Topic-2 Shallow Foundations [Compatibility Mode]

8/10/2019 Topic-2 Shallow Foundations [Compatibility Mode]

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29/09/2011

1

D M B

C E, N M C

Topic 2

Module: H23GGE

A

. A ,

, .

F :

D ()

D

E

N:

C =

AFE

ALLABLE

1.

2. A

F : D

:

*

,

(, , ), , ,

, .

* G

,

, , /

. , ,

. I ,

.


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2

The objectives of this study are:

Calculate the safe bearing capacity of soils

Estimate the settlement of shallow foundations

Estimates the size of shallow foundations to satisfybearing capacity and settlement criteria

Forces on a foundation

B

F

L

Foundation: Structure transmits loads to the underlying

ground (soil).

Footing: Slab element that transmit load from superstructureto ground.

Embedment depth, Df : The depth below ground surface

where the base of the footing rests.

Bearing pressure: The normal stress imposed by

the footing on the supporting ground.

Load

Df

D

B

5.2BD 5.2>BD

B:

D:

F

I

C

,

,


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3

F

:

I

,

:

,

C:

A

B C (B8004:1986)D

.

A .

E 7D ( )

.

The solution is usually based on:

a) General bearing capacity failure:

b) Strip footing: width, B, length L =,

and founded at depth, D

c) Soil: homogeneous & isotropic, unit

weight,

d) Soil Strength: parameters c &

I

M C :

tan+= c

Total stress

D

F

.

B

,

0 0000.

() , 2D :

0000

( )


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4

:

,

Where and are the bearing capacity factors can beobtained from data sheet

qcult NDNcNBq .....2

10 ++=

cNN , qN

N:

N:

N:

B

They are derived from plasticity theory and failure theory.

M

0

5

10

15

20

25

30

35

40

45

5.7

7.3

9.6

12.9

17.7

25.1

37.2

57.8

95.7

172

1.0

1.6

2.7

4.4

7.4

12.7

22.5

41.4

81.3

173

0.0

0.5

1.2

2.5

5.0

9.7

19.7

42.4

100

298

5.14

6.49

8.34

11.0

14.8

20.1

30.1

46.4

75.3

134

1.0

1.6

2.5

3.9

6.4

10.7

18.4

33.5

64.1

135

0.0

0.1

0.4

1.2

2.9

6.8

15.1

34.4

79.4

201

0.0

0.1

0.4

1.1

2.9

6.8

15.7

37.6

93.6

262.3

0.0

0.4

1.2

2.6

5.4

12.5

22.4

48.1

109.3

271.3

Nc Nq N Nc Nq N N N

Terzaghis (1943)

Expression

Hansen,Meyerhoff,

and Vesics

Expressions

Hansen

(1970)

Meyerhoff

(1951,1963)

Vesic

(1973,1975)

B

1. :

( )

MC:

H :

0=== uucc

uc=

Footingqult 45+/2

45-/2

0).2( pcq uult ++=

0=N 1=qN

F

N = 5.14


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5

:

/

cN

uultc cpqN /)( 0=

cN

Dp .00 =

1. :

1. :

BL

( B


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Effective unit weight =

Ground WT

Unit weight = d

D

B

E

0 <

F , + ( ) ,

.


Ground WT

Unit weight = d

D

B

E

+

. I ,= . , :

( )( )BDd+=


Ground WT

Unit weight = d

D

B

E

> +

F > + ,= . . .

F ,

. ,

:

( BB):

C ( B):

:

++= N.B..4.0N.pN.c.2.1q q0cult

++= N.B..3.0N.pN.c.2.1q q0cult

N:

( B M. D),

F G

NBLBNpNcq

qcult ..)./2.01(...2.1

0 ++=


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, ..E A (EA) :

, ,

qqqqccccult disNpdisNcdisNBq .............2

10++=

F

= (1 + B.N

/L.N

)

=(1 + B./L)

= (1 0.4B/L)

D B (1970)

:

L Length of foundation > B its width

D

D

= 1 + 0.41(D/B)

=1 + 2(1 )2.1(D/B)

= 1

F

D

= (1 + 0.4D/B)

=(1 + 2(1 )2D/B)

= 1

H (1970)

1/ BD

1/ >BD

D

B

Note: (1) L Length of foundation > B its width(2) tan-1(Df/B) is in radians

I

F

I

(

)

= =(1 /90)2

F =(1 / )2 <

= 0

M (1963)

H &

M (1981)

: Angle of load with vertical

E

F (..

