# 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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NPTEL ADVANCED FOUNDATION ENGINEERING-I

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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NPTEL ADVANCED FOUNDATION ENGINEERING-I

= 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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NPTEL ADVANCED FOUNDATION ENGINEERING-I

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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NPTEL ADVANCED FOUNDATION ENGINEERING-I

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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NPTEL ADVANCED FOUNDATION ENGINEERING-I

, , 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

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