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    University College LondonDepartment of Civil, Environmental & Geomatic

    Engineering

    Advanced Seismic Designof Structures(CEGEG140/CEGEM140)

    ~ Coursework 4 ~

    Student: Carmine Russo 1410!10"#nstru$tor: Carlos Molina %utt

    &$ademi$ 'ear

    01*1"

    L+,-+,.S GL+&L U,#ERS#' 

    https://moodle.ucl.ac.uk/course/view.php?id=20463https://moodle.ucl.ac.uk/course/view.php?id=20463https://moodle.ucl.ac.uk/course/view.php?id=20463https://moodle.ucl.ac.uk/course/view.php?id=20463

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    Sommario

    #,R+-UC#+,22222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222!

    C&LCUL&#+,S 3+R -#33ERE, ELEME,S2222222222222222222222222222222222222222222222222222222222222!

     A) B1035.041: !E"#$!%!'DGE (*"B BEA+"C$+# -$'#% 2222222222222222222222222222222!

    B) B1035.01A: EB* /EA! '#, #$ *$$! BEA+/, '# ( 100 *22222222222222222

    C) C1011.001A: (A A!%'%'$#, %2E: G2/+ ('% +E%A /%D/, *E'G%, *'ED BE$(, *'ED AB$E2222222222222222222222222222222222222222222222222222222222222222222211

    D) C303.001A: //E#DED CE''#G, /DC A, B6 A!EA 7A): A 50, E!% /$!% $#2 22222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222221

    5UES#+, & * 67at $7anges $ould you ma8e to $om9onents 1 t7roug7 4 to redu$edamage under an eual set o; engineering demand 9arameters

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    Introduction

    Using the results obtained from the previous analysis, performed on a moment resisting frame, in

    the following steps we are going to calculate the damages and the economic losses that occur to

    some particular structural components.

    The aim of the exercise is to show the last part of a performance-based design procedure. Should be

    noted that this could be either part of preliminary design, when the designer wants to calculate

    different performances of alternative structural solutions to show to the (future) building owner or 

    can be part of a definitive design procedure when its necessary to assess the costs of eventual

    damages or to label the building.

    The damage levels have been evaluated by using the !"#T ( Performance Assessment Calculation

    Tool ) database that gives some information about different Damage-State (DS) for various building

    components. $ach damage state is associated with some degree of performance of the building.

    %or each element, the probability of reaching a particular damage state has been calculated by using

    the relative fragility curves. "lso, using the conse&uence functions, for each 'amage-State have

     been obtained relevant repairing costs, the time to repair, and finally the total loss costs.

    n the following table are reported the results obtained from the analysis performed

    S9e$trummat$7ed

    a$$elerograms

    Linearly S$aleda$$elerograms

    ? @ ? @

    Aea8 -is9la$ement Bm0.1521138

    330.0405493

    570.144319 0.040948

    -is9la$ement Ratio Brad*#alculated considering that the height of the frame is +.m

    0.043461095

    0.011585531

    0.0412340.0116994

    29

    Aea8 &$$eleration Bg0.4579471

    630.0866762

    260.485565

    070.1003532

    86

    Rotation node Brad0.0364693

    330.0096629

    750.034548

    060.0123033

    39

    Rotation node " Brad 0.042768367

    0.013504671

    0.03984613

    0.011476659

    Calculations for different elements

     A) B1035.041: Pre-Northridge WUF-B beam-colum !oit 

    The fragility curves are derived in !"#T for the storey drift ratio and are showed in the following

    image

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    The mean values and the standard deviation for the analysis are

    3or s9e$trally mat$7edeart7ua8es

    3or linearly s$aled

     μdrift =0.043461

    σ drift =0.01159

    ¿ μdrift +σ drift =0.055047

     μdrift −σ drift =0.031876

     μdrift =0.041234

    σ drift =0.011699505

    ¿ μdrift +σ drift =0 .052933451

     μdrift −σ drift =0.029534441

    This structural component consists of a post-/orthridge with a welded unreinforced flange-bolted

    web and a beam on one side.

    The database considers three damage states, named: DS1, DS2, DS3. The first two of these, are

    subdivided into two mutually exclusive damage states the element can reach a damage, within the

     particular damage-state considered, in only one of the two sub-damage state and both of these two

    are associated with a certain probability.

    -amageStates

    -amageRe9resented

    Mutually eD$lusivedamage state

    Aroaility

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     D S

    Fracture of o!er"eam#ange

    !ed andfaiure of !e" "ots$s%ear ta"

    connection&' !it%

    fracturescon(ned to

    t%e !edregion.

     D S1

    Fracture of o!er "eam#ange !ed

    and faiure of !e" "ots$s%ear ta"

    connection&'!it% fractures

    con(ned to t%e

    !ed region.

