# Flexural Free Vibrations of Multistep Nonuniform Beams

Authors: Guojin Tan; Wensheng Wang; Yubo Jiao
Journal: Mathematical Problems in Engineering
 (2016)

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## Abstract

This paper presents an exact approach to investigate the flexural free vibrations of multistep nonuniform beams. Firstly, one-step beam with moment of inertia and mass per unit length varying asI(x)=α11+βxr+4 and m(x)=α21+βxr was studied. By using appropriate transformations, the differential equation for flexural free vibration of one-step beam with variable cross section is reduced to a four-order differential equation with constant coefficients. According to different types of roots for the characteristic equation of four-order differential equation with constant coefficients, two kinds of modal shape functions are obtained, and the general solutions for flexural free vibration of one-step beam with variable cross section are presented. An exact approach to solve the natural frequencies and modal shapes of multistep beam with variable cross section is presented by using transfer matrix method, the exact general solutions of one-step beam, and iterative method. Numerical examples reveal that the calculated frequencies and modal shapes are in good agreement with the finite element method (FEM), which demonstrates the solutions of present method are exact ones.


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## Body

## 1. Introduction

Beams with nonuniform cross section are widely used in various engineering fields, such as bridges, tall buildings, and helicopter rotor blades. A large number of studies can be found in literature about the free vibrations of nonuniform beams. Vibration problems of beams with nonuniform cross section are often described by partial differential equations and in most cases it is extremely difficult to find their closed form solutions. Consequently, a wide range of approximate and numerical solutions such as Rayleigh-Ritz, Galerkin, finite difference, finite element, and spectral finite element methods have been used to obtain the natural vibration characteristics of variable-section beams [1–4].Huang and Li investigated the free vibration of axially functionally graded beams with variable flexural rigidity and mass density [5]. A novel and simple approach was presented to solve modal shapes and corresponding frequencies through transforming traditional fourth-order governing differential equation into Fredholm integral equation. Koplow et al. proposed an analytical solution for the dynamic response of Euler-Bernoulli beams with step changes in cross section, which was verified by experimental tests and receptance coupling methods [6]. Firouz-Abadi et al. studied the transverse free vibrations of a class of variable-cross section beams using Wentzel-Kramers-Brillouin (WKB) approximation [7]. The governing equation of motion for the Euler-Bernoulli beam including axial force distribution was utilized to obtain a singular differential equation in terms of the natural frequency of vibration and a WKB expansion series was applied to find the solution. Inaudi and Matusevich investigated longitudinal vibration problems of variable-cross section rods using an improved power series method [8]. This method introduced domain partition implementation in matrix formulation, as an alternative to other power series techniques in vibration analysis. Therefore, the method solved linear differential equations efficiently up to a desired degree of accuracy and remedies two limitations of the conventional power series method. The Adomian decomposition method (ADM) is employed to investigate the free vibrations of tapered Euler-Bernoulli beams with a continuously exponential variation of width and a constant thickness along the length under various boundary conditions [9]. Duan and Wang demonstrated the free vibration of beams with multiple step changes using the modified discrete singular convolution (DSC) [10]. The jump conditions at the steps were used to overcome the difficulty in using ordinary DSC for dealing with ill-posed problems. A transfer matrix method and the Frobenius method were adopted by J. W. Lee and J. Y. Lee to solve the free vibration characteristics of a tapered Bernoulli-Euler beam and obtain the power series solution for bending vibrations [11].Nevertheless, besides all advantages of such numerical methods, exact solutions can provide adequate insight into the physics of the problems and convenience for parametric studies. The other advantage of exact solutions is their significance in the field of inverse problems. An exact solution can be more useful than numerical solutions to design the characteristics and damage identification of a structure.Wang derived the closed form solutions for free vibration of a flexural bar with variably distributed stiffness but uniform mass [12]. Using a systematic approach, Abrate obtained a closed form solution of longitudinal vibration for rods whose cross section varies as Ax=A01+ax/L2 [13]. Kumar and Sujith found the exact solutions for longitudinal vibration of nonuniform rods whose cross section varies as A=a+bxn and A=A0sin2⁡(a+bx) [14]. Li presented an exact approach for free longitudinal vibrations of one-step nonuniform rod with classical and nonclassical boundary conditions [15]. The approach assumed that the distribution of mass is arbitrary, and distribution of longitudinal stiffness is expressed as a functional relation with mass distribution and vice versa. Li et al. obtained exact solutions of flexural vibration for beam-like structures whose moment of inertia and mass per unit length vary as EIx=a1+βxn+4, m(x)=α1+βxn and EIx=a·exp⁡(-bx), m(x)=α·exp⁡(-bx), respectively [16, 17].The components whose moment of inertiaI(x) and mass per unit length m(x) satisfy I(x)=α11+βxr+4 and m(x)=α21+βxr are widely used in civil engineering, such as the bridge with cross section height varying as h(x)=h01+βx2. This kind of beam-like structure is usually solved by dividing it into several segments whose I(x) and m(x) satisfy different distributions. The paper derived the exact general solution of one-step nonuniform beam firstly and then obtained the exact solution of multistep nonuniform beam by combining the transfer matrix method and exact solutions of one-step beam. The exact solutions of present method not only can provide convenience for parametric studies, but also are very useful for inverse problems such as damage identification.


## 2. Modal Shape Function of One-Step Beam

### 2.1. Beams with Uniform Cross Section

The governing differential equation for undamped free flexural vibration of beam with uniform cross section can be written as [18](1)EI∂4yx,t∂x4+m∂2yx,t∂t2=0,where E is Young’s modulus, I is moment of inertia, m is mass per unit length, and y(x,t) is transverse displacement at position x and time t.Assuming the beam performs a harmonic free vibration at equilibrium position, that is,(2)yx,t=ϕuxejωt,where ϕu(x) is mode shape function of uniform beam, ω is corresponding natural frequency.Inserting (2) into (1) obtains(3)EId4ϕuxdx4-ω2mϕux=0.Equation (3) leads to the modal shape function of beam with uniform cross section:(4)ϕux=∑i=14BiSiux,where S1u(x)=sinh⁡kx, S2u(x)=cosh⁡kx, S3u(x)=sin⁡kx, S4u(x)=cos⁡kx, are Bi (i=1,2,3,4) are integration constants, k4=ω2m/EI.


