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Communication

New Failure Mechanism for Evaluating the Inclined Failure Load Adjacent to Undrained Soil Slope

1
School of Geomatics and Prospecting Engineering, Jilin Jianzhu University, Changchun 130118, China
2
School of Aerospace, UNNC-NFTZ Blockchain Laboratory, University of Nottingham Ningbo China, 199 Taikang East Road, Ningbo 315100, China
*
Author to whom correspondence should be addressed.
Sustainability 2023, 15(3), 2159; https://doi.org/10.3390/su15032159
Submission received: 6 December 2022 / Revised: 7 January 2023 / Accepted: 9 January 2023 / Published: 23 January 2023
(This article belongs to the Section Environmental Sustainability and Applications)

Abstract

:
The failure mechanism is the key problem for calculating the inclined failure loads adjacent to the undrained soil slope. Based on the undrained slip line field and boundary value problems, a new failure mechanism is proposed to evaluate the inclined failure loads adjacent the undrained slope. The rationality of the proposed method is proved by the definition of the ultimate load and the finite element upper bound analysis (FE-UB). It is not reasonable to assume the failure models since the failure models vary with the inclination load and angle. The proposed method gives a new evaluating indicator, which first calculates the inclined failure load and then determines the failure models.

1. Introduction

Many engineering structures’ (e.g., buildings, bridge abutments, and towers of transmission lines) foundations are required to be placed adjacent to slopes and need to be built quickly. Thus, the bearing capacity of the foundation adjacent to the undrained slope will have a great effect on the stability of the engineering [1]. The numerical simulation of the stability analysis includes the finite element method [2,3], the discrete element model [4], and the fast Langrangian analysis of continua (FLAC) [5]. Compared with vertical loads, inclined loads should take into account the effect of the inclination angle [6,7,8,9,10]. The methods of the ultimate inclined failure load of the foundation on the undrained soil slope include the empirical approach [11,12] and the finite element limit analysis (FELA) [13,14]. Both the foundation bearing capacity and the slope stability are related to the limit state of the system and should be equivalent in terms of failure mechanism [15], e.g., the UB solutions require assumed failure modes, and the optimization search methods are adopted to address the critical failure surface by FELA. The failure modes are diverse and challenging to identify a priori [16]. The determination of the failure model can be regarded as a nonlinear and nonsmooth global optimization problem. It is difficult to optimize the load problem with the existence of multiple local minima [17].
FE combined with strength reduction techniques (SR) does not require prior assumptions about the shape or location of critical failure mechanisms. However, a literature survey reveals that the instability criterion of FE-SR requires human interpretations [18]. A stress field solution proposed by Georgiadis [14] is only applicable to the special case of weightless soil slope. The current method of characteristics proposed by Li et al. [9,19] and Ahmadi et al. [20] can be used to evaluate the ultimate load of the cohesion-frictional slope. The above techniques assume the outermost slip line is the critical slip surface. However, each slip line may be a slip surface based on the Mohr–Coulomb criterion. Only the slip line associated with the minimum safety factor can be considered a critical slip surface. The slope surface under the limit state condition (referred to as critical slope contour) can be calculated by the slip line field [21,22], e.g., Fang et al. [23] proposed a new instability criterion, which only applies to the cohesion-frictional slope when the critical slope contour and the slope surface intersect at the toe of the slope.
This paper proposes a new failure mechanism for evaluating the vertical and horizontal failure loads of a strip footing adjacent to an undrained soil slope. The considered parameters are the slope angle (η), the footing width (B), the slope height (D), the undrained shear strength (c), and the bulk unit weight (γ). The footings are closer to the slope surface, and the slope is closer to unstable. Thus, the distance of the footing from the slope surface is considered to be 0 in this study. The feasibility of the proposed failure mechanism is verified by the definition of ultimate load (i.e., the factor of safety is 1.0 when the ultimate inclined load is imposed at the slope top surface) and comparison with FE-UB.

