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Batter pile design
Batter pile design
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Batter piles are widely used for retaining walls to resist lateral forces. Batter piles have higher lateral load capacity than vertical piles. Bridge abutments, retaining walls, and platforms can benefit from batter piles. •
Usually in most cases, one row of piles is battered as in Fig. 20.1. In some cases it is necessary to have more than one row of batter piles, as in Fig. 20.2.
20.1 Theory 20.1.1 Forces on batter piles Batter piles are capable of resisting significant lateral forces. Lateral resistance of batter piles comes from two sources: 1. horizontal component of the axial reaction 2. horizontal resistance due to soil (H)
Note that horizontal resistance (H) and horizontal component of the pile axial reaction are two different quantities (Plate 20.1).
20.1.2 Negative skin friction Batter piles should be avoided in situations where negative skin frictional forces can be present. Settling soil could induce large bending moments in batter piles.
20.1.3 Force polygon Please refer to Fig. 20.1 for the following force polygon (calculation of loads will be shown later in the chapter):
Pile Design and Construction Rules of Thumb Copyright © 2016 Elsevier Inc. All rights reserved.
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Pile Design and Construction Rules of Thumb
Figure 20.1 Batter piles. (a) Bridge abutments; and (b) typical profile for retaining wall with batter piles.
Figure 20.2 Forces in batter piles.
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Plate 20.1 Batter piles.
Line AB: calculate the total vertical load on piles. Draw the total vertical load. Line BC: draw the total horizontal load (load due to wind or water). Piles should be able to withstand these two loads. Line CD: draw the load on piles along row 3. These piles are vertical. Line DE: draw the load on piles along row 2. Line EF: draw the load on piles along row 1. These are batter piles. Line FA: this line indicates the lateral force required to stabilize the piles. Lateral resistance of pile should be more than the load indicated by line FA. Please refer to Fig. 20.1b for the following force polygon:
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Line AB: calculate the total vertical load on piles. Draw the total vertical load. Line BC: draw the total horizontal load. Line CD: draw the load on piles along row 3. These piles are vertical. Line DE: draw the load on piles along row 2. These piles are batter piles. Line EF: draw the load on piles along row 1. Line FA: this line indicates the lateral force required to stabilize the piles. Lateral resistance of the pile should be more than the load indicated by line FA. Design example 20.1 Compute the loads on piles due to the retaining wall shown. Assume lateral earth pressure coefficient at rest (K0) to be 0.5. Piles in the front are battered at 20 degrees and center piles are battered at 15 degrees to the vertical. Assume the piles have been placed at 3 ft. intervals (Fig. 20.3).
Horizontal stress at bottom of the wall; K0 × 110 × 20 = 1100 psf H, horizontal force = area of the stress triangle = 1,100 × 20/2 = 11,000 lb. Step 1: horizontal force. Horizontal force H, acts 6.666 ft. from the base (y = 20/3). Moment (M) = 11,000 × 6.666 lb ft. = 73,333 lb ft. per linear ft. of wall. Moment in 3 linear ft. section = 3 × 73,333 = 219,999 lb ft. Step 2: weight of soil and concrete (Fig. 20.4). Weight of soil resting on the retaining wall (height = 18, width = 5) = 5 × 18 × 110 = 9900 lb. Weight of concrete = (2 × 9 + 2 × 18) × 160 = 8640 lb (160 pcf = concrete density). Total weight = 18,540 lb per linear feet of wall. Total weight (W) in a 3-ft. section = 55,620 lb. There are three piles in the 3-ft. section. Hence, each pile carries a vertical load of 18,540 lb.
Figure 20.3 Design example.
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Figure 20.4 Horizontal force.
Step 3: overturning moment (Fig. 20.5). Overturning moment = resisting moment Take moments around point “E” R × L = W × L = M = H. y = 219,999 lb ft. (see step 1) L = 219,999/W = 219,999/55,620 = 3.96 ft. M = overturning moment due to H. Step 4: stability (Fig. 20.6). The center of gravity of the piles lies along the center of the middle row. Vertical reaction acts 3.96 ft. from the edge. Distance to the reaction from the center of the pile system = 4.5 − 3.96 = 0.54 ft. Moment around the center line = R × 0.54 = 55,620 × 0.54 = 30,038.4 lb per 3-ft. section. Note that moment around the centerline is different from the moment around the edge (point “E”).