### 2.2. Beams with Variable Cross Section

The governing differential equation for undamped free flexural vibration of beam with variable cross section can be written as [19](5)∂2∂x2EIx∂2yx,t∂x2+mx∂2yx,t∂t2=0,where E is Young’s modulus, I(x) is bending moment of inertia at position x, m(x) is mass per unit length at position x, and y(x,t) is transverse displacement at position x and time t.Assuming the beam performs a harmonic free vibration at equilibrium position, that is,(6)yx,t=ϕxejωt,where ϕ(x) is mode shape function of beam, ω is natural frequency.Substituting (6) into (5) arrives at(7)d2dx2EIxd2ϕxdx2-ω2mxϕx=0.The moment of inertiaI(x) and mass per unit length m(x) of beam are assumed to vary as (8)Ix=α11+βxr+4,mx=α21+βxr,α1,α2,β≠0,where α1, α2, and β are arbitrary constants but not equal to 0 and r is a positive integer.Substituting (8) into (7) yields(9)0=Eα11+βx4dϕ4xdx4+2Eα1βr+41+βx3dϕ3xdx3+Eα1β2r+4r+31+βx2dϕ2xdx2-ω2α2ϕx.Let(10)z=ln1+βx,D=ddz.It can be derived that(11)dϕkxdxk1+βxk=βkDD-1D-2⋯D-k+1ϕz,k=1,2,…,n.Introducing (11) into (9), one arrives at(12)0=Eα1β4DD-1D-2D-3ϕz+2Eα1β4r+4DD-1D-2ϕz+Eα1β4r+4r+3DD-1ϕz-ω2α2ϕz.Equation (12) can be simplified as(13)0=D4+2r+4-6D3+r+4r+3-6r+4+11D2+4r+4-r+4r+3-6D-ω2α2Eα1β4ϕz.Let(14)a1=2r+4-6,a2=r+4r+3-6r+4+11,a3=4r+4-r+4r+3-6,a4=-ω2α2Eα1β4.Equation (13) can be written as(15)D4+a1D3+a2D2+a3D+a4ϕz=0.The characteristic equation of (15) is(16)d4+a1d3+a2d2+a3d+a4=0.The four roots of (16) are(17)d1,2=g-a1/2±a1/2-g2-4t/2-f2,d3,4=-g-a1/2±a1/2+g2-4t/2+f2,where(18)t=3-q2+q24+p327+3-q2-q24+p327+a23,tisarealroot,p=a1a3-13a22-4a4,q=13a1a2a3-227a23-a12a4+83a2a4-a32,f=t24-a41/2,g=a1t/4-a3/2f.The modal shape function of one-step beam with variable cross section can be written as(19)ϕx=c1ed1z+c2ed2z+c3ed3z+c4ed4z,where c1, c2, c3, and c4 are four undetermined coefficients.Sincea4<0, then f>t/2, and t/2-f<0, it can be found that d1 and d2 are two real roots in the first equation of (17). From the second equation in (17), it can be found that when (a1/2+g)2>4(t/2+f), d3 and d4 are also two real roots and when (a1/2+g)2<4(t/2+f), d3 and d4 are two conjugate complex roots.Modal shapes of one-step beam are determined by the roots as follows:(1) Whend1, d2, d3, and d4 are all real roots, substituting z=ln⁡(1+βx) into (19), one obtains (20)ϕx=B11+βxd1+B21+βxd2+B31+βxd3+B41+βxd4.(2) Whend1 and d2 are real roots, while d3 and d4 are two conjugate complex roots, let(21)s1=-g-a1/22,s2=4t/2+f-a1/2+g22;the roots d3 and d4 can be expressed as(22)d3=s1+s2j,d4=s1-s2j,where j=-1.Substitutingz=ln⁡(1+βx) and (22) into (19) yields(23)ϕx=B11+βxd1+B21+βxd2+B31+βxs1cos⁡s2ln⁡1+βx+B41+βxs1sin⁡s2ln⁡1+βx.In (20) and (23), Bi (i=1,2,3,4) have the same meanings as those in (4) and they are integration constants. The solutions of (20) and (23) are the exact general solutions only at the condition of β≠0. This is because z will be equal to zero in (10) if for β=0. And the derivation of constant term for the second equation in (10) has no mathematical meaning. Meanwhile, from the numerical point of view, small value of β (β→0) is adopted and the uniform beam is approached; then, it could be solved by (20) or (23). However, compared to solutions of uniform beam by (4), the solutions of approximate uniform beam are not exact general solutions but numerical solutions.Equations (20) and (23) can be expressed with a uniform style as (24)ϕx=∑i=14BiSix;when the modal shape function is (20), Si(x)=1+βxdi for i=1,2,3,4, and when it is (23), Si(x)=1+βxdi for i=1,2, and S3(x)=1+βxs1cos⁡s2ln⁡(1+βx), S4(x)=1+βxs1sin⁡s2ln⁡(1+βx).


## 2.1. Beams with Uniform Cross Section

The governing differential equation for undamped free flexural vibration of beam with uniform cross section can be written as [18](1)EI∂4yx,t∂x4+m∂2yx,t∂t2=0,where E is Young’s modulus, I is moment of inertia, m is mass per unit length, and y(x,t) is transverse displacement at position x and time t.Assuming the beam performs a harmonic free vibration at equilibrium position, that is,(2)yx,t=ϕuxejωt,where ϕu(x) is mode shape function of uniform beam, ω is corresponding natural frequency.Inserting (2) into (1) obtains(3)EId4ϕuxdx4-ω2mϕux=0.Equation (3) leads to the modal shape function of beam with uniform cross section:(4)ϕux=∑i=14BiSiux,where S1u(x)=sinh⁡kx, S2u(x)=cosh⁡kx, S3u(x)=sin⁡kx, S4u(x)=cos⁡kx, are Bi (i=1,2,3,4) are integration constants, k4=ω2m/EI.


