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.
Use the Nordlund method for cohesionless soils.
For uniform pile diameter (no tapering), w = 0, use Equation (4.9). The ultimate bearing capacity of the driven pile is:
Since the pile penetrates two soil layers, the above equation can be written as:
The perimeter of the pile is:
The cross-sectional area at the pile toe is:
The effective stress at the pile toe:
Since the limiting value of s¢t is 150 kPa, choose s¢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 strata / Given parameters / Figures/Tables used / Derived parametersLayer 1 / ϕ1 = 30°
Displaced soil volume: V = 0.1256 m3/m / Figure 4.5, Table 4.1 / Kz(1) = 0.300logV + 1.459 = 1.189
Displaced soil volume: V = 0.1256 m3/m, ϕ1 = 30°, and precast concrete pile / Figure 4.6, curve (c) / d1/f1 = 0.84, so d1 = 25.2°,
ϕ1 = 30°, d1/f1 = 0.84 / Figure 4.7 (use d/f = 0.8 in the chart) / CK(1)» 0.94
γ1= 18.5kN/m3, H1 = 12 m / N/A / Average s¢z(1)= 18.5´ 6= 111kN/m2
Layer 2 / ϕ2 = 35°
Displaced soil volume: V = 0.1256 m3/m / Figure 4.5, Table 4.1 / Kz(2) = 0.600logV + 2.369 = 1.828
Displaced soil volume: V = 0.1256 m3/m, ϕ2 = 35°, and precast concrete pile / Figure 4.6, curve (c) / d2/f2 = 0.84, so d2 = 29.4°,
ϕ2 = 35°, d2/f2 = 0.84, / Figure 4.7 (use d2/f2 = 0.8 in the chart) / CK(2)» 0.91
g2= 19 kN/m3, H2 = 8 m / N/A / Average s¢z(2)= 18.5´12 +19´4= 298kN/m2
ϕ2 = 35°, L=H1+H2 = 20 m, B = 0.4 m, L/B = 50 / Figure 4.8 / at» 0.65
ϕ2 = 35°, / Figure 4.9 / Nq» 75
The ultimate skin resistance is:
The unit toe resistance is:
The ultimate toe resistance is:
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.
Use the Nordlund method for cohesionless soils.
For uniform pile diameter (no tapering), w = 0, use Equation (4.9). The ultimate bearing capacity of the driven pile is:
The perimeter of the pile is:
The cross-sectional area at the pile toe is:
The effective stress at the pile toe:
Note the limiting value of s¢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 d/f = 0.7, so d = 24.5°.
From Figure 4.7, using d/f = 0.7 and ϕ = 35°, find CK» 0.85
Average s¢z= 18.5 L/2 =9.25L (kN/m2)
at 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:
Use trial-and-error:
Assume L/B = 30, i.e., L = 9 m, find = 0.65, calculate Qu = 4894 kN
Assume L/B = 20, i.e., L = 6 m, find = 0.66, calculate Qu = 3188kN
Assume L/B = 15, i.e., L = 4.5 m, find » 0.66, calculate Qu = 2269 kN
Assume L/B = 16, i.e., L = 4.8 m, find » 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/m3, 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.
Since the subsoil is undrained clay, use the a-method. Since f = 0, the undrained shear strength su= cu.
The unit skin resistance is:
Three methods are used to determine and compare a, as shown in the following table.
Soil Strata / Methods / Input values / Figure or equation used / aLayer 1 / 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) / su= 90 kPa;
C1 = 0.5;
=(18.5-9.81)´10/2=43.4kPa / / 0.36
Layer 2 / Tomlinson (1979) / L/B = 5/0.4=12.5,
su= 120 kPa.
Smooth steel pile. / Figure 4.12 / 0.47
Terzaghi et al. (1996) / su= 120 kPa / Figure 4.12 / 0.42
Sladen (1992) / su= 120 kPa;
C1 = 0.5;
=(18.5-9.81)´10 + (19-9.81) ´5/2=109.9kPa / / 0.48
The a values determined usingTergazhi’s method are used.
