Synthesizing reduced systems by complex receptances

Journal of Sound and Vibration (1994) 179(5), 855–867

SYNTHESIZING REDUCED SYSTEMS BY COMPLEX RECEPTANCES D. T. S†  W. S School of Mechanical Engineering, Purdue University, West Lafayette, Indiana 47907–1288, U.S.A. (Received 26 February 1993, and in final form 20 October 1993) While the addition of linear dynamic systems using receptances is well established and relatively widely used, at times it becomes necessary to subtract systems. For example, when measuring transfer functions at the tire of an automotive suspension system, the system is held in the position of interest by an auxiliary airspring, which requires correction of the experimental data. It is shown that the subtraction is a non-trivial problem, requiring auxiliary experiments or calculations, but if they can be performed then natural frequencies, modes and corrected responses of the reduced system can be obtained from measured or calculated receptances of the total system.

1. INTRODUCTION

When testing the responses of vibrating systems, the measured response is not always the true response. This is due to artificial additions to the system in order to be able to perform experimental response tests. In most cases, it is assumed that the additions can simply be neglected. An example of this is the response testing of a tire on an automobile suspension. In order for the suspension not to sag, an airbag is placed under it in order to support it so that a shaker can be placed under the tire without having to support the weight of the vehicle. The airbag is the addition to the system that engineers typically neglect to account for, since there has been no reasonable way to subtract its influence. The argument which is usually made is that the airbag is a very soft spring which will not influence the response in higher frequency regions. But what about the lower frequency regions? With the airbag, the tire response is not a true response to a given forcing. This paper shows how to obtain the true response by theoretically taking away the addition to the system (theoretically, taking away the airbag from the automobile suspension). The theory can also be used theoretically to remove part of a system to see if this change decreases the response at points of interests, instead of having physically to take part of the system away (if possible) and testing the system again. The method that is used to solve the problem uses the receptance method [1, 2] in a new subtractive way instead of the usual additive way. A receptance is defined as the ratio of a deflection response at a point to a harmonic force or moment input at a point. Receptances have been used in the additions of even relatively complicated harmonic systems with excellent success [3–5]. The method is useful in theoretically obtaining deflection responses at points of interest in systems that are the combinations of two or more subsystems, by knowing the receptances of the subsystems experimentally or theoretically. †Now with Noell Inc., Herndon, Virginia 22071, U.S.A.

855 0022–460X/95/050855 + 13 $08.00/0

7 1995 Academic Press Limited

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. .   . 

The receptance method applies to both discrete and continuous systems [1]. Examples include joining a beam to a beam, a discrete oscillator to a plate or a shell, or a spring mass to a ring. Although much has been done in the additions of systems with the receptance method, there has been little done, if anything at all, in subtracting a subsystem from a system. Subtracting a system is not as simple as substituting a negative receptance (−bij ) into the receptance equation for the system under analysis, but equations for the receptances of interest must be found in terms of known receptances (experimentally or theoretically). In the following, the general theory of subtraction is presented for a system divided into two parts (one being the part to be removed and the other being the rest of the system). It will be pointed out that subtraction may necessitate some auxiliary experiments. Next, two examples will be shown in which the general theory is applied. The first example is a simple two-degree-of-freedom mass–spring–damper system, where one spring–mass combination is to be subtracted. The second is the automobile suspension response test with airbag, where the suspension is treated as a discretized system.

2. GENERAL THEORY

2.1.    Using a generalized system C, which consists of systems A and B, as shown in Figure 1, and using the theory of receptances [1], the equations for the receptances of system A alone can be found. Let us at first pretend that we would like to add systems A and B. In the case of Figure 1, where a force F1 e jvt is applied to the system at location 1 [2], XA1 = XC1 = a11 FA1 + a12 FA2 ,

XA2 = XC2 = a21 FA1 + a22 FA2 ,

(1, 2)

where x = X e jvt

(3)

and where the aij can be complex numbers. Since FA2 = −FB2 ,

(4)

XC2 = XB2 = b22 FB2 ,

(5)

FA2 = −XC2 /b22 .

(6)

FA1 = F1 ,

(7)

and

one obtains

Since

Figure 1. System C, composed of systems A and B.

