# Seismic Response Characteristics of Stabilizing Pile Based on Elastic-Plastic Analysis

Authors: Honglue Qu; Ying Liu; Hao Luo; Huanguo Hu; Qindi Hu
Journal: Shock and Vibration
 (2018)

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## Abstract

Stabilizing pile is widely used in the landslide controlling projects and shows excellent seismic performance under the action of earthquake. Therefore, in order to improve seismic design theory, it is of importance to study the seismic response characteristics of stabilizing pile based on elastic-plastic analysis. In view of this, elastic-plastic constitutive model was established to deduce the plastic zone of stabilizing pile. Based on elastic-plastic analysis, the seismic response characteristics and the influence of different section sizes, material strengths, and peak ground motion acceleration (PGA) were analyzed by ANSYS 3D. Resultantly, the elastic-plastic fourth-order tensorCijklep was deduced, which can be used to calculate plastic strain of stabilizing pile under loading. Compared with Chinese code, the material of stabilizing pile working with elastic-plastic state will be decreased under the same section size and the same property. Furthermore, stabilizing pile is in the elastic stage at the beginning under the action of earthquake. With the increase of ground motion time, the section starts to exhibit elastic-plastic state and then the plastic zone expands gradually. Finally, the plastic zone runs through the whole section, resulting in the performance loss of the pile. In addition, under the different design parameters, pile shows different seismic response characteristics; namely, changing these parameters reasonably can improve the seismic design.


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## Body

## 1. Introduction

Stabilizing pile, as a flexible retaining structure, has advantages of convenient construction, flexible arrangement, and strong stabilizing ability, which is widely used in the treatment of geological hazards in recent years [1–3] and also shows excellent seismic performance in Wenchuan earthquake [4]. Therefore, in order to provide a theoretical basis for optimizing seismic design system, it is necessary to study the seismic response characteristics of stabilizing pile based on elastic-plastic analysis.Nowadays, many researches about stabilizing pile have been carried out [5–8]. On the one hand, many scholars put forward the corresponding calculation formula in view of pile spacing [9], pile length [10], embedded depth [11], and so on to optimize the structure design, which provided theory basis for the engineering practical application of stabilizing pile [12]. On the other hand, as an important measure to control landslides, the seismic response characteristics of slope soil and pile-soil interaction were also research topics in geotechnical engineering. For example, Luan et al. [13] established a consolidated soil-pile model to analyze pile side friction force and seismic response of soil by considering the pile-soil interaction. Fard and Rasoolan [14] investigated soil-pile interaction using the numerical method and provided greater familiarity with force distribution in deformation mechanisms in the soil as well as soil-pile interaction.The above researches mainly focus on the structure design, the pile-soil interaction, and the seismic response characteristics of reinforced soil. However, the research on elastic-plastic analysis of stabilizing pile under the action of earthquake is still rare [15]. In view of this, elastic-plastic constitutive model was established to deduce the plastic zone of stabilizing pile. Based on elastic-plastic analysis, the seismic response characteristics of stabilizing pile and the influence of different design parameters were analyzed, such as section size, material strength, and PGA. The results of numerical analysis can provide a basis for the optimization of stabilizing pile seismic design system [16].


## 2. Elastic-Plastic Theory of Stabilizing Pile

### 2.1. The Establishment of Constitutive Model

During the process of the structural design with elastic theory, the stabilizing pile fails to meet the material damage conditions completely so that the design is conservative and the masonry is wasteful. Therefore, it is necessary to establish the elastic-plastic constitutive model, which is advantageous to make full use of materials and maximize their performance.The elastic constitutive relationship is not applicable when the stress state of stabilizing pile in particular point meets yield condition and goes into plastic deformation stage. Therefore, it is necessary to establish stress-strain relationship for describing the elastic-plastic behavior of that point.The stress state can be simplified down to plane stress state because the length is much greater than the width of stabilizing pile. On account of von Mises yield criterion, stabilizing pile can be viewed as isotropic material in the elastic stage and it will be viewed as isotropic hardening material in the plastic stage. On the basis, the elastic-plastic stress and strain can be calculated according to the external loading of stabilizing pile, and the size of plastic zone can be calculated in the case of plastic hinge. Finally, a numerical simulation with ANSYS was used to verify the correctness of elastic-plastic constitutive model.From the elastic-plastic theory, it can be seen that elastic increment plus plastic increment is strain increment [17], that is, (1)dεij=dεije+dεijpIn (1), εij, εije, and εijp are second-order strain tensor, second-order elastic strain tensor, and second-order plastic strain tensor, respectively. The relationship between elastic strain increment and stress increment is shown in (2), where σij is second-order stress tensor and Cijkl is fourth-order elastic tensor. Additionally, the plastic strain increment can be expressed by plastic potential function and flow rule equation, as shown in (3), where λ is positive scalar.(2)dσij=Cijkldεkle(3)dεijp=dλ∂g∂σijCombining (1), (2), and (3), (4) can be obtained.(4)dσij=Cijkldεkl-dλ∂g∂σij


#### 2.1.1. Initial Yield

Based on the relationship between stress increment and strain increment,f equals g when the associated flow rule is adopted. Taking von Mises criterion as yield condition, the initial yield condition is given in (5), where Kk is hardening function, k is hardening parameter, and J2 is the second invariant of stress deviator. In view of this, subsequent yield function for isotropic hardening is shown in (6) and (7), where sij is second-order stress deviator, σe is elastic stress, ε-p is equivalent plastic strain, and Ee, Ep, and Et are elastic modulus, plastic modulus, and tangent modulus, respectively, which satisfied (8).(5)fσij,k=J2-Kk=0(6)12sijsij-13σε2ε-p=0orJ2-13σε2ε-p=0(7)dσ=Eedεedσ=Epdεpdσ=Etdε(8)1Ee+1Ep=1EtIn von Mises yield criterion, the yield surface is cylindrical surface in principal stress space with the average principal stress axes. The intersecting line among yield plane,π plane, and flat plane is a circle with the radius of 2k, while it is an ellipse in σ2-σ1 coordinate plane, as shown in Figure 1.Figure 1
The von Mises yield criterion inσ2-σ1 coordinate plane.In addition, the equation of the ellipse is shown in (9), where σ1, σ2, and σs0 are first main stress, second main stress, and initial yield stress, respectively. Therefore, the yield function at the time of initial yield is given in(9)σ12+σ22-σ1σ2=σs02(10)f0σij=1212sijsij-13σs02=0


#### 2.1.2. Subsequent Yield under Loading

When the elastic-plastic deformation occurs, the boundary of elastic zone is constantly changing, and plastic state will inevitably follow the change. The yield functions are shown in (11), so (12) can be obtained. Combining (4), (10), and (12), elastic-plastic tensor of isotropic hardening material under the yield condition can be solved in (13).(11)fσij,ε-p=0,fσij+dσij,ε-p+dε-p=0(12)∂f∂σijdσij+∂f∂ε-p=0(13)Cijklep=Cijkl-sijskl1/9G2Ep+3Gσe2Therefore, the elastic-plastic constitutive tensor of stabilizing pile with isotropic hardening on the plane is shown in (14), where μ is Lame coefficient, G is shear modulus, Cijklep is fourth-order elastic-plastic tensor, and sx and sy are stress deviator along the x direction and y direction, respectively. (14)Cijklep=μ+Gμ0μμ+G000G-11/9G2Ep+3Gσe2sxsxsxsysxτxysysxsysysyτxyτyxsxτyxsyτyxτxy


