# Dynamic Analysis of Cracked Plate Subjected to Moving Oscillator by Finite Element Method

Authors: Nguyen Thai Chung; Nguyen Thi Hong; Le Xuan Thuy
Journal: Mathematical Problems in Engineering
 (2019)

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## Abstract

This paper presents the finite element algorithm and results of dynamical analysis of cracked plate subjected to moving oscillator with a constant velocity and any motion orbit. There are many surveys considering the dynamic response of the plate when there is a change in number of cracks and the stiffness of the spring k. The numerical survey results show that the effect of cracks on the plate's vibration is significant. The results of this article can be used as a reference for calculating and designing traffic structures such as road surface and bridge surface panels.


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## Body

## 1. Introduction

There are several types of plate structures affected by the vehicle load: pavement, railway system, and bridge floor, etc. Calculating these types of structures and the means of loading are modeled by different kinds of forces such as force, mass, and moving oscillators. Typically, the tracked vehicle is a moving mass, while the wheeled vehicle is described as a moving oscillator. Accordingly, structural dynamic analysis under the influence of mobile loads has been considered by many scientists. Nguyen Thai Chung and Le Pham Binh [1] analyzed the cracked beam on the elastic foundation under moving mass by using the finite element method (FEM). S.R. Mohebpour and P. Malekzadeh [2], P. Malekzadeh, A.R. Fiouz, H. Razi [3], Qinghua Song, Zhanqiang Liu, Jiahao Shi, Yi Wan [4], Qinghua Song, Jiahao Shi, and Zhanqiang Liu [5] presented a finite element model based on the first order shear deformation theory to investigate the dynamic behavior of laminated composite, FGM plates traversed by a moving oscillator, and a moving mass. Ahmad Mamandi, Ruhollah Mohsenzadeh, and Mohammad H. Kargarnovin [6] used finite element methods and Ansys software to simulate the nonlinear dynamic of rectangular plates subjected to accelerated or decelerated moving load. A.R. Vosoughi, P. Malekzadeh, and H. Razi [7] analyzed the moderately thick laminated composite plates on the elastic foundation subjected to moving load. G.L. Oian, S.N. Gu, J.S. Jiang [8], and Marek Krawczuk [9] analyzed the cracked plate subjected to dynamic loads by FEM. Yin T. and Lam H.F [10, 11] used a new solution method for investigation of the vibration characteristics of finite-length circular cylindrical shells with a circumferential part-through crack with four representative sets of boundary conditions being considered: simply supported, clamped-clamped, clamped-simply supported, and clamped-free. Li D. H., Yang X., Qian R. L., and Xu D. [12, 13] used the extended layer method (XLWM) to analyze the static reaction, free vibration, and transient response of cracked FGM plates.


## 2. Finite Element Simulation and Dominant Equations

Figure1 shows the cracked plate under the moving oscillator on the plate in the general coordinate system (X,Y,Z).Figure 1
Cracked plate subjected to moving oscillator.For finite element model formulation, the following assumptions are made:(i)
The materials of the system are linear elastic.(ii)
The load and pavement are not speared in the activity duration of system.


