A refined beam theory for bending and vibration of functionally graded tube-beams

Journal Pre-proofs A refined beam theory for bending and vibration of functionally graded tubebeams Wei-Li Ma, Zi-Cheng Jiang, Kang-Yong Lee, Xian-Fang Li PII: DOI: Reference:

S0263-8223(19)33583-4 https://doi.org/10.1016/j.compstruct.2020.111878 COST 111878

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Composite Structures

Received Date: Accepted Date:

22 September 2019 3 January 2020

Please cite this article as: Ma, W-L., Jiang, Z-C., Lee, K-Y., Li, X-F., A refined beam theory for bending and vibration of functionally graded tube-beams, Composite Structures (2020), doi: https://doi.org/10.1016/j.compstruct. 2020.111878

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A refined beam theory for bending and vibration of functionally graded tube-beams Wei-Li Ma1 , Zi-Cheng Jiang2 , Kang-Yong Lee3 , Xian-Fang Li1,4∗ 1 School of Civil Engineering, Central South University, Changsha 410075, PR China 2 School of Resources and Safety Engineering, Central South University, Changsha 410075, PR China 3 Department of Aerospace Engineering, San Diego State University, San Diego, CA 92182, United States 4 State Key Laboratory of High Performance Complex Manufacturing, Central South University, Changsha 410083, PR China

Abstract This paper presents a novel approach for analyzing transverse bending and vibration of functionally graded circular cylindrical tubes with radial nonhomogeneity. Different from the Euler-Bernoulli and Timoshenko theories of beams, a refined beam theory or third-order shear deformation beam theory for radially graded tubes is proposed, where warping, shear deformation and rotational moment of inertia of cross-section are all considered. The shear correction coefficient is not needed. Coupled governing equations for the deflection and rotation about the neutral axis of cross-section are derived from equilibrium equations, and then converted to a single fourth-order partial differential governing equation. The deflection and stress distribution for cantilever and simply-supported tubes are derived explicitly. The frequency equations for free flexural vibration of radially graded hollow cylinders with clamped-clamped, pinned-pinned, and clamped-free ends are obtained and the natural frequencies are calculated for different powerlaw gradients and various length/thicknesses ratios. The effects of radial gradient on the stress distribution and the natural frequencies are analyzed in detail. Keywords: Flexure of cylindrical tube; Functionally graded hollow cylinder; Free vibration; Radial gradient. ∗

Corresponding author, E-mail: [email protected]

1

1

Introduction

Hollow cylinders with annular cross-section are widely used in lots of engineering fields from macro scale to micro scale, such as steel pipes, polyvinylchlorid tubes, carbon nanotubes, microtubules, etc. As one of the most important structure units, cylindrical tubes have gained more application in many areas like mechanical, civil, aerospace, marine engineering in recent years. In these applications, the tubes or pipes in service are loaded by complex mechanical loading, which directly affects the structural integrity, reliability, and stability of tube structures. The study of such tube structures has a very long history of research [1, 2]. In particular, the study of the mechanical behavior of tube structures can be divided into two categories in general. One approach is to model it as a three-dimensional realistic structure including shell structures, which involves three spatial variables, and complex manipulation to achieve the response analysis of the structure is inevitable, although it provides more accurate results [3, 4]. The other is to simulate it as a one-dimensional problem based on reasonable hypotheses of cylindrical structures. In other words, the latter is to transform a three-dimensional hollow cylinder or tube to a one-dimensional beam with annular cross-section based on crucial and rational assumptions and neglecting some secondary factors. In the context of classical three-dimensional (3D) elasticity theory, many papers are devoted to bending and free vibration of solid and hollow cylindrical rods. Soldatos and Hadjigeorgiou gave a detailed analysis of the free vibration of homogeneous isotropic thick cylindrical shell or panel subjected to simply supported edge boundary conditions [5]. By employing the Ritz method and expanding three displacement components as a periodic function in the circumferential angle, Leissa and So calculated the natural frequencies of free-free and free-fixed elastic cylinders [6]. A double polynomial series was employed to handle the free vibration problem of a stress-free hollow cylinder of arbitrary cross-section [7]. As compared to the 3D elasticity theory, theories based on the one-dimensional (1D) beam model are quite efficient, especially for slender beams. To date, two recognized models, the Euler–Bernoulli and Timoshenko theories [8], have been widely used. The Euler–Bernoulli theory of beams provides a simplified solution for static bending, vibration and buckling of beams. However, it neglects the shear deformation and rotational inertia, which

2

leads to the overestimates of both the natural frequencies of free vibration and the critical loads of buckling. The Timoshenko theory of beams considers two factors mentioned above and modifies the simplest beam theory [9]. To date, the Timoshenko beam theory has been broadly adopted to treat bending and vibration problems [10–13] as well as the stability problems of elastic beams [14]. It is worth pointing out that, in the Timoshenko beam theory, a shear correction coefficient needs to be introduced, which is an essential part of this theory and cannot be derived by the theory itself. Towards more accurate theoretical models, Levinson [15] proposed a third-order beam theory for rectangular beams, in which the shear correction coefficient is not necessary and the shear-free surface conditions are automatically satisfied. Furthermore, for cylindrical beams of circular cross-section, Huang and Li presented the third-order beam model to treat static bending, vibration and buckling problems [16–18]. Unfortunately, for beams of other cross-section such as annular cross-section, there is little information on the higher-order beam model relating vibration, bending and stability of beams. On the other hand, functionally graded materials (FGMs) have attracted strong scientific and technological interest in recent years due to its outstanding performance in the high temperature environment applications [19]. Usually FGMs are fabricated by ceramic and metal. So they show advantages in suffering from thermal shocks and maintaining the structural integrity since ceramic components are beneficial to achieve higher temperature resistance while metal part prevents the structure from cracking. Considerable research has been done for the manufacture of FGMs. The material properties of FGMs can be designed by changing the volume fraction of two or more components and material properties may vary along arbitrary direction like plates’ or beams’ thickness, radial direction of cylinders, etc. There have been a large number of theoretical analyses on FGM beams or rods. For example, Sankar proposed an elasticity solution for a thickness-wise functionally graded beam subjected to transverse loads [20]. Aydogdu investigated the free vibration of simply supported thickness-wise FGM beam [21]. Li formulated a unified approach to analyze the static and dynamic behavior for thickness-wise functionally graded beams using the Euler-Bernoulli and Timoshenko theories of beams [22]. For axially graded beams, some approximate methods and exact analytic approaches have been put forward to deal with free vibration and buckling problems of 3

