# Vibration Analysis of Hollow Tapered Shaft Rotor

Authors: P. M. G. Bashir Asdaque; R. K. Behera
Journal: Advances in Acoustics and Vibration
 (2014)

---

## Abstract

Shafts or circular cross-section beams are important parts of rotating systems and their geometries play important role in rotor dynamics. Hollow tapered shaft rotors with uniform thickness and uniform bore are considered. Critical speeds or whirling frequency conditions are computed using transfer matrix method and then the results were compared using finite element method. For particular shaft lengths and rotating speeds, response of the hollow tapered shaft-rotor system is determined for the establishment of dynamic characteristics. Nonrotating conditions are also considered and results obtained are plotted.


---

## Body

## 1. Introduction

Shaft is a major component of any rotating system, used to transmit torque and rotation. Hence the study shaft-rotor systems has been the concern of researchers for more than a century and will continue to persist as an active area of research and analysis in near future. Geometry of shaft is of the main concern during the study of any rotating system. Most papers related to shaft-rotor systems consider cylindrical shaft elements for study and analysis of rotating systems. The first idea of transfer matrix method (TMM) was compiled by Holzer for finding natural frequencies of torsional systems and later adapted by Myklestad [1, 2] for computing natural frequencies of airplane wings, coupled in bending and torsion. Gyroscopic moments were first introduced by Prohl [3] for rotor-bearing system analysis. Lund [4] used complex variables as the next significant advancement in the method. An improved method for calculating critical speeds and rotor stability of turbo machinery was investigated by Murphy and Vance [5]. Whalley and Abdul-Ameer [6] used frequency response analysis for particular profiled shafts to study dynamic response of distributed-lumped shaft rotor system. They studied the system behavior in terms of frequency response for the shafts with diameters which are functions of their lengths. They derived an analytical method which uses Euler-Bernoulli beam theory in combination with TMM. On the other hand, there are large numbers of numerical applications of finite element techniques for the calculation of whirling and the computation of maximum dynamic magnitude. In this regard, Ruhl and Booker [7] modeled the distributed parameter turbo rotor systems using finite element method (FEM). Nelson and McVaugh [8] reduced large number of eigenvalues and eigenvectors identified, following finite element analysis, and the erroneous modes of vibration predicted were eliminated. Nelson [9] again formulated the equations of motion for a uniform rotating shaft element using deformation shape functions developed from Timoshenko beam theory including the effects of translational and rotational inertia, gyroscopic moments, bending and shear deformation, and axial load. Greenhill et al. [10] derived equation of motion for a conical beam finite element form Timoshenko beam theory and include effects of translational and rotational inertia, gyroscopic moments, bending and shear deformation, axial load, and internal damping. Genta and Gugliotta [11] also analyzed element with annular cross-section based on Timoshenko beam theory having two degrees of freedom at each node. Mohiuddin and Khulief [12] derived a finite element model of a tapered rotating cracked shaft for modal analysis and dynamic modeling of a rotor-bearing system, based on Timoshenko beam theory, that is, included shear deformation and rotary inertia. Rouch and Kao [13] presented numerically integrated formulation of a tapered beam element for rotor dynamics.In this era of machines, tapered shafts are widely used for rotating systems. Using the approach of Whalley and Abdul-Ameer [6], the dynamic analysis of hollow tapered shaft-rotor has been done. Later the results obtained were compared with that obtained from finite element method. The effect of length and speed on the dynamic analysis of the hollow tapered shaft-rotor system is also clearly shown. Frequency response of the rotor system for an impulse of unit force at the free end is determined in terms of critical speeds for various rotor speeds and shaft lengths.


## 2. Transfer Matrix Method

### 2.1. Shaft Model

The shaft model is derived in matrix form Whalley and Abdul-Ameer [6] as
(1){Y2θ2My2Qy2}=F(s){Y1θ1My1Qy1},
where (2)F(s)=[cos⁡Γ(l)l+cosh⁡Γ(l)l2sinΓ(l)l+sinh⁡Γ(l)l2Γ(l)-Γ(0)(sinΓ(l)l-sinh⁡Γ(l)l)2cos⁡Γ(l)l+cosh⁡Γ(l)l2Γ(0)(cos⁡Γ(l)l-cosh⁡Γ(l)l)2C(0)Γ(0)(sinΓ(l)l-sinh⁡Γ(l)l)2C(0)-Γ(0)Γ(0)(sinΓ(l)l+sinh⁡Γ(l)l)2C(0)Γ(0)(cos⁡Γ(l)l-cosh⁡Γ(l)l)2C(0)C(0)(cos⁡Γ(l)l-cosh⁡Γ(l)l)2Γ(l)C(0)(sinΓ(l)l-sinh⁡Γ(l)l)2Γ(0)Γ(0)-C(0)(sinΓ(l)l+sinh⁡Γ(l)l)2Γ(0)C(0)(cos⁡Γ(l)l-cosh⁡Γ(l)l)2Γ(0)cos⁡Γ(l)l+cosh⁡Γ(l)l2sinΓ(l)l+sinh⁡Γ(l)l2Γ(0)-Γ(0)(sinΓ(l)l-sinh⁡Γ(l)l)2cos⁡Γ(l)l+cosh⁡Γ(l)l2].The complete derivation is present in [6].


### 2.2. Rigid Disk

The output vector from the shaft will become the input for the rigid disk model, as shown in Figure1; that is, for disk model, we have
(3)y3(s)=y2(s),θ3(s)=θ2(s),My3(s)=My2(s)-JΩsθ2(s),Qy3(s)=ms2y2(s)+Qy2(s),
where (y3(s),y1(s)), (θ3(s),θ1(s)), (My3(s),My1(s)), and (Qy3(s),Qy1(s)) are the deflections, slopes, bending moments, and shear forces at the free and fixed end, respectively.Figure 1
Vibrating shaft element.Hence, writing in matrix form, we have(4)(y3(s),θ3(s),My3(s),Qy3(s))T=R(s)(y2(s),θ2(s),My2(s),Qy2(s)),
where
(5)R(s)=[100001000-JΩs10ms2001].
Form transfer matrix method [6]
(6)H(s)=R(s)F(s),
where
(7)H(s)=[H11H12H21H22],
and input-output vectors relationship is given by
(8)[y3θ3My3Qy3]=H(s)[y1θ1My1Qy1].
After applying the boundary conditions for cantilever beam, deflection at the free end is obtained and hence leads to transfer function.


