Giám sát hệ xe cầu có vết nứt dạng đóng mở : Luận văn ThS. Kỹ thuật cơ khí: 60 52 02

ĐẠI HỌC QUỐC GIA HÀ NỘI TRƢỜNG ĐẠI HỌC CÔNG NGHỆ HÀ NỘI PHẠM THỊ MAI HOA GIÁM SÁT HỆ XE CẦU CÓ VẾT NỨT DẠNG ĐÓNG MỞ LUẬN VĂN THẠC SỸ CƠ HỌC KỸ THUẬT Hà Nội – 2012 ĐẠI HỌC QUỐC GIA HÀ NỘI TRƢỜNG ĐẠI HỌC CÔNG NGHỆ HÀ NỘI PHẠM THỊ MAI HOA GIÁM SÁT HỆ XE CẦU CÓ VẾT NỨT DẠNG ĐÓNG MỞ Ngành: Cơ học kỹ thuật Chuyên ngành: Cơ học kỹ thuật Mã số: 60 52 02 LUẬN VĂN THẠC SỸ CƠ HỌC KỸ THUẬT NGƢỜI HƢỚNG DẪN KHOA HỌC: TS - Nguyễn Việt Khoa Hà Nội - 2012 LỜI CAM ĐOAN Tôi xin cam đoan: Luận văn: “Giám sát hệ xe cầu có vết nứt dạng đóng mở” cơng trình nghiên cứu riêng Các số liệu nêu trích dẫn luận văn trung thực Kết nghiên cứu đƣợc trình bày luận văn chƣa đƣợc cơng bố cơng trình khác Hà Nội, ngày 20 tháng năm 2012 Tác giả luận văn Phạm Thị Mai Hoa LỜI CẢM ƠN Lời tác giả xin bày tỏ lòng biết ơn sâu sắc tới TS Nguyễn Việt Khoa - cán hƣớng dẫn Thầy tận tình bảo giúp đỡ nhiều suốt trình làm luận văn Qua đó, tơi học tập tích lũy đƣợc nhiều kiến thức bổ ích Thầy bảo cho phƣơng pháp nghiên cứu khoa học kinh nghiệm vô quý giá Tác giả xin chân thành cảm ơn tới cán Khoa Cơ học kỹ thuật Tự động hóa-Viện Cơ học tạo điều kiện thuận lợi giúp đỡ suốt trình học tập hồn thành luận văn Tơi xin cảm ơn gia đình, bạn bè ngƣời thân động viên, khích lệ tinh thần trình học tập nhƣ thực đề tài Hà Nội, ngày 20 tháng năm 2012 Tác giả luận văn Phạm Thị Mai Hoa MỤC LỤC MỞ ĐẦU CHƢƠNG I: Động lực học hệ xe cầu có vết nứt dạng đóng mở 1.1 1.2 1.3 1.4 1.5 Động lực học hệ xe cầu khơng có vết nứt Động lực học hệ xe cầu có vết nứt mở 11 Động lực học hệ xe cầu có vết nứt đóng mở 14 Cách xác định ma trận độ cứng tổng thể dầm phƣơng pháp phần tử hữu hạn 16 Kết luận 18 CHƢƠNG II: Phép biến đổi Wavelet 2.1 Phép biến đổi Wavelet 19 19 2.1.1 Biến đổi wavelet liên tục CWT (Continuous Wavelet Transform) biến đổi ngƣợc 20 2.1.2 Biến đổi wavelet rời rạc DWT (Discrete Wavelet Transform) 22 2.1.3 Các hàm wavelet 24 2.2 Ví dụ ứng dụng wavelet phát thay đổi đột ngột tín hiệu 26 2.3 Kết luận CHƢƠNG III: Kết mô dao động hệ xe cầu có vết nứt 28 29 3.1 Ảnh hƣởng vết nứt đóng mở tới tần số 29 3.2 Ảnh hƣởng vết nứt đóng mở đến dịch chuyển 31 3.3 Ảnh hƣởng vết nứt đóng mở việc phát vết nứt 34 3.4 Kết luận 37 KẾT LUẬN 38 TÀI LIỆU THAM KHẢO 39 DANH MỤC CÁC CƠNG TRÌNH ĐÃ CƠNG BỐ 42 PHỤ LỤC I 68 DANH MỤC CÁC HÌNH VẼ Hình1.1 Mơ hình cầu dạng dầm dƣới tác động tải trọng di động Hình 1.2 Mơ hình dầm có vết nứt 11 Hình 1.3 Mơ hình vết nứt đóng mở 14 Hình 1.4 Mơ hình bốn phần tử có vết nứt dầm 16 Hình 2.1 Các thành phần wavelet tƣơng ứng với hệ số co giãn vị trí khác 21 Hình 2.2 Minh họa dịch chuyển vị trí wavelet 21 Hình 2.3 Cây phân tích wavelet mức 23 Hình 2.4 Hàm  (t) họ biến đổi wavelet Haar 24 Hình 2.5 Hàm  (t) họ biến đổi wavelet Daubechies 24 Hình 2.6 Hàm  (t) họ biến đổi wavelet Biorthogonal 25 Hình 2.7 Hàm  (t) họ biến đổi wavelet Coiflets 25 Hình 2.8 Hàm  (t) họ biến đổi wavelet Symlets 25 Hình 2.9 Hàm  (t) họ biến đổi wavelet Morlet 26 Hình 2.10 Hàm  (t) họ biến đổi wavelet Mexican Hat 26 Hình 2.11 Hàm  (t) họ biến đổi wavelet Meyer 26 Hình 2.12 Ví dụ phát thay đổi đột ngột tín hiệu 27 Hình 2.12 Ví dụ phân tích tần số hƣ hỏng 28 Hình 3.1 Mơ hình dầm với hai vết nứt 29 Bảng 3.1 Sự khác tần số 30 Hình 3.2 Sự thay đổi tần số phụ thuộc vào độ sâu vết nứt 31 Hình 3.3 Sự dịch chuyển xe độ sâu vết nứt 10% 31 Hình 3.4 Sự dịch chuyển xe độ sâu vết nứt 20% 32 Hình 3.5 Sự dịch chuyển xe độ sâu vết nứt 30% 32 Hình 3.6 Sự dịch chuyển xe độ sâu vết nứt 40% 33 Hình 3.7 Sự dịch chuyển xe độ sâu vết nứt 50% 33 Hình 3.8 Biến đổi Wavelet, độ sâu vết nứt 10% 35 Hình 3.9 Biến đổi Wavelet, độ sâu vết nứt 20% 35 Hình 3.10 Biến đổi Wavelet, độ sâu vết nứt 30% 35 Hình 3.11 Biến đổi Wavelet, độ sâu vết nứt 40% 36 Hình 3.12 Biến đổi Wavelet, độ sâu vết nứt 50% 36 MỞ ĐẦU Sự tác động môi trƣờng nhƣ tải trọng di động, tải trọng sóng, gió, ăn mịn, tập trung ứng suất… gây hƣ hỏng kết cấu cầu điển hình hƣ hỏng dạng vết nứt Sự phát triển vết nứt theo thời gian dẫn tới phá hủy kết cấu Do