DISPLACEMENT-BASED SEISMIC DESIGN

ISRAELI STRUCTURAL ENGINEERING CONFERENCE
TEL AVIV, JUNE 2009
DISPLACEMENT-BASED SEISMIC
DESIGN
Nigel Priestley
European School for Advanced Studies in
Reduction of Seismic Risk, Pavia, Italy
1

Review of a 15 year research effort
culminating in the preparation of a design text
book:
DISPLACEMENT-BASED SEISMIC
DESIGN OF STRUCTURES
Priestley, Calvi and Kowalsky
IUSS Press (May 2007)
Available online (www.iusspress.it)
2

WITHIN A PERFORMANCE-BASED
ENVIRONMENT:
• WHAT SHOULD BE THE STRUCTURAL
STRENGTH (BASE SHEAR FORCE)
• HOW SHOULD THIS STRENGTH BE
DISTRIBUTED
• HOW CAN DESIGN BE ELEVATED
ABOVE ANALYSIS
• WHAT SHOULD BE THE STRENGTH
OF CAPACITY-PROTECTED ACTIONS
3

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 or Force-Reduction Factors
• Assumption that Increased Strength Reduces
potential for damage
Note: Damage is strain or drift (not strength)
related
4

Concrete Bridge under Longitudinal Seismic Excitation
Spectral Acceleration S A(T)
H A H
H C
B
C
A
B
Structure
Elastic
Ductile
Period T (seconds)
Acceleration Response Spectrum
K
long
Stiffness:
= K + K
A
Period:
T = 2π m /
F
i
B
+
K long
K
C
=
Base shear force:
=
F =
m ⋅
g ⋅ S
R
A ( T )
Pier Shear Force:
F
K
K
i
long
Design Displacement:
2
T
∆
long
= ⋅ S
2 A(
T )
4π
⋅ g
∑
A,
B,
C
5
12EI
H
3

T = 2π m /
K long
Concrete Bridge Under Longitudinal Response
H A H B
H C
C
A
Stiffness:
K
long
B
= K
A
+
K
B
+
K
C
=
∑
A,
B,
C
12EI
H
3
Period:
T = 2π m /
K long
Pier Strength:
F =
i
F
K
K
i
long
What value for EI What force-reduction
factor Is the strength distribution logical
6

STIFFNESS OF CONCRETE ELEMENTS:
Stiffness is assumed to be independent of strength,
and a fixed percentage (often 100% or 50%) of gross
(uncracked) section stiffness
7

Elastic Stiffness
40000
EI=M y /φ y
50000
40000
Nu/f'cAg = 0.4
Nu/f'cAg = 0.3
Nu/f'cAg = 0.2
Nu/f'cAg = 0.1
Moment (kNm)
30000
20000
Nu/f'cAg = 0.4
Nu/f'cAg = 0.3
Nu/f'cAg = 0.2
Nu/f'cAg = 0.1
Nu/f'cAg = 0
Moment (kNm)
30000
20000
Nu/f'cAg = 0
10000
10000
0
0
0 0.002 0.004 0.006
Curvature (1/m)
(a) Reinforcement Ratio = 1%
0 0.002 0.004 0.006
Curvature (1/m)
(b) Reinforcement Ratio = 3%
SELECTED MOMENT-CURVATURE CURVES FOR
CIRCULAR COLUMNS (D=2m,f’ c = 35MPa, f y = 450MPa)
8

Dimensionless Moment (M N
/f'cD 3 )
0.2
0.16
0.12
0.08
0.04
0
ρ l
= 0.04
ρ l
= 0.03
ρ l
= 0.02
ρ l
= 0.01
ρ l = 0.005
Dimensionless Curvature (φ y
D/ε y
)
2.5
2
1.5
1
0.5
0
Ave. +10%
ρ l
= 0.04
Ave. -10%
ρ l
= 0.005
Average φ y
D/ε y
= 2.25
Average φ y
D/ε y
= 2.25
0 0.1 0.2 0.3 0.4
0 0.1 0.2 0.3 0.4
Axial Load Ratio (N u
/f'cA g
)
Axial Load Ratio (N u
/f' c
A g
)
(a) Nominal Moment
(b) Yield Curvature
DIMENSIONLESS NOMINAL MOMENT AND YIELD
CURVATURE FOR CIRCULAR COLUMNS
STRENGTH VARIES; YIELD CURVATURE IS CONSTANT
9