Euler-Bernoulli and Timoshenko beams [23–25]. For a functionally graded beam with bi-directional gradient such as the thickness- and length-directions, some works have been conducted. Simsek [26] dealt with free and forced vibration of Timoshenko beams with various ends. Nguyen et al. analyzed the vibration of a bi-dimensional functionally graded Timoshenko beam excited by a moving load, [27]. Furthermore, Ghayesh [28] gave an analysis of nonlinear vibration analysis of axially graded shear-deformable tapered beams. Sun and Li [29] coped with free vibration of axially loaded functionally graded Timoshenko beams with nonuniform cross section. Tang and Ding [30] tackled nonlinear vibration of a bi-directional functionally graded beam under hygro-thermal loads. Using isogeometric analysis, Nguyen et al provided an efficient computational approach for analyzing functionally graded plate structures [31–33]. Although many works on functionally graded beams with gradient along the axial and/or radial/thickness directions have been reported, these works are limited to rectangular functionally graded beams, for which some higher-order beam theories have been established. For functionally graded cylindrical tube-beams, the theories available are either the Euler-Bernoulli beam theory or the Timoshenko beam theory. To date, there is no work focusing on a higher-order beam theory for functionally graded beams with annular cross-section. To adequately describe the mechanical behavior of circular cylindrical radially graded tubes, this paper aims at presenting a third-order beam theory to analyze their bending and vibration. Using the shear-stress-free condition at the inner and outer surface of the cylinder, appropriate expressions for the axial and vertical displacements are constructed. Then, two coupled equations for the deflection and rotation are deduced, and they are further converted to a single governing equation. Bending deflection and free vibration of straight radially graded tubes are tackled. Evaluated results are consistent with FEM results, and it indicates the efficiency of the method suggested here. The effects of gradient variation on the deflection, stress distribution, and natural frequencies are analyzed.

2

Theoretical formulation

The bending and free vibration of a straight cylindrical tube with radially varying material properties over annular cross-section is considered in the present paper, and the geometry of a radially 4

graded hollow cylinder is shown in Fig. 1. We denote its inner and outer radii as Ri and Ro , respectively, and Young’s modulus E(r) and shear modulus G(r) of the tube are dependent on r, r being radius, i.e. Ri ≤ r ≤ Ro . In the following, we choose a Cartesian coordinate system and a cylindrical polar coordinate system (see Fig. 1), where the x-axis is taken along the centroidal axis of the tube, y, z are related to r, θ by y = r cos θ,

z = r sin θ.

(1)

Consider a straight hollow tube subjected to a transverse force or distributed loading in the oxz-plane and a bending moment perpendicular to the oxz-plane, the tube is deflected in the xoz-plane. We denote three displacement components by (u, ur , uθ ) according to the coordinates (x, r, θ), and (u, v, w) according to the coordinates (x, y, z), respectively. It is clear that they are related with each other through the relationship ur = v cos θ + w sin θ.

(2)

In polar coordinates, because of (1) and (2), the relationship between the shear strain and the displacement vector can be expressed as γxr

∂ur ∂u 1 = + = ∂x ∂r r

( ) ∂v ∂w ∂u ∂u y +z +y +z . ∂x ∂x ∂y ∂z

(3)

Consider a simple case, that is, the displacement component v in the y-direction is independent of x. The shear strain can be rewritten as γxr

[ ( ) ] 1 ∂w ∂u ∂u = z + +y . r ∂x ∂z ∂y

If taking the shear strain γxr of the form ( ) ∂u ∂w ∂u (y 2 + z 2 − Ri2 )(y 2 + z 2 − Ro2 ) y +z + =z ζ(x) ∂y ∂x ∂z y2 + z2

(4)

(5)

where ζ(x) is an unknown function, it is evident that the shear-stress-free boundary conditions τxr (x, r) = 0, r = Ri , Ro .

(6)

are automatically satisfied at the inner surface and the outer surface of the tube owing to the shear Hooke’s law τxr (x, r) = G (r) γxr (x, r) . 5

(7)

To get the expression for ζ(x), we separate u into two parts ( u(x, y, z) = u1 (x, z) +

1 2 R2 R2 y − 2 i o2 3 y +z

) zζ(x).

(8)

After inserting (8) into (5), we find ] ∂u1 ∂w [ 2 =− + z − (Ri2 + Ro2 ) ζ(x). ∂z ∂x

(9)

By performing integration with respect to z to both sides of Eq. (9), we find [ ] 1 3 ∂w 2 2 u1 = z ζ(x) − z (Ri + Ro )ζ(x) + 3 ∂x

(10)

Then, we introduce a new function ψ representing the rotation angle of u1 , ψ can be expressed as ψ(x) = −(Ri2 + Ro2 )ζ(x) −

∂w . ∂x

(11)

After combining (11) and (8), ζ(x) can be eliminated immediately, then we get an expression for the displacement component u below 3Ri2 Ro2 − (y 2 + z 2 )2 u = zψ + z 3(Ri2 + Ro2 )(y 2 + z 2 )

(

) ∂w +ψ . ∂x

(12)

With the above expression for u, one immediately gets the normal strain εxx and shear strains γxy , γxz , γxr . They are εxx γxy γxz γxr

( ) ∂ψ 3Ri2 Ro2 − (y 2 + z 2 )2 ∂ 2 w ∂ψ =z +z + , ∂x ∂x 3(Ri2 + Ro2 )(y 2 + z 2 ) ∂x2 ( ) 3Ri2 Ro2 + (y 2 + z 2 )2 ∂w = −2yz + ψ , 3(Ri2 + Ro2 )(y 2 + z 2 )2 ∂x [ ]( ) 3Ri2 Ro2 (y 2 − z 2 ) − (y 2 + 3z 2 )(y 2 + z 2 )2 ∂w = 1+ +ψ , ∂x 3(Ri2 + Ro2 )(y 2 + z 2 )2 [ ]( ) z Ri2 Ro2 + (y 2 + z 2 )2 ∂w = 1− 2 +ψ r ∂x (Ri + Ro2 )(y 2 + z 2 )

(13) (14) (15) (16)

Based on the assumption of beams, the stress components σyy = σzz = τyz = 0 vanish in the whole beam [3]. In addition, on the inner and outer surfaces of a tube, the shear stresses are nil.

