The General Bearing Capacity Equation

8/10/2019 The General Bearing Capacity Equation

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NPTEL ADVANCED FOUNDATION ENGINEERING-I

Module 3

Lecture 10

SHALLOW FOUNDATIONS: ULTIMATE BEARINGCAPACITY

Topics

1.1THE GENERAL BEARING CAPACITY EQUATION

Bearing Capacity Factors

General Comments

1.2 EFFECT OF SOIL COMPRESSIBILITY

1.3 ECCENTRICALLY LOADED FOUNDATIONS

1.3.1 Foundation with Two-Way Eccentricity


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THE GENERAL BEARING CAPACITY EQUATION

The ultimate bearing capacity equations presented in equations (3, 7 and 8) are for

continuous, square, and circular foundations only. The do not address the case of

rectangular foundations (0


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=2 45 +2 tan [3.26]

=

1

cot

[3.27]

The equation for given by equation (27) was originally derived by Prandtl (1921), andthe relation for [equation (26)] was presented by Reissner (1924). Caquot and Kerisel(1953) and Vesic (1973) gave the relation for as = 2 + 1 tan [3.28]Table 4 shows the variation of the preceding bearing capacity factors with soil friction

angles.

In many texts and reference books, the relationship for

may be different from that in

equation (28). The reason is that there is still some controversy about the variation of with the soil friction angle, .In this text, equation (28) is used.Other relationships for generally cited are those given by Meyerhof (1963), Hansen(1970), and Lundgren and Mortensen (1953). They values for various soil frictionangles are given in appendix B (table B-1, B-2, B-3).

Table 4 Bearing Capacity Factors / tan / tan

0 5.14 1.00 0.00 0.20 0.00 26 22.25 11.85 12.54 0.53 0.49

1 5.38 1.09 0.07 0.20 0.02 27 23.94 13.20 14.47 0.55 0.51

2 5.63 1.20 0.15 0.21 0.03 28 25.80 14.72 16.72 0.57 0.53

3 5.90 1.31 0.24 0.22 0.05 29 27.86 16.44 19.34 0.59 0.55

4 6.19 1.43 0.34 0.23 0.07 30 30.14 18.40 22.40 0.61 0.58

5 6.49 1.57 0.45 0.24 0.09 31 32.67 20.63 25.99 0.63 0.60

6 6.81 1.72 0.57 0.25 0.11 32 35.49 23.18 30.22 0.65 0.62

7 7.16 1.88 0.71 0.26 0.12 33 38.64 26.09 35.19 0.68 0.65

8 7.53 2.06 0.86 0.27 0.14 34 42.16 29.44 41.06 0.70 0.67

9 7.92 2.25 1.03 0.28 0.16 35 46.12 33.30 48.03 0.72 0.70

10 8.35 2.47 1.22 0.30 0.18 36 50.59 37.75 56.31 0.75 0.73

11 8.80 2.71 1.44 0.31 0.19 37 55.63 42.92 66.19 0.77 0.7512 9.28 2.97 1.69 0.32 0.21 38 61.35 48.93 78.03 0.80 0.78

13 9.81 3.26 1.97 0.33 0.23 39 67.87 55.96 92.25 0.82 0.81

14 10.37 3.59 2.29 0.35 0.25 40 75.31 64.20 109.41 0.85 0.84

15 10.98 3.94 2.65 0.36 0.27 41 83.86 73.90 130.22 0.88 0.87

16 11.63 4.34 3.06 0.37 0.29 42 93.71 85.38 155.55 0.91 0.90

17 12.34 4.77 3.53 0.39 0.31 43 105.11 99.02 186.54 0.94 0.93

18 13.10 5.26 4.07 0.40 0.32 44 118.37 115.31 224.64 0.97 0.97


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19 13.93 5.80 4.68 0.42 0.34 45 133.88 134.88 271.76 1.01 1.00

20 14.83 6.40 5.39 0.43 0.36 46 152.10 158.51 330.35 1.04 1.04

21 15.82 7.07 6.20 0.45 0.38 47 173.64 187.21 403.67 1.08 1.07

22 16.88 7.82 7.13 0.46 0.40 48 199.26 222.31 496.01 1.12 1.11

23 18.05 8.66 8.20 0.48 0.42 49 229.93 265.51 613.16 1.15 1.15

24 19.32 9.60 9.44 0.50 0.45 50 266.89 319.07 762.89 1.20 1.1925 20.72 10.66 10.88 0.51 0.47aAfter Vesic (1973)

Shape, Depth, and Inclination Factors

The relationships for the shape factors, depth factors, and inclination factors

recommended for use are shown in table 5. Other relationships generally found in many

texts and references are shown in table B-4 (appendix B).