    0.75

     D S1

    Simiar to DS1'e)ce*t t%at

    fracture*ro*agatesinto coumn

    #anges.

    0.25

     D S

    Fracture of u**er "eam #ange !ed'

    !it%out DS1 t+*e damage. Fracture iscon(ned to "eam #ange region.

     D S2

    Fracture of u**er "eam#ange !ed'!it%out DS1

    t+*e damage.Fracture iscon(ned to

    "eam #angeregion.

    0.75

     D S2

    Simiar to DS3'e)ce*t t%at

    fracture*ro*agatesinto coumn

    #anges.

    0.25

     D S

    Fracture initiating at !ed access %oeand *ro*agating t%roug% "eam #ange'*ossi"+ accom*anied "+ oca "uc,ing

    deformations of !e" and #ange.

    none

    0eading the values from the fragility curves we have

    CasesS9e$trally mat$7ed

    eart7ua8esLinearly s$aled

    eart7ua8es

     μdrift −σ drift 

     P DS3

    =0.53

     P DS2

    =0.71−0.53=0.18

     P DS1

    =0.93−0.71=0.22

     Pundmaged=1.0−0.93=0.07

     P DS3

    =0.49

     P DS2

    =0.68−0.49=0.19

     P DS1

    =0.92−0.68=0.24

     Pundmaged=1.0−0.92=0.08

     μdrift 

     P DS3=0.82

     P DS2

    =0.92−0.82=0.1

     P DS1

    =0.99−0.92=0.07

     Pundmaged=1.0−0.99=0.01

     P DS3=0.79

     P DS2

    =0.89−0.79=0.1

     P DS1

    =0.99−0.89=0.1

     Pundmaged=1.0−0.99=0.01

     μdrift +σ drift 

     P DS3

    =0.93

     P DS2

    =0.96−0.93=0.03

     P DS1

    =1−0.96=0.04

     Pundmaged=0

     P DS3

    =0.91

     P DS2

    =0.95−0.91=0.04

     P DS1

    =1−0.95=0.05

     Pundmaged=0

    The costs related to the different limit states considered have been extrapolated from the diagramsgiven in !"#T. The following figures show that costs decrease as a number of parts to be repaired

    increases since fixed costs affect thousands more if the amounts are minor. %or this element costing

    is on a per bay basis and does not include fireproofing removal or reapplication cost.

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    Re9air Cost Conseuen$e ;or

     DS1 I 

    Re9air Cost Conseuen$e ;or

     DS1 II 

    Re9air Cost Conseuen$e ;or

     DS2 I 

    Re9air Cost Conseuen$e ;or

     DS2 II 

    Re9air Cost Conseuen$e ;or  DS3

    #onsidering that the cost is the higher possible, we have

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    DS1

    -e*air !i t+*ica+reuire gouging out andre/!eding of t%e "eam#ange !ed' re*air of 

    s%ear ta"' and re*acings%ear "ots.

    Cost  DS1 I =14472$

    re*airs tocoumn !i "enecessar+ t%at

    !i invovere*acing a

    *ortion of t%e!eded *ates

    Cost  DS 1 II =16272 $

    DS2

    -e*airs !i "e simiar tot%ose reuired for DS1'e)ce*t t%at access to!ed !i i,e+ reuireremova of a *ortion of t%e #oor sa" a"ove t%e

    !ed.

    Cost  DS2 I =20472$

    n addition tocoumn

    measures forDS3' re*airs tocoumn !i "enecessar+ t%at

    !i invovere*acing a

    *ortion of t%ecoumn #ange.

    Cost  DS2 II =21120$

    DS3

    -e*air is simiar to t%at for DS1 e)ce*t t%at a *ortion of t%e "eam !e"and #ange ma+ need to "e %eat straig%tened or re*aced.

      Cost  DS3=17472$

    The total repairing cost can be calculated ta1ing into account the different probabilities related to

    every damage state (also those related to the mutually exclusive damage states).Cost tot = P DS

    1

    ∙ ( P DS1 I ∨ D S1

    ∙Cost  DS1 I + P DS1 II ∨ D S 1∙Cost  DS1 II )+¿+ P DS

    2

    ∙ ( P DS2 I ∨ DS 2

    ∙Cost  DS2 I + P DS2 II ∨ D S2 ∙Cost  DS2 II  )+ P DS3 ∙Cost  DS3

    2here we too1 in account also the conditional probabilities for mutual exclusive 'amage State

    obtained from !"#T

    -amage state 1   P DS1 I ∨ D S1

    =0.75   P DS1 II ∨ D S1

    =0.25

    -amage State   P DS2 I ∨ DS 2

    =0.75   P DS2 II ∨ D S2

    =0.25

    Cases

    analysedotal $ost ;or s9e$trally mat$7ed eart7ua8es otal

     μdrift −σ drift    Cost tot =0.22 ∙ (0.75 ∙14472$+0.25 ∙16272$ )+0.18 ∙ (0.75 ∙2047   16257$