Figure 20.5 Weight of soil and concrete.
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Figure 20.6 Overturning moment.
Step 5: additional load due to bending moments. • Each pile carries a vertical load of 18,540 lb (see step 2). • Due to the moment, the base of the retaining wall would undergo a bending moment. This bending moment would create additional stresses on piles. Stress developed on piles due to bending is given by the following equation: M σ = , I y where M = bending moment, σ = bending stress, I = moment of area, and y = distance. Find the moment of area of piles by taking moments around the centerline of the footing. Moment of area of row 1 = A × 32 = 9A (A is area of the piles). Moment of area of row 2 = A × 0 = 0 (distance is taken from the center of gravity of piles). Moment of area of row 3 = A × 32 = 9A. Total moment of area = 18 A. Bending load on piles in row 1 (σ ) =
M 30,034.8 5005.8 ⋅y= ×3= lb/ft 2 A I 18 A
M = 30,034.8 lb per 3-ft. section (see step 4) y = 3 ft. (distance from center line to corner row of piles) Bending load per pile =
5005.8 × A = 5005.8 lb per pile A
Step 6: total loads on piles. Total vertical load on pile = vertical load + load due to bending. Total vertical load on piles in row 1 = 18,540 + 5005.8 = 23,545.8 lb. Bending load is positive for piles in row 1. Total vertical load on piles in row 2 = 18,540 + 0 = 18,540 lb (since y = 0). Load due to bending is zero for row 2. Total vertical load on piles in row 3 = 18,540 − 5005.8 = 13,534.2 lb. Bending load is negative for piles in row 3.
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Step 7: force polygon. Draw the force polygon for the 3-ft. section selected. Total horizontal load for a 3-ft. s ection is 33,000 lb (11,000 lb per linear foot. See step 1). Total vertical load for a 3-ft. s ection is 55,620 lb. See step 4. Start drawing the force polygon from I to J. HI is the final leg.
EF = 18,540 × tan 15 = 4451 lb (See Fig. 20.6 for EF) GH = 23,545.8 tan 20 = 7,650.5 lb. Total horizontal load resisted = 4451 + 7650.5 = 12,101.5 lb. Additional load that need to be resisted = 33,000 – 12,101.5 = 20,898.5 lb. This load needs to be resisted by the soil pressure acting on the piles laterally (see Fig. 20.7). The lateral pile resistance should not be confused with axial pile resistance. In the case of a vertical pile, axial load is vertical and there is no horizontal component to the axial load. On the other hand, when a horizontal load is applied to a vertical pile, it would resist. Hence, all piles (including vertical piles) have a lateral resistance (Fig. 20.7). Axial load “P” can be broken down into “X” and “Y” components. Computation of lateral resistance of the pile due to soil pressure is given in a further chapter (see chapter: Lateral Pile Resistance). There are three piles in the considered 3-ft. section. Each pile should be able to resist a load of 6,966.5 lb. Use the principles given in chapter: Lateral Pile Resistance to compute the lateral pile capacity due to soil pressure.
Figure 20.7 Lateral resistance.
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Factor of safety (FOS) of 2.5–3.0 should be used. For instance, the required vertical pile capacity for piles in row 1 is 23,545 lb. The piles should have an ultimate pile capacity of 70,635 lb assuming a FOS of 3.0. Note: no credit should be given to the resistance due to soil friction acting on the base of the retaining wall. Piles would stress the soil due to the horizontal load acting on them. Stressed soil underneath the retaining wall will not be able to provide any frictional resistance to the retaining wall. If piles were to fail, piles would pull the soil with them. Hence, soil would not be able to provide any frictional resistance to the retaining structure. Design example 20.2 Compute the loads on piles due to the retaining wall shown. Assume lateral earth pressure coefficient at rest (K0) to be 0.5. Piles in the front are battered at 20 degrees and center piles are battered at 15 degrees to the vertical (this problem is similar to the previous example, except for the pile configuration, Fig. 20.8).
Step 1: horizontal force H acts 6.666 ft. from the base. Moment (M) = 11,000 × 6.666 lb ft. = 73,333 lb ft. per linear foot of wall. Moment in 6 linear ft. section = 6 × 73,333 = 439,998 lb ft. Assume the piles have been placed at 3-ft. intervals (Fig. 20.9).
