11 Bearing Capacity

7/31/2019 11 Bearing Capacity

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CHAPTER 11. BEARING CAPACITY

In fig. 11.1 is shown a stripfooting, which is a shallow foundation supporting a load-bearing wall. When establishing the area A of contact between the foundation andthe soil, two fundamental requirements must be satisfied:

- to ensure safety against the risk of shear failure of the supporting soil (fig. 11.1a),

- to limit the settlement s of the foundation to values allowable for the structureand for its normal exploitation (fig. 11.1 b).

a. b.Fig. 11.1

The problem ofbearingcapacity, this chapter is dealing with, refers to the first ofthe two above outlined requirements.

Bearingcapacityrepresents the ability of a soil to carry a load.The allowablebearingcapacityis defined as the maximumpressure which may be

applied to the soil such that the two fundamental requirements are satisfied.

The ultimatebearingcapacityis defined as the leastpressure which would causeshear failure of the supporting soil immediately below and adjacent to a foundation.

As shown in the chapter 7, the problem ofultimatebearingcapacity is a specialcase of limiting or plastic equilibrium in a soil mass.

In the following paragraphs, the particular problem of the ultimatebearingcapacityof shallow foundations will be considered.

11.1 Failure modes

Present knowledge concerning the way in which failure of the soil supporting shallowfoundations takes place is based on analysis of both causes of accidents in whichvarious structures lost stability and interpretation of experimental data. Theexperiments were conducted, in general, at small scale in installations allowing tovisualize the trajects followed by soil particles during the process of gradual loadinguntil the failure condition was reached.

On that basis, three main modes of failure were recognized, depending, in essence,on the ground conditions.


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a. general shear failure

Continuous failure surfaces develop between the edges of the footing and the groundsurface (fig. 11.2 a). As the pressure is increased towards the value of the ultimatebearing capacity pf, the state of plastic equilibrium is reached initially in the soilaround the edges of the footing then gradually spreads downwards and outwards.Ultimately, the state of plastic equilibrium is fully developed throughout the soil abovethe failure surfaces. Heave of the ground surface occur on both sides of the footing,although the final slip movement would occur only on one side, accompanied bytilting of the footing, as shown in fig. 11.1 a. The load-settlement diagram, whichaccompanies this mode of failure, shown in the diagram a in fig. 11.3, puts intoevidence clearly the values of the ultimate bearing capacity pf for which deformationsincrease indefinitely. The transition from the initial, quasi-linear, part of the diagramand the point corresponding to pf is a short one.

Fig. 11.2

The general shear failure (sometimes named complete shear failure) is typical forsoils of low compressibility (dense sands, stiff clays) and for rocks.

b. local shear failure

In this mode of failure, there is significant compression of the soil under the footingand only partial development of the state of plastic equilibrium. The failure surfacesdo not reach the ground surface and tilting of the foundation is unlikely to occur. The

load-settlement diagram (b in the fig. 11.3) shows that the ultimate bearing capacityis not clearly defined and is characterized by the occurrence of relatively largesettlements. This mode of failure is associated with soils of medium to high


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compressibility, (non-cohesive soils of medium relative density, cohesive soils ofmedium consistency).

c. punching shear failure

This mode of failure occurs when there is compression of the soil under the footing,accompanied by shearing in the vertical direction around the edges of the footing. Asthe pressure is increased, the foundation penetrates into the soil like a piston. Thereis no heave of the ground surface away from the edges and no tilting of the footing.The load-settlement diagram (c in fig. 11.3) shows that large settlements are alsocharacteristics to this mode of failure and the ultimate bearing capacity, like in thecase b, is not well defined. Punching shear failure, is associated with soils of veryhigh compressibility such as loose sands and soft clays.

