The role of a shallow foundation is to spread the column load (from a super-structure) on a wider area in a uniform manner. Thus, instead of applying the
concentrated column load directly to the “weak” soil, the shallow foundation will apply a much gentler uniform pressure to the soil.
Consider a uniformly loaded rectangular area with length L and width B as shown in Figure 3.28a. Note that L is always greater than B in a rectangle, and L is equal toB in a square. The uniform load q is expressed in force per unit area (pressure units). Equation (3.13) can be used to calculate the increase of vertical stress under the corner of a loaded rectangle as illustrated in Figure 3.28a.
σz= q 4π
2mn√
m2+ n2+ 1 m2+ n2+ m2n2+ 1
m2+ n2+ 2
m2+ n2+ 1+ tan−1 2mn√
m2+ n2+ 1 m2+ n2− m2n2+ 1
(3.13) wherem= B/z and n = L/z.
X
Y
q (kN/m2)
∆σz
Z z B L
(a)
A
B L
∆σz
Z z
(b)
FIGURE 3.28 Stresses caused by a uniformly loaded rectangular area.
In most practical cases an increase in vertical stress under the center (point A in Figure 3.28b) of a uniformly loaded rectangular area is required. Remember that (3.13) is only for the stress increase below the corner of a rectangle. To use (3.13) to calculate the stress increase under the center of a loaded rectangle, we can divide the rectangle into four identical “small” rectangles each of which has point A as its corner. The increase in the vertical stress under the corner A of each small rectangle can be calculated using (3.13) and assuming thatL and B are the length and width of a small rectangle (see Figure 3.28b). The total increase in vertical stress is then calculated by adding the four stress increases of the four identical small rectangles.
Example 3.10 (a) Using (3.13), calculate the increase in vertical stress under the center of a 4 m× 2 m rectangle that is loaded uniformly with q = 10 kPa. Assume that the soil layer underlying the loaded area is very thick and linear elastic with E= 1 × 107 kPa and ν = 0.3. (b) Calculate the increase in vertical stress under the center using the finite element method. Compare the two answers.
SOLUTION: (a) Since we need to calculate the stress increase under the center using (3.13), we will have to divide the 4 m× 2 m rectangle into four 2 m × 1 m small rectangles. Then for the small rectangles we havem= B/z = 1/z and n = L/z= 2/z. Let’s vary z from 0.01 to 20 m. For each z we calculate m and n and substitute those into (3.13) to get the stress increase σz under the corner of the small rectangles. To calculate the stress increase under the center of the 4 m× 2 m rectangle, multiply σz by 4 to account for the four identical small rectangles. The stress increase with depth calculated is listed in Table 3.2 and plotted in Figure 3.29. Note that near the surface (at z= 0.01 m), σz is equal to q(= 10 kPa). The figure also shows that the stress declines very rapidly. At z= 4B = 8 m, σzis only 0.56 kPa (i.e., less than 6% of the applied pressureq).
(b) Finite element solution (filename: Chapter3 Example10.cae) This is a three-dimensional geometry that needs to be treated as such in a finite element analysis. The three-dimensional model analyzed is shown in Figure 3.30. The soil layer is 50 m deep and 100 m× 100 m in plan. The loaded area is 4 m × 2 m.
The model considers only one-fourth of the soil layer and the loaded area, taking
TABLE 3.2
z (m) σz (kPa) z (m) σz(kPa)
0.01 10.0 6 0.952
1 7.996 8 0.56
2 4.808 10 0.368
3 2.928 15 0.168
4 1.9 20 0.096
5 1.312
0 1 2 3 4 5 6 7 8 9 10 Vertical Stress (kPa)
0 2 4 6 8 10 12 14 16 18 20
Depth (m)
FEM, Center Boussinesq, Center
X
Y q (kN/m2)
Z L B
FIGURE 3.29 Stresses caused by a uniformly loaded rectangular area: FEM versus ana-lytical solution.