2. Approach

2.1. Governing Equations

According to the Mohr–Coulomb criterion, the expressions of normal stress and shear stress [21] are:
σ x = σ ( 1 + sin φ · cos 2 θ ) c · cot φ
σ y = σ ( 1 sin φ · cos 2 θ ) c · cot φ
τ x y = τ y x = σ · sin φ · sin 2 θ
where σ is characteristic stress, c is cohesion, φ is the internal friction angle, and θ is the angle between the maximum principal stress σ1 and the x-axis.
The formula of σ is introduced:
σ = S + c · cot φ
where S is σ 1 + σ 3 2 , and σ3 is the minimum principal stress.
For undrained shear strength c > 0 and φ = 0, and substituting Equation (2) into Equation (1):
σ x = S + c · cos 2 θ
σ y = S c · cos 2 θ
τ x y = τ y x = c · sin 2 θ
The differential equations are given as follows:
σ x x + τ x y y = 0
τ y x x + σ y y = γ
where γ represents the unit weight.
Partial differential equations of pure clay slip lines can be obtained by substituting Equation (3) into Equation (4):
S x 2 c ( sin 2 θ θ x cos 2 θ θ y ) = 0
S y + 2 c ( sin 2 θ θ y + cos 2 θ θ x ) = γ
The undrained differential equation of the α and β families can be obtained using the finite difference method as shown in Appendix A:
d y d x = tan ( θ π 4 ) d S 2 c · d θ = γ · d y
d y d x = tan ( θ + π 4 ) d S + 2 c · d θ = γ · d y
The finite difference method (FDM) is used to solve (6a) and (6b) approximately:
y y α x x α = tan ( θ α π 4 ) ( S S α ) 2 c · ( θ θ α ) = γ ( y y α )
y y β x x β = tan ( θ β + π 4 ) ( S S β ) + 2 c · ( θ θ β ) = γ ( y y β )
where Mα(xα, yα, θα, σα) is the point of the α family, Mβ(xβ, yβ, θβ, σβ) is the point of the β family, and (x, y) is the coordinate value.
The point M(x, y, θ, σ) on the slip line is assessed using Formula (7a,b):
x = x α · tan ( θ α π 4 ) x β · tan ( θ β + π 4 ) ( y α y β ) tan ( θ α π 4 ) tan ( θ β + π 4 )
y = ( x x α ) · tan ( θ α π 4 ) + y α y = ( x x β ) · tan ( θ β + π 4 ) + y β
θ = S β S α + 2 c ( θ β + θ α ) + γ ( y α y β ) 4 c
S = S β + S α 2 + c ( θ β θ α ) + γ ( 2 y y α y β 2 )
When σ3 is 0 and σ1 is 2c, S ij = c can be obtained in the critical slope contour as shown in Figure 1. The differential equation of the critical slope contour is d y d x = tan θ [22]. The coordinate points Mij (xij, yij, θij, σij) of the critical slope contour can be assessed by d y d x = tan θ in conjunction with the β family slip line:
x i j = x b · tan θ b x β · tan ( θ β + π 4 ) ( y b y β ) tan θ b tan ( θ β + π 4 )
y i j = ( x x b ) · tan θ b + y b y i j = ( x x β ) · tan ( θ β + π 4 ) + y β
θ i j = S β S b + 2 c ( θ β + θ b ) + γ ( y b y β ) 4 c
S ij = c
where Mb(xb, yb, θb, σb) and Mβ(xβ, yβ, θβ, σβ) are known points of the critical slope contour and the β family slip line.

2.2. Boundary Value Problems

The loads toward the slope surface are more dangerous than those deviating from the slope surface. Thus, the first condition is considered in this study. The computational mode is shown in Figure 2a. The horizontal and vertical failure loads are H and V. The inclination angle is δ. The ultimate failure inclined load qu is H 2 + V 2 . The slope angle is η. The minimum coordinate of the critical slope contour is (xmin, ymin), and the y′ axis is parallel to the ordinate, i.e., y0 = xmin·tanη. There are three boundary value problems, i.e., the Cauchy boundary condition, the degenerative Riemann boundary condition, and the mixed boundary condition, which are shown in Figure 2b–d.
(1)
Cauchy boundary OAB
As shown in Figure 2a, the known points at Cauchy boundary OA are Mα and Mβ, whose abscissa is x = D tan η + Δ x · i , D is the slope height, i is an integer between 0 and N1, N1 is the step number, and Δx is the calculation step, e.g., N1 = 20 and Δx = 0.1 in Figure 1a. The number of nodes is Count 1 = (N2 + 1)·(N2 + 2)/2. The footing width is B = Δx·N1, and the ordinate is D. Undrained stress Mohr’s circle of the Cauchy boundary is shown in Figure 1. The radius of the stress Mohr’s circle of undrained soil is equal to cohesion (i.e., O 1 F = c ), O 1 F sin ( 2 ψ ) = F G = τ 0 , and 2 ψ = 2 θ 1 π , where τ 0 = P · sin δ , and P is the load at the slope top. The intersection angle (θ1) between σ1 and the x-axis can be derived as 2 θ 1 π = arcsin ( τ 0 c ) , i.e., :
θ 1 = π 2 + 1 2 arcsin ( P · sin δ c )
As shown in Figure 1, O 2 E = O 2 O 1 sin δ = O 1 F sin ( 2 ψ δ ) . Thus, the characteristic stress (S1 = O1O2) of OA is derived:
S 1 = c · sin ( 2 θ 1 π δ ) sin δ
(2)
Mixed boundary OCD
The characteristic stress (Sb) of the known point Mb in the slope crest can be obtained by Formula (15):
S b = S 3 = c
According to the characteristic of the β family slip line integral equation (i.e., S + 2 c θ = c o n s t . ), the intersection angle (θ3) can be obtained:
θ b = θ 3 = ( S 1 + 2 c θ 1 c ) / 2 c
The number of nodes is Count 2 = N2 · (N2 + 1)/2.
(3)
Degenerative Riemann boundary OBC
The known point O at the degenerative Riemann boundary is the slope crest, and the characteristic stress is:
S 2 = S 1 + 2 c ( θ 1 θ 2 )
where θ 2 = θ 1 + k · Δ θ N 2 , k is an integer between 0 to N2, Δ θ = θ 3 θ 1 , and N2 is the point partition of the Riemann boundary, e.g., N2 = 5 in Figure 2a. The number of nodes is Count 3 = (N1 + 1)·N2. The total number of nodes is Count = Count 1 + Count 2 + Count 3, e.g., Count = 546 is shown in Figure 2a.
Boundary value problems under vertical only loading (δ = 0) were given by Zhao [24]:
(1)
The Cauchy boundary OAB: θ 1 = π / 2 and S 1 = P 0 c ;
(2)
The mixed boundary OCD: θ b = θ 3 = P 0 2 c + π 2 1 and S b = S 3 = c ;
(3)
The degenerative Riemann boundary OBC: θ 2 = θ 1 + k · Δ θ N 2 and S 2 = P 0 c ( 2 θ 2 π + 1 ) , where Δ θ = θ 3 θ 1 = P 0 2 c 1 .

