chap 7 shallow static bearing strength

7/30/2019 Chap 7 Shallow Static Bearing Strength

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77SHALLOW FOUNDATION

STATIC BEARINGSTRENGTH

7.1 ULTIMATE LIMIT STATE METHODS FOR SHALLOW FOUNDATIONS

There are three approaches to ultimate limit state design. Two of them, LRFDand Partial Factor methods, have already been discussed in chapter 6. Forshallow foundation design the total factor of safety approach has now mostlybeen superseded, herein it is needed for comparative purposes.

The equations used by the three approaches:

Tot a l fac to r of sa fety:

i

RF

FoS (7.1)

where: R is the ideal1 resistance (or strength).FoS is the total factor of safety.Fi are the applied loads.

1This resistance is commonly referred to in several ways. Among them are theoreticalresistance, nominal resistance as well as ideal resistance.


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Loa d a nd resista nc e fa c to r m eth od (LRFD):

i iR F (6.3)

Pa rt ia l fac tor me thod:

i iF R (6.4)

In the context of this chapter the ideal strengths, R and R , are the bearingstrength of the shallow foundation.

7.2 SHALLOW FOUNDATION BEARING STRENGTH EQUATION

7.2.1 Terzaghi bearing capacity equation

The usual equation for the bearing capacity of a strip foundation at the groundsurface is:

c qu

1= c + q + Bq N N N

2gg (7.2)

where: Nc, Nq, & N are bearing capacity factors (functions of),qu is the ultimate bearing pressure,q is the surcharge pressure adjacent to the foundation,

is the unit weight of the soil,B is the width of the foundation.

The form of the above equation is due to Terzaghi, hence the title of thissection. Terzaghi (1943) made the assumption that the three separate bearingcapacity components can simply be added as expressed in equation (7.2). Theequation is a general one in the sense that it is based on a simple linear shear

strength relationship with f = c + ntan. It needs to be written in formsspecific to short term total stress analysis or long term effective stress analysis.

These two common design situations are handled as special cases as follows:

(i) Short term, total stress bearing capacity analysis ( = 0)For this case equation (7.2) reduces to (as Nq= 1 for = 0):

u u cq =s N +q (7.3)

(ii) Long term, effective stress bearing capacity analysis

For this case equation (7.2) is written as:

u c v q

1q = c N + N + BN

2 (7.4)

where: is the appropriate unit weight of the soil to give effective stressesbeneath the foundation,


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Chapter 7: Shallow foundation static bearing strength

179

)2

+(45tane=N2tan

q

0=for5.14but0,>forcot1)-N(=N qc

tan1)N2(=N q

10 20 30 40

Angle of shearing resistance (degrees)

1

10

100

1000

2

3

4

5

678

2

3

4

5

678

2

3

4

5

678

Bearingcapacityfactors

Nc

Nq

N

N

For = 0: Nc = 5.14, Nq = 1, N

Figure 7.1Bea ring stren g th fac tors.

and v is the vertical effective stress in the soil adjacent to the foundation ata depth corresponding to the underside of the foundation.

We need to look at equation (7.2) carefully; it is written with the expectation

that the bearing capacity is derived from effective stresses, so c appears in thefirst term, and is used to evaluate Nc, Nq, and N, as shown in Figure 7.1.Note that q must also be in terms of effective stress, hence the notation vforthis term in equation (7.4). Similarly with the N term, we substitute for ifthe water table is at or above the base of the footing and if it is deeper thanthe zone involved in the bearing capacity failure mechanism (say deeper thantwice the foundation width).


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The usual way equation(7.2) is presented is in terms of an ultimate bearingpressure. However, equations (6.3) and (6.4) are expressed in terms of actionsrather than bearing pressure. Consequently, in ultimate limit state approaches itis the bearing strength (units of force) that is referred to rather than bearing

capacity (units of pressure). The ultimate bearing strength is then the ultimatebearing pressure times the effective area of contact between the underside ofthe foundation and the soil.

The above equations are for footings subject only to vertical load. Consider,now, a footing subject to vertical load as well as moment. The underside of thefooting can provide compressive normal stress only. As the footing is notattached to the material below, the only way moment equilibrium can besatisfied is by offsetting the centroid of the pressure distribution block so that acouple is generated. This is a standard operation in elementary statics, theoffset distance for the location of the centroid of the pressure block, the

eccentricity, usually denoted with the letter e, is:

e = M/V

where: M is the moment applied to the footingand V is the vertical load.

If moments are applied about more than one axis then there are eccentricitiesin different directions. A complex loading situation can be reduced tocomponent moments about axes in two orthogonal directions, thus:

x y y xe = M /V and e = M /V (7.5)

where: ex is the eccentricity in the direction of the x axisand Mx is the moment applied about the x axis, similarly for eyand My.