     μdrift    Cost tot =0.07 ∙ (0.75 ∙14472 $+0.25 ∙16272$)+0.1 ∙ (0.75 ∙20472   17434$

     μdrift +σ drift    Cost tot =0.04 ∙ (0.75 ∙14472$+0.25 ∙16272$)+0.03 ∙ (0.75 ∙2047   17465$

    Cases

    analysed

    otal $ost ;or linearly s$aled eart7ua8es otal

     μdrift −σ drift    Cost tot =0.24 ∙ (0.75 ∙14472$+0.25 ∙16272$)+0.19 ∙ (0.75 ∙2047   16063$

     μdrift    Cost tot =0.1∙ (0.75 ∙14472$+0.25 ∙16272$ )+0.2∙ (0.75 ∙20472$   17358$

     μdrift +σ drift    Cost tot =0.05 ∙ (0.75 ∙14472$+0.25 ∙16272$ )+0.04 ∙ (0.75 ∙2047   17471$

    The estimated repairing time has been derived from similar diagrams from !"#T database

    considering the highest possible values

    -amage State 1   T  D S1 I =38.59days   T  D S

    1 II =43.39days

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    -amage State   T  D S2 I =54.59 days T   D S

    2 II =56.32days

    -amage State !   T  D S3

    =46.6 days

    2ith the same methodology used for the repairing costs, we have calculated the time

    Cases

    analysed

    otal re9air time ;or s9e$trally mat$7ed eart7ua8es otal

     μdrift −σ drift    Timetot =0.22 ∙ (0.75 ∙38.59d+0.25 ∙43.39d )+0.18 ∙ (0.75 ∙54.59   43days

     μdrift    Timetot =0.07 ∙ (0.75 ∙38.59d+0.25∙43.39d )+0.1 ∙ (0.75 ∙54.59d   46.5days

     μdrift +σ drift    Timetot =0.04 ∙ (0.75 ∙38.59d+0.25 ∙43.39d )+0.03∙ (0.75 ∙54.59   46.6days

    Casesanalysed otal re9air time ;or linearly s$aled eart7ua8es otal

     μdrift −σ drift    Timetot =0.24 ∙ (0.75 ∙38.59d+0.25 ∙43.39d )+0.18∙ (0.75 ∙54.59   42.3days

     μdrift    Timetot =0.1∙ (0.75 ∙38.59d+0.25 ∙43.39d )+0.2 ∙ (0.75 ∙54.59d+   51.79days

     μdrift +σ drift    Timetot =0.05∙ (0.75 ∙38.59d+0.25 ∙43.39d )+0.04 ∙ (0.75 ∙54.59   46.6days

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     B) B1035.0"1a: #BF $hear %i&' o (loor beam' li& * + 100 P%F 

    The component analysed is an eccentrically braced frame lin1. $ccentrically braced steel frames

    ($3%s) are very efficient structures for resisting earth&ua1es as they combine the ductility of that is

    characteristic of moment frames and the stiffness associated with braced frames. n the $3%s

    inelastic activity is confined to a small length of the floor beams which yields mostly in shear 

    (therefore called the shear lin1).

    n this case, the fragility curves are derived for the plastic rotation hinges

    3or s9e$trally mat$7edeart7ua8es

    3or linearly s$aled

     μrotation=0.036469333

    σ rotation=0.009662975

    ¿ μrotation+σ rotation=0.046132

     μrotation−σ rotation=0.026806

     μrotation=0.034548067

    σ rotation=0.012303339

    ¿ μrotation+σ rotation=0.046851

     μrotation−σ rotation=0.022245

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    -amageStates

    -amage Re9resentedMutually eD$lusive damage

    state

     D S

    Damage toconcrete

    sa" a"ovet%e in,"eam.

    none

     D S

    e" oca"uc,ing'

    #ange oca"uc,ing.

    none

     D S

    nitiation of fracture in

    t%e in,"eam andin, #ange.

