# A Computational Method of Active Earth Pressure from Finite Soil Body

Authors: Yi Tang; Jiangong Chen
Journal: Mathematical Problems in Engineering
 (2018)

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## Abstract

Nowadays, Coulomb and Rankine earth pressure theories have been widely applied to solve the earth pressure on a retaining structure. However, both of the theories established on the basis of the semi-infinite space assumption are not suitable for calculating the earth pressure from finite soil body. Therefore, this paper focuses on a theoretical study about the active earth pressure from finite soil body. Firstly, a common calculation model of finite soil body is established according to the results of previous studies. And then, based on Coulomb’s theory and the wedge element method, an analytical solution of the unit active earth pressure from finite soil body is deduced without an assumption of its linear distribution in advance. Meanwhile, formulas of the active earth pressure strength coefficient and the application point of the resultant force are also deduced. Finally, the influence of parameters such as the frictional angle between the retaining wall back and backfill, slope angle of backfill, dip angle of the retaining wall back, the frictional angle between backfill and rock slope, and uniformly applied load on the backfill surface on the distribution of the unit active earth pressure and the application point of the resultant force is analyzed in detail.


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## Body

## 1. Introduction

With wide applications of retaining walls in engineering, there have been more and more researches on the retaining walls. The methods to determine the active pressure on a retaining wall constitute a classical topic of soil mechanics [1]. Many scholars have studied this topic, and Coulomb and Rankine earth pressure theory are the most classical theories [2–4]. However, both of the theories are established on the basis of semi-infinite space assumption, which does not conform to actual conditions of many projects. When the retaining walls built in mountainous areas are often close to rock slope, which leads backfill behind the wall to form finite soil body, it is not suitable to calculate the earth pressure from finite soil body by Coulomb’s and Rankine’s theory. There are more and more problems about finite soil body in the design and construction of actual projects, but only a small number of scholars have conducted a preliminary study on them. According to the previous soil arch theory, Take and Valsangkar do an extensive series of centrifuge model tests to evaluate the use of flexible subminiature pressure cells and measure lateral earth pressures behind retaining walls of narrow backfill width [5]. Kniss et al. investigate the earth pressure against walls in narrow spaces using the finite element method [6]. Yang and Liu present finite element analyses of earth pressures in narrow retaining walls for both at-rest and active conditions, and the predicted data show a favorable agreement with measured data from centrifuge tests [7]. Fan and Fang present a numerical study on the behaviour of active earth pressures behind a rigid retaining wall with limited backfill space of various geometries [8]. On the basis of the limit equilibrium method with planar slip surfaces, Greco presents an analytical method to obtain a solution for the active thrust exerted by backfill of narrow width on gravity retaining walls [9]. Greco presents a limit equilibrium method, for calculating the active thrust on fascia retaining walls, where common methods cannot be used owing to the narrowness of the backfill [10].However, the above achievements are limited and insufficient. Therefore, based on Coulomb’s theory and wedge element method, analytical solutions of the unit active earth pressure from finite soil body and the application point of the resultant force are deduced. Finally, the influence of parameters on the distribution of the unit active earth pressure from finite soil body and the application point of the resultant force is analyzed in detail.


## 2. Method of Analysis

A retaining wall model of finite soil body is shown in Figure1. It is a common retaining wall near rock slope, which leads backfill behind the wall to form finite soil body. The retaining wall is a rigid retaining wall of height H and inclination α with the vertical. The cohesionless backfill surface is inclined at an angle β with the horizontal. q0 is the uniform surcharge acting on the ground surface. Rock slope is inclined at an angle θ0 with the horizontal.Figure 1
Schematic drawing of a rigid retaining wall near rock slope.


