Federal Highway Administration Design Manual: Deep Mixing for Embankment and Foundation Support

CHAPTER 7. DESIGN EXAMPLE

7.1 introduction

This chapter provides an example of DMM to support an embankment for a transportation application. The problem background is described and step-by-step analysis and design calculations are presented.

A new approach embankment is to be constructed over a 25-ft (7.6-m)-thick deposit of soft clay underlain by a dense sand layer. The ground water table (GWT) is located 3 ft (0.9 m) below the native ground surface. Preliminary analyses determined that without some type of ground improvement, both the factor of safety against slope stability failure (F_s = 0.77) and the predicted settlement (about 2.3 ft (0.7 m)) are unacceptable. DMM has been proposed to stabilize the soft clay layer. The DMM design is performed using steps 1-7 in the following sections, based on the design guidance presented in chapter 6.

7.1.1 Step 1—Establish Project Requirements

The geometry and soil properties for the proposed embankment are shown in figure 76. The slope of the embankment is 1.5 horizontal to 1 vertical (1.5H:1V). A uniform traffic surcharge load, q_s, of 200 lbf/ft² (10 kPa) is included across the entire width of the embankment crest.

1 ft = 0.305 m
1 lbf/ft² = 0.0479 kPa
1 lbf/ft3 = 0.157 kN/m³
Figure 76. Illustration. Example problem—embankment cross section.

For this project, a thorough site investigation was conducted, and a customary degree of conservatism was used when the soil strength parameters were selected. Therefore, based on the recommendations in section 6.1.1, the following factors of safety (defined in table 11) were selected for design:

• F_cc = 1.3.
• F_s = 1.5.
• F_o = 1.3.
• F_c = 1.3.
• F_v = 1.3.
• F_e = 1.3.

The maximum allowable settlement of the embankment was 2 inches (51 mm).

7.1.2 Step 2—Establish Representative Subsurface Conditions

The soil material property values to be used in the geotechnical analysis and design of the deep mixed ground are shown in figure 76.

7.1.3 Step 3—Establish Trial Deep Mixed Ground Property Values

The procedure for step 3 is as follows:

1. Assume a value of the 28-day q_dm,spec. Typical values range from about 75 to 150 psi (517 to 1,034 kPa) for soft ground conditions. For this example, q_dm,spec of 125 psi (862 kPa) is assumed.

2. Determine f_c, as shown in figure 77, using estimated t in days between mixing and application of 75 percent of the proposed embankment height. For this example, t equals 60 days, which means that the embankment height will not be above about 13 ft (4 m) until about 60 days after mixing, which is about 1 month after the 28-day strength has been verified. Generally, a significant height of embankment fill will not be placed until after the 28-day strength has been verified.

Figure 77. Equation. Example problem—curing factor.

3. Determine s_dm according to figure 33. An fr equal to 0.8 is used for this example, as shown in figure 78.

1 lbf/ft² = 0.0479 kPa
Figure 78. Equation. Example problem—shear strength of the deep mixed ground.

4. Determine f_v from table 12. For this example, the estimated V_dm is 0.5, and the estimated pdm is 80 percent. Based on these values, f_v is equal to 0.83 for slope stability analyses (factor of safety equals 1.5) and f_v is equal to 0.95 when considering the other failure modes (factor of safety equals 1.3) that involve the strength of the deep mixed ground.

5. Determine the Young's modulus of the deep mixed ground, E_dm, according to figure 34 for wet mixing or figure 35 for dry mixing. For this example, it is assumed that wet mixing will be used (see figure 79).

1 lbf/ft² = 0.0479 kPa
Figure 79. Equation. Example problem—determine E_dm according to figure 34.

The unit weights of the deep mixed zones are also necessary for stability analyses. For this example, it is assumed that the unit weights of the deep mixed zones are approximately equal to the unit weight of the soil prior to mixing, as discussed in section 6.1.3.

7.1.4 Step 4—Establish Trial Deep Mixed Geometry

In step 4, a trial geometry of deep mixing to support the embankment is established.

