# A New Form of the General Solution of the Elastic Space Axisymmetric Problem in Pavement Mechanics

Authors: Jia Liang; Guo Jian; He Shikai
Journal: Mathematical Problems in Engineering
 (2016)

---

## Abstract

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.


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## Body

## 1. 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 Section2, 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 thez-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)
∂
σ
r
∂
r
+
∂
τ
z
r
∂
z
+
σ
r
-
σ
θ
r
=
0
,
∂
σ
z
∂
z
+
∂
τ
z
r
∂
r
+
τ
z
r
r
=
0
.Geometric equations are as follows:(2)
ε
r
=
∂
u
∂
r
,
ε
θ
=
u
r
,
ε
z
=
∂
w
∂
z
,
γ
z
r
=
∂
w
∂
r
+
∂
u
∂
z
.Constitutive equations are as follows:(3)
ε
r
=
1
E
σ
r
-
μ
σ
θ
+
σ
z
,
ε
θ
=
1
E
σ
θ
-
μ
σ
r
+
σ
z
,
ε
z
=
1
E
σ
z
-
μ
σ
r
+
σ
θ
,
γ
z
r
=
1
G
τ
z
r
,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)
∇
2
σ
r
-
2
r
2
σ
r
-
σ
θ
+
1
1
+
μ
∂
2
Θ
∂
r
2
=
0
,
∇
2
σ
θ
-
2
r
2
σ
r
-
σ
θ
+
1
1
+
μ
1
r
∂
Θ
∂
r
=
0
,
∇
2
σ
z
+
1
1
+
μ
∂
2
Θ
∂
r
2
=
0
,
∇
2
τ
z
r
-
τ
z
r
r
2
+
1
1
+
μ
∂
2
Θ
∂
r
∂
z
=
0
,where ∇
2
=
∂
2
/
∂
r
2
+
1
/
r
∂
/
∂
r
+
∂
2
/
∂
z
2, Θ
=
σ
r
+
σ
θ
+
σ
z.We introducedχ
(
r
,
z
) of the Southwell displacement function [1]; the stress can be obtained as follows:(5)
σ
r
=
∂
∂
z
μ
∇
2
χ
-
∂
2
χ
∂
r
2
,
σ
θ
=
∂
∂
z
μ
∇
2
χ
-
1
r
∂
χ
∂
r
,
σ
z
=
∂
∂
z
2
-
μ
∇
2
χ
-
∂
2
χ
∂
z
2
,
τ
z
r
=
τ
r
z
=
∂
∂
r
1
-
μ
∇
2
χ
-
∂
2
χ
∂
z
2
.Substituting (5) into (1) and (4) yields(6)
∇
2
∇
2
χ
=
0
.According to (2), (3), and (5), the displacement components can be expressed as(7)
u
=
1
+
μ
E
1
r
∂
2
χ
∂
z
2
-
2
μ
υ
2
χ
,
w
=
1
+
μ
E
1
r
∂
∂
z
∂
2
χ
∂
z
2
-
2
υ
2
χ
,where υ
2
=
∂
/
∂
r
2
-
1
/
r
∂
/
∂
r
+
∂
/
∂
z
2 and υ
2 is the Southwell operator.From (2), (3), and (7), the expression of displacement component and stress component can be obtained as follows:(8)
σ
r
=
1
r
∂
∂
r
∂
2
χ
∂
z
2
-
μ
υ
2
χ
-
1
r
2
∂
2
χ
∂
z
2
-
2
μ
υ
2
χ
,
(9)
σ
θ
=
μ
r
∂
υ
2
χ
∂
r
+
1
r
2
∂
2
χ
∂
z
2
-
2
μ
υ
2
χ
,
(10)
σ
z
=
-
1
r
∂
∂
r
∂
2
χ
∂
z
2
-
1
+
μ
υ
2
χ
,
(11)
τ
z
r
=
1
r
∂
∂
z
∂
2
χ
∂
z
2
-
1
+
μ
υ
2
χ
