Chapter 4. Introduction to Deep Foundation Design
1.
A 20-m long concrete pile is driven into cohesionless soil of two strata. The topsoil
stratum has unit weight of 18.5 kN/m3, internal friction angle of 30, and thickness
of 12 m. The second stratum has unit weight of 19.0 kN/m3, internal friction angle
of 35, and thickness of 50 m. The groundwater table is found to be at 42 m below
the ground surface. The concrete pile is circular in cross section with a diameter of
40 cm. Determine the ultimate bearing load of the pile.
Solution:
The pile and the subsoil condition are show in the figure below. The pile length is 20 m,
and GWT is 42 m below ground surface, so GWT is 22 m below the tip of the pile and
has no effect on bearing capacity.
z
H1 = 12 m
Sand layer 1:
γ1 = 18.5 kN/m3
c1 = 0; ϕ1 = 30°
H2 = 8 m
Sand layer 2:
γ2 = 19 kN/m3
c2 = 0; ϕ2 = 35°
Use the Nordlund method for cohesionless soils.
For uniform pile diameter (no tapering),  = 0, use Equation (4.9). The ultimate bearing
capacity of the driven pile is:
z=L
Qu = Qs + Qt = å éë( K z CKs ¢z sin d ) lDz ùû + a t N q Ats t¢
z=0
Since the pile penetrates two soil layers, the above equation can be written as:
(
)
(
)
Qu = Qs + Qt = K z(1)CK (1)s ¢z(1) sin d1 l H1 + K z(2)CK (2)s ¢z(2) sin d 2 l H2 + a t N q Ats t¢
The perimeter of the pile is: l = p B = 3.14 ´ 40 cm = 125.6 cm=1.256 m
1
The cross-sectional area at the pile toe is: At = p B2 = 1256 cm 2 = 0.1256 m 2
4
The effective stress at the pile toe:
s t¢ = g 1H1 + g 2 H 2 = 18.5 ´12 +19 ´ 8 = 374 kN/m2
Since the limiting value of t is 150 kPa, choose t = 150 kPa
The following table is developed to obtain the parameters in the ultimate bearing capacity
equation.
Parameters in the ultimate bearing capacity equation
Soil
Figures/Tables
Given parameters
strata
used
ϕ1 = 30°
Figure 4.5,
Displaced soil volume: V =
Table 4.1
0.1256 m3/m
Displaced soil volume: V =
Figure 4.6,
0.1256 m3/m, ϕ1 = 30°, and
curve (c)
Layer 1 precast concrete pile
Figure 4.7 (use
ϕ1 = 30°, 1/1 = 0.84
/ = 0.8 in the
chart)
γ1= 18.5kN/m3, H1 = 12 m
ϕ2 = 35°
Displaced soil volume: V =
0.1256 m3/m
Displaced soil volume: V =
0.1256 m3/m, ϕ2 = 35°, and
precast concrete pile
Layer 2
Derived parameters
Kz(1) = 0.300logV +
1.459 = 1.189
1/1 = 0.84, so 1 =
25.2°,
CK(1) 0.94
N/A
Average z(1)= 18.5 6=
111kN/m2
Figure 4.5,
Table 4.1
Kz(2) = 0.600logV +
2.369 = 1.828
Figure 4.6,
curve (c)
2/2 = 0.84, so 2 =
29.4°,
ϕ2 = 35°, 2/2 = 0.84,
Figure 4.7 (use
2/2 = 0.8 in
the chart)
CK(2) 0.91
2= 19 kN/m3, H2 = 8 m
N/A
Average z(2)= 18.512
+194= 298kN/m2
ϕ2 = 35°, L=H1+H2 = 20 m, B
Figure 4.8
= 0.4 m, L/B = 50
ϕ2 = 35°,
Figure 4.9
The ultimate skin resistance is:
t 0.65
Nq 75
The unit toe resistance is:
qt = a t Nqs t¢ = 0.65 ´ 75 ´150 = 7312.5 kN/m 2
The ultimate toe resistance is: Qt = qt At = 5000 ´ 0.1256 = 628 kN
The total ultimate bearing capacity of the driven concrete pile is:
Qu = Qs + Qt =3241.3 + 628 = 3869.3 kN
2.
A concrete pile is driven into a homogeneous cohesionless soil. The soil’s unit
weight is 18.5 kN/m3, and its internal friction angle is 35. The groundwater table is
not found during the subsoil exploration. The pile is subjected to a load of 800 kN.
Using a factor of safety of 3 and pile diameter of 30 cm, determine the required pile
length.
Solution:
The pile and the subsoil condition are illustrated in the figure below. The pile
length is L.