)

I

B L.

I

M (1963)

B L :

H:

B L

B B

L

B

eB

eL

=

=

L

B

eLL

eBB

2

2


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:

B

qcult NDNcNBq .....2

10 ++=

eBB 2=

E

:1 C ,

.

2 M .

E

F .

.

,

:

I , 2. 3.0.

.

00 p

F

pqq ulta +

=

0

0 2/.)1(

pF

BNpNq

q

a +

+=

E 1A 2.25 1.5 ,

=0 =38. D

() I ,

() I .

=18 N/3 : =20 N/

3.


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1

For a square footing, the ultimate bearing capacity:

qult = 0.4BN + DNq

For =38 Nq=49 , N=67

qult = 0.4x18x2.25x67 + 18x1.5x49qult = 2408 kN/m

2

When the water table is at the surface:

qult = 0.4(sat w)BN + (sat w)DNqqult = 0.4x10.2x2.25x67 + 10.2x1.5x49qult = 1365 kN/m

2

E 2D 4.5 2.25 3.5

3

. 20 N/3

C=135 N/2 =0.

2

qult = Cu.Nc.(0.84+0.16xB/L)+D=135x8.1(0.84+0.16x2.25/4.5) + 20x3.5=1076 kN/m2

qnet = (qult D)/FS=(1076 20x3.5)/3=335 kN/m2

qall = qnet + D =335 +20x3.5=405 kN/m2

Allowable load = 405x2.25x4.5=4101 kN

E 3A 1.2 3500 N.

20 . D

3.

20

3500 kN

1.2m

B

c = 0 = 40= 18 kN/m3


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10

3

C = 0 therefore C.Nc=0

qult =0.5BN.s.d.i + DNq.Sq.dq.iq

For 40

N=95 Nq= 65

Correction factors:

Shape:

Depth:

20

3500 kN

1.2m

B

c = 0 = 40= 18 kN/m3

s= (1 0.4B/L)=0.6sq =(1 + B.tan/L)=1.84

d = 1dq =(1 + 2.tan(1 sin)

2Df/B)=1+0.26/B

3

20

3500 kN

1.2m

B

c = 0 = 40= 18 kN/m3

Correction factors:

Inclination:

Then

qult

=0.5x18xBx95x0.6x1x0.25 + 18x1.2x65x1.84x(1+0.26/B)x0.605

qult =128.25B + 406.4/B + 1562.9

i =(1 /90)2=0.605

i=(1 / )2=0.25

3

So qnet= qult Dqnet=128.25B + 406.4/B + 1562.9 18x1.2

= 128.25B + 406.4/B + 1541.3and

qall =qnet/FS +D= (128.25B + 406.4/B + 1541.3)/3+21.6

qall = 42.75B + 135.47/B + 535.37

Sinceqall = q/B

2 = 3500/B2

3500/B2 = 42.75B + 135.47/B + 535.3742.75B3 + 535.37B2 + 135.47B 3500 = 0 B=2.25 so D/B


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/C/

0. pNcq cult +=

0)/16.84.0.( pNLBcq cult ++=

)/( BDfNc =

C

Strip footing (dry soil):

Strip footing (soil saturated)

qcult NpNcNBq .....5.0 0++=


With:


eBB 2=

:

D :

0pF

pqq

oultall +

=

0

.p

F

Ncq call +=

0

0 ..5.0)1(.p

F

BNpNNcq

qc

all +++

=

FAC AFE,

0

0

pq

pq

pressurefoundationnet

capacitybearingultimatenetF ult

==

(): (.. ).

N ():

.

A ():

.

: A

.

: A

, ,

.

oultnet pqq =

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Topic-2 Shallow Foundations [Compatibility Mode]