application of s.i.d. to non-linear engine mounts...

Application of S.I.D. to non-linear engine mounts design

U. Tornar1

1 PSA Peugeot Citroen, Research and Development

18 Rue des Fauvelles, 92250, La Garenne Colombes, France

e-mail: [email protected]

Abstract Forces are generally defined in physics as functions of position (Newton: gravity) or velocity (Laplace:

magnetic force on a moving electric charge). Damping forces are little known even today and represent

one of the most intriguing subjects of physics. Maxwell elements and fractional derivatives are used to

modelize time domain hysteretic damping. The resulting models are comparatively complicated and have

a limited domain of validity especially when strong non-linearity is involved. The mathematical model we

use is based on the introduction of a new state variable and is particularly suitable in the non-linear

vibration case. S.I.D. (Strain Integral Damping: see ref. [2]) is a very suitable mean to modelize hysteretic

damping in the time domain and for engine mounts in particular [3]. In the present paper some very

important technical advice is given to possible SID users. The theory is better explained and enriched in

comparison with previous papers. The effectiveness of SID is shown by some examples one of them

concerning a strongly non-linear model. Two ‘’Scilab’’ scripts are provided to better explain.

1 Introduction

Natural damping is only seldom viscous. Hysteretic damping is much more common and can be described

as follows in the frequency domain.

If:

[ ( ( ))]

(1)

Where an imaginary part of the stiffness matrix is introduced (tg(φ)). We shall call this I.S.D. (Imaginary

Stiffness Damping) in the following.

Such a formulation is much used in the frequency domain because it is simple and practical to use and not

because there is a real physical theory behind it. S.I.D. (Strain Integral Damping) wants to be as simple

and practical to use for hysteretic damping modeling in the time domain.

4123

2 SID Formulation

Figure 1: hysteresis of a spring-damper

Let us consider the spring-damper of FIG. 1. If ( ) is the displacement at time ‘’t’’ and we apply a

sinusoidal force we shall obtain:

( ) ( ) (2)

Velocity ‘’v’’ and acceleration ‘’a’’:

( )

( ) (3)

( )

( ) (4)

The applied force will be:

( ) ( ) (5)

‘’F’’ being the force amplitude. We can rewrite:

( ) ( ) ( ) ( ) ( ) (6)

Following equation (1) the springer-damper stiffness ‘’k’’ is defined by:

( ) (7)

Force F Force

Displacement

4124 PROCEEDINGS OF ISMA2014 INCLUDING USD2014

We can then write equation (6) in the form:

( ) ( ) ( ) ( ) (8)

By substituting equations (2) and (3) in equation (8) we obtain:

( ) ( ( ) ( )

( )

) (9)

Where we have introduced the same ( ) factor of eq. (1).

Now we must express the ‘’ω’’ of eq. (9) as a function of state variables only. We may think of expressing

the ‘’1/‘’ factor of eq. (9) as the ratio:

| ( )

( )|

(10)

But as it is shown in reference [1], forces cannot in general be expressed as functions of the accelerations

and this leads us to define a new state variable which is the solution of the differential eq.:

( ) (11)

The solution is:

( ) ∫ ( ) ( )

(12)

The constant ‘’ω1’’ is introduced to define as ‘‘remote past’’ all events for which:

( ) (13)

Such events will have negligible effect on ‘’y’’ (strain integral) and, as a consequence, on the damping

force. We must remark that if ‘’ω1’’ is zero, ‘’y’’ goes to infinity for all x(t) whose average is not zero

(spring preloading). This of course wouldn’t be physical. So ‘’ω1’’ can be seen as a high pass filter

parameter: it has the same physical dimensions as a frequency and it must be set well lower than the

frequencies of interest but it must not be negligible in comparison with the frequencies of interest to

avoid ‘’y’’ to go to infinity. We can better understand this by writing eq. (11) in the frequency domain:

(14)

VIBRATION ISOLATION AND CONTROL 4125

Where X and Y are the complex amplitudes of ‘’x’’ and ‘’y’’ respectively. We can see from this formula

that if ω is an angular frequency of interest, it must be for ‘’y’’ to be close to the integral of ‘’x’’.

