DIRECT DISPLACEMENT-BASED SEISMIC DESIGN OF...
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DIRECT DISPLACEMENT-BASED SEISMIC DESIGN OF
STRUCTURES
Nigel Priestley
European School for Advanced Studies in Reduction of Seismic Risk
(“Rose”
School), Pavia, Italy
ENG.PRI.0003.1
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•
WHAT SHOULD BE THE STRUCTURAL STRENGTH? (BASE SHEAR FORCE)?
•
HOW SHOULD THIS STRENGTH BE DISTRIBUTED?
•
IS CURRENT DESIGN PHILOSOPHY ADEQUATE, PARTICULARLY WITH REFERENCE TO NEW TECHNOLOGY SYSTEMS?
WITHIN A PERFORMANCE-BASED ENVIRONMENT:
ENG.PRI.0003.2
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1. The Need for DDBD
(Problems with Force-based Design)
• Estimating Elastic Stiffness
•
Distribution of Required Strength based on Elastic Stiffness
• Displacement-Equivalence Rules (equal displacement)
• Specified Ductility of Force-Reduction Factors
•
Assumption that Increased Strength Reduces potential for damage
Note: Damage is strain or drift (not strength) related
ENG.PRI.0003.3
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HA HB HC
A
B
C
Period T (seconds)
Spec
tral
Acc
eler
atio
n S
A(T
)
Elastic
Ductile
Structure
Acceleration Response Spectrum
Concrete Bridge under Longitudinal Seismic Excitation
CBA
CBAlong HEIKKKK
,,3
12Stiffness:
Period:
longKmT /2
Base shear force:
RSgm
F TA )(
Pier Shear Force:
long
ii K
KFF
Design Displacement:
gSTTAlong )(2
2
4
ENG.PRI.0003.4
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HA HB HC
A
B
C
Concrete Bridge Under Longitudinal Response
CBA
CBAlong HEIKKKK
,,3
12
longKmT /2
longKmT /2
Stiffness:
Period:
Pier Strength:long
ii K
KFF
What value for EI? What force-reduction factor? Is the strength distribution logical?
ENG.PRI.0003.5
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0 0.002 0.004 0.006Curvature (1/m)
0
10000
20000
30000
40000M
omen
t (k
Nm
)
Nu/f'cAg = 0
Nu/f'cAg = 0.1
Nu/f'cAg = 0.2
Nu/f'cAg = 0.3
Nu/f'cAg = 0.4
(a) Reinforcement Ratio = 1%
0 0.002 0.004 0.006Curvature (1/m)
0
10000
20000
30000
40000
50000
Mom
ent
(kN
m)
Nu/f'cAg = 0
Nu/f'cAg = 0.1
Nu/f'cAg = 0.2
Nu/f'cAg = 0.3
Nu/f'cAg = 0.4
(b) Reinforcement Ratio = 3%
SELECTED MOMENT-CURVATURE CURVES FOR CIRCULAR COLUMNS (D=2m,f’c
= 35MPa, fy
= 450MPa)
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0 0.1 0.2 0.3 0.4Axial Load Ratio (Nu/f'cAg)
0
0.2
0.4
0.6
0.8
1
Stif
fnes
s R
atio
(E
I/E
I gro
ss)
l = 0.04
l = 0.03
l = 0.02
l = 0.01
l = 0.005
EFFECTIVE STIFFNESS RATIO FOR CIRCULAR COLUMNS
EIeff
= MN
/y
EIeff
/EIgross
=MN
/y
EIgross
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INTERDEPENDENCY OF STRENGTH AND STIFFNESS
Stiffness EI = M/
M M
y3
y2
y1
y
M1
M2
M3
M1
M2
M3
(a) Design assumption (constant stiffness)
(b) Realistic assumption (constant yield curvature)
INFLUENCE OF STRENGTH ON MOMENT-CURVATURE RESPONSE
ENG.PRI.0003.8
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HA HB HC
A
B
C
Bridge Under Longitudinal Response
CBA
CBAlong HEIKKKK
,,3
12
longKmT /2
longKmT /2
Stiffness:
Period:
Pier Strength:long
ii K
KFF
What value to use for EI? What ductility?