## 2.1. Shaft Model

The shaft model is derived in matrix form Whalley and Abdul-Ameer [6] as
(1){Y2θ2My2Qy2}=F(s){Y1θ1My1Qy1},
where (2)F(s)=[cos⁡Γ(l)l+cosh⁡Γ(l)l2sinΓ(l)l+sinh⁡Γ(l)l2Γ(l)-Γ(0)(sinΓ(l)l-sinh⁡Γ(l)l)2cos⁡Γ(l)l+cosh⁡Γ(l)l2Γ(0)(cos⁡Γ(l)l-cosh⁡Γ(l)l)2C(0)Γ(0)(sinΓ(l)l-sinh⁡Γ(l)l)2C(0)-Γ(0)Γ(0)(sinΓ(l)l+sinh⁡Γ(l)l)2C(0)Γ(0)(cos⁡Γ(l)l-cosh⁡Γ(l)l)2C(0)C(0)(cos⁡Γ(l)l-cosh⁡Γ(l)l)2Γ(l)C(0)(sinΓ(l)l-sinh⁡Γ(l)l)2Γ(0)Γ(0)-C(0)(sinΓ(l)l+sinh⁡Γ(l)l)2Γ(0)C(0)(cos⁡Γ(l)l-cosh⁡Γ(l)l)2Γ(0)cos⁡Γ(l)l+cosh⁡Γ(l)l2sinΓ(l)l+sinh⁡Γ(l)l2Γ(0)-Γ(0)(sinΓ(l)l-sinh⁡Γ(l)l)2cos⁡Γ(l)l+cosh⁡Γ(l)l2].The complete derivation is present in [6].


## 2.2. Rigid Disk

The output vector from the shaft will become the input for the rigid disk model, as shown in Figure1; that is, for disk model, we have
(3)y3(s)=y2(s),θ3(s)=θ2(s),My3(s)=My2(s)-JΩsθ2(s),Qy3(s)=ms2y2(s)+Qy2(s),
where (y3(s),y1(s)), (θ3(s),θ1(s)), (My3(s),My1(s)), and (Qy3(s),Qy1(s)) are the deflections, slopes, bending moments, and shear forces at the free and fixed end, respectively.Figure 1
Vibrating shaft element.Hence, writing in matrix form, we have(4)(y3(s),θ3(s),My3(s),Qy3(s))T=R(s)(y2(s),θ2(s),My2(s),Qy2(s)),
where
(5)R(s)=[100001000-JΩs10ms2001].
Form transfer matrix method [6]
(6)H(s)=R(s)F(s),
where
(7)H(s)=[H11H12H21H22],
and input-output vectors relationship is given by
(8)[y3θ3My3Qy3]=H(s)[y1θ1My1Qy1].
After applying the boundary conditions for cantilever beam, deflection at the free end is obtained and hence leads to transfer function.


## 3. Finite Element Method

The vector of nodal displacements is given by(9){q}={VWβΓ}.
So, each element is having eight degrees of freedom.


### 3.1. Rigid Disc

Rigid disk is having two translations and two rotations inY and Z direction, respectively (considering X coordinate in axial direction). For constant spin condition, the Lagrangian equation of motion is given by
(10)([MdT]+[MdR]){qd′′}-Ω[Gd]{qd′}={Qd},
where
(11)[MdT]=[md0000md0000000000],[MdR]=[0000000000ID0000ID],[Gd]=[00000000000-IP00IP0].
The forcing term may include mass unbalance and other external forces.


### 3.2. Finite Shaft-Rotor Element

The rotor-shaft element considered here has eight degrees of freedom, that is, four degrees of freedom per node as in Nelson and McVaugh [8]. For constant spin condition, the Lagrangian equation of motion is given by
(12)([MeT]+[MeR]){qe′′}-Ω[Ge]{qe′}+([Ke]){qe}={Qe},
where
(13)[MeT]=∫0lρA(x)[N]T[N]dx,[MeR]=∫0lID(x)[ϕ]T[ϕ]dx,[Ne]=∫0lIP(x)[ϕΓ]T[ϕβ]dx,[Ke]=∫0lEI(x)[N′′]T[N′′]dx,[Ge]=[Ne]-[Ne]T.
Except skew-symmetric gyroscopic matrix [Ge], others are symmetric matrices. Since the element is linearly tapered, area and inertias are the function of the shaft-rotor length. The translational shape function is given by
(14)[N]=[N100N2N300N40N1-N200N3-N40],
where
(15)N1=1-3x2l2+2x3l3,N2=x-2x2l+x3l2,N3=3x2l2-2x3l3,N4=-x2l+x3l2.
The rotational shape function is given by
(16)[Φ]=[⌊Φβ⌋⌊ΦΓ⌋]=[0-N1′N2′00-N3′N4′0N1′00N2′N3′00N4′].
The element matrices are assembled together to get the equation of motion for the complete system.


## 3.1. Rigid Disc

Rigid disk is having two translations and two rotations inY and Z direction, respectively (considering X coordinate in axial direction). For constant spin condition, the Lagrangian equation of motion is given by
(10)([MdT]+[MdR]){qd′′}-Ω[Gd]{qd′}={Qd},
where
(11)[MdT]=[md0000md0000000000],[MdR]=[0000000000ID0000ID],[Gd]=[00000000000-IP00IP0].
The forcing term may include mass unbalance and other external forces.


## 3.2. Finite Shaft-Rotor Element

The rotor-shaft element considered here has eight degrees of freedom, that is, four degrees of freedom per node as in Nelson and McVaugh [8]. For constant spin condition, the Lagrangian equation of motion is given by
(12)([MeT]+[MeR]){qe′′}-Ω[Ge]{qe′}+([Ke]){qe}={Qe},
where
(13)[MeT]=∫0lρA(x)[N]T[N]dx,[MeR]=∫0lID(x)[ϕ]T[ϕ]dx,[Ne]=∫0lIP(x)[ϕΓ]T[ϕβ]dx,[Ke]=∫0lEI(x)[N′′]T[N′′]dx,[Ge]=[Ne]-[Ne]T.
Except skew-symmetric gyroscopic matrix [Ge], others are symmetric matrices. Since the element is linearly tapered, area and inertias are the function of the shaft-rotor length. The translational shape function is given by
(14)[N]=[N100N2N300N40N1-N200N3-N40],
where
(15)N1=1-3x2l2+2x3l3,N2=x-2x2l+x3l2,N3=3x2l2-2x3l3,N4=-x2l+x3l2.
The rotational shape function is given by
(16)[Φ]=[⌊Φβ⌋⌊ΦΓ⌋]=[0-N1′N2′00-N3′N4′0N1′00N2′N3′00N4′].
The element matrices are assembled together to get the equation of motion for the complete system.


