3

BFC 4043

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

Vesic (1973) proposed a relationship for the bearing capacity failure on sands in terms of relative density, Dr depth of foundation, Df and B*, Figure 2.2 Where :

and B width, L length of foundation

NOTE : L IS ALWAYS GREATER THAN B For square; B=Land for circular; B=L=Diameter of foundation and B* = B

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

2.2 Terzaghis Bearing Capacity Terzaghi suggested for a continuous or strip foundation with failure surface as in Figure 2.3

Figure 2.3 Bearing capacity failure in soil under rough rigid continuous 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 :

Where : c cohesion of soil

- unit weight of soil

q = Df

Nc, Nq, N- bearing capacity factors

And

where - passive pressure coefficient Table 2.1 summarizes values for Nc, Nq, and N

Table 2.1 Terzaghis Bearing Capacitys Factors

NcNqN

NcNqN

0

1

2

3

4

5

6

7

8

9

10

11

12

13

1415

16

17

18

19

20

21

22

23

24

255.706.00

6.30

6.62

6.97

7.34

7.73

8.15

8.60

9.09

9.6110.16

10.76

11.41

12.11

12.86

13.68

14.60

15.12

16.56

17.69

18.92

20.27

21.75

23.36

25.131.001.10

1.22

1.35

1.49

1.64

1.81

2.00

2.21

2.44

2.69

2.98

3.29

3.63

4.02

4.45

4.92

5.45

6.04

6.70

7.44

8.26

9.19

10.23

11.40

12.720.000.01

0.04

0.06

0.10

0.14

0.20

0.27

0.35

0.44

0.56

0.69

0.85

1.04

1.26

1.52

1.82

2.18

2.59

3.07

3.64

4.31

5.09

6.00

7.08

8.342627

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47

48

49

5027.0929.24

31.61

34.24

37.16

40.41

44.04

48.09

52.6457.75

63.53

70.01

77.50

85.97

95.66

106.81

119.67

134.58

151.95

172.28

196.22

224.55

258.28

298.71

347.50

14.2115.90

17.81

19.98

22.46

25.28

28.52

32.23

36.50

41.44

47.16

53.80

61.55

70.61

81.27

93.85

108.75

126.50

147.74

173.28

204.19

241.80

287.85

344.63

415.149.8411.60

13.70

16.18

19.13

22.65

26.87

31.94

38.04

45.41

54.36

65.27

78.61

95.03

115.31

140.51

171.99

211.56

261.60

325.34

407.11

512.84

650.67

831.99

1072.80

From Kumbhojkar (1993)

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

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

Table 2.2 Terzaghis Modified Bearing Capacitys Factors

NcNqN

NcNqN

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

255.70

5.906.106.306.516.746.977.227.477.748.028.328.638.969.319.6710.0610.4710.9011.3611.8512.3712.9213.5114.1414.801.00

1.071.141.221.301.391.491.591.701.821.942.082.222.382.552.732.923.133.363.613.884.174.484.82

5.205.600.00

0.0050.020.040.055

0.0740.100.1280.160.200.240.300.350.420.480.570.670.760.881.031.121.351.551.741.972.25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47

48

49

5015.5316.3017.1318.0318.9920.0321.1622.3923.7225.1826.7728.5130.4332.5334.8737.4540.3343.5447.1351.1755.7360.9166.8073.5581.31

6.056.547.077.668.319.039.8210.6911.6712.75

13.9715.3216.8518.5620.5022.7025.2128.0631.3435.1139.4844.4550.4657.4165.602.592.883.293.764.394.835.516.327.22

8.359.4110.9012.7514.7117.2219.7522.5026.2530.4036.00

41.70

49.30

59.2571.4585.75

Example 2.1

Given : A square foundation, 1.5m x 1.5m in plan view

Soil parameters :

= 20, c = 15.2 kN/m2, =17.8 kN/m3Assume : FS = 4, general shear failure condition and Df= 1 m

Find : Allowable gross load on the foundation

Solution :

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

Thus

Allowable bearing capacity :

Thus total allowable gross load, Q

Example 2.2

Given : Repeat example 2.1Assume : Local shear failure condition

Solution:

For = 20, (Table 2.2); Nc = 11.85, Nq = 3.88, N= 1.12

Allowable load :

;

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

Figure 2.4 Modification of bearing capacity for water table

Case I : 0 D1 Df q(effective surcharge) =

where :

- effective unit weight =

- saturated unit weight of soil

- 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

Case III : d B water has no effect on the quNote : the values of bearing capacity factors used strictly depending on whether the condition is general or local shear failure.2.4 Factor of Safety, FS

, 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

, and :

;

Where : qall(net) net allowable bearing capacity

qu(net) net ultimate bearing capacity

Procedure for FSsheara. Find developed cohesion,cd and angle of friction,d;

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

With : Nc, Nq, N- bearing capacity factors for dc. Thus, the net allowable bearing capacity :

Example 2.3

Using ; and FS = 5; find net allowable load for the foundation in example 2.1 with qu = 521 kN/m2

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

HenceQall(net) = 100.64(1.5x1.5) = 226.4 kN

Example 2.4

Using Example 3.1, and Terzaghis equation

with FSshear = 1.5;

Find net allowable load for the foundation

For c=15.2 kN/m2, = 20 and

cd =

d = tan-1[] = tan-1[] = 13.64

With:

From Table 2.1 : =13.6 ; ; ; (estimation)Hence :

2.5 The General Bearing Capacity Equation The need to address for rectangular shape foundation where :(0 1

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

Where : inclination of load from vertical For undrained condition ( = 0)

Skemptons :

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

Nq/ NcTan NcNqN

Nq/ NcTan

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

255.145.38

5.63

5.90

6.19

6.49

6.81

7.16

7.53

7.92

8.35

8.80

9.28

9.81

10.37

10.98

11.63

12.34

13.10

13.93

14.83

15.82

16.88

18.05

19.32

20.721.001.09

1.20

1.311.43

1.57

1.72

1.88

2.06

2.25

2.47

2.71

2.97

3.26

3.59

3.94

4.34

4.77

5.26

5.80

6.40

7.07

7.82

8.66

9.60

10.660.000.07

0.15

0.24

0.34

0.45

0.57

0.71

0.86

1.03

1.22

1.44

1.69

1.97

2.29

2.65

3.06

3.53

4.07

4.68

5.39

6.20

7.13

8.20

9.44

10.880.200.20

0.21

0.22

0.23

0.24

0.25

0.26

0.27

0.28

0.30

0.31

0.32

0.33

0.35

0.36

0.37

0.39

0.40

0.42

0.43

0.45

0.46

0.48

0.50

0.510.000.02

0.03

0.05

0.07

0.09

0.11

0.12

0.14

0.16

0.18

0.19

0.21

0.23

0.25

0.27

0.29

0.31

0.32

0.34

0.36

0.38

0.40

0.42

0.45

0.4726

27

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47

48

49

5022.2523.94

25.80

27.86

30.14

32.67

35.49

38.64

42.16

46.12

50.59

55.63

61.35

67.87

75.31

83.86

93.71

105.11

118.37

133.88

152.10

173.64

199.26

229.93

266.89

11.8513.20

14.72

16.44

18.40

20.67

23.18

26.09

29.44

33.30

37.75

42.92

48.93

55.96

64.20

73.90

85.38

99.02

115.31

134.88

158.51

187.21

222.31

265.51

319.07

12.5414.47

16.72

19.34

22.40

25.99

30.22

35.19

41.06

48.03

56.31

66.1978.03

92.25

109.41

130.22

155.55

186.54

224.64

271.76

330.35

403.67

496.01

613.16

762.890.530.55

0.57

0.59

0.61

0.63

0.65

0.68

0.70

0.72

0.75

0.77

0.80

0.82

0.85

0.88

0.91

0.94

0.97

1.01

1.04

1.08

1.12

1.15

1.200.490.51

0.53

0.55

0.58

0.60

0.62

0.65

0.67

0.70

0.73

0.75

0.78

0.81

0.84

0.87

0.90

0.93

0.97

1.00

1.04

1.07

1.11

1.15

1.19

Example 2.5

Figure 2.5Given : 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 factorsFind

: The width of foundation B

;

