Terzaghi bearing capacity

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    Content

    Bearing Capacity Concepts

    Analysis for Shallow Foundations

    2

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    Bearing Capacity Failures

    Modes of Soil Failure:

    General shear failure

    Local shear failure

    Punching shear failure

    3

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    Terzaghi Bearing Capacity Formulas

    4

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    General Shear Failure

    5

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    Local Shear Failure

    6

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    Punching Shear Failure

    7

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    Model Tests - Vesic (1973)

    8

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    General Guidelines

    Footings on clays - general shear

    Footings on dense sands (> 70%) - general

    shear

    Footings on loose to medium dense - local

    shear

    Footings on very loose sand (< 35%) -

    punching shear

    9

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    Terzaghis Basic Assumptions

    D B

    No sliding between footing and soil

    Soil is a homogeneous semi-infinite mass

    General shear failure

    Footing is very rigid compared to soil

    10

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    Terzaghi Bearing Capacity Formulas

    General Shear Failure

    BNqNcNq qcult 5.0

    For Square foundations:

    For Continuous foundations:

    BNqNcNq qcult 4.03.1

    For Circular foundations:

    BNqNcNq qcult 3.03.1

    11

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    Terzaghi Bearing Capacity Factors

    General Shear Failure

    12

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    13

    Terzaghi Bearing Capacity Formulas

    Local Shear Failure

    cc

    3

    2

    )tan3

    2(tan

    tan3

    2tan

    1

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    Modified General Ultimate Bearing

    Capacity Formula

    ids

    qiqdqsq

    cicdcscult

    FFFBN

    FFFqN

    FFFcNq

    5.0

    14

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    15

    Modified General Ultimate Bearing

    Capacity Formula

    1

    4.01

    tan1

    1

    L

    B

    L

    BF

    L

    BF

    N

    N

    L

    BF

    s

    qs

    c

    q

    cs

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    16

    Modified General Ultimate Bearing

    Capacity Formula

    1

    1

    tan)sin1(21

    4.01

    2

    B

    D

    F

    B

    DF

    B

    DF

    f

    d

    f

    qd

    f

    cd

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    17

    Modified General Ultimate Bearing

    Capacity Formula

    1

    1

    tantan)sin1(21

    tan4.01

    12

    1

    B

    D

    F

    B

    DF

    B

    DF

    f

    d

    f

    qd

    f

    cd

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    18

    Modified General Ultimate Bearing

    Capacity Formula

    calthe vertirespect toload with

    iedf the appllination ois the inc

    F

    FF

    i

    qici

    )(

    1

    901

    2

    2

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    19

    Effect of GWT Level

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    Case I: 0 D1D

    f

    '21 DDq

    '

    20

    Effect of GWT Level

    wsat

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    Case II: 0 d B

    fDq

    B

    d'

    21

    Effect of GWT Level

    wsat

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    Case III: d > B

    22

    Effect of GWT Level

    fDq

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    Selection of Soil Strength

    Parameters

    Use saturated strength parameters

    Use undrained strength in clays (cu)

    Use drained strength in sands

    For intermediate soils with partially drained

    conditions, undrained shear strength can be

    used but it will be conservative

    23

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    Other Approaching Methods

    Skempton (1951)

    Meyerhof (1953, 1963)

    De Beer and Ladanyi (1961)

    Brinch Hanson (1961, 1970)

    Vesic(1973, 1975) Others

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    Factor of Safety

    Depends on:

    Type of soil

    Level of uncertainty in soil strength

    Importance of structure

    Consequences of failure

    Likelihood of design load occurrence

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    Gross Allowable Bearing Capacity

    Net Allowable Bearing Capacity

    26

    Factor of Safety

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    27

    Factor of Safety

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    28

    Design Factor of Safety

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    Use Terzaghis formula to determine the gross

    allowable load for the square footing shown

    below at a general shear failure condition with

    no GWT effect. Take Nc = 17.69, Nq = 7.44, N =3.64, and FS= 3.0.

