On the reliability of timber structures the role of timber grading

Transcription

1 Institute of Structural Engineering[ 1 COST E53 WG III Meeting, Oslo, On the reliability of timber structures the role of timber grading Jochen Köhler ETH Zürich, Institute for Structural Engineering, Group Risk & Safety Bild: Jussi Tiainen, NTC/LIGNUM

2 Institute of Structural Engineering[ 2 Bild: Frédéric Beaud/LIGNUM Overview Lifecycle and Performance of timber structures Structural Reliability Theory a brief introduction Modelling timber material properties Timber Grading the multiple challenge

3 Institute of Structural Engineering[ 3 Timber is advantorous in constructions: Timber..can be sustainably used in construction..provides a pleasant indoor environment..is a efficient building material

4 Institute of Structural Engineering[ 4 However timber is not associated as a high performance building material: Failure and/or collapse: Is it a concrete or steel structure -> the engineers fault Is it a timber structure? -> it is timber! In the general perception timber is not associated with properties as durability, strength, reliability. Siemens Arena, Copenhagen DK,

5 Institute of Structural Engineering[ 5 This is not due to the properties of timber,. Timber..is a natural grown material and its properties can not be designed by some recipe...is a complex building material, its proper use in construction requires expertise. Siemens Arena, Copenhagen DK,

6 Institute of Structural Engineering[ 6 This is not due to the properties of timber,.... but due to our incomplete knowledge about its properties!! Siemens Arena, Copenhagen DK,

7 Institute of Structural Engineering[ 7 The estimation of the performance of timber structures is associated with large uncertainties!! Siemens Arena, Copenhagen DK,

8 Institute of Structural Engineering[ 8 The lifecycle of (timber)-structures

9 Institute of Structural Engineering[ 9 The lifecycle of (timber)-structures

10 Institute of Structural Engineering[ 10 The lifecycle of (timber)-structures

11 Institute of Structural Engineering[ 11 The lifecycle of (timber)-structures Uncertainties Traffic volume Loads Resistances (material, soil,..) Degradation processes Service life Manufacturing costs Execution costs Decommissioning costs

12 Institute of Structural Engineering[ 12 The lifecycle of (timber)-structures Uncertainties Traffic volume Loads Resistances (material, soil,..) Degradation processes Service life Manufacturing costs Execution costs Decommissioning costs

13 Institute of Structural Engineering[ 13 Structural Engineering Decision Problem: Design Process:

14 Institute of Structural Engineering[ 14 Structural Engineering Decision Problem: Optimal Design: Expected Benefit of the Structure Benefit of the Structure in Service E Reliability [ B] P ( C ) Risk [ ] E B = I ( 1 F D ) CD CF PF ( CD ) = 0 C D

15 Institute of Structural Engineering[ 15 Structural Engineering Decision Problem: Optimal Design: Expected Benefit of the Structure Benefit of the Structure in Service E Reliability [ B] P ( C ) Risk [ ] E B = I ( 1 F D ) CD CF PF ( CD ) = 0 C Probability of Failure: f ( 0) P = P R S ( ( ) 0) ( ) Pf = P g X = fx x dx g D ( x) 0

16 Institute of Structural Engineering[ 16 Reliability assessment: Probability of Failure: f ( 0) P = P R S Consider the simple case where the resistance R and the load S are modelled as normal distributed random variables: ( ) R ~ N µ, σ R R ( ) S ~ N µ, σ S S

17 Institute of Structural Engineering[ 17 Reliability assessment: Probability of Failure: f ( 0) P = P R S Consider the simple case where the resistance R and the load S are modelled as normal distributed random variables: ( ) R ~ N µ, σ R R ( ) S ~ N µ, σ S S The probability of failure can be estimated by considering the so called safety margin: M = R S

18 Institute of Structural Engineering[ 18 Reliability assessment: Probability of Failure: f ( 0) P = P R S Consider the simple case where the resistance R and the load S are modelled as normal distributed random variables: ( ) R ~ N µ, σ R R ( ) S ~ N µ, σ S S The probability of failure can be estimated by considering the so called safety margin: M = R S The failure probability can now be controlled by, e.g., variation of the resistance and a combination with acceptable failure probability might be identified.

