seismic design and detailing of reinforced concrete structures based on csa a23.3 - 2004
DESCRIPTION
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. - PowerPoint PPT PresentationTRANSCRIPT
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Seismic Design and Detailing of Seismic Design and Detailing of Reinforced Concrete Structures Reinforced Concrete Structures Based on CSA A23.3 - 2004Based 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
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Reinforced concrete structures are designed to dissipate seismic induced energy through
inelastic deformations
Basic Principles of DesignBasic Principles of Design
Ve = S(Ta) Mv IE W / (Rd Ro)Ve
Ve /Rd Ro
Ve /Rd
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Basic Principles of DesignBasic Principles of Design
Inelasticity results softening in the structure, elongating structural period
S(T)
TT1 T2
S1
S2
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Basic Principles of DesignBasic 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.
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To reduce seismic demands…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
![Page 6: Seismic Design and Detailing of Reinforced Concrete Structures Based on CSA A23.3 - 2004](https://reader038.vdocument.in/reader038/viewer/2022102511/5681395d550346895da0fbe9/html5/thumbnails/6.jpg)
Seismic Amplification due to Soft SoilSeismic Amplification due to Soft Soil
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LiquefactionLiquefaction
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LiquefactionLiquefaction
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LiquefactionLiquefaction
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To reduce seismic demands…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
![Page 11: Seismic Design and Detailing of Reinforced Concrete Structures Based on CSA A23.3 - 2004](https://reader038.vdocument.in/reader038/viewer/2022102511/5681395d550346895da0fbe9/html5/thumbnails/11.jpg)
Use of Unnecessary MassUse of Unnecessary Mass
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Use of Unnecessary MassUse of Unnecessary Mass
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Use of Unnecessary MassUse of Unnecessary Mass
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Use of Unnecessary MassUse of Unnecessary Mass
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To reduce seismic demands…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
![Page 16: Seismic Design and Detailing of Reinforced Concrete Structures Based on CSA A23.3 - 2004](https://reader038.vdocument.in/reader038/viewer/2022102511/5681395d550346895da0fbe9/html5/thumbnails/16.jpg)
Effect of TorsionEffect of Torsion
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Effect of TorsionEffect of Torsion
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Effect of TorsionEffect of Torsion
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Effect of TorsionEffect of Torsion
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Effect of TorsionEffect of Torsion
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Effect of TorsionEffect of Torsion
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Effect of TorsionEffect of Torsion
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Effect of TorsionEffect of Torsion
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To reduce seismic demands…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
![Page 25: Seismic Design and Detailing of Reinforced Concrete Structures Based on CSA A23.3 - 2004](https://reader038.vdocument.in/reader038/viewer/2022102511/5681395d550346895da0fbe9/html5/thumbnails/25.jpg)
Effect of Vertical DiscontinuityEffect of Vertical Discontinuity
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Effect of Vertical DiscontinuityEffect of Vertical Discontinuity
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To reduce seismic demands…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
![Page 28: Seismic Design and Detailing of Reinforced Concrete Structures Based on CSA A23.3 - 2004](https://reader038.vdocument.in/reader038/viewer/2022102511/5681395d550346895da0fbe9/html5/thumbnails/28.jpg)
Effect of Soft StoreyEffect of Soft Storey
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Effect of Soft StoreyEffect of Soft Storey
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Effect of Soft StoreyEffect of Soft Storey
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Effect of Soft StoreyEffect of Soft Storey
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To reduce seismic demands…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
![Page 33: Seismic Design and Detailing of Reinforced Concrete Structures Based on CSA A23.3 - 2004](https://reader038.vdocument.in/reader038/viewer/2022102511/5681395d550346895da0fbe9/html5/thumbnails/33.jpg)
R/C Frame Buildings without Drift ControlR/C Frame Buildings without Drift Control
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Buildings Stiffened by Structural WallsBuildings Stiffened by Structural Walls
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To reduce seismic demands…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
![Page 36: Seismic Design and Detailing of Reinforced Concrete Structures Based on CSA A23.3 - 2004](https://reader038.vdocument.in/reader038/viewer/2022102511/5681395d550346895da0fbe9/html5/thumbnails/36.jpg)
Short Column EffectShort Column Effect
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Short Column EffectShort Column Effect
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Seismic Design Requirements of Seismic Design Requirements of CSA A23.3 - 2004CSA A23.3 - 2004
Capacity design is employed…..
