Geo Tech

Foundation Types and Selection Selection of an appropriate foundation system is dependent upon many factors. These factors may include: · · · · · ·

soil conditions groundwater conditions surface conditions structural loads structural function (i.e. basement, cold structure, etc) economy

Foundations are typically constructed utilizing either shallow or deep foundation structures. Foundations are sometimes considered shallow if the footing depth divided by the footing width is less than 1. Shallow foundations include mat (raft, pad), continuous footings and isolated (spread) footings. Isolated footings could be square or circular. Deep foundations include piles, micropiles (minipiles, pin piles), mats and piers. Piles are designed to provide bearing either at the bottom tip of the pile, friction between the pile sidewalls and subgrade, or a combination of both. Piles are typically driven (hammered) into the subgrade, or piles can be drilled or jetted into the ground. In addition, piles may be wood, steel or concrete. Micropiles are usually used in lieu of piles in areas with difficult or restricted access. Micropiles are usually drilled with a small machine, and grouted in place. The definition of piers varies greatly, and ranges from a prefabricated rebar cage and wet concrete placed in a pre-bored hole to a larger diameter pile. Analysis of a pier is similar to a pile. Floating foundations are typically deep mats, in which enough soil is excavated so that the weight of soil is equal to the weight of the structure. Directly from a technical manual produced by the Department of Army (TM 5-818-1), the following foundation possibilities for various subsurface conditions are feasible Foundation Possibilities for Different Subsoil Conditions, TM 5-818-1 Subsoil Conditions Light, Flexible Structure Heavy, Rigid Structure Deep compact or stiff soils

Shallow footing Shallow footing, shallow mat Shallow footing overlying compacted granular fill, Deep compressible strata Shallow mat, Friction piles Deep mat, Friction piles End bearing piles or piers, Shallow footing on Soft or Loose strata overlying compacted granular fill, End bearing piles or piers, firm strata Shallow mat Deep mat Compact or stiff layer overlying Deep mat (floating a soft or compressible strata Shallow footing, Shallow mat foundation), Piles Alternating soft and stiff layers Shallow footing, Shallow mat Deep mat, Piles

General Foundation Considerations 1. The allowable soil pressure should not be exceeded. 2. Footings should be located below the frost line. Although you could analytically estimate the frost depth, your local City or County Building Department provides these values. 3. The applied load should be within the middle third of the footing. 4. Foundations should have an adequate factor of safety against uplift, sliding and overturning.

Bearing Capacity Factors Bearing capacity is the ability of the underlying soil to support the foundation loads without shear failure. Bearing capacity factors are empirically derived factors used in a bearing capacity equation that usually correlates with the angle of internal friction of the soil.

See the bearing capacity technical guidance for equations and detailed calculations for applying the following bearing capacity factors. This link also explains each BC factor with relation to the bearing capacity component. Each BC factor below is related to the angle of internal friction. Terzaghi’s Bearing Capacity Factors f Nc Nq Ng 0 5 10 15

5.7 7.3 9.6 12.9

1 1.6 2.7 4.4

0 0.5 1.2 2.5

20 25 30 34 35 40 45

17.7 25.1 37.2 52.6 57.8 95.7 172.3

7.4 12.7 22.5 36.5 41.4 81.3 173.3

5 9.7 19.7 35.0 42.4 100.4 297.5

Meyerhof Bearing Capacity Factors f Nq Nc Ng 0 5 10 15 20 25 30 32 34 36 38 40 42 44

5.14 6.5 8.3 11.0 14.8 20.7 30.1 35.5 42.4 50.6 61.4 75.3 93.7 118.4

1.0 1.6 2.5 3.9 6.4 10.7 18.4 23.2 29.4 37.7 48.9 64.2 85.4 115.3

Vesic Bearing Capacity Factors f Nc Nq 0 5 10 15 20

5.14 6.5 8.3 11.0 14.8

1.0 1.6 2.5 3.9 6.4

0.0 0.07 0.37 1.1 2.9 6.8 15.7 22.0 31.2 44.4 64.1 93.7 139.3 211.4

Ng 0.0 0.5 1.2 2.6 5.4

25 30 32 34 36 38 40 42 44

20.7 30.1 35.5 42.4 50.6 61.4 75.3 93.7 118.4

10.7 18.4 23.2 29.4 37.7 48.9 64.2 85.4 115.3

10.8 22.4 30.2 41.1 56.3 78.0 109.4 155.6 224.6

Hansen Bearing Capacity Factors f Nq Nc Ng 0 6 10 16 20 26 30 32 34 36 38 40 42 44

5.14 6.81 8.3 11.63 14.8 22.25 30.1 35.5 42.4 50.6 61.4 75.3 93.7 118.4

1.0 1.23 2.5 4.34 6.4 11.85 18.4 23.2 29.4 37.7 48.9 64.2 85.4 115.3

0.0 0.11 0.39 1.43 2.95 7.94 15.07 20.79 28.77 40.05 56.17 79.54 113.95 165.48

Foundation Engineering Handbook f Nc Nq Ng 0 5 10

5.14 6.5 8.35

1.0 1.57 2.47

0.0 0.45 1.22

15 20 25 30 32 34 36 38 40 42 44

10.98 14.83 20.72 30.14 35.49 42.16 50.59 61.35 75.31 93.71 118.37

3.94 6.4 10.66 18.4 23.18 29.44 37.75 48.93 64.20 85.38 115.31

2.65 5.39 10.88 22.4 30.22 41.06 56.31 78.03 109.41 155.55 224.64

Bearing Capacity Factors for Deep Foundations

Meyerhof Values of Nq For Driven and Drilled Piles f Driven Drilled 20 25 28 30 32 34 36 38 40 42 45

8 12 20 25 35 45 60 80 120 160 230

4 5 8 12 17 22 30 40 60 80 115

Angle of Internal Friction Angle of internal friction for a given soil is the angle on the graph (Mohr's Circle) of the shear stress and normal effective stresses at which shear failure occurs. Angle of Internal Friction, f, can be determined in the laboratory by the Direct Shear Test or the Triaxial Stress Test.

