ch-04

Introduction

The ultimate bearing capacity problems described in Chapter 3 assume that the soilsupporting the foundation is homogeneous and extends to a great depth below thebottom of the foundation. They also assume that the ground surface is horizontal.However, that is not true in all cases: It is possible to encounter a rigid layer at ashallow depth, or the soil may be layered and have different shear strength parameters. In some instances, it may be necessary to construct foundations on or near aslope. This chapter discusses bearing capacity problems relating to these specialcases, as well as the seismic bearing capacity of shallow foundations.

Foundation Supported by a Soil With a Rigid Baseat Shallow Depth

Figure 4.1(a) shows a shallow, rough continuous foundation supported by a soil thatextends to a great depth. Neglecting the depth factor, for vertical loading Eq. (3.21)will take the form

(4.1)

164

The general approach for obtaining expressions for Nc, Nq , and Ny was outlined in Chapter 3. The extent of the failure zone in soil, D, at ultimate load obtainedin the derivation of Nc and Nq by Prandtl (1921) and Reissner (1924) is given in Figure 4.1(b). Similarly, the magnitude of D obtained by Lundgren and Mortensen(1953).in evaluating Ny is given in tIle figure.

Now, if a rigid, rough base is located at a depth of H < D below tIle bottom ofthe foundation, full development of the failure surface in soil will be restricted. Insuch a case, the soil failure zone and the development of slip lines at ultimate load

4.2 Foundation Supported by a Soil With a Rigid Base at Shallow Depth 165

y¢/c '

(a)

3.------r--------,-----r------,----,----,

21-----+-----t---+-------:~--+--------l

11------+::,,.......,,,:::::.--f---+----:::;>'.L::---+--------l

Figure 4. 1 (a) Failuresurface under a roughcontinuous foundation;(b) variation of D / Bwith soil friction angle <P'

5040

(b)

20 30

Soil friction angle, <P' (deg)

10OL..-----'----------l---....I..------'--------'

o

will be as shown in Figure 4.2. Mandel and Salencon (1972) determined the bearingcapacity factors applicable to this case by numerical integration, using the theory ofplasticity. According to their theory, the ultimate bearing capacity of a rough continuous foundation with a rigid, rough base located at a shallow depth can be given bythe relation

(4.2)

where N;, N;, N; = modified bearing capacity factorsB = width of foundationy= unit weight of soil

Note that, for H ~ D, N; = Np N; = Nq , and N; = Ny (Lundgren and Mortensen,1953). The variations of N;, N:, and N~ with H/B and the soil friction angle </>' aregiven in Figures 4.3,4.4, and 4.5, respectively.

166 Chapter 4 Ultimate Bearing Capacity of Shallow Foundations: Special Cases

~-'--·r~ ...H

:' .' '." .. ". :' '." "... ... L ····1 q ~;DfJ...-.-.---:- --;-:--:---:-------,;----:--:-- ,-:-:-

y¢/c'

Rough rigid base

Figure 4.2 Failure surface under a rough, continuous foundation with a rigid, rough baselocated at a shallow depth

10,000

5000

2000

1000

500

200

N* 100c

50

20

10

5

2

10 10 20

1>' (deg)

30 40

D/B=2.4

Figure 4.3 Mandel and Salencon's bearing capacity factor Nt [Eq. (4.2)]

~,I

I

I


Figure 4.4 Mandel and Salencon's bearing capacity factor N: [Eq. (4.2)]

500 I-----+-----j--+---t--f----t-t-----i

5000 I----+---+--+-f-----+-+----i

2000 I----+---+-+--f-----,t----i

1000 1----+---+---f----1---+--+----i

10,000 .------.----,--------.~-_____r-___,____,

200D/B=3.0

N* 100q

50

20

10

51.2

2

120 25 30 35 40 45

tP' (deg)

Neglecting the depth factors, the ultimate bearing capacity of rough circularand rectangular foundations on a sand layer (c' = 0) with a rough, rigid base located at a shallow depth can be given as

