# Study on the Mixed Materials Proportion of Stratum Based on the Modelling Experiment

Authors: Yijun Zhou; Tao Xu
Journal: Advances in Civil Engineering
 (2021)

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## Abstract

It is highly significant to select similar materials as well as the proportion of mixed materials in the model test. The mixed materials are used to simulate the stratum of the model test, including the iron ore powder, natural sands, gypsum, and lime. The stratum contains silty clay and silt soil. First of all, the symmetry coefficient of model mechanics parameters are calculated by the equation, and the symmetry ratio is 16 : 1. Second, calculate the proportion of compositions in mixed materials by the orthogonal test. The deviation method is used to analyze the mixed materials and how to influence the elastic modulus, cohesion, and friction angle. Finally, get the mixed materials which meet the symmetry theory and control factors.


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## Body

## 1. Introduction

The model test is a method that uses the symmetry theory to reduce the size of the prototype. The symmetrical material is highly significant to the model test, and it usually contains some kinds of materials called mixed materials. Choosing suitable mixed materials can determine the model test whether success or not [1–3].The mixed materials have been researched by some researchers. In abroad, Fumagalli [4] researched the model test of engineering geology initially in the 1960s. He used gypsum, powder of PbO, expansive soil, and water to simulate the stratum. Han et al. [5] researched the materials of MIB to study the rock and soil. Wang [6] selected the barite, quartz, and vaseline to study proportion of mixed materials in the model test. He found that different proportions of mixed materials lead to different results of the test. Chen and Zuo [7, 8] introduced several materials to study the influence for proportion of stratum, including PbO, gypsum, expansive soil, sands, starch, hardener, and so on.The symmetry theory is mainly used to guide the model test to determine the proportion for the model and prototype [9–12].The geology model test is highly complex and is affected by lots of factors, such as density of soil, cohesion of soil, friction angle of soil, elastic modulus of soil, and so on. Therefore, the much more important factor must be controlled, ignoring the less important factors [13–17].To measure the proportion of mixed materials, some kinds of methods are introduced, including the direct shear test, orthogonal test, deviation analysis method, three axes test, and so on [18–26].


## 2. Materials and Methods

### 2.1. Determination of Symmetry Coefficient of Stratum

#### 2.1.1. Symmetry Ratio

In the process of the model test, the symmetry ratio is a crucial step and also determines the model test whether it can correctly react to objective laws or not. The symmetry ratio is the ratio between the prototype and the model and markedC. The definitions of model test parameters are as follows: L is the length, r is the density, δ is the displacement, σ is the stress, ε is the strain, σt is the tensile strength, σc is the compressive strength, c is the cohesion, ϕ is the friction angle, μ is Poisson’s ratio, and f is the coefficient of friction. All of the parameters of symmetry ratio are given in Table 1.Table 1
Symmetry coefficient calculation formula.
ParameterLengthStrainDensityVolume forceDisplacementStressElastic modulusPoisson’s ratioFriction angleCohesionSymmetry ratioCL=LP/LmCε=εP/εmCγ=γP/γmCX=XP/XmCδ=δP/δmCσ=σtP/σtm=σcP/σcm=σP/σmCE=Ep/EmCμ=μp/μmCϕ=ϕP/ϕmCc=cP/cm


#### 2.1.2. Establishment of Symmetry Equation

According to the symmetry theory, establish the equation of the prototype and model, including the equilibrium equation, geometric equation, and physical equation.(1)
Establish a symmetrical condition by the equilibrium equation:(1)Equilibrium equation of prototype:∂σx∂xp+∂τyx∂yp+∂τzx∂zp+XP=0∂σy∂yp+∂τzy∂zp+∂τxy∂xp+YP=0∂σz∂zp+∂τxz∂xp+∂τyz∂yp+ZP=0.(2)Equilibrium equation of model:∂σx∂xm+∂τyx∂ym+∂τzx∂zm+Xm=0∂σy∂ym+∂τzy∂zm+∂τxy∂xm+Ym=0∂σz∂zm+∂τxz∂xm+∂τyz∂ym+Zm=0.Substitute the symmetry coefficientcσ,cL,cX into formula (1), and the following formula is obtained:(3)∂σx∂xm+∂τyx∂ym+∂τzx∂zm+cγcLcσXm=0∂σy∂ym+∂τzy∂zm+∂τxy∂xm+cγcLcσYm=0∂σz∂zm+∂τxz∂xm+∂τyz∂ym+cγcLcσZm=0.According to formulas (2) and (3), we can get the equation for cγ,cL,cσ.(4)cγcLcσ=1.(2)
Establish a symmetrical condition by the geometry equation:(5)Geometry equation of prototype:εxp=∂u∂xpεyp=∂v∂ypεzp=∂w∂zpγxyp=∂u∂yp+∂v∂xpγyzp=∂v∂zp+∂w∂ypγzxp=∂u∂zp+∂w∂xp.(6)Geometry equation of model:εxm=∂u∂xmεym=∂v∂ymεzm=∂w∂zmγxym=∂u∂ym+∂v∂xmγyzm=∂v∂zm+∂w∂ymγzxm=∂u∂zm+∂w∂xm.Substitute the symmetry coefficientcε,cδ,cL into formula (5), and the following formula is obtained:(7)εxmcξcLcδ=∂u∂xmεymcξcLcδ=∂v∂ymεzmcξcLcδ=∂w∂zmγxymcξcLcδ=∂u∂ym+∂v∂xmγyzmcξcLcδ=∂v∂zm+∂w∂ymγzxmcξcLcδ=∂u∂zm+∂w∂xm.According to formulas (6) and (7), we can get the equation for cξ,cδ,cL.(8)cξcLcδ=1.(3)
Establish a symmetrical condition by the physical equation:(9)Physical equation of prototype:εxp=1EPσx−μσy+σzpεyp=1EPσy−μσx+σzpεzp=1EPσz−μσx+σypγxyp=21+μEpτxypγxzp=21+μEpτxzpγyzp=21+μEpτyzp(10)Physical equation of model:εxm=1EPσx−μσy+σzmεym=1EPσy−μσx+σzmεzm=1EPσz−μσx+σymγxym=21+μEpτxymγxzm=21+μEpτxzmγyzm=21+μEpτyzm.Substitute the symmetrical coefficientcε,cσ,cμ,cE into formula (9), and the following formula is obtained.(11)εxm=cσcεcE1Emσx−cμμσy+σzmεym=cσcεcE1Emσy−cμμσx+σzmεzm=cσcεcE1Emσz−cμμσx+σymγxym=cσcεcE21+μEpτxymγxzm=cσcεcE21+μEpτxzmγyzm=cσcεcE21+μEpτyzm.According to formulas (10) and (11), we get the equation for cσ,cε,cE. The symmetry coefficient of Poisson’s is cμ=1.(12)cσcεcE=1.