đó, việc giám sát nhằm phát sớm vết nứt kết cấu vấn đề quan trọng Trong thực tế, vết nứt khơng có trạng thái đóng mở mà đóng mở liên tục tùy thuộc vào tải trọng tác dụng vào vết nứt (tải trọng, trọng lƣợng vết nứt, v.v), rung động Đây đƣợc gọi vết nứt đóng mở đƣợc cơng bố Chondros [1] Các phản ứng động hệ để phát vết nứt đóng mở đƣợc phân tích Ruotolo Surace [2], Rizzo Scalea [3] Trong nghiên cứu họ, tần số riêng dầm với vết nứt đóng mở khơng liên tục q trình rung động, mà thay đổi theo thời gian, tần số riêng nhỏ nhiều so với tần số riêng dầm với vết nứt mở hoàn toàn Douka Hadjileontiadis [4] đề xuất phƣơng pháp gọi phƣơng pháp phân tích thực nghiệm để phân tích tần số tức thời Họ tần số tức thời thay đổi từ trạng thái mở trạng thái đóng cho thấy đóng mở vết nứt Sự có mặt tƣợng phi tuyến hệ có vết nứt đóng mở đƣợc nghiên cứu Sundermeyer Weaver [5] Trong nghiên cứu này, có phản ứng với tần số nằm hai tần số kích động Phản ứng tính phi tuyến đáp ứng dầm Bovsunovsky Matveev [6] trình bày khái niệm dạng riêng song hành xảy thời điểm vết nứt đóng mở để giải thích cho tính phi tuyến gây vết nứt đóng mở Qian [7] Ariaei [8] cho khác biệt phản ứng động hệ khơng có vết nứt có vết nứt đóng mở nhỏ so với hệ khơng có vết nứt có vết nứt mở hồn tồn Các phân tích hệ đàn hồi chủ đề đƣợc quan tâm nhiều lĩnh vực đa dạng nhƣ: xây dựng dân dụng hàng không vũ trụ kỷ qua Vấn đề phát sinh thiết kế cầu đƣờng sắt, cầu đƣờng bộ, đƣờng hầm cầu cống Đặc biệt kỹ thuật cầu đƣờng, nhiều ứng dụng đƣợc phát triển từ nghiên cứu chủ đề Parhi Behera [9] trình bày phƣơng pháp phân tích với kiểm tra thực nghiệm để nghiên cứu rung động dầm có vết nứt chịu tác động khối lƣợng di động Tƣơng tác hệ xe - cầu đƣợc tính tốn Piombo [10] cách coi cầu nhƣ ba nhịp chịu tác dụng hệ nhiều vật bảy bậc tự với hệ giảm xóc tuyến tính lốp xe không tuyệt đối cứng Trong nghiên cứu khác, Mahmoud Zaid [11] trình bày phƣơng pháp nghiên cứu ảnh hƣởng vết nứt nằm ngang lên phản ứng động dầm cơngxon EulerBernoulli khơng cản có điều kiện biên khớp hai đầu chịu tác dụng khối lƣợng di động Trong Lee [12] đề xuất quy trình để xác định đặc trƣng động lực học xác định vị trí mức độ hƣ hỏng chúng kết cấu Bilello Bergman [13] nghiên cứu dầm có vết nứt đƣợc mơ nhƣ lị xo quay chịu tải trọng di động Gần đây, Zhu Law [14] phân tích độ võng động theo thời gian cầu chịu tải trọng di động sử dụng biến đổi wavelet cho việc phát vết nứt Tuy nhiên, hầu hết phƣơng pháp tiếp cận để phát hƣ hỏng hệ xe -cầu sử dụng đáp ứng động cầu Các tác giả báo gần sử dụng phản ứng động đƣợc đo trực tiếp xe di chuyển cầu với vết nứt mở hoàn toàn[15] Tuy nhiên ảnh hƣởng vết nứt dạng đóng mở chƣa đƣợc nghiên cứu nhiều việc giám sát kết cấu cầu chịu tải trọng di động Do vậy, luận văn nghiên cứu ảnh hƣởng vết nứt đóng mở lên phản ứng hệ xe cầu đƣợc đo trực tiếp xe sau xem xét ảnh hƣởng việc phát hƣ hỏng cách sử dụng biến đổi wavelet, cơng cụ hiệu cho xử lý tín hiệu [16, 17, 18, 19, 20, 21, 22] Bố cục luận văn bao gồm ba chƣơng Chƣơng thứ xây dựng mơ hình phần tử hữu hạn hệ xe-cầu, xe đƣợc mơ hình hóa nhƣ hệ bậc tự do, cầu đƣợc mơ hình hóa nhƣ dầm Euler- Bernoulli Từ đó, hệ phƣơng trình tƣơng tác hệ xe cầu đƣợc thiết lập Xây dựng mô hình dầm đƣợc chia thành Q phần tử, có vết nứt đóng mở nằm phần tử thứ i Xác định ma trận tổng thể khối lƣợng M, ma trận cản C ma trận độ cứng K giải hệ phƣơng trình phƣơng pháp Newmark ta thu đƣợc phản ứng động xe dầm Cách xác định ma trận độ cứng tổng thể dầm chia thành bốn phần tử, phƣơng pháp phần tử hữu hạn, vết nứt nằm dầm Chƣơng thứ hai giới thiêu sở toán học phép biến đổi wavelet Một ví dụ minh họa cho việc sử dụng phân tích wavelet để phát nhƣ đánh giá thay đổi đột ngột tín hiệu Chƣơng thứ ba phân tích ảnh hƣởng vết nứt đóng mở tới thay đổi tần số riêng phản ứng xe di chuyển cầu có vết nứt đóng mở Phân tích ảnh hƣởng vết nứt đóng mở lên phƣơng pháp phát vết nứt wavelet Từ xác định đƣợc vị trí vết nứt cầu 56 Detection of a breathing crack of a beam-like bridge subjected to a moving vehicle using wavelet technique Khoa Viet Nguyen, Hoa