Allowable Bearing Capacity



Allowable Bearing Capacity

Qa =

Qu F.S.

Qa = Allowable bearing capacity (kN/m2) or (lb/ft2)

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 +  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 below D = 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 factors N=(1/2) tan(kp /cos2  - 1), *see typical bearing capacity factors e = 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 10

(tons/ft2)

(standard)

+ ND 10

For footing widths of 1.2 meters (4 feet) or less Qa = 11,970N

(kN/m2)

(metric)

Qa = 1.25N 10

(tons/ft2)

(standard)

For footing widths of 3 meters (10 feet) or more Qa = 9,576N

(kN/m2)

(metric)

Qa = N 10

(tons/ft2)

(standard)

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 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 = pqfL

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 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  = 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. ************ 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 50

standard

Theoretical unit friction-bearing capacity for drilled piles in sand: qf = N tons/ft2 100

standard

Where: D = pile embedment depth, ft B = pile diameter, ft Nc = 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 values Cohesion, c = 0 *from soil testing, see typical cvalues

 

angle of Internal Friction,  = 32 degrees footing depth, D = 0.6 m (2 ft)

*from soil testing, see typical  values *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) Qu = 362 kN/m2

metric

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

standard

Solve for allowable bearing capacity, Qa = Qu F.S. Qa = 362 kN/m2 = 121 kN/m2 3 Qa = 7577lbs/ft2 = 2526 lbs/ft2 3

not o.k. not o.k.

metric standard

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) Qu = 438 kN/m2

metric

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

standard

Qa = 438 kN/m2 = 146 kN/m2 3

Qa > 144 kN/m2

o.k.

metric

Qa = 9029 lbs/ft2 = 3010 lbs/ft2 3

Qa > 3000 lbs/ft2

o.k.

standard

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/ft 2 unit weight of saturated soil, sat= 20.3 kN/m3 (129 lbs/ft3) *see typical values 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 cvalues angle of Internal Friction,  = 0 degrees *from soil testing, see typical  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, Nc and Nq. See typical bearing capacity factors relating to the soils' angle of internal friction.

  

N = 0 Nc = 5.7 Nq = 1

Solve for ultimate bearing capacity, 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 > 48.9 kN/m2 Qa > 1000 lbs/ft2

o.k. o.k.

metric standard

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 10

(tons/ft2)

(standard)

+ ND 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 10 10

(standard)

Solve for allowable bearing capacity,

Qa = Qu F.S. Qa = 1029 kN/m2 = 343 kN/m2 Qa > 144 kN/m2 3 Qa = 10.8 tons/ft2 = 3.6 tons/ft2 Qa > 1.5 tons/ft2 3

o.k. o.k.

(metric) (standard)

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 o unit weight,  = 19.6 kN/m3 (125 lbs/ft3) *from soil testing, see typical values o cohesion, c = 47.9 kN/m2 (1000 lb/ft2) *from soil testing, see typical cvalues o angle of internal friction,  = 30 degrees *from soil testing, see typical  values Pile Information o driven o steel o 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 Ap  (B/2)2(2ft/2)2 = 3.14 ft2 qp = DNq

metric standard

 = 19.6 kN/m3 (125 lbs/ft3); given soil unit weight  = 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  qp = 19.6 kN/m3(1.5 m)25 = 735 kN/m2 qp = 125 lb/ft3(5 ft)25 = 15,625 lb/ft2

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

metric standard

metric standard

Determine ultimate friction capacity of pile, Qf = Afqf Af = pL p = 2(0.61m/2) = 1.92 m metric p = 2(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 Af = 6.28 ft(5 ft) = 31.4 ft2

metric standard

qf = cA + k tan  = cA + kD 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 > D c, then use Dc  = 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 qf = 500 lb/ft2 + 0.5(125 lb/ft3)(5ft)tan 20 = 614 lb/ft2 Qf = Afqf = 2.88 m2(29.4 kN/m2) = 84.7 kN Qf = Afqf = 31.4 ft2(614 lb/ft2) = 19,280 lb

metric standard metric standard

Determine ultimate pile capacity, Qult = Qp + Qf Qult = 214.6 kN + 84.7 kN = 299.3 kN Qult = 49,063 lb + 19,280 lb = 68,343 lb

metric 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 68,343 lbs = 22,781 lbs Qa > applied load (15 kips) o.k. standard 3

