Chapter 7
Reliability-Based Design Methods of
Structures
Chapter 7: Reliability-Based Design Methods of Structures
Contents
7.1 Reliability-Based Design Codes
7.2 Reliability-Based Design Formulas
7.3 Calibration for Deterministic Codes
7.4 Target Reliability Index in Chinese Codes
7.5 Practical LRFD Formulas in Current Codes
Chapter 7
Reliability-Based Design Methods of
Structures
7.1 Reliability-Based Design Codes
7.1 Reliability-Based Design Codes …1
7.1.1 Role of a Code in the Building Process
–
The building process includes planning, design, manufacturing of
materials, transportation, construction, operation/use, and demolition.
–
The role of a design code is to establish the requirements needed to
ensure an acceptable level of reliability for a structure.
–
The central role of a code is diagrammed in the following figure:
Designer
Owner
Code
Contractor
User
7.1 Reliability-Based Design Codes …2
7.1.2 Code Levels
–
Level Ⅰ Codes: Use deterministic formulas
K (SGk  SQk ) ≤ Rk
–
Level Ⅱ Codes: Use approximate probability limit state design formula
–
Level Ⅲ Codes: Use full probability analysis and design formula
–
Level Ⅳ Codes: Use the total expected life cycle cost of the design as the
optimization criterion
7.1 Reliability-Based Design Codes …2
7.1.3 Reliability-Based Design Codes
1. International Standard
General Principles on Reliability for Structures
(ISO2394: 1998)
2. Chinese Codes
Unified Standard for Reliability Design of
Engineering Structures
(GB50153 — 92)
7.1 Reliability-Based Design Codes …3
1. Unified Standard for Reliability Design of
Building Structures (GB50068 — 2001)
2. Unified Standard for Reliability Design of
Highway Engineering Structures (GB/T50283 — 1999)
3. Unified Standard for Reliability Design of
Railway Engineering Structures (GB50216 — 94)
4. Unified Standard for Reliability Design of
Hydraulic Engineering Structures (GB50119 — 94)
5. Unified Standard for Reliability Design of
Harbor Engineering Structures (GB50158 — 92)
Chapter 7
Reliability-Based Design Methods of
Structures
7.2 Reliability-Based Design Formulas
7.2 Reliability-Based Design Formulas …1
7.2.1 Formulas of Reliability Checking
–
There are three kinds of reliability checking formulas:
Ps ≥[Ps ]
… … … … …(1)
Pf ≤[Pf ]
… … … … …(2)
 ≥[ ]
… … … … …(3)
where, [ P ] , [ Pf ] , [  ] are called target safety probability,
s
target failure probability, or target reliability index.
–
The third formula is generally used in practical engineering.
Given: the probability distribution and digital characteristic

of the loads and resistance


Find: design vector x

Subjected to:  ( x ) ≥[ ]
7.2 Reliability-Based Design Formulas …2
7.2.2 Single Factor Design Formulas
–
The single factor formula based on mean values is as following:
K 0 S   R
where,
S
R
is the mean value of load effect
is the mean value of resistance
K0 is the central safety factor
1    R2   S2 (1   2 R2 )
K0 
1   2 R2
K 0  exp(   R2   S2 )
–
–
R & S are normal distributions
R & S are lognormal distributions
7.2 Reliability-Based Design Formulas …2
7.2.2 Single Factor Design Formulas
–
The single factor formula based on characteristic values is as following:
KSk  Rk
where,
Sk is the characteristic value of load effect
Rk is the characteristic value of resistance
K is the characteristic safety factor
Rk  R (1  kR R )
Sk  S (1  kS S )
1  k R R
K  K0
1  kS  S
7.2 Reliability-Based Design Formulas …3
Relationships among nominal load, mean load, and factored load
Frequency
S , load effect
0
S
Mean load
Sn  S n
Factored load
Nominal load
S
7.2 Reliability-Based Design Formulas …4
Relationships among nominal resistance, mean resistance, and
factored resistance
Frequency
R , Resistance
0
 Rn Rn R
Factored resistance
Mean resistance
Nominal resistance
R
7.2 Reliability-Based Design Formulas …2
7.2.3 Multiple Factor Design Formulas
(Load and Resistance Factor Design, LRFD)
–
The LRFD formula is as following:
Total factored nominal load effect

