settlement of shallow foundation final

43
1 Shallow Foundation Settlement

Upload: others

Post on 18-Dec-2021

13 views

Category:

Documents


0 download

TRANSCRIPT

Page 1: Settlement of Shallow Foundation Final

1

Shallow Foundation Settlement���� ��� � ��

Page 2: Settlement of Shallow Foundation Final

2

Settlement

Immediate Settlement: Occurs immediately after the construction. This is computed using elasticity theory (Important for Granular soil)

Primary Consolidation: Due to gradual dissipation of pore pressure induced by external loading and consequently expulsion of water from the soil mass, hence volume change. (Important for Inorganic clays)

Secondary Consolidation: Occurs at constant effective stress with volume change due to rearrangement of particles. (Important for Organic soils)________________________________________________________________For any of the above mentioned settlement calculations, we first need vertical stress increase in soil mass due to net load applied on the foundation

Page 3: Settlement of Shallow Foundation Final

3

Page 4: Settlement of Shallow Foundation Final

4

Page 5: Settlement of Shallow Foundation Final

5

Page 6: Settlement of Shallow Foundation Final

6

Page 7: Settlement of Shallow Foundation Final
Page 8: Settlement of Shallow Foundation Final

8

Stress Due to a Concentrated Load

Page 9: Settlement of Shallow Foundation Final

9

Stress due to a Circularly Loaded Area

Page 10: Settlement of Shallow Foundation Final

10

Distribution of Stress(from a vertical line load)

Page 11: Settlement of Shallow Foundation Final

11

Stress below a Rectangular Area

Page 12: Settlement of Shallow Foundation Final

12

Stress Influence Chart

Page 13: Settlement of Shallow Foundation Final

13

2:1 method

Approximate methods

Page 14: Settlement of Shallow Foundation Final

14

Page 15: Settlement of Shallow Foundation Final

15

Page 16: Settlement of Shallow Foundation Final
Page 17: Settlement of Shallow Foundation Final

17

Page 18: Settlement of Shallow Foundation Final

18

Elastic Settlement

Page 19: Settlement of Shallow Foundation Final

19

Elastic settlement Based on theTheory of Elasticity

Page 20: Settlement of Shallow Foundation Final

20

Elastic Settlement of Rectangular footings

�� �� ��� � � �� ��

��� ������ ����

�������� ���� ��� �

�������� !"�# ��� �

$%" �&���

'() �&���

Es = ��� !&*�&�+� ,-��

Page 21: Settlement of Shallow Foundation Final

21

Page 22: Settlement of Shallow Foundation Final

IMMEDIATE SETTLEMENT

General Equation (Harr, 1966)• Flexibel Foundation

– At the corner of foundation

– At the center of foundation

– Average

• Rigid Foundation

(((( ))))2

1Eq.B

S 2s

s

oe

ααααµµµµ−−−−====

Es = Modulus of elasticity of soil

B = Foundation width L = Foundation length

(((( ))))ααααµµµµ−−−−==== 2s

s

oe 1

Eq.B

S

(((( )))) r2s

s

oe 1

Eq.B

S ααααµµµµ−−−−====

−−−−++++++++++++++++

−−−−++++++++++++

ππππ====αααα

1m1

1m1ln.m

mm1

mm1ln

12

2

2

2

(((( )))) av2s

s

oe 1

Eq.B

S ααααµµµµ−−−−====

B

Lm =; ; H = ∞∞∞∞

Page 23: Settlement of Shallow Foundation Final

23

Page 24: Settlement of Shallow Foundation Final

IMMEDIATE SETTLEMENT

If Df = 0 and H < ∞, the elastic settlement of foundation can be determined from the following formula:

( ) ( ) ( )[ ]

( ) ( ) ( )[ ]22

122

22

12

2

2111.

2

2111

.