3. Failure Mechanism

3.1. Evaluation Indicator

A new failure mechanism proposed for the evaluation of H and V is shown in Figure 3a. The results are presented in terms of the load interaction diagrams of the normalized failure load H/H0 and V/V0, where H0 and V0 are the maximum ultimate horizontal or vertical failure loads, respectively. As shown in Figure 2, H0 is c according to the undrained stress circle of the Cauchy boundary, i.e., H0 is the maximum shear stress. V0 can be obtained from the proposed method when δ is zero. The critical slope contour assessed by the undrained slip line field varies with Pi and δi as shown in Section 2.2 Boundary value problems:
P i = H i 2 + V i 2
δ i = arctan ( H i V i )
where i = 1, 2 n.
The proposed failure mechanism for predicting undrained H/H0 and V/V0 of a shallow strip footing placed adjacent to an undrained slope is: (1) the slope is in a stable state and Pi < qu when ymin < y0; (2) the slope is in a limited equilibrium state and Pi = qu when ymin = y0; (3) the slope is in an unstable state and Pi > qu when ymin > y0. Figure 3a gives the case of ymin > 0 or y0 > 0, and the proposed failure mechanism is still valid when ymin ≤ 0 or y0 ≤ 0. Fang et al. [23] proposed an instability criterion for the critical slope contours and the slopes intersecting at the foot of the slope (i.e., ymin = y0 = 0), which can be seen as a special case of the proposed method. In this study, the right-most β slip line (i.e., the curve ABCD in Figure 1) is not the critical slip surface. The calculation flow chart is shown in Figure 3b. See the attachment Appendix B for the Matlab calculation program.

