Exact receptance function of a tapered afg beam with nonlinearly varying ratios of beam properties carrying concentrated masses

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This paper presents a method for establishing the exact receptance function of a tapered axially functionally graded (AFG) beam with nonlinear ratios of properties using the Adomian method. In current papers, the Adomian method was applied for linearly tapered beams where the geometric series was used conveniently.

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Nội dung Text: Exact receptance function of a tapered afg beam with nonlinearly varying ratios of beam properties carrying concentrated masses

Vietnam Journal of Mechanics, Vol. 45, No. 4 (2023), pp. 296 – 317 DOI: https:/ /doi.org/10.15625/0866-7136/19697 EXACT RECEPTANCE FUNCTION OF A TAPERED AFG BEAM WITH NONLINEARLY VARYING RATIOS OF BEAM PROPERTIES CARRYING CONCENTRATED MASSES Thao Thi Bich Dao1,2 , Khoa Viet Nguyen1,2,3,∗ , Quang Van Nguyen1 1 Institute of Mechanics, VAST, Hanoi, Vietnam 2 Graduate University of Science and Technology, VAST, Hanoi, Vietnam 3 VNU University of Engineering and Technology, Hanoi, Vietnam ∗ E-mail: nvkhoa@imech.vast.vn Received: 15 October 2023 / Revised: 27 December 2023 / Accepted: 29 December 2023 Published online: 31 December 2023 Abstract. This paper presents a method for establishing the exact receptance function of a tapered axially functionally graded (AFG) beam with nonlinear ratios of properties using the Adomian method. In current papers, the Adomian method was applied for linearly tapered beams where the geometric series was used conveniently. However, for nonuni- form AFG beams with nonlinearly varying ratios of properties, the geometric series can- not be used, thus the other type of power series needs to be established and applied. In this paper, the derivation of the power series applied for obtaining the exact receptance function of a nonuniform AFG beam with nonlinearly varying ratios of properties is pre- sented. Numerical simulation results of the receptance function of a tapered AFG beam with nonlinearly varying ratios of beam properties carrying concentrated masses are con- ducted and provided. The influences of the concentrated masses and the varying ratios of properties of beam on the receptance matrix are also investigated and presented. Keywords: receptance, frequency response function, concentrated mass, functionally graded material beam. 1. INTRODUCTION Frequency response function is used in many problems such as finite element model updating, vibration and noise control, system identification, structural damage detec- tion, dynamic optimization, numerous mechanical, aerospace and civil engineering sys- tems, etc. Mottershead et al. [1, 2] described the theory and practical application of the receptance method for vibration suppression in structures by multi-input partial pole placement and studied the measured zeros form frequency response functions and its
Exact receptance function of a tapered AFG beam with nonlinearly varying ratios . . . 297 application to model assessment and updating. A new damage detection algorithm is formulated to utilize an original analytical model and FRF data measured prior and pos- terior to damage for structural damage detection is presented by Wang et al. [3]. Based on nonlinear perturbation equations of FRF data, an algorithm has been derived which can be used to determine a damage vector indicating both location and magnitude of damage from perturbation equations of FRF data. The frequency response function of a cantilevered beam, which is simply supported in-span is determined by Gürgöze and Erol [4]. In this work, the frequency response function is obtained through a formula, which was established for the receptance matrix of discrete systems subjected to linear constraint equations. Huang et al. [5] presented a new method for system identifica- tion and damage detection of controlled building structures equipped with semi-active friction dampers through model updating based on frequency response functions. The influence of the higher-order modes on the frequency response functions (FRFs) of non- proportionally viscously damped systems is eliminated by Li et al. [6]. In this study, two power-series expansions in terms of eigenpairs and system matrices are derived to obtain the FRF matrix based on the Neumann expansion theorem. Failla [7] concerned the frequency response analysis of beams and plane frames with an arbitrary number of Kelvin–Voigt viscoelastic dampers. The theory of generalised functions within a 1D for- mulation of equations of motion are used to derive the exact closed-form expressions for beam dynamic Green’s functions and frequency response functions under arbitrary poly- nomial load, with arbitrary number of dampers. Nguyen [8] presented the general form of receptance functions of the isotropic homogeneous and axially functionally graded beams carrying concentrated masses and then compared the matrices of them. Singh et al. [9] presented a method for feedback control design using the receptance method is presented which can utilize the available partial measurements for control gain compu- tation. Non-uniform beams and concentrated mass are the two subjects concerned in study- ing the vibrations of the beam. Lenci et al. [10] used the asymptotic development method to obtain approximate analytical expressions for the natural frequencies of non-uniform cables and beams. In this work, some examples are reported to illustrate the effectiveness and simplicity of the proposed formulas. Eberle and Oberguggenberger [11] presented the bending stiffness curve of a non-uniform Euler-Bernoulli beam based on measured data from a static bending test and the calculus of variations. Hadian Jazi et al. [12] in- troduced an exact closed-form explicit solution for the transverse displacement of a non- uniform multi-cracked beam with any type of boundary conditions. A good agreement is evident when the obtained results are compared. An exact approach for free vibration analysis of a non-uniform beam with an arbitrary number of cracks and concentrated masses is proposed by Li [13]. By using the fundamental solutions and recurrence for- mulas, the mode shape function of vibration of a non-uniform beam with an arbitrary
