Chapter 3 Mod

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Foundation EngineeringChapter (3) Bearing capacity of shallow foundation

Foundation EngineeringECIV 4352

Lecture (3) Bearing capacity of shallow foundation Ultimate bearing capacity (qu)

Shear failure Allowable bearing capacity (qall)

Shear failure qallshould be adequate to prevent excessive settlement and shear failure

Types of shear failure:1- General shear failure: For Dense sand.

2- Local shear failure: For Medium compaction soil.

3- Punching shear failure: For loose soil

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Calculation of ultimate bearing capacity of shallow foundations without eccentricity:

1- Terzaghis theory:Assumption for Terzaghis theory: The foundation is considered to be shallow if ( )BDf , in recent studies the

foundation is considered to be shallow if ( 4/ BDf . Other wise it is consideredto be deep foundation.

Foundation is considered to be strip if ( )00.0/ LB . The soil from ground surface ( ) to the bottom of the foundation (

) is replaced by stress fDq = .

For General shear failure:Type of foundation Ultimate bearing capacity qu

Strip Footing BNqNcNq qcu2

1++=

Square footing BNqNcNq qcu 4.03.1 ++=

Circular footing BNqNcNq qcu 3.03.1 ++=

c: Cohesive.fDq =

B: Foundation width (Diameter if circular).NNN qc ,, : Bearing capacity factors given from table 3.1 P.158 as function of angle of

friction .

For Local shear failure:Type of foundation Ultimate bearing capacity qu

Strip Footing '''2

1

3

2

BNqNcNq qcu ++=

Square footing ''' 4.0867.0 BNqNcNq qcu ++=

Circular footing ''' 3.0867.0 BNqNcNq qcu ++=

''' ,,

NNN qc : Factors for bearing capacity given from table 3.2 P.160Or from table 3.1 P158 but replace ' instead of :

= tan

3

2tan 1'

2- Meyerhofs equations (General bearing capacity equation):Terzagi equations neglect:

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Rectangular footings. Inclination of loads. Shear strength of soil above the foundation.

Meyerhofs equation takes in consideration theses variables:idsqiqdqsqcicdcscu

FFFBNFFFqNFFFcNq

5.0++=

NNN qc ,, : Table 3.4 P.168

b eS h o

f a c t o r s .nI n c l i n a t i o,,

f a c t o r s .D e p t h,,

f a c t o r s .S h a p e,,

iq ic i

dq dc d

sq sc s

FFF

FFF

FFF

Shape Factors:

L

BF

L

BF

N

N

L

BF

s

qs

c

q

cs

4.01

tan1

1

=

+=

+=

Depth Factors: Case I: 1/ BDf

( )( )

1

sin1tan21

4.01

2

=

+=

+=

d

f

qd

f

cd

F

B

DF

B

DF

Depth Factors: Case II: 1/ >BDf

( )( )

1

tansin1tan21

tan4.01

12

1

=

+=

+=

d

f

qd

f

cd

F

B

D

F

B

DF

Inclination Factors

3

The Term is in radian


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2

2

1

901

=

==

i

qici

F

FF

Effect of water table in bearing capacity equations:

4

Case I: 0Dw1Df Case II: 0Dw2B


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Case I) Water table is located at depth Dw1 so that 0 Dw1 Df:

( )1

'

1 wfwDDDq += ( )

wsat =='

( )

Case II) Water table is located at depth Dw2 below the foundation so that 0 Dw2 B:fDq = .( )

( )'2' +==B

Dw ( (

Case III) Water table is located at depth Dw2 below the foundation so that Dw2 > B:No changes in equations.

Factor of safety:Ultimate bearing capacity

.

( )( )

( )

load.Ultimate

capacitybearingallowableNet

capacitybearingallowableGross

capacitybearingultimateNet

capacitybearingultimateGross

GrossQ

q

q

qqq

q

u

netall

all

unetu

u

=

( )( )

FS

qq

FS

qq

FS

qq

FS

qq

allunetu

netall

uall

=

==

=

FS = (3 4) for bearing capacity

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Example 1)Determine the size of square footings to carry gross allowable load (295 KN) given that:

00.0

.35

./15.18

00.1

3

3

=

=

=

=

=

C

mKN

D

FS

f

Use Terzagi equations assuming general shear failure.295KN

Df=1.0

0

B

C=0.00

=35

=18.15KN/m3

Solution

92cmB:errorandBy trial684.22814.2

41.4515.184.044.41115.180885

41.45,44.41,75.5735At

4.03.1

:footingsquareFor

.885

3295

.295

23

2

22

2

==+

++=

====

++=

===

===

BB

BB

NNN

BNqNcNq

BBFSqq

BA

Qq

AreaqQ

qc

qcu

allu

allall

allall

Example 2)

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Determine the net allowable load that foundation can carry (no inclination), UseMeyerhof equation given that:

./50

.25

./10

00.2

4

2

3

mKNC

mKN

D

FS

w

f

== =

==

Solutionidsqiqdqsqcicdcscu FFFBNFFFqNFFFcNq 5.0++=

88.10,66.10,72.2025At ====

NNN qc

The water table is at depth = 1m


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( )( )

1

313.1

2

225sin125tan21

4.12

24.01

12/2/

2

=

=+=

=

+=

==

d

qd

cd

f

F

F

F

BD

Inclination factors:Due to absence of inclined load, the inclination factor is 1 every where.

( )

( )( )

( ) ( ) KNAqQ

mKNFS

qq

mKNqqq

mKNq

q

netallnetall

netu

netall

unetu

u

u

1.3716635.619

/35.6194

4.2477

/4.24772.266.2503

./6.2503

733.088.1024.95.01313.1311.166.102.2614.1343.172.2050

2

2

2

===

===

===

=

++=

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P

e

BL

Pe

B

Foundation EngineeringChapter 3: Bearing capacity of shallow foundation


Lecture (4) Bearing capacity of shallow foundation (cont.) Eccentrically loaded foundation:

** :

P

q = P /( B x L )

B x L

** :

One way eccentricity

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P

B

M = P e

q m a x

q m i n


1- For e


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3- For e > B/6: There will be tension stresses on the foundation which is prohibitedin design, so we will neglect the tension stress and calculate qmax as follow:

qmax qmax

3(B/2- e)

)2(3

4max

eBL

Pq

=

Derivation:

2........................2

32323

00.0

00.0

:

1................................2

1max

=

=

=

=

==

=

eB

XeBX

eB

PX

N

M

PNF

Stabilityfor

LXqN

edgeright

y

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)2(3

4

2

23*

2

1

2

1

:2&1

maxmax

max

eBL

Pq

eBLqP

LXqPN

From

=

=

==

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B

L

e

B / 2 - eB / 2 - e

B - 2 e2 e


:

( ) ( )

BB

eLL

LBB

LBA

LL

eBB

used

=

=

=

=

=

=

'

'

'''

'''

'

'

2

:(L)ofplanein theismomenttheIf

,.min

2

The equation used to calculate the bearing capacity is Meyerhof's or Terzagi's equation: idsqiqdqsqcicdcscu FFFNBFFFqNFFFcNq

''5.0++=

To find shape factors: use '' , LBTo find depth factors: use B, Lin last term is related to soil below the foundation.To find the gross ultimate load Qu:

''AqQ uu =

max

'

max :qforadequateissafetyoffactor

capacity)bearingAgainst(

q

qFSthatCheck

Q

QFS

u

all

u

=

=

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Example 3)Determine the size of square footing (B x B) that subjected to vertical load (100,000Ib)and moment (25,000Ib.ft), the soil profile is given below:

Use FS=6, w=62.4 Ib/ft3Qal l=100,000Ib

B

M=25,000Ib.f td=100Ib/f t3

C=0.00

= 3 0

sat=120Ib/f t3

C = 0 . 0 0

= 30

4ft

Solution1- Find the eccentricity (e):

ftP

Me 25.0

000,100

000,25===

2- Find the effective areaA':

( )

( )( ) BBBBABL

BLBB

BL

BB

LBA

used

5.05.0

.