Figure 20.8 Compute pile loads.
Figure 20.9 Assume piles are placed at 3 ft. intervals.
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Consider a section of 6 ft. for computational purposes. There are six half piles and one full pile in this section. Rows 1 and 2 are batter piles. Row 1 has twice as more piles than rows 2 and 3. Weight of soil resting on the retaining wall = 5 × 18 × 110 = 9900 lb. Weight of concrete = (2 × 9 + 2 × 18) × 160 = 8640 lb (160 pcf = concrete d ensity). Total weight = 18,540 lb per linear foot of wall. Total weight (W) in a 6-ft. section = 111,240 lb. There are four piles in the 6-ft. section. Hence, each pile carries a load of 27,810 lb. Take moments around point “E” R × L = W × L = M = H · k = 439,998 lb ft. L = 439,998/111,240 = 3.96 ft.
20.1.4 Center of gravity of piles There are six half piles and one full pile in this section (Fig. 20.10 and Fig. 20.11). There are two half piles and one full pile in row 1.
Figure 20.10 Taking moments around point “E”.
Figure 20.11 Center of gravity of piles.
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There are two half piles in row 2 and two half piles in row 3 (total piles in the section = 4). Take moments around row 3. (2 × 6 + 1 × 3)/4 = 3.75 ft. • •
As per these calculations, the center of gravity of piles is located 3.75 ft. from row 3. Moment of area of the system has to be computed from the center of gravity of piles. Distance to row 1 from center of gravity = (6 − 3.75) ft. = 2.25 ft. Distance to row 2 from center of gravity = 0.75 ft. Distance to row 3 from center of gravity = 3.75 ft. Step 2: Each pile carries a load of 27,810 lb (see step 1). • Stress developed on piles M σ = . I y Here, M = bending moment, σ = bending stress, I = moment of area, and y = distance.
Find the moment of area of piles.
20.1.5 Row 1 Moment of area of row 1 = A · r2 = A × 2.252 = 10.12 A (A = area of piles). Moment of area of row 2 = A × 0.752 = 0.56 A. (Distance is taken from the center of gravity). Moment of area of row 3 = A × 3.752 = 14.06 A. Total moment of area = 24.74 A. Bending moment (M) = 439,998 ft. lb (see step 1). Bending load on piles in row 1(σ ) = Bending load on first row piles =
M 439,998 40, 016 ⋅y = × 2.25 = lb. A I 24.74 A
40, 016 × 2 A = 80, 032 lb. A
There are two piles in the first row. Two half piles and one full pile is equivalent to two piles. Total load on pile = vertical load + load due to bending. Total load on piles in row 1 = 27,810 × 2 + 80,032 = 135,652 lb. Vertical load per pile is 27,810 lb. Since there are two piles in row 1, 27,810 lb is multiplied by 2. Step 3: row 2. Bending load on piles in row 2(σ ) = Bending load per pile =
13,339 M 439,998 ⋅y= × 0.75 = lb. A I 24.74 A
13, 339 × A = 13, 339 lb per pile. A
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There are two half piles in row 2, which amounts to one pile. Multiply by area of the pile to convert the load from lb per ft.2 to lb per pile. Total load on piles in row 2 = 27,810 – 13,339 = 14,471 lb. (Since center of gravity is on the opposite side, bending moment creates a tensile force on the pile.) Step 4: row 3. Bending load on piles in row 3(σ ) = Bending load per pile =
M 439,998 66,693 ⋅y= × 3.75 = lb. A I 24.74 A
66, 693 × A = 66, 693 lb per pile. A
Row 3 has two half piles, which amounts to one pile. Total load on piles in row 3 = 27,810 − 66,693 = −38,883 lb per pile The value 27,810 lb is the vertical load on piles and 66,693 lb is the load due to bending, which is tensile. Total force on piles in row 3 is tensile. Step 5: draw the force polygon. NM = 14,471 × tan 15 = 3,877 lb. GH = 135,652 × tan 20 = 49,373 lb (this load accounts for two piles in row 1).
Total horizontal load resisted by axial forces of piles = 49,373 + 3,877 = 53,250 lb. Additional load that needs to be resisted = 66,000 – 53,250 = 12,750 lb. This load needs to be resisted by the soil pressure acting on piles laterally. Each pile should have a lateral pile capacity of 12,750/4 lb = 3,188 lb.
