CH2 Shallowfoundation (SEM2 200910)

8/20/2019 CH2 Shallowfoundation (SEM2 200910)

1/35

BFC 4043

AKS & PM I R AZI ZAN SEM 2 200910 1

2.0 SHALLOW FOUNDATION :

2.1

General Concept

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


2/35

BFC 4043


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 : L B BL B 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)


3/35

BFC 4043


2.2 Terzaghi’s Bearing Capacity

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 DFas arcs of logarithmic spiral

- Rankine passive zones AFH and CEG

CAD and ACD are assume to equal friction angle, Ø

Thus ultimate bearing capacity, qu for general shear failure can

be expressed as :

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

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

)........(5.0

foundationcircular BN qN cN q

foundation square BN qN cN q

foundation strip BN qN cN q

qcu

qcu

qcu


4/35

BFC 4043


Where : c – cohesion of soil - unit weight of soil

q = Df Nc, Nq, N - bearing capacity factors

And

1cot1

24cos2

cot2

tan2/4/32

qc N e

N

245cos2 2

tan2/4/32

e N q

tan1cos2

12

p K N

where p K - passive pressure coefficient

Table 2.1 summarizes values for Nc, Nq, and N

Table 2.1 Terzaghi’s Bearing Capacity’s Factors

Ø Nc Nq N Ø Nc Nq N

01234567891011121314

1516171819202122232425

5.706.006.306.626.977.347.738.158.609.099.6110.1610.7611.4112.11

12.8613.6814.6015.1216.5617.6918.9220.2721.7523.3625.13

1.001.101.221.351.491.641.812.002.212.442.692.983.293.634.02

4.454.925.456.046.707.448.269.1910.2311.4012.72

0.000.010.040.060.100.140.200.270.350.440.560.690.851.041.26

1.521.822.182.593.073.644.315.096.007.088.34

262728293031323334353637383940

41424344454647484950

27.0929.2431.6134.2437.1640.4144.0448.0952.6457.7563.5370.0177.5085.9795.66

106.81119.67134.58151.95172.28196.22224.55258.28298.71347.50

14.2115.9017.8119.9822.4625.2828.5232.2336.5041.4447.1653.8061.5570.6181.27

93.85108.75126.50147.74173.28204.19241.80287.85344.63415.14

9.8411.6013.7016.1819.1322.6526.8731.9438.0445.4154.3665.2778.6195.03115.31

140.51171.99211.56261.60325.34407.11512.84650.67831.991072.80

From Kumbhojkar (1993)


5/35

BFC 4043


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

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

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

)........('5.0''32




qcu

qcu

qcu

Where : N’c, N’q, N’ (see Table 2.2) are reduced bearing

capacity factors can be calculated by using N’c, N’q, N’ -

bearing capacity factors with

tan

3

2tan' 1

Table 2.2 Terzaghi’s Modified Bearing Capacity’s Factors

Ø N’ c N’ q N’ Ø N’ c N’ q N’

012345678

9101112131415161718192021222324

25

5.705.906.106.306.516.746.977.227.47

7.748.028.328.638.969.319.6710.0610.4710.9011.3611.8512.3712.9213.5114.14

14.80

1.001.071.141.221.301.391.491.591.70

1.821.942.082.222.382.552.732.923.133.363.613.884.174.484.825.20

5.60

0.000.0050.020.040.0550.0740.100.1280.16

0.200.240.300.350.420.480.570.670.760.881.031.121.351.551.741.97

2.25

262728293031323334

35363738394041424344454647484950

15.5316.3017.1318.0318.9920.0321.1622.3923.72

25.1826.7728.5130.4332.5334.8737.4540.3343.5447.1351.1755.7360.9166.8073.5581.31

6.056.547.077.668.319.039.8210.6911.67

12.7513.9715.3216.8518.5620.5022.7025.2128.0631.3435.1139.4844.4550.4657.4165.60

2.592.883.293.764.394.835.516.327.22

8.359.4110.9012.7514.7117.2219.7522.5026.2530.4036.0041.7049.3059.2571.4585.75


6/35

BFC 4043


Example 2.1

Given : A square foundation, 1.5m x 1.5m in plan viewSoil parameters :

