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DEEP FOUNDATIONS
Example: (i) How may the load carrying capacity of a single driven pile and a single bored pile be assessed? Discuss the differences in behaviour of a single and a group of similar piles and distinguish between block failure and local failure of a group. (ii) A large diameter straight-shafted bored pile is installed in a thick layer of overconsolidated clay whose shear strength at depth z is excess 3m below ground level is given by c = [ 75 + 1.5 ( z F 02 D 3)] kPa. The pile is 1.25 m in diameter and has an embedded length of 15 m. Calculate the ultimate pile capacity stating the shaft adhesion factor used, and making an appropriate allowance for loss of adhesion between pile and clay due to seasonal moisture movement near the ground surface. What would you suggest as the working load of this pile? Give your reasons. Solution: A driven pile is one which is hammered into the ground and in the process displaces a volume of soil equal to the volume of the pile. A bored pile is one which is formed by the removal of soil by boring. The load-bearing capacity of a single pile may be assessed by considering the measured soil properties adjacent to the pile. A general formula is: Qf = Q (^) b + Qs where Qf = ultimate load-carrying capacity of pile Qb = ultimate base resistance due to bearing capacity of ground at toe Qs = ultimate shaft resistance due to skin friction over length of pile For a cohesionless soil: Qb may be calculated using Terzaghi’s formula for a circular or square footing q b =^ F 07 3F 0A 2 v N^ q +0.3^ F 06 7sub BN^ F 0 6 7 where F 07 3F 0A 2 v is the effective overburden pressure at the pile toe. It is generally assumed for deep foundations that the last term is small and may be neglected, that the weight of soil removed is equal to the weight of the pile replacing it and that the water table is at ground level. Qb = F 07 3F 0A 2 v N (^) q Ab where A (^) b = area of pile base Meyerhof has suggested that the ultimate skin friction on the side of a pile qs =^ K^ s F 07 3F 0A 2 vm tan^ F 06 4 where K (^) s = coefficient of lateral earth pressure acting horizontally on side of pile F 07 3F 0A 2vm = average effective overburden pressure on side of pile
F 06 4 = angle of friction between soil and pile Qs = K (^) s F 07 3F 0A 2 vm As tan F 06 4 where As = surface area of pile in the ground Typical values of K (^) s are 1.0 for loose sand and 2.0 for dense sand; F 06 4 may be taken as 0.75. The values of N (^) q shown in Figure below should be used.
For a cohesive soil: Qb = cb N (^) c Ab Where cb = undrained shear strength of soil at base of pile N (^) c = bearing-capacity factor (usually taken as 9) Qs = cs F 06 1 As Wher e c (^) s = average undrained shear strength of soil adjacent to pile shaft F 06 1= shaft adhesion factor The most difficult value to assign is often the shaft adhesion factor F 06 1. In the absence of particular evidence, values of 0.8 for a driven concrete pile and 0.45 for a bored concrete pile may be assumed. Both types of pile may be subjected to loading tests, when a maximum allowable settlement is specified under a load of, say, 1½ times the expected working load. The load-carrying capacity of a group of piles will not be the same as the load-carrying capacity of a single pile times the number of piles in the group. That is because the pressure bulbs for each pile tend to overlap, indicating a greater stress concentration on the surrounding soil. Figure (a) shows this and also that the influence of a group of piles is more extensive than that of a single pile, and this leads to greater settlement of the pile group. Excessive settlement of the pile group may cause local failure. If, however, a pile group is overloaded, it may fail as a block by breaking into the ground as on Figure (b). For the given pile, the shear strength of the soil adjacent to the shaft is shown on Figure (c).
Neglecting the difference between the weight of the pile and the excavated soil: Qu = Q (^) b + Q (^) sT
Qs = cs α A (^) s and taking the shaft adhesion factor as 0. = neglecting the adhesion along the top 3 m of pile to allow for seasonal moisture movement.
