Resistance vs. Load Reliability Analysis

Transcription

1 esistance vs. oad eliability Analysis et be the load acting on a system (e.g. footing) and let be the resistance (e.g. soil) Then we are interested in controlling such that the probability that > (i.e. the reliability) is acceptably high or, equivalently, that < (i.e. the failure probability) is acceptably low, where P [ < ] = f ( r, l) drdl r< l and where f (,) r l is the joint (bivariate) distribution of and. f r l drdl r r dr l l dl (,) P = [ < + < + ]

2 Bivariate Distributions [ ] P l1 < l2 r1 < r = 2 f ( r, l) drdl l r l r 1 1

3 esistance vs. oad eliability Analysis The estimation of f (,) r l typically requires vast amounts of data, which is generally impractical. Simplifications: 1. assume and are independent so that f (,) r l = f () r f () l and [ ] P < = f ( r) f ( l) drdl = f ( l) f ( r) drdl r< l r< l

4 esistance vs. oad eliability Analysis 2. Assume and are either normally or lognormally distributed. The event { < } is the same as the events i) { < 0} ii) {/ < 1} If both and are normally distributed, then X = is also normally distributed with parameters μ = μ μ X σ = σ + σ 2 X (assuming and are independent).

5 eliability Index the reliability index, β, is the number of standard deviations the mean is from failure. superior to the Factor-of-Safety approach because it depends on both the mean and the standard deviation. failure occurs if X < 0 (normal) or ln X < 0 (lognormal). Defining μ X β =, (normal) σ X μln X β =, (lognormal) σ ln X μ X P[ failure] = P[ X < 0 ] =Φ =Φ( β ), (normal) σ X μ ln X P[ failure] = P[ ln X < 0 ] =Φ =Φ( β ), (lognormal) σ ln X

6 esistance vs. oad eliability Analysis Suppose load is normally distributed with mean 10 and standard deviation 3 Suppose resistance is normally distributed with mean 20 and standard deviation 4. Then X = has mean and variance μ = μ μ = = 10 Mean FS = X σ = σ + σ = = 25 2 X μ /μ = 20/10 = 2 eliability index = β = μ X /σ X = 10/5 = Probability of failure = P[ < ] = P[ X < 0] =Φ 5 =Φ( 2) = 0.023

7 esistance vs. oad eliability Analysis μ X P < = P < 0 = P < 0 =Φ σ X where Φ( x) is the standard normal cumulative distribution function. Now [ ] [ ] [ X ]

8 esistance vs. oad eliability Analysis Alternatively, if and are lognormally distributed, then X = is also lognormally distributed with parameters μ = μ μ ln X ln ln σ = σ + σ (assuming independence) 2 ln X ln ln so that P[ < ] = P [ / < 1] = P[ X < 1] = P[ lnx < 0] μ =Φ σ ln X ln X

9 esistance vs. oad eliability Analysis Suppose load is lognormally distributed with mean 10 and standard deviation 3 Suppose resistance is lognormally distributed with mean 20 and standard deviation 4. Then X = / is lognormally distributed μ μ = 20, σ = 4 = 10, σ = 3 σ σ = ln 1+ = ln 2 μ ( ) μ = ln μ 0.5σ = σ 2 ln ln σ = ln 1+ = ln 2 μ ( ) μ = ln μ 0.5σ = ln ln

10 β esistance vs. oad eliability Analysis X = / ln X = ln ln μ = μ μ = ln X ln ln σ = σ + σ = σ = μ σ 2 ln X ln ln ln X [ ] ln X = = = < =Φ = ln X P / 1 ( 2.02) 0.022

11 eliability Index

12 eliability Index More generally, system failure can be defined in terms of a failure or limit state function. Also called the safety margin M = gz (, Z, ) 1 2 Failure occurs when M = g(z 1, Z 2, ) < 0. In this case, the reliability index is defined as β = μ σ M M Problem: different choices of the function M lead to different reliability indices (e.g. M = or M = ln(/) both imply failure when M < 0, but lead to different values of β in first order approximations).

13 eliability Index Example 1: suppose that M = and and are normally distributed. Then μ = μ μ and M σ = σ + σ M (assuming independence) so that β = μ σ μ + σ and Pfailure [ ] = P[ M < 0] =Φ( β ) This is exact, so long as and are normally distributed and independent (if not independent then must involve covariances in computation of σ M ).

14 eliability Index Example 2: suppose that M = ln( / ) = ln ln and and are lognormally distributed. Then μ = μ μ and M ln ln σ = σ + σ M ln ln (assuming independence) μln μln so that β = σ + σ ln ln and Pfailure [ ] = P[ M < 0] =Φ( β ) This is exact, so long as and are lognormally distributed and independent (if not independent then must involve covariances in computation of σ M ).

