chapter 7 reliability-based design methods of structures

Chapter 7Chapter 7

Reliability-Based Design Methods of

Structures

Chapter 7: Reliability-Based Design Methods of StructuresChapter 7: Reliability-Based Design Methods of Structures

7.2 Reliability-Based Design Formulas

7.5 Practical LRFD Formulas in Current Codes

7.1 Reliability-Based Design Codes

Contents

7.3 Calibration for Deterministic Codes

7.4 Target Reliability Index in Chinese Codes

7.1 Reliability-Based Design Codes

Chapter 7Chapter 7 Reliability-Based Design Methods of Reliability-Based Design Methods of StructuresStructures

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:

Owner

Designer

Contractor

UserCode


7.1.2 Code Levels

– Level CodesⅠ : Use deterministic formulas

( )k kG Q kK S S R ≤

– Level CodesⅡ : Use approximate probability limit state design formula

– Level CodesⅢ : Use full probability analysis and design formula

– Level CodesⅣ : Use the total expected life cycle cost of the design as the optimization criterion


7.1.3 Reliability-Based Design Codes

General Principles on Reliability for Structures (ISO2394: 1998)

1. International Standard

2. Chinese Codes

Unified Standard for Reliability Design of Engineering Structures

(GB50153 — 92)


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)

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:

[ ]sPwhere,target failure probability, or target reliability index.

[ ]fP [ ], , are called target safety probability,

s sP P≥[ ]

f fP P≤[ ]

≥[ ]

… … … … …(1)

… … … … …(2)

… … … … …(3)

– The third formula is generally used in practical engineering.

Given: the probability distribution and digital characteristic

of the loads and resistance

Find: design vector

Subjected to: ( )

x

x

≥[ ]


Swhere,

is the mean value of load effect

is the mean value of resistanceRis the central safety factor0K

7.2.2 Single Factor Design Formulas

– The single factor formula based on mean values is as following:

– R & S are normal distributions

0 S RK

2 2 2 2

0 2 2

1 (1 )

1R S R

R

K

2 20 exp( )R SK – R & S are lognormal distributions


kSwhere,

is the characteristic value of load effect

is the characteristic value of resistancekR

is the characteristic safety factorK

7.2.2 Single Factor Design Formulas

– The single factor formula based on characteristic values is as following:

k kKS R

0

1

1R R

S S

kK K

k

(1 )k R R RR k

(1 )k S S SS k


0S

Frequency

S , load effect

S nS nSMean load

Nominal load

Factored load

Relationships among nominal load, mean load, and factored load


Relationships among nominal resistance, mean resistance, and factored resistance

0R

Frequency

R , Resistance

RnRnRMean resistance

Nominal resistance

Factored resistance


niSwhere,

is the nominal (design) value of load effect component,

is the load partial factor for load component,Siis the nominal (design) value of resistance or capacity,nR

is the resistance partial factor.R

7.2.3 Multiple Factor Design Formulas

(Load and Resistance Factor Design, LRFD)

– The LRFD formula is as following:

1ni n

R

S R

Si

Factored nominal resistance Total factored nominal load effect


7.2.3 Multiple Factor Design Formulas

(Load and Resistance Factor Design, LRFD)

* * *1 2( , , , ) 0ng X X X

– The partial safety factors and must be calibrated based on the target index adopted by the code.

R Si

* 1

1S S

Sk S S

S

S k

* (1 )S S S S S SS

(1 )k S S SS k

*

1

1k R R

RR R

R k

R

* (1 )R R R R R RR

(1 )k R R RR k

7.3 Calibration for Deterministic Codes


7.3 Calibration for Deterministic Codes …1

7.3.1 Calibration of Target Reliability Index

1. Basic Principles

( ) 0k Gk QkR K S S

where, — safety factor,K

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:

kR — characteristic value of member resistance ,

GkS , — characteristic value of permanent load effect and

variant load effect designed according to the

deterministic code .

QkS


0G QR S S

Now, the problem can be re-formulated as follows:

How much is the reliability implicit in the original deterministic structural design code (Level Code)?Ⅰ

– When the calibration method is used, the limit state equation in simple load combination condition can be formulated as:

where, — structural member resistance,R

GS — dead load effect,

QS — 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.