6

It is clearly seen from the following relationships [

σxx τxy τxz τxr

( )] ∂ψ 3Ri2 Ro2 − (y 2 + z 2 )2 ∂ 2 w ∂ψ = E(r)εxx = E(r) z +z + , ∂x ∂x 3(Ri2 + Ro2 )(y 2 + z 2 ) ∂x2 ( ) 3Ri2 Ro2 + (y 2 + z 2 )2 ∂w = G(r)γxy = −2G(r)yz +ψ , 3(Ri2 + Ro2 )(y 2 + z 2 )2 ∂x [ ]( ) 3Ri2 Ro2 (y 2 − z 2 ) − (y 2 + 3z 2 )(y 2 + z 2 )2 ∂w = G(r)γxz = G(r) 1 + +ψ , ∂x 3(Ri2 + Ro2 )(y 2 + z 2 )2 [ ]( ) z Ri2 Ro2 + (y 2 + z 2 )2 ∂w = G (r) 1− 2 +ψ . r ∂x (Ri + Ro2 )(y 2 + z 2 )

(17) (18) (19) (20)

It indicates that the above relations allow us to analyze flexural behavior as well as flexural vibration of straight radially graded cylindrical tubes, where the shear stresses on the inner and outer surfaces of the tube automatically vanish. In the Timoshenko beam theory, the shear stress on the cross section is assumed to be uniform, and a shear correction coefficient must be introduced to relax the traction-free condition on the surface of the beam [9]. The shear correction coefficient is an inherent part of this theory and cannot be directly determined by the theory itself. Here we abandon the uniform stress assumption. The shear stress is assumed to meet the shear-stress-free surface condition. The shear correction coefficient is not necessary. Due to the equilibrium of internal force, the bending moment M and shear force Q at any position x are ∫ M=

σxx zdA,

(21)

τxz dA,

(22)

∫

A

Q= A

respectively, where A is the cross-sectional area. Plugging (17) and (19) into the above integrals, one gets 2 ˜2 ∂ψ − E4 ∂ w , M =E 2 ∂x ∂x ( ) ˜ 0 ψ + ∂w , Q=G ∂x

7

(23) (24)

where ˜2 = E2 − E4 , ˜ 0 = G0 − G2 , E G ∫ Ro ∫ Ro π 3 E2 = π E(r)r dr, E4 = E(r)(r4 − 3Ri2 Ro2 )rdr, 2 + R2 ) 3(R Ri Ri o i ∫ Ro ∫ Ro 4π G0 = 2π G(r)rdr, G2 = G(r)r3 dr. 2 + R2 ) 3(R Ri Ri o i

(25) (26) (27)

Since our attention focuses on a hollow tube subjected to a transverse force or distributed loading in the oxz-plane and a bending moment perpendicular to the oxz-plane, the bending of the tube takes place in the xoz-plane. Thus, the motion in the y-direction is neglected, so the remaining two equations of motion read ∂σxx ∂τxy ∂τxz ∂2u + + = ρ(r) 2 , ∂x ∂y ∂z ∂t ∂τyz ∂τxz ∂σzz ∂2w + + = ρ(r) 2 , ∂x ∂y ∂z ∂t

(28) (29)

where ρ (r) stands for r-dependent mass density. Now we insert (17)-(19) into Eq. (28), multiply both sides by z and then integrate both sides over cross-section. After some manipulations we get 2 3 ˜2 ∂ ψ − E4 ∂ w − G ˜0 E ∂x2 ∂x3

(

∂w +ψ ∂x

) = ρ˜2

∂2ψ ∂3w − ρ 4 ∂t2 ∂x∂t2

(30)

where ρ˜2 = ρ2 − ρ4 , ∫ Ro ρ2 = π ρ(r)r3 dr, ρ4 = Ri

π 2 3(Ri + Ro2 )

∫

(31) Ro

ρ(r)(r4 − 3Ri2 Ro2 )rdr.

(32)

Ri

Then we perform integration to both sides of Eq. (29) over cross-section, giving ( ˜0 G

∂ 2 w ∂ψ + ∂x2 ∂x

where

∫

) = ρ0

∂2w − q, ∂t2

(33)

Ro

ρ0 = 2π

ρ(r)rdr,

(34)

Ri

and q is a distributed lateral loading along the centroidal line in the oxz-plane. Thus, one obtains two coupled equations (30) and (33) governing the dynamic behaviors of radially graded cylindrical tubes. 8

A further simplification can be achieved by introducing a new auxiliary function of F with length dimension, which satisfies ∂4F ∂2F E2 4 + ρ0 2 − ∂x ∂t

(

˜2 ρ0 E ρ2 + ˜0 G

)

∂4F ρ˜2 ρ0 ∂ 4 F + = q. ˜ 0 ∂t4 ∂x2 ∂t2 G

(35)

For this purpose, if choosing the deflection w and the rotation ψ in the form ˜2 ∂ 2 F E ˜ 0 ∂x2 G ρ4 ∂ 3 F ψ= − ˜ 0 ∂x∂t2 G

w=F −

ρ˜2 ∂ 2 F , ˜ 0 ∂t2 G E4 ∂ 3 F ∂F − , 3 ˜ ∂x ∂x G0 +

(36) (37)

we find that after inserting (36) and (37) into Eqs. (30) and (33), they are identically fulfilled. Therefore, we only need to determine the auxiliary function F through appropriate boundary conditions. For convenience, we also give bending moment M and shear force Q in terms of F . It is done by plugging (36) and (37) into (23) and (24). The bending moment M and shear force Q can be expressed as ∂2F E 2 ρ4 − E 4 ρ2 ∂ 4 F + , ˜0 ∂x2 ∂x2 ∂t2 G ∂3F ∂3F Q = ρ2 − E2 3 . 2 ∂x∂t ∂x

M = −E2

(38) (39)

The remaining task is to solve the governing equation (35) for the unknown F under suitable boundary conditions, which can be provided by the deflection w, rotation ψ, bending moment M , and shear force Q. It is noted that for static bending of a tube, we solely remove all the term related to time t.

