# An Analytical Model for a Pneumatic Vibration Isolator with the Stiffness Effect of the Elastomeric Diaphragm

Authors: Dengfeng Xu; Qiang Yu; Fei Shen; Yu Zhu; Gaofeng Guan
Journal: Shock and Vibration
 (2018)

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## Abstract

The elastomeric diaphragm is widely used in pneumatic vibration isolators, and the relevant model is often ignored due to its complexity. Considering that the ignored model of the elastomeric diaphragm in pneumatic vibration isolators plays an important role in the discrepancy between the predicted and practical behavior, this paper develops an analytical model for the elastomeric diaphragm using the Mooney-Rivlin modeling method and elastomeric theory. Specifically, the elastomeric diaphragm consists of several segments in the axial section. After considering the structural restriction, each segment can be simplified as uniaxial stretching, and the force-strain equation can be established for each segment. By combining the equations of all segments, an analytical model of the elastomeric diaphragm can be built and solved via numerical methods. The developed model is added to the standard model of pneumatic cylinders to supply a complete analytical model for pneumatic vibration isolators. The experimental results demonstrate that the analytically predicted behavior is similar to the practical behavior. The proposed analytical model can be used as a guide for the parameter design of pneumatic isolators in practice.


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## Body

## 1. Introduction

Pneumatic vibration isolators are widely applied in optics, semiconductors, and other fields [1, 2]. Many efforts have been devoted to the design of active control systems, which require an accurate mathematical model of the pneumatic vibration isolator [3, 4]. Harris [5] assumed that the airflow in the damping hole was laminar and first offered a linear model for the pneumatic vibration isolators. In 1984, Debra and Bryan [6] established a simpler linear model of the volume ratio of the upper chamber to the lower chamber and the inherent frequency. Erin and Wilson [7] proposed an analytical modeling method for the pneumatic vibration isolator based on the ideal gas state equation. In all previous studies, the discrepancies between the model-predicted behavior and the practically observed behavior of pneumatic vibration isolators exceeded 50% because the effect of the elastomeric diaphragm was ignored.To obtain a more accurate model of the pneumatic isolator, Erin and Wilson proposed an improved model for pneumatic isolators by incorporating the stiffness and damping effect of the elastomeric diaphragm based on experimental investigation. Lee and Kim [8, 9] reported a nonlinear complex stiffness model that reflected the frequency and excitation amplitude, which was validated by comparison with experimental measurements. Pu et al. [10] obtained the stiffness and damping characteristics of the elastomeric diaphragm indirectly by comparing the experimental and theoretical results. Additionally, Pu established the stiffness and damping model using a similar approach in other research. Since the 1990s, analog simulation has become an attractive method for examining the pneumatic spring, and the stiffness characteristic is often studied using the finite element method (FEM) [11]. However, experimental modification or the FEM is inconvenient for the parameter optimization design of a pneumatic isolator. Significant error typically still exists between the actual behavior and the presented model after the experimental modification or FEM study. Therefore, a direct theoretical derivation method must be established for the development of an analytical model for a pneumatic vibration isolator.Zhu et al. [12] and Li [13] proposed an analytic model of an elastomeric belt in a magnetic-levitation vibration isolator using the Mooney-Rivlin model, and the model was validated experimentally. In that research, the elastomeric belt was chosen as the research object, and an accurate model was obtained according to direct theoretical derivation. Inspired by this work, we use the same method and theory to study and analyze the elastomeric diaphragm in pneumatic vibration absorbers and develop an analytical model of the elastomeric diaphragm in this paper. Specifically, because the elastomeric diaphragm is composed of several segments in the axial section, each segment can be simplified as uniaxial stretching after considering the structural restriction, and a force-strain equation can be established for each segment. The analytical model of the elastomeric diaphragm can be built by combining all segments, and the model can be resolved by numerical solution. Experimental tests were conducted, and the results demonstrate that the analytically predicted behavior is similar to the practically observed behavior. The error in this method can be reduced to less than 10% after the elastomeric diaphragm model is added to the standard pneumatic cylinder model. Although certain scholars have proposed several improved models that also imitate the practical behavior, this paper supplies a direct theoretical derivation method with a notably small error. The proposed analytical model facilitates the practical parameter optimization design of pneumatic isolators.


## 2. Mathematic Model of the Elastomeric Diaphragm

### 2.1. Configuration of the Pneumatic Vibration Isolator

The configuration of a typical pneumatic vibration isolator is shown in Figure1(a), which includes an air-filled cylinder, elastomeric diaphragm, and piston. The cylinder consists of a top chamber and bottom chamber that are both filled with compressed air. The elastomeric diaphragm is located between the piston and cylinder, which is the seal component of the cylinder. The payload is connected to the piston, and the flow restrictor supplies damping for the vibration isolator.Figure 1
Pneumatic vibration isolator.
(a)
Schematic of a pneumatic vibration isolator (b)
Simplification of a pneumatic vibration isolatorFigure1(b) illustrates that the payload is supported by two components: the pressurized air in the pneumatic cylinder and the elastomeric diaphragm, the model of which is often ignored. This paper only analyzes the stiffness effect of the elastomeric diaphragm without considering the damping effect because the vibration damping effect of the elastomeric diaphragm is a highly complex phenomenon. Therefore, the total stiffness of the pneumatic spring isolators can be obtained by adding the vertical stiffness of the elastomeric diaphragm (the model typically ignored in previous analysis and research) to the vertical stiffness of the pressurized air. The total stiffness of the pneumatic spring can be expressed as follows [14]:(1)Ktotal=Km+Kawhere Ktotal represents the total stiffness of the pneumatic spring isolators, Ka represents the vertical stiffness of the compressed air, and Km represents the vertical stiffness of the elastomeric diaphragm. Km has typically been neglected in previous analyses of the pneumatic spring, which causes a certain error between the theory and experiments. This error in the total vertical stiffness can result in a series of problems. For example, the actual inherent frequency of a pneumatic vibration absorber cannot be easily estimated with high accuracy. After considering the effect of elastomeric diaphragm, a complete analytical model for a pneumatic vibration isolator is developed, and the developed analytical model can be used in the parameter optimization design of a pneumatic vibration isolator when the total stiffness and inherent frequency of the pneumatic vibration absorber are both known.Figure2 shows a cross-sectional view of an elastomeric diaphragm, and the parameters of the elastomeric diaphragm are listed in Table 1. After the elastomeric diaphragm is divided into four segments, the vertical stiffness of the segments, expressed as K1, K2, K3, and K4, is deduced such that the total vertical stiffness of the elastomeric diaphragm can be determined.Table 1
Parameters of the elastomeric diaphragm.
Symbol Description R Radius of the piston (radius of the inner circle of the elastomeric diaphragm) r Radius of the arc of the elastomeric diaphragm h1 Height of the inner circle of the elastomeric diaphragm h3 Height of the outer circle d0 Thickness of the elastomeric diaphragm “1” Inner circle of the elastomeric diaphragm “2” Inner arc of the elastomeric diaphragm “3” Outer circle of the elastomeric diaphragm “4” Outer arc of the elastomeric diaphragmFigure 2
Elastomeric diaphragm.
(a)
Size definition of the elastomeric diaphragm (b)
Parameters of the elastomeric diaphragm