Qa =

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. ********************************


 





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 values o cohesion, c = 0 *from soil testing, see typical cvalues o angle of internal friction,  = 30 degrees *from soil testing, see typical  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 Information o driven o wood o 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 = (B/2)2(0.61m/2)2 = 0.292 m2 Ap  (B/2)2(2ft/2)2 = 3.14 ft2

metric standard

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

metric standard

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

metric standard

Determine ultimate friction capacity of pile, Qf = pqfL p = 2(0.61m/2) = 1.92 m p = 2(2 ft/2) = 6.28 ft

metric standard

upper 1.5 m (5 ft) of soil qfL = [k tan ]L = [kD 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 > D c, then use Dc  = (2/3) = 20 deg; external friction angle, equation chosen from Broms timber piles B = 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 qfL = [1.5(125 lb/ft3)(5ft)tan (20)]5 ft = 1706 lb/ft

metric standard

soils below 1.5 m (5 ft) of subgrade qfL = Su 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)  = 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 qfL = [0.3(2000 lb/ft2)]3 ft = 1800 lb/ft

metric standard

ultimate friction capacity of combined soil layers Qf = pqfL Qf = 1.92 m(24.1 kN/m + 26.2 kN/m) = 96.6 kN Qf = 6.28 ft(1706 lb/ft + 1800 lb/ft) = 22,018 lb

metric standard


metric standard

Solve for allowable bearing capacity, Qa = Qult F.S. Qa =


Qa =

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 F.S.

Qa = Allowable bearing capacity (kN/m2) or (lb/ft2)

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 +  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 below D = 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 factors N=(1/2) tan(kp /cos2  - 1), *see typical bearing capacity factors e = 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 10

(tons/ft2)

(standard)

+ ND 10

For footing widths of 1.2 meters (4 feet) or less Qa = 11,970N

(kN/m2)

(metric)

Qa = 1.25N 10

(tons/ft2)

(standard)

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

Qa = 9,576N

(kN/m2)

(metric)

Qa = N 10

(tons/ft2)

(standard)

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 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 = pqfL

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 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  = 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. ************ 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 50

standard

Theoretical unit friction-bearing capacity for drilled piles in sand: qf = N tons/ft2 100

standard

Where: D = pile embedment depth, ft B = pile diameter, ft Nc = 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 values Cohesion, c = 0 *from soil testing, see typical cvalues angle of Internal Friction,  = 32 degrees *from soil testing, see typical  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) Qu = 362 kN/m2

metric

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

standard


not o.k. not o.k.

metric standard

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) Qu = 438 kN/m2

metric

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

standard

Qa = 438 kN/m2 = 146 kN/m2 3 Qa = 9029 lbs/ft2 = 3010 lbs/ft2 3

Qa > 144 kN/m2

Qa > 3000 lbs/ft2

o.k.

o.k.

metric

standard

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/ft 2 unit weight of saturated soil, sat= 20.3 kN/m3 (129 lbs/ft3) *see typical values 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 cvalues angle of Internal Friction,  = 0 degrees *from soil testing, see typical  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, Nc and Nq. See typical bearing capacity factors relating to the soils' angle of internal friction.

  

N = 0 Nc = 5.7 Nq = 1

Solve for ultimate bearing capacity, 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 Solve for allowable bearing capacity, Qa = Qu F.S. Qa = 163 kN/m2 = 54 kN/m2 3 Qa = 3394lbs/ft2 = 1130 lbs/ft2 3 Conclusion

Qa > 48.9 kN/m2 Qa > 1000 lbs/ft2

o.k. o.k.

metric standard

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

   

2(

2

bearing pressure from building = 144 kN/m 1.5 tons/ft ) 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 10

(tons/ft2)

(standard)

+ ND 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 10 10

(standard)

Solve for allowable bearing capacity, Qa = Qu F.S. Qa = 1029 kN/m2 = 343 kN/m2 Qa > 144 kN/m2 3 Qa = 10.8 tons/ft2 = 3.6 tons/ft2 Qa > 1.5 tons/ft2 3 Conclusion

o.k. o.k.