## 2.2. Beams with Variable Cross Section

The governing differential equation for undamped free flexural vibration of beam with variable cross section can be written as [19](5)∂2∂x2EIx∂2yx,t∂x2+mx∂2yx,t∂t2=0,where E is Young’s modulus, I(x) is bending moment of inertia at position x, m(x) is mass per unit length at position x, and y(x,t) is transverse displacement at position x and time t.Assuming the beam performs a harmonic free vibration at equilibrium position, that is,(6)yx,t=ϕxejωt,where ϕ(x) is mode shape function of beam, ω is natural frequency.Substituting (6) into (5) arrives at(7)d2dx2EIxd2ϕxdx2-ω2mxϕx=0.The moment of inertiaI(x) and mass per unit length m(x) of beam are assumed to vary as (8)Ix=α11+βxr+4,mx=α21+βxr,α1,α2,β≠0,where α1, α2, and β are arbitrary constants but not equal to 0 and r is a positive integer.Substituting (8) into (7) yields(9)0=Eα11+βx4dϕ4xdx4+2Eα1βr+41+βx3dϕ3xdx3+Eα1β2r+4r+31+βx2dϕ2xdx2-ω2α2ϕx.Let(10)z=ln1+βx,D=ddz.It can be derived that(11)dϕkxdxk1+βxk=βkDD-1D-2⋯D-k+1ϕz,k=1,2,…,n.Introducing (11) into (9), one arrives at(12)0=Eα1β4DD-1D-2D-3ϕz+2Eα1β4r+4DD-1D-2ϕz+Eα1β4r+4r+3DD-1ϕz-ω2α2ϕz.Equation (12) can be simplified as(13)0=D4+2r+4-6D3+r+4r+3-6r+4+11D2+4r+4-r+4r+3-6D-ω2α2Eα1β4ϕz.Let(14)a1=2r+4-6,a2=r+4r+3-6r+4+11,a3=4r+4-r+4r+3-6,a4=-ω2α2Eα1β4.Equation (13) can be written as(15)D4+a1D3+a2D2+a3D+a4ϕz=0.The characteristic equation of (15) is(16)d4+a1d3+a2d2+a3d+a4=0.The four roots of (16) are(17)d1,2=g-a1/2±a1/2-g2-4t/2-f2,d3,4=-g-a1/2±a1/2+g2-4t/2+f2,where(18)t=3-q2+q24+p327+3-q2-q24+p327+a23,tisarealroot,p=a1a3-13a22-4a4,q=13a1a2a3-227a23-a12a4+83a2a4-a32,f=t24-a41/2,g=a1t/4-a3/2f.The modal shape function of one-step beam with variable cross section can be written as(19)ϕx=c1ed1z+c2ed2z+c3ed3z+c4ed4z,where c1, c2, c3, and c4 are four undetermined coefficients.Sincea4<0, then f>t/2, and t/2-f<0, it can be found that d1 and d2 are two real roots in the first equation of (17). From the second equation in (17), it can be found that when (a1/2+g)2>4(t/2+f), d3 and d4 are also two real roots and when (a1/2+g)2<4(t/2+f), d3 and d4 are two conjugate complex roots.Modal shapes of one-step beam are determined by the roots as follows:(1) Whend1, d2, d3, and d4 are all real roots, substituting z=ln⁡(1+βx) into (19), one obtains (20)ϕx=B11+βxd1+B21+βxd2+B31+βxd3+B41+βxd4.(2) Whend1 and d2 are real roots, while d3 and d4 are two conjugate complex roots, let(21)s1=-g-a1/22,s2=4t/2+f-a1/2+g22;the roots d3 and d4 can be expressed as(22)d3=s1+s2j,d4=s1-s2j,where j=-1.Substitutingz=ln⁡(1+βx) and (22) into (19) yields(23)ϕx=B11+βxd1+B21+βxd2+B31+βxs1cos⁡s2ln⁡1+βx+B41+βxs1sin⁡s2ln⁡1+βx.In (20) and (23), Bi (i=1,2,3,4) have the same meanings as those in (4) and they are integration constants. The solutions of (20) and (23) are the exact general solutions only at the condition of β≠0. This is because z will be equal to zero in (10) if for β=0. And the derivation of constant term for the second equation in (10) has no mathematical meaning. Meanwhile, from the numerical point of view, small value of β (β→0) is adopted and the uniform beam is approached; then, it could be solved by (20) or (23). However, compared to solutions of uniform beam by (4), the solutions of approximate uniform beam are not exact general solutions but numerical solutions.Equations (20) and (23) can be expressed with a uniform style as (24)ϕx=∑i=14BiSix;when the modal shape function is (20), Si(x)=1+βxdi for i=1,2,3,4, and when it is (23), Si(x)=1+βxdi for i=1,2, and S3(x)=1+βxs1cos⁡s2ln⁡(1+βx), S4(x)=1+βxs1sin⁡s2ln⁡(1+βx).


## 3. Transfer Relationship for Undetermined Coefficients of Multistep Beam Modal Shapes

As shown in Figure1, the multistep beam is divided into m segments. For an arbitrary beam segment k (k=1,2,…,m) with length lk, a local Cartesian coordinated system is established with the origin locating at the left end of segment, Ik(x) is bending moment of inertia for the kth segment, and ϕk(x) is mode shape function for the kth segment, where x is defined as belonging to the kth segment in the local Cartesian coordinated system.Figure 1
Multistep beam.Taking thekth segment, for example, the continuity of deformations, equilibrium of moments, and shear forces are satisfied at the right end of the kth segment and left end of the (k+1)th segment; that is,(25)ϕkxx=lk=ϕk+1xx=0,ϕk′xx=lk=ϕk+1′xx=0,EIkx·ϕk′′xx=lk=EIk+1x·ϕk+1′′xx=0,dEIkx·ϕk′′xdxx=lk=dEIk+1x·ϕk+1′′xdxx=0,where prime denotes the derivative with respect to local coordinate x.Substituting modal shape functions of thekth and (k+1)th segments ((4) or (24)) into (25) yields(26)∑i=14BikSiklk=∑i=14Bik+1Sik+10,∑i=14BikSi′klk=∑i=14Bik+1Si′k+10,Iklk∑i=14BikSi′′klk=Ik+10∑i=14Bik+1Si′′k+10,Ik′lk∑i=14BikSi′′klk+Iklk∑i=14BikSi′′′klk=Ik+1′0∑i=14Bik+1Si′′k+10+Ik+10∑i=14Bik+1Si′′′k+10,where Bik denotes the ith undetermined coefficient of modal shape function of the kth segment and Sik is the ith term in modal shape function of the kth segment.Equation (26) can be rewritten into a matrix form:(27)HkUk=Hk+1Uk+1,where(28)Uk=B1kB2kB3kB4kT,Uk+1=B1k+1B2k+1B3k+1B4k+1T,Hk=Hk1Hk2Hk3Hk4,Hk+1=Hk+11Hk+12Hk+13Hk+14,in which(29)Hki=SiklkSi′klkIklkSi′′klkIk′lkSi′′klk+IklkSi′′′klk,Hk+1i=Sik+10Si′k+10Ik+10Si′′k+10Ik+1′0Si′′k+10+Ik+10Si′′′k+10.Let(30)Tk=Hk+1-1Hk;then,(31)Uk+1=Hk+1-1HkUk=TkUk.Equation (31) represents the transfer relationship of four undetermined coefficients for modal shapes of the kth and (k+1)th segments. Tk is the transfer matrix. In accordance with (31), transfer relationship between the first segment and the last segment can be expressed as(32)Um=Tm-1Tm-2···T2T1U1=TU1,where T=Tm-1Tm-2···T2T1.