Q = 800 kN
z
Homogeneous sand:
γ1 = 18.5 kN/m3
c1 = 0; ϕ1 = 35°
L
D = 0.3 m
Use the Nordlund method for cohesionless soils.
For uniform pile diameter (no tapering),  = 0, use Equation (4.9). The ultimate bearing
capacity of the driven pile is:
Qu = K zCKs z¢ sin (d ) lL + a t N qs t¢
The perimeter of the pile is: l = p B = 3.14 ´ 30 cm = 94.2 cm=0.942 m
The cross-sectional area at the pile toe is: At =
1 2
p B = 706 cm 2 » 0.071 m 2
4
The effective stress at the pile toe:
s t¢ = 18.5L
Note the limiting value of t is 150 kPa.
Given: ϕ = 35°, displaced soil volume: V = 0.071 m3/m, find Kz from Table 4.1:
Kz = 0.600logV + 2.369=1.680
From Figure 4.6, curve “c”, find / = 0.7, so  = 24.5°.
From Figure 4.7, using / = 0.7 and ϕ = 35°, find CK 0.85
Average z= 18.5 L/2 =9.25L (kN/m2)
t depends on pile length L and can be determined from Figure 4.8.
From Figure 4.9 and use ϕ = 35°, find Nq 75
So:
To satisfy FS = 3:
Qu = 3Q = 3´ 800 = 2400 (kN)
Use trial-and-error:
Assume L/B = 30, i.e., L = 9 m, find a t = 0.65, calculate Qu = 4894 kN
Assume L/B = 20, i.e., L = 6 m, find a t = 0.66, calculate Qu = 3188kN
Assume L/B = 15, i.e., L = 4.5 m, find a t  0.66, calculate Qu = 2269 kN
Assume L/B = 16, i.e., L = 4.8 m, find a t  0.66, calculate Qu = 2435 kN
So, L = 4.8 m to satisfy FS = 3.0
3.
A 15-m closed-end steel pipe pile is driven into layered undrained clay. The top
layer has unit weight of 18.5 kN/m3, undrained cohesion of 90 kN/m2, and a
thickness of 10 m. The second layer has unit weight of 19.5 kN/m 3, undrained
cohesion of 120 kN/m2, and it extends to a great depth. The groundwater table is at
the ground surface. The pile diameter is 40 cm. Determine the ultimate bearing load
of the pile.
Solution:
Solution using allowable stress design:
The pile and subsoil condition are shown in the following figure.
z
10 m
Undrained clay layer 1:
γsat(1) = 18.5 kN/m3
cu1 = 90 kPa; ϕ1 = 0°
5m
Undrained clay layer 2:
γsat(2) = 19.5 kN/m3
cu2 = 120 kPa; ϕ2 = 0°
D = 0.4 m
Since the subsoil is undrained clay, use the -method. Since  = 0, the undrained shear
strength su= cu.
The unit skin resistance is:
fs = ca = a su
Three methods are used to determine and compare , as shown in the following table.
Soil
Strata
Layer 1
Methods
Figure or
equation used

Tomlinson
(1979)
L/B = 10/0.4=25,
su= 90kPa.
Smooth steel pile.
Figure 4.12
0.75 (use
interpolation)
Terzaghi et
al. (1996)
su= 90 kPa
Figure 4.12
0.50
Sladen
(1992)
Tomlinson
(1979)
Layer 2
Input values
Terzaghi et
al. (1996)
Sladen
(1992)
su= 90 kPa;
C1 = 0.5;
q =(18.59.81)10/2=43.4kPa
L/B = 5/0.4=12.5,
su= 120 kPa.
Smooth steel pile.
su= 120 kPa
su= 120 kPa;
C1 = 0.5;
q =(18.5-9.81)10 + (199.81) 5/2=109.9kPa
æqö
a = C1 ç ÷
è su ø
0.45
0.36
Figure 4.12
0.47
Figure 4.12
0.42
æqö
a = C1 ç ÷
è su ø
0.45
0.48
The  values determined usingTergazhi’s method are used.