20
Batter piles are widely used for retaining walls to resist lateral forces. Batter piles have higher lateral load capacity than vertical piles. Bridge abutments, retaining walls, and platforms can benefit from batter piles. •
Usually in most cases, one row of piles is battered as in Fig. 20.1. In some cases it is necessary to have more than one row of batter piles, as in Fig. 20.2.
20.1 Theory 20.1.1 Forces on batter piles Batter piles are capable of resisting significant lateral forces. Lateral resistance of batter piles comes from two sources: 1. horizontal component of the axial reaction 2. horizontal resistance due to soil (H)
Note that horizontal resistance (H) and horizontal component of the pile axial reaction are two different quantities (Plate 20.1).
20.1.2 Negative skin friction Batter piles should be avoided in situations where negative skin frictional forces can be present. Settling soil could induce large bending moments in batter piles.
20.1.3 Force polygon Please refer to Fig. 20.1 for the following force polygon (calculation of loads will be shown later in the chapter):
Pile Design and Construction Rules of Thumb Copyright © 2016 Elsevier Inc. All rights reserved.
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Pile Design and Construction Rules of Thumb
Figure 20.1 Batter piles. (a) Bridge abutments; and (b) typical profile for retaining wall with batter piles.
Figure 20.2 Forces in batter piles.
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Plate 20.1 Batter piles.
Line AB: calculate the total vertical load on piles. Draw the total vertical load. Line BC: draw the total horizontal load (load due to wind or water). Piles should be able to withstand these two loads. Line CD: draw the load on piles along row 3. These piles are vertical. Line DE: draw the load on piles along row 2. Line EF: draw the load on piles along row 1. These are batter piles. Line FA: this line indicates the lateral force required to stabilize the piles. Lateral resistance of pile should be more than the load indicated by line FA. Please refer to Fig. 20.1b for the following force polygon:
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Line AB: calculate the total vertical load on piles. Draw the total vertical load. Line BC: draw the total horizontal load. Line CD: draw the load on piles along row 3. These piles are vertical. Line DE: draw the load on piles along row 2. These piles are batter piles. Line EF: draw the load on piles along row 1. Line FA: this line indicates the lateral force required to stabilize the piles. Lateral resistance of the pile should be more than the load indicated by line FA. Design example 20.1 Compute the loads on piles due to the retaining wall shown. Assume lateral earth pressure coefficient at rest (K0) to be 0.5. Piles in the front are battered at 20 degrees and center piles are battered at 15 degrees to the vertical. Assume the piles have been placed at 3 ft. intervals (Fig. 20.3).
Horizontal stress at bottom of the wall; K0 × 110 × 20 = 1100 psf H, horizontal force = area of the stress triangle = 1,100 × 20/2 = 11,000 lb. Step 1: horizontal force. Horizontal force H, acts 6.666 ft. from the base (y = 20/3). Moment (M) = 11,000 × 6.666 lb ft. = 73,333 lb ft. per linear ft. of wall. Moment in 3 linear ft. section = 3 × 73,333 = 219,999 lb ft. Step 2: weight of soil and concrete (Fig. 20.4). Weight of soil resting on the retaining wall (height = 18, width = 5) = 5 × 18 × 110 = 9900 lb. Weight of concrete = (2 × 9 + 2 × 18) × 160 = 8640 lb (160 pcf = concrete density). Total weight = 18,540 lb per linear feet of wall. Total weight (W) in a 3-ft. section = 55,620 lb. There are three piles in the 3-ft. section. Hence, each pile carries a vertical load of 18,540 lb.
Figure 20.3 Design example.
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Figure 20.4 Horizontal force.
Step 3: overturning moment (Fig. 20.5). Overturning moment = resisting moment Take moments around point “E” R × L = W × L = M = H. y = 219,999 lb ft. (see step 1) L = 219,999/W = 219,999/55,620 = 3.96 ft. M = overturning moment due to H. Step 4: stability (Fig. 20.6). The center of gravity of the piles lies along the center of the middle row. Vertical reaction acts 3.96 ft. from the edge. Distance to the reaction from the center of the pile system = 4.5 − 3.96 = 0.54 ft. Moment around the center line = R × 0.54 = 55,620 × 0.54 = 30,038.4 lb per 3-ft. section. Note that moment around the centerline is different from the moment around the edge (point “E”).
