# The Spreading Residue Harmonic Balance Method for Strongly Nonlinear Vibrations of a Restrained Cantilever Beam

Authors: Y. H. Qian; J. L. Pan; S. P. Chen; M. H. Yao
Journal: Advances in Mathematical Physics
 (2017)

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## Abstract

The exact solutions of the nonlinear vibration systems are extremely complicated to be received, so it is crucial to analyze their approximate solutions. This paper employs the spreading residue harmonic balance method (SRHBM) to derive analytical approximate solutions for the fifth-order nonlinear problem, which corresponds to the strongly nonlinear vibration of an elastically restrained beam with a lumped mass. When the SRHBM is used, the residual terms are added to improve the accuracy of approximate solutions. Illustrative examples are provided along with verifying the accuracy of the present method and are compared with the HAM solutions, the EBM solutions, and exact solutions in tables. At the same time, the phase diagrams and time history curves are drawn by the mathematical software. Through analysis and discussion, the results obtained here demonstrate that the SRHBM is an effective and robust technique for nonlinear dynamical systems. In addition, the SRHBM can be widely applied to a variety of nonlinear dynamic systems.


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## Body

## 1. Introduction

A lot of problems in physical, mechanical, and aeronautical technology and even in structural applications are essentially nonlinear. Majority of the nonlinear dynamical models are mainly composed of a group of differential equations and auxiliary conditions for modeling processes [1]. In general, it is difficult to obtain the exact solution for strongly nonlinear high dimensional dynamic systems. Hence, the analytical approximate solution of the nonlinear problem has become the research object of many scholars in recent years [2–28].Generally speaking, the fifth-order Duffing type problem with the inertial and static nonlinear terms is sophisticated all the better [29]. Recently, some scholars have tried to study this kind of nonlinear problem. For instance, Telli and Kopmaz [30] and Lai and Lim [31] used the harmonic balance method to study the linear and nonlinear springs. S.-S. Chen and C.-K. Chen [32] dealt with this fifth-order nonlinear problem by applying the differential transformation approach. Subsequently, Ganji et al.[33] and Mehdipour et al.[34], respectively, brought in the homotopy perturbation method, amplitude-frequency formulation, and the energy balance method. They used these methods to solve this strongly nonlinear problem, and lower-order approximate solutions are yielded. Qian et al. [35] studied the nonlinear vibrations of cantilever beam by the HAM. Latterly, Guo et al. [36, 37] have presented the residue harmonic balance solution procedure to approximate the periodic behavior of different oscillation systems and they have obtained some more accurate results. Ju and Xue [38, 39] proposed the global residue harmonic balance method to study strongly nonlinear systems. Comparing the obtained solutions with the exact one, they discovered that the approximate results excellently agree with the exact one. Lee [40] used the multilevel residue harmonic balance method to solve a nonlinear panel coupled with extended cavity.The principal intention of this paper is to investigate the utility of the spreading residue harmonic balance method (SRHBM) [36] for the fifth-order strongly nonlinear problem. The paper consists of the following several parts. Section 2 describes how the strongly nonlinear equation is educed from the governing equations of the cantilever beam model in a nutshell. In Section 3, the SRHBM is introduced and the solution process of different order solutions will be presented. The numerical examples of the SRHBM are rendered and compared with other solutions in Section 4. Finally, conclusion of the paper is drawn in Section 5.


## 2. Mathematical Formulation

An isotropic slender beam with uniform lengthl and mass m per unit length is considered, as shown in Figure 1 [25]. It is assumed that the beam thickness is much smaller than the beam length, so the effects of shear deformation and rotary inertia can be ignored. The angle of inclination is θ and the beam displacement is a=b/l. For the boundary condition constraints, one of the conditions is hinged at the bottom of a rotational spring with stiffness K, and the other condition is independent. Moreover, the intermediate lumped mass M is also connected in s=d along the beam span. By the Euler-Lagrange differential equation, the fifth-order Duffing type temporal problem with strongly inertial and static nonlinearities is able to be derived as follows [25]:(1)x¨+x+ε1x2x¨+ε1xx˙2+ε2x4x¨+2ε2x3x˙2+ε3x3+ε4x5=0,x0=A,x˙0=0,where x is the dimensionless deflection at the tip of the beam, A is the maximum amplitude, the overdot indicates the derivative relative to t, and ε1, ε2, ε3, and ε4 are parameters. For the complete formulation of (1), readers are referred to [25] for details.Figure 1
Geometry and coordinate system for a beam with a lumped mass.