  

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equations (1) and (2) become XC1 = a11 F1 −

a12 X , b22 C2

XC2 = a21 F1 −

a22 X . b22 C2

(8, 9)

Solving these two equations results in g11 =

XC1 a a = a11 − 21 12 , F1 a22 + b22

g21 =

XC2 b a = 22 21 , F1 a22 + b22

(10, 11)

where g designates system C (A and B) receptances, a designates system A receptances, and b designates system B receptances. We assume next that we know all g receptances and the b22 receptance from theory or from measurements. From Maxwell’s reciprocity theorem, aij = aji .

(12)

Therefore, there are two equations with three unknowns (a11 , a12 = a21 , a22 ), since the responses of system A alone are of interest here. This shows that we need to generate a third equation. This third equation is found by placing a force at the junction of system A and B. Similar to the above analysis, one obtains as the third equation: g22 = a22 b22 /(a22 + b22 ).

(13)

Solving equations (10), (11), and (13) for a11 , a21 and a22 gives a22 =

−g22 b22 , g22 − b22

a11 =

2 g11 (g22 − b22 ) − g21 , g22 − b22

a21 =

−g21 b22 . g22 − b22

(14–16)

Therefore, for any system that fits the system of Figure 1, one can obtain the system A response (after system B has been taken away) knowing receptances for system C (g11 , g21 , g22 ) and the receptance for system B (b22 ). If g22 is to be obtained experimentally, this means that the shaker will have to be attached to point 2 of system C. This is unexpected in case of the tire-suspension example, because uncorrected experiments are usually conducted by locating the shaker at point 1 only, but it is necessary and is termed, in the following, an ‘auxiliary’ experiment. 2.2.     If only the natural frequencies of the reduced system are to be found, one obtains from equations (14)–(16), for zero damping, the frequency equation g22 − b22 = 0.

(17)

The values of v which satisfy this equation are the natural frequencies vk of the reduced system. On an intuitive level, it seems possible to think of the system subtraction in terms of adding a ‘‘negative’’ receptance (−b22 ) to the total system receptance g22 , very much like the standard frequency equation for system C, which in terms of its component receptances a22 and b22 is a22 + b22 = 0 [1]. This is a speculation which was shared by the present authors and the authors of references [3–5] some time ago. This is not unreasonable for simple systems B, such as a mass void (a negative mass) or a missing linear spring (a negative spring rate). However, this way of thinking encounters interpretation difficulties for more complicated systems B. Also, to obtain the natural modes of the reduced system by such an intuitive approach is not possible.

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. .   . 

To find natural modes, it is therefore necessary to formulate the response at a point 3 of system A, which requires auxiliary measurements or calculations of receptances g31 and g32 . In terms of the subsystem receptances, g31 = a31 −

a32 a21 , a22 + b22

g32 =

a32 b22 . a22 + b22

(18, 19)

g32 b22 g22 − b22

(20, 21)

These equations are solved for a31 and a32 : a31 = g31 −

g32 g21 , g22 − b22

a32 = −

Equation (20) defines the natural modes. It is evaluated at various points 3 for each of the natural frequencies vk . In practice, this is done for negligibly small damping to avoid the singularity at v = vk . Equation (20) is valid for discrete as well as continuous systems. 3. EXAMPLE 1: SUBTRACTING A SPRING–DAMPER–MASS SYSTEM FROM A TWO-DEGREE-OF-FREEDOM SYSTEM

The system to be analyzed is shown in Figure 2. System C is the original system, system B is the system to be subtracted, and system A is the system of interest. The receptances necessary to synthesize the reduced system were found theoretically (see Appendix A), but typically they would be measured when the method is applied to a real system. Since all the receptances for system C turn out to be equal (g11 = g12 = g22 ), equations (14), (15) and (16) are equal (a11 = a21 = a22 ). Substitution of g11 , g21 , g22 and b22 into equation (16) gives a21 = −g21 b22 /(g22 − b22 ).

(22)

This is the receptance of system A. The forced response of system A at point 1 is then simply xA2 = a21 F1 e jvt.

(23)

The receptance a11 is a complex number. In terms of its magnitude and phase angle it is a21 = A21 e−jf,

Figure 2. Reduction of a standard two-degree-of-freedom system.

(24)

  

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A21 = =g21 ==b22 =/z=b22 = 2 + =g21 = 2 − 2=b22 ==g21 = cos (fB2 − fA ),

(25)

f = fA + fB2 − tan−1 ((=g21 = sin fA − =b22 = sin fB2 )/(=g21 = cos fA − =b22 = cos fB2 )).