Ø’ = 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 foundation square BN qN N cq qcu

For Ø’ = 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 BN qN cN q qcu 2/52187.3843.13255.349 mkN

Allowable bearing capacity : 2/1304

521mkN

FS

qq uall

Thus total allowable gross load, Q

kN B AqQ all all 5.292)5.15.1(130130 2

Example 2.2

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

).........('4.0''867.0 foundation square BN qN cN q qcu

For Ø’ = 20°, (Table 2.2); Nc = 11.85, N

q = 3.88, N

= 1.12

)12.1)(5.1)(8.17)(4.0(88.38.17185.112.15867.0'4.0'''867.0

BN qN N cqqcu

2/3.2370.121.692.156 mkN

Allowable load :2/3.59

4

3.237mkN

FS

qq uall ; kN AqQ all all 4.133)5.15.1(3.59


7/35

BFC 4043


2.3

Effect of Water Table on Bearing Capacity

All equations mentioned before are based on the location of

water table well below the foundation; if otherwise, somemodification 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 D D

-

where :- ' - effective unit weight = w sat

- sa t - 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 f Dq

-

'' Bd

Case III : d ≥ B -

water has no effect on the qu

Note : the values of bearing capacity factors used strictly depending onwhether the condition is general or local shear failure.


8/35

BFC 4043


2.4 Factor of Safety, FS

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 unet all

)( , and :

f

unet u

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,Ød;

shear

d

shear

d FS

and FS

cc

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

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

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

)........(5.0




qcu

qcu

qcu

b. Terzaghi’s equations become (with cd and Ød):

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

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

)........(5.0

foundationcircular BN qN N cq

foundation square BN qN N cq

foundation strip BN qN N cq

qcd u

qcd u

qcd u

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

c. Thus, the net allowable bearing capacity :

BN N q N cqqq qcd all net all 2

11)(


9/35

BFC 4043


Example 2.3

Using FS

qqq unet all

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

foundation in example 2.1 with qu = 521 kN/m

2

With qu = 521 kN/m2; q = 1(17.8) = 17.8 kN/m2

2

)( /64.1005

8.17521mkN

FS

qqq unet all

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

Example 2.4

Using Example 3.1, and Terzaghi’s equation

).........(4.03.1 foundation square BN qN cN q qcu with FSshear = 1.5;

Find net allowable load for the foundation

For c=15.2 kN/m2, Ø = 20° and

shear

d

shear

d

FS

and

FS

cc

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

cd =2/13.10

5.1

2.15mkN

FS

c

shear

Ød = tan-1[

shear FS

tan] = tan-1[

5.1

20tan] = 13.64°

With : BN N q N cq qcd net all 4.013.1)(

From Table 2.1 : Ø=13.6° ; 2.1 N ; 8.3q N ; 12c N (estimation)

Hence :

2

)(

/2202.128.490.158

2.15.18.174.018.38.171213.103.1

mkN

q net all

kN Q net all 4955.15.1220)(


10/35

BFC 4043


2.5 The General Bearing Capacity Equation

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

tan2

245tan e N q


11/35

BFC 4043


B

D F

f

cd

1tan4.01

B

D F

f

qd

12 tansin1tan21 1d F

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

-

inclination2

901

qici F F

2

1

i F

Where : β – inclination of load from vertical

For undrained condition (Ø = 0)

q F F N cq cd cscuu

cd cscuuunet F F N cqqq )(

Skempton’s :

L

B

B

Dcq f

unet 2.012.015)(

Table 2.3 Vesic’s Bearing Capacity Factors for General Equation (1973)

Ø Nc Nq N Nq / Nc Tan Ø Ø Nc Nq N Nq / Nc Tan Ø012345678910111213141516171819202122232425

5.145.385.635.906.196.496.817.167.537.928.358.809.289.8110.3710.9811.6312.3413.1013.9314.8315.8216.8818.0519.3220.72

1.001.091.201.311.431.571.721.882.062.252.472.712.973.263.593.944.344.775.265.806.407.077.828.669.6010.66

0.000.070.150.240.340.450.570.710.861.031.221.441.691.972.292.653.063.534.074.685.396.207.138.209.4410.88