#### 2.1.3. Symmetry Ratio Determination

According to the symmetry theory and the size of model box, finally the symmetry ratio in the model test isCL=16:1. It is assumed that the symmetry ratio of density is 1 : 1, and the dimensionless parameter is 1 : 1. Therefore, all of the parameters of symmetry ratio in the model test are given in Table 2.Table 2
Each of parameter for symmetry ratio.
ParameterGeometryDisplacementStressStrainCohesionFriction angleElastic modulusPoisson’s ratioBending momentPrototype1616161161161164Model111111111


## 2.1. Determination of Symmetry Coefficient of Stratum

### 2.1.1. Symmetry Ratio

In the process of the model test, the symmetry ratio is a crucial step and also determines the model test whether it can correctly react to objective laws or not. The symmetry ratio is the ratio between the prototype and the model and markedC. The definitions of model test parameters are as follows: L is the length, r is the density, δ is the displacement, σ is the stress, ε is the strain, σt is the tensile strength, σc is the compressive strength, c is the cohesion, ϕ is the friction angle, μ is Poisson’s ratio, and f is the coefficient of friction. All of the parameters of symmetry ratio are given in Table 1.Table 1
Symmetry coefficient calculation formula.
ParameterLengthStrainDensityVolume forceDisplacementStressElastic modulusPoisson’s ratioFriction angleCohesionSymmetry ratioCL=LP/LmCε=εP/εmCγ=γP/γmCX=XP/XmCδ=δP/δmCσ=σtP/σtm=σcP/σcm=σP/σmCE=Ep/EmCμ=μp/μmCϕ=ϕP/ϕmCc=cP/cm


### 2.1.2. Establishment of Symmetry Equation

According to the symmetry theory, establish the equation of the prototype and model, including the equilibrium equation, geometric equation, and physical equation.(1)
Establish a symmetrical condition by the equilibrium equation:(1)Equilibrium equation of prototype:∂σx∂xp+∂τyx∂yp+∂τzx∂zp+XP=0∂σy∂yp+∂τzy∂zp+∂τxy∂xp+YP=0∂σz∂zp+∂τxz∂xp+∂τyz∂yp+ZP=0.(2)Equilibrium equation of model:∂σx∂xm+∂τyx∂ym+∂τzx∂zm+Xm=0∂σy∂ym+∂τzy∂zm+∂τxy∂xm+Ym=0∂σz∂zm+∂τxz∂xm+∂τyz∂ym+Zm=0.Substitute the symmetry coefficientcσ,cL,cX into formula (1), and the following formula is obtained:(3)∂σx∂xm+∂τyx∂ym+∂τzx∂zm+cγcLcσXm=0∂σy∂ym+∂τzy∂zm+∂τxy∂xm+cγcLcσYm=0∂σz∂zm+∂τxz∂xm+∂τyz∂ym+cγcLcσZm=0.According to formulas (2) and (3), we can get the equation for cγ,cL,cσ.(4)cγcLcσ=1.(2)
Establish a symmetrical condition by the geometry equation:(5)Geometry equation of prototype:εxp=∂u∂xpεyp=∂v∂ypεzp=∂w∂zpγxyp=∂u∂yp+∂v∂xpγyzp=∂v∂zp+∂w∂ypγzxp=∂u∂zp+∂w∂xp.(6)Geometry equation of model:εxm=∂u∂xmεym=∂v∂ymεzm=∂w∂zmγxym=∂u∂ym+∂v∂xmγyzm=∂v∂zm+∂w∂ymγzxm=∂u∂zm+∂w∂xm.Substitute the symmetry coefficientcε,cδ,cL into formula (5), and the following formula is obtained:(7)εxmcξcLcδ=∂u∂xmεymcξcLcδ=∂v∂ymεzmcξcLcδ=∂w∂zmγxymcξcLcδ=∂u∂ym+∂v∂xmγyzmcξcLcδ=∂v∂zm+∂w∂ymγzxmcξcLcδ=∂u∂zm+∂w∂xm.According to formulas (6) and (7), we can get the equation for cξ,cδ,cL.(8)cξcLcδ=1.(3)
Establish a symmetrical condition by the physical equation:(9)Physical equation of prototype:εxp=1EPσx−μσy+σzpεyp=1EPσy−μσx+σzpεzp=1EPσz−μσx+σypγxyp=21+μEpτxypγxzp=21+μEpτxzpγyzp=21+μEpτyzp(10)Physical equation of model:εxm=1EPσx−μσy+σzmεym=1EPσy−μσx+σzmεzm=1EPσz−μσx+σymγxym=21+μEpτxymγxzm=21+μEpτxzmγyzm=21+μEpτyzm.Substitute the symmetrical coefficientcε,cσ,cμ,cE into formula (9), and the following formula is obtained.(11)εxm=cσcεcE1Emσx−cμμσy+σzmεym=cσcεcE1Emσy−cμμσx+σzmεzm=cσcεcE1Emσz−cμμσx+σymγxym=cσcεcE21+μEpτxymγxzm=cσcεcE21+μEpτxzmγyzm=cσcεcE21+μEpτyzm.According to formulas (10) and (11), we get the equation for cσ,cε,cE. The symmetry coefficient of Poisson’s is cμ=1.(12)cσcεcE=1.