3

Bending of cylindrical tube-beams

According to the governing equation (35) in the preceding section, for static bending of tubes we solely remove all the terms related to time t. On this occasion, the governing equation (35) reduces to E2

d4 F = q(x). dx4

(40)

A general solution to Eq. (40) is 1 F (x) = 6E2

∫

x

(x − s)3 q (s) ds + A1 x3 + A2 x2 + A3 x + A4 ,

0

9

(41)

where Aj (j = 1, . . . , 4) are unknown constants to be obtained through appropriate boundary conditions. In the following, we consider two typical cases. For other cases, the exact solution is also easily obtained and omitted here.

3.1

Cantilever tube-beam subjected to uniform distributed loading

Here we consider a cantilever tube-beam subjected to uniform distributed loading q. In this case, the boundary conditions for a cantilever tube-beam with the free end at x = L and the clamped end at x = 0 are stated as w(0) = 0,

ψ(0) = 0,

M (L) = 0,

Q(L) = 0,

(42)

where L is the length of the tube-beam. Using the above boundary conditions, it is a simple matter to determine Aj in (41). Omitting the detail, one gets the final expression for F (x) below ( ) ˜2 L2 q 24E L 12 E 4 F (x) = x4 − 4Lx3 + 6L2 x2 + x+ . ˜0 ˜0 24E2 G G

(43)

With F (x) at hand, we obtain all physical quantities of interest as follows: [ ] qz 6E4 x 6Ri2 Ro2 − 2(y 2 + z 2 )2 E2 (L − x) 3 2 2 u=− x − 3Lx + 3L x + − , ˜0 ˜0 6E2 (Ri2 + Ro2 )(y 2 + z 2 ) G G [ ] q 12 4 3 2 2 ˜ w= x − 4Lx + 6L x + (2E2 L − E2 x)x , ˜0 24E2 G ( ) q 6E4 x 3 2 2 ψ=− x − 3Lx + 3L x + , ˜0 6E2 G

(44) (45) (46)

and σxx τxy τxz τxr

[ ] qzE(r) 2 2E4 3Ri2 Ro2 − (y 2 + z 2 )2 2E2 2 =− x − 2Lx + L + + , ˜0 ˜0 2E2 3(Ri2 + Ro2 )(y 2 + z 2 ) G G 2qG(r)yz 3Ri2 Ro2 + (y 2 + z 2 )2 = (x − L) , ˜0 3(Ri2 + Ro2 )(y 2 + z 2 )2 G [ ] qG(r) 3Ri2 Ro2 (y 2 − z 2 ) − (y 2 + 3z 2 )(y 2 + z 2 )2 =− 1+ (x − L) , ˜0 3(Ri2 + Ro2 )(y 2 + z 2 )2 G [ ] qG (r) z Ri2 Ro2 + (y 2 + z 2 )2 =− 1− 2 (x − L) . ˜ 0r (Ri + Ro2 )(y 2 + z 2 ) G

(47) (48) (49) (50)

From the above, it is observed that the contribution of flexural behavior is composed of two ˜ 0 tend to parts, one arising from bending moment and the other from shear force. If letting G 10

infinity, equivalently to the fact that the beam has sufficiently high reduced shear rigidity, it means shear locking. Thus, the above results reduce to those for Euler-Bernoulli radially graded beams of annular cross-section, in which there is uniform vanishing shear stress, as expected. On the other hand, if setting Ri = 0, a hollow tube reduces to a solid cylinder. In this case, the above stresses collapse to those for a solid cylinder subjected to uniform transverse loading [16].

3.2

Simply supported tube-beam subjected to a concentrated force

Next we consider a simply supported tube-beam subjected to a concentrated force (Py = 0, Pz = P ) at the mid-span of the beam. Owing to symmetry, one can write the following conditions w(0) = 0,

M (0) = 0,

Q

( ) L P = , 2 2

ψ

( ) L = 0. 2

(51)

L . 2

(52)

Using these conditions, we obtain F (x) as follows: P P F (x) = − x3 + 12E2 2E2

(

L2 E4 + ˜0 8 G

) x, 0 ≤ x ≤

With this F (x), we obtain all physical quantities of interest. For saving space, we only give expressions in x ∈ [0, L/2] below and those in x ∈ [L/2, L] can be directly written due to symmetry 3Ri2 Ro2 − (y 2 + z 2 )2 u = zψ + z 3(Ri2 + Ro2 )(y 2 + z 2 )

(

) ∂w +ψ . ∂x

( ) L2 P z 3Ri2 Ro2 − (y 2 + z 2 )2 x2 − + ˜ 0 (Ri2 + Ro2 )(y 2 + z 2 ) 4 6G ( 2 ) P P L E2 3 w=− x + + x, ˜0 12E2 2E2 8 G ( ) P L2 2 ψ= x − , 4E2 4 u=

Pz 4E2

(53)

(54) (55) (56)

and xzP E(r) , 2E2 yzP G(r) 3Ri2 Ro2 + (y 2 + z 2 )2 =− , ˜0 3(Ri2 + Ro2 )(y 2 + z 2 )2 G [ ] P G(r) 3Ri2 Ro2 (y 2 − z 2 ) − (y 2 + 3z 2 )(y 2 + z 2 )2 = 1+ , ˜0 3(Ri2 + Ro2 )(y 2 + z 2 )2 2G [ ] zP G (r) R2 R2 + (y 2 + z 2 )2 = 1 − i2 o 2 2 . ˜0r (Ri + Ro )(y + z 2 ) 2G

σxx =

(57)

τxy

(58)

τxz τxr

11

(59) (60)