The perimeter of the pile is: l = B = 1.25 m
1
The cross-sectional area at the pile toe: At = p B 2 = 0.125 m2
4
Layer 1: f s(1) = a su = 0.50´ 90 = 45 kN/m2
Layer 2: f s(2) = a su = 0.42´120 = 50.4 kN/m2
The total skin resistance is:
Qs = f s(1) A1 + f s(2) A2
= 45´1.25´10+50.4 ´1.25´5
= 877.5 kN
The unit toe resistance is: qt = cu N c = 120 ´ 9 = 1080 kPa
The total toe resistance is:
Qt = qt At = 1080´0.125 = 135 kN
The total ultimate bearing capacity of the driven concrete pile is:
Qu = Qs+ Qt = 877.5 + 135 = 1012.5kN
Solution using limit state design:
1
The cross-sectional area at the pile toe:𝐴𝑡 = 4 𝜋𝐵 2 = 0.126m2
The characteristic value of the bearing resistance is the minimum value of:
𝑅𝑐;𝑘 =
(𝑅𝑡𝑜𝑒;𝑐𝑎𝑙 +𝑅𝑠𝑘𝑖𝑛;𝑐𝑎𝑙 )𝑎𝑣𝑒𝑟𝑎𝑔𝑒
𝜉3
=
𝑐𝑢 𝑁𝑐 𝐴𝑡 +𝛼𝑠𝑢 𝜋𝐵𝑙
𝜉3
and
𝑅𝑐;𝑘 =
(𝑅𝑡𝑜𝑒;𝑐𝑎𝑙 +𝑅𝑠𝑘𝑖𝑛;𝑐𝑎𝑙 )𝑚𝑖𝑛𝑖𝑚𝑢𝑚
𝜉4
=
𝑐𝑢 𝑁𝑐 𝐴𝑡 +𝛼𝑠𝑢 𝜋𝐵𝑙
𝜉4
Assuming the geotechnical parameters are the result of only one ground test per clay
layer,ξ3 and ξ4 are assumed both equal to 1.4 as suggested in EN-1997-1:2004 (Design
approach 2). Hence, using Terzaghi´s method for and assuming also that the material
properties are design values (as if they were already multiplied by their corresponding
partial factor of safety):
𝑅𝑐;𝑘 =
𝑅𝑐;𝑘 =
𝑐𝑢 𝑁𝑐 𝐴𝑡 +𝛼𝑠𝑢 𝜋𝐵𝑙
𝜉4
136.08+893.47
1.4
=
120×9×0.126+(0.51×90×𝜋×0.4×10+0.42×120×𝜋×0.4×5)
1.4
= 735.39kN
and assuming that toe = skin = 1.1 (Note that these might change locally and according to
the chosen design approach)
𝑅𝑐;𝑑 =
4.
𝑅𝑡𝑜𝑒;𝑘
𝛾𝑡𝑜𝑒
+
𝑅𝑠𝑘𝑖𝑛;𝑘
𝛾𝑠𝑘𝑖𝑛
=
136.08
1.1
+
893.47
1.1
= 𝟗𝟑𝟓. 𝟗𝟓 𝒌𝑵
A subsoil profile is shown in Figure 4.23. The concrete pile’s diameter is 50
cm.Determine the total length of the concrete pile to take a load of 250 kN with a
factor of safety of 3.
Groundwater table
2m
z
8m
Sand, scour zone. γsat = 17.5 kN/m3, c = 0, ϕ = 30°
Clay layer 1:
γsat(1) = 18 kN/m3
cu1 = 100 kPa; ϕ1 = 0°
Clay layer 2:
γsat(2) = 19 kN/m3
cu2 = 120 kPa; ϕ2 = 0°
To significant
depth
Figure 4.23 Subsoil profile for Problem 4
Solution:
Solution using allowable stress design:
The scour zone is not considered in bearing capacity calculation.
Since the subsoil is clay and is beneath the groundwater table, use the -method. Since 
= 0, the undrained shear strength su= cu.
Also given: concrete pile diameter B = 0.5 m. Total load Q = 250 kN
Assume the length of the pile in the second layer is L.
fs = ca = a su
The unit skin resistance is:
To avoid using trial-and-error, use the Sladen method (1992) to directly calculate L.
In layer 1:
su= 100 kPa; C1 = 0.5; q =(18-9.81)8/2=33 kPa
æqö
a = C1 ç ÷
è su ø
In layer 2:
0.45
=0.30
f s(1) = a su = 0.30´100 = 30 kPa
su= 120 kPa; C1 = 0.5;
q =(18-9.81)8 + (19-9.81) L/2=65.5 + 4.6L(kPa)
æqö
a = C1 ç ÷
è su ø
0.45
æ 65.5+ 4.6L ö
= 0.5ç
120 ÷ø
è
0.45
f s(2) = a su = 120a kPa
The perimeter of the pile is: l = B = 1.25 m
1 2
2
The cross-sectional area at the pile toe: At = p B = 0.196 m
4
The total skin resistance is:
Qs = f s(1) A1 + f s(2) A2
= 30´1.25´ 8 +120a ´1.25´ L
= 300+150a L (kN)
The unit toe resistance is: qt = cu N c = 120 ´ 9 = 1080 kPa
The total toe resistance is:
Qt = qt At = 1080´0.196 = 212 kN
The total ultimate bearing capacity of the driven concrete pile is:
Qu = Qs + Qt= 300 + 150L + 212 = 512 + 150L(kN)
FS =
Qu
Q
=
512+150a L
=3
250
0.45
0.45
æqö
æ 65.5+ 4.6L ö
a = C1 ç ÷ = 0.5ç
120 ÷ø
è
è su ø
Solve the above equation and find L = 3.75 m
L = 4.0 m can be chosen.