298 Thao Thi Bich Dao, Khoa Viet Nguyen, Quang Van Nguyen number of cracks and concentrated masses are determined. Mahmoud [14, 15] presented a general solution for the free transverse vibration of non-uniform, axially functionally graded cantilevers loaded at the tips with point masses. The free vibration analysis of non-uniform and stepped axially functionally graded (AFG) beams carrying arbitrary numbers of point masses is studied by using the Myklestad method. Tan et al. [16] pre- sented an approach for free vibration analysis of the cracked non-uniform beam with general boundary conditions, whose mass per unit length and bending moment of iner- tia varying as polynomial functions. In current published works, the receptance function of beam can be established using the Adomian method in which the receptance function is expressed as a geometric series - a special case of power series. In these publications the terms ( EI )′ /EI and ( EI )′′ /EI, where EI is a polynomial function, can be expressed easily as a geometric series. How- ever, for nonuniform AFG beams with nonlinear ratios of properties these terms might not be expressed as a geometric series. In order to overcome this problem, we need to express these terms by another type of power series, not the geometric series. The aim of this paper is to present a method for establishing the receptance func- tion of a nonuniform AFG beam with nonlinear ratios of properties using the Adomian method where EI is an arbitrary polynomial function. In this paper, the derivation of the receptance function of a nonuniform AFG beam with nonlinearly ratios of properties is given in detail. The receptance function of a nonuniform AFG beam with nonlinearly ratios of properties and the influence of concentrated masses on that of the beam is inves- tigated by numerical simulations. Numerical results show that the receptance matrices are changed when masses are attached on beam. When masses are attached at peak po- sitions of the receptance matrices, these peaks will decrease significantly. The peaks and nodes of the receptance of the AFG beam move to the mass positions. These results can be useful for controlling the vibration amplitude of along the beam by using concentrated masses. Some new simulations of beam with nonlinearly ratios of properties of beam in this paper can be used for validating future works using other techniques. 2. THEORETICAL BACKGROUND The modal of an axially tapered functionally graded beam is shown in Fig. 1. Assume that the elasticity modulus E ( x ), inertia moment I ( x ) and the mass density ρ ( x ) of the beam are defined by E ( x ) = E0 (1 − α1 x n1 ) , b ( x ) = b0 (1 − α2 x n2 ) , (1) h ( x ) = h 0 (1 − α 3 x n3 ) , ρ ( x ) = ρ 0 (1 − α 4 x n4 ) ,
( ) h ( x ) = h0 1 - a 3 x n3 r ( x ) = r (1 - a x ) 0 4 n4 Where, E0, b0, h0 and r0 are Young’s modulus, the width, the depth and mass density of beam at x = 0 , respectively; 0 < a i < 1; n1 , n2 are the varying ratios of material properties. The governing equation of bending vibration of an axially tapered functionally graded beam carrying concentrated masses can be presented as follows: Exact receptance function of a tapered AFG beam with nonlinearly varying ratios . . . 299 é n ù ë û y ( é E ( x ) I ( x ) y¢¢ù¢¢ + ê µ ( x ) + å mk d ( x - xmk ) ú  = d x - x f f ( t ) (2) ) ë û where, E0 , b0 , h0 and ρ0 are Young’s modulus, the width, the depth and mass density k =1 where m the kth concentrated mass < α of beam at kxis = 0, respectively; 0 locatedi at
300 Thao Thi Bich Dao, Khoa Viet Nguyen, Quang Van Nguyen  1  2 EIϕ′′ 1 dξ 0 ... 0  0   1  2 1  0 EIϕ′′ 2 dξ ... 0  K= 4 , (5)   0 L  ... ...     1  2 0 0 ... EIϕ′′ N dξ 0 Φ (ξ ) = [ϕ1 (ξ ) , ..., ϕN (ξ )] , q (t) = [q1 (t) , ..., q N (t)] , q (t) = [q1 (t) , ..., q N (t)] T , ¨ T ¨ ¨ T and the term Φ T (ξ ) q and f¯ are the amplitudes of the response at ξ and the force at ξ f , ¯ respectively. In order to derive formula (4), the mode shape ϕ of the beam without attached masses needs to be determined. Any mode shape ϕ is determined by solving the eigen- value problem of Eq. (3) ′′ E (ξ ) I (ξ ) ϕi′′ (ξ ) − L4 ω 2 µ (ξ ) ϕi (ξ ) = 0. (6) Using the rules of differentiation, yields 2 ( EI )′ ′′′ ( EI )′′ ′′ L4 ω 2 µ ϕi′′′′ = − ϕi − ϕ + ϕi , EI EI i EI or 2 ( EI )′ ′′′ ( EI )′′ ′′ λ2 (1 − α3 ξ n2 ) ϕi′′′′ = − ϕi − ϕ + ϕi , (7) EI EI i 1 − α 1 ξ n1 ω 2 µ0 L4 where λ2 = . E0 I0 Applying the Adomian decomposition method, any mode shape ϕ is decomposed into the infinite sum of convergent series ∞ ϕi (ξ ) = ∑ Ck ξ k . (8) k =0 d4 ξ ξ ξ ξ By using the linear operators ℓ = and ℓ−1 = (...)dξdξdξdξ, we dξ 4 0 0 0 0 have ϕi (ξ ) = C0 + C1 ξ + C2 ξ 2 + C3 ξ 3 + l −1 ϕi′′′′ , or, 3 2 ( EI )′ ′′′ ( EI )′′ ′′ λ2 (1 − α3 ξ n2 ) ϕi = ∑ Ck ξ k + l −1 − EI ϕi − EI i ϕ + 1 − α 1 ξ n1 ϕi . (9) k =0
Exact receptance function of a tapered AFG beam with nonlinearly varying ratios . . . 301 ( EI )′ ( EI )′′ Since EI is the polynomial function so the terms and are the rational EI EI fraction. From the partial fraction decomposition theory, a fraction of two polynomials P (x) can be expressed in general form as follows Q (x) P (ξ ) Ak Bk x + Ck = ∑ k +∑ . (10) Q (ξ ) k (ξ − a) k (ξ 2 + bξ + c)k ( EI )′ ( EI )′′ Ak If the partial fractions of terms and consist of only ∑ k then these EI EI k (ξ − a) partial fractions can be expressed by geometric series ∞ Ak ξ k = ∑ Ak ξ˜i , ˜ ˜ where ξ = . a (11) (ξ − a) i =0 Bk x + Ck However, if those partial fractions include ∑ then they cannot be ex- (ξ 2 + bξ + c)k k pressed simply as a geometric series. Let us consider an axially tapered AFG beam with nonlinearly varying width as follows E (ξ ) = E0 (1 − α1 ξ n1 ) , b = b0 1 − α2 ξ 3 , (12) h = h0 , ρ = ρ 0 (1 − α 4 ξ n4 ) . In this case, the moment of inertia is expressed as the third polynomial function I = I0 1 − α2 ξ 3 , then ( EI )′ 1 = − n1 α1 ξ n1 −1 + 3α2 ξ 2 − (n1 + 3) α1 α2 ξ n1 +2 n1 ) (1 − α ξ 3 ) , EI (1 − α1 ξ 2 ( EI )′′ 1 = − n1 (n1 − 1) α1 ξ n1 −2 + 6α2 ξ − (n1 + 3) (n1 + 2) α1 α2 ξ n1 +1 , EI (1 − α 1 ξ n1 ) (1 − α 2 ξ 3 ) (13) in which, 1 1 1 1 β2 ξ + 2 = + , (14) (1 − α1 ξ n1 ) (1 − α 2 ξ3) 3 1 − ( β 1 ξ ) n1 1 − β2 ξ 1 + β 2 ξ + β2 ξ 2 2 √ √ where β 1 = n1 α1 , β 2 = 3 α2 with the assumption that β i ξ < 1, i = 1, 2, 3, 4.