ENG.PRI.0003.9
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Column C
Column A
Column B
Force
yC
Displacement 4yC
,yA yB
Column height ratios: hA
:hB
:hC
= 2.0:2.75:1.0
Yield displacement ratio: yA
:yB
:yC
= 4.0:7.56:1.0
Strength distribution: VA
:VB
:VC
= 0.125:0.048:1.0
y
= y
h2/3
=2.25y
h2/(3D)
Force-Based Design of Bridge
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BRIDGE WITH UNEQUAL COLUMN HEIGHTS
•
Shortest piers are stiffest. Shear is in proportion to 1/h3. Therefore moments (and, approximately reinforcement ratio) are in proportion to 1/h2. Increasing the reinforcement ratio of the short piers relative to the long piers increases the relative stiffness of the short piers, attracting still higher shear in elastic response.
•
Allocating high shear to the short column increases its susceptibility to shear failure.
•
Displacement capacity of the short pier is DECREASED by increasing its reinforcement ratio
•
It doesn’t make sense to have a higher reinforcement ratio for the shorter columns than the longest columns (and in practice, they would probably have the same reinforcement ratio, for simplicity of construction)
ENG.PRI.0003.11
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YIELD DRIFTS OF CONCRETE FRAMES
ELASTIC DEFORMATION CONTRIBUTIONS TO DRIFT
OF A BEAM/COLUMN JOINT SUBASSEMBLAGE
Beam flexure and shear deform.Column flexure and shear deform.Joint shear deformation
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Column height/beam length aspect ratio (lc/lb) : 0.4 – 0.86 Concrete compression strength (f’c) : 22.5MPa – 88MPa Beam Reinforcement yield strength (fy) : 276MPa – 611MPa Maximum beam reinforcement ratio (A’s/bwd) : 0.53% – 3.9% Column axial load ratio (Nu/f’cAg) : 0 – 0.483 Beam aspect ratio (lb/hb) : 5.4 – 12.6
CONCRETE FRAME DRIFT EQUATION
y = 0.5y (lb /hb )
Equation Checked against test data with:
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y = 0.5y (lb /hb )
EXPERIMENTAL DRIFTS OF BEAM/COLUMN TEST UNITS COMPARED WITH EQUATION
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Circular column: Dyy /25.2
Rectangular column: cyy h/10.2
Rectangular cantilever walls: wyy l/00.2
T-Section Beams: byy h/70.1
b
byy h
l 5.0Concrete Frames:
DIMENSIONLESS YIELD CURVATURES AND DRIFTS
Steel Frames:b
byy h
l 65.0
ENG.PRI.0003.15
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FRAME DUCTILITY LIMITS
NOTE: CODES LIMIT DRIFT TO 0.02-0.025
b
byy h
l 5.0Yield Drift:
Example: lb
= 6m, hb
=0.6m, fy
= 500MPa
y = 500/200,000 = 0.0025;
y = 0.5*0.0025*6.0/0.6 = 0.0125
Thus ductility limit is 1.6 to 2.0
Note: using high-strength reinforcement provides no benefit in reduced steel content!
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FORCE-BASED DESIGN – ASSUMPTIONS OF SYSTEM DUCTILITY
AND CONSEQUENT FORCE-REDUCTION FACTOR
•
In current force-based design it is assumed that structural systems have a unique ductility capacity, and hence a unique force-reduction factor
e.g.
Concrete Frame Building: R
= 6 (depends on country)
Concrete Wall Building: R
= 4
(::)
Concrete Bridge:
R
= 3
(::)
ENG.PRI.0003.17
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Structural Type and Material
US West Coast Japan New** Zealand
Europe
Concrete Frame 8 1.8-3.3 9 5.85 Conc. Struct. Wall 5 1.8-3.3 7.5 4.4
Steel Frame 8 2.0-4.0 9 6.3 Steel EBF* 8 2.0-4.0 9 6.0
Masonry Walls 3.5 6 3.0 Timber (struct. Wall) 2.0-4.0 6 5.0
Prestressed Wall 1.5 - - - Dual Wall/Frame 8 1.8-3.3 6 5.85
Bridges 3-4 3.0 6 3.5 * Eccentrically Braced Frame **SP factor of 0.67 incorporated.