1
0.90
0.8
ρ l
= 0.04
Stiffness Ratio (EI/EI gross
)
0.6
0.4
ρ l
= 0.03
ρ l
= 0.02
ρ l
= 0.01
ρ l
= 0.005
EI eff = M N /φ y
EI eff /EI gross
=M N /φ y EI gross
0.2
0
0.12
0 0.1 0.2 0.3 0.4
Axial Load Ratio (N u
/f' c
A g
)
EFFECTIVE STIFFNESS RATIO FOR
CIRCULAR COLUMNS
10

INTERDEPENDENCY OF STRENGTH AND STIFFNESS
Stiffness EI = M/φ
M
M 1
M
M 1
M 2
M 2
M 3
M 3
φ y3
φ y2
φ y1
φ φ y
φ
(a) Design assumption
(constant stiffness)
(b) Realistic assumption
(constant yield curvature)
INFLUENCE OF STRENGTH ON MOMENT-CURVATURE RESPONSE
11

Bridge Under Longitudinal Response: Elastic Strength
Distribution
Force-Based Design: Column Stiffness
Force-Based Design: Column Moments:
K = C EI
i
M = C V h =
Bi
h
3
1 i.
eff
/
i
C C
EI
h
2
2 i i 1 2 i.
eff
/
i
Short columns need highest long.rebar: stiffness increases: higher moment! 12

BRIDGE WITH UNEQUAL COLUMN HEIGHTS
• Shortest piers are stiffest. Shear is in proportion to 1/h 3 .
Therefore moments (and, approximately reinforcement ratio)
are in proportion to 1/h 2 . 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)
13

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
14

CONCRETE FRAME DRIFT EQUATION
θ y = 0.5ε y (l b /h b )
Equation Checked against test data with:
• Column height/beam length aspect ratio (l c /l b ) : 0.4 – 0.86
• Concrete compression strength (f’ c ) : 22.5MPa – 88MPa
• Beam Reinforcement yield strength (f y ) : 276MPa – 611MPa
• Maximum beam reinforcement ratio (A’ s /b w d) : 0.53% – 3.9%
• Column axial load ratio (N u /f’ c A g ) : 0 – 0.483
• Beam aspect ratio (l b /h b ) : 5.4 – 12.6
15

θ y = 0.5ε y (l b /h b )
EXPERIMENTAL DRIFTS OF BEAM/COLUMN
TEST UNITS COMPARED WITH EQUATION
16

DIMENSIONLESS YIELD CURVATURES
AND DRIFTS
• Circular column: φ = 2.25ε
D
y y
/
• Rectangular column: φ
y
= 2.10ε
y
/ hc
• Rectangular cantilever walls: φ
y
= 2.00ε
y
/ lw
• T-Section Beams: φ
y
= 1.70ε
y
/ hb
Concrete Frames:
Steel Frames:
θ = 0. 5ε
y
y
θ = 0. 65ε
y
y
l
h
b
b
l
h
b
b
17

FORCE-REDUCTION FACTORS IN DIFFERENT
COUNTRIES
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 **S P factor of 0.67 incorporated.
18

FRAME DUCTILITY LIMITS
NOTE: CODES LIMIT DRIFT TO 0.02-0.025
Yield Drift:
θ = 0. 5ε
Example: l b = 6m, h b =0.6m, f y = 500MPa
ε y = 500/200,000 = 0.0025;
y
y
l
h
b
b
θ 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!
19

2. DIRECT DISPLACEMENT -
BASED SEISMIC DESIGN
A RATIONAL SEISMIC DESIGN
APPROACH
20

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 (nonstructural
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
21

ELASTIC RESPONSE OF STRUCTURES
1
0.5
Acceleration a(T) (g)
0.8
0.6
0.4
0.2
Force-based design
Displacement ∆(T) (m)
0.4
0.3
0.2
0.1
DDBD
0
0 1 2 3 4 5
Period T (seconds)
(a) Acceleration Spectrum for 5% damping
0
0 1 2 3 4 5
Period T (seconds)
(b Displacement Spectrum for 5% damping
Elastic, Force Based: F = m.a(T).g; D = F/K
Elastic, Displ. Based: F = K.D(T)
22