Selected elements are designed to yield while critical elements remain elastic
Design for
Strength and Deformability
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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
Load CombinationsLoad Combinations
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Stiffness Properties for AnalysisStiffness 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.
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Moment
Curvature
My
Mcr
Mn
Elastic rigidity
Post-cracking rigidity
Post-yield rigidity
ActualIdealized
Flexural Behaviour of R/CFlexural Behaviour of R/C
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Flexural Behaviour of R/CFlexural Behaviour of R/CMoment
Curvature
Mn
0.75Mn
Actual
Idealized(bi-linear)
Effective elastic rigidity
y u
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Section Properties for Analysis Section Properties for Analysis as per CSA A23.3-04as per CSA A23.3-04
Beams Ie = 0.40 Ig
Columns Ie = cIg
Coupling Beams
without diagonal reinforcement Ave = 0.15Ag
Ie = 0.40 Ig
with diagonal reinforcement Ave = 0.45Ag
Ie = 0.25 Ig
Slab-Frame Element Ie = 0.20 Ig
Walls Axe = wAg
Ie = w Ig
1.0AfP
0.60.5αg
'c
sc
1.0AfP
0.6αg
'c
sw
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Seismic Design Requirements of Seismic Design Requirements of CSA A23.3 - 2004CSA 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
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Ductile Moment Resisting Frame Ductile Moment Resisting Frame Members Subjected to FlexureMembers Subjected to Flexure
Rd = 4.0 Pf ≤ Agf’c /10
h d
bw
h0.3bw
mm250bw
d4n yx
yxcb 2w
h3/4x
h3/4y
c2
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Mr Mr
Mr > 1/2 Mr + -
-
Mr > 1/2 Mr + -
-
Top and Botom 2 bars continuous
Mr > 1/4 Mr - -
Mr > 1/4 Mr + -
Top and Bottom: 1.4bwd / fy ≤ r ≤ 0.025
Beam Longitudinal ReinforcementBeam Longitudinal Reinforcement
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Beam Transverse ReinforcementBeam Transverse Reinforcementc1
h
n
s1 50 mm2/ds2
2d
4/ds1
mm300s1
bar.longb1 )d(8s
hoopb1 )d(24s
Hoops HoopsSirrups with seismic hooks
db
No lap splicing within this region
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Formation of Plastic HingesFormation of Plastic Hinges
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Beam Shear StrengthBeam Shear Strength
Wf
M-pr
M+pr
(Ve)left (Ve)right
Ve =M-
pr M+pr
ln
ln
Wf ln
2
++-
Plastic Hinge
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Beam Shear StrengthBeam Shear Strength
The factored shear need not exceed that
obtained from structural analysis under
factored load combinations with RdRo = 1.0
The values of = 45o and = 0 shall be used
in shear design within plastic hinge regions
The transverse reinforcement shall be
seismic hoops
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Ductile Moment Resisting Frame Ductile Moment Resisting Frame Members Subjected to Flexure and Members Subjected to Flexure and
Significant Axial LoadSignificant Axial Load
Rd = 4.0 Pf > Agf’c /10
hshort
hlong D
hshort ≥ 300 mm D ≥ 300 mmhshort / hlong ≥ 0.4
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Longitudinal ReinforcementLongitudinal Reinforcement
r min = 1% r max = 6%
Design for factored axial forces and moments using Interaction Diagrams
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Strong Beam-Weak Column DesignStrong Beam-Weak Column Design
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Strong Beam-Weak Column DesignStrong Beam-Weak Column Design
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Manc
Mbnc
M lpb M r
pb
Strong Column-Weak Beam DesignStrong Column-Weak Beam Design
pbnc MM
Nominal moment resistance of columns under factored axial loads
Probable moment resistance of beams
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Column Confinement Column Confinement ReinforcementReinforcement