## 4. Solutions of Natural Frequencies and Modal Shapes

Based on the modal shape function of one-step beam ((4) or (24)) and transfer relationships for undetermined coefficients of modal shapes for each segment ((32)), natural frequencies and modal shapes of multistep beam can be determined by boundary conditions of the beam. The cantilever and simply supported boundaries are considered in this paper, respectively.For cantilever multistep beam, the boundary conditions are(33)ϕ10=0,ϕ1′xx=0=0,EImx·ϕm′′xx=lm=0,EdImx·ϕm′′xdxx=lm=0.Inserting the modal shape functions into (33) and utilizing transfer relationship in (32), one arrives at(34)C1CmTU1=0,where(35)C1=S110S210S310S410S1′10S2′10S3′10S4′10,Cm=S1′′mlmS2′′mlmS3′′mlmS4′′mlmS1′′′mlmS2′′′mlmS3′′′mlmS4′′′mlm.For simply supported multistep beam, the boundary conditions are(36)ϕ10=0,EI1x·ϕ1′′xx=0=0,ϕmlm=0,EImx·ϕm′′xx=lm=0.Similar to (34), one obtains(37)C1′Cm′TU1=0,where(38)C1′=S110S210S310S410S1′′10S2′′10S3′′10S4′′10,Cm′=S1mlmS2mlmS3mlmS4mlmS1′′mlmS2′′mlmS3′′mlmS4′′mlm.Equations (34) and (37) can be written into a uniform style(39)CU1=c11c12c13c14c21c22c23c24c31c32c33c34c41c42c43c44B11B21B31B41=0,where the elements cij (i,j=1,2,3,4) in C are functions of natural frequencies ω for multistep beam.Existence of nontrivial solutions of (39) leads to the following frequency equation(40)C=0.If the frequency region [ωai, ωbi] for the ith natural frequency ωi of multistep beam and the expressions of modal shape functions for each segment ((20) or (23)) can be determined, then the expressions of elements in matrix C can be uniquely determined. The ith natural frequency ωi of multistep beam can be calculated by (40) directly, and undetermined coefficients in modal shape function of the first segment can be solved by (39); then, the undetermined coefficients for other segments can be obtained by using transfer relationship in (32).If the frequency region or expressions of modal shape function for each segment ((20) or (23)) cannot be determined, the natural frequencies and modal shapes can be solved by the half-interval method [20]. The detailed process for determining natural frequencies is as follows. Firstly, an initial value of the ith natural frequency ωai is assumed, and the expressions for modal shape function of each segment corresponding to ωai can be determined ((20) or (23)); then, the coefficient matrix C corresponding to ωai is obtained and the determinant of matrix C is calculated (denoted as Dai=|C(ωai)|). Then, a new value ωbi=ωai+Δω with Δω (e.g., Δω=0.5) representing the increment of ω is assumed, and the same calculations are repeated to determine the new determinant corresponding to ωbi (denoted as Dbi=|C(ωbi)|). If Dai and Dbi have opposite signs, there is at least one natural frequency in the interval [ωai, ωbi]; otherwise, let ωai=ωbi and continue the above procedures until Dai and Dbi have opposite signs. For the next step, let ωci=(ωai+ωbi)/2; if Dai and Dci have the same sign, let ωai=ωci; otherwise, let ωbi=ωci; a new interval [ωai, ωbi] is obtained. The same calculations are repeated to determine the new interval [ωai, ωbi] until the rank of matrix C is equal to three. The accurate values of ω are obtained, respectively, using the half-interval method.


## 5. Numerical Examples

To verify the correctness of present analytical method, several numerical examples are studied in this paper. The materials used in these examples are all the same; that is, Yong’s modulus is 3.25 × 1010 Pa and density is 2500 kg/m3. Poisson’s ratio is ν=0.3 in a two-dimensional FEM (2D-FEM) analysis.


### 5.1. Reliability of the Proposed Method

#### 5.1.1. Validation by Finite Element Method

In order to illustrate the proposed method, a two-step beam with 10 m span, as shown in Figure2, is used as numerical model and solved by one-dimensional FEM (1D-FEM), 2D-FEM, and the present analytical method, respectively.Figure 2
Two-step beam.Cross section of this beam is rectangular with constant width 0.3 m. The heights of sections atA, B, and C are expressed by hA, hB, and hC, respectively. hA=hC, hB=0.7hA. And the height of each step varies nonlinearly asA to B:(41)hABx=hA1+βABx2,0≤x≤5,βAB=hB/hA-15=7/10-15;B to C:(42)hBCξ=hB1+βBCξ2,ξ=x-5,5≤x≤10,βBC=hC/hB-15=10/7-15.For 1D-FEM. The finite element method with two-nodes beam element. Each node of the beam element has two degree-of-freedoms, that is, transverse displacement and rotation angle. The cubic Hermite polynomial is adopted as shape function for beam element. Therefore, the shape functions Ni (i=1,2,3,4) can be expressed as(43)N1=l3-3lx2+2x3l3,N2=l2x-2lx2+x3l2,N3=3lx2-2x3l3,N4=x3-lx2l2,0≤x≤l.In (43), l is the length of beam element and l = 0.5 m.LetN=N1N2N3N4; stiffness matrix and mass matrix of the beam element can be evaluated as(44)Ke=∫0lBTEIxBdx,Me=∫0lmxNTNdx,where B=d2N/dx2 is the strain matrix of beam element. Using above stiffness matrix and mass matrix, finite element analysis is carried on by the codes written in MATLAB.For 2D-FEM. A finite element analysis is carried out using ANSYS software. A 2D-FEM model of two-step beam is built with 8-node PLANE183 elements as shown in Figure 3.Figure 3
A two-dimensional ANSYS model for 2D-FEM analysis (L/hA = 10).The ratios of span and height atA (L/hA) vary from 10 to 100. hA, hB, and hC are variables with L/hA, and βAB and βBC are constants. Relative errors of the first two-order natural frequencies obtained by 1D-FEM and 2D-FEM with respect to present analytical method are shown in Figure 4.Figure 4
Relative error of 1D-FEM and 2D-FEM.As can be seen from Figure4, the relative errors between 1D-FEM and present method remain at a very small lever all the time no matter how L/hA changes, and the relative errors between 2D-FEM and the present method decrease rapidly with L/hA increasing. This is because 1D-FEM and the present method are employed without considering the effect of shear deformation and the stiffness and mass matrixes are formed by integration method for the variable moment of inertia and mass in 1D-FEM modeling. However, the shear deformation is inevitable for a two-dimensional FEM model. When L/hA reaches a certain lever, the error between 2D-FEM and the present method becomes smaller. For example, when L/hA = 60, the frequencies obtained by 2D-FEM and the present method are very close and relative error of the first two order frequencies is about −0.1%. With regard to beam-like structural free vibration, it can be treated as plane stress problem, so a 2D-FEM analysis is more accurate than a 1D-FEM analysis in theory. For easy of calculation, the beam-like structure can be treated as Euler-Bernoulli beam when L/hA reaches a certain lever and the effect of shear deformation could be ignored. And the object of this paper is not considering the effect of shear deformation, so the solutions of the proposed method are exact ones based on Euler-Bernoulli beam theory. This leads to different relative errors of 1D-FEM and 2D-FEM shown in Figure 4. It also reveals that the present method possesses favorable accuracy for the larger ratio of span and sectional height.