    none

    0eading the values from the fragility curves, we have

    CasesS9e$trally mat$7ed

    eart7ua8esLinearly s$aled

    eart7ua8es

     μrotation−σ rotation

     P DS3

    =0

     P DS2

    =0

     P DS 1=0.05 Pundmaged=1.0−0.05=0.95

     P DS3

    =0

     P DS2

    =0

     P DS1=0.045 Pundmaged=1.0−0.045=0.955

     μrotation

     P DS3

    =0.005

     P DS2

    =0.025−0.005=0.002

     P DS1

    =0.33−0.025=0.305

     Pundmaged=1.0−0.33=0.67

     P DS3

    =0.004

     P DS2

    =0.023−0.004=0.019

     P DS1

    =0.3−0.023=0.277

     Pundmaged=1.0−0.3=0.7

     μrotation+σ rotation

     P DS3

    =0.025

     P DS2

    =0.19−0.025=0.165

     P DS1

    =0.75−0.19=0.56

     Pundmaged=1−0.75=0.25

     P DS3

    =0.03

     P DS2

    =0.19−0.03=0.16

     P DS1

    =0.76−0.19=0.57

     Pundmaged=1−0.76=0.24

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    #onsidering that the cost is the higher possible, from the we have

    -amage State 1   Cost  DS1=123900$ -e*ace concrete sa".

    -amage State   Cost  DS2=132900 $eat straig%tening of "uc,edeements.

    -amage State !   Cost  DS3=188400 $ -e*ace F in,.

    The total repairing cost can be calculated ta1ing into account the different probabilities related to

    every damage state.

    Cost tot = P DS1

    ∙Cost  DS1+ P DS2

    ∙Cost  DS2+ P DS3

    ∙Cost  DS3

    Cases analysed otal $ost ;or s9e$trally mat$7ed eart7ua8es otal

     μrotation−σ rotation   Cost tot =0.05∙123900$+0 ∙132900 $+0 ∙188400 $   6195$

     μrotation   Cost tot =0.305 ∙123900$+0.002 ∙132900 $+0.005 ∙188400   38997$

     μrotation+σ rotation   Cost tot =0.56∙123900$+0.165 ∙132900$+0.025 ∙188400 $   96022$

    Cases analysed otal $ost ;or linearly s$aled eart7ua8es otal

     μrotation−σ rotation   Cost tot =0.045∙123900 $+0 ∙132900$+0 ∙188400$   5576$

     μrotation   Cost tot =0.277 ∙123900$+0.019 ∙132900 $+0.004 ∙188400   37599$

     μrotation+σ rotation   Cost tot =0.57∙123900$+0.16 ∙132900 $+0.03 ∙188400$   97539$

    The estimated repairing for each damage state are

    -amageState 1

      T  DS1=340.37 days -e*ace concrete sa".

    -amageState

      T  DS2=364.61dayseat straig%tening of "uc,edeements.

    -amageState !

      T  DS3=517.39days -e*ace F in,.

    2ith the same procedure, we can calculate the total time repair

    Cases analysedotal re9air time ;or s9e$trally mat$7ed

    eart7ua8esotal

     μrotation−σ rotation   Timetot =0.05∙340.37d+0 ∙364.61d+0 ∙517.39 d   17days

     μrotation   Timetot =0.305 ∙340.37d+0.002 ∙364.61d+0.005 ∙517.39d   107days

     μrotation+σ rotation   Timetot =0.56 ∙340.37 d+0.165 ∙364.61d+0.025 ∙517.39d   264days

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    Cases analysed otal re9air time ;or linearly s$aled eart7ua8es otal

     μrotation−σ rotation   Timetot =0.045 ∙340.37d+0 ∙364.61d+0 ∙517.39d   15days

     μrotation   Timetot =0.277 ∙340.37d+0.019 ∙364.61d+0.004 ∙517.39 d   103days

     μrotation+σ rotation   Timetot =0.57 ∙340.37 d+0.16 ∙364.61 d+0.03 ∙517.39d   268days

    ,) ,1011.001a: Wall Partitio' /e: /um *ith metal tud' Full eight' Fi2ed Belo*' Fi2ed 

     Aboe

    4ypsum walls (illustrated in the picture below) are a type of lightweight, non-loadbearing,

     partitions. 5etal framed partitions could be used in all types of new and existing buildings,

    covering all applications, from simple space division, through to high-performance walls designed

    to meet the most demanding fire resistance, sound insulation, impact and height re&uirements. This

    1ind of wall partitions is easy to reconfigure with minimal impact to both building and occupants

    resulting in less disruption, optimising the transformation process.

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    The demand parameter for this (non-structural) component is the storey drift ratio

    3or s9e$trally mat$7edeart7ua8es

    3or linearly s$aled

     μdrift =0.043461

    σ drift =0.01159

    ¿ μdrift +σ drift =0.055047

     μdrift −σ drift =0.031876

     μdrift =0.041234

    σ drift =0.011699505

    ¿ μdrift +σ drift =0.052933451

     μdrift −σ drift =0.029534441

    The database considers three damage states the second and the third of these are subdivided into

    two mutually exclusive damage states.