Equation (7.5) locates the centroid of the contact pressure distribution on theunderside of the foundation. Consider the case of a footing of length L andbreadth B, subject to moments Mx and My. The origin of coordinates is at thecentre of the footing, thus the coordinates of the resultant vertical load are exand ey. The distances from the centroid of the block to the nearest edges of thefooting are: L/2 - ey and B/2 - ex. The dimensions of the stress block are then:

y x x y L = L - 2e and B = B - 2e (if e and e are such

that B > L then interchange L and B ) (7.6)

7.2.2 General bearing capacity equation

The above paragraphs describe the modification made to the original bearingcapacity equation to handle moment applied to a strip foundation. In additionthere are four further effects that need to be considered. These handle theshape of the foundation (that is rectangular foundations are common whereas

strip foundations are not), the depth to the underside of the foundation (that isthe shear resistance of the soil adjacent to and above the foundation may also


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contribute to the bearing strength), the effect horizontal shear being applied tothe foundation (the resultant to vertical and horizontal forces is inclined, hencethe term inclined load factors); and fourth the underside of the foundation maynot be horizontal so we need another factor to handle base inclination. The

general ultimate bearing capacity equation now becomes:1

2cs cd ci c qs qd qi q s d ic qu = c + q + Bq N N Ng g g ga a gal l g ll l l l l l l l l (7.7)

where: cs, qs, and s are shape factors;cd, qd, and d are depth factors;ci, qi and i are inclined load factors,

and c, q, and are base inclination factors.

A suite of equations for the factors is given in Appendix I. Notice that theeffective footing dimensions, LandB are used in all these factors and thatthe cohesion terms are associated with the effective area A, that is LB. Theshape, inclined load and base inclination factors given in Appendix I are thesame as or similar to those in EuroCode EC7 (2001). Depth factors areincluded here although EC7 eschews these. Other sources of equations forthese factors are Vesic (1975) and Brinch Hansen (1970).

The reader may wonder about the origin of these factors. Mostly they areempirical, they certainly do not have the theoretical basis underlying the

bearing capacity factors Nc, Nq and N (although, in saying this, we are notdenying that there are many different theoretical analyses for these parameters,

particularly for N, which give widely differing values), yet they are thought tobe qualitatively reasonable. In addition model testing and back analysis ofobserved foundation failure provide insight into these values.

Equations (7.2) and (7.7) and the associated equations in Appendix I are thestandard approach to estimating the static bearing strength of shallowfoundations. This approach was originally developed for static vertical loadingonly. One limitation of the equation is that it does not provide much insightinto how the various actions, vertical load, horizontal shear and momentinteract to affect the bearing strength; that is the equations work as a kind ofblack box. Deeper insight is obtained by considering the bearing strength

surface, which represents all the combinations of vertical load, horizontal shearand moment that induce bearing failure. This is done below, but prior to thatthe idea of a compensated foundation needs to be considered with respect tothe various ultimate limit state approaches.

7.2.3 Compensated foundations

In total factor of safety design the definition of the bearing capacity factor ofsafety and the idea of a compensated shallow foundation are closely relatedideas. The definition of bearing capacity factor of safety is in terms of netbearing pressures.


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Consider the construction and subsequent loading of a foundation at somedepth beneath the ground surface. Firstly the excavation has to be made forthe placing of the footing. The digging of this means that there is a reductionin the vertical stress, for example at the depth of the base of the footing the

vertical stress before any excavation is Df and at the end of the diggingprocess it is zero. When the footing is loaded there is a certain vertical load atwhich the contact stress between the footing and the underlying soil is again

equal to Df. At this stage the soil beneath the foundation does not know theorigin of the vertical stress; does it come from foundation loading or fromoverlying soil? In other words only when the foundation loading applies

vertical stresses above the in situ vertical stress is the material beneath thefoundation being loaded towards failure. Thus the stress that contributes tofailure of the soil beneath the footing is that in excess of the in situ vertical

stress Df.

More generally we need to distinguish the total stress ( = 0) and effectivestress cases. In terms of total stresses q = Df, for the effective stress case q =v= Df- u(z = Df).

Thus we define the following terms:

Gross bearing pressure = Vertical load carried by the footing/effectivefooting area

Net bearing pressure = gross bearing pressure - vertical stress in the soiladjacent to the foundation at the depth of the underside of the footing.