### 2.1. Cracked Plate Element Subjected to Moving Oscillator

#### 2.1.1. Cracked Plate Element Subjected to Dynamic Loads

Plate is described by bending rectangular four-node elements (Figure2). Arbitrary point in the element has positions (x, y) in global coordinate and positions (r, s) in local coordinate [14]. We assume that the thickness of plate element h is a constant and the conditions of Mindlin–Reissner plate theory are satisfied.Figure 2
Model of 4-node plate element and the coordinate system.The displacement fields are written as [15](1)ux,y,z,t=u0x,y,t+zθyx,y,t,vx,y,z,t=v0x,y,t-zθxx,y,t,wx,y,z,t=w0x,y,t,where u0, v0, w0 are the displacements of the mid plane and θx, θy are rotations of normal about, respectively, the y and x axes.The strain vector is presented in the form(2)εp=εxεyγxyγxzγyzT=εbTεsTT,where(3)εb=∂u0∂x∂v0∂y∂u0∂y+∂v0∂xT+z∂θy∂x-∂θx∂y∂θy∂x-∂θx∂yT=ε0+zκ,(4)εs=γxzγyzT=∂w0∂x+θy∂w0∂y-θxT,(5)κ=kxkykxyT=∂θy∂x-∂θx∂y∂θy∂x-∂θx∂yT.The constitutive equation can be written as(6)
σ
=
σ
b
σ
s
=
D
b
0
0
D
s
ε
b
γ
s
,where {σb} is stress vector without shear deformation:(7)σb=σxσyτxy=E1-ν21ν0ν10001-ν2εxεyγxy=Dbεb=Dbε0+zκ,{σs} is stress vector of shear stress:(8)σs=τxzτyz=Gγxzγyz=E21+ν1001γxzγyz=Dsεs,with E being elastic modulus of longitudinal deformation and ν being Poisson ratio.Using (7) and (8), the components of internal force vector {Fif} are determined as(9)MxMyMxyT=∫-h/2h/2zσxσyτxydz=Db∫-h/2h/2zε0+zκdz=h312Dbκ,QxQyT=∫-h/2h/2Dsεsdz=αhDsεs,so that one obtains(10)Fif=Dcsεcs,where Dcs=(h3/12)Db00αhDs- strain matrix, εcs=kxkykxyγxzγyzT is the vector of curvatures and shear strains, and α is the shear strain correction factor, usually equal to α = 5/6.According to the FEM procedure, the displacement of a point of the element is represented as [14, 16](11)w=∑i=14Niwi,θx=∑i=14Niθxi,θy=∑i=14Niθyi,where wi, θxi, θyi are displacements w, θx, θy at ith node, respectively, and Ni are shape functions, which allows us to obtain (12)εcse=Bqe=∑i=14Biqi,where Be is a matrix for the internal force determination, and qe=q1Tq2Tq3Tq4TeT is a vector of the node displacement, with qi=wiθxiθyiT, (i = 1,2,3,4).Substituting (12) into (10) leads to(13)Fif=∑i=14DcsBiqi,(14)whereDcsBi=DcsBib+DcsBis,DcsBib,DcsBis are matrices corresponding to bending moment and shear force, respectively, [10, 11].The dynamic equation of plate element can be derived by using Hamilton’s principle [14, 17]:(15)δ∫t1t2Te-Πedt=0,where Te, Πe are kinetic energy and total potential energy of the element, respectively.The kinetic energy of the element level is defined as [14](16)Πe=12∫AeFifeTDcsFifedAe-∫AewpdAe=12qeTK0eqe-qeTfe,with K0e=∫AeBTDcsBdAe,fe=∫AeNTpdAe being stiffness matrix and node loading vector of the element, respectively, [N] is mode shape function matrix of element, p is pressure of intensity, w=Nqe, and Ae is the surface area of the plate elements.Kinetic energyTe of element is determined by [14](17)Te=12∫Veρu˙eTu˙edVe=12q˙eT∫VeρNTNdVeq˙e=12q˙eTM0eq˙e,where M0e=∫VeρNTNdVe - mass matrix, ρ - mass density and q˙e-velocity vector.Substituting (16) and (17) into (15), the dynamic matrix equation of the plate element without damping can be written as(18)M0eq¨e+K0eqe=fe.In the case of cracked plate element, the stiffness matrixKce of the element can be written as [8, 18](19)Kce=TTCf-1T,where T is the transformation matrix, given in Appendix A, Cf=C0f+C1f in which C0f is the flexibility matrix of the noncracked element, given in Appendix B, and C1f is the flexibility matrix due to the presence of the crack, given in Appendix C [8].Now, the dynamic matrix equations of the cracked plate element subjected to dynamic loads become(20)M0eq¨e+Kceqe=fe.


#### 2.1.2. Cracked Plate Element Subjected to Moving Oscillator

The force of the moving oscillator on the plate at the time t is determined as follows:(21)Rt=-m1d2wx,y,tdt2-m2u¨-m1+m2g+Qtx=ξy=η,where g is acceleration due to gravity, u¨ is acceleration of the mass m2, and d2wx,y,t/dt2=w¨ is acceleration of the plate at the force set point given by(22)d2wdt2=∂2w∂x2x˙2+∂2w∂y2y˙2+∂2w∂t2+2∂2w∂x∂yx˙y˙+2x˙∂2w∂x∂t+2y˙∂2w∂y∂t+x¨∂w∂x+y¨∂w∂y.Substitutingw=Nqe into (22) yields(23)d2wdt2=Nxxx˙2qe+Nyyy˙2qe+Nq¨e+2x˙y˙Nxyqe+2x˙Nxq˙e+2y˙Nyq˙e+x¨Nxqe+y¨Nyqe,where Nx=∂N/∂x,Nxx=∂2N/∂x2,Nxy=∂2N/∂x∂y,Ny=∂N/∂y,Nyy=∂2N/∂y2,x˙,y˙ and x¨,y¨ are the velocity and acceleration of the loads along x, y axes, respectively.By substituting (23) into (22), the force of the moving oscillator on the plate at the time t can be written as(24)Rt=Qt-m1Nq¨e-2m1x˙Nx+y˙Nyq˙e-m1Nxxx˙2+Nyyy˙2+2x˙y˙Nxy+x¨Nx+y¨Nyqe-m2u¨-m1+m2g.Concentrated force (24) is described by the uniformly distributed load as follows [12–14]:(25)px,y,t=δx-ξδy-ηRx,y,t,where δ(.) is the Dirac’s delta function, and(26)∫-∞∞δx-afxdx=fa,(27)∫abδx-ξfxdx=0ifξ<a<bfξifa<ξ<b0ifa<b<ξ.Substituting (24) into (25) leads to(28)p=Qδx-ξδy-η-m1Nδx-ξδy-ηq¨e-2m1x˙Nx+y˙Nyδx-ξδy-ηq˙e-m1Nxxx˙2+Nyyy˙2+2x˙y˙Nxy++x¨Nx+y¨Nyδx-ξδy-ηqe-m2u¨δx-ξδy-η-m1+m2gδx-ξδy-η.The element nodal load vector is [14](29)fe=∫0b∫0aNTpx,y,tdxdy=∫0b∫0aNTδx-ξδy-ηRx,y,tdxdy.Substituting (28) into (29) leads to the nodal load vector equation(30)fe=Pe-Mpm1eq¨e-Mpm2eu¨-Cpeq˙e-Kpeqe,where(31)Pte=Nξ,ηTQ-m1+m2g,(32)Mpm1e=m1Nξ,ηTNξ,η,(33)Mpm2e=m2Nξ,ηT,(34)Cpe=2m1Nξ,ηTx˙Nxξ,η+y˙Nyξ,η,(35)Kpe=m1Nξ,ηTNxxξ,ηx˙2+Nyyξ,ηy˙2+2x˙y˙Nxyξ,η+x¨Nxξ,η+y¨Nyξ,η,Substituting (30) into (20) leads to the dynamic equation of the cracked plate element subjected to moving oscillator, which is(36)M0e+Mpm1eq¨e+Mpm2eu¨+Cpeq˙e+Kce+Kpeqe=Pe.The dynamic equation of mass m2 can be written as(37)m2u¨+cu˙+ku-cNq˙e-kNqe=Qt,By combining (1) and (2), the dynamic system of equations of the cracked plate and mass m2 are presented as follows:(38)M0e+Mpm1eq¨e+Mpm2eu¨+Cpeq˙e+Kce+Kpeqe=Pem2u¨+cu˙+ku-cNq˙e-kNqe=Qt,Or(39)M0e+Mpm1eMpm2e0m2q¨eu¨+Cpe0-cNcq˙eu˙+Kce+Kpe0-kNkqeu=PeQt.