advantage of symmetry as indicated in Figure 3.30a. The loaded area simulates a foundation with perfect contact with the soil. Reduced-integration eight-node linear brick elements are used for the soil layer. The base of the soil layer is fixed in all directions. All vertical boundaries are fixed in the horizontal direction but free in the vertical direction. The finite element mesh used in the analysis is shown in Figure 3.30b. It is noted that the mesh is finer in the vicinity of the foundation since that zone is a zone of stress concentration. No mesh convergence studies have been performed. However, the dimensions of the soil layer are chosen such that the boundary effect on foundation behavior is minimized. The increase in vertical stress under the center of the rectangular area is plotted as a function of depth in Figure 3.29. The figure shows excellent agreement between the stresses calculated using the analytical solution (3.13) and those calculated using the finite element method.
Note that the finite element analysis presented above provides stresses, strains, and displacements at all nodal points within the loaded semi-infinite soil mass —the finite element results are not limited to the stresses under the center of the loaded rectangle. Also note that this analysis assumes linear elastic soil behavior that is not suited for failure analysis. The present finite element analysis was carried out only to calculate the stress increase within the soil. In Chapter 6 we use elastoplastic soil models to predict the bearing capacity of various types of shallow foundations in a realistic manner.
Plane of Symmetry Plane of Symmetry
1/4 of the 4 m × 2 m Rectangular Foundation
50 m 50 m
50 m
(b)
Symmetry Axis 4 m × 2 m Loaded Area
100 m 100 m
Quarter Analyzed Symmetry Axis
50 m
(a)
FIGURE 3.30 Finite element discretization of the loaded rectangle problem.
PROBLEMS
3.1 Plot the vertical stress, pore water pressure, and effective-stress distributions for the soil strata shown in Figure 3.31. The water table is located 5 m below the ground surface.
3.2 By the end of the summer, the water table in Problem 3.1 dropped 2 m.
Plot the new distribution of the vertical stress, pore water pressure, and effective stress for the soil strata knowing that the bulk unit weight of the clay layer (above the water table) is 18 kN/m3 and its saturated unit weight is 19.7 kN/m3, as shown in Figure 3.31.
3.3 Refer to Figure 3.8. In the winter, the water table is located at a distance H1 = 10 m below the ground surface, as shown in Figure 3.8a. In the spring, the water table rises a distanceh= 5 m above the winter level, due to snow thaw (Figure 3.8b). Calculate the change in effective stress at the bottom of the soil layer from winter to spring. The soil has a dry unit weight of γd = 17 kN/m3 and a saturated unit weight ofγsat= 18.7 kN/m3.
3.4 A 3.5-m-thick silt layer underlain by a 3-m-thick clay layer is shown in Figure 3.11a (refer to Example 3.3). Calculate the total stress, pore water pressure, and effective stress at points A, B, C, D, and E. The water table is located 2.5 m below the ground surface. Ignore the capillary rise in the silt layer and assume that the soil above the water table is dry. Compare the effective stresses at points A, B, C, D, and E with those calculated in Example 3.3.
3.5 Two vertical point loads of 100 kN each are applied at the surface of a semi-infinite soil mass. The horizontal distance between the two point loads is 0.5 m. Using a Boussinesq solution, calculate the increase in vertical stress directly under one of the applied loads atz= 1 m. If you were to solve this problem using the finite element method, what type of geometry would you use: axisymmetric, plane strain, or three-dimensional?
Water Table 3 m
2 m
3 m Sand
Clay
Sand γsat= 20 kN/m3 γsat= 19.7 kN/m3
3 m γ = 18 kN/m3
γ = 19 kN/m3
FIGURE 3.31
q1= 100 kN/m q2= 100 kN/m
1.5 m B 1 m
A
E D C 0.75 m
0.75 m
0.75 m
0.75 m
FIGURE 3.32
3.6 Two parallel line loads, 100 kN/m each, are applied at the surface of the backfill soil behind a rigid basement wall as shown in Figure 3.32. This type of loading can be thought of as the load caused by the rails of a train while the train is standing still or moving at a constant speed. (a) Calculate the increase in vertical stress at points A, B, C, D, and E. (b) Repeat your solution using the finite element method and assuming that the soil is linear elastic with E= 1 × 107 kPa and ν = 0.3. What is the horizontal stress increase at points A, B, C, D, and E?
