Part4-Settlement2.pdf

Part 4

Settlement of Shallow

Foundations

1

2

Settlement of Shallow Foundations

It is practically impossible to prevent settlement of

foundations. At least, elastic settlement will occur.

The task of the geotechnical engineer is to prevent

the foundation system from reaching a serviceability

limit state rather than by ultimate limit state.

However, both limit states must be satisfied.

Hooke’s Law: Elasticity

The stress s is directly proportional to the
strain e, and

s = E e

The coefficient E is called the modulus of elasticity

of the material involved, or also Young’s modulus

The ratio of the radial (or lateral) strain

to the vertical strain is called Poisson’s

ratio, n, defined as

l

a

Lateral strain

Axial strain

e
n

e
  

4

The distribution of stresses within a soil from applied surface loads or stresses is determined by assuming that the soil is a semi-

infinite, homogeneous, linear, isotropic, elastic material. A semi-infinite mass is bounded on one side and extends infinitely in

all other directions; this is also called an “elastic half space.” For soils, the horizontal surface is the bounding side. Because of

the assumption of a linear elastic soil mass, we can use the principle of superposition. That is, the stress increase at a given point

in a soil mass in a certain direction from different loads can be added together.

Stresses From Surface Loads

Boussinesq (1885)

5

Under Corner of Rectangular Foundation

∆𝜎𝑧 = 𝑞𝐼𝜎

A rectangular concrete slab, 3 m x 4.5 m, rests on the surface of a soil

mass. The load on the slab is 2025 kN. Determine the vertical stress

increase at a depth of 3 m (a) under the center of the slab, point A; (b)

under point B; and (c) at a distance of 1.5 m from a corner, point C.

Example 4-1

6

Find the vertical stress increase for one rectangle
of size 1.5mx2.25m and multiply the results by 4

find the vertical stress increase for one rectangle of
size 2.25mx3m and multiply the result by 2.

find the vertical stress increase for large rectangle
of size 4.5mx4.5m and deduct from it the stress
increase under the small rectangle of size 1.5mx4.5.

This will be left for the student to practice using the
chart

Example 4-1

7

Two rectangular concrete slabs, 3 m x 4.5 m each, rest on the surface of a

soil mass as shown. The load on each slab is 2025 kN. Determine the

vertical stress increase at a depth of 3 m (a) under point A; (b) and under

point B

Example 4-2
Set up for stress program

8

9

Under Circular Foundation

∆𝜎𝑧 = 𝑞𝐼𝜎

10

Find the vertical stress increase at A, B, and C

Point A (r/a = 0/6 = 0, z/a = 6/6 = 1 -> Is = 0.7 -> Dsz = 0.7(450) = 315 kPa

Point B (r/a = 3/6 = 0.5, z/a = 6/6 = 1 -> Is = 0.6 -> Dsz = 0.6(450) = 270 kPa

Point C (r/a = 6/6 = 1, z/a = 6/6 = 1 -> Is = 0.35 -> Dsz = 0.35(450) = 158 kPa

Example 4-3

solution

Under Continuous Foundation

11

Take B = 2 m, qs = 200 kN/m
2, find the vertical stress increase when

x = 1 m and 2.5 m with depth z = 1 m

When x = 1m, z = 1m
b = – tan-1(1/1) = – 0.785 , a = /2 = 1.571
Dsz = 200/3.14[1.571+sin(1.571)cos(1.571+2(- 0.785))] = 163.7 kN/m

2

When x = 2.5m, z = 1m
b = tan-1(0.5/1) = 0.464, a = tan-1(2.5/1) – 0.464 = 0.727
Dsz = 200/3.14[0.727+sin(0.727)cos(0.727+2(0.464)] = 42.7 kN/m

2

Example 4-4

solution

12

Total Settlement

The settlement of a foundation can have three components:

(a) elastic settlement (Se)

(b) primary consolidation settlement (Sp)

(c) secondary consolidation settlement (Ss)

St = Se + Sp + Ss

13

Differential Settlement

The differential settlement, D, is the difference in total settlement between two foundations or between
two points in a single foundation. Differential settlements are generally more troublesome than total

settlements because they distort the structure causing damage to the structure and could threaten the

integrity of the structure

D = qd S

where
qd is allowable angular distortion
S is column spacing

14

Settlement Based on the Theory of Elasticity

15

Elastic Settlement in Granular soils

Bowles 1987

Below center of flexible foundation

16

17

18

19

A rigid shallow foundation 1 m  2 m is as shown. Calculate the
elastic settlement at the center of the foundation.

