Active earth pressure against flexible retaining wall for finite soils under the drum deformation mode

A reasonable method is proposed to calculate the active earth pressure of finite soils based on the drum deformation mode of the flexible retaining wall close to the basement’s outer wall. The flexible retaining wall with cohesionless sand is studied, and the ultimate failure angle of finite soils close to the basement’s outer wall is obtained using the Coulomb theory. Soil arch theory is led to get the earth pressure coefficient in the subarea using the trace line of minor principal stress of circular arc after stress deflection. The soil layers at the top and bottom part of the retaining wall are restrained when the drum deformation occurs, and the soil layers are in a non-limit state. The linear relationship between the wall movement’s magnitude and the mobilization of the internal friction angle and the wall friction anger is presented. The level layer analysis method is modified to propose the resultant force of active earth pressure, the action point’s height, and the pressure distribution. Model tests are carried out to emulate the process of drum deformation and soil rupture with limited width. Through image analysis, it is found that the failure angle of soil within the limited width is larger than that of infinite soil. With the increase of the aspect ratio, the failure angle gradually reduces and tends to be constant. Compared with the test results, it is shown that the horizontal earth pressure reduces with the reduction of the aspect ratio within critical width, and the resultant force decreases with the increase of the limit state region under the same ratio. The middle part of the distribution curve is concave. The active earth pressure strength decreases less than Coulomb’s value, the upper and lower soil layers are in the non-limit state, and the active earth pressure strength is more than Coulomb’s value.