3.2. Results and Discussion

3.2.1. Cases Studied

The case studied by Georgiadis (see Figure 10 in [14]) with strength ratio c/γB = 1.0 (i.e., c = 40 kPa, γ = 20 kN/m3, B = 2 m), η = 45°, and D/B = 0.5 (D = 1 m) was adopted to validate the feasibility of the proposed failure mechanism. The finite element program Plaxis Version 8.6 was used by Georgiadis [14], and the details of the upper bound analysis (UB) were shown in [14]. This case is re-studied in the proposed method.
V0 (i.e., δ = 0 and H/H0 = 0) needs to be solved first. As shown in Figure 4a–c, ymin changes from less than y0 to larger than y0 as V/V0 increases from 0.92 to 1.08, and ymin = y0 when V/V0 = 1.0 (V0 = 131 kPa). Figure 4d shows that the critical slope contour shifts from the interior to the exterior of the slope as V/V0 increases. As shown in Figure 3a, V0 = 131 kPa is the ultimate vertical failure load according to the proposed method. Figure 4e,f show that the factor of safety (FS) calculated using the Bishop method (using SLIDE7.0 with auto refine search) [25] and FLAC (using strength reduction technology) are 1.04 and 1.03, respectively, when V0 = 131 kPa is imposed at the slope top surface, and the failure model is not the foundation width on the top of the slope. Thus, it is unreasonable for the current method of characteristics [9,20] to assume that the outermost slip line is the failure model.
Figure 5a–c show that ymin changes from less than y0 to larger than y0 as V/V0 increases from 0.73 to 0.93 when H/H0 = 0.5 (i.e., H0 = c = 40 kPa, H = 20 kPa) and V/V0 = 0.83 (i.e., V0 = 131 kPa, V = 108.73 Pa) when ymin = y0. The FS calculated using the Bishop method and FLAC are 1.0 and 1.0 when q u = H 2 + V 2 = 20 2 + 108.7 2 = 110.5 kPa and δ = arctan ( H V ) = arctan ( 20 108.7 ) = 10.4 0 are imposed at the slope top surface.
Figure 6a–c show that ymin changes from less than y0 to larger than y0 as V/V0 increases from 0.45 to 0.65 when H/H0 = 0.9 (i.e., H0 = c = 40 kPa, H = 36 kPa) and V/V0 = 0.55 (i.e., V0 = 131 kPa, V = 72 kPa) when ymin = y0. The FS calculated using the Bishop method and FLAC are 1.0 and 1.01, respectively, when q u = H 2 + V 2 = 36 2 + 72 2 = 80.5 kPa and δ = arctan ( H V ) = arctan ( 36 72 ) = 26.5 0 are imposed at the slope top surface.
When H/H0 = 0.5 and 0.9, Figure 5d and Figure 6d show that as V/V0 increases, the critical slop contour shifts from the interior to the exterior of the slope, as is the case for H/H0 = 0 (as shown in Figure 4d). By comparing Figure 4d, Figure 5d, and Figure 6d, the critical slope contour becomes shorter as δ increases from 0° to 26.5°. Thus, there may be three cases of the comparison between the critical slope contour length and the slope surface length, namely, the critical slope contour length is larger than, equal to, or less than the slope surface length when the footings are loaded at the slope top surface. The proposed indicators for calculating the inclined failure load can be applied to the above three cases, e.g., V/V0 = 1.0, 0.83, 0.55 correspond to H/H0 = 0, 0.5, 0.9, which is consistent with FE-UB (see c/γB = 1.0 in Figure 10 by Georgiadis [14]). Figure 5e,f and Figure 6e,f show that the FSs calculated using the Bishop method and FLAC are close to 1.0 when qu = 110.5 kPa (δ = 10.4°) and 80.5 kPa (δ = 26.5°) are imposed at the slope top surface, and the failure models move up from the toe of the slope to the face as δ increases from 10.4° to 26.5°. Thus, it is not reasonable to assume a failure model before calculating the inclined failure load.

3.2.2. Parametric Analyses

When c/γB = 2.5, η = 30°, and D/B = 1.0, the load interaction diagrams of V/V0 and H/H0 assessed by the different methods are shown in Figure 7. The solutions of the proposed method are consistent with the FE-UB [14], and V/V0 increases as H/H0 decreases, namely, the smaller the horizontal load is, the larger the vertical load is. Vesic’s solution underestimates and overestimates V/V0 in the case of lower and higher H/H0. Figure 8 shows that nearly identical load interaction diagrams can be obtained by the proposed method and FE-UB. The variation of c/γB does not affect the load interaction diagram of V/V0 and H/H0. Figure 9 shows the load interaction diagrams of η = 15°, 30°, and 45°. It is evident that the shape of the load interaction diagram depends on the slope angle. The V/V0 reduces with η increasing, and the failure loads and the load interaction curves obtained from the proposed method and FE-UB are in excellent agreement. The maximum error is 2.3%.

4. Conclusions

(1)
The slip line field of the undrained soil slope was derived based on the Mohr–Coulomb criterion and the undrained differential equations. Three inclined load boundary value problems of the undrained slope, i.e., the Cauchy, Riemann, and mixed boundary problems, were derived based on the undrained stress Mohr’s circle. The expressions of the horizontal or vertical failure loads were given according to undrained shear strength.
(2)
A new failure mechanism was proposed to evaluate the ultimate inclination failure load of undrained soil slopes. The critical slope contour shifts from the interior to the exterior of the slope as the inclined failure load increases. The critical slope contour length changes from larger than the slope surface length to less than the slope surface length as the inclination angle increases. The proposed evaluation indicators apply to a wide range of conditions, and it is not needed to assume or search the failure models.
(3)
The rationality of the proposed method is verified by the definition of the ultimate load. The solutions of the proposed method are consistent with those of FE-UB. Under the vertical ultimate load, the current method of characteristics assumes the outermost slip line is the failure model, which is inconsistent with those of the Bishop method and FLAC. The failure model becomes shallow with the inclination load decreasing and the inclination angle increasing. The proposed method calculates the inclination load first and then determines the failure models. The setback distance of footing to the slope surface is not 0 and the physical model test of the proposed method will be carried out in the future.

Author Contributions

Conceptualization, software, validation, formal analysis, H.F.; writing—original draft preparation, review and editing, Y.X. All authors have read and agreed to the published version of the manuscript.