    -amageStates

    -amageRe9resented

    Mutually eD$lusivedamage state

    Aroaility

     D S Scre!s *o*/out' minor crac,ing of !a"oard' !ar*ing or crac,ing of ta*e.

    none none

     D S

    oderate crac,ing or crus%ing of g+*sum !a "oards $t+*ica+ in cornersand in corners of o*enings&.

     D S2

    oderatecrac,ing orcrus%ing of 

    g+*sum !a"oards

    $t+*ica+ incorners and in

    corners of o*enings&.

    0.8

     D S2

    oderatecrac,ing orcrus%ing of 

    g+*sum !a"oards

    $t+*ica+ incorners and incorners of o*enings&..

    0.2

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     D S

    Signi(cant crac,ing andor crus%ing of 

    g+*sum !a "oards/ "uc,ing of studsand tearing of trac,s.

     D S3

    Signi(cantcrac,ing andor

    crus%ing of g+*sum !a

    "oards/"uc,ing of studs andtearing of 

    trac,s.

    0.8

     D S3

    Signi(cantcrac,ing andor

    crus%ing of g+*sum !a

    "oards/"uc,ing of studs andtearing of 

    trac,s.

    0.2

    0eading the values from the fragility curves, we have

    CasesS9e$trally mat$7ed

    eart7ua8esLinearly s$aled

    eart7ua8es

     μdrift −σ drift 

     P DS3

    =0.98

     P DS2

    =1−0.98=0.02

     P DS1

    =0

     Pundmaged=0

     P DS3

    =0.96

     P DS2

    =1−0.96=0.04

     P DS1

    =0

     Pundmaged=0

     μdrift 

     P DS3

    =1

     P DS2

    =0

     P DS1

    =0

     Pundmaged=0

     P DS3

    =1

     P DS2

    =0

     P DS1

    =0

     Pundmaged=0

     μdrift +σ drift 

     P DS3

    =1

     P DS2

    =0

     P DS1

    =0

     Pundmaged=0

     P DS3

    =1

     P DS2

    =0

     P DS1

    =0

     Pundmaged=0

    The costs related to the different limit states considered have been extrapolated from the diagrams

    given in !"#T. #onsidering that the cost is the higher possible, we have

    -S1

    -eta*e oints' *aste and re*aint "ot% sides of t%e 50/foot engt% of !a"oard.

    Cost  DS3=2730$

    -S

    -emove 25/foot engt% of !a"oard$"ot% sides&' insta ne! !a "oard $"ot%

    sides&' ta*e' *aste and re*aint.

    -emove fu 100/foot engt% of !a"oard$"ot% sides&' insta ne! !a "oard $"ot%

    sides&' ta*e' *aste and re*aint.Cost  DS2 I =5190$ Cost   DS 2 II =19800$

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    -S!

    -emove and re*ace a 25/foot engt% of t%e meta stud !a' "ot% sides of t%eg+*sum !a"oard and an+ em"eddedutiities' and ta*e' *aste and re*aint.

    -emove and re*ace a fu 100/foot engt%of t%e meta stud !a' "ot% sides of t%eg+*sum !a"oard and an+ em"eddedutiities' and ta*e' *aste and re*aint.

    Cost  DS3 I =7940$ Cost   DS3 II =31100$

    The total repairing cost can be calculated ta1ing into account the different probabilities related toevery damage state (also those related to the mutually exclusive damage states).

    Cost tot = P DS1

    ∙Cost  DS1+ P DS2

    ∙( P DS2 I ∨ D S2

    ∙Cost  DS2 I + P DS2 II ∨ D S2∙Cost  DS2 II )+ P DS

    3

    ∙ ( P DS3 I ∨ DS 3

    ∙Cost  DS3 I + P DS3 II ∨ DS 1 ∙Cost  DS3 II )

    2here we too1 in account also the conditional probabilities for mutual exclusive 'amage State

