Helical piles are proposed to support a new structure. The proposed pile layout is shown on the foundation plan along with a design working load of 30 kips per pile with a Factor of Safety (FOS) = 2. Preliminary product selection suggests that the HP288 helical pile is the best fit for this load condition with an ultimate torque rated capacity of 71.1 kips. The allowable torque rated capacity would then be 35.5 kips with a FOS = 2. A geotechnical investigation was completed for the project and the soil profile is shown in Figure 35.
The helical piles will penetrate the upper fill and medium stiff clay to bear within the deeper very stiff clay. With the helix blades bearing entirely within the very stiff clay soil below a depth of 15 feet, we can use the equation from Section 2.7.1 for purely cohesive soils with Ф = 0:
Qu = ∑Ah(9c)
Solve for the required helix plate area:Ah = Qu/9c
Qu = Design Working Load (30,000 lb) x FOS (2) = 60,000 lbc = 3,000 lb/ft2Ah = 60,000 / (9)(3,000)
Ah = 2.22 ft2
Helix plate areas for the various shaft sizes can be found in Appendix A. For the HP288 shaft (2.875-inch O.D.), a total helix plate area of 2.22 ft2 can be most efficiently achieved with a 10/12/14 triple-helix plate configuration.
A10” = 0.50 ft2
A12” = 0.74 ft2
A14” = 1.02 ft2
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∑Ah = 2.26 ft2
Solve for the ultimate and allowable pile capacities:
Qu = (2.26)(9)(3,000) = 61,000 lb = 61 kips
The allowable pile capacity,
Qa = Qu / FOS
Qa = 61,000 / 2 = 30,500 lb = 30.5 kips…….OK
Determine the required final installation torque in accordance with the equations and procedures of Section 2.7.3:
Qu = KtT
The equation can be rewritten to solve for torque:
T = Qu / Kt
Without site-specific load testing and determination of Kt, we use the default value from ICC-ES AC358 for a 2.875-inch O.D. shaft, Kt = 9 ft-1:
T = 60,000 / 9 = 6,667 ft-lb
Install the helical piles to a final installation torque of at least 6,700 ft-lb.
Grain conveyor towers will be constructed at an ethanol facility. The towers will be designed with four support legs, each leg designed for working loads of 40 kips in compression and 15 kips in tension/uplift. A FOS = 2 is required for both the compression and uplift pile capacities. A geotechnical exploration was completed for the project and the soil profile is shown in Figure 36. Groundwater was encountered at a depth of 10 feet below the surface. Preliminary product selection suggests that the HP349 helical pile is best suited to support the proposed loads. The HP349 has an ultimate torque rated capacity of 91 kips and an allowable torque rated capacity of 45.5 kips with a FOS = 2. The helical piles will be embedded into the dense sand as shown in figure on the right.
For purely granular (frictional) soils with c = 0, the ultimate pile capacity can be determined from equation:
Qu = ∑Ah(q’Nq)
Solve for the required helix plate area:Ah = Qu/q’Nq
The helix plates should be embedded several plate diameters into the dense sand to provide uplift resistance. This depth depends upon the pile load. We can fine tune the embedment depth at a later point, but for an uplift load of 15 kips, we’ll consider a minimum helix plate embedment of 3 diameters. A pile with an ultimate capacity of 80 kips often has three helix plates on the lead section. A 10/12/14 lead has a distance of 5.5 feet between the uppermost and bottommost plates. With these parameters in mind, we’ll choose a trial depth of:
13 feet + 3.5 feet (depth of 14-inch plate into dense sand)+2.75 feet (half the distance between bearing plates) = 19.5 feet.
The vertical effective overburden stress, q’, at 19.5 feet:
q’ = (110 lb/ft3)(10 ft) + ((115-62.4) lb/ft3)(3 ft) + ((130-62.4) lb/ft3)(6.5 ft) = 1,697 lb/ft2
Qu = Design Working Load (40,000 lb) x FOS (2) = 80,000 lb
Nq = 1+0.56(12Ф)Ф/54 = 42.6 (for Ф = 38 ̊)
Ah = 80,000 / (1,697)(42.6)
Ah = 1.11 ft2
For the HP349 shaft (3.5-inch O.D.), a total helix plate area of at least 1.11 ft2 can be achieved with a 10/12 double-helix plate configuration.
A10” = 0.47 ft2
A12” = 0.71 ft2
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∑Ah = 1.18 ft2
Solve for the ultimate and allowable pile capacities:
Qu = (1.18)(1,697)(42.6) = 85,000 lb = 85 kips
Qa, compression = 85000 / 2 = 42,500 lb = 42.5 kips……….OK
To maintain the average vertical effective overburden stress at a depth of 19.5 feet, the 12-inch blade would be installed to a depth of 18.25 feet and the 10-inch blade would be installed to a depth of 20.75 feet. The upper helix plate is now 5.25 feet below the loose sand to dense sand interface. With this depth of embedment, we would expect the allowable uplift capacity to be similar to the allowable compressive capacity.
To be very conservative and consider that the loose sand above the 12-inch plate could have some effect on the uplift capacity, we could model the soil strength (friction angle) above the 12-inch plate to represent the loose sand.
Qu = ∑Ah(q’Nq)
q’12” = (110)(10) + (115-62.4)(3) + (130-62.4)(5.25) = 1612 lb/ft2
q’10” = (110)(10) + (115-62.4)(3) + (130-62.4)(7.75) = 1781 lb/ft2
Nq, 12” = 15.7 (for Ф = 30 ̊)
Nq, 10” = 42.6 (for Ф = 38 ̊)
Qu = (0.71)(1612)(15.7) + (0.47)(1781)(42.6) = 53,600 lb = 53.6 kips
Qa,uplift = 53,600 / 2 = 26,800 lb or 26.8 kips……...OK
Determine the required final installation torque in accordance with the equations and procedures of Section 2.7.3:
Qu = KtT
The equation can be rewritten to solve for torque:
T = Qu / Kt
Without site-specific load testing and determination of Kt, we use the default value from ICC-ES AC358 for a 3.5-inch O.D. shaft, Kt = 7 ft-1:
T = 80,000 / 7 = 11,428 ft-lb
Install the helical piles to a final installation torque of at least 11,500 ft-lb.