Fig. 11.3

In cases of local shearand punching shear failures, the ultimate bearing capacityshould be defined based on a deformation criterion. Available experimental datashow that settlements of shallow foundations corresponding to a failure load are ofthe order of (3%...7%) B for clay soils and of (5%...15 %) B for sands where B is thewidth of the foundation. Hence, a settlement of 10% B could be adopted as adeformation criterion for any soil condition in order to define pf (fig. 11.4). It followsalso that plate load tests on compressible soils should be conducted to settlementsequal to at least 0.25 B, to be able to define the ultimate load from the load-settlement diagram.

Fig. 11.4


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Besides the nature of the soil, the mode of failure depends also on other factors suchas:

- the depth of the foundation; punching shear failure will occur in a soil of lowcompressibility, for instance dense sands, if the foundation is located atconsiderable depth (deep foundation);

- the kind of loading; a dense sand subjected to cyclic loading will exhibitpunching shear failure;- the rhythm of loading; a saturated, normally consolidated clay, exhibits a general

shear failure under a sudden loading, when no volume change takes place, andapunching shear failure when the rhythm of applying the load is slow and aftereach load stage the time required for the consolidation of the soil is provided.

11.2 General hypothesis adopted for computing the ultimate bearingcapacity

For the computation of the ultimate bearing capacity pf the following hypothesis areadopted:

- a continuous failure surface characteristic for the general shear failure mode(fig. 11.5);

Fig. 11.5

- the failure condition ctanf += is fulfilled in each point of the failure surface;- the shear strength of the soil between the level of the foundation and the ground

surface (part CD of the failure surface) is neglected;- the friction between the soil above the level of the foundation and the lateralface of the foundation (EB) is neglected;

- the friction between the soil located above and below the foundation level (onthe line BC) is neglected;

- the friction between the base of the foundation (AB) and the soil to which it c.. incontact, is neglected.

With these hypothesis, the soil located above the foundation level is replaced by asurcharge q = D, where D is the foundation depth.


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11.3 Ultimate bearing capacity in the case of a failure surface made bytwo planes

The two failure planes (fig. 11.6) have the inclinations in respect to the horizontal of

)

2

45( o

+ and )

2

45( o

, corresponding to the development in the mass of soil under

the footing of two Rankine zones on both sides of a imaginary, fictitious, perfectlysmooth (frictionless) wall BD, namely the active zone on the left of the wall and thepassive zone on the right of the wall.

Fig. 11.6

Computing pf is based on expressing the active earth thrust Pa behind a vertical wallBD limited by an horizontal ground surface, on which a surcharge pf is applied, andthe passive resistance Pp in front of the same wall, limited by an horizontal groundsurface on which a surcharge q = D is applied (fig. 11.7).

Fig. 11.7

aafa2

a KHc2KHpKH2

1P += (11.1 a)

ppp2

p KHc2KHqKH2

1P ++= (11.1 b)

To find pf, the condition Pa = Pp is written, considering that:


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po KB)

245(tanBH =

+=

pa K

1K =

ppppp

f

p

Kc2KqKH2

1

K

1c2

K

p

K

1H

2

1++=+

[ ] [ ]

)KK(c2Kq)KK(B2

1

KKKc2KqKKKB21

K

Kc2H

2

1KKc2KqKH

2

1p

21

p23

p2p

21

p25

p

ppp2pp

2pp

p

ppp

2p

2pf

+++=

=+++=

=+++=

(11.2)

The expression (11.2) can be put into the form:

++= NB2

1NcNqp cqf (11.3)

where N,N,N cq , named bearing capacity factors, are depending on the angle of

internal friction , and have the following expressions:

)KK(N);KK(2N;KN 21

p25

p21

p25

pc2pq =+==

(11.4)

11.4 Ultimate bearing capacity in the case of a curved failure surface

The problem is solved in three phases, corresponding to the following conditions:

a. cohesionless, weightless soil ( )0;0c;0 ==

b. frictionless, weightless soil ( 0;0;0c == )

c. soil with weight ( 0 )

a. In the case of a soil without cohesion and weight, a suitable failure mechanism fora strip footing is shown in fig. 11.8. The footing, of width B and infinite length, carriesa uniform pressure on the surface of a mass of homogeneous, isotropic soil. Whenthe pressure becomes equal to the ultimate bearing capacity pf the footing will bepushed downwards into the soil mass, producing a state of plastic equilibrium, in theform of an active Rankine zone, below the footing, the angles ABC and BAC being (