FORCE-REDUCTION FACTORS IN DIFFERENT COUNTRIES
ENG.PRI.0003.18
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A B C
STRUCTURE Column Moment-Curvature Relationships
INFLUENCE OF AXIAL LOAD ON COLUMN STIFFNESS
REFINED ANALYSIS MYTH
A
C
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INFLUENCE OF ELASTIC DAMPING ON DISPLACEMENT-EQUIVALENCE RULE
Displacement-equivalence rules are based on Inelastic Time History Analysis (ITHA). The representation of the initial response has a considerable influence on the results:
“Elastic damping” is typically added to represent the initial stages,and expressed as a % of critical damping – typically 5%. There are two main ways this could be modelled: as initial-stiffness proportional damping, or tangent-stiffness proportional damping:
Initial-stiffness: damping always proportional to this slope
Tangent stiffness: damping force reduces when stiffness reduces; increases when stiffness increases
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IS HIGH STRENGTH NECESSARY OR EVEN DESIRABLE?
High Strength:
•
SESOC recommends high strength to reduce damage to frame buildings (mu
= 1.25)
• Reduces damage in small or moderate earthquakes
•
Does not necessarily reduce damage in Design or Extreme EQ
• Does not necessarily reduce displacements
•
Subjects contents to higher accelerations (increased contents damage)
•Increases foundation forces and cost
ENG.PRI.0003.21
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0 1 2 3 4Reinforcement Ratio (%)
0
0.5
1
1.5
2
Par
amet
er R
atio
Displacement
Stre
ngt
h
uy1 y2
u/y
Strength
StiffnessDuctilityCapacityDispl.
Capacityu Demand/Capacity
(a) Strength vs Ductility (b) Influence of Rebar % on Parameters
S1
S2
INFLUENCE OF STRENGTH ON DAMAGE
ENG.PRI.0003.22
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2. DIRECT DISPLACEMENT - BASED SEISMIC DESIGN
A RATIONAL SEISMIC DESIGN APPROACH
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Force
Displacement
Ki Ks
First cycle
Subsequent cycle
Secant Elastic
CHARACTERIZING STRUCTURE BY ELASTIC OR SECANT STIFFNESS
Maximum
Residual
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FORMULATION OF THE DIRECT DISPLACEMENT-BASED (DDBD)
APPROACH
DDBD is based on the observation that damage is directly related to strain (structural effects) or drift (non-
structural effects), and both can be integrated to obtain displacements. Hence damage and displacement can be directly related. The design approach ACHIEVES a specified damage limit state.
It is not possible to formulate an equivalent relationship between strength (force) and damage. This is one of the major deficiencies of current force-based seismic design). The level of damage is uncertain
ENG.PRI.0003.25
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DIRECT DISPLACEMENT-BASED
SEISMIC DESIGN FUNDAMENTALS
• Based on fundamental inelastic modes of response
•
Hysteretic response represented by equivalent viscous damping
•
Higher modes considered by new capacity design provisions
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me Fu
F Fn rKi
He Ki Ke
y d
(a) SDOF Simulation (b) Effective Stiffness Ke
0 2 4 6Displacement Ductility
0
0.05
0.1
0.15
0.2
0.25
Dam
pin
g R
atio
,
0 1 2 3 4 5Period (seconds)
0
0.1
0.2
0.3
0.4
0.5
Dis
pla
cem
ent
(m)
0.05
=0.10
=0.15
=0.20
=0.30
d
Te
Elasto-Plastic
Steel Frame
Concrete Frame
Hybrid Prestress
(c) Equivalent damping vs. ductility (d) Design Displacement Spectra
Concrete Bridge
FUNDAMENTALS OF DDBDENG.PRI.0003.27
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Fme
He
Fu
Fn
Ki
rKi
Ke
y D
(a) SDOF Simulation (b) Initial and Secant Stiffness
INITIAL AND SECANT STIFFNESS
Initial Stiffness: Ki with (5%) elastic damping (traditional)
Secant Stiffness: Ke with effective (e.g.16%) damping (new)
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0 2 4 6Displacement Ductility
0
0.05
0.1
0.15
0.2
0.25
Dam
pin
g (
%)
0 1 2 3 4 5Period (seconds)
0
0.1
0.2
0.3
0.4
0.5
Dis
pla
cem
ent
(m)
5%
10%
15%
20%
30%
d
Te
Elasto-Plastic
Steel Frame
Concrete Frame
Hybrid Prestress
(c) Equivalent damping vs. ductility (d) Design Displacement Spectra
Concrete Bridge
(c) Equivalent Damping vs Ductility (d) Design Displacement Spectra
ASPECTS OF DDBD
Ductility Damping; Damping+Displacement Period Te
Fu =KeD224
ee T
MK
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MDOF STRUCTURES
DDBD is a SDOF Design method, based on fundamental inelastic mode. With MDOF structure:
• Determine displaced shape, and characteristic disp.