Typical relationships for estimating the angle of internal friction, f, are as follows: Empirical values for f, of granular soils based on the standard penetration number, (from Bowels, Foundation Analysis). SPT Penetration, NValue (blows/ foot) f (degrees) 0 4 10 30 50

25 - 30 27 - 32 30 - 35 35 - 40 38 - 43

Relationship between f, and standard penetration number for sands, (from Peck 1974, Foundation Engineering Handbook). SPT Penetration, N-Value Density of (blows/ foot) f (degrees) Sand 50

Very loose Loose Medium Dense Very dense

41

Relationship between f, and standard penetration number for sands, (from Meyerhof 1956, Foundation Engineering Handbook). SPT Penetration, N-Value Density of (blows/ foot) f (degrees) Sand 50

Very loose Loose Medium Dense Very dense

45

External Friction Angle The external friction angle, d, or friction between a soil medium and a material such as the composition from a retaining wall or pile may be expressed in degrees as the following:

Piles 20 degrees for steel piles (NAVFAC) 0.67f - 0.83f (USACE)

20 degrees for steel (Broms) f3/4 for concrete (Broms) f2/3 for timber (Broms) 0.67f (Lindeburg) Nordlund attempts to more precisely quantify the external friction angle, d, by using various charts based on the angle of internal friction, f, and volume of pile where f = angle of internal friction of the soil (degrees)

Retaining Walls d = 2f 3

for concrete walls (Coulomb) where f = angle of internal friction of the soil (degrees)

Unit Weight of Soil Unit weight of a soil mass is the ratio of the total weight of soil to the total volume of soil. Unit Weight, g, is usually determined in the laboratory by measuring the weight and volume of a relatively undisturbed soil sample obtained from a brass ring. Measuring unit weight of soil in the field may consist of a sand cone test, rubber balloon or nuclear densiometer. Empirical values for g, of granular soils based on the standard penetration number, (from Bowels, Foundation Analysis). SPT Penetration, N3 Value (blows/ foot) g (lb/ft ) 0-4 4 - 10 10 - 30 30 - 50 >50

70 - 100 90 - 115 110 - 130 110 - 140 130 - 150

Empirical values for g, of cohesive soils based on the standard penetration number, (from Bowels, Foundation Analysis). SPT Penetration, N3 Value (blows/ foot) gsat (lb/ft ) 0-4 4-8 8 - 32

100 - 120 110 - 130 120 - 140

Typical Soil Characteristics (from Lindeburg, Civil Engineering Reference Manual for the PE Exam, 8th ed.) 3 3 Soil Type g (lb/ft ) gsat (lb/ft ) Sand, loose and uniform Sand, dense and uniform sand, loose and well graded Sand, dense and well graded glacial clay, soft glacial clay, stiff

90

118

109

130

99

124

116

135

76

110

106

125

Typical Values of Soil Index Properties (from NAVFAC 7.01) 3 3 Soil Type g (lb/ft ) gsub (lb/ft ) Sand; clean, uniform, fine or medium Silt; uniform, inorganic Silty Sand Sand; Wellgraded Silty Sand and Gravel Sandy or Silty Clay Silty Clay with Gravel;

84 - 136

52 - 73

81 - 136 88 - 142

51 - 73 54 - 79

86 - 148

53 - 86

90 - 155

56 - 92

100 - 147

38 - 85

115 - 151

53 - 89

uniform Well-graded Gravel, Sand, Silt and Clay Clay Colloidal Clay Organic Silt Organic Clay

125 - 156 94 - 133

62 - 94 31 - 71

71 - 128 87 - 131 81 - 125

8 - 66 25 - 69 18 - 62

Typical Soil Characteristics (from Lindeburg, Civil Engineering Reference Manual for the PE Exam, 8th ed.) 3 3 Soil Type g (lb/ft ) gsat (lb/ft ) Sand, loose and uniform Sand, dense and uniform sand, loose and well graded Sand, dense and well graded glacial clay, soft glacial clay, stiff

90

118

109

130

99

124

116

135

76

110

106

125

Young's Modulus of Soil The modulus of elasticity or Young's modulus of a soil is an elastic soil parameter most commonly used in the estimation of settlement from static loads. Young's soil modulus, Es, may be estimated from empirical correlations, laboratory test results on undisturbed specimens and results of field tests. Laboratory tests that may be used to estimate the soil modulus are the triaxial unconsolidated undrained compression or the triaxial consolidated undrained compression tests. Field tests include the plate load test, cone penetration test, standard penetration test (SPT) and the pressuremeter test. Empirical correlations summarized from USACE EM 1110-1-1904 is presented below:

Es = KcCu where: Es = Young's soil modulus (tsf) Kc = correlation factor Cu = undrained shear strength, tsf

Typical Elastic Moduli of soils based on soil type and consistency/ density, (from USACE, Settlement Analysis). Soil Es (tsf) very soft clay soft clay medium clay stiff clay, silty clay sandy clay clay shale loose sand dense sand dense sand and gravel silty sand