5.0),.(min

5.0

2'

'

'''

'

'

'''

==

=

==

=

=

=

3- Find 'uq :

idsqiqdqsqcicdcscu FFFNBFFFqNFFFcNq ''

5.0++=

4.22,4.1803At ===

NNq

./6.574.62120

./4001004

3'

2

ftIb

ftIbq

===

==

Shape factors:

BB

BF

BB

BF

s

qs

2.06.0

5.04.01

2886.0577.130tan

5.01

+==

=+=

Depth factors:

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1

154.11

4)30sin1(30tan21

1/DAssume

2

f

=

+=+=

==

>==

( )

( )

mL

AB

mLBA

mLBL

mL

mB

675.0908.1

2879.1

2879.1908.135.15.02

1

908.1),max(

908.1)182.0(35.12

35.1)2.0(35.15.1

'

''

2

11

'

11

'

1

1

===

===

==

====

idsqiqdqsqcicdcscu FFFNBFFFqNFFFcNq ''

5.0++=

Foe sand: c=0.0003.48,3.3353At ===

NNq

Shape factors:

858.0908.1

675.04.01

248.135tan908.1

675.01

==

=+=

s

qs

F

F

Depth factors:

1

167.15.1

1)35sin1(35tan21

166.05.1/1/D

2

f

=

=+=


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Lecture (5) Bearing capacity of shallow foundation (Cont.)Bearing capacity of layered soilIn practice the foundation may be based on layered soil profile, or in some times we haveto replace adequate thickness of weak soil by stronger one.Calculating the ultimate bearing capacity for such case have some different as that will beshown below.

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For continuous foundation:

11

12

1

tan1

2qH

B

K

H

DH

B

Hcqq s

fa

bu

+++=

( ) ( ) ( ) fqc DqBNNqNcq 11111111 5.0 =++=

( )1cN , ( )1qN and ( )1N are given from the same tables but use 1=

uq : Ultimate bearing capacity.

bq : bearing capacity of bottom layer.

In this case ( ) ( ) ( ) 1222222 5.0 HDqBNNqNcq fqcb +=++=

( )2cN , ( )2qN and ( )2N are given from the same tables but use 2=

ac : AdhesionH: Thickness of top soil layer below bottom of foundation.

B: Width of foundation.sK : Punching shear coefficient.

To find sK To find ac

1- find 12 / qq

( ) ( )

( ) ( )22222

11111

5.0

5.0

BNNcq

BNNcq

c

c

+=

+=

2- Go to figure 3.21 P.190 to find sK

1- find 12 / qq2- Go to figure 3.21 P.190 to find

1/ cca

When the top layer height (H) is relatively large, the failure surface will occur in the toplayer and so uq = 1qOtherwise there will be punching shear failure in the top layer and then general shearfailure in the bottom layer.

For rectangular foundation:

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( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )1111111111

222221222

11

12

1

5.0

5.0

tan211

21

sqsqfcsc

sqsqfcscb

sfa

bu

FBNFNDFNCq

FBNFNHDFNCq

qHB

k

H

D

L

BH

B

Hc

L

Bqq

++=

+++=

+

++

++=

Read the special cases from text book P.191 192

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Example 1)Rectangular footing 1.5m x 1m based on the shown soil profile; find the grossallowable load that foundation can carry.

Special case III

112

2114.52.01 qD

B

Hc

L

BC

L

Bq

f

a

u+

++

+=

In this case 9.0/4.0120/48// 11212 === CcCCqq a from figure 3.22./1081209.0 2mKNca ==

2/4.65618.161

110825.1

114814.55.1

12.01 mKNqu =+

++

+=

2

111 /8.71518.1612014.55.1

12.0114.52.01 mKNDC

L

Bq f =+

+=+

+=

uq < OKq 1 , otherwise use uq = 1q

( ) KNQ

mKNq

all

all

15.2465.111.164

/1.1644

4.656 2

==

==

Read Example 3.9 P.194

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Bearing Capacity of foundation on top of slopeFor continuous shallow foundation:

qcqu BNCNq 5.0+=

For pure sand C = 0.00 qu BNq 5.0=

For pure clay = 00.0 cqu CNq =

To find cqN :

1- Find stability numberC

HNs

=

2- Find b/B and BDf /

3- Go to figure A shown below: If B < H: use curves with 00.0=sN

If HB : use curves with calculated sN .To find qN :

1- Find b/B and BDf /2- Go to figure B shown below to find qN .

b: Distance from edge of foundation tothe top of slope.H: Height of top of slope.

: Angle of slope with horizontal

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Fig.A Finding Ncq

Fig.B Finding N q

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Example 2)For the shown soil profile determine the ultimate bearing capacity of the continuousfoundation.

Solution= 00.0C qu BNq 5.0=

B=1.5m

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