The total pile length is: 2m (scour zone) + 8m + 4m = 14 m.
Solution using limit state design:
In limit state design approaches using partial factors of safety as demonstrated in this
book, the concept of the global factor of safety suggested in this problem is not
applicable. Hence, an alternative solution in which the pile length required to satisfy that
Rc;d ≥ Fc;d (where Rc;d is the design resistance and Fc;d is the design value of all forces
imposed on the pile) is proposed.
1
The cross-sectional area at the pile toe: At = p B2 = 0.196 m 2
4
The forces imposed on the pile should include both the forces and the self-weight of the
pile. In the calculation below it is assumed that both of these forces are permanent and
unfavorable, as well as included in the load of 250 kN. Hence the partial factor of safety
G =1.35. Assuming that the pile will have to penetrate into the third soil layer by a length
L, and noting that the scour zone is neglected in the calculation, then
Fc;d = 𝑉𝐺 × 𝛾𝐺
𝐹𝑐;𝑑 = 250 × 1.35
Fc;d = 337.50 kN
Noting that partial factors of safety for geotechnical parameters are all equal to 1.00 for
the design approach used here. Also assuming the geotechnical parameters are the result
of only one ground test per layer,it can be found that the characteristic value of the
bearing capacity of the pile is the minimum of:
𝑅𝑐;𝑘 =
(𝑅𝑡𝑜𝑒;𝑐𝑎𝑙 +𝑅𝑠𝑘𝑖𝑛;𝑐𝑎𝑙 )𝑎𝑣𝑒𝑟𝑎𝑔𝑒
𝜉3
and
𝑅𝑐;𝑘 =
(𝑅𝑡𝑜𝑒;𝑐𝑎𝑙 +𝑅𝑠𝑘𝑖𝑛;𝑐𝑎𝑙 )𝑚𝑖𝑛𝑖𝑚𝑢𝑚
𝜉4
But because ξ3 and ξ4 are assumed both equal to 1.4 when only one test is available, as
suggested in EN-1997-1:2004 then,
𝑅𝑐;𝑘 =
(𝑅𝑡𝑜𝑒;𝑐𝑎𝑙 +𝑅𝑠𝑘𝑖𝑛;𝑐𝑎𝑙 )
1.4
The skin resistance for each layer can then be calculated. The aim is to leave the total
skin resistance as a function of the penetration of the pile into layer 3 (L)
In layer 2 (clay):
su= 100 kPa; C1 = 0.5; q =(18-9.81)8/2=33 kPa
æqö
a = C1 ç ÷
è su ø
0.45
=0.30
𝑓𝑠(2) = 𝛼𝑠𝑢 = 0.30 × 100 = 30 kPa
In layer 3 (clay):
𝑅𝑠𝑘𝑖𝑛;𝑐𝑎𝑙(3) = 𝛼𝑠𝑢 𝜋𝐵𝑙 = 0.42×120×π×0.5×L = 79.17L[kN]
su= 120 kPa; C1 = 0.5;
q =(18-9.81)8 + (19-9.81) L/2=65.5 + 4.6L(kPa)
0.45
æqö
æ 65.5+ 4.6L ö
a = C1 ç ÷ = 0.5ç
120 ÷ø
è
è su ø
𝑓𝑠(3) = 𝛼𝑠𝑢 = 120𝛼
0.45
The perimeter of the pile is: l = B = 1.25 m
The characteristic skin resistance is:
𝑄𝑠;𝑘 = 𝑅𝑠𝑘𝑖𝑛;𝑐𝑎𝑙(2) + 𝑅𝑠𝑘𝑖𝑛;𝑐𝑎𝑙(3)
𝑄𝑠;𝑘 = 30 × 1.25 × 8 + 120𝛼 × 1.25 × 𝐿
𝑄𝑠;𝑘 = 300 + 150𝛼𝐿
The toe bearing capacity:
𝑅𝑡𝑜𝑒;𝑐𝑎𝑙 = 𝑐𝑢 𝑁𝑐 𝐴𝑡 = 120 × 9 × 0.196 = 211.68kN
Hence
𝑅𝑐;𝑘 =
(𝑅𝑡𝑜𝑒;𝑐𝑎𝑙 +𝑅𝑠𝑘𝑖𝑛;𝑐𝑎𝑙 )
1.4
=
(211.68+300+150𝛼𝐿)
1.4
=
511.68+150𝛼𝐿
1.4
𝑅𝑐;𝑘 = 365.49 + 107𝛼𝐿
and assuming that toe = skin = 1.1 (Note that these might change locally and according to
the chosen design approach)
𝑅𝑐;𝑑 =
𝑅𝑡𝑜𝑒;𝑘
𝛾𝑡𝑜𝑒
+
𝑅𝑠𝑘𝑖𝑛;𝑘
𝛾𝑠𝑘𝑖𝑛
=
211.68
1.1
+
300+150𝛼𝐿
1.1
= 192.44 + 136.36𝛼𝐿
Finally, it must be satisfied that Rc;d ≥ Fc;d