5 - 50 50 - 200 200 - 500 500 - 1000 250 - 2000 1000 - 2000 100 - 250 250 - 1000 1000 - 2000 250 - 2000

Factor of Safety

Foundatation Analysis by Bowels has good recommendations for safety factors. He evaluates uncertainties and assigns a factor of safety by taking into account the following: 1. Magnitude of damages (loss of life and property damage) 2. Relative cost of increasing or decreasing the factor of safety 3. Relative change in probability of failure by changing the factor of safety 4. Reliability of soil data 5. Construction tolerances 6. Changes in soil properties due to construction operations 7. Accuracy (or approximations used) in developing design/ analysis methods

Typical values of customary safety factors, F.S., as presented by Bowels. Failure Foundation Mode Type F.S. Shear Shear Shear

Shear Shear Shear Shear Seepage Seepage

Earthwork for Dams, Fills, etc. Retaining Walls Sheetpiling, Cofferdams Braced Excavations (Temporary) Spread Footings Mat Footings Uplift for Footings Uplift, heaving Piping

1.2 - 1.6 1.5 - 2.0 1.2 - 1.6

1.2 - 1.5 2-3 1.7 - 2.5 1.7 - 2.5 1.5 - 2.5 3-5

Other customary factors of safety, F.S., used are: 1.5 for retaining walls overturning with granular backfill 2.0 for retaining walls overturning with cohesive backfill 1.5 for retaining walls sliding with active earth pressures 2.0 for retaining walls sliding with passive earth pressures

Cohesion of Soil Cohesive soils are clay type soils. Cohesion is the force that holds together molecules or like particles within a soil. Cohesion, c, is usually determined in the laboratory from the Direct Shear Test. Unconfined Compressive Strength, Suc, can be determined in the laboratory using the Triaxial Test or the Unconfined Compressive Strength Test. There are also correlations for Suc with shear strength as estimated from the field using Vane Shear Tests.

c = Suc/2 Where: c = cohesion, kN/m2 (lb/ft2), and Suc = unconfined compressive strength, kN/m2 (lb/ft2).

Guide for Consistency of Fine-Grained Soil, NAVFAC 7.02 SPT Penetration Estimated (blows/ foot) Consistency Suc (tons/ft2) 30

Very Soft Soft Medium Stiff Very Stiff Hard

4

Empirical Values for Consistency of Cohesive Soil, (from Foundation Analysis,

Bowels) SPT Penetration (blows/ foot) 0-2 2-4 4-8 8 - 16 16 - 32 >32

Estimated Consistency Very Soft Soft Medium Stiff Very Stiff Hard

2

Suc (kips/ft ) 0 - 0.5 0.5 - 1.0 1.0 - 2.0 2.0 - 4.0 4.0 - 8.0 >8

Typical Strength Characteristics (from Lindeburg, Civil Engineering Reference Manual for the PE Exam, 8th ed.) USCS Soil c, as compacted c, saturated 2 2 Group (lb/ft ) (lb/ft ) GW GP GM GC SW SP SM SM-SC SC ML ML-CL CL OL MH CH

0 0 1050 1050 1550 1400 1350 1800 1500 2150

0 0 420 300 230 190 460 270 420 230

·

Ultimate Bearing Capacity for Shallow Foundations Terzaghi Ultimate Bearing Capacity Theory

Qu = c Nc + g D Nq + 0.5 g B Ng = Ultimate bearing capacity equation for shallow strip footings, (kN/m2) (lb/ft2) Qu = 1.3 c Nc + g D Nq + 0.4 g B Ng = Ultimate bearing capacity equation for shallow square footings, (kN/m2) (lb/ft2) Qu = 1.3 c Nc + g D Nq + 0.3 g B Ng = Ultimate bearing capacity equation for shallow circular footings, (kN/m2) (lb/ft2) Where: c = Cohesion of soil (kN/m2) (lb/ft2), g = effective unit weight of soil (kN/m3) (lb/ft3), *see note below D = depth of footing (m) (ft), B = width of footing (m) (ft), Nc=cotf(Nq – 1), *see typical bearing capacity factors 2 Nq=e (3p/4-f/2)tanf / [2 cos2(45+f/2)], *see typical bearing capacity factors 2 N g=(1/2) tanf(kp /cos f - 1), *see typical bearing capacity factors e = Napier's constant = 2.718..., kp = passive pressure coefficient, and f = angle of internal friction (degrees). Notes: Effective unit weight, g, is the unit weight of the soil for soils above the water table and capillary rise. For saturated soils, the effective unit weight is the unit

weight of water, gw, 9.81 kN/m3 (62.4 lb/ft3), subtracted from the saturated unit weight of soil. Find more information in the foundations section.

Meyerhof Bearing Capacity Theory Based on Standard Penetration Test Values Qu = 31.417(NB + ND)

(kN/m2)

(metric)

Qu = NB 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).

These equations are for shallow footings, where D = 10, or 1.0 otherwise s-q = 1 + 0.1*Kp*B/L for PHI >= 10, or 1.0 otherwise i-q = i-c = (1 - theta/90)^2 Ngam = wedge weight multiplier = (Nq - 1)*tan(1.4*PHI) d-gam = d-q = 1 + 0.1*sqrt(Kp)*D/B for PHI >= 10, or 1.0 otherwise s-gam = s-q = 1 + 0.1*Kp*B/L for PHI >= 10, or 1.0 otherwise i-gam = (1 - theta/PHI)^2 where PHI is in degrees or i-gam = 0 if PHI = 0 or theta > PHI Input values are obtained from tests of soil samples.