## 4. Numerical Results

### 4.1. Rotating Condition

By using the transfer matrix approach as in the paper of Whalley and Abdul-Ameer, we will ultimately get the transfer function which will be plotted. The FEM will be applied and then is compared with the TMM approach of Whalley and Abdul-Ameer [6].Example 1.
Let us consider a cantilever tubular shaft with uniform thickness and a disc at the free end with downward unit force,P, on the disc as shown in Figure 2. The default values of various parameters are tabulated in Table 1.Table 1
Various parameters of the system shown in Figure2 and their default values.
Parameters
Values
Length of the shaft rotor,l (m)
0.1
Mass of the disk,m (Kg)
0.7443
Diameter of the disk,D (m)
0.09
Young’s modulus of elasticity,E (GPa)
209
Density of the material,ρ (Kg/m3)
7800
Rotational speed,N (rpm)
10000
Thickness of the hollow shaft,t (m)
0.002
Beginning radius,R0 (m)
0.0050
End radius,R1 (m)
0.0037Figure 2
Hollow tapered shaft disc with uniform thickness.Transfer function for tubular shaft with constant thickness as shown in Figure2 for default values is given by
(17)1.344s+1.016×104s3+7564s2+8.401×105s+1.589×109.
Bode plots for different lengths and rotating speed have been plotted using MATLAB software, as shown in Figures 3 and 4.Figure 3
Bode plot for different lengths with TMM.Figure 4
Bode plot for various speeds with TMM.Applying FEM on the same system, we get mass, gyroscopic, and stiffness matrices. A finite hollow tapered shaft element is shown in Figure5.Figure 5
Hollow tapered shaft finite element with uniform thickness.The stiffness matrix for hollow tapered shaft element with uniform thickness is given by is given by(18)Kn=[kn11kn21kn22kn31kn32kn33sym.kn41kn42kn43kn44kn51kn52kn53kn54kn55kn61kn62kn63kn64kn65kn66kn71kn72kn73kn74kn75kn76kn77kn81kn82kn83kn84kn85kn86kn87kn88],where elements of the stiffness matrices are
(19)kn11=3πEt(7R03+3R02R1-12R02t+3R0R12-6R0R1ts+10R0t2+7R13-12R12t+10R1t2-5t3)×(5l3)-1,kn41=πEt(30R03+12R02R1-51R02t+6R0R12-18R0R1ts+40R0t2+12R13-21R12t+20R1t2-15t3)×(10l2)-1,kn81=+20R0t2+30R13-51R12t+40R1t2-15t3))πEt(12R03+6R02R1-21R02t+12R0R12-18R0R1ts+20R0t2+30R13-51R12t+40R1t2-15t3)×(10l2)-1,kn33=πEt(11R03+5R02R1-19R02t+2R0R12-7R0R1ts+15R0t2+2R13-4R12t+5R1t2-5t3)×(5l)-1,kn73=πEt(8R03+2R02R1-13R02t+2R0R12-4R0R1ts+10R0t2+8R13-13R12t+10R1t2-5t3)×(10l)-1,kn77=πEt(2R03+2R02R1-4R02t+5R0R12-7R0R1ts+5R0t2+11R13-19R12t+15R1t2-5t3)×(5l)-1,kn21=kn42=kn82=kn31=0,kn43=kn53=kn83=kn64=kn74=0,kn52=kn53=kn65=kn75=kn61=0,kn65=kn86=kn71=kn87=0,kn22=kn55=kn66=kn11,kn51=kn62=-kn11,kn63=kn41,kn32=kn54=-kn41,kn76=kn81,kn72=kn85=-kn81,kn44=kn33,kn84=kn73.
Translational mass matrix is given by(20)Mnt=[mnt11mnt21mnt22mnt31mnt32mnt33sym.mnt41mnt42mnt43mnt44mnt51mnt52mnt53mnt54mnt55mnt61mnt62mnt63mnt64mnt65mnt66mnt71mnt72mnt73mnt74mnt75mnt76mnt77mnt81mnt82mnt83mnt84mnt85mnt86mnt87mnt88],where elements of translational mass matrix are given by
(21)mnt11=πρlt(20R0+6R1-13t)35,mnt41=πρtl2(15R0+7R1-11t)210,mnt51=9πρlt(R0+R1-t)70,mnt81=-(πρtl2(14R0+12R1-13t))420,mnt33=πρl3t(5R0+3R1-4t)420,mnt63=-(πρl2t(12R0+14R1-13t))420,mnt73=-(πρl3t(R0+R1-t))140,mnt76=πρl2t(7R0+15R1-11t)210,mnt77=πρl3t(3R0+5R1-4t)420,mnt21=mnt42=mnt82=mnt31=0,mnt43=mnt53=mnt83=mnt64=mnt74=0,mnt52=mnt53=mnt75=mnt61=0,mnt65=mnt86=mnt71=mnt87=0,mnt22=mnt55=mnt66=mnt11,mnt51=mnt62,mnt32=-mnt41,mnt63=mnt54,mnt72=-mnt81,mnt44=mnt33,mnt84=mnt73,mnt85=-mnt76.Rotational mass matrix is given by(22)Mnr=[mnr11mnr21mnr22mnr31mnr32mnr33sym.mnr41mnr42mnr43mnr44mnr51mnr52mnr53mnr54mnr55mnr61mnr62mnr63mnr64mnr65mnr66mnr71mnr72mnr73mnr74mnr75mnr76mnr77mnr81mnr82mnr83mnr84mnr85mnr86mnr87mnr88],where elements of rotational mass matrix are given by
(23)mnr11=3πρt2(8R02+12R0R1-14R0t+8R12-14R1t+7t2)×(70l)-1,mnr41=πρt2(-8R02+6R0R1+20R12-28R1t+7t2)×(280)-1,mnr51=-(3πρt2(8R02+12R0R1-14R0t+8R12-14R1t+7t2))×(70l)-1,mnr33=πρt2l(18R02+6R0R1-21R0t+4R12-7R1t+7t2)×(210)-1,mnr63=πρt2(-8R02+16R0R1+20R12-28R1t+7t2)×(280)-1,mnr73=-(πρt2l(12R02+4R0R1-14R0t+12R12-14R1t+7t2))×(840)-1,mnr76=πρt2(20R02+16R0R1-28R0t-8R12+7t2)×(280)-1,mnr77=πρt2l(4R02+6R0R1-7R0t+18R12-21R1t+7t2)×(210)-1,mnr21=mnr42=mnr82=mnr31=mnr43=0,mnr53=mnr83=mnr64=mnr74=0,mnr52=mnr53=mnr75=mnr61=0,mnr65=mnr86=mnr71=mnr87=0,mnr22=mnr55=mnr66=mnr11,mnr51=mnr62,mnr32=-mnr41,mnr63=mnr54,mnr81=mnr76,mnr72=-mnr76,mnr44=mnr33,mnr84=mnr73,mnr85=-mnr76.Gyroscopic matrix is given by(24)Gne=[gne11gne21gne22gne31gne32gne33skewsym.gne41gne42gne43gne44gne51gne52gne53gne54gne55gne61gne62gne63gne64gne65gne66gne71gne72gne73gne74gne75gne76gne77gne81gne82gne83gne84gne85gne86gne87gne88],where the elements of the gyroscopic matrix are given by
(25)gne21=3πρt2(8R02+12R0R1-14R0tsss+8R12-14R1t+7t2)×(35l)-1,gne31=-28R1t+7t2)))-(πρt2(-8R02+16R0R1+20R12sss-28R1t+7t2))×(140)-1,gne71=-8R12+7t2)))-(πρt2(20R02+16R0R1-28R0tsss-8R12+7t2))×(140)-1,gne43=πρlt2(18R02+6R0R1-21R0tss+4R12-7R1t+7t2)×(105)-1,gne83=-(πρlt2(12R02+4R0R1-14R0tss+12R12-14R1t+7t2))×(420)-1,gne87=πρlt2(4R02+6R0R1-7R0tss+18R12-21R1t+7t2)×(105)-1,gne52=gne65=gne21,gne61=-gne21,gne42=gne53=gne64=gne31,gne82=gne71,gne86=gne75=-gne71,gne75=-gne83,gne11=gne41=gne51=gne81=gne22=gne32=0,gne62=gne72=gne33=gne63=0,gne73=gne44=gne54=gne84=gne55=gne85=0,gne66=gne76=gne77=gne88=0.
Discretizing the tapered shaft into six elements as shown in Figure 6 and then assembling, we get the assembled equation of motion
(26)[Mns]{q¨s}-Ω[Gns]{q˙s}+[Kns]{qs}={Qns}.
where [Mns] is the assembled mass matrix containing both the translational and rotational mass matrices.Figure 6
Discretized shaft element.The assembled equation of motion is arranged in the first order state vector form(27)[[0][Mns][Mns]-Ω[Gns]]{z˙}+[[-Mns][0][0][Kns]]{z}={Z},
where
(28){z}={{q˙}{q}},{Z}={{0}{Qns}}.
The shaft rotor has been discretized into six elements of equal length. Hence the order of assembled matrices, after applying the fixed-free boundary condition, is 7×4-4=24.MATLAB program is used to find the bode plot for different values of shaft length and rotor speed as shown in Figures7 and 8 and are found to be in good agreement with bode plots found using TMM as shown in Figures 3 and 4.Figure 7
