Seismic Design and Detailing of
Reinforced Concrete Structures
Based on CSA A23.3 - 2004
Murat Saatcioglu PhD,P.Eng.
Professor and University Research Chair
Department of Civil Engineering
The University of Ottawa
Ottawa, ON
Basic Principles of Design
Reinforced concrete structures are designed to
dissipate seismic induced energy through
inelastic deformations
Ve
Ve = S(Ta) Mv IE W / (Rd Ro)
Ve /Rd
Ve /Rd Ro
D
Basic Principles of Design
Inelasticity results softening in the structure,
elongating structural period
S(T)
S1
S2
T1
T2
T
Basic Principles of Design
Capacity  Demand
It is a good practice to reduce seismic
demands, to the extent possible….
This can be done at the conceptual stage
by selecting a suitable structural system.
To reduce seismic demands…
 Select a suitable site with favorable soil conditions
 Avoid using unnecessary mass
 Use a simple structural layout with minimum
torsional effects
 Avoid strength and stiffness taper along the height
 Avoid soft storeys
 Provide sufficient lateral bracing and drift control by
using concrete structural walls
 Isolate non-structural elements
Seismic Amplification due to Soft Soil
Liquefaction
Liquefaction
Liquefaction
To reduce seismic demands…
 Select a suitable site with favorable soil conditions
 Avoid using unnecessary mass
 Use a simple structural layout with minimum
torsional effects
 Avoid strength and stiffness taper along the height
 Avoid soft storeys
 Provide sufficient lateral bracing and drift control by
using concrete structural walls
 Isolate non-structural elements
Use of Unnecessary Mass
Use of Unnecessary Mass
Use of Unnecessary Mass
Use of Unnecessary Mass
To reduce seismic demands…
 Select a suitable site with favorable soil conditions
 Avoid using unnecessary mass
 Use a simple structural layout with minimum
torsional effects
 Avoid strength and stiffness taper along the height
 Avoid soft storeys
 Provide sufficient lateral bracing and drift control by
using concrete structural walls
 Isolate non-structural elements
Effect of Torsion
Effect of Torsion
Effect of Torsion
Effect of Torsion
Effect of Torsion
Effect of Torsion
Effect of Torsion
Effect of Torsion
To reduce seismic demands…
 Select a suitable site with favorable soil conditions
 Avoid using unnecessary mass
 Use a simple structural layout with minimum
torsional effects
 Avoid strength and stiffness taper along the height
 Avoid soft storeys
 Provide sufficient lateral bracing and drift control by
using concrete structural walls
 Isolate non-structural elements
Effect of Vertical Discontinuity
Effect of Vertical Discontinuity
To reduce seismic demands…
 Select a suitable site with favorable soil conditions
 Avoid using unnecessary mass
 Use a simple structural layout with minimum
torsional effects
 Avoid strength and stiffness taper along the height
 Avoid soft storeys
 Provide sufficient lateral bracing and drift control by
using concrete structural walls
 Isolate non-structural elements
Effect of Soft Storey
Effect of Soft Storey
Effect of Soft Storey
Effect of Soft Storey
To reduce seismic demands…
 Select a suitable site with favorable soil conditions
 Avoid using unnecessary mass
 Use a simple structural layout with minimum