Hansen's Bearing-Capacity Equations For footings that support eccentric loads. Hansen (1970) proposed bearing-capacity equations that are an extension of the earlier Meyerhof work. Hansen's equations allow any D/B and thus can be used for shallow footings and deep bases. The spreadsheet is based on Chapter 4-3 of Foundation Analysis and Design, Ed. 5 by Joseph E. Bowles (permission granted). The equations for vertical load with moments on a horizontal base are: PHI>0: q-ult = c*Nc*d-c*s-c + q-bar*Nq*d-q*s-q + 0.5*gamma*B'*Ngam*d-gam*sgam PHI=0: q-ult = 5.14*su*(1 + s'c + d'c) + q-bar where: c = soil cohesion in kPa su = undrained shear strength in kPa Nc = cohesion multiplier = (Nq - 1)*cot(PHI) d-c = depth factor for cohesion = 1 + 0.4*k d'c = depth factor for PHI = 0, d'c = 0.4*k where k = D/B for D/B 144 kN/m2 Qa > 3000 lbs/ft2

o.k.

o.k.

metric

standard

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 (12inch) square isolated footing bearing on saturated cohesive soil. The frost penetration depth is 0.61 meter (2 feet). Structural parameters require the foundation to withstand 4.4 kN (1000 lbs) of force on a 0.3 meter (12-inch) square column. Given · · · · · · ·

bearing pressure from building column = 4.4 kN/ (0.3 m x 0.3 m) = 48.9 kN/m2 bearing pressure from building column = 1000 lbs/ (1 ft x 1 ft) = 1000 lbs/ft2 unit weight of saturated soil, gsat= 20.3 kN/m3 (129 lbs/ft3) *see typical g values unit weight of water, gw= 9.81 kN/m3 (62.4 lbs/ft3) *constant Cohesion, c = 21.1 kN/m2 (440 lbs/ft2) *from soil testing, see typical c values angle of Internal Friction, f = 0 degrees *from soil testing, see typical f values footing width, B = 0.3 m (1 ft)

Solution Try a footing depth, D = 0.61 meters (2 feet), because foundation should be below frost depth. Use a factor of safety, F.S = 3. See factor of safety for more information. Determine bearing capacity factors Ng, Nc and Nq. See typical bearing capacity factors relating to the soils' angle of internal friction. · ·

Ng = 0 Nc = 5.7

·

Nq = 1

Solve for ultimate bearing capacity, Qu = 1.3c Nc + g D Nq + 0.4 g B Ng

*square footing eq.

Qu =1.3(21.1kN/m2)5.7+(20.3kN/m3-9.81kN/m3)(0.61m)1+0.4(20.3kN/m39.81kN/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

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 N Value, N = 36 at 0.61 m (2 ft) depth N Value, N = 50 at 1.5 m (5 ft) depth

*from SPT soil testing *from SPT soil testing *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 3 3 o unit weight, g = 19.6 kN/m (125 lbs/ft ) *from soil testing, see typical g values 2 2 o cohesion, c = 47.9 kN/m (1000 lb/ft ) *from soil testing, see typical c values o angle of internal friction, f = 30 degrees *from soil testing, see typical 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 regional soils. Determine ultimate end bearing of pile, Qp = Apqp Ap = p(B/2)2 = p(0.61m/2)2 = 0.292 m2 Ap = p(B/2)2 = p(2ft/2)2 = 3.14 ft2

metric standard

qp = gDNq g = 19.6 kN/m3 (125 lbs/ft3); given soil unit weight f = 30 degrees; given soil angle of internal friction B = 0.61 m (2 ft); trial pile width D = 1.5 m (5 ft); trial depth, may need to increase or decrease depending on capacity check to see if D < Dc Dc = 15B = 9.2 m (30 ft); critical depth for medium dense silts.

If D > Dc, then use Dc Nq = 25; Meyerhof bearing capacity factor for driven piles, based on f qp = 19.6 kN/m3(1.5 m)25 = 735 kN/m2 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 = 2p(0.61m/2) = 1.92 m metric p = 2p(2 ft/2) = 6.28 ft standard L = D = 1.5 m (5 ft); length and depth used interchangeably. check Dc as above Af = 1.92 m(1.5 m) = 2.88 m2 Af = 6.28 ft(5 ft) = 31.4 ft2

metric standard

qf = cA + ks tan d = cA + kgD tan d k = 0.5; lateral earth pressure coefficient for piles, value chosen from Broms low density steel g = 19.6 kN/m3 (125 lb/ft3); given effective soil unit weight. If water table, then g - gw D = L = 1.5 m (5 ft); pile length. Check to see if D < Dc Dc = 15B = 9.2 m (30 ft); critical depth for medium dense silts. If D > Dc, then use Dc d = 20 deg; external friction angle, equation chosen from Broms steel piles B = 0.61 m (2 ft); selected pile diameter cA = 0.5c; for clean steel. See adhesion in pile theories above. = 24 kN/m2 (500 lb/ft2) qf = 24 kN/m2 + 0.5(19.6 kN/m3)(1.5m)tan 20 = 29.4 kN/m2 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

Determine ultimate pile capacity,

metric standard metric standard

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 = 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

Conclusion A 0.61 m (2 ft) steel pile shall be plugged and driven 1.5 m (5 feet) below the ground surface. Many engineers neglect the skin friction within the upper 1 to 5 feet of subgrade due to seasonal variations or soil disturbance. Seasonal variations may include freeze/ thaw or effects from water. The end bearing alone (neglect skin friction) is sufficient for this case. Typical methods for increasing the pile capacity are increasing the pile diameter or increasing the embedment depth of the pile. ********************************

