# Displacement and Stress Analysis of Thin Plate for Cement Concrete Pavement

Authors: Liang Jia; Li Zhang; Jian Guo; Kai Yao; Shikai He
Journal: Mathematical Problems in Engineering
 (2018)

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## Abstract

In order to analyze the displacement and stress of thin plate for cement concrete pavement with rectangle shape and resting on Winkler soil foundation, an analysis model was set up based on the theory of elastic thin plate on Winkler foundation. According to elasticity Kirchhoff theory of thin plates and Winkler soil foundation model, the expressions of displacement and stress were yielded by using the inverse Hankel transform. Numerical method was employed to calculate displacement and stress of the thin plate. Results showed that displacement and stress corresponded with using the method fromCode for Highway Design-Pavement. Results showed that the method can provide a new computation method for displacement and stress of the thin plate about cement concrete pavement.


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## Body

## 1. Introduction

The model of a thin plate on elastic foundation was mainly used in structural engineering in last years. Currently, thin plates or films of metal, ceramic, or synthetic materials were bonded in the surface of machine structural parts or electronic devices to improve their mechanical, thermal, electronic, or tribological properties [1]. At these applications, the subgrade of the thin plate can be simulated as a Winkler foundation, which reacted with pressure proportional to the deflection of the plate at each point. The plate in the above applications was loaded by vertical loads or bending moments. For cement concrete pavement, the applied load is mainly from the vehicles, which is vertically applied on pavement. So that the width and thickness of the pavement are usually determined by the traffic volume. And the thickness will increase with the designed traffic volume. A large number of research works have been published to solve problem of a classical thin plate [2–4] or a thin plate on an elastic foundation [5–9]. Fisher investigated the elastic impact of a sphere on a thin plate which is continuous contact with a foundation [10]. Jędrysiak investigated the dynamic response of thin elastic plates in Winkler foundation [11].Some researches introduced an efficient boundary element approach for the analysis of thin plate resting on an elastic Winkler foundation with different boundary conditions [12–15]. Shao combined the Fourier spectral method and differential quadrature method for solving problems for thin plates resting on Winkler foundations with irregular domains [16].However, few research works have been published concerning the stress and displacement of a thin plate on Winkler foundation. The present work is aimed to obtain an analytical solution of a finite rectangle thin plate on an elastic foundation under vertical forces. For this purpose, both the Hankel transform and inverse Hankel transform were adopted.


## 2. Basic Assumption and Governing Equations

According to the theory of elastic thin plate on Winkler soil foundation, a rectangle thin plate with thicknessh was considered to rest on a Winkler soil foundation (Figure 1), the thickness of plate was less than one-fifth the size of plane, the plate was loaded by the vertical pressure qx,y and the foundation reaction pressurep assumed to be proportional to the vertical deflection ωx,y of the plate [17], and the rectangular coordinate system was placed on midplane.Figure 1
Model of the thin plate.Kirchhoff proposed the basic assumption of the thin plate as follows [17].( 1 ) Length of normal line of the plate is unchanged before and after deformation, that is,(1)εz=0where εz is normal strain.( 2 ) Before deformation, the normal line of the plate is perpendicular to midplane; after deformation, the normal line is still perpendicular to midplane, that is,(2)γzx=0,γzy=0where γzx,γzy are shear strain.( 3 ) The plane of parallel to midplane does not squeeze, that is,(3)σz=0where σz is stress.( 4 ) Horizontal displacement of the midplane does not occur during the process thin plate bending, which are(4)u0=0,v0=0where u0,v0 are displacement.According to the hypothesis (2), it yieldsγ z x = 0, namely, ∂w/∂x+∂u/∂z=0,γ z y = 0, namely, ∂w/∂y+∂v/∂z=0.Thus,(5)u=-z∂ω∂x+u0v=-z∂ω∂y+v0According to the hypothesis (4), the displacement of thin plate can be obtained as follows:(6)u=-z∂ω∂xv=-z∂ω∂yAccording to (6), the strain of thin plate can be obtained as follows:(7)εx=∂u∂x=-z∂2ω∂x2εy=∂v∂y=-z∂2ω∂y2γxy=∂u∂y+∂v∂x=-2z∂2ω∂x∂yAccording to general Hook's law, we can obtain(8)εx=1Eσx-μσy+σzεy=1Eσy-μσx+σzγxy=1Eσz-μσx+σySubstituting (7) into (8), the stress of thin plate can be obtained as follows:(9)σx=-Ez1-μ2∂2ω∂x2+μ∂2ω∂y2σy=-Ez1-μ2∂2ω∂y2+μ∂2ω∂x2τxy=-Ez1+μ∂2ω∂x∂ywhereE is elasticity modulus,μis Poisson's ratio,u andv are displacement,ωis vertical deflection of the plate, and ω=ωx,y.


