Allowable Bearing Capacity

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 theoryF.S. = Factor of Safety *See information on factor of safety

Ultimate Bearing Capacity for Shallow Foundations

Terzaghi Ultimate Bearing Capacity Theory

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

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

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

Where:

c = Cohesion of soil (kN/m2) (lb/ft2), = effective unit weight of soil (kN/m3) (lb/ft3), *see note belowD = depth of footing (m) (ft), B = width of footing (m) (ft), Nc=cot(Nq – 1), *see typical bearing capacity factors Nq=e2(3/4-/2)tan / [2 cos2(45+/2)], *see typical bearing capacity factorsN=(1/2) tan(kp /cos2 - 1), *see typical bearing capacity factorse = Napier's constant = 2.718..., kp = passive pressure coefficient, and = angle of internal friction (degrees).

Notes: Effective unit weight, , 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, w, 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 = (B/2)2 m2 (ft2)qp = DNq = 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) = effective unit weight of soil, kN/m3 (lb/ft3), *See notes belowD = 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), andDc = 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 = pqfL 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 = k tan = Theoretical unit friction capacity for cohesionless soils, kN/m2 (lb/ft2) qf = cA + k tan = Theoretical unit friction capacity for silts, kN/m2 (lb/ft2) qf = Su = Theoretical unit friction capacity for cohesive soils, kN/m2 (lb/ft2)p = perimeter of pile cross-section, m (ft) for a circular pile; p = (B/2) for a square pile; p = 4B L = Effective length of pile, m (ft) *See Notes below = 1 - 0.1(Suc)2 = adhesion factor, kN/m2 (ksf), where Suc < 48 kN/m2 (1 ksf) = 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) = external friction angle of soil and wall contact (deg) = angle of internal friction (deg) = D = effective overburden pressure, kN/m2, (lb/ft2) k = lateral earth pressure coefficient for piles = effective unit weight of soil, kN/m3 (lb/ft3) *See notes belowB = 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 = 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, , 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, w, 9.81 kN/m3 (62.4 lb/ft3), subtracted from the saturated unit weight of soil.

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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, ftB = pile diameter, ftNc = Cn(N)Cn = 0.77 log 20 N = N-Value from SPT test = D = effective overburden stress at pile embedment depth, tons/ft2

= ( - w)D = effective stress if below water table, tons/ft2

= effective unit weight of soil, tons/ft3 w = 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, = 21 kN/m3 (132 lbs/ft3) *from soil testing, see typical gvalues Cohesion, c = 0 *from soil testing, see typical cvalues angle of Internal Friction, = 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, Nc and Nq. See typical bearing capacity factors relating to the soils' angle of internal friction.

N = 22 Nc = 35.5 Nq = 23.2

Solve for ultimate bearing capacity,

Qu = c Nc + D Nq + 0.5 B N *strip footing eq. Qu = 0(35.5) + 21 kN/m3(0.6m)(23.2) + 0.5(21 kN/m3)(0.3 m)(22) metricQu = 362 kN/m2

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

Solve for allowable bearing capacity,

Qa = Qu F.S.

Qa = 362 kN/m 2 = 121 kN/m2 not o.k. metric 3 Qa = 7577lbs/ft 2 = 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) metricQu = 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/m 2 = 146 kN/m2 Qa > 144 kN/m2 o.k. metric 3

Qa = 9029 lbs/ft 2 = 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

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, sat= 20.3 kN/m3 (129 lbs/ft3) *see typical gvalues unit weight of water, w= 9.81 kN/m3 (62.4 lbs/ft3) *constant Cohesion, c = 21.1 kN/m2 (440 lbs/ft2) *from soil testing, see typical cvalues angle of Internal Friction, = 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.


N = 0 Nc = 5.7 Nq = 1


Qu = 1.3c Nc + D Nq + 0.4 B N *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


Qa = Qu F.S.

Qa = 163 kN/m 2 = 54 kN/m2 Qa > 48.9 kN/m2 o.k. metric 3 Qa = 3394lbs/ft 2 = 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

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).


Solve for ultimate bearing capacity



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


Qa = Qu F.S.

