2

I

**Practical Response Analysis of a Simply Supported Beam with Clearance and Hysteresis **

**Damping **

Shigeru Aoki 1) and Takeshi Watanabe 2~

1)Department of Mechanical Engineering, Tokyo Metropolitan College of Technology, Higashi-Ohi, Shinagawa-ku, Tokyo, Japan 2)Faculty of Education and Human Sciences, Yamanashi University, Takeda, Kofu-City, Japan

ABSTRACT

Collision characteristics or hysteresis damping are typical nonlinear characteristics which are observed at supports and joints in the piping systems of the nuclear power plant. In this paper, a beam simply supported at one end is used as one span model of the piping system. Nonlinear boundary condition at the other end of the beam is assumed to have clearance and hysteresis damping. Considering the energy loss in a collision with a stop and duration of collision, an analytical method of approximate solution for the beam which collides with the stop having quadrilateral hysteresis loop characteristics is presented. Quadrilateral hysteresis loop characteristics is more practical model for collision characteristics. Some numerical results of the approximate solution for the resonance curves and the mode shapes are shown. Proposed method is applied to the system with no clearance and with hysteresis damping. Some numerical results of the approximate solution are shown.

I N T R O D U C T I O N

Forced response of continuous system with boundary conditions is of great importance for several nuclear engineering applications[i]. Piping systems in nuclear power plants consists of many nonlinear components[2]. For example, piping systems have supports and joints. Collision characteristics is one of the typical nonlinear characteristics which are observed at supports and joints[3]. In this paper, a beam simply supported at one end is used as one span model of the piping system. And the beam at the other end collides with the stop in half period of its vibration. Considering the energy loss in a collision with a stop and duration of collision, an analytical method of approximate solution for the steady-state response of the beam which collides with the stop having quadrilateral hysteresis loop characteristics is presented. In quadrilateral hysteresis loop characteristics, the coefficient of restitution depends on velocity response at beam end[4]. Quadrilateral hysteresis loop characteristics is a~ore practical model for collision characteristics. Some numerical results of the approximate solution for the resonance curves and the mode shapes are shown. It is well known that vibration of the system with collision is a complicated phenomenon and resonance curves are sometimes shown as discontinuous line[5]. In this paper, force of restitution is expanded into Fourier series and neglecting the terms higher orders than n=l, only fundamental terms of the series are considered. And then, the approximate solutions are given, the resonance curves are obtained.

In the case of no clearance, the system with only hysteresis damping also is one of the typical nonliner systems[6]. The proposed method is applied to the beam simply supported at one end ,with supports having no clearance and hysteresis damping at the other end. The approximate solutions are easily obtainedby simple procedure and are more practical.

ANALYTICAL M E T H O D F O R T H E SYSTEM W I T H C O I J JSION

A simplified dynamical model of the piping system with symmetrical collision characteristics is shown in Fig.1. Namely, this model consists of a beam with one end simply supported, with two-sided amplitude constrained by the hysteresis loop characteristics (Fig.2) at the other end. Let 9 be the mass density, A the cross-sectional area and E1 the modulus of flexural rigidity. The equation of motion for transverse free vibration of a beam can be written as follows:

### 02y

### EI

### 04y

### (1)

### 0t2

pA### 0X4

The relations between y and z as shown in Fig.1 are given by

y - z + Y0 cos00t

### (2)

Hence we have

### 02z

### EI 04z

**0t2 ** **9 A 0X4 **- - -

### y

000 2 COS e0t### (3)

The solution of Eq.(1) is assumed **as: **

y = n~ 1-- X n (x)cosn00t

### (4)

A formal solution of Eq.(3) can be expressed as follows:

o o

~ ( A )

z = - y 0 cosmt + n cosh)vnX + Bn sinh )VnX + Cn COS)VnX + Dn sin)vnX cosnmt (5)

=

where A n , Bn, C n and D n are constants to be determined in each particular case from the boundary conditions ofthe beam, where

~n 4~4= Zl 4n 2~"~12, ~"~1 = 03/0) 1 (6)

and

Z12 / E I

0)1 --~-'--77-.~1"--7, Zl = 3.927 (7)

g'- VpA

The boundary conditions for this case are as follows:

1)x =0, z = 0 (8)

2)x = 0, ~ d2z - 0 (9)

dx 2

3)X -/?, ~ d2z - 0 (10)

dx 2

4)x **= **g, E I ~ **d 3 z = f(z e ze) **

dx 3

### (n)

where ze is the transverse displacement at the beam end (x= g ) and the force of restitution f (z e, ~ e ) is defined by the piecewise-linear characteristics as shown in Fig.2:

" KI(Z e - e o ~ e o ~ z e ~(e o +6o),~ e > 0 (I)

=K16 o + K 2 ( z e - e o - 6 o ~ e o + 6 o ~ z e ~ z e m a x , ~ e a 0 (II)

= K 3 (z e - Ze3 ~Ze3 ~ z e ~ Zemax,~ e ~ 0 (III)

**- 0 ; - e o ~ z e ~ z e B , k e < 0 ** **(IV) **

(12) f(ze'2:e)" = K l ( Z e + e o ~ - 6 0 ~ z e ~ 0 , z e < 0 (V)

**= **-Ka60 + K2 (Ze + eo + 60 ~-Zemax ~ Ze ~ -(eo + 60),ze ~ 0 (VI)

= K 3 (z e + Ze3 },-Zemax ~ z e ~ -ze3,2 e > 0 (VII)

**= 0 ; - z e 3 ~ z e ~ e o , 2 ; e > 0 ** (VIII)

where z e max denotes the maximum dispalcement at the end (x= g ). K 1 , K 2 and K 3 are spring constants of nonlinear support as shown in Fig.2 and 6 o is the maximum deformation by the spring constant K 1 . And then, z e3 is written as follows:

ze3 = e° + ( 1 -K3)K1 ]6° + ( 1 - K2 ) ( z - ~ 3 gmax - e 0 - 60) (13)

In quadrilateral hysteresis loop characteristics, the coefficient of restitution depends on velocity response at the beam end. The coefficient of restitution e is the ratio of the retreating velocity v 2 to the approaching velocity Vl and is given as the following equation.

E = V 2 / V 1 = 1.0,v I ~V ]

(14) E = V2/V 1 = ~ / K 2 / K 3 + 0 - K 2 / K l X K 1 / K 3 X v / v 1 ) 2

which V denotes the critical velocity distinguishing a perfectly elastic collision from an imperfectly elastic one, which is

V -- 6 0 ~/K 1 / m (15)

Relations between the coefficient restitution and velocity are shown in Fig.3.

From Eqs.(5) and the boundary condition Eq.(8), A n and C n are expressed as

Then Eq.(5) is written as

A n = C n = 0 (n = 2,3,. .... ) (17)

oo

(1 1 ) n~l(Bnsinh)~n + D sin)~ x ) c o s n c o t = (18)

z - Yo ~c°sh)~lX

### +--COS)~IX-1

cos~t + x n nUsing the boundary condition Eq.(9), the following equations are obtained.

1 (COS~lg - cosh)~lg)

B 1 sinh ~1 ~ -- D 1 sin ~1,~ = -~ Y0 (n =1) (19)

B n sinh)~ng-D n sin)~ng = 0 (n = 2,3,. .... ) (20)

In this paper, the steady-state vibration is dealt with. Once the vibration of the beam becomes steady and periodic, the nonlinear force of restitution f (z e, ~ e ) becomes also periodic and can be represented as a periodic function g(0) of 0 with the period 2a. And 0 is defined by the following equation:

0 = cot-ct (21)

where ot is the phase lag angle.

