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Bearing Capacity Technical Guidance

Bearing capacity of soil is the value of the average contact pressure between the foundation and the soil which will produce shear failure in the soil. Ultimate bearing capacity is the theoretical maximum pressure which can be supported without failure. Allowable bearing capacity is what is used in geotechnical design, and is the ultimate bearing capacity divided by a factor of safety.

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

Qa   =    Qu                                 Qa = Allowable bearing capacity  (kN/m2) or (lb/ft2)
F.S.

Where:

Qu = ultimate bearing capacity (kN/m2) or (lb/ft2)                *See below for theory
F.S. = Factor of Safety                                    *See information on factor of safety

## Ultimate Bearing Capacity for Shallow Foundations

#### Terzaghi Ultimate Bearing Capacity Theory

Qu = c Nc + g D Nq + 0.5 g B Ng
= Ultimate bearing capacity equation for shallow strip footings, (kN/m2) (lb/ft2)

Qu = 1.3 c Nc + g D Nq + 0.4 g B Ng
= Ultimate bearing capacity equation for shallow square footings, (kN/m2) (lb/ft2)

Qu = 1.3 c Nc + g D Nq + 0.3 g B Ng
= Ultimate bearing capacity equation for shallow circular footings, (kN/m2) (lb/ft2)

Where:

c = Cohesion of soil (kN/m2) (lb/ft2),
g = effective unit weight of soil (kN/m3) (lb/ft3),  *see note below
D = depth of footing (m) (ft),
B = width of footing (m) (ft),
Nc=cotf(Nq � 1),                                             *see typical bearing capacity factors
Nq=e2(3p/4-f/2)tanf / [2 cos2(45+f/2)],         *see typical bearing capacity factors
N g=(1/2) tanf(kp /cos2 f - 1),                         *see typical bearing capacity factors
e = Napier's constant = 2.718...,
kp = 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, gw, 9.81 kN/m3 (62.4 lb/ft3), 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

Qu = 31.417(NB + ND)      (kN/m2)                        (metric)

Qu =   NB    +   ND            (tons/ft2)                       (standard)
10           10

For footing widths of 1.2 meters (4 feet) or less

Qa =   11,970N               (kN/m2)                        (metric)

Qa =   1.25N                   (tons/ft2)                       (standard)
10

For footing widths of 3 meters (10 feet) or more

Qa =   9,576N                 (kN/m2)                        (metric)

Qa =   N                          (tons/ft2)                       (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).

## Ultimate Bearing Capacity for Deep Foundations (Pile)

Qult = Qp + Qf

Where:

Qult = Ultimate bearing capacity of pile, kN (lb)
Qp = Theoretical bearing capacity for tip of foundation, or end bearing, kN (lb)
Qf = Theoretical bearing capacity due to shaft friction, or adhesion between foundation shaft and soil, kN (lb)

#### End Bearing (Tip) Capacity of Pile Foundation

Qp = Apqp

Where:

Qp = Theoretical bearing capacity for tip of foundation, or end bearing, kN (lb)
Ap = Effective area of the tip of the pile, m2 (ft2)
For a circular closed end pile or circular plugged pile; Ap = p(B/2)2 m2 (ft2)
qp = gDNq
= Theoretical unit tip-bearing capacity for cohesionless and silt soils, kN/m2 (lb/ft2)
qp = 9c
= Theoretical unit tip-bearing capacity for cohesive soils, kN/m2 (lb/ft2)
g = effective unit weight of soil, kN/m3 (lb/ft3),                                *See notes below
D = Effective depth of pile, m (ft), where D < Dc,
Nq = Bearing capacity factor for piles,
c = cohesion of soil, kN/m2 (lb/ft2),
B = diameter of pile, m (ft), and
Dc = critical depth for piles in loose silts or sands m (ft).
Dc = 10B, for loose silts and sands
Dc = 15B, for medium dense silts and sands
Dc = 20B, for dense silts and sands

#### Skin (Shaft) Friction Capacity of Pile Foundation

Qf = Afqf       for one homogeneous layer of soil

Qf = pSqfL    for multi-layers of soil

Where:

Qf = Theoretical bearing capacity due to shaft friction, or adhesion between foundation shaft and soil, kN (lb)
Af = pL; Effective surface area of the pile shaft, m2 (ft2)
qf = ks tan d = Theoretical unit friction capacity for cohesionless soils, kN/m2 (lb/ft2)
qf = cA + ks tan d = Theoretical unit friction capacity for silts, kN/m2 (lb/ft2)
qf = aSu = Theoretical unit friction capacity for cohesive soils, kN/m2 (lb/ft2)
p = perimeter of pile cross-section, 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(Suc)2 = adhesion factor, kN/m2 (ksf), where Suc < 48 kN/m2 (1 ksf)
a =    1    [0.9 + 0.3(Suc - 1)] kN/m2, (ksf) where Suc > 48 kN/m2, (1 ksf)
Suc
Suc = 2c = Unconfined compressive strength , kN/m2 (lb/ft2)
= c for rough concrete, rusty steel, corrugated metal
0.8c < cA < c for smooth concrete
0.5c < cA < 0.9c for clean steel
c = cohesion of soil, kN/m2 (lb/ft2)
d = external friction angle of soil and wall contact (deg)
f = angle of internal friction (deg)
s = gD = effective overburden pressure, kN/m2, (lb/ft2)
k = lateral earth pressure coefficient for piles
g = effective unit weight of soil, kN/m3 (lb/ft3)                    *See notes below
B = diameter or width of pile, m (ft)
D = Effective depth of pile, m (ft), where D < Dc
Dc = critical depth for piles in loose silts or sands m (ft).
Dc = 10B, for loose silts and sands
Dc = 15B, for medium dense silts and sands
Dc = 20B, for dense silts and sands
S = summation of differing soil layers (i.e. a1 + a2 + .... + an)

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, gw, 9.81 kN/m3 (62.4 lb/ft3), subtracted from the saturated unit weight of soil.

************

Meyerhof Method for Determining  qp and qf in Sand

Theoretical unit tip-bearing capacity for driven piles in sand, when  D  > 10:
B
qp = 4Nc  tons/ft2                      standard

Theoretical unit tip-bearing capacity for drilled piles in sand:

qp = 1.2Nc  tons/ft2                   standard

Theoretical unit friction-bearing capacity for driven piles in sand:

qf =  N   tons/ft2                         standard
50

Theoretical unit friction-bearing capacity for drilled piles in sand:

qf =  N   tons/ft2                         standard
100

Where:

D = pile embedment depth, ft
B = pile diameter, ft
Nc = Cn(N)
Cn = 0.77 log  20
s

N = N-Value from SPT test
s = gD = effective overburden stress at pile embedment depth,  tons/ft2
= (g - gw)D = effective stress if below water table,  tons/ft2
g = effective unit weight of soil,  tons/ft3
gw = 0.0312 tons/ft3 = unit weight of water

## Examples for determining allowable bearing capacity

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/m2 (3000 lb/ft2) building pressure.

Given

• bearing pressure from building = 144 kN/m2 (3000 lbs/ft2)

• unit weight of soil, g = 21 kN/m3 (132 lbs/ft3)  *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 Ng, Nc and Nq. See typical bearing capacity factors relating to the soils' angle of internal friction.

• Ng = 22
• Nc = 35.5
• Nq = 23.2

Solve for ultimate bearing capacity,

Qu = c Nc + g D Nq + 0.5 g B Ng                                *strip footing eq.

Qu = 0(35.5) + 21 kN/m3(0.6m)(23.2) + 0.5(21 kN/m3)(0.3 m)(22)                 metric
Qu = 362 kN/m2

Qu = 0(35.5) + 132lbs/ft3(2ft)(23.2) + 0.5(132lbs/ft3)(1ft)(22)                          standard
Qu = 7577 lbs/ft2

Solve for allowable bearing capacity,

Qa =   Qu
F.S.

Qa =  362 kN/m2  = 121 kN/m2                                          not o.k.                 metric
3
Qa =  7577lbs/ft2  = 2526 lbs/ft2                                          not o.k.                 standard
3

Since Qa < required 144 kN/m2 (3000 lbs/ft2) bearing pressure, increase footing width, B or foundation depth, D to increase bearing capacity.

Try footing width, B = 0.61 m (B = 2 ft).

Qu = 0 + 21 kN/m3(0.61 m)(23.2) + 0.5(21 kN/m3)(0.61 m)(22)                      metric
Qu = 438 kN/m2

Qu = 0 + 132 lbs/ft3(2 ft)(23.2) + 0.5(132 lbs/ft3)(2 ft)(22)                                standard
Qu = 9029 lbs/ft2

Qa =   438 kN/m2   = 146 kN/m2          Qa > 144 kN/m2            o.k.                   metric
3

Qa =   9029 lbs/ft2 = 3010 lbs/ft2           Qa > 3000 lbs/ft2           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 Nq = 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 (12-inch) 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 (12-inch) square column.