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

;

EMBED Equation.3 ;

;

So

EMBED Equation.3 By trial and error : B=1.3m

2.6 Eccentrically Loaded Foundations Eccentrically loaded foundations give non-uniform distribution of pressure, Figure 2.6

Figure 2.6 Eccentrically loaded foundations

Eccentricity,

qmax and qmin is given by :

and

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

Factor of safety against bearing capacity failure; effective area method, 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 :

use L and B to find

use B to find

c. Total ultimate load, ; where A effective area

d. Factor of safety,

e. Check FS against qmax ;

Example 2.6

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

Figure 2.7

Find : Ultimate load, Qult,

assume one way load eccentricity, e = 0.15mSolution : with c = 0;

Where :

q = 0.7(18) = 12.6 kN/m2for = 30, from Table 2.3 : Nq=18.4 and N=22.4

B = 1.5 2(0.15) = 1.2m

L = 1.5m

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

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

Qall = 988/3 = 330kN with FS=32.7 Load on strip footingExample 2.7 :

Given : The strip footing shown below is to be constructed in a uniform deposit of stiff clay and must support a wall that imposes a loading of 152 kN/m of wall length. Use general bearing capacity factors.Find : The width of footing with FS of 3.

Figure 2.8Solution :

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

B required is 1.5 meter to be conservative

2.8 Dimension of loaded square pad footing

Example 2.8 :

Soil deposit has the following ;=20.44 kN/m3, =30, c=38.3kN/m2Square footing located 1.52 m below surface, carries 2670 kN and groundwater is negligible. Use Terzaghis values, (Table 2.1).

Find : The right dimension B. Use Terzaghis equation

With =30; Nc=37.16, Nq=22.46, N=19.13Assume B=3 m;

Assume B=1 m;

Assume B=2m;

Assume B=1.8m;

Assume B=1.7m;

Therefore use 1.7m x 1.7m 2.9 Contact Pressure and stability check. Can be computed by using flexural formula of :

Where :

q contact pressure

Q total axial vertical load

A area of footing

Mx, My total moment about respective x and y axes

Ix, Iy moment of inertia about respective x and y axes

x, y distance from centroid to the outer most point at which the contact pressure is computed along respective x and

y axes.

Example 2.9A pad footing with dimension of 1.52 x 1.52m acted upon by the load of 222.4kN. Estimate soil contact pressure and FS against bearing capacity.

Given :

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

=24 kN/m3; qu = 143.64 kN/m2Find :

a. Soil contact pressure

b. FS against bearing capacity pressure

Solution :

a. ; Mx=My=0; since load on centroid

Total load calculation, Q :

Column load, P = 222.4kN

Weight of footing base = (1.52m)(1.52m)0.31m(24kN/m3) = 17.19 kN

Weight of footing pedestal = (0.14m)(0.14m)(0.91m)(24kN/m3) = 0.43 kN

Weight 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 = 279.32kN

Area, A = 1.52mx1.52m = 2.31m2

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

b.

Assuming cohesive soil has : =0 and c>0; thus :

Nc=5.14, Nq=1.0, N=0, Df=1.22m

Since FS > 3.0; thus ok.

Given : 2.29m by 1.52m rectangular footing

Find : Contact pressure and soil pressure diagram

Solution :

Using flexural formula;

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;

Take V = 0 and Mc = 0 will produce :

V = 0 :

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

Example 2.11 Checking stability on shallow foundation

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

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, = 0.404. FS against overturning.

Solution :

1.

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

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

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

2. FDG and EDH are similar triangles; so

3.

4.

2.10 Settlement of shallow foundation

Foundation settlement under load can be classified according to two major types :

(a) immediate or elastic settlement, Se(b) consolidation settlement, ScElastic settlement, Se takes place immediately during or after construction of structure.

Consolidation settlement, Sc is time dependent comprises of two phases; 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 Janbu et al., (1956) using the equation :

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 as shown in Figure 2.14.