– It is assumed that the following parameters of the basic random variables are known:

, , QG

G Q

SSRR S S

k Gk QkR S S

bias factor:

, , QG

G Q

G Q

SSRR S S

R S S

V V V

variation factor:

– It is assumed that is linearly related with and .kR GkS QkS

Let Qk

Gk

S

S

then ( ) ( )

(1 )

k Gk Qk Gk Gk

Gk

R K S S K S S

K S

, is called load effect ratio,


(1) Assume one value of the load effect ratio ;

kR

2. Calculation Procedure

(2) Determine the characteristic value of member resistance :

(3) Determine the mean values and standard deviations of the basic variables :

, ,G G Q QR R k S S Gk S S QkR S S mean values:

, ,G G G Q Q QR R R S S S S S SV V V standard deviations:

(1 )k GkR K S

(4) Determine the limit state equation:

0G QR S S (5) Solve the reliability index by the JC method. (6) Adjust the load effect ratio, calculate the mean value of different reliability indexes.


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 following parameters are known:

Assume that the ratio of live load to dead load ,

Try to calibrate the reliability index of the ultimate limit state in TJ9-74 code.

/ 1.0Qk GkS S

1.33R R is lognormal 0.17RV

1.06GS

GS is normal 0.07GS

V

0.70LS

LS is Extreme Ⅰ 0.29LS

V

1.55K 10LkS kN m


(1) Determine

1.0

Solution

kR

/ 10 /1 10k kG LS S

( ) 1.55 (10 10) 31k Gk QkR K S S

(2) Determine the means and standard deviations

0.742G G GS S SV

1.33 31 41.23R R kR

0.17 41.23 7.009R R RV

10.6G GS S GkS

2.03L L LS S SV

7.0L LS S LkS


(3) Determine the ultimate limit state equation

0G LR S S

(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 , then2.0 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!


7.3.2 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.

– In determining partial factors, the problem is reversed compared with reliability analysis context introduced in Chapter3.

Reliability analysis

iX

iXVKnown: ,

Find: ,

Partial factor calibration

iX

Known: ,

Find: ,

[ ] iX

V

*

i

di iX

ri ri

X x

X X

*ix

*ix


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 , the other corresponds a variant load effect . Load effect ratios are used to relate the means of the load effects.

RiS

*ix(2) Obtain an initial design point by assuming mean values.

For the first iteration, we can use the limit state equation evaluated at the mean values to get a relationship between the two unknown means.

0Z

i

eX

(3) For each of the design point values corresponding to a non- normal distribution, determine the equivalent normal mean and standard deviation by using equivalent normalization.

*ix

i

eX

i i

eX X

i i

eX X


*ix(5) Calculate the n values of design point

* [ ]i ii X i Xx ( 1,2, , )i n

(6) Update the relationship between the two unknown mean values by solving the limit state function.

* * *1 2( , , , ) 0ng x x x

(7) Repeat Steps 3-6 until converge.{ }i

i(4) Calculate the n values of direction cosine

*

*

2

1

i

i

Xi P

in

Xi i P

gX

gX

( 1,2, , )i n

(8) Once convergence is achieved, calculate the partial factors.* /

iX i rix X


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!

0.1RV 0.12QV [ ] 3.0

Z R Q

R R Q Q ≥The limit state function:

The design equation:

Known parameters:

Probability information: R and Q are all normal and uncorrelated.