### 2.2. Governing Differential Equations for Total System

By assembling all elements matrices and nodal force vectors, the governing equations of motions of the total system can be derived as(40)M0+Mpm1Mpm20m2q¨u¨+Cp+CR0-cNξ,ηcq˙u˙+K0+Kc+Kp0-kNξ,ηkqu=PQt,where M0=∑N0M0e is the overall structural mass matrix, K0=∑N0-NcK0e is the overall structural stiffness matrix with total uncracked elements, Kc=∑NcKce is the overall structural stiffness matrix with total cracked elements, Mpm1=∑Nem1Mpm1e is the overall mass matrix due to moving mass m1, Mpm2=∑Nem1Mpm2e is the overall mass matrix due to moving mass m2, Cp=∑Nem1∑Cpe is the overall damping matrix due to moving mass m1, and CR=αRM0+βRK0+Kc is the overall structural damping matrix [14, 17].This is a linear differential equation system with time dependence coefficient, which can be solved by using direct integration Newmark’s method. A MATLAB program named Cracked_Plates_Moving_2019 was conducted to solve (40).


## 2.1. Cracked Plate Element Subjected to Moving Oscillator

### 2.1.1. Cracked Plate Element Subjected to Dynamic Loads

Plate is described by bending rectangular four-node elements (Figure2). Arbitrary point in the element has positions (x, y) in global coordinate and positions (r, s) in local coordinate [14]. We assume that the thickness of plate element h is a constant and the conditions of Mindlin–Reissner plate theory are satisfied.Figure 2
Model of 4-node plate element and the coordinate system.The displacement fields are written as [15](1)ux,y,z,t=u0x,y,t+zθyx,y,t,vx,y,z,t=v0x,y,t-zθxx,y,t,wx,y,z,t=w0x,y,t,where u0, v0, w0 are the displacements of the mid plane and θx, θy are rotations of normal about, respectively, the y and x axes.The strain vector is presented in the form(2)εp=εxεyγxyγxzγyzT=εbTεsTT,where(3)εb=∂u0∂x∂v0∂y∂u0∂y+∂v0∂xT+z∂θy∂x-∂θx∂y∂θy∂x-∂θx∂yT=ε0+zκ,(4)εs=γxzγyzT=∂w0∂x+θy∂w0∂y-θxT,(5)κ=kxkykxyT=∂θy∂x-∂θx∂y∂θy∂x-∂θx∂yT.The constitutive equation can be written as(6)
σ
=
σ
b
σ
s
=
D
b
0
0
D
s
ε
b
γ
s
,where {σb} is stress vector without shear deformation:(7)σb=σxσyτxy=E1-ν21ν0ν10001-ν2εxεyγxy=Dbεb=Dbε0+zκ,{σs} is stress vector of shear stress:(8)σs=τxzτyz=Gγxzγyz=E21+ν1001γxzγyz=Dsεs,with E being elastic modulus of longitudinal deformation and ν being Poisson ratio.Using (7) and (8), the components of internal force vector {Fif} are determined as(9)MxMyMxyT=∫-h/2h/2zσxσyτxydz=Db∫-h/2h/2zε0+zκdz=h312Dbκ,QxQyT=∫-h/2h/2Dsεsdz=αhDsεs,so that one obtains(10)Fif=Dcsεcs,where Dcs=(h3/12)Db00αhDs- strain matrix, εcs=kxkykxyγxzγyzT is the vector of curvatures and shear strains, and α is the shear strain correction factor, usually equal to α = 5/6.According to the FEM procedure, the displacement of a point of the element is represented as [14, 16](11)w=∑i=14Niwi,θx=∑i=14Niθxi,θy=∑i=14Niθyi,where wi, θxi, θyi are displacements w, θx, θy at ith node, respectively, and Ni are shape functions, which allows us to obtain (12)εcse=Bqe=∑i=14Biqi,where Be is a matrix for the internal force determination, and qe=q1Tq2Tq3Tq4TeT is a vector of the node displacement, with qi=wiθxiθyiT, (i = 1,2,3,4).Substituting (12) into (10) leads to(13)Fif=∑i=14DcsBiqi,(14)whereDcsBi=DcsBib+DcsBis,DcsBib,DcsBis are matrices corresponding to bending moment and shear force, respectively, [10, 11].The dynamic equation of plate element can be derived by using Hamilton’s principle [14, 17]:(15)δ∫t1t2Te-Πedt=0,where Te, Πe are kinetic energy and total potential energy of the element, respectively.The kinetic energy of the element level is defined as [14](16)Πe=12∫AeFifeTDcsFifedAe-∫AewpdAe=12qeTK0eqe-qeTfe,with K0e=∫AeBTDcsBdAe,fe=∫AeNTpdAe being stiffness matrix and node loading vector of the element, respectively, [N] is mode shape function matrix of element, p is pressure of intensity, w=Nqe, and Ae is the surface area of the plate elements.Kinetic energyTe of element is determined by [14](17)Te=12∫Veρu˙eTu˙edVe=12q˙eT∫VeρNTNdVeq˙e=12q˙eTM0eq˙e,where M0e=∫VeρNTNdVe - mass matrix, ρ - mass density and q˙e-velocity vector.Substituting (16) and (17) into (15), the dynamic matrix equation of the plate element without damping can be written as(18)M0eq¨e+K0eqe=fe.In the case of cracked plate element, the stiffness matrixKce of the element can be written as [8, 18](19)Kce=TTCf-1T,where T is the transformation matrix, given in Appendix A, Cf=C0f+C1f in which C0f is the flexibility matrix of the noncracked element, given in Appendix B, and C1f is the flexibility matrix due to the presence of the crack, given in Appendix C [8].Now, the dynamic matrix equations of the cracked plate element subjected to dynamic loads become(20)M0eq¨e+Kceqe=fe.