- Qf = 1027 + 1780 = 2807 kN
s = 0.01 m
Example: A cast in–situ bored pile, formed in a saturated clay stratum, has a shaft length of 14.5 m and a shaft diameter of 0.94 m. The undrained shear strength of the clay at base level and for several metres beyond is 150 kN/m 2 and the average shear strength value over the shaft length is 128 kN/m^2 Estimate the vertical failure load and working load that may be applied externally to the pile for: i) this parallel–sided pile, ii) this pile if it were extended by under-reaming (at 1:3) to a base diameter of 1.86 m. Assume the values of factors of safety F to be F (^) overall = 2, F (^) skin friction = 1.5 and F (^) end bearing = 3. The skin friction coefficient F 06 1 for a bored concrete pile is 0.45 and Skempton’s bearing- capacity coefficient N (^) c for a deep circular foundation is 9. Solution: i) The vertical failure load is Qf =^ Q^ b +^ Qs = c (^) b. N (^) c. A (^) b + c (^) s. F 06 1. As = = 937 + 2466 = 3403 kN This is the vertical failure load. Applying the given factors of safety: Overall Qa = 3403/2 = 1701 kN End bearing Q (^) ab = 937/3 = 312 kN Skin friction Q (^) as = 2466/1.5 = 1644 kN F 05 C Qab + Q (^) as = 1956 kN
The working load will be the lower of these two figures, i.e.1701 kN ii) If the pile is extended by under-reaming, the size of the base is increased but, because of the disturbance caused, it is usual to neglect the skin friction above the under-reaming for a distance of 2 F 0B 4 diameter of base. Qf = 150 F 0B 4 9 F 0B 4+ 128 F 0B 4 0.45 F 0B 4 ( F 07 0^ F 0B 4 0.94 F 0B 49.40) = 3668 + 1599 = 5267 kN = failure load Overall Qa = 5267/2 = 2634 kN
End bearing Q (^) ab = 3668/3 = 1223 kN Skin friction Q (^) as = 1599/1.5 = 1066 kN 2289 kN Working load = 2289 kN
Example: What do you understand by the following terms, in relation to piling works: negative skin friction; efficiency and settlement ratio of a pile group; shaft adhesion factor? A group of concrete piles, square in plan consists of sixteen piles, each 18 m long and 0.45 m in diameter. They are installed at 1.10 m centres in a deep deposit of estuarine clay having a mean unconfined compressive strength over the pile length of cu = 30 kN/ m^2. At and below the level of the pile bases the mean strength is^ c^ u = 40 kN/m^2. The mean unit weights of the soil and concrete are respectively 19.2 kN/m^3 and 24.0 kN/m 3. Stating clearly your assumptions concerning bearing-capacity and shaft adhesion factors estimate the superimposed load required to cause failure of a single pile if installed ( a ) by boring, ( b ) by driving. What total working load would you recommend for the pile group immediately after its installation by driving? Solution: Negative skin friction : In general, friction between the pile shaft and the surrounding soil tends to increase the load-carrying capacity of the pile. In special circumstances, such as piles driven through soil which tends to consolidate under its own weight or compressible fills, a drag-down force will act on the pile and reduces its bearing capacity. Such a force is known as negative skin friction. Efficiency ratio = Settlement ratio = Shaft adhesion factor = (a) Single bored pile: Qf = c (^) b N (^) c Ab + cs F 06 1 As and using a shaft adhesion factor of 0. Qf = 409+ = 57+343 = 400 kN If the difference in weight between the pile and the displaced soil is taken into account, this figure is reduced by (24.0 F 02 D 19.2) = 14 kN, i.e.3.4% (b) Single driven pile: A shaft adhesion factor of 0.8 may be used, so
Bored piles should be spaced at about 3 times their diameter and a suggested arrangement is shown on Figure below:
The action of the group must be checked for stability. Consider a pier 2.35m square: Qfg = 125+115). = 6212+4378 = 10590 kN Qfg = = 3530 kN Since this is more than the load of 2000 kN, the arrangement should be satisfactory from a load-carrying point of view. If the piles had been spaced at 1.5diameter, the block would measure (0.35 = 1.4 m each side. F 05 C Qfg = 125+ = 4813 kN Qfa = = 1604 kN which is inadequate. Spacing should be at least 2diameter of piles. Example: A mass concrete pier of circular section is to carry a load, including its own weight, of 2500 kN. The pier base is to be located 4.5 m below ground level in a deep saturated clay. The bulk density of the clay is 1.84 Mg/m 3 and cu average = 50 kPa. Choosing a suitable value for the bearing capacity factor N (^) c , determine the diameter of the pier to give FOS = 3 against shear failure. Assume that cohesion 0.3 cu can be mobilized on the cylindrical surface of the pier below ground level. Solution: Net applied load = 2500 – weight of soil excavated = 2500 – (1.84) F 0B 49.81 F 0B 44.5 F 0B 4 = 2500 – 63.80 D^2 (i)
• 294.5 D^2 + 212 D = 7500 – 191 D^2
or 485.5 D^2 + 212 D = 7500 (ii)
Example:
A group of nine bearing piles of 300 mm diameter are driven in a square grid with the lines of the piles at 900 mm centres. The piles are cast-in-situ and are 6.5 m long. The soil is a deep deposit of clay having the following properties: cu = 72 kN/m^2 , φ (^) u = 0, γ = 2.00 Mg/m^3. Determine the bearing capacity of the pile group if the minimum factor of safety is 3. Solution: Using 0.8 as adhesion factor for a driven pile
F 05 C Qfg = 2857.68 + 3144.96 = 6002.64 kN F 05 C Qfa gross = Example: A series of undrained triaxial tests on clay taken from the site of a proposed building in central London gave the following results: O.D level Shearing resistance (kN/ m^2 )
Bulk density (Mg/m^3 ) Ground surface +6m - 2. 0 165 1. -6 210 1. -12 250 1. -18 295 2. -24 330 2.