Given

• bearing pressure from building column = 4.4 kN/ (0.3 m x 0.3 m) = 48.9 kN/m2

• bearing pressure from building column = 1000 lbs/ (1 ft x 1 ft) = 1000 lbs/ft2

• unit weight of saturated soil, gsat= 20.3 kN/m3 (129 lbs/ft3)            *see typical g values

• unit weight of water, gw= 9.81 kN/m3 (62.4 lbs/ft3)                        *constant

• Cohesion, c = 21.1 kN/m2 (440 lbs/ft2)                *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 Ng, Nc and Nq. See typical bearing capacity factors relating to the soils' angle of internal friction.

• Ng = 0
• Nc = 5.7
• Nq = 1

Solve for ultimate bearing capacity,

Qu = 1.3c Nc + g D Nq + 0.4 g B Ng                                  *square footing eq.

Qu =1.3(21.1kN/m2)5.7+(20.3kN/m3-9.81kN/m3)(0.61m)1+0.4(20.3kN/m3-9.81kN/m3)(0.3m)0
Qu = 163 kN/m2                                                                                       metric

Qu = 1.3(440lbs/ft2)(5.7) + (129lbs/ft3 - 62.4lbs/ft3)(2ft)(1) + 0.4(129lbs/ft3 - 62.4lbs/ft3)(1ft)(0)
Qu = 3394 lbs/ft2                                                                                      standard

Solve for allowable bearing capacity,

Qa =   Qu
F.S.

Qa =   163 kN/m2   = 54 kN/m2             Qa > 48.9 kN/m2         o.k.          metric
3
Qa =    3394lbs/ft2   = 1130 lbs/ft2         Qa > 1000 lbs/ft2         o.k.          standard
3

Conclusion

The 0.3 meter (12-inch) 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/m2 (1.5 tons/ft2) building pressure.

Given

• bearing pressure from building = 144 kN/m2 (1.5 tons/ft2)
• 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 N-value) 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

Qu = 31.417(NB + ND)      (kN/m2)                          (metric)

Qu =   NB    +   ND            (tons/ft2)                         (standard)
10           10

Qu = 31.417(36(0.3m) + 36(0.61m)) = 1029 kN/m2    (metric)

Qu =   36(1 ft)   +   36(2 ft)   = 10.8 tons/ft2                 (standard)
10                10

Solve for allowable bearing capacity,

Qa =   Qu
F.S
.

Qa =  1029 kN/m2  = 343 kN/m2    Qa > 144 kN/m2      o.k.    (metric)
3
Qa =  10.8 tons/ft2  = 3.6 tons/ft2     Qa > 1.5 tons/ft2      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/m3 (125 lbs/ft3) *from soil testing, see typical g values
• cohesion, c = 47.9 kN/m2 (1000 lb/ft2)   *from soil testing, see typical c values
• angle of internal friction, f = 30 degrees   *from soil testing, see typical f values

• Pile Information
• driven
• steel
• plugged end

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,

Qp = Apqp

Ap = p(B/2)2 = p(0.61m/2)2 = 0.292 m2                                      metric
Ap = p(B/2)2 = p(2ft/2)2 = 3.14 ft2                                                        standard

qp = gDNq

g = 19.6 kN/m3 (125 lbs/ft3); 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 < Dc
Dc = 15B = 9.2 m (30 ft); critical depth for medium dense silts.
If D > Dc, then use Dc
Nq = 25; Meyerhof bearing capacity factor for driven piles, based on f

qp = 19.6 kN/m3(1.5 m)25 = 735 kN/m2                                    metric
qp = 125 lb/ft3(5 ft)25 = 15,625 lb/ft2                                          standard

Qp = Apqp = (0.292 m2)(735 kN/m2) = 214.6 kN                     metric
Qp = Apqp = (3.14 ft2)(15,625 lb/ft2) = 49,063 lb                      standard