General Comments

When the water table is present at or near the foundations, the factors and given inthe general bearing capacity equations, equation (25), will need modifications. The

procedure for modifying them is the same.

For undrianed loading conditions ( = 0 concept) in clayey soils, the general load-bearing capacity equation [equation (25)] takes the form (vertical load)

= + [3.29]Table 5 Shape, Depth, and Inclination Factors Recommended for Use

Factor Relationship Source

Shapea = 1 + = + tan = 1 0.4

Where

= length of the foundation (

>

)

De Beer

(1970Hansen

(1970)

Depth Condition (a):/ 1 = 1 + 0.4 = 1 + 2 tan(1 sin)2 = 1Condition (b):/ > 1

Hansen(1970


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= 1 + 2 tan(1 sin)tan 1 = 1Inclinatio

n

=

=

1

90

2

Where =inclination of the load on the foundation with respect to the ver

Meyerhof (1963);

Hannaand

Meyerho

f (1981)aThese shape factors are empirical relations based on extensive laboratory tests.

bThe factors 1(/) is in radians.

Hence the ultimate baring capacity (vertical load) is

net () = = [3.30]Skempton (1951) proposed an equation for the net ultimate baring capacity for clayeysoils ( = condition), which is similar to equation (30) :net() = 5 1+ 0.2 1+ 0.2 [3.31]Example 2

A square foundation ( )has to be constructed as shown in figure 3.9. Assume that = 105 lb/ft3, sat = 118 lb/ft3, = 4 ft,and 1 = 2 ft . The gross allowable load,all , with = 3 is 150,000 lb. The field standard penetration resistance, values areas follows:

Depth (ft) (blow/ft)5 4

10 6

15 6

20 10

25 5

Determine the size of the footing. Use equation (25).

Solution

Using equation (7 from chapter 2) and the Liao and Whitman relationship (table 4 fromchaper 2), the correct standard penetration number can be determined.


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Depth(ft)

(ton/ft2) cor =15 4 1

2000[2 1 0 5 + 3 (118 62.4)]

= 0.188

12

10 60.188 +

1

2000(5)(118 62.4)= 0.327

11

15 60.327 +

1

2000(5)(118 62.4)= 0.466

9

20 100.466 +

1

2000(5)(118 62.4)= 0.605

13

25 50.605 +

1

2000(5)(118

62.4)

= 0.744

6

Figure 3.9

The average cor can be taken to be about 11.From equation 11 (from chapter 2), 35. Given

all =Qall

B2 =

150,000

B2 lb/ft2

[a]

From equation (25) (note: = 0),all = = 13 + 12 For = 35,from table 4, = 33.3, = 48.03. From table 5,


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= 1 + tan = 1 + tan 35 = 1.7 = 1 0.4 = 1 0.4 = 0.6

= 1 + 2 tan(1 sin)2

= 1 + 2 tan35(1 sin 35)2 4

= 1 +1

= 1 = (2)(105) + 2(118 62.4) = 321.2 lb/ft2So

all = 13 (321.2)(33.3)(1.7) 1 + 1B+ 12 (118 62.4)(B)(48.03)(0.6)(10= 6061.04 +

6061.04

B+ 267.05

[b]

Combining equations (a) and (b)

150,0002 = 6061.04 + 6061.04B + 267.05By trial and error, 4.2 ftExample 3

Refer to example1. Use the definition of factor of safety given by equation (20) and

= 5to determine the net allowable load for the foundation.Solution

From example 1,

= 10,736 lb/ft2 = (3)(115) = 345 lb/ft2

all (net ) =

10,7363455

2078 lb/ft2

Hence

all (net ) = (2078)(5)(5) = 51,950 lbExample 4


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Refer to example 1. Use equation (7) and shear = 1.5determine the net allowable loadfor the foundation.

Solution

For

= 320 lb/ft2 and

= 20,

= shear = 3201.5 213 lb/ft2 =1 tanshear =1 tan 201.5 = 13.64From equation (7),

all (net ) = 1.3 + ( 1) + 0.4 For

= 13.64, the values of the bearing capacity factors from table 1 are

1.2, 3.8, 12Hence

all (net ) = 1.3(213)(12) + (345)(3.8 1) + (0.4)(115)(5)(1.2) = 4565 lb/ft2And

all (net ) = (4565)(5)(5) = 114,125 lb 57 tonNote: There appears to be a large discrepancy between the results of examples 3 (or 1)and 4. The use of trial and error shows that, when shear is about 1.2, the results areapproximated equal.