Example #5: Determine allowable bearing capacity and diameter of a single driven pile. Pile must withstand a 66.7 kN (15 kips) vertical load. Given · ·

·

vertical column load = 66.7 kN (15 kips or 15,000 lb) upper 1.5 m (5 ft) of soil is a medium dense gravelly sand 3 3 o unit weight, g = 19.6 kN/m (125 lbs/ft ) *from soil testing, see typical g values o cohesion, c = 0 *from soil testing, see typical c values o angle of internal friction, f = 30 degrees *from soil testing, see typical f values soils below 1.5 m (5 ft) of soil is a stiff silty clay

unit weight, g = 18.9 kN/m3 (120 lbs/ft3) cohesion, c = 47.9 kN/m2 (1000 lb/ft2) angle of internal friction, f = 0 degrees Pile Information o driven o wood o closed end o o o

·

Solution Try a pile depth, D = 2.4 meters (8 feet) Try pile diameter, B = 0.61 m (2 ft) Use a factor of safety, F.S = 3. Smaller factors of safety are sometimes used if piles are load tested, or the engineer has sufficient experience with the regional soils. Determine ultimate end bearing of pile, Qp = Apqp Ap = p(B/2)2 = p(0.61m/2)2 = 0.292 m2 Ap = p(B/2)2 = p(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 = pSqfL p = 2p(0.61m/2) = 1.92 m p = 2p(2 ft/2) = 6.28 ft

upper 1.5 m (5 ft) of soil

metric standard

qfL = [ks tan d]L = [kgD tan d]L k = 1.5; lateral earth pressure coefficient for piles, value chosen from Broms low density timber g = 19.6 kN/m3 (125 lb/ft3); given effective soil unit weight. If water table, then g - gw D = L = 1.5 m (5 ft); segment of pile within this soil strata. Check to see if D < Dc Dc = 15B = 9.2 m (30 ft); critical depth for medium dense sands. This assumption is conservative, because the soil is gravelly, and this much soil unit weight for a sand would indicate dense soils. If D > Dc, then use Dc d = f(2/3) = 20 deg; external friction angle, equation chosen from Broms timber piles B = 0.61 m (2 ft); selected pile diameter f = 30 deg; given soil angle of internal friction qfL = [1.5(19.6 kN/m3)(1.5m)tan (20)]1.5 m = 24.1 kN/m 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 = aSu Suc = 2c = 95.8 kN/m2 (2000 lb/ft2); unconfined compressive strength c = 47.9 kN/m2 (1000 lb/ft2); cohesion from soil testing (given) a = 1 [0.9 + 0.3(Suc - 1)] = 0.3; because Suc > 48 kN/m2, (1 ksf) Suc L = 0.91 m (3 ft); segment of pile within this soil strata qfL = [0.3(95.8 kN/m2)]0.91 m = 26.2 kN/m qfL = [0.3(2000 lb/ft2)]3 ft = 1800 lb/ft

metric standard

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

Determine ultimate pile capacity, Qult = Qp + Qf

metric standard

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

metric standard

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

222.5 kN = 74.2 kN; Qa > applied load (66.7 kN) o.k. metric 3 50,275 lbs = 16,758 lbs Qa > applied load (15 kips) o.k. standard 3

Conclusion Wood pile shall be driven 8 feet below the ground surface. Many engineers neglect the skin friction within the upper 1 to 5 feet of subgrade due to seasonal variations or soil disturbance. Seasonal variations may include freeze/ thaw or effects from water. Notice how the soil properties within the pile tip location is used in the end bearing calculations. End bearing should also consider the soil layer(s) directly beneath this layer. Engineering judgment or a change in design is warranted if subsequent soil layers are weaker than the soils within the vicinity of the pile tip. Typical methods for increasing the pile capacity are increasing the pile diameter or increasing the embedment depth of the pile. ********************************

Settlement Analysis, Pressure & Stress Guidance Settlement of Coarse-Grained Granular Soils for Static Loads

Settlement analysis of non-cohesive, coarse-grained soils is usually limited to the immediate settlement analysis. Settlement of these soil types primarily occur from the re-arrangement of soil particles due to the immediate compression from the applied load. Typically, engineers justify ignoring the dissipation of pore water pressure based on the assumption that these soil types have a large permeability rate, thus releasing the excess pore water as the load(s) is applied. Earthquakes and other ground movement such as machinery or blasting, may also cause settlements in non-cohesive soils, which may be immediate settlements in the future. To learn more about settlements due to earthquakes

and vibrations, review "Settlement Analysis" by the USACE in the settlement section of the geotechnical publications page.

Immediate Settlement of Non-Cohesive Soils, NAVFAC Method

DHi =

4qB2 Kvi(B + 1)2

for isolated shallow footings and B < 20 ft

DHi =

2qB2 Kvi(B + 1)2

for isolated shallow footings and B > 40 ft

DHi =

2qB2 Kvi(B + 1)2

for deep isolated foundations and B < 20 ft

Where: DHi = immediate settlement of footing (ft) q = footing unit load, bearing pressure (tsf) B = footing width (ft) Kvi = Modulus of vertical subgrade reaction (tons/ft3) Notes: 1. For continuous footings, multiply the computed settlement by 2. 2. Non-cohesive soils include gravels, sands and non-plastic silts. 3. Multiply Kvi by 0.5 if groundwater is at base of footing or above, multiply Kvi by 1.0 if groundwater is at least 1.5B below base of footing, and interpolate multiplication factor for groundwater between base of footing and 1.5B below the footing base.

