Foundation Settlement and Soil Compression - Elements of Soil Mechanics - page 340

Alternative method for determining ρ c

In one-dimensional consolidation the volumetric strain must be equal to the axial strain,
i.e.

dH
H

ρ de

e

c

= =

H 1 + 1

hence:

de
e H

= 1 1
In the example:

ρ c

0 680 0 666
1 680

. − .

ρ c = × 20

.

= 0 008 834 20 . × = 0 167 . m

= 167 mm

Example 11.7: Total settlement using SPT results

The plan of a proposed raft foundation is shown in Fig. 11.12a . The uniform bearing
pressure from the foundation will be 350 kPa and a site investigation has shown that the
upper 7.62 m of the subsoil is a saturated coarse sand of unit weight 19.2 kN/m 3 with
groundwater level occurring at a depth of 3.05 m below the top of the sand. The result
from a standard penetration test taken at a depth of 4.57 m below the top of the sand
gave N = 20. Below the sand there is a 30.5 m thick layer of clay (A = 0.75, E = 16.1 MPa,
E swelling = 64.4 MPa). The clay rests on hard sandstone (Fig. 11.12b ).
Determine the total settlement under the centre of the foundation.

Solution

Using the De Beer and Martens' approach:
Vertical pressure increments

Gross foundation pressure = 350 kPa
Relief due to excavation of sand = 1.52 × 19.2 = 29 kPa
Net foundation pressure increase, Δ p = 350 − 29 = 321 kPa

The foundation is split into four rectangles, as shown in Fig. 11.12a , and Fig. 3.10 is
then used to determine values for I σ .

Depth below
foundation (m) B/z L/z I σ 4I σ Δ σ z (kPa)

3.05 3.0 9.0 0.247 0.988 317

9.15 1.0 3.0 0.203 0.812 261

15.25 0.6 1.8 0.152 0.608 195

21.35 0.43 1.29 0.113 0.452 145

27.45 0.33 1.00 0.086 0.344 110

33.55 0.27 0.82 0.067 0.268 86