Thi Mai Pham Institute of Mechanics, Vietnam Academy of Science and Technology Abstract In this paper a wavelet based technique for detection of a breathing crack of a beam- like bridge subjected to a moving vehicle is presented The stiffness matrix of the beam is modeled as time dependent stiffness matrix using finite element method The stiffness matrix of the cracked element at each moment is calculated from the curvature of the structure at the crack position The time history displacement of the vehicle is obtained by solving simultaneously the dynamic equations of the vehicle-bridge system Numerical simulation results show that natural frequencies of the system increase in comparison with the case of fully open crack Meanwhile, the amplitude of the vibration in the case of breathing crack is smaller than the case of fully open crack This is a warning that the small amplitude of the dynamic displacement does not necessarily corresponds to the small crack size The wavelet technique for crack detection is then carried out by analysing the dynamic response The result of the crack detection in the case of breathing crack is compared to the result in the case of fully open crack It is concluded that the peaks in the wavelet transform of the displacement in case of breathing crack are larger in comparison with the case of non-breathing crack This implies the wavelet method for crack detection is more efficient with the presence of the breathing crack Introduction In practice, depending on loading conditions on the cracked structure (residual loads, body weight of a structure, etc.), and the vibration effect a crack may be fully or partly open, fully closed at all times, or it can open and close regularly This crack which closes and opens during vibration was termed a „„breathing crack‟‟ and was discussed by Chondros et al [1] The dynamic response to harmonic excitation of a beam with several breathing cracks was analyzed by Ruotolo and Surace [2], Rizzo and Scalea [3] In their study, natural frequencies of a beam with a breathing crack are shown to be not constant during vibration but it is changing in time, and the relative decrease in natural frequencies found is much smaller than the decrease due to an open crack Douka and Hadjileontiadis [4] proposed a method called empirical mode decomposition to analyse the instantaneous frequency It is shown that the instantaneous frequency oscillates between frequencies corresponding to the open and closed states revealing the breathing of the crack The presence of non-linear phenomenon of a beam with breathing crack has been studied by Sundermeyer and Weaver [5] In these studies, a response at a frequency that is different from the driving frequencies was discovered This new response is due to the nonlinearity in the beam response Bovsunovsky and Matveev [6] presented a concept of concomitant mode shapes that occur at the time the crack closes and opens was presented to explain the nonlinearity caused by the breathing crack Qian et al [7] and Ariaei et al [8] stated that the difference between the displacement response of the intact beam and the cracked beam due to breathing crack is smaller than that between an open-cracked beam and intact beam The analysis of continuous elastic systems subjected to moving subsystems has been a topic of interest in many diverse fields such as civil and aerospace engineering for well over a century The problem arose in the design of railway bridges, roadways, tunnels and bridges ect Especially in bridge engineering many applications have been developed from the study of this subject Parhi and Behera [9] presented an analytical method along with experimental verification to investigate the vibration behavior of a cracked beam under a moving mass The vehicle–bridge interaction system was calculated by Piombo et al [10] by considering it as a three-span orthotropic plate subject to a seven degrees-of-freedom multi-body system with linear suspensions and tires flexibility In other studies, Mahmoud