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2.0 SHALLOW FOUNDATION :

A shallow foundation must :

- be safe against overall shear failure in the soil- not undergo excessive settlement

Nature of bearing capacity failure are : (as shown in Figure 2.1)

- general shear failure (for stiff clay or dense sand)- local shear failure(for medium dense sand or clayey soil)

- punching shear failure(loose sand or soft clay)

Figure 2.1 Nature of bearing capacity failure : (a) general shear (b)local shear (c) punching shear

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Vesic (1973) proposed a relationship for the bearing capacityfailure on sands in terms of relative density, Dr depth offoundation, Df and B*, Figure 2.2

Where : LBBLB += 2* and B width, L length of foundation

NOTE : L IS ALWAYS GREATER THAN B

For square; B=L and for circular; B=L=Diameter of foundationand B* = B

Figure 2.2 Modes of foundation failure in sand, (Vesic, 1973)

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Terzaghi suggested for a continuous or strip foundation withfailure surface as in Figure 2.3

Figure 2.3 Bearing capacity failure in soil under rough rigidcontinuous foundation

Soil above the bottom of foundation is surcharge, q = Df The failure zone under the foundation is separated into three

parts namely;- triangular ACD under the foundation- radial shear zones ADF and CDE with curves DE and DF

as arcs of logarithmic spiral- Rankine passive zones AFH and CEG

CAD and ACD are assume to equal friction angle,

Thus ultimate bearing capacity, qu for general shear failure can

be expressed as :

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1819202122232425

15.1216.5617.6918.9220.2721.7523.3625.13

6.046.707.448.269.1910.2311.4012.72

2.593.073.644.315.096.007.088.34

44454647484950

151.95172.28196.22224.55258.28298.71347.50

147.74173.28204.19241.80287.85344.63415.14

261.60325.34407.11512.84650.67831.991072.80

From Kumbhojkar (1993)

And ultimate bearing capacity, qufor local shear failure can beexpressed as :

)..........('3.0''867.0

).........('4.0''867.0

)........('5.0''3

2

foundationcircularBNqNcNq

foundationsquareBNqNcNq

foundationstripBNqNcNq

qcu

qcu

qcu

++=

++=

++=

Where :Nc, Nq, N (see Table 2.2) are reduced bearing capacity

factors can be calculated by using Nc, Nq, N - bearing capacity

factors with

= tan

3

2tan' 1

Table 2.2 Terzaghis Modified Bearing Capacitys Factors

Nc Nq N Nc Nq N01

234567891011121314151617

1819202122232425

5.705.90

6.106.306.516.746.977.227.477.748.028.328.638.969.319.6710.0610.47

10.9011.3611.8512.3712.9213.5114.1414.80

1.001.07

1.141.221.301.391.491.591.701.821.942.082.222.382.552.732.923.13

3.363.613.884.174.484.825.205.60

0.000.005

0.020.040.0550.0740.100.1280.160.200.240.300.350.420.480.570.670.76

0.881.031.121.351.551.741.972.25

2627

28293031323334353637383940414243

44454647484950

15.5316.30

17.1318.0318.9920.0321.1622.3923.7225.1826.7728.5130.4332.5334.8737.4540.3343.54

47.1351.1755.7360.9166.8073.5581.31

6.056.54

7.077.668.319.039.8210.6911.6712.7513.9715.3216.8518.5620.5022.7025.2128.06

31.3435.1139.4844.4550.4657.4165.60

2.592.88

3.293.764.394.835.516.327.228.359.4110.9012.7514.7117.2219.7522.5026.25

30.4036.0041.7049.3059.2571.4585.75

Example 2.1

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Given : A square foundation, 1.5m x 1.5m in plan viewSoil parameters : = 20, c = 15.2 kN/m2, =17.8 kN/m3

Assume : FS = 4, general shear failure condition and Df= 1 mFind : Allowable gross load on the foundation

Solution : ).........(4.0'3.1 foundationsquareBNqNNcq qcu ++=

For = 20, (Table 2.1); Nc = 17.69, Nq = 7.44, N= 3.64

Thus( ) ( ) ( ) ( ) )64.3)(5.1)(8.17)(4.0(44.78.17169.172.153.14.03.1 ++=++=

BNqNcNq

qcu

2/52187.3843.13255.349 mkN=++=

Allowable bearing capacity : 2/1304

521mkN

FS

qq uall ===

Thus total allowable gross load, Q

kNBAqQ allall 5.292)5.15.1(1301302 ====

Example 2.2

Given : Repeat example 2.1Assume : Local shear failure conditionSolution :

