PBEE-2 simulation exercise

Transcription

1 PBEE-2 simulation exercise CVEN SPTPS: Nonlinear Structural Analysis; Theory and Applications 21 Apr 2011 Keith Porter, Associate Research Professor Civil, Environmental, and Architectural Engineering University of Colorado at Boulder

2 Today s objectives Recap 5 stages of PBEE-2 Repair costs Exercise Define the asset Hazard analysis Structural analysis Damage analysis Loss analysis

3 Recap PEER-style PBEE p[im A] A Hazard G(IM) Rupture: M, mech r Fault A Building Site, Vs 30 IM: intensity meas. Sa, g Cumulative probability Undamaged Repair T = 0 T = 0.3 T = 1.0 Demolish & replace Peak transient drift Distance r, km

4 Repair costs Component & damage state repair effort Construction cost estimators judge the unit repair cost for each component & damage state They consider economies of scale, uncertain cost An example: Repair cost $250,000 $200,000 $150,000 $100,000 C Component: D b, "Chiller - Capacity: 100 to <350 Ton - Unanchored equipment that is anchored or if vibration isolated has seismic snubbers - Equipment fragility only" Damage state: "Damaged, inoperative" $50,000 Mean Coeff of Variation 0.1 Repair effort: "Repair chiller and attached piping." $ Quantity to repair

5 Exercise Given: A 20x20x20-ft 1-story woodframe structure built this year next to the Caltech cogen plant, with 7/8-in stucco over 7/16-in oriented strandboard (OSB), a steel door, no windows, a mission-style ceramic tile roof over OSB roof sheathing, no interior finishes, housing two 200-ton chillers hardmounted on a concrete slab on grade. Construction cost = $350,000. Required: repair cost with 10% exceedance probability given shaking with 1/475-yr mean recurrence frequency

6 1. Hazard analysis, location N, E

7 Site class w/o soil boring ±0.01N, ±0.01E

8 Site class w/o soil boring Site class C Vs m/sec Accuracy similar to Wills et al. (2000) A nearby soil boring would be preferred

9 Hazard deaggregation

10 Hazard deaggregation R=5 km M = 6.6 epsilon = 0.3

11 Ground-motion time histories

12 Ground-motion time histories

13 Ground-motion time histories

14 Ground-motion time histories

15 Ground-motion time histories

16 2. Structural analysis M = roof + ½ of wall mass Roof: W/sf = 150 tiles/100 sf x 7lb/tile + 10 psf for roof, framing, etc. = 20.5 psf Roof mass = 450 sf x 20.5 psf / 32.2 lbm/slug = 287 slug ½ of wall mass: 15 psf 10 ft (high) 80 ft (perimeter)/ 32.2 lbm/slug = 373 slug M = 287 slug slug = 660 slug

17 Stucco sheathing springs from UCI COLA tests Elastic properties K 0 = 2.5 kip/in/lf wall Wayne Stewart degrading stiffness parameter values F Y = ±0.168 kip/lf wall r = 0.16 F U = kip/lf wall P TRI = F I = 0.25 FY, as did Wayne Stewart P UNL = 1.45 Beta = 1.09 Alpha = 0.38

18 Structural analysis Create springs for each wall, each material E.g., each stucco wall : 20 ft x K 0, F Y, & F U Repeat for OSB sheathing Create roof diaphragm element Perform nonlinear dynamic structural analyses, 1 for each ground-motion time history i {1, 2, 20} Response R i = [PTD Ei, PTD Wi, PTD Ni, PTD Si, PGA i, PRA i ] T Response R = [R 1 ; R 2 ; R 20 ]

19 Structural response R = Sim PTDE, in/in PTDW, in/in PTDN, in/in PTDS, in/in PGA, g PRA, g Sim PTDE, in/in PTDW, in/in PTDN, in/in PTDS, in/in PGA, g PRA, g

20 Damage analysis Chillers: N D = 1; damaged, inoperative; repair chiller and connections θ 1 = 0.72, β 1 = 0.2; Stucco on 7/16-in OSB: N D = 2 sequential states fracture sheathing-framing connection; remove damaged stucco, renail, replace stucco: θ 1 = 0.005, β 1 = 0.3, tearout connectors on 2 edges; dem & replace: θ 2 = 0.011, β 2 = 0.5 Roof tile: N D = 2 sequential states Minor damage, tiles dislodged; reinstall tiles θ 1 = 1.8g, β 1 = 0.3 Major portion (>10-20%?) of tiles dislodged; replace θ 2 = 2.2g, β 2 = 0.3

21 Damage analysis Sim chiller 1 θ β Pf u f chiller 2 θ β Pf u f Wall E θ β θ β Pf Pf u f Wall W θ β θ β Pf Pf u f Wall N θ β θ β Pf Pf u f Wall S θ β θ β Pf Pf u f R f