Using these definitions it is apparent that the bearing pressure given byequations 7.2 and 7.7 is the gross ultimate bearing pressure. The net ultimatebearing pressure is obtained by replacing the second term in the equation (7.2)

with: q(Nq - 1). With the above two definitions in mind it is apparent that asuitable definition of the factor of safety against bearing capacity failure is (thedefinition needs to be interpreted in terms of effective or total stresses asappropriate):

bearing capacity

net ultimate bearing pressureF =

net applied bearing pressure

(7.8)

An interesting question to ask is what is the value for the factor of safety given

by the above definition when v = Df? At this point the net applied bearingpressure is zero and so the applied bearing pressure makes no demand on theshear strength of the material beneath the foundation. It is here that theconcept of a compensated foundation arises. If we desire to construct ashallow foundation on soft ground then the excavation of a basement reducesthe load transferred to the underlying soil. The only requirement is that theexcavated material is replaced with material substantially lighter than the soilremoved. A building with beneath ground car parking fills this requirement.

So to does a building with a so-called cellular foundation that is not put to anypost-construction use other than remaining filled with air. Generally the above


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ground parts of buildings are mostly empty. The total of dead load and liveload is typically about 8 kN/m2 per floor. The excavated material for onebasement level 3 m deep is equivalent to a vertical load of about 3*18 8 = 46kN/m2. Thus one level of basement is equivalent to several above ground

floors; it is only floors above this number that call on the shear strength of thesoil to contribute to the bearing resistance.

The LRFD and partial factor approaches, as discussed in chapter 6, do not usethe concept of net bearing pressure and as such do not take advantage of thecompensated foundation idea. The following paragraphs show that all threeultimate limit state approaches to shallow foundation design are consistent ifthe calculations are done in terms of net actions.

Consider first the short term capacity of a foundation on saturated clay.Writing and manipulating the definition of the bearing capacity for the total

factor of safety in terms of net bearing pressures gives:

:

u u c

a a

u ca

q q s NFoS

q q q q

from which

s Nq q

FoS

(7.9)

where: qa is the gross applied bearing pressure.

Now performing similar manipulations using the LRFD equation, but using

net actions (net bearing pressure times foundation area) we have:

:

u a

u

u c u

a u c

q q Area q q Area

which gives

q q q q a

Substituting s N qforq gives:

q s N q

(7.10)

Finally, the partial factor approach and a similar argument leads to the sameconclusion:

:

a

a

uu c u

pf

a u c a u cpf

q Area q q Area

which gives

1q q q

sSubstituting s N q for where s is gives:

1 1q s N q q s N q

R

R

R

(7.11)


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184

Comparing the last equations in the above three blocks reveals that, for the

short term undrained case when = 0, when the calculations are done in termsof net actions we achieve the same understanding of compensation for all threeapproaches.

The above argument has been expressed for undrained conditions only. Forthe drained case the difference between net and gross pressures is much

smaller than for the undrained situation, because for = 0 Nq=1, so Nq 1 =0, whereas for > 0, say 30 degrees, Nq= 19 and Nq 1 18.

7.3 EC7 BEARING STRENGTH SURFACE

The bearing strength theory gives us a way of estimating what combinations ofvertical load, horizontal shear, and moment mobilise all the available shear

strength of the soil underlying and surrounding a shallow foundation. The sumtotal of these combinations forms a three dimensional bearing strength surface.In this section we examine the features of this surface. There are two surfacesto consider: that for the undrained case and that for the drained case, generatedusing appropriate rearrangements of equation (7.7).

A convenient of way of presenting the surfaces is to use axes defined in termsof dimensionless parameters, one for vertical load, another for horizontalshear, and a third for moment applied to the foundation. Starting with the

vertical load, this is normalised with respect to the ultimate strength of thefoundation subject to vertical load only, the suite of dimensionless parameters

is then:

, ,uo uo uo

V H MV H M

V V V B (7.12)

where: B is the width of the foundation,Vuo is the ultimate vertical that may be applied to the foundation in the

absence of shear and moment loading evaluated with theexpressions given in Appendix II,

V, H and M are a combination of actions that induce an ultimate limitstate, ie the coordinates of a point on the bearing strength surface,

V , H and M are the normalized foundation actions.

7.3.1 Dimensionless bearing strength surface for strip

foundations at the ground surface

For a strip foundation at the surface of a clay layer, subject to vertical load,moment and shear, the undrained static bearing strength surface derived fromthe equation (7.7) is given by:

22

undrained c

2 M 2 M 2 M7 V, H, M 1 2V 1 N H 1 0

V V V

f

(7.13)


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This equation is developed in Appendix III. A view of the surface is given in

Figure 7.2. The2

1

M

Vterms in the above and following equations are

because moment acts about the longitudinal axis and the width B contributesto the bearing strength.

For a strip foundation at the surface of a layer of cohesionless soil, subject tovertical load, moment and shear, the drained static bearing strength surfacederived from the equation (7.7) is given by:

2 3

27 1 1 0drained

M HV, H, M V

V V

f (7.14)

This equation is developed in Appendix III. A view of the surface is given inFigure 7.3.