COMPARISON OF ELASTIC FORCE AND
DISPLACEMENT-BASED DESIGN
We require to know force (F) and displacement (∆)
FORCE BASED: F =m.a (T) .g
Calculate K,T,a (T) , F,∆
∆= F/K
(5 steps)
DISPLACEMENT BASED: F = K.∆ (T)
Calculate K,T,∆ (T) ,F
(4 steps)
23

FUNDAMENTALS OF DDBD
F
m e
h e
F u
F n
K e
K i
∆y ∆d
(a) SDOF Simulation (b) Effective Stiffness K e
Damping (%)
0.25
0.2
0.15
0.1
0.05
←
Elasto-Plastic
Steel Frame
Concrete Frame
Concrete Bridge
Hybrid Prestress
Displacement (m)
0.5
0.4
0.3
0.2
0.1
∆ d
→
5%
10%
15%
20%
30%
0
0 2 4 6
Displacement Ductility
↑
(c) Equivalent damping vs. ductility
0
↓
T e
0 1 2 3 4 5
Period (seconds)
(d) Design Displacement Spectra
24

Damping (%)
0.25
0.2
0.15
0.1
0.05
←
Elasto-Plastic
Steel Frame
Concrete Frame
Concrete Bridge
Hybrid Prestress
Displacement (m)
0.5
0.4
0.3
0.2
0.1
∆ d
→
5%
10%
15%
20%
30%
0
↑
0
↓
T e
0 2 4 6
Displacement Ductility
0 1 2 3 4 5
Period (seconds)
(c) Equivalent damping vs. ductility
(d) Design Displacement Spectra
Limit State Displacement; Ductility Damping
Damping+Displacement T e ; K=4π 2 m e /T e
2
25

m e F u
F F n rK i
H e K i K e
Limit State Displacement; Ductility Damping
Damping+Displacement Te; K=4π 2 me/T e
2
∆ y ∆ d
F u = K e ∆ d
26

ELEMENTS OF DDBD: Multi-storey Building
Effective Stiffness:
K
e
=
2 2
4π
me
/ Te
Base Shear Force:
F
=
V Base
= K e
∆
d
MDOF Design Displacement:
∆
d
=
n
∑
i=
1
n
( m ∆
2 ) ∑( ∆ )
i i
/ mi
i
i=
1
Effective Mass:
m
n
= ∑
e
m i
i=
1
( ∆ )
i
/ ∆d
Effective Height:
H
e
=
n
∑
i=
1
( m ) ∑
i∆iH
i
/ ( mi∆i
)
n
i=
1
27

3. EQUIVALENT VISCOUS DAMPING
Relationships between displacement ductility and
equivalent viscous damping have been determined by
extensive inelastic time-history analyses, using two
different approaches:
• “Real” accelerograms, matching displacement of
equivalent elastic system to inelastic displacement,
on a record-by-record approach, then averaging the
EVD over a very large number of records.
• Spectrum compatible accelerograms averaging the
inelastic response, and matching average equivalent
elastic displacement.
Results were nearly identical
28

F F y
βF y
BASIC HYSTERESIS RULES CONSIDERED
F
F
rk i
k i k i
∆
∆
(a) Elasto-plastic (EPP)
(b) Bi-linear, r = 0.2 (BI)
F
k i
k i µ −α
∆
F
k i
∆
k i µ −α
(c) Takeda “Thin” (TT) (d) Takeda “Fat” (TF)
k i
∆
k i
∆
(e) Ramberg-Osgood (RO) (f) Flag Shaped (FS)
29