#### 5.1.2. Numerical Simulation for Small Value ofβ

From the numerical point of view, small value ofβ (β→0) is adopted, and the uniform beam is approached. In order to verify the explanations, a one-step beam with small value of β (shown in Figure 5) and a one-step uniform beam (shown in Figure 6) are used for numerical simulation by the present method ((20) and (23) for nonuniform beams, (4) for uniform beams).Figure 5
One-step beam with small value ofβ.Figure 6
One-step uniform beam.One-Step Beam with Small Value of β . The span is 15 m. Cross section of this beam is rectangular with constant width 0.3 m.The heights of sections atA and B are hA = 0.25 m and hB = 0.249 m, respectively, and the height varies nonlinearly from A to B as(45)hx=hA1+βx2,0≤x≤15,β=hB/hA-115=249/250-115,β⟶0.Therefore,(46)Ix=bh3x12=bhA31+βx612,mx=ρbhx=ρbhA1+βx2,0≤x≤15.One-Step Uniform Beam. The beam is also 15 m span with constant width 0.3 m, and the height is 0.249 m.The one-step beams with different boundary conditions have been analyzed.Case 1.
A cantilever beam model.Case 2.
A simply supported beam model.The first three natural frequencies of one-step beam with small value ofβ (β→0) calculated by (20) or (23) are compared with those of one-step uniform beam calculated by (4); the results are listed in Tables 1 and 2.Table 1
Calculation results of natural frequencies for Case1.
Beam type One-step beam with small value ofβ One-step uniform beam with height 0.249 m Relative error (%) 1st (rad/s) 4.0680 4.0499 −0.4451 2nd (rad/s) 25.4507 25.3805 −0.2755 3rd (rad/s) 71.2273 71.0663 −0.2261Table 2
Calculation results of natural frequencies for Case2.
Beam type One-step beam with small value ofβ One-step uniform beam with height 0.249 m Relative error (%) 1st (rad/s) 11.3912 11.3684 −0.2002 2nd (rad/s) 45.5647 45.4734 −0.2003 3rd (rad/s) 102.5205 102.3152 −0.2003As can be seen from the results, the relative errors between nonuniform beam with small value ofβ and uniform beam are very small. It reveals that the uniform beam is approached when β is small enough (β→0). However, for small value of β representing a uniform beam, the solutions are not exact general solutions but numerical solutions.


### 5.2. One-Step Beam

A one-step beam with 15 m span, as shown in Figure5, is used for numerical simulation. Cross section of this beam is rectangular with constant width 0.3 m. The heights of sections at A and B are hA = 0.25 m and hB = 0.175 m, respectively, and the sectional height varies nonlinearly from A to B as(47)hx=hA1+βx2,0≤x≤15,β=hB/hA-115=7/10-115.Therefore,(48)Ix=bh3x12=bhA31+βx612,mx=ρbhx=ρbhA1+βx2,0≤x≤15.The one-step beam with different boundary conditions has been analyzed.Case 3.
A cantilever beam model.Case 4.
A simply supported beam model.The first three natural frequencies of one-step beam with different boundary conditions calculated by the proposed method and FEM are listed in Tables3 and 4; modal shapes are shown in Figures 7 and 8.Table 3
Calculation results of natural frequencies for Case3.
Methods Present (rad/s) 1D-FEM (rad/s) 2D-FEM (rad/s) Relative error (%) 1D-FEM 2D-FEM 1st 4.2234 4.2229 4.2232 −0.0120 −0.0047 2nd 22.8615 22.8680 22.8406 0.0283 −0.0914 3rd 61.3271 61.3522 61.1800 0.0410 −0.2399Table 4
Calculation results of natural frequencies for Case4.
Methods Present (rad/s) 1D-FEM (rad/s) 2D-FEM (rad/s) Relative error (%) 1D-FEM 2D-FEM 1st 9.5349 9.5398 9.5316 0.0511 −0.0350 2nd 38.3529 38.3716 38.3029 0.0487 −0.1303 3rd 86.2536 86.2964 85.9980 0.0495 −0.2964Figure 7
Modal shapes of Case3.Figure 8
Modal shapes of Case4.As can be seen from the results calculated by the proposed method and FEM, relative errors for the first three natural frequencies of the one-step beam with different boundary conditions are very small. It reveals that the solutions for one-step beam with variable cross section by the proposed method are exact ones. The reasons lie in that only the differential equations of motion for one-step beam with variable cross section are used to make the solutions, and no other assumptions are introduced.


### 5.3. Multistep Beam

To verify the correctness of present method for free vibration analysis of multistep beams, a two-step simply supported beam with variable cross section (as shown in Figure9) is used for numerical simulation firstly. The cross section of this beam is rectangular. The segments AB and BC have variable heights and widths. In the two-dimensional FEM analysis, a fine mesh of elements is used and the average thickness of element is adopted for element thickness.Figure 9
Two-step simply supported beam: (a) elevation graph; (b) plane graph.
(a) (b)For segmentAB, the heights of sections A and B are hA = 0.36 m and hB = 0.24 m, respectively; the widths of sections A and B are bA = 0.3 m and bB = 0.2449 m, respectively. The height and width from A to B vary as(49)hABx=hA1+βABx2,bABx=bA1+βABx,0≤x≤10,βAB=hB/hA-110=2/3-110.Therefore,(50)IABx=bABx·hAB3x12=bAhA31+βABx712,mABx=ρbABx·hABx=ρbAhA1+βABx3,0≤x≤10.For segmentBC, the heights of sections B and C are hB = 0.24 m and hC = 0.38 m, respectively; the widths of sections B and C are bB = 0.2449 m and bC = 0.3082 m, respectively. The height and width from B to C vary as(51)hBCξ=hB1+βBCξ2,bBCξ=bB1+βBCξ,ξ=x-10,10≤x≤22,βBC=hC/hB-112=19/12-112.Therefore,(52)IBCξ=bBCξ·hBC3ξ12=bBhB31+βBCξ712,mBCξ=ρbBCξ·hBCξ=ρbBhB1+βBCξ3,0≤ξ=x-10≤12.The first three natural frequencies of two-step simply supported beam are calculated by the proposed method and FEM, which are listed in Table5. The modal shapes are shown in Figure 10.Table 5
Calculation results of natural frequencies of two-step simply supported beam.
Methods Present (rad/s) 1D-FEM (rad/s) 2D-FEM (rad/s) Relative error (%) 1D-FEM 2D-FEM 1st 5.7666 5.7712 5.7570 0.0805 −0.1656 2nd 25.1551 25.1761 25.1157 0.0838 −0.1566 3rd 57.1504 57.2114 56.9501 0.1068 −0.3490Figure 10
Modal shapes of two-step simply supported beam.Secondly, a three-step simply supported beam with variable cross section, as shown in Figure11, is used for numerical simulation. Cross section of this beam is rectangular with constant width 0.3 m. The segments A-B and C-D are variable sections, and B-C is uniform cross section.Figure 11
Three-step simply supported beam.For segmentAB, the heights of sections A and B are hA = 0.3 m and hB = 0.2 m, respectively. The height of section from A to B varies as(53)hABx=hA1+βABx2,0≤x≤5,βAB=hB/hA-15=2/3-15.Therefore,(54)IABx=bhAB3x12=bhA31+βABx612,mABx=ρbhABx=ρbhA1+βABx2,0≤x≤5.For segmentCD, the heights of sections C and D are hC = 0.2 m and hD = 0.3 m. The height of section from C to D varies as(55)hCDξ=hC1+βCDξ2,0≤ξ=x-15≤5,βCD=hD/hC-15=3/2-15.Therefore,(56)ICDξ=bhCD3ξ12=bhC31+βCDξ612,mCDξ=ρbhCDξ=ρbhC1+βCDξ2,0≤ξ=x-15≤5.The first three natural frequencies of three-step simply supported beam calculated by the proposed method and FEM are listed in Table6, and the modal shapes are shown in Figure 12.Table 6
Calculation results of natural frequencies of three-step simply supported beam.
Methods Present (rad/s) 1D-FEM (rad/s) 2D-FEM (rad/s) Relative error (%) 1D-FEM 2D-FEM 1st 5.2110 5.2132 5.2089 0.0414 −0.0402 2nd 21.6022 21.5989 21.5683 −0.0155 −0.1571 3rd 50.2561 50.3032 50.1738 0.0938 −0.1638Figure 12
Modal shapes of three-step simply supported beam.From Tables5 and 6 and Figures 10 and 12, for two-step and three-step simply supported beams, it can be found that the first three natural frequencies and modal shapes calculated by the proposed method are in good agreement with those calculated by FEM. It reveals that the solutions for multistep beam with variable cross section by the proposed method are exact ones. This is because only transfer matrix method and exact solutions of one-step beam with variable cross section are used to solve the equations of vibration for multistep beam with variable cross section, and no other assumptions are introduced.


