Foundation on Layered

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128 Shallow Foundations: Bearing Capacity and Settlement

4.4 Foundation on layered cf Soil

StronGer Soil underlain by weaKer Soil

Meyerho and Hanna17 developed a theory to estimate the ultimate bearing capacity o

a shallow rough continuous oundation supported by a strong soil layer underlain by a

weaker soil layer as shown in Figure 4.15. According to their theory, at ultimate load

per unit area qu,the ailure surace in soil will be as shown in Figure 4.15. I the ratio

H/B is relatively small, a punching shear ailure will occur in the top (stronger) soil layerollowed by a general shear ailure in the bottom (weaker) layer. Considering the unit

length o the continuous oundation, the ultimate bearing capacity can be given as

q q

C P

BH

u b

a p= ++

-2

1

( sin ) (4.28)

where

B= width o the oundationg1= unit weight o the stronger soil layerCa= adhesive orce along aa and bbPp= passive orce on aces aa and bbqb= bearing capacity o the bottom soil layerd= inclination o the passive orce Pp with the horizontal

FiGure 4.15 Rough continuous oundation on layered soilstronger over weaker.

Df

H

Pp

Ca Ca

a b

qu

Pp

B

Stronger soil11c1

Weaker soil22c2

a b


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Special Cases of Shallow Foundations 129

Note that, in equation (4.28),

C c H

a a= (4.29)

where

ca= unit adhesion

P HK

D HK

p

pH

f

pH=

+

12 1

21

cos

( )( )cos

= +

1

21

21

2

HD

H

Kf pH

cos (4.30)

where

KpH= horizontal component o the passive earth pressure coefcient

Also,

q c N D H N BN

b c f q= + + +

2 2 1 212 2 2( ) ( ) ( )

( ) (4.31)

where

c2= cohesion o the bottom (weaker) layer o soilg2= unit weight o bottom soil layer

Nc(2),Nq(2),Ng(2)= bearing capacity actors or the bottom soil layer (that is, withrespect to the soil riction angle o the bottom soil layer f2)

Combining equations (4.28), (4.29), and (4.30),

q qc H

BH

D

H

Ku b

a f pH= + + +

22

1

21

21

2coos

sin

-

= + + +

BH

qc H

BH

Db

a

1

12

21

2f pH

H

K

BH

-

tan

1

(4.32)

Let

K KpH stan tan = 1 (4.33)where

Ks= punching shear coefcient

So,

q qc H

BH

D

H

K

BH

u ba f s= + + +

-

21

21

2 11

tan

(4.34)

The punching shear coefcient can be determined using the passive earth pres-

sure coefcient charts proposed by Caquot and Kerisel.18 Figure 4.16 gives the varia-

tion oKs with q2/q1 and f1. Note that q1 and q2 are the ultimate bearing capacities oa continuous surace oundation o widthB under vertical load on homogeneous beds

o upper and lower soils, respectively, or

q c N BN

c1 1 112 1 1

= +( ) ( )

(4.35)


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where

Nc(1),Ng(1)= bearing capacity actors corresponding to soil riction angle f1

q c N BN

c2 2 212 2 2

= +( ) ( )

(4.36)

I the height H is large compared to the width B (Figure 4.15), then the ail-

ure surace will be completely located in the upper stronger soil layer, as shown inFigure 4.17. In such a case, the upper limit or qu will be o the ollowing orm:

q q c N qN BN

u t c q= = + +

1 1 112 1 1( ) ( ) ( )

(4.37)

Hence, combining equations (4.34) and (4.37),

q qc H

BH

D

H

K

BH q

u ba f s

t= + + +

-

21

21

2 11

tan

(4.38)

For rectangular oundations, the preceding equation can be modifed as

q qB

L

c H

B

B

LH

u ba

a= + +

+ +

12

11

l 22 11

12

+

- D

H

K

BH q

f ss t

tanl

(4.39)

FiGure 4.16 Meyerho and Hannas theoryvariation oKs with f1 and q2/q1.

20 30 40 50

0

0.20.4

q2/q1 = 1.0

0

10

20

30

40

Ks

1 (deg)


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where

la, ls= shape actors

q c N D H N BN

b c cs f q qs= + + +

2 2 2 1 2 212 2( ) ( ) ( ) ( )

( )l l l( ) ( )2 2s (4.40)

q c N D N BN

t c cs f q qs= + +1 1 1 1 1 11

2 1 1( ) ( ) ( ) ( ) ( )

l l

lls( )1 (4.41)

lcs(1), lqs(1), lgs(1)= shape actors or the top soil layer (riction angle =f1; seeTable 2.6)

lcs(2), lqs(2), lgs(2)= shape actors or the bottom soil layer (riction angle =f2; seeTable 2.6)