## 3. Solution Methodology

In the following, the spreading residue harmonic is used to solve (1). Firstly, by introducing a new variable τ=ωt and substituting it into (1), we can get (2)ω2x′′+x+ε1ω2x2x′′+ε1ω2xx′2+ε2ω2x4x′′+2ε2ω2x3x′2+ε3x3+ε4x5=0,x0=A,x′0=0,where x′ represents the first-order derivative with respect to τ and ω is the unknown angular frequency of (1).Since we discuss the existence of a periodic solution, we usually choose the base functions(3)cos⁡2k-1τ∣k=1,2,3,…,and we find the expression of the steady state solutions(4)xτ=x0τ+px1τ+p2x2τ+⋯,ω2=ω02+pω1+p2ω2+⋯,where p is an order parameter and ωii=0,1,2,… are unknown.Next, we mainly analyze the zeroth-order harmonic approximation, the first-order harmonic approximation, and the second-order harmonic approximation.


### 3.1. TheZeroth-Order Harmonic Approximation

To meet the initial conditions in (2), we can set the following initial guess solution of xτ; that is,(5)x0τ=Acos⁡τ,τ=ω0t.Substituting (5) into (2), the equation is yielded:(6)R0τ=ω02x0′′τ+x0τ+ε1ω02x02τx0′′τ+ε1ω02x0τx0′2τ+ε2ω02x04τx0′′τ+2ε2ω02x03τx0′2τ+ε3x03τ+ε4x05τ=A1-ω02+34A2ε3-12ε1A2ω02+58A4ε4-38A4ε2ω02cos⁡τ+A314ε3-12ε1ω02+516A2ε4-716A2ε2ω02cos⁡3τ+116A5ε4-3ε2ω02cos⁡5τ.R
0
τ denotes the zeroth-order residual term. When R0τ=0, x0τ is the exact solution.Based on the Galerkin procedure, the secular termcos⁡τ cannot appear on the right hand side of (6). Equating the term’s coefficient to zero, we obtain a linear equation containing an unknown ω0. Through solving that equation, we are able to work out the unknown frequency ω0: (7)ω0=8+6A2ε3+5A4ε48+4A2ε1+3A4ε2.In addition, the zeroth-order approximation solution can be obtained as follows:(8)x0τ=Acos⁡ω0t.When the obtained zeroth-order solution is substituted into (6), the terms of cos⁡3τ and cos⁡5τ generally are not zero.


### 3.2. TheFirst-Order Harmonic Approximations

Substituting (4) into (2), the coefficient of parameter p is put forward, and we can obtain(9)ϕ1τ=ω02x1′′+ε1x1x0′2+2x0x0′x1′+2x0x1x0′′+x02x1′′+ε26x02x1x0′2+4x03x0′x1′+4x03x1x0′′+x04x1′′+ω1x0′′+ε1x0x0′2+x02x0′′+ε22x03x0′2+x04x0′′+x1+3ε3x02x1+5ε4x04x1.According to (3), we choose the following equation as the solution of the representation:(10)x1τ=a3,1cos⁡τ-cos⁡3τ,where a3,1 is the unknown.Combining (10) with (9), we consider(11)R1τ=ϕ1τ+R0τ.We can obtain(12)R1τ=a3,11-ω02-Aω1+32A2a3,1ε3-12A3ω1ε1+516A4a3,15ε4+ε2ω02-38A5ε2ω1cos⁡τ+9a3,1ω02-a3,1+A272a3,1ε1ω02-34a3,1ε3+A314ε3-12ε1ω02+ω1+A43116a3,1ω02ε2-516a3,1ε4+A5516ε4-716ε2ω02+ω1cos⁡3τ+A292a3,1ε1ω02-34a3,1ε3+A4a3,16116ε2ω02-1516ε4+A5116ε4-316ε2ω02+ω1cos⁡5τ+A4a3,13116ε2ω02-516ε4cos⁡7τ.To increase the accuracy, (11) should be added in the zeroth-order residual term R0τ.Based on the Galerkin procedure, (12) should not contain secular terms. Letting coefficients of cos⁡τ and cos⁡3τ be zeros, hence, we obtain a linear equation set containing two unknowns ω1 and a3,1. Through solving the equation set, we can get (13)ω1=A216+24A2ε3+25A4ε4+5A4ε2-16ω02NM,a3,1=2A38+4A2ε1+3A4ε2NM,where(14)M=2561-9ω02+A8ε2205ε4-151ε2ω02+16A615ε2ε3+15ε1ε4-34ε1ε2ω02-16A4-13ε2-18ε1ε3-5ε4+47ε12+23ε2ω02+64A23ε3+ε14-34ω02,N=4ε3-8ε1ω02+A25ε4-7ε2ω02.Therefore, the first-order harmonic approximation can be procured:(15)ω1=ω02+ω1,x1τ=x0τ+x1τ=A+a3,1cos⁡τ-a3,1cos⁡3τ,τ=ω1t.When the obtained first-order solution is substituted into (12), the terms of cos⁡5τ and cos⁡7τ commonly are not zero.


