some absorption still occurs
Absorber Zone
Two Degree of Freedom – Dynamic AbsorberAbsorber Zone
This illustrate that the useful operating range of absorber design is
(0.908 𝜔? < 𝜔 < 1.118 𝜔?) .
Hence if the driving frequency drifts within this range, the absorber design still offers some protection to the primary system by reducing its steady state vibration amplitude.
Two Degree of Freedom – Dynamic Absorber
Two Degree of Freedom – Dynamic AbsorberExample
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
a) Static Coupling
a) Static +Dynamic Couplings
Two Degree of Freedom – Modal Analysis
a) Static Coupling
a) Static +Dynamic Couplings
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Example 4.2.1 (Inman)
Calculate the Matrix 𝐾A For M = 9 00 1 K= 27 −3
−3 3Example 4.2.2 (Inman)
Calculate the Eigen values and vectors of Matrix 𝐾A andnormalize the Eigen vectors and check if they areorthogonal
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
P = 𝑣" 𝑣" = &"1 11 −1
So that PT P becomes
P = &"
&"
1 11 −1
1 11 −1
= &"1 + 1 1 − 11 − 1 1 + 1 = 1 0
0 1 = I
Calculate the Matrix PT 𝐾AP for the previous two degree of freedom example
𝑃H𝐾A𝑃 =121 11 −1
3 −1−1 3
121 11 −1
= &"1 11 −1
2 42 −4
= &"4 00 8 = 2 0
0 4 =ΛNote that the diagonal elements of the spectral matrix Λ are the natural frequencies 𝜔&and 𝜔". That is, from previous calculations, 𝜔&"=2 and 𝜔"" = 4, so that Λ=diag(𝜔&" 𝜔"") =diag(2 4) = 2 0
0 4
Two Degree of Freedom – Modal Analysis
𝑆 = 𝑀N& "⁄ 𝑃 = &"
&P
00 1
1 11 −1
𝑆 = &"
&P
&P
1 −1and
𝑆N& = 𝑃H𝑀N& "⁄ = &"1 11 −1
&P
00 1
𝑆 = &"3 13 −1
Two Degree of Freedom – Modal Analysis
The given initial conditions are𝑥a =
10 and �̇�a =
00
The initial conditions in the generalizedcoordinate 𝑟 will be
𝑟 0 = 𝑆N&𝑥a =&"3 13 −1
10 =
P"P"
�̇� 0 = 𝑆N&�̇�a =&"3 13 −1
00 = 0
0
Two Degree of Freedom – Modal Analysis
So that 𝑟& 0 = 𝑟" 0 = 3 2⁄ and �̇�& 0 = �̇�" 0 = 0𝑥a =
10 and �̇�a =
00
The response in the physical coordinate r 𝑡is calculated from
𝑟& 𝑡 =32sin 2 𝑡 +
𝜋2 =
32cos 2 𝑡
𝑟" 𝑡 =32sin 2 𝑡 +
𝜋2 =
32cos 2 𝑡
Two Degree of Freedom – Modal Analysis
The response in the physical coordinate 𝑥 𝑡is calculated from
𝑥 𝑡 = 𝑆 𝑟 𝑡 =12
13
13
1 −1
32cos 2 𝑡
32cos 2 𝑡
𝑥 𝑡 =0.5 cos 2 𝑡 + cos 2 𝑡1.5 cos 2 𝑡 + cos 2 𝑡
which is identical to the solution obtained in Example 4.1.7
Two Degree of Freedom – Modal Analysis The response in the generalized coordinate 𝑟 𝑡and the physical coordinate 𝑥 𝑡 are plotted as follows
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis
Two Degree of Freedom – Modal Analysis