    obtained from !"#T

    -amage state   P DS2 I ∨ D S2

    =0.80   P DS2 II ∨ D S2

    =0.20

    -amage State !   P DS3 I ∨ DS 3=0.80   P DS3 II ∨ DS 1=0.20

    Casesanalysed

    otal $ost ;or s9e$trally mat$7ed eart7ua8es otal

     μdrift −σ drift Cost tot =0.98 ∙2730$+0.02∙ (0.8 ∙5190$+0.2∙19800$ )+¿

    +0 ∙ (0.8∙7940$+0.2∙31100$ )  2838$

     μdrift Cost tot =1 ∙2730$+0∙ (0.8 ∙5190$+0.2∙19800$ )+¿

    +0 ∙ (0.8∙7940$+0.2∙31100$ )  2730$

     μdrift +σ drift Cost tot =1 ∙2730$+0∙ (0.8 ∙5190$+0.2∙19800$ )+¿

    +0 ∙ (0.8∙7940$+0.2∙31100$ )  2730$

    Casesanalysed

    otal $ost ;or linearly s$aled eart7ua8es otal

     μdrift −σ drift Cost tot =0.96∙2730$+0.04 ∙ (0.8 ∙5190 $+0.2 ∙19800 $ )+¿

    +0 ∙ (0.8∙7940$+0.2∙31100$ )  2945$

     μdrift  Cost tot =1 ∙2730$+0∙ (0.8 ∙5190$+0.2∙19800$ )+¿

    +0 ∙ (0.8∙7940$+0.2∙31100$ )  2730$

     μdrift +σ drift Cost tot =1 ∙2730$+0∙ (0.8 ∙5190$+0.2∙19800$ )+¿

    +0 ∙ (0.8∙7940$+0.2∙31100$ )  2730$

    The estimated repairing time has been derived from similar diagrams from !"#T database

    considering the highest possible values

    -amage State 1   T  D S1

    =8.04days

    -amage State   T  D S2 I =15.3days T   D S2 II =58.1days

    -amage State !   T  D S3 I =23.4 days T   D S

    3 II =91.5days

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    2ith the same methodology used for the repairing costs, we have calculated the timeCases

    analysedotal re9air time ;or s9e$trally mat$7ed eart7ua8es otal

     μdrift −σ drift Timetot =0.98∙8.04 d+0.02 ∙ (0.8 ∙15.3d+0.2 ∙58.1d )+¿

    +0 ∙ (0.8∙23.4d+0.2 ∙91.5d )  8.35days

     μdrift  Timetot =1 ∙8.04d+0 ∙ (0.8∙15.3d+0.2 ∙58.1d )+¿

    +0 ∙ (0.8∙23.4d+0.2 ∙91.5d )  8.04days

     μdrift +σ drift Timetot =1 ∙8.04d+0 ∙ (0.8∙15.3d+0.2 ∙58.1d )+¿

    +0 ∙ (0.8∙23.4d+0.2 ∙91.5d )  8.04days

    Casesanalysed

    otal re9air time ;or linearly s$aled eart7ua8es otal

     μdrift −σ drift Timetot =0.96 ∙8.04 d+0.04 ∙ (0.8 ∙15.3d+0.2∙58.1d )+¿

    +0 ∙ (0.8∙23.4d+0.2 ∙91.5d )  8.67days

     μdrift  Timetot =1

    ∙8.04

    d+0

    ∙ (0.8

    ∙15.3

    d+0.2

    ∙58.1

    d )+¿

    +0 ∙ (0.8∙23.4d+0.2 ∙91.5d )   8.04days

     μdrift +σ drift Timetot =1 ∙8.04d+0 ∙ (0.8∙15.3d+0.2 ∙58.1d )+¿

    +0 ∙ (0.8∙23.4d+0.2 ∙91.5d )  8.04days

     ) ,303.001a: $u/eded ,eilig' $, A' B6 Area 7A): A + 50' 8ert u//ort ol

    The ceiling structure is a non-structural element, in this case, supported by vertical hanging wires

    only. The demand parameter is the acceleration and the values obtained from the analysis are

    3or s9e$trally mat$7edeart7ua8es

    3or linearly s$aled

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     μacc=0.457947163

    σ acc=0.086676226

    ¿ μacc+σ acc=0.544623

     μacc−σ acc=0.371271

     μacc=0.48556507

    σ acc=0.100353286

    ¿ μacc+σ acc=0.58585

     μacc−σ acc=0.38515

    -amage

    States-amage Re9resented

    Mutually eD$lusive damage

    state

     D S1 5 of ties disodge and fa. none

     D S2

    30 of ties disodge and fa andt/"ar grid damaged.

    none

     D S3  ota ceiing coa*se. none

    0eading the values from the fragility curves, we have

    Cases S9e$trally mat$7edeart7ua8es

    Linearly s$aledeart7ua8es

     μdrift −σ drift 

     P DS3

    =0

     P DS2

    =0

     P DS1

    =0

     Pundmaged=1

     P DS3

    =0

     P DS2

    =0

     P DS1

    =0

     Pundmaged=1

     μdrift 

     P DS3

    =0

     P DS2

    =0.01

     P DS1

    =0.03−0.01=0.02

     Pundmaged=1−0.03=0.97

     P DS3

    =0

     P DS2

    =0.01

     P DS1

    =0.03−0.02=0.01

     Pundmaged=1−0.03=0.97

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     μdrift +σ drift 

     P DS3

    =0.009

     P DS2

    =0.03−0.009=0.021

     P DS1

    =0.25−0.03=0.22

     Pundmaged=1−0.25=0.75

     P DS3

    =0.0099

     P DS2

    =0.03−0.009=0.021

     P DS1

    =0.25−0.03=0.22

     Pundmaged=1−0.25=0.75

    #onsidering that the cost is the higher possible, we have

    -amage State 1   Cost  DS1=471.25 $ -einsta ne! acoustic tie for 5 or area.