302 Thao Thi Bich Dao, Khoa Viet Nguyen, Quang Van Nguyen The following terms in (14) can be expressed by geometric series ∞ 1 n1 = ∑ α 1 ξ kn1 , 1 − ( β1 ξ ) k =0 ∞ (15) 1 = ∑ α2 ξ k . 1 − β2 ξ k =0 However, the second term in the bracket cannot be expressed in term of geometric series. Thus, we will express this term as a power series using another method as follows. Since β i ξ < 1, i = 1, 2, using the power series one can obtain  1 ∞  A0 = 1 = ∑ Au ξ u with A1 = − β 2 (16) 1 + β 2 ξ + β2 ξ 2 2 u =0  A u = − β 2 A u −1 − β 2 A u −2 , u ≥ 2 2 The coefficients of this can be expressed as A0 = 1, A1 = − β 2 , A2 = 0, A3 = β3 , A4 = − β4 , A5 = 0, A6 = β6 , ... 2 2 2 Therefore, ∞ 1 = ∑ Cu ξ u = ξ 0 − β1 ξ 1 + β3 ξ 3 − β4 ξ 4 + β6 ξ 6 − β7 ξ 7 + ... 2 2 2 2 2 1 + β 2 ξ + β2 ξ 2 2 u =0 ∞ (17) = ∑ (−1) n bn , 0 ≤ bn < 1. n =0 This is a converged alternating series since it satisfies the convergence conditions of an alternating series bn + 1 < bn , ∀ n lim bn = 0 (18) n→∞ Applying Cauchy product, one has ∞ ∞ ∞ 1 u1 = ∑ β n1 ξ n1 ∑ ( β 2 ξ ) v + ( β 2 ξ + 2) ∑ Av ξ v (1 − α 1 ξ n1 ) (1 − α 2 ξ 3 ) u =0 1 3 v =0 v =0 (19) ∞ ∞ 1 = 3 ∑ B1i ξ i + ( β 2 ξ + 2) ∑ B1i ξ i , i =0 i =0 where m1 m1 i ∑ β1 ∑ β1 n1 j i − n1 j n1 j B1i = β2 , B2i = A i − n1 j , m1 = , (20) j =0 j =0 n1
Exact receptance function of a tapered AFG beam with nonlinearly varying ratios . . . 303 i i in which m1 = is the largest integer that is less than or equal to , p is the remain- n1 n1 i der of , and n1 ∞ ∞ ∞ ϕi ′′′ 1 (1 − α1 ξ n1 ) (1 − α 2 ξ3) = 3 ∑ ( β2 ξ )v + ( β2 ξ + 2) ∑ Av ξ v . ∑ (l + 1) (l + 2) (l + 3) Cl+3 ξ l v =0 v =0 l =0 ∞ i 1 = 3 ∑ ξ i ∑ B1i (i − j + 1) (i − j + 2) (i − j + 3) Ci− j+3 i =0 j =0 ∞ i + ( β 2 ξ + 2) ∑ ξ i ∑ B2i (i − j + 1) (i − j + 2) (i − j + 3) Ci− j+3 , i =0 j =0 ∞ ∞ ∞ ϕi ′′ 1 (1 − α1 ξ n1 ) (1 − α 2 ξ3) = 3 ∑ ( β2 ξ )v + ( β2 ξ + 2) ∑ Av ξ v . ∑ (l + 1) (l + 2) Cl+2 ξ l v =0 v =0 l =0 ∞ i 1 = 3 ∑ ξ i ∑ B1i (i − j + 1) (i − j + 2) Ci− j+2 i =0 j =0 ∞ i (21) + ( β 2 ξ + 2) ∑ ξ i ∑ B2i (i − j + 1) (i − j + 2) Ci− j+2 , i =0 j =0 ∞ ∞ ∞ m1 ϕi = ∑ ( β 1 ξ )n1 u . ∑ Cv ξ v = ∑ ξ i ∑ β 1 Ci−nj . nj 1 − ( β 1 ξ ) n1 u =0 v =0 i =0 j =0 Substituting Eqs. (21) and (13) into Eq. (7), yields ∞ i 2 ϕi ′′′′ = 3 ∑ ξ i+n1 −1 ∑ ( B1i + 2B2i ) n1 α1 (i − j + 1) (i − j + 2) (i − j + 3) Ci− j+3 i =0 j =0 ∞ i + ∑ ξ i + n1 ∑ B2i n1 α1 β2 (i − j + 1) (i − j + 2) (i − j + 3) Ci− j+3 i =0 j =0 ∞ i + ∑ ξ i +2 ∑ ( B1i + 2B2i ) 3α2 (i − j + 1) (i − j + 2) (i − j + 3) Ci− j+3 i =0 j =0 (22) ∞ i +∑ξ i +3 ∑ B2i 3α2 β2 (i − j + 1) (i − j + 2) (i − j + 3) Ci− j+3 i =0 j =0 ∞ i − ∑ ξ i + n1 +2 ∑ ( B1i + 2B2i ) (n1 + 3) α1 α2 (i − j + 1) (i − j + 2) (i − j + 3) Ci− j+3 i =0 j =0 ∞ i − ∑ ξ i + n1 +3 ∑ B2i (n1 + 3) α1 α2 β2 (i − j + 1) (i − j + 2) (i − j + 3) Ci− j+3 i =0 j =0
304 Thao Thi Bich Dao, Khoa Viet Nguyen, Quang Van Nguyen ∞ i 1 + 3 ∑ ξ i+n1 −2 ∑ ( B1i + 2B2i ) α1 n1 (n1 − 1) (i − j + 1) (i − j + 2) Ci− j+2 i =0 j =0 ∞ i + ∑ ξ i + n1 −1 ∑ B2i α1 β2 n1 (n1 − 1) (i − j + 1) (i − j + 2) Ci− j+2 i =0 j =0 ∞ i + ∑ ξ i +1 ∑ ( B1i + 2B2i ) 6α2 (i − j + 1) (i − j + 2) Ci− j+2 i =0 j =0 ∞ i + ∑ ξ i +2 ∑ B2i 6α2 β2 (i − j + 1) (i − j + 2) Ci− j+2 i =0 j =0 ∞ i − ∑ ξ i + n1 +1 ∑ ( B1i + 2B2i ) α1 α2 (n1 + 2) (n1 + 3) (i − j + 1) (i − j + 2) Ci− j+2 i =0 j =0 ∞ i − ∑ ξ i + n1 +2 ∑ B2i α1 α2 β2 (n1 + 2) (n1 + 3) (i − j + 1) (i − j + 2) Ci− j+2 , i =0 j =0 i i ∞ n1 ∞ n1 − ∑ ξ i + n2 ∑ + ∑ ξi ∑ n j n j λ2 α3 β 1 1 Ci−n1 j λ2 β 1 1 Ci−n1 j . i =0 j =0 i =0 j =0 Plugging Eq. (22) into Eq. (9) and implementing the integration operator in Eq. (9), we have: i 3 ∞ ∑ ( B1i + 2B2i ) (k − i + 1) (k − i + 2) (k − i + 3) Ck−i+3 2 ∑ Ck ξ k + 3 n1 α1 ∑ ξ k+n1 +3 i =0 ϕi = k =0 k =0 ( k + n1 ) ( k + n1 + 1) ( k + n1 + 2) ( k + n1 + 3) i ∞ ∑ B2i (k − i + 1) (k − i + 2) (k − i + 3) Ck−i+3 2 + n 1 α 1 β 2 ∑ ξ k + n1 +4 i =0 3 k =0 ( k + n1 + 1) ( k + n1 + 2) ( k + n1 + 3) ( k + n1 + 4) i ∞ ∑ ( B1i + 2B2i ) (k − i + 1) (k − i + 2) (k − i + 3) Ck−i+3 ∑ ξ k +6 i =0 + 2α2 k =0 ( k + 3) ( k + 4) ( k + 5) ( k + 6) (23) i ∞ ∑ B2i (k − i + 1) (k − i + 2) (k − i + 3) Ck−i+3 + 2α2 β 2 ∑ξ k +7 i =0 ( k + 4) ( k + 5) ( k + 6) ( k + 7) k =0 i ∞ ∑ ( B1i + 2B2i ) (k − i + 1) (k − i + 2) (k − i + 3) Ck−i+3 2 − ( n + 3 ) α 1 α 2 ∑ ξ k + n1 +6 i =0 3 1 k =0 ( k + n1 + 3) ( k + n1 + 4) ( k + n1 + 5) ( k + n1 + 6) i ∞ ∑ B2i (k − i + 1) (k − i + 2) (k − i + 3) Ck−i+3 2 − ( n + 3 ) α 1 α 2 β 2 ∑ ξ k + n1 +7 i =0 3 1 k =0 ( k + n1 + 4) ( k + n1 + 5) ( k + n1 + 6) ( k + n1 + 7)
Exact receptance function of a tapered AFG beam with nonlinearly varying ratios . . . 305 k ∞ ∑ ( B1i + 2B2i ) (k − i + 1) (k − i + 2) Ck−i+2 1 + α 1 n 1 ( n 1 − 1 ) ∑ ξ k + n1 +2 i =0 3 k =0 ( k + n1 − 1) ( k + n1 ) ( k + n1 + 1) ( k + n1 + 2) k ∞ ∑ B2i (k − i + 1) (k − i + 2) Ck−i+2 1 + α 1 β 2 n 1 ( n 1 − 1 ) ∑ ξ k + n1 +3 i =0 3 k =0 ( k + n1 ) ( k + n1 + 1) ( k + n1 + 2) ( k + n1 + 3) k ∞ ∑ ( B1i + 2B2i ) (k − i + 1) (k − i + 2) Ck−i+2 ∑ ξ k +5 i =0 + 2α2 k =0 ( k + 2) ( k + 3) ( k + 4) ( k + 5) i ∞ ∑ B2i (k − i + 1) (k − i + 2) Ck−i+2 j =0 + 2α2 β 2 ∑ ξ k +6 ( k + 3) ( k + 4) ( k + 5) ( k + 6) k =0 i ∞ ∑ ( B1i + 2B2i ) (k − i + 1) (k − i + 2) Ck−i+2 1 j =0 − α 1 α 2 ( n 1 + 2 ) ( n 1 + 3 ) ∑ ξ k + n1 +5 3 k =0 ( k + n1 + 2) ( k + n1 + 3) ( k + n1 + 4) ( k + n1 + 5) i ∞ ∑ B2i (k − i + 1) (k − i + 2) Ck−i+2 1 j =0 − α 1 α 2 β 2 ( n 1 + 2 ) ( n 1 + 3 ) ∑ ξ k + n1 +6 3 k =0 ( k + n1 + 3) ( k + n1 + 4) ( k + n1 + 5) ( k + n1 + 6) k n1 n i ∞ ∑ β 1 1 Ck−n1 i ∑ ξ k + n +4 ( k + n 2 + 1 ) ( k + n 2 + 2 ) ( k + n 2 + 3 ) ( k + n 2 + 4 ) i =0 − λ2 α3 2 k =0 k n1 n i ∞ ∑ β 1 1 Ck−n1 i ∑ ξ k +4 ( k + 1 ) ( k + 2 ) ( k + 3 ) ( k + 4 ) . i =0 + λ2 k =0 The coefficients C k where k < 4 can be determined from the boundary conditions. Let us consider a simply supported beam, the boundary conditions can be expressed as: ϕ(0) = 0, ϕ′′ (0) = 0, (24) ′′ ϕ(1) = 0, ϕ (1) = 0. (25) From Eqs. (24) and (25) we have C0 = 0, C1 ̸= 0, C2 = 0, C3 ̸= 0. (26) For k ≥4 the coefficients Ck can be calculated from the recurrent relations depending on the value of n1 as follows