19 Institute of Structural Engineering[ 19 Reliability assessment: Probability of Failure: f ( 0) P = P R S Consider the simple case where the resistance R and the load S are modelled as normal distributed random variables: ( ) R ~ N µ, σ R R ( ) S ~ N µ, σ S S The probability of failure can be estimated by considering the so called safety margin: M = R S The failure probability can now be controlled by, e.g., variation of the resistance and a combination with acceptable failure probability might be identified.

20 Institute of Structural Engineering[ 20 Reliability assessment: Probability of Failure: f ( 0) P = P R S Consider the simple case where the resistance R and the load S are modelled as normal distributed random variables: ( ) R ~ N µ, σ R R ( ) S ~ N µ, σ S S The probability of failure can be estimated by considering the so called safety margin: M = R S The failure probability can now be controlled by, e.g., variation of the resistance and a combination with acceptable failure probability might be identified.

21 Institute of Structural Engineering[ 21 Reliability assessment: Probability of Failure: f ( 0) P = P R S Consider the simple case where the resistance R and the load S are modelled as normal distributed random variables: ( ) R ~ N µ, σ R R ( ) S ~ N µ, σ S S The probability of failure can be estimated by considering the so called safety margin: M = R S The failure probability can now be controlled by, e.g., variation of the resistance and a combination with acceptable failure probability might be identified.

22 Institute of Structural Engineering[ 22 Structural Engineering Decision Problem: Design Process: E [ B] P ( C ) [ ] E B = I ( 1 F D ) CD CF PF ( CD ) = 0 C D! Design strategies might be quantified by explicit cost benefit consideration!!!

23 Institute of Structural Engineering[ 23 Structural Engineering Decision Problem: Design Process: Codes and regulations: -Deterministic limit state equations -Loads, resistance, safety factors --> Decision support tool for engineers Derived by code authorities Traditionally based on experience and judgement AND : recently calibrated and refined by explicit probabilistic assessment

24 Institute of Structural Engineering[ 24 Modern design codes are load and resistance factor design LRFD formats (and these are referred to as semiprobabilistic): LRF Design Equation: modification factor on resistance k γ characteristic values for resistance and load r ( γ ) s mod z k S k = rd sd = 0 M partial safety factors on resistance and load

25 Institute of Structural Engineering[ 25 Modern design codes are load and resistance factor design LRFD formats (and these are referred to as semiprobabilistic): LRF Design Equation: modification factor on resistance k γ characteristic values for resistance and load r ( γ ) s mod z k S k = rd sd = 0 M partial safety factors on resistance and load Note that the design equation above corresponds to the probabilistic formulation from the example before!

26 Institute of Structural Engineering[ 26 Modern design codes are load and resistance factor design LRFD formats (and these are referred to as semiprobabilistic): LRF Design Equation: modification factor on resistance k γ characteristic values for resistance and load r ( γ ) s mod z k S k = rd sd = 0 M partial safety factors on resistance and load

27 Institute of Structural Engineering[ 27 Modern design codes are load and resistance factor design LRFD formats (and these are referred to as semiprobabilistic): LRF Design Equation: modification factor on resistance k γ characteristic values for resistance and load r ( γ ) s mod z k S k = rd sd = 0 M partial safety factors on resistance and load

28 Institute of Structural Engineering[ 28 Modern design codes are load and resistance factor design LRFD formats (and these are referred to as semiprobabilistic): LRF Design Equation: modification factor on resistance k γ characteristic values for resistance and load r ( γ ) s mod z k S k = rd sd = 0 M partial safety factors on resistance and load

29 Institute of Structural Engineering[ 29 Modern design codes are load and resistance factor design LRFD formats (and these are referred to as semiprobabilistic): LRF Design Equation: modification factor on resistance k γ characteristic values for resistance and load r ( γ ) s mod z k S k = rd sd = 0 M partial safety factors on resistance and load Partial safety factors are calibrated to provide similar failure probability over different design equations. The calibration is sensitive to the representation of load and resistance.

