Problems On Piles

PROBLEMS ON PILES/DEEP FOOTING Problem # 01  A cast in place concrete pile in medium sand is 0.5 m in dia and 10 m long. The water table is 2 m below the top of the pile. The bulk density of sand above the water table is 1.8 t/m3 . and below the water table the density is 1.95 t/m3 . Angle of shearing resistance is 38o . Assume Nqp = 60 and ks tan  = 0.5, Dc /B = 8. Determine the ultimate load and the safe load for a pile with F.S = 3. 

Solution 

Qf = Qb + Qs



(1) Base resistence (Qb )

   

 



2 m

Dc = 4 m

B = 0.5 m Lp= 10 m Nqp = 60 Dc= ? From table for medium sand, Dc = 8 x B = 8 x 0.5 = 4m Lp > Dc . fb = o  Nqp Where o = Effective vertical pressure at Dc . o = (1.8 x 9.8 x 2) + ( (1.95-1) x 9.8 x 2) = 53.9 kPa.

• (ii) Frictional resistence, Qs   



 







Qs = Asx fs

fs =? The frictional resistence is calculted by considering the pile in three parts, 0-2m, 2 m - 4 m (Dc ) and 4 m – 10 m. The frictional resistence will increase upto Dc only = 4 m.  at 2 m = 2 x 1.8 x 9.8 = 35.28 kPa.  at 4 m = (35.28 + ( 1.95- 1) x 9.8 x 2) = 53.9 kPa. Average vertical stress for 0-2 m = 17.64 kPa. ( 0+ 35.28/2)  Av. fs1 =17.64 x ks tan  = 17.64 x 0.5 = 8.82 kPa.

• Vertical stress below Dc is assumed constant and equal to 53.9 kPa. • fs3 = 53.9 x 0.5 = 26.95 ( unit skin friction) • Qs = fs1 x B x D1 + fs2 x B (Dc – D1) + fs3 B (Lp - Dc ) • = 8.82 x  x 0.5 x 2 + 22.3 x  x 0.5 x 2 +26.95 x  0.5 x 6 = 352 kN. • Qf = Qb + Qs = 635 + 352 = 987 kN. • Q = Q /3 = 987/3 = 329 kN.

Problem # 02 • A cased concrete pile, 0.42 m outside diameter, is driven through 11 m of loose sand and for a depth of 3 m into underlying medium dense sand. The skin friction is measured with a static cone fitted with a friction sleeve; fs = fc(av) = 20 kPa in loose sand and fs = fc(av) = 55 kPa in medium dense sand. The cone resistence at base level is 12000 kPa. Determine the safe load for a pile, using F.S = 2.5.

Solution • Qf = Qb + Qs

•

fb = qc = • Qb = /4 B2 x fb 12000 kPa but limit is 11000 kPa • Qb = /4 (0.42)2 x 11000 = 1524 kN. • Qs = Qs1 + Qs2 ,Qs1 = through loose sand = 11m • Qs2 = through medium dense sand =3 m. • Qs = B x 11 x 20 + B x 3 x 55 • =  x 0.42 x 11 x 20 +  x 0.42 x 3 X 55 = 290.28 + 217.71 • Q = Q + Q = 1524 + 290.28 + 217.71=

Problem # 03 • Determine the penetration depth for a 40 cm dia. Concrete pile which is required to carry load of 900 kN at a safety factor of 2.5. The soil consists of silty sand upto a depth of 10 m in which the average cone resistance (qc)av = 4000 kPa. Silty sand is underlain by dense sand. The average cone resistance increases to 13000 kPa within the depth 10 m to 10.5 m and to 18000 kPa below 10.5 m. to reach full base resistance the pile must be driven to at least 5B, i-e 2 m, in bearing stratum of dense sand.

Solution • Try the allowable load for a pile, embedded length 12m. Min. • Qf = Qb + Qs • Qb = fb x Ab • fb = qc= 18000 kPa But limit of fb =11000 kPa • Qb = fb x /4 B2 = 11000 x (0.42)2 x /4 = 1381.6 kN. • Now Qs = Qs1 +Qs2 +Qs3 .  0-10m 10-10.5m + > 10.5.