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245o

+ ). The downward movement of the wedge ABC forces the adjoining soil

sideways, producing outward lateral forces on both sides of the wedge. PassiveRankine zones ADE and BGF develop on both sides of the wedge ABC, the angles

DEA and GFB being (2

45o

). The transition between the downward movement of

the wedge ABC and the lateral movement of the wedges ADE and BGF takes placethrough zones of radial shear ACD and BCG. In his solution, Prandtl admits that thesurfaces CD and CG are logarithmic spirals, to which BC and ED, or AC and FG, are

tangential. The equation of the spiral is = tanoerr where is the angle between

the initial radius ro and the one corresponding to a point on the spiral; is the angle

made by the radius with the normal in any point of the spiral. A state of plasticequilibrium exists above the surface EDCGF, the remainder of the soil mass being ina state of elastic equilibrium.

Fig. 11.8

To find pf, first the equilibrium of the wedges ABC and BDE, as equilibrium of forceson vertical direction, will be considered. Then, the equilibrium of the transition zoneBCD, as equilibrium of moments toward the point B, will be written.

On the conjugated failure planes AC and CB are acting the reactions RI, making an

angle with the normal (fig. 11.9 a).

The equation of projection of forces on the vertical direction:

)2

45(cosR2ABp oIf

=

)2

45(cosr2AB oo

+=

)2

45(cosR2)2

45(cosr2p oIo

of

=

+


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)2

45(tanrp

)2

45(cos

)2

45(cos

rpRo

ofo

o

ofI

=

+

+

= (11.5)

On the conjugated failure planes BD and DE are acting the reactions RIII, making anangle with the normal (fig. 11.9 b).

The equation of projection of forces on the vertical direction:

)2

45(cosR2)2

45(cosr2q

)2

45(cosr2BE

)2

45(cosR2BEq

oIII

o1

o1

oIII

+=

=

+=

)2

45(tanrq

)2

45(cos

)2

45(cos

rqR o1o

o

1III

+=

+

= (11.6)

The equilibrium of the transition zone II (fig. 11.9 c) is expressed in terms of themoment around the point B.

Fig. 11.9

The arc of the spiral CD belongs to the failure surface, therefore the reaction RIImakes an angle with the normal to the arc. Hence, the direction of RII coincideswith the direction of the radius and RII produces no moment in respect to B. Themoment equation becomes:


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)2

45(tanrq)2

45(tanrp

2

rcosR

2

rcosR

o21

o2of

1III

oI

+=

=

But r1 = ro tan

2e

)2

45(tanerq)2

45(tanrp otan2oo2

of

+=

)2

45(taneqp o2tanf

+= (11.7)

By writing: qo2tan N)2

45(tane =

+

equation (11.7) becomes:

pf= q Nq (11.8)

From (11.8) follows that, in the case of a cohesionless and weightless material, thereis a bearing capacity only if there is a surcharge q.

To consider the effect of the cohesion, a normal stress equal to c cot isadded to the normal stresses p and q. The equation (11.8) becomes:

qfN)cotcq(cotcp +=+

)1N(cotcNqp qqf+=

(11.9)

By writing cq N)1N(cot =

equation (11.9) becomes:

pf= q Nq + c Nc

Nq and Nc are bearing capacity factors depending on .

An additional term should be added to equation (11.10) to take into account the self-

weight of the soil. Experimental observations showed that a wedge of soil remainingin elastic state, with faces making an angle with the horizontal, is developedbelow the foundation and moves downwards together the foundation, tending to


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produce the lateral movement of the soil along the failure surfaces CDE and CFG(fig. 11.10). The passive resistance of the soil mass above the failure surfaces ismobilized. The problem consists on computing the passive resistance force Pp of a

mass of soil ( 0,0 ), limited by a horizontal ground surface, behind a wall BC

with inclination and height H = tan

2

B.