Figures 7.2 and 7.3 have plots of the upper halves of the undrained anddrained bearing strength surfaces. The diagrams are drawn to the same scale toaid comparison and to indicate that the drained surface is so much skinnierthan the undrained surface. Note that the vertical load is restricted to positive

values whereas the moment and shear can be positive or negative. Note alsothat for very small vertical loads or vertical loads near Vuo only small values forshear and moment are possible, whereas the largest possible values of shear

and moment are possible when V is about 0.5.

Figure 7.4 compares the H-V and M V sections of the surfaces. These showthat in terms of dimensionless parameters the shallow foundation on clay hasgreater shear and moment capacity than a foundation on sand. However, whenthe dimensionless parameter values are converted to actual actions this may notbe the case as Vuo for the foundation on sand maybe greater than Vuo for thefoundation on clay.

7.3.3 Effect of foundation shape and embedment on the

bearing strength surface

The bearing strength expressions presented in EC7 are for strip foundations.Clearly, some adjustment needs to be made for rectangular foundations. These

will have greater bearing strength because of the shearing of the soil adjacent tothe ends. The informative appendix in EC7 includes shape factors.

Depth factors are also standard additions to bearing strength calculations. It sohappens that EC7 does not include these, but they have been included in

Appendix I herein. These can also be applied to the bearing strengthexpressions given above. In the case of the EC7 expressions moment loading

is accommodated by using a reduced area defined by B and L.


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0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

-0.2-0.1

00.1

0.2

0

0.1

V

/Vuo

H/Vuo

M/BVuo

Vuo: Bearing strength under

.......vertical load only

B: ... Foundation width

Figure 7.2EC7 und rained be ar ing strength surfac e for a surfac e strip found at ion.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

-0.2-0.1

00.1

0.2

0

0.1

V/V

uo

H/Vuo

M/BVuo

Vuo

: Bearing strength under

.......vertical load only

B: ... Foundation width

Figure 7.3EC7 d rained be ar ing strength surfac e for a surfac e strip found at ion.


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187

Figure 7.4 Co m pa rison of the H V, (a), a nd M V, (b), sec tions of the EC7

be ar ing streng th sur fac es for c lay (und rained ) and sand (d rained) .

The shape and depth factors are applied to Vuo. In this way they will scale thebearing strength surface, so increasing the actual values of all components ofbearing strength.

In the above equation the normalisation parameter is the bearing strength perunit length of a strip of width B. The drained bearing strength of a rectangular

foundation in the absence of shear or moment is:

( , ) .uo H 0 M 0 uoB

V 1 0 3 VL

(7.15)

The -0.3 factor in the above indicates that s reduces and the foundationmoves from a strip to square. This is different from all the other factors

which increase as the foundation tends towards a square shape. The differencebetween triaxial and plane strain angle of shearing resistance of sand, beingabout 10% greater for plane strain shearing than axial symmetry, might be a

possible explanation. N for 35 degrees is about 42 and for 32 degrees it is

about 28 (ie roughly 70% of 42). If we assume that for a strip foundation planestrain conditions prevail and for a square foundation that axial symmetry is areasonable approximation, these differences provide, perhaps, a plausible

explanation for the -0.3 factor in s.

Figures 7.5 and 7.6 provide comparative views of M = 0 and H = 0 sections ofthe two EC7 bearing strength surfaces for strip foundations at the groundsurface, for square foundations at the ground surface, and for squarefoundations embedded such that (Df/B = 0.2).

0 0.2 0.4 0.6 0.8 10

0.1

0.2

V/Vuo

H/Vuo Clay

Sand

a

0 0.2 0.4 0.6 0.8 10

0.1

0.2

V/Vuo

M/Vuo

B

Sand

Clay

b


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0 0.5 10

0.1

0.2

V/Vuo

H

/Vuo

Square, surface

Strip, surface

aSquare, embedded

0 0.5 10

0.1

0.2

V/Vuo

M/Vuo

B

Strip, surface

Square, surfaceb

Square, embedded

Figure 7.5 Com pa rison of the H V, (a), a nd M V, (b), sec tions of the EC7 b ea ring

st reng th surfac e for sha l low found at ions em be dd ed in c lay (Df/ B = 0.2) .

0 0.5 10

0.1

0.2

V/Vuo

H/Vuo

Strip, surface

Square, surface

a

Square, embedded

0 0.5 10

0.1

0.2

V/Vuo

M/Vuo

B

Square, s urface

Strip, surface

b

Square, embedded

Figure7.6 Co m pa rison of the H V, (a), a nd M V, (b), sec tions of the EC7 b ea ringstreng th sur fac es for sha l low founda t ions em be dd ed in sa nd (Df/ B = 0.2) .