Analysis model of flexible retaining wall
The retaining wall is close to the outer basement wall or vertical rock slope, and the height of the retaining wall is H, as shown in Fig. 1. Cohesionless sand is filled behind the retaining wall, and the narrow width is l = n•H (n is the ratio of width to height).
The middle abdomen of the flexible retaining wall protrudes into the excavation under the top strut's support and the constraint of the embedded end at the bottom, forming a drum deformation. The only rotation occurs at the top and bottom of the retaining wall, and its horizontal displacement is assumed to be zero. It is assumed that the midpoint H/2 is the place of maximum deformation and horizontal displacement. Because of the limited width of retaining sand, the slip plane is cut off by the outer wall or rock slope and cannot fully develop to the sand top surface. Thus, the height of the wall is divided into H 1 and H 2 . The sliding surface is assumed to be a plane, passing through the bottom of the wall and forming an angle β with the level plane.
Due to the different widths of the limited soil, the intersection point c of slip surface and vertical external wall or rock slope may be higher or lower than the maximum horizontal displacement of the midpoint, including two cases, as shown in Fig. 1. When H 1 ≤ H/2, the soil mass is divided into upper and lower areas with ce as the boundary. The area above the ce boundary is zone I, and the area below the ce boundary is divided into zones II, III, and IV from top to bottom. The up thin layer at the maximum horizontal displacement is the intermediate www.nature.com/scientificreports/ transition zone III (Fig. 1a). When the width is minimal, H 1 > H/2, the area below the ce boundary is zone IV, and the area above the ce boundary is divided into zones I, II, and III from top to bottom. The up thin layer at the maximum horizontal displacement is the intermediate transition zone II (Fig. 1b). Based on the wall friction and the existence of force N, for simplifying the calculation, it is approximately considered that the earth pressure distribution on the wall meets the triangular distribution along with the height. Therefore, it is the same at the same depth 16,31,32 . By introducing the parameter m, then: E a is the resultant force of active earth pressure acting below the normal, and its direction is δ angle from the normal of the back of the wall.
Based on the Coulomb method, the vertical and horizontal equilibrium function on the sliding surface is derived.
The extreme value of E a can be solved ( dE a /dβ = 0 ) to obtain the value of β of the most dangerous sliding surface as the soil enters an active limit state. The value of the extreme thrust E a can be obtained by using Eq. (2) 16 .
(2) E a =  The upper zone I is located between the backs of the retaining wall and the outer wall. In the process of drum deformation and ground subsidence, because of friction between two vertical parallel walls, the stress deflection occurs due to the soil arch, and the horizontal stress on the retaining wall is no longer minor principal stress. Therefore, the horizontal layer unit at depth z (in zone I) is taken for analysis. Each point's minor principal stress trajectories are connected to form a continuous arch curve, as shown in Fig. 2. Although the stress and boundary conditions of the retaining soil in the lower zones II, III, and IV are different from those in the upper zone I, the vertical and lateral deformation are also limited by frictions (the interface friction between the retaining wall and the soil and the soil friction on the failure plane). Therefore, the direction of the principal stress deflects, and its magnitude remains unchanged along the arch. Taking the layer unit at depth z (in zones II, III, and IV) for analysis shown in Fig. 3, each point's minor principal stress trajectories on the horizontal unit form a half arch. Moreover, the horizontal shearing stress exists on the level unit's surface objectively, and the shearing stress of each point is not equal because of the unequal deflection angle of each point.
The circular arch is generally employed for analysis by many scholars 16,26,33,34 . Since the circular arch's calculation results are close to those of other shapes of arch curves, the circular arch stress trajectory is used to establish the calculation model in this paper.   www.nature.com/scientificreports/ After stress deflection occurs, an arched curve with radius r is formed at point f. The center of the circle is point O. The vertical stress distribution on the layer unit at depth z is uneven considering the soil arching effect. Herein, according to the study by Handy 15 and Paik and Salgado 14 , the lateral active earth pressure coefficient K awn is defined as where σ h is the normal earth pressure on the interface between retaining wall and soil at the depth z, σ v is the average erect pressure on the level at the same height.
The stress of point Q on the arch line is expressed as follows.
where σ 1 is the major principal stress and θ is the deflection angle between the major principal stress and the level at point Q. The deflection angles at point f and g are indicated as: Considering the symmetry of soil mass in zone I, half of the circular arch trajectory can be taken for analysis. The horizontal span B z is l/2, the deflection angle at point j is θ j = π/2, and the average vertical stress along the arch in zone I is In the formula, the curve radius of the minor principal stress arch curve is r = B z /(cos θ f − cos θ j ) . The lateral coefficient of active earth pressure K awn1 in zone I can be presented from Eqs. (4) and (6), The mean shearing stress of soil arching line in zone I is The average shearing stress coefficient k is defined as average shearing stress ratio to average vertical stress on the level layer unit, which should be less than tanφ.
Thus, the average shearing stress coefficient k 1 of soil in zone I can be given To the lower soil in zones II, III and IV, the average vertical stress on the track line of minor principal stress arching can be expressed as where, the radius r = B z /(cos θ f − cos θ g ) . Similarly, the lateral coefficient of active earth pressure K awn2 = K awn3 = K awn4 and the coefficient of average shearing stress k 2 = k 3 = k 4 can be obtained in zones II, III and IV www.nature.com/scientificreports/ Suppose the ultimate rupture angle is β = π/4 + φ/2, K awn1 = K awn2 , k 1 = k 2 . If β = π/4 + φ/2 and δ = 0, Eqs. (7) and (12) is able to transform into K awn1,2 = tan 2 (π/4 − φ/2), that is Rankine coefficient. Furthermore, taking the situation shown in Fig. 1b as the research object, the coefficient of earth pressure and soil arching are analyzed according to the same method above. Then, we can get the lateral active earth pressure coefficients K awn1 , K awn2 , K awn3 , and average shearing stress coefficients k 1 , k 2 , k 3 in zones I, II, and III.
The coefficient of lateral active earth pressure Kawn4 and average shearing stress k4 in zone IV are obtained.

