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The Shock Response of a VibrationpIsolator with Nonlinear Stiffness

Mike Brennan (UNESP)Mike Brennan (UNESP)Gianluca Gatti (University of Calabria, Italy)Bin Tang (Dalian University of Technology, China)g ( y gy, )

Motivation

V l tiffStatic equilibrium position

Very low stiffness (natural frequency)

Geometrically Nonlinear Stiffness

ef

m

eftx

k cvk c

ex

tf

e

tf

Geometrically Nonlinear Stiffness

efhk hk

m

eftx

k c

l

vk c

ex

tf

e

tf

High Static Low Dynamic Stiffness (HSLDS) isolatorMany engineering applicationsLow frequency isolation

Frequency Response

55

45

50

55

0

1ll

0

0.8ll

B) 

35

400

0.7llility| (dB

---- Displacement trans….  Force trans

20

25

30 0l

smissibi

Shift natural frequency to a very low frequency

10

15

0

0.667ll|T

ran Shift natural frequency to a very low frequency

0 0.2 0.4 0.6 0.8 1 1.20

5

fNormalised frequency

Objective

Shock isolation of HSLDS BUMP AHEAD

m

kvcv

khkh

Engine

Seat isolators

Objective

To to determine whether the HSLDS systemTo to determine whether the HSLDS system described in (Carrella, et al. 2007, 2008, 2009 2012) is better at isolating shock than2009, 2012) is better at isolating shock than a linear isolator.

HSLDS System

k khk hk

m

x

k c

lz x y

vk c

y

y

2 202 1v hk l l zkmz cz z my

HSLDS System

02 1 lkk

0

2 22 1v hkmz cz

l zk z my

Non-dimensional Equation

1ˆ ˆ2 2 1u u u k u y 2 2 2 2

max

2 2 1ˆ ˆˆ 1

u u u k u yl u y l

ˆ ˆ ˆhkl y zl k2 20 max0

, , , h

v

yl k y ul k yl l

Types of Shock Input Applied

1.2VI

Versed-sine impulse

0 6

0.8

1.0

III I

V

IV II

VI

ˆRy 1 cos 2

ˆ , 02

VV py

0 2

0.4

0.6Ry 2

0 2 4 6 8 10 120.0

0.2

τ 1.0

1.2III I

VIV II

VI

Rounded-step displacement 0.6

0.8ˆVy

0 0

0.2

0.4

ˆ 1 1 , 0RR Ry e

0 1 2 3 4 5 60.0

τ

Shock Isolation Performance Indices

Shock Acceleration Ratio (SAR)

max| |SAR| |xy

Shock Displacement Ratio (SDR)

max| |y

Shock Displacement Ratio (SDR)max| |SDR x

y

Relative Displacement Ratio (RDR)maxy

p ( )max

max| |RDR = | |z u

maxy

max| |SAR| |x

Shock Acceleration Ratio

101101101

l = 0.8 l = 0.7 l = 0.667maxˆ 5.0y

max| |y

SAR 10-1

100

SAR 10-1

100

SAR 10-1

100l 0.8 l 0.7 l 0.667max 5.0y

maxˆ 1.0y Linear

0 1 1 10 10010-3

10-2

0 1 1 10 10010-3

10-2

0 1 1 10 10010-3

10-2maxˆ 0.1y

101101101

l 0 8 l 0 7 l 0 667

R

0.1 1 10 100R

0.1 1 10 100R

0.1 1 10 100

Rounded step Displacement ζ = 0.1RRR

SAR 10-1

100

SAR 10-1

100

SAR 10-1

100l = 0.8 l = 0.7 l = 0.667

S

10-3

10-2

S

10-3

10-2

S

10-3

10-2

V

0.1 1 10 10010

V

0.1 1 10 10010

V

0.1 1 10 10010

Versed sine-shape displacementVV V

Shock Isolation Performance Indices

Base displacement is a Heaviside function (βR >>1) Impulse Input (βV >>1)