## 3. Stress Fields of the Elastic Thin Plate

A stress field is a region in a body for which the stress is defined at every point [18]. Stress fields were widely used in materials science. The stress field, which was composed of these generalized stresses, was called the generalized stress field.A hexahedron element with thicknessh was taken out from the thin plate (Figure 2); stress components acting on the four sides of hexahedron element wereMxdy,Mydx,Mxydy,Myxdx,Qxdy, andQydx, whereMx,My,Mxy,Myx,Qx, andQy are called the generalized stress. Generalized stress can be expressed as follows:(10)Mx=-∫-h/2h/2σxzdzMy=-∫-h/2h/2σyzdzMxy=Myx=-∫-h/2h/2τxyzdzQx=-∫-h/2h/2τzxdzQy=-∫-h/2h/2τzydzFigure 2
Stress of the plane.Substituting (9) into (10) yields(11)Mx=-D∂2ω∂x2+μ∂2ω∂y2My=-D∂2ω∂y2+μ∂2ω∂x2Mxy=-D1-μ∂2ω∂x∂yQx=-D∂∇2ω∂xQy=-D∂∇2ω∂ywhere∇2=∂2/∂x2+∂2/∂y2,∇2is the Laplacian operator, D=Eh3/121-μ2, andDis relative stiffness of the plate.Combining (9) and (11) gives(12)σx=12Mxzh3σy=12Myzh3τxy=12Mxyzh3τzx=-6Qxh3h24-z2τzy=-6Qyh3h24-z2Under the condition of axisymmetric loading, stress of the plane can be obtained(13)Mr=-Dd2wdr2+μrdwdrMθ=-Dμd2wdr2+1rdwdr


## 4. Bending Equations of the Thin Plate

As illustrated in Figure3, uniform load q(x,y) is applied on hexahedron element, while Mx, My, Mxy, Myx, Qx, and Qy are generalized stress on the four sides of the element.Figure 3
Internal force of the plane.The loading added up to zero in the direction of thez axis, and solving bending moments of parallel to thex axis and they axis yields(14)Qx=∂Mx∂x+∂Mxy∂y(15)Qy=∂My∂y+∂Mxy∂x(16)∂Qx∂x+∂Qy∂y+q=0Substituting (14) and (15) into (16) yields(17)∂2Mx∂x2+2∂2Mxy∂x∂y+∂2My∂y2+q=0Substituting (11) into (17) yields(18)D∇2∇2ωx,y=qx,ywhere ∇2=∂2/∂x2+∂2/∂y2,∇2 is the Laplace operator.According to the view of Winkler [1], a rectangle thin plate was considered to rest on a Winkler soil foundation, and the foundation reaction pressurep was only related to the vertical deflectionω of the plate.(19)p=kωwherek is reaction modulus of the foundation.The plate was loaded as follows:(20)qx,y-kωx,yA finite elastic thin plate on Winkler soil foundation; the bending equation was obtained as follows:(21)D∇2∇2ωx,y=qx,y-kωx,y


## 5. The Equations of Stress and Displacement

Under the axisymmetric loading, the bending equation of the thin plate on Winkler soil foundation in cylindrical can be obtained as follows:(22)D∇2∇2ωr=qr-kwrwhere ∇2=d2/dr2+1/rd/dr.The application of the first class of zero order Hankel transform to equation (22) yields(23)Dξ4ω-ξ+kω-ξ=q-ξwhere ω-ξ=∫0∞ωrJ0ξrrdr,q-(ξ)=∫0∞qrJ0(ξr)rdr.Solution for (23) yields [17](24)ω-ξ=q-ξDξ4+kwhere D=Eh3/12(1-μ2).Setl4=D/k; l is the relative stiffness radius of the plate on Winkler soil foundation. The application of Hankel transform to (24) and expression of deflection can be obtained as follows.(25)ωr=1k∫0∞q-ξJ0ξr1+l4ξ4ξdξSubstituting (25) into (13) yields(26)Mr=-Dd2ωrdr2+μrdωrdr=l4∫0∞q-ξJ0ξr1+l4ξ4ξJ0ξr-1-μrJ1ξrξ2dξ(27)Mθ=-Dμd2ωdr2+1rdωdr=l4∫0∞q-ξJ0ξr1+l4ξ4μξJ0ξr+1-μrJ1ξrξ2dξThe maximum radial stressσr and tangential stress σθ can be obtained as follows:(28)σr=6Mrh2(29)σθ=6Mθh2On the positionr=0, (25)-(27)can be transformed as follows:(30)ωr=1k∫0∞q-ξ1+l4ξ4ξdξ(31)Mr=l4∫0∞q-ξ1+l4ξ4ξ3dξ(32)Mθ=l4∫0∞q-ξ1+l4ξ4μξ3dξEquations (30)-(32) can be solved by numerical method. Combining equations (31) and (32) gives(33)Mθ=μMrThe thin plate for cement concrete pavement,μ, was less than 1. Therefore, σr was greater than σθ on the position r=0. The maximum stress σr was calculated in the following example.