### 2.2. Force Analysis of the Elastomeric Diaphragm

The forces of the piston and elastomeric diaphragm in the vertical direction are shown in Figure3; then,(2)Fa+F1=G+F0(3)F1=F1′=F3=F3′=FP2where F1 is the pulling force of part “2” acting on part “1” and F3 is the pulling force of part “4” acting on part “3”. In addition, F′1 and F′3 are the reaction forces of F1 and F3, respectively, andFa, F0 and G can be rewritten as(4)Fa=Pa·πR+d0λ312(5)F0=P0·πR+d0λ312(6)G=mgwhere Fa is the vertical force of the air in the chamber acting on the effective piston area, F0 is the vertical force of the air outside the chamber acting on the effective piston area, G is the gravity of the load, P0 is the atmospheric pressure, Pa is the air pressure in the chamber, m is the payload mass, g is the gravitational acceleration, and λ31 is the corresponding stretch ratio of a unit of part “1”, which is explained later. The stretch ratios in this paper are defined as the ratios of the stretched length to the unstretched length or original length of a small-volume element, which is a cube when it is unstretched or in the original state.Figure 3
Force analysis of the piston and elastomeric diaphragm.
(a)
Force analysis of the piston (b)
Force analysis of the elastomeric diaphragmAs shown in Figure3(b), Fp is the force acting on the arc of the elastomeric diaphragm in the vertical direction and is expressed as(7)FP=Pa-P0·stwhere st is the projection area of the arc of the elastomeric diaphragm, as shown in Figure 4. Then, st can be expressed as(8)st=π·R+2d0+2r-d0λ332-π·d0λ31+R2where λ33 is the corresponding stretch ratio of a unit of part “3”, which is explained later.Figure 4
Cross-section of the dual-chamber pneumatic spring.Substituting (8) into (7) leads to(9)FP=πPa-P0R+2d0+2r-d0λ332-d0λ31+R2.By substituting (9) into (3) and combining the result with (2), (4), (5), and (6), the air pressure in the chamber can be obtained as(10)Pa=2mgπR+2d0+2r-d0λ332+d0λ31+R2+P0.By substituting (10) into (9) and noting (3), one can obtain(11)F1=F3=FP2=mgR+2d0+2r-d0λ332-d0λ31+R2R+2d0+2r-d0λ332+d0λ31+R2.


### 2.3. Mooney-Rivlin Model

Mooney-Rivlin [15, 16] assumed that the elastic behavior was isotropic and that simple stretching satisfied Hooke’s law; then, they obtained the rubber energy density function W under the premise of not changing the rubber state and compressing its volume; i.e.,(12)W=WJ1,J2,J3and(13)J1=λ12+λ22+λ32-3J2=λ1λ22+λ2λ32+λ1λ32-3J3=λ12λ22λ32-1where λ1, λ2, and λ3 represent the stretch ratios in three different directions and are defined as the ratios of the stretched length to the unstretched length of a small-volume element, which is a cube when it is unstretched. These three directions, which are related to λ1, λ2, and λ3, are expressed under generalized XYZ coordinates. If εi is defined as a corresponding stretched length, the stretch ratio can be expressed as λi=1+εi/1. Herein, the rubber satisfies the incompressibility condition, i.e., λ1λ2λ3=1, which means that J3=0. The corresponding deformation energy density W only depends on J1 and J2. Under the condition of sufficiently small deformation, the form of deformation energy function can be expressed as(14)W=C1J1+C2J2(15)C1=∂W∂J1,C2=∂W∂J2where C1 and C2 represent different material parameters.


### 2.4. Stiffness Analysis of the Elastomeric Diaphragm

#### 2.4.1. Part “1” Stiffness of the Elastomeric Diaphragm

As shown in Figure5(a), when an elastomeric diaphragm is stretched, the tension F1 acting on part “1” can be expressed as(16)F1=δ11st1where st1 is the sectional area of part “1”, which is shown in Figure 4, and st1 can be expressed as(17)st1=πR+d0λ312-R2where λ31 is the corresponding stretch ratio of a unit of part “1”. In this paper, the letters of the subscript contain two numbers i and j, where i(1,2,3) represents the corresponding principal stress surface and j(1,2,3,4) represents the corresponding component of the elastomeric diaphragm, i.e., the inner circle of the elastomeric diaphragm, the inner arc of the elastomeric diaphragm, the outer circle of the elastomeric diaphragm, and the outer arc of the elastomeric diaphragm, respectively.Figure 5
Force analysis of part “1” of the elastomeric diaphragm.
(a)
Schematic diagram and force analysis of part “1” (b)
Parameters and force analysis of “1” unit of part “1”Figure5(b) shows the unit of part “1”, and the unit stress in three directions can be expressed as(18)δ11=2C1λ112-C2λ11-2+P1(19)δ21=2C1λ212-C2λ21-2+P1(20)δ31=2C1λ312-C2λ31-2+P1=Pawhere λij is the corresponding stretch ratio and P1 is the unknown pressure. The stretch ratio can be expressed as λ11=(h1+ε1)/h1, where ε1 is the length variable in the vertical direction of part “1”.The following equation can be obtained by combining (18) with (20):(21)δ11=Pa+2C1λ112-C2λ11-2-2C1λ312-C2λ31-2.Substituting (17) and (21) into (16) yields(22)F1=πPa+2C1λ112-C2λ11-2-2C1λ312-C2λ31-2R+d0λ312-R2.The force on the unit in the cross-section is shown in Figure6. The force of the piston acting on the unit of part “1” can be expressed as(23)FN1=FP1+2F21·sin⁡γ12where Fp1 is the third principal stress surface of part “1” subjected to the force of the gas chamber,F21 is the second principal stress surface of part “1” subjected to the force, and γ1 is the angle of the unit in the cross-section of the dual-chamber pneumatic spring.Figure 6
Cross-section of one unit of part “1”.F P 1 and F21 are defined as(24)FP1=Pa·h1λ11·b1λ21(25)F21=δ21·h1λ11·d0λ31where h1λ11·b1λ21 is the area of the third principal stress surface of part “1”, δ21 is the stress of the second principal stress surface of part “1”, and h1λ11·d0λ31 is the area of the second principal stress surface of part “1”, as shown in Figure 5(b).Substituting (24) and (25) into (23) yields(26)FN1=Pa·h1λ11·b1λ21+2δ21·h1λ11·d0λ31·sin⁡γ12which indicates that FN1>0. Therefore, part “1” of the elastomeric diaphragm remains in contact with the piston wall, and the size of part “1” of the elastomeric diaphragm does not change when it is stretched in the circumferential direction; i.e., λ21=1. The incompressibility of rubber leads to λ11λ21λ31=1, and the following expression can be obtained:(27)λ31=λ11-1=h1+ε1h1-1.Substituting (10) and (27) into (22) yields(28)F1=π2mgπR+2d0+2r-d0h3/h3+ε32+d0h1/h1+ε1+R2+P0+2C1+C2h1+ε1h12-h1h1+ε12R+h1d0h1+ε12-R2.To obtain a function for the stiffnessK1 of part “1” with respect to variable ε1, the following equation can be obtained by differentiating (28) with respect to ε1:(29)K1=dF1dε1.The complete equation(29) is extremely long and is provided in Appendix B.


#### 2.4.2. Part “2” Stiffness of the Elastomeric Diaphragm

As shown in Figure7, the tension F2 acting on part “2” can be expressed as(30)F2=∫0π/2∫02πδ2dst2·dαwhere F2 originates from the pressure difference of the top chamber and ambient environment in the vertical direction.Figure 7
Force analysis of part “2” of the elastomeric diaphragm.
(a)
Schematic diagram of part “2” (b)
Parameters of “2” unit of part “2” (c)
Force analysis of “2” unit of part “2” (d)
Force analysis of part “2”For the projected area of the unit,dst2 can be written as(31)dst2=R+r1-cos⁡αdθ·d0λ32·cos⁡αwhere R+r(1-cos⁡α) is the length l2 from the analysis point to the center axis, (R+r(1-cos⁡α))dθ is the side length d2 of the unit in part “2”, and d0λ32 is the thickness of the unit in part “2”. θ is shown in Figure 7(a).The vertical stressδ2 of the unit of part “2” can be expressed as(32)δ2=δ12cos⁡α+δ32sin⁡αwhere δ12 and δ32 can be written as(33)δ12=2C1λ122-C2λ12-2+P2(34)δ32=2C1λ322-C2λ32-2+P2=Pa-P0where δ12 and δ32 are the stresses of the corresponding principal stress surface and P2 is the unknown pressure. The stretch ratio can be expressed as λ12=(r+ε2)/r, where ε2 is the length variable in the vertical direction. The size of part “2” does not change when it is stretched in the circumferential direction; i.e., λ22=1. The incompressibility of rubber leads to λ12λ22λ32=1. Thus, the following equation can be obtained:(35)λ32=λ12-1=r+ε2r-1The following equation can be obtained by transferring (32) to (35):(36)δ2=Pa-P0+2C1+C2r+ε2r2-rr+ε22cos⁡α+Pa-P0sin⁡α.Substituting (10), (30), (31), and (36) into (30) yields(37)F2=2π2mgπR+2d0+2r-d0h3/h3+ε32+d0h1/h1+ε1+R2+2C1+C2r+ε2r2-rr+ε22πd0rR+r4r+ε2-2d0r23r+ε2+2mgπR+2d0+2r-d0h3/h3+ε32+d0h1/h1+ε1+R2d0rR+rr+ε2-d0r23r+ε2.To obtain a function of the stiffnessK2 of part “2” with respect to variable ε2, the following equation can be obtained by differentiating (37) with respect to ε2:(38)K2=dF2dε2.The complete equation(38) is extremely long and is provided in Appendix B.


