Capturing the worst-case SOF with n and Г 2

The phenomenon of the worst-case SOF was reviewed in Section 2.10. It is now important to ask whether the WPM is capable of capturing this phenomenon. Although the detailed answer to this question is presented in the next chapter, this section explores the question from a theoretical point of view. In the WPM, the difference between the

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PDF of X and X^mob is handled by n and Γ². As an example, consider n = {1, 10, 30} and Γ² = {0.1, 0.5, 1}. Figure 4.4a shows how 9 combinations of (n, Γ²) influences X^mob.

(a)

(b)

Figure 4.4 (a) Effect of n and Γ² on PDF of X^mob (solid line). The dashed line is point X with µ = 50 and σ = 15; (b) the phenomenon of the worst-case SOF

The top-left panel (n = 1, Γ² = 1) in Figure 4.4a represents the homogenous case, i.e., SOF = ∞. In general, as SOF decreases, n may either remain constant or increase;

Γ² may either remain constant or decrease (see next chapter for details). Therefore, at least three different paths can be considered as SOF decreases:

1. A-B-C (top panels in Figure 4.4a): n remains constant (n = 1), while Γ² decreases. In this case, the mean of X^mob is unchanged (dashed line in Figure 4.4b), and there is no phenomenon of the worst-case SOF. As will be shown in the next chapter, the behavior of a friction pile problem follows this path.

2. A-D-F (left panels in Figure 4.4a): Γ² remains constant (Γ² = 1), while n increases. In this case, the mean of X^mob is reducing toward zero (dotted-dashed line in Figure 4.4b). This path does not exhibit a worst-case SOF. Ching et al.

(2014) showed that a soil column problem under pure shear exhibits this behavior, when its horizontal SOF is very large.

3. A-E-G (diagonal panels in Figure 4.4a): n increases, while Γ² decreases. In this case, the mean of X^mob reaches a minimum at a certain SOF (solid line in Figure 4.4b). The phenomenon of the worst-case SOF is pronounced. As will be shown in the next chapter, most of investigated problems behave in this manner, and the WPM is capable of capturing their worst-case SOFs by adjusting n and Γ².

Note that when SOF = ∞ (homogenous soil), the PDF of X^mob is identical to the PDF of point X (top-left panel in Figure 4.4a). Hence, the characteristic value of X can be simply calculated as 5%-fractile of point X. In other words, one may not need to use the WPM if the soil is homogenous. Nevertheless, the WPM still works for homogenous soils by adopting n = 1 and Γ² = 1 (i.e., no weak-zone seeking and no variance reduction).

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4.5 Description of investigated problems

The current study estimates the two parameters (n, Γ²) of the WPM for six geotechnical ultimate limit state problems:

(a) Strip footing subjected to vertical loading. As will be described in Section 4.5.1, response of interest = bearing capacity qu;

(b) Retaining wall. As will be described in Section 4.5.2, response of interest = active lateral force Pa;

(c) Basal heave for excavation in clay. As will be described in Section 4.5.3, response of interest = Factor of safety, FS, for basal heave;

(d) Soil column subjected to axial compression. As will be described in Section 4.5.4, response of interest = unconfined compressive strength Qu;

(e) Laterally loaded pile. As will be Section 4.5.5, response of interest = lateral load resistance Rlat.

(f) Friction pile under compression. As will be described in Section 4.5.6, response of interest = shaft resistance Rs.

For the first five problems, the X^mob samples simulated by the 2D RFEM are employed to calibrate the WPM. To control the discretization error in the RFEM, the findings of the previous chapter is considered. The midpoint method (e.g., Der Kiureghian and Ke 1988) as opposed to local averaging method (e.g., Fenton and Vanmarcke 1990) is adopted to discretize the random field (i.e., the value at the centroid of each element is adopted). Moreover, the ratio of element size to SOF is taken to less than 1/8 for all problems. For the sixth problem (i.e., friction pile), however, a simple model rather than the RFEM is adopted to simulate X^mob samples to calibrate the WPM.

Three ground scenarios are considered:

(a) A clay ground with spatially variable su (input property X = su) modeled by a stationary lognormal random field. This scenario is hereafter called as

“stationary-su”. Table 4.1 lists the parameters for the problems analyzed under this scenario.

(b) A clay ground with spatially variable normalized su/σ'v (input property X = su/σ'v; σ'v is the vertical effective stress) modeled by a stationary lognormal random field. This scenario is hereafter called “stationary-su/σ'v”. This scenario is suitable for normally consolidated clay. The parameters for the problems analyzed under this scenario are listed in Table 4.2.

