Mätosäkerhet och kundlaster

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1 Kurs i Lastanalys för Utmattning SP Bygg och Mekanik Pär Johannesson Par.Johannesson@sp.se PJ/ Uncertainty of Customer Loads Two scales: Small scale: individual customers (or measurement). Two measurements from the same customer: Uncertainty in the calculated damage value? How long to measure? Large scale: populations of customers (markets/applications). What is the scatter in the population? Uncertainty in estimation? These problems need to be treated separately, using different methods. PJ/

2 Uncertainty of Customer Loads Two scales: Small scale: individual customers (or measurement). Two measurements from the same customer: Uncertainty in the calculated damage value? How long to measure? Large scale: populations of customers (markets/applications). What is the scatter in the population? Uncertainty in estimation? These problems need to be treated separately, using different methods. PJ/ Example: Uncertainty in Measured Stress Measured stress for 3 minutes rough conditions representing test track Questions: Uncertainty in calculated damage? How long to measure? PJ/

3 Example: Scatter in Test Track Loads Measurements of the 15 laps on the test track (3 drivers, 5 laps each): Lap 1 Lap 2 Lap 3 Lap 4 Lap5 Driver 1 Driver 2 Driver 3 PJ/ Example: Scatter in Test Track Loads Severity of the 15 laps on the test track (3 drivers, 5 laps each): The severity is the equivalent amplitude (of N 0 =10 6 cycles) giving the same damage as the measurement (repeated K=1000 times). A eq K = d N0 1/ β Questions: How large is the scatter between laps? Any difference between drivers? PJ/

4 Coefficient of Variation Relative Variation The severity is a positive quantity, therefore the relative variation is natural to work with, and can be interpreted as percentage of variation. The mean, E[x i ] and variance, Var[x i ] can be estimated by standards methods, i.e. x = N i= 1 x i and 2 1 s = n 1 The coefficient of variation is s 1 s R( x i ) = and R( x) = x N x for a 1-lap-measurement and for the mean of N=5 laps. The x-values can be the damage, x i =d i, or the equivalent amplitude, x i =A eq,i. N i= 1 ( x x) i 2 PJ/ Example: Uncertainty in Equivalent Amplitude The equivalent load of 5 laps for driver 1: A eq K = d N0 1/ β Mean, x i E[x i ] = 2.73 Standard deviation, x i s[x i ] = Coefficient of variation, x i R[x i ] = s[x i ] / E[x i ] = 0.052/2.73 = Coefficient of variation, x R[x] = 0.019/ 5 = Conclusion: The uncertainty in the equivalent amplitude is 1.9% for 1 lap and 0.85% for the average of 5 laps. K = 1000 N 0 = 10 6 β = 6 How many laps to measure if 1% uncertainty is required? R( x) 1% N N 0.01 = 3.6 Conclusion: 4 laps! PJ/

5 Övning: Osäkerhet i ekvivalent amplitud Förare 3 har kört 5 varv på provbanan. Man har mätt lasten och beräknat ekvivalenta amplituden för varje varv: Beräkna variationskoefficienten för ekvivalenta amplituden. Redovisa också medelvärde och standardavvikelse. Hur många varv krävs för att få osäkerhet på mindre än 1%? K = 1000 N 0 = 10 6 β = 6 PJ/ Övning: Osäkerhet i ekvivalent amplitud Lösning med Matlab >> Aeq = [ ]; >> mean(aeq) 3.04 >> Aeq-mean(Aeq) >>(Aeq-mean(Aeq)).^ >> sum((aeq-mean(aeq)).^2) >> sum((aeq-mean(aeq)).^2)/(5-1) >> sqrt(sum((aeq-mean(aeq)).^2)/(5-1)) >> std(aeq) >> R = std(aeq)./mean(aeq) >> std(aeq)./mean(aeq)/sqrt(5) >> (R/0.01)^ Resultat: Variationskoefficienten för ett varv är 2.7%, och för 5 varv blir den 1.2%. Det krävs 8 varv för att nå under 1% osäkerhet. PJ/

6 Example: Uncertainty in measured stress Measured stress for 3 minutes rough conditions representing test track Questions: Uncertainty in calculated damage? How long to measure? PJ/ Splitting into subloads Measured stress split into 9 subloads Observations: Same conditions, but Different calculated damage Questions: Uncertainty in calculated damage? How long to measure? PJ/

7 Coefficient of Variation Relative Variation Splitting the Measured Signal into Parts The values of E[d i ] and Var[d i ] can be estimated by standards methods, i.e. giving PJ/ Example: Uncertainty in damage Measured stress split into 9 subloads. Damage values for β=5 [d i /1e12]: Mean, d i E[d i ] = 3.25 Standard deviation, d i s[d i ] = 2.48 Coefficient of variation, d i R[d i ] = s[d i ] / E[d i ] = 2.48/3.25 = Coefficient of variation, d R[d] =0.763/3 = Conclusion: The uncertainty in the calculated damage is 25%! How long to measure if 10% uncertainty is required? PJ/