• fs2 = 10 – 10.5 = qc av /200 = 13000/200 =65 kPa. • fs3 = 10.5 – 12 = qc av /200 = 18000/200 = 90 kPa. (< 100 kPa) • Qs =  B [(10x20) + (0.5x65) + (1.5x90)]  D1 x fs1 D2 x fs2 D3 x fs3 =  x 0.4 ( 200 + 32.5 + 135) = 461.81 kN. Q = Q f b + Qs = 1381.6 + 461.81 = 1 843.41 f s3 x  B x D = 407 

D = 407/90 x x 0.4 = 3.6m.  12 + 3.6 = 15.6 m Q a = Qf/F.S. 1843.41/2.5 =737.36 Q = Q f a x F.S = 900 x 2.5 = 2250. 

 

Receiving Load = 2250 – 1843 = 407 kN

Problem # 04 • A bored concrete pile 30 cm in dia. And 6.5m long, passes through stiff fissured clay, subjected to seasonal shrinkage and swelling upto a depth of 1.5m. The average undrainedstrength of clay varies linearly from 50 kPa at 1.5m to 186 kPa at 10m. Find the ultimate load capacity. Assume  = 0.3.

Solution • 1.5m top is ignored as the shrinkage may cause all adhesion to be lost. • Cu at 1.5m depth = 50 kPa. • Now cu at base = 50 + 186- 50/8.5 x 5 = 130 kPa •  Average cu = 50 + 130 /2 = 90 kPa. • Qf = fb x Ab + fs As • 1170 x  /4 (0.3)2 + 27 x  x 0.3 x 5 • =82.7 + 127.27 = 210 kN.

Problem # 05 (uplift Resistance) • Find the uplift resistance and the safe uplift resistance (F.S = 2.5) of the under reamed pile installed in clay with the following data. (if the top 2m clay is active and subjected to swelling) • Dia of shaft = 40 cm, dia. of bulb at base = 100cm. • Length of shaft above bulb = 5m cu av along shaft = 115 kPa. Cu at base = 150 kPa and adhesion factor  = 0.4

Solution • Projected area of bulb = /4 (Bb – B2)= 0.659 m2 . • Neglecting top 2m of activeness • As =  B x 3 = 3.77 m2 [ fb =Nc • Qup = 0.659 fb + 3.77 fs x cu =9x150=135



] = 0.659x 1350 + 3.77 x 46 = 1063 kN Safe uplift resistance = fs =  cu = 0.4x 115 = 46





Problem # 06 • Find the penetration depth for a 35cm square R.C pile driven through sea bed for a jetty structure to carry a maximum compressive load of 500 kN and a net uplift load of 300 kN. The soil below sea bed consists of a 10m thick layer of saturated medium dense sand overlying dense sand and gravel. The average corrected SPT blows in the top 10m sand are 12 and on bottom sand gravel layer 40. No erosion is expected. The minimum safety factor should be 2.5 on the uplift load and 4 on the compressive

Solution • Nav for top10m = 12 (sandy layer) • N in sand –gravel = 40 • Total ultimate uplift resistance required = 2.5x 300 = 750 kN • Now fs (medium dense sand)= 2 Nav = 2x12=24 kPa. • Qs (uplift ) in top 10m = (4x0.35x10)24 = fs x As =336 kN. • Remaiing uplift resistance = 750 – 336 = 44 kN. • Friction resistance per meter length in sand gravel = fs xAs (one meter) • fs (dense gravel) = 2 Nav = 2x 40 = 80 kPa. Friction resistance /meter length = 80 x 4 x 0.35x1 =112 kN/m. Further penetration required in dense gravel to resist the uplift of 414 kN.  = 414/112 = 3.7m Total Penetration = 10 + 3.7 = 13.7m for uplift. 

Check for compressive load. Q =Q f b + Qs. Q b = fb x Ab.  fb (dense gravel) = 40 N Lp /B ≤ 400 N. ≤400x40 ≤ 16000 kPa. =40x4x 3.7/0.35= 16914.3 ≤ 16000 kPa. But maximum value of f b which could be adopted = 11000 kPa.   fb = 11000. 2 Now Q b = 11000 x (0.35) = 1347.5 kN. Q s is same for compression as for uplift. ∴ Qf = 1347 + 750 = 2097 kN. ∴ F = Qf /Qa= 2097/500 = 4.2 > 4 safe.  this indicates that the required penetration is governed by uplift resistance. 