and Zaid [11] presented iterative methods for the effect of single transverse cracks on the dynamic behavior of simply supported and cantilever undamped Euler-Bernoulli beams subject to a moving mass While Lee et al [12] proposed a procedure for identification of the operational modal properties and the assessment of damage locations and their associated severities Bilello and Bergman [13] studied beams with damages modeled by rotational springs subject to a moving load Recently, Zhu and Law [14] analysed the cracked bridge subject to a moving vehicular load by analyzing the operational deflection time history of a bridge and used the wavelet transform for crack detection However, most of the current approaches for damage detection of the vehicle-bridge system used dynamic responses the bridge, only the authors of this paper recently used the dynamic responses obtained directly from the vehicle moving on bridge with fully open cracks [15] To this end, this study will first investigate the effect of breathing crack on the response of the vehicle-bridge system measured directly from the vehicle and then 57 consider its influence on the damage detection using the wavelet transform which is a very efficient tool for signal processing [16, 17, 18, 19, 20, 21, 22] Intact beam like structure Consider the bridge–vehicle system shown in Fig The vehicle is modelled as two DOFs system with sprung vehicle body and sprung tyre The bridge is considered approximately as an Euler–Bernoulli beam It is assumed that the surface unevenness of the bridge can be ignored and the tyre is always in contact with the supported beam Under these assumptions the governing equation of motion for the bridge–vehicle system in finite element method can be shown as follows: m1 y1  c1 ( y1  y )  k ( y1  y2 )  (1) 0)  m2 y2  c1 ( y1  y )  k ( y1  y2 )  k ( y2  w0 )  c2 ( y  w (2)   Cd  Kd  f  N T f Md o (3) f o  m1  m2 g  m2 y2 (4) Where m1, m2, k, c are vehicle parameters; y is vertical deflection of the vehicle body m1; uo is the vertical deflection of m2 and is equal to the deflection of the beam u at the contact position; M, C, K are structural mass, damping and stiffness matrices; NT denotes the transpose of the shape functions at the position x of the force; f0 is the magnitude of the interaction force between the vehicle and the beam; d is a nodal displacement column vector of the beam The displacement of the beam u at the arbitrary position x can be interpolated from the shape functions N and the nodal displacement d as: u=Nd (5) y1 y2 k1 c1 k2 m2 c2 v x w0, f0 L Fig A beam-like bridge subject to moving vehicle The components of shape function of an element can be obtained as:  x  x x  N1   3   2  ; N  x  1 ; l l l   x  x   x  x N  3   2  ; N  x     l  l l  l  (6) with l is the length of the element The time derivatives of uo are: u o ( x, t )  u u x  x t (7) 58 uo ( x, t )   2u  2u u  2u     x 2 x x xt x x t (8) Because N is spatial function while d is time dependent, from (4) we have:  2u u  2u  2u    N xx d ;  N xd ;  N xd ;  Nd x xt x t (9) where the subscript x implies the differentiation with respect to x Substituting (7), (8) and (9) into equations (1), (2), and (3) yields:    C M  d     M  y   C M  M2 K  d   - N T m1  m2 g        K  y   OT  C1  d   K     C3  y   K (10) Here O is a row zero matrix, and:  y1  y   y2   M1  O T N T m2  O M2    O m1  M3    0 m2  (11)  OT   O  C2     Nc2   c1   c1 C3    (12)  c1 c1  c2   OT  O   K2    Nk  N x c2 x   k1   k1 K3    (13)  k1 k1  k  C1  O T K  OT 2.2 Multi breathing cracks Figure shows a uniform beam-like structure divided into Q elements with R cracks situated in R different elements It is assumed that the cracks only affect the stiffness, not affect the mass and damping coefficient of the beam Neglecting shear action, the strain energy of an element without a crack can be written as [7]: LcR Lci xi xR x i, ui y i+1, ui+1 Mi+1 Pi+1 Mi Pi Fig Model of beam with R cracks W (o)  EI  P 2l   M l  MPl     (11) where P and M are the shear and bending