2010 EN1998 Bridges BKolias

8/4/2019 2010 EN1998 Bridges BKolias

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EUROCODESBridges: Background and applications

Dissemination of information for training Vienna, 4-6 October 2010 1

Overview of Seismic issues for bridge

es gn

Basil KoliasDENCO SA

Athens


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Indian Concrete Institute Workshop on Eurocode 8 - New Delhi, 17-19 September 2009 2


Overview of the presentation

1. Example of ductile piers

concrete deck rigidly connected to piers designedfor ductile behaviour

2. Example of limited ductile piers

Seismic desi n of the eneral exam le: Brid e onhigh piers designed for limited ductile behaviour

3. Example of seismic isolation

Seismic design of the general example: Bridge on

squat piers designed with seismic isolation


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1 Example of ductile piers

Bridge description

3 span voided slab bridge (overpass), withspans 23.0m, 35.0m and 23.0m. Total length of

82.50m.

Piers: single cylindrical columns D=1.20m,. .

for M1 and 8.5m for M2.

Sim l su orted to abutments throu h a air of

sliding bearings.

Foundation of piers and abutments through

piles.


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Example of ductile piers

Longitudinal section

A1 M1 M2 A2


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Plan view


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Cross section of deck d = 1.65m

Pier cross section D = 1.20m


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Linear Analysis Methods

Equivalent Linear Analysis:Elastic force analysis with forces from

unlimited elastic response divided by theglobal behaviour factor = q.=

(except for very low periods T< TB)

esponse spec rum ana ys s:Multi-mode spectrum analysis

Fundamental mode analysisqu va en s a c oa ana ys s

With several models of varying simplification


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Basic prerequisite:

Stiffness of Ductile Elements iers :

secant stiffness at the theoretical yield,based on bilinear fit of the actual curve

My

= My/ yYield of first bar

My = Rd

Guidance for estimation of stiffness in Annex C


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Stiffness of concrete deck

Bending stiffness: uncracked stiffness both

-

open sections or slabs: may be ignored

restressed box sections: 50% of

uncracked

reinforced box sections: 30% of

uncracked


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Ductility Classes

Limited Ductile Behaviour:

q 1.50 (behaviour factor)

min= 7 (curvature ductility)

Ductile Behaviour:

1.50 < q 3.50min = 13


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Ductility Classes

BehaviourBehaviourBehaviour

DuctileDuctileDuctile

Displacement


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Indian Concrete Institute Workshop on Eurocode 8 - New Delhi, 17-19 September 2009 13Dissemination of information for training Vienna, 4-6 October 2010 13

Regular / Irregular bridges

Criterion based on the variation of local required

ri = qMEd i/MRd I =

qx Seismic moment / Section resistance

A bridge is considered regular when the

irregularity index:

ir i i 0

Piers contributing less than 20% of the average

force per pier are not considered


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Regular / Irregular bridges

For regular bridges: equivalent elastic-

specified, without checking of local ductilitydemands

Irregular bridges are: either designed with reduced behaviour

factor:

qr= qo ir .

-

dynamic (time history) analysis


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Design Seismic Action to EN 1998

Two types of elastic response spectra:.

5 types of soil: A, B, C, D, E. (+ S1,S2)

4 period ranges:

short < TB < constant acceleration < TC


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Seismic action: Elastic response spectra Se(T)

Ground motion: Dependence on ground type (A, B, C, D, E)

max. values:AAgg == aaggS, vS, vgg == aaggSTSTCC/(2/(2) ), d, dgg = 0.025= 0.025 aaggSTSTCCTTDDRes onse s ectrum: AccelerationRes onse s ectrum: Acceleration SS TT as a function ofas a function ofTT

15,21:0 ge

TSaTSTT B

ag: P.G.A.S: soil factor

, .

Se(T) /g S, geCB

T

SaTSTTT CeDC 5,2:

2.5

acceleration velocity displacement

(1/) (1/ 2)

DCTT1.0

2geD,

T

C D

q


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Seismic action

Response spectrum type 1= . ,

TC=0.60s and TD=2.50s. Soil factorS=1.15.

e sm c zone : Reference peak groundacceleration agR = 0.16g

Im ortance factor = 1.0 Lower bound factor = 0.20

e sm c ac ons n or zon a rec onsag =

.agR = 1.0.0.16g=0.16g

Behaviour factors 4.1.6 3 of EN 1998-2 :qx=3.5 (s=3.3>3, (s)=1.0) and

qy=3.5 (s=6.7>3, (s)=1.0)


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Methods of analysis

Multi mode spectrum response analysis

- .

- Combination of responses in 3 directions. . - .

of EN1998-1 (1 + 0.3 + 0.3 rule).

>,

Program: SOPHISTIK

Fundamental mode method

In the longitudinal direction

(hand calculation for comparison/check)


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Deck weights and other actions

1. Self wei ht G : 6.89m2.73.5m + 9.97m2.9.0m .25kN/m3 =14903kN

2. Additional dead (G2): qG2 = 43.65kN/m =sidewalks safety barriers road pavement2.25kN/m3.0.50m2 + 2.0.70kN/m + 7.5m. 23kN/m3.0.10m

3. Effective seismic live load (LE): (20% of uniformly distributedtraffic load) qLE= 0.2 x 45.2 kN/m = 9.04kN/m

4. Temperature action (T)*: +52.5o

C / -45o

C5. Creep & Shrinkage (CS)*: Total strain: -32.0x10-5

* Actions 4, 5 applicable only for bearing displacements

Deck seismic weight:

WE=14903kN+(43.65+9.04)kN/m x 82.5m = 19250kN


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Horizontal Stiffness in longitudinal directionFor cylindrical column of diameter 1.2m, Jun =

.1.24/64 = 0.1018m4

ssume e ec ve s ness o p ers eff un = . o e c ec e

later).Concrete grade C30/37 with Ecm = 33GPa

ssum ng o en s o e p ers xe , e or zon a s ness o

each pier in longitudinal direction is:

K1 = 12EJeff/H3

= 12.