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0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0

0.1

0

0.1

V/Vuo

H/Vuo

M

/BVuo

a

b

FoS = b/a

Current state

tan -1(M/HB)

Figure 7.7 In terpretat ion of the be ar ing streng th fa c tor of safety in term s ofthe b ea ring strength sur fac e.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0

0.1

0

0.1

V/Vuo

H/Vuo

M/BVuo

a

b

FoS = b/a

What is the relation between the FoS

and the reserve of bearing strength

along these load paths?

Figure 7.8 Loa d pa ths hav ing reserves of be ar ing streng th di f ferent from that

from whic h the fac tor of safety is de termined .

7.4 ULTIMATE LIMIT STATE CALCULATIONS AND THE BEARING

STRENGTH SURFACE

Equation (7.8) defines the bearing capacity factor of safety (total factor ofsafety) in terms of net pressures. It is instructive to seek an interpretation ofthis factor in terms of the bearing strength surface. As shown in Figure 7.7, thedefinition in equation (7.8) is found to have a very specific and limited meaningassociated with only one point on the surface. The factor of safety is intendedto give an indication of the reserve of strength available, that is by how much


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190

the applied actions would need to be increased to generate limiting equilibrium.However, in Figure 7.7 it is seen that the definition of the factor of safety is

with reference to only one loading path. No insight is given for the myriad ofother load, or state, paths that are possible from the current point shown in

Figure 7.7. This is particularly important when it is possible that increases inactions other than vertical load might occur as shown in Figure 7.8; for thesepaths the definition gives little insight into the reserve of bearing strengthavailable, a point made first by Butterfield and Gottardi (1994). Under staticconditions the state point will, typically, be somewhere in the range V/Vuo 0.2to 0.3, so a large increase in vertical load would be needed to induce a bearingstrength failure, but a much smaller increase in horizontal shear or moment. Inthis sense the bearing strength factor of safety defined by equation (7.8) ismisleading and may engender a false notion of security. When verticalfoundation loading is accompanied by shear and moment loading, this is apowerful argument for the abandonment of the traditional bearing capacity

factor of safety concept.

In terms of Figures 7.7 and 7.8, how are we to understand the LRFD approachto ultimate limit state design of a shallow foundation. Figure 7.9 shows howthe LRFD inequality works for the case where only vertical load is applied tothe foundation. If the strength reduction factor has a value of 0.5, then theLRFD inequality demands that the factored vertical load must remain less than0.5Vuo.

Things are more complicated if the foundation is subject to moment and shear.Then the critical factor may be where the moment and shear lie in relation to

the bearing strength surface. In the case where the vertical load remainsconstant and it is the moment and shear that are "driving" the bearingbehaviour of the footing so a constant vertical load section of the bearingstrength surface, as shown in Figure 7.10, needs to be considered where theLRFD inequality is satisfied inside a smaller version of the bearing strengthsurface.

Figure 7.10 shows how the LRFD inequality works for a shallow foundationunder constant vertical load. What happens when the vertical load is notconstant? This case is handled using the bearing strength surface discussedbelow in section 7.5. The LRFD inequality then operates as shown in Figure

7.11. Note that permissible values for the vertical load are restricted to V/Vuo.

7.5 BEARING STRENGTH SURFACE INCLUDING SHAPE AND DEPTH

FACTORS

Equation 7.13 for the undrained clay case, when extended to include shape anddepth factors, becomes:

, , c2undrainedH N

7 V H M 2 02 M

1 V

f

7.16


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compressive

vertical load

Vu

o-

bearing

str

ength

und

er

vertic

al

load

only

Vu

o

V-

facto

red

vertic

al load

V-

applied

vertic

al

load

0

LRFD is satisfied as long as

Vuo is greater than V

tensile

vertic

al

load

not

possible

,

found

ationuplifts

Figure 7.9 LRFD bearing strength evaluation for a foundation subject to

ver tica l loa d o n ly .

0 0.05 0.1 0.150

0.05

0.1

0.15

Normalised shear

Normalisedmoment

BSS

reduced BSS

Figure 7.10Co nstant ver t ic a l loa d sec t ion o f the b ea ring strength surfac e

(V /Vuo= 0.3).


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0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0

0.1

0

0.1

V/Vuo

H/Vuo

M/BVuo

a b

Satisfactory if b _> a

Current state

State pointdefined by

factored actions

Reduced bearing strength surface because

of (LRFD) or su/partial factor ( fsu)

Figure 7.11 Interpretat ion of the LRFD inequality for shallow foundation

be ar ing streng th.

where:

..