Bode plot for different shaft lengths (FEM).Figure 8
Bode plot for various rotor speeds (FEM).Example 2.
Let us consider a cantilever hollow shaft with uniform bore and a disc at the free end, as shown in Figure9. The values of various parameters are tabulated in Table 2.Table 2
Various parameters of the system shown in Figure9 and their default values.
Parameters
Values
Length of the shaft rotor,l (m)
0.1
Mass of the disk,m (Kg)
0.7443
Diameter of the disk,D (m)
0.09
Young’s modulus of elasticity,E (GPa)
209
Density of the material,ρ (Kg/m3)
7800
Rotational speed,N (rpm)
10000
Inner radius of  hollow shaft,Ri (m)
0.001
Beginning radius,R0 (m)
0.0050
End radius,R1 (m)
0.0037Figure 9
Hollow tapered shaft disc with uniform bore and vertically downward forceP on the disc.The transfer function for hollow shaft with constant thickness for default values is given by(29)1.344s+1.217×104s3+9059s2+1.097×106s+2.483×109.
The bode plots for different lengths and rotating speeds have been plotted using MATLAB software, as shown in Figures 10 and 11.Figure 10
Bode plot for changing length for hollow tapered shaft rotor with uniform bore.Figure 11
Bode plot for different speeds for hollow tapered shaft rotor with uniform bore.Applying FEM in the same system, we get mass, gyroscopic, and stiffness matrices. A finite hollow tapered shaft element with uniform bore is shown in Figure12.Figure 12
Hollow tapered shaft finite element with uniform bore.The stiffness matrix for hollow tapered shaft with uniform bore is given by(30)Ku=[ku11ku21ku22ku31ku32ku33sym.ku41ku42ku43ku44ku51ku52ku53ku54ku55ku61ku62ku63ku64ku65ku66ku71ku72ku73ku74ku75ku76ku77ku81ku82ku83ku84ku85ku86ku87ku88],where elements of the stiffness matrices are
(31)ku11=3πE(11R04+5R03R1+3R02R12ss+5R0R13+11R14-35Ri4)×(35l3)-1,ku41=πE(47R04+22R03R1+9R02R12s+8R0R13+19R14-105*Ri4)×(70l2)-1,ku81=πE(19R04+8R03R1+9R02R12s+22R0R13+47R14-1105Ri4)×(70l2)-1,ku33=πE(17R04+9R03R1+4R02R12s+2R0R13+3R14-35Ri4)×(35l)-1,ku73=πE(13R04+4R03R1+R02R12s+4R0R13+13R14-35Ri4)×(70l)-1,ku77=πE(3R04+2R03R1+4R02R12s+9R0R13+17R14-35Ri4)×(35l)-1,ku21=ku42=ku82=ku31=0,ku43=ku53=ku83=ku64=ku74=0,ku52=ku53=ku65=ku75=0,ku61=ku65=ku86=ku71=ku87=0,ku22=ku55=ku66=ku11,ku51=ku62=-ku11,ku63=ku41,ku32=ku54=-ku41,ku76=ku81,ku72=ku85=ku72=-ku81,ku44=ku33,ku84=ku73.Translational mass matrix is given by(32)Mut=[mut11mut21mut22mut31mut32mut33sym.mut41mut42mut43mut44mut51mut52mut53mut54mut55mut61mut62mut63mut64mut65mut66mut71mut72mut73mut74mut75mut76mut77mut81mut82mut83mut84mut85mut86mut87mut88],where elements of translational mass matrix are given by
(33)mut11=πρl(145R02+70R0R1+19R12-234Ri2)630,mut41=πρl2(65R02+50R0R1+17R12-132Ri2)2520,mut51=πρl(23R02+35R0R1+23R12-81Ri2)630,mut81=-(πρl2(25R02+34R0R1+19R12-78Ri2))2520,mut33=πρl3(5R02+5R0R1+2R12-12Ri2)1260,mut63=-(πρl2(19R02+34R0R1+25R12-78Ri2))2520,mut73=-(πρl3(5R02+8R0R1+5R12-18Ri2))2520,mut76=πρl2(17R02+50R0R1+65R12-132Ri2)2520,mut77=πρl3(2R02+5R0R1+5R12-12Ri2)1260,mut21=mut42=mut82=mut31=mut43=0,mut53=mut83=mut64=mut74=0,mut52=mut53=mut75=mut61=0,mut65=mut86=mut71=mut87=0,mut22=mut55=mut66=mut11,mut51=mut62,mut32=-mut41,mut63=mut54,mut72=-mut81,mut44=mut33,mut84=mut73,mut85=-mut76.Rotational mass matrix is given by(34)Mur=[mur11mur21mur22mur31mur32mur33sym.mur41mur42mur43mur44mur51mur52mur53mur54mur55mur61mur62mur63mur64mur65mur66mur71mur72mur73mur74mur75mur76mur77mur81mur82mur83mur84mur85mur86mur87mur88],where the elements of the rotational mass matrix are given by
(35)mur11=πρ(5R04+10R03R1+12R02R12-24R02Ri2+10R0R13ss-36R0R1Ri2+5R14-24R12Ri2+42Ri4)×(140l)-1,mur41=πρ(-5R04+6R02R12+8R02Ri2+8R0R13ss-16R0R1Ri2+5R14-20R12Ri2+14Ri4)×(560)-1,mur51=-(πρ(5R04+10R03R1+12R02R12-24R02Ri2ss+10R0R13-36R0R1Ri2+5R14ss-24R12Ri2+42Ri4))×(140l)-1,mur81=+8R12Ri2+14Ri4))πρ(5R04+8R03R1+6R02R12-20R02Ri2ss-16R0R1Ri2-5R14+8R12Ri2+14Ri4)×(560)-1,mur33=+2R14-8R12Ri2+28Ri4))πρl(15R04+5R03R1+3R02R12-36R02Ri2ss+3R0R13-12R0R1Ri2+2R14-8R12Ri2+28Ri4)×(840)-1,mur63=-20R12Ri2+14Ri4))πρ(-5R04+6R02R12+8R02Ri2+8R0R13ss-16R0R1Ri2+5R14-20R12Ri2+14Ri4)×(560)-1,mur73=-(πρl(5R04+2R03R1-12R02Ri2+2R0R13ssss-4R0R1Ri2+5R14-12R12Ri2+14Ri4))×(1680)-1,mur76=πρ(5R04+8R03R1+6R02R12-20R02Ri2ss-16R0R1Ri2-5R14+8R12Ri2+14Ri4)×(560)-1,mur77=πρl(2R04+3R03R1+3R02R12ss-8R02Ri2+5R0R13-12R0R1Ri2ss+15R14-36R12Ri2+28Ri4)×(840)-1,mur21=mur42=mur82=mur31=mur43,=mur53=mur83=mur64=mur74=0,mur52=mur53=mur75=mur61=0,mur65=mur86=mur71=mur87=0,mur22=mur55=mur66=mur11,mur51=mur62,mur32=-mur41,mur63=mur54,mur72=-mur81,mur44=mur33,mur84=mur73,mur85=-mur76.
The gyroscopic matrix is given by(36)Gue=[gue11gue21gue22gue31gue32gue33skewsym.gue41gue42gue43gue44gue51gue52gue53gue54gue55gue61gue62gue63gue64gue65gue66gue71gue72gue73gue74gue75gue76gue77gue81gue82gue83gue84gue85gue86gue87gue88],where the elements of the gyroscopic matrix are
(37)gue21=πρ(5R04+10R03R1+12R02R12-24R02Ri2+10R0R13sss-36R0R1Ri2+5R14-24R12Ri2+42Ri4)×(70l)-1,gue31=-(πρ(-5R04+6R02R12+8R02Ri2+8R0R13sssd-16R0R1Ri2+5R14-20R12Ri2+14Ri4))×(280)-1,gue71=-(πρ(5R04+8R03R1+6R02R12-20R02Ri2sd-16R0R1Ri2-5R14+8R12Ri2+14Ri4))×(280)-1,gue43=πρl(15R04+5R03R1+3R02R12-36R02Ri2+3R0R13s-12R0R1Ri2+2R14-8R12Ri2+28Ri4)×(420)-1,gue83=-(πρl(5R04+2R03R1-12R02Ri2+2R0R13sd-4R0R1Ri2+5R14-12R12Ri2+14Ri4))×(840)-1,gue87=πρl(2R04+3R03R1+3R02R12-8R02Ri2+5R0R13sd-12R0R1Ri2+15R14-36R12Ri2+28Ri4)×(420)-1,gue52=gue65=gue21,gue61=-gue21,gue42=gue53=gue64=gue31,gue82=gue71,gue86=gue75=-gue71,gue75=-gue83,gue11=gue41=gue51=gue81=gue22=0,gue32=gue62=gue72=gue33=gue63=0,gue73=gue44=gue54=gue84=gue55=0,gue85=gue66=gue76=gue77=gue88=0,
As proceeded in Example 1, bode plots are obtained for various shaft lengths and rotor speeds as shown in Figures 13 and 14 and are found to be in good agreement with bode plots found using TMM as shown in Figures 10 and 11.Figure 13
Bode plot for different shaft lengths (FEM).Figure 14
Bode plot for different rotating speeds (FEM).