Funding

The research was supported by The Science and Technology Research Project of the Education Department of Jilin province (No. JJKH20220280KJ).

Data Availability Statement

All data included in this study are available upon request by contact with the corresponding author.

Conflicts of Interest

The authors declare no conflict of interest.

Appendix A

The partial differential equations of the undrained method of characteristics are:
S x 2 c ( sin 2 θ θ x cos 2 θ θ y ) = 0
S y + 2 c ( sin 2 θ θ y + cos 2 θ θ x ) = γ
Formula (A1a) multiplied by cos ( θ π 4 ) , and Formula (A1b) multiplied by sin ( θ π 4 ) can obtain:
S x · cos ( θ π 4 ) 2 c ( sin 2 θ θ x cos 2 θ θ y ) · cos ( θ π 4 ) = 0
S y · sin ( θ π 4 ) + 2 c ( sin 2 θ θ y + cos 2 θ θ x ) · sin ( θ π 4 ) γ · sin ( θ π 4 ) = 0
Another form of Equation (A2c) can be obtained by formula adding Equation (A2a) to Equation (A2b):
( S x 2 c · θ x ) cos ( θ π 4 ) + ( S y 2 c · θ y γ ) sin ( θ π 4 ) = 0
The following equations can be obtained by multiplying Equation (A2c) by dx and dy:
( S x 2 c · θ x ) cos ( θ π 4 ) d x + ( S y 2 c · θ y γ ) sin ( θ π 4 ) d x = 0
( S x 2 c · θ x ) cos ( θ π 4 ) d y + ( S y 2 c · θ y γ ) sin ( θ π 4 ) d y = 0
Transforming Equation (A2d,e):
( S x 2 c · θ x ) d x = Z 1 tan ( θ π 4 ) d x = Z 2 d x
( S y 2 c · θ y γ ) d y = Z 2 cot ( θ π 4 ) d y = Z 1 d y
where Z 1 = ( S y 2 c · θ y γ ) , Z 2 = ( S x 2 c · θ x ) .
According to spatial total differential equations: d S = S x d x + S y d y and d θ = θ x d x + θ y d y , and adding Equation (A2f) to Equation (A2g):
d S 2 c · d θ γ d y = Z 1 tan ( θ π 4 ) d x + Z 2 cot ( θ π 4 ) d y
The following equations can be obtained by multiplying Equation (A3a) by sin ( θ π 4 ) and cos ( θ π 4 ) :
( d S 2 c · d θ γ d y ) sin ( θ π 4 ) = Z 1 tan ( θ π 4 ) sin ( θ π 4 ) d x + Z 2 cos ( θ π 4 ) d y
( d S 2 c · d θ γ d y ) cos ( θ π 4 ) = Z 1 sin ( θ π 4 ) d x + Z 2 cot ( θ π 4 ) cos ( θ π 4 ) d y
Substituting Equation (A2f) into Equation (A3b), and substituting Equation (A2g) into Equation (A3c):
( d S 2 c · d θ γ d y ) sin ( θ π 4 ) = Z 2 sin ( θ π 4 ) d x + Z 2 cos ( θ π 4 ) d y
( d S 2 c · d θ γ d y ) cos ( θ π 4 ) = Z 1 sin ( θ π 4 ) d x Z 1 cos ( θ π 4 ) d y
Formula (A3d,e) are transformed into:
Z 2 = d S 2 c · d θ γ d y cos ( θ π 4 ) d y sin ( θ π 4 ) d x sin ( θ π 4 )
Z 1 = d S 2 c · d θ γ d y sin ( θ π 4 ) d x cos ( θ π 4 ) d y cos ( θ π 4 )
The differential equation of two families of slip lines can be obtained according to the concept of characteristic line or slip line:
d y d x = tan ( θ π 4 ) d S 2 c · d θ = γ · d y
d y d x = tan ( θ + π 4 ) d S + 2 c · d θ = γ · d y