2 V

2 M 2 M 0 41 1 0 2 1 1V V 2 M

1V

and:

is B/L is Df/B is (1+0.2B/L)(1+0.4Df/B)

The example below shows two ways of doing an LRFD evaluation of a shallow

foundation subject to constant vertical load. The first inserts the various

factors into equation 7.7 and finds where the moment and shear values on theBSS. The second uses equation 7.16 to find where the LRFD inequality is justsatisfied.

Example 7.1

Consider a shallow foundation 3 m wide and 3 m long resting on the surface ofsaturated clay with an undrained shear strength of 50 kPa. The underside of thefoundation is 1 m beneath the ground surface. Find the moment and shearforce that will induce bearing failure when V/Vuo = 0.3 and H = 0.33M/B.


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File: Undrained bearing strength surface and function Example 8-1 16/08/2012

ORIGIN 1

su 50

B 3

L 3

Df 1

Vuo 5.14 su 1 0.2B

L

1 0.4Df

B

B L Vuo 3.146 10

3 (kN) (Net value)

Let's assume that the vertical load on the foundation is constant a Vuo/3 and the ratio of

moment to shear is 1.0 and that the moment is applied about the longitudinal axis of the

foundation.

Vn 0.3 V Vuo Vn V 943.704

So we need to find what values of shear and moment will induce bearing failure for that vertical

load.

Working in terms of net actions we need to solve for the moment:

Mn 150 Start value for the root finder. Assume Hn = 0.33Mn

f Mn( ) B 2Mn

V A function to reduce the spread of the equation for solving.

Mu root su 5.14 1 0.2f Mn( )

L

10.4 Df

f Mn( )

1

21 1

0.33 Mn

f Mn( ) L su

L f Mn( ) V Mn

Mu 687.333

Now express these in terms of dimensionless variables:

MnMu

Vuo B Mn 0.073 Hn

0.33Mu

Vuo

Hn 0.072

Bearing strength sur face for a rectangular foundation at a depth Df beneath the ground

surface

1.0 B/L ratio 0.33 Df/B ratio

1 0.2 (1+ 0.2B/L) 1 0.4 1 + 0.4*Df/B

1.358 VuoD 5.14 su 1 0.2 ( ) 1 0.4 ( ) B B

First calculate the longitudinal boundary curves of the BSS (that is Hn when Mn is zero and Mn

when Hn is zero).

i 1 11


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Set this to a small postive value as zero gives divide by

zero problems.Vn

ii 1( ) 0.1 Vn

10.001

HnDi

4 Vni

1 Vni

5.14

0 0.2 0.4 0.6 0.8 10

0.05

0.1

0.15

HnD

Vn

Mn1 0.050

MnDi

root Vni

1 0.2 12 Mn1

Vni

1

2 Mn1

Vni

0.4

Mn1

0 0.2 0.4 0.6 0.80

0.05

0.1

MnD

Vn

Now plot some cross-sections:

MM 0.11 Trial value for the root finder

Define a function that enables a more compact solving of equation 7.16:

f1 MM VNN( )2 VNN

1 2MM

VNN

1 0.2 1 2MM

VNN

10.4

1 2MM

VNN

Cross-section with Vn = 0.3:

HnD4i

i 1( )HnD

4

10 HnD4

10.001 Vn

40.3

MnD4i

root f1 MM Vn4

2

2 f1 MM Vn4

HnD4

i 5.14

1 2MM

Vn4

MM


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0 0.05 0.1 0.150

0.05

0.1

0.15

MnD4

HnD4

Now use the undrained bearing strength equation to check on LRFD:

Set-up the function representing the BSS:

f1 Vb Mb Hb( )2 Vb

12 Mb

Vb

1 0.2 12 Mb

Vb

10.4

1 2 MbVb

f2 Vb Mb Hb( ) Hb 5.14

12 Mb

Vb

Now equation 7.16:

fBSS Vb Mb Hb( ) f1 Vb Mb Hb( )2

2 f1 Vb Mb Hb( ) f2 Vb Mb Hb( )

fBSS 0.3 MnD4i

HnD4i

-112.54210

-131.68510

-131.32910

-106.910

-127.84110

-93.5610

-114.04310

-132.510

-104.8610

-113.99510

-114.99910

Thus BSS function takes a value of zero on the BSS.


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0.5 Strength reduction factor for bearing

MnD4 MnD4 HnD4 HnD4 Line0

0.073 1.3

0

0.072 1.3

0 0.05 0.1 0.150

0.05

0.1

0.15

MnD4

MnD4

Line1

HnD4 HnD4 Line2

Note that 0.073 and 0.072 were the pair of values obtained in the first part of this example.

fBSS 0.3 0.073 0.072( ) 4.543 104

fBSS 0.3 0.073 0.99 0.072 0.99( ) 0.021

fBSS 0.3 0.073 1.01 0.072 1.01( ) 0.023

The above calculations show that for combinations of shear and moment dead the BSS, then the

function has a value of zero; negative for points inside, and positive for points outside.