Rc;d ≥ Fc;d
192.44 + 136.36𝛼𝐿 ≥ 337.50
And solving the equation with
æqö
a = C1 ç ÷
è su ø
0.45
æ 65.5+ 4.6L ö
= 0.5ç
120 ÷ø
è
0.45
65.5+4.6𝐿 0.45
192.44 + 136.36 (0.5 (
120
)
) 𝐿 ≥ 337.50
192.44 + 136.36(0.56 + 0.17𝐿)𝐿 ≥ 337.50
192.44 + 76.36𝐿 + 23.18𝐿2 ≥ 337.50
𝐿 ≥ 1.34 m
This means that the length of the pile should be at least 2 m + 8 m + 1.34m ≅ 11.5 m in
length.
5.
A concrete pile is designed to support a load of 4600 kN. The pile is driven into a
homogeneous drained clayey sand with c = 50 kN/m2 and  = 32. The unit weight
of the subsoil is 19 kN/m3. The concrete pile is square in cross section with a width
of 30 cm. Use FS = 3. Determine the minimum length of the pile.
Solution:
Since the subsoil is drained clay, the  -method is used.
Assume the minimum length of the pile is L in meter.
The unit skin resistance is: fs = b × s ¢
s 0¢ = 19(L/2) = 9.5L (kN/m2)
Use Table 4.4, and given clay soil with = 32, select upper limit of = 0.4.
f s = bs ¢ = 0.4 ´ 9.5L = 3.8L
The perimeter of the pile is 0.3  4 = 1.2 m
The cross-sectional area at the pile toe is At = 0.3  0.3 = 0.09 m2
The ultimate total skin resistance is:
Qs = fs As = 3.8L 1.2L = 4.56L2
The unit toe bearing capacity is: qt = N t × s t¢
Using Table 4.4, and given the clay soil with = 32, select upper limit ofNt = 30.
The effective overburden stress at the toe is:
s t¢ = 19L(kN/m2)
qt = Nts t¢ = 30´19L = 570L
The ultimate toe bearing capacity is:
Qt = At qt = 0.09´570L = 51.3L
The total ultimate bearing capacity of the driven concrete pile is:
Qu = Qt +Qs = 51.3L+ 4.56L2
Qu
51.3L + 4.56L2
FS = =
=3
Q
460
Solve L and find: L = 12.66 m
6.
As shown in Figure 4.24, a concrete pile is driven into the top two layers of subsoil
strata. The subsoil profile and properties are shown in the figure. The pile’s
diameter is 50 cm throughout the pile. Determine the ultimate bearing load of the
pile.
z
5m
10 m
Sand, layer 2:
γ1 = 18 kN/m3
c1 = 0; ϕ1 = 35°
Clay, layer 1:
γ2 = 19 kN/m3
c2 = 100 kN/m2; ϕ2 = 25°
Figure 4.24 Subsoil profile for Problem 6
Solution:
Ultimate bearing capacity of the pile: Qu = Qs(sand) + Qs(clay) + Qt(in clay)
Since the groundwater table is not present, assume the subsoil is drained and use the  method.
The perimeter of the pile is: l = p B = 1.57m
1
The cross-sectional area at the pile toe: At = p B2 = 0.196 m 2
4
Determine the skin resistance in the top sand layer.
s 0¢ = 185/2 = 45 kN/m2.
Use Figure 4.14, and given sandy soil with = 35, select  = 0.40.
Qs(sand ) = f s(sand ) As(sand ) = bs 0¢ As(sand ) = 0.40´ 45´1.57´5 = 141.3 kN
Determine the skin resistance in the bottom clay layer.
s 0¢ = 185+ 19(10/2)= 185 kN/m2.
Use Table 4.4 and given clay soil with = 25, select minimum  = 0.23.