### 3.3. TheSecond-Order Harmonic Approximations

Substituting (4) to (2) and presenting the coefficient of parameter p2 yield(16)ϕ2τ=ω026ε2x0x12x0′2+ε1x2x0′2+6ε2x2x02x0′2+2ε1x1x0′x1′+12ε2x02x1x0′x1′+ε1x0x12+2ε2x03x1′2+2ε1x0x0′x2′+4ε2x03x0′x2′+ε1x12x0′′+6ε2x02x12x0′′+2ε1x0x2x0′′+4ε2x03x2x0′′+2ε1x0x1x1′′+4ε2x03x1x1′′+x2′′+ε1x02x2′′+ε2x04x2′′+ω1ε1x1x0′2+6ε2x02x1x0′2+2ε1x0x0′x1′+4ε2x03x0′x1′+2ε1x0x1x0′′+4ε2x03x1x0′′+x1′′+ε1x02x1′′+ε2x04x1′′+ω2ε1x0x0′2+2ε2x03x0′2+x0′′+ε1x02x0′′+ε2x04x0′′+3ε3x0x12+10ε4x03x12+x2+3ε3x02x2+5ε4x04x2.Observing (16), we know it is linear with respect to ω2 and x2τ. In accordance with the form of (3), we apply (17)x2τ=a3,2cos⁡τ-cos⁡3τ+a5,2cos⁡τ-cos⁡5τ,where a3,2 and a5,2 are unknowns.Combining (17) with (16) and calculating the equation(18)R2τ=ϕ2τ+R1τ,we can obtain(19)R2τ=A94a3,12ε3-72a3,12ε1ω02-ω2+A232a3,2ε3+94a5,2ε3-32a5,2ε1ω02+A3154a3,12ε4-134a3,12ε2ω02-12ε1ω2+116A425a3,2ε4+45a5,2ε4+5a3,2ε2ω02-15a5,2ε2ω02+5a3,1ε2ω1-38A5ε2ω2+a3,2+a5,2-a3,2ω02-a5,2ω02-a3,1ω1cos⁡τ+Aa3,12172ε1ω02-94ε3+A272a3,2ε1ω02-34a3,2ε3+ε13a5,2ω02+72a3,1ω1-12A3a3,125ε4-13ε2ω02+ε1ω2+116A45a5,2ε4-5a3,2ε4+31a3,2ε2ω02+41a5,2ε2ω02+31a3,1ε2ω1-716A5ε2ω2+9a3,2ω02+9a3,1ω1-a3,2cos⁡3τ+Aa31272ε1ω02-34ε3+A213a5,2ε1ω02-32a5,2ε3-34a3,1+a3,2ε3-6ε1ω02+92a3,1ε1ω1+12A3a3,1217ε2ω02-5ε4+116A4147a5,2ε2ω02-25a5,2ε4-a3,1+a3,215ε4-61ε2ω02+61a3,1ε2ω1+116A5ε4-3ε2ω02+ω1+ω2+25a5,2ω02-a5,2cos5τ+A34a3,12ε3-172a3,12ε1ω02-A2a5,234ε3-192ε1ω02+18A3a3,125ε4-43ε2ω02+A4-516a3,2ε4-54a5,2ε4+3116a3,2ε2ω02+394a5,2ε2ω02-516a3,2ε4+116a3,131ε2ω02+ω1cos⁡7τ+18A3a3,125ε4-51ε2ω02-516A4a5,2ε4-11ε2ω02cos⁡9τ.With the purpose of increasing the accuracy, (18) should be put on the first-order residual term R1τ.In order to prevent the right hand side of (19) from exhibiting the secular terms cos⁡τ, cos⁡3τ, and cos⁡5τ, we make all their coefficients equal to zero. Then, we can get three linear equations containing three unknown parameters a3,2, a5,2, and ω2. According to the three linear equations, we can solve the three unknowns. Thus, thesecond-order harmonic approximation is shown:(20)ω2=ω02+ω1+ω2,x2τ=x0τ+x1τ+x2τ,τ=ω2t.In conclusion, we draw thekth-order harmonic approximation k=2,3,4,…(21)xkτ=xk-1τ+xkτ,ωk=ωk-12+ωkk=2,3,4,…,xk-1τ=xk-2τ+xk-1τ,ωk-1=ωk-22+ωk-1,xkτ=∑i=1ka2i+1,kcos⁡τ-cos⁡2i+1τ,x0=Acos⁡τ,ω0=ω0,k=2,3,4,….