### 2.1.2. Cracked Plate Element Subjected to Moving Oscillator

The force of the moving oscillator on the plate at the time t is determined as follows:(21)Rt=-m1d2wx,y,tdt2-m2u¨-m1+m2g+Qtx=ξy=η,where g is acceleration due to gravity, u¨ is acceleration of the mass m2, and d2wx,y,t/dt2=w¨ is acceleration of the plate at the force set point given by(22)d2wdt2=∂2w∂x2x˙2+∂2w∂y2y˙2+∂2w∂t2+2∂2w∂x∂yx˙y˙+2x˙∂2w∂x∂t+2y˙∂2w∂y∂t+x¨∂w∂x+y¨∂w∂y.Substitutingw=Nqe into (22) yields(23)d2wdt2=Nxxx˙2qe+Nyyy˙2qe+Nq¨e+2x˙y˙Nxyqe+2x˙Nxq˙e+2y˙Nyq˙e+x¨Nxqe+y¨Nyqe,where Nx=∂N/∂x,Nxx=∂2N/∂x2,Nxy=∂2N/∂x∂y,Ny=∂N/∂y,Nyy=∂2N/∂y2,x˙,y˙ and x¨,y¨ are the velocity and acceleration of the loads along x, y axes, respectively.By substituting (23) into (22), the force of the moving oscillator on the plate at the time t can be written as(24)Rt=Qt-m1Nq¨e-2m1x˙Nx+y˙Nyq˙e-m1Nxxx˙2+Nyyy˙2+2x˙y˙Nxy+x¨Nx+y¨Nyqe-m2u¨-m1+m2g.Concentrated force (24) is described by the uniformly distributed load as follows [12–14]:(25)px,y,t=δx-ξδy-ηRx,y,t,where δ(.) is the Dirac’s delta function, and(26)∫-∞∞δx-afxdx=fa,(27)∫abδx-ξfxdx=0ifξ<a<bfξifa<ξ<b0ifa<b<ξ.Substituting (24) into (25) leads to(28)p=Qδx-ξδy-η-m1Nδx-ξδy-ηq¨e-2m1x˙Nx+y˙Nyδx-ξδy-ηq˙e-m1Nxxx˙2+Nyyy˙2+2x˙y˙Nxy++x¨Nx+y¨Nyδx-ξδy-ηqe-m2u¨δx-ξδy-η-m1+m2gδx-ξδy-η.The element nodal load vector is [14](29)fe=∫0b∫0aNTpx,y,tdxdy=∫0b∫0aNTδx-ξδy-ηRx,y,tdxdy.Substituting (28) into (29) leads to the nodal load vector equation(30)fe=Pe-Mpm1eq¨e-Mpm2eu¨-Cpeq˙e-Kpeqe,where(31)Pte=Nξ,ηTQ-m1+m2g,(32)Mpm1e=m1Nξ,ηTNξ,η,(33)Mpm2e=m2Nξ,ηT,(34)Cpe=2m1Nξ,ηTx˙Nxξ,η+y˙Nyξ,η,(35)Kpe=m1Nξ,ηTNxxξ,ηx˙2+Nyyξ,ηy˙2+2x˙y˙Nxyξ,η+x¨Nxξ,η+y¨Nyξ,η,Substituting (30) into (20) leads to the dynamic equation of the cracked plate element subjected to moving oscillator, which is(36)M0e+Mpm1eq¨e+Mpm2eu¨+Cpeq˙e+Kce+Kpeqe=Pe.The dynamic equation of mass m2 can be written as(37)m2u¨+cu˙+ku-cNq˙e-kNqe=Qt,By combining (1) and (2), the dynamic system of equations of the cracked plate and mass m2 are presented as follows:(38)M0e+Mpm1eq¨e+Mpm2eu¨+Cpeq˙e+Kce+Kpeqe=Pem2u¨+cu˙+ku-cNq˙e-kNqe=Qt,Or(39)M0e+Mpm1eMpm2e0m2q¨eu¨+Cpe0-cNcq˙eu˙+Kce+Kpe0-kNkqeu=PeQt.