### 2.2. The Determination of Plastic Zone

Slip mass behind cantilever section of stabilizing pile is simplified as known external force. In order to facilitate the mechanical analysis for embedded segment, stress analysis coordinate system of stabilizing pile is established when the top of embedded segment is taken as the origin [18], as shown in Figure 2.Figure 2
Internal force calculation of stabilizing pile.Based on the balance of microsection, force relationship is given in (15), where Qy is the shear force, qy is the loading in the y direction, and My is bending moment. Furthermore, (16) and (17) can be obtained based on the principle of material mechanics, where fy is horizontal displacement along y-axis,Eis elastic modulus, and I is cross-sectional moment of inertia of pile. Subgrade reaction of pile meets the relationship qy=Cfy. Therefore, subgrade reaction beard by cross-section of pile meets the relationship of (18), where C is foundation coefficient, Bp is calculative width of pile. In view of the above, and (19) can be obtained.(15)dQydy=qydMydy=Qy(16)xy=fy,φy=dxydy(17)MyEI=-d2fydy2,QyEI=-d3fydy3(18)qy=CBpfy(19)EId4fydy4+CBpfy=0The stress of stabilizing pile at any position is shown in Figure3, and the normal stress and shear stress can be solved by (20) and (21), where h is cross-sectional height, b is cross-sectional width of pile, and Iz is inertia moment with z as the symmetry axis. In addition, principal stress in the plane and stress deviator along the x and y direction can be obtained, which are described in (22) and (23).(20)σx=MyxIz,τxy=QySz∗Izb(21)Sz∗=∫xh/2xbdx=b2h24-x2(22)σ1=Myx2Iz+My2x24Iz2+Qy2Sz∗2Iz2b2,σ2=Myx2Iz-My2x24Iz2+Qy2Sz∗2Iz2b2(23)sx=2Myx3Iz,sy=-Myx3IzFigure 3
Internal force of pile. (a) Shear stress; (b) normal stress.
(a) (b)When the initial yield happens, the yield function equals 0, as shown in (24). Furthermore, elastic and plastic ultimate loading can be given by (25). With the increase of loading, plastic zone begins to extend from the position with the maximum stress. Supposing that the x-coordinate of the intersection between plastic zone and elastic zone on y cross-section is s(y), these points on the cross-section are still in elastic state when the value range of x axis is -s(y)≤x≤s(y) and the value of σx is a linear distribution, as shown in (26), where σx is yield stress.(24)f0σij=2My2x2Iz2+6Qy2Sz∗2Iz2b2-σs023=0(25)My=∫-hhxσx2bdx(26)σx=xsyσs-sy≤x≤syTherefore, in the plastic zone, the stress relationships are given in (27) and (28) and the function of the plastic zone s(y) can be obtained, as shown in (29).(27)σx=σssy≤x≤h(28)σx=-σs-h≤x≤-sy(29)sy=±3h2-3My2σsb


### 2.3. Verification of Constitutive Model and Elastic-Plastic Zone

The elastic-plastic zone of simply supported beam under uniform distributed loading was analyzed by ANSYS, and then the correctness of elastic-plastic constitutive relationship and plastic zone of isotropic hardening material satisfying von Mises yield condition was verified according to the results of numerical simulation. The material is supposed to be elastic-plastic and bilinear isotropic hardening material, which meets von Mises yield criterion. The sectional dimension was taken asb×h= 1 m × 2 m and the length was 5 m, as shown in Figure 4. In addition, the initial yield was assumed to happen when the strain was 0.01 and began to damage when the strain reached 0.2. In the process of loading, the material property was assumed to meet the requirement of linear hardening. Uniform linear loading was applied at the top of the pile, which increases from 0 to 30×106 N/m.Figure 4
The numerical model and material property in ANSYS.


#### 2.3.1. Validation of Elastic-Plastic Limit State

According to the stress analysis, it is known that the stress of cross-section in the middle of beam is the maximum. Therefore, elastic-plastic limited state of the cross-section should be considered firstly. Whens(0)=h, through (29), the bending moment of cross-section reaches yield stage firstly and the elastic limited bending moment Me can be obtained. When the cross-section reaches the yield stage absolutely, the corresponding bending moment is named plastic limit bending moment Mp. At this time, s(0)=0, (30) and (31) can be obtained.(30)Me=43σsbh2Mp=2σsbh2(31)qe=83σsbhl2qp=4σsbhl2In (30) and (31), Me and Mp are elastic limited bending moment and plastic limited bending moment, respectively, and qe and qp is elastic limited loading and plastic limited loading, respectively.Through (31), it can be seen that the beam is in the elastic limited state when qe=16×106 N/m; namely, the yield of the middle-beam just happens when q=qe. At that moment, the strain is ε=0.01, as shown in Figure 5. In addition, the beam is in the plastic limited state when qp=24×106 N/m according to the calculation, namely, q=qp. At the same time, the cross-section in the middle of beam reaches complete yield state and the plastic hinge is developed in the numerical simulation analysis, which satisfied the results of (31), as shown in Figure 6.Figure 5
Elastic limited state analyzed by ANSYS.Figure 6
Plastic limited state analyzed by ANSYS.


#### 2.3.2. Validation of Plastic Zone

The comparison of equivalent stress and yield strength of the material itself under the external loading can be used as a basis for judging whether the structure shows plastic zone. This means that when the two values are equal at a certain point, the pile reaches the critical point of yielding at that point. Therefore, the range of plastic zone can be obtained according to the position of the critical point, and the subsequent yield equivalent stress of each part of plastic zone can be obtained on the basis of plastic hardening criterion.Through (29), the plastic zone of simply supported beam can be obtained, as shown in Figure 7.Figure 7
Plastic zone of simply supported beam.Whenq=21×106 N/m, plastic zone happened in the interval -L/3,L/3, which is consistent with the results analyzed of numerical simulation, as shown in Figure 8.Figure 8
The plastic zone analyzed by ANSYS.


#### 2.3.3. Validation of Plastic Hardening under Loading

Whenq=21×106 N/m, there is qe<q<qp. At this moment, the middle of beam is in elastic-plastic state. Putting the value above into (8), (14), (23), and (29), then the value of total strain can be obtained, namely εij=εije+εijp=0.0102+0.0055=0.01577, which is also consistent with the result of numerical simulation, as shown in Figure 9.Figure 9
Elastic strain and plastic strain analyzed by ANSYS.The numerical simulation was used to analyze the simply supported beam subjected to uniformly distributed loading to verify the correctness of the elastic-plastic constitutive model and plastic zone mentioned above. The results show that the derived plastic zone and plastic strain are consistent with that of numerical simulation; namely, the formulas of elastic-plastic constitutive model and plastic zone proposed in this paper can be applied to the elastic-plastic analysis of isotropic hardening materials which meet the von Mises yield criterion.However, the design of stabilizing pile is carried out by static method in Chinese code for seismic design of railway [19] and highway engineering [20] at present; namely, the effect of earthquake on the pile is expressed by equivalent loading, and then the internal force and displacement are calculated by static analysis with the equivalent loading, so as to check the seismic bearing capacity and deformation of the pile. Certainly, the static method, as an accepted and frequently used method, can satisfy the requirement of seismic design and has many advantages, such as simply calculation and rich experience. However, the code is mainly for prevention so that the stabilizing pile designed by static method is more abundant than the actual one. According to Figure 4, under the same section size and same property, the ultimate stress of the pile with elastic-plastic analysis is 500 N/mm and the pile reaches the state of complete yielding at this time, while the stress is only 300 N/mm with static method, as the code assumes that the pile only exists in elastic state under the action of earthquake. If the pile has to bear the ultimate stress of 500 N/mm, it is necessary to increase the section size, which will undoubtedly increase material consumption and cause waste. Therefore, the establishment of elastic-plastic constitutive model is beneficial to make full use of materials and maximize their performance.


