Module 8 (Lecture 33) PILE FOUNDATIONS Topics

Module 8 (Lecture 33) PILE FOUNDATIONS Topics 1.1 PILE-DRIVING FORMULAS 1.2 NEGATIVE SKIN FRICTION Clay Fill over Granular Soil Granular Soil Fill over Clay 1.3 GROUP PILES 1.4 GROUP EFFICIENCY

PILE-DRIVING FORMULAS To develop the desired load-carrying capacity, a point bearing pile must penetrate the dense soil layer sufficiently or have sufficient contact with a layer of rock. This requirement cannot always be satisfied by driving a pile to a predetermined depth because soil profiles vary. For that reason, several equations have been developed to calculate the ultimate capacity of a pile during driving. These dynamic equations are widely used in the field to determine whether the pile as reached satisfactory bearing value at the predetermined depth. One of the earliest of these dynamic equationscommonly referred to as the Engineering News Record (ENR) formula-is derived from the work-energy theory. That is, Energy imparted by the hammer per blow = (pile resistance) (penetration per hammer blow) According to the ENR formula, the pile resistance is the ultimate load QQ uu, expressed as QQ uu = WW RRh SS+CC [8.118] Where WW RR = wwwwwwwwhtt oooo tthee rrrrrr (ffffff eeeeeeeeeeeeee, ssssss tttttttttt DD. 4AAAAAAAAAAAAAAAA DD) h = heeeeeehtt oooo ffffffff oooo tthee rrrrrr SS = pppppppppppppppppppppp oooo pppppppp pppppp haaaaaaaaaa bbbbbbbb CC = aa cccccccccccccccc The pile penetration, S, is usually based on the average value obtained from the last few driving blows. In the equations original form, the following values of C were recommended. For drop hammers: C = 1 in. (if the units of S and h are in inches) For steam hammers: C = 0.1 in. (if the units of S and h are in inches) Also, a factor of safety, FFFF = 6, was recommended to estimate the allowable pile capacity. Note that, for single- and double-acting hammers, the term WW RR h can be replaced

by EEEE EE (where EE = haaaaaaaaaa eeeeeeiiiiiiiiiiiiii aaaaaa HH EE = rrrrrrrrrr eeeeeeeeeeee oooo haaaaaaaaaa). Thus QQ uu = EEEE EE SS+CC [8.119] The ENR formula has been revised several times over the years, and other pile-driving formulas also have been suggested. Some of them are tabulated in table 1. The maximum stress developed on a pile during the driving operation can be estimated from the pile-driving formulas presented in table 11. To illustrate, we use the modified ENR formula: QQ uu = EEEE RRh SS+CC WW RR +nn 2 WW pp WW RR +WW pp In this equation, S equals the average penetration per hammer blow, which can also be expressed as SS = 1 NN [8.120] Where SS iiii iiii iiiiiiheeee NN = nnnnnnnnnnnn oooo haaaaaaaaaa bbbbbbbbbb pppppp iiiiiih oooo pppppppppppppppppppppp Table 11 Pile-Driving formulas Name Formula Modified formula ENR QQ uu = EEEE RRh WW RR + nn 2 WW pp SS + CC WW RR + WW pp Where EE = haaaaaaaaaa eeeeeeeeeeeeeeeeeeee CC = 0.1 iiii., iiii tthee uuuuuuuu oooo SS aaaaaa h aaaaaa iiii iiiiiiheeee WW pp = wwwwwwwwhtt oooo tthee pppppppp nn = cccccccccccccccccccccc oooo rrrrrrrrrrrrrrrrrrrrrr bbbbbbbbbbbbbb tthee rrrrrr aaaaaa tthee pppppppp ccaaaa Typical values for E