Qa = 1029 kN/m 2 = 343 kN/m2 Qa > 144 kN/m2 o.k. (metric) 3 Qa = 10.8 tons/ft 2 = 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

vertical column load = 66.7 kN (15 kips or 15,000 lb) homogeneous soils in upper 15.2 m (50 ft); silty soil

o unit weight, = 19.6 kN/m3 (125 lbs/ft3) *from soil testing, see typical gvalueso cohesion, c = 47.9 kN/m2 (1000 lb/ft2) *from soil testing, see typical cvalueso angle of internal friction, = 30 degrees *from soil testing, see typical f values

Pile Informationo driveno steelo 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 = (B/2)2(0.61m/2)2 = 0.292 m2 metricAp (B/2)2(2ft/2)2 = 3.14 ft2 standard

qp = DNq

= 19.6 kN/m3 (125 lbs/ft3); given soil unit weight = 30 degrees; given soil angle of internal frictionB = 0.61 m (2 ft); trial pile widthD = 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

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

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

Determine ultimate friction capacity of pile,

Qf = Afqf

Af = pL

p = 2(0.61m/2) = 1.92 m metricp = 2(2 ft/2) = 6.28 ft standardL = 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 + k tan = cA + kD tan

k = 0.5; lateral earth pressure coefficient for piles, value chosen from Broms low density steel = 19.6 kN/m3 (125 lb/ft3); given effective soil unit weight. If water table, then - w

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 = 20 deg; external friction angle, equation chosen from Broms steel pilesB = 0.61 m (2 ft); selected pile diametercA = 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 metricqf = 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 metricQf = 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 metricQult = 49,063 lb + 19,280 lb = 68,343 lb standard


Qa = Qult F.S.

Qa = 299.3 kN = 99.8 kN; Qa > applied load (66.7 kN) o.k. metric 3Qa = 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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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

o unit weight, = 19.6 kN/m3 (125 lbs/ft3) *from soil testing, see typical gvalueso cohesion, c = 0 *from soil testing, see typical cvalueso angle of internal friction, = 30 degrees *from soil testing, see typical f values

soils below 1.5 m (5 ft) of soil is a stiff silty clay o unit weight, = 18.9 kN/m3 (120 lbs/ft3) o cohesion, c = 47.9 kN/m2 (1000 lb/ft2)o angle of internal friction, = 0 degrees

Pile Informationo driveno woodo closed end

Solution




Qp = Apqp


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

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


Qf = pqfL

p = 2(0.61m/2) = 1.92 m metricp = 2(2 ft/2) = 6.28 ft standard

upper 1.5 m (5 ft) of soil

qfL = [k tan ]L = [kD tan ]L

k = 1.5; lateral earth pressure coefficient for piles, value chosen from Broms low density timber = 19.6 kN/m3 (125 lb/ft3); given effective soil unit weight. If water table, then - w

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 = (2/3) = 20 deg; external friction angle, equation chosen from Broms timber pilesB = 0.61 m (2 ft); selected pile diameter = 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 metricqfL = [1.5(125 lb/ft3)(5ft)tan (20)]5 ft = 1706 lb/ft standard

soils below 1.5 m (5 ft) of subgrade

qfL = Su

Suc = 2c = 95.8 kN/m2 (2000 lb/ft2); u nconfined compressive strength c = 47.9 kN/m2 (1000 lb/ft2); cohesion from soil testing (given) = 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 metricqfL = [0.3(2000 lb/ft2)]3 ft = 1800 lb/ft standard

ultimate friction capacity of combined soil layers

Qf = pqfL

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


Qult = Qp + Qf



Qa = Qult F.S.


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.

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 theoryF.S. = Factor of Safety *See information on factor of safety

http://www.geotechnicalinfo.com/factor_of_safety.html

Ultimate Bearing Capacity for Shallow Foundations

Terzaghi Ultimate Bearing Capacity Theory

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

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

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

Where:

c = Cohesion of soil (kN/m2) (lb/ft2), = effective unit weight of soil (kN/m3) (lb/ft3), *see note belowD = depth of footing (m) (ft), B = width of footing (m) (ft), Nc=cot(Nq – 1), *see typical bearing capacity factors Nq=e2(3/4-/2)tan / [2 cos2(45+/2)], *see typical bearing capacity factorsN=(1/2) tan(kp /cos2 - 1), *see typical bearing capacity factorse = Napier's constant = 2.718..., kp = passive pressure coefficient, and = angle of internal friction (degrees).