.Substituting (8)–(11) into (1), displacement must satisfy the following equation:(12)
υ
4
χ
=
0
.We can obtain (13) based on Hankel transform:(13)
χ
r
,
z
=
∫
0
∞
χ
¯
ξ
,
z
J
1
ξ
r
r
d
ξ
.Equation (13) can be rewritten as follows based on inverse Hankel transform:(14)
χ
¯
ξ
,
z
=
∫
0
∞
χ
r
,
z
J
1
ξ
r
ξ
d
r
.A differentiation action is performed on (13), on both sides ofr andz, and we can get the following equations:(15)
∂
χ
r
,
z
∂
r
=
∫
0
∞
χ
¯
ξ
,
z
J
0
ξ
r
ξ
r
d
ξ
,
(16)
∂
2
χ
r
,
z
∂
r
2
=
∫
0
∞
χ
¯
ξ
,
z
J
0
ξ
r
ξ
d
ξ
+
∫
0
∞
-
χ
¯
ξ
,
z
ξ
2
J
1
ξ
r
r
d
ξ
,
(17)
∂
2
χ
r
,
z
∂
z
2
=
∫
0
∞
∂
2
χ
¯
ξ
,
z
∂
z
2
J
1
ξ
r
r
d
ξ
.According to (15)–(17), the displacement can be expressed as follows:(18)
υ
2
χ
=
∂
∂
r
2
-
1
r
∂
∂
r
+
∂
∂
z
2
χ
r
,
z
=
∫
0
∞
d
2
χ
¯
ξ
,
z
d
z
2
-
ξ
2
χ
¯
ξ
,
z
J
1
ξ
r
r
d
ξ
,
(19)
υ
2
υ
2
χ
=
∫
0
∞
d
2
d
z
2
-
ξ
2
2
χ
¯
ξ
,
z
J
1
ξ
r
r
d
ξ
.Combining (12) and (19) gives(20)
∫
0
∞
d
2
d
z
2
-
ξ
2
2
χ
¯
ξ
,
z
J
1
ξ
r
r
d
ξ
=
0
.Equation (20) can be rewritten as follows based on Hankel transform:(21)
d
2
d
z
2
-
ξ
2
2
χ
¯
ξ
,
z
=
0
.For a solution of the differential equation (21) yields the following general solution:(22)
χ
¯
ξ
,
z
=
A
ξ
+
B
ξ
z
e
-
ξ
z
+
C
ξ
+
D
ξ
z
e
ξ
z
.Substituting (22) into (13) yields(23)
χ
r
,
z
=
∫
0
∞
A
ξ
+
B
ξ
z
e
-
ξ
z
+
C
ξ
+
D
ξ
z
e
ξ
z
J
1
ξ
r
r
d
ξ
.Substituting (22) into (17) and (18) yields(24)
∂
2
χ
r
,
z
∂
z
2
=
∫
0
∞
ξ
2
A
ξ
+
-
2
ξ
+
ξ
2
z
B
ξ
e
-
ξ
z
+
ξ
2
C
ξ
+
2
ξ
+
ξ
2
z
D
ξ
e
ξ
z
J
1
ξ
r
r
d
ξ
,
(25)
υ
2
χ
=
∫
0
∞
-
2
ξ
B
ξ
e
-
ξ
z
+
2
ξ
D
ξ
e
ξ
z
J
1
ξ
r
r
d
ξ
.Substituting (24)-(25) into (7)–(10), denoted by A
=
ξ
2
A
ξ, B
=
ξ
B
ξ, C
=
ξ
2
C
ξ, and D
=
ξ
D
ξ, yields(26)
σ
r
=
∫
0
∞
ξ
A
-
2
-
2
μ
-
ξ
z
B
e
-
ξ
z
+
C
+
2
-
2
μ
+
ξ
z
D
e
ξ
z
J
0
ξ
r
d
ξ
-
1
r
U
,
σ
θ
=
-
2
μ
∫
0
∞
ξ
B
e
-
ξ
z
-
D
e
ξ
z
J
0
ξ
r
d
ξ
+
1
r
U
,
σ
z
=
-
∫
0
∞
ξ
A
+
2
μ
+
ξ
z
B
e
-
ξ
z
+
C
-
2
μ
-
ξ
z
D
e
ξ
z
J
0
ξ
r
d
ξ
,
τ
z
r
=
-
∫
0
∞
ξ
A
-
1
-
2
μ
-
ξ
z
B
e
-
ξ
z
-
C
+
1
-
2
μ
+
ξ
z
D
e
ξ
z
J
1
ξ
r
d
ξ
,
u
=
1
+
μ
E
U
,
w
=
1
+
μ
E
∫
0
∞
A
+
1
+
ξ
z
B
e
-
ξ
z
-
C
-
1
-
ξ
z
D
e
ξ
z
J
0
ξ
r
d
ξ
,where U
=
∫
0
∞
{
A
-
2
-
4
μ
-
ξ
z
B
e
-
ξ
z
+
C
+
2
-
4
μ
+
ξ
z
D
e
ξ
z
}
J
1
ξ
r
d
ξ.