Based on the general equations [equations (4.39), (4.40), and (4.41)], some special

cases may be developed. They are as ollows:

Case i: stRongeR sanD layeRoveR weakeR satURateD Clay (f2= 0)

For this case, c1= 0; hence, ca= 0. Also or f2= 0,Nc(2)= 5.14,Ng(2)= 0,Nq(2)= 1, lcs=1 + 0.2(B/L), lqs= 1 (shape actors are Meyerhos values as given in Table 2.6). So,

q cB

L

B

LH

u= +

+ +

5 14 1 0 2 12 1

2. . 112

11

+

+ D

H

K

BD q

f ss f t

tanl

(4.42)

where

q D NB

Lt f q= +

+

1 1

2 11 0 1 452

( ). tan

+ +

+

12

1 0 1 452

1 12 1

BN

B

L( )

. tan

(4.43)

FiGure 4.17 Continuous rough oundation on layered soilH/B is relatively small.

Df

H

qu

B

Stronger soil11c1

Weaker soil22c2


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In equation (4.43) the relationships or the shape actors lqs and lgs are those givenby Meyerho19 as shown in Table 2.6. Note that Ks is a unction oq2/q1 [equations

(4.35) and (4.36)]. For this case,

q

q

c N

BN

c

BN

c2

1

2 2

12 1 1

2

1 1

5 14

0 5= =( )

( ) ( )

.

. (4.44)

Once q2/q1 is known, the magnitude o Ks can be obtained rom Figure 4.16,

which, in turn, can be used in equation (4.42) to determine the ultimate bearing

capacity o the oundation qu. The value o the shape actor ls or a strip oundationcan be taken as one. As per the experimental work o Hanna and Meyerho,20 the

magnitude ols appears to vary between 1.1 and 1.27 or square or circular ounda-

tions. For conservative designs, it may be taken as one.Based on this concept, Hanna and Meyerho20 developed some alternative design charts

to determine the punching shear coefcientKs, and these charts are shown in Figures 4.18

and 4.19. In order to use these charts, the ensuing steps need to be ollowed.

1. Determine q2/q1.

2. With known values of1 and q2/q1, determine the magnitude od/f1 romFigure 4.18.

3. With known values of1, d/f1, and c2, determine Ks rom Figure 4.19.

FiGure 4.18 Hanna and Meyerhos analysisvariation od/f1 with f1 and q2/q1strongersand over weaker clay.

0 0.2 0.4 0.6 0.8 1.0q2/q1

0

0.2

0.4

0.6

0.8

1.0

/1 1 =50

40

30


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FiGure 4.19 Hanna and Meyerhos analysis or coefcient o punching shearstronger

sand over weaker clay.

50

10

20

30

40

50

10 15 20

0.3

0.4

0.5

0.6

0.7

0.3

0.4

0.5

0.6

0.7

25 30 35

Ks

0

10

20

30

Ks

c2 (kN/m2)

5 10 15 20 25 30 35

c2 (kN/m2)

/1 = 0.8

/1 = 0.8

(a) 1 = 50

(b) 1 = 45

0.30.4

0.50.6

0.710

0

20

Ks

5 10 15 20 25 30 35

c2 (kN/m2)

/1 = 0.8

(c) 1

= 40


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Case ii: stRongeR sanD layeRoveR weakeR sanD layeR

For this case, c1= 0 and ca= 0. Hence, reerring to equation (4.39),

q q

B

L H

D

H

K

Bu b

f s

= + +

+

1 1

21

2 1

tan - l s tH q1 (4.45)

where

q D H N BN b f q qs s

= + + l l 1 2 212 2 2 2

( )( ) ( ) ( ) ( )

(4.46)

q D N BN

t f q qs s= + l l 1 1 1

12 1 1 1( ) ( ) ( ) ( )

(4.47)

Using Meyerhos shape actors given in Table 2.6,

l l

qs s

B

L( ) ( ) . tan1 1

2 1

1 0 1 45 2= = +

+

(4.48)

and

l l

qs s

B

L( ) ( )

. tan2 2

2 21 0 1 452

= = +

+

(4.49)

For conservative designs, or allB/L ratios, the magnitude ols can be taken asone. For this case

qq

BNBN

NN

2

1

2 2

1 1

2 2

1 1

0 50 5

= =..

( )

( )

( )

( )

(4.50)

Once the magnitude oq2/q1 is determined, the value o the punching shear coef-

cient Ks can be obtained rom Figure 4.16. Hanna21 suggested that the riction angles

obtained rom direct shear tests should be used.