Example 4-5

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Elastic Settlement of Shallow Foundation on Saturated Clay (ms = 0.5)

21

Example 4-6

22

Settlement in Sand Based on SPT test

N60 is the average SPT blow counts 2B below the foundation level.

23

Example 4-7

24

Schmertmann et al. (1978)

Settlement in Sand Based on CPT test`

25

Example 4-8

26

27

Settlement based on stiffness Variation with Stress Level

Berardi and Lancellotta (1991)

after Lancellotta, 2009

B

Square foundation, H = B

Strip foundation (L/B = 10), H = 2B

Rectangular foundation, 𝐻 = 𝐵 1 + 𝑙𝑜𝑔
𝐿

𝐵

𝜎𝑜
′ = effective stress at middle of influnce depth

∆𝜎 = stress increase below the center of the foundation at middle of influnce depth

𝑝𝑎 = atmospheric pressure ~100 kPa

28

ത𝜎′𝑜 = 1.875 17 = 31.9 Τ𝑘𝑁 𝑚
3

𝐶𝑁 =
95.8

31.9
= 1.73 ഥ𝑁1,60 = 1.73 10 = 17

𝐷𝑟 =
17

60
× 100 = 53%

𝐾𝐸,0.1% = 9.1 53 + 92.5 = 575

∆𝜎 = 4 120 0.175 = 84 𝑘𝑁/𝑚2

𝐸𝑠,0.1% = 575 100
31.9 + 0.5(84)

100
= 49400 𝑘𝑁/𝑚2

( ൗ𝑆𝑒 1.75)
0.3 =

125 120 1−0.32 0.609

49400
→ 𝑆𝑒 = 0.005 𝑜𝑟 5 𝑚𝑚

Example 4-9

solution

Liao and Whitman (1985)

Skempton (1986)

29

Consolidation is a time dependent
deformation or volume change. The volume

change take place as a result of expulsion of

water from the void spaces due to excessive

pressure.

Consolidation Analogy

is the primary consolidation settlement

30

Present effective stress ( 𝝈𝒐
′ ): The calculated effective

stress (pressure) at the present soil condition.

Preconsolidation stress (𝝈𝒄
′ ): Maximum effective past stress

(pressure) that the soil was subjected to.

Compression index ( Cc ): Slope of the in-situ compression

line in e vs. log P plot.

Swelling index ( Cs ): Slope of the rebound curve in the e

vs. log p plot.

Over Consolidation Ratio:

OCR ≤ 1 for normally consolidated soil

OCR > 1 for over consolidated soil

Definitions

𝑂𝐶𝑅 =
𝜎𝑐
′

𝜎𝑜
′

31

For normally consolidated soil

For over consolidated soil when ( sʹo + Dsʹ ) ≤ sʹc

For over consolidated soil ( sʹo + Dsʹ )> sʹc

Primary Consolidation Settlement

32

Calculate the settlement of the 3-m-thick clay layer as shown that will result from the load carried by a 1.5 m square footing. The clay
is normally consolidated.

𝑚 =
𝐵

𝑥
=
0.75

6
= 0.125, 𝑛 =

𝐿

𝑧
=
0.75

6
= 0.125 → 𝐼~0.008

∆𝜎 = 4
890

1.5 × 1.5
0.008 = 12.7 Τ𝑘𝑁 𝑚2

𝑆𝑝 =
3

1 + 1
0.27 log

85.58 + 12.7

85.58
= 0.024 𝑚 𝑜𝑟 24 𝑚𝑚

Example 4-10

33

The ultimate load-bearing capacity of a foundation, as well as the allowable bearing capacity based on
tolerable settlement considerations, can be effectively determined from the field load test, generally referred
to as the plate load test.

The plates that are used for tests in the field
are usually made of steel and are 25 mm thick
and 150 mm to 762 mm in diameter.
Occasionally, square plates that are 305 mm ×

305 mm are
also used.

