Uncertainty budget In many situations we have uncertainties come from several sources. When the total uncertainty is too large, we look for ways of reducing it. 1 One way costs $100 Another way $50

Uncertainty budget The tolerable total uncertainty is our uncertainty budget, and we need to achieve it by reducing individual uncertainties in the most cost effective way. This will be illustrated by a study by Chanyoung Park on deciding between reducing uncertainties at the material level or at the structural level. 2

Chanyoung Park, Raphael T. Haftka, and Nam-Ho Kim Modeling the Effect of Structural Tests on Uncertainty in Estimated Failure Stress (Strength) 3

Multistage testing for design acceptance  Building-block process -Detect failures in early stage of design -Reduce uncertainty and estimate material properties -A large number of tests in lower pyramid (reducing uncertainty) -System-level probability of failure controlled in upper pyramid (certification) ELEMENTS DETAILS COMPONENTS COUPONS SYSTEM DATA BASE STRUCTURAL FEATURES GENERIC SPECIMENS NON-GENERIC SPECIMENS 4

5 Structural elements are under multi-axial stress and element strength has variability (aleatory uncertainty) Element strength is estimated from material strengths in different directions using failure theory, which is not perfectly accurate (epistemic uncertainty) Estimating distribution of element strength based on material strengths and a failure theory Coupon tests are done to characterize the material strength and its aleatory uncertainty ELEMENT COUPON Uncertainty in element strength estimates

6 With finite number of coupon tests we are left with errors in prameters (epistemic uncertainty) Element tests reduce the uncertainty in the failure theory. If we can tolerate a certain total uncertainty we need to decide on number of coupon and element tests. Uncertainty in element strength estimates ELEMENT COUPON Mean and STD of material strength Mean and STD of element strength Failure theory

Estimating mean and STD of material strength Goal: Estimate distribution of material strength from n c samples Assumption: true material strength:  c,true ~ N(  c,true,  c,true ) Sample mean & STD: (  c,test,  c,test ) Predicted mean & STD 7  c,true  c,true  c,test  c,test  c,P Distribution of distributions!!  c,P ~ N(  c,P,  c,P )

Question In tests of 50 samples, the mean strength was 100 and the standard deviation of the strength was 7. What is the typical distance between the red and purple curves in the figure –7 –1 –0.7 8  c,true  c,true  c,test  c,test  c,P

 c,P ~ N(  c,P,  c,P ) Obtaining the predictive strength distribution by sampling?  Predictive distribution of material strength (Double roof MCS) 1.Predictive mean & STD 2.Samples of possible material strength distributions 3.Predictive true material strength distribution How do we decide how many samples? 9 Sampling  c,P  c,P  c,P

Example (predictive material strength)  Predictive distribution of material strength w.r.t. # of specimens 1. τ c,P is biased but it compensates by wider distribution 2.Are any of these results extreme compared to the expected scatter? # of samples 3080 μ c,test σ c,test Std. μ c,P Std. σ c,P Mean τ c,P Std. τ c,P μ c,true 1.1 σ c,true  c,P n c = 30  c,P n c = 80  c,true  c,P ~ N(  c,P,  c,P )

Estimating element strength Assumption: true element strength  e,true ~ N(  e,true,  e,true ) Error in failure theory Element tests used to reduce errors using Bayesian updating 11 Failure theory Coupon strengthElement strength 22 11  c,true  e,true Multiaxial loading Failure envelope k true  e,true = k true  c,true  e,true =   e,P = (1 – e k )k calc  c,P  e,P = (1 – e  )  c,P Errors

Combined uncertainty (mean element strength) Uniform distribution for error in failure theory (±10%) Used Bayesian network to calculate PDF of  e,Ptrue Similar calculation for element std. 12  e,P = (1 – e k )k calc  c,P ncnc  c,test  c,test  c,P ekek  e,P

Bayesian statistics for modeling the effect of tests Combined uncertainty in element failure strength predictions based on coupon tests and failure theory The combined uncertainty represents “current state of knowledge” The “current state of knowledge” can be improved by carrying out element tests Uncertainty in element tests should also be considered Bayesian statistics is a probabilistic method to model the effect of tests 13  How to incorporate the effect of tests?

Bayesian update Element test ~ N(  e,true,  e,true ) Update the joint PDF (  e,P,  e,P ) using n e element tests 14  e,P  e,true f M (  | test) = L(test |  )f M (  ) Reduce epistemic uncertainty in e k

Illustration of convergence of coupon  c,P &  c,P  Estimated mean of (  c,P &  c,P ) (single set cumulative) TestDistributionParameters Coupon testNormal  c,true = 1.1,  c,true =  True distribution of material strength 15

Illustrative example (coupon tests)  Estimated STD of (  c,P &  c,P ) -Increasing n c reduces uncertainties in the estimated parameters -Effectiveness of reducing uncertainty is high at low n c 16

Question Why does the convergence on Slide 16 (uncertainty estimates) look so much better than the convergence on Slide 15 (estimates of mean and standard deviations) – Noise to signal ratio – Difference in scales – Both 17

The effect of element tests for 10 coupon tests  RMS error (500K instances of tests) vs. uncertainty in mean and standard deviation from a single set of tests 18 STD of meanSTD of STD TestDistributionParameters Element testNormal  e,true = 1.1,  e,true = 0.099

The effect of element tests for 90 coupon tests 19  RMS error (500K MCS) vs. estimated uncertainty in means and standard deviation from a single set of tests. STD of meanSTD of STD

The effect of number of tests on error in element strength prediction -First element test has a substantial effect to reduce uncertainty in estimated parameters of element strength 20 Mean of element strengthSTD of element strength

Summary 21  Uncertainty Budget for reducing uncertainty  Reducing uncertainty with multistage testing (building block test)  Modeling uncertainty reduction with tests and uncertainty propagation between stages  Increasing the number of element tests is more efficient than increasing the number of coupon tests for reducing cumulative uncertainty  Uncertainty in failure theory (model) is much larger than uncertainty from coupon tests (characterization)