(4.3)

where F;s, F~s = modified shape factors

The shape factors F;s and F~s are functions of H / Band <p'. On the basis of thework of Meyerhof and Chaplin (1953), and simplifying the assumption that, in radialplanes, the stresses and shear zones are identical to those in transverse planes, Meyerhof (1974) proposed that


10,000

5000

2000

1000

500

200

N* 100y

50

20

10

5

2

/ I1/ /I

/ /H/B~ ~.2 )/ // / l/.;4

/ 1/ JIv7/ / V/V

1.2

1/ ~V;:.o

7 y '/0.8

//V0.6

/0.5

D/B=1.5

120 25 30 35

¢/ (deg)

40 45Figure 4.5 Mandel and Salencon's bearing capacity factor N~ [Eq. (4.2)]

and

F* = 1 - m (B)'}'s 2 L

(4.4)

(4.5)

where L = length of the foundation

The variations of ml and m2 with H / Band 4>' are shown in Figure 4.6.For saturated clay (i.e., under the undrained condition, or 4> = 0), Eq. (4.2) will

simplify to the form

(4.6)


1.0 ,.---...,....------r----r----...,....-----,

454030 35

¢'(deg)

25

0.6 \-----+-~--I----t-~--+------;:l

0.8 \-----+----::;7'"--=---I----t---::;7'""""'---+----1

0.4 \-----+--~.e.-r---t--r---+--____;;,.L-j

0.2 1--.",L---1----:?~---t--::>-'''----1----",I

1.0 r---..,----,----y-----..,-----,

HIB= 0.1

0.8 1------f-:::,.,....:=:;...---+---t---:",......"---1-------!

0.6 \----."l£----f----J-:::,..-<::.---+--------:,.....:!

0.4

0.2 1------+-----+--------1f-----+-----r

Figure 4.6 Variation of m1

and m2 with HIB and 4>'(Meyerhof)

454030 35

¢'(deg)

25OL.-.--....l.-------l...---L.-__....l.-__----I

20

Mandel and Salencon (1972) performed calculations to evaluate N; for continuousfoundations. Similarly, Buisman (1940) gave the following relationship for obtainingthe ultimate bearing capacity of square foundations:

(for~ - v2 ~ 0) (4.7)2H 2


Table 4.1 Values of N~ for Continuous andSquare Foundations (<!> = 0)

N*c

I

B

H

234568

10

Square8

5.435.936.446.947.438.439.43

Continuousb

5.245.716.226.687.208.179.05

aBuisman's analysis (1940)bMandel and Salencon's analysis (1972)

In this equation, Cu is the undrained shear strength.Equation (4.7) can be rewritten as

(

O.S.!!.- - 0.707)qu(square) = S.14 1 + H

S.14

CU + q

Table 4.1 gives the values of N; for continuous and square foundations.

(4.8)

I

I

I

I

I

I

~I

I

4.3 Bearing Capacity of Layered Soils: Stronger Soil Underlain by Weaker Soil 171

From Figure 4.6(a), for 4>' = 35°, H / B = 0.6, and.!be value of mt = 0.34, so

F;s = 1 - (0.34)(1/1) = 0.66

Similarly,

F;s = 1 - m2(B/L)

From Figure 4.6(~);m2 = 0.45, so

F;s = 1 - (0.45)(1/1) = 0.55

(0.66) + (1/2) (17.3).(1) ($Q}(0.55) == 1009 kN/m2

•

Bearing Capacity of Layered Soils: Stronger SoilUnderlain by Weaker Soil

The bearing capacity equations presented in Chapter 3 involve cases in which thesoil supporting the foundation is homogeneous and extends to a considerable depth.The cohesion, angle of friction, and unit weight of soil were assumed to remain constant for the bearing capacity analysis. However, in practice, layered soil profiles areoften encountered. In such instances, the failure surface at ultimate load may extendthrough two or more soil layers, and a determination of the ultimate bearing capacity in layered soils can be made in only a limited number of cases. This section features the procedure for estimating the bearing capacity for layered soils proposedby Meyerhof and Hanna (1978) and Meyerhof (1974).

