Bearing capacity of soil is the value of the average contact pressure between th e foundation and the soil which will produce shear failure in the soil. Ultimate bearing capacity is the theoretical maximum pressure which can be supported wit hout failure.Allowable bearing capacity is what is used in geotechnical design, and is the ultimate bearing capacity divided by a factor of safety. • Allowable Bearing Capacity ________________________________________ Qa = Qu Qa = Allowable bearing capacity (k N/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 f actor 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 ca pacity 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 fa ctors 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 + 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) 10 For footing widths of 3 meters (10 feet) or more
(standard)
Qa =
9,576N
Qa =
N
(kN/m2) (tons/ft2)
(metric) (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 fi rst equation is for ultimate bearing capacity, while the second two are factored within the equation in order to provide an allowable bearing capacity. Linear i nterpolation 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 t o 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 fou ndation 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 fou ndation 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 be low 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 soi ls, 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 weig ht, , is the unit weight of the soil for soils above the water table and capillar y rise. For saturated soils, the effective unit weight is the unit weight of wat er, w, 9.81 kN/m3 (62.4 lb/ft3), subtracted from the saturated unit weight of soi l. ************ 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 100
Where: D = pile embedment depth, ft B = pile diameter, ft
standard
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 f ooting 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 k N/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 g v alues • Cohesion, c = 0 *from soil testing , see typical c values • 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 fac tors 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) m etric 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) be low 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 c onditions that may negatively effect the depth factor are foundations establishe d 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-inc h) square isolated footing bearing on saturated cohesive soil. The frost penetra tion 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 *see typi • unit weight of saturated soil, sat= 20.3 kN/m3 (129 lbs/ft3) cal g values • unit weight of water, w= 9.81 kN/m3 (62.4 lbs/ft3) *consta nt • Cohesion, c = 21.1 kN/m2 (440 lbs/ft2) *from soil testing, see ty pical c values • angle of Internal Friction, = 0 degrees *from soil testing, see t ypical 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 belo w 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 fac tors 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)(0.3 m)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) be low the ground surface. Other considerations may be required for foundations bea ring 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 he avier 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 usin g the Meyerhof Method. Soils consist of poorly graded sand. Footing must withsta nd 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 te sting • N Value, N = 36 at 0.61 m (2 ft) depth *from SPT soil tes ting • N Value, N = 50 at 1.5 m (5 ft) depth *from SPT soil te sting 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 0.61 m (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) bel ow the ground surface. A footing width of only 0.3 m (1 ft) is most likely insuf ficient 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 g 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 typic al f 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 regio nal soils. Determine ultimate end bearing of pile, Qp = Apqp Ap = (B/2)2 (0.61m/2)2 = 0.292 m2 metric Ap (B/2)2 (2ft/2)2 = 3.14 ft2 tandard 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 cap acity 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 metric qp = 125 lb/ft3(5 ft)25 = 15,625 lb/ft2 standard 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
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 metric Af = 6.28 ft(5 ft) = 31.4 ft2 standard qf = cA + k tan = cA + k D tan k = 0.5; lateral earth pressure coefficient for piles, value chosen from Broms l ow density steel = 19.6 kN/m3 (125 lb/ft3); given effective soil unit weight. If water table, the
s
n - 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 metric qf = 500 lb/ft2 + 0.5(125 lb/ft3)(5ft)tan 20 = 614 lb/ft2 stand ard Qf = Afqf = 2.88 m2(29.4 kN/m2) = 84.7 kN metric Qf = Afqf = 31.4 ft2(614 lb/ft2) = 19,280 lb st andard Determine ultimate pile capacity, Qult = Qp + Qf Qult = 214.6 kN + 84.7 kN = 299.3 kN c Qult = 49,063 lb + 19,280 lb = 68,343 lb andard Solve for allowable bearing capacity, Qa = Qult F.S. Qa = 299.3 kN = 99.8 kN; Qa > applied load (66.7 kN) o.k. 3 Qa = 68,343 lbs = 22,781 lbs Qa > applied load (15 kips) o.k. dard 3
metri st
metric stan
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 varia tions may include freeze/ thaw or effects from water. The end bearing alone (neg lect skin friction) is sufficient for this case. Typical methods for increasing the pile capacity are increasing the pile diameter or increasing the embedment d epth of the pile. ******************************** Example #5: Determine allowable bearing capacity and diameter of a single driven pile. Pile must withstand a 66.7 kN (15 kips) vertical load. Given • vertical column load = 66.7 kN (15 kips or 15,000 lb) • upper 1.5 m (5 ft) of soil is a medium dense gravelly sand o unit weight, = 19.6 kN/m3 (125 lbs/ft3) *from soil testing, see typical g values o cohesion, c = 0 *from soil testin g, see typical c values o angle of internal friction, = 30 degrees *from soil testing, see typic al 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 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 regio nal soils. Determine ultimate end bearing of pile, Qp = Apqp Ap = (B/2)2 Ap (B/2)2 tandard
(0.61m/2)2 = 0.292 m2 (2ft/2)2 = 3.14 ft2
metric s
qp = 9c = 9(47.9 kN/m2) = 431.1 kN/m2 qp = 9c = 9(1000 lb/ft2) = 9000 lb/ft2 standard Qp = Apqp = (0.292 m2)(431.1 kN/m2) = 125.9 kN Qp = Apqp = (3.14 ft2)(9000 lb/ft2) = 28,260 lb rd
metric
metric standa
Determine ultimate friction capacity of pile, Qf = p qfL p = 2 (0.61m/2) = 1.92 m metric p = 2 (2 ft/2) = 6.28 ft 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 l ow density timber = 19.6 kN/m3 (125 lb/ft3); given effective soil unit weight. If water table, the n - 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 f or a sand would indicate dense soils. If D > Dc, then use Dc = (2/3) = 20 deg; external friction angle, equation chosen from Broms timber pile s 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 metric qfL = [1.5(125 lb/ft3)(5ft)tan (20)]5 ft = 1706 lb/ft st andard 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 metric qfL = [0.3(2000 lb/ft2)]3 ft = 1800 lb/ft 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 ndard
metric
Determine ultimate pile capacity, Qult = Qp + Qf Qult = 125.9 kN + 96.6 kN = 222.5 kN c Qult = 28,260 lb + 22,018 lb = 50,278 lb andard Solve for allowable bearing capacity, Qa = Qult F.S. Qa = 222.5 kN = 74.2 kN; Qa > applied load (66.7 kN) o.k. 3 Qa = 50,275 lbs = 16,758 lbs Qa > applied load (15 kips) o.k. dard 3
sta
metri st
metric stan
Conclusion Wood pile shall be driven 8 feet below the ground surface. Many engineers neglec t the skin friction within the upper 1 to 5 feet of subgrade due to seasonal var iations or soil disturbance. Seasonal variations may include freeze/ thaw or eff ects 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 vic inity of the pile tip. Typical methods for increasing the pile capacity are incr easing the pile diameter or increasing the embedment depth of the pile. ********************************