306 Thao Thi Bich Dao, Khoa Viet Nguyen, Quang Van Nguyen k − n1 −3 ∑ ( B1i + 2B2i ) (k − n1 − i − 2) (k − n1 − i − 1) (k − n1 − i ) Ck−n1 −i 2 i =0 Ck = n1 α1 3 ( k − 3) ( k − 2) ( k − 1) k k − n1 −4 ∑ B2i (k − n1 − i − 3) (k − n1 − i − 2) (k − n1 − i − 1) Ck−n1 −i−1 2 i =0 + n1 α1 β 2 3 ( k − 3) ( k − 2) ( k − 1) k k −6 ∑ ( B1i + 2B2i ) (k − i − 5) (k − i − 4) (k − i − 3) Ck−i−3 i =0 + 2α2 ( k − 3) ( k − 2) ( k − 1) k k −7 ∑ B2i (k − i − 6) (k − i − 5) (k − i − 4) Ck−i−4 + 2α2 β 2 i=0 ( k − 3) ( k − 2) ( k − 1) k k − n1 −6 ∑ ( B1i + 2B2i ) (k − n1 − i − 5) (k − n1 − i − 4) (k − n1 − i − 3) Ck−n1 −i−3 2 i =0 − ( n1 + 3) α1 α2 3 ( k − 3) ( k − 2) ( k − 1) k k − n1 −7 ∑ B2i (k − n1 − i − 6) (k − n1 − i − 5) (k − n1 − i − 4) Ck−n1 −i−4 2 i =0 − ( n1 + 3) α1 α2 β 2 3 ( k − 3) ( k − 2) ( k − 1) k k − n1 −2 ∑ ( B1i + 2B2i ) (k − n1 − i − 1) (k − n1 − i ) Ck−n1 −i 1 i =0 + α1 n1 ( n1 − 1) 3 ( k − 3) ( k − 2) ( k − 1) k k − n1 −3 ∑ B2i (k − n1 − i − 2) (k − n1 − i − 1) Ck−n1 −i−1 1 i =0 + α1 β 2 n1 ( n1 − 1) 3 ( k − 3) ( k − 2) ( k − 1) k k −5 ∑ ( B1i + 2B2i ) (k − i − 4) (k − i − 3) Ck−i−3 + 2α2 i=0 ( k − 3) ( k − 2) ( k − 1) k k −6 ∑ B2i (k − i − 5) (k − i − 4) Ck−i−4 + 2α2 β 2 i=0 ( k − 3) ( k − 2) ( k − 1) k k − n1 −5 ∑ ( B1i + 2B2i ) (k − n1 − i − 4) (k − n1 − i − 3) Ck−n1 −i−3 1 i =0 − α1 α2 ( n1 + 2) ( n1 + 3) 3 ( k − 3) ( k − 2) ( k − 1) k k − n1 −6 ∑ ( B1i + 2B2i ) (k − n1 − i − 5) (k − n1 − i − 4) Ck−n1 −i−4 1 i =0 − α1 α2 β 2 ( n1 + 2) ( n1 + 3) 3 ( k − 3) ( k − 2) ( k − 1) k k − n2 −4 k −4 n1 n1 ∑ βn1 i Ck−n2 −n1 i−4 1 n i ∑ β 1 1 Ck−n1 i−4 i =0 i =0 − λ2 α3 + λ2 . (27) ( k − 3) ( k − 2) ( k − 1) k ( k − 3) ( k − 2) ( k − 1) k
Exact receptance function of a tapered AFG beam with nonlinearly varying ratios . . . 307 Therefore, ϕi in Eq. (23) can be expressed as a linear function of C1 and C3 . Since coefficients C k are determined from the recurrent equation (27), the coefficients C k are the functions of C1 , C3 and λ, thus ϕi is also a function of C1 , C3 and λ. By substituting Eq. (27) into Eq. (24)–(25), we have ∞ ∑ Ck = 0 ⇔ f 11 (λ)C1 + f 12 (λ)C3 = 0, (28) k =0 ∞ ∑ (k + 1) Ck+1 = 0 ⇔ f 21 (λ)C1 + f 22 (λ)C3 = 0. (29) k =0 In order to have non-trivial C1 and C3 of Eqs. (28) and (29), the following condition must be satisfied f 11 (λ) f 22 (λ) − f 12 (λ) f 21 (λ) = 0. (30) Solving Eq. (30), the dimensionless frequency λi will be determined. Substituting the solution λi into Eqs. (28), C3 can be calculated as the function of a given C1 f 11 (λ) C3 = − C1 . (31) f 12 (λ) The nth mode shape ϕn corresponding to λn is calculated from Eq. (23). Once the mode shape ϕi is determined, the following relations is derived ∞ ∞ ∞ k ϕi2 (ξ ) = ∑ Ck ξ k ∑ Ck ξ k = ∑ ξ k ∑ Ck−i Ci , (32) k =0 k =0 k =0 i =0 ∞ ∞ ∑ Ck ξ k (k + 1) (k + 2) ∑ Ck ξ k (k + 1) (k + 2) 2 ϕ′′ i (ξ ) = k =0 k =0 ∞ k (33) = ∑ ξ ∑ Ck−i+2 Ci+2 (i + 1) (i + 2) (k − i + 1) (k − i + 2) k k =0 i =0 1 1 ∞ k 0 µϕi2 dξ = 0 µ0 (1 − α2 ξ n ) ∑ ξ k ∑ Ck−i Ci dξ k =0 i =0 (34) ∞ µ0 [(k + 1) (1 − α2 ) + n + 1] k =∑ ∑ Ck−i Ci , k =1 ( k + 1) ( k + n + 1) i =0
308 Thao Thi Bich Dao, Khoa Viet Nguyen, Quang Van Nguyen 1 2 IEϕ′′ i dξ = 0 1 ∞ k = 0 IE0 (1 − α1 ξ n ) ∑ ξ k ∑ Ck−i+2 Ci+2 (i + 1) (i + 2) (k − i + 1) (k − i + 2) dξ k =0 i =0 ∞ IE0 [(k + 1) (1 − α1 ) + n + 1] k = ∑ ( k + 1) ( k + n + 1) ∑ Ck−i+2 Ci+2 (i + 1) (i + 2) (k − i + 1) (k − i + 2). k =1 i =0 (35) Substituting Eqs. (32)–(35) into Eq. (5) the matrices M and K and the receptance formula (4) will be determined. 3. NUMERICAL SIMULATION 3.1. Reliability of the theory To justify the formula of receptance function for the AFG beam, a cantilever beam consisting of two constituent materials of aluminum and zirconia presented in [14] is ap- plied. The material properties are E Al = 70 GPa, ρ Al = 2702 kg/m3 for aluminum, and Ez = 200 GPa, and ρz = 5700 kg/m3 for zirconia. Frequency parameters were obtained using ωi L2 12ρs Ωi = . To examine the convergence of the solution, a cantilever beam with h2 Es the material properties varying linearly when ratio n = 1 is considered. The coefficients C k can be approximated by the N-term truncated series, Eq. (17) can be expressed as N ϕ (ξ ) = ∑ Ck ξ k . (36) k =0 Calculations show that when the 100-term truncated series in Eq. (36) is applied the following condition is met Ω1 − Ω1 −1 < ε = 6.055E−10. N N (37) Table 1. Frequency parameter of the cantilever AFG beam Case Frequency parameter Ref. [14] Present paper Error (%) Alumina-Zirconia Ω1 5.888 5.888 0 Ω2 26.060 26.060 0 Ω3 64.939 64.941 0.003080 Ω4 122.532 122.531 0.000816 Ω5 199.075 199.176 0.050710