SolutionIteration cycle 1

(1) Assume iteration initial values*

Rr *Qq

* * 0r q R Q

(2) Calculate direction cosine

*

0.1R R R R R QP

ZG V

R

*

0.12Q Q Q Q Q Q

P

ZG V

Q

2 20.6402R

R

R S

G

G G

2 20.7682Q

Q

R S

G

G G


(3) Calculate design points

* * 0r q 1.5801R Q


2 20.7964R

R

R S

G

G G

2 20.6048Q

Q

R S

G

G G

* [ ] 0.6402 3.0 0.1 0.8079R R R R R Rr * [ ] 0.7682 3.0 0.12 1.2766Q Q Q Q Q Qq

(4) Update the relationship between the two unknown means

Iteration cycle 2

0.1 0.15801R R QG

0.12Q QG



* * 0r q 1.5999R Q


2 20.8000R

R

R S

G

G G

2 20.6000Q

Q

R S

G

G G

* [ ] 0.7964 3.0 0.1 0.7611R R R R R Rr * [ ] 0.6048 3.0 0.12 1.2177Q Q Q Q Q Qq


Iteration cycle 3

0.1 0.15999R R QG

0.12Q QG



* * 0r q 1.6000R Q


2 20.8000R

R

R S

G

G G

2 20.6000Q

Q

R S

G

G G

* [ ] 0.8 3.0 0.1 0.7600R R R R R Rr * [ ] 0.6 3.0 0.12 1.2160Q Q Q Q Q Qq


Iteration cycle 4

0.1 0.1600R R QG 0.12Q QG

have converge. The iteration stop.{ }i


Assuming the mean values are the nominal design values, then the partial factors are :

R

Numbers of Iteration

1 2 3 4

-0.6402 -0.7964 -0.8000 -0.8000

0.7682 0.6048 0.6000 0.6000Q

Table 7.1 Convergence process for Example 7.2

*

0.7600RR

r

*

1.22QQ

q

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.

Safety

Class

Consequences of Damage

Types of

Buildings

Importance factor

Class one Very severe Important buildings 1.1

Class two Severe Common buildings 1.0

Class three not severe Unimportant buildings 0.9

– The safety class is considered through the importance factor 0

0

Table 7.2 Safety class of building structures


7.4.2 Target Reliability Index for Ultimate Limit State

Types of

damage

Safety class

Class one Class two Class three

Ductile 3.7 3.2 2.7

Brittle 4.2 3.7 3.2

Table 7.3 Target reliability index for ULS of structural member[ ]


7.4.3 Target Reliability Index for Serviceability Limit State

Irreversible Limit State

Reversible Limit State

Table 7.4 Target reliability index for SLS of structural member[ ]

1.5≥

0≥

3. How are these target reliability indexes determined ?

2. Why are the target reliability indexes for ultimate limit state and serviceability limit state different ?

1. What are the rules of target reliability indexes ?

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

where, — structural importance factor,0

G — partial factor for dead load,

1Q , — partial factors for the 1st and ith variant load,

iQ

1 102

( ) ( , , )i i

n

G Gk Q Q k Q ci Q k k k ki

S S S R f a

R

1≤

01

( ) ( , , )i i

n

G Gk Q ci Q k k k ki

S S R f a

R

1≤

GkS — effect of permanent load characteristic value

1Q kS — effect of variant load characteristic value which

dominates the load effect combination.


iQ kS — effect of the ith variant load characteristic value

ic — combination factor of the ith variant load

( )R — function of structural member

R — partial factor for structural member resistance,

kf — characteristic value of material behavior,

ka — 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.


7.5.2 Serviceability Limit State Design Formulas

1 12

[ ]i

n

Gk Q k ci Q ki

S S S f

≤

1. Design Formula for Characteristic Values

1 1 22

[ ]i

n

Gk f Q k qi Q ki

S S S f

≤

2. Design Formula for Frequent Values

31

[ ]i

n

Gk qi Q ki

S S f

≤

3. Design Formula for Quasi-Permanent Values


where, 1 1f Q kS — effect of a variant load frequent value which

dominates the frequent load combination.

iqi Q kS — effect of quasi-permanent value of a variant load.

1[ ]f — the deformation or crack limit value corresponding

to characteristic value combination.


to frequent value combination.


to quasi-permanent value combination.

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

Chapter 7: Homework 7

End of

Chapter 7Chapter 7

End of

This CourseThis Course

Thank you

Very Much!

chapter 7 reliability-based design methods of structures

Documents

formulas of reliability

kinds of reliability

single factor design

target reliability index

deterministic codes

factored load slide

current codes

chinese codes unified