### 4.2. Nonrotating Conditions

Bode plot for nonrotating (1 or 2 rpm) tapered shaft-rotor system is slightly different from rotating conditions in terms of amplitude. Hollow shaft with uniform thickness is considered. Bode plots are obtained for zero rpm as shown in Figure15 with TMM.Figure 15
Bode plot at zero rpm with TMM for uniform thickness hollow tapered shaft-rotor system.Applying FEM, then for zero rpm we get the bode plot as shown in Figure16.Figure 16
Bode plot at zero rpm with FEM for uniform thickness hollow tapered shaft-rotor system.


## 4.1. Rotating Condition

By using the transfer matrix approach as in the paper of Whalley and Abdul-Ameer, we will ultimately get the transfer function which will be plotted. The FEM will be applied and then is compared with the TMM approach of Whalley and Abdul-Ameer [6].Example 1.
Let us consider a cantilever tubular shaft with uniform thickness and a disc at the free end with downward unit force,P, on the disc as shown in Figure 2. The default values of various parameters are tabulated in Table 1.Table 1
Various parameters of the system shown in Figure2 and their default values.
Parameters
Values
Length of the shaft rotor,l (m)
0.1
Mass of the disk,m (Kg)
0.7443
Diameter of the disk,D (m)
0.09
Young’s modulus of elasticity,E (GPa)
209
Density of the material,ρ (Kg/m3)
7800
Rotational speed,N (rpm)
10000
Thickness of the hollow shaft,t (m)
0.002
Beginning radius,R0 (m)
0.0050
End radius,R1 (m)
0.0037Figure 2
Hollow tapered shaft disc with uniform thickness.Transfer function for tubular shaft with constant thickness as shown in Figure2 for default values is given by
(17)1.344s+1.016×104s3+7564s2+8.401×105s+1.589×109.
Bode plots for different lengths and rotating speed have been plotted using MATLAB software, as shown in Figures 3 and 4.Figure 3
Bode plot for different lengths with TMM.Figure 4
Bode plot for various speeds with TMM.Applying FEM on the same system, we get mass, gyroscopic, and stiffness matrices. A finite hollow tapered shaft element is shown in Figure5.Figure 5
Hollow tapered shaft finite element with uniform thickness.The stiffness matrix for hollow tapered shaft element with uniform thickness is given by is given by(18)Kn=[kn11kn21kn22kn31kn32kn33sym.kn41kn42kn43kn44kn51kn52kn53kn54kn55kn61kn62kn63kn64kn65kn66kn71kn72kn73kn74kn75kn76kn77kn81kn82kn83kn84kn85kn86kn87kn88],where elements of the stiffness matrices are
(19)kn11=3πEt(7R03+3R02R1-12R02t+3R0R12-6R0R1ts+10R0t2+7R13-12R12t+10R1t2-5t3)×(5l3)-1,kn41=πEt(30R03+12R02R1-51R02t+6R0R12-18R0R1ts+40R0t2+12R13-21R12t+20R1t2-15t3)×(10l2)-1,kn81=+20R0t2+30R13-51R12t+40R1t2-15t3))πEt(12R03+6R02R1-21R02t+12R0R12-18R0R1ts+20R0t2+30R13-51R12t+40R1t2-15t3)×(10l2)-1,kn33=πEt(11R03+5R02R1-19R02t+2R0R12-7R0R1ts+15R0t2+2R13-4R12t+5R1t2-5t3)×(5l)-1,kn73=πEt(8R03+2R02R1-13R02t+2R0R12-4R0R1ts+10R0t2+8R13-13R12t+10R1t2-5t3)×(10l)-1,kn77=πEt(2R03+2R02R1-4R02t+5R0R12-7R0R1ts+5R0t2+11R13-19R12t+15R1t2-5t3)×(5l)-1,kn21=kn42=kn82=kn31=0,kn43=kn53=kn83=kn64=kn74=0,kn52=kn53=kn65=kn75=kn61=0,kn65=kn86=kn71=kn87=0,kn22=kn55=kn66=kn11,kn51=kn62=-kn11,kn63=kn41,kn32=kn54=-kn41,kn76=kn81,kn72=kn85=-kn81,kn44=kn33,kn84=kn73.
Translational mass matrix is given by(20)Mnt=[mnt11mnt21mnt22mnt31mnt32mnt33sym.mnt41mnt42mnt43mnt44mnt51mnt52mnt53mnt54mnt55mnt61mnt62mnt63mnt64mnt65mnt66mnt71mnt72mnt73mnt74mnt75mnt76mnt77mnt81mnt82mnt83mnt84mnt85mnt86mnt87mnt88],where elements of translational mass matrix are given by
(21)mnt11=πρlt(20R0+6R1-13t)35,mnt41=πρtl2(15R0+7R1-11t)210,mnt51=9πρlt(R0+R1-t)70,mnt81=-(πρtl2(14R0+12R1-13t))420,mnt33=πρl3t(5R0+3R1-4t)420,mnt63=-(πρl2t(12R0+14R1-13t))420,mnt73=-(πρl3t(R0+R1-t))140,mnt76=πρl2t(7R0+15R1-11t)210,mnt77=πρl3t(3R0+5R1-4t)420,mnt21=mnt42=mnt82=mnt31=0,mnt43=mnt53=mnt83=mnt64=mnt74=0,mnt52=mnt53=mnt75=mnt61=0,mnt65=mnt86=mnt71=mnt87=0,mnt22=mnt55=mnt66=mnt11,mnt51=mnt62,mnt32=-mnt41,mnt63=mnt54,mnt72=-mnt81,mnt44=mnt33,mnt84=mnt73,mnt85=-mnt76.Rotational mass matrix is given by(22)Mnr=[mnr11mnr21mnr22mnr31mnr32mnr33sym.mnr41mnr42mnr43mnr44mnr51mnr52mnr53mnr54mnr55mnr61mnr62mnr63mnr64mnr65mnr66mnr71mnr72mnr73mnr74mnr75mnr76mnr77mnr81mnr82mnr83mnr84mnr85mnr86mnr87mnr88],where elements of rotational mass matrix are given by