The perimeter of the pile is: l = pB = 1.25 m
The cross-sectional area at the pile toe:
Layer 1:
Layer 2:
The total skin resistance is:
The unit toe resistance is:
The total toe resistance is:
The total ultimate bearing capacity of the driven concrete pile is:
Qu = Qs+ Qt = 877.5 + 135 = 1012.5kN
Solution using limit state design:
The cross-sectional area at the pile toe:At=14πB2=0.126m2
The characteristic value of the bearing resistance is the minimum value of:
Rc;k=Rtoe;cal+Rskin;calaverageξ3=cuNcAt+αsuπBlξ3
and
Rc;k=Rtoe;cal+Rskin;calminimumξ4=cuNcAt+αsuπBlξ4
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 a, 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=cuNcAt+αsuπBlξ4=120×9×0.126+(0.51×90×π×0.4×10+0.42×120×π×0.4×5)1.4
Rc;k=136.08+893.471.4=735.39kN
and assuming that gtoe = gskin = 1.1 (Note that these might change locally and according to the chosen design approach)
Rc;d=Rtoe;kγtoe+Rskin;kγskin= 136.081.1+893.471.1=935.95 kN
4. 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.
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 a-method. Since f = 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.
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; =(18-9.81)´8/2=33 kPa
=0.30
In layer 2: su= 120 kPa; C1 = 0.5;
=(18-9.81)´8 + (19-9.81) ´L/2=65.5 + 4.6L(kPa)
The perimeter of the pile is: l = pB = 1.25 m
The cross-sectional area at the pile toe:
The total skin resistance is:
The unit toe resistance is:
The total toe resistance is:
The total ultimate bearing capacity of the driven concrete pile is:
Qu = Qs + Qt= 300 + 150aL + 212 = 512 + 150aL(kN)
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.
The cross-sectional area at the pile toe:
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 gG =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;calaverageξ3and Rc;k=Rtoe;cal+Rskin;calminimumξ4
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; =(18-9.81)´8/2=33 kPa
=0.30
fs2= αsu=0.30×100=30 kPa
In layer 3 (clay):
Rskin;cal(3)=αsuπBl = 0.42×120×π×0.5×L = 79.17L[kN]
su= 120 kPa; C1 = 0.5;
=(18-9.81)´8 + (19-9.81) ´L/2=65.5 + 4.6L(kPa)
fs3= αsu=120α
The perimeter of the pile is: l = pB = 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=cuNcAt=120×9×0.196=211.68kN
Hence
Rc;k=Rtoe;cal+Rskin;cal1.4=(211.68+300+150αL)1.4=511.68+150αL1.4
Rc;k=365.49+107αL
and assuming that gtoe = gskin = 1.1 (Note that these might change locally and according to the chosen design approach)
Rc;d=Rtoe;kγtoe+Rskin;kγskin= 211.681.1+300+150αL1.1=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
192.44+136.360.565.5+4.6L1200.45L≥337.50
192.44+136.360.56+0.17LL≥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 f¢ = 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 b -method is used.
Assume the minimum length of the pile is L in meter.
The unit skin resistance is:
= 19´(L/2) = 9.5L (kN/m2)
Use Table 4.4, and given clay soil with f= 32°, select upper limit ofb = 0.4.
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:
Using Table 4.4, and given the clay soil with f= 32°, select upper limit ofNt = 30.
The effective overburden stress at the toe is:
= 19L(kN/m2)
The ultimate toe bearing capacity is:
The total ultimate bearing capacity of the driven concrete pile is:
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.
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 b -method.
The perimeter of the pile is:
The cross-sectional area at the pile toe:
Determine the skin resistance in the top sand layer.
= 18´5/2 = 45 kN/m2.
Use Figure 4.14, and given sandy soil with f= 35°, select b = 0.40.
Determine the skin resistance in the bottom clay layer.