What we have in the above diagram is the section of the BSS for Vn = 0.3 (red). We also have a

reduced section for a strength reduction factor of 0.5 (blue).

The next question is how do we apply the LRFD relation to the BSS equation? What we need to do

is decide whether the factored actions, moment and shear, are on or inside the the blue curve in the

above diagram. The 0.071 and 0.079 values lie on the BSS. The values 0.071 and 0.079 lie on the

blue curve above.

Sometimes in LRFD calculations we know what value is required to satisfy the inequality. Then

we can take that value and divide by to find the required theoretical strength of the member or

system. So what we do is take the value(s) that we want to check against the LRFD inequality,

divide these values by , and so get the theoretical strength, we can use the BSS function to

see where the theoretical strengths lie in relation to the BSS.

Let's say our test values are 0.0365 and 0.0360, then:

This means that when divided by (that is doubling

them) the theoretical strengths lie on the BSS, so the

LRFD relation is just satisfied.fBSS 0.3

0.0365

0.036

4.543 104


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194

REFERENCES

Butterfield, R and Gottardi, G (1994) A complete three-dimensional failureenvelope for shallow footings on sand. Geotechnique: 44(1): 181-184.

Brinch Hansen J. (1970) "A revised and extended formula for bearingcapacity", Bulletin No. 28, Danish Geotechnical Institute, Copenhagen,pp. 5-11.

EC7 Drafting Committee (2001) Eurocode 7, part 1: Geotechnical Design,General Rules. Comit Europen de Normalisation, Final Draft, October2001.

Frank, R, Baudin, C, Driscoll, R, Kavvadas, M, Krebs Ovesen, N, Orr, T andSchuppener, B. (2004) Eurocode 7: Geotechnical design General Rules.

Thomas Telford, London.Gourvenec, S. (2004) Bearing capacity under combined loading a study of

the effect of shear strength heterogeneity. Proc. 9th Australia-New

Zealand conference on Geomechanics, Auckland. (2): 527-533.Meyerhof, G. G. (1993) "Development of Geotechnical Limit state design",

Keynote Address: International Symposium on Limit State Design inGeotechnical Engineering, Copenhagen. Danish Geotechnical Society,

Vol. 1, pp. 1-12.NZS 1170.5 (2004) Structural design actions Part 5: Earthquake Design

Actions New Zealand. Standards New Zealand, Wellington.Orr, T. L. L (1993) "Partial safety factors in geotechnical design", Paper

presented to the Geotechnical Society of Ireland, 9th November 1993.Salenon, J and Pecker, A (1994a) Ultimate capacity strength of shallow

foundations under inclined and eccentric loads. Part I: purely cohesive

soil European Journal of Mechanics A/Solids, 14(3), pp. 349-375.Salenon, J and Pecker, A (1994b) Ultimate bearing capacity of shallow

foundations under inclined and eccentric loads. Part II: purely cohesivesoil without tensile strength European Journal of Mechanics A/Solids,14(3), pp. 377-396.

Simpson, B. and Driscoll, R. (1998) Eurocode 7 a commentary. ConstructionResearch Communications, London.

Terzaghi (1943) Theoretical Soil Mechanics, Wiley.Vesic, A. S. (1975) Bearing capacity of shallow foundations, In: Foundation

Engineering Handbook edited by H. F. Winterkorn and Hsai-Yang Fang.Van Notstrand, Reinhold, New York.


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195

Appendix I: EC7 shallow foundation bearing strength

formulae

The general ultimate bearing capacity equation now becomes:

cs cd ci c qs qd qi q s d ic qu

1= c + q + Bq N N N

2g g g ga a gal l g ll l l l l l l l l (IX.1)

where: tan 2q = (45 + )N e tan2

p f f

c q= ( - 1)cot for > 0, but 5.14 for = 0N N f f f

q= 2( 1)tanN Ng - f ,

and cs, qs, and s are the shape factors;cd, qd, and d are the depth factors;ci, qi and i are the inclined load factors,

and c, q, and are the base inclination factors.

If the foundation is subject to moment loading then B and L are used in thefollowing equations. The shape, inclined load and base inclination factors below arethe same as or similar to those in EC7 (2001). Depth factors are included herealthough EC7 eschews these. Other sources of equations for these factors are Vesic(1975) and Brinch Hansen (1970).