Figure 4.7 shows a shallow continuous foundation supported by a stronger soillayer, underlain by a weaker soil that extends to a great depth. For the two soil layers, the physical parameters are as follows:

Soil frictionLayer Unit weight angle Cohesion

Top 1'1 </>i ciBottom 1'2 </>2 C2

At ultimate load per unit area (qu), the failure surface in soil will be as shown in thefigure. If the depth H is relatively small compared with the foundation width B, apunching shear failure will occur in the top soil layer, followed by a general shearfailure in the bottom soil layer. This is shown in Figure 4.7(a). However, if the depth His relatively large, then the failure surface will be completely located in the top soil

.~--~_._._----


(a)

Stronger soil

Y1<p{c{

b '. :.... :·.··.,.i<:··~:::r~·/~/i.:J~;::,'.":·i::.::~.::·.\:~

Weaker soil

Y2<Pzcz

~II

Ii

t'.~I

..•....:.'.

';-!J

'.;1',,_.

Weaker soil

Y2<Pzcz(b)

Stronger soil

H Yl<Pl

; ','C,'i;:'i),(,):;;,yy:ji'?:i:?'/ii/\':'iji::':A:;;\"ei,,:,l;

Figure 4.7 Bearing capacity of a continuous foundation on layered soil i

Ilayer, which is the upper limit for the ultimate bearing capacity. This is shown in Figure 4.7b.

The ultimate bearing capacity for this problem, as shown in Figure 4.7a, canbe given as

I

where B = width of the foundationCa = adhesive forcePp = passive force per unit length of the faces aa' and bb'qb = bearing capacity of the bottom soil layer8 = inclination of the passive force Pp with the horizontal

(4.9)

'·1'.

;;:


Note that, in Eq. (4.7),

where c~ == adhesion

Equation (4.7) can be simplified to the fornl

2c~H 2( 2Dt )KpH tan 8qu = qh +~ + 'YI H 1 + H B - 'YI H

where K pH == horizontal component of passive earth pressure coefficient

However, let

K pH tan D == K s tan cPr

where K s == punching shear coefficient

Then

(4.8)

(4.9)

(4.10)

(4.11)

The punching shear coefficient, K s , is a function of q2/ql and 4Jl, or, specifically,

K s = f(:> ~1)Note that ql anel q2 are the ultimate bearil1g capacities of a continuous founda

tion of width B under vertical load on the surfaces of homogeneous thick beds ofupper and lower soil, or

(4.12)

and

q2 = C2Nc(2) + !'Y2BNY(2) (4.13)

where Nc(l) , Ny(l) == bearing capacity factors for friction angle <Pi (Table 3.4)Nc(2) , Ny (2) == bearing capacity factors for friction angle <P2 (Table 3.4)

Observe that, for the top layer to be a stronger soil, q2!ql should be less than unity.The variation of K s with q2/ql and <Pi is shown in Figure 4.8.The variation of c~/cl

with qJql is shown in Figure 4.9. If the height H is relatively large, then the failure surface in soil will be cOITlpletely located in the stronger upper-soil layer (Figure 4.7b). Forthis case,

where N q(i) == bearing capacity factor for <p' == <Pi (Table 3.4) and q == "liD!

Combining Eqs. (4.11) and (4.14) yields

_ 2c~H 2( 2Dt ) Ks tan <Piqu - qb +~ + 'YIH 1 + H B - 'YIH ~ qt

(4.14)

(4.15)

174 Chapter 4 Ultimate Bearing Capacity of Shallow Foundations: Special Cases I

I

:.;.u....•...• 'Ij I

..t.~.::~;,\ ,

,,)~ i

I.,:' ~

Figure 4.9 Variation of c~/cl with qz/qlbased on the theory of Meyerhof andHanna (1978)

1.00.8

50 Figure 4.8 Meyerhof and Hanna's punching shearcoefficient Ks

30 40

¢{ (deg)