## 2.1.1. Cracked Plate Element Subjected to Dynamic Loads

Plate is described by bending rectangular four-node elements (Figure2). Arbitrary point in the element has positions (x, y) in global coordinate and positions (r, s) in local coordinate [14]. We assume that the thickness of plate element h is a constant and the conditions of Mindlin–Reissner plate theory are satisfied.Figure 2
Model of 4-node plate element and the coordinate system.The displacement fields are written as [15](1)ux,y,z,t=u0x,y,t+zθyx,y,t,vx,y,z,t=v0x,y,t-zθxx,y,t,wx,y,z,t=w0x,y,t,where u0, v0, w0 are the displacements of the mid plane and θx, θy are rotations of normal about, respectively, the y and x axes.The strain vector is presented in the form(2)εp=εxεyγxyγxzγyzT=εbTεsTT,where(3)εb=∂u0∂x∂v0∂y∂u0∂y+∂v0∂xT+z∂θy∂x-∂θx∂y∂θy∂x-∂θx∂yT=ε0+zκ,(4)εs=γxzγyzT=∂w0∂x+θy∂w0∂y-θxT,(5)κ=kxkykxyT=∂θy∂x-∂θx∂y∂θy∂x-∂θx∂yT.The constitutive equation can be written as(6)
σ
=
σ
b
σ
s
=
D
b
0
0
D
s
ε
b
γ
s
,where {σb} is stress vector without shear deformation:(7)σb=σxσyτxy=E1-ν21ν0ν10001-ν2εxεyγxy=Dbεb=Dbε0+zκ,{σs} is stress vector of shear stress:(8)σs=τxzτyz=Gγxzγyz=E21+ν1001γxzγyz=Dsεs,with E being elastic modulus of longitudinal deformation and ν being Poisson ratio.Using (7) and (8), the components of internal force vector {Fif} are determined as(9)MxMyMxyT=∫-h/2h/2zσxσyτxydz=Db∫-h/2h/2zε0+zκdz=h312Dbκ,QxQyT=∫-h/2h/2Dsεsdz=αhDsεs,so that one obtains(10)Fif=Dcsεcs,where Dcs=(h3/12)Db00αhDs- strain matrix, εcs=kxkykxyγxzγyzT is the vector of curvatures and shear strains, and α is the shear strain correction factor, usually equal to α = 5/6.According to the FEM procedure, the displacement of a point of the element is represented as [14, 16](11)w=∑i=14Niwi,θx=∑i=14Niθxi,θy=∑i=14Niθyi,where wi, θxi, θyi are displacements w, θx, θy at ith node, respectively, and Ni are shape functions, which allows us to obtain (12)εcse=Bqe=∑i=14Biqi,where Be is a matrix for the internal force determination, and qe=q1Tq2Tq3Tq4TeT is a vector of the node displacement, with qi=wiθxiθyiT, (i = 1,2,3,4).Substituting (12) into (10) leads to(13)Fif=∑i=14DcsBiqi,(14)whereDcsBi=DcsBib+DcsBis,DcsBib,DcsBis are matrices corresponding to bending moment and shear force, respectively, [10, 11].The dynamic equation of plate element can be derived by using Hamilton’s principle [14, 17]:(15)δ∫t1t2Te-Πedt=0,where Te, Πe are kinetic energy and total potential energy of the element, respectively.The kinetic energy of the element level is defined as [14](16)Πe=12∫AeFifeTDcsFifedAe-∫AewpdAe=12qeTK0eqe-qeTfe,with K0e=∫AeBTDcsBdAe,fe=∫AeNTpdAe being stiffness matrix and node loading vector of the element, respectively, [N] is mode shape function matrix of element, p is pressure of intensity, w=Nqe, and Ae is the surface area of the plate elements.Kinetic energyTe of element is determined by [14](17)Te=12∫Veρu˙eTu˙edVe=12q˙eT∫VeρNTNdVeq˙e=12q˙eTM0eq˙e,where M0e=∫VeρNTNdVe - mass matrix, ρ - mass density and q˙e-velocity vector.Substituting (16) and (17) into (15), the dynamic matrix equation of the plate element without damping can be written as(18)M0eq¨e+K0eqe=fe.In the case of cracked plate element, the stiffness matrixKce of the element can be written as [8, 18](19)Kce=TTCf-1T,where T is the transformation matrix, given in Appendix A, Cf=C0f+C1f in which C0f is the flexibility matrix of the noncracked element, given in Appendix B, and C1f is the flexibility matrix due to the presence of the crack, given in Appendix C [8].Now, the dynamic matrix equations of the cracked plate element subjected to dynamic loads become(20)M0eq¨e+Kceqe=fe.