• Determine effective Mass
• Determine effective Ductility and Damping
• Determine Effective Stiffness & Base shear
• Distribute Base shear to mass locations
•
Analyze structure to determine moments at plastic hinges
• Use capacity design to determine forces elsewhere
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1. DESIGN DISPLACEMENT FOR S.D.O.F STRUCTURES
• Depends on Design Limit State
• Structural displacement limit: Strain related
• Non-structural displacement limit: Drift related
•
chose critical of structural and non-structural limit displacements
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EXAMPLE OF STRAIN LIMIT STATES
Curvature from concrete compression:
mc
= cm
/cCurvature from reinforcement tension:
ms
= sm
/(d-c)
Chose lesser of mc
and ms
,Design Displacement is:
ds
=
y
+
p
= y
H2/3 + (m
-y
)Lp
H
Lp
= plastic hinge length.
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DRIFT LIMIT STATE
Code-specified non-structural drift limit = d
Displacement Limit d
=
d
.H
The lesser of the structural and non-structural displacement is chosen as the design value.
NOTE: Design approach will often be to design for a specified drift limit, and detail the critical sections to ensure strain limits are satisfied (this is particularly appropriate at the damage-control limit state).
NOTE: If the relationship between maximum and residual drift is known, the DDBD approach can be adapted to design for residual drift limits.
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CRITICAL DISPLACEMENTS
•Frames:
normally governed by structural or non- structural drift in beams of lowest storey
•
Cantilever Wall buildings: normally governed by plastic rotation at the wall base (for longest wall), or drift in top storey
•
Bridges: normally governed by plastic rotation, or drift limit, of the shortest column
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35
22 /4 eee TmK Effective Stiffness:
deBase KVF Base Shear Force:
n
iii
n
iiid mm
11
2 /MDOF Design Displacement:
d
n
iiie mm
/1
ELEMENTS OF DDBD: Multi-storey Building
Effective Mass:
n
iii
n
iiiie mHmH
11
/Effective Height:
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3.EQUIVALENT VISCOUS DAMPING
The design procedure requires relationships between displacement ductility and equivalent viscous damping.
Damping is the sum of elastic and hysteretic damping, and is determined by inelastic time-history analysis for different hysteretic rules:
hysteleq
Elastic damping, and the way that this has been modelled
in time-history analysis in the past:
refer
Ch.4 pp 203-210 for details
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37
Concrete Wall Building, Bridges (TT):
1444.005.0eq
Concrete Frame Building (TF):
1565.005.0eq
Steel Frame Building (RO):
1577.005.0eq
Hybrid Prestressed Frame (FS,=0.35):
1186.005.0eq
Friction Slider (EPP):
1670.005.0eq
Bilinear Isolation System (BI, r=0.2):
1519.005.0eq
RELATIONSHIPS FOR TANGENT-STIFFNESS DAMPING
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DAMPING FOR WALL BUILDING
Walls have different ductilities
(d
yA
, d
/yB
), hence damping.