## 5.1. Reliability of the Proposed Method

### 5.1.1. Validation by Finite Element Method

In order to illustrate the proposed method, a two-step beam with 10 m span, as shown in Figure2, is used as numerical model and solved by one-dimensional FEM (1D-FEM), 2D-FEM, and the present analytical method, respectively.Figure 2
Two-step beam.Cross section of this beam is rectangular with constant width 0.3 m. The heights of sections atA, B, and C are expressed by hA, hB, and hC, respectively. hA=hC, hB=0.7hA. And the height of each step varies nonlinearly asA to B:(41)hABx=hA1+βABx2,0≤x≤5,βAB=hB/hA-15=7/10-15;B to C:(42)hBCξ=hB1+βBCξ2,ξ=x-5,5≤x≤10,βBC=hC/hB-15=10/7-15.For 1D-FEM. The finite element method with two-nodes beam element. Each node of the beam element has two degree-of-freedoms, that is, transverse displacement and rotation angle. The cubic Hermite polynomial is adopted as shape function for beam element. Therefore, the shape functions Ni (i=1,2,3,4) can be expressed as(43)N1=l3-3lx2+2x3l3,N2=l2x-2lx2+x3l2,N3=3lx2-2x3l3,N4=x3-lx2l2,0≤x≤l.In (43), l is the length of beam element and l = 0.5 m.LetN=N1N2N3N4; stiffness matrix and mass matrix of the beam element can be evaluated as(44)Ke=∫0lBTEIxBdx,Me=∫0lmxNTNdx,where B=d2N/dx2 is the strain matrix of beam element. Using above stiffness matrix and mass matrix, finite element analysis is carried on by the codes written in MATLAB.For 2D-FEM. A finite element analysis is carried out using ANSYS software. A 2D-FEM model of two-step beam is built with 8-node PLANE183 elements as shown in Figure 3.Figure 3
A two-dimensional ANSYS model for 2D-FEM analysis (L/hA = 10).The ratios of span and height atA (L/hA) vary from 10 to 100. hA, hB, and hC are variables with L/hA, and βAB and βBC are constants. Relative errors of the first two-order natural frequencies obtained by 1D-FEM and 2D-FEM with respect to present analytical method are shown in Figure 4.Figure 4
Relative error of 1D-FEM and 2D-FEM.As can be seen from Figure4, the relative errors between 1D-FEM and present method remain at a very small lever all the time no matter how L/hA changes, and the relative errors between 2D-FEM and the present method decrease rapidly with L/hA increasing. This is because 1D-FEM and the present method are employed without considering the effect of shear deformation and the stiffness and mass matrixes are formed by integration method for the variable moment of inertia and mass in 1D-FEM modeling. However, the shear deformation is inevitable for a two-dimensional FEM model. When L/hA reaches a certain lever, the error between 2D-FEM and the present method becomes smaller. For example, when L/hA = 60, the frequencies obtained by 2D-FEM and the present method are very close and relative error of the first two order frequencies is about −0.1%. With regard to beam-like structural free vibration, it can be treated as plane stress problem, so a 2D-FEM analysis is more accurate than a 1D-FEM analysis in theory. For easy of calculation, the beam-like structure can be treated as Euler-Bernoulli beam when L/hA reaches a certain lever and the effect of shear deformation could be ignored. And the object of this paper is not considering the effect of shear deformation, so the solutions of the proposed method are exact ones based on Euler-Bernoulli beam theory. This leads to different relative errors of 1D-FEM and 2D-FEM shown in Figure 4. It also reveals that the present method possesses favorable accuracy for the larger ratio of span and sectional height.


### 5.1.2. Numerical Simulation for Small Value ofβ

From the numerical point of view, small value ofβ (β→0) is adopted, and the uniform beam is approached. In order to verify the explanations, a one-step beam with small value of β (shown in Figure 5) and a one-step uniform beam (shown in Figure 6) are used for numerical simulation by the present method ((20) and (23) for nonuniform beams, (4) for uniform beams).Figure 5
One-step beam with small value ofβ.Figure 6
One-step uniform beam.One-Step Beam with Small Value of β . The span is 15 m. Cross section of this beam is rectangular with constant width 0.3 m.The heights of sections atA and B are hA = 0.25 m and hB = 0.249 m, respectively, and the height varies nonlinearly from A to B as(45)hx=hA1+βx2,0≤x≤15,β=hB/hA-115=249/250-115,β⟶0.Therefore,(46)Ix=bh3x12=bhA31+βx612,mx=ρbhx=ρbhA1+βx2,0≤x≤15.One-Step Uniform Beam. The beam is also 15 m span with constant width 0.3 m, and the height is 0.249 m.The one-step beams with different boundary conditions have been analyzed.Case 1.
A cantilever beam model.Case 2.
A simply supported beam model.The first three natural frequencies of one-step beam with small value ofβ (β→0) calculated by (20) or (23) are compared with those of one-step uniform beam calculated by (4); the results are listed in Tables 1 and 2.Table 1
Calculation results of natural frequencies for Case1.
Beam type One-step beam with small value ofβ One-step uniform beam with height 0.249 m Relative error (%) 1st (rad/s) 4.0680 4.0499 −0.4451 2nd (rad/s) 25.4507 25.3805 −0.2755 3rd (rad/s) 71.2273 71.0663 −0.2261Table 2
Calculation results of natural frequencies for Case2.
Beam type One-step beam with small value ofβ One-step uniform beam with height 0.249 m Relative error (%) 1st (rad/s) 11.3912 11.3684 −0.2002 2nd (rad/s) 45.5647 45.4734 −0.2003 3rd (rad/s) 102.5205 102.3152 −0.2003As can be seen from the results, the relative errors between nonuniform beam with small value ofβ and uniform beam are very small. It reveals that the uniform beam is approached when β is small enough (β→0). However, for small value of β representing a uniform beam, the solutions are not exact general solutions but numerical solutions.