Appendix B

MaxPoint = 2000;
MaxValue = 2000;
for i=1:1:MaxPoint
    for j=1:1:MaxPoint
      PointValue{i,j} = [MaxValue,MaxValue,MaxValue,0,0,0];
    end
end
% Enter initial value
Gamma=20;                                                  % unit weight
C0=80;                                                           % cohesion
Alpha0=(45/180*pi);                                    % slope angl
H=4;                                                               % slope height
kh=-0.1
xi=0.0;
kv=xi*(abs(kh))    ;
delta=atan(abs(kh)/(1-kv)) ;
delta0=delta/pi*180;
F=1   ;
C1=C0/F      ;
Alpha1=(Alpha0/pi*180) ;
BuchangX=0.01;                                        % calculation step
N1=100;                                                      % step number
N2=10;                                                        % the point partition of the Riemann boundary
B=BuchangX*N1                                        % foundation width
P1=36                                                          % load
P=P1/(Gamma*B)
X_3=-H/tan(Alpha0);
Y_3=H;
Y_3_0 =H ;
X_3_2_1=0;
Y_3_2_1=0;
Y_3_2_1_0 = 0;
Mu=pi/4;
Point_0_0 = {0,0,0,0,0,0};
 Count1 = (N1+1)*(N1+2)/2;
 Count2 = (N1+1)*N2;
 Count3 = N1*(N1+1)/2 ;
 Count = Count1+Count2+Count3;
  %%Cauchy boundary
     h0=C1;                                     %% ultimate horizontal failure load
     v0=131;                                    %% ultimate vertical failure load when delta=0
     xi1=0.1
     h=xi1*h0;                                 %% the horizontal failure load
     xi2=0.92
     v=xi2*v0;                                   %% the vertical failure load
     delta=atan(h/v)                        %% the inclination angle
     delta0=delta/pi*180
     P1=(h^2+v^2)^(1/2)                  %% ultimate inclined failure load
Theta1 = pi/2+(1/2)*asin(P1*sin(delta)/C1) ;
Sigma1 =(C1*sin(2*Theta1-pi-delta))/sin(delta) ;
        I=0;
        for i=1:1:(N1+1)
            for j=i:-1:1
                   if(j==i)
                    PointValue{i,j} = [X_3_2_1+(N1+1-i)*BuchangX, Y_3_2_1, Y_3_2_1,Theta1,Sigma1,Sigma1];
                    I=I+1;
                    Point{I} = [i,j];
                  else
                    p1=PointValue{i,j+1};
                    p2=PointValue{i-1,j};
                    x1 = p1(1);
                    y1 = p1(2);
                    o1 = p1(4);
                    q1 = p1(5);
                    x2 = p2(1);
                    y2 = p2(2);
                    o2 = p2(4);
                    q2 = p2(5);
                    dd=callfun(x1,y1,o1,q1,x2,y2,o2,q2,Mu,C1,Gamma);
                    PointValue{i,j} = [dd(1), dd(2), dd(3),dd(4),dd(5),dd(6)];
                    I=I+1;
                    Point{I} = [i,j];
               end
            end
       end
      %%Degenerative Riemann boundary
     DetaXita=(Sigma1+2*C1*Theta1-C1)/(2*C1)-Theta1;
        for i=(N1+1+1):1:(N1+1+N2)
            ii = i-(N1+1);
            Theta2=Theta1+ii*DetaXita/N2;
           Sigma2=Sigma1+2*C1*(Theta1-Theta2);
            for j=(1+N1):-1:1
                   if(j==(1+N1))
                      PointValue{i,j} = [0, 0, 0,Theta2,Sigma2,Sigma2];
                      I=I+1;
                      Point{I} = [i,j];
                  else
                      p1=PointValue{i,j+1};
                      p2=PointValue{i-1,j};
                      x1 = p1(1);
                      y1 = p1(2);
                      o1 = p1(4);
                      q1 = p1(5);
                      x2 = p2(1);
                      y2 = p2(2);
                      o2 = p2(4);
                      q2 = p2(5);
                      dd=callfun(x1,y1,o1,q1,x2,y2,o2,q2,Mu,C1,Gamma);
                      PointValue{i,j} = [dd(1), dd(2), dd(3),dd(4),dd(5),dd(6)];
                          I=I+1;
                      Point{I} = [i,j];
                end
            end
        end
 %% Mixed boundary
        Sigma3 = C1;
        for i=(N1+1+N2+1):1:(N1+1+N2+N1)
          for j=(N1+1+N2+N1+1-i):-1:1
              if(j==(N1+1+N2+N1+1-i))
                 p1=PointValue{i-1,j+1};
                 p2=PointValue{i-1,j};
                 x1 = p1(1);