Step ImpulseSAR 0 0SDRSDRRDR

max| |SDR xShock Displacement Ratio

1 8

2.0

1 8

2.0

1 8

2.0

l = 0.8 l = 0.7 l = 0.667maxˆ 5.0y

maxy

SDR

1.4

1.6

1.8

SDR

1.4

1.6

1.8

SDR

1.4

1.6

1.8

maxˆ 1.0y

Linear

0 1 1 10 100

1.0

1.2

0.1 1 10 100

1.0

1.2

0.1 1 10 100

1.0

1.2maxˆ 0.1y

1.61.61.6

l 0 8 l 0 7 l 0 667

R

0.1 1 10 100R

0.1 1 10 100R

0.1 1 10 100

Rounded step Displacement ζ = 0.1R R R

SDR 0.8

1.01.21.4

SDR 0.8

1.01.21.4

SDR

0.81.01.21.4 l = 0.8 l = 0.7 l = 0.667

S

0.00.20.40.6

S

0 00.20.40.6

S

0 00.20.40.6

V

0.1 1 10 1000.0

V

0.1 1 10 1000.0

V

0.1 1 10 1000.0

Versed sine-shape displacementV V V

Shock Isolation Performance Indices

Low Amplitude InputLow Amplitude Input

Base displacement is a Heaviside function (βR >>1) Impulse Input (βV >>1) Impulse Input (βV 1)

Step ImpulseSAR 0 0SAR 0 0SDR ? 0RDR

max| |RDR zRelative Displacement Ratio

0 8

1.0

0 8

1.0

0 8

1.0maxˆ 0.1y

maxy

RD

R

0.4

0.6

0.8

RD

R

0.4

0.6

0.8

RD

R

0.4

0.6

0.8

maxˆ 5.0y

maxˆ 1.0y Linear

0.1 1 10 1000.0

0.2

0.1 1 10 1000.0

0.2

0.1 1 10 1000.0

0.2 l = 0.8 l = 0.7 l = 0.667

1.41.41.4 l = 0 8 l = 0 7 l = 0 667

RRR

Rounded step Displacement ζ = 0.1R R R

RD

R

0 6

0.8

1.0

1.2

RD

R

0 6

0.8

1.0

1.2

RD

R

0 6

0.8

1.0

1.2 l = 0.8 l = 0.7 l = 0.667

R

0.0

0.2

0.4

0.6R

0.0

0.2

0.4

0.6R

0.0

0.2

0.4

0.6

V

0.1 1 10 100V

0.1 1 10 100V

0.1 1 10 1000.0

Versed sine-shape displacementV V V

Shock Isolation Performance Indices

Base displacement is a Heaviside function (βR >>1) Impulse Input (βV >>1)

Step ImpulseSAR 0 0SDR ? 0SDR ? 0RDR 1 1

Low Amplitude Input

0l

0

2 22 1v h

lkmz czl z

k z my

When the relative displacement z < 40 %*l,

l z p ,

where l is the length of the horizontal spring.

3

h

31 3mz cz k z k z my

• where

02lk k k 03 h

lk k1 1v hk k kl

3 3 hk kl

Low Amplitude Input

31 3mz cz k z k z my

Non-dimensional Equation

1 3 y

q3 ˆ ˆ2 ( ) (or y ( ))R Vu u u u y

where ˆ ˆ 2

2max 3 ,

ˆ2 1 ˆˆ1 ˆ

1ˆˆ ˆ

lk

ly k

ll

is the parameter that controls the linear

3ll

l̂ is the parameter that controls the linear stiffness and the cubic nonlinearityl

Low Amplitude Input

= 0 8 0 5 0 703125l̂

= 0.8 0.5 0.703125

= 0.7 0.14286 1.48688l̂l

= 0.667 0.001499 1.87069l̂

• When the linear stiffness term theˆ 2 / 3l 0 • When the linear stiffness term, the system becomes a Quasi-Zero-Stiffness (QZS) system

