Instrumentation, measurement and control of bio process parameters ( Temperat...
Problems on piles and deep footing
1. 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 δ = 0.5, Dc /B = 8.
Determine the ultimate load and the safe load for
a pile with F.S = 3.
2. Solution
Qf = Qb + Qs
(1) Base resistence (Qb )
B = 0.5 m
Lp= 10 m
Nqp = 60 Dc = ? From table
for medium sand, Dc = 8 x B = 8 x 0.5 = 4 m
Lp > Dc .
fb = σo ′ Nqp Where σo ′ = Effective vertical pressure at
Dc .
σo ′= (1.8 x 9.8 x 2) + ( (1.95-1) x 9.8 x 2) = 53.9 kPa.
fb = 53.9 x 60 = 3234 kPa.
Qb = Ab x fb = π/4 B2 x fb . = π/4 x (0.5)2 x 3234 = 635 kN.
2 m
Dc = 4 m
3. • (ii) Frictional resistence, Qs
Qs = As x 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.
σ′ at 2 m = 2 x 1.8 x 9.8 = 35.28 kPa.
σ′ 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 δ = 17.64 x 0.5 = 8.82 kPa.
Av. Stress for 2 m to 4 m = 53.9 + 35.28/2 = 44.59 kPa.
∴ Av. fs2 = σ′s x ks tanδ = 44.59 x 0.5 = 22.3 kPa.
4. • 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 πB x D1 + fs2 x πB (Dc – D1) + fs3 πB (Lp -
Dc )
• = 8.82 x π x 0.5 x 2 + 22.3 x π x 0.5 x 2
+26.95 x π 0.5 x 6 = 352 kN.
• Qf = Qb + Qs = 635 + 352 = 987 kN.
• Qa = Qf /3 = 987/3 = 329 kN.
5. 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.
6. Solution
• Qf = Qb + Qs
• Qb = π/4 B2 x fb fb = qc = 12000 kPa but
limit is 11000 kPa
• ∴Qb = π/4 (0.42)2 x 11000 = 1524 kN.
• Qs = Qs1 + Qs2 ,Qs1 = through loose sand = 11m
• Qs2 = through medium dense sand =3 m.
• Qs = πB x 11 x 20 + πB x 3 x 55
• = π x 0.42 x 11 x 20 + π x 0.42 x 3 X 55 = 290.28 + 217.71
• Qf = Qb + Qs = 1524 + 290.28 + 217.71= 2032.0 kN.
• Qa = Qf /2.5 = 2032/2.5 = 812.8 kN.
7. 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.
8. 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 π/4 B2 = 11000 x (0.42)2 x π/4 =
1381.6 kN.
• Now Qs = Qs1 +Qs2 +Qs3.
0-10m 10-10.5m + > 10.5.
∴fs1 = 0-10m = qc av /200 = 4000/200 =20 kPa.
9. • 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 = π B [(10x20) + (0.5x65) + (1.5x90)]
D1 x fs1 D2 x fs2 D3 x fs3
= π x 0.4 ( 200 + 32.5 + 135) = 461.81 kN.
Qf = Qb + Qs = 1381.6 + 461.81 = 1 843.41
fs3 x π B x D = 407
D = 407/90 x πx 0.4 = 3.6m.
12 + 3.6 = 15.6 m
Qa = Qf /F.S. 1843.41/2.5 =737.36
Qf = Qa x F.S = 900 x 2.5 = 2250.
Receiving Load = 2250 – 1843 = 407 kN
10. 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 undrained
strength of clay varies linearly from 50 kPa at
1.5m to 186 kPa at 10m. Find the ultimate
load capacity. Assume α = 0.3.
11. 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 π/4 (0.3)2 + 27 x π x 0.3 x 5
• =82.7 + 127.27 = 210 kN.
12. 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 α = 0.4
13. Solution
• Projected area of bulb = π/4 (Bb – B2)= 0.659
m2 .
• Neglecting top 2m of activeness
• As = π B x 3 = 3.77 m2
• Qup = 0.659 fb + 3.77 fs [ fb =Nc x cu =9x150=135
fs = α cu = 0.4x 115 = 46]
= 0.659x 1350 + 3.77 x 46 = 1063 kN
Safe uplift resistance = 1063/2.5 = 425 kN
14. 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 load.
15. 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.
16. Check for compressive load.
Qf =Qb + Qs.
Qb = 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 fb which could be adopted = 11000 kPa.
∴ fb = 11000.
Now Qb = 11000 x (0.35)2 = 1347.5 kN.
Qs 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.