### 2.1. Inclinationθ of the Sliding Surface

According to Coulomb’s theory based on semi-infinite space assumption, the inclinationθ1 of the Coulomb sliding surface is what makes the earth pressure reach the maximum, and it can be obtained when the Coulomb active earth pressure coefficient reaches the maximum as well. For the situation of finite soil body in Figure 1, the inclination θ of the sliding surface is determined as follows: when θ0 is bigger than θ1, backfill slides along the rock slope surface and the inclination θ of the sliding surface are taken equal to θ0. When θ0 is smaller than θ1, the sliding surface is within backfill and the inclinationθ of the sliding surface is taken equal to θ1. θ1 can be acquired from the following equation.The expression of the Coulomb active earth pressure coefficient is shown as(1)Ka=sin⁡θ1-φcos⁡α-θ1cos⁡α-βcos⁡θ1-α-φ-δcos2⁡αsin⁡θ1-β,where φ is the internal friction angle of backfill and δ is the frictional angle between the back of the wall and backfill. To obtain the critical value of θ1 which yields the maximum value of Ka, dKa/dθ is set equal to 0.


### 2.2. Calculation Model

According to Coulomb’s theory, the earth pressure against a retaining wall is due to the thrust exerted by a sliding soil body between the back of the wall and the sliding surface. The sliding soil body is shown as Figure2. θ is the inclination of the sliding surface, and its value which is equal to θ0 or θ1 is determined as the above-mentioned method. ABD, a part of the sliding soil body at a depth y above the heel, is taken as an isolated unit for analysis and discussion. Its plane AD is parallel with the ground surface.Figure 2
Sliding soil body.Forces analysis on the part wedge ABD of the thrust wedge is shown in Figure3. G(y) is the weight of wedge ABD, and q(y) is the uniform vertical pressure on the top of wedge ABD. Pa(y) is the force on the wall, namely, the active earth pressure, and the angle between Pa(y) and the normal of the wall back is the frictional angle δ between the retaining wall back and backfill. R(y) is the force on sliding surface. When the inclination θ of the sliding surface is equal to θ1, the angle between R(y) and the normal of the sliding surface is the internal frictional angle φ of backfill. When the inclination θ of the sliding surface is equal to θ0, the angle between R(y) and the normal of the sliding surface is the frictional angle δr between backfill and rock slope. If there is no test data about δr, δr is set to be 0.33 times as φ [11].Figure 3
Force analysis of the partial sliding wedge ABD.


### 2.3. Establishment of Fundamental Equations

Figure4 shows the force polygon considered in wedge ABD. Q(y) is the resultant force acting on plane AD. According to sine theorem,(2)Rycos⁡α+δ=Paysin⁡θ-φ=Gy+Qysin⁡α+δ+φ-θ.Figure 4
Static equilibrium of the partial sliding wedge ABD.Using the formula of Coulomb’s active earth pressure,(3)Pay=12Kaγy2+Kaqycos⁡α·cos⁡βcos⁡α-βy,where γ is the unit weight of backfill. Ka is the active earth pressure coefficient, which is written as(4)Ka=sin⁡θ-φcos⁡α-θcos⁡α-βcos⁡θ-α-φ-δcos2⁡αsin⁡θ-β.Introducing two variablespa(y) and r(y), pa(y) is the unit active earth pressure, and r(y) is the unit force on failure surface. There are two equations shown as(5)Pay=∫0ypaydy,Ry=∫0yDrydy.Putting (3) into the first equation of (5) and, meanwhile, taking a derivative from both sides of the equation with respect to y, it can be obtained that(6)pay=Kaγy+Kaqy+q′yycos⁡α·cos⁡βcos⁡α-β(7)Letpay=Kqy,where K is the unit active earth pressure coefficient. Putting (7) into (6), it can be obtained that(8)q′y=KKa·cos⁡α-βcos⁡α·cos⁡β-1·qyy-cos⁡α-βcos⁡α·cos⁡β·γ.Taking the moment equilibrium equation around the B point on the wedge ABD, it can be obtained that(9)∫0ypay·cos⁡δ·ycos⁡αdy+y·cos⁡θ-α·cos⁡βcos⁡α·sin⁡θ-β·qy·12cos⁡θ-αcos⁡α·sin⁡θ-β-sin⁡αcos⁡β·cos⁡α·cos⁡β·y+12γ·y2cos2⁡α·cos⁡θ-α·cos⁡α-βsin⁡θ-β·23·12cos⁡θ-αcos⁡α·sin⁡θ-β-sin⁡αcos⁡β·cos⁡α·cos⁡β·y=∫0yDry·cos⁡φ·ysin⁡θdy.Putting (2), the second equation of (5), and (7) into (9) and taking a derivative from both sides of the equation with respect to y, it can be obtained that (10)q′y=A1K-A2K-2·qyy-cos⁡α-βcos⁡α·cos⁡β·γ,where(11)A1=cos⁡α+δ·cos⁡φ·cos⁡α-βsin⁡θ-φ·cos⁡θ-α·cos2⁡β·12cos⁡θ-αcos⁡α·sin⁡θ-β-sin⁡αcos⁡β·cos⁡α,A2=cos⁡δ·sin⁡θ-βcos⁡θ-α·cos2⁡β·12cos⁡θ-αcos⁡α·sin⁡θ-β-sin⁡αcos⁡β·cos⁡α.


