Sanaz Rezaeian Postdoctoral Research Fellow, PEER, UC Berkeley Group Members: I.M. Idriss

Sanaz Rezaeian Postdoctoral Research Fellow, PEER, UC Berkeley

Group Members:I.M. IdrissYousef BozorgniaKen CampbellWalter SilvaNorm Abrahamson

NGA-West 2 Project: Scaling of NGA Models for Damping

(December 7, 2010)

Outline Review (Literature, Trends in data) Database Regression for given T and Residual plots Functional form Regression coefficients (and sigma) in terms of

http://peer.berkeley.edu/ngawest2_wg/workshops_damping_scaling.htmlReview

Review Definition:

Existing models have looked at dependence of DSF on : damping T: period (frequency)

D: duration M: magnitude R: distance Site class Tectonic setting (WUS, CEUS) PGA Very few models Common factors in modeling

Review Two common formulations: Newmark & Hall, 1982 (1978)

American building codes Idriss, 19931990 French, 1994 Spanish code2001 CaltransEuroCode 8, 2004Tolis & Faccioli, 1999Priestley, 2003

Review Random Vibrations Theory: Assuming White-Noise

Review (formulas with M or Dur) Abrahamson and Silva 1996:

Stafford et al. 2008:

x is a measure of duration Recommended R.V. method by McGuire et al., 2001 (NUREG/CR-6728)0.2< T

Lin and Chang, 2003:

Lin and Chang, 2004: Site class can be neglected if 5% error is acceptable in structural design and if < 20%. a is a function of site and .b and c depend on the site class. DSF(T,) = 1 + ( 5) [1+c1 ln()+c2 (ln())2] [c3+c4 ln(T)+c5 (ln(T ))2] Hatzigeorgiou, 2010:Review (nonlinear regression, thousands of mathematical eqns)

Review (DSF is Tabulated / Plotted) Cameron & Green, 2007:

frequency content and duration M, distance, and tectonic setting.

For >=2%: DSF depends on T, , M, WUS/EUS, Rock/Soil For =1%: Distance (significantly influences duration)

Bommer & Mendis, 2005

Atkinson & Pierre, 2004

Reviewln(DSF) = (,T,M,) + ; ~ N( 0 , ln(DSF) )

Assumed DSF is lognormally distributed: Note:Linear combination of independent normally distributed random variables is normal. Normal in GMPENormalNormal in GMPE

Distribution of DSF: (at a given T)Fitted pdf to ln(DSF) is normal testing the lognormality of DSFT=0.2 sT=1 sT=7.5 s

Distribution of DSF: (at a given T and 2% damping)Fitted pdf to ln(DSF) is normal testing the lognormality of DSFT=0.2 sT=1 sT=7.5 s

Distribution of DSF: (at a given T and 20% damping)Fitted pdf to ln(DSF) is normal testing the lognormality of DSFT=0.2 sT=1 sT=7.5 s

Exception At:T = 0.1 s & Very low dampingor Very high damping

ln(DSF) = (,T,M,) + ; ~ N(0,)

Functional Form ?

ReviewInfluence of Period, T:

Review35Influence of damping, :

Review There is pattern with ln(Duration). Increases at long periods. Looked at data pattern with Duration, Magnitude, Distance, and Vs30: There is significant dependence on M (linear or quadratic). Increases at long periods. Insignificant dependence on Rrup. Insignificant dependence on Vs30.

Database 6,056 Records (5,819 have PSA at T=0)

4.265 M 7.9 (reported for 5,856) 0.07 Rrup 1529 km (reported for 4,878) 93.7 VS30 2016.1 m/s (reported for 5,845)

GMRotI50 is used to calculate DSF

Data for Duration: Average (arithmetic) duration of H1 and H2 components.

Database:

RegressionStep 0: ln(DSF) = c0+ ; ~ N(0,)

Step 01: ln(DSF) = c0+ c1M+

Step 02: ln(DSF) = c0+ c1M+ c2M2+

Step 03: ln(DSF) = c0+ c1M+ c2ln(R)+ ln(DSF) = c0+ c1M+ c2ln{(R2+c32)1/2}+

Step 04: ln(DSF) = c0+ c1ln(D5-95)+ c2ln(D5-95)2+

(Given T and )

Step 0(All data are used.)