## 2.1.2. Cracked Plate Element Subjected to Moving Oscillator

The force of the moving oscillator on the plate at the time t is determined as follows:(21)Rt=-m1d2wx,y,tdt2-m2u¨-m1+m2g+Qtx=ξy=η,where g is acceleration due to gravity, u¨ is acceleration of the mass m2, and d2wx,y,t/dt2=w¨ is acceleration of the plate at the force set point given by(22)d2wdt2=∂2w∂x2x˙2+∂2w∂y2y˙2+∂2w∂t2+2∂2w∂x∂yx˙y˙+2x˙∂2w∂x∂t+2y˙∂2w∂y∂t+x¨∂w∂x+y¨∂w∂y.Substitutingw=Nqe into (22) yields(23)d2wdt2=Nxxx˙2qe+Nyyy˙2qe+Nq¨e+2x˙y˙Nxyqe+2x˙Nxq˙e+2y˙Nyq˙e+x¨Nxqe+y¨Nyqe,where Nx=∂N/∂x,Nxx=∂2N/∂x2,Nxy=∂2N/∂x∂y,Ny=∂N/∂y,Nyy=∂2N/∂y2,x˙,y˙ and x¨,y¨ are the velocity and acceleration of the loads along x, y axes, respectively.By substituting (23) into (22), the force of the moving oscillator on the plate at the time t can be written as(24)Rt=Qt-m1Nq¨e-2m1x˙Nx+y˙Nyq˙e-m1Nxxx˙2+Nyyy˙2+2x˙y˙Nxy+x¨Nx+y¨Nyqe-m2u¨-m1+m2g.Concentrated force (24) is described by the uniformly distributed load as follows [12–14]:(25)px,y,t=δx-ξδy-ηRx,y,t,where δ(.) is the Dirac’s delta function, and(26)∫-∞∞δx-afxdx=fa,(27)∫abδx-ξfxdx=0ifξ<a<bfξifa<ξ<b0ifa<b<ξ.Substituting (24) into (25) leads to(28)p=Qδx-ξδy-η-m1Nδx-ξδy-ηq¨e-2m1x˙Nx+y˙Nyδx-ξδy-ηq˙e-m1Nxxx˙2+Nyyy˙2+2x˙y˙Nxy++x¨Nx+y¨Nyδx-ξδy-ηqe-m2u¨δx-ξδy-η-m1+m2gδx-ξδy-η.The element nodal load vector is [14](29)fe=∫0b∫0aNTpx,y,tdxdy=∫0b∫0aNTδx-ξδy-ηRx,y,tdxdy.Substituting (28) into (29) leads to the nodal load vector equation(30)fe=Pe-Mpm1eq¨e-Mpm2eu¨-Cpeq˙e-Kpeqe,where(31)Pte=Nξ,ηTQ-m1+m2g,(32)Mpm1e=m1Nξ,ηTNξ,η,(33)Mpm2e=m2Nξ,ηT,(34)Cpe=2m1Nξ,ηTx˙Nxξ,η+y˙Nyξ,η,(35)Kpe=m1Nξ,ηTNxxξ,ηx˙2+Nyyξ,ηy˙2+2x˙y˙Nxyξ,η+x¨Nxξ,η+y¨Nyξ,η,Substituting (30) into (20) leads to the dynamic equation of the cracked plate element subjected to moving oscillator, which is(36)M0e+Mpm1eq¨e+Mpm2eu¨+Cpeq˙e+Kce+Kpeqe=Pe.The dynamic equation of mass m2 can be written as(37)m2u¨+cu˙+ku-cNq˙e-kNqe=Qt,By combining (1) and (2), the dynamic system of equations of the cracked plate and mass m2 are presented as follows:(38)M0e+Mpm1eq¨e+Mpm2eu¨+Cpeq˙e+Kce+Kpeqe=Pem2u¨+cu˙+ku-cNq˙e-kNqe=Qt,Or(39)M0e+Mpm1eMpm2e0m2q¨eu¨+Cpe0-cNcq˙eu˙+Kce+Kpe0-kNkqeu=PeQt.


## 2.2. Governing Differential Equations for Total System

By assembling all elements matrices and nodal force vectors, the governing equations of motions of the total system can be derived as(40)M0+Mpm1Mpm20m2q¨u¨+Cp+CR0-cNξ,ηcq˙u˙+K0+Kc+Kp0-kNξ,ηkqu=PQt,where M0=∑N0M0e is the overall structural mass matrix, K0=∑N0-NcK0e is the overall structural stiffness matrix with total uncracked elements, Kc=∑NcKce is the overall structural stiffness matrix with total cracked elements, Mpm1=∑Nem1Mpm1e is the overall mass matrix due to moving mass m1, Mpm2=∑Nem1Mpm2e is the overall mass matrix due to moving mass m2, Cp=∑Nem1∑Cpe is the overall damping matrix due to moving mass m1, and CR=αRM0+βRK0+Kc is the overall structural damping matrix [14, 17].This is a linear differential equation system with time dependence coefficient, which can be solved by using direct integration Newmark’s method. A MATLAB program named Cracked_Plates_Moving_2019 was conducted to solve (40).