Notes: Effective unit weight, , 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, w, 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



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

http://www.geotechnicalinfo.com/foundations.html

http://www.geotechnicalinfo.com/angle_of_internal_friction.html

http://www.geotechnicalinfo.com/lateral_earth_pressure_coefficient.html

http://www.geotechnicalinfo.com/bearing_capacity_factors.html



http://www.geotechnicalinfo.com/soil_unit_weight.html

http://www.geotechnicalinfo.com/cohesion.html

http://www.geotechnicalinfo.com/geotechnical_glossary.html



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 = (B/2)2 m2 (ft2)qp = DNq = 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) = effective unit weight of soil, kN/m3 (lb/ft3), *See notes belowD = 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), andDc = 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 = pqfL 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 = k tan = Theoretical unit friction capacity for cohesionless soils, kN/m2 (lb/ft2) qf = cA + k tan = Theoretical unit friction capacity for silts, kN/m2 (lb/ft2) qf = Su = Theoretical unit friction capacity for cohesive soils, kN/m2 (lb/ft2)p = perimeter of pile cross-section, m (ft) for a circular pile; p = (B/2) for a square pile; p = 4B L = Effective length of pile, m (ft) *See Notes below = 1 - 0.1(Suc)2 = adhesion factor, kN/m2 (ksf), where Suc < 48 kN/m2 (1 ksf) = 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) = external friction angle of soil and wall contact (deg) = angle of internal friction (deg) = D = effective overburden pressure, kN/m2, (lb/ft2) k = lateral earth pressure coefficient for piles = effective unit weight of soil, kN/m3 (lb/ft3) *See notes belowB = 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 = 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, , 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, w, 9.81 kN/m3 (62.4 lb/ft3), subtracted from the saturated unit weight of soil.

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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







http://www.geotechnicalinfo.com/external_friction_angle.html



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, ftB = pile diameter, ftNc = Cn(N)Cn = 0.77 log 20 N = N-Value from SPT test = D = effective overburden stress at pile embedment depth, tons/ft2

= ( - w)D = effective stress if below water table, tons/ft2

= effective unit weight of soil, tons/ft3 w = 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, = 21 kN/m3 (132 lbs/ft3) *from soil testing, see typical gvalues Cohesion, c = 0 *from soil testing, see typical cvalues angle of Internal Friction, = 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).



N = 22 Nc = 35.5







Nq = 23.2


Qu = c Nc + D Nq + 0.5 B N *strip footing eq. Qu = 0(35.5) + 21 kN/m3(0.6m)(23.2) + 0.5(21 kN/m3)(0.3 m)(22) metricQu = 362 kN/m2

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


Qa = Qu F.S.

Qa = 362 kN/m 2 = 121 kN/m2 not o.k. metric 3 Qa = 7577lbs/ft 2 = 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) metricQu = 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/m 2 = 146 kN/m2 Qa > 144 kN/m2 o.k. metric 3

Qa = 9029 lbs/ft 2 = 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

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, sat= 20.3 kN/m3 (129 lbs/ft3) *see typical gvalues unit weight of water, w= 9.81 kN/m3 (62.4 lbs/ft3) *constant Cohesion, c = 21.1 kN/m2 (440 lbs/ft2) *from soil testing, see typical cvalues angle of Internal Friction, = 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.


N = 0 Nc = 5.7 Nq = 1


Qu = 1.3c Nc + D Nq + 0.4 B N *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


Qa = Qu F.S.

Qa = 163 kN/m 2 = 54 kN/m2 Qa > 48.9 kN/m2 o.k. metric 3 Qa = 3394lbs/ft 2 = 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

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).


Solve for ultimate bearing capacity



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


Qa = Qu F.S.

Qa = 1029 kN/m 2 = 343 kN/m2 Qa > 144 kN/m2 o.k. (metric) 3 Qa = 10.8 tons/ft 2 = 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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Given

vertical column load = 66.7 kN (15 kips or 15,000 lb) homogeneous soils in upper 15.2 m (50 ft); silty soil

o unit weight, = 19.6 kN/m3 (125 lbs/ft3) *from soil testing, see typical gvalueso cohesion, c = 47.9 kN/m2 (1000 lb/ft2) *from soil testing, see typical cvalueso angle of internal friction, = 30 degrees *from soil testing, see typical f values

Pile Informationo driveno steelo plugged end

Solution




Qp = Apqp


qp = DNq

= 19.6 kN/m3 (125 lbs/ft3); given soil unit weight = 30 degrees; given soil angle of internal frictionB = 0.61 m (2 ft); trial pile widthD = 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

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

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






Qf = Afqf

Af = pL

p = 2(0.61m/2) = 1.92 m metricp = 2(2 ft/2) = 6.28 ft standardL = 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 + k tan = cA + kD tan

k = 0.5; lateral earth pressure coefficient for piles, value chosen from Broms low density steel = 19.6 kN/m3 (125 lb/ft3); given effective soil unit weight. If water table, then - w

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 = 20 deg; external friction angle, equation chosen from Broms steel pilesB = 0.61 m (2 ft); selected pile diametercA = 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 metricqf = 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 metricQf = Afqf = 31.4 ft2(614 lb/ft2) = 19,280 lb standard


Qult = Qp + Qf



Qa = Qult F.S.