FFE

qBS

FF

E

qBS

sssss

oe

ssss

s

oe

µµµµ

µµµµ

−−+−−=

−−+−−= (corner of rigid foundation)

(corner of flexible foundation)

The variations of F1 and F2 with H/B are given in the graphs of next slide

Page 25: Settlement of Shallow Foundation Final

IMMEDIATE SETTLEMENT

Page 26: Settlement of Shallow Foundation Final

IMMEDIATE SETTLEMENT

Page 27: Settlement of Shallow Foundation Final

EXAMPLE

Problem:A foundation is 1 m x 2 m in plan and carries a net load per unit area, qo = 150 kN/m2. Given, for the soil, Es = 10,000 kN/m2, µs0.3. Assuming the foundation to be flexible, estimate the elastic settlement at the center of the foundation for the following conditions:

a. Df = 0 and H = ∞b. Df = 0 and H = 5 m

Page 28: Settlement of Shallow Foundation Final

EXAMPLE

Solution:Part a.

Part b.

(((( ))))ααααµµµµ−−−−==== 2s

s

oe 1

Eq.B

S

For L/B = 2/1 = 2 � α ≈ 1.53, so

( ) mmmSe 9.200209.0)53.1(3.01000,10

)150)(1( 2 ==−=

( ) ( ) ( )[ ]22

122 2111

'.FF

E

qBS ssss

s

oe µµµµ −−+−−=

For L’/B’ = 2, and H/B’ = 10 � F1 ≈ 0.638 and F2 ≈ 0.033, so

( ) ( ) ( )[ ] mmmxSe 3.160163.04)033.0()3.0(23.01)638.0(3.013.01000,10

)150)(5.0( 222 ==−−+−−=

Page 29: Settlement of Shallow Foundation Final

29

Page 30: Settlement of Shallow Foundation Final

30

Elastic Settlement on Saturated Clay

Page 31: Settlement of Shallow Foundation Final

31

Page 32: Settlement of Shallow Foundation Final

32

Elastic Settlement Using the Strain Influence Factor: [Schmertman & Hartman Method (1978)]

Page 33: Settlement of Shallow Foundation Final

33

Page 34: Settlement of Shallow Foundation Final

34

Elastic Settlement Using the Strain Influence Factor: [Schmertman Method (1978)]

Page 35: Settlement of Shallow Foundation Final

35

Page 36: Settlement of Shallow Foundation Final

36

Procedure for Schmertman Method (1978)

Page 37: Settlement of Shallow Foundation Final

37

Procedure for Schmertman Method (1978)

Page 38: Settlement of Shallow Foundation Final

38

Procedure for Schmertman Method (1978)

Page 39: Settlement of Shallow Foundation Final

39

Procedure for Schmertman Method (1978)

Page 40: Settlement of Shallow Foundation Final

40

Notes on Schmertmann Method

Page 41: Settlement of Shallow Foundation Final

EXAMPLE

A shallow foundation 3 m x 3 m (as shown in the following drawing). The subgrade is sandy soil with Young modulus varies based on N-SPT value (use the following correlation: Es = 766N)

Determine the settlement occur in 5 years (use strain influence method)

Page 42: Settlement of Shallow Foundation Final

EXAMPLE

Page 43: Settlement of Shallow Foundation Final

EXAMPLE

Depth(m)

∆z(m)

Es

(kN/m2)Iz

(average) (m3/kN)

0.0 – 1.0 1.0 8000 0.233 0.291 x 10-4

1.0 – 1.5 0.5 10000 0.433 0.217 x 10-4

1.5 – 4.0 2.5 10000 0.361 0.903 x 10-4

4.0 – 6.0 2.0 16000 0.111 0.139 x 10-4

Σ 1.55 x 10-4

zEI

s

z ∆∆∆∆

( ) 9.05.18.17160

5.18.175.015.011 =

−−=

−−=

x

x

qq

qC 34.1

1.0

5log.2.01

1.0log.2.012 =

+=

+= tC

( )

mmS

xxS

zE

IqqCCS

e

e

B

s

ze

8.24

)1055.1)(5.18.17160)(34.1)(9.0(

...

4

2

021

=−=

∆−=