).........('4.0''867.0 foundationsquareBNqNcNq qcu ++=

For = 20, (Table 2.2); Nc = 11.85, Nq = 3.88, N= 1.12( ) ( ) ( ) ( ) )12.1)(5.1)(8.17)(4.0(88.38.17185.112.15867.0'4.0'''867.0 ++=++=

BNqNNcq

qcu

2/3.2370.121.692.156 mkN=++=

Allowable load :2/3.59

4

3.237mkN

FS

qq uall === ; kNAqQ allall 4.133)5.15.1(3.59 ===

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All equations mentioned before are based on the location ofwater table well below the foundation; if otherwise, some

modification should be made according to the location of thewater table, see Figure 2.4

Figure 2.4 Modification of bearing capacity for water table

Case I : 0 D1 Df

- q(effective surcharge) = '21 DD +

- where :- '- effective unit weight = wsat - sat - saturated unit weight of soil

- w - unit weight of water = 9.81kN/m3 or 62.4 lb/ft3

- '= in the last term of the equation Case II : 0 d B

- the value fDq =

- ( )'' +==B

d

Case III : d B- water has no effect on the qu

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FS

qq uall = , where :

- qall - gross allowable load-bearing capacity,- qu gross ultimate bearing capacity,

- FS factor of safety

Values of FS against bearing capacity failure is 2.5 to 3.0. Net stress increase on soil = net ultimate bearing capacity/FS

FS

qqq unetall

=)( , and :

f

unetu

Dq

qqq

=

=)( ;

Where : qall(net) net allowable bearing capacityqu(net) net ultimate bearing capacity

Procedure for FSshear

a. Find developed cohesion,cd and angle of friction,d;

==

shear

d

shear

dFS

andFS

cc

tantan............. 1

)..........(3.03.1

).........(4.03.1

)........(5.0

foundationcircularBNqNcNq

foundationsquareBNqNcNq

foundationstripBNqNcNq

qcu

qcu

qcu

++=

++=

++=

b. Terzaghis equations become (with cd and d):

)..........(3.03.1

).........(4.03.1

)........(5.0

foundationcircularBNqNNcq

foundationsquareBNqNNcq

foundationstripBNqNNcq

qcdu

qcdu

qcdu

++=

++=

++=

With :Nc, Nq, N - bearing capacity factors for d

c. Thus, the net allowable bearing capacity :

( ) BNNqNcqqq qcdallnetall2

11)( ++==

Example 2.3

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UsingFS

qqq unetall

=)( ; and FS = 5; find net allowable load for the

foundation in example 2.1 with qu = 521 kN/m2

With qu = 521 kN/m

2

; q = 1(17.8) = 17.8 kN/m

2

2

)( /64.1005

8.17521mkN

FS

qqq unetall =

=

=

Hence Qall(net) = 100.64(1.5x1.5) = 226.4 kN

Example 2.4

Using Example 3.1, and Terzaghis equation

).........(4.03.1 foundationsquareBNqNcNq qcu ++= with FSshear = 1.5;

Find net allowable load for the foundation

For c=15.2 kN/m2, = 20 and

==

shear

d

shear

dFS

andFS

cc

tantan............. 1

cd =2/13.10

5.1

2.15mkN

FS

c

shear

==

d = tan-1[shearFS

tan] = tan-1[

5.1

20tan] = 13.64

With : BNNqNcq qcdnetall 4.013.1)( ++=

From Table 2.1 : =13.6 ; 2.1N ; 8.3qN ; 12cN (estimation)

Hence :

( ) ( ) ( ) ( ) ( ) ( )2

)(

/2202.128.490.158

2.15.18.174.018.38.171213.103.1

mkN

q netall

=++=

++=

( ) ( ) kNQ netall 4955.15.1220)( ==

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The need to address for rectangular shape foundation where :(0 1

10

tan2

245tan eNq

+=

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( )

+=

B

DF

f

cd

1tan4.01 , ( )

