# Nonlinear Vibration Analysis of Moving Strip with Inertial Boundary Condition

Authors: Chong-yi Gao; Guo-jun Du; Yan Feng; Jian-xiong Li
Journal: Mathematical Problems in Engineering
 (2015)

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## Abstract

According to the movement mechanism of strip and rollers in tandem mill, the strip between two stands was simplified to axially moving Euler beam and the rollers were simplified to the inertial component on the fixed axis rotation, namely, inertial boundary. Nonlinear vibration mechanical model of Euler beam with inertial boundary conditions was established. The transverse and longitudinal motion equations were derived based on Hamilton’s principle. Kantorovich averaging method was employed to discretize the motion equations and the inertial boundary equations, and the solutions were obtained using the modified iteration method. Depending on numerical calculation, the amplitude-frequency responses of Euler beam were determined. The axial velocity, tension, and rotational inertia have strong influences on the vibration characteristics. The results would provide an important theoretical reference to control and analyze the vertical vibration of moving strip in continuous rolling process.


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## Body

## 1. Introduction

Vibration phenomenon of moving strip is extremely complicated in continuous rolling process. It may lead to strip quality decline, equipment trouble, and production efficiency reduction [1, 2]. For rolling mill vibrations, there are more studies on vertical, axial, torsional, horizontal, cross, and swing vibration modes of roller [3]. However, the strip vibration exists as well in practical production and scholars provided a little focus for this aspect. Therefore, it is significant to study the vibration of moving strip between two stands during rolling.The moving strip during continuous rolling process can be similar to an axially moving beam if the vertical vibration of mill can be neglected [4]. Using the theory of moving beam, the vertical vibration of strip can be equivalent to transverse vibration of axially moving beam. Some studies have been already conducted. Sun et al. [5, 6] established two-dimensional dynamic model and three-dimensional dynamic model of moving strip between two mills with time-dependent tension, and the stability of principle parametric resonances was analyzed. Huang and Chen [7, 8] investigated nonlinear vibration of the axially moving beam with coupled transverse and longitudinal motions, and the incremental harmonic balance method was used for analysis. Świa̧toniowski and Bar [9, 10] analyzed chatter vibration phenomena in tandem rolling mill with use of mathematical models of parametrical self-excited vibration and the nonlinear mathematical model of oscillated system was given: the continuous group of the rolling stands coupled by transferring strip. Miranker [11] put forward differential equations of transverse vibration on moving strings firstly, and the corresponding moving string was analyzed. Hu and Zhang [12] investigated an axially accelerating rectangular thin plate subjected to parametric excitations resulting from the axial time-varying tension and axial time-varying speed in the magnetic field. The parametric resonance of plates with axis tension during zinc coasting process was studied by Kim et al. [13]. Stability of transverse vibration in axial variable motion system was studied by Mote Jr. [14]. The mathematical model of a strip-roller-flexible-supporting hybrid system in coating section of a continuous hot-dip galvanizing line was established by Wang et al. [15]. The above studies considered some aspects of moving strip but did not give the influences of rotational inertia and other factors on the strip. It will not be able to analyze the vibration characteristics of strip accurately.Based on the movement mechanism of rolling and theory of axially moving beam, the vertical vibration of moving strip can be similar to a transverse vibration of axially moving Euler beam with tension [16–18]. The author established a dynamic model of interaction between inertial component and Euler beam. When the rotational inertia is considered, the nonlinear vibration mechanical model of the Euler beam with inertial boundary conditions is established. The modified iteration method is employed to solve the motion equations and used Matlab for simulation and analysis [19, 20]. Hereby, the relationships of axial velocity, tension, and rotational inertia are investigated, and the results may provide an important theoretical reference to control and analyze the vertical vibration of strip during continuous rolling process.


