# A New General Solution of Axisymmetric Elastic Space Problem in Pavement Engineering

Authors: Sun Lei
Journal: Mathematical Problems in Engineering
 (2023)

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## Abstract

The general solution of the axial symmetry elastic space problem in cement concrete pavement is an elementary question. Combining the Love method and the Southwell method, the Southwell operator was used in the variable selection process, the displacement function was introduced to express the displacement component by the Love method, the expression of stress indicated by the displacement function was obtained by combining the displacement components, geometric equations, and physical equations, then the stress was substituted into the equilibrium equation, the biharmonic equation of displacement function was obtained by mathematical operation, and finally, a new general solution of the axisymmetric elastic space problem in cement concrete pavement was obtained. According to the proposed general solution, mechanical calculation for a cement concrete pavement slab on Winkler foundation was carried out, the displacement curve at the top surface of the slab and the stress curve at the top, middle, and bottom surface of the slab were derived, and the results showed that the method is feasible in the calculation of cement concrete pavement. The general solution provided a new method for solving the axisymmetric elastic space problem of cement concrete pavement.


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## Body

## 1. Introduction

The problem of axisymmetric elastic space refers to the elastic body that the geometry, constraint condition, and load are symmetric to a certain axis, and the stress, strain, and displacement of the axisymmetric elastic space problem are also symmetrical to the axis [1–3]. In fact, an axisymmetric problem is a special form of spatial problem where all variables are independent of θ and only related to r and z. Therefore, the shear stress, shear strain, and displacement along the θ direction are all zero [4, 5].The common method for solving the general solution of axisymmetric elastic space in the pavement is usually the Love method or the Southwell method [6, 7]. Both the Love method [8] and the Southwell method [9] use the introduction of the displacement function to represent the displacement component, and by combining the displacement component represented by the displacement function, and the geometric and physical equation, then the expression for stress represented by the displacement function is obtained. Substituting the obtained stress into the equilibrium equation, a harmonic equation about the displacement function is obtained, and solving this equation gives the expression of the displacement function; substituting the obtained expression into the displacement function to represent the displacement and stress, the general solution of the axisymmetric elastic space problem is obtained [10–12].The solution of displacement and stress of the cement concrete pavement belongs to the axisymmetric elastic space problem [13]. Scholars have achieved many results in this area [14, 15]; however, it is difficult to find the stress function for pavement [16, 17]. The author adopted the Southwell operator without variable substitution but directly adopted the idea of the Love method to obtain a general solution of the axisymmetric elastic space problem, and this general solution undergoes certain form changes to derive the general solutions of the Southwell and Love methods. According to the proposed general solution, the mechanical calculation was carried out for a cement concrete pavement slab on Winkler foundation, and the displacement curve at the top surface of the slab and the stress curve at the top, middle, and bottom surfaces of the slab were obtained to verify the feasibility of the method in the cement concrete pavement. The general solution provided a new method for solving the axisymmetric elastic space problem.