(26)

where

and

Therefore, the response of system A can finally be written as xA2 = A21 F1 e j(vt − f).

(27)

In Figure 3 are shown the magnitude and phase of g11 = g22 = g12 of the original system C, as they might have been measured in a practical application. The figure shows the characteristic two resonances of a two-degree-of-freedom system. Figure 4 is a plot of the magnitude and phase of b22 for system B, the system that is to be subtracted from system C. Again, this could be measured input to the procedure. Figure 5 is a plot of the result when system B is subtracted from C (in terms of the magnitude and phase of a11 ). This is, of course, the classical one-degree-of-freedom response curve that one would expect. Needless to say, it was verified by direct calculation. The procedure was found to be robust for any value of damping.

Figure 3. Receptance g11 of system C in terms of (a) magnitude =g11 = and (b) phase fA .

. .   . 

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Figure 4. Receptance b22 of system B in terms of (a) magnitude =b22 = and (b) phase fB2 .

4. EXAMPLE 2: SUBTRACTING A SUPPORTING AIRSPRING FROM AUTOMOTIVE SUSPENSION MEASUREMENTS

The automotive suspension test system to be analyzed is modelled in a simplified way in Figure 6. System C is the system with the airspring (modelled as a spring–damper system), system B is the airspring model, and system A is the subtracted system, namely the quarter-vehicle model without an airspring. Again, as for Example 1, the receptances necessary to synthesize the reduced system were found theoretically (see Appendix B), but typically they would be measured when dealing with a real suspension. Since it is of interest to obtain the response of system A at point 1 due to forcing at point 1, (system C without the airspring), as shown in Figure 6, the receptance a11 must be obtained. From equation (15), one obtains a11 =

2 g11 (g22 − b22 ) − g21 , g22 − b22

(28)

which can be written a11 = =a11 = e−jf,

(29)

where 2 2 2 2 =a11 = = zA11 + B11 /zH11 + G11 ,

f = tan−1 (H11 /G11 ) − tan−1 (B11 /A11 ),

(30, 31)

  

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Figure 5. Receptance a11 of system A (result) in terms of (a) magnitude =A21 = and (b) phase f. The result was verified by direct calculation, and agreement was such that curves were on top of each other, with errors of the order of the line width.

Figure 6. Reduction of an airspring supported vehicle model to a degenerated vehicle model without airspring.

. .   . 

862 and where

A11 = =g11 ==g22 = cos (fA + fB2 ) − =g11 ==b22 = cos (fA + fB22 ) − (b/L)2 =g11 = 2(cos2 fA − sin2 fA ),

(32)

B11 = −=g11 ==g22 = sin (fA + fB2 ) + =g11 ==b22 = sin (fA + fB22 ) + (b/L)2 =g11 = 22 cos fA sin fA , G11 = =g11 = cos fB2 − =b22 = cos fB22 ,

H11 = =b22 = sin fB22 − =g22 = sin fB2 .

(33) (34, 35)

Figure 7 is a plot of the magnitude and phase of g11 of system C, the automative suspension test model. This corresponds to the response measurement at the tire with a shaker attached to the tire and the supporting airspring in place. In Figure 8 are shown the magnitude and phase of b22 for system B, the airspring that is to be subtracted from system C. In this example, mass effects of the airspring are not considered, but could easily be added. Figure 9 is the plot of the result when system B is subtracted from C (in terms of magnitude and phase of a11 ). Removing the airspring creates a degenerate two-degree-offreedom system with a rigid body motion and a corresponding zero natural frequency, as expected.

Figure 7. Receptance g11 of system C (automobile suspension with airspring) in terms of (a) magnitude =g11 = and (b) phase fA . The receptances g22 and g21 are of similar character.

  

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Figure 8. Receptance b22 of system B (airspring) in terms of (a) magnitude =b22 = and (b) phase fB22 .