    29

    Example 1

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    30

    Example 1 - Solution

    lbAqQ

    ftlbF

    qq

    ftlb

    BNqNNcq

    allall

    S

    ultall

    qcult

    28.41041)4)(4)(08.2565(

    /08.25653

    24.7659

    /24.7659

    )64.3)(4)(110(4.0)44.7)(110)(3()69.17)(200(3.1

    4.0'3.1

    2

    2

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    Redo the previuos example under a local shear

    failure condition with no GWT effect. Use

    Terzaghisformula. Take Nc= 11.85, Nq= 3.88, N= 1.12, and FS= 3.0.

    31

    Example 2

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    32

    Example 2 - Solution

    lbAqQ

    ftlbF

    qq

    ftlb

    BNqNNcq

    ftlbcc

    allall

    S

    ultall

    qcult

    72.18834)4)(4)(17.1177(

    /17.11773

    52.3531

    /52.3531

    )12.1)(4)(110(4.0)88.3)(110)(3()85.11)(200)(3

    2(3.1

    4.0'3.1

    64.13))20tan(

    3

    2(tan)'tan

    3

    2(tan'

    /)200(3

    2'

    3

    2'

    2

    2

    11

    2

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    Refer to the previous example to determine the

    net allowable load. Use Terzaghisformula. Take

    FS= 3.0.

    33

    Example 3

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    34

    Example 3 - Solution

    lbAqQ

    ftlbF

    qq

    ftlb

    qqq

    netallnetall

    S

    netnetall

    ultnet

    72.17074)4)(4)(17.1067(

    /17.10673

    52.3201

    /52.3201

    )110)(3(52.3531

    )()(

    2

    )(

    2

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    Determine the breadth of the square footing

    shown below using Terzaghis formula to

    withstand a gross mass of 30 tons with no GWT

    level effects. Nc = 57.75, Nq = 41.44, N = 45.41,and FS= 3.0.

    35

    Example 4

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    36

    Example 4 - Solution

    cmB

    BforSolving

    BB

    B

    BNqNNcq

    BAQq

    kNFQQ

    mkN

    qcult

    ultult

    Sallult

    50.91

    /90.88268.32914.752

    )41.45)()(15.18(4.0)44.41)(15.18)(1()75.57)(0(3.1

    4.0'3.1

    /90.882/

    90.882)1000/()3)(81.9)(30000(

    /15.18)1000/()81.9)(1850(

    2

    2

    3

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    Determine the required breadth of a square

    footing using the general bearing capacity

    formula to withstand a gross load of 150 kN at adepth of 0.7 m with no GWT level effects. The

    applied load is inclined at angle of 20 to the

    vertical. c= 0, = 30, = 18 kN/m3, Nq = 18.40,

    N = 22.40, and FS= 3.0.

    37

    Example 5

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    38

    Example 5 - Solution

    60.04.014.01

    58.130tan1'tan1

    5.0

    '

    B

    B

    L

    BF

    B

    B

    L

    BF

    FFFBN

    FFFqN

    FFFNcq

    s

    qs

    ids

    qiqdqsq

    cicdcscult

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    39

    Example 5 - Solution

    1

    1

    2021.0

    1

    30tan)30sin1(7.0

    21

    'tan)'sin1(21

    2

    2

    B

    D

    F

    B

    B

    B

    DF

    f

    d

    f

    qd

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    40

    Example 5 - Solution

    2

    22

    22

    /450/

    450)3)(150(

    111.030

    2011

    605.0

    90

    201

    90

    1

    BAQq

    kNFQQ

    F

    FF

    ultult

    Sallult

    i

    qici

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    41

    Example 5 - Solution

    OKB

    D

    cmB

    BforSolving

    BBB

    B

    Bq

    f

    ult

    1282.1

    7.0

    20.128

    /45044.13)2021.0

    1)(62.221(

    )111.0)(1)(60.0)(4.22)()(18)(5.0(

    )605.0)(2021.0

    1)(58.1)(4.18)(18)(7.0(

    2

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