8 Example: Uncertainty in damage How long to measure if 10% uncertainty is required? The load was measured for T 0 =180 s, and the coefficient of variation is R[d T0 ] = The variance is inversely proportional to the square-root of the time, T: T0 R ( dt ) = R( d T ) 0 T We want uncertainty less than 10% T0 R( dt ) = 10% R( dt 0 T T 2 ) = R( dt ) = 0 = 3 min = min = 19.4 min 0.10 T 0.10 Conclusion: We need to measure 20 minutes instead of 3 minutes! PJ/ Distribution of Severity log-normal Why log-normal distribution? Agrees well with observations. Positive quantity. Mathematically convenient. Damage and eq.amplitude log-normal. In accordance with load-strength model. Percentage of variation. Std[ d] Std[ln d] is approximately the Standard deviation of log-damage: E[ d] ln A ln d ~ N( µ, σ ) eq coefficient of variation. d d ~ N( µ, σ A eq A eq ) Note: Natural logarithm ln (not the logarithm with base 10). The interpretation of the natural logarithm is very practical for engineering applications, where uncertainties are often judged in percentage of variation. PJ/

9 Confidence Interval for Severity A 95% confidence interval for ln d: A 95% confidence interval for d: Example: Test Track Loads PJ/ Övning: Konfidensintervall för ekvivalent amplitud Förare 3 har kört 5 varv på provbanan. Man har mätt lasten och beräknat ekvivalenta amplituden för varje varv: Beräkna konfidensintervall för ln(a eq ) och A eq. (Använd t ex räknedosa eller Excel.) K = 1000 N 0 = 10 6 β = 6 PJ/

10 Comparison of Severity A 95% confidence interval for difference ln d 2 - ln d 1 : Example: Test Track Loads PJ/ Comparing Severity of many Customers Analysis of Variance (ANOVA) model: y ij = ln(a eq,ij ) y ij = µ + x i + e ij, i = driver, j = lap Severity Mean between = + severity drivers + scatter within drivers scatter µ+x 3 =ln(3.04) µ+x 2 =ln(2.94) µ=ln(2.9) µ+x 1 =ln(2.73) 20 PJ/

11 Quantifying the Scatter Analysis of Variance (ANOVA) model: y ij = ln(a eq,ij ) y ij = µ + x i + e ij, i = driver, j = lap Severity Mean between = + severity drivers + scatter within drivers scatter Result of ANOVA p-value < Standard deviation between/within driver σ x (kn) σ e (kn) Estimate Confidence interval (0.026 ; 0.35) (0.019 ; 0.037) µ+x 3 µ+x 2 µ µ+x 1 PJ/ Results on the Scatter in Severity There is a between drivers effect (with 99.99% certainty). The between and within drivers scatter is of the same order. How long do we need to measure to know the severity accurately for one single driver? Relative error 8% 4% 2% 1% Number of laps to know the severity accurately for the population of drivers? Relative error 8% 4% 2% 1% Number of drivers Half the relative error with four times as many measurements! How much do we gain by letting the drivers go many laps? Three drivers, 1 lap each: 8.1%, 5 laps each: 7.1%, laps each: 6.9% Four drivers, 1 lap each: 6.8% PJ/

12 Example: Measured Service Loads Vertical wheel force measured on the front left wheel of a truck. Three road types: City, Highway and Country. 1. City (21 km) 2. Highway (12 km) 3. Country (14 km) PJ/ Example: Measured Service Loads Split loads into N=5 parts (subloads), damage with β=5. d d2 = , s2 = = , s1 = d3 = , s3 = ~ ~ ~ ρ = R( d ) = ρ2 = R( d2) = ρ3 = R( d 3) = Coefficient of variation for damage PJ/

13 How long do we need to measure? Uncertainty in measured damage!!! City (21 km): 31%, Highway (12 km): 54%, Country (7 km): 25% Uncertainty in load (equivalent amplitude, A eq )!!! City (21 km): 6%, Highway (12 km): 11%, Country (7 km): 5% What is small uncertainty? What is large uncertainty? BIKUPA What is acceptable uncertainty? PJ/ What is acceptable uncertainty? Depend on the application!!! Customer distribution 15% Fatigue test Compare to the uncertainty in the test setup. Measured 10% load 2% Rule of thumb: Less than 20% of the load variation for the population of interest. Typically: 2-4% load variation acceptable. Typically: 1% load variation acceptable. PJ/

14 Uncertainty of Customer Loads Two scales: Small scale: individual customers (or measurement). Two measurements from the same customer: Uncertainty in the calculated damage value? How long to measure? Large scale: populations of customers (markets/applications). What is the scatter in the population? Uncertainty in estimation? These problems need to be treated separately, using different methods. PJ/ Customer Load Distribution Investigate populations of customers (markets/applications). What is the scatter in the population? Uncertainty in estimation? Measuring the customer distribution Three approaches Sample customers Customer usage + sample road types Vehicle independent load description + customer usage + vehicle model Customer usage Load Environment Customer load distribution Vehicle model PJ/