## 5.1.1. Validation by Finite Element Method

In order to illustrate the proposed method, a two-step beam with 10 m span, as shown in Figure2, is used as numerical model and solved by one-dimensional FEM (1D-FEM), 2D-FEM, and the present analytical method, respectively.Figure 2
Two-step beam.Cross section of this beam is rectangular with constant width 0.3 m. The heights of sections atA, B, and C are expressed by hA, hB, and hC, respectively. hA=hC, hB=0.7hA. And the height of each step varies nonlinearly asA to B:(41)hABx=hA1+βABx2,0≤x≤5,βAB=hB/hA-15=7/10-15;B to C:(42)hBCξ=hB1+βBCξ2,ξ=x-5,5≤x≤10,βBC=hC/hB-15=10/7-15.For 1D-FEM. The finite element method with two-nodes beam element. Each node of the beam element has two degree-of-freedoms, that is, transverse displacement and rotation angle. The cubic Hermite polynomial is adopted as shape function for beam element. Therefore, the shape functions Ni (i=1,2,3,4) can be expressed as(43)N1=l3-3lx2+2x3l3,N2=l2x-2lx2+x3l2,N3=3lx2-2x3l3,N4=x3-lx2l2,0≤x≤l.In (43), l is the length of beam element and l = 0.5 m.LetN=N1N2N3N4; stiffness matrix and mass matrix of the beam element can be evaluated as(44)Ke=∫0lBTEIxBdx,Me=∫0lmxNTNdx,where B=d2N/dx2 is the strain matrix of beam element. Using above stiffness matrix and mass matrix, finite element analysis is carried on by the codes written in MATLAB.For 2D-FEM. A finite element analysis is carried out using ANSYS software. A 2D-FEM model of two-step beam is built with 8-node PLANE183 elements as shown in Figure 3.Figure 3
A two-dimensional ANSYS model for 2D-FEM analysis (L/hA = 10).The ratios of span and height atA (L/hA) vary from 10 to 100. hA, hB, and hC are variables with L/hA, and βAB and βBC are constants. Relative errors of the first two-order natural frequencies obtained by 1D-FEM and 2D-FEM with respect to present analytical method are shown in Figure 4.Figure 4
Relative error of 1D-FEM and 2D-FEM.As can be seen from Figure4, the relative errors between 1D-FEM and present method remain at a very small lever all the time no matter how L/hA changes, and the relative errors between 2D-FEM and the present method decrease rapidly with L/hA increasing. This is because 1D-FEM and the present method are employed without considering the effect of shear deformation and the stiffness and mass matrixes are formed by integration method for the variable moment of inertia and mass in 1D-FEM modeling. However, the shear deformation is inevitable for a two-dimensional FEM model. When L/hA reaches a certain lever, the error between 2D-FEM and the present method becomes smaller. For example, when L/hA = 60, the frequencies obtained by 2D-FEM and the present method are very close and relative error of the first two order frequencies is about −0.1%. With regard to beam-like structural free vibration, it can be treated as plane stress problem, so a 2D-FEM analysis is more accurate than a 1D-FEM analysis in theory. For easy of calculation, the beam-like structure can be treated as Euler-Bernoulli beam when L/hA reaches a certain lever and the effect of shear deformation could be ignored. And the object of this paper is not considering the effect of shear deformation, so the solutions of the proposed method are exact ones based on Euler-Bernoulli beam theory. This leads to different relative errors of 1D-FEM and 2D-FEM shown in Figure 4. It also reveals that the present method possesses favorable accuracy for the larger ratio of span and sectional height.


## 5.1.2. Numerical Simulation for Small Value ofβ

From the numerical point of view, small value ofβ (β→0) is adopted, and the uniform beam is approached. In order to verify the explanations, a one-step beam with small value of β (shown in Figure 5) and a one-step uniform beam (shown in Figure 6) are used for numerical simulation by the present method ((20) and (23) for nonuniform beams, (4) for uniform beams).Figure 5
One-step beam with small value ofβ.Figure 6
One-step uniform beam.One-Step Beam with Small Value of β . The span is 15 m. Cross section of this beam is rectangular with constant width 0.3 m.The heights of sections atA and B are hA = 0.25 m and hB = 0.249 m, respectively, and the height varies nonlinearly from A to B as(45)hx=hA1+βx2,0≤x≤15,β=hB/hA-115=249/250-115,β⟶0.Therefore,(46)Ix=bh3x12=bhA31+βx612,mx=ρbhx=ρbhA1+βx2,0≤x≤15.One-Step Uniform Beam. The beam is also 15 m span with constant width 0.3 m, and the height is 0.249 m.The one-step beams with different boundary conditions have been analyzed.Case 1.
A cantilever beam model.Case 2.
A simply supported beam model.The first three natural frequencies of one-step beam with small value ofβ (β→0) calculated by (20) or (23) are compared with those of one-step uniform beam calculated by (4); the results are listed in Tables 1 and 2.Table 1
Calculation results of natural frequencies for Case1.
Beam type One-step beam with small value ofβ One-step uniform beam with height 0.249 m Relative error (%) 1st (rad/s) 4.0680 4.0499 −0.4451 2nd (rad/s) 25.4507 25.3805 −0.2755 3rd (rad/s) 71.2273 71.0663 −0.2261Table 2
Calculation results of natural frequencies for Case2.
Beam type One-step beam with small value ofβ One-step uniform beam with height 0.249 m Relative error (%) 1st (rad/s) 11.3912 11.3684 −0.2002 2nd (rad/s) 45.5647 45.4734 −0.2003 3rd (rad/s) 102.5205 102.3152 −0.2003As can be seen from the results, the relative errors between nonuniform beam with small value ofβ and uniform beam are very small. It reveals that the uniform beam is approached when β is small enough (β→0). However, for small value of β representing a uniform beam, the solutions are not exact general solutions but numerical solutions.


## 5.2. One-Step Beam

A one-step beam with 15 m span, as shown in Figure5, is used for numerical simulation. Cross section of this beam is rectangular with constant width 0.3 m. The heights of sections at A and B are hA = 0.25 m and hB = 0.175 m, respectively, and the sectional height varies nonlinearly from A to B as(47)hx=hA1+βx2,0≤x≤15,β=hB/hA-115=7/10-115.Therefore,(48)Ix=bh3x12=bhA31+βx612,mx=ρbhx=ρbhA1+βx2,0≤x≤15.The one-step beam with different boundary conditions has been analyzed.Case 3.
A cantilever beam model.Case 4.
A simply supported beam model.The first three natural frequencies of one-step beam with different boundary conditions calculated by the proposed method and FEM are listed in Tables3 and 4; modal shapes are shown in Figures 7 and 8.Table 3
Calculation results of natural frequencies for Case3.
Methods Present (rad/s) 1D-FEM (rad/s) 2D-FEM (rad/s) Relative error (%) 1D-FEM 2D-FEM 1st 4.2234 4.2229 4.2232 −0.0120 −0.0047 2nd 22.8615 22.8680 22.8406 0.0283 −0.0914 3rd 61.3271 61.3522 61.1800 0.0410 −0.2399Table 4
Calculation results of natural frequencies for Case4.
Methods Present (rad/s) 1D-FEM (rad/s) 2D-FEM (rad/s) Relative error (%) 1D-FEM 2D-FEM 1st 9.5349 9.5398 9.5316 0.0511 −0.0350 2nd 38.3529 38.3716 38.3029 0.0487 −0.1303 3rd 86.2536 86.2964 85.9980 0.0495 −0.2964Figure 7
Modal shapes of Case3.Figure 8
Modal shapes of Case4.As can be seen from the results calculated by the proposed method and FEM, relative errors for the first three natural frequencies of the one-step beam with different boundary conditions are very small. It reveals that the solutions for one-step beam with variable cross section by the proposed method are exact ones. The reasons lie in that only the differential equations of motion for one-step beam with variable cross section are used to make the solutions, and no other assumptions are introduced.