Qs(clay ) = f s(clay ) As(clay ) = bs 0¢ As(clay ) = 0.23´185´1.57´10 = 668.0 kN
Determine the toe bearing capacity:
Using Table 4.4, and given clay soil with = 25, use minimum Nt = 3.
The effective overburden stress at the toe is:
s t¢ = 185 + 1910 = 280kN/m2.
qt = Nts t¢ = 3´280 = 840 kN/m2
The ultimate toe bearing capacity is:
Qt = qt At = 840´0.196 = 164.6 kN
The total ultimate bearing capacity of the driven concrete pile is:
Qu = Qs(sand) + Qs(clay) + Qt(in clay) = 141.3+668.0+164.6 = 973.9 kN
7.
A pile group is comprised of four circular concrete piles. The diameter of each pile
is 40 cm. The spacing between two adjacent piles is 120 cm. The pile group is
driven into a homogeneous sandy riverbed to support a bridge pier. It is assumed
the river flows year-round. The saturated unit weight of subsoil is 19 kN/m3, the
cohesion is zero, and the internal friction angle is 36. The pile length is decided to
be 12 m. Determine the ultimate bearing capacity and pile group efficiency of the
pile group.
Solution:
Given parameters: b = 0.4 m, s = 1.2 m, L = 12 m, sat = 19 kN/m3, c = 0, = 36.
The plan view of the pile group is show in the figure below.
Assume 0.4 m
b = 0.4 m
s = 1.2 m
Assume 0.4 m
The subsoil is sandy (cohesionless) soil. The unit skin resistance is fs, the unit toe
resistance qt.
The ultimate bearing capacity of pile group: Qu( g) = As( g) f s + At( g)qt
(
)
= (0.4 +1.2+0.4 ) = 4m
As( g) = 0.4 +1.2+0.4 ´ 4 ´12= 96m
2
At( g)
The ultimate bearing capacity of an individual pile in the pile group:
Qu = As f s + At qt
As = p ´0.4 ´12= 15.07 m2
At = p ´0.22 = 0.1256 m2
Use Figure 4.14 and given sandy soil with = 36, find  = 0.41
Average effective stress: s 0¢ = (19-9.81)12/2 = 55.1 kN/m2
f s = bs ¢ = 0.41´55.1 = 22.6 kN/m2
Use Figure 4.15, and given sandy soil with = 36, find Nt67.
The effective overburden stress at the toe is:
s t¢ = (19-9.81)12 = 110.2 kN/m2.
qt = Nts t¢ = 67´110.2= 7388.7 kN/m2
So: Qu( g) = As( g) f s + At( g)qt = 96´22.6+ 4 ´7388.7 = 31724.4 kN
Qu = As f s + At qt = 15.07´22.6+0.1256´7388.7 = 1268.5 kN
The pile group efficiency: h g =
8.
Qu( g)
nQu
=
31724.4
= 6.25
4 ´1268.5
A 30-m long closed-end steel pipe pile group is driven into layered undrained clay.
The pile cap is square and the nine piles are evenly spaced. The layout of the pile
group is shown in Figure 4.25. The topsoil layer has a unit weight of 18 kN/m 3,
undrained cohesion of 100 kN/m2, and a thickness of 10 m. The second layer has
unit weight of 19 kN/m3, undrained cohesion of 150 kN/m2, and it extends to great
depth. The groundwater table is at the ground surface. Determine the ultimate
bearing capacity and pile group efficiency of the pile group.
b = 0.5 m
s = 1.5 m
0.5 m
Figure 4.25 Layout of pile group for Problem 8.
Solution:
The section view of the pile group and the subsoil condition is shown below.
L = Bg = 4.0 m
Ground surface
γsat(1) = 18 kN/m3
cu1 = 100 kN/m2
Groundwater
table
L1 = 10 m
γsat(2) = 19 kN/m3
cu1 = 150 kN/m2
L2 = 20 m
Lg = Bg = 3.5 m
Ultimate bearing of pile group: Qu( g) = Qu(p) +Qu(s )
Ultimate point bearing capacity:
Qu(p) = LgB gcu(2)Nc * = 3.5´3.5´150´ 9 = 16537.5 kN
Nc* = 9 for deep foundation in undrained clay
Ultimate skin resistance:
(
Qu( s ) = å pcu DL
(
)
)
(
)
= 2 Lg + B g cu(1)L1 + 2 Lg + B g cu(2)L2
(
)
(
)
= 2´ 3.5+ 3.5 ´100´10+ 2´ 3.5+ 3.5 ´150´ 20
= 56000 kN
Qu( g) = Qu(p) +Qu(s ) = 16537.5+56000 = 72537.5 kN
Ultimate bearing capacity of each single pile:
Qu = Qp +Qs
Qp = Apcu(2)Nc * = p ´0.252 ´150´ 9 = 264.9 kN
Use -method to calculate the friction resistance of single pile. The  values are
determined usingTergazhi’s method (Figure 4.12).