33000MPa.

(0.40.

0.1018m4

)/(8.0m)3

= 31.5MN/m 2 = eff =

. . . . . . = .

Total horizontal stiffness: K= 31.5 + 26.3 = 57.8MN/m

Total seismic weight: WE = 19250kN (see loads)

=).81m/s(19250kN/9

==2m

un amen a per o :

.57800kN/mK


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Spectral acceleration in longitudinal direction:

S = a S / T /T = 0.16 .1.15. 2.5/3.5 0.60/1.16 = 0.068

Total seismic shear force in piers:VE = SeWE/g = 0.068g.19250kN/g = 1309kN

Distribution to piers M1 and M2 proportional to their stiffness:

V1 = (31.5/57.8).1309kN = 713kNV2 = 1309 713 = 596kN

Seismic moments My):

full fixit of ier columns assumed at to and bottom

My1 V1.H1/2 = 713kN.8.0m/2 = 2852kNmMy2 V2

.H2/2 = 596kN.8.5m/2 = 2533kNm


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Multimode response spectrum analysis

1st mode (T=1.75s)

(rotational MMy =3%)

2nd mode (T=1.43s)

(transverse MMy=95%)

3rd mode (T=1.20s)

(longitudinal MMx = 99%)

4th mode (T=0.32s)

(vertical MMz= 9%)


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Comparison in longidudinal direction

Fundamental

mode

Multimode

response spectrum

Effective period Teff 1.16s 1.20s

for longitudinal

direction

(3rd mode)

, zM2 596kN 556kN

Seismic moment, My M1 2852kNm 2605...2672kNm

M2 2533kNm 2327...2381kNm(values at top and

bottom)


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Required Longitudinal Reinforcement in Piers

Pier concrete C30/37, fck=30MPa, Ec=33000MPa

Reinforcing steel S500, fyk=500MPa

.

Cover to center of reinforcement c=8.2cm

Combination N My Mz As2

maxMy + Mz -7159 4576 -1270 198.7

minMy + Mz -7500 -3720 1296 134.9

maxMz + My -7238 713 4355 172.4

minMz + My -7082 456 -4355 170.0

Final reinforcement: 2532 (201.0cm2)

MRD = 4779kNm


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Verification of Longitudinal Reinforcement in Piers

Moment Axial force interaction diagram for the

bottom section of Pier M1-

-25000

-

-15000

Section with 2532

Design combinations

-

-5000-6000 -4000 -2000 0 2000 4000 6000

N(

KN

5000

15000M (KNm)


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Required Longitudinal Reinforcement in Piers

Pier concrete C30/37, fck=30MPa, Ec=33000MPa

Reinforcing steel S500, fyk=500MPa

.

Cover to center of reinforcement c=8.2cm

Combination N My Mz As2

maxMy + Mz -7528 3370 -1072 103.2

minMy + Mz -7145 -4227 1042 168.0

maxMz + My -7317 -465 3324 89.8minMz + My -7320 -674 -3324 92.5

Final reinforcement: 2132 (168.8cm2)

MRD = 4366kNm


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Verification of sections

Ductile Behaviour: Flexural resistance of lastic hin e re ions with

design seismic effectsAEd (only in piers).AEd ARd-

(shear of elements & joints and soil) are checked

with capacity design effectsACd. For non-ductile failure modes:

ACd ARd/Bd.

Bd.

Cd Local ductility () ensured by special detailing

rules (confinement, restraining of compressed barsetc), i.e. without direct assessment of.


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Capacity Design Effects

AACdCd correspond to the section forces under

assumed pattern of plastic hinges, where theflexural over-strength:

Mo = oMRdhas developed with:

o= 1.35 and

Bd = 1.0

However AACdCd qAqAEdEd==CdCd EdEd . .

Simplifications for AACdCd satisfying the equilibrium

conditions are allowed.

Guidance is given in normative Annex G


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Shear verification of piers (Capacity effects)

Over strength moment Mo = o.MRd

Over strength factoro = 1.35 is increased due to k = 0.22 > 0.1

. .

o = 1.35.[1+2.(0.22-0.1)2] = 1.35.1.029 = 1.39,

Over strength moments:

Mo1 = 1.39. 4779 = 6643kNm

M = 1.39 . 4366 = 6069kNm

Longitudinal direction (seismic actions in negative direction)

VC1 = 2Mo1/H1 = 2x6643/8.0 = 1661kNVC2 = 2Mo2/H2 = 2x6069/8.5 = 1428kN


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Shear verification of piers (Capacity effects)

Capacity design in transverse direction

VC2=1251kN

E1

VE1=680.3kN

=

Mo1=6643kNm

VC1=1476kN

Mo2=6069kNm VE2=450.2kN

The base shear on each pier is calculated by VCi=(Mo/MEi). VEi

according to simplifications of Annex G of EN 1998-2.Design for shear

Design shear force VC1 = 1661kN and VC2 = 1428kN. Bd = 1.0For circular section effective depth: de = 0.60 + 2.0.52/ = 0.93m,

Internal lever arm: z= 0.90 . de = 0.75. 0.93m = 0.84m

Rd,s = sw s . z. ywd

. co Bd, co = Pier M1: A

sw

/s = 1661kN / (0.84m . 50kN/cm2/1.15) = 45.5cm2/mPier M2: Asw/s = 1428kN / (0.84m

. 50kN/cm2/1.15) = 39.1cm2/m


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Confinement reinforcement

Typical arrangements

oops cross es

Overlapping hoops


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Mechanical confinement ratio wd

wd = wfsd/fcd

Geometric reinforcement ratio wRectangular Circular

w = Asw/(sLb) 4Asp/(sLDw)

Requirement

0,01)(0,13 Lcd

k

cc

creqw,

f

A

2

minw,reqw,rwd, 3;max

minw,reqw,wd.c,

Seismic Behaviour w,minDuctile 0,37 0,18

Limited ductile 0,28 0,12


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Special detailing rules: Confinement reinforcement