RELATIONSHIPS FOR TANGENT-STIFFNESS
DAMPING
Concrete Wall Building, Bridges (TT):
Concrete Frame Building (TF):
Steel Frame Building (RO):
Hybrid Prestressed Frame (FS,β=0.35):
Friction Slider (EPP):
Bilinear Isolation System (BI, r=0.2):
ξ
ξ
ξ
ξ
ξ
ξ
eq
eq
eq
eq
eq
eq
⎛ µ −1⎞
= 0.05 + 0.444⎜
⎟
⎝ µπ ⎠
⎛ µ −1⎞
= 0.05 + 0.565⎜
⎟
⎝ µπ ⎠
⎛ µ −1⎞
= 0.05 + 0.577⎜
⎟
⎝ µπ ⎠
⎛ µ −1⎞
= 0.05 + 0.186⎜
⎟
⎝ µπ ⎠
⎛ µ −1⎞
= 0.05 + 0.670⎜
⎟
⎝ µπ ⎠
⎛ µ −1⎞
= 0.05 + 0.519⎜
⎟
⎝ µπ ⎠
30

4
GENERAL FORM OF ELASTIC 5% DISPLACEMENT
SPECTRUM, FROM EC8
∆ max
Plateau
∆ max
T C Period T D T E
∆ PG
Displacement
Corner
Period
Linear
∆ PG
T C Period T D T E
31

.DESIGN DISPLACEMENT SPECTRA (1)
• Can be approximately generated from design
acceleration spectra (5% damping) using accelerationdisplacement
relationships:
∆ (T,5) = (T 2 /4π 2 ).g.a (T,5)
• Values for damping other than 5% can be generated
from relationships such as:
∆
T , ξ
= ∆
T ,5
⋅
⎛
⎜
⎝
2
7 ⎞ ⎟⎠
+ ξ
0.5
“Normal” records, EC8 1998
∆
T , ξ
= ∆
T ,5
⋅
⎛
⎜
⎝
2
7 ⎞ ⎟⎠
+ ξ
0.25
“Velocity pulse” records (Forward
directionality) (tentative!)
32

DESIGN DISPLACEMENT SPECTRA (4)
• Based on Faccioli’s observations,the corner period T c
appears to increase almost linearly with moment
magnitude. For earthquakes with M W > 5.7, the
following expression seems conservative:
( 5.7)
T 1.0
+ 2.5 M −
c
=
w
seconds
• Peak displacement at the corner period can be
estimated from the following expression (firm ground):
δ
max
=
10
( M −3.2)
W
r
mm
r = nearest
distance to fault
plane (km)
33

Spectral Displacement (mm)
2000
1600
1200
800
400
0
M = 7.5
M = 7.0
M = 6.5
M = 6.0
0 2 4 6 8 10
Period (seconds)
5% Damped Spectra Resulting from the
Equations, at r = 10km
34

Spectral Displacement (mm)
2000
1600
1200
800
400
0
M = 7.5
M = 7.0
M = 6.5
0 2 4 6 8 10
Period (seconds)
5% Damped spectra, Resulting from the
Equations, at r = 20 km
35

Spectral Displacement (mm)
2000
1600
1200
800
400
0
M = 7.5
M = 7.0
M = 6.5
0 2 4 6 8 10
Period (seconds)
5% Damped spectra Resulting from the
Equations, at r = 40 km
36

INFLUENCE OF DAMPING AND DUCTILITY ON
SPECTRAL DISPLACEMENT RESPONSE
Current EC8:
1998 EC8:
Newmark and Hall 1987:
∆
∆
∆
T , ξ
T , ξ
=
∆
= ∆
T ,5
T ,5
⋅
⋅
⎛
⎜
⎝
⎛
⎜
⎝
10 ⎞ ⎟⎠
5 + ξ
2
7 ⎞ ⎟⎠
+ ξ
0.5
0.5
( 1.31 ln )
.ξ
ξ
= ∆T
,5
T ,
−
Independent analyses by Priestley et al, Kowalsky et al, Sullivan et
al, support the 1998 EC8, not the 2003 Eqn,nor Newmark and Hall.
Tentative Expression
for Forward Directivity,
Near field
∆
T , ξ
= ∆
T ,5
⋅
⎛
⎜
⎝
2
7 ⎞ ⎟⎠
+ ξ
0.25
37