## 3.1. TheZeroth-Order Harmonic Approximation

To meet the initial conditions in (2), we can set the following initial guess solution of xτ; that is,(5)x0τ=Acos⁡τ,τ=ω0t.Substituting (5) into (2), the equation is yielded:(6)R0τ=ω02x0′′τ+x0τ+ε1ω02x02τx0′′τ+ε1ω02x0τx0′2τ+ε2ω02x04τx0′′τ+2ε2ω02x03τx0′2τ+ε3x03τ+ε4x05τ=A1-ω02+34A2ε3-12ε1A2ω02+58A4ε4-38A4ε2ω02cos⁡τ+A314ε3-12ε1ω02+516A2ε4-716A2ε2ω02cos⁡3τ+116A5ε4-3ε2ω02cos⁡5τ.R
0
τ denotes the zeroth-order residual term. When R0τ=0, x0τ is the exact solution.Based on the Galerkin procedure, the secular termcos⁡τ cannot appear on the right hand side of (6). Equating the term’s coefficient to zero, we obtain a linear equation containing an unknown ω0. Through solving that equation, we are able to work out the unknown frequency ω0: (7)ω0=8+6A2ε3+5A4ε48+4A2ε1+3A4ε2.In addition, the zeroth-order approximation solution can be obtained as follows:(8)x0τ=Acos⁡ω0t.When the obtained zeroth-order solution is substituted into (6), the terms of cos⁡3τ and cos⁡5τ generally are not zero.


## 3.2. TheFirst-Order Harmonic Approximations

Substituting (4) into (2), the coefficient of parameter p is put forward, and we can obtain(9)ϕ1τ=ω02x1′′+ε1x1x0′2+2x0x0′x1′+2x0x1x0′′+x02x1′′+ε26x02x1x0′2+4x03x0′x1′+4x03x1x0′′+x04x1′′+ω1x0′′+ε1x0x0′2+x02x0′′+ε22x03x0′2+x04x0′′+x1+3ε3x02x1+5ε4x04x1.According to (3), we choose the following equation as the solution of the representation:(10)x1τ=a3,1cos⁡τ-cos⁡3τ,where a3,1 is the unknown.Combining (10) with (9), we consider(11)R1τ=ϕ1τ+R0τ.We can obtain(12)R1τ=a3,11-ω02-Aω1+32A2a3,1ε3-12A3ω1ε1+516A4a3,15ε4+ε2ω02-38A5ε2ω1cos⁡τ+9a3,1ω02-a3,1+A272a3,1ε1ω02-34a3,1ε3+A314ε3-12ε1ω02+ω1+A43116a3,1ω02ε2-516a3,1ε4+A5516ε4-716ε2ω02+ω1cos⁡3τ+A292a3,1ε1ω02-34a3,1ε3+A4a3,16116ε2ω02-1516ε4+A5116ε4-316ε2ω02+ω1cos⁡5τ+A4a3,13116ε2ω02-516ε4cos⁡7τ.To increase the accuracy, (11) should be added in the zeroth-order residual term R0τ.Based on the Galerkin procedure, (12) should not contain secular terms. Letting coefficients of cos⁡τ and cos⁡3τ be zeros, hence, we obtain a linear equation set containing two unknowns ω1 and a3,1. Through solving the equation set, we can get (13)ω1=A216+24A2ε3+25A4ε4+5A4ε2-16ω02NM,a3,1=2A38+4A2ε1+3A4ε2NM,where(14)M=2561-9ω02+A8ε2205ε4-151ε2ω02+16A615ε2ε3+15ε1ε4-34ε1ε2ω02-16A4-13ε2-18ε1ε3-5ε4+47ε12+23ε2ω02+64A23ε3+ε14-34ω02,N=4ε3-8ε1ω02+A25ε4-7ε2ω02.Therefore, the first-order harmonic approximation can be procured:(15)ω1=ω02+ω1,x1τ=x0τ+x1τ=A+a3,1cos⁡τ-a3,1cos⁡3τ,τ=ω1t.When the obtained first-order solution is substituted into (12), the terms of cos⁡5τ and cos⁡7τ commonly are not zero.