Confinement reinforcement according to 6.2.1 of EN 1998-2

Normalized axial force: k = Ed c. ck = . m . a = . > . Confinement of compression zone is required

. w,min .Longitudinal reinforcement ratio:- Pier M1:

L

=201.0cm2/11300cm2=0.0178- Pier M2: =168.8cm2/11300cm2=0.0149

Distance to spiral centerline c=5.8cm (Dsp=1.084m) Acc=0.923m

2

equ re mec an ca re n orcemen ra o w,req :- Pier M1: w,req = (Ac/Acc)..k+0.13

.(fyd/fcd)(L-0.01) =(1.13/0.923).0.37.0.22+0.13.(500/1.15)/(0.85.30/1.5).(0.0178-

. .- Pier M2: w,req = 0.116


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Confinement Reinforcement

For circular spirals

wd,c = max(1.4.

w,req; w,min) = max(1.4.

0.126; 0.18) = 0.18

w = wd,c

.(fcd/fyd) = 0.18.(0.85.30/1.5)/(500/1.15) = 0.0070

Asp/sL = w

.Dsp/4 = 0.0070.1.084m/4 = 0.00190m2/m =

19.0cm2/m

Req. spacing for16 spirals sLreq = 2.01/19.0= 0.106m

.sL

allowed = min(6.3.2cm; 108.4cm/5) = min(19.2cm; 21.7cm) =

19.2cm > 10.6cm


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Buckling of longitudinal bars

Avoidance of buckling of longitudinal bars

-

For steel S500 the ratio ftk/fyk = 1.15

= 2.5.(ftk/fyk) + 2.25 = 2.5.1.15 + 2.25 = 5.125

Maximum required spacing of spirals

L

L

. . . .

f f


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Transverse reinforcement of piers

Comparison of requirements for16 spiral

Requirement Confinement Bucklingof bars Shear design

At/sL

(cm2

/m)

2x19.0=38 -M1: 45.5

M2: 39.1

maxsL(cm)

19.2 16.4 8.5

The transverse reinforcement is governed by the shear design.

Reinforcement selected for both piers is one spiral of16/8.5

(47.3cm2/m)

C l l ti f it ff t


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Calculation of capacity effects

General procedure: combination G + Mo-G (acc. to G1 ofAnnex G of EN 1998-2)

Alternative: G + Mo on continuous deck articulated to iers

G G

Permanent Load GMG1 MG2

M M

+ (1) (2)

MG1 MG2

Ac: Over strength GMO1-MG1 MO1 MO2 MG1

MG2MO2-MG2

-MO1-MG1

MO1-MG1

MO1

MO1

MO2

MO2

MG1 MG2

MG1 MG2MO2-MG2

MO2-MG2

general procedure alternative procedure


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Calculation of capacity effects

G loading Overstrength Mo(+x direction)

Mo1=6643kNm Mo2=6069kNm

Capacity effectsG + Mo(+x direction)G

2430 228523402195M (kNm)

3692 4296M (kNm) 6122 6491

M (kNm)

4491 3104

6643 6069 6643 60692206 764

40523207

V (kN)V (kN)

(+)

6643 6069

V (kN)

6643 6069

37953045

(+)

3211 4047 256.7187.4 263.8 161.9(-)

(-)3398 4311

(-)

Reversed signs for x direction

1661(+) 1428(+) 1661(+) 1428(+)

alternative procedure isapplied


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Verification of deck section for capacity effects

Top layer: 4620 + 3316 (210.8cm2)Deck section, reinforcement and tendons

Bottom layer: 2x3316 (182.9cm2)

4 groups of 3 tendons of type DYWIDAG

6815 (area of 2250mm

2

each)

-200000

Combination / location My (kNm) N (kN)Moment Axial force interaction diagram

for deck section

-150000

- -

Pier M1 right side (+x) 2206 -28300

Pier M2 left side (+x) -6491 -29500

Design combinations

-

-50000

-60000 -40000 -20000 0 20000 40000

N(K

N)

Pier M2 right side (+x) 764 -28100Pier M1 left side (-x) 1262 -28100

Pier M1 ri ht side -x -6776 -295000

50000M (KNm)

Pier M2 left side (-x) 2101 -28300

Pier M2 right side (-x) -5444 -29900

C it ff t i f d ti


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Capacity effects on pier foundation

=

ong u na rec on

(pier M1 foundation for +x direction)

Transverse direction(pier M1 foundation for +/- direction)

Vz=1661kN Vy=1476kNVz=730kN

M =124kNmy

Mz=6643kNmX

Di l t i li l i


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Displacements in linear analysis

Control of Displacements

ssessmen o se sm c sp acemen E

dE = ddEe.dEe = result of elastic analysis. = damping correction factor for 0.05.

= dis lacement ductilit as follows:

0.05

.

when TT0=1.25TC : d = q

< = - + -

Di l t i li l i


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Displacements in linear analysis

Provision of ade uate structure clearancefor the total seismic design displacements:

Ed = E G 2 T w 2= .

dG due to permanent and quasi-permanent actions(mainly shrinkage + creep).dT due to thermal actions.

Roadway joint displacements:

= + +

Roadway joints over abutments


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y j

Roadway joint: dEd,J = 0.4dE + dG + 2dT, where 2 = 0.5

Structure clearance: dEd = dE + dG + 2dT

Displacements

(mm)

dGShrinkage

+Creep

dTTemper.variation

dEEarthquake

dEd,JRoadway

joint

dEdclearance

opening +18,7 +10,7 +76,0 +54,5 +100,7Longitudinal closure 0 -8,5 -76,0 -34,7 -80,3

, , ,

Roadwa oint t e: T120

Capacity in longitudinal direction: 60mmCapacity in transverse direction: 50mm

Roadway joints over abutments


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y j

Detailing of back-wall for predictable (controlled) damage.