3.3

Effects of the gradient parameter on flexure

To study the effects of the gradient parameter on flexure, we take a cantilever tube-beam subjected to uniform distributed loading as an example. In the following calculation, the material properties of the tube are assumed to obey the power law ( Y (r) = Yi + (Yo − Yi )

r − Ri Ro − Ri

)N ,

(61)

where Yi and Yo are the corresponding material properties on the inner and the outer surfaces, respectively, and N is the power-law gradient index describing the volume fraction change of both constituents involved. When N = 1, the material properties exhibit a linear variation on the radial coordinate. For other N values, Fig. 2 shows the variation of material properties for Ri /Ro = 0.5. In the present study, a radially graded tube with the ratio of length to radius L/Ro = 10, v = 0.3 is used, unless otherwise stated, and aluminum and zirconia are selected as the inner surface material and the outer surface material, respective, the material properties of which are [34] Al: Ei = 70 GPa, Gi = 27 GPa, ρi = 2702 kg/m3 , ZrO2 : Eo = 200 GPa, Go = 77 GPa, ρo = 5700 kg/m3 . As a representative, consider a radially graded cantilever beam with annular cross-section subjected to uniform distributed loading q. The variation of dimensionless deflection w(x)Ei I/qL4 with the dimensionless axial coordinate x/L is plotted in Fig. 3 for different N values, where I ( ) denotes the rotational moment of inertia of cross-section, i.e. I = π Ro4 − Ri4 /4. It is observed that the deflection increases with the gradient index N rising, implying that different volume fractions can produce different deflections, although the chosen two materials remain unchanged. For comparison, we also display the variation of the deflection for a homogeneous cantilever tube-beam made of aluminum. Clearly, due to the contribution of zirconia with larger Young’s modulus in a functionally graded tube, the radially graded tube-beam seems to become stiffer, and the deflection is lower. When taking N = 10 or more, the deflection is gradually close to that for homogeneous cantilever aluminum tube-beam. It is easily understood since the volume fraction of aluminum rich in the inner surface of a radially graded tube with larger N becomes larger, and the properties of aluminum play a dominant role, as seen in Fig. 2. 12

Furthermore, the distribution of the axial stress σxx (0, 0, z) at the clamped end versus z/Ro is displayed in Fig. 4. The stress σxx in magnitude reaches maximum at z = ±Ro , which coincides with the classical results. However, the maximum σxx in magnitude at z = ±Ro is increased when N becomes large. The top and bottom positions correspond to the maximum compressive and tensile stress, respectively. For shear stress τxz , we plot the distribution profile of τxz L/q over the cross-section of the clamped end in Fig. 5(a-c). Clearly, the shear stress σxz arrives at its maximum at the midplane, in agreement with the classical result. The location of the maximum shear stress in the midplane depends on the gradient index. For example, when the gradient index N is raised, the location of the maximum shear stress is shifted from the inner surface to the outer surface, as seen in Fig. 5(a-c). For comparison, for a homogeneous cantilever tube-beam made of aluminum, the distribution of the shear stress is also presented in Fig. 5d. The gradient index has a remarkable effect on the stress distribution. Finally, we point out that the shear stress τxz does not vanish on the inner and outer surfaces except the top and bottom positions. Contrary to that, the shear stress τxr is always zero on the inner and outer surfaces, which is seen from (50) and (60), but not displayed. To verify the effectiveness of the present approach, the finite element method (FEM) is used to examine the accuracy of our numerical results. In the analysis of the FEM, the commercial software MSC. Patran/Nastran is chosen, and 3D Hex8 element with 8 nodes was applied. Table 1 shows the numerical results of a dimensionless deflection w(x)Ei I/qL4 based on the FEM and the present approach. When adopting the FEM, three different meshes with the numbers of elements being 9000, 16000 and 64800, respectivly, are taken for the same model, and they are denoted as FEM-1, FEM-2, and FEM-3 in Table 1, respectively. As a representative, the meshing model of FEM-2 is shown in Fig. 6a, and those of FEM-1 and FEM-3 are not displayed here for saving space. Additionally, to model the gradient variation, the Al/ZrO2 tube was partitioned into k equal parts along the radial direction, k being a positive integer (see Fig. 6b). In each part, the material is assumed to be homogeneous and has same material properties, and the total properties are piecewise constants, where the elastic displacement and stresses are required to be continuous across the interface of two adjacent parts. 13

From the relative errors listed in Table 1, one sees that as the number of elements grows, the FEM results converge well. All the relative errors between the latter two FEM results and present ones are less than 3%. It validates that our results are accurate and agree with the finite element results. Due to this reason, in what follows the subsequent FEM model in this article is meshed by the same element size as in FEM-2. It is interesting to point out that based on FEM-2, the maximum dimensionless lateral displacement vEi I/qL4 is found to be 0.000264, which is far less than the maximum of dimensionless deflection w(x)Ei I/qL4 as 0.06034. Thus, comparing with the flexural deflection w, one can directly ignore the displacement component v. It also indiates that the assumption of the lateral displacment v being independent of the variable x holds true. Table 1. Dimensionless deflection w(x)Ei I/qL4 for functionally graded cantilever tube-beam with L/Ro = 10subjected to uniformly distributed loading

x/L Present FEM-1 errors-1 FEM-2 errors-2 FEM-3 errors-3

4

0.1 0.001629 0.001398 14.18% 0.001589 2.46% 0.001594 2.15%

0.2 0.005062 0.004852 4.15% 0.004978 1.66% 0.004987 1.48%

0.3 0.009857 0.009705 1.54% 0.009730 1.29% 0.009755 1.03%

0.4 0.015690 0.015660 0.19% 0.015690 0.00% 0.015610 0.51%

0.5 0.022280 0.022430 0.67% 0.022350 0.31% 0.022250 0.13%

0.6 0.028640 0.029670 3.60% 0.029470 2.90% 0.029420 2.72%

0.7 0.036740 0.03721 1.28% 0.037210 1.28% 0.036960 0.60%

0.8 0.044260 0.045000 1.67% 0.044570 0.70% 0.044450 0.43%

0.9 0.051810 0.052800 1.91% 0.052550 1.43% 0.05204 0.44%

Free vibration of cylindrical tube-beams

Here let us analyze free vibration of radially graded cylindrical beams with annular cross-section. For this purpose, the introduced auxiliary function F may take the form F (x, t) = f (x)eiωt ,

(62)

where ω is the circular frequency of free vibration. Putting the above into Eq. (35), in the absence of applied loading, viz. q = 0, one can get ( ) ( ) 2 ˜2 d4 f ρ0 E ρ˜2 2 2d f 2 E2 4 + ρ2 + ω − ρ0 ω 1 − ω f = 0. ˜0 ˜0 dx dx2 G G

(63)