System Damping:
Note: chose wall strength
Proportional to lw2:
m
jj
m
jjje VV
11
/
m
jwj
m
jjwje ll
1
2
1
2 /
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F1 F2 F3 F4 F5
V1
V2 V3
V4
V5
displaced shape
H2 H3
H4
1444.005.0pier damping:
Proportion of forces carried to abutments = x, hence:
Pier shear forces:
5
151
iiFxVV
4
2
5
1
1/11i iii
ii HHFxV Assumes all
piers yield
Inertia forcespier shears
abutment shear
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40
4. SEISMIC INPUT FOR DDBD
0 1 2 3 4 5Period T (seconds)
0
0.2
0.4
0.6
0.8
1
Acc
eler
atio
n a
(T)
(g)
0 1 2 3 4 5Period T (seconds)
0
0.1
0.2
0.3
0.4
0.5
Dis
pla
cem
ent
(T
) (
m)
(a) Acceleration Spectrum for 5% damping (b Displacement Spectrum for 5% damping
Force-based design
DDBD
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41
max
TC Period TD TE
PG
max
PG
TC Period TD TE
Displac
emen
t
GENERAL FORM OF ELASTIC 5% DISPLACEMENT SPECTRUM, FROM EC8
Corner Period
Linear
Plateau
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42
DESIGN DISPLACEMENT SPECTRA (4)
• Based on Faccioli’s observations,the corner period Tcappears to increase almost linearly with moment magnitude. For earthquakes with MW > 5.7, the following expression seems conservative:
7.55.20.1 wc MT
• Peak displacement at the corner period can be estimated from the following expression (firm ground):
r
WM )2.3(
max10
seconds
mm r = nearest distance to fault plane (km)
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43
0 2 4 6 8 10Period (seconds)
0
400
800
1200
1600
2000
Spec
tral
Dis
pla
cem
ent
(mm
)
M = 7.5
M = 7.0
M = 6.5M = 6.0
5% Damped Spectra Resulting from the Equations, at r = 10km
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44
0 1 2 3 4 5 6Period (seconds)
0
400
800
1200
1600
2000
Spec
tral
Dis
pla
cem
ent
(mm
)
0 1 2 3 4 5 6Period (seconds)
0
0.4
0.8
1.2
1.6
2
Spec
tral
Acc
eler
atio
n (
xg)Life
safety
Damagecontrol
Serviceability
Lifesafety
Damagecontrol
Serviceability
(a) Displacement Spectra (b) Acceleration Spectra
Elastic Spectra for Different Limit States
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45
0 1 2 3 4 5Period T (seconds)
0
0.2
0.4
0.6
0.8
1
Rel
ativ
e D
isp
lace
men
t
T,
0 1 2 3 4 5Period T (seconds)
0
0.2
0.4
0.6
0.8
1
Rel
ativ
e D
isp
lace
men
t
T,
4,5%
0.05
0.10
0.15
0.20
0.30
0.30
(a) "Normal" Conditions (b) Velocity Pulse Conditions
0.20 0.15
0.10
0.05
(Eq.(2.8)) (Eq.(2.11))
Recommended Spectral correction for Damping
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46
Note that designing for “Velocity Pulse”
spectra of the type in the previous slide using Direct Displacement-Based Design automatically results in a reduced effective period, and hence an increased seismic design force to limit the displacements to the design value.
This is one of the advantages of Direct Displacement-Based Design
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F4 VB4 VB4
Level 4 Base Overturning Moment F3 VB3 VB3 Level 3 F2 Level 2 H3
VB1 VB1
F1 Level 1 Level 0 MC1 MC2 MC3
T C Lbase
CjbaseBi
Cjb
ii
MLVOTM
MLTOTM
HFOTM
VB2 VB2
VC1 VC2 VC3
ANALYSIS OF FRAMES UNDER LATERAL FORCES
n
iiiiiBi mmVF
1
/
Lateral force distribution
0.6H1
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0.6H1
H1
0.15H1
VC
0.325VCH1
0.6VCH1
Level 0
Level 1
FIRST STOREY COLUMN MOMENTS
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49
COMBINATION OF GRAVITY AND SEISMIC MOMENTS AT FRAME PLASTIC HINGES
0 100 200 300Seismic Moment (kNm)
0
100
200
300
Des
ign
Mom
ent
(kN
m)
Gravity Moment
Seismic alone
RedistributedSeismic+Gravity
Design for higher of factored gravity moments OR design seismic moments (not a combination of both). Displacement demand is not influenced.