#### 2.4.3. Part “3” Stiffness of the Elastomeric Diaphragm

As shown in Figure8, when an elastomeric diaphragm is stretched, the tension F3 acting on part “3” can be expressed as(39)F3=δ13st3Figure 8
Force analysis of part “3” of the elastomeric diaphragm.
(a)
Schematic diagram and force analysis of part “3” (b)
Parameters and force analysis of “3” unit of part “3”The analysis of part “3” is similar to that of part “1”. Therefore, based on the Mooney-Rivlin model, the vertical stressδ13 can be expressed as(40)δ13=Pa+2C1λ132-C2λ13-2-2C1λ332-C2λ33-2.where the stretch ratio can be expressed as λ13=(h3+ε3)/h3 and ε3 is the variable length in the vertical stretch of part “3”.Additionally,st3 is the cross-sectional area of part “3”, as shown in Figure 4. Herein, st3 can be expressed as(41)st3=πR+2d0+2r2-R+2d0+2r-d0λ332.Substituting (40) and (41) into (39) yields(42)F3=πPa+2C1λ132-C2λ13-2-2C1λ332-C2λ33-2·R+2d0+2r2-R+2d0+2r-d0λ332.The force on the unit in the cross-section is shown in Figure9. The force of the piston acting on the unit of part “3” can be expressed as(43)FN3=FP3-2F23·sin⁡γ32where Fp3 is the third principal stress surface of part “3” subjected to the force of the gas chamber, F23 is the second principal stress surface of part “3” subjected to the force, and γ3 is the angle of the unit in cross-section of the dual-chamber pneumatic spring.Figure 9
Cross-section of one unit of part “3”.F P 3 and F23 are written as(44)FP3=Pa·h3λ13·b3λ23(45)F23=δ23·h3λ13·d0λ33where h3λ13·b3λ23 is the area of the third principal stress surface of part “3”, δ23 is the stress of the second principal stress surface of part “3”, and h3λ13·d0λ33 is the area of the second principal stress surface of part “3”, as shown in Figure 8(b).Substituting (44) and (45) into (43) yields(46)FN3=Pa·h3λ13·b3λ23-2δ23·h3λ13·d0λ33·sin⁡γ32where sin⁡(γ3/2)=b3/2(R+2r+2d0-d0λ33) and b3 is the arc length of the unit of part “3”, which is smaller than 2(R+2r+2d0-d0λ33). Thus, the first polynomial is larger than the second polynomial of (46); i.e., FN3>0. Therefore, part “3” remains in contact with the fixed ring, and the size of part “3” does not change when it is stretched in the circumferential direction; i.e., λ23=1. However, the incompressibility of rubber leads to λ13λ23λ33=1. Thus, the following equation can be obtained:(47)λ33=λ13-1=h3+ε3h3-1.Substituting (10) and (47) into (42) yields(48)F3=π2mgπR+2d0+2r-d0h3/h3+ε32+d0h1/h1+ε1+R2+P0+2C1+C2h3+ε3h32-h3h3+ε32R+2d0+2r2-R+2d0+2r-d0h3h3+ε32.To obtain a function of the stiffnessK3 of part “3” with respect to variable ε3, the following equation can be obtained by differentiating (48) with respect to ε3:(49)K3=dF3dε3.The complete equation(49) is extremely long and is provided in Appendix B.


#### 2.4.4. Part “4” Stiffness of the Elastomeric Diaphragm

As shown in Figure10, the tension F4 acting on part “4” can be expressed as(50)F4=∫0π/2∫02πδ4dst4·dβwhere F4 originates from the pressure difference of the top chamber and the ambient environment in the vertical direction.Figure 10
Force analysis of part “4” of the elastomeric diaphragm.
(a)
Schematic diagram of part “4” (b)
Parameters of the “4” unit of part “4” (c)
Force analysis of the “4” unit of part “4” (d)
Force analysis of part “4”For the projected area of the unit,dst4 can be written as(51)dst4=R+r1+cos⁡βdθ·d0λ34·cos⁡βwhere R+r(1+cos⁡β) is the length l4 from the analysis point to the center axis, (R+r(1+cos⁡β))dθ is the side length d4 of the unit in part “4”, d0λ34 is the thickness of the unit in part “4”, and θ is shown in Figure 10(a).The vertical stressδ4 of the partial unit in part “4” can be expressed as(52)δ4=δ14cos⁡β+δ34sin⁡βandδ14 andδ34 can be rewritten as(53)δ14=2C1λ142-C2λ14-2+P4(54)δ34=2C1λ342-C2λ34-2+P4=Pa-P0where δ14 and δ34 are the stresses corresponding to the principal stress surface and P4 is the unknown pressure. The stretch ratio can be expressed as λ14=(r+ε4)/r, where ε4 is the variable length in the vertical stretch of part “4”. The size of part “4” does not change when it is stretched in the circumferential direction; i.e., λ24=1. The incompressibility of rubber leads to λ14λ24λ34=1. Thus, the following equation can be obtained:(55)λ34=λ14-1=r+ε4r-1.The following equation can be obtained by transferring (52) to (55):(56)δ4=Pa-P0+2C1+C2r+ε4r2-r+ε4r-2cos⁡β+Pa-P0sin⁡β.Substituting (10), (51), (55), and (56) into (50) yields(57)F4=2π2mgπR+2d0+2r-d0h3/h3+ε32+d0h1/h1+ε1+R2+2C1+C2r+ε4r2-rr+ε42πd0rR+r4r+ε4+2d0r23r+ε4+2mgπR+2d0+2r-d0h3/h3+ε32+d0h1/h1+ε1+R212d0r+ε4r-1R+r+13d0r+ε4r-1r.To obtain a function of the stiffnessK4 of part “4” with respect to variable ε4, the following equation can be obtained by differentiating (57) with respect to ε4:(58)K4=dF4dε4.The entire equation(58) is extremely long and is provided in Appendix B.


#### 2.4.5. Total Stiffness of the Elastomeric Diaphragm

As shown in Figure11, part “1” and part “2” are connected in series such that the equivalent stiffness K12 can be expressed as(59)1K12=1K1+1K2.Figure 11
Total stiffness of the elastomeric diaphragm.Part “3” and part “4” are connected in series such that the equivalent stiffnessK34 can be expressed as(60)1K34=1K3+1K4.The total stiffness can be expressed as(61)Km=K12+K34=K1K2K1+K2+K3K4K3+K4.


### 2.5. Numerical Solution of the Vertical Stiffness for the Elastomeric Diaphragm

When the load of the dual-chamber pneumatic spring ism, each part of the elastomeric diaphragm is stretched to a certain length. The stretched lengths of each corresponding part are defined as ε10, ε20, ε30, and ε40. By analyzing (3) and (11), each component of the force can be expressed as(62)F1=F1′=F2=F3=F3′=F4=FP2=mgR+2d0+2r-d0h3/h3+ε32-d0h1/h1+ε1+R2R+2d0+2r-d0h3/h3+ε32+d0h1/h1+ε1+R2.The analysis process to obtain the stretched length of each corresponding part with respect to the other variables is as follows. Substitutingε1=ε10 into (28) and noting (62), we can obtain (B.5) of the stretched length ε10. (B.6), (B.7), and (B.8) can also be obtained in a similar manner. Equations (B.5) to (B.8) are rather long and are provided in Appendix B.By solving (B.5) to (B.8) in the numerical root of ε10, ε20, ε30, and ε40 and substituting ε1=ε10, ε2=ε20, ε3=ε30, and ε4=ε40 into (29), (38), (49), and (58), we obtain the numerical solutions of stiffnessK1,K2,K3, andK4, respectively. As a result, the total stiffness Km can be obtained from (61).