This is an ordinary differential equation with constant coefficients. Use of a standard approach for solving the above equation permits us to give a general solution to Eq. (63) below f (x) = C1 cos(λ1 x) + C2 sin(λ1 x) + C3 cosh(λ2 x) + C4 sinh(λ2 x), 14

(64)

1.0 0.059310 0.060590 2.16% 0.060340 1.74% 0.059840 0.89%

for frequencies ω less than the cut-off frequency ωc , and f (x) = C1 cos(λ3 x) + C2 sin(λ3 x) + C3 cos(λ4 x) + C4 sin(λ4 x),

(65) √

for frequencies ω larger than the cut-off frequency ωc , where the cut-off frequency ωc = Cj (j = 1, ..., 4) are unknown constants, and v   v( u )2 ( ) u u ( ) u u1  ˜2 ˜2  4ρ0 1 1 ρ2 ρ0 E t ρ2 + ρ0 E λ1,2 = ω u + − 2 ± + , t2  2 ˜ 0 E2 ˜ 0 E2  E2 G E2 ω ωc E2 G and λ3,4

v   v( u )2 ( ) u u ( ) u 1  u ρ2 ˜2 ˜2  ρ0 E 4ρ0 1 1 ρ2 ρ0 E t = ωu + + − + + . t 2 ± ˜ 0 E2 ˜ 0 E2  E2 G E2 ω 2 ωc2 E2 G

˜ 0 /˜ G ρ2 ,

(66)

(67)

An expression for f (x) for ω = ωc may be provided and omitted here. In the above expressions for f (x), the sine and cosine terms in Eqs. (64) and (65) correspond to flexural waves propagating in the hollow tube, while the other terms represent nonpropagating fields or evanescent components. Combined the above formulas with the relevant end conditions, it is easy to determine the natural frequencies of free vibration. In what follows we will discuss three frequently-encountered typical cases: simply supported beams, doubly clamped beams, and cantilever beams. Moreover, for convenience, we only focus our attention on free vibration of lower frequencies. Or rather, we only study the case of Eq. (64). For f (x) given by Eq. (65), an analogous treatment can be done. To this end, it is expedient to give expressions for w, ψ, M, and Q in terms of the λ1,2 . This is done by substituting (64) into (36)-(39). After some algebra, one finally obtains w = (1 + Π1 ) [C1 cos(λ1 x) + C2 sin(λ1 x)] + (1 − Π2 ) [C3 cosh(λ2 x) + C4 sinh(λ2 x)] ,

(68)

ψ = λ1 (1 − Λ1 ) [C1 sin(λ1 x) − C2 cos(λ1 x)] − λ2 (1 + Λ2 ) [C3 sinh(λ2 x) + C4 cosh(λ2 x)] , (69) ( ) } E 2 ρ4 − E 4 ρ2 { 2 M = E2 + λ1 [C1 cos(λ1 x) + C2 sin(λ1 x)] − λ22 [C3 cosh(λ2 x) + C4 sinh(λ2 x)] , ˜0 G (70) { } ( ) λ2 (Π2 + Λ2 ) 2 2 Q = −λ1 E2 λ1 − ρ2 ω C1 sin(λ1 x) − C2 cos(λ1 x) + [C3 cosh(λ2 x) + C4 sinh(λ2 x)] , λ1 (Π1 + Λ1 ) (71) 15

where Π1 =

˜2 λ2 − ρ˜2 ω 2 ˜2 λ2 + ρ˜2 ω 2 E E 1 2 , Π2 = , ˜0 ˜0 G G

(72)

Λ1 =

E4 λ21 − ρ4 ω 2 , ˜0 G

(73)

Λ2 =

E4 λ22 + ρ4 ω 2 ˜0 G

Using the above-resulting results for the deflection, rotation, bending moment and shear force, (68)-(71), from appropriate boundary conditions at both ends of a tube-beam, we can easily determine the natural frequencies of the free vibration of a tube-beam.

4.1

Simply supported tube-beams

First, let us consider a simply supported tube-beam of length L with the boundary conditions: w(0) = 0,

M (0) = 0,

w(L) = 0,

M (L) = 0.

(74)

By applying the boundary conditions (74) in the expressions (68) and (70), respectively, one easily gets C1 = C3 = 0, and C3 and C4 satisfy the following algebraic equations (1 + Π1 ) C2 sin(λ1 L) + (1 − Π2 ) C4 sinh(λ2 L) = 0,

(75)

λ21 C2 sin(λ1 L) − λ22 C4 sinh(λ2 L) = 0.

(76)

Due to the existence of a nontrivial solution, the determinant of the coefficient matrix has to vanish, which results in sin(λ1 L) sinh(λ2 L) = 0.

(77)

Solving the above equation, recalling (66) we derive the natural frequencies, which meet the follow equation v   v( u )2 ( ) u u ( ) u u1  ˜2 ˜2  4ρ0 1 1 ρ2 ρ0 E t ρ2 + ρ0 E ωLu + − 2 + + = nπ, n = 1, 2, 3, ..., t2  2 ˜ 0 E2 ˜0  E2 G E2 ω ωc E2 E2 G The natural frequencies can be given explicitly by looking for the roots of Eq. (78) v √[ u ( nπ )2 ( nπ )2 ]2 4E ρ˜ ( nπ )4 ωc u 2 2 t ωn = √ 1+η − 1+η − ˜ L L L 2 G 0 ρ0 16

(78)

(79)

where η=

˜2 ρ2 E + . ˜0 ρ0 G

(80)

The natural frequencies can be exactly determined by calculating (79). Note that the lower positive roots are chosen. With the natural frequencies, the mode shape of a simply-supported tube-beam corresponding to the n-th natural frequency reads w = sin

( nπx ) L

.