Single- and double-acting hammers 0.7-0.85 Diesel hammers 0.8-0.9 Drop hammers 0.7-0.9 Typical values for n Cast iron hammer and concrete pile (without cap) 0.4-0.5 Wood cushion on steel piles 0.3-0.4 Wooden pile 0.25-0.3 Michigan State Highway Commission formula (1965) QQ uu = 1.25EEEE EE SS + CC WW RR + nn 2 WW pp WW RR + WW pp Where HH EE = mmmmmmmmmmmmmmmmmmmmmmmm ssssssssssssssss rrrrrrrrrr haaaaaaaaaa eeeeeeeeeeee (llll iiii. ) EE = haaaaaaaaaa eeeeeeeeeeeeeeeeeeee CC = 0.1 iiii. A factor of safety of 6 is recommended. Danish formula (Olson and Flaate, 1967) QQ uu = EEEE EE SS + EEEE EELL 2AA pp EE pp Where

EE = haaaaaaaaaa eeeeeeeeeeeeeeeeeeee HH EE = rrrrrrrrrr haaaaaaaaaa eeeeeeeeeeee EE pp = mmmmmmmmmmmmmm oooo eeeeeeeeeeeeeeeeeeee oooo tthee pppppppp mmmmmmmmmmmmmmmm LL = llllllllllh oooo pppppppp AA pp = aaaaaaaa oooo tthee pppppppp cccccccccc ssssssssssssss Pacific Coast Uniform Building Code formula (International Conference of Building Officials, 1982) Janbu s formula (Janbu, 1953) QQ uu = (EEEE EE ) WW RR + nnww pp WW RR + WW pp SS + QQ uull AAAA pp The value of n should be 0.25 for steel piles and 0.1 a for all other piles. A factor of safety of 4 is generally recommended. QQ uu = EEEE EE KK uu SS Where KK uu = CC dd 1 + 1 + λλ CC dd CC dd = 0.75 + 0.14 WW pp WW RR λλ = EEEE EELL AA pp EE pp SS 2 Gates formula (Gates, 1957) QQ uu = aa EEEE EE (bb llllll SS) If QQ uu is in kkkkkkkk, then S is iiii iiii. aa = 27, bb = 1, aaaaaa HH EE is in kkkkkk ffff.

If QQ uu is in kkkk, then S is in mm, aa = 104.5, bb = 2.4, aaaaaa HH EE is in kkkk mm EE = 0.75 for drop hammer; EE = 0.85 for all other hammer Use a factor of safety of 3. Navy-McKay formula QQ uu = EEEE EE SS 1 + 0.3 WW pp WW RR Use a factor of safety of 6. Thus QQ uu = EEEE RRh (1/NN)+0.1 WW RR +nn 2 WW pp WW RR +WW pp [8.121] Different values of N may be assumed for a given hammer and pile and QQ uu calculated. The driving stress can then be calculated for each value of N and QQ uu /AA pp. AA pp = 100 iiii 2 The weight of the pile is 100 iiii 2 AA pp LLγγ cc = (80 ffff)(150 144 llll/ffff3 ) = 8.33 kkkkkk If the weight of the cap is 0.67 kkkkkk, WW pp = 8.33 + 0.67 = 9 kkkkkk Again, from for an 11B3 hammer, Rated energy= 19.2 kkkkkk ffff = HH EE = WW RR h Weight of ram= 5 kkkkkk Assume that the hammer efficiency is 0.85 and that nn = 0.35. Substituting these values in equation (121) yields QQ uu = (0.85)(19.2 2) 1 5+(0.35)2(9) NN +0.1 5+9 = 85.37 1 kkkkkk +0.1 NN Now the following table can be prepared:

NN QQ uu (kkkkkk) AA pp (iiii 2 ) QQ uu /AA pp (kkkkkk/iiii 2 ) 0 0 100 0 2 142.3 100 1.42 4 243.9 100 2.44 6 320.1 100 3.20 8 379.4 100 3.79 10 426.9 100 4.27 12 465.7 100 4.66 20 569.1 100 5.69 Both the number of hammer blows per inch and the stress can now be plotted in a graph, as shown in figure 8.47. If such a curve is prepared, the number of blows per inch of pile penetration corresponding to the allowable pile-driving stress can be easily determined. Figure 8.47 Actual driving stresses in wooden piles are limited to about 0.7ff uu. Similarly, for concrete and steel piles, driving stresses are limited to about 0.6ff cc aaaaaa 0.85ff yy, respectively.