(2.3.6.3 (5), EN 1998-2))

Clearance of

roadway joint

Approach slab

Structure

clearance

Minimum overlapping length


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Minimum overlapping length

Minimum overlapping (seating) length at moveable

joint (according to 6.6.4 of EN 1998-2)

lm = 0.50m > 0.40m

. . . . . . 2. . .g . g C D . . . . . . .

Lg = 400m (Ground type C)

Leff= 82.50/2 = 41.25m

Consequently deg = (2

.dg/Lg)Leff= (2.0.068/400).41.25 = 0.014m < 2dg

=lov

es .

lov = lm + deg + des == 0.50 + 0. 014 + 0.101 = 0.615m

Available seating length: 1.25m > lov

Conclusions for design concept


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Conclusions for design concept

Optimal cost effectiveness of a ductile system is

achieved when all ductile elements (piers) have

dimensions that lead to a seismic demand that is

critical for the main reinforcement of all critical

sec ons an excee s e m n mum e.g:min=

This is difficult to achieve when the piers

- have substantial height differences, or

- have section larger than required.

In such cases it may be economical to use:- limited ductile behaviour for low agR values

- ex e connect on to t e ec se sm c

isolation)

2 E ample of limited d ctile piers


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2 Example of limited ductile piers

high piers designed for limited ductile behaviour

Bridge description

Com osite steel & concrete deckThree spans: 60m+80m+60m

Pier dimensions:

Height 40 m,External/internal diameter 4.0 m/3.2 m

. .

Bridge elevation and arrangement of bearings


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Bridge elevation and arrangement of bearings

Articulated Articulatedconnection

Sliding Longitudinal,Articulated Transverse

Y

P1R P2R

X

C0L P1L P2L C3L

Example of limited ductile piers


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Example of limited ductile piers

Design Concept

Due to the hi h ier flexibilit :

Hinged connection of the deck to both piersboth iers resist earth uake without

excessive restraints

High fundamental period Low spectral acceleration

No need of high ductility or seismic isolation

q= 1.5 (limited ductility)

Design seismic action


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Design seismic action

Soil type B

Importance factorI = 1.00

Reference peak ground acceleration: aGr= . g

Soil factor: S= 1.20, aGrS= 0.36g

q= 1.50

=

Design seismic spectrum


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Design seismic spectrum

0.6

0.7

Ag=0.3,GroundType:BSoilFactor:1.2

0.5(g)b= . , c= . , d= . , = .

0.4

e

leratio

0.2

Ac

0.1

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6

Period(sec)

Seismic analysis


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Seismic analysis

Quasi permanent traffic load for the

. .

For bridges with severe traffic=, , ,

Q is the characteristic load of UDL s stem of,Model 1 .For the 4 lanes of the deck:

3 m x 9 + 3 m x 2.5 + 3 m x 2.5 + 2 m x 2.5

Qk,1= 47.0 kN/m2,1Qk,1 = 9.4 kN/m

Seismic analysis


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Seismic analysis

Structural Model

XY X

Z

Y

Z

Seismic analysis


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Seismic analysis

A beam finite element model of the bridge is

.

Each composite steel concrete beam and

The effective width of the main beams is

assumed e ual to the total eometric width. The bending stiffness about the vertical axis

of the two main beam sections is modified

so that the sum of the stiffnesses is equal tothe relevant stiffness of the entire composite

deck.

Seismic analysis


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y

Effective Pier Stiffness

2.3.6.1 1 of EN1998-2 defines the stiffness

of the ductile elements (piers) to correspond

to the secant stiffness at the theoretical yield

point.

This value depends on the axial force and on

the final reinforcement of the element.

It is estimated from the moment-curvature

an e momen e gross ra o curves o

the piers for the estimated final reinforcement

. ,

Jeff/Jgross 0.30

Seismic analysis


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y

80000.0

85000.0

90000.0

95000.0

Bottom Section

55000.0

60000.0

65000.0

70000.0

75000.0

Nm)

Top Section

30000.0

35000.0

40000.0

45000.0

50000.0

Moment( .

0.0

5000.0

10000.0

15000.0

20000.0

.

N=15000kN

N=20000kN

0.0

0E+

00

1.0

0E03

2.0

0E03

3.0

0E03

4.0

0E03

5.0

0E03

6.0

0E03

Curvature

Moment-Curvature M- curve of Pier Section for = 1.5%

Seismic analysis


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y

85000.0

90000.0

95000.0

N=15000kN

N=20000kNBottom Section

60000.0

65000.0

70000.0

75000.0

80000.0

Nm)

35000.0

40000.0

45000.0

50000.0

55000.0

Moment(k

=1.5%

10000.0

15000.0

20000.0

25000.0

30000.0

Top Section

0.0

5000.0

0.0

0

0.1

0

0.2

0

0.3

0

0.4

0

0.5

0

0.6

0

0.7

0

0.8

0

0.9

0

1.0

0

J/Jgross

=e .

()eff/(EI) = (M/)/()

Seismic analysis


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y

Eigenmodes Response Sectrum Analysis The first 30 eigenmodes of the structure

were calculated and considered in the

response spectrum analysis. e sum o mo a masses o ese mo es

amounts to: 97.1% in the X and 97.2% in the

. Combination of modal responses was carried

.

Following table shows the characteristics of the.

Seismic analysis


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y

Modal Mass %

First 10 eigenmodes

NoSec X Y Z

1 5.03 92.5% 0.0% 0.0%

2 3.84 0.0% 76.8% 0.0%

3 1.49 0.0% 0.0% 0.0%

. . . .

5 0.71 0.0% 0.5% 0.0%

6 0.66 0.0% 8.4% 0.0%

7 0.52 0.0% 0.0% 0.0%8 0.50 0.0% 0.0% 0.0%

. . . .