3.7 A pressure of 100 kPa is distributed uniformly on a circular area withR= 2 m. (a) Calculate the increase in vertical stress directly under the center of the applied load for z= 0 to 12 m. (b) Repeat your solution using the finite element method and assuming that the soil is linear elastic with E= 1× 107 kPa andν = 0.3.
3.8 Consider the three-layer system shown in Figure 3.33. A pressure of 100 kPa is distributed uniformly on a circular area withR= 2 m. Using the finite ele-ment method, calculate the increase in vertical stress directly under the center of the circular area forz= 0 to 12 m. Compare the stress increase at points A, B, and C with that obtained in Problem 3.7 for a single homogeneous soil layer.
3.9 A 2-m-wide strip load of 100 kN/m2 is applied at the surface of the backfill soil behind a rigid basement wall as shown in Figure 3.34. (a) Calculate the increase in vertical stress directly under the center of the applied load for z= 0 to 3 m, (b) Calculate the increase in vertical stress at points A, B, C, D, and E. (c) Repeat your solution using the finite element method and assuming that the soil is linear elastic withE= 1 × 107 kPa andν = 0.3.
3.10 Using (3.13), calculate the increase in vertical stress under the center of a 4 m× 4 m foundation that is loaded uniformly with q = 100 kPa. Assume
that the soil layer underlying the loaded area is very thick and linear elastic withE= 1 × 107 kPa andν = 0.3. Also calculate the increase in vertical stress under the center using the finite element method. Compare the two answers.
q= 100 kPa 2 m
B A
C 4 m
4 m 4 m
E= 5 × 105 kPa E= 107kPa
Sand Sand
Clay
E= 107kPa
FIGURE 3.33
q= 100 kPa
1 m 2 m B A
E D C 0.75 m
0.75 m
0.75 m
0.75 m
FIGURE 3.34
CONSOLIDATION
4.1 INTRODUCTION
When a saturated soil is loaded, its pore pressure increases. This pore pressure increase, called excess pore pressure,u, dissipates from the boundaries of the soil layer as time goes by, resulting in consolidation settlement. This process is time dependent and is a function of the permeability of the soil, the length of the drainage path (defined later), and the compressibility of the soil.
When saturated sands and gravels are loaded slowly, volume changes occur, resulting in excess pore pressures that dissipate rapidly due to high permeability.
This is called drained loading. On the other hand, when silts and clays are loaded, they generate excess pore pressures that remain entrapped inside the pores because these soils have very low permeabilities. This is called undrained loading. Conse-quently, the excess pore pressures generated by undrained loading dissipate slowly from the soil layer boundaries, causing consolidation settlement.
Consider a saturated clay layer sandwiched between two sand layers, with a groundwater table close to its top surface as shown in Figure 4.1. A uniform sur-charge pressure ofσ = 10 kPa is suddenly applied to the top surface. This loading is undrained and an excess pore pressure in the clay layer is generated instanta-neously:u= σ = 10 kPa. The water level in a piezometer (standpipe) positioned at point A will rise a distance h= u/γw= 10 kPa/9.81 kN/m3 ≈1 m above the level of the groundwater table. As time goes by, the excess pore pressure gradu-ally dissipates from the top and bottom boundaries of the clay layer, and the water level in the piezometer drops accordingly. This loss of water from the boundaries is associated with consolidation settlements that take place gradually until the excess pore pressure is totally dissipated and the water level in the piezometer drops to the level of the groundwater table.
124
Applied Soil Mechanics: with ABAQUS Applications. Sam Helwany
© 2007 John Wiley & Sons, Inc. ISBN: 978-0-471-79107-2
Water Level
FIGURE 4.1 One-dimensional consolidation.