Theoretical (Ultimate) and allowable bearing capacity can be assessed for the following:
For comprehensive examples of bearing capacity problems see:
 Bearing Capacity Examples
Allowable Bearing Capacity
Q_{a} = Qu
Q_{a}
= Allowable bearing capacity
(kN/m^{2) }or (lb/ft^{2}) F.S.
Where:
Q_{u} = ultimate bearing capacity (kN/m^{2})
or (lb/ft^{2})
*See below for theory F.S. = Factor of Safety
*See information on factor of safety
Terzaghi Ultimate Bearing Capacity Theory
Q_{u} = c N_{c} +
g D N_{q} + 0.5 g B N_{g}
= Ultimate bearing capacity equation for shallow
strip footings,
(kN/m^{2})
(lb/ft^{2})
Q_{u} = 1.3 c N_{c} + g D N_{q} + 0.4 g B N_{g}
= Ultimate bearing capacity equation for shallow
square footings,
(kN/m^{2})
(lb/ft^{2})
Q_{u} = 1.3 c N_{c} + g D N_{q} + 0.3 g B N_{g}
= Ultimate bearing capacity equation for shallow
circular footings,
(kN/m^{2})
(lb/ft^{2})
Where:
c = Cohesion of soil
(kN/m^{2}) (lb/ft^{2}),
g = effective
unit weight of soil
(kN/m^{3}) (lb/ft^{3}), *see note below D = depth of footing
(m) (ft),
B = width of footing
(m) (ft),
N_{c}=cotf(N_{q} – 1),
*see typical bearing capacity factors
N_{q}=e^{2}(3p/4f/2)tanf / [2
cos2(45+f/2)],
*see typical bearing capacity factors
N
_{g}=(1/2) tanf(k_{p} /cos^{2}
f  1),
*see typical bearing capacity factors
e = Napier's constant = 2.718...,
k_{p} =
passive pressure coefficient, and
f =
angle of internal
friction (degrees).
Notes: Effective unit weight, g, is the unit weight of the soil for soils above
the water table and capillary rise. For saturated soils, the effective unit
weight is the unit weight of water,
g_{w}, 9.81 kN/m^{3} (62.4 lb/ft^{3}), subtracted from the
saturated unit weight of soil.
Find more information in the foundations
section.
Meyerhof Bearing Capacity Theory Based on Standard Penetration Test Values
Q_{u} = 31.417(NB + ND)
(kN/m^{2})
(metric)
Q_{u} = NB + ND
(tons/ft^{2}) (standard)
10
10
For footing widths of 1.2 meters (4 feet) or less
Q_{a} = 11,970N
(kN/m^{2})
(metric)
Q_{a} = 1.25N
(tons/ft^{2}) (standard)
10
For footing widths of 3 meters (10 feet) or more
Q_{a} = 9,576N
(kN/m^{2})
(metric)
Q_{a} = N
(tons/ft^{2}) (standard)
10
Where:
N
= N value derived from Standard Penetration Test (SPT)
D = depth of footing (m) (ft), and
B = width of footing (m) (ft).
Note: All Meyerhof equations are for foundations bearing on clean
sands. The first equation is for ultimate bearing capacity, while the second
two are factored within the equation in order to provide an allowable
bearing capacity. Linear interpolation can be performed for footing widths
between 1.2 meters (4 feet) and 3 meters (10 feet). Meyerhof equations are
based on limiting total settlement to 25 cm (1 inch), and differential
settlement to 19 cm (3/4 inch).
Q_{ult} = Q_{p} + Q_{f}
Where:
Q_{ult} = Ultimate bearing capacity of pile,
kN (lb)
Q_{p} = Theoretical bearing capacity for tip of foundation, or end bearing,
kN (lb)
Q_{f} = Theoretical bearing capacity due to shaft friction, or adhesion between
foundation shaft and soil, kN (lb)
End Bearing (Tip) Capacity of Pile Foundation
Q_{p} = A_{p}q_{p}
Where:
Q_{p} = Theoretical bearing capacity for tip of foundation, or end bearing,
kN (lb)
A_{p} = Effective area of the tip of the pile,
m^{2 }(ft^{2})
For a circular closed end pile or circular plugged pile;
A_{p} = p(B/2)^{2}
m^{2} (ft^{2}) q_{p} = gDN_{q}
= Theoretical unit tipbearing capacity for cohesionless and silt soils, kN/m^{2 }(lb/ft^{2})
q_{p} = 9c = Theoretical unit tipbearing capacity for
cohesive soils, kN/m^{2 }(lb/ft^{2})
g = effective unit weight of soil, kN/m^{3} (lb/ft^{3}),
*See notes below D = Effective depth of pile, m (ft), where D < D_{c}, N_{q} = Bearing capacity factor for piles,
c = cohesion of soil, kN/m^{2} (lb/ft^{2}), B = diameter of pile, m (ft),