## 2.1. Configuration of the Pneumatic Vibration Isolator

The configuration of a typical pneumatic vibration isolator is shown in Figure1(a), which includes an air-filled cylinder, elastomeric diaphragm, and piston. The cylinder consists of a top chamber and bottom chamber that are both filled with compressed air. The elastomeric diaphragm is located between the piston and cylinder, which is the seal component of the cylinder. The payload is connected to the piston, and the flow restrictor supplies damping for the vibration isolator.Figure 1
Pneumatic vibration isolator.
(a)
Schematic of a pneumatic vibration isolator (b)
Simplification of a pneumatic vibration isolatorFigure1(b) illustrates that the payload is supported by two components: the pressurized air in the pneumatic cylinder and the elastomeric diaphragm, the model of which is often ignored. This paper only analyzes the stiffness effect of the elastomeric diaphragm without considering the damping effect because the vibration damping effect of the elastomeric diaphragm is a highly complex phenomenon. Therefore, the total stiffness of the pneumatic spring isolators can be obtained by adding the vertical stiffness of the elastomeric diaphragm (the model typically ignored in previous analysis and research) to the vertical stiffness of the pressurized air. The total stiffness of the pneumatic spring can be expressed as follows [14]:(1)Ktotal=Km+Kawhere Ktotal represents the total stiffness of the pneumatic spring isolators, Ka represents the vertical stiffness of the compressed air, and Km represents the vertical stiffness of the elastomeric diaphragm. Km has typically been neglected in previous analyses of the pneumatic spring, which causes a certain error between the theory and experiments. This error in the total vertical stiffness can result in a series of problems. For example, the actual inherent frequency of a pneumatic vibration absorber cannot be easily estimated with high accuracy. After considering the effect of elastomeric diaphragm, a complete analytical model for a pneumatic vibration isolator is developed, and the developed analytical model can be used in the parameter optimization design of a pneumatic vibration isolator when the total stiffness and inherent frequency of the pneumatic vibration absorber are both known.Figure2 shows a cross-sectional view of an elastomeric diaphragm, and the parameters of the elastomeric diaphragm are listed in Table 1. After the elastomeric diaphragm is divided into four segments, the vertical stiffness of the segments, expressed as K1, K2, K3, and K4, is deduced such that the total vertical stiffness of the elastomeric diaphragm can be determined.Table 1
Parameters of the elastomeric diaphragm.
Symbol Description R Radius of the piston (radius of the inner circle of the elastomeric diaphragm) r Radius of the arc of the elastomeric diaphragm h1 Height of the inner circle of the elastomeric diaphragm h3 Height of the outer circle d0 Thickness of the elastomeric diaphragm “1” Inner circle of the elastomeric diaphragm “2” Inner arc of the elastomeric diaphragm “3” Outer circle of the elastomeric diaphragm “4” Outer arc of the elastomeric diaphragmFigure 2
Elastomeric diaphragm.
(a)
Size definition of the elastomeric diaphragm (b)
Parameters of the elastomeric diaphragm


## 2.2. Force Analysis of the Elastomeric Diaphragm

The forces of the piston and elastomeric diaphragm in the vertical direction are shown in Figure3; then,(2)Fa+F1=G+F0(3)F1=F1′=F3=F3′=FP2where F1 is the pulling force of part “2” acting on part “1” and F3 is the pulling force of part “4” acting on part “3”. In addition, F′1 and F′3 are the reaction forces of F1 and F3, respectively, andFa, F0 and G can be rewritten as(4)Fa=Pa·πR+d0λ312(5)F0=P0·πR+d0λ312(6)G=mgwhere Fa is the vertical force of the air in the chamber acting on the effective piston area, F0 is the vertical force of the air outside the chamber acting on the effective piston area, G is the gravity of the load, P0 is the atmospheric pressure, Pa is the air pressure in the chamber, m is the payload mass, g is the gravitational acceleration, and λ31 is the corresponding stretch ratio of a unit of part “1”, which is explained later. The stretch ratios in this paper are defined as the ratios of the stretched length to the unstretched length or original length of a small-volume element, which is a cube when it is unstretched or in the original state.Figure 3
Force analysis of the piston and elastomeric diaphragm.
(a)
Force analysis of the piston (b)
Force analysis of the elastomeric diaphragmAs shown in Figure3(b), Fp is the force acting on the arc of the elastomeric diaphragm in the vertical direction and is expressed as(7)FP=Pa-P0·stwhere st is the projection area of the arc of the elastomeric diaphragm, as shown in Figure 4. Then, st can be expressed as(8)st=π·R+2d0+2r-d0λ332-π·d0λ31+R2where λ33 is the corresponding stretch ratio of a unit of part “3”, which is explained later.Figure 4
Cross-section of the dual-chamber pneumatic spring.Substituting (8) into (7) leads to(9)FP=πPa-P0R+2d0+2r-d0λ332-d0λ31+R2.By substituting (9) into (3) and combining the result with (2), (4), (5), and (6), the air pressure in the chamber can be obtained as(10)Pa=2mgπR+2d0+2r-d0λ332+d0λ31+R2+P0.By substituting (10) into (9) and noting (3), one can obtain(11)F1=F3=FP2=mgR+2d0+2r-d0λ332-d0λ31+R2R+2d0+2r-d0λ332+d0λ31+R2.


## 2.3. Mooney-Rivlin Model

Mooney-Rivlin [15, 16] assumed that the elastic behavior was isotropic and that simple stretching satisfied Hooke’s law; then, they obtained the rubber energy density function W under the premise of not changing the rubber state and compressing its volume; i.e.,(12)W=WJ1,J2,J3and(13)J1=λ12+λ22+λ32-3J2=λ1λ22+λ2λ32+λ1λ32-3J3=λ12λ22λ32-1where λ1, λ2, and λ3 represent the stretch ratios in three different directions and are defined as the ratios of the stretched length to the unstretched length of a small-volume element, which is a cube when it is unstretched. These three directions, which are related to λ1, λ2, and λ3, are expressed under generalized XYZ coordinates. If εi is defined as a corresponding stretched length, the stretch ratio can be expressed as λi=1+εi/1. Herein, the rubber satisfies the incompressibility condition, i.e., λ1λ2λ3=1, which means that J3=0. The corresponding deformation energy density W only depends on J1 and J2. Under the condition of sufficiently small deformation, the form of deformation energy function can be expressed as(14)W=C1J1+C2J2(15)C1=∂W∂J1,C2=∂W∂J2where C1 and C2 represent different material parameters.


