Foundations with Df equal to 3 to 4 times the width may be defined as shallow foundations

TWO MAIN CHARACTERISTICS ◦ Safe against overall shear failure ◦ Cannot undergo excessive displacement, or

settlement

ULTIMATE BEARING CAPACITY

the load per unit area of the foundation at which shear failure in soil occurs

GENERAL SHEAR FAILURE

Long rectangular model footing of width B at the surface of a

dense sand.

The triangular wedge-shaped zone of soil marked I is pushed down and in turn presses the zones marked II and III sideways. The soil on the both sides of the foundation will bulge out and the slip surface will extend to the ground surface.

LOCAL SHEAR FAILURE

Medium dense sand or clayey soil of medium compaction

Movement of the foundation will be accompanied by sudden jerks

The triangular wedge-shaped zone of soil marked I is pushed down but unlike in general shear failure, the slip surface end somewhere inside the soil.

PUNCHING SHEAR FAILURE

Fairly loose soil

Soil will not extend to the ground surface

GENERAL SHEAR FAILURE Applies to dense granular soil and to firmer saturated

cohesive soils subject to undrained loading (the U-U and C-U shearing conditions apply)

PUNCHING SHEAR FAILURE Applies to compressible soils, such as sands having low-to-

medium relative density, and for cohesive soils subject to slow loading (the C-D shearing conditions apply)

Vesic1963

Laboratory load-bearing tests on circular and rectangular plates supported by a sand at various relative densities of compaction, Dr.

Dr≥ about 70%

(general shear failure)

Vesic 1973

Relationship for the mode of bearing capacity failure of foundations resting on sands.

Vesic1973

Dr = relative density of sand Df = depth of foundation measured from the ground surface

B* = 2𝐵𝐿

𝐵+𝐿

where B= width of foundation L= length of foundation

General range of S/B with the relative density of compaction of sand.

General shear failure – ultimate load may occur at settlements of 4 to 10% of B.

Local or punching shear failure – ultimate load may occur at settlements of 15 to 25% of B.

Bearing capacity failure in soil under a rough rigid continuous (strip) foundation

1. The triangular zone ACD immediately under the foundation

2. The radial shear zones ADF and CDE, with the curves DE and DF being arcs of a logarithmic spiral

3. Two triangular Rankine passive zones AFH and CEG

Continuous or Strip foundation

𝑞𝑢 = 𝑐′𝑁𝐶 + 𝑞𝑁𝑞 +1

2𝛾𝐵𝑁𝛾 Eq.(3.3)

where 𝑐′= cohesion

𝛾 = unit weight of soil

q = 𝛾Df

𝑁𝐶, 𝑁𝑞, 𝑁𝛾= bearing capacity factors that are nondimensional and are functions only of the soil friction angle ɸ’

Eq.(3.4)

Eq.(3.5)

Eq.(3.6)

where Kp𝛾= passive pressure coefficient

Square foundation

𝑞𝑢 = 1.3𝑐′𝑁𝐶 + 𝑞𝑁𝑞 + 0.4𝛾𝐵𝑁𝛾 Eq.(3.7)

Circular foundation

𝑞𝑢 = 1.3𝑐′𝑁𝐶 + 𝑞𝑁𝑞 + 0.3𝛾𝐵𝑁𝛾 Eq.(3.8)

LOCAL SHEAR FAILURE Strip foundation

𝑞𝑢 =2

3𝑐′𝑁′𝐶 + 𝑞𝑁′𝑞 +

1

2𝛾𝐵𝑁′𝛾 Eq.(3.9)

Square foundation 𝑞𝑢 = 0.867𝑐′𝑁′𝐶 + 𝑞𝑁′𝑞 + 0.4𝛾𝐵𝑁′𝛾 Eq.(3.10) Circular foundation 𝑞𝑢 = 0.867𝑐′𝑁′𝐶 + 𝑞𝑁′𝑞 + 0.3𝛾𝐵𝑁′𝛾 Eq.(3.11)

ɸ′

= tan−1(2

3tan ɸ

′)

𝑞𝑎𝑙𝑙 =𝑞𝑢

𝐹𝑆 Eq.(3.12)