and D_{c} = critical depth for piles in loose silts or sands
m (ft). D_{c}
= 10B, for loose silts and sands D_{c}
= 15B, for medium dense silts and sands D_{c}
= 20B, for dense silts and sands
Skin (Shaft) Friction Capacity of Pile Foundation
Q_{f} = A_{f}q_{f}
for one homogeneous layer of soil
Q_{f} = pSq_{f}L for multilayers of soil
Where:
Q_{f} = Theoretical bearing capacity due to shaft friction, or adhesion between
foundation shaft and soil, kN (lb)
A_{f} = pL; Effective surface area of the pile shaft,
m^{2 }(ft^{2})
q_{f} = ks tan d = Theoretical unit friction capacity for
cohesionless soils, kN/m^{2 }(lb/ft^{2})
q_{f} = c_{A} + ks tan d = Theoretical unit friction capacity for silts, kN/m^{2 }(lb/ft^{2})
q_{f} = aS_{u} = Theoretical unit friction capacity
for cohesive soils, kN/m^{2 }(lb/ft^{2}) p = perimeter of pile
crosssection, m (ft)
for a circular pile;
p = 2p(B/2) for a square pile; p = 4B
L = Effective length of pile, m (ft)
*See Notes below
a = 1  0.1(S_{uc})^{2}
= adhesion factor, kN/m^{2} (ksf), where S_{uc}
< 48 kN/m^{2} (1 ksf)
a = 1 [0.9 + 0.3(S_{uc}  1)] kN/m^{2}, (ksf) where
S_{uc} > 48 kN/m^{2}, (1 ksf)
S_{uc} S_{uc}
= 2c = Unconfined compressive strength
, kN/m^{2} (lb/ft^{2})
c_{A} = adhesion
= c for rough concrete, rusty steel, corrugated metal
0.8c < c_{A} < c for smooth concrete
0.5c < c_{A} < 0.9c for clean steel
c = cohesion of soil,
kN/m^{2} (lb/ft^{2})
d
= external friction angle
of soil and wall contact (deg)
f
= angle of internal
friction (deg)
s = gD = effective
overburden pressure, kN/m^{2},
(lb/ft^{2})
k = lateral earth pressure coefficient
for piles
g =
effective unit weight of soil, kN/m^{3} (lb/ft^{3})
*See notes below B = diameter or width of pile, m (ft) D
= Effective depth of pile, m (ft), where D < D_{c}
D_{c} = critical depth for piles in loose silts or sands
m (ft). D_{c}
= 10B, for loose silts and sands D_{c}
= 15B, for medium dense silts and sands D_{c}
= 20B, for dense silts and sands
S = summation of differing soil layers (i.e. a_{1}
+ a_{2} + .... + a_{n})
Notes: Determining effective length requires engineering judgment. The effective length can be the pile depth minus any disturbed surface soils,
soft/ loose soils, or seasonal variation.
The effective length may also be the length of a pile segment within a
single soil layer of a multi layered soil. Effective unit weight, g, is the unit weight of the soil for soils above
the water table and capillary rise. For saturated soils, the effective unit
weight is the unit weight of water,
g_{w}, 9.81 kN/m^{3} (62.4 lb/ft^{3}),
subtracted from the saturated unit weight of soil.
************
Meyerhof Method for Determining qp and qf in Sand
Theoretical unit tipbearing capacity for driven piles in sand, when
D > 10:
B
q_{p} = 4N_{c}
tons/ft^{2} standard
Theoretical unit tipbearing capacity for drilled piles in sand:
q_{p} = 1.2N_{c}
tons/ft^{2} standard
Theoretical unit frictionbearing capacity for driven piles in sand:
q_{f} = N
tons/ft^{2} standard
50
Theoretical unit frictionbearing capacity for drilled piles in sand:
q_{f} = N
tons/ft^{2} standard
100
Where:
D = pile embedment depth, ft
B = pile diameter, ft
N_{c} = C_{n}(N)
C_{n} = 0.77 log 20
s
N = NValue from SPT test
s = gD =
effective overburden stress at pile embedment depth,
tons/ft^{2}
= (g  g_{w})D
= effective stress if below water table,
tons/ft^{2}
g =
effective unit weight of soil,
tons/ft^{3}
g_{w }=
0.0312 tons/ft^{3} = unit weight of water
Example #1: Determine allowable bearing capacity and width for a
shallow strip footing on cohesionless silty sand and gravel soil. Loose
soils were encountered in the upper 0.6 m (2 feet) of building subgrade. Footing must withstand a
144 kN/m^{2} (3000 lb/ft^{2}) building pressure.
Given