Problem # 07 (Pile load test) • The following data refers to a cyclic pile load test carried out on a 300mm dia. pile 10m long. • Load on pile top (kN) 150 200 250 300 400 500 600 Total settlement on pile top 1.45 2.25 2.75 3.6 5.75 10.75 30.00 • (mm) Net settlement on pile top 0.40 0.65 0.80 1.0 1.70 5.25 22.80 (mm) • • Plot the load settlement curves (total and net) and estimate the allowable load by different methods/criterians.

Solution

Settlement

• Plot the load vs settlement ( Arithmatic graph) • Plot the load vs Net settlement(Arth. graph) • (a) Two third of the load causing 12 mm settlement, that is • Qa = 2/3 x 500 = 333.33 kN • (b) 50% of the load causing • Total settlement equal to 10% Load • Of the pile dia. i.e settlement of • 30mm (i.e 600 kN) • Qa = 1/2 x 600 = 300 kN. • (c) Net settlement 6mm 2/3 rd of that • 2/3 x 550 = 1100/3 = 366.66 kN • (d) Falling to striaght line Qu 600 kN, use F.S =2. • 600/2 = 300 • Least of the above is 300 kN

Problem # 08 • Two independent loading tests on 300mm dia. Short bored piles in clay for which  = 0, yielded the following results. • Embedded length of pile Added load at failure (kN) • 2.15 100 • 2.75 110 • Assuming the adhesion is effective over the whole of the embedded length, estimate the mean cohesion of the soil and the shaft adhesion factor to be used in extrapolating the test results to larger piles. The densities of the soil and concrete are 1.92 Mg/m3 and 2.40 Mg/m respectively.

Solution: • • • • •

P

At point of failure  v =0 Wt. s = 1.92 Mg/m3 Qs 3. c = 2.4 Mg/m Qb + Qs = P + (wt. of pile – wt. of excavated material) Now P = 100 kN Lp = 2.15m.

c • Qb + Qs = c Ncp x Ab +  c As= c Ncp x /4xd2 +Q b x x d x Lp. • = c x 9x /4 x 0.32 +  c x  x 0.3 x 2.15 = 0.636 c + 2.027  c (Lp /d > 4) • Ncp = 9 • P = 100 + wt. of pile – wt. of excavated material • = 100 + (2.4- 1.92) x 9.81 (/4 x 0.32 x 2.15) • = 100 + 0.715

• v=0 •  100.715 = 0.636 c + 2.027  c -----------(i) • Similarly for 2nd test results  v = 0 • Qb + Qs =9x c x /4 (0.3)2 +  c x x 0.3 x2.75 • = 0.636 c + 2.592  c • Load = 110 + (2.4 – 1.92) x 9.81 (/4 x 0.32 x 2.75) • = 110 + 0.915 = 110.915 • 110.915 = 0.636 c + 2.592  c ---------------------(ii) • Solve equation simultaneously

Problem # 09 • A free standing, square group of 16 piles extends to a depth of 12 m in stiff clay which is underlain by rock at 24 m depth. The piles are 0.5 m in diameter and are spaced at 1.5 m centres. cT at base = 140 kPa, average (cT)av along shaft = 90 kPa, adhesion  = 0.45, coefficient of volume decrease mv below base = 0.08 m2/MN, remoulded cT along shaft = 50 kPa. Assume group efficiency Eg = 0.65. Find the allowable load for the group with F = 2.5. Also estimate the

Solution: • (a) Considering individual piles.

fb = cT Ncp = 140 x 9 = 1260 kPa  fs =  cT = 0.45 x 90 = 40.5 kPa  Qf =  /4 B2 fb + B Lp fs  = /4 0.52x 1260 +  x 0.5 x 12 x 40.5 = 247.4 + 763.4 = 1010.8 kN  Qg = Eg .n Qf = 0.65 x 16 x 1010.8 = 10512 kN (b)Considering block failure. QB = Bo Lo fb + 2 Lp ( Bo +Lo) fB ; Bo = Lo = 5 m, Lp = 12 m fb = cT Ncp = 140 x 9 =1260 kPa fs = cT ( remolded) = 50 kPa QB = 43500 kN > Qg (31500 + 12000) Q = 1/2.5 Q = 10512/2.5 = 4205 kN 



   

• For a settlement the load is assumed to act at an equivalent raft (5 m x 5 m) located at 12/3 = 4 m above the base. The compressible zone of 16 m thickness is divided into 4 layers, each 4 m thick. Pressure increase at the middle of each layer is calculated by assuming 2:1 spread of load Qs. The calculations are given in table. The total settlement is 48 mm.