0.200.200.210.220.230.240.250.260.270.280.300.310.320.330.350.360.370.390.400.420.430.450.460.480.500.51

0.000.020.030.050.070.090.110.120.140.160.180.190.210.230.250.270.290.310.320.340.360.380.400.420.450.47

26272829303132333435363738394041424344454647484950

22.2523.9425.8027.8630.1432.6735.4938.6442.1646.1250.5955.6361.3567.8775.3183.8693.71105.11118.37133.88152.10173.64199.26229.93266.89

11.8513.2014.7216.4418.4020.6723.1826.0929.4433.3037.7542.9248.9355.9664.2073.9085.3899.02115.31134.88158.51187.21222.31265.51319.07

12.5414.4716.7219.3422.4025.9930.2235.1941.0648.0356.3166.1978.0392.25109.41130.22155.55186.54224.64271.76330.35403.67496.01613.16762.89

0.530.550.570.590.610.630.650.680.700.720.750.770.800.820.850.880.910.940.971.011.041.081.121.151.20

0.490.510.530.550.580.600.620.650.670.700.730.750.780.810.840.870.900.930.971.001.041.071.111.151.19


12/35

BFC 4043


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( id sqiqd qsqu F F F BN F F F qN q ;

2/6.12187.0 mkN q

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

58.158.01tan1 B

B

L

B F qs ; 6.04.014.01

B

B

L

B F s

B B B

D F

f

qd

202.01

7.030sin158.021sin1tan21

22 ; 1d F

605.090

201

901

22

qici F F ; 11.0

30

2011

22

i F

So

B B

B B

F F F BN F F F qN q id sqiqd qsqu

3.1368.44

2.22111.016.04.22182

1605.0

202.0158.14.186.12

)'2

1(

B B B

set thusq

q uall 43.489.14

73.73150:

3 2

By trial and error : B=1.3m


13/35

BFC 4043


2.6

Eccentrically Loaded Foundations

Eccentrically loaded foundations give non-uniform distribution

of pressure, Figure 2.6

Figure 2.6 Eccentrically loaded foundations

Eccentricity,Q

M e

qmax and qmin is given by :

B

e

BL

Qq

61max and

B

e

BL

Qq

61min

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

e B LQ

q23

4max


14/35

BFC 4043


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 :

id sqiqd qsqcicd cscu F F F N B F F F qN F F F cN q '2

1'

- use L’ and B’ to find sqscs F and F F ..,

-

use B to find d qd cd F and F F ..,

c. Total ultimate load, ''' '' L Bq AqQ uuult ; where A’ – effective

area

d.

Factor of safety,Q

Q FS ult

e.

Check FS against qmax ;max

'

q

q FS u

Example 2.6

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

Figure 2.7

Sand :

0

30/18 3

c

mkN

1.5m x 1.5 m

0.7 m


15/35

BFC 4043


Find : Ultimate load, Qult,assume one way load eccentricity, e = 0.15m

Solution : with c = 0;id sqiqd qsqu F F F N B F F F qN q '

2

1'

Where :

q = 0.7(18) = 12.6 kN/m2

for Ø = 30°, from Table 2.3 : Nq=18.4 and N =22.4

B’ = 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

B F qs

135.15.1

7.0289.01sin1tan21

2

B

D F

f

qd

68.05.1

2.14.01'

'4.01

L

B F s

1d F

2

21

/2.5495.1647.384

168.04.222.118135.1462.14.186.12'

mkN

q u

Qult = q’ u X A’ = 549.2 X (1.5X1.2) = 988kN

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


16/35

BFC 4043


2.7 Load on strip footing

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 capacityfactors.

Find : The width of footing with FS of 3.