bearing pressure from building = 144 kN/m^{2 (}3000 lbs/ft^{2})

unit weight of soil,
g = 21 kN/m^{3} (132 lbs/ft^{3}) *from soil testing,
see typical
g values

Cohesion, c = 0 *from soil testing,
see typical c values

angle of Internal Friction,
f = 32 degrees *from soil testing,
see typical
f values

footing depth, D = 0.6 m (2 ft)
*because loose soils in upper soil strata
Solution
Try a minimal footing width, B = 0.3 m (B = 1 foot).
Use a factor of safety, F.S = 3. Three is typical for this type of application. See
factor of safety for more
information.
Determine bearing capacity factors N_{g}, N_{c} and
N_{q}. See typical bearing capacity
factors relating to the soils' angle of internal friction.
 N_{g} = 22
 N_{c} = 35.5
 N_{q} = 23.2
Solve for ultimate bearing capacity,
Q_{u} = c N_{c} +
g D N_{q} + 0.5
g B N_{g} *strip footing eq.
Q_{u} = 0(35.5) + 21 kN/m^{3}(0.6m)(23.2) + 0.5(21
kN/m^{3})(0.3 m)(22)
metric Q_{u} = 362 kN/m^{2}
Q_{u} = 0(35.5) + 132lbs/ft^{3}(2ft)(23.2) + 0.5(132lbs/ft^{3})(1ft)(22)
standard Q_{u} = 7577 lbs/ft^{2}
Solve for allowable bearing capacity,
Q_{a} = Qu F.S.
Q_{a} = 362 kN/m^{2} = 121 kN/m^{2}
not o.k.
metric
3 Q_{a} = 7577lbs/ft^{2} = 2526
lbs/ft^{2}
not o.k.
standard
3
Since Q_{a} < required 144 kN/m^{2} (3000 lbs/ft^{2}) bearing pressure, increase footing width, B or
foundation depth, D to increase bearing capacity.
Try footing width, B = 0.61 m (B = 2 ft).
Q_{u} = 0 + 21 kN/m^{3}(0.61 m)(23.2) + 0.5(21 kN/m^{3})(0.61
m)(22)
metric Q_{u} = 438 kN/m^{2}
Q_{u} = 0 + 132 lbs/ft^{3}(2 ft)(23.2) + 0.5(132 lbs/ft^{3})(2 ft)(22)
standard Q_{u} = 9029 lbs/ft^{2}
Q_{a} = 438 kN/m^{2} = 146 kN/m^{2}
Q_{a} > 144 kN/m^{2
}o.k.
metric
3
Q_{a} = 9029 lbs/ft^{2} = 3010 lbs/ft^{2}
Q_{a} > 3000 lbs/ft^{2
}o.k.
standard
3
Conclusion
Footing shall be 0.61 meters (2 feet) wide at a depth of 0.61 meters (2 feet) below ground surface.
Many engineers neglect the depth factor (i.e. D N_{q}
= 0) for shallow foundations. This inherently increases the factor of
safety. Some site conditions that may negatively effect the depth factor are
foundations established at depths equal to or less than 0.3 meters (1 feet)
below the ground surface, placement of foundations on fill, and disturbed/
fill soils located above or to the sides of foundations.
********************************
Example #2: Determine allowable bearing capacity of a shallow,
0.3 meter (12inch) square
isolated footing bearing on saturated cohesive soil. The frost penetration
depth is 0.61 meter (2 feet). Structural parameters require the foundation
to withstand 4.4 kN (1000 lbs) of force on a 0.3 meter (12inch) square
column.
Given