𝑁𝑒𝑡 𝑠𝑡𝑟𝑒𝑠𝑠 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑜𝑛 𝑠𝑜𝑖𝑙 =𝑛𝑒𝑡 𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦

𝐹𝑆 Eq.(3.13)

𝑞𝑛𝑒𝑡(𝑢) = 𝑞𝑢 − 𝑞 Eq.(3.14)

where

𝑞𝑛𝑒𝑡(𝑢) = net ultimate bearing capacity

𝑞 = 𝛾Df

So,

𝑞𝑎𝑙𝑙(𝑛𝑒𝑡) =𝑞𝑢−𝑞

𝐹𝑆 Eq.(3.15)

The factor of safety as defined by Eq.(3.15) should be at least 3 in all cases.

A square foundation is 2mx2m in plan. The soil supporting the foundation has a friction angle of ɸ’ = 25˚ and 𝑐′= 20kN/m2. The unit weight of soil, 𝛾, is 16.5kN/m3. Determine the allowable gross load on the foundation with a factor of safety (FS) of 3. Assume that the depth of the foundation (Df) is 1.5m and that general shear failure occurs in the soil.

Solution

From Eq.(3.7) 𝑞𝑢 = 1.3𝑐′𝑁𝐶 + 𝑞𝑁𝑞 + 0.4𝛾𝐵𝑁𝛾

From Table 3.1, for ɸ’ = 25˚, 𝑁𝐶 = 25.13

𝑁𝑞 = 12.72

𝑁𝛾 = 8.34

Thus,

𝑞𝑢 = 1.3 20 (25.13) + (1.5𝑥16.5)(12.72) + 0.4(16.5)(2)(8.34) = 1078.29kN/m2

So, the allowable load per unit area of the foundation is

𝑞𝑎𝑙𝑙 =𝑞𝑢

𝐹𝑆=

1078.29

3= 359.5KN/m2

Thus, the total allowable gross load is

Q= 𝑞𝑎𝑙𝑙(B2) = 359.5(2x2) = 1438KN

The Bearing Capacity Equation is modified when the water table is in the proximity of the foundation.

Bearing Capacity Equation

Modified Bearing Capacity Equation ◦ Case I

◦ Case II

◦ Case III

next

back

BNqNNcq

BNqNNcq

BNqNNcq

qcu

qcu

qcu

3.0'3.1

4.0'3.1

2

1'

GENERAL SHEAR FAILURE (Continuous or strip foundation) (square foundation) (circular foundation)

If 0≤D1≤Df,

q=D1γ+D2(γsat-γw)

Where:

γsat = sat unit wt of soil

γw = unit wt of water

γ in ½γBNγ becomes γ’

where γ’= γsat-γw

back

q= γDf

BNqNNcq

qcu

2

1'

If 0≤d≤B,

q=γDf

γ in the last term becomes

* The preceding modifications are based on the assumption that there is no seepage force in the soil.

back

BNqNNcq

qcu

2

1'

)'(' B

d

If d≥B,

*The water will have

no effect on the

ultimate bearing

capacity.

back

BNqNNcq

qcu

2

1'

SHAPE: The bearing capacity eqns

do not address the case of

rectangular foundations

(0<B/L<1). Wherein L>B.

DEPTH: The eqns also do not take

into account the shearing

resistance along the failure surface

in soil above the bottom of the

foundation.

LOAD INCLINATION: The load on

the foundation may be inclined.

Wherein:

c’= cohesion

q= effective stress at the level of the bottom of the foundation

γ= unit weight of soil

B= width of foundation (or diameter for circular foundation),

Fcs, Fqs, Fγs = shape factors

Fcd, Fqd, Fγd = depth factors

Fci, Fqi, Fγi = load inclination factors

Nc, Nq, Nγ = bearing capacity factors

idsqiqdqsqcicdcscFFFBNFFFqNFFFNcqu

2

1'

α=45 + ϕ’/2

Nq = tan2 (45 + ϕ’/2) eπtan ϕ’

◦ Reissner (1924)

Nc = (Nq – 1) cot ϕ’

◦ Prandtl (1921)

Nγ = 2(Nq + 1) tan ϕ’