Recommended Spectral correction for Damping
Relative Displacement ∆ T,ξ
/∆ 4,5%
1
0.8
0.6
0.4
0.2
0
(Eq.(2.8))
ξ= 0.05
ξ= 0.10
ξ= 0.15
ξ= 0.20
ξ= 0.30
Relative Displacement ∆ T,ξ
/∆ 4,5%
1
0.8
0.6
0.4
0.2
0
(Eq.(2.11))
ξ= 0.05
ξ= 0.10
ξ= 0.15
ξ= 0.20
ξ= 0.30
0 1 2 3 4 5
Period T (seconds)
(a) "Normal" Conditions
0 1 2 3 4 5
Period T (seconds)
(b) Velocity Pulse Conditions
K e =4π 2 m e /T e
2
; V base =K e ∆ d
38

SCOPE OF RESEARCH AND BOOK
The research study (and the text book) considered
the following structural systems:
• Frames (concrete, steel, infilled, braced, precast)
• Structural walls (cantilever and coupled)
• Dual Wall/Frame structures
• Timber structures
• Unreinforced masonry buildings
• Structures with Seismic Isolation and added
damping
• Bridges
• Marginal wharves
39

CHAPTER 4
ANALYSIS TOOLS FOR DIRECT
DISPLACEMENT-BASED DESIGN
Deals with a number of particularly relevant issues:
• Moment-curvature analysis (Cumbia on CD)
• Equilibrium considerations
• Section stiffness
• Shear strength
• Capacity design analyses
• Inelastic time-history analyses (Ruaumoko, SeismoStruct
on CD)
• Pushover analyses (Ruaumoko, SeismoStruct)
40

CHAPTER 5
FRAME BUILDINGS
Considers issues specially relevant to design of frames:
• design displacement profiles
• one way/two way frames
• irregular frames
• elastic response
• combination with gravity-load effects (consider
separately!
• capacity design for columns, and beams
• hybrid precast frames
41

CHAPTER 6
STRUCTURAL WALL BUILDINGS
Issues specially relevant to wall buildings considered:
• Yield displacement and elastically responding walls
• Torsional response
• Capacity design effects
• Foundation rotational stiffness
• Coupled wall design (based on equilibrium, not elastic
stiffness)
42

CHAPTER 7
DUAL WALL/FRAME BUILDINGS
(a) Plan View
(b) Long Direction Model
Issues: How to distribute lateral
force between walls and frames
(a designer’s choice –not based
on elastic stiffness)
How to distribute shear up the
frame (also a designer’s choice)
Analysis for design requires no
information about relative
stiffnesses, which are, in fact
irrelevant.
(c) Short Direction Model
43

CHAPTER 8
MASONRY BUILDINGS
This chapter essentially deals with unreinforced, or very
lightly reinforced buildings, such as are used extensively in
Europe and Central and South America. Displacementbased
design is a much more rational way of designing such
structures than current force-based approaches. Coupling
between walls by floor slabs and spandrels is rationally
considered with an innovative design approach
44

Chapter 9: DUCTILE TIMBER STRUCTURES
F
Joint
Framing
Shear
Panel
End View
(a) Ductile Moment-Resisting Frame (b) Timber Shear Panel Bracing
Ductile
connectors
Post-tensioning
P
P
(c) Post-tensioned Timber Frame
45

CHAPTER 10
BRIDGES
Consideration of aspects specially applicable to bridges:
• Longitudinal/transverse design
• Elastically responding bridges
• Influence of bearings and foundation flexibility
• P-∆ Effects
• Different foundation conditions
• Capacity design (Effective modal superposition)
46

Isolator
Isolated rigid
structure
Braced frame
with added
damping
Isolated bridge
Partially isolated bridge
Foundation
sliding on gravel
layer
Pier
Gravel
layer Soil
Concrete
Chapter 11
ISOLATION
Deck
Rocking bridge bent
47

CHAPTER 12
WHARVES AND PIERS
Design is dominated by soil/structure interaction effects
Displacement-based design, to specified limit-strains in
piles (two-level design)
Torsional response is critical
This chapter is the basis of the POLA seismic design code
48

INELASTIC FORCE-DISPLACEMENT RESPONSE
FROM PUSHOVER ANALYSIS
Wharf Deck
F E D C B A
T F T E T D T C T B T A
F L
L Cl
Ground
L sp
IG F
IG E
IG D
IG C
IG B
IG A
Piles
Inelastic pile
members
Inelastic Soil
Springs
x F
x E
x D
x C
x B
x A
49

