There are numerous cases where the soil near the ground surface is not homogeneous, but is layered. If the layers are in the zone where the soil would move up and out as a wedge, some modifications would be needed in the method to compute the ultimate soil resistance pu, and consequently modifications would be needed in the p-y curves.
The problem of the layered soil has been given intensive study by Allen (1985); however, Allen’s formulations require the use of several computer programs. Integrating the methods developed by Allen with the methods shown herein must be delayed until a later date when this research can be put in a readily usable form.
3-11-1 Layering Correction Method of Georgiadis
The method of Georgiadis (1983) is based on the determination of the “equivalent” depth of all the layers existing below the upper layer. The p-y curves of the upper layer are determined according to the methods presented herein for homogeneous soils. To compute the p-y curves of the second layer, the equivalent depth H2 to the top of the second layer has to be determined by summing the ultimate resistances of the upper layer and equating that value to the summation as if the upper layer had been composed of the same material as in the second layer. The values of pu are computed according to the equations given earlier. Thus, the following two equations are solved simultaneously for H2.
∫
=
1
0 1 1
H u dH p
F ... (3-124)
and
∫
=
2
0 2 1
H u dH p
F ...(3-125)
The equivalent thickness H2 of the upper layer along with the soil properties of the second layer, are used to compute the p-y curves for the second layer.
The concepts presented above can be used to get the equivalent thickness of two or more dissimilar layers of soil overlying the layer for whom the p-y curves are desired. One possible consequence is that the equivalent depths may be either smaller or greater than the actual depths of the soil layers, depending on the relative strengths of the layers of the soil profiles. This is illustrated in Figure 3-56.
Soft Soil (Layer 3) Soft Soil (Layer 1)
Stiff Soil (Layer 2)
F1= Total force acting on pile above point i at the time of soil failure
hi= Equivalent depth of top of layer i
h1
h2 h3
F2 F1
Fi Groundline
Soft Soil (Layer 3) Soft Soil (Layer 1)
Stiff Soil (Layer 2)
F1= Total force acting on pile above point i at the time of soil failure
hi= Equivalent depth of top of layer i
h1
h2 h3
F2 F1
Fi Groundline
Figure 3-56 Illustration of Equivalent Depths in a Multi-layer Soil Profile
3-11-2 Example p-y Curves in Layered Soils
The example problem to demonstrate the manner in which layered soils are modeled is shown in Figure 3-57. As seen in the sketch, a pile with a diameter of 610 mm (24 in.) is embedded in soil consisting of an upper layer of soft clay, overlying a layer of loose sand, which in turn overlays a layer of stiff clay. The water table is at the ground surface, and the loading is static.
Four p-y curves for the case of layered soil are shown in Figure 3-58. The curves are for points A, B, C and D as shown in the sketch in Figure 3-59, at depths of 0.92 m (36 in.), 1.83 m (72 in.), 3.66 m (144 in.), and 7.32 m (288 in.), respectively. The curve at a depth of 0.92 m (36 in.) falls in the upper zone of soft clay; the curve for the depth of 1.83 m (72 in.) falls in the sand just below the soft clay; and the curves for depths of 3.66 m (144 in.) and 7.32 m (288 in.) fall in the lower zone of stiff clay.
Soft Clay
Loose Sand
Stiff Clay
Static Loading
c = 23.9 kPa ε50= 0.02 γ′ = 7.9 kN/m3 φ= 30 deg.
γ′= 7.9 kN/m3
c = 95.8 kPa ε50= 0.005 γ′= 9.4 kN/m3 k = 20,400 kPa 1.73 m
1.32 m
6.1 m
0.61 m
Soft Clay
Loose Sand
Stiff Clay
Static Loading
c = 23.9 kPa ε50= 0.02 γ′ = 7.9 kN/m3 φ= 30 deg.
γ′= 7.9 kN/m3
c = 95.8 kPa ε50= 0.005 γ′= 9.4 kN/m3 k = 20,400 kPa 1.73 m
1.32 m
6.1 m
0.61 m
Figure 3-57 Soil Profile for Example of Layered Soils
Following the method suggested by Georgiadis, the p-y curve for soft clay can be computed as if the profile consists altogether of that soil. When dealing with the sand, an equivalent depth of sand is found such that the integrals of the ultimate soil resistance of an equivalent sand layer and for the soft clay are equal at the interface. The equivalent thickness of loose sand to replace the 1.73 m (68 in.) of soft clay was found to be 1.88 meters (74 in.). Thus, the equivalent depth to point B in loose sand is 1.98 meters (78 in.). A plot of the integrals of ultimate soil resistance and equivalent depths is presented in Figure 3-59.
An equivalent depth of stiff clay was found such that the sum of the ultimate soil resistance for the stiff clay is equal to the sum of the ultimate soil resistance of the loose sand and soft clay. In making the computation, the equivalent and actual thicknesses of the loose sand, 1.88 m (74 in.) and 1.32 m (52 in.), respectively, were replaced by 1.14 m (45 in.) of stiff clay.
Thus, the actual thicknesses of the soft clay and loose sand of 3.05 m (120 in.) were reduced by 1.91 m (75 in.), leading to equivalent depths in the stiff clay of points C and D of 1.75 m (69 in.) and 5.41 m (213 in.), respectively (Figure 3-59).
Lateral Deflection y, meters
Load Intensity p, kN/m
400
Figure 3-58 Example p-y Curves for Layered Soil
Soft
Figure 3-59 Equivalent Depths of Soil Layers Used for Computing p-y Curves
Another point of considerable interest is that the recommendations for p-y curves for stiff clay in the presence of no free water were used for the stiff clay. This decision is based on the assumption that the sand above the stiff clay can move downward and fill any gap that develops between the clay and the pile. Furthermore, in the stiff-clay experiment where free water was present, the free water moved upward along the face of the pile with each cycle of loading. The presence of soft clay and sand to a depth of 3.05 m (120 in.) above the stiff clay is believed to suppress the hydraulic action of free water even though the sand did not serve to close the potential gaps in the stiff clay.
The equations used to compute lateral load transfer at failure are the ultimate values.
Soft Clay static loading
cb Soft Clay cyclic loading
Jc
Stiff Clay with Free Water Static
pct = 2cab + γ′bx + 2.83 cax... (3-26) pcd = 11cb ... (3-27)
(
1.225 −0.75 −0.411)
= pc As As
p ... (3-34)
Stiff Clay with Free Water Cyclic
p
Stiff Clay without Free Water static and cyclic loading cb
b