Parameter value in non-limit state
The horizontal displacement of the retaining wall is s under the drum movement mode, and the magnitude of displacement at the middle point is the largest, which value is s max . Assuming that the horizontal displacement required for the soil to enter the full limit state is s a , the area with the displacement s ≥ s a is the entire limit state area, where the internal friction angle of the fill and the external wall friction angle are fully mobilized. In this paper, the region is defined as the intermediate transition region.
The area with horizontal displacement s < s a of retaining wall is a non-limit state area. In the non-limit state, because of the small magnitude of displacement, the soil friction angle φ′ and the wall friction angle δ′ partial mobilize that their values are between the initial state values φ 0 , δ 0 , and the ultimate state values φ, δ, respectively. Considering that the mobilization of φ′ and δ′ are affected by the magnitude of horizontal displacement of the retaining wall, it is assumed that φ′ and δ′ increase linearly with the rise of horizontal displacement, the following expression is given by and where, In which, In general, δ = 2φ/3 and δ 0 = φ/2. The variation of φ′ and δ′ along the retaining wall under drum deformation mode is shown in Fig. 4.
The height of the intermediate transition zone is set as ∆z = x·H, the soil mass within the height ∆z reaches the active limit equilibrium state. x is the ratio of soil layer height entering the limit state along the retaining wall. For the first case of H 1 ≤ H/2, as shown in Fig. 5a, the depth from the top surface of the transition zone to (13) .

Solution for active earth pressure
The shearing stress on the level unit surface usually is not considered under the translation mode (T) because the soil mass moves as a whole, and there is no relative movement between the horizontal soil layers. However, under the drum movement mode, the flexible retaining wall can be regarded as the upper retaining wall rotating about the top (RT) and the lower retaining wall rotating about the bottom of the wall (RB). Each soil layer produces relative motion in the rotation direction with respect to the below layer. Therefore, there must be level shearing stress between the upper and lower soil. The distribution of shearing stress is very complex, and it will affect the moment balance condition. If the moment equilibrium condition is not involved in the derivation, then the specific distribution of shear stress is not concerned 29,35 . Nevertheless, different stress distribution assumptions on the horizontal plane will not affect two static equilibrium of force along with the horizontal and vertical directions.
In this paper, the differential level layer method is introduced. According to the relative movement trend of the horizontal layer unit of the soil wedge behind the wall, the action direction of the friction shearing stress between the level layer units in each zone is determined 16,25,35,36 . Under the condition of satisfying the balance of forces, the active earth pressure differential equation in a non-limit state is established, and then its distribution is discussed.
Zone I. The level layer unit in zone I is shown in Fig. 6, and the static equilibrium equations are established.
In which, the second order differentiation has been omitted, σ v1 is the vertical stress on the layer unit at depth z, and σ h1 is the lateral active earth pressure. τ 1 is the horizontal shearing stress on the surface of the layer unit, assuming it is average distribution, it can be obtained.
where τ w1 is the shearing stress on the interface, and the magnitude of τ w1 is: σ n1 is the horizontal lateral pressure on the outer wall at depth z, and τn1 is the shearing stress. τ n1 can be expressed by dw 1 is the self-weight of the level unit in zone I, and its magnitude is obtained as: In general, there are www.nature.com/scientificreports/ When z = 0, σ v1 = 0 is regarded as the boundary condition of zone I, the first-order linear differential equation (Eq. 29) can be solved as follows.