wo methods are listedthe Table 1. Therefore, in series summation limit N in Eq. (37) will be truncated to N=100 in all the calculations in this work. Five lowest frequency parameters of the cantilever AFG beam obtained by two methods are listed in1. Frequency parameter of the cantilever AFG beam Table Table 1. Case Table 1. Frequency parameter of the cantilever AFG beam Error Frequency Ref. [17] Present paper parameter (%) Case Frequency Alumina-Zirconia Exact receptance 1 Ref.5.888 [17] Present5.888 paper Ω function of a tapered AFG beam with nonlinearly varying ratios . . . Error 0 309 parameter Ω2 26.060 26.060 (%) 0 Alumina-Zirconia Ω1 5.888 5.888 0 Therefore, the series Ω 3 summation limit N64.939 (37) will be truncated to N 0.003080 in Eq. 64.941 = 100 in Ω2 26.060 26.060 0 Ω4 122.532 122.531 all the calculations in this work. Five lowest frequency parameters of the cantilever AFG 0.000816 Ω3 64.939 64.941 0.003080 Ω5 beam obtained by two methods are listed in Table 1. 199.075 199.176 0.050710 Ω4 122.532 122.531 0.000816 The excellentexcellent agreementΩ 5 the frequency 199.075 between the present work and 199.176 The agreement of the frequency parameters between the present work and Ref. [17] of parameters 0.050710 presented in table 1 Ref. [14] verifies the reliability of the proposed method. presented in Table 1 verifies the reliability of the proposed method. The excellent agreement of the frequency parameters between the present work and Ref. [17] 3.2. Influence inInfluence of the the reliability of theof on the receptance presented of tablevarying propertiesproperties proposed method.receptance 3.2. the 1 verifies varying of material material on the 3.2.Fig. 2 presents the 3D theproperties of material on thethe simply supported beam incases: n2 = 1 Influence of the varying 3D graphs of receptance of receptance Fig. 2 presents graphs of receptance of the simply supported beam in two two and n2 = 3 when the = 1 and the= 3 when the forcing the first, the secondthe first, the second andcases: n2 = 1 cases:2npresents nfrequencies equal to frequencies equal tosupported third natural frequencies Fig. 2 forcing 2 3D graphs of receptance of the simply and the beam in two the of beam, n2 = 3 when the frequencies 2e beam,equal to the first, the second 2(e)nshow the receptance 2f show and third natural forcing frequencies respectively. Figs. beam when = third natural frequencies respectively. Figs. 2a, 2c, of show the receptance of 2(a), 2(c), and 2the 1. Figs. 2b, 2d, he receptancerespectively.nFigs. 2a, 2c, 2e ratios2(f) material properties n2 = 3.nAs can the varying 2f Figs. of beam, beam when 2 =theFigs. 2(b), show the show the receptance when 2 when be seen2d, show of of beam when 1. varying 2(d), of receptance of beam of beam = 1. Figs. 2b, from 2a and 2b, when the material frequency is2equal As thematerial properties n2(a) 3. Asreceptance the Figs. the receptance of beam when the varying 3. to can first natural frequency, and 2(b), be seen from in ratios of forcing properties n = ratios of be seen from Figs. 2 = the can when matrices both 2a andare maximumforcingequal to of the beam. When the forcing frequency is equal to the second cases 2b, when the at the middle theis equal to the first natural frequency, the receptance matrices in forcing frequency is frequency first natural frequency, the receptance matrices in both natural frequency, the receptance the middle of maximum When the forcingof L/4 and equal from the left both cases are maximum at at matricesof the beam. Whenthe positions frequency is equal to second cases are maximum the middle are the beam. at the forcing frequency is 3L/4 to the end of the beam as can the observed from Figs.are maximum at the positions of at matrices are minimum natural the second natural frequency, the receptance2d. While, the receptance the positions of the left frequency, be receptance matrices 2c and matrices are maximum L/4 and 3L/4 from at theend of the beam 3L/4 fromobserved from the beamandn2d. beand nthe receptance matrices difference of middle of and asHowever, the peaksof Figs. 2c as can While, 2 = 3fromdifferent: and are minimum L/4 beam. can be the left end in two cases 2 = 1 observed are Figs. 2(c) the 2(d). wo peaks inWhile,of beam.1However, the peaks case ofcases 3.2middle ofbe explained thatthe peaks section at the middle thenreceptance matrices are minimum at= n Thisand n2beam. However, the cross the case 2 = are larger than the in two n2 the = 1 can = 3 are different: the difference of at the two peaks the case n= 1 1 = 1 = 3 far morecase of difference of two n2 = in the case 2cross larger two peakstwoin the n2 2 = n2aren2 are are different: then2 =than the casepeaks 3 resultingnin thesection in in cases case and larger than the different 3. This can be explained that the = 1 difference between the two peaks 2 =n1 = case nmore be explained that the cross = 3 n = at the two 2e and at the two larger than the case of theare far 2 can different than the case n2case resulting in the larger are peaks in the case n of 2 3. This = 1 in comparison with the section 3. Graphs 2f show thepeaks in the case n2 peaks farcases when 1 in comparison with the equalin= 3. larger 2e and difference between the two= 1 are of the case n2 =the forcingcase n2 = 3 resulting ntothe Graphsnatural receptance matrices in two more different than the frequency is case the third 2f show the receptance matrices inpeakscases when the=forcing frequencywith the case n = 3. natural two ofmaximum at the positions of about L/6, the third 5L/6, requency. As presented, the receptances are the case n2 1 in comparison difference between the two is equal to 3L/6 and frequency. As presented,showreceptances are matrices in two cases when the about L/6, 3L/6 and 5L/6, Figs. 2(e) and 2(f) the the receptance maximum at the positions of forcing frequency while they are minimum at the positions of 2L/6 and 4L/6. Similar to the case when the forcing frequency while they are minimum at natural frequency. Asand 4L/6. Similar to the case when the forcing frequency is second natural the positions of 2L/6 of the receptance in the are of the linearly s equal to theequal to the thirdfrequency, the peaks presented, the receptancescasemaximum at the varying is equalpositions of about L/6,frequency,5L/6, while they are minimum at the positions of 2L/6 varying to the second natural the peaks of the receptance in the case of the linearly parameters is smaller thanthan that3L/6 and varying parameters. It It is concludedthat, the positions of parameters is smaller that of nonlinearly varying parameters. is concluded that, the positions of of nonlinearly maxima andand 4L/6.of the receptances obtained at any natural frequency coincide with the positions of maxima minima Similar to the case when the forcing frequency is equal to the second natural and minima of the receptances obtained at any natural frequency coincide with the positions of maxima andfrequency, theof corresponding mode shape. case of the linearly varying parameters is maxima minima of the the corresponding mode shape. and minima peaks of the receptance in the (a) n2 = 1, ω = ω1 (b) n2 = 3, ω = ω1