lo ≥ 1.5h
lo ≥ 1/6 of clear col. height
If Pf ≤ 0.5 c f’c Ag ;
lo ≥ 2.0hIf Pf > 0.5 c f’c Ag ;
Columns will be confined for improved inelastic deformability
lo
lo
Columns connected to rigid members such as foundations and discontinuous walls, or columns at the base will be confined along the entire height
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Poorly Confined ColumnsPoorly Confined Columns
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Poorly Confined ColumnsPoorly Confined Columns
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Well-Confined Well-Confined ColumnColumn
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Column Confinement ReinforcementColumn Confinement Reinforcement
yh
cps f
f'0.4kρ
o
fp P
Pk
yh
c
c
gs f
f'1)
A
A0.45(ρ
Circular Spirals
MPa500yhf
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Column Confinement ReinforcementColumn Confinement Reinforcement
cch
g shA
A
yh
cpnsh f
f'k0.2kA
o
fp P
Pk
cshyh
csh f
f'0.09A
Rectilinear Ties
MPa500yhf)2n/(n nk
n : No. of laterally supported bars
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Spacing of Confinement Spacing of Confinement ReinforcementReinforcement
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
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Vcol =M
acol Mb
col
lu
+
M-prM+
pr
lu
M+pr M -
pr
Ma
col
Mbcol
Vcol
Vcol
Column Shear Column Shear StrengthStrength
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Column Shear StrengthColumn Shear Strength
The factored shear need not exceed that
obtained from structural analysis under
factored load combinations with RdRo = 1.0
The values of ≥ 45o and ≤ 0.10 shall be
used in shear design in regions where the
confinement reinforcement is needed
The transverse reinforcement shall be
seismic hoops
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Shear Deficient ColumnsShear Deficient Columns
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Shear Deficient ColumnsShear Deficient Columns
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Beam-Column JointsBeam-Column Joints
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Poor Joint PerformancePoor Joint Performance
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As
A's
C1 = T1
C2 = T2 T1 = 1. 25 A's fy
T2 = 1. 25 As fy
xx
Ve
Ve
Vx-x = Ve - T2 - C1
Computation of Joint ShearComputation of Joint Shear
Vx-x ≤ that obtained from frame analysis using RdRo = 1.0
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jccj A'f2.2V
jccj A'f6.1V
jccj A'f3.1V
Shear Resistance of Joints Shear Resistance of Joints
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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
Transverse Reinforcement in Joints Transverse Reinforcement in Joints
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Design Example Design Example
Six-Storey Ductile Moment Resisting Frame in Vancouver
Chapter 11
By D. Mitchell and P. Paultre
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•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
Six-Storey Ductile Moment Resisting Frame in Vancouver
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Material PropertiesConcrete: normal density concrete with 30 MPaReinforcement: 400 MPaLive loadsFloor live loads:2.4 kN/m2 on typical office floors4.8 kN/m2 on 6 m wide corridor bayRoof load2.2 kN/m2 snow load, accounting for parapets and equipment projections1.6 kN/m2 mechanical services loading in 6 m wide strip over corridor bayDead loadsself-weight of reinforced concrete members calculated as 24 kN/m3
1.0 kN/m2 partition loading on all floors0.5 kN/m2 mechanical services loading on all floors0.5 kN/m2 roofingWind loading1.84 kN/m2 net lateral pressure for top 4 storeys1.75 kN/m2 net lateral pressure for bottom 2 storeysThe fire-resistance rating of the building is assumed to be 1 hour.