In most cases, wooden piles are driven with hammer energy of less than 45 kkkkkk ffff ( 60 kkkk mm). Driving resistances are limited are limited mostly to 4-5 blows per inch of pile penetration. For concrete and steel piles, the usual N values are 6-8 and 12-14, respectively. Example 11 A precast concrete pile 12 iiii. 12 iiii. in cross sections in driven by a hammer. Given: MMMMMMMMMMMMMM rrrrrrrrrr haaaaaaaaaa eeeeeeeeeeee = 30 kkkkkk ffff HHHHHHHHHHHH eeeeeeeeeeeeeeeeeeee = 0.8 WWWWWWWWhtt oooo rrrrrr = 7.5 kkkkkk PPPPPPPP llllllllllh = 80 ffff CCCCCCCCCCCCCCCCCCCCCC oooo rrrrrrrrrrrrrrrrrrrrrr = 0.4 WWWWWWWWhtt oooo pppppppp cccccc = 550 llll EE pp = 3 10 6 kkkkkk/iiii 2 Number of blow for last 1 in. of penetration = 8 Estimate the allowable pile capacity by the a. Modified ENR formula (use FFFF = 6) b. Danish formula (use FFFF = 4) c. Gates formula (use FFFF = 3) Solution Part a QQ uu = (0.8)(30 12 kkkkkk iiii.) 1 7.5+(0.4)2 (12.55) = 607 kkkkkk 8 +0.1 7.5+12.55 QQ aaaaaa = QQ uu = 607 101 kkkkkk FFFF 6 Part b QQ uu = EEEE EE SS+ EEEE EE LL 2AApp EEpp Use EE pp = 3 10 6 llll/iiii 2.

EEEE EELL (0.8)(30 12)(80 12) = 2AA pp EE pp = 0.566 iiii. 2(12 12) 3 106 1000 kkkkkk /iiii 2 QQ uu = (0.8)(30 12) 1 8 +0.566 417 kkkkkk QQ aaaaaa = 417 4 104 kkkkkk Part c QQ uu = aa EEEE EE (bb llllll SS) = 27 (0.8)(30) [1 llllll 1 ] 252 kkkkkk 8 QQ aaaaaa = 252 3 = 84 kkkkkk NEGATIVE SKIN FRICTION Negative skin friction is a downward drag force exerted on the pile by the soil surrounding it. This action can occur under conditions such as the following: 1. If a fill of clay soil is placed over a granular soil layer into which a pile is driven, the fill will gradually consolidate. This consolidation process will exert a downward drag force on the pile (figure 8.48a) during the period of consolidation. 2. If a fill of granular soil is placed over a layer of soft clay, as shown in figure 8. 48b, it will induce the process of consolidation in the clay layer and thus exert a downward drag on the pile. 3. Lowering of the water table will increase the vertical effective stress on the soil at any depth, which will induce consolidation settlement in clay. If a pile is located in the clay layer, it will be subjected to a downward drag force. In some cases, the downward drag force may be excessive and cause foundation failure. This section outlines two tentative methods for the calculation of negative skin friction.

Figure 8.48 Negative skin friction Clay Fill over Granular Soil (Figure 8.48a) Similar to the ββ method presented in section 12, the negative (downward) skin stress on the pile is ff nn = KK σσ vv tttttt δδ [8.122] Where KK = eeeeeeeeh pppppppppppppppp cccccccccccccccccccccc = KK oo = 1 ssssssss σσ vv = vvvvvvvvvvvvvvvv eeeeeeeeeeeeeeeeee ssssssssssss aaaa aaaaaa ddddddddh zz = γγ ff zz γγ ff = eeeeeeeeeeeeeeeeee uuuuuuuu wwwwwwwwhtt oooo ffffffff δδ = ssssssss pppppppp ffffffffffffffff aaaaaaaaaa 0.5 0.7φφ Hence the total downward drag force, QQ nn, on a pile is QQ nn = HH ff 0 (ppkk γγ ff tttttt δδ)zz dddd = ppkk γγ ff HH ff 2 tttttt δδ 2 [8.123] Where HH ff = heeeeeehtt oooo tthee ffffffff If the fill is above the water table, the effective unit weight, γγ ff, should be replaced by the moist unit weight.

