Bearing Capacity of Anisotropic Soils (BM Das - Shallow Foundations)

8/13/2019 Bearing Capacity of Anisotropic Soils (BM Das - Shallow Foundations)

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cc qcqc

qN

=

=

=1

03531 0 353

10 66 250228

tan.

.

. tan.

=90

)(211

i

Also

Equation (2.84):

qu= (48)(20.72)(1.257)(1.4)(0.228)

+ (0.6)(18) (10.66) (1.233) (1.155)(0.353)

+ 12

(18)(0.6)(10.88)(0.8)(1)(0.353)

= 399.05 + 57.88 + 16.59 474 kN/m2

2.11 BEARING CAPACITY OF FOUNDATIONS

ON ANISOTROPIC SOIL

Foundation on Sand (c= 0)

Most natural deposits of cohesionless soil have an inherent anisotropic

structure due to their nature of deposition in horizontal layers. Initial deposition

of the granular soil and subsequent compaction in the vertical direction causes

the soil particles to take a preferred orientation. For a granular soil of this type

Meyerhof suggested that, if the direction of application of deviator stress

makes an angle i with the direction of deposition of soil (Fig. 2.25), then thesoil friction anglecan be approximated in a form

(2.90)

where1= soil friction angle with i= 0

2= soil friction angle with i= 90

Figure 2.26shows a continuous (strip)roughfoundation on an anisotropic

sand deposit. The failure zone in the soil at ultimate load is also shown in thefigure. In the triangular zone (Zone 1) the soil friction angle will be = 1 .

However, the magnitude ofwill vary between the limits of1and2in Zone

2. In Zone 3 the effective friction angle of the soil will be equal to 2 .

Meyerhof [21] suggested that the ultimate bearing capacity of a continuous

foundation on an anisotropic sand could be calculated by assuming an equiva-

lent friction angle = eq , or

1999 by CRC Press LLC


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FIGURE 2.25 Aniostropy in sand deposit

FIGURE 2.26 Continuous rough foundation on anisotropic sand deposit

eq = +

= +( ) ( )2

3

2

3

1 2 1n

where friction ratio = 2n=

1

(2.91)

(2.92)

Once the equivalent friction angle is determined, the ultimate bearing capacity

for vertical loading conditions on the foundation can be expressed as

(neglecting the depth factors)

qu= q Nq(eq)qs+ 12B N(eq)s (2.93)



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FIGURE 2.27 Variation ofN(eq)[Eq. (2.93)]



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FIGURE 2.28 Variation ofNq(eq)[Eq. (2.93)]

whereNq(eq) ,N(eq)= equivalent bearing capacity factors corresponding to the

friction angle = eqIn most cases the value of 1 will be known. Figures 2.27 and 2.28

present the plots ofNq(eq) andN(eq)in terms of nand 1 . Note that the soil



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+

+= L

BBN

L

BqNq

qu4.01

2

1tan1

(eq)eq(eq)

b

a

c

c c

u i

uV uH

= = ( )( )( )

45

+=

2)(

uHuV

icu

ccNq

friction angle =eqwas used in Eqs. (2.66) and (2.72) to prepare the graphs.

So combining the relationships for shape factors (Table 2.5) given by DeBeer

[19]

(2.94)

Foundations on Saturated Clay (= 0 concept)

As in the case of sand discussed above, saturated clay deposits also exhibit

anisotropic undrained shear strength properties. Figures 2.29aand 2.29b show

the nature of variation of the undrained shear strength of clays, cu , with respect

to the direction of principal stress application [22]. Note that the undrained

shear strength plot shown in Fig. 2.29b is elliptical. However, the center of the

ellipse does not match the origin. The geometry of the ellipse leads to the

equation

(2.95)

where cuV= undrained shear strength with i= 0

cuH= undrained shear strength with i= 90

A continuous foundation on a saturated clay layer ( = 0) whose

directional strength variation follows Eq. (2.95) is shown in Fig. 2.29c. The

failure surface in the soil at ultimate load is also shown in the figure. Note that,

in Zone I, the major principal stress direction is vertical. The direction of themajor principal stress is horizontal in Zone III; however, it gradually changes

from vertical to horizontal in Zone II. Using the stress characteristic solution,

Davis and Christian [22] determined the bearing capacity factorNc(i)for the

foundation. For a surface foundation

(2.96)

The variation ofNc(i) with the ratio ofa/b(Fig. 2.29b) is shown in Fig. 2.30.

Note that, when a= b,Nc(i)becomes equal toNc= 5.14 [isotropic case; Eq.(2.67)].

In many practical conditions, the magnitudes of cuVand cuHmay be known,

but not the magnitude of cu(i= 45) . If such is the case, the magnitude of a/b[Eq.

