Hasofer and Lind Method

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    4.4 Hasofer-Lind Reliability Index (1974)

    A modified reliability index: it did not exhibit

    the invariance problem. The correction is to

    evaluate the limit state function at a point

    known as the design point instead of the

    mean values. The design point is generally not

    known a priori, an iteration technique must be

    used (in general).

    Consider a limit state function ),...,,( 21 nXXXg

    where the random variables Xi are all

    uncorrelated. The limit state function is

    rewritten in terms of the standard form of the

    variables using

    i

    i

    X

    Xi

    i

    XZ

    =

    As before, the Hasofer-Lind Reliability Indexis defined as the shortest distance from the

    origin of the reduced variable space to the

    limit state function g=0.

    An iteration is required to find the design

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    point { }**2*1 ,...,, nZZZ in reduced variable spacesuch that still corresponds to the shortest

    distance (shown in Figure).

    The iterative procedure requires us to solve a

    set of (2n+1) simultaneous equations with(2n+1) unknowns:

    **

    2

    *

    121 ,...,,,,...,,, nn ZZZ ,

    where

    Z2

    Z10

    Z*2

    Z*

    1

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    =

    =n

    k

    podesignatevaluated

    k

    podesignaievaluated

    ii

    Zg

    Z

    g

    1

    2

    int

    int

    )(

    (a)

    iX

    ii

    i

    ii X

    g

    Z

    X

    X

    g

    Z

    g

    =

    =

    (b)

    1)( 2

    1

    ==

    n

    i

    i (c )

    iiZ =*

    (d)

    0),...,,( **2*

    1 =nZZZg (e)

    The simultaneous equation procedure:

    (1) Formulate the limit state function andappropriate parameters for all random

    variables involved.

    (2) Express the limit state function in terms ofreduced variables Zi.

    (3) Use Eq. (d) to express the limit statefunction in terms of and i .

    (4) Calculate the n i values. Use Eq.(d)

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    here also to express each i as a function

    of all i and (note the coupling effect

    between i).

    (5) Contact the initial cycle: Assumenumerical values of and all i ,

    noting that thei values must satisfy Eq.

    (c ).

    (6) Use the numerical values of and ion the right-hand sides of the equations

    formed in Steps 3 and 4 above.

    (7) Solve the n+1 simultaneous equations inStep 6 for and i.

    (8) Go back to Step 6 and repeat. Iterate untilthe and i values converge.

    Example:

    Calculate the Hassofer-Lind reliability index

    for the three-span continuous beam shown in

    following Figure.

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    The random variables in the problem are

    distributed load (w), span length (L), modulus

    of elasticity (E), and moment of inertia (I). The

    limit state to be considered is deflection, and

    the allowable deflection is specified as L/360.

    The maximum deflection is 0.0069wL4/EI, and

    it occurs at 0,446L from either and

    (AISC,1986). The limit state function is

    EI

    wLLIELwg

    4

    0069.0360

    ),,,( =

    The means and standard deviations of therandom variables are listed in following Table.

    Variables Mean Standard

    deviation

    w 10 kN/m 0.4 kN/m

    A B C D

    LLL

    W

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    L 5 m ~0

    E 7102 kN/m2 7105.0 kN/m

    2

    I 4108 m4 4105.1 m4

    Solution

    We follow the Steps in the simultaneous

    equation procedure.

    (1) Formulate the limit state function andappropriate parameters for all random

    variables involved. It has been done.

    (2) Express g as a function of reducedvariables. First, substituting some numbers,g can be expressed as

    05.3100)5(0069.0360

    50 4 === wEIwEIg

    Define the reduced variables:

    w

    w

    E

    E

    I

    I w

    Z

    E

    Z

    I

    Z

    =

    =

    =321 ;;

    wwEEII ZwZEZI 321 ;; +=+=+=

    Substitute into g:

    0)(5.310))(( 312 =+++ wwIIEE ZZZ

    0)]4.0(10(5.310)]105.1(108)][105.0(102[ 34

    1

    47

    2

    7 =+++ ZZZ

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    012895)2.124()750()4000()3000( 32121 =+++ ZZZZZ

    (3) Formulate g in terms of andi := iiZ

    *

    0128952.12475040003000 3212

    21 =+++

    32121 2.12475040003000

    12895

    ++

    =

    (4) Calculate i values:22

    1

    2

    2

    21

    )2.124()7504000()7503000(

    )7503000(

    ++++

    +=

    22

    1

    2

    2

    12

    )2.124()7504000()7503000(

    )7504000(

    ++++

    +=

    22

    1

    2

    2

    3

    )2.124()7504000()7503000(

    )2.124(

    ++++

    =

    (5) The iterations start with a guess for321 ,,, . For example, let us start with

    58.0333.0;58.0333.0 321=====

    andlet 0.3=

    (6)~(8) The iterations are summarized in the

    following Table. Notice that between iterations

    5 and 6, the values change very little, so the

    solution has converged. Faster convergence

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    occurs when the correct signs are used for

    each i (+ for load effects,- for resistances).

    Thus the calculated reliability index is

    approximately 3.17

    Table

    Iteration numberInitial guess

    1 2 3 4 5 6

    3 3.644 3.429 3.213 3.175 3.173 3.173

    1 -0.58 -0.532 -0.257 -0.153 -0.168 -0.179 -0.182

    2 -0.58 -0.846 -0.965 -0.988 -0.985 -0.983 -0.983

    3 +0.58 0.039 0.047 0.037 0.034 0.034 0.034

    The matrix procedure:

    (1) Formulate the limit state function andappropriate parameters for all randomvariables ),...,2,1( niXi = involved.