15 eliability Index Example 3: suppose that M = ln( / ) = ln ln and and are normally distributed. Then the distribution of M is complex and we must approximate its moments; μ ln μ ln μ (to first order) M M 2 M 2 M ln + ln μ μ σ σ σ = σ μ σ + = V + V μ now β = ln μ V ln μ + V It was β = μ σ μ + σ in Example 1 These are clearly different.

16 Hasover-ind eliability Index Hasover and ind (1974) solved this ambiguity by mapping the set of system variables, Z, onto a set of standardized (mean zero, unit variance) and uncorrelated variables, X [ ] ( ) X= A Z E Z where the transformation matrix A is the solution of AC A Z T = I where C Z is the matrix of covariances between the system variables, Z, and I is the identity matrix. In terms of Z, ( [ ]) T 1 z Z C z [ Z] ( ) β = min E E z Z Z where Z is the failure surface. The value of z which minimizes this is called the design point, z *.

17 Hasover-ind eliability Index Hasover-ind s reliability index is the minimum distance from the mean to the failure surface in standardized space (figure from Madsen, Krenk, and ind, 1986)

18 Going Beyond Calibration Must move beyond calibration for real benefits of FD Simple probability-based methods take load and resistance distributions into account - nominal or characteristic resistance: n = k μ - nominal or characteristic load: n = k μ Design: ϕ n = γ n P[failure] = P[ < ] = P[ / < 1]

19 Going Beyond Calibration let M = ln(/) then P[failure] = P[ M < 0] β is the reliability index typically β ranges from 2.0 to 3.0

20 To determine both load and resistance factors: We want to produce a design such that the mean and standard deviation of resistance satisfies P < 1 = P[ ln ln< 0 ] =Φ( β ) In detail < = 0 E[ln ln ] < =Φ μ μ =Φ ln ln P[ln ln 0] P Z ( ) SD[ln ln ] β σln + σ ln so that μ σ ln μ + σ ln ln ln Now, since or = β μ = μ + β σ + σ μ ln( μ ) 0.5σ 2 ln ln 2 ln ln ln ln μ βσ ( + σ ) ln ln ln 2 = and μ = exp( μln + 0.5σln ) { ln ln } { ln ln } μ = μ exp 0.5σ βσ exp 0.5σ βσ { ln + ln } = { ln + ln } exp 0.5σ 0.75βσ μ exp 0.5σ 0.75βσ μ we get

21 Writing this in terms of the nominal load and resistance = k μ ( k < 1) n = k μ ( k > 1) n gives us exp{ 0.5σln 0.75βσln } exp{ 0.5σln βσln } n = k k ecalling that our FD has the form ϕ n = γ n implies n esistance factor: oad factor: ϕ = γ = 2 { σln βσln } exp k 2 { σln + βσln } exp k

22 If load factors are known (e.g. from structural codes) then the resistance factor becomes dependent on both the resistance variability and the load variability. In this case, our FD can be written γ γ i ni i ni i i n = in = = i i n kμ ϕ γ ϕ where n = k i μ i i μ V ni μ i i i k = = σ μ i γ iqn i 2 1+ V i = exp + k 2 μ 1 V + { } β σ ln σln = = i i i (assuming loads are independent) V μ μ

23 Going Beyond Calibration thus, for given target reliability index and variances, the resistance factor can be computed as γ in i 2 1+ V i ϕ = exp β σ + σ k 2 μ 1 V + { } ln ln which depends on - coefficient of variation of load ( V ) - coefficient of variation of resistance ( ) V - load factors γ ι - characteristic coefficients, k and k i

24 Problems Implementing FD the coefficient of variation of resistance depends on; - variability in material properties - error in design models - measurement and correlation errors - construction variability with steel and concrete, the material property variability does not change significantly with location (quality controlled materials) with soils, the material property mean and variability change within a site and from site to site

25 Problems Implementing FD there is a dependence between resistance and load which is generally absent (or small) in structural engineering, e.g. shear strength is dependent on stress; τ = c + σ tanφ f

26 Problems Implementing FD No common definition of characteristic value - often defined as a cautious estimate of the mean, but sometimes as a low percentile - we badly need a standard definition (median?) V changes with intensity of site investigation - resistance factor should approach 1.0 as the site is more thoroughly investigated - this would lead to a complex table of resistance factors (however, see, e.g., AS 5100, AS 2159, AS4678, Eurocode 7, NCHP507)

27 Future Directions probabilistic methods generally limited to single random variable models (e.g. vs. ) to consider the effect of spatial variability, random field simulation combined with finite element analysis is necessary (FEM) the random field simulation allows the representation of a soil s spatial variability the finite element analysis allows the soil to fail along weakest paths (decreased model error)

28 Conclusions geotechnical engineers led the way with imit States concepts (1940 s) but have been slow to migrate to reliability-based design methods. the most advanced FD codes currently are AS 5100 and Eurocode 7. all current FD geotechnical codes have load and resistance factors calibrated from older WSD codes, with some adjustments based on engineering judgement and simple probability methods.