Immediate Settlement of Non-Cohesive Soils, Alpan Approximation

p = m'[2B/(1 + B)]2(a/12)q Where: p = immediate settlement (ft) m' = shape factor = (L/B)0.39 L = footing length (ft)

B = footing width (ft) a = Alpan parameter q = average bearing pressure applied by footing on soil (tsf)

Settlement of Cohesive Soils for Static Loads

Total settlement for cohesive soils are generally estimated by the sum of immediate settlement, primary consolidation and secondary compression, where immediate settlement usually constitutes a significant portion of the total settlement.

Immediate Settlement of Cohesive Soils, Janbu Approximation

p = (u0)(u1) qB Es Where: p = immediate settlement (ft) u0 = Influence factor for depth foundation u1 = Influence factor for shape of foundation q = footing unit load, bearing capacity (tsf) B = footing width (ft) Es = Young's modulus of soil (tsf)

Primary Consolidation

Review free and downloadable settlement publications in our Settlement Analysis Publications

Secondary Compression

Review free and downloadable settlement publications in our Settlement Analysis Publications

Settlement of Single Pile (Vesic)

The following calculations were first derived by Vesic, and can be found in the USACE manual. See USACE EM 1110-2-2906 - Design of Pile Foundations. w = ws + wf + wp where, w = vertical settlement of a single pile at the top of pile, m (ft) ws = (Qp + asQf) L m (ft) AE = amount of settlement due to the axial deformation of the pile shaft wf = Cs(Qs) m (ft) Dqp = amount of settlement at the pile tip caused by load transmitted along the pile shaft wp = Cp(Qp) m (ft) Bqp = amount of settlement at the pile tip due to the load transferred at the tip and, Qp = Theoretical bearing capacity for tip of foundation, kN (lb) see deep foundations in bearing capacity technical guidance Qf = Theoretical bearing capacity due to shaft friction, kN (lb) see deep foundations in bearing capacity technical guidance qp = Theoretical unit tip bearing capacity, kN/m2 (lb/ft2) = see deep foundations in bearing capacity technical guidance L = Length of pile, m (ft)

A = Cross-sectional area of pile, m2 (ft2) = p(B/2)2 for closed end piles E = Modulus of elasticity of the pile material, kN/m2 (lb/ft2) as = Alpha approximation. See referenced manual for long piles in dense soils or flexible shafts. D = Embedment depth of the pile, m (ft) B = Diameter of pile, m (ft) Cs = Cp[0.93 + 0.16(D/B)0.5] Cp = empirical coefficient: See referenced manual for different soils within 10B of the pile tip. soil type Driven Piles Bored Piles sand (dense to loose) 0.02 to 0.04 0.09 to 0.18 silt (dense to loose) 0.03 to 0.05 0.09 to 0.12 clay (stiff to soft) 0.02 to 0.03 0.03 to 0.06

Pressure Analysis

Inducing pressure on a soil from structural loads is sometimes important to calculate, especially for settlement analyses. The pressure, or stress, on the soil where the structure is in contact with the soil is simply the structural load. The methods provided below will allow us to determine the additional pressure on a soil at a certain depth below the contact point of the structure. This includes a pressure bulb, and the Boussinesq theory. Pressure Bulb Knowing the amount of applied building loads and the designed footing width, we can use charts to determine the additional stress on a soil at specified depths beneath the footing and points beyond the foundation footprint. See the following source from the NAVFAC manual: ·

Pressure Bulb

Boussinesq Theory The change in soil pressure due to an applied load may be calculated from the following methods:

Circular Foundation Pv =

3q 2pz2

[1/(1+(r/z)2]-2.5

Where: Pv = change in vertical stress at point z below the center of a circularly loaded area, and point r horizontally from the center of the circularly loaded area, lb/ft2 q = applied stress from structural load, lb/ft2 p = 3.1412... z = depth below center of circularly loaded area in which a change in vertical stress is desired, ft r = horizontal distance from the center of a circularly loaded area in which a change in vertical stress is desired, ft

Rectangular Foundation Dsv = SPv Where: Dsv = total change in vertical stress due to an applied stress, kN/m2 (lb/ft2) SPv = summation of all stress components (i.e. Pv1 + Pv2 + .... + Pvn) Pv = qIs = change in vertical stress at point z below the corner of a rectangular loaded area, kN/m2 (lb/ft2) If the vertical stress is desired below the middle of the foundation, where the stress is maximum, the rectangular footing must be divided into 4 sections, or quadrants, so that the corner of each section is located in the middle of the foundation. Thus we calculate the change in vertical stress from 4 different sections. The total change in vertical stress is the summation of the 4 different sections. q = applied stress from structural load, kN/m2 (lb/ft2) Is = Influence value from Boussinesq chart. Is is determined from chart using m and n values. m= x z n= y z z = depth below corner of rectangular loaded area in which a change in vertical stress is desired, m (ft) x = length of foundation, m (ft) y = width of foundation, m (ft)

Stress Analysis of Soil

The following equations will aid in determining pore water pressure, total stress and effective stress. See examples below for sample calculations. s' = s - m Where: s' = s - m, kN/m2 (lb/ft2) s = gD, kN/m2 (lb/ft2) m = gwd, kN/m2 (lb/ft2)

effective stress total stress pore water pressure (see note below)

and, g = unit weight of soil, kN/m2 (lb/ft2) D = depth of overburden, m (ft), or vertical distance between surface of soil to a subsurface depth at which total stress is determined. gw = 9.81 kN/m3 (62.4 lb/ft3) = unit weight of water, constant d = depth of water, m (ft), or vertical distance between surface of water table to a subsurface depth at which pore water pressure is determined. Notes: Capillary rise will increase the effective stress by creating a negative pore water pressure within the capillary zone. Capillary rise above the water table is sometimes calculated as: hc =