14 Thao Thi Bich Dao, Khoa Viet Nguyen, Quang Van Nguyen Thao Thi Bich Dao, Khoa Viet Nguyen, Quang Van Nguyen a) n2=1, =1, ω=ω a) n ω=ω1 b) b)=3, ω=ω1 n2 n =3, ω=ω 2 1 2 1 310 a)a)Thao Thi Bich Dao, Khoa Viet Nguyen, Quang Van Nguyenn22=3,ω=ω11 n2n2=1, ω=ω1 =1, ω=ω1 b) n =3, ω=ω b) c) n2=1,2=1, ω=ω2 c) n ω=ω2 d) d) =3, ω=ω2 2 n2 n2=3, ω=ω (c) n2 = 1,n2=1, ω=ω2 c) ω = ω2 (d) n2 = 3, n2=3, 2 d) ω = ω ω=ω2 c) n2=1, ω=ω2 d) n2=3, ω=ω2 (e) n2 = 1, ω=1,ω3 e) n = ω=ω (f) n2 = 3,f) n =3, ω=ω ω = ω3 e) n2e) n22ω=ω3 3 =1, =1, ω=ω 3 e) n2=1, ω=ω3 f) n =3, ω=ω3 2 ω=ω f) n22=3, ω=ω 3 3 f) n2=3, 3 Fig. Fig. 2. Receptance matrices ofofAFG beam when thewidth varies Fig. 2. Receptance matrices of AFG beam when the width varies Fig. 2. Receptance matrices AFG beam when the width varies 2. Receptance matrices of AFG beam when the width varies Fig. 2. Receptance matrices of AFG beam when the width varies In order to give a clearer comparison of the differences between two peaks in in the two cases n2 In order to give a clearer comparison of the differences between two peaks the two cases n2 In order to give a clearer comparison of the differences between two peaks in the two cases n n2 in the two cases = = 1 andnn2==3, to give a clearer comparison ofparameters.fixedbetween two peaks L/3, are extracted 2and 1 andIn 2order the that of nonlinearly varyingtheforce is fixed atat the position the positions of and when the differences the position of L/3, smaller than receptance curves, when the forcefixed at concluded that, of are extracted and 3, the receptance is It is = 1 and and=n2 =the the receptance curves, when the force is fixed atthe position of L/3, that extracted and = 1 n2 3, in 2Dgraph as shown in Fig. 3. The force position of L/3 isis chosen of ensure that the excitation presented 2Dreceptance curves, when The force is the the chosen toto L/3, are the excitation position ensure are extracted presented in 3, and minima of the receptances force position ofnatural frequency coincide with graph as shown 3. any L/3 presented in applied graph shown of receptance. The Graph 3a shows the chosen to ensure that thethe forcing presented in 2Dat the nodes of receptance. forceGraph 3a shows is receptance curve when excitation 2D graph as maxima in Fig. 3. The force obtained atof L/3 is chosen to ensure that the excitation position isis notthe positions ofas shown in Fig. 3. Theof the corresponding mode not applied at the nodes and minimaThe position of L/3 the receptance curve when the forcing is not frequency isequal to the firstreceptance. The Graph peak of receptance shape.case n2=3=3 moves slightly applied at the the to the first natural frequency. The 3ashows receptance in the case when the slightly is not applied at nodes of of natural frequency. The peak of the receptance curve n moves forcing frequency is equal nodes receptance. The Graph 3a shows the receptance curve when the forcing maxima in the 2 frequency isleftendorder to givecomparision with theThe peakof As can bebetweenFigs. n22=3 moves slightly the frequency isendto thethe in natural frequency.thecase the =1.receptance in the case 3b3b and the slightly toto the equal of beam firstanatural frequency. casenn=1. receptance seen intwo peaksmoves when equal to first comparision with The of 2 2 differences be in in case n =3 in 3c, the left In of beam in clearer comparison peak of As can seen Figs. and 3c, when the to the forcing frequencyis equal to2the3, the with the case 2=1. As can beforce is fixed at3b and two when the the to left left endbeam is 1comparision withand third curves, frequencies, the differencetheof 3c, peaks in forcing frequency2 = equal to the second and thirdnatural when the seen indifferenceand 3c, when the the end of of beam in comparision receptance nn2=1. As can be seen Figs. 3b two peaks in two cases n in and n = second the case natural frequencies, the Figs. of position forcing frequency higherequalthe andsecond as mentioned. frequencies, the difference of two peaks in the forcingnfrequencyequal to thethe presented in 2Dnatural as shown in the difference of two peaks in the case n =1L/3, is is than to case n2=3 and third graph case 2of=1 ishigher than thecase n2=3 as third natural frequencies, Fig. 3. The force position 2 is are extracted second and mentioned. casecase n2=1 L/3 isthan thetheensure 2thatas mentioned. is not applied at the nodes of receptance. n2=1 is higher chosen to case =3=3 mentioned. of is higher than case n2 n as the excitation Fig. 3(a) shows the receptance curve when the forcing frequency is equal to the first nat- ural frequency. The peak of receptance in the case n2 = 3 moves slightly to the left end of beam in comparison with the case n2 = 1. As can be seen in Figs. 3(b) and 3(c), when the forcing frequency is equal to the second and third natural frequencies, the difference of two peaks in the case n2 = 1 is higher than the case n2 = 3 as mentioned.