## 2.1. The Establishment of Constitutive Model

During the process of the structural design with elastic theory, the stabilizing pile fails to meet the material damage conditions completely so that the design is conservative and the masonry is wasteful. Therefore, it is necessary to establish the elastic-plastic constitutive model, which is advantageous to make full use of materials and maximize their performance.The elastic constitutive relationship is not applicable when the stress state of stabilizing pile in particular point meets yield condition and goes into plastic deformation stage. Therefore, it is necessary to establish stress-strain relationship for describing the elastic-plastic behavior of that point.The stress state can be simplified down to plane stress state because the length is much greater than the width of stabilizing pile. On account of von Mises yield criterion, stabilizing pile can be viewed as isotropic material in the elastic stage and it will be viewed as isotropic hardening material in the plastic stage. On the basis, the elastic-plastic stress and strain can be calculated according to the external loading of stabilizing pile, and the size of plastic zone can be calculated in the case of plastic hinge. Finally, a numerical simulation with ANSYS was used to verify the correctness of elastic-plastic constitutive model.From the elastic-plastic theory, it can be seen that elastic increment plus plastic increment is strain increment [17], that is, (1)dεij=dεije+dεijpIn (1), εij, εije, and εijp are second-order strain tensor, second-order elastic strain tensor, and second-order plastic strain tensor, respectively. The relationship between elastic strain increment and stress increment is shown in (2), where σij is second-order stress tensor and Cijkl is fourth-order elastic tensor. Additionally, the plastic strain increment can be expressed by plastic potential function and flow rule equation, as shown in (3), where λ is positive scalar.(2)dσij=Cijkldεkle(3)dεijp=dλ∂g∂σijCombining (1), (2), and (3), (4) can be obtained.(4)dσij=Cijkldεkl-dλ∂g∂σij


### 2.1.1. Initial Yield

Based on the relationship between stress increment and strain increment,f equals g when the associated flow rule is adopted. Taking von Mises criterion as yield condition, the initial yield condition is given in (5), where Kk is hardening function, k is hardening parameter, and J2 is the second invariant of stress deviator. In view of this, subsequent yield function for isotropic hardening is shown in (6) and (7), where sij is second-order stress deviator, σe is elastic stress, ε-p is equivalent plastic strain, and Ee, Ep, and Et are elastic modulus, plastic modulus, and tangent modulus, respectively, which satisfied (8).(5)fσij,k=J2-Kk=0(6)12sijsij-13σε2ε-p=0orJ2-13σε2ε-p=0(7)dσ=Eedεedσ=Epdεpdσ=Etdε(8)1Ee+1Ep=1EtIn von Mises yield criterion, the yield surface is cylindrical surface in principal stress space with the average principal stress axes. The intersecting line among yield plane,π plane, and flat plane is a circle with the radius of 2k, while it is an ellipse in σ2-σ1 coordinate plane, as shown in Figure 1.Figure 1
The von Mises yield criterion inσ2-σ1 coordinate plane.In addition, the equation of the ellipse is shown in (9), where σ1, σ2, and σs0 are first main stress, second main stress, and initial yield stress, respectively. Therefore, the yield function at the time of initial yield is given in(9)σ12+σ22-σ1σ2=σs02(10)f0σij=1212sijsij-13σs02=0


### 2.1.2. Subsequent Yield under Loading

When the elastic-plastic deformation occurs, the boundary of elastic zone is constantly changing, and plastic state will inevitably follow the change. The yield functions are shown in (11), so (12) can be obtained. Combining (4), (10), and (12), elastic-plastic tensor of isotropic hardening material under the yield condition can be solved in (13).(11)fσij,ε-p=0,fσij+dσij,ε-p+dε-p=0(12)∂f∂σijdσij+∂f∂ε-p=0(13)Cijklep=Cijkl-sijskl1/9G2Ep+3Gσe2Therefore, the elastic-plastic constitutive tensor of stabilizing pile with isotropic hardening on the plane is shown in (14), where μ is Lame coefficient, G is shear modulus, Cijklep is fourth-order elastic-plastic tensor, and sx and sy are stress deviator along the x direction and y direction, respectively. (14)Cijklep=μ+Gμ0μμ+G000G-11/9G2Ep+3Gσe2sxsxsxsysxτxysysxsysysyτxyτyxsxτyxsyτyxτxy


## 2.1.1. Initial Yield

Based on the relationship between stress increment and strain increment,f equals g when the associated flow rule is adopted. Taking von Mises criterion as yield condition, the initial yield condition is given in (5), where Kk is hardening function, k is hardening parameter, and J2 is the second invariant of stress deviator. In view of this, subsequent yield function for isotropic hardening is shown in (6) and (7), where sij is second-order stress deviator, σe is elastic stress, ε-p is equivalent plastic strain, and Ee, Ep, and Et are elastic modulus, plastic modulus, and tangent modulus, respectively, which satisfied (8).(5)fσij,k=J2-Kk=0(6)12sijsij-13σε2ε-p=0orJ2-13σε2ε-p=0(7)dσ=Eedεedσ=Epdεpdσ=Etdε(8)1Ee+1Ep=1EtIn von Mises yield criterion, the yield surface is cylindrical surface in principal stress space with the average principal stress axes. The intersecting line among yield plane,π plane, and flat plane is a circle with the radius of 2k, while it is an ellipse in σ2-σ1 coordinate plane, as shown in Figure 1.Figure 1
The von Mises yield criterion inσ2-σ1 coordinate plane.In addition, the equation of the ellipse is shown in (9), where σ1, σ2, and σs0 are first main stress, second main stress, and initial yield stress, respectively. Therefore, the yield function at the time of initial yield is given in(9)σ12+σ22-σ1σ2=σs02(10)f0σij=1212sijsij-13σs02=0


## 2.1.2. Subsequent Yield under Loading

When the elastic-plastic deformation occurs, the boundary of elastic zone is constantly changing, and plastic state will inevitably follow the change. The yield functions are shown in (11), so (12) can be obtained. Combining (4), (10), and (12), elastic-plastic tensor of isotropic hardening material under the yield condition can be solved in (13).(11)fσij,ε-p=0,fσij+dσij,ε-p+dε-p=0(12)∂f∂σijdσij+∂f∂ε-p=0(13)Cijklep=Cijkl-sijskl1/9G2Ep+3Gσe2Therefore, the elastic-plastic constitutive tensor of stabilizing pile with isotropic hardening on the plane is shown in (14), where μ is Lame coefficient, G is shear modulus, Cijklep is fourth-order elastic-plastic tensor, and sx and sy are stress deviator along the x direction and y direction, respectively. (14)Cijklep=μ+Gμ0μμ+G000G-11/9G2Ep+3Gσe2sxsxsxsysxτxysysxsysysyτxyτyxsxτyxsyτyxτxy