Set out below are the factors for use in equation I.1

The shape factors:cs, qs, ands

a for = 0cs qs

B= 1 + 0.2 1L

l =l (IX.2)

b for > 0qs q

cs

q

N 1

N 1

l -l =

- (IX.3)

qs

B= 1 + sin

L

fl (IX.4)

s

B= 1 - 0.3

Lg

l (IX.5)

Why does s reduce as the foundation approaches a square, whereas the other shapefactors give an increase in bearing strength when moving from a strip to a squarefoundation? For the strip foundation plane strain conditions prevail whereas for asquare foundation axisymmetric conditions approximate. It is known that the planestrain angle of shearing resistance is about 10% greater than that for triaxial

conditions. N for 35 degrees is about 40 and for 32 degrees it is about 26 (ie 26 isroughly 70% of 40.)

The depth factors:cd, qd, andd

a cd:

for = 0: f fcd

D Dfor 1 : = 1 + 0.4

B B

l (IX.6)


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196

f f-1cd

D Dfor > 1 : = 1 + 0.4tan

B B

l (IX.7)

for > 0: qdcd qdq

(1 - )= -

tanN

ll l

f

(IX.8)

b qd:

2f fqd

D D)for 1 : = 1 + 2 tan (1 - sin

B B

f fl (IX.9)

2 f-1f qd

D)for /B > 1 : = 1 + 2 tan (1 - sin tanD

B

f fl (IX.10)

c d:d = 1gl (IX.11)

The inclination factors:ci, qi, andi

a for = 0( )ci u qi0.5 1 1 H/As 1l = + - l = (IX.12)

for this equation H is restricted to be less than or equal to As u, H = Asu is thecondition for sliding failure rather than bearing.

b for > 0qi

ci qi

c

1

N tan

-ll = l -

f (IX.13)

m

qi

H1

V Accot

l = - + f (IX.14)

m 1

i

H1

V Accot

+

g

l = - + f (IX.15)

where:

B

2 B/Lm m when H acts in the direction of B

1 B/L

+= =

+ (IX.16)

L

2 L/Bm m when H acts in the direction of L

1 L/B

+= =

+ (IX.17)

In cases where the horizontal load acts in a direction forming and angle with thedirection of L, m may be calculated from:

2 2L L Bm m m cos m sin= = q + q

(IX.18)

The foundation base inclination factors:cq and

a for = 0c q

21 12

a aal = - l =

p + (IX.19)


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197

b for > 0q

c q

c

(1 )

N tan

a

a a

- ll = l -

f (IX.20)

2

q

(1 tan )a ga

l = l = - a f (IX.21)

where: is the angle the underside of the foundation makes with the horizontal(radians).


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198

Appendix II: Estimation of Vuo

The ultimate bearing strength of a shallow foundation under vertical load onlyis:

uo uV q A

where: qu is the ultimate capacity, expressed as a bearing pressure,A is the contact area of the foundation.

Since for Vuo there is no moment or horizontal shear all we need is the bearingstrength for vertical load only but it may be necessary to include shape anddepth factors.

The ultimate bearing capacity equation, for a horizontal foundation underside,becomes:

12cs cd c qs qd q s du

= c + q + Bq N N N

where: Nc, Nqand N and the factors are given in Appendix I.

a Strip foundation

ground surface:

12c quoV = c + q + B AN N N embedded:

12cd c qd q duoV = c + q + B AN N N

b Rec tangular found at ion

ground surface:

12cs c qs q suoV = c + q + B AN N N

embedded:

12cs cd c qs qd q s duoV = c + q + B AN N N


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199

Appendix III: EC7 bearing strength surfaces

The bearing capacity surface for a rectangular foundation at

the ground surface, subject to vertical load, shear and momentabout the length axis.

We need to express M/V in terms ofM V:

uoM MV B , soM MB

V V

If the foundation is subject to moment loading then use B and L given by:2 2M M

L L and B BV V

Also we need the effective width in terms of dimensionless parameters:2 2

1M M

B B BV V

Drained case:3

2

u uo

1 H 1q B N 1 and V B N

2 V 2

Convert qu to the vertical load per unit length of the strip foundation:3

21 HV B N 12 V

Convert to dimensionless parameters by dividing byVuo:

32B H

V 1B V

Substitute for (B/B) and rearrange:

2 32M H

V 1 1 0V V

Undrained case:

u u c ci u c uo u c

u

1 Hq s N s N 1 1 and V s N B

2 B s

Converting the pressure to vertical load by multiplying by B:

1 HV s N B 1 1

u c2 B su

Convert to dimensionless parameters by dividing both sides by uoV

u c

u c u

s N B1 HV 1 12 s N B B s


27/27


Substitute for B:

1 21 1 1

22

1 u

M HV

MV

sV

Now convert the term inside the square root bracket to dimensionless form:

cHNMVV M

V

1 21 1 1

2 21

Rearrange and eliminate the square root:

2 2

cHN2M 2M

2V 1 1 1V V 2M1

V

Final rearranging:

c

M M MV N H

V V V

2 22 2 2

2 1 1 1 0

chap 7 shallow static bearing strength

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