### 2.1.3. Symmetry Ratio Determination

According to the symmetry theory and the size of model box, finally the symmetry ratio in the model test isCL=16:1. It is assumed that the symmetry ratio of density is 1 : 1, and the dimensionless parameter is 1 : 1. Therefore, all of the parameters of symmetry ratio in the model test are given in Table 2.Table 2
Each of parameter for symmetry ratio.
ParameterGeometryDisplacementStressStrainCohesionFriction angleElastic modulusPoisson’s ratioBending momentPrototype1616161161161164Model111111111


## 2.1.1. Symmetry Ratio

In the process of the model test, the symmetry ratio is a crucial step and also determines the model test whether it can correctly react to objective laws or not. The symmetry ratio is the ratio between the prototype and the model and markedC. The definitions of model test parameters are as follows: L is the length, r is the density, δ is the displacement, σ is the stress, ε is the strain, σt is the tensile strength, σc is the compressive strength, c is the cohesion, ϕ is the friction angle, μ is Poisson’s ratio, and f is the coefficient of friction. All of the parameters of symmetry ratio are given in Table 1.Table 1
Symmetry coefficient calculation formula.
ParameterLengthStrainDensityVolume forceDisplacementStressElastic modulusPoisson’s ratioFriction angleCohesionSymmetry ratioCL=LP/LmCε=εP/εmCγ=γP/γmCX=XP/XmCδ=δP/δmCσ=σtP/σtm=σcP/σcm=σP/σmCE=Ep/EmCμ=μp/μmCϕ=ϕP/ϕmCc=cP/cm


## 2.1.2. Establishment of Symmetry Equation

According to the symmetry theory, establish the equation of the prototype and model, including the equilibrium equation, geometric equation, and physical equation.(1)
Establish a symmetrical condition by the equilibrium equation:(1)Equilibrium equation of prototype:∂σx∂xp+∂τyx∂yp+∂τzx∂zp+XP=0∂σy∂yp+∂τzy∂zp+∂τxy∂xp+YP=0∂σz∂zp+∂τxz∂xp+∂τyz∂yp+ZP=0.(2)Equilibrium equation of model:∂σx∂xm+∂τyx∂ym+∂τzx∂zm+Xm=0∂σy∂ym+∂τzy∂zm+∂τxy∂xm+Ym=0∂σz∂zm+∂τxz∂xm+∂τyz∂ym+Zm=0.Substitute the symmetry coefficientcσ,cL,cX into formula (1), and the following formula is obtained:(3)∂σx∂xm+∂τyx∂ym+∂τzx∂zm+cγcLcσXm=0∂σy∂ym+∂τzy∂zm+∂τxy∂xm+cγcLcσYm=0∂σz∂zm+∂τxz∂xm+∂τyz∂ym+cγcLcσZm=0.According to formulas (2) and (3), we can get the equation for cγ,cL,cσ.(4)cγcLcσ=1.(2)
Establish a symmetrical condition by the geometry equation:(5)Geometry equation of prototype:εxp=∂u∂xpεyp=∂v∂ypεzp=∂w∂zpγxyp=∂u∂yp+∂v∂xpγyzp=∂v∂zp+∂w∂ypγzxp=∂u∂zp+∂w∂xp.(6)Geometry equation of model:εxm=∂u∂xmεym=∂v∂ymεzm=∂w∂zmγxym=∂u∂ym+∂v∂xmγyzm=∂v∂zm+∂w∂ymγzxm=∂u∂zm+∂w∂xm.Substitute the symmetry coefficientcε,cδ,cL into formula (5), and the following formula is obtained:(7)εxmcξcLcδ=∂u∂xmεymcξcLcδ=∂v∂ymεzmcξcLcδ=∂w∂zmγxymcξcLcδ=∂u∂ym+∂v∂xmγyzmcξcLcδ=∂v∂zm+∂w∂ymγzxmcξcLcδ=∂u∂zm+∂w∂xm.According to formulas (6) and (7), we can get the equation for cξ,cδ,cL.(8)cξcLcδ=1.(3)
Establish a symmetrical condition by the physical equation:(9)Physical equation of prototype:εxp=1EPσx−μσy+σzpεyp=1EPσy−μσx+σzpεzp=1EPσz−μσx+σypγxyp=21+μEpτxypγxzp=21+μEpτxzpγyzp=21+μEpτyzp(10)Physical equation of model:εxm=1EPσx−μσy+σzmεym=1EPσy−μσx+σzmεzm=1EPσz−μσx+σymγxym=21+μEpτxymγxzm=21+μEpτxzmγyzm=21+μEpτyzm.Substitute the symmetrical coefficientcε,cσ,cμ,cE into formula (9), and the following formula is obtained.(11)εxm=cσcεcE1Emσx−cμμσy+σzmεym=cσcεcE1Emσy−cμμσx+σzmεzm=cσcεcE1Emσz−cμμσx+σymγxym=cσcεcE21+μEpτxymγxzm=cσcεcE21+μEpτxzmγyzm=cσcεcE21+μEpτyzm.According to formulas (10) and (11), we get the equation for cσ,cε,cE. The symmetry coefficient of Poisson’s is cμ=1.(12)cσcεcE=1.


## 2.1.3. Symmetry Ratio Determination

According to the symmetry theory and the size of model box, finally the symmetry ratio in the model test isCL=16:1. It is assumed that the symmetry ratio of density is 1 : 1, and the dimensionless parameter is 1 : 1. Therefore, all of the parameters of symmetry ratio in the model test are given in Table 2.Table 2
Each of parameter for symmetry ratio.
ParameterGeometryDisplacementStressStrainCohesionFriction angleElastic modulusPoisson’s ratioBending momentPrototype1616161161161164Model111111111


## 3. Testing

### 3.1. Selecting of Stratum Symmetry Materials

#### 3.1.1. Prototype Stratum Parameters

In the model test, selecting the soil of Shanghai is considered as the prototype stratum. The information of prototype stratum in Shanghai is given in Table3. According to formula (13), the relationship between compression modulus and elastic modulus is(13)E=ES1−2μ21−μ,where E is the elastic modulus of soil, kPa; ES is the compression modulus of soil, kPa; μ is Poisson’s ratio of soil.Table 3
The physical quantity of prototype stratum in Shanghai.
StratumDensity (kN/m3)Cohesion (kPa)Friction angle (°)Elastic modulus (MPa)Poisson’s ratioThickness of stratum (m)Fill the soil18.5015110.330.64Silty clay19.22017380.361.92Silt soil17.51311100.3117.6Silty clay18.42714420.346.24Based on the exiting conclusion of research, finally, it selected mixed materials to research stratum in the model test, including iron ore powder, natural sands, gypsum, and lime. The iron ore powder and natural sands are aggregate, which have a great density. The gypsum and lime is the adhesive, which has better sticky property and great tension. In addition, the mixed materials are no harm for beings, easily available, low cost, and so on.