## 2.4. Stiffness Analysis of the Elastomeric Diaphragm

### 2.4.1. Part “1” Stiffness of the Elastomeric Diaphragm

As shown in Figure5(a), when an elastomeric diaphragm is stretched, the tension F1 acting on part “1” can be expressed as(16)F1=δ11st1where st1 is the sectional area of part “1”, which is shown in Figure 4, and st1 can be expressed as(17)st1=πR+d0λ312-R2where λ31 is the corresponding stretch ratio of a unit of part “1”. In this paper, the letters of the subscript contain two numbers i and j, where i(1,2,3) represents the corresponding principal stress surface and j(1,2,3,4) represents the corresponding component of the elastomeric diaphragm, i.e., the inner circle of the elastomeric diaphragm, the inner arc of the elastomeric diaphragm, the outer circle of the elastomeric diaphragm, and the outer arc of the elastomeric diaphragm, respectively.Figure 5
Force analysis of part “1” of the elastomeric diaphragm.
(a)
Schematic diagram and force analysis of part “1” (b)
Parameters and force analysis of “1” unit of part “1”Figure5(b) shows the unit of part “1”, and the unit stress in three directions can be expressed as(18)δ11=2C1λ112-C2λ11-2+P1(19)δ21=2C1λ212-C2λ21-2+P1(20)δ31=2C1λ312-C2λ31-2+P1=Pawhere λij is the corresponding stretch ratio and P1 is the unknown pressure. The stretch ratio can be expressed as λ11=(h1+ε1)/h1, where ε1 is the length variable in the vertical direction of part “1”.The following equation can be obtained by combining (18) with (20):(21)δ11=Pa+2C1λ112-C2λ11-2-2C1λ312-C2λ31-2.Substituting (17) and (21) into (16) yields(22)F1=πPa+2C1λ112-C2λ11-2-2C1λ312-C2λ31-2R+d0λ312-R2.The force on the unit in the cross-section is shown in Figure6. The force of the piston acting on the unit of part “1” can be expressed as(23)FN1=FP1+2F21·sin⁡γ12where Fp1 is the third principal stress surface of part “1” subjected to the force of the gas chamber,F21 is the second principal stress surface of part “1” subjected to the force, and γ1 is the angle of the unit in the cross-section of the dual-chamber pneumatic spring.Figure 6
Cross-section of one unit of part “1”.F P 1 and F21 are defined as(24)FP1=Pa·h1λ11·b1λ21(25)F21=δ21·h1λ11·d0λ31where h1λ11·b1λ21 is the area of the third principal stress surface of part “1”, δ21 is the stress of the second principal stress surface of part “1”, and h1λ11·d0λ31 is the area of the second principal stress surface of part “1”, as shown in Figure 5(b).Substituting (24) and (25) into (23) yields(26)FN1=Pa·h1λ11·b1λ21+2δ21·h1λ11·d0λ31·sin⁡γ12which indicates that FN1>0. Therefore, part “1” of the elastomeric diaphragm remains in contact with the piston wall, and the size of part “1” of the elastomeric diaphragm does not change when it is stretched in the circumferential direction; i.e., λ21=1. The incompressibility of rubber leads to λ11λ21λ31=1, and the following expression can be obtained:(27)λ31=λ11-1=h1+ε1h1-1.Substituting (10) and (27) into (22) yields(28)F1=π2mgπR+2d0+2r-d0h3/h3+ε32+d0h1/h1+ε1+R2+P0+2C1+C2h1+ε1h12-h1h1+ε12R+h1d0h1+ε12-R2.To obtain a function for the stiffnessK1 of part “1” with respect to variable ε1, the following equation can be obtained by differentiating (28) with respect to ε1:(29)K1=dF1dε1.The complete equation(29) is extremely long and is provided in Appendix B.


### 2.4.2. Part “2” Stiffness of the Elastomeric Diaphragm

As shown in Figure7, the tension F2 acting on part “2” can be expressed as(30)F2=∫0π/2∫02πδ2dst2·dαwhere F2 originates from the pressure difference of the top chamber and ambient environment in the vertical direction.Figure 7
Force analysis of part “2” of the elastomeric diaphragm.
(a)
Schematic diagram of part “2” (b)
Parameters of “2” unit of part “2” (c)
Force analysis of “2” unit of part “2” (d)
Force analysis of part “2”For the projected area of the unit,dst2 can be written as(31)dst2=R+r1-cos⁡αdθ·d0λ32·cos⁡αwhere R+r(1-cos⁡α) is the length l2 from the analysis point to the center axis, (R+r(1-cos⁡α))dθ is the side length d2 of the unit in part “2”, and d0λ32 is the thickness of the unit in part “2”. θ is shown in Figure 7(a).The vertical stressδ2 of the unit of part “2” can be expressed as(32)δ2=δ12cos⁡α+δ32sin⁡αwhere δ12 and δ32 can be written as(33)δ12=2C1λ122-C2λ12-2+P2(34)δ32=2C1λ322-C2λ32-2+P2=Pa-P0where δ12 and δ32 are the stresses of the corresponding principal stress surface and P2 is the unknown pressure. The stretch ratio can be expressed as λ12=(r+ε2)/r, where ε2 is the length variable in the vertical direction. The size of part “2” does not change when it is stretched in the circumferential direction; i.e., λ22=1. The incompressibility of rubber leads to λ12λ22λ32=1. Thus, the following equation can be obtained:(35)λ32=λ12-1=r+ε2r-1The following equation can be obtained by transferring (32) to (35):(36)δ2=Pa-P0+2C1+C2r+ε2r2-rr+ε22cos⁡α+Pa-P0sin⁡α.Substituting (10), (30), (31), and (36) into (30) yields(37)F2=2π2mgπR+2d0+2r-d0h3/h3+ε32+d0h1/h1+ε1+R2+2C1+C2r+ε2r2-rr+ε22πd0rR+r4r+ε2-2d0r23r+ε2+2mgπR+2d0+2r-d0h3/h3+ε32+d0h1/h1+ε1+R2d0rR+rr+ε2-d0r23r+ε2.To obtain a function of the stiffnessK2 of part “2” with respect to variable ε2, the following equation can be obtained by differentiating (37) with respect to ε2:(38)K2=dF2dε2.The complete equation(38) is extremely long and is provided in Appendix B.


### 2.4.3. Part “3” Stiffness of the Elastomeric Diaphragm

As shown in Figure8, when an elastomeric diaphragm is stretched, the tension F3 acting on part “3” can be expressed as(39)F3=δ13st3Figure 8
Force analysis of part “3” of the elastomeric diaphragm.
(a)
Schematic diagram and force analysis of part “3” (b)
Parameters and force analysis of “3” unit of part “3”The analysis of part “3” is similar to that of part “1”. Therefore, based on the Mooney-Rivlin model, the vertical stressδ13 can be expressed as(40)δ13=Pa+2C1λ132-C2λ13-2-2C1λ332-C2λ33-2.where the stretch ratio can be expressed as λ13=(h3+ε3)/h3 and ε3 is the variable length in the vertical stretch of part “3”.Additionally,st3 is the cross-sectional area of part “3”, as shown in Figure 4. Herein, st3 can be expressed as(41)st3=πR+2d0+2r2-R+2d0+2r-d0λ332.Substituting (40) and (41) into (39) yields(42)F3=πPa+2C1λ132-C2λ13-2-2C1λ332-C2λ33-2·R+2d0+2r2-R+2d0+2r-d0λ332.The force on the unit in the cross-section is shown in Figure9. The force of the piston acting on the unit of part “3” can be expressed as(43)FN3=FP3-2F23·sin⁡γ32where Fp3 is the third principal stress surface of part “3” subjected to the force of the gas chamber, F23 is the second principal stress surface of part “3” subjected to the force, and γ3 is the angle of the unit in cross-section of the dual-chamber pneumatic spring.Figure 9
Cross-section of one unit of part “3”.F P 3 and F23 are written as(44)FP3=Pa·h3λ13·b3λ23(45)F23=δ23·h3λ13·d0λ33where h3λ13·b3λ23 is the area of the third principal stress surface of part “3”, δ23 is the stress of the second principal stress surface of part “3”, and h3λ13·d0λ33 is the area of the second principal stress surface of part “3”, as shown in Figure 8(b).Substituting (44) and (45) into (43) yields(46)FN3=Pa·h3λ13·b3λ23-2δ23·h3λ13·d0λ33·sin⁡γ32where sin⁡(γ3/2)=b3/2(R+2r+2d0-d0λ33) and b3 is the arc length of the unit of part “3”, which is smaller than 2(R+2r+2d0-d0λ33). Thus, the first polynomial is larger than the second polynomial of (46); i.e., FN3>0. Therefore, part “3” remains in contact with the fixed ring, and the size of part “3” does not change when it is stretched in the circumferential direction; i.e., λ23=1. However, the incompressibility of rubber leads to λ13λ23λ33=1. Thus, the following equation can be obtained:(47)λ33=λ13-1=h3+ε3h3-1.Substituting (10) and (47) into (42) yields(48)F3=π2mgπR+2d0+2r-d0h3/h3+ε32+d0h1/h1+ε1+R2+P0+2C1+C2h3+ε3h32-h3h3+ε32R+2d0+2r2-R+2d0+2r-d0h3h3+ε32.To obtain a function of the stiffnessK3 of part “3” with respect to variable ε3, the following equation can be obtained by differentiating (48) with respect to ε3:(49)K3=dF3dε3.The complete equation(49) is extremely long and is provided in Appendix B.