bearing pressure from building column = 4.4 kN/ (0.3 m x 0.3 m) =
48.9 kN/m^{2}

bearing pressure from building column = 1000 lbs/ (1 ft x 1 ft) = 1000 lbs/ft^{2}

unit weight of saturated soil,
g_{sat}= 20.3 kN/m^{3} (129 lbs/ft^{3})
*see typical
g values

unit weight of water,
g_{w}=
9.81 kN/m^{3} (62.4 lbs/ft^{3})
*constant

Cohesion, c = 21.1 kN/m^{2} (440 lbs/ft^{2})
*from soil testing, see typical
c values

angle of Internal Friction,
f = 0 degrees
*from soil testing, see typical
f values

footing width, B = 0.3 m (1 ft)
Solution
Try a footing depth, D = 0.61 meters (2 feet), because foundation should
be below frost depth.
Use a factor of safety, F.S = 3. See
factor of safety for more
information.
Determine bearing capacity factors N_{g}, N_{c} and
N_{q}. See typical bearing capacity
factors relating to the soils' angle of internal friction.
 N_{g} = 0
 N_{c} = 5.7
 N_{q} = 1
Solve for ultimate bearing capacity,
Q_{u }= 1.3c N_{c} +
g D N_{q} + 0.4
g B N_{g} *square footing eq.
Q_{u} =1.3(21.1kN/m^{2})5.7+(20.3kN/m^{3}9.81kN/m^{3})(0.61m)1+0.4(20.3kN/m^{3}9.81kN/m^{3})(0.3m)0
Q_{u} = 163 kN/m^{2 }
metric
Q_{u} = 1.3(440lbs/ft^{2})(5.7) + (129lbs/ft^{3} 
62.4lbs/ft^{3})(2ft)(1) + 0.4(129lbs/ft^{3} 
62.4lbs/ft^{3})(1ft)(0)
Q_{u} = 3394 lbs/ft^{2 }
standard
Solve for allowable bearing capacity,
Q_{a} = Qu F.S.
Q_{a} = 163 kN/m^{2} =
54 kN/m^{2 } Q_{a}
> 48.9 kN/m^{2}
o.k. metric
3 Q_{a} =
3394lbs/ft^{2} = 1130 lbs/ft^{2
} Q_{a} >
1000 lbs/ft^{2}
o.k. standard
3
Conclusion
The 0.3 meter (12inch) isolated square footing shall be 0.61 meters (2 feet) below the ground surface.
Other considerations may be required for foundations bearing
on moisture sensitive clays, especially for lightly loaded structures such as in
this example. Sensitive clays could expand and contract, which could cause
structural damage. Clay used as bearing soils may require mitigation such as
heavier loads, subgrade removal and replacement below the foundation, or
moisture control within the subgrade.
********************************
Example #3: Determine allowable bearing capacity and width for a
foundation using the Meyerhof Method. Soils consist of poorly graded sand. Footing must withstand a
144 kN/m^{2} (1.5 tons/ft^{2}) building pressure.
Given
 bearing pressure from building = 144 kN/m^{2 (}1.5 tons/ft^{2})
 N Value, N = 10 at 0.3 m (1 ft) depth *from
SPT soil testing
 N Value, N = 36 at 0.61 m (2 ft) depth *from
SPT soil testing
 N Value, N = 50 at 1.5 m (5 ft) depth *from
SPT soil testing
Solution
