Foundation Analysis

Part 2

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.

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 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.

back

BNqNNcq qcu2

1'

)'('_

B

d

If d ≥ B,

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

back

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'

Ex 3.3

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

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

General Bearing Capacity Equation

is modified to (Vesic, 1973)

where Fcc, Fqc and Fγc are compressibility factors

The soil compressibility factors were derived by Vesic

(1973) by analogy to the expansion of cavities. According

to that theory, in order to calculate Fcc, Fqc and Fγc, the

following steps should be taken:

idsqiqdqsqcicdcscu FFFBNFFFqNFFFNcq 2

1'

cdsqcqdqsqcccdcscu FFFBNFFFqNFFFNcq 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 is the shear modulus of the soil

q' is the effective overburden pressure at a depth of Df + B/2

'tan'' qc

GI s

r

2

'45cot45.03.3exp

2

1)(

L

BI crr

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 𝐼𝑟 ≥ 𝐼𝑟(𝑐𝑟) then 𝐹𝑐𝑐 = 𝐹𝑞𝑐 = 𝐹𝛾𝑐 = 1.

However if 𝐼𝑟 < 𝐼𝑟(𝑐𝑟), then

Figure 3.12 shows the variation of 𝐹𝛾𝑐 = 𝐹𝑞𝑐 with ∅′ and 𝐼𝑟 .

'sin1

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

rqcc

I

L

BFF

rcc IL

BF log60.012.032.0

'tan

1

c

qc

qcccN

FFF

For ∅ > 0,

For ∅ = 0,

1. For a shallow foundation, B = 0.6 m, L = 1.2 m, and Df = 0.6 m. The known soil characteristics are as follows: ϕ’ = 25°, c’ = 48 kN/m², γ = 18 kN/m³, modulus of elasticity (Es) = 620 kN/m², and Poisson’s ratio (μs) =

0.3. Calculate the ultimate bearing capacity.

Solution:

Rigidity Index

)'tan'')(1(2

)1(2

'tan''

qc

EI

EG

qc

GI

s

sr

s

ss

sr

29.4)25tan2.1648)(3.01(2

620

/2.162

6.06.018

2' 2

rI

mkNB

Dfq

Critical Rigidity Index

41.622

2545cot

2.1

6.045.03.3exp

2

1

2

'45cot45.03.3exp

2

1

)(

)(

crr

crr

I

L

BI

Since 𝐼𝑟(𝑐𝑟) > 𝐼𝑟,

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

cc

c

c

qc

qccc

qcc

rqcc

F

thereforeseeTableNFor

N

FFF

and

xFF

I

L

BFF

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

cdsqcqdqsqcccdcscu FFFBNFFFqNFFFNcq 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

c

qd

qdcd

F

B

DF

N

FFF

From Table 3.3, for ϕ’ = 25°, 𝑁𝑐 = 20.72, 𝑁𝑞 = 10.66, 𝑎𝑛𝑑 𝑁𝛾 = 10.88.

𝑞𝑢 = 48 20.72 1.257 1.343 0.279 + 0.6𝑥18 10.66 1.233 1.311 (0.347) + 0.5 18 0.6 10.88 0.8 1 0.347

𝒒𝒖 = 𝟓𝟒𝟗. 𝟑𝟐 𝒌𝑵/𝒎𝟐

When foundations are

subjected to moments in

addition to the vertical load, the distribution of

pressure on the soil is not

uniform. The nominal

distribution of pressure is,

where Q is the total vertical load and M

is the moment on the foundation.

LB

M

BL

Qq

LB

M

BL

Qq

2

2

6

6

min

max

B

e

BL

Qq

B

e

BL

Qq

Q

Me

LB

M

BL

Qq

LB

M

BL

Qq

61

61

6

6

min

max

min

max

2

2

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 is the ultimate load-carrying capacity

Q

QFS ult

Effective Area Method (Meyerhoff, 1953)

The following is a step-by-step procedure for determining the

ultimate load that the soil can support and the factor of safety against bearing capacity failure:

Step 1: Determine the effective dimensions of the foundation.

B’ = effective width = B – 2e

L’ = effective length = L

Step 2: Use the general bearing capacity equation.

To evaluate the shape factors, use the effective dimensions (B’,

L’) instead of B and L. To determine the depth factors, use B and L.