Step 4.1—General Layout and Definitions

For this example, the deep mixed columns were arranged as shown in figure 36. Isolated columns were used under the central portion of the embankment to control settlement, and continuous shear walls composed of overlapping columns oriented perpendicular to the embankment centerline were used under the embankment side slopes to improve stability. Typical values of e/d for shear panels beneath embankment side slopes range from about 0.2 to 0.35, and a minimum value of 0.3 was selected for this example.

Step 4.2—Establish Center Replacement Ratio

Establish a trial value of a_s,center using figure 44, as shown in figure 80.

Figure 80. Equation. Example problem—area replacement ratio beneath the central portion of the embankment.

Where f_v is the variability factor determined in step 3 corresponding to the design value of
F_cc= 1.3.

Typical values of a_s,center for deep mixing support of embankments range from about 0.2 to 0.4. To satisfy figure 44 and stay within the typical range of values, an a_s,center of 0.2 was selected.

Step 4.3—Estimate the Shear Wall Zone Replacement Ratio

Estimate a minimum value of a_s,shear. Typical values of a_s,shear for deep mixing support of embankments are greater than or equal to a_s,shear and range from about 0.2 to 0.4. For this example, as,shear equal to 0.25 was selected as a trial value. Calculate using figure 40 as shown in figure 81.

Figure 81. Equation. Example problem—chord angle in radians.

Calculate the value of c/s_shear using figure 45 based on the selected values of e/d and a_s,shear, as shown in figure 82.

Figure 82. Equation. Example problem—ratio of chord length to shear wall spacing.

The minimum and/or maximum trial values of the geometric parameters required for design are summarized in table 17.

Table 17. Example problem—geometric parameters.

7.1.5 Step 5—Evaluate Settlement

Evaluate M_comp using figure 46, as shown figure 83.

1 lbf/ft² = 0.0479 kPa
Figure 83. Equation. Example problem—composite modulus.

Where M_soil is assumed to equal 25,000 lbf/ft² (1,196 kPa) for this example and can be determined from oedometer tests in practice. E_dm was determined in step 3.

Calculate ΔH_dm, based on figure 47, as shown in figure 84.

1 inch = 25.4 mm
Figure 84. Equation. Example problem—compression of the treated zone.

For this example, it is assumed that the compression of the dense sand is small and takes place as the embankment is constructed. Therefore, the predicted settlement is equal to the 0.63-inch (16‑mm) compression of the deep mixed zone, which is less than the allowable settlement of 2 inches (51 mm).

H_emb is greater than 2 times the maximum allowed clear spacing between adjacent columns under the central portion of the embankment (i.e., H_emb > 2(s_center - d) = 2(8) = 16 ft (4.9 m)). Therefore, there is little risk of surface expression of differential settlements occurring at the base of a well-constructed embankment, and special provisions for a load transfer platform at the base of the embankment are not necessary. Nevertheless, if spoils from the DMM operation are available, they could be used to construct the lower portion of the embankment to further reduce any risk of differential surface settlements.

On the embankment side slopes, there is a potential for differential settlement of the embankment surface because H_emb is less than 2 times the maximum allowed clear spacing between shear walls (i.e., H_emb < 2(s_shear - d) = 2(12) = 24 ft (7.3 m)). Therefore, it may be necessary to consider settlement control measures such as use of the DMM spoils to create a strong load transfer platform at the base of the embankment beneath the side slopes.

7.1.6 Step 6—Evaluate Stability

In this step, the trial geometry established in step 4 is analyzed for stability.

Step 6.1—Slope Stability

Perform slope stability analyses to determine the critical failure surface and corresponding factor of safety. Determine the composite shear strengths of the deep mixed zones beneath the embankment as calculated in figure 85 and figure 86 and assigned in figure 87.

1 lbf/ft² = 0.0479 kPa
Figure 85. Equation. Example problem—composite shear strength of the deep mixed zone beneath the shear walls.

Where f_v is the variability factor determined in step 3 corresponding to the design value of F_s = 1.5.