Layer Mid-depth Area (m) (m 1 2 49 2 ) 2 6 121 3 10 225 4 14 361

 (kPa) 85.80 34.75 18.68 11.65

mv  H (mm) 27.45 11.12 5.97 3.73 48.28

Problem # 10 • A 40-ft-long precast pile is fully embedded in sand. The cross section of pile measures 12 in x 12 in. The allowable working load for the pile is 80 kip and 54 kip is contributed by skin friction. Determine the elastic settlement of the pile. Ep = 3 x 106 lb/in2 , Es = 4500 lb/in2 , and s = 0.3.

Solution: s1

( Q =

wp

+ ξQws ) L

Ap E p

Let ξ = 0.6, E p = 3 x 106 lb / in 2 = 3 x 103 kip / in 2 , Qws = 54 kip, Qwp = 26 kip Ap = 12 x 12 = 144 in 2 .Then s1 =

[ 26 + (0.6)(54)](40 ×12) = 0.065 in.

(144)(3 × 103 ) qwp D 2 s2 = (1 − µ s ) I wp Es

From Figure 3.27, I wp = 0.82. Also, qwp

Qwp

26 = = = 0.18 kip / in 2 Ap 144

So 0.18 ×12 s2 = 1 − 0.32 ( 0.82 ) = 0.38 in.  4500     1000 

(

)

 Qws  D  (1 − µ s 2 ) I ws s3 =   pL  Es I ws

L 40 = 2 + 0.35 = 2 + 0.35 4.21 D 1.0

So 54 12 s3 = (1 − 0.32 )( 4.21) = 0.287 in. ( 4 ×12)( 40)  4500   1000  Hence the total settlement is s = 0.065 + 0.358 + 0.287 = 0.71in.

Problem # 11 



 

Refer to Figure 8.47 which shows the plan of a pile group. Each pile has a square cross section measuring 12 in x 12 in. the center-to-center spacing of the piles is 30 in. Determine the group efficiency of the piles by using: a. Eq. (8.112) b. Eq. (8.114)

12in

30 in. Fig.

8.47

Solution: • Part a: Using Eq. (8.112) •

2( n1 + n2 − 2 ) d + 4 D η= pn1n2 For n1 = 4, n2 = 3, D = 12 in, and p = 4 D = 48 in.,

2( 4 + 3 − 2 )( 30) + 48 η= = 0.604 = 60.4% (18)( 4)( 3)

• Part b: Using Eq. (8.114) •  ( n1 − 1) n2 + ( n2 − 1) n1  −1  D  η = 1−   tan   90n1n2 d   −1  D  −1  12  tan   = tan   = 21.8o d  30  So  ( 3)( 3) + ( 2 )( 4 )  o η = 1−  21 . 8 = 0.657 = 65.7%.   ( 90 )( 3)( 4 ) 

(

)

Problem #13 • Refer to Figure 8.41a. For this group pile, n1=4,n2= 3, D = 305 mm, d = 1220 mm, and L= 15 m. The piles are square in cross section and are embedded in a homogenous clay with cu = 70 kN/m2. Use FS = 4 and determine the allowable loadbearing capacity of the group pile. Assume  = 0.63

Solution: • From Eq. (8.116), • ΣQu = n1n2 [9 Ap cu ( p ) + Σαpcu ∆L ]

Ap = ( 0.305)( 0.305) = 0.093 m 2 p = ( 4 )( 0.305) = 1.22 m

2

From Figure 8.19 for cu = 70kN / m 2 , α = 0.63. So ΣQu = ( 4 )( 3) [ ( 9 )( 0.093)( 70 ) + ( 0.63)(1.22 )( 70 )(15) ] = 12( 58.59 + 807.03) ≈10387 kN