### 2.4.4. Part “4” Stiffness of the Elastomeric Diaphragm

As shown in Figure10, the tension F4 acting on part “4” can be expressed as(50)F4=∫0π/2∫02πδ4dst4·dβwhere F4 originates from the pressure difference of the top chamber and the ambient environment in the vertical direction.Figure 10
Force analysis of part “4” of the elastomeric diaphragm.
(a)
Schematic diagram of part “4” (b)
Parameters of the “4” unit of part “4” (c)
Force analysis of the “4” unit of part “4” (d)
Force analysis of part “4”For the projected area of the unit,dst4 can be written as(51)dst4=R+r1+cos⁡βdθ·d0λ34·cos⁡βwhere R+r(1+cos⁡β) is the length l4 from the analysis point to the center axis, (R+r(1+cos⁡β))dθ is the side length d4 of the unit in part “4”, d0λ34 is the thickness of the unit in part “4”, and θ is shown in Figure 10(a).The vertical stressδ4 of the partial unit in part “4” can be expressed as(52)δ4=δ14cos⁡β+δ34sin⁡βandδ14 andδ34 can be rewritten as(53)δ14=2C1λ142-C2λ14-2+P4(54)δ34=2C1λ342-C2λ34-2+P4=Pa-P0where δ14 and δ34 are the stresses corresponding to the principal stress surface and P4 is the unknown pressure. The stretch ratio can be expressed as λ14=(r+ε4)/r, where ε4 is the variable length in the vertical stretch of part “4”. The size of part “4” does not change when it is stretched in the circumferential direction; i.e., λ24=1. The incompressibility of rubber leads to λ14λ24λ34=1. Thus, the following equation can be obtained:(55)λ34=λ14-1=r+ε4r-1.The following equation can be obtained by transferring (52) to (55):(56)δ4=Pa-P0+2C1+C2r+ε4r2-r+ε4r-2cos⁡β+Pa-P0sin⁡β.Substituting (10), (51), (55), and (56) into (50) yields(57)F4=2π2mgπR+2d0+2r-d0h3/h3+ε32+d0h1/h1+ε1+R2+2C1+C2r+ε4r2-rr+ε42πd0rR+r4r+ε4+2d0r23r+ε4+2mgπR+2d0+2r-d0h3/h3+ε32+d0h1/h1+ε1+R212d0r+ε4r-1R+r+13d0r+ε4r-1r.To obtain a function of the stiffnessK4 of part “4” with respect to variable ε4, the following equation can be obtained by differentiating (57) with respect to ε4:(58)K4=dF4dε4.The entire equation(58) is extremely long and is provided in Appendix B.


### 2.4.5. Total Stiffness of the Elastomeric Diaphragm

As shown in Figure11, part “1” and part “2” are connected in series such that the equivalent stiffness K12 can be expressed as(59)1K12=1K1+1K2.Figure 11
Total stiffness of the elastomeric diaphragm.Part “3” and part “4” are connected in series such that the equivalent stiffnessK34 can be expressed as(60)1K34=1K3+1K4.The total stiffness can be expressed as(61)Km=K12+K34=K1K2K1+K2+K3K4K3+K4.


## 2.4.1. Part “1” Stiffness of the Elastomeric Diaphragm

As shown in Figure5(a), when an elastomeric diaphragm is stretched, the tension F1 acting on part “1” can be expressed as(16)F1=δ11st1where st1 is the sectional area of part “1”, which is shown in Figure 4, and st1 can be expressed as(17)st1=πR+d0λ312-R2where λ31 is the corresponding stretch ratio of a unit of part “1”. In this paper, the letters of the subscript contain two numbers i and j, where i(1,2,3) represents the corresponding principal stress surface and j(1,2,3,4) represents the corresponding component of the elastomeric diaphragm, i.e., the inner circle of the elastomeric diaphragm, the inner arc of the elastomeric diaphragm, the outer circle of the elastomeric diaphragm, and the outer arc of the elastomeric diaphragm, respectively.Figure 5
Force analysis of part “1” of the elastomeric diaphragm.
(a)
Schematic diagram and force analysis of part “1” (b)
Parameters and force analysis of “1” unit of part “1”Figure5(b) shows the unit of part “1”, and the unit stress in three directions can be expressed as(18)δ11=2C1λ112-C2λ11-2+P1(19)δ21=2C1λ212-C2λ21-2+P1(20)δ31=2C1λ312-C2λ31-2+P1=Pawhere λij is the corresponding stretch ratio and P1 is the unknown pressure. The stretch ratio can be expressed as λ11=(h1+ε1)/h1, where ε1 is the length variable in the vertical direction of part “1”.The following equation can be obtained by combining (18) with (20):(21)δ11=Pa+2C1λ112-C2λ11-2-2C1λ312-C2λ31-2.Substituting (17) and (21) into (16) yields(22)F1=πPa+2C1λ112-C2λ11-2-2C1λ312-C2λ31-2R+d0λ312-R2.The force on the unit in the cross-section is shown in Figure6. The force of the piston acting on the unit of part “1” can be expressed as(23)FN1=FP1+2F21·sin⁡γ12where Fp1 is the third principal stress surface of part “1” subjected to the force of the gas chamber,F21 is the second principal stress surface of part “1” subjected to the force, and γ1 is the angle of the unit in the cross-section of the dual-chamber pneumatic spring.Figure 6
Cross-section of one unit of part “1”.F P 1 and F21 are defined as(24)FP1=Pa·h1λ11·b1λ21(25)F21=δ21·h1λ11·d0λ31where h1λ11·b1λ21 is the area of the third principal stress surface of part “1”, δ21 is the stress of the second principal stress surface of part “1”, and h1λ11·d0λ31 is the area of the second principal stress surface of part “1”, as shown in Figure 5(b).Substituting (24) and (25) into (23) yields(26)FN1=Pa·h1λ11·b1λ21+2δ21·h1λ11·d0λ31·sin⁡γ12which indicates that FN1>0. Therefore, part “1” of the elastomeric diaphragm remains in contact with the piston wall, and the size of part “1” of the elastomeric diaphragm does not change when it is stretched in the circumferential direction; i.e., λ21=1. The incompressibility of rubber leads to λ11λ21λ31=1, and the following expression can be obtained:(27)λ31=λ11-1=h1+ε1h1-1.Substituting (10) and (27) into (22) yields(28)F1=π2mgπR+2d0+2r-d0h3/h3+ε32+d0h1/h1+ε1+R2+P0+2C1+C2h1+ε1h12-h1h1+ε12R+h1d0h1+ε12-R2.To obtain a function for the stiffnessK1 of part “1” with respect to variable ε1, the following equation can be obtained by differentiating (28) with respect to ε1:(29)K1=dF1dε1.The complete equation(29) is extremely long and is provided in Appendix B.


## 2.4.2. Part “2” Stiffness of the Elastomeric Diaphragm

As shown in Figure7, the tension F2 acting on part “2” can be expressed as(30)F2=∫0π/2∫02πδ2dst2·dαwhere F2 originates from the pressure difference of the top chamber and ambient environment in the vertical direction.Figure 7
Force analysis of part “2” of the elastomeric diaphragm.
(a)
Schematic diagram of part “2” (b)
Parameters of “2” unit of part “2” (c)
Force analysis of “2” unit of part “2” (d)
Force analysis of part “2”For the projected area of the unit,dst2 can be written as(31)dst2=R+r1-cos⁡αdθ·d0λ32·cos⁡αwhere R+r(1-cos⁡α) is the length l2 from the analysis point to the center axis, (R+r(1-cos⁡α))dθ is the side length d2 of the unit in part “2”, and d0λ32 is the thickness of the unit in part “2”. θ is shown in Figure 7(a).The vertical stressδ2 of the unit of part “2” can be expressed as(32)δ2=δ12cos⁡α+δ32sin⁡αwhere δ12 and δ32 can be written as(33)δ12=2C1λ122-C2λ12-2+P2(34)δ32=2C1λ322-C2λ32-2+P2=Pa-P0where δ12 and δ32 are the stresses of the corresponding principal stress surface and P2 is the unknown pressure. The stretch ratio can be expressed as λ12=(r+ε2)/r, where ε2 is the length variable in the vertical direction. The size of part “2” does not change when it is stretched in the circumferential direction; i.e., λ22=1. The incompressibility of rubber leads to λ12λ22λ32=1. Thus, the following equation can be obtained:(35)λ32=λ12-1=r+ε2r-1The following equation can be obtained by transferring (32) to (35):(36)δ2=Pa-P0+2C1+C2r+ε2r2-rr+ε22cos⁡α+Pa-P0sin⁡α.Substituting (10), (30), (31), and (36) into (30) yields(37)F2=2π2mgπR+2d0+2r-d0h3/h3+ε32+d0h1/h1+ε1+R2+2C1+C2r+ε2r2-rr+ε22πd0rR+r4r+ε2-2d0r23r+ε2+2mgπR+2d0+2r-d0h3/h3+ε32+d0h1/h1+ε1+R2d0rR+rr+ε2-d0r23r+ε2.To obtain a function of the stiffnessK2 of part “2” with respect to variable ε2, the following equation can be obtained by differentiating (37) with respect to ε2:(38)K2=dF2dε2.The complete equation(38) is extremely long and is provided in Appendix B.