(metric) (standard)

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. ********************************


 



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 values o cohesion, c = 47.9 kN/m2 (1000 lb/ft2) *from soil testing, see typical cvalues o angle of internal friction,  = 30 degrees *from soil testing, see typical  values Pile Information o driven o steel o plugged end


Ap = (B/2)2(0.61m/2)2 = 0.292 m2 Ap  (B/2)2(2ft/2)2 = 3.14 ft2

metric standard

qp = DNq  = 19.6 kN/m3 (125 lbs/ft3); given soil unit weight  = 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 

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

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

metric standard

metric standard

Determine ultimate friction capacity of pile, Qf = Afqf Af = pL p = 2(0.61m/2) = 1.92 m metric p = 2(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 Af = 6.28 ft(5 ft) = 31.4 ft2

metric standard

qf = cA + k tan  = cA + kD 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 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 qf = 500 lb/ft2 + 0.5(125 lb/ft3)(5ft)tan 20 = 614 lb/ft2 Qf = Afqf = 2.88 m2(29.4 kN/m2) = 84.7 kN Qf = Afqf = 31.4 ft2(614 lb/ft2) = 19,280 lb

metric standard metric standard


Solve for allowable bearing capacity, Qa = Qult F.S.

metric standard

Qa =


Qa =

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. ********************************


 





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 values o cohesion, c = 0 *from soil testing, see typical cvalues o angle of internal friction,  = 30 degrees *from soil testing, see typical  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 Information o driven o wood o closed end


Ap = (B/2)2(0.61m/2)2 = 0.292 m2 Ap  (B/2)2(2ft/2)2 = 3.14 ft2

metric standard

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

metric standard

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

metric standard

Determine ultimate friction capacity of pile, Qf = pqfL p = 2(0.61m/2) = 1.92 m p = 2(2 ft/2) = 6.28 ft

metric standard

upper 1.5 m (5 ft) of soil qfL = [k tan ]L = [kD 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 piles B = 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 qfL = [1.5(125 lb/ft3)(5ft)tan (20)]5 ft = 1706 lb/ft

metric standard

soils below 1.5 m (5 ft) of subgrade qfL = Su 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)  = 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 qfL = [0.3(2000 lb/ft2)]3 ft = 1800 lb/ft

ultimate friction capacity of combined soil layers Qf = pqfL

metric standard

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

metric standard


metric standard

Solve for allowable bearing capacity, Qa = Qult F.S. Qa = Qa =


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 OF B EARING C APACITY – D RIVEN P ILE F OUNDATION 8 - 20 D

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 IGURE 8.3 C OHESIONLESS S OIL – F REE H EAD C ONDITION The minimum embedment depth of the pile due to shear force Q is : () eDQKBD5.0FS minp3min +γ = φ+= 245tanK 2p [8.40] where :D min = minimum embedment depthQ = lateral shear forceB = pile diameters u = undrained shear strengthe = eccentricity of lateral loadK p = coefficient of passive lateral earth pressureFS = safety factor (FS = 3)

A NALYSIS OF B EARING C APACITY – D RIVEN P ILE F OUNDATION 8 - 21 The following figure is the pressure diagram proposed by Brom for cohesionless soil for restrained head condition . F IGURE 8.3 C OHESIONLESS S OIL – R ESTRAINED H

EAD C ONDITION The minimum embedment depth of the pile due to shear force Q is : () pmin BK5.1FSQD γ= [8.40] where :D min = minimum embedment depthQ = lateral shear forceB = pile diameters u = undrained shear strengthK p = coefficient of passive lateral earth pressureFS = safety factor (FS = 3) 8.9 GROUP PILE FOUNDATION 8.9.1 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 P ILE C ONFIGURATION The minimum spacing between pile in group pile is : () D5.35.2s −= [8.41]

A NALYSIS OF B EARING C APACITY

– D RIVEN P ILE F OUNDATION 8 - 22 where :s = pile spacingD = pile diameter 8.9.3 L OAD T RANSFER The group pile may be subjected to centric load and eccentric load.The load transfer to the pile due to centric load is :nPP 1 = [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 : ∑ ± ∑ ±= 2x2y1 yyMxxMnPP [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 G ROUP E FFICIENCY 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

Allowable Bearing Capacity

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