Field Load Test

34

( ) ( )

( ) ( )

For tests in clay,

For tests in sandy soil

s

u F u P

F
u F u P

p

Ultimate bearing capacity

q q

B
q q

B





2

(for clayey soil)

2
(for sandy soil)

F
F P

P

F
F P

F P

Settlement considerat

B
S S

B

B
S S

B B

ion



 
  

 

Plate Load Test

35

Example 4-11
A rigid foundation 1.5m x 1.5m is to constructed at depth 1.5 below

ground subjected to 675 kN gross load on a sandy deposit with a

borehole log as shown. Estimate the elastic settlement (Use Bowles)

Assume mavg = 0.3

𝑞𝑜 =
𝑄

𝐴
=

675

1.5 × 1.5
= 300 𝑘𝑁/𝑚2

solution

We will use

Depth (m) N60 Es DH EsDH

0.5-1 8

1-1.5 9 Foundation level

1.5-2 12 13500 0.5 6750

2-2.5 11 13000 0.5 6500

2.5-3 10 12500 0.5 6250

3-3.5 13 14000 0.5 7000

3.5-4 15 15000 0.5 7500

4-4.5 22 18500 0.5 9250

4.5-5 27 21000 0.5 10500

5-6 33 24000 1 24000

6-7 45 30000 1 30000

7-8 50 32500 1 32500

8-9 50 32500 1 32500

9-10 50

Es(wt. avg) = 23033 kN/m
2

5B =7. 5m

Inputs
B 1.5 m

L 1.5 m

Df 1.5 m

qo 300 N/m
2

ms 0.3

Es 23033 kN/m
2

H 7.5 m

Calculated values

m’ 1

n’ 10

F1 0.50

F2 0.02

Is 0.51

If 0.65

Se (flexible) = 11.7 mm

Se (Rigid = 0.93 Se(flex.)) 10.9 mm

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Example 4-12

A rigid foundation 1.5m x 1.5m is to constructed at depth 1.5 below ground subjected to 675 kN gross load on

a sand soil with a borehole log as shown. Estimate the immediate settlement (use Meyerhof/Bowles)

solution

Depth (m) N60

0.5-1 8

1-1.5 9 Foundation level

1.5-2 12

2-2.5 11

2.5-3 10

3-3.5 13

3.5-4 15

4-4.5 22

4.5-5 27 Navg = 14

5-6 33

6-7 45

7-8 50

8-9 50

9-10 50

2B =3 m

𝑞𝑂 =
675

(1.5)(1.5)
= 300 Τ𝑘𝑁 𝑚2

𝑞𝑛𝑒𝑡 = 𝑞𝑂 − 𝑞 = 300 − 1.5 17 = 275 Τ𝑘𝑁 𝑚
2

𝛾𝑑 = 17 Τ𝑘𝑁 𝑚
3

input data

B 1.5 m
Df 1.5 m
qnet 275 kN/m2 (applied)

calculated values

Nave 14

Fd 1.33

Se 20.8 mm
qnet 331 kN/m

2 (for 25 mm settlement)

37

A rigid foundation 1.5m x 1.5m is to constructed at depth 1.5 below ground subjected to 675 kN gross load on a sand soil

with a borehole log as shown. Estimate the immediate settlement (Use Schmertmann)

solution

Depth (m) N60 Es

0.5-1 8

1-1.5 9 Foundation level

1.5-2 12 13500

2-2.5 11 13000

2.5-3 10 12500

3-3.5 13 14000

3.5-4 15 15000

4-4.5 22 18500

4.5-5 27 21000

5-6 33 24000

6-7 45 30000

7-8 50 32500

8-9 50 32500

9-10 50

2B =3 m

Example 4-13

Input Calculated values

B 1.5 m q 25.5 kN/m
2

L 1.5 m C 1 0.954

Df 1.5 m C 2 1.4

g 17 kN/m
3

I o 0.100 Final Results

300 kN/m
2

Z 1 0.750 S e = 32.4 mm

Time 10 years Z 2 3

Dw 5.3 m q z1 38.3 kN/m3

gsat 20 kN/m
3

I z1 0.768

I z2 0

Auto E z I z Δz I z
Depth

(m)
kN/m

2 (m) avg.

0 0.100 0

0.5 13500 0.545 0.5 0.323 1.195E-05

0.75 13000 0.768 0.25 0.657 1.263E-05

1 13000 0.683 0.25 0.725 1.395E-05

1.5 12500 0.512 0.5 0.597 2.389E-05

2 14000 0.341 0.5 0.427 1.524E-05

2.5 15000 0.171 0.5 0.256 8.532E-06

3 18500 0.000 0.5 0.085 2.306E-06

Sum 8.849E-05

0.00

0.50

1.00

1.50

2.00

2.50

3.00

3.50

0.00 0.25 0.50 0.75 1.00

𝑞ത

Iz Variation with Z