Foundation Settlement and Soil Compression - Elements of Soil Mechanics - page 342

Clay: in Fig. 11.2, H 1 = 6.1 m and H 2 = 36.6 m.

L
B

27 44
9 15

.
.

H
B

36 6
9 15

.

For H : = = 3 0 . ; = = 4 0 . 2

.

Hence I p = 0.475.

L
B

H
B

6 1
9 15

.

For H : = 3 0 . ; = = 0 67 . 1

.

Hence I p = 0.18.
Settlement under centre of foundation ( note as heave effects will be allowed for, use
gross contact pressure. If heave is not allowed for net foundation pressure should be
used).

pB
E

( 1

2

) 4 I Rigidity factor

ρ = − ν ×

i p

350
16100

= × 9 15 0 75 4 0 475 . × . × ( . − 0 18 . ) × 0 8 .

= 0 141 . m = 141 mm

Heave effects: relief of pressure due to sand excavation = 29 kPa

29
64 400

⇒ Heave = × 9 15 0 75 4 0 475 0 18 . × . × ( . − . ) × 0 8 .

= 0 0029 . m = 3 mm

Hence ρ i in the clay = 138 mm.
As can be seen from this example the effects of heave are usually only significant
when a great depth of material is excavated.

Consolidation settlement

The clay layer has been divided into five layers of thickness, H, equal to 6.1 m.

m v Δ σ z m v Δ σ z H

0.000145 261 0.231

0.000114 195 0.136

0.0000413 145 0.081

0.000073 110 0.049

0.0000456 86 0.024

0.521 m = 521 mm

This value of settlement can be reduced by the factor

µ = + − A ( 1 A ) α

An approximate value for α can be obtained from Table 11.4.
Hence:

α = 0.26
μ = 0.75 + 0.25 × 0.26 = 0.82
ρ c = 521 × 0.82 = 448 mm
Total settlement = 62 + 138 + 448 = 648 mm