• Again from Eq. (8.117), the ultimate block capacity is Lg Bg cu(p) Nc*+2(Lg +Bg) cu  L, So • Lg = (n1 – 1)d +2(D/2) = (4-1)(1.22)+ 0.305=3.965 m • Bg = (n2-1)d + 2(D/2) = (3-1)(1.22)+ 0.305= 2.745 m • L/Bg = 15/2.745= 5.46 • Lg /Bg = 3.965/2.745 = 1.44 • From Figure 7.16, Nc* 8.6. Thus • Block capacity = (3.965)(2.745)(70)(8.6) + 2(3.965 + 2.745)(70)(15) = 6552 + 14091 = 20643 kN • So

Problem # 14 • A group pile in clay is shown in Fig. Determine the consolidation settlement of the pile groups. All clays are normally consolidated.

Qg = 500 kip sand 6 ft

 = 105 lb/ft3 3 ft

Water table 30 ft 45 ft

z Clay

Clay

2v:1H

27 ft

Po(1), p(1) Po(2), p(2) Po(3), p(3) Rock

21 ft

Clay 48 ft sat = 115 lb/ft3 eo = 0.82 Cc = 0.3 3 2v:1H sat = 120lb/ft 12 ft eo = 0.7 Cc = 0.2

sat = 122 lb/ft3 6 ft e = 0.75 o Cc = 0.25 Pile group: Lg = 9ft; Bg = 6ft

Solution: • The stress distribution pattern is shown in Figure 8.49. Hence Q ( 500)(1000) = 1554 lb / ft ∆p = = • ( L + z )( B + z )  9 + 21  6 + 21  g

2

(1)

g

1

g

1



2 

( 500)(1000) = 421lb / ft 2 ( 9 + 27 )( 6 + 27 ) ( 500)(1000) = 265 lb / ft 2 ∆p( 3) = ( 9 + 36)( 6 + 36)

2

∆p( 2 ) =

 po (1) + ∆p(1)  log   1 + eo (1) p   o (1) 21   = ( 6)(105) +  27 + (115 − 62.4) = 2603 lb / ft 2 2 

∆s1 = po (1)

∆s1 =

Cc (1) H1

( 0.3)( 21) log 2603 + 1554  = 0.7 ft = 8.45 in. 1 + 0.82

 

2603

 

 po ( 2 ) + ∆ p( 2 )  ∆ s2 = log  1 + eo ( 2) po ( 2)   po ( 2) = ( 6)(105) + ( 27 + 21)(115− 62.4) + ( 6)(120− 62.4) = 3500lb / ft 2 Cc ( 2 ) H 2

∆ s2

( 0.2)(12)  3500+ 421 = log = 0.84in

∆ s3

( 3)(122− 62.4) = 4025lb / ft 2 ( 0.25)( 6)  4025+ 265 = log = 0.024 ft ≈ 0.29in

 3500  1 + 0.7 po (3) = ( 6)(105) + ( 48)(115− 62.4) + (12)(120− 62.4) +

1 + 0.75  4025  Total settlement, ∆ s g = 8.45 + 0.84 + 0.29 = 9.58in.

Problem # 15 • It is proposed to carry the load from a column on 350 mm diameter bored piles. The total load to be carried is 2000 kN and the soil consists of a deep deposit of clay having the following properties: Depth Undrained shear strength • If the piles are 0 100 2 (m) (kN/m ) 110 • about 9 m long, 3 6 125 • Estimate the 125 • Number reqired 9 135 • And suggest how 12 • they should be • arranged

Solution: • Consider the ultimate load carried on pile: • Qf = Qb + Qs • Taking average cu on shaft of pile as = 115 kN/m2 and an adhesion factor of 0.45 for a bored pile: Qf = 125 x 9 x (/4 x 0.352) + 115 x 0.45 • x ( x0.35 x 9) • = 108 + 512 = 620 kN • Using a factor of safety of 3: Qa = 620/3 = 207 kN •  No of piles required to carry 2000 kN = 2000/207 = 9.6 • Say 9 piles which reduces the factor of

• Bored piles should be spaced at about 3 times their diameter and a suggested arrangement is shown on Fig. 1 m • The action of the group • Must be checked for 1 m 2.35 m  stability. Consider a pier 2.35 m square: 2 +115 x 0.45 x (4 x Q = 125 x 9 x 2.35 fg 2.35 x 9)  = 6212 + 4378 = 10590 kN

•