#### 3.1.2. Determination of Mixed Materials Ratio Initially

According to the direct shear test for many times, rely on the density of symmetry ratio. The mixed materials ratio is obtained initially, as given in Table4. The mixed materials are made to simulate each stratum, as given in Table 5.Table 4
Mixed materials finished initially.
Proportion of iron and sand in mixture (%)Quality ratio between iron and sandQuality ratio between gypsum and limeVolume ratio between water and gypsum851.5 : 3.51 : 12 : 8Table 5
Simulation materials of each stratigraphic.
StratumThickness of prototype (m)Thickness of the model (cm)Mixed materialsFill the soil0.644Natural sandsSilty clay1.9212Iron ore powder, natural sands, gypsum, and limeSilt soil17.6110Iron ore powder, natural sands, gypsum, and limeSilty clay6.2439Iron ore powder, natural sands, gypsum, and limeFrom what has been researched above, it just studies about the mixed materials ratio for silty clay and silt soil.


### 3.2. Determination of Mechanics Parameters for Stratum Symmetry Materials

#### 3.2.1. Design of the Orthogonal Test

The orthogonal test is used to research the proportion of mixed materials, design three factors and three levels, a total of nine tests, according to the purpose of the test, considering the density, cohesion, friction angle, elastic modulus, and Poisson’s ratio as the control index, as given in Table6.Table 6
Value of control index.
FactorsProportion of iron and sand in mixture (%)Quality ratio between iron and sandQuality ratio between gypsum and limeLevel 1802 : 32 : 1Level 2851.5 : 3.51 : 1Level 3901 : 41 : 2


#### 3.2.2. Parameters of the Orthogonal Test

In order to get the five parameters, that is, density, cohesion, friction angle, elastic modulus, and Poisson’s ratio, the research adopts, respectively, the density test, the direct shear test, elastic modulus test, and Poisson’s ratio test.(1)
Density testIn the density test, the formula of density is given in the following equation, and the instruments are given in Table7.(14)ρ=m1−m2V2,ρ is the density of soil, g/cm3; m1 is the total quality of soil and ring knife, g; m2 is the quality of ring knife, g; V2 is the volume of ring knife, cm3.(2)
Direct shear testThe direct shear test is a common method to measure the shear strength of soil. There is about four times to measure the shear strength in one direct shear test, under different vertical pressures, measuring the shear stress when soil is destroyed. The formula is given as follows:(15)τf=c+σtanϕ,τf is the shear strength of soil, kPa; c is the cohesion of soil, kPa; ϕ is the friction angle of soil; σ is the vertical stress, kPa.(3)
Poisson’s ratio and the elastic modulus testThe value of Poisson’s ratio is measured by two steps:(a)
The lateral pressure coefficientK0 of soil samples is obtained by the static pressure coefficient test(b)
Getting the value of Poisson’s ratio according to the generalized Hooker’s lawTable 7
Instrument of the density test.
NameRing knifeBalanceVernier caliperCompaction meterParametersInternal diameter, 6–8 cm; height 2 cmAccuracy, 0.1 gAccuracy, 0.02 mmDiameter,100 mm; height, 127 mm; volume, 997 cm3The elastic modulus is measured from the lateral compression test of similar materials, as shown in Figure1, and the formula is derived as follows:Figure 1
Stress condition of the confined compress test.According to generalized Hooke’s law,(16)εx+εy=0,(17)εx=σxE−μσy+σzE,(18)εy=σyE−μσx+σzE.Substituting formulas (17) and (18) into formula (16), the following equation is obtained:(19)K0=σxσz=μ1−μ,where μ is Poisson’s ratio; K0 is the side pressure coefficient; E is the elastic modulus, kPa.According to generalized Hooke’s law, the strain ofZ axis is given in the following formula:(20)εz=σzE−μσ+xσyE.Substitutingσz=p, σx=σy=μ/1−μp into formula (20), the following equation is obtained:(21)E=PεZ1−2μ21−μ,(22)εZ=ΔHH=e0−e11+e0.Compression coefficient under confinement conditions is(23)a=ΔeΔσ=e0−e1P.Finally, formula (21) is given as follows:(24)E=1+e0a1−2μ21−μ.Compression modulus under confinement conditions is(25)ES=1+e0a,E=ES1−2μ21−μ,where ES is the compression modulus under confinement conditions, kPa; e0 is the porosity ratio; a is the compression coefficient under confinement conditions, kPa−1; Δe is the change of the amount porosity ratio; Δσ is the change of the amount vertical stress, kPa.


## 3.1. Selecting of Stratum Symmetry Materials

### 3.1.1. Prototype Stratum Parameters

In the model test, selecting the soil of Shanghai is considered as the prototype stratum. The information of prototype stratum in Shanghai is given in Table3. According to formula (13), the relationship between compression modulus and elastic modulus is(13)E=ES1−2μ21−μ,where E is the elastic modulus of soil, kPa; ES is the compression modulus of soil, kPa; μ is Poisson’s ratio of soil.Table 3
The physical quantity of prototype stratum in Shanghai.
StratumDensity (kN/m3)Cohesion (kPa)Friction angle (°)Elastic modulus (MPa)Poisson’s ratioThickness of stratum (m)Fill the soil18.5015110.330.64Silty clay19.22017380.361.92Silt soil17.51311100.3117.6Silty clay18.42714420.346.24Based on the exiting conclusion of research, finally, it selected mixed materials to research stratum in the model test, including iron ore powder, natural sands, gypsum, and lime. The iron ore powder and natural sands are aggregate, which have a great density. The gypsum and lime is the adhesive, which has better sticky property and great tension. In addition, the mixed materials are no harm for beings, easily available, low cost, and so on.