## 2.4.3. Part “3” Stiffness of the Elastomeric Diaphragm

As shown in Figure8, when an elastomeric diaphragm is stretched, the tension F3 acting on part “3” can be expressed as(39)F3=δ13st3Figure 8
Force analysis of part “3” of the elastomeric diaphragm.
(a)
Schematic diagram and force analysis of part “3” (b)
Parameters and force analysis of “3” unit of part “3”The analysis of part “3” is similar to that of part “1”. Therefore, based on the Mooney-Rivlin model, the vertical stressδ13 can be expressed as(40)δ13=Pa+2C1λ132-C2λ13-2-2C1λ332-C2λ33-2.where the stretch ratio can be expressed as λ13=(h3+ε3)/h3 and ε3 is the variable length in the vertical stretch of part “3”.Additionally,st3 is the cross-sectional area of part “3”, as shown in Figure 4. Herein, st3 can be expressed as(41)st3=πR+2d0+2r2-R+2d0+2r-d0λ332.Substituting (40) and (41) into (39) yields(42)F3=πPa+2C1λ132-C2λ13-2-2C1λ332-C2λ33-2·R+2d0+2r2-R+2d0+2r-d0λ332.The force on the unit in the cross-section is shown in Figure9. The force of the piston acting on the unit of part “3” can be expressed as(43)FN3=FP3-2F23·sin⁡γ32where Fp3 is the third principal stress surface of part “3” subjected to the force of the gas chamber, F23 is the second principal stress surface of part “3” subjected to the force, and γ3 is the angle of the unit in cross-section of the dual-chamber pneumatic spring.Figure 9
Cross-section of one unit of part “3”.F P 3 and F23 are written as(44)FP3=Pa·h3λ13·b3λ23(45)F23=δ23·h3λ13·d0λ33where h3λ13·b3λ23 is the area of the third principal stress surface of part “3”, δ23 is the stress of the second principal stress surface of part “3”, and h3λ13·d0λ33 is the area of the second principal stress surface of part “3”, as shown in Figure 8(b).Substituting (44) and (45) into (43) yields(46)FN3=Pa·h3λ13·b3λ23-2δ23·h3λ13·d0λ33·sin⁡γ32where sin⁡(γ3/2)=b3/2(R+2r+2d0-d0λ33) and b3 is the arc length of the unit of part “3”, which is smaller than 2(R+2r+2d0-d0λ33). Thus, the first polynomial is larger than the second polynomial of (46); i.e., FN3>0. Therefore, part “3” remains in contact with the fixed ring, and the size of part “3” does not change when it is stretched in the circumferential direction; i.e., λ23=1. However, the incompressibility of rubber leads to λ13λ23λ33=1. Thus, the following equation can be obtained:(47)λ33=λ13-1=h3+ε3h3-1.Substituting (10) and (47) into (42) yields(48)F3=π2mgπR+2d0+2r-d0h3/h3+ε32+d0h1/h1+ε1+R2+P0+2C1+C2h3+ε3h32-h3h3+ε32R+2d0+2r2-R+2d0+2r-d0h3h3+ε32.To obtain a function of the stiffnessK3 of part “3” with respect to variable ε3, the following equation can be obtained by differentiating (48) with respect to ε3:(49)K3=dF3dε3.The complete equation(49) is extremely long and is provided in Appendix B.


## 2.4.4. Part “4” Stiffness of the Elastomeric Diaphragm

As shown in Figure10, the tension F4 acting on part “4” can be expressed as(50)F4=∫0π/2∫02πδ4dst4·dβwhere F4 originates from the pressure difference of the top chamber and the ambient environment in the vertical direction.Figure 10
Force analysis of part “4” of the elastomeric diaphragm.
(a)
Schematic diagram of part “4” (b)
Parameters of the “4” unit of part “4” (c)
Force analysis of the “4” unit of part “4” (d)
Force analysis of part “4”For the projected area of the unit,dst4 can be written as(51)dst4=R+r1+cos⁡βdθ·d0λ34·cos⁡βwhere R+r(1+cos⁡β) is the length l4 from the analysis point to the center axis, (R+r(1+cos⁡β))dθ is the side length d4 of the unit in part “4”, d0λ34 is the thickness of the unit in part “4”, and θ is shown in Figure 10(a).The vertical stressδ4 of the partial unit in part “4” can be expressed as(52)δ4=δ14cos⁡β+δ34sin⁡βandδ14 andδ34 can be rewritten as(53)δ14=2C1λ142-C2λ14-2+P4(54)δ34=2C1λ342-C2λ34-2+P4=Pa-P0where δ14 and δ34 are the stresses corresponding to the principal stress surface and P4 is the unknown pressure. The stretch ratio can be expressed as λ14=(r+ε4)/r, where ε4 is the variable length in the vertical stretch of part “4”. The size of part “4” does not change when it is stretched in the circumferential direction; i.e., λ24=1. The incompressibility of rubber leads to λ14λ24λ34=1. Thus, the following equation can be obtained:(55)λ34=λ14-1=r+ε4r-1.The following equation can be obtained by transferring (52) to (55):(56)δ4=Pa-P0+2C1+C2r+ε4r2-r+ε4r-2cos⁡β+Pa-P0sin⁡β.Substituting (10), (51), (55), and (56) into (50) yields(57)F4=2π2mgπR+2d0+2r-d0h3/h3+ε32+d0h1/h1+ε1+R2+2C1+C2r+ε4r2-rr+ε42πd0rR+r4r+ε4+2d0r23r+ε4+2mgπR+2d0+2r-d0h3/h3+ε32+d0h1/h1+ε1+R212d0r+ε4r-1R+r+13d0r+ε4r-1r.To obtain a function of the stiffnessK4 of part “4” with respect to variable ε4, the following equation can be obtained by differentiating (57) with respect to ε4:(58)K4=dF4dε4.The entire equation(58) is extremely long and is provided in Appendix B.


## 2.4.5. Total Stiffness of the Elastomeric Diaphragm

As shown in Figure11, part “1” and part “2” are connected in series such that the equivalent stiffness K12 can be expressed as(59)1K12=1K1+1K2.Figure 11
Total stiffness of the elastomeric diaphragm.Part “3” and part “4” are connected in series such that the equivalent stiffnessK34 can be expressed as(60)1K34=1K3+1K4.The total stiffness can be expressed as(61)Km=K12+K34=K1K2K1+K2+K3K4K3+K4.


## 2.5. Numerical Solution of the Vertical Stiffness for the Elastomeric Diaphragm

When the load of the dual-chamber pneumatic spring ism, each part of the elastomeric diaphragm is stretched to a certain length. The stretched lengths of each corresponding part are defined as ε10, ε20, ε30, and ε40. By analyzing (3) and (11), each component of the force can be expressed as(62)F1=F1′=F2=F3=F3′=F4=FP2=mgR+2d0+2r-d0h3/h3+ε32-d0h1/h1+ε1+R2R+2d0+2r-d0h3/h3+ε32+d0h1/h1+ε1+R2.The analysis process to obtain the stretched length of each corresponding part with respect to the other variables is as follows. Substitutingε1=ε10 into (28) and noting (62), we can obtain (B.5) of the stretched length ε10. (B.6), (B.7), and (B.8) can also be obtained in a similar manner. Equations (B.5) to (B.8) are rather long and are provided in Appendix B.By solving (B.5) to (B.8) in the numerical root of ε10, ε20, ε30, and ε40 and substituting ε1=ε10, ε2=ε20, ε3=ε30, and ε4=ε40 into (29), (38), (49), and (58), we obtain the numerical solutions of stiffnessK1,K2,K3, andK4, respectively. As a result, the total stiffness Km can be obtained from (61).