The top 4.5 m is Thames Ballast. Estimate the load which a bored pile of 1.40 m diameter and 30 m in depth below ground surface can carry in this clay if the bottom is belled out at an angle of 60 F 0B 0 to the horizontal to a diameter of 3.5 m.
Solution:
(i) From Figure Effective shaft length, l = 30 – (1.82 + 7.0) = 21.18 m Effective length in clay below El 0.0, l F 0A 2 = 21.18 – 6.0 = 15.18 m Calculate Q (^) s as follows: Shaft length (m) Soil type Average c (^) s (kPa)
γ (kN/m 3 ) α cs α As (kN) 4.5 Thames Ballast -- 20.40 -- --
Qb = mNA (^) b = 400 F 0B 4 16 F 0B 40.0930 = 595.2 kN (Note: m = 400 for driven piles and 120 for bored piles).
Example: A 350 F 0B 4350 mm reinforced concrete pile in a jetty structure is required to carry a maximum compressive load Q (^) u = 400 kN and a net uplift load of 320 kN. The soil conditions are shown below. Estimate the required depth of penetration of the pile. For driven piles, use K/Ko = 2. (Note:^ fs F 0A 3 110 kPa and^ q^ b = 15,000 kPa). Solution: For φ = 31 F 0B 0,^ K^ o = 1 – sinφ^ = 0.485^ F 05 C K^ = 0.484^ F 0B 42 = 0. For concrete ~sand, F 06 4 = 0.75φ = 23.25 F 0B 0
Now Q (^) s 1 = ( f (^) s 1)( As 1) But < 110 kPa F 05 Cuse fs 1 =16.88 kPa
Thus Qs 1 = 16.88 F 0B 4(4 F 0B 40.350 F 0B 49) = 212.70 kN Try 4 m penetration into dense sandy-gravel Now K/K o = 2^ F 05 C K^ = 2^ F 0B 42 = 4^ and
110 kPa F 05 Cuse fs 2 =110 kPa
Thus Qs 2 = 110 F 0B 4(4 F 0B 40.350 F 0B 44) = 616 kN F 05 C Qs = Q (^) s 1 + Qs 2 = 212.7 + 616 = 828.7 kPa F 05 CFactor of safety against uplift = 828.7/320 = 2. Now L (^) b /0.35 = 11.43, for φ = 38.5 F 0B 0, N (^) q = 130 F 05 C qb = 129 F 0B 4130 = 16770 kPa > 15,000 kPa
so use q (^) b = 15,000 kPa F 05 C Qb = (15,000)(0.35 F 0B 40.35) = 1837.5 kN OK Also using Meyerhof (1976) method based on SPT blows:
In each term above, bearing capacity is more than the limiting value, so use limiting values, i.e.
FoS = 1991.5/400 = 4.98 > 4 OK
Example: HP 310 F 0B 41.079 steel pile is driven into sand shown below. Estimate the point capacity of the pile and calculate Qa.
Solution: For HP 310 F 0B 41.079 steel pipe pile Depth of pile section is = 308 mm and width = 310 mm F 05 C Ab = 0.308 F 0B 40.310 = 0.0955 m 2 qb = σ F 0A 2 N (^) q For L/B = 4/0.310 = 12.9 and φ = 40 F 0B 0 N (^) q = 140 F 05 C qb = 224.63 F 0B 4140 = 31448.2 kPa > 15,000 kPa, use q (^) b = 15,000 kPa
and Qb = 15,000 F 0B 40.0955 = 1432.5 kN For steel piles in loose sand K (^) o = 0. For steel piles in dense sand K (^) o = 1. F 05 C
- Qs = ( f (^) s 1 F 0B 4 L (^) 1 + fs 2 F 0B 4 L (^) 2 + f (^) s 3 F 0B 4 L 3) As = (7.1 F 0B 45 + 24.1 F 0B 413 + 74.8 F 0B 44) F 0B 42(0.308 + 0.310) = 801 kN
- Qa =
Example: A bridge pier imposing a total load of 22,000 kN has plan dimensions of 18 F 0B 44.5 m. It is sited on 7.5 m of recently placed loose sand filling ( N = 9 blows/300 mm) overlying 4. m of peaty soft clay ( c = 24 kPa) followed by stiff clay ( c = 60 kPa at 12 m below ground surface increasing to 300 kPa at 25 m below ground surface). Ground water table is at 0. m below ground surface. Design a pile foundation using 0.508 m diameter F 0B 4 20 m long concrete driven cast-in-situ piles. Subsoil parameters are shown in the figure. Solution: It will be desireable to avoid excavating below water level, so we will allow the pile cap to project 0.6 m above the ground surface and we shall have to add the weight of the cap to the working load on the piles.