                 y1 = p1(2);
                 o1 = p1(4);
                 q1 = p1(5);
                 x2 = p2(1);
                 y2 = p2(2);
                 o2 = p2(4);
                 q2 = p2(5);
dd=callfan(x1,y1,o1,q1,x2,y2,o2,q2,Mu,C1,Gamma);
PointValue{i,j} = [dd(1), dd(2), dd(3),dd(4),Sigma3,Sigma3];
                 I=I+1 ;
                 Point{I} = [i,j];
            else
                 p1=PointValue{i,j+1};
                 p2=PointValue{i-1,j};
                 x1 = p1(1);
                 y1 = p1(2);
                 o1 = p1(4);
                 q1 = p1(5);
                 x2 = p2(1);
                 y2 = p2(2);
                 o2 = p2(4);
                 q2 = p2(5);
                 dd=callfun(x1,y1,o1,q1,x2,y2,o2,q2,Mu,C1,Gamma);
                 PointValue{i,j} = [dd(1), dd(2), dd(3),dd(4),dd(5),dd(6)];
                 I=I+1;
                 Point{I} = [i,j];
              end
            end
        end
for k=1:1:I
    i=Point{k}(1);
    j=Point{k}(2);
    p=PointValue{i,j};
    p(1) = p(1)-X_3;
    p(2) = Y_3-p(2);
    p(3) = Y_3-p(3);
    PointValue{i,j}=[p(1), p(2), p(2),p(4),p(5),p(6)];
end
Y_3_0 = Y_3 - Y_3_0;
Y_3_2_1_0 = Y_3 - Y_3_2_1_0;
CountAlpha =2*N1+N2+1;
 CountBeta =N1+1;
%%Alpha line
for i=1:1:CountAlpha
    UN_0 = 0;
    for j=1:1:CountBeta
      p=PointValue{i,j};
      if p(1)~=MaxValue || p(2)~=MaxValue
          UN_0 = UN_0+1;
      end
    end
    x_p = zeros(1,UN_0);
    y_p = zeros(1,UN_0);
         UN_0 = 0;
    for j=1:1:CountBeta
      p=PointValue{i,j};
      if p(1)~=MaxValue || p(2)~=MaxValue
          UN_0 = UN_0+1;
          x_p(1,UN_0) = p(1);
          y_p(1,UN_0) = p(2);
      end
    end
    hold on
    plot(x_p,y_p)
end
%%Beta line
for j=1:1:CountBeta
    UN_0 = 0;
    for i=1:1:CountAlpha
      p=PointValue{i,j};
      if p(1)~=MaxValue || p(2)~=MaxValue
         UN_0 = UN_0+1;
      end
    end
    x_p = zeros(1,UN_0);
    y_p = zeros(1,UN_0);
        UN_0 = 0;
     for i=1:1:CountAlpha
      p=PointValue{i,j};
      if p(1)~=MaxValue || p(2)~=MaxValue
          UN_0 = UN_0+1;
          x_p(1,UN_0) = p(1);
          y_p(1,UN_0) = p(2);
      end
     end
    hold on
    plot(x_p,y_p)
end
Y_3_0;
Y_3_2_1_0;
UN_0 = 0;
for j=1:1:CountBeta
     for i=1:1:CountAlpha
      p=PointValue{i,j};
      if  p(2)==Y_3_2_1_0
          UN_0 = UN_0+1;
      end
    end
end
x_p = zeros(1,UN_0);
y_p = zeros(1,UN_0);
UN_0 = 0;
for j=1:1:CountBeta
     for i=1:1:CountAlpha
    p=PointValue{i,j};
      if  p(2)==Y_3_2_1_0
          UN_0 = UN_0+1;
          x_p(1,UN_0) = p(1);
          y_p(1,UN_0) = p(2);
      end
    end
end
hold on
plot(x_p,y_p,’b’)
X_3=0;
Y_3_0 =0;
X_3_2_1=H/tan(Alpha0);
Y_3_2_1_0=H;
x_po = zeros(1,2);
y_po = zeros(1,2);
x_po(1,1)=X_3;
y_po(1,1) = Y_3_0;
x_po(1,2)=X_3_2_1;
y_po(1,2)=Y_3_2_1_0;
hold on
plot(x_po,y_po,’b’)
x_xie = zeros(1,CountBeta);
y_xie = zeros(1,CountBeta);
UN_0 = 0;
for j=1:1:CountBeta
    for i=1:1:CountAlpha
      p=PointValue{i,j};
      if p(1)~=MaxValue || p(2)~=MaxValue
         x_xie(1,j)=p(1);
         y_xie(1,j) = p(2);
      end
    end
end
 hold on
 plot(x_xie,y_xie,’r’)
x_sj = zeros(1,3);
y_sj = zeros(1,3);
x_sj(1,1) = 0;
y_sj(1,1) = 0;
x_sj(1,2) = H/tan(Alpha0);
y_sj(1,2) = 0;
x_sj(1,3) = H/tan(Alpha0);
y_sj(1,3) = H;
hold on
plot(x_sj,y_sj,’b’)
xlabel(’x/m’); ylabel(’y/m’);
axis equal;
xmin=PointValue{2*N1+N2+1,1}(1);
ymin=PointValue{2*N1+N2+1,1}(2)
y0=tan(Alpha0)*(xmin)      %% y0<ymin: unstable; y0=ymin: limit state; y0>ymin: stable
%%callfun function:Eqs.(8-11)
function dd=callfun(x1,y1,o1,p1,x2,y2,o2,p2,u,c,r)
    dd(1)=(x1*tan(o1-u)-x2*tan(o2+u)-(y1-y2))/(tan(o1-u)-tan(o2+u));
    dd(2)=(dd(1)-x1)*tan(o1-u)+y1;
    dd(3)=(dd(1)-x2)*tan(o2+u)+y2;
    dd(4)=(r*(y1-y2)+(p2-p1)+2*c*(o2+o1))/(4*c);
    dd(5)=c*(o2-o1)+(p1+p2)/2+r*(dd(2)-y1/2-y2/2);
    dd(6)=c*(o2-o1)+(p1+p2)/2+r*(dd(2)-y1/2-y2/2);
end
%%callfan function:Eqs.(12-15)
function dd=callfan(x1,y1,o1,p1,x2,y2,o2,p2,u,c,r)
    dd(1)=(x1*tan(o1)-x2*tan(o2+u)-(y1-y2))/(tan(o1)-tan(o2+u));
    dd(2)=(dd(1)-x1)*tan(o1)+y1;
    dd(3)=(dd(1)-x2)*tan(o2+u)+y2;
    dd(4)=(r*(y1-y2)+(p2-p1)+2*c*(o2+o1))/(4*c);
end