## 2. Mechanical and Mathematical Models

The strip moves between two stands during rolling process; the schematic diagram is shown in Figure1. The moving strip can be simplified to moving Euler beam which is supported by both rollers. The rolling direction can be considered as axial motion direction, and it is assumed that the motion is uniform movement, v0 is axial velocity, and P0 is tension. Rollers can be equivalent to symmetric and rigid inertial components and regarded as the disc of fixed axis rotation along the width of beam; j is rotational inertia. Therefore, Figure 2 shows an axially moving Euler beam at velocity v0 and acted tension P0 with inertial boundary conditions. The origin of coordinate locates as a crossing point of the center line of the first stand and the passing line. The transverse displacement and the longitudinal displacement of Euler beam are w(x0,y0,t) and u(x0,y0,t), respectively. l is the distance of Euler beam between two stands.Figure 1
The schematic diagram of moving strip between two stands.Figure 2
Mechanical model of Euler beam with inertial boundary conditions.Based on Hamilton’s principle, the mathematic model is established. The following equation can be obtained:(1)δ∫t2t1T1+T2-U1-U2dt=0.Kinetic energy T1 of moving Euler beam can be written as (2)T1=12∫0lρAv0+u,t+v0u,x02+w,t+v0w,x02dx0,where ρ is density of strip, A is cross-sectional area of Euler beam, and (·),t=∂(·)/∂t and (·),x0=∂(·)/∂x0 denote the partial derivatives in time t and spatial coordinate x0, respectively.Potential energyU1 of Euler beam can be written as(3)U1=12∫0lEAu,x0+12w,x022+EIw,x0x02dx0,where E is Young’s modulus and I is moment of inertia.Kinetic energyT2 of roller can be written as (4)T2=12jdudt1r2=j2r2u,t+v0u,x02,where r is radius of roller.Potential energyU2 of tension can be written as(5)U2=∫0lP0u,x0+12w,x02dx0.Substitution of (2)~(5) into (1), the motion equations can be expressed as(6)ρAu,tt+ρAv02-EAu,x0x0+2ρAv0u,x0t-EAw,x0w,x0x0=0,(7)ρAw,tt+EIw,x0x0x0x0+ρAv02-P0w,x0x0+2ρAv0w,x0t-EAw,x0x0u,x0+u,x0x0w,x0+32w,x02w,x0x0=0.Boundary conditions can be written as(8)EAru,x0-P0r=-ju,tt+2v0u,x0t+v02u,x0x0+v,tu,x01rwhenx0=0,P0r+EAru,x0=-ju,tt+2v0u,x0t+v02u,x0x0+v,tu,x01rwhenx0=l,w0=wl=w,x0x00=w,x0x0l=0,wmax=φm,where wmax is the maximum transverse displacement.In general case, the kinetic energy of axial movement caused by transverse vibration is relatively small. So, letu,t=u,tt=0; (6) can be simplified as(9)ρAv-02-EAu,x0x0-EAw,x0w,x0x0=0.By substitution ofw=φ-x0cos⁡ω-t into (9), the following equation can be obtained:(10)u=ϕ-x0cos2⁡ω-t,where ω- is vibration frequency.According to Kantorovich averaging method on an interval0,2π/ω-, the time variables can be eliminated. The motion equations and the inertial boundary equations after being discreted are(11)ρAv02-EAϕ-,x0x0-EAφ-,x0φ-,x0x0=0,(12)EIφ-,x0x0x0x0+ρAv02-P0φ-,x0x0-ω02ρAφ--34EAφ-,x0x0ϕ-,x0+φ-,x0ϕ-,x0x0+32φ-,x02φ-,x0x0=0,(13)jv02ϕ-,x0x0+EAr2ϕ-,x0-43jω02ϕ--43P0r2=0whenx0=0,(14)jv02ϕ-,x0x0+EAr2ϕ-,x0-43jω02ϕ-+43P0r2=0whenx0=l,(15)φ-0=φ-l=φ-,x0x00=φ-,x0x0l=0,(16)φ-12=φ-m.The dimensionless quantities are given by(17)x=x0l,ϕ=ϕ-l,φ=φ-l,v=v0ρE,ω=ω0ρAl4EI,S=Al2I,J=jρAr2l,P=P0l2EI.Then, the dimensionless form of (11)~(16) can be obtained, respectively:(18)v2-1ϕ,xx-φ,xφ,xx=0,(19)φ,xxxx+Sv2-Pφ,xx-ω2φ-34Sφ,xxϕ,x+φ,xϕ,xx+32φ,x2φ,xx=0,(20)Jv2ϕ,xx+ϕ,x-4J3Sω2ϕ-4P3S=0whenx=0,(21)Jv2ϕ,xx+ϕ,x-4J3Sω2ϕ+4P3S=0whenx=1,(22)φ0=φ1=φ,xx0=φ,xx1=0,(23)φ12=φm.