where,
Si

Sni 
Factored nominal resistance
1
R
Rn
Sni is the nominal (design) value of load effect component,
 Si
is the load partial factor for load component,
Rn is the nominal (design) value of resistance or capacity,
R
is the resistance partial factor.
7.2 Reliability-Based Design Formulas …2
7.2.3 Multiple Factor Design Formulas
(Load and Resistance Factor Design, LRFD)
g ( X1* , X 2* ,
, X n* )  0
S *  S  S  S  S (1  S  S )
R*  R  R  R  R (1 R  R )
Sk  S (1  kS S )
Rk  R (1  kR R )
S * 1   S  S
S  
Sk
1  kS  S
–
Rk
1  kR R
R  * 
R 1   R  R
The partial safety factors  R and  Si must be calibrated based on the
target index adopted by the code.
Chapter 7
Reliability-Based Design Methods of
Structures
7.3 Calibration for Deterministic Codes
7.3 Calibration for Deterministic Codes …1
7.3.1 Calibration of Target Reliability Index
1. Basic Principles
Consider a structural member which carry a dead load and a variant load.
According to the original deterministic structural design code, the design
formula of ultimate limit state design for this member can be stated as
follows:
Rk  K (SGk  SQk )  0
where, K
— safety factor,
Rk — characteristic value of member resistance ,
SGk , SQk — characteristic value of permanent load effect and
variant load effect designed according to the
deterministic code .
7.3 Calibration for Deterministic Codes …2
Now, the problem can be re-formulated as follows:
How much is the reliability implicit in the original deterministic structural
design code (Level Ⅰ Code)?
–
When the calibration method is used, the limit state equation in simple
load combination condition can be formulated as:
R  SG  SQ  0
where, R
— structural member resistance,
SG — dead load effect,
SQ — live load effect.
–
It is assumed that the parameters and the probability distribution types of
the three basic random variables are known.
–
The calibration method can be implemented by the FORM method, for
example, JC Method.
7.3 Calibration for Deterministic Codes …3
–
It is assumed that the following parameters of the basic random variables
are known:
R
R 
variation factor:
S
S
R
VR 
, VS 
, VS 
R
S
S
G
SGk
, SQ 
S
bias factor:
Rk
, SG 
S
Q
SQk
G
G
Q
G
–
It is assumed that
Let
then

SQk
SGk
Rk is linearly related with SGk and SQk .
,
 is called load effect ratio,
Rk  K ( SGk  SQk )  K ( SGk   SGk )
 K (1   ) SGk
Q
Q
7.3 Calibration for Deterministic Codes …4
2. Calculation Procedure
(1) Assume one value of the load effect ratio
 ;
(2) Determine the characteristic value of member resistance
Rk  K (1   )SGk
Rk :
(3) Determine the mean values and standard deviations of the
basic variables :
mean values:
R  R Rk , S  S SGk , S  S SQk
G
standard deviations:
G
Q
Q
 R  VR R , S  VS S , S  VS S
G
G
G
Q
Q
Q
(4) Determine the limit state equation:
R  SG  SQ  0
(5) Solve the reliability index

by the JC method.
(6) Adjust the load effect ratio, calculate the mean value of different
reliability indexes.
7.3 Calibration for Deterministic Codes …5
Example 7.1
Consider a RC axial compression short column carrying a dead load and an
office live load, the column was designed according to the old “Design Code
of Concrete Structures (TJ9-74)”.
Assume that the ratio of live load to dead load
  SQk / SGk  1.0
,
Try to calibrate the reliability index of the ultimate limit state in TJ9-74 code.
Assume that the following parameters are known:
R is lognormal
R  1.33 VR  0.17
SG is normal
S  1.06 VS  0.07
SL is Extreme Ⅰ
S  0.70 VS  0.29
G
L
K  1.55
G
L
SLk  10kN  m
7.3 Calibration for Deterministic Codes …6
Solution
(1) Determine
Rk
  1.0
SGk  SLk /   10/1  10
Rk  K (SGk  SQk )  1.55  (10 10)  31
(2) Determine the means and standard deviations
R  R Rk  1.33 31  41.23
 R  VRR  0.17  41.23  7.009
S  S SGk  10.6
S  S SLk  7.0
 S  VS S  0.742
 S  VS S  2.03
G
G
G
G
G
L
L
L
L
L
7.3 Calibration for Deterministic Codes …7
(3) Determine the ultimate limit state equation
R  SG  SL  0
(4) Determine the reliability index by the JC method
The solution process of JC method is omitted.
The solution result is :
  3.8082
If the load effect ratio   2.0 , then
  3.5828
Please refer to the reference book “Reliability of
Structures” by Professors Ou and Duan.
Turn to Page 97, look at the table 5.3 carefully!
7.3 Calibration for Deterministic Codes …8
7.3.2 Calibration of Partial Factors
1. Basic Principles
–
The partial factors in the LRFD format must be calibrated based on the
target reliability index adopted by the code.
X
–
i
X di
xi*


X ri X ri
In determining partial factors, the problem is reversed compared with
reliability analysis context introduced in Chapter3.
Reliability analysis
Known:
X
Find:
 , xi*
i
,
VX i
Partial factor calibration
Known:
Find:
  [ ] , VX i
 X i , xi*
7.3 Calibration for Deterministic Codes …9
2. Iteration Algorithm
(1) Formulate the limit state function and design equation.