The perimeter of the pile is: l = B = 1.57 m
Layer 1: f s(1) = a su = 0.49´100 = 49 kN/m2
Layer 2: f s(2) = a su = 0.40´150 = 60 kN/m2
The total skin resistance is:
Qs = f s(1) A1 + f s(2) A2
= 49´1.57 ´10+60´1.57 ´ 20
= 2653.3 kN
Qu = Qp +Qs = 264.9+2543.3= 2918.2 kN
The pile group efficiency: h g =
9.
Qu( g)
nQu
=
72537.5
= 2.76
9´2918.2
The subsoil profile of a riverbed is shown in Figure 4.23. It is determined that a pile
group comprising four piles is needed to support the bridge pier. The four piles are
evenly spaced. The center-to-center spacing is three times of the pile diameter, and
each pile’s outside circumference is assumed to align with the edge of the pile cap.
Each concrete pile’s diameter is 50 cm, and length is 15 m. Determine the ultimate
bearing capacity and pile group efficiency of the pile group.
Solution:
The subsoil condition is shown below:
Groundwater table
2m
z
8m
5m
Sand, scour zone. γsat = 17.5 kN/m3, c = 0, ϕ = 30°
Clay layer 1:
γsat(1) = 18 kN/m3
cu1 = 100 kPa; ϕ1 = 0°
Clay layer 2:
γsat(2) = 19 kN/m3
cu2 = 120 kPa; ϕ2 = 0°
The plan view of the pile group is show in the figure below. Pile length L = 15 m
0.25 m
b = 0.5 m
s = 1.5 m
0.25 m
The scour zone is not considered in bearing capacity calculation.
(1) Determine the ultimate bearing capacity of pile group:
Qu( g) = As( g) f s + At( g)qt
For undrained clayey soil:
Unit skin resistance in layer 1: f s(1) = cu1 = 100 kPa
Unit skin resistance in layer 2: f s(2) = cu2 = 120 kPa
Unit toe resistance: qt = cu2 N c
æ
æ 12 ö æ 2 ö
L öæ B ö
Nc = 5ç 1+ ÷ ç 1+ 1 ÷ = 5´ ç 1+ ÷ ç 1+ ÷ = 70 > 9 , use the limiting value of 9.
B2 ø
2 ø è 2ø
è
è B1 ø è
qt = cu2Nc = 120´ 9 = 1080 kPa
Qu( g) = As( g)(1) f s(1) + As( g)(2) f s(2) + At( g)qt
= 2´ 4 ´ 8 ´100+ 2´ 4 ´8 ´120+2´2´1080
= 18400 kN
(2) Determine the ultimate bearing capacity of an individual pile in the pile group:
Qu = As f s + At qt
Follow the solution in Problem 4, and use the Sladen method (1992)
In layer 1:
su= 100 kPa; C1 = 0.5; q =(18-9.81)8/2=33 kPa
æqö
a = C1 ç ÷
è su ø
0.45
=0.30
f s(1) = a su = 0.30´100 = 30 kPa
In layer 2:
su= 120 kPa; C1 = 0.5;
q =(18-9.81)8 + (19-9.81) 5/2=88.5 kPa
æqö
a = C1 ç ÷
è su ø
0.45
æ 88.5 ö
= 0.5ç
è 120 ÷ø
0.45
= 0.43
f s(2) = a su = 0.43´120 = 51.6 kPa
The perimeter of the pile is: l = B = 1.25 m
1
The cross-sectional area at the pile toe: At = p B 2 = 0.196 m2
4
The total skin resistance is:
Qs = f s(1) A1 + f s(2) A2
= 30´1.25´ 8 +51.6 ´1.25´5
= 622.5 (kN)
The unit toe resistance is: qt = cu N c = 120 ´ 9 = 1080 kPa
The total toe resistance is:
Qt = qt At = 1080´0.196 = 211.6 kN
The total ultimate bearing capacity of the driven concrete pile is:
Qu = Qs + Qt= 622.5 + 211.6 = 834.1 kN
The pile group efficiency: h g =
Qu( g)
nQu
=
18400
= 5.51
4 ´ 834.1
10. A 15-m long closed-end steel pipe pile group is driven into a homogeneous clay.
The pile cap is square and the nine piles are evenly spaced. The layout of the pile
group is shown in Figure 4.25. The pile group is subjected to a vertical load of 5200
kN. The soil has a unit weight of 18.5 kN/m3, cohesion of 100 kN/m2, friction angle
of 10 degrees. The clay layer is 100 m deep and beneath the clay layer is dense
sand. The groundwater table is at the ground surface. Preliminary laboratory testing
found the clay’s void ratio is 0.45, compression index is 0.3, swell index is 0.08,
and the clay is overconsolidated. The preconsolidation pressure is 200
kN/m2.Determine the primary consolidation settlement of the pile group.
