Bearing capacity
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.
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Theoretical
(Ultimate) and allowable bearing capacity can be assessed for the
following:
For
comprehensive examples of bearing capacity problems see:
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) cA = adhesion = 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
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.
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.
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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
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.
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.
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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
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.
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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
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.
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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
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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Wednesday, June 17, 2015
Bearing capacity of soil
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