## 2. The Derivation of General Solution

The cylindrical coordinate system is adopted as solving the axisymmetric elastic space problem in pavement engineering. The basic equations are shown as follows.Equilibrium equations are as follows:(1)∂σr∂r+∂τzr∂z+σr−σθr=0,∂σz∂z+∂τzr∂r+τzrr=0.Geometric equations are as follows:(2)εr=∂u∂r,εθ=ur,εz=∂w∂z,γzr=∂w∂r+∂u∂z.Constitutive equations are as follows:(3)εr=1Eσr−μσθ+σz,εθ=1Eσθ−μσr+σz,εz=1Eσz−μσr+σθ,γzr=1Gτzr,where μ is Poisson’s ratio, E is elasticity modulus, u and w are displacement of pavement, and G is shear modulus.The compatibility equation is as follows:(4)∇2σr−2r2σr−σθ+11+μ∂2Θ∂r2=0,∇2σθ−2r2σr−σθ+11+μ1r∂Θ∂r=0,∇2σz+11+μ∂2Θ∂r2=0,∇2τzr−τzrr2+11+μ∂2Θ∂r∂z=0,where ∇2=∂2/∂r2+1/r∂/∂r+∂2/∂z2,Θ=σr+σθ+σz.The stress can be obtained as follows:(5)σr=∂∂zμ∇2χ−∂2χ∂r2,σθ=∂∂zμ∇2χ−1r∂χ∂r,σz=∂∂z2−μ∇2χ−∂2χ∂z2,τzr=τrz=∂∂r1−μ∇2χ−∂2χ∂z2.Substituting equation (5) into equations (1) and (4) yields(6)∇2∇2χ=0.According to equations (2), (3), and (5), the displacement component can be expressed as follows:(7)u=1+μE1r∂2χ∂z2−2μυ2χ,(8)w=1+μE1r∂∂z∂2χ∂z2−2υ2χ,where υ2=∂/∂r2−1/r∂/∂r+∂/∂z2, υ2 is the Southwell operator.From equations (2), (3), (7), and (8), the expression of the displacement component and stress component is as follows:(9)σr=1r∂∂r∂2χ∂z2−μυ2χ−1r2∂2χ∂z2−2μυ2χ,(10)σθ=μr∂υ2χ∂r+1r2∂2χ∂z2−2μυ2χ,(11)σz=−1r∂∂r∂2χ∂z2−1+μυ2χ,(12)τzr=1r∂∂z∂2χ∂z2−1+μυ2χ.Substituting equations (9)–(12) into equation (1), displacement must be satisfied by the following equation:(13)υ4χ=0.According to Hankel transform, we can obtain the following equation:(14)χr,z=∫0∞χ¯ξ,zJ1ξrrdξ.According to Hankel inverse transform, equation (14) can be rewritten as the following equation:(15)χ¯ξ,z=∫0∞χr,zJ1ξrξdr.The differential operation is performed on both sides of equation (14) r and z, and we can get the following equations:(16)∂χr,z∂r=∫0∞χ¯ξ,zJ0ξrξrdξ,(17)∂2χr,z∂r2=∫0∞χ¯ξ,zJ0ξrξdξ+∫0∞−χ¯ξ,zξ2J1ξrrdξ,(18)∂2χr,z∂z2=∫0∞∂2χ¯ξ,z∂z2J1ξrrdξ.According to equations (16)–(18), the displacement of the cement concrete pavement can be expressed as follows:(19)υ2χ=∂∂r2−1r∂∂r+∂∂z2χr,z=∫0∞d2χ¯ξ,zdz2−ξ2χ¯ξ,zJ1ξrrdξ,(20)υ2υ2χ=∫0∞d2dz2−ξ22χ¯ξ,zJ1ξrrdξ.Combining equations (13) and (20), we can draw the conclusion as the following equation:(21)∫0∞d2dz2−ξ22χ¯ξ,zJ1ξrrdξ=0.According to Hankel transform, equation (21) can be rewritten as the following equation:(22)d2dz2−ξ22χ¯ξ,z=0.General solution is constructed to solve differential equation (22).(23)χ¯ξ,z=Aξ+Bξze−ξz+Cξ+Dξzeξz.Substituting equation (23) into equation (14) yields:(24)χr,z=∫0∞Aξ+Bξze−ξz+Cξ+DξzeξzJ1ξrrdξ.Substituting equation (23) into equations (18) and (19) yields(25)∂2χr,z∂z2=∫0∞ξ2Aξ+−2ξ+ξ2zBξe−ξz+ξ2Cξ+2ξ+ξ2zDξeξzJ1ξrrdξ,(26)υ2χ=∫0∞−2ξBξe−ξz+2ξDξeξzJ1ξrrdξ.Substituting equations (25) and (26) into equations (7)–(11), denoted by A = ξ2Aξ, B = ξBξ, C = ξ2Cξ, D = ξDξ, yield(27)σr=∫0∞ξA−2−2μ−ξzBe−ξz+C+2−2μ+ξzDeξzJ0ξrdξ−1rU,σθ=−2μ∫0∞ξBe−ξz−DeξzJ0ξrdξ+1rU,σz=−∫0∞ξA+2μ+ξzBe−ξz+C−2μ−ξzDeξzJ0ξrdξ,τzr=−∫0∞ξA−1−2μ−ξzBe−ξz−C+1−2μ+ξzDeξzJ1ξrdξ,u=1+μEU,w=1+μE∫0∞A+1+ξzBe−ξz−C−1−ξzDeξzJ0ξrdξ,where U=∫0∞A−2−4μ−ξzBe−ξz+C+2−4μ+ξzDeξzJ1ξrdξ.