torsional effects
 Avoid strength and stiffness taper along the height
 Avoid soft storeys
 Provide sufficient lateral bracing and drift control by
using concrete structural walls
 Isolate non-structural elements
R/C Frame Buildings without Drift Control
Buildings Stiffened by Structural Walls
To reduce seismic demands…
 Select a suitable site with favorable soil conditions
 Avoid using unnecessary mass
 Use a simple structural layout with minimum
torsional effects
 Avoid strength and stiffness taper along the height
 Avoid soft storeys
 Provide sufficient lateral bracing and drift control by
using concrete structural walls
 Isolate non-structural elements
Short Column Effect
Short Column Effect
Seismic Design Requirements of
CSA A23.3 - 2004
Capacity design is employed…..
Selected elements are designed to yield
while critical elements remain elastic
Design for
Strength and Deformability
Load Combinations
Principal loads:
1.0D + 1.0E
And either of the following:
1) For storage occupancies, equipment areas and
service rooms:
1.0D + 1.0E + 1.0L + 0.25S
2) For other occupancies:
1.0D + 1.0E + 0.5L + 0.25S
Stiffness Properties for Analysis
 Concrete cracks under own weight of
structure
 If concrete is not cracked, then the structure
is not reinforced concrete (plain concrete)
 Hence it is important to account for the
softening of structures due to cracking
 Correct assessment of effective member
stiffness is essential for improved accuracy
in establishing the distribution of design
forces among members, as well as in
computing the period of the structure.
Flexural
Moment
Behaviour of R/C
Mn
My
Post-yield rigidity
Post-cracking rigidity
Mcr
Elastic rigidity
Actual
Idealized
Curvature
Flexural
Behaviour
of
R/C
Moment
Mn
0.75Mn
Effective elastic rigidity
Actual
Idealized
(bi-linear)
y
u
Curvature
Section Properties for Analysis as
per CSA A23.3-04
Beams
P
α  0.5 0.6
Columns
f A
Coupling Beams
c
'
c
 1.0
s
g
without diagonal reinforcement
with diagonal reinforcement
Slab-Frame Element
Walls
α w  0.6
Ps
 1.0
'
fc A g
Ie = 0.40 Ig
Ie = a cIg
Ave = 0.15Ag
Ie = 0.40 Ig
Ave = 0.45Ag
Ie = 0.25 Ig
Ie = 0.20 Ig
Axe = awAg
Ie = aw Ig
Seismic Design Requirements of
CSA A23.3 - 2004
Chapter 21 covers:
 Ductile Moment Resisting Frames (MRF)
 Moderately Ductile MRF
 Ductile Shear Walls
 Ductile Coupled Shear Walls
 Ductile Partially Coupled Shear Walls
 Moderately Ductile Shear Walls
Ductile Moment Resisting Frame
Members Subjected to Flexure
Rd = 4.0
Pf ≤ Agf’c /10
c2
h
d
x
y
bw
b w  0.3 h
bw  c2  x  y
bw 250mm
x  3/4 h
n  4 d
y  3/4 h
Beam Longitudinal Reinforcement
-
Mr
+
-
-
-
Mr > 1/4 Mr
-
Mr > 1/2 Mr
Mr
+
-
Mr > 1/2 Mr
+
Mr > 1/4 Mr
Top and Botom 2 bars continuous
Top and Bottom: 1.4bwd / fy ≤ r ≤ 0.025
Beam Transverse Reinforcement
No lap splicing within
this region
2d
s1
c1
s2  d / 2
50 mm
h
Hoops
Sirrups with
seismic hooks
Hoops
n
s1  d / 4
s 1  8(d b ) long .bar
s1  300 mm
s 1  24(d b )hoop
db
Formation of Plastic Hinges
Beam Shear Strength
Plastic Hinge
Wf
M+pr
M pr
ln
(Ve )left
(Ve ) right
-
Ve =
M pr + M+pr +
ln
Wf ln
2
Beam Shear Strength
 The factored shear need not exceed that
obtained from structural analysis under
factored load combinations with RdRo = 1.0