10 0.46 0.0% 0.0% 0.0%

Seismic analysis


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1st Mode - Longitudinal Period 5.02 sec c(Mass Participation Factor Ux:93%)

Seismic analysis


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2nd Mode -Traqnsverse Period 3.84 sec

Mass Partici ation Factor U :77%

Seismic analysis


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3rd Mode - Rotation Period 1.49 sec

Seismic analysis


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11th Mode - Vertical Period 0.42sec

(Mass Participation Factor Uz:63%)

Seismic analysis


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6680.6

1-673

4.83 6282.21-6279.87

13324

.92

-1343

3.36 8637.09

-8633.13

26256

.58

-2647

3.46

19836

.-19

999.

14245.07

-14237.88

11345.81-11340.23

3935

4.93

-396

80.25

3271

9.57

-3299

0.67

20622.12

-20611.70

17322.35

-17313.55

37.13

-536

70.89

4620

8.34-46587.88

27881.8

-27868.23

24158.64-24146.62

5

Max bending moment distribution along pier P1

2nd order effects for the seismic analysis


66/110


Geometric Imperfections of PiersAccording to 5.2 of EN 1992-2:2005:

31 2

1.58 10200i

0i i

le

Imperfection Eccentricities

0 i

X 80 0.063

.

I di C t I tit t W k h E d 8 N D lhi 17 19 S t b 2009 67



67/110


The first and second order effect of imperfectioneccentricities ei,II including creep for = 2.0

1(1 )lle e

where

,

v

B EdNB: buckling load according to 5.8 of EN1992-1-1

Direction ei = /ED ei,II /ei ei,II

x 0.063 19.65 1.161 0.073

y 0.032 78.62 1.039 0.033

Indian Concrete Institute Workshop on Eurocode 8 New Delhi 17 19 September 2009 68



68/110


Second order effects due to seismic first ordereffects

Two alternative approaches:

1. To 5.8 of EN 1992-1-1 (nominal stiffness method)se o nom na s ness me o . . or eff=

0.30 (EI)

MF= 1+[/((NB/NEd)-1)]

2 2B eff 1 0 1

Longitudinal direction MF= 1.154= .


Di i ti f i f ti f t i i Vi 4 6 O t b 2010 69



69/110

Indian Concrete Institute Workshop on Eurocode 8 New Delhi, 17 19 September 2009 69Dissemination of information for training Vienna, 4-6 October 2010 69

2nd order effects due to seismic 1st order effects2. To 5.4 of EN 1998-2 (followed in the example)

Increase of bending moments at the plastic hinge

section= . +q Ed Ed

dEd: seismic displacement of pier top

iCombination EN1992-1-1 EN1998-2

. . .

Ey+0.3x+2nd Ord Mx 27178.8 30508.3

The Eff. Stiffness method of EN1992-1-1 may

be unsafe for 1.5 < q 3.5


Dissemination of information for training Vienna 4 6 October 2010 70

Verification of piers


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p , pDissemination of information for training Vienna, 4-6 October 2010 70

Flexure and axial force for pier base sectionDesign action effects:

NEd = 19568 kN

MEy = 59610 kNm

Ex = m

As,req = 678 cm2

Internal perimeter 4928 (301 cm2) 1.5%


Dissemination of information for training Vienna 4-6 October 2010 71



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20000

30000

40000

=1.5%

=1.0%

30000

2000010000

0

10000

N)

70000

60000

50000

40000

A

xialForce(k

120000

110000

100000

90000

140000

130000

0

10000

20000

30000

40000

50000

60000

70000

80000

Design Interaction Diagram of Pier Base Section





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Dissemination of information for training Vienna, 4 6 October 2010 72

Shear verificationDesign shear: Vx,d = 1887 Vy,d = 281

1/3

2 2, , 1908d x d y d V V V kN

, , 1Rd c Rd c l ck cp w

,0.18 0.18

0.12

1.5

Rd c

c

C

(EN1998-2:2005, 5.6.3.3.(2))

2 2 1.82.0 3.15s

e

rd r

15EdN

1 1 1.253150

k

d

1 ..

4.52cp

c

A

,2670

2136Rd cv

kN VV 2670kN

No shear reinforcement required 1 .Bd

,





73/110

g

Ductility requirements Confining reinforcement

To 6.2.1.4 of EN 1998-2 for limited ductile

, ,max(1.4 ;0.12) max(1.4 0.058;0.12) 0.12wd c w req

, 0.28 0.13 ( 0.01)ydc

w req k l

cc cd

fA

A f

4.52 19580 500000 1.50.28 0.13 (0.015 0.01) 0.058

3.39 35000 4.52 35000 1.15

35000 1.150.12 0.0064500000 1.5

cdw w

yd

ff

16/11sp spwcc l

D A

A s

No buckling of reinforcement sL < 5dbL = 14 cm



3 Example of seismic isolation


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Bridge Elevation:

-general example

bridge

Three spans

(60m+80m+60m)

Com osite steel &

concrete deck



Pier Layout


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Transversesection:

Longitudinalsection:

Section A-A:

ec on - :

Section C-C:

=

Rectangular cross-section 5.0m x 2.5m

Enlarged pier head 9.0m x 2.5m to support bearings



Bridges with Seismic Isolation


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Effect of period shift from Tin to Teff

ad

Period schift

a d

2Taag Displacement spectrum

Acceleration spectrum

T





77/110

Effect of increasing damping (eff)

d = 0.05

reduces displacements

but not necessarily forces

n e

eff

0,05eff dd

Teff 0,40,05

0,10

eff


78/110



Triple FPS bearings


79/110

Photo of Triple PendulumTM Bearing Schematic Cross Section

Concaves and Slider Assembly Concaves and Slider Components



Triple FPS bearings


80/110

Section:

h

4 Stainless steel concave slidin

Plan:

surfaces & articulated slider

Special low-friction sliding

low horizontal stiffness (increasedflexibility)

bearing capacity

Protective seal



Triple FPS Force-Displ. relation


81/110

(b)(c)

a

Adaptive behaviour,

improved recentering (b) design earthquake: softening,

(c) extreme events: stiffening, increased

damping



General Loads


82/110

1. Permanent loads: (from general example)Total

support Self weight Minimum Maximum Total with Total withTime

variation

MN (bothbeams)

construction load load equipment equipmentdue to creep& shrinkage

C0 2.328 0.664 1.020 2.993 3.348 -0.172P1 10.380 2.440 3.744 12.819 14.123 0.206

. . . . . .