## 2.2. The Determination of Plastic Zone

Slip mass behind cantilever section of stabilizing pile is simplified as known external force. In order to facilitate the mechanical analysis for embedded segment, stress analysis coordinate system of stabilizing pile is established when the top of embedded segment is taken as the origin [18], as shown in Figure 2.Figure 2
Internal force calculation of stabilizing pile.Based on the balance of microsection, force relationship is given in (15), where Qy is the shear force, qy is the loading in the y direction, and My is bending moment. Furthermore, (16) and (17) can be obtained based on the principle of material mechanics, where fy is horizontal displacement along y-axis,Eis elastic modulus, and I is cross-sectional moment of inertia of pile. Subgrade reaction of pile meets the relationship qy=Cfy. Therefore, subgrade reaction beard by cross-section of pile meets the relationship of (18), where C is foundation coefficient, Bp is calculative width of pile. In view of the above, and (19) can be obtained.(15)dQydy=qydMydy=Qy(16)xy=fy,φy=dxydy(17)MyEI=-d2fydy2,QyEI=-d3fydy3(18)qy=CBpfy(19)EId4fydy4+CBpfy=0The stress of stabilizing pile at any position is shown in Figure3, and the normal stress and shear stress can be solved by (20) and (21), where h is cross-sectional height, b is cross-sectional width of pile, and Iz is inertia moment with z as the symmetry axis. In addition, principal stress in the plane and stress deviator along the x and y direction can be obtained, which are described in (22) and (23).(20)σx=MyxIz,τxy=QySz∗Izb(21)Sz∗=∫xh/2xbdx=b2h24-x2(22)σ1=Myx2Iz+My2x24Iz2+Qy2Sz∗2Iz2b2,σ2=Myx2Iz-My2x24Iz2+Qy2Sz∗2Iz2b2(23)sx=2Myx3Iz,sy=-Myx3IzFigure 3
Internal force of pile. (a) Shear stress; (b) normal stress.
(a) (b)When the initial yield happens, the yield function equals 0, as shown in (24). Furthermore, elastic and plastic ultimate loading can be given by (25). With the increase of loading, plastic zone begins to extend from the position with the maximum stress. Supposing that the x-coordinate of the intersection between plastic zone and elastic zone on y cross-section is s(y), these points on the cross-section are still in elastic state when the value range of x axis is -s(y)≤x≤s(y) and the value of σx is a linear distribution, as shown in (26), where σx is yield stress.(24)f0σij=2My2x2Iz2+6Qy2Sz∗2Iz2b2-σs023=0(25)My=∫-hhxσx2bdx(26)σx=xsyσs-sy≤x≤syTherefore, in the plastic zone, the stress relationships are given in (27) and (28) and the function of the plastic zone s(y) can be obtained, as shown in (29).(27)σx=σssy≤x≤h(28)σx=-σs-h≤x≤-sy(29)sy=±3h2-3My2σsb


## 2.3. Verification of Constitutive Model and Elastic-Plastic Zone

The elastic-plastic zone of simply supported beam under uniform distributed loading was analyzed by ANSYS, and then the correctness of elastic-plastic constitutive relationship and plastic zone of isotropic hardening material satisfying von Mises yield condition was verified according to the results of numerical simulation. The material is supposed to be elastic-plastic and bilinear isotropic hardening material, which meets von Mises yield criterion. The sectional dimension was taken asb×h= 1 m × 2 m and the length was 5 m, as shown in Figure 4. In addition, the initial yield was assumed to happen when the strain was 0.01 and began to damage when the strain reached 0.2. In the process of loading, the material property was assumed to meet the requirement of linear hardening. Uniform linear loading was applied at the top of the pile, which increases from 0 to 30×106 N/m.Figure 4
The numerical model and material property in ANSYS.


### 2.3.1. Validation of Elastic-Plastic Limit State

According to the stress analysis, it is known that the stress of cross-section in the middle of beam is the maximum. Therefore, elastic-plastic limited state of the cross-section should be considered firstly. Whens(0)=h, through (29), the bending moment of cross-section reaches yield stage firstly and the elastic limited bending moment Me can be obtained. When the cross-section reaches the yield stage absolutely, the corresponding bending moment is named plastic limit bending moment Mp. At this time, s(0)=0, (30) and (31) can be obtained.(30)Me=43σsbh2Mp=2σsbh2(31)qe=83σsbhl2qp=4σsbhl2In (30) and (31), Me and Mp are elastic limited bending moment and plastic limited bending moment, respectively, and qe and qp is elastic limited loading and plastic limited loading, respectively.Through (31), it can be seen that the beam is in the elastic limited state when qe=16×106 N/m; namely, the yield of the middle-beam just happens when q=qe. At that moment, the strain is ε=0.01, as shown in Figure 5. In addition, the beam is in the plastic limited state when qp=24×106 N/m according to the calculation, namely, q=qp. At the same time, the cross-section in the middle of beam reaches complete yield state and the plastic hinge is developed in the numerical simulation analysis, which satisfied the results of (31), as shown in Figure 6.Figure 5
Elastic limited state analyzed by ANSYS.Figure 6
Plastic limited state analyzed by ANSYS.


### 2.3.2. Validation of Plastic Zone

The comparison of equivalent stress and yield strength of the material itself under the external loading can be used as a basis for judging whether the structure shows plastic zone. This means that when the two values are equal at a certain point, the pile reaches the critical point of yielding at that point. Therefore, the range of plastic zone can be obtained according to the position of the critical point, and the subsequent yield equivalent stress of each part of plastic zone can be obtained on the basis of plastic hardening criterion.Through (29), the plastic zone of simply supported beam can be obtained, as shown in Figure 7.Figure 7
Plastic zone of simply supported beam.Whenq=21×106 N/m, plastic zone happened in the interval -L/3,L/3, which is consistent with the results analyzed of numerical simulation, as shown in Figure 8.Figure 8
The plastic zone analyzed by ANSYS.


### 2.3.3. Validation of Plastic Hardening under Loading

Whenq=21×106 N/m, there is qe<q<qp. At this moment, the middle of beam is in elastic-plastic state. Putting the value above into (8), (14), (23), and (29), then the value of total strain can be obtained, namely εij=εije+εijp=0.0102+0.0055=0.01577, which is also consistent with the result of numerical simulation, as shown in Figure 9.Figure 9
Elastic strain and plastic strain analyzed by ANSYS.The numerical simulation was used to analyze the simply supported beam subjected to uniformly distributed loading to verify the correctness of the elastic-plastic constitutive model and plastic zone mentioned above. The results show that the derived plastic zone and plastic strain are consistent with that of numerical simulation; namely, the formulas of elastic-plastic constitutive model and plastic zone proposed in this paper can be applied to the elastic-plastic analysis of isotropic hardening materials which meet the von Mises yield criterion.However, the design of stabilizing pile is carried out by static method in Chinese code for seismic design of railway [19] and highway engineering [20] at present; namely, the effect of earthquake on the pile is expressed by equivalent loading, and then the internal force and displacement are calculated by static analysis with the equivalent loading, so as to check the seismic bearing capacity and deformation of the pile. Certainly, the static method, as an accepted and frequently used method, can satisfy the requirement of seismic design and has many advantages, such as simply calculation and rich experience. However, the code is mainly for prevention so that the stabilizing pile designed by static method is more abundant than the actual one. According to Figure 4, under the same section size and same property, the ultimate stress of the pile with elastic-plastic analysis is 500 N/mm and the pile reaches the state of complete yielding at this time, while the stress is only 300 N/mm with static method, as the code assumes that the pile only exists in elastic state under the action of earthquake. If the pile has to bear the ultimate stress of 500 N/mm, it is necessary to increase the section size, which will undoubtedly increase material consumption and cause waste. Therefore, the establishment of elastic-plastic constitutive model is beneficial to make full use of materials and maximize their performance.