( ) is a typical value.

We can then assume:

| ( )

( )|

(15)

By substituting eq. (15) into eq. (9) we easily obtain:

( ) ( ( ) ( ( )) | ( ) ( )| ) (16)

We remark that the term | ( ) ( )| has the physical dimensions of a displacement but is ‘’phased’’

like a velocity.

We assume as initial condition for ‘’y’’:

( ) ( ) (17)

We can easily see that, with this initial condition, the cycle starts at the origin like the dotted curve shown

in Fig. 1.

Work experience has shown that the introduction of factor ‘’ω1’’ in eq. (11) is not enough to avoid that

‘’y’’ goes to infinity. This problem of course can only exist in case of spring (engine mount) preloading.

The problem is easily solved by the introduction of a moving average in equation (11):

( ( ) ( )) (18)

The moving average ( ) is defined as the solution of the following differential equation:

(19)

The solution is:

( ) ∫ ( ) ( )

(20)


And the corresponding weighted moving average:

( ) ∫ ( ) ( ) ( )

(21)

Where:

( ) (22)

Is the normalization factor. We can see from eq. (21) that such an average has the important property that

all events of the ‘’past’’ that happened at time such that:

( ) (23)

Are ‘’squeezed’’ by the weighting factor:

( ) (24)

So that only the most ‘’recent’’ events are really included in the average.

The fact that factor (22) goes to infinity when t=0 is generally avoided by the substitution:

( ) (25)

In practice we generally assume:

(26)

3 Nonlinearity

SID is most useful in nonlinear problems. To introduce nonlinearity we only need modifying eq. (16) as

follows:

( ) ( ) ( ) ( ( )) | ( ) ( )| (27)

Where ‘’spl’’ is a spline representing the non-linear spring and kS is the secant stiffness (very seldom the

tangent stiffness as written in reference [3]).


4 To better understand SID (Numerical example)

We are now going to present with a numerical example concerning a particularly critical case. In all

calculation that we perform we assume that the zero of displacements coincides with the point of static

equilibrium. In this case instead we consider a mass of 1 kg with an attached stiffness that corresponds to

an Eigen frequency of 10 Hz and we load this mass with a 10 N force constant in time. We define also:

- tg(φ)=0.2 ;

- ω1 = ω2 = 4*π rad/sec = 2 Hz ;

We didn’t do anything to make the zero of displacements to coincide with the working point

corresponding to this static load of 10 N.

4.1 What the results are if everything is done according to SID theory.

Figure 2: displacement

Figure 3: strain integral

We remark that the displacement stabilizes at 2.529 mm which is the static working point. Sometimes

static loads are not really constant. For example the engine torque varies depending on how much the

driver presses on the accelerator and ‘’static’’ loads on the mounts will vary accordingly. The results look

correct.

DEPL SID selon la théorie

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

0 0.2 0.4 0.6 0.8 1 1.2

SEC

mm DEPL

Strain Integral selon la théorie

-0.00002

-0.00001

0

0.00001

0.00002

0.00003

0.00004

0.00005

0.00006

0.00007

0 0.2 0.4 0.6 0.8 1 1.2

SEC

M*S

EC

SI

sec

mm

m

*se

c

sec


4.2 What the results are if the moving average (eq. 21) is not subtracted.



We can see that in this case, the SI doesn’t go to zero (as it should go at the end of the transient) but

stabilizes at a value which depends on the value of ω1. The results are obviously wrong.