Granular Soil Fill over Clay (figure 8.48b) In this case, the evidence indicates that the negative skin stress on the pile may exist from zz = 0 tttt zz = LL 1, which is referred to as the neutral depth (see Vesic, 1977, pp. 25-26, for discussion). The neutral depth may be given as (Bowles, 1982): LL 1 = (LL HH ff ) LL HH ff LL 1 2 + γγ ffhh ff 2γγ ff HH ff γγ γγ [8.124] Where γγ ff aaaaaa γγ = eeeeeeeeeeeeeeeeee uuuuuuuu wwwwwwwwhtttt oooo tthee ffffffff aaaaaa tthee uuuuuuuuuuuuuuuuuuuu cccccccc llllllllll, rreeeeeeeeeeeeeeeeeeeeee For end-bearing piles, the neutral depth may be assumed to be located at the pile tip (that is, LL 1 = LL HH ff ). Once the value of LL 1 is determined, the downward drag force is obtained in the following manner. The unit negative skin friction at any depth from zz = 0 tttt zz = LL 1 is ff nn = KK σσ vv tttttt δδ [8.125] Where KK = KK oo = 1 ssssss φφ σσ vv = γγ ff HH ff + γγ zz δδ = 0.5 0.7φφ LL 1 0 LL 1 0 QQ nn = ppff nn dddd = ppkk γγ ff HH ffγγ zz tttttt δδ dddd = (ppkk γγ ff HH ff tttttt δδ)ll 1 + LL 1 2 ppkk γγ tttttt δδ 2 [8.126] If the soil and the fill are above the water table, the effective unit weights should be replaced by moist unit weights. In some cases, the piles can be coated with bitumen in the downdrag zone to avoid this problem. Baligh et al. (1978) summarized the results of several field tests that were conducted to evaluate the effectiveness of bitumen coating in reducing the negative skin friction. Their results are presented in table 12. A limited number of case studies on negative skin friction is available in the literature. Bjerrum et al. (1969) reported monitoring of downdrag force on a test pile at Sorenga in the harbor of Oslo, Norway (noted as pile G in the original paper). This was also discussed by Wong and The (1995) in terms of the pile being driven to bedrock at 40 m. Figure 8.49 a shows the soil profile and the pile. Wong and The (1995) estimated the following:

Figure 8.49 Negative skin friction on a pile in the harbor of Oslo, Norway [based on Bjerrum et al., (1969); and Wong and The (1995)] FFFFFFFF: MMMMMMMMMM uuuuuuuu wwwwwwwwhtt, γγ ff = 16 kkkk/mm 3 SSSSSSSSSSSSSSSSSS uuuuuuuu wwwwwwwwhtt, γγ ssssss (ff) = 18.5 kkkk/mm 3 So γγ ff = 18.5 9.81 = 8.69 kkkk/mm 3 HH ff = 13 mm CCCCCCCC: KK tttttt δδ 0.22 SSSSSSSSSSSStttttt eeeeeeeeeeeeeeeeee uuuuuuuu wwwwwwwwhtt, γγ = 19 9.81 = 9.19 kkkk/mm 3 PPPPPPPP: LL = 40mm DDDDDDDDDDDDDDDD, DD = 500mm Thus, the maximum downdrag force on the pile can be estimated from equation. (126). Since it is a point bearing pile, the magnitude of LL 1 = 27 mm, so QQ nn = (pp)(kk tttttt δδ)[γγ ff 2 + (13 2)γγ ff ](LL 1) + LL 1 2 ppγγ (KK tttttt δδ) 2