Figure 2.8Solution :

2

2

/9.722

/8.1452

;

........(5.0

mkN mkN qcwith


u

qcu

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

m

mkN

mkN dthofwall requiredwi

mkN mkN

q

mkN BmkN mmkN mkN q

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

B required is 1.5 meter to be conservative


17/35

BFC 4043


2.8 Dimension of loaded square pad footing

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

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

Find : The right dimension B. Use Terzaghi’s equation).........(4.03.1 foundation square BN qN cN q qcu

With Ø=30°; Nc=37.16, Nq=22.46, N =19.13

Assume B=3 m;

m B BmmkN

kN wall of widthrequired

mkN mkN

q

mkN mkN

mmkN mmkN mkN q

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;

m B BmmkN


mkN mkN

q

mkN mkN

mmkN mmkN mkN q

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

22

2

22

22

332

Assume B=2m;

2670 kN

γ = 20.44 kN/m3

1.52m Ø=30˚

c = 38.3 kN/m2

Figure 2.9


18/35

BFC 4043


m B BmmkN


mkN mkN

q

mkN mkN

mmkN mmkN mkN q

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

222

22

22

332

Assume B=1.8m;

m B BmmkN


mkN mkN

q

mkN mkN

mmkN mmkN mkN q

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;

m B BmmkN


mkN mkN

q

mkN mkN

mmkN mmkN mkN q

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

Contact Pressure and stability check.

Can be computed by using flexural formula of :

y

y

x

x

I

x M

I

y M

A

Qq

Where :

q – contact pressureQ – total axial vertical load

A – area of footingMx, My – total moment about respective x and y axesIx, Iy – moment of inertia about respective x and y axes


19/35

BFC 4043


x, y – distance from centroid to the outer most point atwhich the contact pressure is computed along respective x andy 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; so il =18.85kN/m

3

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

2

Find :a.

Soil contact pressureb. FS against bearing capacity pressure

Solution :

a. y

y

x

x

I

x M

I

y M

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

222.4KN

0.14m20.91m

1.22m

0.31m1.52m

Figure 2.10


20/35

BFC 4043


= [(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 = 279.32kN

Area, 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

mkN mkN q

c

BN N DcN q

u

q f cult

Assuming cohesive soil has : Ø=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

q FS

mkN

BN N DcN q

ult

q f cult

Since FS > 3.0; thus ok.

Example 2.10

Draw soil contact pressure forfooting 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)

Figure 2.11


21/35

BFC 4043


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

Solution :

Using flexural formula; y

y

x

x

I x M

I y M

AQq

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

mm I y

22

22

42

/53.53...../47.232

/143/47.8952.1

)145.1)(88.189(

48.3

36.311

mkN qand mkN q

mkN mkN m

mmkN

m

kN q

left right

Take ΣV = 0 and ΣMc = 0 will produce :

ΣV = 0 : )........(36.311)52.1(2

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

AkN mqd

and W P Ld q

ΣMc = 0 : see Figure 2.12 (b) and (c)

03

))((2

))((

d x Ld

qS H M

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

mkN qkN mmq

Ao substituemd d

Bd m

mkN mkN

Figure 2.12 (a) and (b)

1.61m

254.46kN/m2

2.29m2.29m


22/35

BFC 4043


Example 2.11 Checking stability on shallow foundation

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/ft

2.

Find :1.

Contact pressure and soil pressure diagram.2.

Shear and moment at section A-A (in the Figure E3.14)3.

FS against sliding if coefficient of friction, δ = 0.40

4.

FS against overturning.

Solution :

1. y

y

x

x

I

x M

I

y M

A

Qq

Q=P+W1+W2=60+9.3+11.2=80.5kips A=6ftx6ft=36ft2 My=4kipsx4.5ft=18kip-ft (about point C)x=6ft/2=3ft

Iy=6ft(6ft)3

/12=108ft4

; Mx=0; Mxy/Ix=0

22

42 /50.0/24.2

108

)3(.18

36

5.80 ft kip ft kips

ft

ft ft kip

ft

kips

I

x M

I

y M

A

Qq

y

y

x

x

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

qleft = 1.74 kips/ft2 < 3.0 kips/ft2 ; OK

Figure 2.13


23/35

BFC 4043


2. ΔFDG and ΔEDH are similar triangles; so

ft kips ft kips ft

kips A Aat Moment

kipskipskips

ft ft kip ft ft ft kips ft A Aat Shear

ft kip DE ft

ft DE ft FG

ft

ft ft

EH ft kips DF FG

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

forces Horizontal eandsoil betweenbas frictionof t coefficienload vertical Total sliding against FS

4. 4.13)5.4(4

)2/6(5.80

.