## 3. The Relationship between General Solutions

Parameters are identified asAL, BL, CL, and DL. Denoted by A = AL + (4μ − 1)BL, B = −BL, C=CL+4μ−1CL, and D=DL, we will just get the Love solution as follows by the equation of (27):(28)σr=−∫0∞ξAL−1+2μ−ξzBLe−ξz−CL+1+2μ+ξzDLeξzJ0ξrdξ+1rUL,σθ=2μ∫0∞ξBLe−ξz+DLeξzJ0ξrdξ−1rUL,σz=∫0∞ξAL+1−2μ+ξzBLe−ξz−CL−1−2μ−ξzDLeξzJ0ξrdξ,τzr=∫0∞ξAL−2μ−ξzBLe−ξz+CL+2μ+ξzDLeξzJ1ξrdξ,u=−1+μEUL,w=−1+μE∫0∞AL+2−4μ+ξzBLe−ξz+CL−2−4μ−ξzDLeξzJ0ξrdξ,where UL=∫0∞AL−1−ξzBLe−ξz−CL+1+ξzDLeξzJ1ξrdξ.Parameters are identified asAS, BS, CS, and DS. Denoted by A = As + 2μBs, B = −Bs, C = Cs + 2μDs, and D = Ds, we will just get the Southwell solution as follows by the equation of (27):(29)σr=∫0∞ξAS+2−ξzBSe−ξz+CS+2+ξzDSeξzJ0ξrdξ−1rUS,σθ=2μ∫0∞ξBSe−ξz+DSeξzJ0ξrdξ+1rUS,σz=−∫0∞ξAS−ξzBSe−ξz+CS+ξzDSeξzJ0ξrdξ,τzr=−∫0∞ξAS+1−ξzBSe−ξz−CS+1+ξzDSeξzJ1ξrdξ,u=1+μEUS,w=1+μE∫0∞AS−1−2μ+ξzBSe−ξz−CS−1−2μ−ξzDSeξzJ0ξrdξ,where US=∫0∞AS+2−2μ−ξzBSe−ξz+CS+2−2μ+ξzDSeξzJ1ξrdξ.Denoted byAs = −AL − (2μ − 1)BL, Bs = BL, Cs = CL + (2μ − 1)DL, and Ds = DL, we can convert the Southwell solution into Love solution.Combining these results leads to the general transformation relationship between solutions as shown in Table1.Table 1
Conversion relations between general solutions.
Love solutionSouthwell solutionThe present solutionLove solution—AL=−AS+1−2μBS,BL=BSCL=CS+1−2μDS,DL=DSAL=−A+1−4μB,BL=−BCL=C+1−4μD,DL=DSouthwell solutionAS=−AL+2μ−1BL,BS=BLCS=CL+2μ−1DL,DS=DL—AS=A+2μB,BS=−BCS=C−2μD,DS=DThe present solutionA=−AL+4μ−1BL,B=−BLC=CL+4μ−1CL,D=DLA=AS+2μBS,B=−BSC=CS+2μDS,D=DS—


## 4. Application of General Solution on Winkler Foundation

As shown in Figure1, δ is radius of the circle, and p is vertical circular uniform distributed load.Figure 1
Load distribution scheme.Boundary conditions are as follows:(30)z=0,τzr=0,σz=−p,r≤δ,0,r>δ,(31)z⟶∞,σ,u=0.Combining equations (26) and (31), we can get(32)C=D=0.Denoted byz = 0, from equations (30) and (31), we can get(33)σz=−∫0∞ξA+2μBJ0ξrdξ=−p0≤r≤δ,τzr=−∫0∞ξA−1−2μBJ1ξrdξ=0.Combining equations (30) and (33), we can get the following equation:(34)A=1−2μpδJ1ξδξ,B=pδJ1ξδξ.Substituting equations (33) and (34) into equation (26) yield(35)w=2pa1−μ2E∫0∞J1xJ0x/δxdx=2F112,−12;1,r2δ2,r<δ,2π,r=δ,δ2r2F112,−12;1,r2δ2,r>δ,where x = ξδ.