EFFECT OF SOIL COMPRESSIBILITY

In section 3 equation 3, 7, and 8, which were for the case of general shear failure, were

modified to equations 9, 10, and 11 to take into account the change of failure mode in soil(that is, local shear failure). The change in failure mode is due to soil compressibility. In

order to account for soil compressibility, Vesic (1973) proposed the following

modification to equation (25),

= + + 12 [3.32]

Where


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, , and = soil compressibility factorsThe soil compressibility factors were derived by Vesic (1973) from the analogy of the

expansion of cavities. According to that theory, in order to calculate , , and thefollowing steps should be taken:

1. Calculate the rigidity index, , of the soil at a depth approximately /2below thebottom of the foundation, or = + tan [3.33]Where

= shear modulus of the soil

= effective overburden pressure at a depth of

+

/2

2. The critical rigidity index, ( ), can be expressed as( ) = 12 exp 3.30 0.45 cot45 2 [3.34]The variation of ( )for / = 0and / = 1are given in table 6.

3. If ( ), then

=

=

= 1

However, if


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30 152 70

35 283 120

40 592 225

45 1442 482

50 4330 1258

After Vesic (1973)

Figure 3.10shows the variation of = [equation (35)] with and . For = 0, = 0.32 + 0.12 + 0.60log [3.36]For > 0,

=

1

tan

[3.37]

Figure 3.10 Variation of = with and Example 5

For a shallow foundation, the following are given: = 0.6 m, = 1.2 m, = o. 6 m.Soil: = 25 = 48 kN/m2


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= 18 kN/m3Modulus of elasticity, = 620 kN/m2Poissons ratio, = 0.3Calculate the ultimate bearing capacity.

Solution

From equation (33)

= + tanHowever,

=

2(1+)

So

= 2(1+)[+ tan] = +2 = 18 0.6 + 0.62 = 162 kN/m2 = 6202(1+0.3)[48+16.2tan 25] = 4.29From equation (34)

( ) = 12 exp 3.3 0.45 cot45 21

2exp 3.3 0.45 0.6

1.2 cot45 25

2 = 62.46

Since ( ) > , use equations 35 and 37. = = exp 4.4 + 0.6 tan + (3.07 sin)log(2)1+sin = exp 4.4 + 0.6 0.61.2 tan 25 + (3.07 sin 25)log(24.29)1+sin 25 = 0.347 = 1 tan For = 25, = 10.66(table 4),


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= 0.347 10.34710.66tan 25 = 0.216Now, from equation (32),

=

+

+ 1

2

From table 4, for = 25, = 20.72, = 10.66, = 10.88. From table 5, = 1 + = 1 + 10.6620.72 0.61.2 = 1.257 = 1 + tan = 1 + 0.61.2 tan 25 = 1.233 = 1 0.4 = 1 0.4 0.61.2 = 0.8

= 1 + 0.4

= 1 + 0.4

0.6

0.6= 1.4

= 1 + 2 tan(1 sin)2 = 1 + 2tan25(1 sin 25)2 0.60.6 = 1.311 = 1Thus

= (48)(20.72)(1.257)(1.4)(0.216) + (0.6 18)(10.66)(1.233)(1.311)(0.347) +12(18)(0.6)(10.88)(0.8)(1)(0.347) = 459 kN/m2

ECCENTRICALLY LOADED FOUNDATIONS

In several instances, as with the base of a retaining wall, foundations are subjected tomoments in addition to the vertical load, as shown in figure 3.11a. In such cases the

distribution of pressure by the foundation on the soil is not uniform. The distribution of

nominal pressure is

max = + 62 [3.38]Andmax = 62 [3.39]Where

= total vertical load


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= moment on the foundationFigure 3.11b shows force system equivalent to that shown in figure 3.11a. The distance e

is he eccentricity, or

= [3.40]Substituting equation (40) in equations (38) and (39) gives

max = 1 + 6 [3.41a]And

max = 1 6 [3.41b]

Figure 3.11 Eccentrically loaded foundations

Note that, in these equations, when the eccentricity, e,becomes /6, max is zero. for >/6,min will be negative, which means that tension will develop. Because soilcannot take any tension, there will be a separation between the foundation and the soilunderlying it. The nature of the pressure distribution on the soil will be as shown in figure

3.11a. the value of

maxthen is

max = 43(2) [3.42]The exact distribution of pressure is difficult to estimate.


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The factor of safety for such types of loading against baring capacity failure can be

evaluated by using the procedure suggested by Meyerhof (1953), which is generally

referred to as the effective area method. The following is Meyerholf a step-by-stepprocedure for determination of the ultimate load that the soil can support and the factor of

safety against bearing capacity failure.

1.

Determine the effective dimensions of the foundation as

= effective width = 2 = effective length =Note that, if the eccentricity were in the direction of the length of the foundation,

the value of would be equal to 2. The value of would equal . Thesmaller of the two dimensions (that is, and ) is the effective width of thefoundation.

2.