(81)

This result coincides with the classical one, and it turns out that the mode shapes of a simply supported radially graded tube-beam are independent of the gradient index. Nonetheless, the natural frequencies are clearly dependent on the gradient index and material properties, as viewed from (79). To verify the accuracy of the presented theory, the first three natural frequencies of a simply supported tube based on the presented theory are calculated for a radially graded cylindrical tube √ made of Al/ZrO2 . Numerical results of dimensionless natural frequencies Ω = L2 ρo Aω 2 /Eo I based on the present approach and FEM are listed in Table 2 for the simplest linear gradient, N = 1. Here we discuss the free vibration of a simply supported radially graded tube-beam and compare our numerical results with those based on the FEM. From Table 2, our numerical results agree well with those computed through FEM, in which the numbers of the elements and DOFs used in the FEM model are listed in Table 3. A comparison reveals satisfactory agreement. A slight difference is possible to due to the relatively rough partition. If more layers are partitioned, it is anticipated that more accurate results of the FEM can be achieved. Fig. 7 shows the finite element results of the first three mode shapes of a simply supported tube. In addition, from (81) we easily conclude that the mode shapes of simply-supported radially graded tube-beams are independent of gradient index, which is essentially different from the natural frequencies. Table 2. Dimensionless frequencies Ω of a simply supported radially graded tube-beam with Ri /Ro = 0.5, N = 1, v = 0.3 n L/Ro = 20 L/Ro = 10 L/Ro = 5 Present FEM Present FEM Present FEM 1 9.7057 9.7001 9.1405 9.1793 7.6461 7.7656 2 36.5620 36.7170 30.5845 31.0625 21.0083 21.5269 3 75.6586 76.4260 56.6664 57.9680 34.7390 35.2628 17

Table 3. The number of the elements and DOFs of a simply-supported radially graded tube-beam with Ri = 1, Ri /Ro = 0.5 L/Ro = 20 L/Ro = 10 L/Ro = 5 elements 32000 16000 8000 DOFs 105138 52338 25938

4.2

Clamped tube-beams

For a doubly clamped tube-beam of length L, the boundary conditions can be written as w(0) = 0,

ψ(0) = 0,

w(L) = 0,

ψ(L) = 0.

(82)

Plugging the first two boundary conditions in (82) into (36) and (37), one get C3 = −

1 + Π1 C1 , 1 − Π2

C4 = −

λ1 (1 − Λ1 ) C2 . λ2 (1 + Λ2 )

(83)

In addition, using the third boundary condition in (82) along with the above results (83) we have C2 = −

cos(λ1 L) − cosh(λ2 L) C1 sin(λ1 L) − Ψ sinh(λ2 L)

(84)

with Ψ=

λ1 (1 − Λ1 ) (1 − Π2 ) . λ2 (1 + Λ2 ) (1 + Π1 )

(85)

Finally, applying the remaining boundary condition in (82) one obtains the frequency equation to be

( ) 1 2 − 2 cos(λ1 L) cosh(λ2 L) − Ψ − sin(λ1 L) sinh(λ2 L) = 0. Ψ

(86)

Once the natural frequencies are determined by solving the above frequency equation (86), the corresponding mode shape can be given by w = cos(λ1 x) − cosh(λ2 x) −

cos(λ1 L) − cosh(λ2 L) [sin(λ1 x) − Ψ sinh(λ2 x)] sin(λ1 L) − Ψ sinh(λ2 L)

(87)

Table 4 also shows a comparison of the first three dimensionless frequencies Ω between the our results with those according to the FEM for a doubly clamped tube-beam. In Fig. 8, the finite element results of the first three modes shapes Table 4. Dimensionless frequencies Ω of a with Ri /Ro = 0.5, N = 1, v = 0.3 n L/Ro = 20 L/Ro = 10 Present FEM Present FEM 1 20.6660 20.7593 17.0964 17.2690 2 51.9404 52.3566 37.9044 38.6744 3 92.2070 93.3804 62.0131 63.8464

for a clamped tube-beam are displayed. clamped tube-beam L/Ro = 5 Present FEM 11.4105 11.6524 22.3890 23.5231 35.3381 37.2219 18

4.3

Cantilever tube-beams

In this case, the boundary conditions of a cantilever tube-beam read w(0) = 0,

ψ(0) = 0,

M (L) = 0,

Q(L) = 0.

(88)

According to the first two boundary condition in (88), one gets the same C3 and C4 in (83). Furthermore, application of the third boundary condition gives C2 = −

λ1 (1 + Λ2 ) Ψ cos(λ1 L) + λ2 (1 − Λ1 ) cosh(λ2 L) C1 . Ψ [λ1 (1 + Λ2 ) sin(λ1 L) + λ2 (1 − Λ1 ) sinh(λ2 L)]

(89)

With (83), (89) and the last boundary conditions in (88) at hands, we can derive the frequency equation of a radially graded cantilever tube-beam as follows: λ2 (Π2 + Λ2 ) (1 − Λ1 )2 + Ψλ1 (Π1 + Λ1 ) (1 + Λ2 )2 [( ) ( ) ] 1 − Λ1 (Π2 + Λ2 ) λ2 λ2 (Π2 + Λ2 ) + cos(λ1 L) cosh(λ2 L) + − sin(λ1 L) sinh(λ2 L) = 0 1 + Λ2 (Π1 + Λ1 ) λ1 Ψ λ1 (Π1 + Λ1 ) Ψ (90)

1+

The corresponding mode shape can be given by w = cos(λ1 x)−cosh(λ2 x)−

λ1 (1 + Λ2 ) Ψ cos(λ1 L) + λ2 (1 − Λ1 ) cosh(λ2 L) [sin(λ1 x) − Ψ sinh(λ2 x)] . Ψ [λ1 (1 + Λ2 ) sin(λ1 L) + λ2 (1 − Λ1 ) sinh(λ2 L)] (91)

Similarly, in Table 5 we give a comparison of the dimensionless frequencies between our numerical results and those by the FEM for a radially graded cantilever tube with Rin /Rout = 0.5, N = 1, v = 0.3. Fig. 9 shows the finite element results of the first three mode shapes of a radially graded cantilever tube. Table 5. Dimensionless frequencies Ω of a with Ri /Ro = 0.5, N = 1, v = 0.3 n L/Ro = 20 L/Ro = 10 Present FEM Present FEM 1 3.5007 3.5034 3.4047 3.4133 2 20.7216 20.7654 17.6878 17.8208 3 53.6195 53.9399 40.9382 41.5966

4.4

radially graded cantilever tube-beam L/Ro = 5 Present FEM 3.0945 3.3713 12.2091 12.3668 25.5707 26.1074

Effects of the gradient index on natural frequencies

Based on the characteristic equations (79), (86) and (90), dimensionless natural frequencies Ω are calculated for classic end conditions, and obtained results are listed in Tables 6-8. As seen in Tables. 19