## 2.3.1. Validation of Elastic-Plastic Limit State

According to the stress analysis, it is known that the stress of cross-section in the middle of beam is the maximum. Therefore, elastic-plastic limited state of the cross-section should be considered firstly. Whens(0)=h, through (29), the bending moment of cross-section reaches yield stage firstly and the elastic limited bending moment Me can be obtained. When the cross-section reaches the yield stage absolutely, the corresponding bending moment is named plastic limit bending moment Mp. At this time, s(0)=0, (30) and (31) can be obtained.(30)Me=43σsbh2Mp=2σsbh2(31)qe=83σsbhl2qp=4σsbhl2In (30) and (31), Me and Mp are elastic limited bending moment and plastic limited bending moment, respectively, and qe and qp is elastic limited loading and plastic limited loading, respectively.Through (31), it can be seen that the beam is in the elastic limited state when qe=16×106 N/m; namely, the yield of the middle-beam just happens when q=qe. At that moment, the strain is ε=0.01, as shown in Figure 5. In addition, the beam is in the plastic limited state when qp=24×106 N/m according to the calculation, namely, q=qp. At the same time, the cross-section in the middle of beam reaches complete yield state and the plastic hinge is developed in the numerical simulation analysis, which satisfied the results of (31), as shown in Figure 6.Figure 5
Elastic limited state analyzed by ANSYS.Figure 6
Plastic limited state analyzed by ANSYS.


## 2.3.2. Validation of Plastic Zone

The comparison of equivalent stress and yield strength of the material itself under the external loading can be used as a basis for judging whether the structure shows plastic zone. This means that when the two values are equal at a certain point, the pile reaches the critical point of yielding at that point. Therefore, the range of plastic zone can be obtained according to the position of the critical point, and the subsequent yield equivalent stress of each part of plastic zone can be obtained on the basis of plastic hardening criterion.Through (29), the plastic zone of simply supported beam can be obtained, as shown in Figure 7.Figure 7
Plastic zone of simply supported beam.Whenq=21×106 N/m, plastic zone happened in the interval -L/3,L/3, which is consistent with the results analyzed of numerical simulation, as shown in Figure 8.Figure 8
The plastic zone analyzed by ANSYS.


## 2.3.3. Validation of Plastic Hardening under Loading

Whenq=21×106 N/m, there is qe<q<qp. At this moment, the middle of beam is in elastic-plastic state. Putting the value above into (8), (14), (23), and (29), then the value of total strain can be obtained, namely εij=εije+εijp=0.0102+0.0055=0.01577, which is also consistent with the result of numerical simulation, as shown in Figure 9.Figure 9
Elastic strain and plastic strain analyzed by ANSYS.The numerical simulation was used to analyze the simply supported beam subjected to uniformly distributed loading to verify the correctness of the elastic-plastic constitutive model and plastic zone mentioned above. The results show that the derived plastic zone and plastic strain are consistent with that of numerical simulation; namely, the formulas of elastic-plastic constitutive model and plastic zone proposed in this paper can be applied to the elastic-plastic analysis of isotropic hardening materials which meet the von Mises yield criterion.However, the design of stabilizing pile is carried out by static method in Chinese code for seismic design of railway [19] and highway engineering [20] at present; namely, the effect of earthquake on the pile is expressed by equivalent loading, and then the internal force and displacement are calculated by static analysis with the equivalent loading, so as to check the seismic bearing capacity and deformation of the pile. Certainly, the static method, as an accepted and frequently used method, can satisfy the requirement of seismic design and has many advantages, such as simply calculation and rich experience. However, the code is mainly for prevention so that the stabilizing pile designed by static method is more abundant than the actual one. According to Figure 4, under the same section size and same property, the ultimate stress of the pile with elastic-plastic analysis is 500 N/mm and the pile reaches the state of complete yielding at this time, while the stress is only 300 N/mm with static method, as the code assumes that the pile only exists in elastic state under the action of earthquake. If the pile has to bear the ultimate stress of 500 N/mm, it is necessary to increase the section size, which will undoubtedly increase material consumption and cause waste. Therefore, the establishment of elastic-plastic constitutive model is beneficial to make full use of materials and maximize their performance.


## 3. Model Design

### 3.1. Model and Parameters of Numerical Simulation

A model was established by ANASYS 3D to analyze the seismic response characteristics of stabilizing pile and the piles were made of C20 reinforced concrete, as shown in Figure10. In addition, the length of the pile was 25 m, where the embedded depth was 10 m. The section size of a pile was 2 m × 3 m (B×H), and the spacing of pile centers L was 6 m. The calculated model parameters are shown in Table 1.Table 1
The calculation parameters of stabilizing piles.
Materials Elastic modulus [MPa] Poisson ratio Cohesion [kPa] Internal friction angle [°] Yield strength [MPa] Density [kg/m3] Sliding mass 50 0.3 15 28 — 1900 Bedrock 1.5 × 104 0.25 35 30 — 2500 stabilizing pile 1 (C20) 2.0 × 104 0.2 — — 13.4 2400 stabilizing pile 2 (C25) 2.5 × 104 0.2 — — 16.7 2400 stabilizing pile 3 (C30) 3.0 × 104 0.2 — — 20.1 2400Figure 10
Structural design and landslide characteristics of the model. (a) The landslide characteristics; (b) the design parameters of stabilizing pile.
(a) (b)When the numerical calculation was performed with ANSYS and LS-DYNA, the first was implicit static analysis by ANSYS, and then the results of static analysis would be input into LS-DYNA as the initial conditions. Finally, the Wenchuan Wolong seismic wave was input for explicit dynamic analysis. In the simulation, block element was chosen to simulate slide mass, bedrock, and stabilizing pile, where SOLID 185 block element is in implicit analysis with ANSYS and SOLID 164 block element is in explicit analysis with LS-DYNA. In addition, the D-P constitutive model was adopted in the numerical model. The no reflection boundary conditions and Rayleigh damping were selected for dynamic calculation. The mesh generation is shown in Figure11. Furthermore, the monitoring points were arranged behind the piles which were prone to plastic yield with the vertical spacing of 1 m from the bottom to the top of a pile.Figure 11
The calculated model.


### 3.2. Input of Seismic Wave

Wenchuan wolong seismic wave, with time 0~15 s, was input after baseline correction and the direction was horizontal direction which was perpendicular to stabilizing pile [21]. The acceleration curve and acceleration spectrum curve after manual processing are shown in Figure 12.Figure 12
The acceleration and acceleration spectral curve after modification. (a) Acceleration time history curve; (b) Fourier spectral (EW) curve.
(a) (b)


## 3.1. Model and Parameters of Numerical Simulation

A model was established by ANASYS 3D to analyze the seismic response characteristics of stabilizing pile and the piles were made of C20 reinforced concrete, as shown in Figure10. In addition, the length of the pile was 25 m, where the embedded depth was 10 m. The section size of a pile was 2 m × 3 m (B×H), and the spacing of pile centers L was 6 m. The calculated model parameters are shown in Table 1.Table 1
The calculation parameters of stabilizing piles.
Materials Elastic modulus [MPa] Poisson ratio Cohesion [kPa] Internal friction angle [°] Yield strength [MPa] Density [kg/m3] Sliding mass 50 0.3 15 28 — 1900 Bedrock 1.5 × 104 0.25 35 30 — 2500 stabilizing pile 1 (C20) 2.0 × 104 0.2 — — 13.4 2400 stabilizing pile 2 (C25) 2.5 × 104 0.2 — — 16.7 2400 stabilizing pile 3 (C30) 3.0 × 104 0.2 — — 20.1 2400Figure 10
Structural design and landslide characteristics of the model. (a) The landslide characteristics; (b) the design parameters of stabilizing pile.
(a) (b)When the numerical calculation was performed with ANSYS and LS-DYNA, the first was implicit static analysis by ANSYS, and then the results of static analysis would be input into LS-DYNA as the initial conditions. Finally, the Wenchuan Wolong seismic wave was input for explicit dynamic analysis. In the simulation, block element was chosen to simulate slide mass, bedrock, and stabilizing pile, where SOLID 185 block element is in implicit analysis with ANSYS and SOLID 164 block element is in explicit analysis with LS-DYNA. In addition, the D-P constitutive model was adopted in the numerical model. The no reflection boundary conditions and Rayleigh damping were selected for dynamic calculation. The mesh generation is shown in Figure11. Furthermore, the monitoring points were arranged behind the piles which were prone to plastic yield with the vertical spacing of 1 m from the bottom to the top of a pile.Figure 11
The calculated model.