Hanna21 also provided an improved design chart or estimating the punching shear

coefcient Ks in equation (4.45). In this development he assumed that the variation o

d or the assumed ailure surace in the top stronger sand layer will be o the nature

shown in Figure 4.20, or

= + z az2

2(4.51)

where

=q

q

2

1

(4.52)

a

H

q

q=

- ( ) 1 22

2

1(4.53)

So,

=

+

- ( )

z

q

qq

q Hz2

1

2

1 2

2

2

2

1

(4.54)


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The preceding relationship means that atz= 0 (that is, at the interace o the twosoil layers),

=

q

q

2

1 2

(4.55)

and at the level o the oundation, that isz=H,

=

1 (4.56)

FiGure 4.20 Hannas assumption or variation odwith depth or determination oKs.

Df

H

H

Pp

a b

a b

qu

B

Sand11

Sand22

z

z

z

(a)

(b)

2

1


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Equation (4.51) can also be rewritten as

z

q

qq

q HH z=

+

- ( )

-21

2

1 2

2

2

1( )2

(4.57)

where dZ is the angle o inclination o the passive pressure with respect to the hori-zontal at a depthz measured rom the bottom o the oundation. So, the passive orce

per unit length o the vertical surace aa (or bb) is

PK

z D dzp

pH z

z

H

f=

+

1

0

( )

cos( ) (4.58)

whereKpH(z)= horizontal component o the passive earth pressure coefcient at a depthzmeasured rom the bottom o the oundation

The magnitude o Pp expressed by equation (4.58), in combination with the

expression dz given in equation (4.57), can be determined. In order to determine themagnitude o the punching shear coefcient Ks given in equation (4.33), we need to

know an average value od. In order to achieve that, the ollowing steps are taken:

1. Assume an average value od and obtain KpH as given in the tables byCaquot and Kerisel.18

2. Using the average values od and KpH obtained rom step 1, calculate Pprom equation (4.30).

3. Repeat steps 1 and 2 until the magnitude oPp obtained rom equation

(4.30) is the same as that calculated rom equation (4.58).

4. The average value od[or which Pp calculated rom equations (4.30) and (4.58)is the same] is the value that needs to be used in equation (4.33) to calculate Ks.

Figure 4.21 gives the relationship or d/f1 versus f2 or various values o f1

obtained by the above procedure. Using Figure 4.21, Hanna21 gave a design chart or

Ks, and this design chart is shown in Figure 4.22.

Case iii: stRongeR Clay layeR (f1= 0) oveR weakeR Clay (f2= 0)

For this case,Nq(1) andNq(2) are both equal to one andNg(1)=Ng(2)= 0. Also,Nc(1)=Nc(2)= 5.14. So, rom equation (4.39),

qB

Lc N

B

L

c

u c

a= +

+ +

1 0 2 1

2

2 2

.( )

HH

BD q

a f t

+ l

1 (4.59)

where

qB

Lc N D

t c f= +

+1 0 2 1 1 1. ( ) (4.60)


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FiGure 4.22 Hannas analysisvariation oKs or stronger sand over weaker sand.

35

30

25

20

15

Ks

10

5

020 25 30 35 40 45 50

1 = 50

2 (deg)

47.7

45

40 3530

FiGure 4.21 Hannas analysisvariation od/f1.

20 25 30 35 40 45 502 (deg)

1 = 25 30 35 40 45 50

/1

0.5

0.6

0.7

0.8

0.9

1.0


11/14


For conservative design the magnitude o the shape actor la may be taken as one.The magnitude o the adhesion ca is a unction oq2/q1. For this condition,

q

q

c N

c N

c

c

c

c

c

c

2

1

2 2

1 1

2

1

2

1

5 14

5 14= = =( )

( )

.

.(4.61)

Figure 4.23 shows the theoretical variation oca with q2/q1.17

Example 4.2

Reer to Figure 4.15. Let the top layer be sand and the bottom layer saturated clay.

Given:H= 1.5 m. For the top layer (sand): g1= 17.5 kN/m3; f1= 40; c1= 0; or thebottom layer (saturated clay): g2= 16.5 kN/m3; f2= 0; c2= 30 kN/m2; and or the oun-dation (continuous):B= 2 m;Df= 1.2 m. Determine the ultimate bearing capacity qu.Use the results shown in Figures 4.18 and 4.19.

Solution

For the continuous oundation BL

= 0and ls= 1, in equation 4.42 we obtain

q c HD

H

K

Buf s= + +

+5 14 1

22 1

2 11

.tan

DD

HH

K

f

= + +

( . )( ) ( . )( )

( )( . )5 14 30 17 5 1

2 1 22 ss

sH K

H

tan( . )( . )

. . .

40

217 5 1 2

175 2 7 342 1 2 42

+

= + +

(a)

To determine Ks, we need to obtain q2/q1. From equation (4.44),

q

q

c

BN2

1

2

1

5 14

0 5=

.

.( )

FiGure 4.23 Analysis o Meyerho and Hanna or the variation oca/c1 with c2/c1.