Figure 2.14 : Parameters

Figure 2.15 : A2 Versus Df/B

Figure 2.16 : A1 Versus H/B and L/B

Elastic settlement of foundation on sand based on SPT N- value

By Meyerhof

Refer section 5.13 page 263 Equations 5.61 and 5.622.10.2 Elastic settlement of foundations on sandy soil: use of strain influence factor

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

where :

Iz strain influence factor

C1 correction factor due to depth =

C2 correction factor due to soil creep =

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

Figure 2.17 : Calculation of elastic settlement using strain influence factor

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

Iz = 0.1at z = 0

Iz = 0.5atz = z1 = 0.5B

Iz = 0

atz = z2 = 2B

Footing with L/B 10 (rectangular footing) :

Iz = 0.2at z = 0

Iz = 0.5atz = z1 = B

Iz = 0

atz = z2 = 4B

2.10.3 Range of material parameters

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

Table 2.4 : Elastic parameters of various soils

Type of soilModulus of Elasticity, Es(MN/m2)Poissons ratio, s

Loose sand10.5 24.00.20 0.40

Medium dense sand17.25 27.600.25 0.40

Dense sand34.50 55.200.30 0.45

Silty sand10.35 17.250.20 0.40

Sand and gravel69.00 172.500.15 0.35

Soft clay4.1 20.70.20 0.50

Medium clay20.7 41.4

Stiff clay41.4 96.6

2.10.4 Consolidation settlement

(a) Primary consolidation, ScMany methods were developed in estimating the value of consolidation settlement, Sc. Due to simplicity only chart based on Newmarks (1942), Figure 2.18 will be used in estimating the consolidation settlement.

Primary consolidation, Sc calculated as :

where : Cc compression index (given)

H thickness of clay layer

e0 initial void ratio (given)

p = p0 + p, final pressure

p0 overburden pressure

p =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 rigid foundation and isotropic soilExample 2.7

Given :

Figure 2.19A 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, which include all self weight. The initial void ratio, e0 is 1.38 and compression 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/m2Weight of excavation = 19.83(200 - 198) + (19.83 9.81)(198 195.5) = 64.7kN/m2

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

mz = 1.5m

nz = 3.0m

;

From Figure 2.18, the influence coefficient is 0.04

Therefore ;

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

Therefore;

(b) secondary consolidation

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

where :

Ss secondary compression settlement

C coefficient of secondary compression, can be determined from Figure

3.26

H thickness of clay layer that is considered

ts time for which settlement is required

tp time to completion of primary consolidation

Figure 2.20 : Value of C2.11 Allowable bearing pressure in sand based on settlement consideration.

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

The following equations are used :

And

Where :

Fd depth factor =

S tolerable settlement, in mm.

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 SPT numbers from field exploration as shown in the table.Assume that the footing must be 1.5m deep, the tolerable settlement as 25.4mm and the size is > 1.22m.Required :

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

Navg = (7+8+11+11+13+10+9+10+12)/9=10

With S=25.4mm and N=10

By trial and error (set the table for calculation)

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) :

;

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

With no inclination so Fqi=Fi=1.0

From table 2.3 Nq=29.44, N=41.06

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

EMBED Equation.3

1.5m x 1.5 m

0.7 m

Sand :

EMBED Equation.3

222.4KN

0.14m2

0.91m

1.22m

0.31m

1.52m

Pressure diagram

Figure 2.13

Figure 2.9

Example 2.10

Draw soil contact pressure for footing in Figure 2.11

Conversion to SI unit

P=222.4 kN;

H=88.96 kN;

M=81.35kN.m;

W=88.96 kN

Df=1.22m;

B=2.29m (7.5ft);

L=1.52m (5ft)

2670 kN

= 20.44 kN/m3

1.52m=30

c = 38.3 kN/m2

Figure 2.10

Figure 2.12 (a) and (b)

1.61m

254.46kN/m2

2.29m

2.29m

Figure 2.11

PAGE 26AKS & PM IR AZIZAN SEM 2 200910

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_1321777128.unknown

_1321780833.unknown

_1322383563.unknown

_1322406945.unknown

_1323669513.unknown

_1323669858.unknown

_1323671688.unknown

_1323672375.unknown

_1323672467.unknown

_1323671327.unknown

_1323669724.unknown

_1322416799.unknown

_1322417273.unknown

_1322407060.unknown

_1322383739.unknown

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