Determine the probability distributions and appropriate parameters
for basic variables.

There can be at most only two unknown mean values needed to
solve. One is  R , the other corresponds a variant load effect Si .
Load effect ratios are used to relate the means of the load effects.
For the first iteration, we can use the limit state equation Z  0
evaluated at the mean values to get a relationship between the two
unknown means.

*
(2) Obtain an initial design point xi by assuming mean values.
(3) For each of the design point values xi* corresponding to a nonnormal distribution, determine the equivalent normal mean  Xe
i
e
and standard deviation  X by using equivalent normalization.
i
X  Xe
i
i
 X   Xe
i
i
7.3 Calibration for Deterministic Codes …10
(4) Calculate the n values of direction cosine i

i 
g
X i
 Xi
P*
 g


i 1  X i
n

 Xi 

P*

2
(i  1, 2, , n)
(5) Calculate the n values of design point xi*
xi*  Xi  i [ ] Xi
(i  1, 2,
, n)
(6) Update the relationship between the two unknown mean values by
solving the limit state function.
g ( x1* , x2* ,
, xn* )  0
(7) Repeat Steps 3-6 until {i } converge.
(8) Once convergence is achieved, calculate the partial factors.
 X  xi* / X ri
i
7.3 Calibration for Deterministic Codes …11
Example 7.2
Please refer to the textbook
“Reliability of Structures”
by Professor A. S. Nowak.
Turn to Page 231, look at the example 8.1 carefully!
The limit state function: Z  R  Q
The design equation:
 R R ≥ Q Q
Known parameters:
VR  0.1
VQ  0.12
[  ]  3.0
Probability information: R and Q are all normal and uncorrelated.
7.3 Calibration for Deterministic Codes …12
Solution
Iteration cycle 1
(1) Assume iteration initial values
r*  R
q*  Q
r *  q*  0
(2) Calculate direction cosine
 R  Q
Z
GR  
  R   R  VR R  0.1Q
R
P*
Z
GQ  
  Q   Q  VQ Q  0.12Q
Q
P*
GR
GQ
R 
 0.6402
Q 
 0.7682
2
2
2
2
GR  GS
GR  GS
7.3 Calibration for Deterministic Codes …13
(3) Calculate design points
r *  R   R [  ] R  R  0.6402  3.0  0.1 R  0.8079 R
q*  Q  Q [ ] Q  Q  0.7682  3.0  0.12Q  1.2766Q
(4) Update the relationship between the two unknown means
R  1.5801Q
r *  q*  0
Iteration cycle 2
(1) Calculate direction cosine
GR  0.1R  0.15801Q
GQ  0.12Q
R 
GR
G G
2
R
2
S
 0.7964
Q 
GQ
G G
2
R
2
S
 0.6048
7.3 Calibration for Deterministic Codes …14
(2) Calculate design points
r *  R   R [  ] R   R  0.7964  3.0  0.1 R  0.7611 R
q*  Q  Q [ ] Q  Q  0.6048  3.0  0.12Q  1.2177Q
(3) Update the relationship between the two unknown means
 R  1.5999Q
r *  q*  0
Iteration cycle 3
(1) Calculate direction cosine
GR  0.1 R  0.15999Q
GQ  0.12Q
R 
GR
G G
2
R
2
S
 0.8000
Q 
GQ
G G
2
R
2
S
 0.6000
7.3 Calibration for Deterministic Codes …15
(2) Calculate design points
r *  R   R [  ] R   R  0.8  3.0  0.1 R  0.7600 R
q*  Q  Q [ ] Q  Q  0.6  3.0  0.12Q  1.2160Q
(3) Update the relationship between the two unknown means
r *  q*  0
 R  1.6000Q
Iteration cycle 4
(1) Calculate direction cosine
GR  0.1 R  0.1600Q
GR
R 
 0.8000
2
2
GR  GS
GQ  0.12Q
GQ
Q 
 0.6000
2
2
GR  GS
{i } have converge. The iteration stop.
7.3 Calibration for Deterministic Codes …16
Table 7.1 Convergence process for Example 7.2
Numbers of
Iteration
R
Q
1
2
3
4
-0.6402
-0.7964
-0.8000
-0.8000
0.7682
0.6048
0.6000
0.6000
Assuming the mean values are the nominal design values, then the
partial factors are :
R 
Q 
r*
R
q*
Q
 0.7600
 1.22
Chapter 7
Reliability-Based Design Methods of
Structures
7.4 Target Reliability Index in Chinese Codes
7.4 Target Reliability Index in Chinese Codes …1
7.4.1 Safety Class of Building Structures
–
–
According to the importance and the consequences of structural damage,
the safety class of buildings in Unified Standard for Reliability Design of