In which
When z = H 1 , σ v1 = D 1 can be regarded as the boundary condition of equivalent load on the surface of the isolated body in zone II.
Zone II. On the basis of the static equilibrium conditions of horizontal and vertical directions acting on the layer unit (Fig. 7), the equation is established. The second-order differential components are omitted to obtain.
where σ v2 is the mean vertical normal stress on the surface of layer unit, and σ h2 is the lateral active earth pressure. τ 2 is the level shearing stress on the soil layer unit. It is assumed to be uniformly distributed, and its magnitude is expressed as follows: where τ w2 refers to the shearing stress on the interface and its expression is σ n2 is the normal stress distributed uniformly on the rupture surface. τ n2 is the shearing stress distributed uniformly, the formula is In which dw 2 is the self-weight of level layer element in zone II, and its expression is By synthesizing the above formula, the first order differential equation is obtained, in which,  www.nature.com/scientificreports/ in zone IV. The distribution of σ h3 along the height ∆z of the middle transition zone can be approximately considered as a linear distribution, and its expression is When ∆z → 0, h = H/2 is taken for calculation, and zone III is a thin transition layer. In accordance with its equilibrium conditions, the solution can be obtained.
The mean vertical stress at the bottom of the thin transition level unit is Therefore, when ∆z → 0, the mean pressure stress at the bottom of the thin transition layer is taken as the equivalent load on the insulator's top surface in zone IV, and the formula is as follows.
Zone IV. From the static equilibrium conditions of the unit in horizontal and vertical directions (in Fig. 9), we can get: In which, σ v4 is the mean direct stress on the layer unit's surface at depth z, andσ h4 is the horizontal active earth pressure. τ 4 is the level shearing stress on the surface of the layer unit, assuming a mean distribution, and its expression is: where τ w4 is the shearing stress on the contact surface, and its expression is: σ n4 is the normal stress and τ n4 is the friction shearing stress, which is expressed as: In which, dw 4 is the self-weight of level layer unit in zone IV, which can be expressed as: By synthesizing the above formula, the first order differential equation can be get, in which, . .
(53) σ h4 dz − τ 4 dz cot β + dτ 4 (H − z) cot β − σ n4 dz + τ n4 cot βdz = 0, www.nature.com/scientificreports/ By using the boundary condition, i.e. the mean vertical stress σ v4 = D 3 on the top of zone IV, we can solve the differential Eq. (60) and get By substituting the above Eq. (62) with Eq. (55), the lateral active earth pressure in zone IV is generated. For calculating the second case of H 1 > H/2, the same method can be used for analysis. Given the length of the paper, a detailed derivation is omitted. When z = 0, σ v1 = 0 is the boundary condition of zone I, and the vertical stress on the surface of the level unit at depth z in zone I is obtained as: When z = H/2 − ∆z/2, σ v1 = D 1 is regarded as the equivalent load on the top surface of zone II.
Taking into account the overall static balance of the middle transition layer in zone II, we can get in which, Taking σ v2 = D 1 at depth z = H/2 − ∆z/2 as the known loading conditions, σ′ v2 at the bottom of the layer can be obtained. Assuming σ′ v2 = D 2 , it can also be regarded as the equivalent load on the top surface of zone III.
Similarly, the distribution of the earth pressure σ h2 along the height ∆z of the middle transition zone can be considered a linear distribution approximately.
Considering the equivalent load on the top surface of zone III, the vertical stress on the surface of the level unit at depth z is in which, Then we can take σ v3 = D 3 at depth z = H 1 as the known loading conditions of equivalent load on the top surface of zone IV.
The vertical stress in zone IV is

Resultant force and height of action point
When σ h1 , σ h2 , σ h3 , and σ h4 are integrated along with the wall height, the horizontal component E ax of resultant force for earth pressure on the whole retaining wall can be obtained.

The expression of resultant force is given by
The height y of the point of application of the resultant force is as follows In the first case, when ∆z/2 ≥ H/2 − H 1 , the depth of the top surface of the transition zone is h = H 1 , and the calculation height [H 1 , h] of zone II is zero. In the calculation formula of E ax and y, the calculation components in zone II are zero, and the original zone II is canceled. For the second case, when ∆z/2 ≥ H 1 − H/2, the calculated height of zone III is zero, and zone III is actually canceled.
When the aspect ratio n is large, the slip plane slides out from the soil's top surface. In this paper, H 1 = 0, σ h1 in zone I is always zero, and the calculation components of zone I in the calculation formula of E ax and y are all zero. In fact, the original zone I has been canceled, and this problem has developed into the earth pressure problem of infinite soil.