+=

B

DF

f

qd

12 tansin1tan21

1=dF

NOTE : tan-1(Df/B) is in

- inclination2

901

==

qici FF

2

1

=

iF

Where : inclination of load from vertical

For undrained condition ( = 0)

qFFNcq cdcscuu +=

cdcscuuunet FFNcqqq ==)(

Skemptons :

+

+=

L

B

B

Dcq

f

unet 2.012.015)(

Table 2.3Vesics Bearing Capacity Factors for General Equation (1973)

Nc Nq N Nq/ Nc Tan Nc Nq N Nq/ Nc Tan

012345678910

1112131415161718192021

5.145.385.635.906.196.496.817.167.537.928.35

8.809.289.8110.3710.9811.6312.3413.1013.9314.8315.82

1.001.091.201.311.431.571.721.882.062.252.47

2.712.973.263.593.944.344.775.265.806.407.07

0.000.070.150.240.340.450.570.710.861.031.22

1.441.691.972.292.653.063.534.074.685.396.20

0.200.200.210.220.230.240.250.260.270.280.30

0.310.320.330.350.360.370.390.400.420.430.45

0.000.020.030.050.070.090.110.120.140.160.18

0.190.210.230.250.270.290.310.320.340.360.38

2627282930313233343536

3738394041424344454647

22.2523.9425.8027.8630.1432.6735.4938.6442.1646.1250.59

55.6361.3567.8775.3183.8693.71105.11118.37133.88152.10173.64

11.8513.2014.7216.4418.4020.6723.1826.0929.4433.3037.75

42.9248.9355.9664.2073.9085.3899.02115.31134.88158.51187.21

12.5414.4716.7219.3422.4025.9930.2235.1941.0648.0356.31

66.1978.0392.25109.41130.22155.55186.54224.64271.76330.35403.67

0.530.550.570.590.610.630.650.680.700.720.75

0.770.800.820.850.880.910.940.971.011.041.08

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22232425

16.8818.0519.3220.72

7.828.669.6010.66

7.138.209.4410.88

0.460.480.500.51

0.400.420.450.47

484950

199.26229.93266.89

222.31265.51319.07

496.01613.16762.89

1.121.151.20

Example 2.5

Figure 2.5

Given : A square foundation (B x B), Figure 2.5, Q=150 kN.Df = 0.7m, load is inclined at 20 from vertical, FS = 3.Use general bearing capacity factors

Find : The width of foundation B

)'2

1( idsqiqdqsqu FFFBNFFFqNq += ;

( ) ( )

2

/6.12187.0 mkNq ==

From Table 2.3 : For =30: Nq = 18.4, N = 22.4, Nq/ Nc = 0.61,Tan = 0.58

( ) 58.158.01tan1 =+=+=B

B

L

BFqs ;

6.04.014.01 =

==B

B

L

BFs

( ) ( ) ( )BBB

DF

f

qd

202.01

7.030sin158.021sin1tan21

22 +=+=+= ;

1=dF

605.090

201

901

22

=

=

== qici FF ;

11.030

2011

22

=

=

=

iF

12

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So

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) BBB

B

FFFBNFFFqNq idsqiqdqsqu

3.1368.442.22111.016.04.221821605.0202.0158.14.186.12

)'2

1(

++= + +=

+=

BBB

setthusq

q uall 43.489.14

73.73150

:3 2

++==

By trial and error : B=1.3m

Eccentrically loaded foundations give non-uniform distributionof pressure, Figure 2.6

Figure 2.6 Eccentrically loaded foundations

Eccentricity,Q

Me =

qmax and qmin is given by :

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+=

B

e

BL

Qq

61max and

=

B

e

BL

Qq

61min

if e > B/6, and qmin becomes negative then :

( )eBL

Qq

23

4max

=

Factor of safety against bearing capacity failure; effective areamethod, by Meyerhof (1953)

a. Find effective dimensions of the dimensions

- the smaller of B and L is the width- effective width, B = B 2e- effective length, L = L- if e is in the direction of L than L = L 2e

b. Find the ultimate bearing capacity, qu :

idsqiqdqsqcicdcscu FFFNBFFFqNFFFcNq '2

1' ++=

- use L and B to find sqscs FandFF ..,

- use B to find dqdcdFandFF

..,

c. Total ultimate load, ( )''''' LBqAqQ uuult == ; where A effective

area

d. Factor of safety,Q

QFS ult=

e. Check FS against qmax ;max

'

q

qFS u=

Example 2.6

Given : A square foundation as shown in Figure 2.7. Using generalbearing capacity factors, (table 2.3)

