Session 5 – 6 BEARING CAPACITY OF SHALLOW

FOUNDATION

Course : S0484/Foundation Engineering

Year : 2007

Version : 1/0

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

TYPES OF SHALLOW FOUNDATION

TYPES OF SHALLOW FOUNDATION

TERZAGHI MODEL

Assumptions:

• Subsoil below foundation structure is homogenous

• Shallow foundation Df < B

• Continuous, or strip, footing : 2D case

• Rough base

• Equivalent surcharge

TERZAGHI MODEL

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

• 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 FOUNDATION

qult = 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

pyKN

eNq

eNc

Where:

c = cohesion of soil

q = . Df ; Df = the thickness of foundation embedded on subsoil

= unit weight of soil

B = foundation width

Nc, Nq, N = bearing capacity factors

TERZAGHI MODEL (GENERAL FAILURE)

BEARING CAPACITY FACTORS

GENERAL FAILURE

BEARING CAPACITY FACTORS

GENERAL FAILURE

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 foundation embedded on subsoil

= unit weight of soil

B = foundation width

Nc, Nq, N = bearing capacity factors

BEARING CAPACITY FACTORS

LOCAL FAILURE

BEARING CAPACITY FACTORS

GROUND WATER INFLUENCE

GROUND WATER INFLUENCE

• CASE 1

0 D1 < Df q = D1.dry + D2 . ’

• CASE 2

0 d B q = dry.Df

the value of in third part of equation is replaced with

= ’ + (d/B).(dry - ’)

FACTOR OF SAFETY

FS

qq

FS

qq

unetnetall

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

.

)(

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 the shear strength parameters of the soil

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

sheard FS

cc

sheard FS

tantan 1

FSshear = 1.4 – 1.6

Ex.: qall = cd.Nc + q.Nq + ½ .B.N

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

qall(net) = qall - q

EXAMPLE – PROBLEM

A square foundation is 5 ft x 5 ft in plan. The soil supporting 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 the foundation (Df) is 3 ft and the general shear failure 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 with FSshear = 1.5

EXAMPLE – SOLUTION

Foundation Type: Square Foundation

EXAMPLE – SOLUTION

GENERAL BEARING CAPACITY EQUATION

idsqiqdqscicdcsu FFFNBFFFNqqFFFNccq .....).5.0(........

Df

Meyerhof’s Theory

BEARING CAPACITY FACTOR

tan)1(2

cot1

245tan tan.2

NqN

NqNc

eNq

SHAPE, DEPTH AND INCLINATION FACTOR

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 use factor of safety, FS = 3.5.

EXAMPLE 3

DETERMINE THE FACTOR OF SAFETY FOR:

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

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

dry = 13 kN/m3B = 4m

SQUARE FOUNDATION

ECCENTRICALLY LOADED FOUNDATIONS

ECCENTRICALLY LOADED FOUNDATIONS

ONE WAY ECCENTRICITY

Meyerhof’s 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 widthto determine Fcd, Fqd, Fd use B

• The total ultimate load that the foundation can sustain isQult = qu’.B’.L’ ; where B’xL’ = 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........

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

EXAMPLE – SOLUTION

With c = 0, the bearing capacity equation becomes

TWO-WAY ECCENTRICITY

TWO-WAY ECCENTRICITY – CASE 1

TWO-WAY ECCENTRICITY – CASE 2

TWO-WAY ECCENTRICITY – CASE 3

TWO-WAY ECCENTRICITY – CASE 4

BEARING CAPACITY OF LAYERED SOILS

STRONGER SOIL UNDERLAIN BY WEAKER SOIL

BEARING CAPACITY OF LAYERED SOILS

HB

K

H

DH

B

Hcqq sfabu 1

121

tan21

2

)2(2)2(22

)1(1)1(11

2

12

1

BNNcq

BNNcq

c

c

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

1sqsqfcscb FBNFNHDFNcq

BEARING CAPACITY OF LAYERED SOILS

tsfa

bu qHB

K

H

DH

B

Hcqq

1

121

tan21

2

tsfa

bu qHB

K

H

D

L

BH

B

Hc

L

Bqq

1

121

tan211

21

Rectangular Foundation

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

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

2

12

1

sqsqfcsct

sqsqfcscb

FBNFNDFNcq

FBNFNHDFNcq

)2(2)2(22

)1(1)1(11

2

12

1

BNNcq

BNNcq

c

c

BEARING CAPACITY OF LAYERED SOILS

SPECIAL CASES– TOP LAYER IS STRONG SAND AND BOTTOM

LAYER 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 WEAKER SATURATED CLAY (2 = 0)

Find the formula for the above special cases

BEARING CAPACITY FROM N-SPT VALUE

A square foundation BxB has to be constructed as shown in the following figure. Assume that = 105 lb/ft3, sat = 118 lb/ft3, Df = 4 ft and D1 = 2 ft. The gross allowable load, Qall, with FS = 3 is 150,000 lb. The field standard penetration resistance, NF values are as follow:

Determine the size of the foundation

SOLUTION

Correction of standard penetration number

(Liao and Whitman relationship)

SOLUTION