## 5. The Proposed General Solution Apply to Winkler Foundation

### 5.1. Model

The arrangement of the circular uniform load on the Winkler foundation is shown in Figure2.Figure 2
Calculation sketch map.Boundary conditions are as follows:(36)z=0,τzr=0,σz=−qr,r≤δ,0,r>δ,(37)z=h,τzr=0,σz=−kωr.Substituting equation (26) into equations (36) and (37) yields(38)σz=−∫0∞ξA+2μB+C−2μDJ0ξrdξ=−qr,τzr=−∫0∞ξA−1−2μB−C+1−2μDJ1ξrdξ=0,σz=−∫0∞ξA+2μ+ξhBe−ξh+C−2μ−ξhDeξhJ0ξrdξ=−kωr,τzr=−∫0∞ξA−1−2μ−ξhBe−ξh−C+1−2μ+ξhDeξhJ1ξrdξ=0.According to Hankel transform, we can obtain the following equation:(39)A+2μB+C−2μD=−q¯ξ,A−1−2μB−C+1−2μD=0,A+2μ+ξhBe−ξh+C−2μ−ξhDeξh=kω¯ξ,A−1−2μ−ξhBe−ξh−C+1−2μ+ξhDeξh=0,where h is the thickness of the cement concrete slab; k is Winkler foundation model.The solution of equation (39) can obtain the expression for A, B, C, and D about ξ, μ, E, h, and k. Substituting A, B, C, and D into equation (27), the displacement and stress of cement concrete pavement can be obtained.


### 5.2. Examples

The calculation diagram of cement concrete pavement is shown in Figure3, the specific calculation parameters are E = 11500 MPa, h = 0.18 m, μ = 0.15, k = 1.45 × 107 N/m3, p = 700 kN/m2, r = 0.151 m, I choose integral interval 0–10, 0–20, 0–30, 0–40, 0–50, 0–500 and calculate displacement at point a, stress at point a, b, c, and the results are summarized in Tables 1–4.Figure 3
Calculation sketch of cement concrete pavement.Table 2
Displacement at pointa.
Integral interval0–100–200–300–400–500–500Displacement (mm)0.8560910.8640950.8641960.8642950.8643950.864479Table 3
Stress at pointa.
Integral interval0–100–200–300–400–500–500Stress (MPa)−1.15−1.12−1.11−1.12−1.10−1.10Table 4
Stress at pointb.
Integral interval0–100–200–300–400–500–500Stress (MPa)−0.010−0.026−0.028−0.025−0.028−0.028Based on Table2, displacement at point a is about 0.86 mm. According to Tables 3−5, stress at point a is −1.1 MPa, stress at point b is −0.028 MPa, and stress at point c is 1.09 MPa.Table 5
Stress at pointc.
Integral interval0–100–200–300–400–500–1000Stress (MPa)1.111.101.091.091.091.09


### 5.3. Variation of Displacement and Stress withr

The relative stiffness radiusl of cement concrete pavement is as follows:(40)l=Eh3121−μ2k4=0.853m.l ∈ (0.8, 1.5), stresses and displacement vary with cement concrete pavement board size, and the range of finite size slab instead of infinite size slab is referred to Zheng [6]:(41)r=0.853×5+0.1=4.4m.Calculation range ofr on the Winkler foundation is 0 to 4.4 m. According to the former calculation models, the displacement at point a on the top surface of the slab is calculated to vary with r as shown in Figure 4, and the stresses at points a, b, and c vary with r as shown in Figures 5−7, respectively.Figure 4
Displacement varied withr at point a.Figure 5
Stress varied withr at point a.Figure 6
Stress varied withr at point b.Figure 7
Stress varied withr at point c.From Figure4, the displacement at point a decreases rapidly with the increase of r, and the displacement at point a gradually tends to a constant value when the r is greater than 3 m. From Figures 5–7, it can be seen that the stresses at each point occurred great changing amplitude with the increase of r. In particular, the stress of the plate changes from compressive stress to tensile stress (Figure 5) or changes from tensile to compressive stresses (Figures 6 and 7) with the increase of r.