Fig. 11.10

The failure surface CDE is made of the line DE, corresponding to the passiveRankine zone BDE, and by the arc of logarithmic spiral CD.

The passive resistance force Pp can be expressed:

===22

p2

2

p2

p tanB8

1Ktan

4

B

2

1KH

2

1P (11.11)

The equilibrium of the elastic wedge ABC:

)(cosP2Q p = (11.12)

The ultimate bearing capacity is:

==

=

===

BN2

1)(costanB

4

1

B8

)(costanB2)(cos

B

P2

B

Qp

2

22p

f

(11.13)

The following notation was used:

)(costan2

1N

2 =

Terzaghi assumed that = and obtained the value of the passive resistance forcein the hypothesis of a curved failure surface.


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Adding the additional term bringing the effect of the self-weight of the soil, theexpression of the ultimate bearing capacity pfbecomes:

++= NB2

1NcNqp cqf (11.14)

Relations of the kind of (11.14) were established by Terzaghi and other authors. Mostof them differ only with respect of the third component, introducing the influence ofthe self-weight of the soil. These relations are theoretically incorrect for a plasticmaterial since they are superposing terms corresponding to different failure figuressuch as those represented in fig. 11.8 and 11.10. However, the error implied isconsidered to be on the safe side and is accepted in engineering practice.

11.4 Ultimate bearing capacity in the case of a purely cohesive soil

This is a particular case of the problem previously considered. The failure mechanism

shown in fig. 11.8 is transformed, when 0= , in the one shown in fig. 11.11.

Equation (11.10) becomes:

DNcNDNcp cqcf+=+= (11.15)

(For 0= , Nq = 1)

Fig. 11.11

One defines as netto ultimate bearing capacity the difference between the criticalpressure in the geological pressure at the level of the foundation base:

cfnettofNcDpp == (11.16)

The problem is to find the bearing capacity factor Nc for this case ( )0,0,0c == .

An approach similar to the one used for the case ( )0,0c,0 == is adopted:

Equilibrium of forces acting on the prism I (fig. 11.12 a)


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Bc2R2

2

2

2

2

B

c22

2R2BNc IIc +=+=

)1N(Bc2

1R cI = (11.17)

The normal stress acting on the faces AC and BC:

)1N(c

2

2

2

B2

)1N(Bc

2

2

2

B

Rp c

c

II

=

==

(11.18)

Equilibrium of forces acting on the prism III (fig. 11.12 b)

2

2

2

2

2

B

c22

2R2 III =

2

BcRIII = (11.19)

Fig. 11.12

The normal stress acting on the faces BD and DE:

c

2

2

2

B2

Bc

pIII =

= (11.20)


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pf is obtained by writing the condition that the moment of all forces acting onthe failure prism, in respect to the point B, is zero. Normal pressures acting onthe circular are CD having the direction of the radius, do not give momenttoward B.

2

DEDEc

2

ABABN2

2

ACAC)1N(cBCACcRDBD

2cDBEDc

c

c

+=

=++

+

(11.21)

But AC = BC = BD = DE = r

2r2

2r2AB ==

Relation (11.21) becomes:

2

r

r

22rN

2

r)1N(r

2r

2

c

2

c22 +

=+

+ ;

2

N1

22

2

1N

2

1N

22

2

rrN

2

r)1N(r

2r2

c

cc

22

c

2

c22

=

+

+=

+

+

+=+

+

14.52Nc =+= (11.22)

c14.5p

netto

f= (11.23)

Dc14.5pf

+= (11.24)

Skempton has shown that, in fact, the netto ultimate bearing capacityincreases withthe depth D of the foundation until a depth D = 5B (fig. 11.13), reaching a limit value9 for Nc.


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Fig. 11.13

For rectangular foundations B x L, for which 5.2B

D , Skempton proposed the

relation:

)

L

B2.01()

B

D2.01(c5p

nettof

++= (11.25)

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11 Bearing Capacity