## 3. Solution Based on Modified Iteration Method

Due to more complicated solving process of (18)~(23), the modified iteration method will be employed to solve the equations in this section.


### 3.1. The First-Order Approximate Solution

Firstly, all the nonlinear terms will be omitted in (19), and (19) can be transformed into(24)φ1,xxxx-ω2φ1=0.The series solution of (24) is(25)φ1=a0M0x+a1N0x+a2I0x+a3K0x,where(26)M0=∑n=0∞ω2n4n!x4n,N0=∑n=0∞ω2n4n+1!x4n+1,I0=∑n=0∞ω2n4n+2!x4n+2,K0=∑n=0∞ω2n4n+3!x4n+3.By substitution of (25) into (22) and (23), one has ω1=14.709. Then, the coefficients of (25) have(27)a0=0,a1=μ1φm,a2=0,a3=μ2φm,where(28)μ1=-K01N01K01/2-N01/2K01,μ2=N01N01K01/2-N01/2K01.Then,(29)φ1x=φmμ1N0x+μ2K0x=φmμ1∑n=0∞ω12nx4n+14n+1!+μ2∑n=0∞ω12nx4n+34n+3!.And substituting (29) into (18), one can get(30)ϕ1x=12v2-1∫φ1,x2dx+c1x+c2.The coefficientsc1 and c2 in (30) can be determined by (20) and (21), and the results are as follows:(31)c1=12v2-13Sv22ω12φ1,xφ1,xx+3S4Jω12φ1,x2-∫φ1,x2dxx=1+2PJω12c2=3S8Jω12v2-13Sv22ω12φ1,xφ1,xx+3S4Jω12φ1,x2-∫φ1,x2dxx=1+3S2Jω12-1PJω12.


### 3.2. The Second-Order Modified-Iterative Solution

In this subsection, the second-order modified-iterative solution can be given.By substitution of (29) and (30) into (19), one has(32)φ2,xxxx-ω2φ2=αφ1,xx+βφ1,x2φ1,xx,where(33)α=P+34Sc1-Sv2,β=9Sv28v2-1,φ1,x2x=φm2∑n=0∞An1x4n+∑n=0∞Bn1x4n+2,φ1,xxx=φm∑n=1∞Cn1x4n-1+∑n=0∞Dn1x4n+1,where(34)A01=μ12,An1=∑n1=0nμ12ω12n4n1!4n-4n1!+∑n1=0n-1μ22ω12n-24n1+2!4n-4n1-2!n=1,2,…,Bn1=∑n1=0n2μ1μ2ω12n4n1!4n-4n1+2!n=0,1,…,Cn1=μ1ω12n4n-1!,n=1,2,…,Dn1=μ2ω12n4n+1!,n=0,1,….Due to the property of series solution, the solution of (32) can be written as(35)φ2x=φmζ1∑n=0∞ω2nx4n+14n+1!+ζ2∑n=0∞ω2nx4n+34n+3!+φm∑n=1∞Anx4n-1+∑n=0∞Bnx4n+1+φm3∑n=1∞Cnx4n-1+∑n=0∞Dnx4n+1+∑n=0∞Enx4n+3,where ζ1 and ζ2 are undetermined coefficients. Consider(36)A1=B0=C1=D0=E0=0,An+1=ω2An+αCn14n+34n+24n+14n,Bn+1=ω2Bn+αDn14n+54n+44n+34n+2,Cn+1=ω2Cn+βAn24n+34n+24n+14n,Dn+1=ω2Dn+βBn24n+54n+44n+34n+2,En+1=ω2En+βCn24n+74n+64n+54n+4,where(37)An2=∑m=0nAn1Cn-m1,n=1,2,…,Cn2=∑m=0nBn1Dn-m1,n=0,1,…,B02=μ12μ2,Bn2=∑m=0nAn1Dn-m1+Bn1Cn-m1,n=1,2,….By substitution of (35) into (22) and (23), the solvability condition is(38)Dξ=0,where(39)ζ=ζ1ζ21T,D=d11d12d13d21d22d23d31d32d33,where(40)d11=∑n=0∞ω2n4n+1!,d12=∑n=0∞ω2n4n+3!,d13=∑n=1∞An+∑n=0∞Bn+φm2∑n=1∞Cn+∑n=0∞Dn+∑n=0∞En,d21=∑n=0∞ω2n4n+1!124n+1,d22=∑n=0∞ω2n4n+3!124n+3,d23=∑n=1∞An124n-1+∑n=0∞Bn124n+1+φm2∑n=1∞Cn124n-1+∑n=0∞Dn124n+1+∑n=0∞En124n+3-1d31=∑n=1∞ω2n4n-1!,d32=∑n=0∞ω2n4n+1!,d33=∑n=1∞4n-14n-2An+∑n=0∞4n+14nBn+φm2∑n=1∞4n-14n-2Cn+∑n=0∞4n+14nDn+∑n=0∞4n+34n+2En.The frequencyω2 can be obtained by det⁡D=0. The coefficients ζ1 and ζ2 can be determined by (38); then the second-order modified-iterative solution can be obtained.