(23)mnr11=3πρt2(8R02+12R0R1-14R0t+8R12-14R1t+7t2)×(70l)-1,mnr41=πρt2(-8R02+6R0R1+20R12-28R1t+7t2)×(280)-1,mnr51=-(3πρt2(8R02+12R0R1-14R0t+8R12-14R1t+7t2))×(70l)-1,mnr33=πρt2l(18R02+6R0R1-21R0t+4R12-7R1t+7t2)×(210)-1,mnr63=πρt2(-8R02+16R0R1+20R12-28R1t+7t2)×(280)-1,mnr73=-(πρt2l(12R02+4R0R1-14R0t+12R12-14R1t+7t2))×(840)-1,mnr76=πρt2(20R02+16R0R1-28R0t-8R12+7t2)×(280)-1,mnr77=πρt2l(4R02+6R0R1-7R0t+18R12-21R1t+7t2)×(210)-1,mnr21=mnr42=mnr82=mnr31=mnr43=0,mnr53=mnr83=mnr64=mnr74=0,mnr52=mnr53=mnr75=mnr61=0,mnr65=mnr86=mnr71=mnr87=0,mnr22=mnr55=mnr66=mnr11,mnr51=mnr62,mnr32=-mnr41,mnr63=mnr54,mnr81=mnr76,mnr72=-mnr76,mnr44=mnr33,mnr84=mnr73,mnr85=-mnr76.Gyroscopic matrix is given by(24)Gne=[gne11gne21gne22gne31gne32gne33skewsym.gne41gne42gne43gne44gne51gne52gne53gne54gne55gne61gne62gne63gne64gne65gne66gne71gne72gne73gne74gne75gne76gne77gne81gne82gne83gne84gne85gne86gne87gne88],where the elements of the gyroscopic matrix are given by
(25)gne21=3πρt2(8R02+12R0R1-14R0tsss+8R12-14R1t+7t2)×(35l)-1,gne31=-28R1t+7t2)))-(πρt2(-8R02+16R0R1+20R12sss-28R1t+7t2))×(140)-1,gne71=-8R12+7t2)))-(πρt2(20R02+16R0R1-28R0tsss-8R12+7t2))×(140)-1,gne43=πρlt2(18R02+6R0R1-21R0tss+4R12-7R1t+7t2)×(105)-1,gne83=-(πρlt2(12R02+4R0R1-14R0tss+12R12-14R1t+7t2))×(420)-1,gne87=πρlt2(4R02+6R0R1-7R0tss+18R12-21R1t+7t2)×(105)-1,gne52=gne65=gne21,gne61=-gne21,gne42=gne53=gne64=gne31,gne82=gne71,gne86=gne75=-gne71,gne75=-gne83,gne11=gne41=gne51=gne81=gne22=gne32=0,gne62=gne72=gne33=gne63=0,gne73=gne44=gne54=gne84=gne55=gne85=0,gne66=gne76=gne77=gne88=0.
Discretizing the tapered shaft into six elements as shown in Figure 6 and then assembling, we get the assembled equation of motion
(26)[Mns]{q¨s}-Ω[Gns]{q˙s}+[Kns]{qs}={Qns}.
where [Mns] is the assembled mass matrix containing both the translational and rotational mass matrices.Figure 6
Discretized shaft element.The assembled equation of motion is arranged in the first order state vector form(27)[[0][Mns][Mns]-Ω[Gns]]{z˙}+[[-Mns][0][0][Kns]]{z}={Z},
where
(28){z}={{q˙}{q}},{Z}={{0}{Qns}}.
The shaft rotor has been discretized into six elements of equal length. Hence the order of assembled matrices, after applying the fixed-free boundary condition, is 7×4-4=24.MATLAB program is used to find the bode plot for different values of shaft length and rotor speed as shown in Figures7 and 8 and are found to be in good agreement with bode plots found using TMM as shown in Figures 3 and 4.Figure 7
Bode plot for different shaft lengths (FEM).Figure 8
Bode plot for various rotor speeds (FEM).Example 2.
Let us consider a cantilever hollow shaft with uniform bore and a disc at the free end, as shown in Figure9. The values of various parameters are tabulated in Table 2.Table 2
Various parameters of the system shown in Figure9 and their default values.
Parameters
Values
Length of the shaft rotor,l (m)
0.1
Mass of the disk,m (Kg)
0.7443
Diameter of the disk,D (m)
0.09
Young’s modulus of elasticity,E (GPa)
209
Density of the material,ρ (Kg/m3)
7800
Rotational speed,N (rpm)
10000
Inner radius of  hollow shaft,Ri (m)
0.001
Beginning radius,R0 (m)
0.0050
End radius,R1 (m)
0.0037Figure 9
Hollow tapered shaft disc with uniform bore and vertically downward forceP on the disc.The transfer function for hollow shaft with constant thickness for default values is given by(29)1.344s+1.217×104s3+9059s2+1.097×106s+2.483×109.
The bode plots for different lengths and rotating speeds have been plotted using MATLAB software, as shown in Figures 10 and 11.Figure 10
Bode plot for changing length for hollow tapered shaft rotor with uniform bore.Figure 11
Bode plot for different speeds for hollow tapered shaft rotor with uniform bore.Applying FEM in the same system, we get mass, gyroscopic, and stiffness matrices. A finite hollow tapered shaft element with uniform bore is shown in Figure12.Figure 12
Hollow tapered shaft finite element with uniform bore.The stiffness matrix for hollow tapered shaft with uniform bore is given by(30)Ku=[ku11ku21ku22ku31ku32ku33sym.ku41ku42ku43ku44ku51ku52ku53ku54ku55ku61ku62ku63ku64ku65ku66ku71ku72ku73ku74ku75ku76ku77ku81ku82ku83ku84ku85ku86ku87ku88],where elements of the stiffness matrices are
