Recent Shinkansen trains are surprisingly quiet and have very little vibration. Generally, vibration damping measures are taken to reduce noise, but in addition to noise reduction, methods of elastic support have also advanced to improve ride comfort. As I have mentioned before, I will discuss vibration isolation again this time.
To prevent machine vibrations from being transmitted to the floor, elastic materials are often placed between the machine and the floor. The purposes of elastic support can be broadly categorized as follows:
(1) Active vibration isolation
To prevent vibrations generated by the machine from being transmitted to the floor.
(2) Passive vibration isolation
To prevent vibrations from being transmitted from the floor to precision machinery and measuring instruments.
They can be divided into two categories. The vehicle's spring system is intended to prevent vibrations of the axle caused by unevenness in the track or road from being transmitted to the vehicle body, and is an example of (2).
The vibration isolation characteristics of these two elastic supports are shown in Figure 1, and the corresponding equations are equations (1) and (2).
Equation (1) represents the transmission coefficient τ of the force Q transmitted to the floor by the forced vibration P caused by the machine,
Equation (2) represents the transmission coefficient τ of the forced vibration xo due to floor displacement, which is transmitted to the machine as displacement vibration x.
Figure 1. Vibration transmission rate of a damped 1-degree-of-freedom drive system
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Frequency ratio η=ω/ω0
τ＝Q/P＝√{1+(2ζη)^2}/{(1-η^2)^2+(2ζη)^2} ...(1)
τ＝x/xo＝√{1+(2ζη)^2}/{(1-η^2)^2+(2ζη)^2}...(2)
Here,
Damping ratio ζ = C/Cc (C: damping coefficient) ・・・・・・・・(3)
Frequency ratio η=ω/ωo=f/fo ・・・・・・・・・(4)
Critical damping coefficient Cc = 2√mk = 2mω/ωo ... ... (5)
Natural frequency ωo = √k/m fo = wo/2π ・・・・・・(6)
Static deflection δ = mG/k ・・・・・・・・・・・・(7)
Mass m=W/G
W: Load (kg)
G: Gravitational acceleration
(√: The notation following the √ indicates what is inside the square root.)
Furthermore, if there is no damping C, substituting ζ = 0 results in equation (8).
τ=1/｜1-η^2｜ ・・・・・・・・・・・・・・・(8)
(|｜ indicates absolute value)
Vibration isolation measures are designed to minimize this τ.
As can be seen from Figure 1,
(1) When f/fo > √2, the transmitted force is always greater than the excitation force.
In this case, the smaller the value of ζ, the smaller the value of τ, which is desirable from a vibration isolation perspective.
(2) When f/fo < √2, the transmission force is always greater than the excitation force.
In this case, the larger the value of ζ, the smaller the value of τ, which is desirable from a vibration isolation perspective.
It has the following characteristics. Based on these characteristics, the following points should be considered as vibration damping measures.
(1) Vibration isolation should be implemented so that f/fo > 3 if possible. The transmission coefficient will be approximately 12.5% or less from equation (8).
(2) If it is not possible to do (1), set f/fo < 0.4.
(3) When the excitation force frequency gradually increases from 0, a damping of η < 0.2 is applied to pass through the resonance point.
(4) For soundproofing purposes, a frequency of f < 20 Hz or less is desirable.
■Examples of natural frequencies
(1) The natural frequency fo of the spring-load system is
fo＝1/2π√k/m
Let m be the mass of the machine, and let W be the weight of the machine (load in kg), and let G be the acceleration due to gravity.
m＝W/G
The spring constant k may be used in cases where multiple vibration-damping rubbers are used. The spring constant is considered as a composite when used in series and in parallel, as follows:
Figure 2 Combination of spring constants
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series -
parallel
- In the case of series springs
1/𝑘=1/𝑘_1 +1/𝑘_2 +1/𝑘_3 - In the case of parallel springs
k = k1 + k2 + k3
■ Example Problem: An air compressor has mass M and a rotational speed of 540 r/min. The amplitude P of the unbalance force Psinωt of the compressor is 40 kN. This machine is mounted on a concrete base of mass M', and M' is supported by a spring k. The vibration transmission force Q transmitted to the surrounding ground is to be 5 kN. What natural frequency fo should be chosen for the elastically supported system?
Substitute P=40kN and Q=5kN into equation (1)
τ = Q/P = 5kN/40kN = 1/8
From equation (8)
1/8＝1/｜1-η^2｜
Therefore, η^2 = 9, η = 3
Also, from 2πf = 540/60, f = 9 (Hz).
From equation (4)
f/fo = 3
Substituting f = 9,
Natural frequency fo=3 (Hz)
The M' and k will be designed so that the natural frequency fo = 3. Since the foundation, such as concrete, is supported by the ground, the formula for the natural frequency, taking into account the spring constant of this ground, is as follows.
Natural frequency considering fixed foundation
In vertical vibration, if the weight of the machine is W (kg), the weight of the base is Wo (kg), and the area of the base in contact with the ground is A (cm^2), then the natural frequency is...
ωo＝√GkvA/（W+Wo) ・・・（9）
kV is called the soil coefficient, and it varies depending on the properties of the soil. Approximate values are shown in the table below.
| Types of ground | kv（kg/cm^3） |
| Soft clay | >2 |
| Loam genus | 3～5 |
| fine sand | 5～6 |
| sand | 8～10 |
| Sand and gravel | 11～13 |
In the case of horizontal motion, the ground reaction force acts on the base of the foundation, and the natural frequency is
ωo＝√GkhA/(W+Wo ・・・(10)
In the above formula, kh is called the horizontal ground coefficient. The value of kh is not definitive.
kh＝（0.6～0.7）kv
This is used.
References:
| Machine noise and countermeasures | Published by Kyoritsu Shuppan |
| Practical Mechanical Oscillometry | Published by Rikogakusha |
(Excerpt from the email newsletter issued on July 21, 2005)
