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p 43
Chapter 2
Helical Foundation Systems
CHAPTER 2
HELICAL FOUNDATION SYSTEMS
The vertical effective overburden stress, q’, at
19.5 feet:
q’
= (110 lb/ft
3
)(10 ft) + ((115-62.4) lb/ft
3
)(3 ft)
+ ((130-62.4) lb/ft
3
)(6.5 ft) = 1,697 lb/ft
2
Q
u
= Design Working Load (40,000 lb) x FOS
(2) = 80,000 lb
N
q
= 1+0.56(12
Φ
)
Φ
/
54
= 42.6
(
for
Φ
= 38°)
A
h
= 80,000 / (1,697)(42.6)
A
h
= 1.1
1
ft
2
For the HP350 shaft (3.5-inch O.D.), a total helix
plate area of at least 1.11 ft
2
can be achieved
with a 10/12 double-helix plate configuration.
A
10”
= 0.47 ft
2
A
12”
= 0.71 ft
2
∑A
h
= 1.18 ft
2
Solve for the ultimate and allowable pile
capacities:
Q
u
= (1.18)(1,697)(42.6) = 85,000 lb =
85 kips
Q
a, compression
= 85,000 / 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.
Q
u
= ∑A
h
(q’N
q
)
q’
12”
= (110)(10) + (115-62.4)(3) + (130-62.4)
(5.25) = 1612 lb/ft
2
q’
10”
=
(110)(10) + (115-62.4)(3) + (130-62.4)
(7.75) = 1781
lb/ft
2
N
q, 12”
= 15.7 (for
Φ
= 30°)
N
q, 10”
= 42.6 (for
Φ
= 38°)
Q
u
= (0.71)(1612)(15.7) + (0.47)(1781)(42.6) =
53,600 lb = 53.6 kips
Q
a,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:
Q
u
= K
t
T
The equation can be rewritten to solve for torque:
T = Q
u
/ K
t
Without site-specific load testing and
determination of K
t
, we use the default value
from ICC-ES AC358 for a 3.5-inch O.D. shaft,
K
t
= 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.