 The values of q = 45o and b = 0 shall be used
in shear design within plastic hinge regions
 The transverse reinforcement shall be
seismic hoops
Ductile Moment Resisting Frame
Members Subjected to Flexure and
Significant Axial Load
Pf > Agf’c /10
Rd = 4.0
hshort
hlong
hshort ≥ 300 mm
hshort / hlong ≥ 0.4
D
D ≥ 300 mm
Longitudinal Reinforcement
Design for factored axial forces
and moments using Interaction
Diagrams
r min = 1%
r max = 6%
Strong Beam-Weak Column Design
Strong Beam-Weak Column Design
Strong Column-Weak Beam Design
Manc
Nominal moment
resistance of columns
under factored axial loads
M rpb
M lpb
Probable moment
resistance of beams
Mbnc
M
nc
  Mpb
Column Confinement
Reinforcement
Columns will be confined for improved
inelastic deformability
lo
lo ≥ 1/6 of clear col. height
If Pf ≤ 0.5 c f’c Ag ;
If Pf > 0.5 c f’c Ag ;
lo ≥ 1.5h
lo ≥ 2.0h
Columns connected to rigid members such as
foundations and discontinuous walls, or columns
at the base will be confined along the entire height
lo
Poorly Confined Columns
Poorly Confined Columns
Well-Confined
Column
Column Confinement Reinforcement
f'c
ρ s  0.4kp
fyh
Circular Spirals
fyh  500MPa
Pf
kp 
Po
Ag
f'c
ρ s  0.45(
 1)
Ac
fyh
Column Confinement Reinforcement
A sh
Ag f'c
 0.2knkp
shc
Ach fyh
A sh
f'c
 0.09 shc
fyh
Rectilinear Ties
Pf
kp 
P
o
fyh  500MPa
n
kn  n /(n  2)
: No. of laterally supported bars
Spacing of Confinement
Reinforcement
 ¼ of minimum member dimension
 6 x smallest long. bar diameter
 sx = 100 + (350 – hx) / 3
Spacing of laterally supported longitudinal
bars, hx ≤ 200 mm or 1/3 hc
M+pr
-
M pr
a
Column Shear
Strength
Mcol
Vcol
a
M
Vcol=
lu
+
M pr
-
M pr
Vcol
b
M col
col
b
+ M col
lu
Column Shear Strength
 The factored shear need not exceed that
obtained from structural analysis under
factored load combinations with RdRo = 1.0
 The values of q ≥ 45o and b ≤ 0.10 shall be
used in shear design in regions where the
confinement reinforcement is needed
 The transverse reinforcement shall be
seismic hoops
Shear Deficient Columns
Shear Deficient Columns
Beam-Column Joints
Poor Joint Performance
Computation of Joint Shear
Ve
As
T2 = 1. 25 As fy
C1 = T1
x
C2 = T2
x
A's
T1 = 1. 25 A's fy
Ve
Vx-x = Ve - T2 - C1
Vx-x ≤ that obtained from frame analysis using RdRo = 1.0
Shear Resistance of Joints
Vj  2.2c f 'c A j
Vj  1.6 c f 'c A j
Vj  1.3 c f 'c A j
Transverse Reinforcement in Joints
 Continue column confinement
reinforcement into the joint
 If the joint is fully confined by four
beams framing from all four sides,
then eliminate every other hoop. At
these locations sx = 150 mm
Design Example
Six-Storey Ductile Moment Resisting Frame in Vancouver
Chapter 11
By D. Mitchell and P. Paultre
Six-Storey Ductile
Moment Resisting
Frame in Vancouver
•Rd = 4.0 and Ro = 1.7
•Site Classification C
(Fa & Fv = 1.0)
Interior columns: 500 x 500 mm
Exterior columns: 450 x 450 mm
Slab: 110 mm thick
Beams (1-3rd floors): 400 x 600 mm
Beams (4-6th floors): 400 x 550 mm
Material Properties
Concrete: normal density concrete with 30 MPa
Reinforcement: 400 MPa
Live loads
Floor live loads:
2.4 kN/m2 on typical office floors
4.8 kN/m2 on 6 m wide corridor bay
Roof load
2.2 kN/m2 snow load, accounting for parapets