Try a minimal footing width, B = 0.3 m (B = 1 foot) at a depth, D = 0.61
meter (2 feet). Footings for single family residences are typically 0.3m (1
ft) to 0.61m (2ft) wide. This depth was selected because soil density
greatly increases (i.e. higher Nvalue) at a depth of 0.61 m (2 ft).
Use a factor of safety, F.S = 3. Three is typical for this type of application. See
factor of safety for more
information.
Solve for ultimate bearing capacity
Q_{u} = 31.417(NB + ND)
(kN/m^{2}) (metric)
Q_{u} = NB + ND
(tons/ft^{2}) (standard)
10
10
Q_{u} = 31.417(36(0.3m) + 36(0.61m)) = 1029 kN/m^{2}
(metric)
Q_{u} = 36(1 ft) + 36(2 ft)
= 10.8 tons/ft^{2}
(standard) 10
10
Solve for allowable bearing capacity,
Q_{a} = Qu F.S.
Q_{a} = 1029 kN/m^{2} = 343 kN/m^{2} Q_{a}
> 144 kN/m^{2} o.k. (metric)
3 Q_{a} = 10.8 tons/ft^{2} = 3.6
tons/ft^{2} Q_{a} > 1.5 tons/ft^{2} o.k.
(standard)
3
Conclusion
Footing shall be 0.3 meters (1 feet) wide at a depth of 0.61 meters (2 feet) below
the ground surface. A footing width of only 0.3 m (1
ft) is most likely insufficient for the structural engineer when designing
the footing with the building pressure in this problem.
********************************
Example #4: Determine allowable bearing capacity and diameter of a single driven pile. Pile must withstand a
66.7 kN (15 kips)
vertical load.
Given
 vertical column load = 66.7 kN (15 kips or 15,000 lb)
 homogeneous soils in upper 15.2 m (50 ft); silty soil
 unit weight,
g = 19.6 kN/m^{3} (125 lbs/ft^{3})
*from soil testing, see typical
g values
 cohesion, c = 47.9 kN/m^{2} (1000 lb/ft^{2})
*from soil testing, see typical
c values
 angle of internal friction,
f = 30 degrees *from soil testing,
see typical
f values
 Pile Information
Solution
Try a pile depth, D = 1.5 meters (5 feet) Try pile diameter,
B = 0.61 m (2 ft)
Use a factor of safety, F.S = 3. Smaller factors of safety are sometimes used if piles are load tested, or the engineer has sufficient
experience with the regional soils.
Determine ultimate end bearing of pile,
Q_{p} = A_{p}q_{p}
A_{p} =
p(B/2)^{2} =
p(0.61m/2)^{2 }= 0.292 m^{2}
metric
A_{p}
=
p(B/2)^{2} = p(2ft/2)^{2} = 3.14 ft^{2
} standard
q_{p} =
gDN_{q}
g = 19.6 kN/m^{3} (125 lbs/ft^{3});
given soil unit weight
f = 30 degrees; given soil
angle of internal friction B = 0.61 m (2 ft); trial pile width
D = 1.5 m (5 ft); trial depth, may need to increase or decrease depending on
capacity check to see if D < D_{c}
Dc = 15B = 9.2 m (30 ft); critical depth for
medium dense silts. If D > D_{c}, then
use D_{c}
N_{q} = 25; Meyerhof bearing capacity