0.2

/~

//

V

/"~;fql

/ O.}V /

/

:?VV V

/--::-~ f..-- f-"" ~

I----o20

40

30

10

0.6o

~'" 20

1.0

0.9

c'a0.8C'1

0.7

For rectangular foundations, the preceding equation can be extended to the form

(4.16)

where

I,

I(4.17)


and

(4.18)

in which Fcs(l) ' F~s(l)' Fys(l) == shape factors with respect to top soil layer (Eqs. 3.25through 3.27)

Fcs(2) ' Fqs(2) ' Fys(2) == shape factors with respect to bottom soil layer(Eqs. 3.25 through 3.27)

Special Cases

1. Top layer is strong sand and bottom layer is saturated soft clay (4J2 == 0). FromEqs. (4.16), (4.17), and (4.18),

qb = (1 + O.2~)5.14C2 + Yl(Df + H) (4.19)

and

(4.20)

Hence,

(4.21)

where C2 == undrained cohesion

For a determination of K s from Figure 4.8,

q2 C2Nc(2)

ql ~'YIBNY(l)(4.22)

2. Top layer is stronger sand and bottom layer is weaker sand (cl == 0, C2 == 0). Theultimate bearing capacity can be given as

(4.23)


where

Then

(4.24)

q2 !yzBNy(z)

q1 !y1BNY(1)

(4.25)

3. Top layer is stronger saturated clay (</>1 == 0) and bottom layer is weaker saturated clay (4J2 == 0). The ultimate bearing capacity can be given as

(4.26)

where

and Cl and Cz are undrained cohesions. For this case,

qz 5.14cz Cz

ql 5.14cl Cl

(4.27)

(4.28)

178 Chapter 4 Ultirnate Bearing Capacity of Shallow Foundations: Special Cases

Bearing Capacity of Foundations on Top of a Slope

In some instances, shallow foundations need to be constructed on top of a slope. InFigure 4.10, the height of the slope is H, and the slope makes an angle f3 with thehorizontal. The edge of the foundation is located at a distance b from the top of theslope. At ultimate load, qu, the failure surface will be as shown in the figure.

Meyerhof (1957) developed the following theoretical relation for the ultimatebearing capacity for continuous foundations:

qu == c'Ncq + !yBNyq (4.29)

For purely granular soil, c' = 0, thus,

(4.30)

4.4 Bearing Capacity of Foundations on Top of a Slope 179

Tyc'¢/

Figure 4.10 Shallow foundation on top of a slope

Figure 4.11 Meyerhofs6 bearing capacity factor

Nyq for granular soil(e' = 0)

543bB

21

Df Df- =0 ---- - =1B B

f3-, I- 0° .- :1-- ,-

1-------- 1..-----20°._

.--- -;1 400-- .--_.- -- ---- .,,--",- .,,-

40°.,,-.,,- </>' - 40°

v 0°

~::--/

~0° / </>' = 30°_-;L_ -7 -- '"-'---- ----

v: Ziclo-- 40°.,,-/ /

/ 0° </>' = 30°/ ~/~V

105

1

o

25

300

50

400

200

~~ 100

Again, for purely cohesive soil, <P = 0 (the undrained condition); hence,

qu = cNcq (4.31)

where c = undrained cohesion.

The variations of Nyq and Ncq defined by Eqs. (4.30) and (4.31) are shown inFigures 4.11 and 4.12, respectively. In using Ncq in Eq. (4.31) as given in Figure 4.12,the following points need to be kept in mind:

1. The term

yHN=

s c (4.32)

is defined as the stability number.


2. If B < H, use the curves for Ns = O.3. If B ;::. H, use the curves for the calculated stability number Ns •

Stress Characteristics Solutionfor Granular Soil Slopes

For slopes in granular soils, the ultimate bearing capacity of a continuous foundation can be given by Eq. (4.30), or

1qu = 2yBNyq

On the basis of the method of stress characteristics, Graham, Andrews,(1988) provided a solution for the bearing capacity factor Nyq for a shallow continuous foundation on the top of a slope in granular soil. Figure 4.13 shows the schematics of the failure zone in the soil for embedment (DtlB) and setback (biB)assumed for those authors' analysis. The variations of Nyq obtained by this methodare shown in Figures 4.14,4.15, and 4.16.