## 6. Calculation Example

In order to check the validity of the method presented in the paper, an example of thin plate for cement concrete pavement was given. The material parameters were chosen as follows:h = 0.18 m andh = 0.22 m,E= 11000 MPa,μ= 0.15,k= 1.4×107 N/m3,δ= 0.151 mq(x,y) = 0.7 MPa. We selected the integral interval changed from 0-10, 0-20, 0-30, 0-40, 0-50, and 0-1000, respectively, the displacement and stress were calculated, and the results were summarized in Tables 1, 2, 3, and 4.Table 1
Displacement of plate in the middle position (h= 0.18 m).
integration interval 0-10 0-20 0-30 0-40 0-50 0-1000 displacement of the plate in the middle position/mm 0.614963 0.615458 0.615507 0.615473 0.615469 0.615433Table 2
Stresses of plate in the middle position (h= 0.18 m).
integration interval 0-10 0-20 0-30 0-40 0-50 0-1000 stress of the plate in the middle position /MPa 1.46 1.53 1.54 1.53 1.53 1.53Table 3
Displacement of plate in the middle position (h= 0.22 m).
integration interval 0-10 0-20 0-30 0-40 0-50 0-1000 displacement of the plate in the middle position/mm 0.594542 0.595469 0. 595564 0. 595535 0. 595531 0. 595529Table 4
Stresses of plate in the middle position (h= 0.22 m).
integration interval 0-10 0-20 0-30 0-40 0-50 0-1000 stress of the plate in the middle position /MPa 1.35 1.41 1.42 1.41 1.41 1.41According to the result of Table1, integral interval tends to 0-20; when the thicknesshof the thin plate is from 0.18 m toh= 0.22 m, displacement of the thin plate on the positionr= 0 was 0.615 mm and 0.595 mm, respectively. According to the results of Table 2, integral interval tends to 0-40; when the thicknesshof the thin plate is from 0.18 m toh= 0.22 m, stress of the thin plate on the positionr = 0 was 1.53 MPa and 1.41 MPa, respectively.


## 7. Validation of the Method

The material parameters were chosen as the calculation example; according to the literature [19], calculation procedure was as follows.Relative stiffness radiuslof the pavement is(34)lh=0.18m=Eh3121-μ2k4=0.25mlh=0.22m=Eh3121-μ2k4=0.42mBased on this, we can obtain coefficientα:(35)αh=0.18m=δlh=0.18m=0.604αh=0.22m=δlh=0.18m=0.356The deflection of the thin plate on the positionr= 0 can be found in the table [19].(36)α=0.604,ϖ=1.23×10-2α=0.356,ϖ=1.19×10-2Deflection of the plate on the positionr= 0 can be obtained as follows:(37)ωh=0.18m=pkϖ=0.714×1.23×10-2=6.15×10-4mωh=0.22m=pkϖ=0.714×1.19×10-2=5.95×10-4mThe bending moment coefficient on the positionr= 0 can be found in the table [19].(38)α=0.604,Mr¯=0.172α=0.356,Mr¯=0.237Bending moment of the plate on the positionr= 0 can be obtained as follows:(39)Mrh=0.18m=pMr¯=48×103×0.172=8.26×103NMrh=0.22m=pMr¯=48×103×0.237=11.37×103NRadial stress of the plate on the positionr= 0 can be obtained as follows:(40)σrh=0.18m=6Mrh2=6×8.26×1030.182=1.53MPaσrh=0.22m=6Mrh2=6×11.37×1030.222=1.41MPaFrom the above analysis, the stress and deflection based on the method were equal to using the method fromCode for Highway Design-Pavement [19]. It also indicated that the present method was reasonable in the paper.


## 8. Conclusions

( 1 ) The expressions of displacement and stress for cement concrete pavement on elastic foundation were derived.( 2 ) In order to verify the theoretical results, an example of thin plate for cement concrete pavement was given; the stress and deflection correspond with using the method fromCode for Highway Design-Pavement.( 3 ) Future experiment research is needed to be carried out to verify the theoretical results present in this paper.



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