## 3. Experimental Validation

To verify the theoretical analysis, we test the performance and calculate the stiffness of the specific dual-chamber pneumatic spring.The following text describes the method for obtaining 2(C1+C2), whereC1 andC2 are the Mooney-Rivlin material parameters. Classical rubber theory yields the relationship between the stiffness values of different types of rubbers and their Young’s elastic moduli [15]. The elastomeric diaphragms commonly used in design of the pneumatic spring structure consist of chloroprene rubber, nitrile butadiene rubber, butyl rubber, natural rubber, and other materials. A moderate stiffness is typically used to obtain a better overall performance. Therefore, the widely applied nitrile butadiene rubber with hardness (55±5) HA is used in the next two experiments. The Young’s elastic modulus E changes from 2.465 MPa to 3.619 MPa via the relationship between the hardness and Young’s elastic modulus. According to the incompressibility of the rubber material, we can obtain E≈3Gs. Hence, Gs ranges from 0.82 MPa to 1.20 MPa, where Gs is the shear modulus and Gs=2(C1+C2). Finally, 2(C1+C2) also ranges from 0.82 MPa to 1.20 MPa.The vertical total stiffnessKtotal of the dual-chamber pneumatic isolator can be calculated according to the parameters in Table 2, where the stiffness of the gas is obtained based on Erin’s study. Two experiments are shown in Figures 12 and 13. The main components of the isolation platforms are composed of four dual-chamber pneumatic isolators. The microvibration of the ground can be attenuated by the dual-chamber pneumatic isolators. Thus, we can obtain the transmissibility curve using the testing instrument HANDYSAM [17]. The input and output sensors used in this study are the ultralow frequency vibration pickup sensor 991B produced by the Institute of Engineering Mechanics, China Earthquake Administration. The sensitivity of the sensors is 4×10−8 m/s. The load in experiment 1 is 50 kg, and the load in experiment 2 is 200 kg. Ground excitation used in the two experiments is just ambient vibration, whose signal in vertical direction measured by the input sensors is shown in Figure 14.Table 2
Parameters of the dual-chamber pneumatic in the two experiments.
Symbol Description Value 1 Value 2 Unit 2 ( C 1 + C 2 ) C 1 and C2 are the Mooney-Rivlin material parameters 0.82-1.20 0.82-1.20 MPa h 1 Part “1” vertical length 0.009 0.015 m r Part “2” or “4” projection length 0.0034 0.0050 m h 3 Part “3” vertical length 0.014 0.020 m d 0 Diaphragm thickness 0.0006 0.002 m p 0 Atmospheric pressure 101325 101325 MPa g Gravitational acceleration 9.81 9.81 m/s2 M Payload mass 20 375 kg n Polytropic exponent 1.4 1.4 - L Flow restrictor length 4e-4 8e-3 m D Flow restrictor diameter 1e-3 1e-3 m A b Effective piston area 0.0015 0.016 m2 μ Air viscosity 1.824e-5 1.824e-5 Pa⋅s C r Flow restriction constant 3.3640e-06 1.6019e-07 m3/(s⋅Pa) V b Lower chamber volume 3.0000e-04 1.8170e-04 m3 V t Upper chamber volume 2.7298e-05 2.3000e-03 m3Figure 12
Photograph of experiment 1.Figure 13
Photograph of experiment 2.Figure 14
Ground vibration velocity curve in vertical direction versus time.The theoretical expression of the frequency response functions has been inferred without considering the elastomeric diaphragm [10]. In this paper, the theoretical transmissibility curves are obtained as shown in Figures 15 and 16 by substituting the actual parameters of the experimental pneumatic vibration absorber shown in Table 2 into Erin’s analytical model. The first curve is Erin’s result without consideration of the elastomeric diaphragm. The second and third curves are our results with consideration of the elastomeric diaphragm. The last curve is the experimental result with the parameters shown in Table 2. Our model yields two curves, namely, the upper and lower bounds for the inherent frequencies, which correspond to the maximum and minimum values of 2(C1+C2).Figure 15
Transmissibility curves of experiment 1.Figure 16
Transmissibility curves of experiment 2.The results of experiment 1 are shown in Figure15, where fa1=1.89Hz, fC1=6.03-7.70Hz, and fE1=6.75Hz are the inherent frequencies of the double-chamber pneumatic springs based on Erin’s study and the proposed theory in this paper that considers the elastomeric diaphragm stiffness and experiment results, respectively. The results of experiment 2 are shown in Figure 16, where fa2=1.31Hz, fC2=3.38-4.30Hz,and fE2=3.75Hz are the inherent frequencies based on Erin’s study and the proposed theoretical and experiment results, respectively. The value of the damping in our two experiments is related to both the length and diameter of the flow restrictor. In experiment 1, the flow restrictor length is 0.0004, and the flow restrictor diameter is 0.001; thus, the value of the damping ratio is 0.005. In experiment 2, the flow restrictor length is 0.008, and the flow restrictor diameter is 0.016; thus, the value of the damping ratio is 0.1.Figures15 and 16 illustrate that fE1 and fE2 are within the ranges of fC1 and fC2, respectively, demonstrating the effectiveness of the vertical stiffness model for the elastomeric diaphragm.The stiffness values of the experiments can be obtained from the relationship of the inherent frequency, the stiffness, and the payload; i.e.,(63a)KE1=mωE12=m2πfE12,(63b)KE2=mωE22=m2πfE22.The calculation results of the two structures with different parameters of the dual-chamber pneumatic spring are shown in Table3. The vertical stiffness,KE1 andKE2, in the experimental tests is within the corresponding ranges of total stiffnessKtotal obtained via calculations, demonstrating the effectiveness of the vertical stiffness model for the elastomeric diaphragm.Table 3
Stiffnesses from the calculations and experiments.
Symbol Description Value 1 Value 2 Unit ε 10 Part “1” length stretched 15.0e-04 to 9.19e-04 22.0e-04 to 14.0e-04 m ε 20 Part “2” length stretched 8.53e-04 to 5.16e-04 9.90e-04 to 5.85e-04 m ε 30 Part “3” length stretched 13.0e-04 to 8.55e-04 18.0e-04 to 11.0e-04 m ε 40 Part “4” length stretched 6.26e-04 to 3.88e-04 8.23e-04 to 4.94e-04 m  K 1 Part “1” stiffness 1.87e+04 to 3.24e+04 9.97e+04 to 1.75e+05 N/m K 2 Part “2” stiffness 3.54e+04 to 6.32e+04 2.34e+05 to 4.12e+05 N/m K 3 Part “3” stiffness 1.90e+04 to 3.10e+04 1.04e+05 to 1.73e+05 N/m K 4 Part “4” stiffness 4.84e+04 to 8.42e+04 2.52e+05 to 4.54e+05 N/m K 12 Part “1” and “2” equivalent stiffness 1.23e+04 to 2.14e+04 6.99e+04 to 1.23e+05 N/m K 34 Part “3” and “4” equivalent stiffness 1.36e+04 to 2.27e+04 7.35e+04 to 1.25e+05 N/m  K m Rubber diaphragm stiffness 2.59e+04 to 4.41e+04 1.43e+05 to 2.48e+05 N/m K a Air stiffness 0.2850e+04 2.5406e+04 N/m K t o t a l Total stiffness 2.87e+04 to 4.90e+04 1.69e+05 to 2.74e+05 N/m K E Experimental stiffness 3.5975e+04 2.0819e+05 N/m  f C Calculated inherent frequency 6.0332 to 7.7096 3.3769 to 4.2931 Hz f E Experimental inherent frequency 6.625 3.5 Hz


## 4. Conclusions

This work developed an analytical model of the elastomeric diaphragm using the Mooney-Rivlin modeling method and elastomeric theory. The stiffness of the elastomeric diaphragm was ignored in the previous model of pneumatic vibration isolators. Specifically, each segmented elastomeric diaphragm can be simplified as uniaxial stretching, and the Mooney-Rivlin model can be adopted to establish the analytical expression of the vertical stiffness. The analytical model of the elastomeric diaphragm is built by combining all segments and can be resolved by numerical solution. The experimental results demonstrated that the analytically predicted behavior is highly similar to the practically observed behavior in terms of the resonant frequency of the isolator after the elastomeric diaphragm model is added to the standard pneumatic cylinder model. Differing from the prior converse correction and experience design, this paper supplies a direct design method that is helpful for optimizing and designing the pneumatic spring parameters in practice.



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