Building Structures (GB50068 — 2001) is divided into three categories.
The safety class is considered through the importance factor  0
Table 7.2 Safety class of building structures
Safety
Class
Consequences of
Damage
Types of
Buildings
Importance
factor  0
Class one
Very severe
Important buildings
1.1
Class two
Severe
Common buildings
1.0
Class three
not severe
Unimportant buildings
0.9
7.4 Target Reliability Index in Chinese Codes …2
7.4.2 Target Reliability Index for Ultimate Limit State
Table 7.3 Target reliability index [  ] for ULS of structural member
Safety class
Types of damage
Class one
Class two
Class three
Ductile
3.7
3.2
2.7
Brittle
4.2
3.7
3.2
7.4 Target Reliability Index in Chinese Codes …3
7.4.3 Target Reliability Index for Serviceability Limit State
Table 7.4 Target reliability index [  ] for SLS of structural member
Irreversible Limit State
≥1.5
Reversible Limit State
≥0
1. What are the rules of target reliability indexes ?
2. Why are the target reliability indexes for ultimate limit
state and serviceability limit state different ?
3. How are these target reliability indexes determined ?
Chapter 7
Reliability-Based Design Methods of
Structures
7.5 Practical LRFD Formulas in Current Codes
7.5 Practical LRFD Formulas in Current Codes …1
7.5.1 Ultimate Limit State Design Formulas
n
 0 ( G SGk   Q SQ k    Q ci SQ k ) ≤
1
1
i 2
n
i
 0 ( G SGk    Q ci SQ k ) ≤
i 1
where,
0
G
i
i
i
1
R
1
R
Rk ( f k , ak , )
Rk ( f k , ak , )
— structural importance factor,
— partial factor for dead load,
Q
1
,
Q
— partial factors for the 1st and ith variant load,
i
SGk — effect of permanent load characteristic value
SQ1k — effect of variant load characteristic value which
dominates the load effect combination.
7.5 Practical LRFD Formulas in Current Codes …2
SQi k — effect of the ith variant load characteristic value
c
— combination factor of the ith variant load
i
R() — function of structural member
R
— partial factor for structural member resistance,
f k — characteristic value of material behavior,
ak — characteristic value of geometric parameter.
–
The second formula is mainly used in the structures, which is dominated
by permanent load. The most unfavorable one of the above two formulas
should be used in practical design situations.
–
The partial factors in the above two formulas are determined by the
principles introduced in this course and optimization method. You can
refer to the P.98-101 in the reference book.
7.5 Practical LRFD Formulas in Current Codes …3
7.5.2 Serviceability Limit State Design Formulas
1. Design Formula for Characteristic Values
n
SGk  SQ1k  ci SQi k ≤ [ f1 ]
i 2
2. Design Formula for Frequent Values
n
SGk   f1 SQ1k  qi SQi k ≤ [ f 2 ]
i 2
3. Design Formula for Quasi-Permanent Values
n
SGk  qi SQi k ≤ [ f3 ]
i 1
7.5 Practical LRFD Formulas in Current Codes …4
where,
 f SQ k
1
1
— effect of a variant load frequent value which
dominates the frequent load combination.
qi SQ k — effect of quasi-permanent value of a variant load.
i
[ f1 ] — the deformation or crack limit value corresponding
to characteristic value combination.
[ f2 ] — the deformation or crack limit value corresponding
to frequent value combination.
[ f3 ] — the deformation or crack limit value corresponding
to quasi-permanent value combination.
Chapter 7: Homework 7
Homework 7
Programming the above algorithms in MATLAB
environment according to the iteration algorithm proposed
by this course.
(1) By using your own subroutine, re-check the example 7.2
in this course.
(2) By using your own subroutine, re-calculate the example
8.3 in the text book on P.231
End of
Chapter 7
End of
This Course
Thank you
Very Much!