1 lbf/ft² = 0.0479 kPa
Figure 86. Equation. Example problem—composite shear strength of the deep mixed zone beneath the central portion of the embankment.

Figure 87. Illustration. Slope stability results.

As recommended in section 6.1, Spencer's method should be used for the slope stability analyses.⁽⁸⁷⁾ For this example, only failure surfaces that passed through or below the deep mixed shear wall zone were analyzed. Stability of the 1.5H:1V embankment slopes would have to be improved with geosynthetic reinforcement or some other stabilizing method given the embankment strength parameter values of Φ = 35 degrees and c = 0. Relatively steep embankment slopes were selected for this example to illustrate the capability of deep mixing to stabilize a steep embankment on soft clay.

The resulting minimum value of F_s from a comprehensive search for the critical failure surface through and below the deep mixed shear wall was 1.51, which exceeds the design value of 1.5. The critical failure surface for this example, which is shown in figure 87, passes partly through and partly below the deep mixed shear wall zone. For the same embankment configuration on the native soft clay prior to deep mixing, the resulting minimum factor of safety value was 0.77. Thus, a relatively modest amount of deep mixing can create a large improvement in stability for the conditions considered in this example.

Step 6.2—Combined Overturning and Bearing Capacity

1. Select a design value of F_o. For this example, F_o equals 1.3.
2. Determine mobilized shear strength parameter values for each layer of soil beside and beneath the deep mixed shear wall zone using figure 52 and figure 53 for total strength parameters and figure 54 and figure 55 for effective strength parameters.

   The shear strength of the soft clay layer is characterized by total stresses, as shown in figure 88 and figure 89.

   
   1 lbf/ft² = 0.0479 kPa
   Figure 88. Equation. Example problem—shear strength of the soft clay using total normal stresses.

   
   Figure 89. Equation. Example problem—mobilized total stress friction angle of the soft clay.

   The composite shear strength of the center deep mixed zone is characterized by total stresses, as shown in figure 90.

   
   1 lbf/ft² = 0.0479 kPa
   Figure 90. Equation. Example problem—composite shear strength of the center deep mixed zone using total normal stresses.

   The shear strength of the embankment material is characterized by effective stresses, as shown in figure 91 and figure 92.

   
   1 lbf/ft² = 0.0479 kPa
   Figure 91. Equation. Example problem—composite shear strength of the embankment material using effective normal stresses.

   
   Figure 92. Equation. Example problem—mobilized effective stress friction angle of the embankment material.

   The shear strength of the dense sand layer is characterized by effective stresses, as shown in figure 93.

   
   Figure 93. Equation. Example problem—mobilized effective stress friction angle of the dense sand layer.

3. Calculate values of Pa, ha, Va, Pp, hp, and Vp using the mobilized strength parameter values from step 2.

   Calculate the active forces based on the mobilized strength parameters of the embankment material and center deep mixed zone, as shown in figure 94 through figure 104.

   
   Figure 94. Equation. Example problem—effective stress active lateral earth pressure coefficient.

   
   1 lbf/ft = 0.01459 kN/m
   Figure 95. Equation. Example problem—active force component from embankment.

   
   1 ft = 0.305 m
   Figure 96. Equation. Example problem—vertical distance from overturning point to line of action of active force component from embankment.

   
   1 lbf/ft = 0.01459 kN/m
   Figure 97. Equation. Example problem—active force component from surcharge.

   
   1 ft = 0.305 m
   Figure 98. Equation. Example problem—vertical distance from overturning point to line of action of active force component from surcharge.

   
   1 lbf/ft = 0.01459 kN/m
   Figure 99. Equation. Example problem—active force component from clay rectangle.

   
   1 ft = 0.305 m
   Figure 100. Equation. Example problem—vertical distance from overturning point to line of action of active force component from clay rectangle.

   
   1 lbf/ft = 0.01459 kN/m
   Figure 101. Equation. Example problem—active force component from clay triangle.

   
   1 ft = 0.305 m
   Figure 102. Equation. Example problem—vertical distance from overturning point to line of action of active force component from clay triangle.