◦ Caquot and Kerisel (1953), Vesic (1973)

idsqiqdqsqcicdcscFFFBNFFFqNFFFNcqu

2

1'

Ex3.3

Ex 3.4

Shape Factors

Reference: DeBeer (1970)

L

BF

L

BF

N

N

L

BF

s

qs

c

q

cs

4.01

'tan1

1

Depth Factors

Reference: Hansen (1970)

1

)'sin1('tan21

'tan

1

'

1

1

4.01

1

2

d

f

qd

qd

qdcd

d

qd

f

cd

f

F

B

DF

Nc

FFF

For

F

F

B

DF

For

B

D

1

tan)'sin1('tan21

'tan

1

'

1

1

tan4.01

1

12

1

d

f

qd

qd

qdcd

d

qd

f

cd

f

F

radiansB

DF

Nc

FFF

For

F

F

radiansB

DF

For

B

D

Inclination Factors

Reference: Meyerhof (1963);

Hanna and Meyerhof (1981)

'1

901

2

i

qici

F

FF

inclination of the load on the foundation with respect to the vertical

mD

FS

mkN

mkNc

f5.1

3

/5.16

/20'

25'

3

2

Square foundation 2mx2m

General shear failure Reqd: Allowable gross load

idsqiqdqsqcicdcscFFFBNFFFqNFFFNcqu

2

1'

87.1025tan)166.10(2'tan)1(2

72.2025cot)166.10('cot)1(

66.10)2/2545(tan

)2/'45(tan

25tan2

'tan2

q

qc

q

NN

NN

e

eN

Solution

Solution: Bearing Capacity Factors: *table 3.3 can also be used Load inclination Factors Since load is vertical, Fci, Fqi, Fyi=1

Shape Factors

Fcs=1+ (2/2)(10.66/20.72) = 1.514

Fqs=1 + (2/2)tan25 = 1.466

Fγs =1-0.4(2/2)= 0.6

Depth Factors (Df/B = 1.5/2 = 0.75)

L

BF

L

BF

N

N

L

BF

s

qs

c

q

cs

4.01

'tan1

1

1

233.12

5.1)25sin1)(25tan2(1)'sin1('tan21

257.125tan72.20

233.11233.1

'tan

1

'

1

22

d

f

qd

qd

qdcd

f

F

B

DF

Nc

FFF

For

B

D

kNxQ

mkNFS

qq

mkNq

q

FFFBNFFFqNFFFNcq

u

all

u

u

idsqiqdqsqcicdcscu

8.1830)22(7.457

/7.4573

2.1373

/2.13737.1079.4766.788

)116.088.1025.165.0(

)1233.1466.1666.105.165.1(

)1257.1514.172.2020(

2

1'

2

2

sizefootingqd

FS

kNQall

mD

mD

mkN

mkN

BxBsquare

f

sat

:Re

3

2.667

61.0

22.1

34'

/55.18

/5.16

;

1

3

3

idsqiqdqsqcicdcscFFFBNFFFqNFFFNcqu

2

1'

dsqdqsqFFBNFFqNqu

'

2

1

3

1

/2.667 2

22

FS

qq

mkNBB

Qq

u

all

all

all

dsqdqsqFFBNFFqN

'

2

1

Since there is no cohesion, Becomes Eq1 Eq2 Bearing Capacity Factors Table 3.3 For ϕ’ = 34

Nq = 29.44 Ny= 41.06

Case I q=D1γ+D2(γsat-γw) q= (0.61)(16.5) + (0.61)(18.55-9.81)= 15.4KN/m2

1

05.11

4)34sin1(34tan21)'sin1('tan21

6.04.014.01

67.134tan1'tan1

22

d

f

qd

s

qs

F

BBB

DF

L

BF

L

BF

mBerrorandtrialBy

BBB

FS

qq

B

Qq

EqnsCombine

B

B

Bxx

FS

qq

mkNxq

FFBNFFqNFS

qqEqn

u

all

all

all

u

all

dsqdqsq

u

all

123.1,

89.35265

38.2522.667

:2&1

89.35B

265252.38

)1)(6.0)(06.41()81.955.18(2

1

05.1167.144.294.15

3

1

/4.15)81.955.18(61.05.1661.0

)'2

1(

3

1.2

2

2

2

Soil Compressibility Factors

Eqn 3.19

Is modified to

Eqn 3.27

Wherein Fcc, Fqc and Fγc are compressibility factors

idsqiqdqsqcicdcscFFFBNFFFqNFFFNcqu

2

1'

cdsqcqdqsqcccdcscFFFBNFFFqNFFFNcqu

2

1'