8...---------------......Df Df---- B =1 ---Jj =0

r"

III

I

I

\1I

:1;,',.. (,

- i"

1 2 3 4b bIi for Ns = 0; Ii for Ns > 0

5 Figure 4.12 Meyerhof's bearingcapacity factor Ncq for purely cohesive soil


(a)

(b)

Figure 4. j'3 Schematic diagram of failure zones for embedment and setback: (a) Df / B > 0;(b) b/B > 0


-- b/B= 1---bIB=2

1000 1000 ,--------,-----,--------,,----,

--b/B=O--- biB = 0.5

403010 20

13 (deg)

(b)

10 L-__---l. ---l.... ...l-__----....l

o403010 20

13 (deg)

(a)

Figure 4.14 Graham et al.'s theoretical values of Nyq (Dt / B = 0)

1000 r------,-------"T---~---_, 1000 ,-------r----.....,....---....,------,

--bIB=l---bIB=2

--bIB=O--- biB = 0.5

403020

13 (deg)

(b)

1010 10 '--__----'- ...J..-__-----' ---l

o 10 20 30 40 0

j3 (deg)

(a)

Figure 4.15 Graham et al.'s theoretical values of Nyq(Df / B = 0.5)


1000 .------.-------,-------..,---------, 1000 .------r-----.-----,-----,

--bIB=O--- biB = 0.5

4030

(b)

20f3 (deg)

1010 '-- ..L..- ...L- "'"'---__---'

o403020f3 (deg)

(a)

101OL..- -'--- ..L..- ...L-__-----'

o

Figure 4.16 Graham et al.'s theoretical values of Nyq(Df / B = 1)

•


6m

L5ml~1.5 m 1.5 ill

y = 16.8 kN/m3

1/ = 40°c' = 0

4.1 A rectangular foundation is shown in Figure P4.1. Determine the gross allowable load the foundation can carry, given that B = 3 ft, L = 6 ft, Df = 3 ft,H = 2 ft, 4>' = 40°, c' = 0, and 'Y = 115lb/ft3

. Use FS = 4.4.2 Repeat Problem 4.1 with the following data: B = 1.5 m, L = 1.5 m, Df = 1 m,

H = 0.6 m, 4>' = 35°, c' = 0, and 'Y = 15 kN/m3• Use FS = 3.

4.3 A square foundation measuring 4 ft X 4 ft is supported by a saturated claylayer of limited depth underlain by a rock layer. Given that Df = 3 ft,

Problems 185

Sand¢'c'

H y

Td7T/77//77//77//77//77//7T/77T/77Rock Figure P4.1

. " .....

y = 1151b/ft3

¢ = 0°c = 1200 Ib/ft2

FigureP4.4

0.92 m Length, L = 1.22 m

--l- y ~ 17 kN/m'I I I ~=if• B = 0.92m • 'f/

O.76m c = 72kN/m2

~>.:;;H))\'it;{\;?::;tY:ii~:;>i ';':~; kN/m'

c = 43 kN/m2 Figure P4.5

H = 2 ft, Cu = 2400 Ib/ft2, and 'Y = 120 Ib/ft3

, estimate the ultimate bearingcapacity of the foundation.

4.4 A strip foundation in a two-layered clay is shown in Figure P4.4. Find thegross allowable bearing capacity. Use a factor of safety of 3.

4.5 Find the gross ultimate load that the footing shown in Figure P4.5 can carry.

186 Chapter 4 Ultimate Bearing Capacity ofShallow Foundations: Special Cases

0 • • '

1.2 m

-+-H

l..-c--B=2m ----.·1

Sand

1'1 = 18 kN/m3

<PI' = 40°e1' = 0

Figure P4.6

1.5 m

Y1 -18kN~<Pi = 40°ci = 0

I

LOrn

~j~,.~..~... ~..~.. ~1'2 ~·16.7lli/m3 .'<Pz = 32°Cz = 0

Figure P4.7

4.6 Figure P4.6 shows a continuous foundation.a. If H = 1.5 m, determine the ultimate bearing capacity qu.b. At what minimum value of H / B will the clay layer not have any effect on

the ultimate bearing capacity of the foundation?4.7 A square foundation on a layered sand is shown in Figure P4.7. Determine

the net allowable load that the foundation can support. Use FS = 4.