factor for driven piles, based on
f
q_{p} = 19.6 kN/m^{3}(1.5 m)25
= 735 kN/m^{2}
metric
q_{p} =
125 lb/ft^{3}(5 ft)25
= 15,625 lb/ft^{2}
standard
Q_{p} = A_{p}q_{p }= (0.292 m^{2})(735 kN/m^{2})
= 214.6 kN
metric Q_{p} = A_{p}q_{p
}= (3.14 ft^{2})(15,625 lb/ft^{2})
= 49,063 lb
standard
Determine ultimate friction capacity of pile,
Q_{f} = A_{f}q_{f}
A_{f} = pL
p = 2p(0.61m/2) = 1.92 m
metric p = 2p(2 ft/2) = 6.28 ft
standard L = D = 1.5 m (5 ft); length and depth used interchangeably. check
D_{c}
as above
A_{f} = 1.92 m(1.5 m) = 2.88 m^{2}
metric A_{f} = 6.28 ft(5 ft) = 31.4 ft^{2}
standard
q_{f} = c_{A} + ks tan
d = c_{A} + kgD tan
d
k = 0.5; lateral earth pressure
coefficient
for piles, value chosen from Broms low density steel
g = 19.6 kN/m^{3
}(125 lb/ft^{3}); given effective soil unit weight. If water table, then
g 
g_{w}
D = L = 1.5 m (5 ft); pile length. Check to see if D < D_{c}
Dc = 15B = 9.2 m (30 ft); critical depth for medium dense silts.
If D > D_{c}, then use D_{c}
d =
20 deg; external friction angle,
equation chosen from Broms steel piles B = 0.61 m (2 ft); selected pile diameter
c_{A} = 0.5c; for clean steel. See adhesion in
pile theories above. = 24 kN/m^{2} (500 lb/ft^{2})
q_{f} = 24 kN/m^{2} + 0.5(19.6 kN/m^{3})(1.5m)tan
20 = 29.4 kN/m^{2} metric
q_{f} = 500 lb/ft^{2} + 0.5(125 lb/ft^{3})(5ft)tan 20 =
614 lb/ft^{2} standard
Q_{f} =
A_{f}q_{f} = 2.88 m^{2}(29.4 kN/m^{2})
= 84.7 kN
metric Q_{f} =
A_{f}q_{f} = 31.4 ft^{2}(614 lb/ft^{2})
= 19,280 lb
standard
Determine ultimate pile capacity,
Q_{ult} = Q_{p} + Q_{f}
Q_{ult} = 214.6 kN + 84.7 kN
= 299.3 kN
metric Q_{ult} = 49,063 lb + 19,280 lb = 68,343 lb
standard
Solve for allowable bearing capacity,
Q_{a} = Qult F.S.
Q_{a}
= 299.3 kN = 99.8 kN; Q_{a}
> applied load (66.7 kN) o.k. metric
3 Q_{a} = 68,343 lbs =
22,781 lbs Q_{a}
> applied load (15 kips) o.k. standard
3
Conclusion
A 0.61 m (2 ft) steel pile shall be plugged and driven 1.5 m (5 feet) below the ground surface.
Many engineers neglect the skin friction within the upper 1 to 5 feet of subgrade due to seasonal variations or soil disturbance.
Seasonal variations may include freeze/ thaw or effects from water. The end
bearing alone (neglect skin friction) is sufficient for this case. Typical methods for
increasing the pile capacity are increasing the pile diameter or increasing the embedment depth of the pile.
********************************
Example #5: Determine allowable bearing capacity and diameter of a single driven pile. Pile must withstand a
66.7 kN (15 kips)
vertical load.
Given