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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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

o unit weight, = 19.6 kN/m3 (125 lbs/ft3) *from soil testing, see typical gvalueso cohesion, c = 0 *from soil testing, see typical cvalueso angle of internal friction, = 30 degrees *from soil testing, see typical f values

soils below 1.5 m (5 ft) of soil is a stiff silty clay o unit weight, = 18.9 kN/m3 (120 lbs/ft3) o cohesion, c = 47.9 kN/m2 (1000 lb/ft2)o angle of internal friction, = 0 degrees

Pile Informationo driveno woodo closed end

Solution




Qp = Apqp


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

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





Qf = pqfL

p = 2(0.61m/2) = 1.92 m metricp = 2(2 ft/2) = 6.28 ft standard

upper 1.5 m (5 ft) of soil

qfL = [k tan ]L = [kD tan ]L

k = 1.5; lateral earth pressure coefficient for piles, value chosen from Broms low density timber = 19.6 kN/m3 (125 lb/ft3); given effective soil unit weight. If water table, then - w

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 = (2/3) = 20 deg; external friction angle, equation chosen from Broms timber pilesB = 0.61 m (2 ft); selected pile diameter = 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 metricqfL = [1.5(125 lb/ft3)(5ft)tan (20)]5 ft = 1706 lb/ft standard

soils below 1.5 m (5 ft) of subgrade

qfL = Su

Suc = 2c = 95.8 kN/m2 (2000 lb/ft2); u nconfined compressive strength c = 47.9 kN/m2 (1000 lb/ft2); cohesion from soil testing (given) = 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 metricqfL = [0.3(2000 lb/ft2)]3 ft = 1800 lb/ft standard

ultimate friction capacity of combined soil layers

Qf = pqfL

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






Qult = Qp + Qf



Qa = Qult F.S.


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.

A

NALYSIS OFB

EARINGC

APACITY – D

RIVENP

ILEF

OUNDATION

8 - 20D

min= minimum embedment depthQ = lateral shear forceB = pile diameters

u= undrained shear strengthFS = safety factor (FS = 3)

C. Cohesionless Soil The following figure is the pressure diagram proposed by Brom for cohesionless soil for

free head condition .F

IGURE8.3

C

OHESIONLESSS

OIL – F

REEH

EADC

ONDITION

The minimum embedment depth of the pile due to shear force Q is :( )

eDQKBD5.0FSminp3min

+γ =

φ+=

245tanK2p

[8.40]where :D


u= undrained shear strengthe = eccentricity of lateral loadK

p= coefficient of passive lateral earth pressureFS = safety factor (FS = 3)

A

NALYSIS OFB

EARINGC

APACITY – D

RIVENP

ILEF

OUNDATION

8 - 21The following figure is the pressure diagram proposed by Brom for cohesionless soil for

restrained head condition .F

IGURE8.3

C

OHESIONLESSS

OIL – R

ESTRAINEDH

EADC

ONDITION

The minimum embedment depth of the pile due to shear force Q is :( )

pminBK5.1FSQD

γ =

[8.40]where :D


u= undrained shear strengthK

p= coefficient of passive lateral earth pressureFS = safety factor (FS = 3)

8.9 GROUP PILE FOUNDATION8.9.1 G

ENERAL

When the load is becomes bigger the group pile must be used to carry the load. The design of grouppile must consider the

efficiency of the group and the

arrangement of the pile .

8.9.2 PILEC

ONFIGURATION

The minimum spacing between pile in group pile is :( )

D5.35.2s−=

[8.41]

A

NALYSIS OFB

EARINGC

APACITY

– D

RIVENP

ILEF

OUNDATION

8 - 22where :s = pile spacingD = pile diameter

8.9.3 LOAD

TRANSFER

The group pile may be subjected to centric load and eccentric load.The load transfer to the pile due to centric load

is :nPP1=

[8.42]where :P

1= vertical load in one pileP = total centric vertical loadn = number of pileThe load transfer to the pile due to eccentric

load is :∑±∑±=

2x2y1yyMxxMnPP

[8.43]where :P

1= vertical load in one pileP = total centric vertical loadM

x= moment about X axisM

y= moment about Y axisx = x distance from center of pile capy = y distance from center of pile capn = number of pile

8.9.4 GROUP

EFFICIENCY

The group efficiency of group pile can be calculated based on the Converse – Labarre formula, asfollows :

{ } { }

−+−θ−=mn90n1mm1n1E

g sDtan

1−

=θ

[8.44]where :E

g= efficiency of group pilem = number of pile columnsn = number of pile rows

Documents

Allowable Bearing Capacity