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Figure 2.7Find : Ultimate load, Qult,

assume one way load eccentricity, e = 0.15m

Solution : with c = 0; idsqiqdqsqu FFFNBFFFqNq '21

' +=

Where :

q = 0.7(18) = 12.6 kN/m2

for = 30, from Table 2.3 : Nq=18.4 and N=22.4B = 1.5 2(0.15) = 1.2mL = 1.5m

Thus values for general bering capacity equations : (using B and L)

462.130tan5.1

2.11tan

'

'1 =

+=+=

L

BFqs

( )( ) ( )

135.15.1

7.0289.01sin1tan21

2 =+=+=B

DF

f

qd

68.05.1

2.14.01

'

'4.01 =

=

=L

BFs

1=dF

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2

21

/2.5495.1647.384

168.04.222.118135.1462.14.186.12'

mkN

q u

=+=

+=

15

Sand :

0

30

/18 3

==

=

c

mkN

1.5m x 1.5 m

0.7 m

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Qult = qu X A = 549.2 X (1.5X1.2) = 988kN

Qall = 988/3 = 330kN with FS=3

Example 2.7 :

Given : The strip footing shown below is to be constructed in auniform deposit of stiff clay and must support a wall that imposes aloading of 152 kN/m of wall length. Use general bearing capacity

factors.

Find : The width of footing with FS of 3.

Figure 2.8Solution :

22

/9.722

/8.145

2;

........(5.0

mkNmkNq

cwith

foundationstripBNqNcNq

u

qcu

===

++=

And =0; from the Table 2.3 Nc = 5.14, Nq = 1.0 and N

=0

mmkN

mkNdthofwallrequiredwi

mkNmkN

q

mkNBmkNmmkNmkNq

all

ult

15.1/4.132

/0.152

/4.1323

/3.397

/3.397)0)()(/82.18(5.0)0.1)(2.1)(/82.18()14.5)(/9.72(

2

22

2332

==

==

=++=

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B required is 1.5 meter to be conservative

2.8

Example 2.8 :Soil deposit has the following ;=20.44 kN/m3, =30, c=38.3kN/m2

Square footing located 1.52 m below surface, carries 2670 kN andgroundwater is negligible. Use Terzaghis values, (Table 2.1).

Find : The right dimension B. Use Terzaghis equation).........(4.03.1 foundationsquareBNqNcNq qcu ++=

With =30; Nc=37.16, Nq=22.46, N

=19.13

Assume B=3 m;

mBBmmkN

kNwallofwidthrequired

mkNmkN

q

mkNmkN

mmkNmmkNmkNq

all

ult

63.165.2/7.1005

2670

/7.10053

/3017

/3017/4696981850

)13.19)(3)(/44.20(4.0)46.22)(52.1)(/44.20()16.37)(/3.38(3.1

22

2

22

22

332

====

==

=++=

++=

Assume B=1 m;

17

2670 kN

= 20.44 kN/m3

1.52m =30

c = 38.3 kN/m2Figure 2.9

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mBBmmkN

kNwallofwidthrequired

mkNmkN

q

mkNmkN

mmkNmmkNmkNq

all

ult

65.172.2/980

2670

/9803

/2939

/2939/3916981850

)13.19)(1)(/44.20(4.0)46.22)(52.1)(/44.20()16.37)(/3.38(3.1

222

22

22

332

====

==

=++=

++=

Assume B=2m;

mBBmmkN

kNwallofwidthrequired

mkNmkN

q

mkNmkN

mmkNmmkNmkNq

all

ult

67.180.2/954

2670

/9543

/2861

/2861/3136981850

)13.19)(2)(/44.20(4.0)46.22)(52.1)(/44.20()16.37)(/3.38(3.1

22

2

22

22

332

====

==

=++=

++=

Assume B=1.8m;

mBBmmkN

kNwallofwidthrequired

mkNmkN

q

mkNmkN

mmkNmmkNmkNq

all

ult

68.183.2/943

2670

/9433

/2830

/2830/2826981850

)13.19)(8.1)(/44.20(4.0)46.22)(52.1)(/44.20()16.37)(/3.38(3.1

22

2

22

22

332

====

==

=++=

++=

Assume B=1.7m;

mBBmmkN

kNwallofwidthrequired

mkNmkN

q

mkNmkNmmkNmmkNmkNq

all

ult

7.185.2/938

2670

/9383

/2814

/2814/2666981850)13.19)(7.1)(/44.20(4.0)46.22)(52.1)(/44.20()16.37)(/3.38(3.1

22

2

22

22

332

====

==

=++=++=

Therefore use 1.7m x 1.7m

2.9

Can be computed by using flexural formula of :