CHAPTER 13
DISPLACEMENT-BASED SEISMIC ASSESSMENT
The principles of displacement-based design are adapted to
assessment of existing structures. Both buildings and
bridges are considered in developing the topic, and in the
design examples
50

MDOF PLASTIC MECHANISMS
5
4
3
2
1
Height
Limit state
∆ P,c
Yield
Plastic mechanism
Displacement
(a) Beam Sway, S i < 0.85, all levels (b) Beam Sway Displacement Profiles
5
4
3
2
Height
Yield
Column sway
at Level 1
Column sway
at Level 2
1
∆ P
Plastic mechanism
Displacement
(c) Column Sway, S i > 1, Levels 1 or 2 (d) Column Sway Displacement
51

CHAPTER 14
DRAFT DISPLACEMENT-BASED CODE FOR
SEISMIC DESIGN OF BUILDINGS
Intended as a starting point for development of design
codes.
52

DIRECT DISPLACEMEMENT-BASED DESIGN
DESIGN DISPLACEMENT
Damage is strain dependent. (e.g. onset of concrete
crushing: strain = 0.004, residual crack widths require
grouting if peak reinforcement strain exceeds 0.015).
From strains we can calculate curvature. We can
integrate curvature to find displacement.
Thus we can directly relate structural damage to
displacement
Non-structural damage can generally be related to
drift (Drift = interstorey displacement divided by
interstorey height). Drift can be integrated to find
displacement.
Thus we can relate non-structural damage to
displacement
53

EXAMPLE OF STRAIN LIMIT STATES
Curvature from concrete
compression:
φ mc = ε cm /c
Curvature from reinforcement
tension:
φ ms = ε sm /(d-c)
Chose lesser of φ mc and φ ms ,
Design Displacement is:
∆ ds = ∆ y + ∆ p
= φ y H 2 /3 + (φ m -φ y )L p H
L p = plastic hinge length.
54

CRITICAL DISPLACEMENTS
•Frames: normally governed by structural or nonstructural
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
55

FRAMES
1.0
n=4:
δ
i
=
H /
i
H
n
H/H n
n>4:
δ =
i
4
3
⎛
⋅
⎜
⎝
H
H
i
n
⎞ ⎛ ⎟ ⋅ ⎜1
−
⎠ ⎝
Hi
4H
n
⎞
⎟
⎠
Mode displ. 1.0
(Based on results of inelastic timehistory
analysis. Note, in some cases
it may be necessary to review the
displacement profile after initial
analysis)
56

CANTILEVER WALL DESIGN PROFILES:
If roof drift is less than the code drift limit θ c , the
design displacement profile is:
∆
i
= ∆
yi
+ ∆
pi
=
ε y ⎛ H ⎞ ⎛
i
y
Hi
m
lw
H
+ ε
⎜ − ⎟
⎜φ
−
2
2
1
⎝ 3
n ⎠ ⎝ lw
⎟ ⎞
L
⎠
p
H
i
If the code drift limit governs the roof drift, the design
displacement profile is:
∆
i
= ∆
yi
+
( − )
y
i
y n
H =
2
θ θ H ⎜1
⎟ ⎜θc
− Hi
c
yn
i
ε
l
w
i
⎛
⎜ −
⎝
H
3H
n
⎞ ⎛
⎟ + ⎜
⎠ ⎝
ε H
l
w
⎟ ⎞
⎠
57

BRIDGE CHARACTERISTIC MODE SHAPES
∆
∆ 1
∆
∆ 2
∆ 3
∆ 4
∆ 5
∆ 1
∆ 2
∆
∆ 3
∆
∆
(a) Symm., Free abuts. (b) Asymm., Free abuts. (c) Symm., free abuts.
Rigid SS translation Rigid SS translation+rotation Flexible SS
∆ 1 ∆ 1 ∆ 1
∆ 2 ∆ 2 ∆ 2
∆ 3 ∆ 3 ∆ 3
∆ 4
∆ 4 ∆ 4 ∆ 4
∆ 5
∆ 5
∆ 5
∆ 5
(d) Symm,. Restrained abuts. (e) Internal movement joint (f) Free abuts., M.joint
Flexible SS Rigid SS, Restrained abuts. Flexible SS
58