## 3.1. The First-Order Approximate Solution

Firstly, all the nonlinear terms will be omitted in (19), and (19) can be transformed into(24)φ1,xxxx-ω2φ1=0.The series solution of (24) is(25)φ1=a0M0x+a1N0x+a2I0x+a3K0x,where(26)M0=∑n=0∞ω2n4n!x4n,N0=∑n=0∞ω2n4n+1!x4n+1,I0=∑n=0∞ω2n4n+2!x4n+2,K0=∑n=0∞ω2n4n+3!x4n+3.By substitution of (25) into (22) and (23), one has ω1=14.709. Then, the coefficients of (25) have(27)a0=0,a1=μ1φm,a2=0,a3=μ2φm,where(28)μ1=-K01N01K01/2-N01/2K01,μ2=N01N01K01/2-N01/2K01.Then,(29)φ1x=φmμ1N0x+μ2K0x=φmμ1∑n=0∞ω12nx4n+14n+1!+μ2∑n=0∞ω12nx4n+34n+3!.And substituting (29) into (18), one can get(30)ϕ1x=12v2-1∫φ1,x2dx+c1x+c2.The coefficientsc1 and c2 in (30) can be determined by (20) and (21), and the results are as follows:(31)c1=12v2-13Sv22ω12φ1,xφ1,xx+3S4Jω12φ1,x2-∫φ1,x2dxx=1+2PJω12c2=3S8Jω12v2-13Sv22ω12φ1,xφ1,xx+3S4Jω12φ1,x2-∫φ1,x2dxx=1+3S2Jω12-1PJω12.