.Re..

ft kips

ft kips

moment Turning

moment sisting g overturninagainst FS

Pressure diagram


24/35

BFC 4043


2.10 Settlement of shallow foundationFoundation 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.

2.10.1 Elastic settlement of foundations on saturated clay

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

s

e E

Bq A AS 021

where :

A 1 is a function of H/B and L/B and A 2 is a function of Df /B All parameters of H, B and Df (with L into the paper) are asshown in Figure 2.14.

Figure 2.14 : Parameters


25/35

BFC 4043


Figure 2.15 : A 2 Versus Df /B

Figure 2.16 : A 1 Versus H/B and L/B

2.10.2 Elastic settlement of foundations on sandy soil: use ofstrain influence factor

Schmertmann, (1978) proposed that the elastic settlement in sandysoil as :

2

0

21

z

s

z e z

E

I qqC C S

where :


26/35

BFC 4043


Iz – strain influence factor

C1 – correction factor due to depth =

qq

q5.01

C2 – correction factor due to soil creep =

1.0

log2.01 yearsintime

q - stress at the level of foundation (due to loading + self

weight of footing + weight of soil above footing)

f Dq

Figure 2.17 : Calculation of elastic settlement using strain influencefactor

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 = 0


27/35

BFC 4043


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

2.10.3 Range of material parameters

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

Table 2.4 : Elastic parameters of various soils

Type of soil Modulus of Elasticity,Es

(MN/m2

)

Poisson’s ratio, μs

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

2.10.4 Consolidation settlement

(a) Primary consolidation, Sc

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 :


28/35

BFC 4043


00

log1 p

p

e

H C S cc

where : Cc – compression index (given)

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

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

clay layer

Ip – Influence factor (from Figure 2.18)

q0 – net stress increase

Figure 2.18 : Chart for determining stresses below corners of rigidand isotropic.


29/35

BFC 4043


Example 2.7

Given :

Figure 2.19

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

20

/3.2355.19519881.983.1919820083.1963

5400

,

mkN mm

kN

excavationof weight pressureload qincrease stress Net


30/35

BFC 4043


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

mz = 1.5m nz = 3.0m

mmm

m z 7.62

6.1850.1925.195

224.07.6

5.1

m

mm ; 448.0

7.6

0.3

m

mn

From Figure 2.18, the influence coefficient is 0.04

Therefore ; 22 /6.37/3.23504.04 mkN mkN p

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

Therefore; mmkN

mkN m

p

p

e

H C S cc 212.0

/1.123

/7.160log

38.11

4.668.0log

1 2

2

00

(b) secondary consolidation

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

p

s s

t

t H C S log

where :

Ss – secondary compression settlement

Cα – coefficient of secondary compression, can be determinedfrom Figure

3.26H – thickness of clay layer that is consideredts – time for which settlement is requiredtp – time to completion of primary consolidation


31/35

BFC 4043


Figure 2.20 : Value of Cα

2.11 Allowable bearing pressure in sand based onsettlement consideration.

Bowles (1977) proposed a correlation of the net allowablebearing pressure for foundations with SPT (N-values).

The following equations are used :

m B for S

F N mkN q d all net 22.14.25

16.19)/( 2)(

And

m B for S

F B

B N mkN q d all net 22.1

4.2528.3

128.398.11)/(

2

2

)(

Where :

Fd – depth factor = 33.133.01

B

D f

S – tolerable settlement, in mm.


32/35

BFC 4043


Example 2.8

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

d d all net F B

B F

B

BmkN q

22

2

)(28.3

128.38.119

4.25

4.25

28.3

128.31098.11)/(

33.133.01

B

D F

f

d

By trial and error (set the table for calculation)


33/35

BFC 4043


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

unet u

Dq

qqq

)( ; id sqiqd qsqcicd cscu F F F BN F F F qN F F F cN q

2

1

q F F F BN F F F qN qqq id sqiqd qsqult net u

2

1

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


34/35

BFC 4043


With no inclination so Fqi=Fγi=1.0From table 2.3 Nq=29.44, Nγ=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.


35/35

BFC 4043

Documents

CH2 Shallowfoundation (SEM2 200910)