22 Damage analysis: east wall, sim 1 θ β1 0.3 θ β2 0.5 Pf Pf u f 0 R 1 = in/in, θ 1 = 0.005, β 1 = 0.3 P[D 1 R=0.0028] = Φ(ln(0.0028/0.005)/0.3) = P[D 2 R=0.0028] = Φ(ln(0.0028/0.011)/0.5) = rand() = 0.58 > not failed DS1 or DS2

23 Damage summary Damage summary Sim Walls Walls Chiller Roof Roof Sim 1 2 Walls Walls Chiller 0 1 Roof Roof (units of 64 sf) (units of 100 sf)

24 Loss analysis: unit costs i.e., cost to repair 1 unit Chillers: (as shown earlier) Stucco on 7/16-in OSB: N D = 2 damage states q 1 = $131/64 sf; b 1 = 0.2; independent of qty q 2 = $742/64 sf, b 2 = 0.2; independent of qty Tile roof: N D = 2 damage states q 1 = $ 80/100 sf; b 1 = 0.3; independent of qty q 2 = $800/100 sf; b 1 = 0.3; independent of qty

25 Loss analysis 1 Loss analysis Sim Walls damage state 1 1 q, $/64sf $ 131 $ 131 $ 131 $ 131 $ 131 $ 131 $131$ 64 $ sf 131 exp $ $ $ 131 $ 13 b rand() $110 64sf Cost/64sf, $ $ 148 $ 110 $ 151 $ 106 $ 152 $ 127 $ 106 $ 91 $ 119 $ 136 $ 176 $ 9 Count 64 sf Cost, $ $ - $ 2,058 $ 944 $ 661 $ 947 $ C- Units $ - C$ / unit - $ 743 $ 848 $ - $ - Walls damage state 2 q, $/64sf $ 742 $ 742 $ 742 $ 742 $ 742 $ $ 742 $110 $ 742 $ 742 $ 742 $ 742 $ 74 b rand() $2, Cost/64sf, $ $ 772 $ 878 $ 581 $ 861 $ 574 $ 874 $ 708 $ 732 $ 880 $ 806 $ 587 $ 73 Count 64 sf Cost, $ $ - $ - $ - $ - $ - $ - $ - $ - $ - $ - $ - $ - Chiller Sim 1 2 Lower plateau count q lower, $/each $ 160,000 $ 160,000 $ 160,000 $ 160,000 $ 160,000 $ 160,000 $ 160,000 $ 160,000 $ 160,000 $ 160,000 $ 160,000 $ 160,00 Upper plateau count q upper, $/each $ 140,000 $ 140,000 $ 140,000 $ 140,000 $ 140,000 $ 140,000 $ 140,000 $ 140,000 $ 140,000 $ 140,000 $ 140,000 $ 140,00 q, $/64sf $ 131 $ 131 $ Count q at count, $/each $ 160,000 $ 160,000 $ 160,000 $ 160,000 $ 160,000 $ 160,000 $ 153,333 $ 160,000 $ 153,333 $ 160,000 $ 160,000 $ 160,00 b b rand() Cost/each, $ $ 193,965 $ 135,463 $ rand() 175,132 $ 178,635 $ 173,351 $ 155,975 $ 120,603 $ 146, $ 156,008 $ 188, $ 199,842 $ 150,60 Cost, $ $ - $ 135,463 $ - $ - - $ 155,975 $ 241,206 $ - $ 312,016 $ 188,153 $ - $ - Cost/64sf, $ $ 148 Roof $ damage state $ q, $/100 sf $ 80 $ 80 $ 80 $ 80 $ 80 $ 80 $ 80 $ 80 $ 80 $ 80 $ 80 $ 8 b Count sf rand() Cost/100sf $ 93 $ 92 $ Cost, 75 $ 73 $ 75 $ 89 $ $ 59 $ $ $ 69 $ 2, $ $ 68 $ 13 Count 100sf Cost $ - $ - $ - $ - $ - $ 355 $ 235 $ - $ - $ 208 $ - $ - C/ unit q exp b u

26 Results Rank Cost Percentile 1 $ $ $ $ $ $ $ $ $ 2, $ 5, $ 140, $ 143, $ 156, $ 165, $ 189, $ 214, $ 238, $ 241, $ 312, $ 348, Total cost simulations sorted in increasing order 90 th percentile = (241, ,759)/2 = $277,100 Call it $280,000

27 Results NB: a lognormal fits the simulations badly; fails Lilliefors goodness-of-fit test 1.00 Nonexceedance probability CDF of 20 simulations LN: ĉ = $2,850; σ lnc = 5.2; c 90 = $2.1 million $- $250,000 $500,000 Repair cost, C

28 Thanks (626)