The procedure shows little sensitivity to variations in the spring rate and damping coefficient of the airspring. The same final result is obtained, no matter what the parameters of the airspring are. The suspension system response without the airspring emerges from the calculations unchanged. The numerical procedure seems to be robust. Again, the response of the system without the airspring was also calculated directly, in order to verify the numerical procedure. The results are so accurate that they differ by less than the line thickness of Figure 9 and are, therefore, not shown separately. 5. CONCLUSIONS

The method presented here is useful when one wants to obtain deflection responses of the leftover system without actually having physically to remove the undesired part of the system and take measurements. In some cases, as for the automative suspension test (Example 2), it is actually impossible to physically remove the unwanted part of the system. The airspring cannot be removed from the suspension test because the suspension would sag if it were removed, and erroneous results would be obtained. While the two examples are both discrete systems, the receptance subtraction equations will also work for continuous systems that follow the general schema of Figure 1. An example of such a continuous system is a uniform, mass loaded plate with a point mass to be subtracted or a spring on an elastic foundation, with part of the foundation missing. Also, it should be noted that the modelling of the airspring in Example 2 is very simplified. It was not the surpose of this project to obtain exact results for a suspension

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. .   . 

Figure 9. Receptance a11 of system A (suspension system with airspring removed) in terms of (a) magnitude =a11 = and (b) phase f. The result was verified by direct calculation, and agreement was such that all curves were on top of each other, with errors of the order of the line width.

test, but to show that the theory can be applied to this problem. Also, all the receptances required for the aii synthesis were found theoretically for a simplified quarter-vehicle model. As pointed out, measured data would typically be used which would include the continuous vibration components of the entire vehicle. The approach would remain the same. The reduced system synthesis equations (14)–(16) can only be used for systems that follow the general schema of Figure 1, which is that the system to be subtracted is connected by one co-ordinate. However, reduction equations for systems where the subsystem to be subtracted is connected by two or more co-ordinates can be developed following the same philosophy of approach. To use the reduced system synthesis equations experimentally, the magnitudes of the required receptances have to be measured, and also the phase angles.

REFERENCES 1. R. E. D. B and D. C. J 1960 The Mechanics of Vibration. London: Cambridge University Press. 2. W. S 1981 Vibrations of Shells and Plates. New York: Marcel Dekker. 3. L. E. K, W. S and T. Y. Y 1986 Journal of Sound and Vibration 107, 181–194. Free vibration of a pneumatic tire–wheel unit using a ring on a elastic foundation and a finite element model. 4. D. A, W. S and T. Y. Y 1986 Journal of Sound and Vibration 111, 9–27. Natural frequencies and modes of rings that deviate from perfect axisymmetry.

  

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5. S. A, W. S and J. F. H 1986 Journal of Sound and Vibration 109, 79–88. Natural frequencies and modes of cylindrical polygonal ducts using receptance methods.

APPENDIX A: RECEPTANCES FOR EXAMPLE 1

To find the receptances for system C (Figure 2), the equation of motion

$

mB 0 0 mA

%6 7 $

%6 7 $

−cB (cA + cB )

x¨B cB + x¨A −cB

x˙B kB + x˙A −kB

%6 7 6 7

−kB (kB + kA )

xB 0 = xA F1 e jvt

(A1)

is solved for the steady state responses xB = XB e j(vt − fB ),

xA = XA e j(vt − fB ),

(A2)

where XB = zA12 + B12 /zG 2 + H 2, fA = tan−1 (H/G) − tan−1 (B2 /A2 ),

XA = zA22 + B22 /zG 2 + H 2,

(A3, A4)

fB = tan−1 (H/G) − tan−1 (B1 /A1 ),

(A5, A6)

2 B

G = mA mB v − [mA kB + (kA + kB )mB + cB cA ]v + k ,

(A7)

H = vcB (kA − mA v 2) − cB mB v 3 + vcA ((kB − mB v 2),

(A8)

4

A1 = FkB ,

2

B1 = FcB v;

A2 = F(kB − mB v 2),

B2 = FcB v.

(A9, A10)

Therefore, receptance g21 is g21 = =g21 = e−jfA ,

(A11)

=g21 = = XA /F.

(A12)

where

Since the mass of the example is inelastic, the deflection response, xA , is the same at any point on the mass mA . For a forcing at any point on mA , we obtain (for this special example—not in general) g21 = g11 = g22 .