Problems 187

t.-I---b--

H

JFigure P4.10

4.8 A continuous foundation with a width of 1 m is located on a slope madeof clay soil. Refer to Figure 4.10, and let Df = 1 m, H = 4 m, b = 2 m,'Y = 16.8kN/m3 ,c = 68kN/m3 ,4> = O,andf3 = 60°. UsingEq.(4.31),3. Determine the allowable bearing capacity of the foundation. (Let FS = 3.)b. Plot a graph of the ultimate bearing capacity qu if b is changed from 0 to 6 m.

4.9 A continuous foundation is to be constructed near a slope made of granularsoil. (See Figure 4.10). If B = 4 ft, b = 6 ft, H = 15 ft, Df = 4 ft, f3 = 30°,4>' = 40°, and 'Y = 110 lb/ft3

, estimate the allowable bearing capacity of thefoundation. Use Eq. (4.30) and FS = 4.

4.10 Figure P4.10 shows a shallow strip foundation on the top of a slope. We aregiven the following data:

Slope (sand): f3 = 15°c' = 0

4>' = 40°'Y = 15 kN/m3

Foundation: B = 1.5 mDf = 0.75 m

b = 1.5 m

Estimate the allowable bearing capacity. Use the stress characteristics solution and a factor of safety of 4.

4.11 Repeat Problem 4.10 with the following data:

Slope (sand): f3 = 20°c' = 0

4>' = 30°y = 112lb/ft3

Foundation: B = 4 ftDf = 4 ftb=4ft

Use a factor of safety of 3.


Buisman, A. S. K. (1940). Grondmechanica, Waltman, Delft, the Netherlands.Graham, 1., Andrews, M., and Shields, D. H. (1988). "Stress Characteristics for Shallow Footings

in Cohesionless Slopes," Canadian Geotechnical Journal, Vol. 25, No.2, pp. 238-249.Lundgren, H., and Mortensen, K. (1953). "Determination by the Ibeory of Plasticity on the

Bearing Capacity of Continuous Footings on Sand," Proceedings, Third InternationalConference on Soil Mechanics and Foundation Engineering, Zurich, Vol. 1, pp. 409-412.

Mandel, J., and Salencon, J. (1972). "Force portante d'un sol sur une assise rigide (etudetheorique)," Geotechnique, Vol. 22, No.1, pp. 79-93.

Meyerhof, G. G. (1957). "111e Ultimate Bearing Capacity of Foundations on Slopes," Proceedings, Fourth International Conference on Soil Mechanics and Foundation Engineering,London, Vol. 1, pp. 384-387.

Meyerhof, G. G. (1974). "Ultimate Bearing Capacity of Footings on Sand Layer OverlyingClay," Canadian Geotechnical Journal, Vol. 11, No.2, pp. 224-229.

Meyerhof, G. G., and Chaplin, T. K. (1953). "The Compression and Bearing Capacity of Cohesive Soils," British Journal ofApplied Physics, VoL 4, pp. 20-26.

Meyerhof, G. G., and Hanna, A. M. (1978). "Ultimate Bearing Capacity of Foundations onLayered Soil under Inclined Load," Canadian Geotechnical Journal, Vol. 15, No.4,pp.565-572.

Prandt!, L. (1921). "Uber die Eindringungsfestigkeit (Harte) plastischer Baustoffe und die Festigkeit von Schneiden," Zeitschrift fUr angewandte Mathematik und Mechanik, Vol. 1,No.1, pp. 15-20.

Reissner, H. (1924). "Zum Erddruckproblem," Proceedings, First International Congress ofApplied }vfechanics, Delft, the Netherlands, pp. 295-311.

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