6-8, no matter which ends, the first two dimensionless natural frequencies increase with increasing N , arriving at its maximum at a certain N value, and decrease as N continues to rise. Nonetheless, from the third-order dimensionless frequencies, they always drop as the power-law index N is raised. Therefore, material properties play a dominant role in determining natural frequencies and their trends. In other words, natural frequencies are sensitive to the volume fraction of constituents of each phase. Table 6. Dimensionless natural frequencies Ω of a simply supported functionally graded tube-beam with L/Ro = 20, v = 0.3 N 1 2 3 4 5 6 0.1 9.6951 36.6625 76.2160 123.7781 176.0439 230.9249 0.5 9.7303 36.7113 76.1077 123.2787 174.9362 229.0379 1 9.7057 36.5620 75.6586 122.3378 173.3405 226.6658 5 9.2679 34.8820 72.1064 116.4790 164.8996 215.4774 10 8.9657 33.7924 69.9714 113.2095 160.4895 209.9520 Table 7. Dimensionless natural frequencies Ω of a clamped functionally graded tube-beam with L/Ro = 20, v = 0.3 N 1 2 3 4 5 6 0.1 20.7626 52.4913 93.6300 140.4777 190.8816 260.4101 0.5 20.7675 52.3197 93.0550 139.2963 188.9609 240.7467 1 20.6660 51.9404 92.2070 137.8225 186.7600 237.7533 5 19.7020 49.4448 87.6771 130.9443 177.3402 225.6760 10 19.1005 48.0339 85.3216 127.5958 172.9746 220.2788 Table 8. Dimensionless natural frequencies Ω of a cantilever functionally with L/Ro = 20, v = 0.3 N 1 2 3 4 5 0.1 3.4959 20.7747 54.0182 96.7249 145.4473 0.5 3.5080 20.8061 53.9408 96.3141 144.4905 1 3.5007 20.7216 53.6195 95.5628 143.1424 5 3.3437 19.7673 51.0924 90.9604 136.1338 10 3.2325 19.1488 49.5801 88.4164 132.5099

5

graded tube-beam 6 197.6921 196.0227 193.9546 184.3368 179.6305

Conclusions

This paper introduces a new approach for analyzing the bending and free vibration of functionally graded beams with annular cross-section and radial gradient. By introducing a warping crosssection to meet the shear-free conditions at the inner and outer surfaces, coupled governing equations were deduced. Then a single governing differential equation was derived and all physical quantities were expressed in terms of the solution of the resulting equation. In this approach, shear deformation and rotational moment of inertia of cross-section have been both considered, and 20

it does not require knowledge of the shear correction coefficient. The frequencies equations were obtained explicitly. Numerical results were presented for analyzing the static and free vibration behaviors of a radially graded tube-beam. Some conclusions are drawn below. • A third-order beam model for bending and free vibration of straight functionally graded cylindrical tubes with radial gradient was presented. • Explicit expressions for the bending deflection, stresses in a radially graded cylindrical tube were given. • The frequencies equations for free transverse vibration of a hollow cylinder with radial gradient were obtained. • The gradient index has a remarkable effect on the deflection and stress distribution. • There is a maximum dimensionless frequency for the first two dimensionless frequencies as the power-law index N varies, and the third- and higher-order dimensionless frequencies have a declination with N rising.

Acknowledgments This work was supported by the National Natural Science Foundation of China (Grant Nos. 11672336 and 11872379).

Declarations of interest The authors declare that they have no conflict of interest.

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24

List of Figure Captions Fig. 1. Schematic of a circular cylindrical beam with radial gradient together with the corresponding coordinates. Fig. 2. Variation of material properties with Yo /Yi = 2.86 and Ri /Ro = 0.5 when taking different gradient indexes N . Fig. 3. Normalized deflection w(x)Ei I/qL4 versus the dimensionless axial coordinate x/L for different gradient indexes of a cantilever functionally graded tube-beam subjected to uniformly distributed loading with L/Ro = 10. Fig. 4. Effect of the gradient index on the distribution of the normal stress σxx (0, 0, z)I/qL2 Ro at the clamped end of a radially graded cantilever tube-beam subjected to uniformly distributed loading with L/Ro = 10. Fig. 5. The distribution of the shear stress τxz (0, y, z)L/q over the clamped end of a radially graded cantilever tube-beam subjected to uniformly distributed loading; a) N = 0.1, b) N = 1, c) N = 10, d) homogeneous aluminum tube-beam. Fig. 6. a) The meshing model used in the FEM, b) The continuously varying (left) and piecewise constant (right) material properties along the radial direction. Fig. 8. The first three mode shapes for a simply supported tube-beam with L/Ro = 10 in FEM results, a) 1st mode shape, b) 2nd mode shape, c) 3rd mode shape. Fig. 9. The first three mode shapes for a clamped tube-beam with L/Ro = 10 in FEM results, a) 1st mode shape, b) 2nd mode shape, c) 3rd mode shape. Fig. 10. The first three mode shapes for a cantilever tube-beam with L/Ro = 10 in FEM results, a) 1st mode shape, b) 2nd mode shape, c) 3rd mode shape.

25

Fig.1

Ri/Ro=0.5 3.0

Yo/Yi=2.86

N=0.1

0.5

Y/Y i

2.5

1 2.0

2

1.5

10 1.0

0.5

0.6

0.7

0.8

r/R o

Fig.2

26

0.9

1.0

0.14

0.12

Aluminum beam

0.08

0.06

0.04

N=0.1,0.5,1,2,10

0.02

0.00 0.0

0.2

0.4

0.6

0.8

1.0

x/L

Fig.3

1.0

N=0.1,0.5,1,2,10 0.8

0.6

0.4

z/R o

w(x)EiI/qL

4

0.10

-0.4

-0.6

-0.8

-1.0 -4

-3

-2

-1

s

0

(0,0,z) I/qL xx

Fig.4

27

2

1

Ro

2

3

4

Fig.5a,b

Fig.5c,d

28

2 ...

k

Fig.6a,b

Fig.7a,b

Fig.7c

29

1

Fig.8a,b

Fig.8c

Fig.9a,b

30

Fig.9c

31