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y

y

x

x

I

xM

I

yM

A

Qq =

Where :

q contact pressureQ total axial vertical loadA area of footing

Mx, My total moment about respective x and y axesIx, Iy moment of inertia about respective x and y axesx, y distance from centroid to the outer most point at

which the contact pressure is computed along respective x and

y axes.

Example 2.9

A pad footing with dimension of 1.52 x 1.52m acted upon by the loadof 222.4kN. Estimate soil contact pressure and FS against bearingcapacity.

Given :1.52m by 1.52m square footing; P=222.4kN; soil =18.85kN/m3

concrete =24 kN/m3; qu = 143.64 kN/m2

Find :a. Soil contact pressureb. FS against bearing capacity pressure

19

0.14m20.91m

1.22m

0.31m1.52m

Figure 2.10

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Solution :

a.y

y

x

x

I

xM

I

yM

A

Qq = ; Mx=My=0; since load on centroid

Total load calculation, Q :

Column load, P = 222.4kNWeight of footing base= (1.52m)(1.52m)0.31m(24kN/m3) = 17.19 kNWeight of footing pedestal= (0.14m)(0.14m)(0.91m)(24kN/m3) = 0.43 kNWeight of backfill soil

= [(1.52m)(1.52m)-(0.14m)(0.14m)](0.91m) x 18.85kN/m3

= 39.3kN

Q = 222.4 + 17.19 + 0.43 + 39.3 = 279.32kNArea, A = 1.52mx1.52m = 2.31m2

Soil contact pressure or Stress, q = Q/A = 120.9 kN/m2

b.2

2

/82.712

/64.143

2

4.02.1

mkNmkNq

c

BNNDcNq

u

qfcult

===

++=

Assuming cohesive soil has : =0 and c>0; thus :Nc=5.14, Nq=1.0, N=0, Df=1.22m

85.39.120

98.465

/98.465

0)0.1)(22.1(85.18)14.5)(82.71(2.14.02.1

2

===

=

++=++=

q

qFS

mkN

BNNDcNq

ult

qfcult

Since FS > 3.0; thus ok.

20

Example 2.10

Draw soil contact pressure for

footing in Figure 2.11

Conversion to SI unitP=222.4 kN;H=88.96 kN;M=81.35kN.m;W=88.96 kNDf=1.22m;B=2.29m (7.5ft);

L=1.52m (5ft)

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Given : 2.29m by 1.52m rectangular footingFind : Contact pressure and soil pressure diagram

Solution :

Using flexural formula;y

y

x

x

I

xM

I

yM

A

Qq =

Q = P + W = 222.4 kN + 88.96 kN = 311.36 kN.A = 2.29m x 1.52m = 3.48 m2;

Mx=0; My=88.96(1.22)+81.35=189.88kN.m (Moment about point C)

x = 2.29/2 = 1.145m; 43

52.112

)29.2(52.1m

mmIy ==

22

22

42

/53.53...../47.232

/143/47.8952.1

)145.1)(88.189(

48.3

36.311

mkNqandmkNq

mkNmkNm

mmkN

m

kNq

leftright =+=

=

=

Take V = 0 and Mc = 0 will produce :

V = 0 : )........(36.311)52.1(2

......0))((2

AkNmqd

andWPLdq

=

=

Mc = 0 : see Figure 2.12 (b) and (c)

03

))((2

))(( =

+

dxLd

qSHM

21

Figure 2.11

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( ) ( )

2/46.254.,36.311)52.1)(61.1(2

:)(int.

61.1079.10351.35653.10835.81

.....032

29.236.31122.196.88.35.81

mkNqkNmmq

Aosubstitue

mdd

Bdm

mkNmkN

==

==++

=

+

Example 2.11

22

Figure 2.12 (a) and (b)

1.61m

254.46kN/m2

2.29m9m

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Given : A 6 ft x 6 ft footing as shown; load P=60kips; weight ofconcrete footing including pedestal + base pad, W1=9.3kips; backfill,W2=11.2kips; horizontal load = 4kips; qall for soil = 3.0 kips/ft2.