Step 3: The total ultimate load that the foundation can sustain is

𝑄𝑢𝑙𝑡 = 𝑞𝑢(𝐵′𝑥𝐿′) = 𝑞𝑢𝐴′ (A’ is the effective area)

Step 4: The factor of safety against bearing capacity failure is

𝐹𝑆 =𝑄𝑢𝑙𝑡

𝑄

2. A continuous foundation, supported by sand, has a width of 2 m and the depth of foundation is 1.5 m. The known soil characteristics are as follows: ϕ’ = 40°, c’ = 0, and γ = 16.5 kN/m³. If the load

eccentricity is 0.2 m, determine the ultimate load per unit length of the

foundation. (𝑄𝑢𝑙𝑡 = 5,260 𝑘𝑁)

Note that,

𝑒𝐵 =𝑀𝑦

𝑄𝑢𝑙𝑡 𝑎𝑛𝑑 𝑒𝐿 =

𝑀𝑥

𝑄𝑢𝑙𝑡

𝑄𝑢𝑙𝑡 = 𝑞𝑢𝐴′ = 𝑞𝑢(𝐵′𝑥𝐿′)

In determining the effective are A’, effective width B’, and effective

length L’, five possible cases may arise (Highter and Anders, 1985).

idsqiqdqsqcicdcscu FFFBNFFFqNFFFNcq 2

1'

CASE 1: 𝑒𝐿

𝐿≥

1

6 𝑎𝑛𝑑

𝑒𝐵

𝐵≥

1

6

𝐴′ =1

2(𝐵1)(𝐿1)

𝐵1 = 𝐵 1.5 −3𝑒𝐵

𝐵

𝐿1 = 𝐿 1.5 −3𝑒𝐿

𝐿

The effective length L’ is the larger of

the two dimensions 𝐵1 and 𝐿1. So the effective width is B’=A’/L’.

CASE 2: 𝑒𝐿

𝐿<

1

2 𝑎𝑛𝑑 0 <

𝑒𝐵

𝐵<

1

6

𝐴′ =1

2𝐿1 + 𝐿2 𝐵

𝐵′ =𝐴′

𝐿1 𝑜𝑟 𝐿2 (𝑤ℎ𝑖𝑐ℎ𝑒𝑣𝑒𝑟 𝑖𝑠 𝑙𝑎𝑟𝑔𝑒𝑟)

𝐿′ = 𝐿1𝑜𝑟 𝐿2 (𝑤ℎ𝑖𝑐ℎ𝑒𝑣𝑒𝑟 𝑖𝑠 𝑙𝑎𝑟𝑔𝑒𝑟)

The magnitudes of 𝐿1 𝑎𝑛𝑑 𝐿2 can be

determined from Figure 3.21b.

CASE 2: 𝑒𝐿

𝐿<

1

2 𝑎𝑛𝑑 0 <

𝑒𝐵

𝐵<

1

6

CASE 3: 𝑒𝐿

𝐿<

1

6 𝑎𝑛𝑑 0 <

𝑒𝐵

𝐵<

1

2

𝐴′ =1

2𝐵1 + 𝐵2 𝐿

𝐵′ =𝐴′

𝐿

𝐿′ = 𝐿

The magnitudes of 𝐵1 𝑎𝑛𝑑 𝐵2 can be

determined from Figure 3.22b.

CASE 3: 𝑒𝐿

𝐿<

1

6 𝑎𝑛𝑑 0 <

𝑒𝐵

𝐵<

1

2

CASE 4: 𝑒𝐿

𝐿<

1

6 𝑎𝑛𝑑

𝑒𝐵

𝐵<

1

6

𝐴′ = 𝐿2𝐵 +1

2𝐵 + 𝐵2 (𝐿 − 𝐿2)

𝐵′ =𝐴′

𝐿

𝐿′ = 𝐿

The ratio 𝐵2/𝐵 𝑎𝑛𝑑 𝑡ℎ𝑢𝑠 𝐵2 can be determined by using the 𝑒𝐿/𝐿 curves that slope upward. Similarly, the ratio 𝐿2/𝐿 𝑎𝑛𝑑 𝑡ℎ𝑢𝑠 𝐿2 can be determined by using the 𝑒𝐿/𝐿 curves that slope downward.

CASE 4: 𝑒𝐿

𝐿<

1

6 𝑎𝑛𝑑

𝑒𝐵

𝐵<

1

6

CASE 5: Circular Foundation

𝐿′ =𝐴′

𝐵′

3. A square foundation (1.5 m x 1.5 m), supported by sand, has its bottom 0.7 m below the ground level, with 𝑒𝐿 = 0.3 m and 𝑒𝐵= 0.15 m. The known soil characteristics are as follows: ϕ’ = 30°, c’ = 0, and γ = 18

kN/m³. Assume two-way eccentricity, and determine the ultimate load.

(𝑄𝑢𝑙𝑡 = 606 𝑘𝑁)

4. A square foundation (1.5 m x 1.5 m), supported by sand, has its bottom 0.7 m below the ground level, with 𝑒𝐿 = 0.18 m and 𝑒𝐵= 0.12 m. The known soil characteristics are as follows: ϕ’ = 25°, c’ = 25 kN/m², and

γ = 16.5 kN/m³. Assume two-way eccentricity, and determine the

ultimate load. (𝑄𝑢𝑙𝑡 = 1,670 𝑘𝑁)