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Session 5 6

BEARING CAPACITY OF SHALLOWFOUNDATION

Course : S0484/Foundation Engineering

Year : 2007

Version : 1/0

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SHALLOW FOUNDATION

Topic: General

Terzaghi Model

Meyerhoff Model Brinch Hansen Model

Influence of multi layer soil

Influence of ground water elevation Shallow Foundation Bearing by N-SPT

value

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TYPES OF SHALLOW FOUNDATION

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TYPES OF SHALLOW FOUNDATION

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TERZAGHI MODEL

Assumptions: Subsoil below foundation structure is

homogenous

Shallow foundationDf

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TERZAGHI MODEL

FAILURE ZONES:1. ACD : TRIANGULAR ZONES2. ADF & CDE : RADIAL SHEAR ZONES3. AFH & CEG : RANKINE PASSIVE ZONES

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STRIP FOUNDATION

qult = c.Nc + q.Nq + 0.5..B.N SQUARE FOUNDATION

qult = 1.3.c.Nc + q.Nq + 0.4..B.N

CIRCULAR FOUNDATIONqult = 1.3.c.Nc + q.Nq + 0.3..B.N

( )

( )

tan1cos2

1

24cos.2

1

24cos.2

cot

2

2

tan2/4/32

2

tan2/4/32

=

+

=

+

=

pyK

N

eNq

eNc

Where:

c = cohesion of soil

q = . Df ; Df = the thickness of foundationembedded on subsoil

= unit weight of soil

B = foundation width

Nc, Nq, N = bearing capacity factors

TERZAGHI MODEL(GENERAL FAILURE)

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BEARING CAPACITY FACTORS

GENERAL

FAILURE

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GENERALFAILURE

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TERZAGHI MODEL(LOCAL FAILURE)

STRIP FOUNDATION

qult = 2/3.c.Nc + q.Nq + 0.5..B.N SQUARE FOUNDATION

qult = 0.867.c.Nc + q.Nq + 0.4..B.N

CIRCULAR FOUNDATION

qult = 0.867.c.Nc + q.Nq + 0.3..B.N

( )

( )

'tan1'cos2

1

2

'

4cos.2

1

2

'

4cos.2

'cot

2

2

'tan2/'4/32

2

'tan2/'4/32

=

+

=

+

=

pyKN

eNq

eNc

= tan-1 (2/3. tan)

Where:

c = cohesion of soil

q = . Df ; Df = the thickness of foundationembedded on subsoil

= unit weight of soil

B = foundation width

Nc, Nq, N = bearing capacity factors

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LOCAL FAILURE

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GROUND WATER INFLUENCE

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GROUND WATER INFLUENCE

CASE 10 D1 < Df q = D1.dry + D2 .

CASE 20 d B q =dry.Df

the value of in third part of equation is

replaced with

= + (d/B).(dry - )

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FACTOR OF SAFETY

FS

qq

FS

qq

unet

netall

uall

)(

)( =

=

Where:

qu = gross ultimate bearing capacity of shallow foundation

qall = gross allowable bearing capacity of shallow foundation

qnet(u) = net ultimate bearing capacity of shallow foundation

qall = net allowable bearing capacity of shallow foundation

FS = Factor of Safety (FS 3)

f

uunet

Dq

qqq

.

)(

=

=

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NET ALLOWABLE BEARING CAPACITY

PROCEDURE:

1. Find the developed cohesion and the angle of friction

2. Calculate the gross allowable bearing capacity (qall)according to terzaghi equation with cd and d as theshear strength parameters of the soil

3. Find the net allowable bearing capacity (qall(net))

shear

dFS

cc =

=

shear

dFS

tantan 1 FSshear = 1.4 1.6

Ex.: qall = cd.Nc + q.Nq + .B.N

Where Nc, Nq, N = bearing capacity factor for the friction angle, d

qall(net) = qall - q

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EXAMPLE PROBLEM

A square foundation is 5 ft x 5 ft in plan. The soilsupporting the foundation has a friction angle of = 20o and c = 320 lb/ft2. The unit weight of soil,, is 115 lb/ft3. Assume that the depth of thefoundation (Df) is 3 ft and the general shearfailure occurs in the soil.Determine:

- the allowable gross load on the foundation with

a factor of safety (FS) of 4.- the net allowable load for the foundation withFSshear = 1.5

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EXAMPLE SOLUTION

Foundation Type: Square Foundation

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EXAMPLE SOLUTION

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GENERAL BEARING CAPACITY EQUATION

idsqiqdqscicdcsu FFFNBFFFNqqFFFNccq .....).5.0(........ ++=

Df

Meyerhof s Theory

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BEARING CAPACITY FACTOR

( )

tan)1(2

cot1

245tan tan.2

+=

=

+=

NqN

NqNc

eNq

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SHAPE, DEPTH AND INCLINATION FACTOR

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EXAMPLE 2

2 m

GWL

dry = 13 kN/m3

sat = 18 kN/m3

c = 1 kg/cm2

= 20o

P = 73 ton

Tank

Foundation

Determine the size (diameter) circle foundation of tank structure as

shown in the following picture

With P is the load of tank, neglected the weight of foundation and usefactor of safety, FS = 3.5.