## 3. Numerical Analysis

We consider a rectangular cracked plate with size L = 3.0 m, W = 1.6 m, thickness h = 0.025 m, crack lengthWcr = 0.5 m, and it appears in the middle of the plate (Figure 1). Material parameters of the plate: Young Modulus E = 2.0x1011 N/m2, Poisson coefficient ν = 0.28, density ρ = 7800 kg/m3. The boundary condition of the plate is SFSF (simply support, free/ simply support, free). The moving oscillator has mass m1 = 300 kg connected with m2 = 200 kg via spring with stiffness k = 1.5x105 N/m and damping element with resistance coefficient, c = 4.5x103 Ns/m, and k and c are parallel. Moving oscillator moves at velocity v = 10 m/s along the centerline y = W/2 of the plate.The results of the vibration of the plate subjected to moving oscillator (MO) and the effect of moving mass (MM) (M = m1 + m2 = 500 kg) are shown in Table1 and Figures 3, 4, 5, and 6, in which the “int” symbol represents the intensity value (stress intensity and train intensity).Table 1
The extreme value at the points A and B.
Case of load wAmax w ¨ A m a x σintAmax σintBmax [ c m ] [ m / s 2 ] [ N / m 2 ] [ N / m 2 ] MO 1.012 13.217 1.364×107 10.536×107  MM 1.122 42.4458 4.434x107 15.748x107Figure 3
Dynamic vertical response at point A of the plate.Figure 4
Vertical acceleration response at point A of the plate.Figure 5
Stress response at points A and B of the plate.Figure 6
Strain response at points A and B of the plate.Comment: Compared to the case of cracked plate subjected to moving oscillator, the case of cracked plate subjected to moving mass indicates greater response of the plate. Therefore, the destructive capacity of the structure is greater.


### 3.1. The Effect of the Number of Cracks

To evaluate the effect of the number of cracks on vibration of cracked plate under moving oscillator, the three cases were investigated: Case 1: the plate has one crack in the middle (X = L/2, the basic problem); Case 2: the plate has one crack in the middle and one same size crack at X = L/4; Case 3: the plate has 3 cracks at X = L/4, L/2, 3Lp/4. The results of the vibration of the plate are shown in Table 2 and Figures 7, 8, 9, and 10.Table 2
The extreme value at the points A and B.
Case wAmax w ¨ A m a x σintAmax σintBmax [ c m ] [ m / s 2 ] [ N / m 2 ] [ N / m 2 ] 1 1.012 13.217 1.364x107 10.536×107  2 1.029 33.814 2.053x107 10.177×107  3 1.066 33.821 2.062x107 10.265×107Figure 7
Dynamic vertical response at point A of the plate with different numbers of cracks.Figure 8
Vertical acceleration response at point A of the plate with different numbers of cracks.Figure 9
Stress response at point B of the plate with different numbers of cracks.Figure 10
Strain response at point B of the plate with different numbers of cracks.Comment: Strain, displacement, stress, and acceleration at point A increase as the number of cracks increases, but these values fluctuate at the crack edge, sometimes increase and sometimes decrease.


### 3.2. The Effect of the Stiffness of the Spring k

To evaluate the effect of k hardness in the oscillation system on the response of the system, the authors examine the problem when k varies from 1x105 N/m to 9.0x105 N/m. Response of the system at points A and B is shown in Table 3 and Figures 11, 12, 13, 14, 15, and 16.Table 3
The extreme value at the points A and B.
k×105 wAmax w ¨ A m a x σintAmax σintBmax [ N / m ] [ c m ] [ m / s 2 ] [ N / m 2 ] [ N / m 2 ] 1.0 0.081 8.942 0.813×107 8.426×107  1.5 1.012 13.217 1.364×107 10.536×107  2.0 1.075 19.415 2.285×107 11.284×107  2.5 1.078 23.073 3.150×107 11.553×107  3.5 1.033 21.224 4.036×107 11.363×107  6.0 0.905 17.740 2.380×107 11.213×107  9.0 0.783 12.360 1.554×107 8.693×107Figure 11
w A m a x - k relationship.Figure 12
w ¨ A m a x - k relationship.Figure 13
σ i n t A m a x and σintBmax - k relationship.Figure 14
ε i n t A m a x and εintBmax - k relationship.Figure 15
Stress response at point A of the plate with different k.Figure 16
Stress response at point B of the plate with different k.Comment: When the k hardness changes, the oscillation of the system varies considerably. With the parameters of the given plate, the displacement response, acceleration, stress, and strain at the computed points are the greatest when k = 2.5x105 N/m.