    -amage State   Cost  DS2=3688.75 $-einsta ne! acoustic tie and ceiing grids

    for 30 of t%e area.

    -amage State !   Cost  DS3=7588.75 $ ntire+ re*ace ceiing and grid.

    The total repairing cost can be calculated ta1ing into account the different probabilities related to

    every damage state.Cost tot = P DS

    1

    ∙Cost  DS1+ P DS2

    ∙Cost  DS2+ P DS3

    ∙Cost  DS3

    Cases analysed otal $ost ;or s9e$trally mat$7ed eart7ua8es otal

     μrotation−σ rotation   Cost tot =0∙471.25 $+0 ∙3688.75$+0 ∙7588.75 $   0$

     μrotation   Cost tot =0.02 ∙471.25$+0.01 ∙3688.75 $+0 ∙7588.75$   46.32$

     μrotation+σ rotation   Cost tot =0.22∙471.25 $+0.021 ∙3688.75$+0.009 ∙7588.75   249.44$

    Cases analysed otal $ost ;or linearly s$aled eart7ua8es otal

     μrotation−σ rotation   Cost tot =0∙471.25 $+0 ∙3688.75$+0 ∙7588.75 $   0$

     μrotation   Cost tot =0.02 ∙471.25$+0.01 ∙3688.75 $+0 ∙7588.75$   46.32$

     μrotation+σ rotation   Cost tot =0.22∙471.25 $+0.021 ∙3688.75$+0.009 ∙7588.75   249.44$

    The estimated repairing for each damage state are

    -amageState 1

      T  DS1=1.5912 days-einsta ne! acoustic tie for 5 or

    area.

    -amageState

      T  DS2=11.6706 days-einsta ne! acoustic tie and ceiing

    grids for 30 of t%e area.

    -amageState !

      T  DS3=23.7441days ntire+ re*ace ceiing and grid.

    2ith the same procedure, we can calculate the total time repair

    Cases analysedotal re9air time ;or s9e$trally mat$7ed

    eart7ua8esotal

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     μrotation−σ rotation   Timetot =0∙1.5912 d+0 ∙11.6706 d+0∙23.7441 d   0days

     μrotation   Timetot =0.02 ∙1.5912d+0.01∙11.6706d+0 ∙23.7441d   0.14853days

     μrotation+σ rotation   Timetot =0.22 ∙1.5912 d+0.021∙11.6706d+0.009 ∙23.7441   0.8088435da

    Cases analysed otal re9air time ;or linearly s$aled eart7ua8es otal

     μrotation−σ rotation   Timetot =0∙1.5912 d+0 ∙11.6706 d+0∙23.7441 d   0days

     μrotation   Timetot =0.02 ∙1.5912d+0.01∙11.6706d+0 ∙23.7441d   0.14853days

     μrotation+

    σ rotation   Timetot =0.22

    ∙1.5912

    d+0.021

    ∙11.6706

    d+0.009

    ∙23.7441   0.8088435

    da

    Question A - What changes could you mae to com!onents 1 through " to reduce damage

    under an e#ual set of engineering demand !arameters$

    f the set of demanding parameters is the same for all elements, a possible method to improve their 

    resistance to the effects of an earth&ua1e depends on the element considered an accurate analysis of 

    each one is necessary to find the wea1 points and to understand the way they reach the failure.

    The first element analysed, a  pre-Northridge WU-! "eam-column #oint$ the damages occur in the

    weld or web bolts with a possible propagation of the crac1s. To improve the performance of the

    connection we can select a different &uality of welding (li1e total penetration) or a different &uality

    of bolts (increasing the class if possible). 6ther improvements could be a strengthening of thesection by using some supports under the beam (the fracture always begins from the bottom flange),

    whether to change element completely and adopt a post-/orthridge beam connection, with a

    reduced beam section, or, if the structure allows the adoption of hinged connection together with the

    implementation of braces with the function of absorption of the hori7ontal forces.

    The second elements is a shear lin1. n this case, damages occur to the concrete slab (in the first

    damage state ), buc1ling of the web (in the second DS ) and finally the fracture in the web (third

     DS ). " way to improve the section can be to increase the number of stiffeners of the lin1 so that the

    connection would be more rigid. This change can give a better performance also with the respect to

    the buc1ling of the web and increase the element resistance to the cyclic fatigue. "nother possible

    way can be the selection of a lin1 with a thic1er web, flange and vertical stiffeners. ts also possibleto change the lin1 length (this also improves the ductility of the system) increasing the length is

     possible to have a more flexible element with a better performance against brittle failure but in

    charge of higher vertical displacements.