and equipment projections
1.6 kN/m2 mechanical services loading in 6 m
wide strip over corridor bay
Dead loads
self-weight of reinforced concrete members
calculated as 24 kN/m3
1.0 kN/m2 partition loading on all floors
0.5 kN/m2 mechanical services loading on all
floors
0.5 kN/m2 roofing
Wind loading
1.84 kN/m2 net lateral pressure for top 4 storeys
1.75 kN/m2 net lateral pressure for bottom 2
storeys
The fire-resistance rating of the building is
assumed to be 1 hour.
Gravity Loading
Design Spectral Response
Acceleration E-W Direction
Empirical: Ta = 0.075 (hn)3/4 = 0.76 s
Dynamic: T = 1.35 s but not greater than 1.5Ta = 1.14s
Design of Ductile Beam
Design of Ductile Beam
Design of Ductile Beam
Design of Ductile Beam
Design of Ductile Beam
Design of Ductile Beam
Design of Ductile Interior Column
Design of Ductile Interior Column
Design of Ductile Interior Column
Design of Ductile Interior Column
Design of Ductile Interior Column
Design of Ductile Interior Column
Design of Interior Beam-Column Joint
Design of Interior Beam-Column Joint
Design of Interior Beam-Column Joint
Ductile Shear Walls
Rd = 3.5 or 4.0
if hw / ℓw ≤ 2.0; Rd = 2.0
SFRS without irregularities:
Plastic hinge length:1.5 ℓw
 Flexural and shear reinforcement
required for the critical section
will be maintained within the
hinging region
hw
Plastic
Hinge
Length
ℓw
 For elevations above the plastic
hinge region, design values will be
increased by Mr/Mf at the top of
hinging region
Ductile Shear Walls
Wall thickness in the plastic hinge:
tw ≥ ℓu / 14 but may be limited to
hw
ℓu / 10 in high compression regions
ℓu
ℓw
tw
Plastic
Hinge
Length Because walls are relatively thin
members, care must be taken to
prevent possible instability in
plastic hinge regions
Ductile Shear Walls
Ductile Shear Walls
Ductile Shear Walls
Effective flange width:
ℓf
ℓf ≤ ½ distance to adjacent wall web
ℓf ≤ ¼ of wall height above the section
Wall
Reinforcement
Plastic Hinges
Other Regions
Distributed Reinforcement in Each Direction
r ≥ 0.0025
r ≥ 0.0025
Amount
Spacing
≤ 300 mm
≤ 450 mm
Concentrated Reinforcement
Where
@ends and
corners
As ≥ 0.015 bwlw
Amount
(at least 4 bars) As ≤ 0.06 (A)be
@ends
Hoops
Like nonseismic
columns
Confine like
columns
As ≥ 0.001 bwlw
As ≤ 0.06 (A)be
Ductile Shear Walls
 Vertical reinforcement outside the plastic
hinge region will be tied as specified in
7.6.5 if the area of steel is more than
0.005Ag and the maximum bar size is #20
and smaller
 Vertical reinforcement in plastic hinge
regions will be tied as specified in 21.6.6.9 if
the area of steel is more than 0.005Ag and
the maximum bar size is #15 and smaller
Ductile Shear Walls
 At least two curtains of reinforcement will
be used in plastic hinge regions, if:
Vf  0.18c fc' A cv
Where;
Acv : Net area of concrete section bounded by
web thickness and length of section in the
direction of lateral force
Ductile Shear Walls
For buckling prevention, ties shall be provided
in the form of hoops, with spacing not to
exceed:
 6 longitudinal bar diameters
 24 tie diameters
 ½ of the least dimension of of the member
Ductility of Ductile Shear Walls
Rotational Capacity, qic> Inelastic Demand, qid
D f RoRd  D f  w
q id 
 0.004
w 