Solution:
Subsoil parameters: sat = 18.5 kN/m3, c = 100 kN/m2, = 10, e0 = 0.45, cc = 0.3, cs =
0.08, c= 200 kN/m2.
The layout of pile group is shown below. AndB = 4 m, L = 12 m.
b = 0.5 m
s = 1.5 m
0.5 m
The 2:1 method is used to calculate the vertical stress increases due to the pile group
loading. The vertical stress increases are assumed to occur starting at the depth of two
thirds of the pile length.
(1) Determine the depth of the soil layer for which the consolidation settlement should
be calculated. The consolidation settlement should be considered to a depth of z
0.1 0.
pB 2
Use 2:1 method: Ds z ¢ =
2
B+z
(
where: p =
)
Q 5200
= 2 = 325 kPa
B2
4
z starts from the (2/3)L = 8 m
The effective stress: s 0¢ = (18.5- 9.81)(8+ z) = 69.52+8.69z
Let: z = 0.10, and solve z = 12.9 m. Use z = 13.0 m
(2) Since the vertical stress increase is nonlinear with depth, the thick soil layers should
be divided into a number of thin layers, and the consolidation settlement for each thin
layer is calculated, then the individual settlements are summed. Based on Figure 3.24
and for easy calculation, the second clay layer of the subsoil is divided into two
layers, as shown in the figure below.
Groundwater
table
Ground surface
γsat = 18.5 kN/m3
Layer 1: 2 m
8m
z
4m
Layer 2: 4 m
B = 4m
Layer 3: 7 m
The vertical stress increases, the average vertical stress increase, and the average
effective stress in each layer are calculated and listed as follows. Note, the vertical
stress increase is calculated from a depth of 10 m, while the effective stress is
calculated from the ground surface.
The average vertical stress increase uses Equation (3.55):
1
Ds z(av) = Ds z(top) + 4Ds z(mid ) + Ds z(bo)
6
(
z(m)
0
1
2
4
6
9.5
13
Layer
)
Vertical stress
increase
(kN/m2)
325.0
208.0
144.4
81.2
52.0
28.5
18.0
Average
vertical stress
increase,
z(av) (kN/m2)
Average in-situ
effective stress,
0(av) (kN/m2)
Precon-solidation
pressure, c
(kN/m2)
Thickness,
H (m)
#1
216.9
#2
86.9
#3
30.7
(18.5-9.81)9=78.2
(18.59.81)12=104.3
(18.59.81)17.5=152.1
180
2
180
4
180
7
In layer #1: c=200 kN/m2 <0(av) +z(av) = 78.2+216.9=295.1 kN/m2, the soil is
overconsolidated.
s ¢ + Ds (z¢ )av
cH
s¢
cH
Sc = s log c + c log 0(av)
1+ e0
s 0(av)
1+ e0
s c¢
¢
0.08 ´ 2
200 0.3 ´ 2
78.2 + 216.9
log
+
log
1+ 0.45
78.2 1+ 0.45
200
= 0.115 m = 11.5 cm
=
In layer #2: c =200 kN/m2 >0(av) +z(av) = 104.3+86.9=191.2kN/m2,the soil is
overconsolidated.
s ¢ + Ds (z¢ )av
cH
Sc = s log 0(av)
1+ e0
s 0(av)
¢
0.08 ´ 4
104.3 + 86.9
log
1+ 0.45
104.3
= 0.058 m = 5.8 cm
=
In layer #3: c =200 kN/m2 >0(av) +z(av) = 152.1+30.7=182.8 kN/m2,the soil is
overconsolidated.
s ¢ + Ds (z¢ )av
cH
Sc = s log 0(av)
1+ e0
s 0(av)
¢
0.08 ´ 7
152.1 + 30.7
log
1+ 0.45
152.1
= 0.031 m = 3.1 cm
=
The total consolidation settlement is: 20.4 cm.
11. The problem statement is the same as in Problem 9, and the subsoil profile is shown
in Figure 4.23. The total load on the pile cap is 6000 kN. Assume both clay layers
are normally consolidated. Both clay layers have the void ratio of 0.4, compression
index of 0.3. Determine the total settlement of the pile cap.
Solution: Not provided.