(2.95)] cannot be determined. For such conditions, the following approximation

may be used



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FIGURE 2.29 Bearing capacity of continuous foundation on anisotropic

saturated clay

14.5

29.0

=

+ uHuV

cu

ccNq

(2.97)



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FIGURE 2.30 Variation ofNc(i)with a/bbased on the analysis of Davis and

Christian

qdqsqcdcs

iuHiuV

icu qNcc

Nq +

+= ==

2

)90()0(

)(

qdqsfcdcs

uHuV

icu Dcc

Nq +

+=

2)(

The preceding equation, which was suggested by Davis and Christian [22], is

based on the undrained shear strength results of several clays. So, in general,

for a rectangular foundation with vertical loading condition

(2.98)

For= 0 condition,Nq= 1 and q= Df. So

(2.99)

The desired relationships for the shape and depth factors can be taken fromTable 2.5and the magnitude ofqucan be estimated.

Foundation on cSoil

The ultimate bearing capacity of a continuous shallow foundation supported



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FIGURE 2.31 Anisotropic clay soilassumptions for bearing capacity evaluation

by anisotropic csoil was studied by Reddy and Srinivasan [23] using the

method of characteristics. According to this analysis the shear strength of a soil

can be given as

s= tan+ c

However, it is assumed that the soil is anisotropic only with respect tocohesion. As mentioned previously in this section, the direction of the major

principal stress (with respect to the vertical) along a slip surface located below

the foundation changes. In anisotropic soils, this will induce a change in the

shearing resistance to the bearing capacity failure of the foundation. Reddy and

Srinivasan [23] assumed the directional variation of cat a given depthzbelow

the foundation as (Fig. 2.31a)



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Kc

c

V z

H z

=( )

( )

c

V z

V z

l

c

lc

=

=

=

=

( )

( )

0

0where characteristic length =

ci(z)= cH(z)+ [cV(z) cH(z)]cos2i (2.100)

where ci(z)= cohesion at a depthzwhen the major principal stress is inclined

at an anglei to the vertical (Fig. 2.31b)

cV(z)= cohesion at depthzfor i= 0

cH(z)= cohesion at depthzfor i= 90

The preceding equation is of the form suggested by Casagrande and Carrillo[24].

Figure 2.31bshows the nature of variation ofci(z)with i. The anisotropy

coefficient Kis defined as the ratio of cV(z)to cH(z) .

(2.101)

Inoverconsolidatedsoils Kis less than one and, for normally consolidated

soils the magnitude of Kis greater than one.

For many consolidated soils, the cohesion increases linearly with depth

(Fig. 2.31c). Thus

cV(z)= cV(z=0)+ s (2.102)

wherecV(z) ,cV(z=0)= cohesion in the vertical direction (that is, i= 0) at depths

ofzandz= 0, respectively

= the rate of variation with depthz

According to this analysis, the ultimate bearing capacity of a continuous foun-

dation may be given as

qu= cV(z=0)Nc(i)+ qNq(i)+ 1

2BN(i) (2.103)

whereNc(i) ,Nq(i) ,N(i)= bearing capacity factors

q= DfThis equation is similar to Terzaghis bearing capacity equation for continuous

foundations [Eq. (2.31)].

The bearing capacity factors are functions of the parameters cand K. The

term ccan be defined as

(2.104)

(2.105)



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FIGURE 2.32 Reddy and Srinivasans bearing capacity factor,Nc(i)

influence of K(c = 0)

Furthermore, Nc(i) is also a function of the nondimensional width of the

foundation,B

B = B

l(2.106)

The variations of the bearing capacity factors with c

, B, , and K

determined using the method of analysis by Reddy and Srinivasan [23] are

shown in Figs. 2.32to 2.37. This study shows that the rupture surface in soil

at ultimate load extends to a smaller distance below the bottom of the founda-

tion for the case where the anisotropic coefficient Kis greater than one. Also,

when Kchanges from one to two with = 0, the magnitude ofNc(i)is reduced

by about 30% 40%.



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FIGURE 2.33 Reddy and Srinivasans bearing capacity factorNc(i

influence of K(c = 0.2)



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FIGURE 2.34 Reddy and Srinivasans bearing capacity factor,Nc(i)

influence of K(c = 0.4)



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FIGURE 2.35 Reddy and Srinivasans bearing capacity factors, N(i) and Nq(i)( c = 0)

EXAMPLE 2.3

Estimate the ultimate bearing capacity quof a continuous foundation with the

following:B= 9 ft, cV(z=0)= 250 lb/ft2, = 25 lb/ft2/ft,Df= 3 ft, = 110 lb/ft3,and= 20. Assume K= 2.

Solution From Eq. (2.105)



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Characteristic length,

Nondimensional width,

l

c

B B

l

V z

= = =

= = =

=( )

.

..

0 250

110 2 27

9

2 27396

Also

FIGURE 2.36 Reddy and Srinivasans bearing capacity factors, N(i) andNq(i) influence ofK(c = 0)



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FIGURE 2.37 Reddy and Srinivasans bearing capacity factors,N(i) andNq(i) influence of K(c = 0.2)

c

V z

l

c=

= ==( )

( )( . )

.0

25 2 27

250 0227

Now, referring to Figs. 2.33, 2.34, 2.36, and 2.37, for = 20,c= 0.227, K

= 2, andB = 3.96 (by interpolation)

Nc(i) 14.5;Nq(i) 6, andN(i) 4

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Bearing Capacity of Anisotropic Soils (BM Das - Shallow Foundations)