(31)ku11=3πE(11R04+5R03R1+3R02R12ss+5R0R13+11R14-35Ri4)×(35l3)-1,ku41=πE(47R04+22R03R1+9R02R12s+8R0R13+19R14-105*Ri4)×(70l2)-1,ku81=πE(19R04+8R03R1+9R02R12s+22R0R13+47R14-1105Ri4)×(70l2)-1,ku33=πE(17R04+9R03R1+4R02R12s+2R0R13+3R14-35Ri4)×(35l)-1,ku73=πE(13R04+4R03R1+R02R12s+4R0R13+13R14-35Ri4)×(70l)-1,ku77=πE(3R04+2R03R1+4R02R12s+9R0R13+17R14-35Ri4)×(35l)-1,ku21=ku42=ku82=ku31=0,ku43=ku53=ku83=ku64=ku74=0,ku52=ku53=ku65=ku75=0,ku61=ku65=ku86=ku71=ku87=0,ku22=ku55=ku66=ku11,ku51=ku62=-ku11,ku63=ku41,ku32=ku54=-ku41,ku76=ku81,ku72=ku85=ku72=-ku81,ku44=ku33,ku84=ku73.Translational mass matrix is given by(32)Mut=[mut11mut21mut22mut31mut32mut33sym.mut41mut42mut43mut44mut51mut52mut53mut54mut55mut61mut62mut63mut64mut65mut66mut71mut72mut73mut74mut75mut76mut77mut81mut82mut83mut84mut85mut86mut87mut88],where elements of translational mass matrix are given by
(33)mut11=πρl(145R02+70R0R1+19R12-234Ri2)630,mut41=πρl2(65R02+50R0R1+17R12-132Ri2)2520,mut51=πρl(23R02+35R0R1+23R12-81Ri2)630,mut81=-(πρl2(25R02+34R0R1+19R12-78Ri2))2520,mut33=πρl3(5R02+5R0R1+2R12-12Ri2)1260,mut63=-(πρl2(19R02+34R0R1+25R12-78Ri2))2520,mut73=-(πρl3(5R02+8R0R1+5R12-18Ri2))2520,mut76=πρl2(17R02+50R0R1+65R12-132Ri2)2520,mut77=πρl3(2R02+5R0R1+5R12-12Ri2)1260,mut21=mut42=mut82=mut31=mut43=0,mut53=mut83=mut64=mut74=0,mut52=mut53=mut75=mut61=0,mut65=mut86=mut71=mut87=0,mut22=mut55=mut66=mut11,mut51=mut62,mut32=-mut41,mut63=mut54,mut72=-mut81,mut44=mut33,mut84=mut73,mut85=-mut76.Rotational mass matrix is given by(34)Mur=[mur11mur21mur22mur31mur32mur33sym.mur41mur42mur43mur44mur51mur52mur53mur54mur55mur61mur62mur63mur64mur65mur66mur71mur72mur73mur74mur75mur76mur77mur81mur82mur83mur84mur85mur86mur87mur88],where the elements of the rotational mass matrix are given by
(35)mur11=πρ(5R04+10R03R1+12R02R12-24R02Ri2+10R0R13ss-36R0R1Ri2+5R14-24R12Ri2+42Ri4)×(140l)-1,mur41=πρ(-5R04+6R02R12+8R02Ri2+8R0R13ss-16R0R1Ri2+5R14-20R12Ri2+14Ri4)×(560)-1,mur51=-(πρ(5R04+10R03R1+12R02R12-24R02Ri2ss+10R0R13-36R0R1Ri2+5R14ss-24R12Ri2+42Ri4))×(140l)-1,mur81=+8R12Ri2+14Ri4))πρ(5R04+8R03R1+6R02R12-20R02Ri2ss-16R0R1Ri2-5R14+8R12Ri2+14Ri4)×(560)-1,mur33=+2R14-8R12Ri2+28Ri4))πρl(15R04+5R03R1+3R02R12-36R02Ri2ss+3R0R13-12R0R1Ri2+2R14-8R12Ri2+28Ri4)×(840)-1,mur63=-20R12Ri2+14Ri4))πρ(-5R04+6R02R12+8R02Ri2+8R0R13ss-16R0R1Ri2+5R14-20R12Ri2+14Ri4)×(560)-1,mur73=-(πρl(5R04+2R03R1-12R02Ri2+2R0R13ssss-4R0R1Ri2+5R14-12R12Ri2+14Ri4))×(1680)-1,mur76=πρ(5R04+8R03R1+6R02R12-20R02Ri2ss-16R0R1Ri2-5R14+8R12Ri2+14Ri4)×(560)-1,mur77=πρl(2R04+3R03R1+3R02R12ss-8R02Ri2+5R0R13-12R0R1Ri2ss+15R14-36R12Ri2+28Ri4)×(840)-1,mur21=mur42=mur82=mur31=mur43,=mur53=mur83=mur64=mur74=0,mur52=mur53=mur75=mur61=0,mur65=mur86=mur71=mur87=0,mur22=mur55=mur66=mur11,mur51=mur62,mur32=-mur41,mur63=mur54,mur72=-mur81,mur44=mur33,mur84=mur73,mur85=-mur76.
The gyroscopic matrix is given by(36)Gue=[gue11gue21gue22gue31gue32gue33skewsym.gue41gue42gue43gue44gue51gue52gue53gue54gue55gue61gue62gue63gue64gue65gue66gue71gue72gue73gue74gue75gue76gue77gue81gue82gue83gue84gue85gue86gue87gue88],where the elements of the gyroscopic matrix are
(37)gue21=πρ(5R04+10R03R1+12R02R12-24R02Ri2+10R0R13sss-36R0R1Ri2+5R14-24R12Ri2+42Ri4)×(70l)-1,gue31=-(πρ(-5R04+6R02R12+8R02Ri2+8R0R13sssd-16R0R1Ri2+5R14-20R12Ri2+14Ri4))×(280)-1,gue71=-(πρ(5R04+8R03R1+6R02R12-20R02Ri2sd-16R0R1Ri2-5R14+8R12Ri2+14Ri4))×(280)-1,gue43=πρl(15R04+5R03R1+3R02R12-36R02Ri2+3R0R13s-12R0R1Ri2+2R14-8R12Ri2+28Ri4)×(420)-1,gue83=-(πρl(5R04+2R03R1-12R02Ri2+2R0R13sd-4R0R1Ri2+5R14-12R12Ri2+14Ri4))×(840)-1,gue87=πρl(2R04+3R03R1+3R02R12-8R02Ri2+5R0R13sd-12R0R1Ri2+15R14-36R12Ri2+28Ri4)×(420)-1,gue52=gue65=gue21,gue61=-gue21,gue42=gue53=gue64=gue31,gue82=gue71,gue86=gue75=-gue71,gue75=-gue83,gue11=gue41=gue51=gue81=gue22=0,gue32=gue62=gue72=gue33=gue63=0,gue73=gue44=gue54=gue84=gue55=0,gue85=gue66=gue76=gue77=gue88=0,
As proceeded in Example 1, bode plots are obtained for various shaft lengths and rotor speeds as shown in Figures 13 and 14 and are found to be in good agreement with bode plots found using TMM as shown in Figures 10 and 11.Figure 13
Bode plot for different shaft lengths (FEM).Figure 14
Bode plot for different rotating speeds (FEM).