   
   1 lbf/ft = 0.01459 kN/m
   Figure 103. Equation. Example problem—total active force.

   
   
   1 ft = 0.305 m
   Figure 104. Equation. Example problem—vertical distance between overturning point and total active force.

   Calculate the side shear forces based on the mobilized strength parameters of the soft clay layer, as shown in figure 105 and figure 106.

   
   1 lbf/ft = 0.01459 kN/m
   Figure 105. Equation. Example problem—active side shear force from the soft clay.

   Calculate the passive forces based on the mobilized strength parameters of the soft clay layer, as shown in figure 107 through figure 112.

   
   1 lbf/ft = 0.01459 kN/m
   Figure 106. Equation. Example problem—passive side shear force from the soft clay.

   
   1 ft = 0.305 m
   Figure 107. Equation. Example problem—passive lateral force component from the clay rectangle.

   
   1 lbf/ft = 0.01459 kN/m
   Figure 108. Equation. Example problem—vertical distance from overturning point to line of action of passive force component from the clay rectangle.

   
   1 ft = 0.305 m
   Figure 109. Equation. Example problem—passive lateral earth force component from the clay triangle.

   
   1 lbf/ft = 0.01459 kN/m
   Figure 110. Equation. Example problem—vertical distance from the overturning point to the line of action of passive force component from clay triangle.

   
   1 ft = 0.305 m
   Figure 111. Equation. Example problem—total passive lateral earth force.

   
   1 lbf/ft = 0.01459 kN/m
   Figure 112. Equation. Example problem—vertical distance between the overturning point and the total passive force.

4. Determine the resultant force N using figure 56. Determine W and the location of x_W, as shown in figure 113 through figure 119.

   
   1 ft = 0.305 m
   Figure 113. Equation. Example problem—weight of the embankment.

   
   1 ft = 0.305 m
   Figure 114. Equation. Example problem—location of the resultant of the embankment weight.

   
   1 lbf/ft = 0.01459 kN/m
   Figure 115. Equation. Example problem—weight of the deep mixed zone.

   
   1 ft = 0.305 m
   Figure 116. Equation. Example problem—location of the resultant of the deep mixed zone.

   
   1 lbf/ft = 0.01459 kN/m
   Figure 117. Equation. Example problem—total weight.

   
   1 ft = 0.305 m
   Figure 118. Equation. Example problem—location of the resultant of the total weight.

   
   1 lbf/ft = 0.01459 kN/m
   Figure 119. Equation. Example problem—resultant total vertical force.

   The shear strength of the soil beneath the base of the deep mixed zone is characterized by effective normal stresses. Calculate N' using figure 57. Determine U and x_U, as shown in figure 120 through figure 122.

   
   1 lbf/ft = 0.01459 kN/m
   Figure 120. Equation. Example problem—water force acting on the base of the deep mixed zone.

   
   1 ft = 0.305 m
   Figure 121. Equation. Example problem—location of water force acting on the base of the deep mixed zone.

   
   1 lbf/ft = 0.01459 kN/m
   Figure 122. Equation. Example problem—resultant effective vertical force.

5. Determine x_N using figure 58, as shown in figure 123.

   
   1 ft = 0.305 m
   Figure 123. Equation. Example problem—location of total resultant force acting on the base of the deep mixed zone.

   The shear strength of the soil beneath the base of the deep mixed zone is characterized by effective normal stresses. Calculate x_N' using figure 59, as shown in figure 124.

   
   1 ft = 0.305 m
   Figure 124. Equation. Example problem—location of the effective resultant force acting on the base of the deep mixed zone.

6. The shear stress of the soil beneath the deep mixed shear walls is characterized by effective normal stresses. Calculate q_toe using figure 61, as shown in figure 125 and figure 126.

   
   Figure 125. Equation. Example problem—Position of resultant within the base.

   
   1 lbf/ft² = 0.0479 kPa
   Figure 126. Equation. Example problem—bearing pressure at the toe of the deep mixed shear walls.

7. The shear strength of the soil beneath the deep mixed shear walls is characterized by effective normal stresses. Determine q_all using figure 64, as shown in figure 127.