30 Institute of Structural Engineering[ 30 Modern design codes are load and resistance factor design LRFD formats (and these are referred to as semiprobabilistic): LRF Design Equation: modification factor on resistance k γ characteristic values for resistance and load r ( γ ) s mod z k S k = rd sd = 0 M partial safety factors on resistance and load Partial safety factors are calibrated to provide similar failure probability over different design equations. The calibration is sensitive to the representation of load and resistance.

31 Institute of Structural Engineering[ 31 In general reliability assessment is less simplistic, e.g.: Problems are time variant Basic variables are non-normal Basic variables are not independent Limit state functions are complex -> the methods of structural reliability provide a toolbox to cope with these problems.

32 Institute of Structural Engineering[ 32 In general reliability assessment is less simplistic, e.g.: Problems are time variant Basic variables are non-normal Basic variables are not independent Limit state functions are complex -> the methods of structural reliability provide a toolbox to cope with these problems. The models for the basic variables (as load and resistance) should be based on the best knowledge available!!! -> it is clear that the model development is an ongoing procedure!!!

33 Institute of Structural Engineering[ 33 The probabilistic modelling of timber material properties is a challenging task!!! Timber material properties depend on: Loads Q T Q L G Size 25 years t Moisture ω Temperature T 25 years t 25 years t

34 Institute of Structural Engineering[ 34 The probabilistic modelling of timber material properties is a challenging task!!! -Timber is a natural grown material -Timber is orthotroph: --> different material properties for different loading modes

35 Institute of Structural Engineering[ 35 The probabilistic modelling of timber material properties is a challenging task!!! -Timber is a graded material -Grading is performed considering reference material properties -Timber material properties depend on size, time and type

36 Institute of Structural Engineering[ 36 The probabilistic modelling of timber material properties requires a careful and differentiated consideration of uncertainties; Uncertainties due to: Natural variability Model uncertainty Statistical uncertainty

37 Institute of Structural Engineering[ 37 The probabilistic modelling of timber material properties requires a careful and differentiated consideration of uncertainties; Uncertainties due to: Natural variability Model uncertainty Statistical uncertainty The explicit consideration of uncertainties is especially important when the uncertainties are high!!!

38 Institute of Structural Engineering[ 38 Timber Grading is an excellent opportunity to control uncertainties! The objective of timber grading is to divide a population of ungraded timber in sub-populations that fulfill certain requirements in regard to certain material properties. The strength class system specifies these requirements in regard to the bending strength and stiffness and the timber density of test specimen. Other material properties are estimated based on these.

39 Institute of Structural Engineering[ 39 Timber Grading is an excellent opportunity to control uncertainties! Existing rules for the calibration of grading procedures take uncertainties only implicitly into account. -> this means that these procedures can not be used as a basis for the derivation of probabilistic models. (but these are the essence of semiprobabilistic design) that information is not efficiently used.

40 Institute of Structural Engineering[ 40 Timber Grading could be represented simply as regression problem: property of interest (poi) indicating property (IP)

41 Institute of Structural Engineering[ 41 Timber Grading could be represented simply as regression problem: poi grade 3 grade 2 grade 1 IP2 IP1 IP

42 Institute of Structural Engineering[ 42 Timber Grading could be represented simply as regression problem: poi grade 3 grade 2 grade 1 POI = β + β IP + ε 0 1 IP2 IP1 IP regression model

43 Institute of Structural Engineering[ 43 Timber Grading could be represented simply as regression problem: The regression model can be formulated probabilistically; this implies that: Uncertainties, e.g. due to limited data can be represented explicitly, The model can be updated when new information becomes available, The model can be verified to represent sub populations from different origins, The model is capable to consider the machine control and output control approach. Existing data can be easily used to develop regression models.

44 Institute of Structural Engineering[ 44 Summary and outlook Modern design formats are semi-probabilistic. They require that models for loads and resistance are based on our best available knowledge, And that uncertainties are represented complete and consistently. Timber grading has high potential to control and reduce these uncertainties. Future developments for grading rules and regulations should consider the potential of a probabilistic representation.