### 3.1.2. Determination of Mixed Materials Ratio Initially

According to the direct shear test for many times, rely on the density of symmetry ratio. The mixed materials ratio is obtained initially, as given in Table4. The mixed materials are made to simulate each stratum, as given in Table 5.Table 4
Mixed materials finished initially.
Proportion of iron and sand in mixture (%)Quality ratio between iron and sandQuality ratio between gypsum and limeVolume ratio between water and gypsum851.5 : 3.51 : 12 : 8Table 5
Simulation materials of each stratigraphic.
StratumThickness of prototype (m)Thickness of the model (cm)Mixed materialsFill the soil0.644Natural sandsSilty clay1.9212Iron ore powder, natural sands, gypsum, and limeSilt soil17.6110Iron ore powder, natural sands, gypsum, and limeSilty clay6.2439Iron ore powder, natural sands, gypsum, and limeFrom what has been researched above, it just studies about the mixed materials ratio for silty clay and silt soil.


## 3.1.1. Prototype Stratum Parameters

In the model test, selecting the soil of Shanghai is considered as the prototype stratum. The information of prototype stratum in Shanghai is given in Table3. According to formula (13), the relationship between compression modulus and elastic modulus is(13)E=ES1−2μ21−μ,where E is the elastic modulus of soil, kPa; ES is the compression modulus of soil, kPa; μ is Poisson’s ratio of soil.Table 3
The physical quantity of prototype stratum in Shanghai.
StratumDensity (kN/m3)Cohesion (kPa)Friction angle (°)Elastic modulus (MPa)Poisson’s ratioThickness of stratum (m)Fill the soil18.5015110.330.64Silty clay19.22017380.361.92Silt soil17.51311100.3117.6Silty clay18.42714420.346.24Based on the exiting conclusion of research, finally, it selected mixed materials to research stratum in the model test, including iron ore powder, natural sands, gypsum, and lime. The iron ore powder and natural sands are aggregate, which have a great density. The gypsum and lime is the adhesive, which has better sticky property and great tension. In addition, the mixed materials are no harm for beings, easily available, low cost, and so on.


## 3.1.2. Determination of Mixed Materials Ratio Initially

According to the direct shear test for many times, rely on the density of symmetry ratio. The mixed materials ratio is obtained initially, as given in Table4. The mixed materials are made to simulate each stratum, as given in Table 5.Table 4
Mixed materials finished initially.
Proportion of iron and sand in mixture (%)Quality ratio between iron and sandQuality ratio between gypsum and limeVolume ratio between water and gypsum851.5 : 3.51 : 12 : 8Table 5
Simulation materials of each stratigraphic.
StratumThickness of prototype (m)Thickness of the model (cm)Mixed materialsFill the soil0.644Natural sandsSilty clay1.9212Iron ore powder, natural sands, gypsum, and limeSilt soil17.6110Iron ore powder, natural sands, gypsum, and limeSilty clay6.2439Iron ore powder, natural sands, gypsum, and limeFrom what has been researched above, it just studies about the mixed materials ratio for silty clay and silt soil.


## 3.2. Determination of Mechanics Parameters for Stratum Symmetry Materials

### 3.2.1. Design of the Orthogonal Test

The orthogonal test is used to research the proportion of mixed materials, design three factors and three levels, a total of nine tests, according to the purpose of the test, considering the density, cohesion, friction angle, elastic modulus, and Poisson’s ratio as the control index, as given in Table6.Table 6
Value of control index.
FactorsProportion of iron and sand in mixture (%)Quality ratio between iron and sandQuality ratio between gypsum and limeLevel 1802 : 32 : 1Level 2851.5 : 3.51 : 1Level 3901 : 41 : 2


### 3.2.2. Parameters of the Orthogonal Test

In order to get the five parameters, that is, density, cohesion, friction angle, elastic modulus, and Poisson’s ratio, the research adopts, respectively, the density test, the direct shear test, elastic modulus test, and Poisson’s ratio test.(1)
Density testIn the density test, the formula of density is given in the following equation, and the instruments are given in Table7.(14)ρ=m1−m2V2,ρ is the density of soil, g/cm3; m1 is the total quality of soil and ring knife, g; m2 is the quality of ring knife, g; V2 is the volume of ring knife, cm3.(2)
Direct shear testThe direct shear test is a common method to measure the shear strength of soil. There is about four times to measure the shear strength in one direct shear test, under different vertical pressures, measuring the shear stress when soil is destroyed. The formula is given as follows:(15)τf=c+σtanϕ,τf is the shear strength of soil, kPa; c is the cohesion of soil, kPa; ϕ is the friction angle of soil; σ is the vertical stress, kPa.(3)
Poisson’s ratio and the elastic modulus testThe value of Poisson’s ratio is measured by two steps:(a)
The lateral pressure coefficientK0 of soil samples is obtained by the static pressure coefficient test(b)
Getting the value of Poisson’s ratio according to the generalized Hooker’s lawTable 7
Instrument of the density test.
NameRing knifeBalanceVernier caliperCompaction meterParametersInternal diameter, 6–8 cm; height 2 cmAccuracy, 0.1 gAccuracy, 0.02 mmDiameter,100 mm; height, 127 mm; volume, 997 cm3The elastic modulus is measured from the lateral compression test of similar materials, as shown in Figure1, and the formula is derived as follows:Figure 1
Stress condition of the confined compress test.According to generalized Hooke’s law,(16)εx+εy=0,(17)εx=σxE−μσy+σzE,(18)εy=σyE−μσx+σzE.Substituting formulas (17) and (18) into formula (16), the following equation is obtained:(19)K0=σxσz=μ1−μ,where μ is Poisson’s ratio; K0 is the side pressure coefficient; E is the elastic modulus, kPa.According to generalized Hooke’s law, the strain ofZ axis is given in the following formula:(20)εz=σzE−μσ+xσyE.Substitutingσz=p, σx=σy=μ/1−μp into formula (20), the following equation is obtained:(21)E=PεZ1−2μ21−μ,(22)εZ=ΔHH=e0−e11+e0.Compression coefficient under confinement conditions is(23)a=ΔeΔσ=e0−e1P.Finally, formula (21) is given as follows:(24)E=1+e0a1−2μ21−μ.Compression modulus under confinement conditions is(25)ES=1+e0a,E=ES1−2μ21−μ,where ES is the compression modulus under confinement conditions, kPa; e0 is the porosity ratio; a is the compression coefficient under confinement conditions, kPa−1; Δe is the change of the amount porosity ratio; Δσ is the change of the amount vertical stress, kPa.