   
   1 lbf/ft² = 0.0479 kPa
   Figure 127. Equation. Example problem—allowable bearing pressure at the toe of the deep mixed shear walls.

   The value of q_toe = 10,500 lbf/ft² (502 kPa) is less than the value of q_all = 18,400 lb/ft² (880 kPa). Therefore, the design is sufficient to prevent combined overturning and bearing capacity failure of the deep mixed shear walls.

Step 6.3—Crushing of the Deep Mixed Shear Walls at the Outside Toe

The deep mixed ground overlies a hard bearing stratum. Therefore, the design is checked against crushing of the deep mixed ground at the toe of the shear walls.

Because the factor of safety F_o is equal to F_c, use the intermediate values from step 6.2 in the current step, where q_toe = 10,500 lbf/ft² (503 kPa) and Φ'_m = 30.1 degrees.

The shear strength of the soil beneath the deep mixed ground is characterized by effective normal stresses. Use figure 67 to calculate the at-rest σ'_h, as shown in figure 128 through figure 130.

Figure 128. Equation. Example problem—at-rest lateral earth pressure coefficient.

1 lbf/ft² = 0.0479 kPa
Figure 129. Equation. Example problem—effective vertical stress.

1 lbf/ft² = 0.0479 kPa
Figure 130. Equation. Example problem—at-rest effective lateral earth pressure.

Determine q_all based on figure 65, as shown in figure 131.

1 lbf/ft² = 0.0479 kPa
Figure 131. Equation. Example problem—allowable bearing capacity of deep mixed ground.

Where f_v is the variability factor determined in step 3 corresponding to the design value of F_c = 1.3.

The value of q_toe = 10,500 lb/ft² (502 kPa) is less than q_all = 12,400 lb/ft² (593 kPa). Therefore, the design is sufficient to prevent crushing of the deep mixed ground at the toe of the shear walls.

Step 6.4—Shearing on Vertical Planes in the Deep Mixed Shear Walls

1. Determine the values of Vp, N, and xN corresponding to F_v. Because the factor of safety F_o is equal to F_c, the following intermediate values from step 6.2 were used in the current step:

• V_p = 6,730 lb/ft (98 kN/m).
• N = 84,470 lb/ft (1231.9 kN/m).
• x_N = 10.01 ft (3.0 m).

2. Compute τ_v on the critical vertical plane in the deep mixed zone using figure 69, as shown in figure 132 through figure 133.

,
Figure 132. Equation. Example problem—location of the force resultant along the base of the deep mixed zone.

1 lbf/ft² = 0.0479 kPa
Figure 133. Equation. Example problem—average vertical shear stress on the critical vertical plane.

3. Compute τ_v,all in the deep mixed zone using figure 70, as shown in figure 134.

1 lbf/ft² = 0.0479 kPa
Figure 134. Equation. Example problem—allowable vertical shear stress in the deep mixed zone.

Where f_v is the variability factor determined in step 3 corresponding to the design value of F_v = 1.3.

The value of τ_v = 814 lb/ft² (39 kPa) is less than τ_v,all = 1,180 lb/ft² (56 kPa). Therefore, the design is sufficient to prevent shearing on vertical planes in the deep mixed shear wall.

Step 6.5—Extrusion of Soil between the Deep Mixed Shear Walls

Check s_shear - d using figure 71 to assure that extrusion of the soft clay could not occur between shear walls, as shown in figure 135. For this example, the procedure is illustrated using the extrusion of the entire thickness of the soft clay.

1 ft = 0.305 m
Figure 135. Equation. Example problem—maximum clear spacing between shear columns.

Therefore, the allowable maximum clear spacing of 12 ft (3.7 m) between shear wall columns established in step 4 is adequate to prevent extrusion of the soft clay because it is less than the value of 19.6 ft (6.0 m) calculated using figure 135.

7.1.7 Step 7—Prepare Plans and Specifications

Incorporate the final design parameters, including q_dm,spec = 125 psi (862 kPa) and the geometric parameters listed in table 17, in the project plans and specifications.