Find :Contact pressure and soil pressure diagram.Shear and moment at section A-A (in the Figure E3.14)FS against sliding if coefficient of friction, = 0.40FS against overturning.

Solution :

1.y

y

x

x

I

xM

I

yM

A

Qq =

Q=P+W1+W2=60+9.3+11.2=80.5kipsA=6ftx6ft=36ft2

My=4kipsx4.5ft=18kip-ft (about point C)x=6ft/2=3ftIy=6ft(6ft)3/12=108ft4; Mx=0; Mxy/Ix=0

22

42/50.0/24.2

108

)3(.18

36

5.80ftkipftkips

ft

ftftkip

ft

kips

I

xM

I

yM

A

Qq

y

y

x

x ===

So : qright = 2.74 kips/ft2 < 3.0 kips/ft2 ; OKqleft = 1.74 kips/ft2 < 3.0 kips/ft2 ; OK

23

gure 2.13

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2. FDG and EDH are similar triangles; so

( ) ftkipsftkipsft

kipsAAatMoment

kipskipskips

ftftkipftftftkipsftAAatShear

ftkipDEft

ftDEftFG

ft

ftft

EHftkipsDFFG

EH

DF

DE

.7.3925.253.22

25.293.31:...

46.3453.293.31

)6)(/375.0)(25.2()6)(/375.074.2(25.2:...

/375.0;..6

25.2

0.1;...6

25.22

5.1

2

6

;.../0.174.174.2;.....

32

2

212

2

2

=+

=

=+=

+=

===

=====

3.( )

05.84

)40.0(2.113.960

.........

=++

=

=

kips

kipskipskips

forcesHorizontaleandsoilbetweenbasfrictionoftcoefficienloadverticalTotalslidingagainstFS

4. 4.13)5.4(4

)2/6(5.80

.

.Re.. ===

ftkips

ftkips

momentTurning

momentsistinggoverturninagainstFS

24

Pressure diagram

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Foundation settlement under load can be classified according to twomajor types :

(a) immediate or elastic settlement, Se

(b) consolidation settlement, Sc

Elastic settlement, Se takes place immediately during or afterconstruction of structure.

Consolidation settlement, Sc is time dependent comprises of twophases; namely, primary and secondary consolidation settlement.

Elastic settlement of foundations on saturated clay is given by Janbuet al., (1956) using the equation :

s

eE

BqAAS 021=

where :

A1 is a function of H/B and L/B and A2 is a function of Df/B

All parameters of H, B and Df (with L into the paper) are asshown in Figure 2.14.

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Parameters

A2 Versus Df/B

A1 Versus H/B and L/B

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Schmertmann, (1978) proposed that the elastic settlement in sandy

soil as :

( ) =2

0

21

z

s

ze z

E

IqqCCS

where :

Iz strain influence factor

C1 correction factor due to depth =( )

qq

q5.01

C2 correction factor due to soil creep =

+

1.0log2.01

yearsintime

q - stress at the level of foundation (due to loading + selfweight of footing + weight of soil above footing)

fDq =

Calculation of elastic settlement using strain influencefactor

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The variation of Iz with depth below the footing for square orcircular are as below :

Iz = 0.1 at z = 0Iz = 0.5 at z = z1 = 0.5BIz = 0 at z = z2 = 2B

Footing with L/B 10 (rectangular footing) :

Iz = 0.2 at z = 0Iz = 0.5 at z = z1 = BIz = 0 at z = z2 = 4B

Elastic parameters such as Es and s in Table 2.4 can be used if thereal laboratory test results not available.

Elastic parameters of various soils

Type of soil Modulus of Elasticity,Es(MN/m2)

Poissons ratio, s

Loose sand 10.5 24.0 0.20 0.40Medium dense sand 17.25 27.60 0.25 0.40

Dense sand 34.50 55.20 0.30 0.45Silty sand 10.35 17.25 0.20 0.40

Sand and gravel 69.00 172.50 0.15 0.35Soft clay 4.1 20.7

0.20 0.50Medium clay 20.7 41.4Stiff clay 41.4 96.6

(a) Primary consolidation, Sc

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Many methods were developed in estimating the value ofconsolidation settlement, Sc.