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Figure 1. Undrained Mohr’s stress circle.
Figure 1. Undrained Mohr’s stress circle.
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Figure 2. Problem definition: (a) undrained slip line field; (b) Cauchy boundary schematic; (c) degenerative Riemann boundary schematic; (d) mixed boundary schematic.
Figure 2. Problem definition: (a) undrained slip line field; (b) Cauchy boundary schematic; (c) degenerative Riemann boundary schematic; (d) mixed boundary schematic.
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Figure 3. The proposed method: (a) failure mechanism; (b) calculation flow chart.
Figure 3. The proposed method: (a) failure mechanism; (b) calculation flow chart.
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Figure 4. The proposed failure mechanism with H/H0 = 0: (a) V/V0 = 0.92; (b) V/V0 = 1.0 (V0 = 131 kPa); (c) V/V0 = 1.08; (d) comparison of critical slope contours; (e) FS calculated using the Bishop method; (f) FS calculated by FLAC.
Figure 4. The proposed failure mechanism with H/H0 = 0: (a) V/V0 = 0.92; (b) V/V0 = 1.0 (V0 = 131 kPa); (c) V/V0 = 1.08; (d) comparison of critical slope contours; (e) FS calculated using the Bishop method; (f) FS calculated by FLAC.
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Figure 5. The proposed failure mechanism with H/H0 = 0.5: (a) V/V0 = 0.73; (b) V/V0 = 0.83; (c) V/V0 = 0.93; (d) comparison of critical slope contours; (e) FS calculated using the Bishop method; (f) FS calculated by FLAC.
Figure 5. The proposed failure mechanism with H/H0 = 0.5: (a) V/V0 = 0.73; (b) V/V0 = 0.83; (c) V/V0 = 0.93; (d) comparison of critical slope contours; (e) FS calculated using the Bishop method; (f) FS calculated by FLAC.
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Figure 6. The proposed failure mechanism with H/H0 = 0.9: (a) V/V0 = 0.45; (b) V/V0 = 0.55; (c) V/V0 = 0.65; (d) comparison of critical slope contours; (e) FS calculated using the Bishop method; (f) FS calculated by FLAC.
Figure 6. The proposed failure mechanism with H/H0 = 0.9: (a) V/V0 = 0.45; (b) V/V0 = 0.55; (c) V/V0 = 0.65; (d) comparison of critical slope contours; (e) FS calculated using the Bishop method; (f) FS calculated by FLAC.
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Figure 7. Comparison of different methods (c/γB = 2.5, η = 30°).
Figure 7. Comparison of different methods (c/γB = 2.5, η = 30°).
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Figure 8. Variation of c/γB (η = 45°).
Figure 8. Variation of c/γB (η = 45°).
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Figure 9. Variation of η (c/γB = 2.5).
Figure 9. Variation of η (c/γB = 2.5).
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MDPI and ACS Style

Fang, H.; Xu, Y. New Failure Mechanism for Evaluating the Inclined Failure Load Adjacent to Undrained Soil Slope. Sustainability 2023, 15, 2159. https://doi.org/10.3390/su15032159

AMA Style

Fang H, Xu Y. New Failure Mechanism for Evaluating the Inclined Failure Load Adjacent to Undrained Soil Slope. Sustainability. 2023; 15(3):2159. https://doi.org/10.3390/su15032159

Chicago/Turabian Style

Fang, Hongwei, and Yixiang Xu. 2023. "New Failure Mechanism for Evaluating the Inclined Failure Load Adjacent to Undrained Soil Slope" Sustainability 15, no. 3: 2159. https://doi.org/10.3390/su15032159

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