### 2.4. Solution of the Basic Equation

According to (8) with (10), the expression of the unit active earth pressure coefficient K can be obtained:(12)K=1A1-A2-A3,where A3=cos⁡α-β/cos⁡α·cos⁡β·1/Ka.Lettingλ=K/Ka·cos⁡α-β/cos⁡α·cos⁡β-1 and solving (8), it can be obtained that(13)qy=Cyλ+cos⁡α-βλ-1cos⁡α·cos⁡β·γy,whereC is an integration constant; it can be determined by the boundary condition that qy=q0 when y=H. It can be written as(14)C=q0H-λ-cos⁡α-βλ-1cos⁡α·cos⁡β·γ·H1-λ.According to (7), the unit active earth pressure can be obtained as(15)pay=Kq0H-λ-cos⁡α-βλ-1cos⁡α·cos⁡β·γ·H1-λ·yλ+Kcos⁡α-βλ-1cos⁡α·cos⁡β·γy.


### 2.5. Application Point of the Resultant Earth Pressure

The heightHp of application point of the resultant earth pressure from the wall bottom is (16)Hp=∫0Hypaydy∫0Hpaydy=23λ+1λ+23q0+cos⁡α-β/cos⁡α·cos⁡βγH2q0+cos⁡α-β/cos⁡α·cos⁡βγHH.


## 2.1. Inclinationθ of the Sliding Surface

According to Coulomb’s theory based on semi-infinite space assumption, the inclinationθ1 of the Coulomb sliding surface is what makes the earth pressure reach the maximum, and it can be obtained when the Coulomb active earth pressure coefficient reaches the maximum as well. For the situation of finite soil body in Figure 1, the inclination θ of the sliding surface is determined as follows: when θ0 is bigger than θ1, backfill slides along the rock slope surface and the inclination θ of the sliding surface are taken equal to θ0. When θ0 is smaller than θ1, the sliding surface is within backfill and the inclinationθ of the sliding surface is taken equal to θ1. θ1 can be acquired from the following equation.The expression of the Coulomb active earth pressure coefficient is shown as(1)Ka=sin⁡θ1-φcos⁡α-θ1cos⁡α-βcos⁡θ1-α-φ-δcos2⁡αsin⁡θ1-β,where φ is the internal friction angle of backfill and δ is the frictional angle between the back of the wall and backfill. To obtain the critical value of θ1 which yields the maximum value of Ka, dKa/dθ is set equal to 0.


## 2.2. Calculation Model

According to Coulomb’s theory, the earth pressure against a retaining wall is due to the thrust exerted by a sliding soil body between the back of the wall and the sliding surface. The sliding soil body is shown as Figure2. θ is the inclination of the sliding surface, and its value which is equal to θ0 or θ1 is determined as the above-mentioned method. ABD, a part of the sliding soil body at a depth y above the heel, is taken as an isolated unit for analysis and discussion. Its plane AD is parallel with the ground surface.Figure 2
Sliding soil body.Forces analysis on the part wedge ABD of the thrust wedge is shown in Figure3. G(y) is the weight of wedge ABD, and q(y) is the uniform vertical pressure on the top of wedge ABD. Pa(y) is the force on the wall, namely, the active earth pressure, and the angle between Pa(y) and the normal of the wall back is the frictional angle δ between the retaining wall back and backfill. R(y) is the force on sliding surface. When the inclination θ of the sliding surface is equal to θ1, the angle between R(y) and the normal of the sliding surface is the internal frictional angle φ of backfill. When the inclination θ of the sliding surface is equal to θ0, the angle between R(y) and the normal of the sliding surface is the frictional angle δr between backfill and rock slope. If there is no test data about δr, δr is set to be 0.33 times as φ [11].Figure 3
Force analysis of the partial sliding wedge ABD.