At a given period and damping:

ln(DSF) = c0+ ; ~N(0,)

= 0.5, 1, 2, 3, 5, 7, 10, 15, 20, 25, 30 %T = 0.1 , 0.4 , 1 , 7 sec

ln(DSF) = c0 +

Damping, %0.5123571015202530T=0.1 0.46 0.32 0.18 0.10 0 0.59 0.45 0.28 0.16 0 0.55 0.43 0.27 0.16 0 0.34 0.28 0.19 0.12 0-0.07 -0.13 -0.21 -0.26 -0.30 -0.34 -0.11 -0.24 -0.41 -0.53 -0.63 -0.71 -0.11 -0.25 -0.41 -0.54 -0.64 -0.73 -0.10 -0.22 -0.37 -0.50 -0.60 -0.69T=0.4T=1T=7

Damping, %0.5123571015202530T=0.10.26 0.20 0.13 0.08 0 0.20 0.16 0.10 0.07 0 0.20 0.16 0.10 0.07 0 0.16 0.13 0.10 0.07 0 0.06 0.11 0.17 0.21 0.24 0.27 0.05 0.09 0.13 0.16 0.19 0.21 0.05 0.09 0.13 0.15 0.17 0.19 0.05 0.09 0.14 0.18 0.21 0.24T=0.4T=1T=7

ln(DSF) = c0 +

ln(DSF) = c0 + Conclusion:

Residuals have a linear pattern with magnitude.Maybe a break in pattern around M=6.75

Strong pattern with Duration.

Mild pattern with Distance.

Step 01(Data with R < 50km are used.)


ln(DSF) = c0 + c1 M +

= 0.5, 1, 2, 3, 5, 7, 10, 15, 20, 25, 30 %T = 0.1 , 0.4 , 1 , 7 sec

ln(DSF) = c0 + c1 M +

Damping, %0.5123571015202530T=0.1 0.37 0.34 0.25 0.15 0 0.05 0.08 0.08 0.06 0 0.10 0.08 0.08 0.07 0-0.35 -0.28 -0.17 -0.10 0 -0.12 -0.26 -0.45 -0.60 -0.71 -0.80 -0.08 -0.17 -0.32 -0.47 -0.60 -0.71 -0.06 -0.13 -0.25 -0.37 -0.47 -0.58 0.09 0.15 0.20 0.24 0.28 0.28T=0.4T=1T=7

Damping, %0.5123571015202530T=0.1 0.22 0.17 0.11 0.07 0 0.19 0.15 0.10 0.07 0 0.19 0.15 0.10 0.07 0 0.13 0.11 0.08 0.06 0 0.05 0.10 0.15 0.19 0.22 0.24 0.05 0.09 0.13 0.15 0.17 0.19 0.05 0.09 0.13 0.15 0.17 0.19 0.05 0.08 0.13 0.16 0.19 0.22T=0.4T=1T=7

Damping, %0.5123571015202530T=0.1 0.02 0.01 -0.00 -0.00 0 0.08 0.05 0.03 0.02 0 0.06 0.05 0.03 0.01 0 0.09 0.08 0.05 0.03 0 0.01 0.02 0.03 0.04 0.05 0.06 -0.01 -0.01 -0.01 -0.01 -0.01 -0.00 -0.01 -0.02 -0.02 -0.02 -0.02 -0.02 -0.03 -0.05 -0.08 -0.10 -0.12 -0.13T=0.4T=1T=7

ln(DSF) = c0 + c1 M +

ln(DSF) = c0 + c1 M + Conclusion:

No pattern with Magnitude.

There is linear/quadratic pattern with Duration in log space.

Mild pattern with Rrup, looks logarithmic (specially at long T).

Problem at small T and very high or very low damping (suspect nonlognormality).



ln(DSF) = c0 + c1 M + c2 M2 +

= 0.5, 1, 2, 3, 5, 7, 10, 15, 20, 25, 30 %T = 0.1 , 0.4 , 1 , 7 sec

ln(DSF) = c0 + c1 M + c2 M2 +

Recall ln(DSF) = c0 + c1 M +

ln(DSF) = c0 + c1 M + c2 M2 +

ln(DSF) = c0 + c1 M + c2 M2 + Compare to Step 01:

No difference in Residuals vs. M (infinitesimal improvements)

Pattern with Duration did not improve at all.