## 3.2.1. Design of the Orthogonal Test

The orthogonal test is used to research the proportion of mixed materials, design three factors and three levels, a total of nine tests, according to the purpose of the test, considering the density, cohesion, friction angle, elastic modulus, and Poisson’s ratio as the control index, as given in Table6.Table 6
Value of control index.
FactorsProportion of iron and sand in mixture (%)Quality ratio between iron and sandQuality ratio between gypsum and limeLevel 1802 : 32 : 1Level 2851.5 : 3.51 : 1Level 3901 : 41 : 2


## 3.2.2. Parameters of the Orthogonal Test

In order to get the five parameters, that is, density, cohesion, friction angle, elastic modulus, and Poisson’s ratio, the research adopts, respectively, the density test, the direct shear test, elastic modulus test, and Poisson’s ratio test.(1)
Density testIn the density test, the formula of density is given in the following equation, and the instruments are given in Table7.(14)ρ=m1−m2V2,ρ is the density of soil, g/cm3; m1 is the total quality of soil and ring knife, g; m2 is the quality of ring knife, g; V2 is the volume of ring knife, cm3.(2)
Direct shear testThe direct shear test is a common method to measure the shear strength of soil. There is about four times to measure the shear strength in one direct shear test, under different vertical pressures, measuring the shear stress when soil is destroyed. The formula is given as follows:(15)τf=c+σtanϕ,τf is the shear strength of soil, kPa; c is the cohesion of soil, kPa; ϕ is the friction angle of soil; σ is the vertical stress, kPa.(3)
Poisson’s ratio and the elastic modulus testThe value of Poisson’s ratio is measured by two steps:(a)
The lateral pressure coefficientK0 of soil samples is obtained by the static pressure coefficient test(b)
Getting the value of Poisson’s ratio according to the generalized Hooker’s lawTable 7
Instrument of the density test.
NameRing knifeBalanceVernier caliperCompaction meterParametersInternal diameter, 6–8 cm; height 2 cmAccuracy, 0.1 gAccuracy, 0.02 mmDiameter,100 mm; height, 127 mm; volume, 997 cm3The elastic modulus is measured from the lateral compression test of similar materials, as shown in Figure1, and the formula is derived as follows:Figure 1
Stress condition of the confined compress test.According to generalized Hooke’s law,(16)εx+εy=0,(17)εx=σxE−μσy+σzE,(18)εy=σyE−μσx+σzE.Substituting formulas (17) and (18) into formula (16), the following equation is obtained:(19)K0=σxσz=μ1−μ,where μ is Poisson’s ratio; K0 is the side pressure coefficient; E is the elastic modulus, kPa.According to generalized Hooke’s law, the strain ofZ axis is given in the following formula:(20)εz=σzE−μσ+xσyE.Substitutingσz=p, σx=σy=μ/1−μp into formula (20), the following equation is obtained:(21)E=PεZ1−2μ21−μ,(22)εZ=ΔHH=e0−e11+e0.Compression coefficient under confinement conditions is(23)a=ΔeΔσ=e0−e1P.Finally, formula (21) is given as follows:(24)E=1+e0a1−2μ21−μ.Compression modulus under confinement conditions is(25)ES=1+e0a,E=ES1−2μ21−μ,where ES is the compression modulus under confinement conditions, kPa; e0 is the porosity ratio; a is the compression coefficient under confinement conditions, kPa−1; Δe is the change of the amount porosity ratio; Δσ is the change of the amount vertical stress, kPa.


## 4. Results and Analysis

According to design and the performed orthogonal test, the results of the orthogonal test obtained are given in Table8.Table 8
Orthogonal test conclusion.
Serial numberProportion of iron and sand in mixture (%)Quality ratio between iron and sandQuality ratio between gypsum and limeElastic modulus (MPa)Cohesion (kPa)Friction angle (°)Density (g/cm3)Poisson’s ratio1802 : 32 : 19.7616292.240.302801.5 : 3.51 : 117.952.831.282.160.353801 : 41 : 223.7816.0528.802.110.374852 : 31 : 119.9429.6926.602.280.345851.5 : 3.51 : 221.5622.8727.192.170.366851 : 42 : 18.4327.2925.112.050.317902 : 31 : 216.3720.0627.912.250.358901.5 : 3.52 : 12.4824.4827.582.150.309901 : 41 : 113.6810.4328.482.090.33Because the value of density almost has no change, as well as Poisson’s ratio, they are out of consideration in the following test.First of all, to produce the mixed materials, select the cohesive force, the friction angle, and the elastic modulus as the control factor.


### 4.1. Analysis of Cohesive Force as the Control Factor

According to the results of the orthogonal test for nine group data, calculate the relative error of nine group cohesive data. The smaller the value of relative error, the more accurate the results, as given in Table9.Table 9
Cohesive force of relative error analysis.
Serial numberProportion of iron and sand in mixture (%)Quality ratio between iron and sandQuality ratio between gypsum and limeRelative error1802 : 32 : 1152801.5 : 3.51 : 11.83801 : 41 : 215.054852 : 31 : 128.695851.5 : 3.51 : 221.876851 : 42 : 126.297902 : 31 : 219.068901.5 : 3.52 : 123.489901 : 41 : 19.43Ι10.6220.9221.59P = 160.68ΙΙ25.6215.7213.31δ=P/9 = 17.85ΙΙΙ17.3216.9218.66P, the sum of the data for every factor; δ, the average of the data for every factor; Ι, the average of relative error for factor 1; ΙΙ, the average of relative error for factor 2; III, the average of relative error for factor 3.According to the orthogonal test and relative error of cohesive, the dispersion of factors forA, B, and C is(26)A:ΙA−δ2+ΙΙA−δ2+ΙΙΙA−δ2=112.92,B:ΙB−δ2+ΙΙB−δ2+ΙΙΙB−δ2=14.82,C:ΙC−δ2+ΙΙC−δ2+ΙΙΙC−δ2=35.25.From the test results analysis and Figure2, we can see that(a)
The relationship between theA, B, and C is A>C>B(b)
The factor 2 point is inflection point, and the line changes suddenly when through the factor 2 point.(c)
A1,B2,C2 is the closest value, respectively, in each factor compared with the prototype cohesion valueFigure 2
Cohesive force of trend diagram.