15 Random sampling of customer loads 36 customers Example: 36 drivers were instructed to drive on specific roads which covered several types of roads, and the equivalent amplitude for the vertical load direction was evaluated. A eq K = d N0 1/ β PJ/ Random sampling of customer loads 36 customers Assumptions: Random sampling of customers. Log-normal distribution: ln ~ N( µ, σ) A eq PJ/

16 Estimation of customer distribution 36 customers Mean: 1 µ = ( x1 + x2 + K+ x n ) = with n n 1 = n Variance: s ( µ ) = i= 1 x i = ln x i A eq, i Median customer: ( µ ) = exp( 3.31) 27.5 [ kn] θ = exp = PJ/ Estimation of a severe customer Define a severe customer as a high load quantile, A q, with probability 1-q to find a more severe customer: lna q µ + z σ = q The 95%-customer is estimated at: ln A 95% A 95% µ + z = = = 3.47 = exp( µ + z s s) = exp(3.47) = 32.0 q q [ kn] PJ/

17 Uncertainty in estimated customer distribution Bootstrap demonstration: 100 simulated samples of 36 customers. PJ/ Uncertainty in estimated median customer 95% Confidence interval for the mean * s * I µ = µ 2 ; µ + 2 n s n = [ 3.28;3.34] 95% Confidence interval for the median of the A eq * s * I θ = exp µ 2 ; exp µ 2 n s = n [ 26.6; 28.3] ( units: kn) PJ/

18 Uncertainty in an estimated severe customer The variance for lna * 95% is approximatively Var * [ ln A ] q 2 1 z σ + n 2 2 q ( n 1) 2( 36 1) A 95% confidence interval for the 95%-customer is I A 95% * * [ ( ln A 2s / n) ;exp ( ln A + 2s / )] = exp 95% 95% 95% 95% n = [ exp( 3.42) ;exp( 3.51) ] = [ 30.5;33.6] ( units :kn) PJ/ Estimation of customer distribution Summary Example: A customer survey where 36 drivers were instructed to drive on specific roads which covered several types of roads, and the equivalent amplitude for the vertical load direction was evaluated. PJ/

19 Customer usage and load environment Road classes Different types of roads, e.g. city, highway, country. Customer usage How much does the customers drive in the different road classes? Customer load distribution Goal: Characterize the load distribution by its mean and variance. PJ/ Example: Customer Usage Distribution Vertical wheel force measured on the front left wheel of a truck. Three road types: City, Highway and Country. A typical customer load is often defined by combining measurements from different road types. 1. City (21 km) µ 1 ~ d 1 = = 45% Highway (12 km) µ 2 ~ d 2 = = 30% Country µ 3 = 25% (14 km) ~ d = PJ/

20 User Variation & Load Environment Uncertainty Uncertainties: Variation in customer usage Measurement uncertainty in damage Road classes Different types of roads, e.g. city, highway, country. The mean and standard deviation can be computed as PJ/ Customer usage and load environment Road classes Different types of roads, e.g. city, highway, country. Customer usage How much does the customers drive in the different road classes? Customer usage µ d 7.05 σ = 0.18 = life dlife + uncertainties µ d 7.01 σ = 0.34 = life dlife Customer load distribution Goal: Characterize the load distribution by its mean and variance. lnd life ~ N ( µ, σ ) d life d life PJ/

21 Vehicle independent load descriptions Customer usage How much does the customers drive on different types of roads? Vehicle independent load description Describe roads, climate, traffic intensity, legislation, etc. Vehicle model Mechanical model of dynamic load response of the vehicle. Customer load distribution Goal: Characterize the load environment independent of the vehicle. PJ/ Example: Lateral loads Curves Modelling of lateral loads induced by curves (Karlsson, 2007) PJ/

22 Example: Lateral loads Curves Modelling of lateral loads induced by curves (Karlsson, 2007) PJ/ Example: Synthetic Road with Pot Holes Simulated road profile with added pot holes (Bogsjö, 2007). A synthetic road profile: Z(x) = Z (x) + Z 0 (x) : standard Gaussian road profile Z i 1 ( x ) : extra rough parts added to Z 0 (x) 0 i i Z ( x ) 1 PJ/

23 Example: Synthetic Road with Pot Holes Response to simulated road profile with added pot holes. Fatigue damage filter = quarter-vehicle travelling on road profile The simulated force acting on the sprung mass of the quarter-vehicle. Schematic description of the quarter vehicle PJ/ Evaluation of Customer Loads (Chapter 7) Purpose: Describe and estimation of the customer load distribution. Three approaches: 1. Random Sampling of Customer (Sec. 7.4) 2. Customer Usage and Load Environment (Sec. 7.5) 3. Vehicle Independent Load Description (Sec. 7.6) Customer usage Load Environment Vehicle model Customer load distribution PJ/

24 Uncertainty of Customer Loads Two scales: Small scale: individual customers (or measurement). Two measurements from the same customer: Uncertainty in the calculated damage value? How long to measure? Large scale: populations of customers (markets/applications). What is the scatter in the population? Uncertainty in estimation? These problems need to be treated separately, using different methods. PJ/