internal forces at the right node of the element (Fig 2) For a rectangular beam with the thickness h, the width b, and the additional energy due to the crack can be written as W    K I2  K II2 1   K III2   b   E E 0 a (1)  da  (12) 59 where E   E for plane stress, E   E for plane strain and a is the crack depth, and KI, KII, KIII are stress  intensity factor for opening type, sliding type and tearing type cracks, respectively Taking into account only bending, equation (12) leads to a W (1)  b  K IM  K IP   K IIP da E (13) where K IM  6M a FI ( s) 3Pl a FI ( s) P a FII ( s) ; K IP  ; K IIP  2 bh bh bh   s  0.923  0.1991  sin   s     FI ( s)  tg   s    s  cos  2  FII ( s)  3s  2s 1.122  0.561s  0.085s (14)  0.18s 1 s (15) (16) Where s=a/h; a is the crack depth and h is the beam height ~ C of the intact element can be calculated as The generic component of the flexibility matrix  2W ( o ) c~ij( o )  ; i, j  1,2 ; P1  P ; P2  M Pi Pj (17) The additional flexibility coefficient is  2W (1) c~ij(1)  ; i, j  1,2 ; P1  P ; P2  M Pi Pj (18) Therefore, the total flexibility coefficient is: c~ij  c~ij(o)  c~ij(1) (19) From the equilibrium condition the following equation can be derived Pi Mi Pi 1 M i 1   TPi 1 T M i 1  T (20) where    l 0 T  1  0 T (21) By the principle of virtual work the stiffness matrix of the cracked element can be expressed as: ~ K c  T T C 1 T (22) The stiffness matrix and mass matrix for an element without a crack can be obtained as: 6l  12  6l 4l EI  Ke  l  12  6l  2l  6l  12  6l 12  6l 6l  2l   6l   4l  (23) 60  156  ml  22l Me  420  54   13l 22l 4l 13l  3l  13l  13l  3l  156  22l    22l 4l  54 (24) where I is the moment of inertia; E is the Young‟s modulus; m and l are the mass and the length of the element Element mass matrices Me are assembled to form the global mass matrix M, while matrices K e and Kc are assembled to form the global stiffness matrix K of the cracked beam Rayleigh damping in the form of C  M  K is used for the beam When a breathing crack is presented in the bridge, the crack opens and closes gradually leading to the gradual change in the stiffness at the cross section of the crack during vibration It is assumed that the effect of cracks depends on the curvature of beam at crack location and that range from to when the curvature ranges from negative maximum positive maximum As a result, the stiffness matrix K of the element with breathing crack breathing crack can be modelled as follows [8]:  d   K b  K e  ( K c  K e ) 1     d max (25) Where, Kb and Kc are the stiffness of the element with breathing crack, open crack and K e is the stiffness of the  is the maximum intact element d  is the instantaneous curvature of the beam at the crack position and d max curvature of the beam at the crack position during motion of the vehicle Substituting global matrices M, C, and K of the cracked beam into equation (9) and solving this equation by Newmark method, the dynamic responses of the vehicle and the beam will be obtained Wavelet transform The continuous wavelet transform is defined as follows [23]:  W ( a, b)   f (t ) a,b dt (28)  * t  b     , a is a real number called scale or dilation, b is a real number called position, a  a  t b W(a,b) are wavelet coefficients at scale a and position b, f(t) is input signal,    is wavelet function and  a  t b t b  *  is complex conjugate of     a   a  Where  a ,b (t )  In order to be classified as a wavelet a function must satisfy the following mathematical criteria: 1) A wavelet must have finite energy:  E 2) If ˆ   t  dt   (29)    is Fourier transform of  t  , i.e  ˆ     t e it dt (30)  then the following condition must be satisfied: ˆ   Cg   d     (31) This implies that the wavelet has no zero frequency component: ˆ (0)  ,   (t )e  jt dt  when    or in other words, the wavelet must have a zero mean: (32) 61   (t )dt  (33)  3) An additional criterion is that, for complex wavelets, the Fourier transform must both be real and vanish for negative frequencies Simulation results A numerical example of