## 5.3. Multistep Beam

To verify the correctness of present method for free vibration analysis of multistep beams, a two-step simply supported beam with variable cross section (as shown in Figure9) is used for numerical simulation firstly. The cross section of this beam is rectangular. The segments AB and BC have variable heights and widths. In the two-dimensional FEM analysis, a fine mesh of elements is used and the average thickness of element is adopted for element thickness.Figure 9
Two-step simply supported beam: (a) elevation graph; (b) plane graph.
(a) (b)For segmentAB, the heights of sections A and B are hA = 0.36 m and hB = 0.24 m, respectively; the widths of sections A and B are bA = 0.3 m and bB = 0.2449 m, respectively. The height and width from A to B vary as(49)hABx=hA1+βABx2,bABx=bA1+βABx,0≤x≤10,βAB=hB/hA-110=2/3-110.Therefore,(50)IABx=bABx·hAB3x12=bAhA31+βABx712,mABx=ρbABx·hABx=ρbAhA1+βABx3,0≤x≤10.For segmentBC, the heights of sections B and C are hB = 0.24 m and hC = 0.38 m, respectively; the widths of sections B and C are bB = 0.2449 m and bC = 0.3082 m, respectively. The height and width from B to C vary as(51)hBCξ=hB1+βBCξ2,bBCξ=bB1+βBCξ,ξ=x-10,10≤x≤22,βBC=hC/hB-112=19/12-112.Therefore,(52)IBCξ=bBCξ·hBC3ξ12=bBhB31+βBCξ712,mBCξ=ρbBCξ·hBCξ=ρbBhB1+βBCξ3,0≤ξ=x-10≤12.The first three natural frequencies of two-step simply supported beam are calculated by the proposed method and FEM, which are listed in Table5. The modal shapes are shown in Figure 10.Table 5
Calculation results of natural frequencies of two-step simply supported beam.
Methods Present (rad/s) 1D-FEM (rad/s) 2D-FEM (rad/s) Relative error (%) 1D-FEM 2D-FEM 1st 5.7666 5.7712 5.7570 0.0805 −0.1656 2nd 25.1551 25.1761 25.1157 0.0838 −0.1566 3rd 57.1504 57.2114 56.9501 0.1068 −0.3490Figure 10
Modal shapes of two-step simply supported beam.Secondly, a three-step simply supported beam with variable cross section, as shown in Figure11, is used for numerical simulation. Cross section of this beam is rectangular with constant width 0.3 m. The segments A-B and C-D are variable sections, and B-C is uniform cross section.Figure 11
Three-step simply supported beam.For segmentAB, the heights of sections A and B are hA = 0.3 m and hB = 0.2 m, respectively. The height of section from A to B varies as(53)hABx=hA1+βABx2,0≤x≤5,βAB=hB/hA-15=2/3-15.Therefore,(54)IABx=bhAB3x12=bhA31+βABx612,mABx=ρbhABx=ρbhA1+βABx2,0≤x≤5.For segmentCD, the heights of sections C and D are hC = 0.2 m and hD = 0.3 m. The height of section from C to D varies as(55)hCDξ=hC1+βCDξ2,0≤ξ=x-15≤5,βCD=hD/hC-15=3/2-15.Therefore,(56)ICDξ=bhCD3ξ12=bhC31+βCDξ612,mCDξ=ρbhCDξ=ρbhC1+βCDξ2,0≤ξ=x-15≤5.The first three natural frequencies of three-step simply supported beam calculated by the proposed method and FEM are listed in Table6, and the modal shapes are shown in Figure 12.Table 6
Calculation results of natural frequencies of three-step simply supported beam.
Methods Present (rad/s) 1D-FEM (rad/s) 2D-FEM (rad/s) Relative error (%) 1D-FEM 2D-FEM 1st 5.2110 5.2132 5.2089 0.0414 −0.0402 2nd 21.6022 21.5989 21.5683 −0.0155 −0.1571 3rd 50.2561 50.3032 50.1738 0.0938 −0.1638Figure 12
Modal shapes of three-step simply supported beam.From Tables5 and 6 and Figures 10 and 12, for two-step and three-step simply supported beams, it can be found that the first three natural frequencies and modal shapes calculated by the proposed method are in good agreement with those calculated by FEM. It reveals that the solutions for multistep beam with variable cross section by the proposed method are exact ones. This is because only transfer matrix method and exact solutions of one-step beam with variable cross section are used to solve the equations of vibration for multistep beam with variable cross section, and no other assumptions are introduced.


## 6. Conclusions

In this paper, an exact approach to investigate the flexural free vibrations of multistep nonuniform beams is presented. The differential equation for flexural free vibration of one-step beam withI(x)=α11+βxr+4 and m(x)=α21+βxr is reduced to a four-order differential equation with constant coefficients by using appropriate transformations. And the general solutions of one-step beam are obtained, whose modal shape functions are found to have two kinds of expressions. Then, combining transfer method, iterative method, and the general solutions of one-step beam, the solving method for natural frequencies and modal shapes of multistep beam is formulated. Numerical simulations on multistep nonuniform beam are used to verify the feasibility. The following conclusions can be obtained:(1)
The comparison of relative errors among the present analytical method and finite element methods reveals that the solutions of proposed method are exact ones based on Euler-Bernoulli beam theory. The present method is suitable for the larger ratio of span and sectional height and possesses favorable accuracy.(2)
The relative errors between nonuniform beam with small value ofβ and uniform beam are very small. It reveals that for small value of β representing a uniform beam, the solutions are not exact general solutions but numerical solutions.(3)
Numerical studies of one-step cantilever beam and simply supported beam indicate that natural frequencies and modal shapes calculated by the proposed method are very close to the FEM results, which demonstrates the solutions of one-step beam are exact ones. And numerical examples of two-step and three-step simply supported beams show that the calculated frequencies and modal shapes are also in good agreements with FEM results, which also demonstrate the solutions of presented method for multistep beam with variable cross section are exact ones. The reasons lie in that only transfer matrix method and general solution of one-step beam are used to obtain the solution equation of natural vibration characteristics, and no other assumptions are introduced.The exact solutions of proposed method can provide adequate insight into the physics of problems and convenience for parametric studies. Furthermore, they are very useful for inverse problems such as structural damage identification. Therefore, it is always desirable to obtain exact solutions for such problems. However, when the ratio of span and sectional height is small, relative errors between FEM and present method may be rather larger. The deficiency of this paper is not considering shear deformation in vibration analysis of nonuniform beams, leading to larger error for the small ratio of span and sectional height. This also has become the topic which the author further deliberated from now on.



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