Figure 20.5 Weight of soil and concrete.
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Figure 20.6 Overturning moment.
Step 5: additional load due to bending moments. • Each pile carries a vertical load of 18,540 lb (see step 2). • Due to the moment, the base of the retaining wall would undergo a bending moment. This bending moment would create additional stresses on piles. Stress developed on piles due to bending is given by the following equation: M σ = , I y where M = bending moment, σ = bending stress, I = moment of area, and y = distance. Find the moment of area of piles by taking moments around the centerline of the footing. Moment of area of row 1 = A × 32 = 9A (A is area of the piles). Moment of area of row 2 = A × 0 = 0 (distance is taken from the center of gravity of piles). Moment of area of row 3 = A × 32 = 9A. Total moment of area = 18 A. Bending load on piles in row 1 (σ ) =
M 30,034.8 5005.8 ⋅y= ×3= lb/ft 2 A I 18 A
M = 30,034.8 lb per 3-ft. section (see step 4) y = 3 ft. (distance from center line to corner row of piles) Bending load per pile =
5005.8 × A = 5005.8 lb per pile A
Step 6: total loads on piles. Total vertical load on pile = vertical load + load due to bending. Total vertical load on piles in row 1 = 18,540 + 5005.8 = 23,545.8 lb. Bending load is positive for piles in row 1. Total vertical load on piles in row 2 = 18,540 + 0 = 18,540 lb (since y = 0). Load due to bending is zero for row 2. Total vertical load on piles in row 3 = 18,540 − 5005.8 = 13,534.2 lb. Bending load is negative for piles in row 3.
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Step 7: force polygon. Draw the force polygon for the 3-ft. section selected. Total horizontal load for a 3-ft. s ection is 33,000 lb (11,000 lb per linear foot. See step 1). Total vertical load for a 3-ft. s ection is 55,620 lb. See step 4. Start drawing the force polygon from I to J. HI is the final leg.
EF = 18,540 × tan 15 = 4451 lb (See Fig. 20.6 for EF) GH = 23,545.8 tan 20 = 7,650.5 lb. Total horizontal load resisted = 4451 + 7650.5 = 12,101.5 lb. Additional load that need to be resisted = 33,000 – 12,101.5 = 20,898.5 lb. This load needs to be resisted by the soil pressure acting on the piles laterally (see Fig. 20.7). The lateral pile resistance should not be confused with axial pile resistance. In the case of a vertical pile, axial load is vertical and there is no horizontal component to the axial load. On the other hand, when a horizontal load is applied to a vertical pile, it would resist. Hence, all piles (including vertical piles) have a lateral resistance (Fig. 20.7). Axial load “P” can be broken down into “X” and “Y” components. Computation of lateral resistance of the pile due to soil pressure is given in a further chapter (see chapter: Lateral Pile Resistance). There are three piles in the considered 3-ft. section. Each pile should be able to resist a load of 6,966.5 lb. Use the principles given in chapter: Lateral Pile Resistance to compute the lateral pile capacity due to soil pressure.
Figure 20.7 Lateral resistance.
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Factor of safety (FOS) of 2.5–3.0 should be used. For instance, the required vertical pile capacity for piles in row 1 is 23,545 lb. The piles should have an ultimate pile capacity of 70,635 lb assuming a FOS of 3.0. Note: no credit should be given to the resistance due to soil friction acting on the base of the retaining wall. Piles would stress the soil due to the horizontal load acting on them. Stressed soil underneath the retaining wall will not be able to provide any frictional resistance to the retaining wall. If piles were to fail, piles would pull the soil with them. Hence, soil would not be able to provide any frictional resistance to the retaining structure. Design example 20.2 Compute the loads on piles due to the retaining wall shown. Assume lateral earth pressure coefficient at rest (K0) to be 0.5. Piles in the front are battered at 20 degrees and center piles are battered at 15 degrees to the vertical (this problem is similar to the previous example, except for the pile configuration, Fig. 20.8).
Step 1: horizontal force H acts 6.666 ft. from the base. Moment (M) = 11,000 × 6.666 lb ft. = 73,333 lb ft. per linear foot of wall. Moment in 6 linear ft. section = 6 × 73,333 = 439,998 lb ft. Assume the piles have been placed at 3-ft. intervals (Fig. 20.9).