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Gravity Loading
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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
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Design of Ductile Beam
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Design of Ductile Beam
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Design of Ductile Beam
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Design of Ductile Beam
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Design of Ductile Beam
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Design of Ductile Beam
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Design of Ductile Interior Column
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Design of Ductile Interior Column
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Design of Ductile Interior Column
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Design of Ductile Interior Column
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Design of Ductile Interior Column
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Design of Ductile Interior Column
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Design of Interior Beam-Column Joint
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Design of Interior Beam-Column Joint
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Design of Interior Beam-Column Joint
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ℓw
hw
Plastic Hinge Length
Ductile Shear Walls 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
For elevations above the plastic hinge region, design values will be increased by Mr/Mf at the top of
hinging region
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ℓw
hw
Plastic Hinge Length
Ductile Shear Walls Ductile Shear Walls
Wall thickness in the plastic hinge:
tw ≥ ℓu / 14 but may be limited to
ℓu / 10 in high compression regions
tw
ℓu
Because walls are relatively thin members, care must be taken to
prevent possible instability in plastic hinge regions
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Ductile Shear Walls Ductile Shear Walls
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Ductile Shear Walls Ductile Shear Walls
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Ductile Shear Walls Ductile Shear Walls
ℓf
Effective flange width:
ℓf ≤ ½ distance to adjacent wall web
ℓf ≤ ¼ of wall height above the section
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Wall Wall Reinforcement Reinforcement
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Distributed Reinforcement in Each Direction
Amount r ≥ 0.0025 r ≥ 0.0025
Spacing ≤ 300 mm ≤ 450 mm
Concentrated Reinforcement
Where @ends and corners
@ends
Amount
(at least 4 bars)
s ≥ 0.015 bwlw
s ≤ 0.06 (A)be
s ≥ 0.001 bwlw
s ≤ 0.06 (A)be
Hoops Confine like columns
Like non-seismic columns
Plastic Hinges Other Regions
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Ductile Shear Walls 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
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Ductile Shear Walls Ductile Shear Walls
At least two curtains of reinforcement will
be used in plastic hinge regions, if:
cv'ccf Af18.0V
Where;
Acv : Net area of concrete section bounded by web thickness and length of section in the direction of lateral force
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Ductile Shear Walls 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
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Ductility of Ductile Shear Walls Ductility of Ductile Shear Walls Rotational Capacity, ic> Inelastic Demand, id
004.0
2h
RR
ww
wfdofid
ℓw
hw
ycu
ℓw/2025.0002.0
c2wcu
ic
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Ductility of Ductile Ductility of Ductile Shear Walls Shear Walls
004.0
2h
RR
ww
wfdofid
025.0002.0c2
wcuic
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Ductility of Ductile Shear Walls Ductility of Ductile Shear Walls
w'cc11
f'cc1nsns
bf
AfPPPc
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x PP
E.Q.
M2M1
Mtotal = M1 + M2 + P x
If P x 2/3Mtotal Coupled Wall
If P x < 2/3Mtotal Partially Coupled Wall
Ductile Coupled Walls Ductile Coupled Walls
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Ductility of Ductile Coupled Ductility of Ductile Coupled Walls Walls
Rotational Capacity, ic> Inelastic Demand, id
004.0h
RR
w
dofid
025.0002.0c2
wcuic
ℓw: Length of the coupled wall system
ℓw: Lengths of the individual wall segments for partially coupled walls
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Ductility of Coupling Beams Ductility of Coupling Beams Rotational Capacity, ic> Inelastic Demand, id
u
cg
w
dofid h
RR
ic = 0.04 for coupling beams with diagonal reinforcement as per 21.6.8.7
ic = 0.02 for coupling beams without diagonal reinforcement as per 21.6.8.6
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Coupling Beams with Diagonal Coupling Beams with Diagonal Reinforcement Reinforcement
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Wall Capacity @ Ends of Coupling Wall Capacity @ Ends of Coupling Beams 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.