## 3.2. Input of Seismic Wave

Wenchuan wolong seismic wave, with time 0~15 s, was input after baseline correction and the direction was horizontal direction which was perpendicular to stabilizing pile [21]. The acceleration curve and acceleration spectrum curve after manual processing are shown in Figure 12.Figure 12
The acceleration and acceleration spectral curve after modification. (a) Acceleration time history curve; (b) Fourier spectral (EW) curve.
(a) (b)


## 4. Result Analysis

In the simulation, the revised Wenchuan Wolong seismic wave was input along the stress direction of piles at the bottom of the model. In addition, PGA was selected 0.8 g to damage those piles. On the basis, the seismic response characteristics of displacement and equivalent stress during the process of damage were studied.


### 4.1. Time History Response Analysis of Stabilizing Piles

In the time history response analysis of stabilizing pile, the time and position of maximum displacement and maximum stress were considered. The displacement time history curve of the pile top and the equivalent stress time history curve of embedded section are shown in Figure13.Figure 13
Displacement and equivalent stress time history curve of stabilizing piles. (a) The time history curve of displacement; (b) the time history curve of equivalent stress.
(a) (b)Some conclusions can be seen from Figure13. Firstly, as the ground motion time increases, the displacement of pile top shocks to increase. When the time reaches 12 s, the displacement begin to increase slowly and tends to be stable gradually. Furthermore, the displacement after stability reaches 12.7 cm. Secondly, the equivalent stress of stabilizing pile also gradually increases with the increase of vibration time. When the time reaches about 10 s, the plastic yielding occurs firstly so that the model enters plastic state. Since then, the equivalent stress comes up to the maximum and remains constant, whereas the plastic zone is not invariable until it runs through the whole section of stabilizing piles. Finally, according to the performance reaction of the stabilizing pile, the pile is in an elastic state at the beginning and then the plastic zone begins to occur with the increase of vibration time. In addition, as the time continues to increase, the plastic zone continuously spreads to the whole section, losing its working performance. The elastic-plastic states of the stabilizing pile with different times are shown in Figure 14.Figure 14
The plastic strain cloud diagram of stabilizing pile at different vibration times.


### 4.2. Seismic Response Characteristics of Stabilizing Pile

According to the time history response analysis of stabilizing pile, the displacement and equivalent stress increase with the increase of the vibration time. Considering the seismic performance, the 13.63 s was selected as the most severe time of response and the worst time of seismic design, and then the overall displacement and equivalent stress response characteristics of stabilizing pile (that is to the meaning of “seismic response characteristics” in this paper) were studied [22]. Finally, the response curves of horizontal displacement and equivalent stress were obtained, as shown in Figure 15.Figure 15
The dynamic response curve of stabilizing pile. (a) Horizontal displacement; (b) equivalent stress.
(a) (b)As is indicted from Figure15, the displacement response under the seismic dynamic action is much larger than that of static state, where the displacement of the pile top is the largest with the value of 10.1 cm. As for equivalent stress, the static and seismic dynamic response curves have the same tendency; namely, equivalent stress reaches minimum at both ends of the pile, while it is the maximum near the embedded section. Similarly, the equivalent stress of the static state is far less than that of seismic action.In addition, under the action of static state, the maximum of equivalent stress is 3.73 MPa which has not reached the yield strength of the material, so it is still in elastic state. However, the maximum is 13.4 MPa under the action of earthquake, which exceeds the yield strength, so it leads to the development of connected plastic zone. Finally, stabilizing pile starts to damage and loses its working performance gradually. The equivalent stress and plastic strain cloud diagram are shown in Figure16.Figure 16
The plastic strain cloud diagram of stabilizing pile under the action of seismic force. (a) Equivalent stress; (b) plastic strain.
(a) (b)


### 4.3. The Influence of Design Parameters on the Stabilizing Pile Performance

#### 4.3.1. Horizontal Displacement of Stabilizing Pile

The influence of material strength, section size, and PGA on the seismic response characteristics of stabilizing pile was also analyzed by controlling variable method, where the material strength included C20, C25, and C30 of reinforced concrete, the design dimensions includedB×H=2×2.5 m, 2×3 m, and 2×3.5 m, and the PGA included 0.2 g, 0.4 g, and 0.8 g.The displacement response curves of stabilizing pile in different section sizes, PGAs, and material strengths are shown in Figure17, and the corresponding plastic strain is shown in Figure 18.Figure 17
The displacement response curves of stabilizing pile in different working conditions. (a) Different section sizes; (b) different PGAs; (c) different material strengths.
(a) (b) (c)Figure 18
The plastic strain cloud diagram of stabilizing pile in different working conditions. (a) Different section sizes; (b) different PGAs; (c) different material strengths.
(a) (b) (c)It can be seen from Figures17-18 that when the size of structure is larger, the displacement is smaller. Therefore, changing the sectional design size of stabilizing pile can reduce the seismic displacement effectively, leading to the decrease of plastic zone. In addition, the displacement increases with the increase of PGA. Under the action of earthquake, 0.2 g, 0.4 g, and 0.8 g of PGA corresponds to 2.32 cm, 4.55 cm, and 8.68 cm of displacement, respectively. However, the displacement is 1.02 cm with the static force. Additionally, the connected plastic zone exhibits plastic state in severe earthquake region so that the capacity of stabilizing pile will lose.Similarly, when the material strength is larger, the displacement response is smaller, whereas the displacement difference of the stabilizing pile under static and seismic states is not obvious with the same structural material strength. Additionally, the plastic threshold of the material is improved by increasing the strength. Therefore, if the material strength changes, its plastic yield condition and bearing capacity can be enhanced availably.