## 3.2. The Second-Order Modified-Iterative Solution

In this subsection, the second-order modified-iterative solution can be given.By substitution of (29) and (30) into (19), one has(32)φ2,xxxx-ω2φ2=αφ1,xx+βφ1,x2φ1,xx,where(33)α=P+34Sc1-Sv2,β=9Sv28v2-1,φ1,x2x=φm2∑n=0∞An1x4n+∑n=0∞Bn1x4n+2,φ1,xxx=φm∑n=1∞Cn1x4n-1+∑n=0∞Dn1x4n+1,where(34)A01=μ12,An1=∑n1=0nμ12ω12n4n1!4n-4n1!+∑n1=0n-1μ22ω12n-24n1+2!4n-4n1-2!n=1,2,…,Bn1=∑n1=0n2μ1μ2ω12n4n1!4n-4n1+2!n=0,1,…,Cn1=μ1ω12n4n-1!,n=1,2,…,Dn1=μ2ω12n4n+1!,n=0,1,….Due to the property of series solution, the solution of (32) can be written as(35)φ2x=φmζ1∑n=0∞ω2nx4n+14n+1!+ζ2∑n=0∞ω2nx4n+34n+3!+φm∑n=1∞Anx4n-1+∑n=0∞Bnx4n+1+φm3∑n=1∞Cnx4n-1+∑n=0∞Dnx4n+1+∑n=0∞Enx4n+3,where ζ1 and ζ2 are undetermined coefficients. Consider(36)A1=B0=C1=D0=E0=0,An+1=ω2An+αCn14n+34n+24n+14n,Bn+1=ω2Bn+αDn14n+54n+44n+34n+2,Cn+1=ω2Cn+βAn24n+34n+24n+14n,Dn+1=ω2Dn+βBn24n+54n+44n+34n+2,En+1=ω2En+βCn24n+74n+64n+54n+4,where(37)An2=∑m=0nAn1Cn-m1,n=1,2,…,Cn2=∑m=0nBn1Dn-m1,n=0,1,…,B02=μ12μ2,Bn2=∑m=0nAn1Dn-m1+Bn1Cn-m1,n=1,2,….By substitution of (35) into (22) and (23), the solvability condition is(38)Dξ=0,where(39)ζ=ζ1ζ21T,D=d11d12d13d21d22d23d31d32d33,where(40)d11=∑n=0∞ω2n4n+1!,d12=∑n=0∞ω2n4n+3!,d13=∑n=1∞An+∑n=0∞Bn+φm2∑n=1∞Cn+∑n=0∞Dn+∑n=0∞En,d21=∑n=0∞ω2n4n+1!124n+1,d22=∑n=0∞ω2n4n+3!124n+3,d23=∑n=1∞An124n-1+∑n=0∞Bn124n+1+φm2∑n=1∞Cn124n-1+∑n=0∞Dn124n+1+∑n=0∞En124n+3-1d31=∑n=1∞ω2n4n-1!,d32=∑n=0∞ω2n4n+1!,d33=∑n=1∞4n-14n-2An+∑n=0∞4n+14nBn+φm2∑n=1∞4n-14n-2Cn+∑n=0∞4n+14nDn+∑n=0∞4n+34n+2En.The frequencyω2 can be obtained by det⁡D=0. The coefficients ζ1 and ζ2 can be determined by (38); then the second-order modified-iterative solution can be obtained.