1.0

0.9

0.8

0.7

ca

/c1

c2/c1

0.60 0.2 0.4 0.6 0.8 1.0


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From Table 2.3 or f1= 40, Meyerhos value oNg(1) is equal to 93.7. So,

q

q2

1

5 14 30

0 5 17 5 2 93 70 094= =

( . )( )

( . )( . )( )( . ).

Reerring to Figure 4.18 or q2/q1= 0.094 and f1= 40, the value od/f1= 0.42. Withd/f1= 0.42 and c2= 30 kN/m2, Figure 4.19c gives the value oKs= 3.89. Substitutingthis value into equation (a) gives

q HH

qu t

= + +

175 2 28 56 12 4

2. ..

(b)

From equation (4.43),

q D NB

L

t f q= +

+

1 12 11 0 1 45

2

( ). tan

+ +

+

1

2

1 0 1 45

2

1 12 1

BN

B

L

( ). tan

(c)

For the continuous oundationB/L= 0. So,

q D N BN

t f q= + 1 1

12 1 1( ) ( )

For f1= 40, use Meyerhos values oNg(1)= 93.7 andNq(1)= 62.4 (Table 2.3). Hence,

q

t= + =( . )( . )( . ) ( . )( )( . )17 5 1 2 62 4 17 5 2 93 7 13481

2. . .2 1639 75 2987 95+ = kN/m2

IH= 1.5 m is substituted into equation (b),

qu

= + +

=175 2 28 56 1 5 12 4

1 5342 32. ( . )( . )

.

.. kN/m2

Since qu= 342.3


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For f2= 32, Meyerhos bearing capacity actors are Ng(2)= 22.02 and Nq(2)= 23.18(Table 2.3). Also rom Table 2.6, Meyerhos shape actors

l l

qs s

B

L( ) ( ). tan

2 2

2 21 0 1 452= = +

+

= +

+

=1 0 1

1 5

1 545

32

21 3252( . )

.

.tan .

qb

= + +( )( . )( . )( . ) ( . )( . )(18 1 5 1 23 18 1 3251

216 7 1 5 22 02 1 325 1382 1 365 4 1747 5. )( . ) . . .= + = kN/m2

Hence, rom equation (4.45),

qu

= + +

+

1747 5 1

1 5

1 518 1 1

2 1 5

12.

.

.( )( )

.

- = +5 75 40

1 518 1 1747 5 463 2

. tan

.( )( ) . . --

=

18

2192 7. kN/m2

CHECK


q D H N BN

t f q qs s= + + l l 1 1 1

12 1 1 2

( )( ) ( ) ( ) ( )

For f1= 40, Meyerhos bearing capacity actors are Nq(1)= 62.4 and Ng(1)= 93.69(Table 2.3).

l l

qs s

B

L( ) ( )

. tan1 1

2 11 0 1 452

= = +

+

= +

+

1 0 11 5

1 545

40

21 462( . )

.

.tan .

q

t= + +( )( . )( . )( . ) ( )( . )( .18 1 5 1 64 2 1 46 18 1 5 93 61

29 1 46 4217 9 1846 6 6064 5)( . ) . . .= + = kN/m2

So, qu=19.7 kN/.

Example 4.4

Figure 4.24 shows a shallow oundation. Given:H= 1 m; undrained shear strength c1(or f1= 0 condition) = 80 kN/m 2; undrained shear strength c2 (or f2= 0 condition) =32 kN/m 2; g1= 18 kN/m3;Df= 1 m;B= 1.5 m;L= 3 m. Estimate the ultimate bearingcapacity o the oundation.

Solution


q

q

c

c2

1

2

1

32

800 4= = = .

From Figure 4.23 or q2/q1= 0.4, ca/c1= 0.9. So ca= (0.9)(80) = 72 kN/m 2. From equa-tion (4.60),

qB

Lc N D

t c f= +

+ = +1 0 2 1 0 21

1 1 1. .

( )

.( )( . ) ( )( ) .

5

380 5 14 18 1 470 32

+ = kN/m2 .


14/14


With ls= 1, equation (4.59) yields

qB

L

c NB

L

cu c

a= +

+ +

1 0 2 12

2 2.

( )

HH

B

D qa f t

+

= + +

l 1

1 0 1 32 5 14 1 5( . )( )( . ) ( . ) ( )( ) ( )( )

.

2 72 18 1

198 93 216

H

B

H

B

+

= +

= +

198 93 2161

1 5.

. kN/

Gure 4.24 Shallow oundation on layered clay.

1 m

1.5 m 3 mStronger clay

1 = 18 kN/m3

2 = 0c2 = 80 kN/m

2

Weaker clay

1 = 16 kN/m3

1 = 0c1 = 32 kN/m

2

H

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