This periodic function g(0) must satisfy the conditions of the given characteristics of the nonlinear force of restitution equation(12), which is in this case, to be written as the following equations:

g ( 0 ) - KI(Z ~ - e o ~-(01 + 0 2 ) < 0 < - 0 2 (I)

g ( 0 ) = K160 + K 2 ( z e - e o - 6 o } , - 0 2 < 0 < 0 (II) g(0) = K 3 (z e - ze3 ),0 < 0 < 03 (III)

g(0) = 0 ; 0 3 ~ 0 ~ 71; - ( 0 1 + 0 2 ) (IV) (22)

f ( z e ' z e ) g ( 0 ) = K l ( Z e + e o } , Z t - ( 0 1 + 0 2 ) < 0 < ~ t - 0 2 (V)

g ( 0 ) = - K x ( 3 0 + K 2 ( Z e + e 0 + ~ 0 ~ - 0 2 < 0 < ~ (VI) g(0) = K 3 (z e + z e3 ~ ~t < 0 < zt + 03 (VII)

g(0) = 0; 7t + 03 < 0 < 2zt - (01 + 02 ) (VIII)

where 01 , 02 and 03 denote the range of the phase angle 0 . In the forgoing, one period 2zt of the resulting vibration is divided into eight intervals. During the first, the ser.~nd and the third intervals and the fifth, the sixth and the seventh intervals of length 01 , 02 and 03, respectively, the beam end moves with nonlinear force of restitution of spring. And during the fourth and the eighth intervals of 7t - (01 + 02 + 03 ), the beam moves without force of restitution.

A Fourier series expansion for periodic function g(0) is assumed as:

o0

g(0) = ~ (a n Cos no~t + b n sin no, t) (23)

n--1

In this paper, let the function be approximated by

g(O) = a I cos ncot+ b I sin no~t (24)

As shown in reference[5], when g(0) is approximated by only the fundamental terms of the Fourier expansion, approximate solution agree with exact solution for relatively low ratio of nonlinear parameters, K 1 / k , K 2 / k and K 3 / k [5], used in this paper. And k is equivalent stiffness of the beam defined by Eq.(34).

From the boundary condition equation (11), the following equation is obtained.

B1 c o s h ) V l g - D1 cos)vlgg ._ a 1 c o s 0 + b 1 sin0

EI)~ 1 COS(O+ Ct)

### (25)

From Eqs.(10) and (25), B 1 and D 1 are obtained as:

B1 _ a I cos0 + b 1 sin 0 Yo

**- ** sinh)~lg - 0 + sinh )~lg sin )~lg - cosh )~lg cos)~lg)

EI)~13A1 cos(0 + or)

(26)

A 1 = cosh )v lg sin ~1 g - sinh ~1 g cos ~1 g (27)

D 1 = a 1 cosO+ b I sin 0 EI~13A1COS(O + Or)

Equation(18) can be written as:

Yo (sinh)Vlgsin )rig + cosh)Vlg COS~lg -1)

sinh )Vlg - - ~ - 1 (28)

**1 **

**) **

z = Y0 cosh ~1 x + - - c o s ~ 1 x - 1 cos(0 + a ) + (B 1 sinh ~1 x + D 1 sin ~1 x )cos(0 + o~) 2

Substituting Eqs.(26) and (28), z is given as:

(29)

z = YoNx cos(0 + or)+ 1 (sin)vlgsinh)VlX + sinh )Vlgsin )vlxXa I cos0 + b I sin0) (30) EI)~13A1

where

(cosh ~IX + COS )v lx - 2 Xcosh ~1 zp sin )VlZ# - sinh )Vl~ COS )Vl~ )

- (1 - cosh ~1,~ COS ~1,~ + sinh ~1,~ sin ~l,~)sinh ~lX Nx _ - (sinh)~1 ~ sin £1 ~ + cosh )~lg cos)~ag - 1)sin )~1 x

2(cosh ~1 e sin ~1 g -- sinh ~1 g COS ~1 g) (31)

Displacement of beam end Ze is

where

z e = yoNe cos(0 + or)+ M 1 al m c o s 0 + M l b l s i n 0

k k

N ~ --

M 1 =

sin )Vl~ 0 - cosh ~1 ~)+ sinh )Vl~(COS ~1 ~ - 1)]

### --7ooshXl-Tsin

### s-anh

### [

6)Vlg sin ~1 g sinh )Vlg [

### (32)

### 03)

k is equivalent stiffness of the beam and given as:

3EI

k = ~ (34)

,~3

Meanwhile the switching-over conditions from one to another of the eight intervals (I), (II), OH), (IV), (V), (VI), (VII) and (VIH) are expressed as:

0 = -(01 + 02 ), z e = e o (VIII---, I) (35)

0 = -02, z e = e o + 6 o (I ---* II) (36)

0 " 0 , 7 . ~ = 0 , Zg =Zemax ( I I ~ III)

### (37)

0 = 03, z = z e3 (III---* IV) (38)

0 = 71; --(0 1 + 0 2 ~ Z g "- - e 0 (IV --'-" V ) (39)

0 = ~ - 0 2 , z e = - ( e o +60) (V---, VI) (40)

0 = 71;, Ze = 0, Zg = -Zemax (VI ---" VII) (41)

Using Eqs.(32) and (37), z e is given as:

where

And,

z e = F cos 0 (43)

a 1

F = yoNe cosot + r a M 1 (44)

k

1 b 1

sin ct = ~ m M 1 (45)

Yo Ne k

**1( al ) **

** 46, **

cosct = ~ F - - - M 1

YoNe k

And the nondimensional Fourier coefficients are defined as follows:

al bl

Xl =~-~, Yl - kF

From Eqs.(45), (46) and (47), the phase lag angle ot is expressed as:

o r = t a n - l ( MIyI ) l _ M l X l (48)

From Eqs.(35), (43), (45), (46) and (47), the amplitude of response at the beam end F and that of sinusoidal excitation y0 is determined as:

F _ Yo , _ Ne (49)

**eo eo ./a **

**_**

**)2**Mix1)2 + (MlYl

(47)

Mode shape Z x

where

Yo

**~/0-MIX ) **

**21 +(MlY 1)2 **

eo - N e cos(01 + 0 2 )
is obtained by using (6), (30), (34), (47) and (49) and substituting 0 = 0 as:

Z x = F I ~ / 0 - M l X l ) 2 +(MlYl) 2 Nx

**t **

Ne
### (5o)

3)Vlg I }

COSa + ~ x Xl (51)

Zl 4 ~"~ 1 2 A 1

I x = sin £1gsinh)VlX + sinh)Vlgsin )VlX

### (52)

From Eq.(6)

Using Eqs.(46) and (49), Eq.(51) is written as:

From Eqs.(13), (35), (36), (38) and (43),

~14X 4 _-- Zl 4 ~--~12X 4 / g4

Z x = F I N x ( I _ M l X l ) + 3)Vle I x } [Ng Zl 4 ~"~----~ 2 A1 xl

~0 COS(01 + 02 ) COS02 ---- COS(01 + 0 2 ) + ~

e0

**(53) **

### (54)

### (55)

c o s 0 3 = l _ K 1 ~)°c0s(01+02)_ K2

_{eo }

_{eo }

_{ (1-cos02) }

_{ (1-cos02) }

Since g(0) given by Eq.(22) is symmetrical, nondimensional g(6) is expressed as follows using Eqs.(35), (36), (37), (38), (43) and (47).

) tj

t L

g(0) kF g(0)

k r g(0)

k r g(0)

kF

### - X 1 COS 0 + Y l sin

0### = K1 r]COS 0 -- COS(01 + 0 2 \~)~ ;-(01 + 0 2 ")~ 0 ~ --0 2

k

### -- XICOS0+ Ylsin0= Ul {COS02 --COS(01 +02)}+-'~-(COS0--COS02) "--02 ~0~0

k

**-XlCOS0+ylsin0=K3(cos0-cos03) **

; 0 z 0 z 0 3
k
### - XICOSO+ YlSinO=O ;03 ~0~7[-(01 "t-02)

(57)

Applying a technique similar to that for determining Fourier coefficients, namely, multiplying both sides of Eq.(57) by cos0 and sin0 integrating through the whole period of 2~, the nondimensional coefficients xl and Yl are obtained as follows:

### [

### {

### } K2 ( 02 - sin 02 cOs 02 ) K3(03-sin03c°s03)

2 K 1 01 + sin 02 cos02 - sin(01 + 02)cos(01 + 02) + + (58)

### Xl =m

~ T 2 T 2 T 2

### Yl =

** [ {cos o2 _cos (ol +o2 **

- c o s 0 2 + c o s ( 0 1 + 0 2 ### )} _ ~ ( 1

+ s i n 2 0 2 - 1 + c o s 0 2 ---~-### ) K3(1

sin 0 3 - 1 + c o s 0 3### 2

### )]

### (59)

ANALYTICAL METHOD FOR THE SYSTEM WITH HYSTERF_~IS DAMPING

A beam simply supported at one end, with hysteresis damping at the other end is shown in Fig.5. In this case, e o = 0. Hysteresis damping has quadrilateral hysteresis loop characteristics as shown in Fig.6. f (z e, ~e) is shearing force at the other end (x= g ). z e3 is given as:

**z 3_(l_ **

_{K3) emax- K3 K3 }

_{K3) emax- K3 K3 }

**(K1 **

**A **

periodic function g(q) of interval **(I), (1I), (V) **

and interval (VI) are written as:
f ( z e , £ e ) - g ( 0 ) =

### K1ze;-(01 +02)<0<-02

(I) (61)f ( z e , k e ) ~ g ( 0 ) = K16 o + K2(z e

### --(~0~--02 ~0~0

(II) (62)### f(Ze,~e)- g(0)= KlZg ;~-(01 +02)<0<7g--02 (V)

(63)f ( z e , ~ e ) - g(0) =-K16o +K2(z e + 6o);,a,-0 2 ~ O ~ n

Switching-over conditions (VIN-->I), (I--qI), (1V-->V) and (V-->VI) are expressed as:

0 = -(01 + 02 ), z e = 0 (VIII--> I)

(VI) (64)

(65)

O = - 0 2 , Ze =6 o (I ---> II) (66)

### 0--7[--(01 +02~Zg "-0 (IV---~ V)

(67)**0 = n - 0 2 ,z e=-60 (V'-->VI) **

The amplitude at the beam end (x= g ) F and that of sinusoidal excitation yo are determined as:

F Yo Ne

60

_{~0 ~/0-MIX1) 2 +(MlYl) 2 }

(68)

I I,

~/0 - m lXl )2 + (M1 Y l )2

Y o _{(70) }

60 Ne cos0 2

Since g(0) is symmetrical, nondimensional g(0) is expressed as follows using Eqs.(12), (61) and (62).

g(0) _ Xl cos0 + Yl sin 0 - Ka cos0 ;-(01 + 0 2)< 0 < ,02

kF k

g(O) _ Xl cos0 + Yl sin O = K1 cos02 + K2 (cos0 - cos02) ;-02 < 0 < 0

kV k - - ~ (71)

g(0) _ XICOS0.t" Ylsin0= K 3 (cos0_cos03) ; 0 < 0 < 0 3

kF k

g(0)

- X l C O S 0 + y l s i n 0 = 0 ; 0 3 < 0 < a ~ - ( 0 1 + 0 2 ) kF

Applying a technique similar to that for determining Fourier coefficients, namely, multiplying both sides of Eq.(71) by cos0 and sin0 integrating through the whole period of 2~ the nondimensional coefficients Xl and Yl are obtained as follows:

2 01+ sin 02 c°s 02 K2 02 - sin 02 c°s 02 + (72)

X 1 = ~ +

a, 2 ---if- 2 2

= - - COS 02-COS02 ~sin 02-1+COS02 ~-sin 0 3 - 1 + c o s 0 3 Yl a~

From Eqs.(60) and (65)

01 +0 2 = ~ , COS(01 + 0 2 ) = 0

### (73)

(74)

cos03 = 1 - K2 ** _{-- }**( K1 K2 ]

_{C O S 0 2 }

_{(75) }K3 ~K3 K3

### )

NUMERICAL EXAMPLES

Figures 4 and 5 show the resonance curves of the system with collision and hysteresis damping, with the amplitude F / e 0 of the beam end versus frequency ratio f21 for excitation ratio Y0/e0 =1.0. Figures 6 and 7 show the resonance curves of the system with hysteresis damping and without collision for excitation ratio y 0 / 6o =1.0. Fig.8 shows mode shapes.