    (2) Obtain an initial design point { }*iX byassuming values for n-1 of the random

    variables Xi (Mean values are often a

    reasonable initial choice). Solve the limit

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    state equation g=0 for the remaining

    random variable. This ensures that the

    design point is on the failure boundary.(3) Determine the reduced variables { }*iZ

    corresponding to the design point { }*iX

    usingi

    i

    X

    Xi

    i

    XZ

    =

    *

    *

    (f).

    (4) Determine the partial derivatives of thelimit state function with respect to the

    reduced variables using Eq.(b). For

    convenience, define a column vector {G}

    as the vector whose elements are these

    partial derivatives multiplied by 1:

    { }

    =

    nG

    G

    G

    GM

    2

    1

    where intpodesignatevaluatedi

    iZgG

    = (g)

    (5) Calculate an estimate of using thefollowing formula:

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    { } { }{ } { }GG

    zG

    T

    T *

    = where

    { }

    =

    *

    *

    2

    *

    1

    *

    nZ

    Z

    Z

    Z

    M (f)

    If the limit state equation is linear, then Eq.(f)

    reduces to

    =

    =

    +

    =n

    i

    Xi

    n

    i

    Xi

    i

    i

    a

    aa

    1

    2

    1

    0

    )(

    (6) Calculate a column vector containing thesensitivity factor using

    { }{ }

    { } { }GG

    G

    T=

    (g)

    (7) Determine a new design point in reducedvariables for n-1 of the variables using

    Eq. (d).

    (8) Determine the corresponding design pointvalues in original coordinates for the n-1

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    values in Step 7 using Eq. (f) in an

    alterative formii XiXi

    ZX ** += .

    (9) Determine the value of the remainingrandom variable (i.e., the one not found in

    Steps 7 and 8) by solving the limit state

    function g=0.

    (10) Repeat Steps 3 to 9 until and thedesign point { }*iX converge.

    Example:

    Repeat above Example using the matrix

    procedure.

    Solution

    (1) For convenience, let X1=I, X2=E, X3=w.The limit state equation is

    12

    4

    3

    321 0069.0360

    ),,(XX

    LXLXXXg =

    (2) For the first iteration, we will assumex1

    *and x

    *2 are the mean values of X1 and

    X2. The value of x3*

    will be obtained by

    solving the limit state equation g=0. Thus

    == 7*

    2

    4*

    1 102;108 xx

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    53.514026.0])0069.0([

    360 3

    *

    1

    *

    2

    4

    *

    1

    *

    2*

    3 ===L

    xx

    L

    xxLx

    (3) Determine reduced variables Zi* for3,2,1=i :

    0

    1

    1

    *

    1*

    1 =

    =X

    Xxz

    ;

    02

    2

    *

    2*

    2=

    =

    X

    Xx

    z

    ;

    8.103

    3

    3

    *

    3*

    3 =

    =X

    Xxz

    (4) Determine the {G} vector, which involvesthe partial derivatives of g with respect to

    the reduced variables Zi*

    :

    11** 2*

    1

    *

    2

    4*

    3

    }{1

    }{1

    1)(

    0069.0 XXxz xx

    Lx

    X

    g

    Z

    gG

    ii

    =

    =

    =

    22** 2*

    2

    *

    1

    4*

    3

    }{2

    }{2

    2)(

    0069.0 XXxz xx

    Lx

    X

    g

    Z

    gG

    ii

    =

    =

    =

    331** *

    1

    *

    2

    4

    }{3

    }{3

    3 0069.0 XXxz xx

    L

    X

    g

    Z

    gG

    ii

    =

    =

    =

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    Plugging in the information for the first

    iteration, we find the vector {G} to be

    =

    4

    3

    3

    10078.1

    10472.3

    10604.2

    }{G

    (5) Calculate :578.2

    }{}{

    }{}{ *

    == GG

    zG

    T

    T

    (6) Calculate {}:

    ==

    025.0

    800.0

    600.0

    }{}{

    }{}{

    GG

    G

    T

    (7) Determine a new design point in reducedcoordinates for n-1 of the random

    variables. To be consistent with what we

    did in Step 2, we will find z1*

    and z2*:

    546.1)578.2)(600.0(1*1 === z

    062.2)578.2)(800.0(2*

    2 === z

    (8) For the reduced variables from Step (7),determine the corresponding values of the

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    design point in original coordinates:

    444*1*1 1068.5)105.1)(546.1()108(11

    =+=+= XX zx

    677*

    2

    *

    21069.9)105.0)(062.2()102(

    22=+=+= XX zx

    (9) Determine the updated value of x3* fromthe limit state equation g=0. We can use

    the formula presented in the Step 2 for this.

    The result is 7.17*3 =x .

    (10) Repeat Step (3) ~ (9) until convergenceis achieved. The subsequent iterations are

    summarized in the following Table.

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    Table

    Iteration number

    1 2 3 4 5 6 74108 1068.5

    41075.5

    41070.6

    41005.7

    41012.7

    41014.7

    7102 61069.9

    61037.5

    61054.4

    61042.4

    61038.4

    61037.4

    51.5 17.7 9.95 9.80 10.0 10.0 10.0

    2.58 3.29 3.21 3.18 3.18 3.18 3.18

    1068.5

    1075.5

    41070.6

    41005.7

    41012.7

    41014.7

    41014.7

    x1*

    x2*

    x3*

    x1*

    x2*

    61069.9

    61037.5

    61054.4

    61042.4

    61038.4

    61037.4

    61037.4

    x3* 17.7 9.95 9.80 10.0 10.0 10.0 10.0

    Note that the value of (within numerical

    precision) obtaining using this procedure is the

    same value obtained using the simultaneous

    equation procedure.