Model test
As shown in Fig. 10, the self-made model is used for the experimental study [37][38][39] . The movable baffle on the left side of the sandbox is polypropylene plate, the fixed baffle on the right side is steel plate, the front side baffle is tempered glass, and the backside baffle is frame steel plate, simulating the flexible retaining wall close to the outer wall of the basement. The real object of the test model device is shown in Fig. 11.
Three motors are installed on the outer side of the sand loading box, as shown in Fig. 12. During the test, the upper and lower motors do not operate (simulating that the top of the retaining wall is supported and the bottom of the retaining wall is embedded). When the middle motor is running, the transmission shaft rotates slowly. The center of the movable baffle gradually moves horizontally outward, forming a certain horizontal displacement and realizing the drum-shaped deformation displacement mode. Taking the soil with limited width in the test box as the PIV analysis area, the digital camera is used to take photos automatically, and the shooting time interval is 1-2 s. Two light sources are placed on both sides of the sandbox to reduce specular reflection. Finally, the images are processed by PIV analysis software.
The earth pressure is measured by five CYY9 micro earth pressure gauges arranged on the movable retaining wall, with a measuring range of 5 kPa and a size of Ф 22 mm × 13 mm. First, the groove is excavated along the vertical centerline of the movable baffle at different depths, and the micro-earth pressure gauges are embedded. The groove's depth is the same as the gauge's thickness to reduce the influence of the gauges protruding from the baffle. Then, a hole is drilled at the groove side to lead out the wire of the earth pressure gauge from the back of the wall, as shown in Fig. 13.
Four groups of soil pressure tests were carried out in this model test, and the specific test parameters are shown in Table 1. The mechanical parameters of sand samples are cohesion c = 0, internal friction angle φ = 36.5°, wall friction angle δ = 24.3°, and unit weight of the tested sand specimen γ = 15 kN/m 3 . (72) (73) E a = E ax cos δ .
(74) As shown in Fig. 14, the sliding failure surface is a plane developing upward from the bottom of the movable baffle. With the increase of the filled sand's width, the intersection of the slip plane and fixed baffle moves upward until the slip plane slides out from the filled sand's top surface. When the ratio of width to height increases from 0.4 to 0.5, the intersection point of the slip plane gradually changes from fixed retaining wall to sand top surface. Therefore, it can be judged that the critical ratio of width height ratio of the finite soil in the model test is between 0.4 and 0.5. When the sand width is greater than the critical width, the retaining soil is considered semi-infinite.
The ultimate fracture surface inclination angle is measured and compared with that calculated on the basis of the generalized Coulomb method, as shown in Table 2. It can be seen that the model test results are close to the theoretical calculation results in the limited width range. With the increase of aspect ratio, the ultimate fracture angle decreases gradually and becomes stable. The experimental analysis shows that the fracture angle β approaches to π/4 + φ/2 = 63.25° under the infinite width (width height ratio n = 0.5, 0.7).
Earth pressure test results. By using the theoretical method in this paper, the distribution of lateral active earth pressure with different ratios (n = 0.2, 0.3, 0.4, 0.5) in the model test is calculated, as shown in Fig. 15.
The theoretical calculation solution shows that the active earth pressure in the figure is nonlinear along with the height of the wall. The lateral earth pressure reduces with the reduction of ratio within the limited width. When the ratio decreases to n = 0.2, the lateral earth pressure decreases significantly. As the limit state region (x from 0.1 to 0.5) increases, the lateral earth pressure near the retaining wall's top and bottom decreases gradually. In contrast, the lateral earth pressure in the middle of the retaining wall does not change significantly.
The resultant force increases with the increase of width to height and decreases with the rise of limit state area under the same ratio. When the width (reaching and exceeding the critical width) and the limit state region increase, the resultant thrust approach to that of Coulomb's result, which is consistent with the previous study, as shown in Fig. 16a. E' ax is obtained by subtracting the prediction and Coulomb's solution. Figure 16b shows the difference at different x. Figure 17 and Table 3 show the comparison of the theoretical calculation of lateral earth pressure distribution with the test results. It can be seen from the figure that the initial horizontal displacement of the retaining wall is smaller under the same ratio, and the limit state region is also tiny, which is confined to the middle part of the wall. The lateral earth pressure on the upper and bottom of the wall is relatively large and highly nonlinear. With the increase of the drum deflection, most central areas enter the active limit state, and the horizontal displacement increases while the earth pressure decreases. With the increase of the limit state region (x from 0.1 to 0.5), the horizontal lateral earth pressure distribution tends to be linearized gradually, and it is close to Coulomb's distribution of finite soil.
The distribution of the earth pressure on the retaining wall calculated theoretically is consistent with that of the model test, and the distribution of the horizontal earth pressure in the middle area is concave. The drum deformation mode of retaining wall under different aspect ratios can be deemed that the upper part rotates about the top of the wall and the lower part rotates about the bottom of the wall. The supporting anchor structure restrains the upper part of the retaining wall, and the bottom is controlled by the fixed end, making the upper www.nature.com/scientificreports/ and bottom soil layers fail to reach the limit state completely. They are still in the active middle state, that is the non-limit active state, and the shearing strength of the soil is insufficient. Therefore, the earth pressure distribution of the upper and the bottom measuring points on the retaining wall are greater than Coulomb's solution to the finite soil, while the middle area is entirely in the active limit state, and the earth pressure distribution of the intermediate measuring points are very close to the Coulomb's solution to the finite soil. www.nature.com/scientificreports/ The internal friction angle is an essential factor affecting the active earth pressure. When n = 0.3 and x = 0.2, the active earth pressure decreases with the increase of internal friction angle, as shown in Fig. 18. The active earth pressure without considering the soil arching is calculated and compared with the theoretical solution in this paper, as shown in Fig. 19. When the soil arching is not considered, the active earth pressure acting on the upper part of the retaining wall is almost the same as considering the soil arching. However, the earth pressure   www.nature.com/scientificreports/ exerted on the middle and lower part of the retaining wall is obviously less than the earth pressure considering the soil arching. In particular, the earth pressure in the transition zone decreases obviously.