Use equation (25) for the ultimate bearing capacity as = + + 12 [3.43]To evaluate , , and , use table 5 with effective length and effective widthdimensions instead of and , respectively. To determine , , and usetable 5 (do not replace with ).

3. The total ultimate load that the foundation can sustain is

ult = ()()

[3.44]

Where

= effective area4. The factor of safety against bearing capacity failure is

=ult [3.45]5.

Check the factor of safety against max ,or, =/max .Note that eccentricity tends to decrease the load-bearing capacity of a foundation. In such

cases, placing foundation columns off center, as shown in figure 3.12, probably is

advantageous. Doing so, in procedures a centrally loaded foundation with uniformlydistributed pressure.


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Figure 3.12 Foundation of columns with off-center loading

Foundation with Two-Way Eccentricity

Consider a situation in which a foundation is subjected to a vertical ultimate load ultanda moment M as shown in figure 3.13a and b. For this case, the components of the

moment,M, about thexandyaxes can be determined as and respectively (figure3.13). This condition is equivalent to a load ultplaced eccentrically on the foundationwith = and =(figure 3.13d). Note that =ult [3.46]And

=ult [3.47]If ultis needed, it can be obtained as follows [equation (44)]:ult =Where, from equation (43)

= + + 12 [3.48]And

= effective area =


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Figure 3.13

Figure 3.14 Effective area for the case of / 16and eB/B 16Where

1 =

1.5

3

[3.49a]

1 = 1.5 3 [3.49b]The effective length, L, is the larger of the two dimensions, that is, 1or 1. So, theeffective width is

= [3.50]


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Case II

/ < 0.5 and 0 < eB/B < 16. The effective area for this condition is shown in figure3.15a.

=

1

2(1 + 2) [3.51]

Figure 3.15 Effective area for the case of / < 0.5 and 0


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/ < 16 and 0 < eB/B < 0.5. The effective area for this condition is shown in figure3.16a:

= 12(1 + 2) [3.54]

Figure 3.16 Effective area for the case of / < 16and 0


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Case IV

/ < 16and eB/B < 16. Figure 3.17a shows the effective area for this case. The ratio2/ and thus 2 can be determined by using the / curves that slope upward.Similarly, the ratio

2/

and thus

2 can be determined by using the

/

curves that

slope downward. The effective area is then

Figure 3.17 Effective area for the case of / < 16and 0


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Example 6

A square foundation is shown in figure 3.18. Assume that the one-way load

eccentricity = 0.15 m. Determine the ultimate load, ult.

Figure 3.18

Solution

With = 0, equation (43) becomes = + 12 = (0.7)(18) = 12.6 kN/m2For = 30, from table 4, = 18.4 and = 22.4 = 1.5 2(0.15) = 1.2 m

= 1.5 m

From table 5

= 1 + tan = 1 + 1.21.5 tan 30 = 1.462 = 1 + 2 tan(1 sin)2 = 1 + (0.289)(0.7)1.5 = 1.135


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= 1 0.4 = 1 0.4 1.21.5 = 0.68 = 1So

= (12.6)(18.4)(1.462)(1.135) + 12(18)(1.2)(22.4)(0.68)(1) = 384.7 + 164.50 =549.2 kN/m2

Hence

ult = = (1.2)(1.5)(549.2) 988kNExample 7

Refer to example 6. Other quantities remaining the same, assume that the load has a two-

way eccentricity. Given: = 0.3 m, and = 0.15 m (figure 3.19). Determine theultimate load, ult.

Figure 3.19

Solution

=0.3

1.5

= 0.2

= 0.151.5 = 0.1This case is similar to that shown in figure 3.15a. From figure 3.15b, for / =0.2 and eB/B = 0.1

1 0.85; 1 = (0.85)(1.5) = 1.275 m


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And

2 0.21; 2 = (0.21)(1.5) = 1.315 mFrom equation (51)

= 12(1 + 2) = 12(1.275 + 0.315)(1.5) = 1.193 m2From equation (53)

=1 = 1.275 mFrom equation (52)

= = 1.1931.275 = 0.936 mNote, from equation (43) for = 0 = + 12 = (0.7)(18) = 12.6 kN/m2For = 30, from table 4, = 18.4 and = 22.4. Thus = 1 + tan = 1 + 1.9361.275 tan 30 = 1.424

= 1

0.4

= 1

0.4

1.936

1.275= 0.706

= 1 + 2 tan(1 sin)2 = 1 + (0.289)(0.7)1.5 = 1.135 = 1So

ult = =( + 12 =(1.193)[(12.6)(18.4)(1.424)(1.135) + (0.5)(18)(0.936)(22.4)(0.706)(1)] =605.95 kN

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The General Bearing Capacity Equation