Foundation Analysis

Foundations with Df equal to 3 to 4 times the width may

be defined as shallow foundations.

TWO MAIN CHARACTERISTICS

o Safe against overall shear failure

o Cannot undergo excessive displacement, or settlement

ULTIMATE BEARING CAPACITY

o 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 UU and

CU 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 CD shearing condition apply)

Vesic, 1963

Laboratory plate 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.

Vesic, 1973

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 Shear 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

where,

𝑐′ is the cohesion

is the unit weight of soil

q is the equivalent surcharge load equal to 𝛾Df

𝑁𝐶, 𝑁𝑞, 𝑁𝛾 are bearing capacity factors that are nondimensional and are

functions only of the soil friction angle ɸ’.

where,

Kp𝛾 is the passive pressure coefficient

Modified for:

Square Foundation 𝑞𝑢 = 1.3𝑐′𝑁𝐶 + 𝑞𝑁𝑞 + 0.4𝛾𝐵𝑁𝛾

Circular Foundation 𝑞𝑢 = 1.3𝑐′𝑁𝐶 + 𝑞𝑁𝑞 + 0.3𝛾𝐵𝑁𝛾

LOCAL SHEAR FAILURE

Strip Foundation

𝑞𝑢 =2

3𝑐′𝑁′𝐶 + 𝑞𝑁′𝑞 +

1

2𝛾𝐵𝑁′𝛾

Square Foundation 𝑞𝑢 = 0.867𝑐′𝑁′𝐶 + 𝑞𝑁′𝑞 + 0.4𝛾𝐵𝑁′𝛾

Circular Foundation 𝑞𝑢 = 0.867𝑐′𝑁′𝐶 + 𝑞𝑁′𝑞 + 0.3𝛾𝐵𝑁′𝛾

ɸ′

= tan−1(2

3tanɸ

′)

𝑞𝑎𝑙𝑙 =𝑞𝑢

𝐹𝑆

𝑁𝑒𝑡 𝑠𝑡𝑟𝑒𝑠𝑠 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑜𝑛 𝑠𝑜𝑖𝑙 =𝑛𝑒𝑡 𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦

𝐹𝑆

𝑞𝑛𝑒𝑡(𝑢) = 𝑞𝑢 − 𝑞

where,

𝑞𝑛𝑒𝑡(𝑢) is the net ultimate bearing capacity

𝑞 = 𝛾𝐷𝑓

So,

𝑞𝑎𝑙𝑙(𝑛𝑒𝑡) =𝑞𝑢 − 𝑞

𝐹𝑆

The factor of safety should be at least 3 in all cases.

1. A square foundation is 2 m x 2 m in plan. The soil supporting the foundation has a friction angle of ɸ’ = 25˚ and 𝑐′= 20 kN/m2. The unit

weight of soil, 𝛾, is 16.5 kN/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 1.5 m and that general shear failure

occurs in the soil.

Solution 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) = 1078.29 kN/m2

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

𝑞𝑎𝑙𝑙 =𝑞𝑢

𝐹𝑆=

1078.29

3= 359.5 kN/m2

Thus, the total allowable gross load is

𝑄 = 𝑞𝑎𝑙𝑙(𝐵2) = 359.5(2x2) = 1438 kN

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

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

q = γDf

BNqNNcq qcu2

1'

If 0 ≤ d ≤ B,

q = γDf

γ in the last term is

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

BNqNNcq qcu2

1'

)'('_

B

d

If d ≥ B,

*The water will have no effect on the ultimate bearing capacity.

BNqNNcq qcu2

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.

where,

c’ is the cohesion

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

γ is the unit weight of soil

B is the width of foundation (or diameter for circular foundation)

Fcs, Fqs, Fγs are shape factors

Fcd, Fqd, Fγd are depth factors

Fci, Fqi, Fγi are load inclination factors

Nc, Nq, Nγ are bearing capacity factors

idsqiqdqsqcicdcscu FFFBNFFFqNFFFNcq 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)

idsqiqdqsqcicdcscu FFFBNFFFqNFFFNcq 2

1'

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

c

qd

qdcd

d

qd

f

cd

f

F

B

DF

N

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

c

qd

qdcd

d

qd

f

cd

f

F

radiansB

DF

N

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

2. Solve Problem 1 using the general bearing capacity equation.

Square foundation

2m x 2m

General shear failure

Reqd: Allowable gross load

mD

FS

mkN

mkNc

f 5.1

3

/5.16

/20'

25'

3

2

idsqiqdqsqcicdcscu FFFBNFFFqNFFFNcq 2

1'

Solution:

Bearing Capacity Factors:

*Table 3.3 can also be used.

Load Inclination Factors Since load is vertical, Fci , Fqi , Fγi =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

eeN

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

idsqiqdqsqcicdcscu FFFBNFFFqNFFFNcq 2

1'

𝑞𝑢 = 20 20.72 1.514 1.257 1 + 1.5𝑥16.5 10.66 1.466 1.233 1 + 0.5 16.5 2 10.88 0.6 1 1

𝑞𝑢 = 1,373.2 𝑘𝑁/𝑚2

𝑞𝑎𝑙𝑙 =𝑞𝑢

𝐹𝑆=

1,373.2

3= 457.7 𝑘𝑁/𝑚2

𝑄 = 457.7 2𝑥2 = 1,830.8 𝑘𝑁

3. A square foundation (B x B) has to be constructed as shown below.

Assume that 𝛾 = 16.5 𝑘𝑁/𝑚3, 𝛾𝑠𝑎𝑡 = 18.55 𝑘𝑁/𝑚3, ∅′ = 34°, 𝐷𝑓 = 1.22 𝑚,

and 𝐷1 = 0.61 𝑚. The gross allowable load, 𝑄𝑎𝑙𝑙 , with FS = 3 is 667.2 𝑘𝑁. Determine the size of the footing using the general bearing capacity

equation.

idsqiqdqsqcicdcscu FFFBNFFFqNFFFNcq 2

1'

dsqdqsqu FFBNFFqNq '2

1

3

1

/2.667 2

22

FS

qq

mkNBB

Qq

u

all

allall

dsqdqsq FFBNFFqN '

2

1

Since there is no cohesion,

Becomes

Eqn. 1

Eqn. 2

Bearing Capacity Factors From Table 3.3, for ϕ’ = 34

Nq = 29.44

Nγ = 41.06

Case I

q = D1γ + D2(γsat - γw)

q = (0.61)(16.5) + (0.61)(18.55-9.81)= 15.4 kN/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

uq

allq

B

allQ

allq

EqnsCombine

B

B

Bxx

FS

uq

allq

mkNxq

dF

sFBN

qdF

qsF

qqN

FS

uq

allqEqn

3.1,

89.35265

38.2522

2.667

2

:2&1

89.35B

265252.38

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

1

05.1167.144.294.15

3

1

2/4.15)81.955.18(61.05.1661.0

)'2

1(

3

1.2