the beam with two cracks at locations of Lc1=L/3 and Lc2=2L/3 is carried out The crack depths of two cracks are the same Parameters of the beam are: mass density is 7855 kg/m 3; modulus of elasticity E=2.1x1011 N/m2; L=50 m; b=0.5 m; h=1 m Vehicle parameters are adopted from [24] as follows: m1=m2=50000 N; k1=k2=1.0x106 N/m; c1=c2=5.0x102 Ns/m The displacement-time history of the moving vehicle is obtained to investigate the influence of the cracks 4.1 Influence of the breathing crack on frequencies When there are breathing cracks, the natural frequency of the system is larger than the case of fully open cracks which is in agreement with the discussion in the Introduction section This is because the crack remains not fully open but it is changing between close and open states which cause the beam stiffer Table presents the comparison of the first two natural frequencies in two cases: fully open and breathing cracks with five levels of cracks It can be seen from this Table and Figure that the differences of the two natural frequencies increase when the depth of crack increase This suggests that when the crack depth is smaller than 20 %, the beam with breathing crack behaves similar to one with fully open crack 15 10 -5 10 20 30 40 Crack depth (%) a) 1nd frequency 50 Breathing crack 3.38 3.36 3.32 3.28 3.26 Change in frequency (%) Change in frequency (%) Table Difference of frequencies Crack 1st frequency depth Breathing Open Difference (%) crack crack (%) 10 0.90 0.90 20 0.90 0.90 30 0.90 0.88 2.27 40 0.90 0.84 7.14 50 0.90 0.80 12.50 2nd frequency Open Difference (%) crack 3.36 0.60 3.26 3.07 3.04 9.21 2.64 24.24 2.16 50.93 60 50 40 30 20 10 10 20 30 40 50 Crack depth (%) b) 2rd frequency Fig Frequency difference vs crack depth 4.2 Influence of the breathing crack on displacement Figs from to show the dynamic displacements of the vehicle moving on the bridge without a crack, with a fully open crack, and with a breathing crack calculated from five different levels of the cracks As can be seen from these Figs, the amplitude of the vertical displacement of vehicle moving on the bridge with breathing cracks is smaller than that with open cracks and larger than the case of intact beam This means that it is more difficult when using the dynamic displacement to detect the crack if it behaves as a breathing crack Another conclusion can be drawn from these Figs is that, when the crack depth is smaller than 30 %, the difference between the response of vehicle moving on the beam without a crack and beam with a breathing crack is smaller than that of the same beam and one with a fully open crack However, when the crack depth increases, this difference decreases This again suggests that the system with a breathing crack behaves similar to a fully open crack when the crack depth is small, but when the crack depth is larger the effect of the breathing crack can primitively be considered as the effect of a fully open crack with much smaller depth 62 -3 x 10 open crack breathing crack uncrack -1 -1.5 -2 -2.5 -3 -3.5 10 15 20 25 30 35 40 45 50 45 50 Position of vehicle (m) Fig Vertical displacement of the vehicle, crack depth is 10% x 10 -3 open crack breathing crack uncrack -0.5 Displacement of the body (m) Displacement of the body (m) -0.5 -1 -1.5 -2 -2.5 -3 -3.5 10 15 20 25 30 35 40 Position of vehicle (m) Fig Vertical displacement of the vehicle, crack depth is 20% 63 x 10 -3 open crack breathing crack uncrack Displacement of the body (m) -0.5 -1 -1.5 -2 -2.5 -3 -3.5 10 15 20 25 30 35 40 45 50 Position of vehicle (m) Fig Vertical displacement of the vehicle, crack depth is 30% x 10 -3 open crack breathing crack uncrack Displacement of the body (m) -0.5 -1 -1.5 -2 -2.5 -3 -3.5 10 15 20 25 30 35 40 Position of vehicle (m) Fig Vertical displacement of the vehicle, crack depth is 40% 45 50 64 -3 x 10 open crack breathing crack uncrack Displacement of the body (m) -0.5 -1 -1.5 -2 -2.5 -3 -3.5 -4 10 15 20 25 30 35 40 45 50 Position of vehicle (m) Fig Vertical displacement of the vehicle, crack