Due to simplicity only chart based on Newmarks (1942), Figure 2.18will be used in estimating the consolidation settlement.

Primary consolidation, Sc calculated as :

00

log1 p

p

e

HCS cc

+

=

where : Cc compression index (given)

H thickness of clay layere0 initial void ratio (given)p = p0 + p, final pressure

p0 overburden pressurep =4(Ip)q0 net consolidation pressure at mid-height of

clay layer

Ip Influence factor (from Figure 2.18)

q0 net stress increase

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Chart for determining stresses below corners of rigidand isotropic.

Example 2.7

Given :

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A foundation to be constructed as in Figure 2.19. The base of the

foundation is 3m by 6m, and it exerts a total load of 5400 kN, whichinclude all self weight. The initial void ratio, e0 is 1.38 andcompression index, Cc is 0.68.

Required :

Expected primary consolidation settlement of clay layer.

Solution :

p0 = 19.83(200 - 198) + (19.83 9.81)(198 - 192) + (17.1 9.81)(192 185.6)/2 = 123.1 kN/m2

Weight of excavation = 19.83(200 - 198) + (19.83 9.81)(198 195.5) = 64.7kN/m2

( ) ( ) ( )( )[ ] 2

0

/3.2355.19519881.983.1919820083.19

63

5400

,

mkN

mm

kN

excavationofweightpressureloadqincreasestressNet

=+

=

=

By dividing the base into 4 equal size of 1.5m by 3.0m :

mz = 1.5m nz = 3.0m

mmm

mz 7.62

6.1850.1925.195 =

+=

224.07.6

5.1

== mm

m ; 448.07.60.3

== mm

n

From Figure 2.18, the influence coefficient is 0.04

Therefore ; ( ) 22 /6.37/3.23504.04 mkNmkNp ==

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Final pressure, p = p0 + p = 123.1 + 37.6 = 160.7 kN/m2.

Therefore; mmkN

mkNm

p

p

e

HCS cc 212.0

/1.123

/7.160log

38.11

4.668.0log

1 2

2

00

=

+=

+

=

Secondary settlement, Ss is computed from the following calculation(U.S. Department of the Navy, 1971)

=

p

ss

t

tHCS log

where :

Ss secondary compression settlementC coefficient of secondary compression, can be determined

from Figure3.26H thickness of clay layer that is consideredts time for which settlement is requiredtp time to completion of primary consolidation

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Value of C

Bowles (1977) proposed a correlation of the net allowable

bearing pressure for foundations with SPT (N-values). The following equations are used :

( )( ) mBforS

FNmkNq dallnet 22.14.25

16.19)/( 2)(

=

And

( ) ( ) mBforS

FB

BNmkNq dallnet 22.1

4.2528.3

128.398.11)/(

2

2

)( >

+=

Where :

Fd depth factor = 33.133.01

+ B

Df

S tolerable settlement, in mm.

Example 2.8

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Given:

A shallow square footing for a column is to be constructed.

Design load is 1000 kN. The foundation soil is sand. The SPTnumbers from field exploration as shown in the table.

Assume that the footing must be 1.5m deep, the tolerablesettlement as 25.4mm and the size is > 1.22m.

Required :

(a) The exact size of the footing (b) safety factor for foundation

Solution :

Navg = (7+8+11+11+13+10+9+10+12)/9=10With S=25.4mm and N=10

( ) ( ) ddallnet FB

BF

B

BmkNq

22

2

)(28.3

128.38.119

4.25

4.25

28.3

128.31098.11)/(

+=

+=

33.133.01

+=

B

DF

f

d

By trial and error (set the table for calculation)

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From the table it is seen that the appropriate B=2.4

Setting general equation and equation for net ultimate with c=0(for sandy soil) :

f

unetu

Dq

qqq

=

=)(

; idsqiqdqsqcicdcscu FFFBNFFFqNFFFcNq 21++=

( ) qFFFBNFFFqNqqq idsqiqdqsqultnetu +== 2

1

For N=10; friction angle of=34 is considered (from table onSI)

With no inclination so Fqi=Fi=1.0

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From table 2.3 Nq=29.44, N=41.06

So for a tolerable settlement of 25.4mm, the SF required iscalculated as : SF=Qnet(u)/Q= 10,322kN/1000kN = 10.3 whichis OK, therefore most design controlled by tolerable criterion.

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