Or QQ nn = (ππ 0.5)(0.22)[(16 2) + (8.69 11)](27) + (27)2 (ππ 0.5)(9.19)(0.22) 2 = 2348 kkkk The measured value of maximum QQ nn was about 2500 kkkk (figure 8. 49b), which is in good agreement with the calculated value. Table 12 Summary of Case Studies of Bitumen-Coated Piles(After Baligh et al. (1978)) Downward drag Test loadings Case number 1 2 3 4 5 6 7 Soil type Fill, sand, and clay Fill and silty clay Fill and clay Sand and silty clay Silty clay Silty clay Sand fill, clay, and peat Pile type Pile cross section (mm) Cast-inplace concrete DD = 530 Steel pipe DD = 300 Steel pipe DD = 500 Steel pile DD = 760 6 RC piles 300 300 6 RC piles 300 300 Precast concrete 380 450 Length in contact with settling soil (m) 25 26 40 25 7-17 9-16 24 Installation method Predriven casing Enlarged tip and slurry Enlarge tip and casing Driving Driving Driving Driving Bitumen coating Type 25 CC) Coating thickness (mm) (pen 20/30 10 80/100 1.2 80/10 1.2 60/70 1.5 60/70 1.2 80-100 RC-0 cutback 1.2 43 special grade 10

Measured shaft resistance Uncoated pile (ton) 70-80 120 300 180 31-40 31-40 160 Coated pile (ton) 5-7 10 15 3 10-33 20-42 Coating effectiveness (%) 92 92 95 98 30-80 30-80 Predicted downdrag Coated pile (ton) 0.1 2-11 5 0-23 Coating Effectiveness (%) 100 91-98 98 87-100 Example 12 Refer to figure 8. 48a; HH ff = 3 mm. The pile is circular in cross section with a diameter of 0.5 m. For the fill that is above the water table, γγ ff = 17.2 kkkk/mm 3 aaaaaa φφ = 36. Determine the total drag force. Use δδ = 0.7 φφ. Solution From equation. (123), QQ nn = ppkk γγ ff HH ff 2 tttttt δδ 2 pp = ππ(0.5) = 1.57 mm KK = 1 ssssss φφ = 1 ssssss 36 = 0.41 δδ = (0.7)(36) = 25.2 QQ nn = (1.57)(0.41)(17.2)(3)2 tttttt 25.2 2 = 23.4 kkkk

Example 13 Refer to figure 8. 48b. Here, HH ff = 2 mm, pppppppp dddddddddddddddd = 0.305 mm, γγ ff = 16.5 kkkk/mm 3, φφ cccccccc = 34, γγ ssssss = 17.2 kkkk/mm 3, aaaaaa LL = 20 mm. The water table coincides with the top of the clay layer. Determine the downward drag force. Assume δδ = 0.6φφ cccccccc. Solution The depth of the neutral plane in given in equation (124) as LL 1 = LL HH ff LL 1 LL HH ff 2 + γγ ffhh ff 2γγ ff HH ff γγ γγ Note that γγ ff in equation (124) has been replaced by γγ ff because the fill is above the water table, so LL 1 = (20 2) (20 2) + (16.5)(2) (2)(16.5)(2) LL 1 2 (17.2 9.81) (17.2 9.81) LL 1 = 242.4 LL 1 8.93; LL 1 = 11.75 mm Now, referring to equation (126), we have QQ nn = (ppkk γγ ff HH ff tttttt δδ)ll 1 + LL 1 2 KK γγ tttttt δδ 2 pp = ππ(0.305) = 0.958 mm KK = 1 ssssss 34 = 0.44 QQ nn = (0.958)(0.44)(16.5)(2)[tttttt(0.6 34)](11.75) + (11.75)2 (0.958)(0.44)(17.2 9.81)[tttttt (0.6 34)] 2 = 60.78 + 79.97 = 140.75 kkkk GROUP PILES GROUP EFFICIENCY In many cases, piles are used in groups, as shown in figure 8.50, to transmit the structural load to the soil. A pile cap is constructed over group piles. The pile cap can be contact with the ground, as in most cases (figure 8.50a), or well above the ground, as in the case of offshore platforms (figure 8.50b).