Figure 20.8 Compute pile loads.
Figure 20.9 Assume piles are placed at 3 ft. intervals.
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Consider a section of 6 ft. for computational purposes. There are six half piles and one full pile in this section. Rows 1 and 2 are batter piles. Row 1 has twice as more piles than rows 2 and 3. Weight of soil resting on the retaining wall = 5 × 18 × 110 = 9900 lb. Weight of concrete = (2 × 9 + 2 × 18) × 160 = 8640 lb (160 pcf = concrete d ensity). Total weight = 18,540 lb per linear foot of wall. Total weight (W) in a 6-ft. section = 111,240 lb. There are four piles in the 6-ft. section. Hence, each pile carries a load of 27,810 lb. Take moments around point “E” R × L = W × L = M = H · k = 439,998 lb ft. L = 439,998/111,240 = 3.96 ft.
20.1.4 Center of gravity of piles There are six half piles and one full pile in this section (Fig. 20.10 and Fig. 20.11). There are two half piles and one full pile in row 1.
Figure 20.10 Taking moments around point “E”.
Figure 20.11 Center of gravity of piles.
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There are two half piles in row 2 and two half piles in row 3 (total piles in the section = 4). Take moments around row 3. (2 × 6 + 1 × 3)/4 = 3.75 ft. • •
As per these calculations, the center of gravity of piles is located 3.75 ft. from row 3. Moment of area of the system has to be computed from the center of gravity of piles. Distance to row 1 from center of gravity = (6 − 3.75) ft. = 2.25 ft. Distance to row 2 from center of gravity = 0.75 ft. Distance to row 3 from center of gravity = 3.75 ft. Step 2: Each pile carries a load of 27,810 lb (see step 1). • Stress developed on piles M σ = . I y Here, M = bending moment, σ = bending stress, I = moment of area, and y = distance.
Find the moment of area of piles.
20.1.5 Row 1 Moment of area of row 1 = A · r2 = A × 2.252 = 10.12 A (A = area of piles). Moment of area of row 2 = A × 0.752 = 0.56 A. (Distance is taken from the center of gravity). Moment of area of row 3 = A × 3.752 = 14.06 A. Total moment of area = 24.74 A. Bending moment (M) = 439,998 ft. lb (see step 1). Bending load on piles in row 1(σ ) = Bending load on first row piles =
M 439,998 40, 016 ⋅y = × 2.25 = lb. A I 24.74 A
40, 016 × 2 A = 80, 032 lb. A
There are two piles in the first row. Two half piles and one full pile is equivalent to two piles. Total load on pile = vertical load + load due to bending. Total load on piles in row 1 = 27,810 × 2 + 80,032 = 135,652 lb. Vertical load per pile is 27,810 lb. Since there are two piles in row 1, 27,810 lb is multiplied by 2. Step 3: row 2. Bending load on piles in row 2(σ ) = Bending load per pile =
13,339 M 439,998 ⋅y= × 0.75 = lb. A I 24.74 A
13, 339 × A = 13, 339 lb per pile. A
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There are two half piles in row 2, which amounts to one pile. Multiply by area of the pile to convert the load from lb per ft.2 to lb per pile. Total load on piles in row 2 = 27,810 – 13,339 = 14,471 lb. (Since center of gravity is on the opposite side, bending moment creates a tensile force on the pile.) Step 4: row 3. Bending load on piles in row 3(σ ) = Bending load per pile =
M 439,998 66,693 ⋅y= × 3.75 = lb. A I 24.74 A
66, 693 × A = 66, 693 lb per pile. A
Row 3 has two half piles, which amounts to one pile. Total load on piles in row 3 = 27,810 − 66,693 = −38,883 lb per pile The value 27,810 lb is the vertical load on piles and 66,693 lb is the load due to bending, which is tensile. Total force on piles in row 3 is tensile. Step 5: draw the force polygon. NM = 14,471 × tan 15 = 3,877 lb. GH = 135,652 × tan 20 = 49,373 lb (this load accounts for two piles in row 1).
Total horizontal load resisted by axial forces of piles = 49,373 + 3,877 = 53,250 lb. Additional load that needs to be resisted = 66,000 – 53,250 = 12,750 lb. This load needs to be resisted by the soil pressure acting on piles laterally. Each pile should have a lateral pile capacity of 12,750/4 lb = 3,188 lb.