ELASTIC RESPONSE OF FRAMES
The shape of the displacement spectrum, with a peak displacement
plateau, implies that in regions of moderate seismicity, tall frames
will respond elastically
Assuming start of the constant velocity slope of acceleration
spectrum at T = 0.5 sec, and peak spectral acceleration = 2.5
times PGA, then the plateau displacement can be written as:
∆
TC
=
0.031×
PGA(1
+ 2.5( M −1))
W
Taking a typical frame of n storeys with beam length/depth =10,
f ye = 400MPa the equivalent yield displacement can be approximated
as
∆
y
= θ
y.( 0.7×
3.5n)
= 0.01×
0.7×
3.5n
= 0. 0245n
59

1.6
Displacement (m)
1.2
0.8
0.4
0
Yield
displacement
Peak elastic spectral
Displacement
0.7g
0.6g
0.5g
0.4g
0.3g
0.2g
0.1g
24 storeys
20 storeys
16 stories
12 storeys
8 storeys
4 storeys
6 6.4 6.8 7.2 7.6 8
Moment Magnitude, M W
60

DRIFT-LIMITED DUCTILITY FOR WALLS
Maximum Design Ductility (µ)
10
8
6
4
2
0
θ c =0.02
Elastic
θ c =0.025
2 4 6 8 10
Wall Aspect Ratio (A r =H n /l w )
Maximum Design Ductility (µ)
10
8
6
4
2
0
θ c =0.02
Elastic
θ c =0.02
2 4
Wall Aspec
(a) Yield Drift = ε y A r (b) Yield Drift =
61

ANALYSIS OF FRAMES UNDER LATERAL FORCES
F 4 V B4 V B4
Level 4
Base Overturning Moment
F 2
F 3 V B3 V B3
V B2 V B2
Level 3
Level 2
OTM
OTM
OTM
∑
= Fi
⋅ H
i
= T ⋅ Lb
+
= V ⋅ L
∑
Bi
∑
base
M
+
Cj
∑
M
Cj
V B1 V B1
F 1 Level 1
V C1 V C2 V C3
0.6H 1
M C1 M C2 M C3
T C
L base
Level 0
H 3
Lateral force distribution
F
i
=
V
B
( m ∆ ) ∑( ∆ )
i i
/ mi
i
n
i=
1
62

PLASTIC DISPLACEMENT OF CANTILEVER WALLS
l W
( )
Plastic
Hinge
H e
∆ p
L P
L SP
θ p
0.5L P -L SP
P
φ p φ y
(a) Wall (b) Curvature (c) Plastic Displacement
P
P
( H − 0.5L
− L ) = θ ( H − ( 0. L − L ))
∆ = φ L
5
∆
P
P
P
e
= φ L H = θ H
P
e
P
P
e
e
SP
63
plastic hinge length: L = k ⋅ H + 0. 1l
+ L k = 0.2(fu/fy – 1) =0.08
W
P
SP
e
P
SP
(acceptable approximation)

LIMIT STATE CURVATURES FOR WALLS:
From analyses:
• Serviceability curvature:
• Damage Control curvature:
φ s
⋅l
w
= 0.0175
φ dc
⋅l
w
= 0.072
In both cases steel strain limit dominates. A general equation can be
written:
• General limit curvature:
φ ⋅ =1.2ε
ls
l
w s,
ls
Design Displacement
profile is thus:
ε H
2
y i ⎛ ∆
i = ∆
yi + ∆
pi =
l
⎜1
−
w
3
⎝
H
i
H
n
⎞
⎟ + θ
pH
⎠
i
64

TORSIONAL RESPONSE OF WALL BUILDINGS WITH
STRENGTH AND STIFFNESS ECCENTRICITY
Z
Wall 4
k trans
L Z =15m
k Z1,
C R
C V
Wall C M
Wall 2
V 1
e VX Strength V2
e RX
V
eccentricity
Stiffness
B
k
eccentricity
trans
Wall 3
L X = 25 m
k Z2
X
Both strength and stiffness eccentricity affect torsional response.
In DDBD the design displacement at centre of mass is reduced to
account for torsional response
65