Pattern with Distance did not change.



ln(DSF) = c0 + c1 M + c2 ln(R) +

= 0.5, 1, 2, 3, 5, 7, 10, 15, 20, 25, 30 %T = 0.1 , 0.4 , 1 , 7 sec

ln(DSF) = c0 + c1 M + c2 ln(R) +

Damping, %0.5123571015202530T=0.1 0.26 0.28 0.22 0.14 0 -0.06 -0.00 0.05 0.04 0-0.09 -0.06 -0.00 0.04 0-0.53 -0.44 -0.27 -0.17 0 -0.12 -0.27 -0.46 -0.61 -0.73 -0.83 -0.06 -0.13 -0.26 -0.39 -0.50 -0.61 -0.04 -0.08 -0.18 -0.27 -0.38 -0.49 0.15 0.28 0.42 0.52 0.62 0.66T=0.4T=1T=7

Damping, %0.5123571015202530T=0.1 0.03 0.02 0.01 0.00 0 0.03 0.02 0.01 0.01 0 0.06 0.04 0.02 0.01 0 0.04 0.03 0.02 0.01 0 -0.00 0.00 0.00 0.00 0.01 0.01 -0.01 -0.01 -0.02 -0.02 -0.03 -0.03 -0.01 -0.01 -0.02 -0.03 -0.03 -0.03 -0.01 -0.03 -0.04 -0.06 -0.07 -0.08T=0.4T=1T=7

Damping, %0.5123571015202530T=0.1 0.03 0.01 -0.00 -0.00 00.08 0.06 0.03 0.02 00.06 0.05 0.03 0.01 00.10 0.08 0.05 0.03 0 0.01 0.02 0.03 0.04 0.05 0.06 -0.01 -0.01 -0.01 -0.01 -0.01 -0.00 -0.01 -0.02 -0.02 -0.03 -0.03 -0.02 -0.03 -0.06 -0.09 -0.12 -0.14 -0.15T=0.4T=1T=7

ln(DSF) = c0 + c1 M + c2 ln(R) +

Damping, %0.5123571015202530T=0.1 0.22 0.17 0.11 0.07 00.18 0.15 0.10 0.07 00.18 0.14 0.10 0.07 00.12 0.10 0.08 0.06 0 0.05 0.10 0.15 0.19 0.22 0.240.05 0.09 0.13 0.15 0.17 0.190.05 0.08 0.12 0.15 0.17 0.190.05 0.08 0.12 0.16 0.19 0.21T=0.4T=1T=7

ln(DSF) = c0 + c1 M + c2 ln(R) +

ln(DSF) = c0 + c1 M + c2 ln(R) + Conclusion:

c1 and sigma are very similar to those of Step01

No pattern with Magnitude (similar to step01).

Some improvement in pattern with Duration (compared to step01).

Step 03nlin(Data with R < 50km are used.)


ln(DSF) = c0 + c1 M + c2ln{(R2+c32)1/2} +

= 0.5, 1, 2, 3, 5, 7, 10, 15, 20, 25, 30 %T = 0.1 , 0.4 , 1 , 7 sec

Damping, %0.5123571015202530T=0.1 0.25 0.28 0.22 0.14 0-0.47 -0.69 -1.25 0.04 0-0.23 -0.16 -0.04 -0.02 0-0.58 -0.47 -0.31 -0.19 0 -0.12 -1.94 -2.25 -4.55 -3.83 -6.47 -0.06 -0.13 -0.26 -0.39 -0.50 -0.61 -0.03 -0.06 -0.13 -0.22 -0.31 -0.36 0.16 0.31 0.46 0.58 0.69 0.75T=0.4T=1T=7

Damping, %0.5123571015202530T=0.1 0.00 -0.00 0.00 0.00 1.0026.55 45.05 91.87 1.00 1.00 7.86 8.03 6.60 13.75 1.00 3.07 2.26 4.17 4.54 1.00-0.00 160.60 150.62 184.50 140.91 166.42-0.00 -0.00 -0.00 -0.00 0.00 -0.00 5.16 4.91 6.43 6.23 8.12 13.67 3.30 2.53 2.26 2.51 2.56 3.02T=0.4T=1T=7

Damping, %0.5123571015202530T=0.1 0.03 0.01 -0.00 -0.00 0 0.08 0.06 0.03 0.02 0 0.06 0.05 0.03 0.01 0 0.10 0.08 0.05 0.03 0 0.01 0.02 0.03 0.04 0.05 0.06 -0.01 -0.01 -0.01 -0.01 -0.01 -0.00 -0.01 -0.02 -0.02 -0.03 -0.03 -0.02 -0.03 -0.06 -0.09 -0.12 -0.14 -0.16T=0.4T=1T=7

Damping, %0.5123571015202530T=0.1 0.22 0.17 0.11 0.07 0 0.18 0.15 0.10 0.07 0 0.18 0.14 0.10 0.07 0 0.12 0.10 0.08 0.06 00.05 0.10 0.15 0.19 0.22 0.240.05 0.09 0.13 0.15 0.17 0.190.05 0.08 0.12 0.15 0.17 0.190.05 0.08 0.12 0.16 0.19 0.21T=0.4T=1T=7