vertical column load = 66.7 kN (15 kips or 15,000 lb)

upper 1.5 m (5 ft) of soil is a medium dense gravelly sand
 unit weight,
g = 19.6 kN/m^{3} (125 lbs/ft^{3})
*from soil testing, see typical
g values
 cohesion, c = 0
*from soil testing, see typical
c values
 angle of internal friction,
f = 30 degrees *from soil testing,
see typical
f values

soils below 1.5 m (5 ft) of soil is a stiff silty clay
 unit weight,
g = 18.9 kN/m^{3 }(120 lbs/ft^{3})
 cohesion, c = 47.9 kN/m^{2} (1000 lb/ft^{2})
 angle of internal friction,
f = 0 degrees

Pile Information
Solution
Try a pile depth, D = 2.4 meters (8 feet) Try pile diameter,
B = 0.61 m (2 ft)
Use a factor of safety, F.S = 3. Smaller factors of safety are sometimes used if piles are load tested, or the engineer has sufficient
experience with the regional soils.
Determine ultimate end bearing of pile,
Q_{p} = A_{p}q_{p}
A_{p} =
p(B/2)^{2} =
p(0.61m/2)^{2 }= 0.292 m^{2}
metric A_{p}
=
p(B/2)^{2} = p(2ft/2)^{2} = 3.14 ft^{2
} standard
q_{p} = 9c = 9(47.9 kN/m^{2})
= 431.1 kN/m^{2}
metric q_{p} =
9c = 9(1000 lb/ft^{2})
= 9000 lb/ft^{2}
standard
Q_{p} = A_{p}q_{p }= (0.292 m^{2})(431.1 kN/m^{2})
= 125.9 kN
metric Q_{p} = A_{p}q_{p
}= (3.14 ft^{2})(9000 lb/ft^{2})
= 28,260 lb
standard
Determine ultimate friction capacity of pile,
Q_{f} = pSq_{f}L
p = 2p(0.61m/2) = 1.92 m
metric p = 2p(2 ft/2) = 6.28 ft
standard
upper 1.5 m (5 ft) of soil
q_{f}L = [ks tan
d]L
= [kgD tan
d]L
k = 1.5; lateral earth pressure
coefficient
for piles, value chosen from Broms low density timber
g = 19.6 kN/m^{3
}(125 lb/ft^{3}); given effective soil unit weight. If water table, then
g 
g_{w}
D = L = 1.5 m (5 ft); segment of pile within this soil strata. Check to see if
D < D_{c} D_{c} = 15B = 9.2 m (30 ft); critical depth for medium
dense sands. This assumption is conservative, because the soil is
gravelly, and this much soil unit weight for a sand would indicate dense
soils. If D > D_{c}, then use D_{c}
d =
f(2/3) = 20 deg;
external friction angle,
equation chosen from Broms timber piles B = 0.61 m (2 ft); selected pile diameter
f = 30 deg; given soil angle of
internal friction
q_{f}L = [1.5(19.6 kN/m^{3})(1.5m)tan (20)]1.5 m = 24.1 kN/m
metric q_{f}L = [1.5(125 lb/ft^{3})(5ft)tan (20)]5 ft =
1706 lb/ft standard
soils below 1.5 m (5 ft) of subgrade
q_{f}L =
aS_{u}
S_{uc}
= 2c = 95.8 kN/m^{2 }(2000 lb/ft^{2});
unconfined compressive strength
c = 47.9 kN/m^{2 }(1000 lb/ft^{2});
cohesion from soil testing (given)
a = 1 [0.9 + 0.3(S_{uc}  1)]
= 0.3; because S_{uc} > 48 kN/m^{2}, (1 ksf) S_{uc}
L = 0.91 m (3 ft); segment of pile within this soil strata
q_{f}L = [0.3(95.8 kN/m^{2})]0.91 m = 26.2 kN/m
metric q_{f}L = [0.3(2000 lb/ft^{2})]3 ft =
1800 lb/ft
standard
ultimate friction capacity of combined soil layers
Q_{f} = pSq_{f}L
Q_{f} = 1.92 m(24.1 kN/m + 26.2 kN/m) = 96.6 kN
metric Q_{f} = 6.28 ft(1706 lb/ft + 1800 lb/ft) =
22,018 lb
standard
Determine ultimate pile capacity,
Q_{ult} = Q_{p} + Q_{f}
Q_{ult} = 125.9 kN + 96.6 kN
= 222.5 kN
metric Q_{ult} = 28,260 lb + 22,018 lb = 50,278 lb
standard
Solve for allowable bearing capacity,
Q_{a} = Qult F.S.
Q_{a}
= 222.5 kN = 74.2 kN; Q_{a}
> applied load (66.7 kN) o.k. metric
3 Q_{a} = 50,275 lbs =
16,758 lbs Q_{a}
> applied load (15 kips) o.k. standard
3
Conclusion
Wood pile shall be driven 8 feet below the ground surface.
Many engineers neglect the skin friction within the upper 1 to 5 feet of subgrade due to seasonal variations or soil disturbance.
Seasonal variations may include freeze/ thaw or effects from water. Notice how the soil properties within the pile tip location is used in the end
bearing calculations. End bearing should also consider the soil layer(s) directly beneath this layer. Engineering judgment or a change in design is warranted
if subsequent soil layers are weaker than the soils within the vicinity of
the pile tip. Typical methods for
increasing the pile capacity are increasing the pile diameter or increasing the embedment depth of the pile.
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