DEPLACEMENT SID sans soustraction de la moyenne mobile

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

0 0.2 0.4 0.6 0.8 1 1.2

SEC

mm DEPL

Strain Integral sans soustraction de la moyenne mobile

-0.00005

0

0.00005

0.0001

0.00015

0.0002

0.00025

0 0.2 0.4 0.6 0.8 1 1.2

SEC

M*S

EC

SID

sec

mm

sec

m*

sec


4.3 What the results are if the moving average is not subtracted and ω1 is set to zero



We see that the SI goes to infinity and the resulting displacement is wrong.

DEPLACEMENT SID sans soustraction de la moyenne mobile ni omega1

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

0 0.2 0.4 0.6 0.8 1 1.2

SEC

mm DEPL

Strain Integral SID sans soustraction de la moyenne mobile ni omega1

-0.0005

0

0.0005

0.001

0.0015

0.002

0.0025

0.003

0 0.2 0.4 0.6 0.8 1 1.2

SEC

M*S

EC

SID

sec

sec

m*

sec

mm


4.4 What the results are with ω1=0



We can see here that subtracting the moving average is not enough: the SI stabilizes around a constant

value instead of going to zero and the resulting displacement is again wrong. ω1 ≠ 0 in equation (18) is

always necessary.

DEPLACEMENT SID sans omega1

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

0 0.2 0.4 0.6 0.8 1 1.2

SEC

mm DEPL

Strain Integral SID sans omega1

0

0.00001

0.00002

0.00003

0.00004

0.00005

0.00006

0.00007

0.00008

0.00009

0 0.2 0.4 0.6 0.8 1 1.2

SEC

M*S

EC

SID

mm

sec

sec

m*

sec


5 Comparison between SID (Strain Integral Damping) and ISD (Imaginary Stiffness Damping) of equation (1)

Figure 10: two masses system

Script number 1 was used to solve the 2 masses system in figure (10) both in the frequency domain (with

ISD) and in the time domain (with SID).

In script 1:

X1 = vertical displacement of mass m1

X2 = vertical displacement of mass m2

F1 = blocked base Eigen-frequency of spring-mass system 1

F2 = blocked base Eigen-frequency of spring-mass system 2

K1 = stiffness of spring-damper 1

K2 = stiffness of spring-damper 2

Csi1 = tg(φ) of spring-damper 1

Csi2 = tg(φ) of spring-damper 2

Figure 11 shows:

The square wave base excitation (blue)

Displacement x1 (black)

The difference (x2 – x1) (red)

Mass m2

Mass m1

Spring damper 2

Spring damper 1

Ground

contact

point


Figure 11: time domain signals

Figure 12 shows :

- (X1 / base excitation) transfer function in the frequency domain (green)

- (X2 / base excitation) transfer function in the frequency domain (red)

- abs((fft of X1) / (fft of base excitation)) in time domain (black)

- abs((fft of X2) / (fft of base excitation)) in time domain (blue)

Figure 12: transfert functions

Please note the strong similarity of the results. Similarity would be even greater if the boxcar effect had

been corrected. Script 1 can also be used to perform other tests of SID that require dealing with the

‘’BOXCAR’’ effect (ref. [4]). The reader should not be appalled by the computational time: Scilab was


chosen just because it is free software. Other software (e.g. ADAMS or VlabMotion) is much more

targeted to time step integration and solution would be much quicker.

5.1 Nonlinear calculation

As we already said, the main use of SID is in non-linear calculations. Script 2 was used to solve a one

mass-spring system of the same kind of the two shown in figure 2.The spring is strongly non-linear and

the base excitation is pseudo-random:

Figure 13: the pseudo-random base excitation (sec,meters)

Figure 14 shows the corresponding displacement-force hysteresis cycles of the non-linear spring-damper

(meters,Newtons).

Figure 14: displacement-force hysteresis cycles

All engineers who design engine mounts are very familiar with the kind of cycles shown in figure 14. In

most time-step integration codes (e.g. ADAMS, VlabMotion) the tg(φ) term of equation (27) can be made

a function of displacement enabling the user to modelize a great variety of different cycle shapes.