## 4.2. Nonrotating Conditions

Bode plot for nonrotating (1 or 2 rpm) tapered shaft-rotor system is slightly different from rotating conditions in terms of amplitude. Hollow shaft with uniform thickness is considered. Bode plots are obtained for zero rpm as shown in Figure15 with TMM.Figure 15
Bode plot at zero rpm with TMM for uniform thickness hollow tapered shaft-rotor system.Applying FEM, then for zero rpm we get the bode plot as shown in Figure16.Figure 16
Bode plot at zero rpm with FEM for uniform thickness hollow tapered shaft-rotor system.


## 5. Conclusions

Shaft geometry plays one of the important roles in dynamic characteristics of rotating systems. Vibration analysis with the help of bode plots has been done for hollow tapered shaft-rotor system. Both TMM and FEM have been used for the purpose. The equation of motion for a tapered beam finite element has been developed using Euler-Bernoulli beam theory. Mass, stiffness, and gyroscopic matrices are found and values of all these elements are stated in a systematic manner for ease of understanding. The results obtained from both methods are compared and are found to be in good agreement. However, the above procedures show that the method of TMM is simpler in calculations. Two types of hollow tapered shafts have been analyzed, that is, one with uniform thickness and another with uniform bore.The length of the shaft is a vital parameter that affects the frequency response of the shaft-rotor system. As shown in Figures3, 7, 10, and 13 there is increase in amplitude of vibration for increased value of shaft length, while reducing the whirling speed of shaft. The system exhibits this behaviour due to the bending effect of the shaft and stiffness change. For changing lengths, there are large differences in frequencies even for small increase in lengths. Bode plots obtained in Figures 4, 8, 11, and 14 show that rotating speeds have very little effect on the critical frequencies; however with increasing speed, the amplitude is lowered due to gyroscopic couple. The phase angle changes abruptly for lower value of shaft speeds than higher speeds. This is due to the fact that there is reduction in gyroscopic couple as the rotational speed of the shaft decreases thereby giving larger amplitude of vibration. Nonrotating conditions are also shown in Figures 15 and 16 as bode plots which show that any rotating system at a very low speed vibrates with high amplitudes due to lack of gyroscopic couple.Effects of bearing may be included in the problem. Geared systems and other rotary elements can be mounted instead of discs and further calculations can be made. Multidiscs and other complex problems can be solved using these methods.



---

*Source: 410851-2014-04-28.xml*