Verification by comparison
Taking Lu's model test 4 as an example, dry sand is used in the test, γ = 16 kN/m 3 , φ = 31°, δ = 2φ/3, and the height of the flexible retaining wall with a single anchor is 2 m. Thus, the displacement of the wall under each excavation condition is typical drum deformation. By the theoretical method in this paper, the lateral earth pressure distribution is calculated for soils with infinite width (the ratio of width to height is taken as n = 0.5).
The distribution of earth pressure at different excavation depths obtained by this method, Ying's method 5 , and Lu's test 4 is shown in Fig. 20.
The calculated results of the proposed method are close to the calculated results of Ying 5 and the measured values of the model test by Lu 4 , and the earth pressure distribution law is the same. The results show that the earth pressure on the middle part of the retaining wall decreases with the drum deformation and the horizontal displacement, even less than the Coulomb earth pressure strength. On the other hand, the earth pressure strength on the upper and lower part of the retaining wall is greater than Coulomb's solution due to soils being in a non-limit state.

Conclusion
(1) Based on the characteristics of the drum deformation mode of the flexible retaining wall close to the outer wall of the basement, four zones are divided to establish the mechanical analysis model for the solution of the active earth pressure. The analysis model takes account of the relative movement trend of the fill with the limited width. (2) The active earth pressure coefficient is obtained using the soil arch theory and considering the horizontal shearing stress between differential layers. Considering the drum deformation of the retaining wall and the non-limit state of upper and lower soil layers, the linear relationship between the mobilization of internal friction angle and external friction angle and the magnitude of displacement is presented, and the differential layer analysis method is modified.  www.nature.com/scientificreports/ (3) The model tests are conducted, it is found that the failure angle reduces gradually and becomes stable with the increase of the ratio of width to height. When the ratio rises to infinite soil, the failure angle approaches π/4 + φ/2. (4) The test results show that the active earth pressure of soils with finite width is nonlinear, and the lateral earth pressure reduces with the reduction of the ratio of width to height in the critical width range. Furthermore, as the limit state region increases, the resultant force of earth pressure decreases under the same ratio of width to height. (5) The earth pressure strength on the upper and bottom parts of the retaining wall is greater than the Coulomb solution for finite soil. The earth pressure strength on the middle part of the retaining wall decreases continuously, less than the Coulomb earth pressure strength. As a result, the concave in the middle of the distribution curve is close to a linear line, and the lower part of the distribution curve has higher nonlinearity.  www.nature.com/scientificreports/ www.nature.com/scientificreports/

Data availability
All data, models, and code generated or used during the study appear in the submitted article.