## 3.3. TheSecond-Order Harmonic Approximations

Substituting (4) to (2) and presenting the coefficient of parameter p2 yield(16)ϕ2τ=ω026ε2x0x12x0′2+ε1x2x0′2+6ε2x2x02x0′2+2ε1x1x0′x1′+12ε2x02x1x0′x1′+ε1x0x12+2ε2x03x1′2+2ε1x0x0′x2′+4ε2x03x0′x2′+ε1x12x0′′+6ε2x02x12x0′′+2ε1x0x2x0′′+4ε2x03x2x0′′+2ε1x0x1x1′′+4ε2x03x1x1′′+x2′′+ε1x02x2′′+ε2x04x2′′+ω1ε1x1x0′2+6ε2x02x1x0′2+2ε1x0x0′x1′+4ε2x03x0′x1′+2ε1x0x1x0′′+4ε2x03x1x0′′+x1′′+ε1x02x1′′+ε2x04x1′′+ω2ε1x0x0′2+2ε2x03x0′2+x0′′+ε1x02x0′′+ε2x04x0′′+3ε3x0x12+10ε4x03x12+x2+3ε3x02x2+5ε4x04x2.Observing (16), we know it is linear with respect to ω2 and x2τ. In accordance with the form of (3), we apply (17)x2τ=a3,2cos⁡τ-cos⁡3τ+a5,2cos⁡τ-cos⁡5τ,where a3,2 and a5,2 are unknowns.Combining (17) with (16) and calculating the equation(18)R2τ=ϕ2τ+R1τ,we can obtain(19)R2τ=A94a3,12ε3-72a3,12ε1ω02-ω2+A232a3,2ε3+94a5,2ε3-32a5,2ε1ω02+A3154a3,12ε4-134a3,12ε2ω02-12ε1ω2+116A425a3,2ε4+45a5,2ε4+5a3,2ε2ω02-15a5,2ε2ω02+5a3,1ε2ω1-38A5ε2ω2+a3,2+a5,2-a3,2ω02-a5,2ω02-a3,1ω1cos⁡τ+Aa3,12172ε1ω02-94ε3+A272a3,2ε1ω02-34a3,2ε3+ε13a5,2ω02+72a3,1ω1-12A3a3,125ε4-13ε2ω02+ε1ω2+116A45a5,2ε4-5a3,2ε4+31a3,2ε2ω02+41a5,2ε2ω02+31a3,1ε2ω1-716A5ε2ω2+9a3,2ω02+9a3,1ω1-a3,2cos⁡3τ+Aa31272ε1ω02-34ε3+A213a5,2ε1ω02-32a5,2ε3-34a3,1+a3,2ε3-6ε1ω02+92a3,1ε1ω1+12A3a3,1217ε2ω02-5ε4+116A4147a5,2ε2ω02-25a5,2ε4-a3,1+a3,215ε4-61ε2ω02+61a3,1ε2ω1+116A5ε4-3ε2ω02+ω1+ω2+25a5,2ω02-a5,2cos5τ+A34a3,12ε3-172a3,12ε1ω02-A2a5,234ε3-192ε1ω02+18A3a3,125ε4-43ε2ω02+A4-516a3,2ε4-54a5,2ε4+3116a3,2ε2ω02+394a5,2ε2ω02-516a3,2ε4+116a3,131ε2ω02+ω1cos⁡7τ+18A3a3,125ε4-51ε2ω02-516A4a5,2ε4-11ε2ω02cos⁡9τ.With the purpose of increasing the accuracy, (18) should be put on the first-order residual term R1τ.In order to prevent the right hand side of (19) from exhibiting the secular terms cos⁡τ, cos⁡3τ, and cos⁡5τ, we make all their coefficients equal to zero. Then, we can get three linear equations containing three unknown parameters a3,2, a5,2, and ω2. According to the three linear equations, we can solve the three unknowns. Thus, thesecond-order harmonic approximation is shown:(20)ω2=ω02+ω1+ω2,x2τ=x0τ+x1τ+x2τ,τ=ω2t.In conclusion, we draw thekth-order harmonic approximation k=2,3,4,…(21)xkτ=xk-1τ+xkτ,ωk=ωk-12+ωkk=2,3,4,…,xk-1τ=xk-2τ+xk-1τ,ωk-1=ωk-22+ωk-1,xkτ=∑i=1ka2i+1,kcos⁡τ-cos⁡2i+1τ,x0=Acos⁡τ,ω0=ω0,k=2,3,4,….


