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8/13/2019 Tugas Desain Pondasi - I

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DISUSUN OLEH :

FRANS PARLINDUNGAN SIREGAR (080404026)

BRAM P. S. (080404079)

M. HAFIZ (080404081)

ARAN G. SIMARMATA (080404095)


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3.10 BEARING CAPACITY OF LAYERED SOILS

STRONGER SOIL UNDERLAIN BY WEAKER SOIL

The bearing capacity equations presented in the precedingsections involve cases in which the soil supporting thefoundation is homogeneous and extends to a considerabledepth.

Cohesion, angle of friction, and unit weight of soil wereassumed to remain constant for the bearing capacity analysis.

In such instances, the failure surface at ultimate load mayextend through two or more soil layers. Determination of

ultimate bearing capacity in layered soils can be made in onlya limited number of cases.

This section features the procedure for estimating bearingcapacity for layered soils proposed by Meyerhof and Hanna(1978) and Meyerhof (1974)


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Figure 3.20 shows a shallow continuous foundation supported

by a stronger soil layer underlain by a weaker soil, which

extends to a great depth. For the two soil layers, the physicalparameters are as follows :


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If the depth H is relatively small compared to the foundation width B, a

punching shear failure will occur in the top soil layer followed by a general

shear failure in the bottom soil layer. This is shown in Figure 3.20a.

However, if the depth H is relatively large, then the failure surface will be

completely located in the top soil layer, which is the upper limit for the

ultimate bearing capacity. This is shown in Figure 3.20b.

The ultimate bearing capacity, qu, for this problem as shown in Figure

3.20a can be given as

(3.60)

Where

B = width of the foundation

Ca = adhesive forcePp= passive force per unit length of the faces and bb

qb= bearing capacity of the bottom soil layer

= inclination of the passive force Ppwith the horizontal

HB

PCqq

pa

bu 1

)sin(2


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Note that, in Eq. (3.60),

(3.61)

where = adhesionEquation (3.60) can be simplified to the form

(3.62)

Where, = horizontal component of passive earth pressurecoefficient

However, let

= (3.63)

Where, Ks = punching shear coefficientSo,

(3.64)


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The punching shear coefficient, Ks, is a function of q2/ q1and

, or

Note that q1 and q2 are the ultimate bearing capacities of aconditinuous foundation of width B under vertical load on thesurfaces of homogeneous thick beds of upper and lower soil,or

(3.65)and

(3.66)

where = bearing capacity factors for frictionangle (Table 3.4)

= bearing capacity factors for frictionangle (Table 3.4)

It is important to note that, for the top layer to be a stronger soil,

q2/ q1should be less than one.


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The variation of Ks, with q2/ q1 and is shown in Figure 3.21.


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The variation ca/ c1with q2/ q1 is shown in Figure 3.22


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If the height H is relatively large, then the failure

surface in soil will be completely located in the

stronger upper-soil layer (Figure 3.20 b).

For this case,

where = bearing capacity factor for

(Table 3.4)

and q =

)1(qN


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Now, combining Eqs. (3.64) and (3.67)

(3.68)

For rectangular foundations, the preceding equation can be

extended to the form

(3.68)

where

(3.69)

(3.70)

where

= shape factors with respect to top soil layer (Table 3.5)

= shape factors with respect to bottom soil layer (Table 3.5)


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Special Cases

1. Top layer is strong sand and bottom layer is saturated soft clay (2=0). From eqs.

(3.68), (3.69), and (3.70),

(3.71)

(3.72)

Hence,

(3.73)

For determination ofKsfrom figure 3.21,

(3.74)

)1()1(1)1()1(1

12

2

1

)(14.52.01

sqsqft

fb

FBNFNDq

HDcLBq

)1()1(1)1()1(11

12

12

2

1

tan21114.52.01

sqsqff

sf

b

FBNFNDD

B

K

H

D

L

BHc

L

Bq

)1(1

2

)1(1

)2(2

1

2

5.0

14.5

2

1

BN

c

BN

Nc

q

q c


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2. Top layer is strong sand and bottom layer is weaker sand (c1=0, c2=0).