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TORSIONAL RESPONSE OF WALL BUILDINGS WITH STRENGTH AND STIFFNESS ECCENTRICITY
Wall Wall 2
Wall 3
Wall 4 Z
X kZ1, kZ2
ktrans
ktrans
CM CV CR
eVX eRX
LX = 25 m
LZ=15m
VB
V1 V2 Strength eccentricity
Stiffness eccentricity
Both strength and stiffness eccentricity affect torsional
response. In DDBD the design displacement at centre of mass is reduced to account for torsional
response
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51
HIGHER MODE EFFECTS
Two effects in DDBD
1. Drift amplification –
A reduction to the design drift is applied to frames higher than about 10 storeys
2. Moment and Shear amplification for capacity- protected actions and members. These have been
extensively researched by inelastic time-history analysis. Results are found to depend on the effective structural displacement ductility demand. New design equations and simplified design approaches have been developed for different structural systems.
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52
(e.g.) Design Column Moments in Frames
f,t
f,c
0.75H
0.25H
1.0
H
first storey
Efo
Nf MM
113.015.1, ocf
1 oo
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53
0 2000 4000 6000 8000 10000Moment (kNm)
0
5
10
15
20
Hei
gh
t (m
)
IR= 0.5 1.0 1.5 2.0
SRSS/
Cap.Des
0 4000 8000 12000 16000 20000Moment (kNm)
0
10
20
30
Hei
ght(
m)
IR=0.5 1.0 1.5 2.0
Cap.Des
SRSS/
30
40
t (m
)
IR=0 5 1 0 1 5 2 0
40
60t
(m)
IR=0 5 1 0 1 5 2 0
( ) y ( ) y
(c) Eight-Storey Wall (d) Twelve-Storey Wall
Capacity Design Moments and Time History Results for Different Seismic Intensity Ratios (IR=1=Design Intensity)
CANTILEVER WALL MOMENTS
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54
0 400 800 1200 1600 2000Shear Force (kN)
0
5
10
15
20
Hei
gh
t (m
)
IR= 0.5 1.0 1.5 2.0
SRSS/
Cap.Des
0 500 1000 1500 2000 2500Shear Force (kN)
0
10
20
30
Hei
gh
t (m
)
IR= 0.5 1.0 1.5 2.0
SRSS/
Cap.Des
30
40
t (m
) 40
60t
(m)
(c) Eight-Storey Wall (d) Twelve-Storey Wall
Capacity Design Shears and Time History Results for Different Seismic Intensity Ratios (IR=1=Design Intensity)
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78 Park Tce. Typical internal shear wall.
2 layers rebar, high %, compression failure? Lack of cross linking. Higher mode shear forces?
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56
MODIFIED MODAL SUPERPOSITION (MMS)
Extension to Method suggested by Eibl
and Keintzel,1988
SHEAR FORCE PROFILES5.02
32
22
1 ...)( EiEii VVVViV1i
= Inelastic
first-mode shear force at level i (from DBD forces) or elastic shear force, if lower
V2Ei
(etc) = elastic
2nd (etc) mode shear force at level i
THUS FORCE REDUCTION FOR DUCTILITY IS ONLY APPLIED TO FIRST MODE
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57
0 4000 8000 12000 16000 20000Moment (kNm)
0
10
20
30
Hei
gh
t (m
)
0 4000 8000 12000 16000 20000Moment (kNm)
0
10
20
30
Hei
gh
t (m
)
0 10000 20000 30000Moment (kNm)
0
10
20
30
40
Hei
gh
t (m
)
0 10000 20000 30000Moment (kNm)
0
10
20
30
40
Hei
gh
t (m
)
0 10000 20000 30000 40000Moment (kNm)
0
20
40
60
Hei
gh
t (m
)
0 10000 20000 30000 40000Moment (kNm)
0
20
40
60
Hei
gh
t (m
)
THA MMS THA MMSTHA MMS
(n) Twelve-Storey Wall, IR=1.0
(o) Twelve-Storey Wall, IR=1.5
(p) Twelve-Storey Wall, IR=2.0
(r) Sixteen Storey Wall, IR 1.0 (v) Twenty Storey Wall, IR 1.0
(s) Sixteen-Storey Wall, IR=1.5 (w) Twenty-Storey Wall, IR=1.5
(t) Sixteen-Storey Wall, IR=2.0 (x) Twenty-Storey Wall, IR=2.0
THA MMS
THA MMS
THA MMS