## 4. Numerical Simulation

To verify the proposed method, some numerical simulations will be given. The main parameters of a tandem rolling mill are listed as below: the distance between stand F2 and stand F3 is 3000 mm, the thickness of strip is 20 mm, rolling velocity of mill is 3 m/s, the nondimensional velocity is6×10-4, and the nondimensional tension is 5×104.Figure3 illustrates the influence of nondimensional rotational inertia J on nondimensional frequency ω under the condition of constant velocity and constant tension. It can be found that the rotational inertia of roller has great effect on amplitude-frequency response characteristics, and the vibration performance will be changed. From Figure 3(a), one can see that the vibration performance is changed from sclerotic type to intenerate type gradually and the nondimensional maximum amplitude about φm is 0.02, 0.06, 0.10, 0.14, and 0.18, respectively. When φm≤0.10, the nondimensional frequencies increase with increasing of rotational inertia, but the hardening degree decreases with increasing of φm. When φm≥0.14, with increasing of rotational inertia, the nondimensional frequencies decrease, and the softening degree increases with increasing of φm. It is obvious that Figure 3(b) shows the same vibration performance. With increasing of φm, the frequencies decrease, and vibration performance present uniformly changes. When φm=0.12, the point is limit between sclerotic type and intenerate type. In addition, when rotational inertia increases to infinity, the boundary condition tends to be fixed. And the bigger the rotational inertia, the greater the impact of amplitude on the frequency.Figure 3
Influence of rotational inertia on frequency.
(a)
Different values of maximum amplitude
(b)
Different values of rotational inertiaFigure4 shows the amplitude-frequency responses under the condition of different nondimensional velocities, which are 2×10-4, 4×10-4, 6×10-4, 8×10-4, and 10×10-4, respectively. The nondimensional rotational inertia J=500; the nondimensional tension P=5×104. From Figure 4, one can see that the nondimensional frequencies decrease with the increasing of amplitude. With the decrease of velocity, the influence of amplitude acts the less on frequency, and the downtrend of amplitude-frequency responses becomes more gradual. That is, the larger velocity results in the greater influence of the amplitude on frequency.Figure 4
Amplitude-frequency responses under the condition of different velocities.Figure5 depicts the amplitude-frequency responses with different nondimensional tensions: 1×104, 2×104, 3×104, 4×104, and 5×104. The nondimensional rotational inertia J=500; the nondimensional velocity v=6×10-4. As can be seen from Figure 5, with increasing of amplitude, the nondimensional frequencies decrease. And the smaller is the tension, the faster declines the amplitude-frequency response. When φm≤0.11, the larger tension leads to less frequency, whereas an adverse result is presented from φm>0.11. That is to say, the smaller tension has greater effect on vibration frequency of strip.Figure 5
Amplitude-frequency responses under the condition of different tensions.Figure6 displays the relationship between nondimensional velocity and nondimensional frequency under the condition of different nondimensional rotational inertia, which is 100, 200, 300, 400, and 500, respectively. The amplitude φm=0.12; the nondimensional tension P=5×104. From Figure 6, one can see that, with increasing of velocity, the nondimensional frequencies decrease. And the bigger is the rotational inertia, the faster declines the trend of velocity-frequency curve. When v≤5.7×10-4, the bigger is the rotational inertia, the greater is the frequency, whereas when v>5.7×10-4, the result shows the opposite. So, the velocity has strong influence on frequency, while the rotational inertia is bigger than others. It also can be known that the velocity-frequency curves are closer with bigger rotational inertia. In other words, a further proof is given to the rule of boundary condition that tends to be fixed at that time.Figure 6
Relationship between velocity and frequency.Figure7 demonstrates relationship between nondimensional tension and nondimensional frequency under the condition of different nondimensional rotational inertia, which is 100, 200, 300, 400, and 500, respectively. The nondimensional amplitude φm is 0.12; the nondimensional velocity v is 6×10-4. From Figure 7, one can see that, with increasing of tension, the nondimensional frequencies increase, and the rising tendencies keep consistent completely. Also it can be seen that the bigger is the rotational inertia, the less is the frequency and the more approximate is the tension-frequency curves. Moreover, it is shown that the less rotational inertia has much influence on frequency.Figure 7
Relationship between tension and frequency.In Figures6 and 7, it can be seen that the rotational inertia of roller has much influence on vertical vibration. When the rotational inertia is bigger, the velocity variables have greater influence on vibration, but the tension variables have less influence. Thus, rotational inertia is not to be ignored in researching of strip vibration. To improve the strip quality and prolong the service life of roller, harmful vibration phenomenon of strip can be depressed during rolling process by adjusting the above parameters.


## 5. Conclusions

(1) Based on the movement mechanism of rolling and the theory of axially moving beam, the mechanical model of the Euler beam with inertial boundary conditions is established. The Kantorovich averaging method is employed to separate the time variables and space variables of equations. Using modified iteration method, the solutions of nonlinear vibration equations are obtained.(2) The simulation results indicate that axial velocity, tension, and rotational inertia of the roller have strong influences on the vibration characteristics. Under the case that tension and velocity are constants, the vibration performances are changed by the maximum amplitude while the rotational inertia is considered. The pointφm=0.12 is limit between sclerotic type and intenerate type. When φm≤0.12, with increasing of rotational inertia, the vibration frequency increases. The vibration performance shows sclerotic type and hardening degree is decreased along with amplitude increase gradually. With further increasing the amplitude, that is, φm>0.12, the curves present downward trend with rotational inertia increases, namely, intenerate type. And with increasing of φm, the softening degree increases. In addition, when rotational inertia increases infinitely, the boundary condition tends to be fixed, and the bigger the rotational inertia, the greater the impact of amplitude on the frequency.(3) Depending on the variation of velocity and tension, distinct cases arise. When the velocity is faster or the tension is smaller, the influences of amplitude on frequency are strong. Moreover, when the rotational inertia is bigger, the velocity variables have greater influence on vibration, but the tension variables have less influence.(4) The study on vibration characteristics of axially moving Euler beam with inertial boundary conditions accords with rolling actuality. The simulation results can provide an important theoretical reference to control and analyze the vertical vibration of strip during rolling process of tandem mill.



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