## 3. The Relationship between General Solutions

Parameters of Love solution are identified asA
L, B
L, C
L, and D
L, denoted by A
=
A
L
+
(
4
μ
-
1
)
B
L, B
=
-
B
L, C
=
C
L
+
4
μ
-
1
C
L, and D
=
D
L; we will just get the general Love solution as follows:(27)
σ
r
=
-
∫
0
∞
ξ
A
L
-
1
+
2
μ
-
ξ
z
B
L
e
-
ξ
z
-
C
L
+
1
+
2
μ
+
ξ
z
D
L
e
ξ
z
J
0
ξ
r
d
ξ
+
1
r
U
L
,
σ
θ
=
2
μ
∫
0
∞
ξ
B
L
e
-
ξ
z
+
D
L
e
ξ
z
J
0
ξ
r
d
ξ
-
1
r
U
L
,
σ
z
=
∫
0
∞
ξ
A
L
+
1
-
2
μ
+
ξ
z
B
L
e
-
ξ
z
-
C
L
-
1
-
2
μ
-
ξ
z
D
L
e
ξ
z
J
0
ξ
r
d
ξ
,
τ
z
r
=
∫
0
∞
ξ
A
L
-
2
μ
-
ξ
z
B
L
e
-
ξ
z
+
C
L
+
2
μ
+
ξ
z
D
L
e
ξ
z
J
1
ξ
r
d
ξ
,
u
=
-
1
+
μ
E
U
L
,
w
=
-
1
+
μ
E
∫
0
∞
A
L
+
2
-
4
μ
+
ξ
z
B
L
e
-
ξ
z
+
C
L
-
2
-
4
μ
-
ξ
z
D
L
e
ξ
z
J
0
ξ
r
d
ξ
,where U
L
=
∫
0
∞
A
L
-
1
-
ξ
z
B
L
e
-
ξ
z
-
C
L
+
1
+
ξ
z
D
L
e
ξ
z
J
1
ξ
r
d
ξ.Parameters of Southwell solution are identified asA
S, B
S, C
S, and D
S, denoted by A
=
A
S
+
2
μ
B
S, B
=
-
B
S, C
=
C
S
+
2
μ
D
S, and D
=
D
S; we will just get the general Southwell solution as follows:(28)
σ
r
=
∫
0
∞
ξ
A
S
+
2
-
ξ
z
B
S
e
-
ξ
z
+
C
S
+
2
+
ξ
z
D
S
e
ξ
z
J
0
ξ
r
d
ξ
-
1
r
U
S
,
σ
θ
=
2
μ
∫
0
∞
ξ
B
S
e
-
ξ
z
+
D
S
e
ξ
z
J
0
ξ
r
d
ξ
+
1
r
U
S
,
σ
z
=
-
∫
0
∞
ξ
A
S
-
ξ
z
B
S
e
-
ξ
z
+
C
S
+
ξ
z
D
S
e
ξ
z
J
0
ξ
r
d
ξ
,
τ
z
r
=
-
∫
0
∞
ξ
A
S
+
1
-
ξ
z
B
S
e
-
ξ
z
-
C
S
+
1
+
ξ
z
D
S
e
ξ
z
J
1
ξ
r
d
ξ
,
u
=
1
+
μ
E
U
S
,
w
=
1
+
μ
E
∫
0
∞
A
S
-
1
-
2
μ
+
ξ
z
B
S
e
-
ξ
z
-
C
S
-
1
-
2
μ
-
ξ
z
D
S
e
ξ
z
J
0
ξ
r
d
ξ
,where U
S
=
∫
0
∞
A
S
+
2
-
2
μ
-
ξ
z
B
S
e
-
ξ
z
+
C
S
+
2
-
2
μ
+
ξ
z
D
S
e
ξ
z
J
1
ξ
r
d
ξ.We can convert Southwell solution into Love solution, denoted byA
S
=
-
A
L
-
(
2
μ
-
1
)
B
L, B
S
=
B
L, C
S
=
C
L
+
(
2
μ
-
1
)
D
L, and D
S
=
D
L.