Damping, %0.5123571015202530T=0.1 0.03 0.02 0.01 0.00 0 0.14 0.19 0.29 0.01 0 0.09 0.07 0.03 0.03 0 0.05 0.04 0.03 0.02 0 -0.00 0.33 0.36 0.76 0.63 1.10 -0.01 -0.01 -0.02 -0.02 -0.03 -0.03-0.01 -0.02 -0.03 -0.04 -0.05 -0.06 -0.02 -0.03 -0.05 -0.07 -0.09 -0.10T=0.4T=1T=7

Conclusion:

c1 and sigma almost identical to Step03.

No change in residual plots for Magnitude, Duration, or Distance compared to Step03.



ln(DSF) = c0 + c1ln(D5-75)+ c2ln(D5-75)2 +

= 0.5, 1, 2, 3, 5, 7, 10, 15, 20, 25, 30 %T = 0.1 , 0.4 , 1 , 7 sec

ln(DSF) = c0 + c1 ln(D5-95) + c2 ln(D5-95)2 +

Damping, %0.5123571015202530T=0.1 0.15 0.09 0.03 0.01 0 0.14 0.11 0.06 0.03 0 0.06 0.06 0.05 0.03 0 -0.13 -0.10 -0.07 -0.05 0 0.00 -0.01 -0.01 0.00 0.01 0.01-0.03 -0.04 -0.07 -0.07 -0.09 -0.08-0.05 -0.09 -0.13 -0.15 -0.17 -0.19 0.02 0.01 0.03 0.02 0.01 -0.01T=0.4T=1T=7

Damping, %0.5123571015202530T=0.1-0.02 -0.01 -0.00 -0.00 0-0.01 -0.01 -0.00 -0.00 0 0.01 0.01 0.00 -0.00 0 0.05 0.04 0.03 0.02 0 -0.00 0.00 0.00 0.00 0.00 0.01 0.00 0.01 0.01 0.01 0.02 0.02 0.01 0.01 0.02 0.02 0.03 0.03 -0.01 -0.01 -0.02 -0.03 -0.03 -0.03T=0.4T=1T=7

Damping, %0.5123571015202530T=0.1 0.32 0.26 0.17 0.10 0 0.31 0.25 0.17 0.11 0 0.31 0.23 0.15 0.09 0 0.30 0.25 0.17 0.12 0-0.08 -0.16 -0.26 -0.34 -0.40 -0.46 -0.08 -0.19 -0.32 -0.44 -0.54 -0.63 -0.05 -0.12 -0.23 -0.32 -0.41 -0.48 -0.07 -0.14 -0.26 -0.33 -0.39 -0.43T=0.4T=1T=7

ln(DSF) = c0 + c1 ln(D5-95) + c2 ln(D5-95)2 +

Damping, %0.5123571015202530T=0.10.22 0.17 0.11 0.07 0 0.17 0.14 0.10 0.06 0 0.17 0.14 0.10 0.07 0 0.12 0.10 0.08 0.06 00.05 0.10 0.15 0.19 0.22 0.24 0.05 0.09 0.13 0.15 0.17 0.19 0.05 0.08 0.12 0.15 0.17 0.18 0.05 0.08 0.13 0.16 0.19 0.21T=0.4T=1T=7

ln(DSF) = c0 + c1 ln(D5-95) + c2 ln(D5-95)2 +

ln(DSF) = c0 + c1 ln(D5-95) + c2 ln(D5-95)2 + Conclusion:

No pattern with Duration.

Some pattern with Mag (@T=7).

Mild pattern with Distance.

What else contributes to the pattern with Duration?All Data:corrcoef( D5t75 , M ) = 0.48corrcoef( D5t75 , R ) = 0.73corrcoef( D5t75 , Vs30 ) = - 0.17corrcoef( D5t75 , depth ) = 0.13

Data with R < 50 :corrcoef( D5t75 , M ) = 0.52corrcoef( D5t75 , R ) = 0.32corrcoef( D5t75 , Vs30 ) = - 0.24corrcoef( D5t75 , depth ) = 0.14

Selected Form Step 01: ln(DSF) = c0+ c1M+ Step 03: ln(DSF) = c0+ c1M+ c2ln(R)+

Regression Coefficients: c0 Step 01:ln(DSF) = c0+ c1M+ Step 03:ln(DSF) = c0+ c1M+ c2ln(R)+