depth is 50% 4.3 Influence of the breathing crack on crack detection When the beam is cracked, there are distortions in the dynamic response of the vehicle at crack locations However, these local distortions are generally small and difficult to be detected visually Therefore, in this work the CWT with its special property is applied for data processing After trying different wavelet functions for signal processing, the wavelet function “Symlet” is chosen as the most suitable one for this study Figs from to 12 present wavelet transforms of the vehicle displacement in five levels of the crack depth As can be seen in these Figs, there are clear peaks in the wavelet transform at crack positions These peaks become more significant when the depth of crack is larger Therefore, the peaks are indicators of the present of cracks It is interesting that, although the amplitude of the displacement of the system with the breathing crack is smaller than that with open crack, peaks in the wavelet transform of the displacement in case of breathing crack are larger This can be explained as follows: when moving on the beam the vehicle loading causes local maximum influences on the breathing crack phenomenon in the area adjacent to the cracks This leads to the significant local changes in the displacement at crack positions As a result, these local changes will be amplified as significant peaks in wavelet transform This implies that the wavelet method for crack detection is more efficient with the present of the breathing crack 65 -5 Wavelet coefficient Wavelet coefficient -5 x 10 0.5 -0.5 -1 10 20 30 40 50 x 10 0.5 -0.5 -1 10 Time (s) 20 30 40 50 Time (s) b) b) Fig Wavelet transform of y, crack depth is 10%: a) Open crack; b) Breathing crack -5 x 10 Wavelet coefficient Wavelet coefficient -5 x 10 0.5 -0.5 -1 10 20 30 40 -1 50 10 20 30 40 50 Time (s) Time (s) a) b) Fig Wavelet transform of y, crack depth is 20%: a) Open crack; b) Breathing crack -5 -1 10 20 30 Time (s) 40 50 Wavelet coefficient Wavelet coefficient -5 x 10 x 10 -5 -10 10 20 30 40 50 Time (s) b) b) Fig 10 Wavelet transform of y, crack depth is 30%: a) Open crack; b) Breathing crack 66 -5 Wavelet coefficient Wavelet coefficient -5 x 10 -2 -4 10 20 30 40 50 x 10 -5 -10 10 20 30 40 50 Time (s) Time (s) b) b) Fig 11 Wavelet transform of y, crack depth is 40%: a) Open crack; b) Breathing crack -5 -5 x 10 x 10 Wavelet coefficient Wavelet coefficient -2 -4 -6 -8 10 20 30 Time (s) 40 50 -5 -10 10 20 30 40 50 Time (s) b) b) Fig 12 Wavelet transform of y, crack depth is 50%: a) Open crack; b) Breathing crack Conclusion In this paper, effects of the breathing crack phenomenon on the dynamic response of a vehicle-bridge system obtained directly from the moving vehicle, and the influence of the breathing crack on crack detection have been investigated The main conclusions can be drawn from this work as follows: - Natural frequency of the vehicle-bridge with the breathing crack is larger in comparison with the case of open crack, which is in agree with other reports [2, 3, 8] - The vertical displacement of the vehicle moving on the bridge with breathing cracks is smaller than that with open cracks This is a warning when using the dynamic displacement to estimate the crack size: the small amplitude of dynamic response does not correspond to a small crack size if there is a presence of the breathing crack - When the crack depth is small the responses of vehicle moving on the beam with a breathing crack and beam with a fully open crack are close together However, when the crack depth increases, the beam with a breathing crack behaves similar to the beam without a crack in comparison with the case of open crack - Although the dynamic displacement of the system with the breathing crack is smaller than that of open crack, but it is interesting that the peaks in the wavelet transform of the displacement in case of breathing crack are larger This implies that the wavelet method for crack detection is more efficient with the present of