0.15 D10

where D10 = soil grain size diameter at the 10% finer of the particle size distribution

Examples for calculating settlement, stress & pressures Example #1: The soil profile indicates a soil unit weight of 17.28 kN/m3 (110 lb/ft3) to a depth of 6.1 m (20 ft) below the ground surface. The water table is at 6.1 m 20 ft

below the ground surface. A saturated soil unit weight of 20.42 kN/m3 (130 lb/ft3) extends to a depth of 7.6 m (25 ft) below the water table. A capillary rise of 1.5 m (5 ft) was determined to exist above the water table. a) Calculate the effective stress at a depth of 3.0 m (10 ft) below the ground surface. b) Calculate the effective stress at a depth of 5.5 m (18 ft) below the ground surface. c) Calculate the effective stress at a depth of 6.1 m (20 ft) below the ground surface. d) Calculate the effective stress at a depth of 13.7 m (45 ft) below the ground surface.

Given · · · · · ·

upper soil profile is 6.1 m (20 ft) deep upper soil profile has a unit weight, g, of 17.28 kN/m3 (110 lbs/ft3) lower soil profile is 7.6 m (25 ft) thick lower soil saturated unit weight, g, is 20.42 kN/m3 (130 lbs/ft3) water table is at 6.1 m (20 ft) below the ground surface capillary rise is 1.5 m (5 ft) above the water table

Solution a) s = gD = (17.28 kN/m3)(3.0 m) = 51.8 kN/m2 = (110 lb/ft3)(10 ft) = 1100 lb/ft2 m = g wd =0

metric standard

because depth is above influence of the water table

s' = s - m = 51.8 kN/m2 - 0 = 51.8 kN/m2 = 1100 lb/ft2 - 0 = 1100 lb/ft2

metric standard

b) s = gD = (17.28 kN/m3)(5.5 m) = 95.0 kN/m2 = (110 lb/ft3)(18 ft) = 1980 lb/ft2

metric standard

m = g wd = (9.81 kN/m3)(- 0.71 m) = - 6.97 kN/m2 = (62.4 lb/ft3)(- 2 ft) = - 124.8 lb/ft2

metric standard

s' = s - m = 95.0 kN/m2 - (- 6.97 kN/m2) = 102.0 kN/m2 = 1980 lb/ft2 - (- 124.8 lb/ft2) = 2105 lb/ft2

metric standard

c) s = gD = (17.28 kN/m3)(6.1 m) = 105.4 kN/m2 = (110 lb/ft3)(20 ft) = 2200 lb/ft2

metric standard

m = g wd = (9.81 kN/m3)(0 m) = 0 = (62.4 lb/ft3)(0 ft) = 0

metric standard

s' = s - m = 105.4 kN/m2 - 0 = 105.4 kN/m2 = 2200 lb/ft2 - 0 = 2200 lb/ft2

metric standard

d) s = gD = (17.3 kN/m3)(6.1 m) + (20.4 kN/m3)(7.6 m) = 260.6 kN/m2 = (110 lb/ft3)(20 ft) + (130 lb/ft3)(25 ft) = 5450 lb/ft2 m = g wd = (9.81 kN/m3)(7.6 m) = 74.6 kN/m2 = (62.4 lb/ft3)(25 ft) = 1560 lb/ft2

metric standard

s' = s - m = 260.6 kN/m2 - 74.6 kN/m2 = 186.0 kN/m2 = 5450 lb/ft2 - 1560 lb/ft2 = 3890 lb/ft2

metric standard

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

metric standard

Example #2: Using the Boussinesq theory, calculate the change in vertical stress at 0.61 m (2 ft) below the middle of a 1.2 m x 1.8 m (4 ft x 6 ft) rectangular foundation. The applied building load on this foundation is 167.6 kN/m2 (3500 lb/ft2).

Given · · ·

z = 0.61 m (2 ft) *See the theory, equations and definitions provided above q = 167.6 kN/m2 (3500 lb/ft2) 1.2 m x 1.8 m (4 ft x 6 ft) rectangular footing

Solution Dsv = SPv SPv = summation of all stress components (i.e. Pv1 + Pv2 + .... + Pvn). In this case, we analyze the foundation in 4 equal but separate quadrants. Instead of a single 1.2 m x 1.8 m (4 ft x 6 ft) foundation, we have 4 separate 0.61 m x 0.91 m (2 ft x 3 ft) quadrants. This is done so that one corner of each quadrant is located in the center of the footing.

4Pv = 4qIs Since the quadrants have equal dimensions with the same applied load, we simply multiply the equation by 4 (4 quadrants). If footing had varying applied loads or unequal quadrant shapes, then each stress summation must be done separately.

m= x z m= x z

=

0.91 m = 1.5 0.61 m = 3.0 ft = 1.5 2.0 ft

n= y z n= y z

= =

0.61 m = 1.0 0.61 m 2.0 ft = 1.0 2.0 ft

metric standard

metric standard

Is = 0.2 Influence value from Boussinesq chart, where m = 1.5 and n= 1.0. Dsv = SPv = 4Pv = 4qIs

= 4(167.6 kN/m2)(0.2) = 134.1 kN/m2 = 4(3500 lb/ft2)(0.2) = 2800 lb/ft2

metric standard

Conclusion Additional pressure on the soil at a distance of 0.61 m (2 ft) below the center of the footing due to an applied building load of 167.6 kN/m2 (3500 lb/ft2) is 134.1 kN/m2 (2800 lb/ft2).