Regression Coefficients: c0 Step 01:ln(DSF) = c0+ c1M+ Step 03:ln(DSF) = c0+ c1M+ c2ln(R)+ c0 = b0 + b1 ln() + b2 ln()2 + 0


b0b1b20T=0.10.34-0.10-0.070.007T=0.40.110.07-0.090.029T=10.120.03-0.070.029T=7-0.260.160.000.025

b0b1b20T=0.10.28-0.04-0.090.008T=0.40.030.12-0.090.028T=1-0.010.12-0.070.036T=7-0.420.220.030.024


b3B4b51T=0.10.01-0.020.010.001T=0.40.05-0.040.010.004T=10.04-0.030.000.004T=70.07-0.03-0.010.002

b3b4b51T=0.10.01-0.020.010.001T=0.40.05-0.040.010.004T=10.04-0.030.000.004T=70.08-0.04-0.010.002


b6b7b82T=0.10.02-0.020.000.001T=0.40.02-0.01-0.000.001T=10.04-0.030.000.002T=70.03-0.01-0.010.001

Standard Deviation: Step 01:ln(DSF) = c0+ c1M+ Step 03:ln(DSF) = c0+ c1M+ c2ln(R)+

Standard Deviation (for given T and ): Step 01:ln(DSF) = c0+ c1M+ Step 03:ln(DSF) = c0+ c1M+ c2ln(R)+

Final Form:For given T and :ln(DSF) = c0+ c1M+ c2ln(R)+ For given T:ln(DSF) = {b0 + b1 ln() + b2 ln()2 + 0} + {b3 + b4 ln() + b5 ln()2 + 1} M + {b6 + b7 ln() + b8 ln()2 + 2} ln(R) +

Final Form:For given T:ln(DSF) = {b0 + b1 ln() + b2 ln()2 + 0 } + {b3 + b4 ln() + b5 ln()2 + 1} M + {b6 + b7 ln() + b8 ln()2 + 2} ln(R) +

mean(ln(DSF)) = {b0 + b1 ln() + b2 ln()2 } + {b3 + b4 ln() + b5 ln()2 } M + {b6 + b7 ln() + b8 ln()2 } ln(R)

Std(ln(DSF)) ~= sqrt { 02 + M212 + ln(R) 222 + 2 }

Simplify?

Step 05:ln(DSF) = {c0 + c1 ln() + c2 ln()2 } + c3 M + c4 ln(R) +

Step 06:ln(DSF) = c0 + c1 M + c2 ln(R) + {c3 ln() + c4 ln()2 } {1 + M + ln(R)} + (for given T)

Step 05 (for given T):ln(DSF) = {c0 + c1 ln() + c2 ln()2 } + c3 M + c4 ln(R) + 95% confidence intervals:R< 50km

c0c1c2c3c4T=0.1 0.21 -0.23 -0.00 0.02 0.01 0.16 0.37 -0.20 -0.04 0.01 -0.00 0.13 0.37 -0.17 -0.05 0.00 0.00 0.13 0.48 -0.06 -0.06 -0.03 -0.02 0.13T=0.4T=1T=7

c0c1c2c3c4T=0.10.18 0.250.34 0.400.34 0.400.44 0.52-0.24 -0.23-0.21 -0.20-0.17 -0.16-0.07 -0.06-0.00 0.00-0.04 -0.04-0.05 -0.04-0.06 -0.050.02 0.030.01 0.02-0.00 0.01-0.03 -0.020.00 0.01-0.01 -0.00-0.00 0.00-0.02 -0.01T=0.4T=1T=7

Step 05 (for given T):ln(DSF) = {c0 + c1 ln() + c2 ln()2 } + c3 M + c4 ln(R) +

Remaining Issues: Distribution at T=0.1s

Extend beyond 50 km

Model for mean and std of DSF? OR Model for DSF of mean(PSA) and mean+std (PSA)?

Approximation at high damping ratio due to shift in period

Correlation between PSA() and PSA(5%)

x

*******The first four are the basis for building codes***********Record 29: magnitude given, PSA not given (-999)*****An additive form makes sense because of the additive form of PSA :DAF= PSA(beta)-PSA(5)************************************************************************************************

Sanaz Rezaeian Postdoctoral Research Fellow, PEER, UC Berkeley Group Members: I.M. Idriss

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Transcript of Sanaz Rezaeian Postdoctoral Research Fellow, PEER, UC Berkeley Group Members: I.M. Idriss