The ultimate bearing capacity can be given as

(3.75)

Where(3.76)

(3.77)

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

1sqsqft FBNFNDq

tsf

sqsqfu

qHB

K

H

D

L

BH

FBNFNHDq

112

1

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

tan211

21)(

)1(1

)2(2

)1(1

)2(2

1

2

2

12

1

N

N

BN

BN

q

q


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3. Top layer is stronger saturated clay (2=0) and bottom layer is weaker

saturated clay (2=0). The ultimate bearing capacity can be given as

(3.78)

(3.79)

For this case

(3.80)

ft

tfa

u

DcL

Bq

qDB

Hc

L

Bc

L

Bq

11

12

14.52.01

2114.52.01

1

2

1

2

1

2

14.5

14.5

c

c

c

c

q

q


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EXAMPLE 3.8

A foundation 1.5 m x 1 m is located at a depth,Df of 1 m in a stronger clay. A softer

clay layer is located at a depth H of 1 m, measured from the bottom of the foundation.

For the top clay layer,Undrained shear strength = 120 kN / m2

Unit Weight = 16.8 kN / m3

And for the bottom clay layer,

Undrained shear strength = 48 kN / m2

Unit Weight = 16.2 kN / m3

Determine the gross allowable load for the foundation with an FS of 4.

SolutionFor this problem, Eqs. (3.78), (3.79), and (3.80) will apply, or

Given :

B = 1 m H = 1 m Df= 1 m

L = 1.5 m 1= 16.8 kN / m3

f

fa

b

DcL

B

DB

Hc

L

Bc

L

Bq

11

12

14.52.01

2114.52.01


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From figure 3.22, c2/ c1= 48 / 120 = 0.4, the value of ca/ c10.9, so

Ca= (0.9)(120) = 108 kN/m2

Check : From Eq. (3.79),

Thus qu= 656.4 kN/m2(that is, the smaller of the two values calculated above)

and

The total allowable load is

( qall) (1 x 1.5) = 246.15 kN

2kN/m656.416.8360279.6

)1)(8.16(1

)1)(108)(2(5.111)48)(14.5(

5.112.01

uq

2kN/m8.7158.16699

)1)(8.16()48)(14.5(5.112.01

tq

2kN/m1.1644

4.656

FS

qq uall


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EXAMPLE 3.9

Refer to Figure 3.20. Assume that the tp layer is sand and the bottom layer is soft

saturated clay. Given :

For the sand : 1= 117 lb / ft2

; 1= 400

For the soft clay ( bottom layer ) : c2= 400 lb / ft2; 2= 0

For the foundation : B = 3 ft; Df=3 ft ; L = 4.5 ft ; H =4 ft

Determine the gross ultimate bearing capacity of the foundation.

SolutionFor this case Eqs. (3.73) and (3.74) apply. For 1= 400, from Table 3.4, N=

109.41 and

From Figure 3.21, for and 1 = 400, the value of

Ks2.5 . Equation (3.73) gives

107.0)41.109)(3)(117)(5.0(

)14.5)(400(

5.0 )1(1

)2(2

1

2 BN

Nc

q

q c

107.05.0/ )1(1)2(2 BNNc c

2

2

112

12

lb/ft813535154542330

)3)(117(3

40tan)5.2(

4

)3)(2(1)4)(117(

5.4

31)400)(14.5(

5.4

32.01

tan211)14.5(2.01

fs

f

u DB

KH

DH

L

Bc

L

Bq


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Again, from Eq. (3.73)

From Table 3.4, for 1= 400, N= 109.41, Nq= 64.20From Table 3.5

Hence,

2

)1(

1)1(

lb/ft622.45)733.0)(4.109)(3)(117(2

1)4.1)(2.64)(3)(117(

733.05.4

3)4.0(14.01

4.140tan5.4

31tan1

t

s

qs

q

L

BF

L

BF

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

1sqsqft FBNFNDq

2lb/ft8135u q


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