In order to analyze the stress and displacement of pavement, a new form of the general solution of the elastic space axisymmetric problem is proposed by the method of mathematics reasoning. Depending on the displacement function put forward by Southwell, displacement function is derived based on Hankel transform and inverse Hankel transform. A new form of the general solution of the elastic space axisymmetric problem has been set up according to a few basic equations as the geometric equations, constitutive equations, and equilibrium equations. The present solution applies to elastic half-space foundation and Winkler foundation; the stress and displacement of pavement are obtained by mathematical deduction. The example results show that the proposed method is practically feasible.
National Natural Science Foundation of China515680441. Introduction
Winkler foundation and elastic half-space foundation have been widely used since Boussinseq represented the solution of the elastic space axisymmetric problem [1, 2]. Study of solutions for elastic space axisymmetric problem is of great interest for a number of researchers. Westergard gave the stress analysis of concrete pavement [3]. Love has obtained the approximate solution in the elastic half-space [4]. Cerruti presented solution of stress and displacement of elastic half-space [5]. Bagisbaev [6] and Rizzo and Shippy [7] made a study of the fundamental solution of axisymmetric elasticity problems. The common solutions of axisymmetric elastic space are Love solution and Southwell solution [8–11]. The idea of generalized images is applied to solve a contact problem [12]. Solution of a thin layer bonded on a viscoelastic medium is presented [13]. Green’s functions are obtained for an infinite prestressed thin plate on an elastic foundation under axisymmetric loading [14].
Great research achievements have been obtained for axisymmetric half-space contact problems. In this paper the potential method will be developed from the stage to which it has been carried previously, and a new form of the general solution has been set up. The relationship between the traditional and the present method is discussed in the last section.
This paper is organized as follows. In Section 2, a brief description is given of the fundamental equations and the new form of general solution. Section 3 is aimed at deriving the formulas for the relationship between general solutions. Sections 4 and 5 apply the new form of the general solution to elastic half-space foundation and Winkler foundation, and in Section 6 we finish with some concluding remarks.
2. General Solutions
For the elastic space axisymmetric contact problems, in the cylindrical coordinates (with the z-axis being positioned normal to the plane of isotropy), the fundamental equations can be rewritten in the following manner.
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 E is elasticity modulus; μ is Poisson’s ratio; u and w are displacement; G is shear modulus.
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.
We introduced χ(r,z) of the Southwell displacement function [1]; 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 (5) into (1) and (4) yields(6)∇2∇2χ=0.
According to (2), (3), and (5), the displacement components can be expressed as(7)u=1+μE1r∂2χ∂z2-2μυ2χ,w=1+μE1r∂∂z∂2χ∂z2-2υ2χ,where υ2=∂/∂r2-1/r∂/∂r+∂/∂z2 and υ2 is the Southwell operator.
From (2), (3), and (7), the expression of displacement component and stress component can be obtained as follows:(8)σr=1r∂∂r∂2χ∂z2-μυ2χ-1r2∂2χ∂z2-2μυ2χ,(9)σθ=μr∂υ2χ∂r+1r2∂2χ∂z2-2μυ2χ,(10)σz=-1r∂∂r∂2χ∂z2-1+μυ2χ,(11)τzr=1r∂∂z∂2χ∂z2-1+μυ2χ.
Substituting (8)–(11) into (1), displacement must satisfy the following equation:(12)υ4χ=0.
We can obtain (13) based on Hankel transform:(13)χr,z=∫0∞χ¯ξ,zJ1ξrrdξ.
Equation (13) can be rewritten as follows based on inverse Hankel transform:(14)χ¯ξ,z=∫0∞χr,zJ1ξrξdr.
A differentiation action is performed on (13), on both sides of r and z, and we can get the following equations:(15)∂χr,z∂r=∫0∞χ¯ξ,zJ0ξrξrdξ,(16)∂2χr,z∂r2=∫0∞χ¯ξ,zJ0ξrξdξ+∫0∞-χ¯ξ,zξ2J1ξrrdξ,(17)∂2χr,z∂z2=∫0∞∂2χ¯ξ,z∂z2J1ξrrdξ.
According to (15)–(17), the displacement can be expressed as follows:(18)υ2χ=∂∂r2-1r∂∂r+∂∂z2χr,z=∫0∞d2χ¯ξ,zdz2-ξ2χ¯ξ,zJ1ξrrdξ,(19)υ2υ2χ=∫0∞d2dz2-ξ22χ¯ξ,zJ1ξrrdξ.
Combining (12) and (19) gives(20)∫0∞d2dz2-ξ22χ¯ξ,zJ1ξrrdξ=0.
Equation (20) can be rewritten as follows based on Hankel transform:(21)d2dz2-ξ22χ¯ξ,z=0.
For a solution of the differential equation (21) yields the following general solution:(22)χ¯ξ,z=Aξ+Bξze-ξz+Cξ+Dξzeξz.