6 Conclusions

The validity of a theory can only be proved by its agreement with reliable experimental results like the

well-known result of eq. (5). In this sense SID has been shown to give the kind of results we are used to

(see figure 12). We don’t know whether SID is a ‘’beable’’ which is what the physicists call something

that has a real link with physical reality: only the future can say. We can say however that it is a very

practical and easy method that corresponds, in the time domain, to the imaginary stiffness damping of eq.

(1) in the frequency domain: no physical base to it but everybody uses it because it is simple and practical.

SID only needs two parameters to be defined. SID is a suitable mathematical description of hysteretic

damping and gives fairly physical results when applied to non-linear problems (see figure 14).

Acknowledgements

I wish to thank Madame N. Collot of PSA Peugeot Citroen for authorizing me to publish my theory in the

present paper.

References

[1] L.A. Pars, A Treatise on Analytical Dynamics, Ox Bow Press.

[2] U. Tornar, Modelizing Structural Hysteretic Damping by Means of MDI ADAMS,

Proceedings of the ADAMS 1997 Users Conference, Marburg, Germany

[3] U. Tornar, Application of Strain Integral Damping to Engine Mounts,

ISMA 2006, Leuven, Belgium

[4] R.K. Otnes and L. Enochson, Applied Time Series Analysis, Wiley-Interscience


Annex

SCILAB SCRIPT 1 clear;

tapering = 0; movingav = 1; gg = 0;

tt=(1:(4096*16))/4096; ff = 0.5*ones(tt)+tt/4;

aa = round(tt.*ff)-floor(tt.*ff);

aa = (aa-0.5*ones(aa))/1000;

aa(length(aa)+1) = aa(length(aa));

csi1 = 0.1; csi2 = 0.05;

m1 = 400; m2 = 70; f1 = 2; f2 = 4;

k1 = (2.*%pi*f1)*(2.*%pi*f1)*m1;

k2 = (2.*%pi*f2)*(2.*%pi*f2)*m2;

dt = 1/4096; x1 = -gg * (m1+m2) / k1;

x2 = x1 - gg * m2 / k2; v1 = 0; v2 = 0; a1 = 0; a2 = 0;

ss1 = 0; ss2 =0; h1 = 0.5*2*%pi;

h2 = 1*2*%pi; z1 = 0; z2 = 0;

zh1 = 0.25*2*%pi; zh2 = 0.5*2*%pi;

for kk = 1 : 4096*16;

va = (aa(kk+1) - aa(kk))/dt;

z1av = 0; z2av = 0;

if movingav == 1 then;

z1 = z1 + (-z1 * zh1 + aa(kk)-x1) * dt;

z2 = z2 + (-z2 * zh2 + x1-x2) * dt;

z1av = z1 * zh1 / max(0.0001,1-exp(-zh1*tt(kk)));


end;

ss1 = ss1 + (-ss1 * h1 + aa(kk)-x1-z1av) * dt;

ss2 = ss2 + (-ss2 * h2 + x1-x2-z2av) * dt;

a1 = ((aa(kk) -x1)*k1 + (x2-x1)*k2)/m1 - gg;

a1 = a1+sign(va-v1)*(abs(ss1)*abs(va-v1))**0.5 * k1 * csi1 /m1;

a1 = a1+sign(v2-v1)*(abs(ss2)*abs(v2-v1))**0.5 * k2 * csi2 /m1;

a2 = ((x1-x2)*k2 + sign(v1-v2)*(abs(ss2)*abs(v1-v2))**0.5 * k2 * csi2)/m2 - gg;

v1 = v1 + a1 * dt; v2 = v2 + a2 * dt; x1 = x1 + v1 * dt; x2 = x2 + v2 * dt;

xx1(kk) = x1; xx2(kk) = x2; vv1(kk) = v1; vv2(kk) = v2;

end;

aa = aa(1:length(aa) - 1);

if tapering == 1 then;

na = length(aa); alfa = %pi * (1:na/10) / (na/10); nalf = length(alfa);