Figure 8.50 Pile groups Determining the load-bearing capacity of group piles is extremely complicated and has not yet been fully resolved. When the piles are placed close to each other, a reasonable assumption is that the stresses transmitted by the piles to the soil will overlap (figure 8. 50c), reducing the load-bearing capacity of the piles. Ideally, the piles in a group should be spaced so that the load-bearing capacity of the group should not be less than the sum of the bearing capacity of the individual piles. In practice, the minimum center-to-center pile spacing, dd, iiii 2.5 DD, and in ordinary situations, is actually about 3 3.5DD. The efficiency of the load-bearing capacity of a group pile may be defined as ηη = QQ gg(uu ) ΣΣQQ uu [8.127] Where ηη = gggggggggg eeeeeeeeeeeeeeeeeeee

QQ gg(uu) = uuuuuuuuuuuuuuuu llllllll bbbbbbbbbbbbbb cccccccccccccccc oooo tthee gggggggggg pppppppp QQ uu = uuuuuuuuuuuuuuuu llllllll bbbbbbbbbbbbbb ccccccccccccccyy oooo eeeeeeh pppppppp wwwwwwhoooooo tthee gggggggggg eeeeeeeeeeee Many structural engineers use a simplified analysis to obtain the group efficiency for friction piles, particularly in sand. This type of analysis can be explained with the aid of figure 8. 50a. Depending on their spacing within the group, the piles may act in one of two ways: (1) as a block with dimensions LL gg BB gg LL, or (2) as individual piles. If the piles act as a block, the frictional capacity is ff aaaa pp gg LL QQ gg(uu). [Note: pp gg = perimeter of the cross section of block= 2(nn 1 + nn 2 2)dd + 4 DD, aaaaaa ff aaaa = average unit frictional resistance.] Similarly, for each pile acting individually, QQ uu ppppff aaaa. (Note: pp = perimeter of the cross section of each pile.) Thus ηη = QQ gg(uu ) ΣΣ QQ uu = ff aaaa [2(nn 1+nn 2 2)dd+4DD]LL nn 1 nn 2 ppppff aaaa = 2(nn 1+nn 2 2)dd+4DD ppnn 1 nn 2 [8.128] Hence QQ gg(uu) = 2(nn 1+nn 2 2)dd+4DD ppnn 1 nn 2 ΣΣ QQ uu [8.129] From equation (129), if the center-to-center spacing, dd, s large enough, ηη > 1. In that case, the piles will behave as individual piles. Thus, in practice, if ηη < 1, QQ gg(uu) = ηη ΣΣ QQ uu And, if ηη 1, QQ gg(uu) = ΣΣ QQ uu There are several other equations like equation (129) for the group efficiency of friction piles. Some of these are given in table 13. Feld (1943) suggested a method by which the load capacity of individual piles (friction) in a group embedded in sand could be assigned. According to this method, the ultimate capacity of a pile is reduced by one-sixteenth by each adjacent diagonal or row pile. The technique can be explained by referring to figure 8.51, which shows the plan of a group pile. For pile type A, there are eight adjacent piles; for pile type B, there are five adjacent piles; and for pile type C, there are three adjacent piles. Now the following table can be prepared:

Figure 8.51 Feld s method for estimation of group capacity of friction piles Table 13 Equations for Group Efficiency of Friction Piles Converse-Labarre equation ηη = 1 (nn 1 1)nn 2 + (nn 2 1)nn 1 θθ 90nn 1 nn 2 wwheeeeee θθ (dddddd) = tttttt 1 (DD/dd) Los Angles Group Action equation ηη = 1 DD ππππnn 1 nn 2 [nn 1 (nn 2 1)] + nn 2 (nn 1 1) + 2(nn 1 1)(nn 2 1)] Seiler and Keeney equation (Seiler and Keeney, 1944) 111dd ηη = 1 7(dd 2 1) nn 1 + nn 2 2 0.3 + nn 1 + nn 2 1 nn 1 + nn 2 wwheeeeee dd iiii iiii ffff Pile type No. of Piles No. of adjacent piles/pile Reduction factor for each pile Ultimate capacity A 1 8 B 4 5 C 4 3 1 8 16 1 5 16 1 3 16 0.5QQ uu 2.75QQ uu 3.25QQ uu

ΣΣ 6.5 QQ uu = QQ gg(uu) (No. of piles)(qq uu ) (reduction factor) QQ uu = uuuuuuuuuuuuuuuu cccccccccccccccc ffffff aaaa iiiiiiiiiiiiiiii pppppppppp Hence ηη = QQ gg(uu ) ΣΣ QQ uu = 6.5 QQ uu 9QQ uu = 72% Figure 8.52 shows a comparison of field test results in clay with the theoretical group efficiency calculated from the Converse-Labarre equation (table 13). Reported by Brand et al. (1972), these tests were conducted in soil for which the details are given in figure 8. 7 from chapter 3. Other test details include LLLLLLLLLLh oooo pppppppppp = 6 mm DDDDDDDDDDDDDDDD oooo pppppppppp = 150 mmmm PPPPPPPP gggggggggggg tttttttttttt = 2 2 LLLLLLLLLLLLLLLL oooo pppppppp heeeeee = 1.5 mm bbbbbbbbbb tthee gggggggggggg ssssssssssssss Figure 8.52 Variation of group efficiency with dd/dd (after Brand et al., 1972)

Pile tests were conducted with and without a cap (free-standing group). Note that for dd/dd 2, the magnitude of ηη was greater than 1.0. Also for similar values of dd/dd the group efficiency was greater with the pile cap than without the cap. Figure 8.53 shows the pile group settlement at various stages of the load test. Figure 8.53 Variation of group pile settlement at various stages of load (after Brand et al., 1972) Figure 8.54 Variation of efficiency of pile group in sand (based on Kishida and Meyerhof, 1965)

Figure 8.55 Behavior of low-set ad high-set pile groups in terms of average skin friction (based on Liu et al., 1985)

Figure 8. 55 (Continued) Figure 8.54 shows the variation of group efficiency (ηη) ffffff aa 3 3 group pile in sand (Kishida and Meyerhof, 1965). It can be seen that, for loose and medium sands, the magnitude of group efficiency is larger than one. This is primarily due to the densification of sand surrounding the pile. Liu et al. (1985) reported the results of field tests on 58 pile groups and 23 single piles embedded in granular soil. Test details include PPPPPPPP llllllllllh, LL = 8DD 23DD PPPPPPPP dddddddddddddddd, DD = 125 mmmm 330 mmmm TTTTTTTT oooo pppppppp iiiiiiiiiiiiiiiiiiiiiiii = bbbbbbbbbb SSSSSSSSSSSSSS oooo ppppppeeee iiii gggggggggg, dd = 2DD 6DD Figure 8. 55 shows the behavior of 3 3 pile groups with low-set and high-set pile caps in terms of average skin friction, ff aaaa. Figure 8.56 shows the variation of average skin friction based on the location of a pile in the group. Figure 8.56 Average skin friction based on pile location (based on Liu et al., 1985)

Based on eh experimental observations of the behavior of group piles in sand to date, the following general conclusions may be drawn. 1. For driven group piles in sand with dd 3DD, QQ gg(uu) may be taken to be ΣΣ QQ uu, which includes the frictional and the point bearing capacities of individual piles. 2. For bored group piles in sand at conventional spacings (dd 3DD), QQ gg(uu) may be taken to be 2 3 tttt 3 4 tttttttttt ΣΣ QQ uu (frictional and point bearing capacities of individual piles).