Substituting (22) into (13) yields(23)χr,z=∫0∞Aξ+Bξze-ξz+Cξ+DξzeξzJ1ξrrdξ.
Substituting (22) into (17) and (18) yields(24)∂2χr,z∂z2=∫0∞ξ2Aξ+-2ξ+ξ2zBξe-ξz+ξ2Cξ+2ξ+ξ2zDξeξzJ1ξrrdξ,(25)υ2χ=∫0∞-2ξBξe-ξz+2ξDξeξzJ1ξrrdξ.
Substituting (24)-(25) into (7)–(10), denoted by A=ξ2Aξ, B=ξBξ, C=ξ2Cξ, and D=ξDξ, yields(26)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+ξzDeξz}J1ξrdξ.
3. The Relationship between General Solutions
Parameters of Love solution are identified as AL, 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 general Love solution as follows:(27)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 of Southwell solution are identified as AS, BS, CS, and DS, denoted by A=AS+2μBS, B=-BS, C=CS+2μDS, and D=DS; we will just get the general Southwell solution as follows:(28)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.
We can convert Southwell solution into Love solution, denoted by AS=-AL-(2μ-1)BL, BS=BL, CS=CL+(2μ-1)DL, and DS=DL.
4. The Present Solution Applies to Elastic Half-Space Foundation
As shown in Figure 1, p is vertical circular uniform distributed load; δ is radius of the circle. Displacement can be obtained under the load.
Load scheme.
Boundary conditions are as follows:(29)z=0,τzr=0,σz=-pr≤δ0r>δ,(30)z⟶∞,σz,u=0.
Combining (25) and (30), we can get(31)C=D=0.
From (29) and (30), denoted by z=0, we can get(32)σz=-∫0∞ξA+2μBJ0ξrdξ=-p0≤r≤δ,τzr=-∫0∞ξA-1-2μBJ1ξrdξ=0.
Combining (29) and (32), we can get (33) based on Hankel transform:(33)A=1-2μpδJ1ξδξ,B=pδJ1ξδξ.
Substituting (32) and (33) into (25) yields(34)w=2pa1-μ2E∫0∞J1xJ0x/δxdx=F1212,-12;1,r2δ2r<δ2πr=δδ2rF1212,-12;1,r2δ2r>δ,where x=ξδ.
The result agreed with Love solution [1].
5. The Present Solution Applies to Winkler Foundation5.1. Model
Circular uniform distributed load on Winkler foundation is as shown in Figure 2.
Calculation sketch map.
Boundary conditions are as follows:(35)z=0,τzr=0,σz=-qrr≤δ0r>δ,z=h,τzr=0,σz=-kωr.
Substituting (25) into (35) yields(36)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.
We can obtain (37) as follows based on Hankel transform:(37)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 k is modulus of foundation reaction; h is the thickness of the pavement.
The solution of (37) can obtain the expression for A, B, C, and D about ξ, μ, E, h, and k. Substituting A, B, C, and D into (26), the stress and displacement of pavement can be obtained.
5.2. Examples
h = 0.2 m, E = 10000 MPa, μ= 0.15, k = 1.5 × 107 N/m3, δ= 0.151 m, and q = 700 KN/m2, as shown in Figure 3. We selected the integral intervals 0–10, 0–20, 0–30, 0–40, 0–50, and 0–1000, respectively, and calculated displacement at point A and stress at points A, B, and C; the results are summarized in Tables 1–4.
Displacement at point A.
Integral interval
0–10
0–20
0–30
0–40
0–50
0–1000
Displacement/mm
0.844091
0.852095
0.85396
0.852095
0.852095
0.852479
Stress at point A.
Integral interval
0–10
0–20
0–30
0–40
0–50
0–1000
Stress/MPa
−1.39
−1.83
−1.86
−1.74
−1.69
−1.69
Stress at point B.
Integral interval
0–10
0–20
0–30
0–40
0–50
0–1000
Stress/MPa
−0.008
0.007
0.009
−0.002
−0.001
−0.001
Stress at point C.
Integral interval
0–10
0–20
0–30
0–40
0–50
0–1000
Stress/MPa
1.33
1.58
1.59
1.58
1.58
1.58
Calculation sketch map.
According to Tables 1, 2, 3, and 4, displacement at point A is about 0.85 mm, stress at point A is −1.69 MPa, stress at point B is −0.01 MPa, and stress at point C is 1.58 MPa.
6. Conclusions
A new form of the general solution of elastic space axisymmetric problem was obtained based on mathematics reasoning. The present solution can provide a new method for the elastic space axisymmetric contact problems, and Love solution and Southwell solution can be obtained by using variable substitution.
According to the boundary condition and characteristics, the present solution can be divided into solving boundary solution, stress solution, and displacement solution. Thus, the present solution would make a very nice complement to the elastic space axisymmetric contact problems.
Competing Interests
The authors declare that they have no competing interests.
Acknowledgments
This work was supported by the National Natural Science Foundation of China (Grant no. 51568044).
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