## 2.3. Establishment of Fundamental Equations

Figure4 shows the force polygon considered in wedge ABD. Q(y) is the resultant force acting on plane AD. According to sine theorem,(2)Rycos⁡α+δ=Paysin⁡θ-φ=Gy+Qysin⁡α+δ+φ-θ.Figure 4
Static equilibrium of the partial sliding wedge ABD.Using the formula of Coulomb’s active earth pressure,(3)Pay=12Kaγy2+Kaqycos⁡α·cos⁡βcos⁡α-βy,where γ is the unit weight of backfill. Ka is the active earth pressure coefficient, which is written as(4)Ka=sin⁡θ-φcos⁡α-θcos⁡α-βcos⁡θ-α-φ-δcos2⁡αsin⁡θ-β.Introducing two variablespa(y) and r(y), pa(y) is the unit active earth pressure, and r(y) is the unit force on failure surface. There are two equations shown as(5)Pay=∫0ypaydy,Ry=∫0yDrydy.Putting (3) into the first equation of (5) and, meanwhile, taking a derivative from both sides of the equation with respect to y, it can be obtained that(6)pay=Kaγy+Kaqy+q′yycos⁡α·cos⁡βcos⁡α-β(7)Letpay=Kqy,where K is the unit active earth pressure coefficient. Putting (7) into (6), it can be obtained that(8)q′y=KKa·cos⁡α-βcos⁡α·cos⁡β-1·qyy-cos⁡α-βcos⁡α·cos⁡β·γ.Taking the moment equilibrium equation around the B point on the wedge ABD, it can be obtained that(9)∫0ypay·cos⁡δ·ycos⁡αdy+y·cos⁡θ-α·cos⁡βcos⁡α·sin⁡θ-β·qy·12cos⁡θ-αcos⁡α·sin⁡θ-β-sin⁡αcos⁡β·cos⁡α·cos⁡β·y+12γ·y2cos2⁡α·cos⁡θ-α·cos⁡α-βsin⁡θ-β·23·12cos⁡θ-αcos⁡α·sin⁡θ-β-sin⁡αcos⁡β·cos⁡α·cos⁡β·y=∫0yDry·cos⁡φ·ysin⁡θdy.Putting (2), the second equation of (5), and (7) into (9) and taking a derivative from both sides of the equation with respect to y, it can be obtained that (10)q′y=A1K-A2K-2·qyy-cos⁡α-βcos⁡α·cos⁡β·γ,where(11)A1=cos⁡α+δ·cos⁡φ·cos⁡α-βsin⁡θ-φ·cos⁡θ-α·cos2⁡β·12cos⁡θ-αcos⁡α·sin⁡θ-β-sin⁡αcos⁡β·cos⁡α,A2=cos⁡δ·sin⁡θ-βcos⁡θ-α·cos2⁡β·12cos⁡θ-αcos⁡α·sin⁡θ-β-sin⁡αcos⁡β·cos⁡α.


## 2.4. Solution of the Basic Equation

According to (8) with (10), the expression of the unit active earth pressure coefficient K can be obtained:(12)K=1A1-A2-A3,where A3=cos⁡α-β/cos⁡α·cos⁡β·1/Ka.Lettingλ=K/Ka·cos⁡α-β/cos⁡α·cos⁡β-1 and solving (8), it can be obtained that(13)qy=Cyλ+cos⁡α-βλ-1cos⁡α·cos⁡β·γy,whereC is an integration constant; it can be determined by the boundary condition that qy=q0 when y=H. It can be written as(14)C=q0H-λ-cos⁡α-βλ-1cos⁡α·cos⁡β·γ·H1-λ.According to (7), the unit active earth pressure can be obtained as(15)pay=Kq0H-λ-cos⁡α-βλ-1cos⁡α·cos⁡β·γ·H1-λ·yλ+Kcos⁡α-βλ-1cos⁡α·cos⁡β·γy.