### 3.3. The Effect of Loading Velocity

The authors analyzed the dynamics of the plate with the speed of the oscillator system varying from 6 m/s to 14 m/s; the results are shown in Table4 and Figures 17, 18, and 19.Table 4
The extreme value at the points A and B.
v w A m a x w ¨ A m a x σintAmax σintBmax [ m / s ] [cm] [ m / s 2 ] [ N / m 2 ] [N/m2] 6 1.211 15.429 4.352×107 16.323×107  8 1.210 13.816 2.931×107 13.124×107  10 1.012 13.217 1.364×107 10.536×107  12 0.724 10.201 0.970×107 7.718×107  14 0.537 11.897 1.137×107 6.048×107Figure 17
Vertical displacement response at point A of the plate with different v.Figure 18
Stress response at point A of the plate with different v.Figure 19
Stress response at point B of the plate with different v.Comment: When the speed of the oscillation system increases, the displacement and stress of the plate decrease, but there is no clear rule. According to the authors, the main reason may be due to the influence of the plate's free vibration frequency and the moving oscillator; this is the difference with the case of the plate subjected to moving mass.


## 3.1. The Effect of the Number of Cracks

To evaluate the effect of the number of cracks on vibration of cracked plate under moving oscillator, the three cases were investigated: Case 1: the plate has one crack in the middle (X = L/2, the basic problem); Case 2: the plate has one crack in the middle and one same size crack at X = L/4; Case 3: the plate has 3 cracks at X = L/4, L/2, 3Lp/4. The results of the vibration of the plate are shown in Table 2 and Figures 7, 8, 9, and 10.Table 2
The extreme value at the points A and B.
Case wAmax w ¨ A m a x σintAmax σintBmax [ c m ] [ m / s 2 ] [ N / m 2 ] [ N / m 2 ] 1 1.012 13.217 1.364x107 10.536×107  2 1.029 33.814 2.053x107 10.177×107  3 1.066 33.821 2.062x107 10.265×107Figure 7
Dynamic vertical response at point A of the plate with different numbers of cracks.Figure 8
Vertical acceleration response at point A of the plate with different numbers of cracks.Figure 9
Stress response at point B of the plate with different numbers of cracks.Figure 10
Strain response at point B of the plate with different numbers of cracks.Comment: Strain, displacement, stress, and acceleration at point A increase as the number of cracks increases, but these values fluctuate at the crack edge, sometimes increase and sometimes decrease.


## 3.2. The Effect of the Stiffness of the Spring k

To evaluate the effect of k hardness in the oscillation system on the response of the system, the authors examine the problem when k varies from 1x105 N/m to 9.0x105 N/m. Response of the system at points A and B is shown in Table 3 and Figures 11, 12, 13, 14, 15, and 16.Table 3
The extreme value at the points A and B.
k×105 wAmax w ¨ A m a x σintAmax σintBmax [ N / m ] [ c m ] [ m / s 2 ] [ N / m 2 ] [ N / m 2 ] 1.0 0.081 8.942 0.813×107 8.426×107  1.5 1.012 13.217 1.364×107 10.536×107  2.0 1.075 19.415 2.285×107 11.284×107  2.5 1.078 23.073 3.150×107 11.553×107  3.5 1.033 21.224 4.036×107 11.363×107  6.0 0.905 17.740 2.380×107 11.213×107  9.0 0.783 12.360 1.554×107 8.693×107Figure 11
w A m a x - k relationship.Figure 12
w ¨ A m a x - k relationship.Figure 13
σ i n t A m a x and σintBmax - k relationship.Figure 14
ε i n t A m a x and εintBmax - k relationship.Figure 15
Stress response at point A of the plate with different k.Figure 16
Stress response at point B of the plate with different k.Comment: When the k hardness changes, the oscillation of the system varies considerably. With the parameters of the given plate, the displacement response, acceleration, stress, and strain at the computed points are the greatest when k = 2.5x105 N/m.


## 3.3. The Effect of Loading Velocity

The authors analyzed the dynamics of the plate with the speed of the oscillator system varying from 6 m/s to 14 m/s; the results are shown in Table4 and Figures 17, 18, and 19.Table 4
The extreme value at the points A and B.
v w A m a x w ¨ A m a x σintAmax σintBmax [ m / s ] [cm] [ m / s 2 ] [ N / m 2 ] [N/m2] 6 1.211 15.429 4.352×107 16.323×107  8 1.210 13.816 2.931×107 13.124×107  10 1.012 13.217 1.364×107 10.536×107  12 0.724 10.201 0.970×107 7.718×107  14 0.537 11.897 1.137×107 6.048×107Figure 17
Vertical displacement response at point A of the plate with different v.Figure 18
Stress response at point A of the plate with different v.Figure 19
Stress response at point B of the plate with different v.Comment: When the speed of the oscillation system increases, the displacement and stress of the plate decrease, but there is no clear rule. According to the authors, the main reason may be due to the influence of the plate's free vibration frequency and the moving oscillator; this is the difference with the case of the plate subjected to moving mass.


## 4. Conclusions

In the end, with the set of survey parameters, the case of the cracked plate under the moving mass is more dangerous than the case of moving oscillators operates. However, the problem of texture affected by the oscillation system is complex. In each case, a reevaluation is needed. The response of the system depends on the interrelation between the frequency of the stimulus and the natural frequency of the system.The results show that stress and strain at the crack head are much larger than they are at other sites. These values vary considerably when the number of cracks and k hardness are changed.



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*Source: 6528251-2019-07-25.xml*