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    The third element is a gypsum wall partition. Unfortunately, due to the brittle nature of the gypsum ,

    is not easy to have better performances, also because the failure is mostly influenced by the

    deformation of the metallic framing members that induce crac1 on the panel, especially in the

    corners or in the openings border where most li1ely, there concentration of stresses (due to

    singularity).

    " possible solution can be the adoption of a material alternative to gypsum, with a more ductile

     behaviour (li1e a fibred layered material) or to coat the partition with a net and then to cover it with

    some special plaster (reinforced plaster). 6bviously, in this second case, the price of this

    improvement has to be considered, against the cost of substitution of the partition wall.

    The fourth element is the ceiling, made by tiles. n this case, for both sets of earth&ua1e records, the

    damages that have been found are small same as the repairing costs. The fragility curves of this

    element are particular because the ceiling is made to be uninstalled or inspected, and therefore has

    all the tiles not connected to the supporting frame. The only possible improvement can be to have a

    larger surface on which the tiles are bac1ed or to use some internally flexible connection that allows

    the tiles to move but not to fall.

    Question % - What changes could you mae to reduce damage to all com!onents$

    " possible method to reduce the damage to all components could be the improvements in the

     performance of the building against the seismic forces, li1e the adoption of braces to reduce the

    drift or base isolation. n some cases, also, the adoption of a tuned mass absorber or dampers can

    give an effective reduction of the seismic demand or a redistribution of masses (if possible) or a

    change of layout (if possible) to improve regularity in plan and elevation.

    Question C - &o' are fragility functions de(elo!ed$

    n a damage analysis, whose input is the engineering demand parameters ($'!) calculated in the

    structural analysis, and whose output is the damage measure ('5) of each damageable structural

    and non-structural component in the facility, we ma1e use of fragility functions.

    %ragility functions are probability distributions that are used to indicate the probability that a

    component, element or system will be damaged to a given or more severe damage state as a

    function of a single predictive demand parameter such as story drift or floor acceleration. %ragility

    functions usually are in the form of lognormal cumulative distribution functions, having a median

    value θ   and logarithmic standard deviation,   β . The mathematical expression for such afragility function is

     F i ( D )=Φ (ln(

     D

    θi ) βi )

    2here  F i ( D )  is the conditional probability that the component will be damaged to damage state8   i 9 or a more severe damage state as a function of demand parameter,  D : Φ  denotes the

    standard normal (4aussian) cumulative distribution function, θi  denotes the median value of the

     probability distribution, and  β i  denotes the logarithmic standard deviation. 3oth θ  and  β

    are established for each component type and damage state using one of the methods presented later.

    The probability that a component will be damaged to damage state 8   i 9 and not to a more or less

    severe level given that it experiences demand, ' is given by P (i∨ D )= F i ( D )− F i+1 ( D )

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    2here   F i+1 ( D )  is the conditional probability that the component will be damaged to damage state“i+1”  or a more severe state and  F i ( D )  is as previously defined.

    " typical form of fragility function plotted in the form of a cumulative distribution function is

    showed in the picture below on the left-hand side: on the right is showed the calculation of the

     probability that a component will be in damage state “i ”   at a particular level of demand, D=d .

    Question D - Why does )AC*+s loss assessment methodology follo' a onte Carlo simulation

    a!!roach$

    "ssessing uncertainty and explore variability in building performance is a process that would

    ideally involve performing a large number of structural analyses, using a large suite of input ground

    motions, and analytical models with properties randomly varied. 5odels would include all

    structural and non-structural components and systems, and would be able to predict explicitly

    damage to each component of each system as it occurs. The results of each single analysis would

    represent one possible outcome, and the results of the large suite (thousands) of analyses would

     produce a smoothed distribution for probabilistic evaluation of earth&ua1e conse&uences.

    4iven the current state of modelling capability, such an approach would be impractical for 

    implementation in practice. nstead, a 5onte #arlo procedure is used to assess a range of possible

    outcomes given a limited set of inputs.

    0ather than re&uiring a large number of structural analyses to develop these demands, the results

    from a limited suite of analyses are mathematically transformed, fictitiously, into a large series of simulated demand sets. %rom this limited number of analyses, one can derive a statistical

    distribution of demands from a series of building response states for a particular intensity of motion.

    %rom this distribution, statistically, consistent demand sets are generated representing a large

    number of possible building response states. These demand sets, together with fragility and

    conse&uence functions, are used to determine a building damage state and compute conse&uences

    associated with that damage. (%$5" !-;