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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    NPTEL ADVANCED FOUNDATION ENGINEERING-I

    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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    NPTEL ADVANCED FOUNDATION ENGINEERING-I

    =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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    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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    NPTEL ADVANCED FOUNDATION ENGINEERING-I

    = 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