## 4. The Present Solution Applies to Elastic Half-Space Foundation

As shown in Figure1,p is vertical circular uniform distributed load; δ is radius of the circle. Displacement can be obtained under the load.Figure 1
Load scheme.Boundary conditions are as follows:(29)
z
=
0
,
τ
z
r
=
0
,
σ
z
=
-
p
r
≤
δ
0
r
>
δ
,
(30)
z
⟶
∞
,
σ
z
,
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)
σ
z
=
-
∫
0
∞
ξ
A
+
2
μ
B
J
0
ξ
r
d
ξ
=
-
p
0
≤
r
≤
δ
,
τ
z
r
=
-
∫
0
∞
ξ
A
-
1
-
2
μ
B
J
1
ξ
r
d
ξ
=
0
.Combining (29) and (32), we can get (33) based on Hankel transform:(33)
A
=
1
-
2
μ
p
δ
J
1
ξ
δ
ξ
,
B
=
p
δ
J
1
ξ
δ
ξ
.Substituting (32) and (33) into (25) yields(34)
w
=
2
p
a
1
-
μ
2
E
∫
0
∞
J
1
x
J
0
x
/
δ
x
d
x
=
F
1
2
1
2
,
-
1
2
;
1
,
r
2
δ
2
r
<
δ
2
π
r
=
δ
δ
2
r
F
1
2
1
2
,
-
1
2
;
1
,
r
2
δ
2
r
>
δ
,where x
=
ξ
δ.The result agreed with Love solution [1].


