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8/3/2019 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.