## 2.5. Application Point of the Resultant Earth Pressure

The heightHp of application point of the resultant earth pressure from the wall bottom is (16)Hp=∫0Hypaydy∫0Hpaydy=23λ+1λ+23q0+cos⁡α-β/cos⁡α·cos⁡βγH2q0+cos⁡α-β/cos⁡α·cos⁡βγHH.


## 3. Parametric Study

In this section, when the inclination of the sliding surface isθ0, the influence of dip angle α of the retaining wall back, slope angle β of backfill, the frictional angle δ between the retaining wall back and backfill, the frictional angle δr between backfill and rock slope, and uniformly applied load q0 on the backfill surface on the distribution of the unit active earth pressure and the application point of the resultant force is discussed as follows. θ0, the inclination of the rock slope, is assumed to be 75°. Another condition can refer to [12].WhenH=4.0 m, r=18Kn/m3, and θ0=75°, the active earth pressure distribution and the height Hp of application point of the resultant earth pressure with change in δr, in δ, in α, in β, and in q0 are shown in Figures 5–14, respectively. As shown in Figure 5, the active earth pressure increases as δr increases near the heel of the wall; the active earth pressure increases as δr decreases in other places. As shown in Figure 7, the active earth pressure decreases as δ increases near the heel of the wall; the active earth pressure increases as δ increases in other places. From Figures 9 and 11, it can be seen that the active earth pressure increases as α and β increase, respectively. As shown in Figure 13, the active earth pressure increases in proportion asq0 increases. From Figures 6, 10, and 12, it can be seen that the height Hp of application point of resultant earth pressure decreases as δr, α, and β increase, respectively. From Figures 8 and 14, it can be seen that the height Hp of application point of the resultant earth pressure decreases as δ and q0 increase, respectively.Figure 5
Distribution of active earth pressure with change inδr.Figure 6
Application point of resultant earth pressure with change inδr.Figure 7
Distribution of active earth pressure with change inδ.Figure 8
Application point of resultant earth pressure with change inδ.Figure 9
Distribution of active earth pressure with change inα.Figure 10
Application point of resultant earth change pressure with change inα.Figure 11
Distribution of active earth pressure with change inβ.Figure 12
Application point of resultant earth pressure with change inβ.Figure 13
Distribution of active earth pressure with change inq0.Figure 14
Application point of resultant earth pressure with change inq0.From Figures5, 7, 9, and 11, it is obvious that there is a critical value existing in dip angle α of the retaining wall back, slope angle β of backfill, the frictional angle δ between the retaining wall back and backfill, and the frictional angle δr between backfill and rock slope. The critical values divide the earth pressure distribution curves into two types. One type is a crooked nonlinear curve and another type is close to a line [12].


## 4. Conclusions

( 1 ) Based on Coulomb’s theory and wedge element method, an analytical solution of the unit active earth pressure from finite soil body is deduced. Meanwhile, the formula of the application point of the resultant force is also deduced. When these formulas are applied, we firstly estimate whether backfill slides along rock slope or the sliding surface happens inside of backfill. And then whether the angle between R(y) and the normal of the sliding surface is the internal frictional angle φ of backfill, or the frictional angle δr between backfill and rock slope is determined.( 2 ) By parametric study, it is found that the frictional angle between the retaining wall back and backfill, slope angle of backfill, dip angle of the retaining wall back, the frictional angle between backfill and rock slope, and uniformly applied load on the backfill surface have great influence on the active earth pressure distribution and the application point of the resultant earth pressure.( 3 ) According to the method of this paper, we can study the earth pressure from finite clayey soil body in the future.



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