the breathing crack 67 References [1] Chondros T.G., Dimarogonas A.D., and Yao J., Vibration of a Beam with a Breathing Crack Journal of Sound and Vibration 2001, Vol 239 (1), 57-67 [2] Pugno N., 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1992 [24] Majumder L and Manohar C.S., A time-domain approach for damage detection in beam structures using vibration data with a moving oscillator as an excitation source Journal of Sound and Vibration 268 (2003) 699– 716 68 PHỤ LỤC I: Chƣơng trình máy tính %FEM matrice construction ne=50; %Number of elements nnode=ne+1; %Number of nodes L=50; %Beam length l=L/ne; %Element length a=l/2; %Half of element length nn=2*(ne+1); %Total degrees of freedom g=10; %Gravity acceleration m1=500; %Mass of the car body m2=500; %Mass of the tyre k1=1e6; %Stiffness of the first floor k2=1e6; %Stiffness of the second floor c1=5e2; %Damping coefficent of the first leg, first store c2=5e2; %Damping coefficent of the first leg, second store aa=0.2; %crack depth b=1; %Element width h=2; %Element height rho=7855; nu=0.3; %Poisson coefficient A=b*h; E=2.1e11; I=b*h^3/12; %Inertial bending moment v=2; %velocity of moving load N=1000; %Number of time step dt=L/v/N; %Time step freq_number=5; %Number of frequencies Lc1=L/3; Lc2=2*L/3; boundary=3; curve_max1=9.2686e-006; curve_max2=9.2686e-006; curve_min1=0; curve_min2=0; overcrack=0; ke=zeros; me=zeros; C=zeros(nn); KK=zeros(nn); KKc=zeros(nn); MM=zeros(nn); %Element stiffness matrix ke=zeros; for i=1:ne ke(i,1,1)=3; ke(i,1,2)=3*a; ke(i,1,3)=-3; ke(i,1,4)=3*a; ke(i,2,1)=3*a; ke(i,2,2)=4*a*a; ke(i,2,3)=-3*a; ke(i,2,4)=2*a*a; 69 ke(i,3,1)=-3; ke(i,3,2)=-3*a; ke(i,3,3)=3; ke(i,3,4)=-3*a; ke(i,4,1)=3*a; ke(i,4,2)=2*a*a; ke(i,4,3)=-3*a; ke(i,4,4)=4*a*a; end ke=ke*E*I/2/a/a/a; nec1=fix(Lc1/l)+1; if nec1>ne nec1=ne; end nec2=fix(Lc2/l)+1; if nec2>ne nec2=ne; end %Determine the crack element number %Determine the crack element number ke_intact1=zeros; ke_intact2=zeros; for i=1:4 for j=1:4 ke_intact1(i,j)=ke(nec1,i,j); %Store ke_intact at Lc1 ke_intact2(i,j)=ke(nec2,i,j); %Store ke_intact at Lc2 end end kec=ke; kec(nec1,:,:)=kc; %The element nec th is now cracked kec(nec2,:,:)=kc; %The element nec th is now cracked for i=1:ne %Assembled stiffness matrix for j=1:4 for k=1:4 KK(2*i-2+j,2*i-2+k)=KK(2*i-2+j,2*i-2+k)+ke(i,j,k); KKc(2*i-2+j,2*i-2+k)=KKc(2*i-2+j,2*i-2+k)+kec(i,j,k); end end end %Element mass matrix -for i=1:ne me(i,1,1)=78; me(i,1,2)=22*a; me(i,1,3)=27; me(i,1,4)=-13*a; me(i,2,1)=22*a; me(i,2,2)=8*a*a; me(i,2,3)=13*a; me(i,2,4)=-6*a*a; me(i,3,1)=27; me(i,3,2)=13*a; me(i,3,3)=78; me(i,3,4)=-22*a; me(i,4,1)=-13*a; me(i,4,2)=-6*a*a; me(i,4,3)=-22*a; me(i,4,4)=8*a*a; end me=me*rho*A*a/105; 70 for i=1:ne %Assembled masss matrix for j=1:4 for k=1:4 MM(2*i-2+j,2*i-2+k)=MM(2*i-2+j,2*i-2+k)+me(i,j,k); end end end %Boundary condition: % Simply supported beam: Fixed translation; free rotation if boundary==3 ndof=2*nnode-2; K=zeros(ndof); M=zeros(ndof); KK(:,nn-1)=KK(:,nn); % Remove translation at right end KK(nn-1,:)=KK(nn,:); % Remove translation at right end KKc(:,nn-1)=KKc(:,nn); % Remove translation at right end KKc(nn-1,:)=KKc(nn,:); % Remove translation at right end MM(:,nn-1)=MM(:,nn); MM(nn-1,:)=MM(nn,:); % Remove translation at right end % Remove translation at right end Ks=KK(2:nn-1,2:nn-1); Ksc=KKc(2:nn-1,2:nn-1); Ms=MM(2:nn-1,2:nn-1); % Remove translation at left end; % Remove translation at left end; % Remove translation at left end; end %End of boundary condition %End of FEM matrice construction function h=Ns(x) global l a b1 b2 %Shape function- Referece: The Finite Element Methode, Liu and Quek xi=x/a; h(1)=1/4*(2-3*xi+xi^3); h(2)=1/4*a*(1-xi-xi^2+xi^3); h(3)=1/4*(2+3*xi-xi^3); h(4)=1/4*a*(-1-xi+xi^2+xi^3); %End of shape function function h=Nsx(x) global l a b1 b2 %Derivative of Shape function- Referece: The Finite Element Methode, Liu and Quek xi=x/a; h(1)=1/4*(-3+3*xi^2); h(2)=1/4*a*(-1-2*xi+3*xi^2); h(3)=1/4*(3-3*xi^2); h(4)=1/4*a*(-1+2*xi+3*xi^2); %End of first derivative of shape function