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EXAMPLE 3

DETERMINE THE FACTOR OF SAFETY FOR:

-CASE 1 : GWL LOCATED AT 0.3m (MEASURED FROM THE SURFACEOF SOIL)

-CASE 2 : GWL LOCATED AT 1.5m (MEASURED FROM THE SURFACEOF SOIL)

dry= 13 kN/m3B = 4m

SQUARE FOUNDATION

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ECCENTRICALLY LOADED FOUNDATIONS

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ECCENTRICALLY LOADED FOUNDATIONS

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ONE WAY ECCENTRICITY

Meyerhofs step by step procedure: Determine the effective dimensions of the foundation as :

B = effective width = B 2eL = effective length = L

Note: if the eccentricity were in the direction of the length of the foundation, the value

of L would be equal to L-2e and the value of B would be B. The smaller of the two dimensions (L and B) is the effective width of the

foundation Determine the ultimate bearing capacity

to determine Fcs, Fqs, Fs use effective length and effective width

to determine Fcd, Fqd, Fd use B

The total ultimate load that the foundation can sustain isQult = qu.B.L ; where BxL = A (effective area)

The factor of safety against bearing capacity failure isFS = Qult/Q

Check the factor of safety against qmax, or,

FS = qu/qmax

idsqiqdqscicdcsu FFFNBFFFNqqFFFNccq ......5,0........ ++=

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EXAMPLE PROBLEM

A Square foundation is shown in the following figure.

Assume that the one- way load eccentricity e = 0.15m.Determine the ultimate load, Qult

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EXAMPLE SOLUTION

With c = 0, the bearing capacity equation becomes

TWO WAY ECCENTRICITY

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TWO-WAY ECCENTRICITY

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TWO-WAY ECCENTRICITY CASE 1

O CC C C S 2

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TWO WAY ECCENTRICITY CASE 3

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TWO WAY ECCENTRICITY CASE 4

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BEARING CAPACITY OF LAYERED SOILS

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STRONGER SOIL

UNDERLAIN BYWEAKER SOIL


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H

B

K

H

DH

B

Hcqq s

fabu 1

12

1

tan21

2

++

+=

)2(2)2(22

)1(1)1(11

21

2

1

BNNcq

BNNcq

c

c

+=

+=

( ) )2()2(2)2()2(1)2()2(22

1sqsqfcscb FBNFNHDFNcq +++=


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tsfa

bu qHB

K

H

DH

B

Hcqq

++

+= 1

12

1

tan21

2

tsfa

bu qHB

K

H

D

L

BH

B

Hc

L

Bqq

+

++

++= 1

12

1

tan211

21

Rectangular Foundation

( )

)1()1(1)1()1(1)1()1(1

)2()2(2)2()2(1)2()2(2

2

1

2

1

sqsqfcsct

sqsqfcscb

FBNFNDFNcq

FBNFNHDFNcq

++=

+++=

)2(2)2(22

)1(1)1(11

21

2

1

BNNcq

BNNcq

c

c

+=

+=


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SPECIAL CASES TOP LAYER IS STRONG SAND AND BOTTOMLAYER IS SATURATED SOFT CLAY (2 = 0)

TOP LAYER IS STRONGER SAND AND BOTTOM

LAYER IS WEAKER SAND (c1 = 0 , c2 = 0) TOP LAYER IS STRONGER SATURATED CLAY (1

= 0) AND BOTTOM LAYER IS WEAKERSATURATED CLAY (2 = 0)

Find the formula for the abovespecial cases

BEARING CAPACITY FROM N SPT VALUE

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BEARING CAPACITY FROM N-SPT VALUE

A square foundation BxB has to

be constructed as shown in thefollowing figure. Assume that =105 lb/ft3, sat = 118 lb/ft

3, Df= 4ft and D1 = 2 ft. The gross

allowable load, Qall, with FS = 3is 150,000 lb. The field standardpenetration resistance, NF valuesare as follow:

Determine the sizeof the foundation

SOLUTION

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SOLUTION

Correction of standard penetration number(Liao and Whitman relationship)

SOLUTION

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SOLUTION

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