#### 4.3.2. Equivalent Stress of Stabilizing Pile

The equivalent stress response curves of stabilizing pile in different section sizes, PGAs, and material strengths are shown in Figure19, and the corresponding plastic strain are shown in Figure 20.Figure 19
The equivalent stress curve of stabilizing pile in different working conditions. (a) Different section sizes; (b) different PGAs; (c) different material strengths.
(a) (b) (c)Figure 20
The equivalent stress cloud diagram of stabilizing pile in different working conditions. (a) Different section sizes; (b) different PGAs; (c) different material strengths.
(a) (b) (c)As is indicated from Figures19 and 20, equivalent stress of the stabilizing pile increases with the increase of PGA and decreases with the increase of section size and material strength. In addition, the influence of section size is extremely obvious when the pile is affected by earthquake, while the influence of material stress is inconspicuous.The variation of section size can efficiently reduce the equivalent stress of stabilizing pile. To be specific, under the action of earthquake, the maximum of equivalent stress near the embedded section is 16.7 MPa, 13.6 MPa, and 11.3 MPa when the section size isB×H=2×2.5 m, 2×3 m, and 2×3.5 m, respectively. In the static force, the maximum of different design sizes all roughly equals 3.88 MPa. Hence, according to the analysis results of equivalent stress, the stabilizing pile with section size of B×H=2×2.5 m reaches the yielding condition and starts to exhibit the plastic state; namely, the plastic zone is developing continuously. Nevertheless, the stabilizing piles with the remaining two sectional sizes are still in elastic state without yielding. Additionally, 0.2 g, 0.4 g, and 0.8 g of PGA correspond to 6.52 MPa, 13.6 MPa, and 16.7 MPa of the equivalent stress, respectively, while the value is 3.88 MPa with different PGAs in the static force. Furthermore, plastic zone in stabilizing pile also develops when PGA equals 0.8 g, whereas the stabilizing piles are still in elastic state when PGA is 0.2 g or 0.4 g. When studying the influence of material strength, it is concluded that the equivalent stress with three materials (C20, C25, and C30) is 13.4 MPa, 13.6 MPa, and 14.1 MPa, respectively, while it is about 3.07 MPa in the static force. Accordingly, the pressure stress of the stabilizing pile with C20 exceeds its yield strength, leading to the development of plastic zone. Likewise, the stabilizing piles with the other working conditions are still in elastic state.Additionally, the plastic threshold of the material is improved by increasing the strength. Therefore, if the material strength of stabilizing pile increases, its plastic yield condition and bearing capacity can be enhanced availably.


## 4.1. Time History Response Analysis of Stabilizing Piles

In the time history response analysis of stabilizing pile, the time and position of maximum displacement and maximum stress were considered. The displacement time history curve of the pile top and the equivalent stress time history curve of embedded section are shown in Figure13.Figure 13
Displacement and equivalent stress time history curve of stabilizing piles. (a) The time history curve of displacement; (b) the time history curve of equivalent stress.
(a) (b)Some conclusions can be seen from Figure13. Firstly, as the ground motion time increases, the displacement of pile top shocks to increase. When the time reaches 12 s, the displacement begin to increase slowly and tends to be stable gradually. Furthermore, the displacement after stability reaches 12.7 cm. Secondly, the equivalent stress of stabilizing pile also gradually increases with the increase of vibration time. When the time reaches about 10 s, the plastic yielding occurs firstly so that the model enters plastic state. Since then, the equivalent stress comes up to the maximum and remains constant, whereas the plastic zone is not invariable until it runs through the whole section of stabilizing piles. Finally, according to the performance reaction of the stabilizing pile, the pile is in an elastic state at the beginning and then the plastic zone begins to occur with the increase of vibration time. In addition, as the time continues to increase, the plastic zone continuously spreads to the whole section, losing its working performance. The elastic-plastic states of the stabilizing pile with different times are shown in Figure 14.Figure 14
The plastic strain cloud diagram of stabilizing pile at different vibration times.


## 4.2. Seismic Response Characteristics of Stabilizing Pile

According to the time history response analysis of stabilizing pile, the displacement and equivalent stress increase with the increase of the vibration time. Considering the seismic performance, the 13.63 s was selected as the most severe time of response and the worst time of seismic design, and then the overall displacement and equivalent stress response characteristics of stabilizing pile (that is to the meaning of “seismic response characteristics” in this paper) were studied [22]. Finally, the response curves of horizontal displacement and equivalent stress were obtained, as shown in Figure 15.Figure 15
The dynamic response curve of stabilizing pile. (a) Horizontal displacement; (b) equivalent stress.
(a) (b)As is indicted from Figure15, the displacement response under the seismic dynamic action is much larger than that of static state, where the displacement of the pile top is the largest with the value of 10.1 cm. As for equivalent stress, the static and seismic dynamic response curves have the same tendency; namely, equivalent stress reaches minimum at both ends of the pile, while it is the maximum near the embedded section. Similarly, the equivalent stress of the static state is far less than that of seismic action.In addition, under the action of static state, the maximum of equivalent stress is 3.73 MPa which has not reached the yield strength of the material, so it is still in elastic state. However, the maximum is 13.4 MPa under the action of earthquake, which exceeds the yield strength, so it leads to the development of connected plastic zone. Finally, stabilizing pile starts to damage and loses its working performance gradually. The equivalent stress and plastic strain cloud diagram are shown in Figure16.Figure 16
The plastic strain cloud diagram of stabilizing pile under the action of seismic force. (a) Equivalent stress; (b) plastic strain.
(a) (b)


## 4.3. The Influence of Design Parameters on the Stabilizing Pile Performance

### 4.3.1. Horizontal Displacement of Stabilizing Pile

The influence of material strength, section size, and PGA on the seismic response characteristics of stabilizing pile was also analyzed by controlling variable method, where the material strength included C20, C25, and C30 of reinforced concrete, the design dimensions includedB×H=2×2.5 m, 2×3 m, and 2×3.5 m, and the PGA included 0.2 g, 0.4 g, and 0.8 g.The displacement response curves of stabilizing pile in different section sizes, PGAs, and material strengths are shown in Figure17, and the corresponding plastic strain is shown in Figure 18.Figure 17
The displacement response curves of stabilizing pile in different working conditions. (a) Different section sizes; (b) different PGAs; (c) different material strengths.
(a) (b) (c)Figure 18
The plastic strain cloud diagram of stabilizing pile in different working conditions. (a) Different section sizes; (b) different PGAs; (c) different material strengths.
(a) (b) (c)It can be seen from Figures17-18 that when the size of structure is larger, the displacement is smaller. Therefore, changing the sectional design size of stabilizing pile can reduce the seismic displacement effectively, leading to the decrease of plastic zone. In addition, the displacement increases with the increase of PGA. Under the action of earthquake, 0.2 g, 0.4 g, and 0.8 g of PGA corresponds to 2.32 cm, 4.55 cm, and 8.68 cm of displacement, respectively. However, the displacement is 1.02 cm with the static force. Additionally, the connected plastic zone exhibits plastic state in severe earthquake region so that the capacity of stabilizing pile will lose.Similarly, when the material strength is larger, the displacement response is smaller, whereas the displacement difference of the stabilizing pile under static and seismic states is not obvious with the same structural material strength. Additionally, the plastic threshold of the material is improved by increasing the strength. Therefore, if the material strength changes, its plastic yield condition and bearing capacity can be enhanced availably.