C3 2.377 0.664 1.019 3.041 3.396 -0.126

Sum ofreactions

25.343 6.209 9.528 31.552 34.871 0.000

. -Lane Number 1: qq1,k = 3 m x 9 kN/m

2 = 27.0 kN/m

Lane Number 2: qq2,k =3 m x 2.5 kN/m2 = 7.5 kN/m

Lane Number 3: qq3,k =3 m x 2.5 kN/m2 = 7.5 kN/m

Residual area: qqr,k =2 m x 2.5 kN/m2 = 5.0 kN/m

Total load = 47.0 kN/m

3. 50% of thermal action: +25oC / -35oC



Seismic Action Design Spectra


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Se(T) (g) Constant velocitybranch (1/T)

0agS

Constant displacement

branch (1/T2

)agS

0agS(TC/T)

2

TB TC TD T(s)

g

- . . ,EN1998-1 3.2.2.2 and 3.2.2.3

Average importance: =1.00 = = High Seismicity Zone: agR=0.40g

Ground Type B: S=1.20, TB=0.15s, TC=0.5s, TD=2.5s

Horizontal s ectrum: s ectral am lification =2.5

g g .

Vertical spectrum: 0=3.0, S=1.00, avg=0.9ag,

TB=0.05s, TC=0.15s, TD=1.0s



Seismic Action Ground Motions


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7 Ground Motions: 2 horizontal & 1 vertical components Semi-artificial accelerograms from modified records

EN 1998-2, 3.2.3(6): 2.00

1.40

1.60

1.80

on

(g)

Damping 5%

of ensemble of earthquakes

1.3 x Elastic spectrum

0.80

1.00

1.20

ctralaccelerati

0.20

0.40

0.60Spe

0.00

.

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

Period (sec)



Triple FPS - Equivalent bilinear model


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F

F0 =0.061W

d:

o - , . . . .

d = Fo/W = effective friction coefficient = 0.061 16% R = effective pendulum length = 1.83m

y = e ec ve y e sp acemen = . m

F = shear force

d = displacement



Design properties of isolators


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Design properties of the isolating system Nominal design properties (NDP) assessed by

prototype tests, confirming the range accepted

by the Designer.

factors (aging, temperature, contamination,

Design is required for: Upper Bound design properties (UBDP). ower oun es gn proper es .

Bounds of Design Properties result either from-

JJ) .



Isolator design properties

Eff ti d l l th R 1 83


87/110

Effective pendulum length: R=1.83m Yield displacement: Dy=0.2=0.005m

properties (UBDP) and Lower bound design properties

(LBDP) in accordance with EN 1998-2, 7.5.2.4 om na va ue range: 0 = . = . ~ .

LBDP: (F0/W)min= minDPnom = 0.051

UBDP: According to EN 1998-2 Annexes J and JJ Minimum isolator temperature for seismic design:

Tmin,b=2Tmin+1 = 0.5 x (-20oC) + 5.0oC = -5.0oC where

=0.5 is the combination factor for thermal actions,

Tmin=-20oC the minimum shade air temperature at the site,

1=+5.0oC for composite deck.

factors: f1 - ageing: max,f1=1.1 (Table JJ.1, for normal environment,

unlubricated PTFE, protective seal)



Isolator design properties

f2 t t f2 1 15 (T bl JJ 2 f T i b 10 0o

C


88/110

f2 - temperature: max,f2=1.15 (Table JJ.2 for Tmin,b=-10.0 C,unlubricated PTFE)

f3 - contamination max,f3=1.1 (Table JJ.3 for unlubricated

PTFE and sliding surface facing both upwards and

downwards)

f4 cumulative travel max,f4=1.0 (Table JJ.4 for

un u r ca e an cumu a ve rave . m

Combination factorfi: fi=0.70 for Importance class II

(Table J.2)

om na on va ue o max ac ors: U,fi= max,fi- fi(J.5) f1 - ageing: U,f1 = 1 + ( 1.1 - 1) x 0.7 = 1.07

- U,f2 . . .

f3 - contamination U,f3 = 1 + (1.1 1) x 0.7 = 1.07 f4 cumulative travel U,f4 = 1 + (1.0 1) x 0.7 = 1.0

UBDP = maxDPnomU,f1U,f2U,f3U,f4 (J.4)

(F0/W)max = 0.071 x 1.07 x 1.105 x 1.07 x 1.0 = 0.09



Fundamental mode spectrum analysis

E ti ti f d d d f d i ilib i


89/110

Assume dcd,a Keff

Estimation ofdmax= dcd from dynamic equilibrium

rom s a c - re a on : max

Keff= Fmax/dcd,a

Fmax

dcd,a

eff

eff 2K

MT

2

cdeff

iD,

eff2

1

dK

E

Displacement spectrum= 0.05

=2Tad

dcd,r

,,2 0,05eff

0,4

0,10

Iterations for dcd,r= deff= dcd,aTeffTDTC

e



Fundamental Mode analysis (LBDP)

F d t l d l i (EN1998 2 7 5 4)


90/110

Fundamental mode analysis (EN1998-2, 7.5.4) Seismic weight: W= 36751kN (see loads)

Assume dcd

=0.15m

Effective Stiffness of Isolation System (ignore piers): Keff= F / dcd = W x [ (F0/W) + dcd / R ] / dcd =

36751kN x [0.051+0.15m/1.83m] / 0.15m =

= Effective period of Isolation System: eq. (7.6)

2.13s).81m/s(36751kN/9

2m

2T2

eff

Dissipated energy per cycle: EN1998-2, 7.5.2.3.5(4) E = 4 x W x F /W x d -D =

eff

4 x 36751kN x (0.051) x (0.15m-0.005m)

ED

= 1087.09 kNm




Effective damping: eq (7 5) (7 9)


91/110

Effective damping: eq. (7.5), (7.9) eff = ED,i / [2 x x Keff x dcd

2] = 1087.09kNm / [2 x x

. .

eff= [0.10 / (0.05 + eff)]0.5 = 0.591

Calculate design displacement dcd (EN 1998-2 Table7.1) dcd=(0.625/

2) x ag x S x effx Teffx TC =

0.625/

2

x 0.40 x 9.81m/s

2

x 1.20 x 0.591 x 2.13s x0.50s = 0.188m

Check assumed displacement.