Step 1. Calculate the rigidity index, Ir, of the soil at a depth approximately B/2 below the bottom of the foundation, or

where Gs = shear modulus of the soil q = effective overburden pressure at a depth of Df + B/2

'tan'' qc

GsIr

2

'45cot45.03.3exp

2

1)(

L

BcrIr

Step 2. The critical rigidity index, Ir(cr), can be expressed as The variations of Ir(cr) with B/L are given in Table 3.6.

Step 3. If then However if Ir< Ir(cr), then Figure 3.12 shows the variation of

)(crIrIr

'sin1

)2)(log'sin07.3('tan6.04.4exp

Ir

L

BFqccF

'tan

1

log60.012.032.0

,0

.'/))30.3((

Nq

FqcFqcFcc

IrL

BFcc

For

andIrwseeEqFqccF

1 cFFqcFcc

Given: ◦ Shallow foundation ◦ B= 0.6m ◦ L= 1.2m ◦ Df=0.6m ◦ Soil characteristics:

ϕ'=25 c’=48Kn/m2

γ=18KN/m3

Modulus of elasticity, Es= 620 KN/m2

Poisson’s ratio, μs=0.3

Required: ◦ Calculate the ultimate

bearing capacity.

Solution: Rigidity Index

)'tan'')(1(2

)1(2

'tan''

qc

EsIr

EsGs

qc

GsIr

s

s

29.4)25tan2.168.4)(3.01(2

620

/2.162

6.06.018

2'

2

Ir

mkNB

Dfq

Critical Rigidity Index

41.622

2545cot

2.1

6.045.03.3exp

2

1)(

2

'45cot45.03.3exp

2

1)(

crIr

L

BcrIr

Since Ir(cr)> Ir, use Eqs/ 3.30 and 3.32

279.025tan72.20

347.01347.0

,);3.3(72.20,25'

'tan

1

347.025sin1

))29.42)(log(25sin07.3(25tan

2.1

6.06.04.4exp

'sin1

)2)(log'sin07.3('tan6.04.4exp

Fcc

thereforeseeTableNcFor

Nc

FqcFqcFcc

and

xFqccF

Ir

L

BFqccF

Shape Factors Depth Factors

Table 3.3

8.02.1

6.04.014.01

233.125tan2.1

6.01'tan1

257.172.20

66.10

2.1

6.011

L

BF

L

BF

N

N

L

BF

s

qs

c

q

cs

cdsqcqdqsqcccdcscFFFBNFFFqNFFFNcqu

2

1'

1

311.16.0

6.0)25sin1(25tan21

)'sin1('tan21

343.125tan72.20

311.11311.1

'tan

1

2

2

d

f

qd

qd

qdcd

F

B

DF

Nc

FFF

cdsqcqdqsqcccdcscFFFBNFFFqNFFFNcqu

2

1'

232.549347.018.088.106.0182

1

]347.0311.1233.166.10)186.0[(

)279.0343.1257.172.2048(

m

kN

qu

When foundations are subjected to moments in addition to the vertical load, the distribution of pressure on the soil is not uniform.

Where Q is the total vertical load and M is the moment on the foundation.

LB

M

BL

Qq

LB

M

BL

Qq

2min

2max

6

6

B

e

BL

Qq

B

e

BL

Qq

Q

Me

LB

M

BL

Qq

LB

M

BL

Qq

61

61

6

6

min

max

2min

2max

When e= B/6, qmin=0

When e> B/6 , qmin <0

Which means tension will

develop.

Soil cannot take any tension

There will be a separation of the foundation and the soil underlying it.

qmax = 4Q/ 3L(B-2e)

The exact distribution of failure

is difficult to estimate.

The factor of safety for such type of loading against bearing capacity failure can be evaluated as

Where Qult = ultimate load-carrying capacity

Q

QFS

ult