(A13)

To find the receptance b22 , a force, FB2 e jvt, is applied at point two of system B (in a practical application, this may have to be done experimentally). Solving the equations of motion,

$

mB 0 0 0

%6 7 $

y¨ cb + x¨B2 −cb

%6 7 $

−cB cB

y˙ kB + x˙B2 −kB

−kB kB

%6 7 6

7

y 0 = , xB2 FB2 e jvt

(A14)

the solution is xB2 = XB2 e j(vt − fB 2 ),

y = Y e j(vt − fy ),

(A15)

where XB2 = FB2 z(kB − v 2mB )2 + v 2cB2 /zv 4mB2 kB2 + v 6cB2 mB2 , fB2 = tan

−1

(vcB /kB ) − tan

−1

(vcB /(kB − mB v )). 2

(A16) (A17)

The receptance b22 is, therefore, b22 = =b22 = e−jfB 2

(A18)

. .   . 

866 where

=b22 = = XB2 /FB2 .

(A19)

APPENDIX B: RECEPTANCES FOR EXAMPLE 2

The equations of motion of system C (Figure 6) are [M]{x¨ } + [C]{x˙ } + [K]{x} = {F }, {x}T = {xa , xc }, [M] =

$

mT 0

%

0 , mc

[K] =

[C] =

{F }T = {F1 e jvt, 0},

$

[cB (b/L)2 + cA (a/L)2] −cA (a/L)

$

%

−cA (a/L) , cA

(B1)

%

[kB (b/L)2 + kA (a/L)2] −kA (a/L) . −kA (a/L) kA

Solving for the steady state response, we obtain xA = XA e j(vt − fA ),

(B2)

where XA = zA12 + B12 /zG 2 + H 2,

fA = tan−1 (H/G) − tan−1 (B1 /A1 ),

(B3, B4)

G = [kB (b/L)2 − mT v 2](kA − mC v 2) − kA (a/L)2mC v 2 − cB (b/L)2cA v 2,

(B5)

H = −mC v 3[cB (b/L)2 + cA (a/L)2] − cA mT v 3 + v(b/L)2(cB kA + cA kB ),

(B6)

and where

A1 = F1 (kA − mC v ),

B1 = F1 vcA .

2

(B7)

The system C receptance g11 is, therefore, g11 = =g11 = e−jfA ,

=g11 = = XA /F1 ,

(B8, B9)

In the real application, g11 is what is measured. In this particular example, g21 is related to g11 geometrically by g21 = g11 (b/L) = (b/L)=g11 = e−jfA .

(B10)

To obtain the required receptance g22 , the response at point 2, xB2 , must be found when a force, FB2 e jvt, is applied to system C at point 2, as shown in Figure 6. This is the required auxiliary experiment in the real application. Here, we solve [M']{x¨ '} + [C']{x˙ '} + [K]{x'} = {F '}, {x'}T = {xB2 , xc }, [M] =

$

mT (L/b)2 0

%

0 , mC

[K] =

{F '}T = {FB2 e jvt, 0}, [C] =

$

%

[cB + cA (a/b)2] −cA (a/b) , −cA (a/b) cA

$

(B11)

%

[kB + kA (a/b)2] −kA (a/b) , −kA (a/b) kA

and obtain the steady state solution xB2 = XB2 e j(vt − fB 2 ),

(B12)

  

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where XB2 = zA'1 2 + B'1 2 /zG'1 + H'2,

fB2 = tan−1 (H'/G') − tan−1 (B'1 /A'1 ),

(B13, B14)

G' = (kB − mT (L/b)2v 2)(kA − mC v 2) − mC v 2kA (a/b)2 − v 2cB cA ,

(B15)

H' = vcB (kA − mC v ) − mC v cA (a/b) + vcA (kB − mT (L/b) v ),

(B16)

and where

2

3

A'1 = FB2 (ka − mC v 2),

2

2

B'1 = FB2 vcA .

2

(B17)

The receptance g22 is, therefore, g22 = =g22 = e−jfB 2,

(B18)

where =g22 = = XB2 /FB2 .

(B19)

To find the receptance for system B, b22 , a force, FB22 e , is applied at point 2 of system B. This would be done experimentally, in the real application. Here, we solve the equation of motion jvt

cB x˙B22 + kB xB22 = FB22 e jvt,

(B20)

xB22 = XB22 e j(vt − fB 22 ),

(B21)

where

and where XB22 = FB22 /zkB2 + v 2cB2 ,

fB22 = tan−1 (vcB /kB ).

(B22)

The receptance b22 is, therefore, b22 = =b22 = e−jfB 22,

(B23)

=b22 = = XB2 /FB22 .

(B24)

where