Calculated displacement 0.188m

o ano er era on





92/110

Assume value for design displacement:=c .

Effective Stiffness of Isolation System (ignore piers): Keff= F / dcd = W x [ (F0/W) + dcd / R ] / dcd =

x . + . m . m . m =

Keff= 28602 kN/m

Effective period of Isolation System: eq. (7.6)

smkN

smkN

K

mT

eff

eff 27.2/28602

)/81.9/36751(22

2

- , . . . .

ED = 4 x W x (F0/W) x (dcd-Dy) =4 x 36751kN x (0.051) x (0.22m-0.005m)

ED = 1611.90 kNm




Effective damping: eq (7 5) (7 9)


93/110

Effective damping: eq. (7.5), (7.9) eff = ED,i / [2 x x Keff x dcd

2] = 1611.90kNm / [2 x x2 =. .

eff= [0.10 / (0.05 + eff)]0.5 = 0.652

Calculate design displ. dcd (EN1998-2 Table 7.1) cd= . x ag x x effx effx C = . x .

x 9.81m/s2) x 1.20 x 0.652 x 2.27s x 0.5s = 0.22m

Check assumed displacement Assumed displacement 0.22m

Calculated displacement 0.22m

Conver ence achieved

Spectral acceleration Se (EN 1998-2 Table 7.1) Se= 2.5 x (TC/Teff) x eff x ag x S = 2.5 x (0.5s/2.27s) x

. . . .

Isolation system shear force (EN 1998-2 eq. 7.10)

Vd= Keffx dcd = 28602 kN/m x 0.22m = 6292kN



Fundamental Mode analysis (UBDP)

Fundamental mode analysis: Summary of results


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Fundamental mode analysis: Summary of results Seismic weight: W= 36751kN (see loads)

cd .

Effective stiffness Keff= 435410 kN/m

Effective period Teff= 1.84s

Dissipated energy per cycle ED = 1799.32 kNm Effective damping eff= 0.331, eff= 0.512

S ectral acceleration S = 0.166

Isolation system shear force Vd= 6096kN

Pier shear forces for averaged vertical load, d

Piers P1, P2: Vd= 2480kN



Comparison with non-isolated bridge

Non-isolated bridge with fixed bearings at pier top


95/110

Non-isolated bridge with fixed bearings at pier top Period: longitudinal Tx = 0.33s, transverse Ty= 0.17s

- , .

q = 3 . 5 x (s), where s=Ls/h the shear ratio

Long. qx=3.5x1.0=3.5, Transverse: qy=3.5x0.82= 2.87

Spectral acceleration in constant acceleration branch of

spectrum Se= 2.5Sag/q

Lon itudinal: S = 2.5x1.2x0.40 /3.5 = 0.34

V=12495kNTransverse: Se = 2.5x1.2x0.40g/2.87 = 0.42gV=15435kN Isolated bridge with UBDP

e .

Reduction of forces with respect to non-isolated:- at 49% in longitudinal direction

- at 40% in transverse direction



Non linear Time-History Analysis

Newmark constant acceleration method


96/110

Newmark constant acceleration method Rayleigh damping

-

Direction X Direction Y600N

)400N

)

Force-Displacement Loops:

-200

0

200

400

-0.300 -0.200 -0.100 0.000 0.100 0.200 0.300

Force(

-300

-200

-100

0

100

200

300

-0.200 -0.100 0.000 0.100 0.200 0.300

Force(

EQ2

- .

0

200

400

600

Force(KN)

- .

0

200

400

600

Force(KN)

EQ3

-400

-200-0.300 -0.200 -0.100 0.000 0.100 0.200 0.300

Displ. (m) -400

-200-0.300 -0.200 -0.100 0.000 0.100 0.200 0.300

Displ. (m)

200

400

rce(KN)

400

600

rce(KN)

-400

-200

0

-0.300 -0.200 -0.100 0.000 0.100 0.200

Displ. (m)

Fo

-400

-200

0

200

-0.300 -0.200 -0.100 0.000 0.100 0.200 0.300

Displ. (m)

Fo



Non linear Time-History Analysis

Force-Displacement Loops (continued):


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EQ4100

200

300

orce(KN)

100

200

300

400

orce(KN)Force Displacement Loops (continued):

-400

-300

-200

-100-0.200 -0.150 -0.100 -0.050 0.000 0.050 0.100 0.150 0.200

Displ. (m)

600N)

-400

-300

-200

-100

0

-0.300 -0.200 -0.100 0.000 0.100 0.200

Displ. (m)

400N)

-400

-200

0

200

400

-0.300 -0.200 -0.100 0.000 0.100 0.200

Force(

-200

-100

0

100

200

300

-0.200 -0.100 0.000 0.100 0.200

Force(

EQ6

- .

0

100

200

300

-0 150 -0 100 -0 050 0 000 0 050 0 100 0 150

Force(KN)

- .

0

200

400

600

Force(KN)

EQ7

-300

-200

-. . . . . . .