F 05 CTotal load on pile shaft at surface of stiff clay layer = 34,145/52 = 657 kN Calculate the ultimate load carrying capacity of a 0.508 dia. F 0B 4 20 m long pile below ground surface i.e. 8 m penetration into stiff clay. c at 20 below ground surface = Average cohesion along 8 m shaft = For α = 0.65, unit shaft adhesion = 0.65 F 0B 4134 = 87 kPa F 05 C Qb = end resistance of pile = 208 F 0B 4 9 F 0B 4F 0B 40.508 (^2) = 379 kN F 05 C Qs = 87 F 0B 4( F 07 0F 0B 40.508 F 0B 48) = 1111 kN F 05 C Qu = Q (^) b + Qs = 379 + 1111 = 1490 kN
Thus FOS on structural load = 1490/461 = 3.2 OK FOS on total load = 1490/657 = 2.3 OK The above factors of safety are satisfactory since the group is relatively narrow and the shear strength of the soil increasing rapidly, therefore the group settlement should be relatively small. Check also FOS for block failure: Qu of block = 2 D ( B + L ) α c + 1.3 c Nc B L = 2 F 0B 48 (19.71 + 5.31) F 0B 4 134 F 0B 40.65 + 1.3 F 0B 4 208 F 0B 4 9 F 0B 419.71 F 0B 45. = 262.112 MN FOS = 262.112/34.145 = 7.7 OK
Example: A Volcam drive hammer of following characteristics is used to install a 12 F 0B 2F 0B 4 12 F 0B 2 pre- cast concrete pile: Rated energy, E = 26 kips-ft Efficiency = 80% No. of blows for the last 1 inch penetration = 5 Constant, C = 0.1 inch Estimate the ultimate pile capacity and the working pile load if FOS = 6 Solution: Using the ENR formula for steam hammer
where W (^) r = weight of hammer h = height of fall of hammer s = amount of point penetration per blow F 05 C
(Note: If h = ft and s = inches, multiply Qu by 12) F 05 C
F 05 C qb = 129 F 0B 4 130 = 16,770 kPa > 15,000 kPa use 15,000 kPa, the limiting value F 05 C Qb = 15000 F 0B 4 A (^) b = 15000 F 0B 4 (0.35) 2 = 1837.5 kN Ultimate bearing capacity in compression, Q (^) c =^ Q^ s1 +^ Q^ b = 835.24 + 1837.5 = 2672.74 kPa Factor of safety against compression = 2672.74/400 = 6.7 > 3 OK As factor of safety against uplift is less than that for compression and close to the minimum recommended (i.e. 2), the uplift pressure will control the pile length, so use L = 13 m.
Example Make a preliminary design for a drilled pier to be founded on the firm clay at depth – m. The top 3.5 m of depth is in a water-bearing sand-gravel stratum. The pier is to carry 10,500 kN and = 35 MPa, use FOS = 2 on the skin resistance. Use bell if required. Assume cu = 120 kPa upto – 25 m depth and 145 kPa below 25 m, E (^) c = 4700 MPa, γ = 19.8 kN/m^3 , γ´ = 10 kN/m^3. Solution:
(i) (^) Diameter of shaft ( D ) for concrete requirement alone as per ACI
F 0D E D F 04 0 1.53 m use D = 1.5 m
(ii) Ultimate skin friction (shaft resistance, Q (^) s ) Neglecting 3.5 m length L = 27 – 3.5 = 23.5 m = 21.5 m + 2.0 m Assume adhesion factor, α = 0. F 05 C Qs = = (0.5 F 0B 4 120 F 0B 4F 07 0F 0B 41.5 F 0B 421.5) + (0.5 F 0B 4 145 F 0B 4F 07 0F 0B 41.5 F 0B 42) = 6,762.3 kN allowable skin friction for FOS 2 = 3,381 kN < 10,500 kN Thus bell will be required (iii) Also check end-bearing, Qb for D = 1.5 m kN allowable end-bearing for FOS 3 = 769 kN Thus without bell, the pier capacity = < 10,500 kN Bell is definitely required. Qb required = 10,500 – 3,381 = 7,119 kN Let B be the diameter of the bell
F 0D E B F 04 0 4.56 m use B = 4.75 m
(iv) check the pier capacity
kN = 3104 kN
