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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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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Example of ductile piers
Plan view
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Example of ductile piers
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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Linear Analysis Methods
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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Linear Analysis Methods
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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Linear Analysis Methods
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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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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Fundamental mode method
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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Fundamental mode method
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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Indian Concrete Institute Workshop on Eurocode 8 - New Delhi, 17-19 September 2009 46Dissemination of information for training Vienna, 4-6 October 2010 46
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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Indian Concrete Institute Workshop on Eurocode 8 - New Delhi, 17-19 September 2009 47Dissemination of information for training Vienna, 4-6 October 2010 47
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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Indian Concrete Institute Workshop on Eurocode 8 - New Delhi, 17-19 September 2009 52Dissemination of information for training Vienna, 4-6 October 2010 52
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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Indian Concrete Institute Workshop on Eurocode 8 - New Delhi, 17-19 September 2009 53Dissemination of information for training Vienna, 4-6 October 2010 53
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
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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
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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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Indian Concrete Institute Workshop on Eurocode 8 - New Delhi, 17-19 September 2009 59Dissemination of information for training Vienna, 4-6 October 2010 59
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%
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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
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Indian Concrete Institute Workshop on Eurocode 8 - New Delhi, 17-19 September 2009 66Dissemination of information for training Vienna, 4-6 October 2010 66
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
2nd order effects for the seismic analysis
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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
2nd order effects for the seismic analysis
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Indian Concrete Institute Workshop on Eurocode 8 - New Delhi, 17-19 September 2009 68Dissemination of information for training Vienna, 4-6 October 2010 68
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= .
Indian Concrete Institute Workshop on Eurocode 8 - New Delhi, 17-19 September 2009 69
Di i ti f i f ti f t i i Vi 4 6 O t b 2010 69
2nd order effects for the seismic analysis
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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
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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%
Indian Concrete Institute Workshop on Eurocode 8 - New Delhi, 17-19 September 2009 71
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Verification of piers
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Dissemination of information for training Vienna, 4-6 October 2010 71
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
Indian Concrete Institute Workshop on Eurocode 8 - New Delhi, 17-19 September 2009 72
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Verification of piers
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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
,
Indian Concrete Institute Workshop on Eurocode 8 - New Delhi, 17-19 September 2009 73
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Verification of piers
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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
Indian Concrete Institute Workshop on Eurocode 8 - New Delhi, 17-19 September 2009 74
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3 Example of seismic isolation
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Bridge Elevation:
-general example
bridge
Three spans
(60m+80m+60m)
Com osite steel &
concrete deck
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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
Indian Concrete Institute Workshop on Eurocode 8 - New Delhi, 17-19 September 2009 76
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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
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Bridges with Seismic Isolation
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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
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Indian Concrete Institute Workshop on Eurocode 8 - New Delhi, 17-19 September 2009 79
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Triple FPS bearings
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Photo of Triple PendulumTM Bearing Schematic Cross Section
Concaves and Slider Assembly Concaves and Slider Components
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Triple FPS bearings
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Section:
h
4 Stainless steel concave slidin
Plan:
surfaces & articulated slider
Special low-friction sliding
low horizontal stiffness (increasedflexibility)
bearing capacity
Protective seal
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Triple FPS Force-Displ. relation
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(b)(c)
a
Adaptive behaviour,
improved recentering (b) design earthquake: softening,
(c) extreme events: stiffening, increased
damping
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General Loads
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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
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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
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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)
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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
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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) .
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Isolator design properties
Eff ti d l l th R 1 83
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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)
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Isolator design properties
f2 t t f2 1 15 (T bl JJ 2 f T i b 10 0o
C
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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
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Fundamental mode spectrum analysis
E ti ti f d d d f d i ilib i
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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
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Fundamental Mode analysis (LBDP)
F d t l d l i (EN1998 2 7 5 4)
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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
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Fundamental Mode analysis (LBDP)
Effective damping: eq (7 5) (7 9)
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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
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Fundamental Mode analysis (LBDP)
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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
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Fundamental Mode analysis (LBDP)
Effective damping: eq (7 5) (7 9)
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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
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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
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Comparison with non-isolated bridge
Non-isolated bridge with fixed bearings at pier top
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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
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Non linear Time-History Analysis
Newmark constant acceleration method
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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
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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
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Check of Lower Bound Action Effects
Lower bound of action effects = 80% of
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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
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Bridges with Seismic Isolation
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
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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
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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))
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Bridges with Seismic Isolation
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
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Design of Piers
Action effects for pier design: EN 1998-2, 7.6.3(2)
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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
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Design of Piers
Provided reinforcement:
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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
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Action Effects on Foundation
Action effects for f d ti d i
Fz
M
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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
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Special Features of FPS isolators
I l t h i t l f ti l t th
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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
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Special Features of FPS isolators
Horizontal seismic components cause continuous
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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.
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Comparison of TH and FMM analyses
Seismic displacement and shear demand at the abutments
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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
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Other subjects covered by EN 1998-2
Main design issues covered by EN 1998-2, notd lt ith i th t
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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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