## 4. Results and Discussion

In order to make sure of the effectiveness of the current technique, we compare the results from the second-order spreading residue harmonic balance approachωSRHB with the energy balance method ωEBM [34], the homotopy analysis method ωHAM [35], and the exact solution ωex [33], which are presented in Table 1, for different parameters εii=1,2,3,4 and amplitudes of vibration A, where the exact solution ωex is computed using the numerical technique. The relative errors of vibration frequency are tabulated in Table 2.Table 1
Comparison of the SRHBM frequencies, EBM frequencies, HAM frequencies, and the exact frequencies for various parameters.
Mode
A
ε
1
ε
2
ε
3
ε
4
ω
E
B
M
ω
H
A
M
ω
S
R
H
B
ω
e
x
1
1.0
0.326845
0.129579
0.232598
0.087584
1.01235
1.01232
1.01004
1.01015
2
0.5
1.642033
0.913055
0.313561
0.204297
0.935046
0.938636
0.935072
0.93639
3
0.2
4.051486
1.665232
0.281418
0.149677
0.965613
0.966516
0.965843
0.96664
4
0.3
8.205578
3.145368
0.272313
0.133708
0.860678
0.871382
0.863939
0.86426Table 2
Comparison of the SRHBM relative error, EBM relative error, and HAM relative error with the exact frequencies for various parameters.
Mode
A
ε
1
ε
2
ε
3
ε
4
ω
E
B
M
-
ω
e
x
ω
e
x
×
100
%
ω
H
A
M
-
ω
e
x
ω
e
x
×
100
%
ω
S
R
H
B
-
ω
e
x
ω
e
x
×
100
%
1
1.0
0.326845
0.129579
0.232598
0.087584
0.21789
0.21482
0.010889
2
0.5
1.642033
0.913055
0.313561
0.204297
0.14353
0.23986
0.140753
3
0.2
4.051486
1.665232
0.281418
0.149677
0.106244
0.01283
0.08245
4
0.3
8.205578
3.145368
0.272313
0.133708
0.414459
0.82406
0.037142For Mode 1 in Table2, we observe that the relative error between ωSRHB and ωex is much less than the relative error between ωHAM and ωex. The same goes for Mode 2 and Mode 4. However, the relative error between ωSRHB and ωex for Mode 3 is more than the relative error between ωHAM and ωex. Hence, we conclude that the accuracy of the second-order SRHBM solutions is improved in Mode 1, Mode 2, and Mode 4. Similarly, from Table 2, the SRHBM relative error is smaller than EBM relative error in Mode 1, Mode 2, Mode 3, and Mode 4. These results show that the approximate solutions obtained by the SRHBM are closer to the exact solutions.To further demonstrate the accuracy of the spreading residue harmonic balance approach, the time history responses and the phase portrait are rendered for four different sets of parameters in Figures2–5. From the phase portrait diagram, we obviously discover that the second-order residue harmonic balance solutions are very consistent with the exact solutions. From the phase portrait, we observe that the system is a periodic motion. Moreover, in the whole range, the presented approximate solutions converge to the exact solutions. Extraordinarily, we argue for t in 560,570.Figure 2
Comparison of the approximate and exact solutions of Mode 1.Figure 3
Comparison of the approximate and exact solutions of Mode 2.Figure 4
Comparison of the approximate and exact solutions of Mode 3.Figure 5
Comparison of the approximate and exact solutions of Mode 4.


## 5. Conclusions

In this paper, the spreading residue harmonic balance method is applied to discuss the strongly nonlinear vibration system. Particularly, we take a restrained cantilever beam as an example. The SRHBM does not need to add small parameters in the calculation process. Besides, this approach approximates the exact solution quickly and only the first- or second-order approximations. And by comparing its results with HAM and EBM for various parameters and amplitudes of vibration, it reveals that SRHBM can be used to solve a nonlinear equation with high nonlinearities. According to the figures and tables, it is effective to explain that the presented approximations are more accurate. Therefore, we can conclude that the SRHBM is more available and effective. Ultimately, we consider that the SRHBM can be used to deal with more complex strongly nonlinear vibration problems.



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*Source: 5214616-2017-04-10.xml*