 hw 

2 

hw
  cu w

q ic  
 0.002  0.025
 2c

ℓw/2
ℓw
y
cu
Ductility of Ductile
Shear Walls
D f RoRd  D f  w
q id 
 0.004
w 

 hw 

2 

  cu w

q ic  
 0.002  0.025
 2c

Ductility of Ductile Shear Walls
Ps  Pn  Pns  a  f Af
c
'
a1 b 1cfcb w
'
1 c c
Ductile Coupled Walls
Mtotal = M1 + M2 + P x
E.Q.
If P x  2/3Mtotal
Coupled Wall
If P x < 2/3Mtotal
Partially
Coupled Wall
M1
P
M2
x
P
Ductility of Ductile Coupled
Walls
Rotational Capacity, qic> Inelastic Demand, qid
D f RoRd
q id 
 0.004
hw
  cu w

q ic  
 0.002  0.025
 2c

ℓw: Length of the coupled wall system
ℓw: Lengths of the individual wall segments
for partially coupled walls
Ductility of Coupling Beams
Rotational Capacity, qic> Inelastic Demand, qid
 D f R o R d   cg

q id  
 hw   u
qic = 0.04 for coupling
beams with diagonal
reinforcement as per
21.6.8.7
qic = 0.02 for coupling beams without
diagonal reinforcement as per 21.6.8.6
Coupling Beams with Diagonal
Reinforcement
Wall Capacity @ Ends of Coupling
Beams
 Walls at each end of a coupling beam shall be
designed so that the factored wall moment
resistance at wall centroid exceeds the
moment resulting from the nominal moment
resistance of the coupling beam.
 If the above can not be achieved, the walls
develop plastic hinges at beam levels. This
requires design and detailing of walls at
coupling beam locations as plastic hinge
regions.
Shear Design of Ductile Walls
Design shear forces shall not be less than;
 Shear corresponding to the development of
probable moment capacity of the wall or the
wall system
 Shear resulting from design load combinations
with RdRo = 1.0
 Shear associated with higher mode effects
Shear Design of Ductile Walls
Shear design will conform to the requirements of
Clause 11. In addition, for plastic hinge regions;
 If qid ≥ 0.015
Vf ≤ 0.10c f’cbwdv
 If qid = 0.005
Vf ≤ 0.15c f’cbwdv
 For qid between the above two values, linear
interpolation may be used
Shear Design of Ductile Walls
For plastic hinge regions:
 If qid ≥ 0.015
b0
 If qid ≤ 0.005
b ≤ 0.18
 For qid between the above two values, linear
interpolation may be used
Shear Design of Ductile Walls
For plastic hinge regions:
 If (Ps + Pp) ≤ 0.1 f’cAg
q  45o
 If (Ps + Pp) ≥ 0.2 f’cAg
q ≥ 35o
 For axial compression between the above
two values, linear interpolation may be
used
Moderately Ductile Moment
Resistant Frame Beams
(Rd = 2.5)
-
Mr
+
-
-
-
Mr > 1/5 Mr
-
Mr > 1/3 Mr
Mr
+
-
Mr > 1/3 Mr
+
Mr > 1/5 Mr
Moderately Ductile Moment
Resistant Frame Beams
2h
s1
c1
s2  h / 2
50 mm
h
Stirrups detailed as
hoops
Stirrups
Stirrups detailed as
hoops
n
s1  d / 4
s 1  8(d b ) long .bar
s1  300 mm
s1  24(d b )hoop
db
Moderately Ductile Moment
Resistant Frame Columns
Marc Factored moment
resistance of columns
M rnb
M lnb
Column design forces
need not exceed those
determined from factored
load combinations using
RdRo = 1.0
Nominal moment
resistance of beams
b
M rc
M
rc
 Mnb
Moderately Ductile Moment
Resistant Frame Columns
Columns will be confined for improved
inelastic deformability
lo
lo ≥ 1/6 of clear col. height
lo ≥ h
lo ≥ 450 mm
lo
Spacing of Confinement
Reinforcement
 1/2 of minimum column dimension
 8 x long. bar diameter
 24 x tie diameters
Crossties or legs of overlapping hoops shall
not have centre-to-centre spacing exceeding
350 mm
Column Confinement Reinforcement
f'c
ρ s  0.3kp
fyh
Circular Hoops
fyh  500MPa
Pf
kp 
Po
Ag
f'c
ρ s  0.45(
 1)
Ac
fyh
Column Confinement Reinforcement
A sh
Ag f'c
 0.15knkp
shc
Ach fyh
A sh
f'c
 0.09 shc
fyh
Rectilinear Ties
Pf
kp 
P
o
fyh  500MPa
n
kn  n /(n  2)
: No. of laterally supported bars
Beam Shear Strength
Wf
M+n
-
Mn
ln
(Ve )left
Ve =
(Ve ) right
Mn
+
ln
+
Mn
+
Wf ln
2
Beam Shear Strength
The factored shear need not exceed
that obtained from structural analysis
under factored load combinations with
RdRo = 1.0
Computation of Joint Shear
Ve
As
T2 = As fy
C1 = T1
x
C2 = T2
Joint shear
associated with
nominal resistance
of beams
x
A's