= 1 and and n3,= 3, receptance curves, when the the forcefixed at the the position L/3, areare extracted an = 1 n2 = 2 the the receptance curves, when force is is fixed at position of of L/3, extracted and presented in 2D graph as shown in Fig.Fig. 3. The force positionL/3L/3 is chosenensure that thethe excitatio presented in 2D graph as shown in 3. The force position of of is chosen to to ensure that excitation is not applied at the nodes of receptance. TheThe Graph shows the the receptance curve when the forcin is not applied at the nodes of receptance. Graph 3a 3a shows receptance curve when the forcing frequency is equal to the firstfirst natural frequency. The peakreceptance in thethe case=32=3 moves slightl frequency is equal to the natural frequency. The peak of of receptance in case n2 n moves slightly to the left end end of beam in comparision with the case n2=1. As can seen in Figs. 3b 3b and 3c, when th to the left of beam in comparision with the case n2=1. As can be be seen in Figs. and 3c, when the forcing frequency is equal to the second andand third natural frequencies, the difference two peaks in theth forcing frequency is equal to the second third natural frequencies, the difference of of two peaks in casecase nis higher thanthan case n2=3 2as mentioned. beam with nonlinearly varying ratios . . . n2=1 2=1 is higher the the case function asa mentioned. Exact receptance n =3 of tapered AFG 311 15 Exact receptance function of a tapered AFG beam with nonlinearly varying ratios of beam properties carryin concentrated masses a) ω=ω1ω = ω1 (a) (b) ω = ω2 b) ω=ω2 (c) ω ω=ω c) = ω3 3 Fig. 3. Receptance curves of beams without attached mass without attached mass Fig. 3. Receptance curves of beams (Solid line: n2 = 1; Dotted line: n2 = 3) Solid line: n2=1; Dotted line: n2=3 3.3. Influence of the concentrated masses and the varying properties of material on the 3.3. Influence receptance of the concentrated masses and the varying properties of material on the receptance In thisIn this paper, equal concentrated masses of 0.6 kg are attached on the simply support paper, two two equal concentrated masses of 0.6 kg are attached on the simply AFG beam in the case n2 = 3 in sevenn2scenarios asscenarios asTable in Table 2. The recep- matrices a supported AFG beam in the case = 3 in seven listed in listed 2. The receptance calculated tance matrices are calculated at on the beam while the force moves along these points. at 100 points spaced equally 100 points spaced equally on the beam while the force moves along these points. Table 2. Seven scenarios of attached masses The receptance matrices of the AFG beams are changed when there are concentrated masses. Figs. 4(a) and 4(b) present the 3D and 2D graphs of the receptanceof m2 beam Scenario ω Position of m1 Position of AFG carrying a mass at position 3L/4 when the forcing frequency is equal to the first natural 1 ω=ω1 3L/4 - frequency of the beam-mass system. The influence of the mass on the receptance matrices 2 ω=ω2 3L/4 - 3 ω=ω3 L/6 - 4 ω=ω3 3L/6 -
312 Thao Thi Bich Dao, Khoa Viet Nguyen, Quang Van Nguyen 16 16 ThaoTable 2. Bich Dao, Khoa Viet Nguyen, Quang Van Nguyen Thao BichSeven scenarios of attached masses Thi Thi Dao, Khoa Viet Nguyen, Quang Van Nguyen Scenario ω Position of m1 Position of m2 1 ω = ω1 3L/4 - 2system. As can can= ωseen from these graphs, the peaks the the mass position are dropped beam-mass system. As be be beam-mass 3L/4 - ω seen2 from these graphs, the peaks at at mass position are dropped 3 and peaksthe ω = ω3 move to the theL/6 where the massattached. - remarkably and peaks of of receptance remarkably the receptance move to side where the mass is is attached. side 4 ω = ω3 3L/6 - When two two masses located at the the two peaksreceptance andand the forcing frequencyequal 5When masses are are located at two peaks 5L/6 receptance the forcing frequency is is equal ω = ω3 of of - to the third6natural frequency of the the beam-mass system, the peaks corresponding these positions to the third natural frequency of beam-mass system, the peaks corresponding to to these positions ω = ω3 L/6 3L/6 decrease as 7 as demonstratedω = ω 7. Figs.andand present the the receptancebeam when two masses areare decrease demonstrated in Fig.Fig. in 7. Figs. 7a 7a 7b 7b present receptance of of beam when two masses L/6 5L/6 3 located at the positions of L/6 L/6 and 3L/6. can canobserved from these graphs, thethe peaks corresponding located at the positions of and 3L/6. As As be be observed from these graphs, peaks corresponding to these positions decrease significantly andand peaks and nodes the the receptance curve moves thethe left to these positions decrease significantly peaks and nodes of of receptance curve moves to to left sidesmall so that two masses attached. When twotwo masses are locatedthe the4(a). However, 5L/6, is side where two masses are are attached. When masses 3D located at at positions of L/6L/6 and 5L/6, where the the it is difficult to be observed from are graph in Fig. positions of and the fourfour peaks the mass on these positions decrease significantly as shown in thethe Figs. and 7d.7d.is is influencecorresponding to thethese positions decrease significantly as shown in Figs. 7c 7c and It It peaks of corresponding to receptance can be inspected visually as presented in Fig. 4(b) noted that that firstfirst node of receptance moves with the the first mass position whichclose to that node when the receptance curves are moves clearly to to force applied at which is is close to that node noted the the node of receptanceextracted clearlythe first mass position0.5L. When there is an and and second node 3L/4 thethe second mass position. the the second node moves to the second mass position. attached mass at moves to peak of receptance moves to the right side of the beam where the mass is attached. a) Receptance matric (a) Receptance matric a) Receptance matric b) Receptance curve (b) Receptance curve b) Receptance curve Fig.Fig. 4. Receptance of beambeam when ω=ω1, mass position: 3L/4 AFG Fig. 4. 4. Receptance of AFG beam when ω=ω11 ,mass position: 3L/4 Receptance of AFG when ω = ω , mass position: 3L/4 Dotted line: without mass; Dashed line: attached mass (Dotted line: without mass; Dashed line: attached mass Dotted line: without mass; Dashed attached mass) When the mass is attached at the position of 3L/4 and the forcing frequency is equal to the second natural frequency, the peak at the mass position is pressed down signif- icantly. The node of the receptance curve moves toward the right of beam where the mass is attached as can be observed from Figs. 5(a) and 5(b). Figs. 6 illustrate the 2D and 3D graphs of the receptance of AFG beam when the mass is attached at L/6, L/2 and 5L/6 and the forcing frequency is equal to the third natural frequency of the beam-mass system. As can be seen from these graphs, the peaks at the mass position are dropped remarkably and peaks of the receptance move to the side where the mass is attached. a. Receptance matric b. Receptance curve a. Receptance matric b. Receptance curve Fig. 5. Receptance of AFG beam when ω=ω2, mass position: 3L/4 Fig. 5. Receptance of AFG beam when ω=ω2, mass position: 3L/4 Dotted line: without mass; Dashed line: attached mass