2 / 3l 0

Low Amplitude Input

3 ˆ ˆ2 ( ) (or ( ))R Vu u u u y y

ˆ 1y 3u u max 1y u u

ˆ ˆ2 ( ) (or y ( ))R Vu u u y

Low Amplitude Input

ˆ ˆ2 ( ) (or y ( ))R Vu u u y R V

2 under damped under damped

22 over damped

2 v. over damped

max| |SDR xShock Displacement Ratio

maxy

1.8

2.0

1.8

2.0

1.8

2.0

l = 0.8 l = 0.7 l = 0.667maxˆ 5.0y ˆ 1 0

SDR

1 2

1.4

1.6

SDR

1 2

1.4

1.6

SDR

1 2

1.4

1.6

ˆ

maxˆ 1.0y

Linear

0.1 1 10 100

1.0

1.2

0.1 1 10 100

1.0

1.2

0.1 1 10 100

1.0

1.2maxˆ 0.1y

r r RRR

Rounded step Displacement ζ = 0.1r r r

Low Amplitude Input

Base displacement is a Heaviside function (βR >>1) Impulse Input (βV >>1)

Step ImpulseSAR 0 0SDR 0 1

SDR 0RDR 1 1

1 e

Concluding Remarks

For both types of excitation the effects of increasing thenonlinearity is beneficial from the point of view of shocknonlinearity is beneficial from the point of view of shockisolation, provided that the amplitude of shock input isrelatively small.y

Nonlinearity has two main effects. The first is it reduces the Nonlinearity has two main effects. The first is it reduces the natural frequency, and the second is that the stiffness of the system is increased for large relative displacements across the isolator.

The first of these is beneficial and is the reason why the nonlinear isolator outperforms the linear isolator. The second of these is not beneficial and has a detrimental effect whenof these is not beneficial, and has a detrimental effect when the amplitude of the shock input is large.

References[1] C.M. Harris, A.G. Piersol, Shock and Vibration Handbook, fifth ed. McGraw Hill, New York, 2002., ,[2] C.W. De Silva, Vibration and Shock Handbook, CRC Press, Boca Raton, 2005.[3] J.C. Snowdon, Vibration and Shock in Damped Mechanical Systems, Wiley, New York, 1968.[4] P. Alabuzhev, A. Gritchin, L. Kim, G. Migirenko, V. Chon, P. Stepanov, Vibration Protecting and Measuring Systems with Q i Z Stiff H i h P bli hi N Y k 1989Quasi‐Zero Stiffness, Hemisphere Publishing, New York, 1989.

References[5] N. Chandra Shekhar, H. Hatwal, A.K. Mallik, Response of non‐linear dissipative shock isolators, Journal of Sound and Vibrationp , f214(4) (1998) 589‐603.[6] N. Chandra Shekhar, H. Hatwal, A.K. Mallik, Performance of non‐linear isolators and absorbers to shock excitations, Journal of Sound and Vibration 227(2) (1999) 293‐307.[7] A. Carrella, M.J. Brennan, T.P. Waters, V. Lopes Jr., Force and displacement transmissibility of a nonlinear isolator with high‐static low dynamic stiffness International Journal of Mechanicalstatic‐low‐dynamic‐stiffness, International Journal of Mechanical Sciences 55(1) (2012) 22‐29.[8] Bin Tang M J Brennan On the shock performance of a[8] Bin Tang, M.J. Brennan. On the shock performance of a nonlinear vibration isolator with high‐static‐low‐dynamic‐stiffness, International Journal of Mechanical Sciences 81(1) (2014) 207‐214.

Low Amplitude Input

Base displacement is a Heaviside function (βR >>1) SAR

The acceleration of the base is infinite

Relative disp cosu e

SAR 0

Relative disp

α > ζ 2, under-damped

1cos dd

u e

α < ζ 2, over-damped

1cos du e d 2( )

2di

du e

i

SDR

SDR 1 1u e

SDR

d

SDR 1

RDR umax ≈ -1 RDR 1 RDR RDR 1

Thank You for Your Attention!!!Thank You for Your Attention!!!Any Questions are welcome!y

谢谢 (Xièxiè)！( )

Bin TangInstitute of Internal Combustion Engine, Dalian University of Technology, China.