## 5.1. Model

The arrangement of the circular uniform load on the Winkler foundation is shown in Figure2.Figure 2
Calculation sketch map.Boundary conditions are as follows:(36)z=0,τzr=0,σz=−qr,r≤δ,0,r>δ,(37)z=h,τzr=0,σz=−kωr.Substituting equation (26) into equations (36) and (37) yields(38)σz=−∫0∞ξA+2μB+C−2μDJ0ξrdξ=−qr,τzr=−∫0∞ξA−1−2μB−C+1−2μDJ1ξrdξ=0,σz=−∫0∞ξA+2μ+ξhBe−ξh+C−2μ−ξhDeξhJ0ξrdξ=−kωr,τzr=−∫0∞ξA−1−2μ−ξhBe−ξh−C+1−2μ+ξhDeξhJ1ξrdξ=0.According to Hankel transform, we can obtain the following equation:(39)A+2μB+C−2μD=−q¯ξ,A−1−2μB−C+1−2μD=0,A+2μ+ξhBe−ξh+C−2μ−ξhDeξh=kω¯ξ,A−1−2μ−ξhBe−ξh−C+1−2μ+ξhDeξh=0,where h is the thickness of the cement concrete slab; k is Winkler foundation model.The solution of equation (39) can obtain the expression for A, B, C, and D about ξ, μ, E, h, and k. Substituting A, B, C, and D into equation (27), the displacement and stress of cement concrete pavement can be obtained.


## 5.2. Examples

The calculation diagram of cement concrete pavement is shown in Figure3, the specific calculation parameters are E = 11500 MPa, h = 0.18 m, μ = 0.15, k = 1.45 × 107 N/m3, p = 700 kN/m2, r = 0.151 m, I choose integral interval 0–10, 0–20, 0–30, 0–40, 0–50, 0–500 and calculate displacement at point a, stress at point a, b, c, and the results are summarized in Tables 1–4.Figure 3
Calculation sketch of cement concrete pavement.Table 2
Displacement at pointa.
Integral interval0–100–200–300–400–500–500Displacement (mm)0.8560910.8640950.8641960.8642950.8643950.864479Table 3
Stress at pointa.
Integral interval0–100–200–300–400–500–500Stress (MPa)−1.15−1.12−1.11−1.12−1.10−1.10Table 4
Stress at pointb.
Integral interval0–100–200–300–400–500–500Stress (MPa)−0.010−0.026−0.028−0.025−0.028−0.028Based on Table2, displacement at point a is about 0.86 mm. According to Tables 3−5, stress at point a is −1.1 MPa, stress at point b is −0.028 MPa, and stress at point c is 1.09 MPa.Table 5
Stress at pointc.
Integral interval0–100–200–300–400–500–1000Stress (MPa)1.111.101.091.091.091.09


## 5.3. Variation of Displacement and Stress withr

The relative stiffness radiusl of cement concrete pavement is as follows:(40)l=Eh3121−μ2k4=0.853m.l ∈ (0.8, 1.5), stresses and displacement vary with cement concrete pavement board size, and the range of finite size slab instead of infinite size slab is referred to Zheng [6]:(41)r=0.853×5+0.1=4.4m.Calculation range ofr on the Winkler foundation is 0 to 4.4 m. According to the former calculation models, the displacement at point a on the top surface of the slab is calculated to vary with r as shown in Figure 4, and the stresses at points a, b, and c vary with r as shown in Figures 5−7, respectively.Figure 4
Displacement varied withr at point a.Figure 5
Stress varied withr at point a.Figure 6
Stress varied withr at point b.Figure 7
Stress varied withr at point c.From Figure4, the displacement at point a decreases rapidly with the increase of r, and the displacement at point a gradually tends to a constant value when the r is greater than 3 m. From Figures 5–7, it can be seen that the stresses at each point occurred great changing amplitude with the increase of r. In particular, the stress of the plate changes from compressive stress to tensile stress (Figure 5) or changes from tensile to compressive stresses (Figures 6 and 7) with the increase of r.


## 6. Conclusions

Using the compatibility equation of the Southwell method and the calculating idea of the Love method, a new general solution of the axisymmetric elastic space problem was derived in pavement engineering. By variable transform, the general solution can be transformed into the Southwell solution or Love solution.According to the boundary conditions of the pavement and the characteristics of the general solution, the general solution is divided into the general solution for solving the composite boundary, the general solution for the stress problem, and the general solution for the displacement problem, and these three types of problem are represented in the form of the Love general solution, the Southwell general solution, and the proposed general solution in the paper. The general solution provides a broader idea for analyzing the axisymmetric elastic space problem of cement concrete pavement.Using the general solution proposed by the author, the mechanical calculation of a cement concrete pavement slab on Winkler foundation as an illustration is carried out, the displacement curve at the top surface of the slab and the stress curves at the top, middle, and bottom surface of the slab were obtained, and the result showed that the method is feasible in the calculation of cement concrete pavement.



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