## 5. The Present Solution Applies to Winkler Foundation

### 5.1. Model

Circular uniform distributed load on Winkler foundation is as shown in Figure2.Figure 2
Calculation sketch map.Boundary conditions are as follows:(35)
z
=
0
,
τ
z
r
=
0
,
σ
z
=
-
q
r
r
≤
δ
0
r
>
δ
,
z
=
h
,
τ
z
r
=
0
,
σ
z
=
-
k
ω
r
.Substituting (25) into (35) yields(36)
σ
z
=
-
∫
0
∞
ξ
A
+
2
μ
B
+
C
-
2
μ
D
J
0
ξ
r
d
ξ
=
-
q
r
,
τ
z
r
=
-
∫
0
∞
ξ
A
-
1
-
2
μ
B
-
C
+
1
-
2
μ
D
J
1
ξ
r
d
ξ
=
0
,
σ
z
=
-
∫
0
∞
ξ
A
+
2
μ
+
ξ
h
B
e
-
ξ
h
+
C
-
2
μ
-
ξ
h
D
e
ξ
h
J
0
ξ
r
d
ξ
=
-
k
ω
r
,
τ
z
r
=
-
∫
0
∞
ξ
A
-
1
-
2
μ
-
ξ
h
B
e
-
ξ
h
-
C
+
1
-
2
μ
+
ξ
h
D
e
ξ
h
J
1
ξ
r
d
ξ
=
0
.We can obtain (37) as follows based on Hankel transform:(37)
A
+
2
μ
B
+
C
-
2
μ
D
=
-
q
¯
ξ
,
A
-
1
-
2
μ
B
-
C
+
1
-
2
μ
D
=
0
,
A
+
2
μ
+
ξ
h
B
e
-
ξ
h
+
C
-
2
μ
-
ξ
h
D
e
ξ
h
=
k
ω
¯
ξ
,
A
-
1
-
2
μ
-
ξ
h
B
e
-
ξ
h
-
C
+
1
-
2
μ
+
ξ
h
D
e
ξ
h
=
0
,wherek is modulus of foundation reaction;h is the thickness of the pavement.The solution of (37) can obtain the expression forA,B,C, andD about ξ, μ,E,h, andk. SubstitutingA,B,C, andD 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, andq= 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.Table 1
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
0.852479Table 2
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
−1.69Table 3
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
−0.001Table 4
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
1.58Figure 3
Calculation sketch map.According to Tables1, 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.


## 5.1. Model

Circular uniform distributed load on Winkler foundation is as shown in Figure2.Figure 2
Calculation sketch map.Boundary conditions are as follows:(35)
z
=
0
,
τ
z
r
=
0
,
σ
z
=
-
q
r
r
≤
δ
0
r
>
δ
,
z
=
h
,
τ
z
r
=
0
,
σ
z
=
-
k
ω
r
.Substituting (25) into (35) yields(36)
σ
z
=
-
∫
0
∞
ξ
A
+
2
μ
B
+
C
-
2
μ
D
J
0
ξ
r
d
ξ
=
-
q
r
,
τ
z
r
=
-
∫
0
∞
ξ
A
-
1
-
2
μ
B
-
C
+
1
-
2
μ
D
J
1
ξ
r
d
ξ
=
0
,
σ
z
=
-
∫
0
∞
ξ
A
+
2
μ
+
ξ
h
B
e
-
ξ
h
+
C
-
2
μ
-
ξ
h
D
e
ξ
h
J
0
ξ
r
d
ξ
=
-
k
ω
r
,
τ
z
r
=
-
∫
0
∞
ξ
A
-
1
-
2
μ
-
ξ
h
B
e
-
ξ
h
-
C
+
1
-
2
μ
+
ξ
h
D
e
ξ
h
J
1
ξ
r
d
ξ
=
0
.We can obtain (37) as follows based on Hankel transform:(37)
A
+
2
μ
B
+
C
-
2
μ
D
=
-
q
¯
ξ
,
A
-
1
-
2
μ
B
-
C
+
1
-
2
μ
D
=
0
,
A
+
2
μ
+
ξ
h
B
e
-
ξ
h
+
C
-
2
μ
-
ξ
h
D
e
ξ
h
=
k
ω
¯
ξ
,
A
-
1
-
2
μ
-
ξ
h
B
e
-
ξ
h
-
C
+
1
-
2
μ
+
ξ
h
D
e
ξ
h
=
0
,wherek is modulus of foundation reaction;h is the thickness of the pavement.The solution of (37) can obtain the expression forA,B,C, andD about ξ, μ,E,h, andk. SubstitutingA,B,C, andD 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, andq= 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.Table 1
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
0.852479Table 2
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
−1.69Table 3
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
−0.001Table 4
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
1.58Figure 3
Calculation sketch map.According to Tables1, 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.



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*Source: 6295204-2016-07-31.xml*