T1 = A's fy
Ve
Vx-x = Ve - T2 - C1
Joint Shear
Joint shear associated with nominal
resistances of the beams and the
columns will be computed and the
smaller of the two values will be used
The joint shear need not exceed that
obtained from structural analysis under
factored load combinations with
RdRo = 1.0
Shear Resistance of Joints in
Moderately Ductile Frames
Vj  2.2c f 'c A j
Vj  1.6 c f 'c A j
Vj  1.3 c f 'c A j
Transverse Reinforcement in Joints
 Longitudinal reinforcement shall have a
centre-to-centre distance not exceeding
300 mm and shall not be cranked within
the joint
 Transverse reinforcement shall be
provided with a maximum spacing of 150
mm
Moderately Ductile Shear Walls
 Wall thicknesses will be similar to those of
ductile shear walls, except;
ℓu / 10
ℓu / 14
ℓu / 14
ℓu / 20
 Ductility limitation will be similar to that
for ductile walls with minimum rotational
demand as 0.003.
Moderately Ductile Shear Walls
 Distributed horizontal reinforcement ratio
shall not be less than 0.0025 in the vertical
and horizontal directions
 Concentrated reinforcement in plastic
hinge regions shall be the same as that for
ductile walls, except the tie requirements
are relaxed to those in Chapter 7
Shear Design of Moderately Ductile
Walls
Design shear forces shall not be less than the
smaller of;
 Shear corresponding to the development of
nominal moment capacity of the wall or the
wall system
 Shear resulting from design load combinations
with RdRo = 1.0
Shear Design of Moderately Ductile
Walls
 Vf ≤ 0.1 cf’cbwdv
 b  0.1
 q = 45o
Design Example
Ductile Core-Wall Structure in Montreal
Chapter 11
By D. Mitchell and P. Paultre
Twelve-Storey Ductile
Core Wall Structure
in Montreal
•E-W: Rd = 4.0 and Ro = 1.7
•N-S: Rd = 3.5 and Ro = 1.6
•Site Classification D
(Fa = 1.124 & Fv = 1.360)
Design Spectral Response
Acceleration N-S Direction
Empirical: Ta = 0.05 (hn)3/4 = 0.87 s
Dynamic:
T = 1.83 s but not greater than 2Ta = 1.74s
Torsion of Core Wall
Torsional Sensitivity
Bx   max /  ave
Max BNS = 1.80
Max BEW = 1.66
Max B > 1.7
irregularity
type 7
Seismic and Wind Loading
Diagonally Reinforced Coupling Beam
Wall Reinforcement Details
Factored Moment Resistance E-W
Factored Moment Resistance N-S
Squat Shear Walls
hw / ℓw ≤ 2.0; Rd = 2.0
 The foundation and diaphragm components
of the SFRS shall have factored resistances
greater than the nominal wall capacity.
 The walls will dissipate energy either;
 through flexural mechanism, i.e., V @
Mn is less than Vr,
 or, through shear mechanism, i.e., V @
Mn is more than Vr.
In this case: Vr  0.2 f'c b wd v
Squat Shear Walls
The distributed reinforcement:
 rh ≥ 0.003
rv ≥ 0.003
 Use two curtains of reinforcement if
Vf  0.18λφc f'c b wd v
 At least 4 vertical bars will be tied with
seismic hooks and placed at the ends
and at junctions of intersecting walls
over 300 mm wall length with r ≥ 0.005.
Squat Shear Walls
Shear Design
 Vf ≤ 0.15 c f’cbwdv
 b=0
q = 300 to 450
 Vertical reinforcement required for shear:
Ps
ρ v  ρhcot θ 
φsf y A g
2
where; rh : required horizontal steel
Conventional Construction
Rd = 1.5
Buildings with Rd = 1.5 can be designed as
conventional buildings. However, detailing
required for nominally ductile columns will be
used unless;
 Factored resistances of columns are more
than those for framing beams
 Factored resistances of columns are greater
than factored loads based on RdRo =1.0
 IEFaSa(0.2) < 0.2
Walls of Conventional Construction
Walls can be designed as conventional walls.
However, the shear resistance will be greater
than the smaller of;
 the shear corresponding to factored
moment resistance,
 the shear computed from factored loads
based on RdRo =1.0.
Frame Members not Considered Part
of the SFRS
Frames that are not part of SFRS, but “go for
the ride” during an earthquake shall be
designed to accommodate forces and
deformations resulting from seismic
deformations.
Thank You…..
Questions or Comments?