CONCLUSIONS

An analytical method of approximate solution for steady-state response of a beam simply supported at one end and colliding with the stop in half period of its vibration at the other end is presented. Considering the energy loss in a collision with a stop and duration of collision, the dynamical characteristics of the stop is assumed to have quadrilateral hysteresis loop characteristics. Moreover, an analytical method of approximate solution for steady-state response of a beam having no clearance and hysteresis damping is also presented. Some numerical results of the approximate solution are shown.

**REFERENCES **

1. Watanabe,T., "Forced Vibration of Continuous System with Nonlinear Boundary Conditions", Trans. of ASME, Journal of Mechanical Design, Vol.100,1978, pp.487-491

2. Baisley, D.E and Johnson, M.A., "Reload Fuel Mixed Core Seismic and Pipe Break Analysis", Trans. of SMiRT-15, VoHI, 1999, pp.453-460

3. Salih, M.H.S. and McLachlan,N., "Modlling Techniques for the Prediction of the Movements of Gas Cooled Reactor Graphite Core Bricks", Trans. of SMiRT-15, Vol.II, 1999, pp.77-84

4. Watanabe,T., ''Forced Response of Continuous System with Collision Characteristics", Proceedings of the 2nd European Nonlinear Conference, Vol.1,1996, pp.489-494

5. Watanabe,T. and Shibata, H., *"On *Nonlinear Vibration of a Beam-Response of a Beam with a Gap at One End-, Report of the Institute of
Industrial Science The University of Tokyo, Vol.36, No.l, 1991, pp.1-25

I

*/ / / / / / / / / / *

**/ / / I 11 **

**/ / / I 11**

*% *

*/ / / *

*i / / / / / / / / / *

*Y *

**Fig.1 Analytical model of simply supported beam with collision **

f ( z ( ,9.( )

K

**6 o **

< > z g 3 " e 0

Z e m a x ' e 0

z g - e 0

**Fig.2 Quadrilateral hysteresis loop charactersitcis **

**20 **

**1 5 ** "

**lO **

**0 ** **2 ** **4 ** **6 ** **8 ** **10 **

**Fig.5 Resonance curve of simply supported beam with collision **
**(K 1/k -- 40,K 2 / k = 20,K 3/k - 100, Yo/e0 = 1.0) **

**20 **

**15 **

**/ **

**lO **

**A **

**II **

** o,k **

**\ **

**0 ** **2 ** **4 ** **6 ** **8 ** **10 **

~"~1

**Fig.6 Resonance curve of simply supported beam **
**with hysteresis damping **

**(K1/k = 4 , K 2 / k = 2 , K 3 / k =10,Yo/6o =1.0) **

0 q i i

0 2 4 6 8 **1 0 **

**v t / 7 - **

**Fig.3 Relation between coetficient of restitution and velocity **
**response at beam end **

**(K 1/k --- 4,K 2 / k = 2,K 3 / k = 10) **

**20 **

**15 **

g

**~ i o : **

## )

0 - '

**0 ** **2 ** **4 ** **6 ** **8 **

**fal **

**Fig.4 Resonance curve of simply supported beam with collision **
**(K 1/k - 4,K 2 / k = 2,K 3 / k = 10, Yo/eo = 1.0) **

**2 0 **

**1 5 ** **- **

**i 0 ** **- **

**0 ** **2 ** **4 ** **6 ** **8 ** **10 **

**Dx **

**Fig.7 Resonance curve of simply supported beam **
**with hysteresis damping **

**(K 1/k = 40,K 2 / k = 20,K 3 / k = 100, Yo/60 = 1.0) **

**10 **

n,_-.o 3

8 . . . n,=l.O

**- - - ** n,=3.4 ,.'

. . . . n,=6.6 .,"" 6

4 . - . . . -.. , " f ' " ' - " ". .'"'" ,,/~' /

2 0

• , . . . , -..<"

~2 *" x . * *, . , . , . ~ , . I * *, . . . . * *,' *
*. . . * *• "" *

*-4 * **T ** **i ** **r ** **r **

**0 ** **0.2 ** **0.4 ** **0.6 ** **0.8 **

x/l