### 4.2. Analysis of Friction Angle as the Control Factor

According to the results of the orthogonal test for nine group data, calculate the relative error of nine group friction angle data. The smaller the value of relative error, the more accurate the results, as given in Table10.Table 10
Friction angle of relative error analysis.
Serial numberProportion of iron and sand in mixture (%)Quality ratio between iron and sandQuality ratio between gypsum and limeRelative error1802 : 32 : 11.642801.5 : 3.51 : 11.843801 : 41 : 21.624852 : 31 : 11.425851.5 : 3.51 : 21.476851 : 42 : 11.287902 : 31 : 21.548901.5 : 3.52 : 11.519901 : 41 : 11.59Ι1.71.531.48P = 13.92ΙΙ1.391.611.62δ=P/9 = 1.55ΙΙΙ1.551.501.54P, the sum of the data for every factor; δ, the average of the data for every factor; Ι, the average of relative error for factor 1; ΙΙ, the average of relative error for factor 2; ΙΙΙ, the average of relative error for factor 3.According to the orthogonal test and relative error of friction angle, the dispersion of factors forA, B, and C is(27)A:ΙA−δ2+ΙΙA−δ2+ΙΙΙA−δ2=0.0481,B:ΙB−δ2+ΙΙB−δ2+ΙΙΙB−δ2=0.0065,C:ΙC−δ2+ΙΙC−δ2+ΙΙΙC−δ2=0.0099.From the test results analysis and Figure3, we can see that(a)
The relationship between theA, B, and C is A>B>C(b)
The factor 2 point is inflection point, and the line changes suddenly when through the factor 2 point.(c)
A2,B3,C1 is the closest value, respectively, in each factor for the prototype friction angle valueFigure 3
Friction angle of trend diagram.


### 4.3. Analysis of Elastic Modulus as the Control Factor

According to the results of the orthogonal test for nine group data, calculate the relative error of nine group elastic modulus data. The smaller the value of relative error, the more accurate the results, as given in Table11.Table 11
Elastic modulus of relative error analysis.
Serial numberProportion of iron and sand in mixture (%)Quality ratio between iron and sandQuality ratio between gypsum and limeRelative error1802 : 32 : 112.942801.5 : 3.51 : 124.643801 : 41 : 232.974852 : 31 : 127.495851.5 : 3.51 : 229.86851 : 42 : 111.047902 : 31 : 222.398901.5 : 3.52 : 12.549901 : 41 : 118.54Ι23.5220.948.84P = 182.36ΙΙ22.7818.9923.56δ=P/9 = 20.26ΙΙΙ14.4920.8528.39P, the sum of the data for every factor; δ, the average of the data for every factor; Ι, the average of relative error for factor 1; ΙΙ, the average of relative error for factor 2; ΙΙΙ, the average of relative error for factor 3.According to the orthogonal test and relative error of elastic modulus, the dispersion of factors forA, B, and C is(28)A:ΙA−δ2+ΙΙA−δ2+ΙΙΙA−δ2=50.27,B:ΙB−δ2+ΙΙB−δ2+ΙΙΙB−δ2=9.14,C:ΙC−δ2+ΙΙC−δ2+ΙΙΙC−δ2=207.40.From the test results analysis and Figure4, we can see that(a)
The relationship between theA, B, and C is C>A>B(b)
The factor 2 point is inflection point, and the line changes suddenly when through the factor 2 point.(c)
A3,B2,C1 is the closest value, respectively, in each factor for the prototype elastic modulus valueFigure 4
Elastic modulus of trend diagram.As shown in Figures5 and 6, according to the standard, the samples are damaged when the displacement of samples has no change obviously in the direct shear test. Finally, the proportion of mixed materials is shown in Figure 7. Based on the above three control factors, use the direct shear test to get the proportion of mixed materials, as given in Table 12.Figure 5
Action cycle of direct shear apparatus.Figure 6
Failure model of test piece.Figure 7
Proportion of mixed materials.
(a)(b)Table 12
Mixed material proportion table.
Mixed materialsProportion of iron and sand in mixture (%)Quality ratio between iron and sandQuality ratio between gypsum and limeSilt soil802 : 32 : 1Silty clay901.5 : 3.51 : 2Use the parameters of mixed materials to compare with the parameters of the prototype and model. It is proved that the proportion of mixed materials is reasonable and meets the requirements of symmetry ratio, as given in Tables13 and 14.Table 13
Comparison of physical mechanical parameters for silty clay.
MaterialsElastic modulus (MPa)Density (kN/m3)Cohesion (kPa)Friction angle (°)Poisson’s ratioPrototype30.819.829150.36Model1.9219.81.81150.36Test2.0821.031.9820.360.33Table 14
Comparison of physical mechanical parameters for silt soil.
MaterialsElastic modulus (MPa)Density (kN/m3)Cohesion (kPa)Friction angle (°)Poisson’s ratioPrototype1017.513110.31Model0.717.51110.31Test0.9518.72.1112.210.34


## 4.1. Analysis of Cohesive Force as the Control Factor

According to the results of the orthogonal test for nine group data, calculate the relative error of nine group cohesive data. The smaller the value of relative error, the more accurate the results, as given in Table9.Table 9
Cohesive force of relative error analysis.
Serial numberProportion of iron and sand in mixture (%)Quality ratio between iron and sandQuality ratio between gypsum and limeRelative error1802 : 32 : 1152801.5 : 3.51 : 11.83801 : 41 : 215.054852 : 31 : 128.695851.5 : 3.51 : 221.876851 : 42 : 126.297902 : 31 : 219.068901.5 : 3.52 : 123.489901 : 41 : 19.43Ι10.6220.9221.59P = 160.68ΙΙ25.6215.7213.31δ=P/9 = 17.85ΙΙΙ17.3216.9218.66P, the sum of the data for every factor; δ, the average of the data for every factor; Ι, the average of relative error for factor 1; ΙΙ, the average of relative error for factor 2; III, the average of relative error for factor 3.According to the orthogonal test and relative error of cohesive, the dispersion of factors forA, B, and C is(26)A:ΙA−δ2+ΙΙA−δ2+ΙΙΙA−δ2=112.92,B:ΙB−δ2+ΙΙB−δ2+ΙΙΙB−δ2=14.82,C:ΙC−δ2+ΙΙC−δ2+ΙΙΙC−δ2=35.25.From the test results analysis and Figure2, we can see that(a)
The relationship between theA, B, and C is A>C>B(b)
The factor 2 point is inflection point, and the line changes suddenly when through the factor 2 point.(c)
A1,B2,C2 is the closest value, respectively, in each factor compared with the prototype cohesion valueFigure 2
Cohesive force of trend diagram.