COUPLED WALLS -TYPICAL ELEVATION
F i
H i
Gravity
Frame
1 2
Coupled
Walls
(a) Structure and Lateral Forces
66

COUPLING BEAM DRIFTS
reinforcement
Conventional
θ W
θ CB
0.5l W L CB 0.5l W
Diagonal
reinforcement
α
α
(a) Drift Geometry (b) Coupling Beam
θ
CB
= θ
W
( 1+ l / L )
W
CB
67

CAPACITY DESIGN
• Overstrength: Material strengths will generally
exceed dependable strengths used for design;
confinement and strain-hardening further
increase flexural strength above design strength
• Higher mode effects: Either not considered
(equivalent lateral force approach), or incorrectly
considered (elastic modal superposition) in current
design.
68

STRUCTURAL WALL
FAILURE
Inadequate transverse
reinforcement in plastic
hinge region
69
KOBE, 1995

INADEQUATE CONSIDER-
ATION OF HIGHER MODE
EFFECTS
Hinge forms above base
70
VALPARAISO, 1985

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 capacityprotected
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.
71

EFFECTIVE MODAL SUPERPOSITION FOR
HIGHER MODE EFFECTS
A modal analysis procedure combining the inelastic
first mode and ELASTIC higher modes by a SRSS
approach provides good simulation of higher mode
effects. The agreement is improved if the stiffness
of plastic hinges is reduced to reflect the design
ductility
i.e. use EI/µ for members with plastic hinges.
72

60
30
40
MMS THA MMS THA MMS THA
Height (m)
20
10
Height (m)
30
20
10
Height (m)
40
20
0
0 500 1000 1500 2000 2500 3000
0
0
0 1000 2000 3000 4000
0 1000 2000 3000 4000
Shear Force (kN)
Shear Force (kN)
Shear Force (kN)
(m) Twelve-Storey Wall, IR=0.5 (q) Sixteen-Storey Wall, IR=0.5 (u) Twenty-Storey Wall, IR=0.5
60
30
40
Height (m)
20
10
MMS THA
Height (m)
30
20
10
THA MMS
Height (m)
40
20
THA MMS
0
0 500 1000 1500 2000 2500 3000
0
0
0 1000 2000 3000 4000
0 1000 2000 3000 4000
Shear Force (kN)
Shear Force (kN)
Shear Force (kN)
(n) Twelve-Storey Wall, IR=1.0 (r) Sixteen-Storey Wall, IR=1.0 (v) Twenty-Storey Wall, IR=1.0
60
MMS Shear Envelopes Compared with Time History Results
30
40
for Different Seismic Intensity Ratios (IR=1= Design)
ht (m)
20
ht (m)
30
ht (m)
40
83 73

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; 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.
74

INFLUENCE OF SEISMIC INTENSITY ON DESIGN
FORCES
Force-Based Design Displacement-Based Design
V
Z
e 2
=
e1
, but
2
b 2
= V
Z2
b1
Z
Hence: K
1
e
If Z 2
= 0.5Z 1
, V b2
= 0.5V b1
Hence:
T
T
Z
1
V
=
K
K
e
⎛ Z
⎜
=
⎞
2
2 1 ⎟ e ⎜
⎝ Z1
⎠
2
= V
⎛ Z
⎜
⎝
⎞
2
2 1 ⎟ B B ⎜ Z1
⎠
2 2
4π
me
/ Te
2
: V 75 b2
= 0.25V b1

BRIDGE COLUMNS OF UNEQUAL HEIGHT
`
F H c
H A H B H C
H A
C
H B
A
B
Force-Based Design Displacement-Based Design
∆
Stiffness: prop. to 1/H 3 prop. to 1/H
Shear Force: prop. to 1/H 3 prop. to 1/H
Moments: prop. to 1/H 2
Rebar: prop. to 1/H 2
equal
equal
Ductility: equal (!) prop. to 1/H 2
76

CONCLUSIONS
DOES DDBD MAKE A DIFFERENCE
YES!
Thank you
77