MMS Moment Envelopes Compared with Time History Results for Different Seismic Intensity Ratios (IR=1=Design)
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20
30
ht
(m)
0 500 1000 1500 2000 2500 3000Shear Force (kN)
0
10
20
30
Hei
ght
(m)
0 500 1000 1500 2000 2500 3000Shear Force (kN)
0
10
20
30
Hei
gh
t (m
)
30
40
ht
(m)
0 1000 2000 3000 4000Shear Force (kN)
0
10
20
30
40
Hei
gh
t (m
)
0 1000 2000 3000 4000Shear Force (kN)
0
10
20
30
40
Hei
gh
t (m
)
40
60
ht (
m)
0 1000 2000 3000 4000Shear Force (kN)
0
20
40
60
Hei
ght
(m)
0 1000 2000 3000 4000Shear Force (kN)
0
20
40
60
Hei
gh
t (m
)
MMS THA MMS THA MMS THA
MMS THA
THA MMS
THA MMS
(m) Twelve-Storey Wall, IR=0.5 (q) Sixteen-Storey Wall, IR=0.5 (u) Twenty-Storey Wall, IR=0.5
(n) Twelve-Storey Wall, IR=1.0 (r) Sixteen-Storey Wall, IR=1.0 (v) Twenty-Storey Wall, IR=1.0
MMS Shear Envelopes Compared with Time History Results for Different Seismic Intensity Ratios (IR=1= Design)
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Hei
ght
MF oMF
oMB
Mo0.5Hn
Tension shift
VF oVF
VoBase=oVVBase
Von
Capacity envelope
Capacity envelope
(a) Moment Capacity Envelope (b) Shear Force Capacity
Bo
To
Hn MCM ,15.0 4.01075.04.0,1
oiT TC
Moment: where
SIMPLIFIED WALL CAPACITY-DESIGN EQUATIONS
,BaseVoo
Base VV ToV C ,21 15.1)5.0(4.0067.0,2 iT TC
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60
DOES DDBD MAKE A DIFFERENCE?
•
Determining moment demand by elastic modal analysis is inappropriate, resulting in poor distributions of lateral strength;
•
Force-based design uses elastic stiffness which is not known at the start of the design. DDBD uses yield displacement or drift which IS known at the start of the design.
•
DDBD achieves a specified limit state at the design intensity (uniform risk); force-based design, at best, is bounded by the limit state, and vulnerability to damage is variable.
•
The design effort with DDBD is less than with force-based design.
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61
INFLUENCE OF SEISMIC INTENSITY ON DESIGN FORCES
Force-Based Design Displacement-Based Design
1
212 Z
ZVV bb 2
112 Z
ZTT ee , but22 /4 eee TmK
Hence:2
1
212
ZZKK ee
Hence:2
1
212
ZZVV BB
If Z2
= 0.5Z1
, Vb2
= 0.5Vb1 : Vb2
= 0.25Vb1
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62
BASE SHEAR AS FUNCTION OF BUILDING HEIGHT
• Assume shape function is linear (i.e. dual wall/frame)
• Assume story mass constant at m per storey.
• Effective mass:
me
= k1
nm
• Design displacement: d
= k2
n
• Effective period:
Te
= k3
d
= k2
k3
n
• Effective Stiffness: !Ke
= 42me
/Te2
= k4
m/n
• Base Shear: VB
= ke
d
= k4
m/n.k2
n = k5
m
•Thus base shear force is independent of building height (!)
Heff,1
Heff,2
m
m
m
m
m
m
m
m
m
VB1 VB2
d2
d1
(a) Building 1 (b) Building 2 (c) Design Displacements
ENG.PRI.0003.62
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HA HB HC
C A
B
Force-Based Design Displacement-Based Design
Stiffness: prop. to 1/H3
prop. to 1/H
Shear Force: prop. to 1/H3
prop. to 1/H
Moments: prop. to 1/H2
equal
Rebar: prop. to 1/H2
equal
Ductility: equal (!)
prop. to 1/H2
BRIDGE COLUMNS OF UNEQUAL HEIGHT
` F Hc HA HB
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CONCLUSIONNew Zealand needs a measured and rational response to the Christchurch earthquake experience. A knee-
jerk band-aid repair to the existing force-based design approach is, in my view, inadequate.
Displacement-based design codes have been developed, and should be considered as a preferred alternative for NZ design.
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