∴ Penetration depth = 13.7 m below sea bed level
17. Problem # 07
(Pile load test)
• The following data refers to a cyclic pile load
test carried out on a 300mm dia. pile 10m
long.
• Plot the load settlement curves (total and net)
and estimate the allowable load by different
methods/criterians.
Load on pile top (kN) 150 200 250 300 400 500 600
Total settlement on pile
top (mm)
1.45 2.25 2.75 3.6 5.75 10.75 30.00
Net settlement on pile
top (mm)
0.40 0.65 0.80 1.0 1.70 5.25 22.80
18. Solution
• 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%
• 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
Load
Settlement
19. Problem # 08
• Two independent loading tests on 300mm dia.
Short bored piles in clay for which φ = 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/m3 respectively.
20. Solution:
• At point of failure ∑ v =0
• γs = 1.92 Mg/m3
• γc = 2.4 Mg/m3. Qb + Qs = P +
• (wt. of pile – wt. of excavated material)
• Now P = 100 kN Lp = 2.15m.
• Qb + Qs = c Ncp x Ab + α c As = c Ncp x π/4xd2 + α c xπ x d x Lp.
• = c x 9x π/4 x 0.32 + α c x π x 0.3 x 2.15 = 0.636 c + 2.027 α c
• Ncp = 9 (Lp /d > 4)
• P = 100 + wt. of pile – wt. of excavated material
• = 100 + (2.4- 1.92) x 9.81 (π/4 x 0.32 x 2.15)
• = 100 + 0.715
Wt.
P
Qs
Qb
21. • ∑ v = 0
• ∴ 100.715 = 0.636 c + 2.027 α c -----------(i)
• Similarly for 2nd test results ∑ v = 0
• Qb + Qs =9x c x π/4 (0.3)2 + α c xπ x 0.3 x2.75
• = 0.636 c + 2.592 α c
• Load = 110 + (2.4 – 1.92) x 9.81 (π/4 x 0.32 x 2.75)
• = 110 + 0.915 = 110.915
• 110.915 = 0.636 c + 2.592 α c ---------------------(ii)
• Solve equation simultaneously
• ∴ α c = 18.06 , c = 100 and αααα = 0.18
22. 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 centre's. cT at
base = 140 kPa, average (cT)av along shaft = 90
kPa, adhesion α = 0.45, coefficient of volume
decrease mv below base = 0.08 m2/MN,
remolded 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
settlement under the allowable load.
23. Solution:
• (a) Considering individual piles.
fb = cT Ncp = 140 x 9 = 1260 kPa
fs = α cT = 0.45 x 90 = 40.5 kPa
Qf = π/4 B2 fb + B Lp fs
= π/4 0.52 x 1260 + π 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)
Qa = 1/2.5 Qg = 10512/2.5 = 4205 kN
24. • 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.
26. 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 µs = 0.3.
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29. 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)
34. Solution:
• From Eq. (8.116),
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35. • 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
• Qg(u) 10,387 kN < 20,643 kN
• Qg(all) = Qg(u) /FS = 10,387/4 ≈ 2597 kN.
36. Problem # 14
• A group pile in clay is shown in Fig. Determine
the consolidation settlement of the pile
groups. All clays are normally consolidated.
37. 6 ft
48 ft
3 ft
27 ft
21 ft
12 ft
6 ft
2v:1H2v:1H
z
45 ft
30 ft
Water table
Qg = 500 kip
Clay
γsat = 115 lb/ft3
eo = 0.82
Cc = 0.3
γsat = 120lb/ft3
eo = 0.7
Cc = 0.2
γsat = 122 lb/ft3
eo = 0.75
Cc = 0.25
Clay
Clay
sand
γ = 105 lb/ft3
Rock
Po(1),
∆p(1)
Po(2),
∆p(2)
Po(3),
∆p(3)
Pile group: Lg = 9ft; Bg = 6ft
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40. 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:
• If the piles are
• about 9 m long,
• Estimate the
• Number reqired
• And suggest how
• they should be
• arranged
Depth
(m)
Undrained shear strength (kN/m2)
0 100
3 110
6 125
9 125
12 135
41. 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 (π/4 x 0.352) + 115 x 0.45 x (π 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 safety slightly.
•
42. • Bored piles should be spaced at about 3 times
their diameter and a suggested arrangement is
shown on Fig.
• The action of the group
• Must be checked for
stability.
Consider a pier 2.35 m square:
Qfg = 125 x 9 x 2.352 +115 x 0.45 x (4 x 2.35 x 9)
= 6212 + 4378 = 10590 kN
1 m
2.35 m
1 m