(c) A sand ground with a spatially variable tangent friction angle (input property X

= tan ϕ) modeled by a stationary lognormal random field. This scenario is hereafter called “stationary-tan ”. The parameters for the problems analyzed under this scenario are listed in Table 4.3.

The friction angle is zero for the two clay scenarios (a) and (b), whereas the cohesion is zero for the sand scenario (c). For scenarios (b) and (c), the shear strength tends to increase with depth because the effective stress increases with depth. The current study only considers a single characteristic value. The problem of addressing multiple characteristic values is treated elsewhere (Ching et al. 2020).

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Table 4.1 Parameters for the stationary-su scenario Soil

Table 4.2 Parameters for the stationary-su/σ'v scenario

Table 4.3 Parameters for the stationary-tan ϕ scenario Soil

79 4.5.1 Strip footing subjected to vertical loading

Figure 4.5 shows the finite element model for a strip footing of width B = 4 m with an embedment depth of Df. The bottom boundary of the model is fixed, and the left and right boundaries are composed of rollers. The finite element mesh consists of eight-node plane strain elements with reduced integration. The element size is 0.05 m × 0.05 m.

Since the smallest SOF considered for the footing problem is 0.1×B = 0.4 m (Table 4.1 to Table 4.3), the ratio of element size to SOF is equal to 0.05/0.4 = 0.125, which satisfies the recommendation obtained in Chapter 3. Obviously, the larger SOFs satisfy the recommendation as well.

Figure 4.5 Strip footing subjected to vertical loading

An elastic perfectly plastic Mohr-Coulomb constitutive model is used. The footing is modeled as a rigid plate with a rough base. The nodes right below the footing are subjected to a monotonically increasing vertical displacement while the horizontal movement of these nodes is restrained. At each displacement increment, the value of total reacting force on the nodes is calculated. Failure is said to have occurred when either (a) the value of the total reacting force at two successive displacement increments

remains unchanged within a prescribed accuracy, which implies that a maximum is reached, or (b) the finite element algorithm is unable to converge within a predefined number of iterations, which implies that no stress distribution can be found that satisfies both the Mohr-Coulomb failure criterion and global equilibrium. The bearing capacity (qu) of the footing is then calculated by dividing the maximum value of the reacting force by B, and is considered to be the response of interest. Note that, for the stationary-su scenario, the footing rests on the ground surface (Df =0). However, for the stationary-su/σ'v and stationary-tan  scenarios, Df = 2 m is adopted, and the 2-m thick soil is modeled by a surcharge pressure in the RFEM. This embedment depth of 2 m ensures that the shear strength is not zero at the footing base, so numerical issues do not occur.

4.5.2 Retaining wall

The backfill for the retaining wall is a rectangular area of size 16 m  8 m, restrained by a wall of height H = 5 m (Figure 4.6). The wall is not explicitly modeled by the RFEM.

Instead, the nodes next to the wall are translated horizontally away from the backfill.

The bottom boundary of the model is fixed. The right boundary is composed of rollers.

The left boundary is unrestrained, with the exception of the nodes adjacent to the wall, which have fixed horizontal displacement and are free to move vertically. The finite element mesh consists of eight-node plane strain elements with reduced integration. The element size is 0.03125 m × 0.03125 m. Since the smallest SOF considered for the retaining wall problem is 0.05×H = 0.25 m (Table 4.1 to Table 4.3), the ratio of element size to SOF is equal to 0.03125/0.25 = 0.125, which satisfies the recommendation obtained in Chapter 3. The active lateral force (Pa) is calculated according to the procedure proposed by Fenton et al. (2005), where the wall nodes are incrementally displaced in the horizontal direction until the total reacting force on the nodes reaches a

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minimum value, and Pa is taken to be this minimum value. Similar to the footing problem, failure is said to have occurred when either (a) the value of the total reacting force flattens out and reaches a minimum, or (b) the finite element algorithm is unable to converge within a predefined number of iterations. The retaining wall problem is modeled by the 2D RFEM for all three ground scenarios.

Figure 4.6 Retaining wall

4.5.3 Basal heave for excavation in clay

Ching et al. (2017a) adopted the 2D RFEM to model the basal heave stability of an excavation in clay. The same RFEM model is adopted here (Figure 4.7). A diaphragm wall with penetration depth Hp = 24 m is used to retain the soil. The bottom boundary of the entire model is fixed, and the left and right boundaries are composed of rollers. The finite element mesh consists of eight-node plane strain elements. The element size is small (0.5 m × 0.5 m) near the excavation region and becomes large (2 m × 2 m) in the remote region. Since the smallest SOF considered for the basal heave problem is 0.21×Hp = 5 m (Table 4.1 to Table 4.3), the ratio of element size to SOF is equal to