Displ. (m)

100

200

300

rce(KN)

-400

-200-0.200 -0.100 0.000 0.100 0.200

Displ. (m)

200

400

rce(KN)

-400

-300

-200

-100

0

-0.200 -0.100 0.000 0.100 0.200

Displ. (m)

Fo

-600

-400

-200

0

-0.300 -0.200 -0.100 0.000 0.100 0.200 0.300

Displ. (m)

Fo



Check of Lower Bound Action Effects

Lower bound of action effects = 80% of


98/110

Lower bound of action effects 80% of corresponding Fundamental Mode method

results (EN 1998-2, 7.5.6(1) & 7.5.5(6):

Applicable for design displacement dcd and totald

o Displacement in X direction: d = dcd / df= 0,193m / 0,22m = 0,88 > 0,80 ok

o Displacement in Y direction: d = dcd / df= 0,207m / 0,22m = 0,94 > 0,80 ok

o Total shear in X direction: v = Vd / Vf= 6929,3kN / 6292kN = 1,10 > 0,80 ok

o o a s ear n rec on: v = d f= , = , > , o




Compliance criteria


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Compliance criteria Isolating system:

Increased reliability is required for the

isolating systemSeismic displacements increased by factor:

IS = 1.50

Sufficient lateral rigidity under service

conditions is required.

Adequate self-restoring capability

Substructure

Design for limited ductile behaviour: q 1.50



Displacement demand of isolators

Max. total displacement EN 1998-2, 7.6.2(1) &(2):


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Max. total displacement EN 1998 2, 7.6.2(1) &(2):dtot = IS x dbi,d + do,i bi,d = es gn sp acement o so ator

IS

= 1.50 seismic displacement amplification factor

d = offset dis lacement due to ermanent actions,,long term actions, 50% of thermal action

Longitudinal: dm = 1.50 x 193mm + 25.5mm = 315mm

Transverse: dm = 1.50 x 207mm = 311mm

Pier P1 bearings:= + =m . .

Transverse: dm = 1.50 x 193mm = 290mm



Lateral restoring capability

EN 1998 2:2005 + Force dcd


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EN 1998-2:2005 + Kp, . .

Check ratio dcd/dr 0.5 (7.7.1) Displdd UPDP give most unfavorable results

Post-elastic stiffness Kp = W/R Force at zero dis lacement F = W x F /W

.

Maximum static residual displacement drdr= F0 / Kp = W x (F0/W) / (W/R) = (F0/W) x R = 0.09 x

=. .

Check ratio dcd/dr= 0.139m / 0.165m = 0.84 > 0.5

equa e a era res or ng capa y w ou ncreaseof displacement demand (EN1998-2, 7.7.1(2))




Lateral restoring capability


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Lateral restoring capabilityd

dy/dcd

0.6cd/1 35 d mi 0i du bi d

1.rdd /01.35 d 1.20du du

dcd/dr0.5



Design of Piers

Action effects for pier design: EN 1998-2, 7.6.3(2)


103/110

p g , ( ) Flexural design: q-factor corresponding to limited

ductile / essentially elastic behaviour, i.e. q 1,50

Confinement reinforcement not required when,

Shear design with q= 1 and additional safety factor

Bd1=1,25 (EN 1998-2, 5.6.2(2)P)

Minimum longitudinal reinforcement to avoid brittlefailure: 0.5% in total



Design of Piers

Provided reinforcement:


104/110

5,0m

0 53 0 53 0 53 0 53 0 53 0 53 0 53 0 53 0 53

2,5m

Stirrups : 4 two-legged 12/15 = 4 x 2 x 7,54cm2/m = 60,3 cm2/m

Longitudinal reinforcement: 1 layer28/13,5 = 45,6 cm /m

Perimetric Hoop: 1 two-legged 16/15 = 2 x 13,40cm2/m = 26.8 cm2/m



Action Effects on Foundation

Action effects for f d ti d i

Fz

M


105/110

foundation design:Fx My

- , . . an

5.8.2(2)P for bridges with

Foundation PadLongitudinal Direction

Analysis results multiplied

by the q-factor used (i.e.

Fy

Foundation Pad

Mx

Tranverse Direction

effectively using q= 1).

Location EnvelopeFx Fy Fz Mx My Mz

Max Fx envelope 783 111 4242 329 78 57C0, C3

Max Fy envelope 470 695 4124 2003 47 191

P1, P2Max Mx envelope 1095 2624 16394 33331 10950 747



Special Features of FPS isolators

I l t h i t l f ti l t th


106/110

Isolator horizontal forces are proportional to the

ver ca so a or oa .

Minimization of horizontal eccentricities

are proportional to the mass.

Motion characteristics (period, displacement

acceleration etc) are ~ independent of the mass Vertical seismic motion causes short period

load May lead to an increase of maximum forces



Special Features of FPS isolators

Horizontal seismic components cause continuous


107/110

Horizontal seismic components cause continuous

Coupled isolator model should be used for T.H. non

linear analysis Increase displacement demand of FMM as a stand

alone analysis

Assuming 0.5dcd to occur simultaneously

transverse to max dcd, as estimated by the FMM,

the displacement demand should be 1.15dcd Increase max forces b the same factor.



Comparison of TH and FMM analyses

Seismic displacement and shear demand at the abutments


108/110

Seismic displacement and shear demand at the abutments

Method of analysisDisplacement

demand

longitudinal

direction

transverse

direction

Time-history

analysis393 783 695

FundamentalMode Method

FMM

405 683 683



Other subjects covered by EN 1998-2

Main design issues covered by EN 1998-2, notd lt ith i th t


109/110

dealt with in the present:

Non linear analysis of bridges: Static (push-over)

and dynamic (time history) Spatial variability of the seismic action (for long

bridges)

y ro ynam c nteract on or mmerse p ers Verification of joints adjacent to plastic hinges

u , w v

and shock transmission units

EUROCODESBridges: Background and applications



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2010 EN1998 Bridges BKolias

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Transcript of 2010 EN1998 Bridges BKolias