COMPACTION

Example #1: A project requires fill to be compacted to 95% relative density with relation to the standard Proctor (ASTM D698). Laboratory results for the standard Proctor indicated that the soil has a maximum dry density of 19.0 kN/m3 (121 lb/ft3), and an optimum moisture content of 8.9%. After compaction of the fill soils with a vibratory roller, field testing with a sand cone, nuclear densiometer, or other appropriate method indicated that the compacted fill soils have an in-place unit weight of 18.76 kN/m3 (124.4 lb/ft3), and a moisture content of 7.5%. Calculate the relative compaction, and does the compacted fill exceed project requirements?

Given gm = 19.0 kN/m3 (121 lbs/ft3) mo = 8.9% g = 19.54 kN/m3 (124.4 lbs/ft3) m = 7.5% Rd = 95%

maximum dry density optimum moisture content in-situ density in-situ moisture content required relative compaction per project specifications

Solution Verify that compacted fill meets or exceeds compaction requirements, Rd > 95% Rd = gd  gm

gd = g - g(m) dry density of the in-situ soil 100 gd =19.54 kN/m3 - 19.54 kN/m3(7.5%) = 18.07 kN/m3 100 3 gd =124.4 lb/ft - 124.4 lb/ft3(7.5%) = 115.1 lb/ft3 100

metric

Rd = 18.07 kN/m3 = 95.1% > 95%  19.0 kN/m3

o.k.

metric

Rd = 115.1 lb/ft3 = 95.1% > 95%  121 lb/ft3

o.k.

standard

standard

Conclusion The compacted fill exceeds project requirements of at least 95% relative density. *****************************

Example #2: A project requires fill to be compacted to 100% relative density with relation to the standard Proctor (ASTM D698). The fill has been vigorously compacted to a relative density of 96.9%. Subsequent compacting does not increase the relative density. What could be the problem?

Solution 1) Check the moisture content of the compacted fill. Depending on the soil type, an insitu moisture content deviating 2% to 4% from the optimum moisture content as determined from the Proctor test, may create impossible conditions to achieve the required compaction. If this is the case, scarify soil and add moisture (or let dry), and recompact at the optimum moisture content. Sometimes, complete removal and replacement of the soil is necessary. 2) Verify the maximum dry density as determined from the Proctor test still holds true for the 'un-compactible' soils. Sometimes the maximum dry density changes as different soils are excavated from the borrow pit. If this is the case, use the new maximum dry density value when determining the relative density. 3) Check compaction methods. Type of equipment used for compaction and the depth of compacted lifts make a difference in the relative compaction.

4) Check for inadequate compaction in underlying lifts. Sometimes achieving adequate relative density is impossible when compacting soils on top of loose or unconsolidated soils. *******************************

EARTHWORK/ COMPACTION/ PHASE DIAGRAM

Example #3: This is in part, a phase diagram problem. A project requires fill to be compacted to 95% relative density with relation to the standard Proctor (ASTM D698). Laboratory results for the standard Proctor indicated that the soil has a maximum dry density of 19.49 kN/m3 (124 lb/ft3), and an optimum moisture content of 9.5%. Borrow soil from another location that will be used as compacted fill for this project has a moisture content of 12%, a void ratio of 0.6, and a specific gravity of 2.65. Assuming that no moisture is lost during transport, what is the volume of borrow required that is needed for 28.32 m3 (1000 ft3) of compacted fill?

Given gm = 19.49 kN/m3 (124 lbs/ft3) mo = 8.9% e = 0.6 Gs = 2.65 m = 12.0% Rd = 95% VT = 28.32 m3 (1000 ft3) gw = 9.81 kN/m3 (62.4 lbs/ft3)

maximum dry density optimum moisture content void ratio of borrow soil specific gravity of soil moisture content of soil required relative compaction per project specifications total soil volume of required fill unit weight of water (constant)

Solution Find dry unit weight, gd, of soil required for 95% compaction. gd = Rd gm 100 = 0.95(19.49 kN/m3) = 18.52 kN/m3 = 0.95(124.0 lb/ft3) = 117.8 lb/ft3

metric standard

Calculate the weight of the soil solids, Ws, required for 95% compaction. The weight of the soil solids will be equal for both the fill and borrow material because only volume changes via compaction. Ws = gd (VT) = 18.52 kN/m3 (28.32 m3) = 524.5 kN = 117.8 lb/ft3 (1000 ft3) = 117,800 lb

*see notes within conclusion metric standard

Determine the volume of soil solids, Vs, required for 95% compaction. Vs =

Ws Gs (gw) = 524.5 kN = 20.18 m3 3 2.65(9.81 kN/m ) = 117,800 lb = 712.4 ft3 3 2.65(62.4 lb/ft )

metric standard

Find the volume of voids, Vv, for the borrow material Vv = e (Vs) = 0.6(20.18 m3) = 12.11 m3 = 0.6(712.4 ft3) = 427.4 ft3

metric standard

Calculate the total volume, VT, of the borrow soil VT = Vv + Vs = 12.11 m3 + 20.18 m3 = 32.3 m3 = 427.4 ft3 + 712.4 ft3 = 1140 ft3

metric standard

Conclusion The volume of soil required from the borrow pit is 32.3 m3 (1140 ft3). Equations used for this problem are standard phase diagram relationships shown here. Other phase diagram equations may be required depending on the situation.