### 4.3.2. Equivalent Stress of Stabilizing Pile

The equivalent stress response curves of stabilizing pile in different section sizes, PGAs, and material strengths are shown in Figure19, and the corresponding plastic strain are shown in Figure 20.Figure 19
The equivalent stress curve of stabilizing pile in different working conditions. (a) Different section sizes; (b) different PGAs; (c) different material strengths.
(a) (b) (c)Figure 20
The equivalent stress cloud diagram of stabilizing pile in different working conditions. (a) Different section sizes; (b) different PGAs; (c) different material strengths.
(a) (b) (c)As is indicated from Figures19 and 20, equivalent stress of the stabilizing pile increases with the increase of PGA and decreases with the increase of section size and material strength. In addition, the influence of section size is extremely obvious when the pile is affected by earthquake, while the influence of material stress is inconspicuous.The variation of section size can efficiently reduce the equivalent stress of stabilizing pile. To be specific, under the action of earthquake, the maximum of equivalent stress near the embedded section is 16.7 MPa, 13.6 MPa, and 11.3 MPa when the section size isB×H=2×2.5 m, 2×3 m, and 2×3.5 m, respectively. In the static force, the maximum of different design sizes all roughly equals 3.88 MPa. Hence, according to the analysis results of equivalent stress, the stabilizing pile with section size of B×H=2×2.5 m reaches the yielding condition and starts to exhibit the plastic state; namely, the plastic zone is developing continuously. Nevertheless, the stabilizing piles with the remaining two sectional sizes are still in elastic state without yielding. Additionally, 0.2 g, 0.4 g, and 0.8 g of PGA correspond to 6.52 MPa, 13.6 MPa, and 16.7 MPa of the equivalent stress, respectively, while the value is 3.88 MPa with different PGAs in the static force. Furthermore, plastic zone in stabilizing pile also develops when PGA equals 0.8 g, whereas the stabilizing piles are still in elastic state when PGA is 0.2 g or 0.4 g. When studying the influence of material strength, it is concluded that the equivalent stress with three materials (C20, C25, and C30) is 13.4 MPa, 13.6 MPa, and 14.1 MPa, respectively, while it is about 3.07 MPa in the static force. Accordingly, the pressure stress of the stabilizing pile with C20 exceeds its yield strength, leading to the development of plastic zone. Likewise, the stabilizing piles with the other working conditions are still in elastic state.Additionally, the plastic threshold of the material is improved by increasing the strength. Therefore, if the material strength of stabilizing pile increases, its plastic yield condition and bearing capacity can be enhanced availably.


## 4.3.1. Horizontal Displacement of Stabilizing Pile

The influence of material strength, section size, and PGA on the seismic response characteristics of stabilizing pile was also analyzed by controlling variable method, where the material strength included C20, C25, and C30 of reinforced concrete, the design dimensions includedB×H=2×2.5 m, 2×3 m, and 2×3.5 m, and the PGA included 0.2 g, 0.4 g, and 0.8 g.The displacement response curves of stabilizing pile in different section sizes, PGAs, and material strengths are shown in Figure17, and the corresponding plastic strain is shown in Figure 18.Figure 17
The displacement response curves of stabilizing pile in different working conditions. (a) Different section sizes; (b) different PGAs; (c) different material strengths.
(a) (b) (c)Figure 18
The plastic strain cloud diagram of stabilizing pile in different working conditions. (a) Different section sizes; (b) different PGAs; (c) different material strengths.
(a) (b) (c)It can be seen from Figures17-18 that when the size of structure is larger, the displacement is smaller. Therefore, changing the sectional design size of stabilizing pile can reduce the seismic displacement effectively, leading to the decrease of plastic zone. In addition, the displacement increases with the increase of PGA. Under the action of earthquake, 0.2 g, 0.4 g, and 0.8 g of PGA corresponds to 2.32 cm, 4.55 cm, and 8.68 cm of displacement, respectively. However, the displacement is 1.02 cm with the static force. Additionally, the connected plastic zone exhibits plastic state in severe earthquake region so that the capacity of stabilizing pile will lose.Similarly, when the material strength is larger, the displacement response is smaller, whereas the displacement difference of the stabilizing pile under static and seismic states is not obvious with the same structural material strength. Additionally, the plastic threshold of the material is improved by increasing the strength. Therefore, if the material strength changes, its plastic yield condition and bearing capacity can be enhanced availably.


## 4.3.2. Equivalent Stress of Stabilizing Pile

The equivalent stress response curves of stabilizing pile in different section sizes, PGAs, and material strengths are shown in Figure19, and the corresponding plastic strain are shown in Figure 20.Figure 19
The equivalent stress curve of stabilizing pile in different working conditions. (a) Different section sizes; (b) different PGAs; (c) different material strengths.
(a) (b) (c)Figure 20
The equivalent stress cloud diagram of stabilizing pile in different working conditions. (a) Different section sizes; (b) different PGAs; (c) different material strengths.
(a) (b) (c)As is indicated from Figures19 and 20, equivalent stress of the stabilizing pile increases with the increase of PGA and decreases with the increase of section size and material strength. In addition, the influence of section size is extremely obvious when the pile is affected by earthquake, while the influence of material stress is inconspicuous.The variation of section size can efficiently reduce the equivalent stress of stabilizing pile. To be specific, under the action of earthquake, the maximum of equivalent stress near the embedded section is 16.7 MPa, 13.6 MPa, and 11.3 MPa when the section size isB×H=2×2.5 m, 2×3 m, and 2×3.5 m, respectively. In the static force, the maximum of different design sizes all roughly equals 3.88 MPa. Hence, according to the analysis results of equivalent stress, the stabilizing pile with section size of B×H=2×2.5 m reaches the yielding condition and starts to exhibit the plastic state; namely, the plastic zone is developing continuously. Nevertheless, the stabilizing piles with the remaining two sectional sizes are still in elastic state without yielding. Additionally, 0.2 g, 0.4 g, and 0.8 g of PGA correspond to 6.52 MPa, 13.6 MPa, and 16.7 MPa of the equivalent stress, respectively, while the value is 3.88 MPa with different PGAs in the static force. Furthermore, plastic zone in stabilizing pile also develops when PGA equals 0.8 g, whereas the stabilizing piles are still in elastic state when PGA is 0.2 g or 0.4 g. When studying the influence of material strength, it is concluded that the equivalent stress with three materials (C20, C25, and C30) is 13.4 MPa, 13.6 MPa, and 14.1 MPa, respectively, while it is about 3.07 MPa in the static force. Accordingly, the pressure stress of the stabilizing pile with C20 exceeds its yield strength, leading to the development of plastic zone. Likewise, the stabilizing piles with the other working conditions are still in elastic state.Additionally, the plastic threshold of the material is improved by increasing the strength. Therefore, if the material strength of stabilizing pile increases, its plastic yield condition and bearing capacity can be enhanced availably.


## 5. Conclusions

In order to analyze the seismic response characteristics of stabilizing pile based on elastic-plastic analysis, the constitutive relationship with isotropic hardening was deduced according to von Mises yield criterion, and then a numerical model was established by ANSYS. In view of this, the seismic response characteristics and the influence of different design parameters on seismic performance were analyzed. The following conclusions can be obtained.(1) The elastic-plastic constitutive relationship of isotropic hardening for stabilizing pile was obtained, and the elastic-plastic fourth-order tensorCijklep was deduced, which can be used to calculate plastic strain under the action of earthquake. In addition, the plastic zone was deduced based on elastic limit state and plastic limit state.(2) As for the piles of same size and same property, the ultimate stress based on elastic-plastic analysis is greater than that of static method. Therefore, the material of stabilizing pile working with elastic-plastic state will be decreased under the action of earthquake.(3) Under the action of earthquake, stabilizing pile is in the elastic stage at the beginning. With the increase of ground motion time, the section starts to exhibit elastic-plastic state and then the plastic zone expands gradually. Finally, the plastic zone runs through the whole section so that the stabilizing pile loses its working performance.(4) Under the different design parameters, stabilizing pile shows different seismic response characteristics. Specifically, with the increase of section size, the displacement and equivalent stress decrease and so does the plastic zone. However, the increase of material strength improves the plastic threshold of pile. In addition, the stabilizing pile in high intensity seismic area may exhibit connected plastic zone so that losing its service capacity.(5) According to the analysis of numerical simulation, it also confirms the feasibly that the elastic-plastic constitutive model deduced in this paper can study those isotropic hardening materials which meet von Mises yield criterion.



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*Source: 7672587-2018-06-25.xml*