aa(1:nalf) = aa(1:nalf) .* (0.5*(ones(alfa)-cos(alfa)));

aa(na-nalf+1:na) = aa(na-nalf+1:na) .* (0.5*(ones(alfa)+cos(alfa)));

xx1(1:nalf) = xx1(1:nalf) .* (0.5*(ones(alfa)'-cos(alfa)'));

xx1(na-nalf+1:na) = xx1(na-nalf+1:na) .* (0.5*(ones(alfa)'+cos(alfa)'));

xx2(1:nalf) = xx2(1:nalf) .* (0.5*(ones(alfa)'-cos(alfa)'));

xx2(na-nalf+1:na) = xx2(na-nalf+1:na).*(0.5*(ones(alfa)'+cos(alfa)'));

end;

figure(1);

plot(tt,aa,'b',tt,xx1,'k',tt,xx2-xx1,'r');

trasfaa = fft(aa,-1); trasfxx1 = fft(xx1,-1); trasfxx2 = fft(xx2,-1);


trasfaa = trasfaa(2:129); trasfxx1 = trasfxx1(2:129);

trasfxx2 = trasfxx2(2:129);

tf1 = trasfxx1./trasfaa'; tf2 = trasfxx2./trasfaa';

for jj = 1 : 128;

om = 2 * %pi* jj / 16; om = om * om;

mm(1,1) = m1 * om - k1 * (1 + csi1 * %i) - k2 * (1 + csi2 * %i);

mm(1,2) = k2 * (1 + csi2 * %i);

mm(2,1) = k2 * (1 + csi2 * %i);

mm(2,2) = m2 * om - k2 * (1 + csi2 * %i);

qq(1) = - k1 * (1 + csi1 * %i); qq(2) = 0;

pp = mm\qq; rr1(jj) = pp(1); rr2(jj) = pp(2);

end;

figure(2);

plot((1:128)/16,log10(abs(tf1)),'k',(1:128)/16,log10(abs(tf2)),'b');

set(gca(),"auto_clear","off");

plot((1:128)/16,log10(abs(rr1)),'g',(1:128)/16,log10(abs(rr2)),'r');

SCILAB SCRIPT 2

clear;

tt=(1:(4096*8))/4096;

ll = 2* %pi;

dd = zeros(tt);

for kk = 9 : 64;

rr = rand(1,1);

dd = dd+(1/kk) * (cos(ll * rr)*cos(2* %pi *kk*tt/8)+sin(ll * rr)*sin(2* %pi *kk*tt/8));

end;

dd = dd/10;

figure(1);

plot(tt,dd,'r');

dd(length(dd)+1) = dd(length(dd));

csi1 = 0.2; m1 = 400; f1 = 2;

k1 = (2.*%pi*f1)*(2.*%pi*f1)*400.; dt = 1/4096; x1 = 0; v1 = 0; a1 = 0;

ss1 = 0; h1 = 0.5*2*%pi; z1 = 0; zh1 = 0.25*2*%pi;

for kk = 1 : 4096*8;

va = (dd(kk+1) - dd(kk))/dt;

z1 = z1 + (-z1 * zh1 + dd(kk)-x1) * dt;


ss1 = ss1 + (-ss1 * h1 + dd(kk)-x1-z1av) * dt;

force1 = (dd(kk) -x1)*k1 + 2*((dd(kk)-x1) > 0.04)*(dd(kk) -x1-0.04)**2*1000000;

secstif = abs(force1/ (dd(kk) -x1));

force1 = force1 + sign(va-v1)*(abs(ss1)*abs(va-v1))**0.5 * secstif * csi1;

a1 = force1 / m1;

v1 = v1 + a1 * dt; x1 = x1 + v1 * dt;

xx1(kk) = x1; vv1(kk) = v1; force(kk) = force1;

end;

dd = dd(1:length(dd)-1);

figure(2);

plot((dd-xx1'),force, 'k');


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