0.5/5 = 0.1, which satisfies the recommendation obtained in Chapter 3. There are two analysis steps. The first step is geostatic equilibrium to build up the in-situ geostatic stress field, and the second step is to simulate the excavation sequence to the final excavation stage. The excavation is conducted in seven stages to the total depth He = 20 m. A surcharge pressure of 10 kN/m² is considered. Failure is defined following the criterion proposed by Do et al. (2013), as demonstrated in Figure 4.8. The horizontal axis is the strength reduction factor (SRF) (e.g., Griffiths and Lane 1999). The vertical axis is the basal heave displacement at the final stage of the excavation. Two straight lines are fitted to the initial and end portions of the SRF-heave relationship. Finally, the FS is obtained by finding the intersection of these two lines. The basal heave problem is analyzed for the two clay scenarios.

Figure 4.7 Basal heave for excavation in clay

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Figure 4.8 Definition of FS for the basal heave problem (after Ching et al. 2017a)

4.5.4 Soil column subjected to axial compression

The 2D RFEM model for the soil column under compression is a rectangular area of size 12.8 m  25.6 m (Figure 4.9). The bottom boundary is supported on rollers, and the lower-left-most node is a hinge. The finite element mesh consists of eight-node plane strain elements with reduced integration. The element size is 0.05 m × 0.05 m. Since the smallest SOF considered for the soil column problem is 0.05×W = 0.64 m (Table 4.1 to Table 4.3), the ratio of element size to SOF is equal to 0.05/0.64 = 0.08, which satisfies the recommendation obtained in Chapter 3. The top nodes are subjected to a monotonically increasing vertical displacement. Following the convergence of stress redistribution at each increment, the total reacting force on the top nodes is calculated.

The maximum value of this total reacting force is found, and the unconfined compressive strength (Qu) is obtained by dividing this maximum value by the column width W = 12.8 m. Similar to the footing and retaining wall problem, failure is said to have occurred when either (a) the value of the total reacting force flattens out and reaches a maximum, or (b) the finite element algorithm is unable to converge within a

predefined number of iterations. Scenarios of stationary-su and stationary-tan  are considered. The confining pressure σ3 is set to 0 for stationary-su, whereas σ3 = 10 kN/m² for stationary-tan .

Figure 4.9 Soil column subjected to axial compression

4.5.5 Laterally loaded pile

Figure 4.10 shows the 2D RFEM model for the laterally loaded pile problem. The bottom boundary is fixed, while the left and right boundaries are composed of rollers.

The finite element mesh consists of eight-node plane strain elements with reduced integration (CPE8R). The element size is small (0.125 m × 0.125 m) near the pile region and becomes large (0.5 m × 0.5 m) in the remote region. Since the smallest SOF considered for the laterally loaded pile problem is 0.2×Dlat = 1 m (Table 4.1), the ratio of element size to SOF is equal to 0.125/1 = 0.125, which satisfies the recommendation obtained in Chapter 3. The top of the pile is horizontally moved to the right. In order to

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minimize the effect of boundary conditions, the soil mass on the right side of the pile is extended to 34.5 m. The soil properties are listed in Table 4.1. The pile properties are as follows: unit weight = 24 kN/m³, Young's modulus = 25450 MN/m³, and Poisson’s ratio

= 0.21. The tangential and normal behaviors of the soil-pile interface is considered. The penalty method with friction coefficient = 0.5 is used for the tangential behavior, and the “hard contact” is adopted for the normal behavior. Moreover, the soil and pile are allowed to be separated during analysis (i.e., “allow separation after contact” is toggled on in Abaqus). Failure is said to have occurred when the finite element algorithm is unable to converge. Note that following Poulos and Davis (1980) criteria, the behavior of this pile is categorized as “short” piles (or rigid piles), in which the lateral capacity is fully dependent on the soil strength.

Figure 4.10 Laterally loaded pile

4.5.6 Friction pile under compression

Unlike the previous five problems, the RFEM is not employed for the friction pile problem (Figure 4.11). Instead, simple models are employed. For the two clay scenarios (a) and (b), the shaft resistance (Rs) is modeled by (Naghibi and Fenton 2011):

s s u

R = A    s (4.8)

where As is the area of the pile shaft; α is an empirical adhesion factor (commonly in the range 0.3 to 1, herein taken to be 0.5); s is the arithmetic average of the spatially _u variable su over the pile shaft. For the sand scenario (c), Rs is modeled by (e.g., Dithinde et al. 2011):

where Ks is the coefficient of earth pressure (herein taken to be 0.8); v is the averaged vertical effective stress along the shaft; b is a reduction factor (herein taken to be 1);

tan(b) is the arithmetic average of the spatially variable tan(b×) over the shaft.

Figure 4.11 Friction pile under compression