a) a) Receptance matric Receptance matric b) Receptance curve b) Receptance curve Fig. 4. 4. Receptance of AFG beam when ω=ω,1,mass position: 3L/4 Fig. Receptance of AFG beam when ω=ω1 mass position: 3L/4 Exact receptance function of a tapered AFG beam with nonlinearly varying ratios . . . 313 Dotted line: without mass; Dashed line: attached mass Dotted line: without mass; Dashed line: attached mass 17 17 17 receptance function of a tapered AFG beam with nonlinearly varying ratios of beam properties carrying Exact 17 Exact receptance function of of tapered AFG beam with nonlinearly varying ratios of beam properties carrying Exact receptance function a a tapered AFGconcentrated masses varying ratios of beam properties carrying beam with nonlinearly Exact receptance function of a tapered AFG beam with nonlinearly varying ratios of beam properties carrying concentrated masses (a) a. Receptance matric concentrated masses (b) Receptance curve Receptance matric a. Receptance matric concentrated masses b. Receptance curve b. Receptance curve Fig. 5. 5. 5. Receptance of AFG beam when ω=ω,2, massposition: 3L/4 Fig. position: 3L/4 Fig. Receptance of AFG beam when ω = ω2 ,2mass position: 3L/4 Receptance of AFG beam when ω=ω mass (Dotted line: without mass; Dashed line: attached mass) Dotted line: without mass; Dashed line: attached mass Dotted line: without mass; Dashed line: attached mass a. Receptance matric, mass position L/6 a. Receptance matric, mass position L/6 b. Receptance curve, mass position L/6 (b) Receptance curve, mass position L/6 L/6 b. Receptance curve, mass position L/6 a. Receptance matric, mass position L/6 (a) Receptance matric, mass b. Receptance curve, mass position a. Receptance matric, mass position L/6 b. Receptance curve, mass position L/6 c. Receptance matric, mass position L/2 L/2 (c) Receptance matric, mass position c. Receptance matric, mass position L/2 (d)d. Receptance curve, mass position L/2 d.Receptance curve, mass position L/2 Receptance curve, mass position L/2 c. Receptance matric, mass position L/2 d. Receptance curve, mass position L/2 c. Receptance matric, mass position L/2 d. Receptance curve, mass position L/2
c.314 Receptance matric, mass position L/2 Khoa Viet Nguyen, Quang Van Nguyen curve, mass position L/2 Receptance matric, mass Thao Thi Bich Dao, c. position L/2 d. Receptance d. Receptance mass position L/2 (e) Receptancematric, mass position 5L/6 e. Receptance matric, mass position 5L/6 f. Receptance, mass position 5L/6 (f) Receptance curve, mass position 5L/6 e. Receptance matric, mass position 5L/6 f. Receptance, mass position 5L/6 Fig. 6. Receptance of AFG beam when ω=ω3 Fig. 6. Receptance of AFG beam when ω=ω Fig. 6. Receptance of AFG beam when ω = ω3 3 Dotted line: without mass; Dashed line: attached mass (Dotted line: without mass; Dashed line: attached mass) Dotted line: without mass; Dashed line: attached mass When two masses are located at the two peaks of receptance and the forcing fre- quency is equal to the third natural frequency of the beam-mass system, the peaks corre- sponding to these positions decrease as demonstrated in Fig. 7. Figs. 7(a) and 7(b) present the receptance of beam when two masses are located at the positions of L/6 and 3L/6. As can be observed from these graphs, the peaks corresponding to these positions decrease significantly and peaks and nodes of the receptance curve moves to the left side where the two masses are attached. When two masses are located at the positions of L/6 and 18 18 5L/6, four peaks corresponding to these Viet Nguyen, QuangQuang Van Nguyenas shown in the Thao Thi Bich Dao, Khoa positionsNguyen, Van Nguyen Thao Thi Bich Dao, Khoa Viet decrease significantly Figs. 7(c) and 7(d). It is noted that the first node of receptance moves clearly to the first mass position which is close to that node and the second node moves to the second mass position. (a) a. Receptance matric, masses at and L/2 L/2 L/2 Receptance, masses at L/6 at L/6 and L/2 Receptance matric, masses at L/6 L/6 at L/6 and b. (b) Receptance, masses at L/6 and L/2 a. Receptance matric, masses and b. Receptance, masses and L/2
a. Receptance matric, masses at aL/6 and L/2 withb. Receptance, ratios . . . at L/6 and L/2 Exact receptance function of tapered AFG beam nonlinearly varying masses 315 a. Receptance matric, masses at L/6 and L/2 b. Receptance, masses at L/6 and L/2 (c) Receptance matric,matric, masses at L/6 and 5L/6 c. Receptance masses at L/6 and 5L/6 (d) Receptance, masses at L/6 at L/6 and 5L/6 d. Receptance, masses and 5L/6 c. Receptance matric, masses at L/6 and 5L/6 d. Receptance, masses at L/6 and 5L/6 Fig. 7. Receptance of AFG beam when ω=ω3, two masses are attached Fig. 7.7. Receptance of AFG beam when ω=ω3, two masses are attached Fig. Receptance of AFG beam when ω = ω3 , two masses are attached (DottedDotted line: without mass; Dashed attached masses) line: without mass; Dashed line: line: attached masses Dotted line: without mass; Dashed line: attached masses 4. CONCLUSION 4. CONCLUSION In this paper, the exact formula CONCLUSION 4. of receptance function of a tapered AFG beam with varying In this paper, the exact formula of receptance function of a tapered AFG beam with In this paper, the exact in of an of receptance polynomialof presented. When beam with )¢ / EI varying parametersingiven formulaform of an arbitrary ordera polynomial is the term ( EI varying parameters given the form the arbitrary order function is tapered AFG presented. When given /in (cannot be expressed)as/EI cannot be expressed as a geometric series ( EI )¢ EI ′ and ( EI )¢¢ EI EI ) /EI and ( EI ′′ parametersthe term the form of an arbitrary geometric series as presented in published the term as will /be a order polynomial is presented. When works, they presented in published works, they will be expressed as a converged alternative power )¢¢ EI numerical simulations of power series. The numerical simulations of receptance ( EIexpressed as a converged alternativereceptance of the presented in published works, they of the and series. /The cannot be expressed as a geometric series as simply supported nonlinear AFG will be simply supported nonlinear AFG beam as an example are provided. expressed as an example are provided.power the exciting frequency is equal to the natural frequencies, the beam as Numerical results show that when series. The numerical simulations of receptance of the a converged alternative simply supported nonlinear AFG beam asof the receptances are the same with the maxima and minima maximum and minimum positions an example are provided. positions of the corresponding mode the the exciting frequencyequal to the the natural fre- Numerical results show that when is equal to Numerical results show that when shapes. The influence of the concentrated masses on the receptance exciting frequency is natural frequencies, the quencies, AFGmaximum and minimum positionsare the receptances are the same with the the beam is also investigated. When there of concentrated massese, the shape of receptance of of the minimum positions of the receptances are the same with the maxima and minima maximum and positions of the corresponding mode of the corresponding mode concentrated attached on of the maxima and minima positions shapes. The influence of the shapes. The masses massesreceptance the tapered AFG beam is changed. The peaks of receptance at the positions of influence the decrease. of the AFG peaks massesinvestigated. Whenof the AFGconcentrated move toward the When there Theof concentrated and nodes of thereceptance of the AFGbeam tendalso investigated. mass positions. The beam is also on the receptance there are beam is to massese, the shape of receptance are concentrated massese, the can be used to validate future work using other methods and they simulation results in this workshape of receptance of the tapered AFG beam is changed. can be the tapered AFG beam is changed. The peaks of receptance at the positions of attached masses decrease. TheThe peaks nodes of the receptance of the AFG beam tend to move toward The mass positions. The peaks and of receptance at the positions of attached masses decrease. the peaks and simulation results in this work the AFG beam tend to move toward the mass positions. they nodes of the receptance ofcan be used to validate future work using other methods andThe can be simulation results in this work can be used to validate future work using other methods and they can be useful for controlling the vibration amplitude at some specific points along the beam by applying concentrated masses. DECLARATION OF COMPETING INTEREST The authors declare that they have no known competing financial interests or per- sonal relationships that could have appeared to influence the work reported in this paper.