Determine ultimate friction capacity of pile,

Qf = Afqf

Af = 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 Dc as above

Af = 1.92 m(1.5 m) = 2.88 m2                                                      metric
Af = 6.28 ft(5 ft) = 31.4 ft2                                                           standard

qf = cA + ks tan d = cA + kgD tan d

k = 0.5; lateral earth pressure coefficient for piles, value chosen from Broms low density steel
g = 19.6 kN/m3 (125 lb/ft3); given effective soil unit weight. If water table, then g - gw
D = L = 1.5 m (5 ft); pile length. Check to see if D < Dc
Dc = 15B = 9.2 m (30 ft); critical depth for medium dense silts. If D > Dc, then use Dc
d = 20 deg; external friction angle, equation chosen from Broms steel piles
B = 0.61 m (2 ft); selected pile diameter
cA = 0.5c; for clean steel. See adhesion in pile theories above.
= 24 kN/m2 (500 lb/ft2

qf = 24 kN/m2 + 0.5(19.6 kN/m3)(1.5m)tan 20 = 29.4 kN/m2      metric
qf = 500 lb/ft2 + 0.5(125 lb/ft3)(5ft)tan 20 = 614 lb/ft2                  standard

Qf = Afqf = 2.88 m2(29.4 kN/m2) = 84.7 kN                               metric
Qf = Afqf = 31.4 ft2(614 lb/ft2) = 19,280 lb                                  standard

Determine ultimate pile capacity,

Qult = Qp + Qf

Qult = 214.6 kN + 84.7 kN = 299.3 kN                                       metric
Qult = 49,063 lb + 19,280 lb = 68,343 lb                                      standard

Solve for allowable bearing capacity,

Qa =  Qult
F.S.

Qa   299.3 kN    = 99.8 kN;  Qa > applied load (66.7 kN)     o.k.     metric
3
Qa   68,343 lbs    = 22,781 lbs  Qa > 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/m3 (125 lbs/ft3) *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/m3 (120 lbs/ft3)
• cohesion, c = 47.9 kN/m2 (1000 lb/ft2)
• angle of internal friction, f = 0 degrees

• Pile Information

• driven
• wood
• closed end

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,

Qp = Apqp

Ap = p(B/2)2 = p(0.61m/2)2 = 0.292 m2                                      metric
Ap = p(B/2)2 = p(2ft/2)2 = 3.14 ft2                                                        standard

qp = 9c = 9(47.9 kN/m2) = 431.1 kN/m2                                     metric
qp = 9c = 9(1000 lb/ft2) = 9000 lb/ft2                                           standard

Qp = Apqp = (0.292 m2)(431.1 kN/m2) = 125.9 kN                    metric
Qp = Apqp = (3.14 ft2)(9000 lb/ft2) = 28,260 lb                           standard

Determine ultimate friction capacity of pile,

Qf = pSqfL

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

qfL = [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/m3 (125 lb/ft3); given effective soil unit weight. If water table, then g - gw
D = L = 1.5 m (5 ft); segment of pile within this soil strata. Check to see if D < Dc
Dc = 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 > Dc, then use Dc
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

qfL = [1.5(19.6 kN/m3)(1.5m)tan (20)]1.5 m = 24.1 kN/m            metric
qfL = [1.5(125 lb/ft3)(5ft)tan (20)]5 ft = 1706 lb/ft                         standard

soils below 1.5 m (5 ft) of subgrade

qfL = aSu

Suc = 2c = 95.8 kN/m2 (2000 lb/ft2); unconfined compressive strength
c = 47.9 kN/m2 (1000 lb/ft2); cohesion from soil testing (given)
a =    1    [0.9 + 0.3(Suc - 1)] = 0.3; because Suc > 48 kN/m2, (1 ksf)
Suc
L = 0.91 m (3 ft); segment of pile within this soil strata

qfL = [0.3(95.8 kN/m2)]0.91 m = 26.2 kN/m                                metric
qfL = [0.3(2000 lb/ft2)]3 ft = 1800 lb/ft                                         standard

ultimate friction capacity of combined soil layers

Qf = pSqfL

Qf = 1.92 m(24.1 kN/m + 26.2 kN/m) = 96.6 kN                         metric
Qf = 6.28 ft(1706 lb/ft + 1800 lb/ft) = 22,018 lb                            standard

Determine ultimate pile capacity,

Qult = Qp + Qf

Qult = 125.9 kN + 96.6 kN = 222.5 kN                                       metric
Qult = 28,260 lb + 22,018 lb = 50,278 lb                                      standard

Solve for allowable bearing capacity,

Qa =  Qult
F.S.

Qa   222.5 kN    = 74.2 kN;  Qa > applied load (66.7 kN)     o.k.     metric
3
Qa   50,275 lbs    = 16,758 lbs  Qa > 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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