## 4.2. Analysis of Friction Angle as the Control Factor

According to the results of the orthogonal test for nine group data, calculate the relative error of nine group friction angle data. The smaller the value of relative error, the more accurate the results, as given in Table10.Table 10
Friction angle of relative error analysis.
Serial numberProportion of iron and sand in mixture (%)Quality ratio between iron and sandQuality ratio between gypsum and limeRelative error1802 : 32 : 11.642801.5 : 3.51 : 11.843801 : 41 : 21.624852 : 31 : 11.425851.5 : 3.51 : 21.476851 : 42 : 11.287902 : 31 : 21.548901.5 : 3.52 : 11.519901 : 41 : 11.59Ι1.71.531.48P = 13.92ΙΙ1.391.611.62δ=P/9 = 1.55ΙΙΙ1.551.501.54P, the sum of the data for every factor; δ, the average of the data for every factor; Ι, the average of relative error for factor 1; ΙΙ, the average of relative error for factor 2; ΙΙΙ, the average of relative error for factor 3.According to the orthogonal test and relative error of friction angle, the dispersion of factors forA, B, and C is(27)A:ΙA−δ2+ΙΙA−δ2+ΙΙΙA−δ2=0.0481,B:ΙB−δ2+ΙΙB−δ2+ΙΙΙB−δ2=0.0065,C:ΙC−δ2+ΙΙC−δ2+ΙΙΙC−δ2=0.0099.From the test results analysis and Figure3, we can see that(a)
The relationship between theA, B, and C is A>B>C(b)
The factor 2 point is inflection point, and the line changes suddenly when through the factor 2 point.(c)
A2,B3,C1 is the closest value, respectively, in each factor for the prototype friction angle valueFigure 3
Friction angle of trend diagram.


## 4.3. Analysis of Elastic Modulus as the Control Factor

According to the results of the orthogonal test for nine group data, calculate the relative error of nine group elastic modulus data. The smaller the value of relative error, the more accurate the results, as given in Table11.Table 11
Elastic modulus of relative error analysis.
Serial numberProportion of iron and sand in mixture (%)Quality ratio between iron and sandQuality ratio between gypsum and limeRelative error1802 : 32 : 112.942801.5 : 3.51 : 124.643801 : 41 : 232.974852 : 31 : 127.495851.5 : 3.51 : 229.86851 : 42 : 111.047902 : 31 : 222.398901.5 : 3.52 : 12.549901 : 41 : 118.54Ι23.5220.948.84P = 182.36ΙΙ22.7818.9923.56δ=P/9 = 20.26ΙΙΙ14.4920.8528.39P, the sum of the data for every factor; δ, the average of the data for every factor; Ι, the average of relative error for factor 1; ΙΙ, the average of relative error for factor 2; ΙΙΙ, the average of relative error for factor 3.According to the orthogonal test and relative error of elastic modulus, the dispersion of factors forA, B, and C is(28)A:ΙA−δ2+ΙΙA−δ2+ΙΙΙA−δ2=50.27,B:ΙB−δ2+ΙΙB−δ2+ΙΙΙB−δ2=9.14,C:ΙC−δ2+ΙΙC−δ2+ΙΙΙC−δ2=207.40.From the test results analysis and Figure4, we can see that(a)
The relationship between theA, B, and C is C>A>B(b)
The factor 2 point is inflection point, and the line changes suddenly when through the factor 2 point.(c)
A3,B2,C1 is the closest value, respectively, in each factor for the prototype elastic modulus valueFigure 4
Elastic modulus of trend diagram.As shown in Figures5 and 6, according to the standard, the samples are damaged when the displacement of samples has no change obviously in the direct shear test. Finally, the proportion of mixed materials is shown in Figure 7. Based on the above three control factors, use the direct shear test to get the proportion of mixed materials, as given in Table 12.Figure 5
Action cycle of direct shear apparatus.Figure 6
Failure model of test piece.Figure 7
Proportion of mixed materials.
(a)(b)Table 12
Mixed material proportion table.
Mixed materialsProportion of iron and sand in mixture (%)Quality ratio between iron and sandQuality ratio between gypsum and limeSilt soil802 : 32 : 1Silty clay901.5 : 3.51 : 2Use the parameters of mixed materials to compare with the parameters of the prototype and model. It is proved that the proportion of mixed materials is reasonable and meets the requirements of symmetry ratio, as given in Tables13 and 14.Table 13
Comparison of physical mechanical parameters for silty clay.
MaterialsElastic modulus (MPa)Density (kN/m3)Cohesion (kPa)Friction angle (°)Poisson’s ratioPrototype30.819.829150.36Model1.9219.81.81150.36Test2.0821.031.9820.360.33Table 14
Comparison of physical mechanical parameters for silt soil.
MaterialsElastic modulus (MPa)Density (kN/m3)Cohesion (kPa)Friction angle (°)Poisson’s ratioPrototype1017.513110.31Model0.717.51110.31Test0.9518.72.1112.210.34


## 5. Conclusion

(1)
According to the symmetry theory, establish the equilibrium equation, geometric equation, and physical equation for the prototype and model. The symmetry ration of mixed materials is 16 : 1.(2)
Select the iron ore powder, natural sands, gypsum, and lime to be the mixed materials for model stratum. Use the orthogonal tests to get the proportion of compositions in mixed materials and analyze the results by the deviation.(3)
The proportion of compositions in mixed materials of silt soil is that the proportion of iron and sand in mixture is 80%, the quality ratio between iron and sand is 2 : 3, and the quality ratio between gypsum and lime is 2 : 1.The proportion of compositions in mixed materials of silty clay is that the proportion of iron and sand in mixture is 90%, the quality ratio between iron and sand is 1.5 : 3.5, and the quality ratio between gypsum and lime is 1 : 2.It is proved that the proportion of compositions in mixed materials is reasonable and meets the requirements of symmetry ratio compared with the parameters of the prototype and model.



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