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Shear Design of Ductile Walls 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
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Shear Design of Ductile Walls Shear Design of Ductile Walls
Shear design will conform to the requirements of
Clause 11. In addition, for plastic hinge regions;
If id ≥ 0.015 Vf ≤ 0.10c f’cbwdv
If id = 0.005 Vf ≤ 0.15c f’cbwdv
For id between the above two values, linear
interpolation may be used
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Shear Design of Ductile Walls Shear Design of Ductile Walls
If id ≥ 0.015
If id ≤ 0.005 ≤
For id between the above two values, linear
interpolation may be used
For plastic hinge regions:
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Shear Design of Ductile Walls Shear Design of Ductile Walls
If (Ps + Pp) ≤ 0.1 f’cAg
If (Ps + Pp) ≥ 0.2 f’cAg ≥
For axial compression between the above
two values, linear interpolation may be
used
For plastic hinge regions:
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Mr Mr
Mr > 1/3 Mr + -
-
Mr > 1/3 Mr + -
-Mr > 1/5 Mr
- -
Mr > 1/5 Mr + -
Moderately Ductile Moment Moderately Ductile Moment Resistant Frame BeamsResistant Frame Beams
(Rd = 2.5)
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c1
h
n
s1 50 mm2/2 hs 2h
4/ds1
mm300s1
bar.longb1 )d(8s
hoopb1 )d(24s
Stirrups
db
Stirrups detailed as hoops
Stirrups detailed as hoops
Moderately Ductile Moment Moderately Ductile Moment Resistant Frame BeamsResistant Frame Beams
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Marc
Mbrc
M lnb M r
nb
nbrc MM
Factored moment resistance of columns
Nominal moment resistance of beams
Moderately Ductile Moment Moderately Ductile Moment Resistant Frame ColumnsResistant Frame Columns
Column design forces need not exceed those determined from factored load combinations using RdRo = 1.0
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lo ≥ h
lo ≥ 1/6 of clear col. height
lo ≥ 450 mm
Columns will be confined for improved inelastic deformability
lo
lo
Moderately Ductile Moment Moderately Ductile Moment Resistant Frame ColumnsResistant Frame Columns
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Spacing of Confinement Spacing of Confinement ReinforcementReinforcement
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
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Column Confinement ReinforcementColumn Confinement Reinforcement
yh
cps f
f'0.3kρ
o
fp P
Pk
yh
c
c
gs f
f'1)
A
A0.45(ρ
Circular Hoops
MPa500yhf
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Column Confinement ReinforcementColumn Confinement Reinforcement
cch
g shA
A
yh
cpnsh f
f'k0.15kA
o
fp P
Pk
cshyh
csh f
f'0.09A
Rectilinear Ties
MPa500yhf)2n/(n nk
n : No. of laterally supported bars
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Beam Shear StrengthBeam Shear Strength
Wf
M-n
M+n
(Ve)left (Ve)right
Ve =M
-n M
+n
ln
ln
Wf ln
2
++-
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The factored shear need not exceed
that obtained from structural analysis
under factored load combinations with
RdRo = 1.0
Beam Shear StrengthBeam Shear Strength
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As
A's
C1 = T1
C2 = T2 T1 = A's fy
T2 = As fy
xx
Ve
Ve
Vx-x = Ve - T2 - C1
Computation of Joint ShearComputation of Joint Shear
Joint shear associated with nominal resistance of beams
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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
Joint Shear Joint Shear
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jccj A'f2.2V
jccj A'f6.1V
jccj A'f3.1V
Shear Resistance of Joints in Shear Resistance of Joints in Moderately Ductile Frames Moderately Ductile Frames
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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
Transverse Reinforcement in Joints Transverse Reinforcement in Joints
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Moderately Ductile Shear Walls 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.
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Moderately Ductile Shear Walls 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
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Shear Design of Moderately Ductile Shear Design of Moderately Ductile Walls 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
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Shear Design of Moderately Ductile Shear Design of Moderately Ductile Walls Walls
Vf ≤ 0.1 cf’cbwdv
= 45o
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Design Example Design Example
Ductile Core-Wall Structure in Montreal
Chapter 11
By D. Mitchell and P. Paultre
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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)
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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
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Torsion of Core Wall
Max BNS = 1.80
Max BEW = 1.66Max B > 1.7irregularity
type 7
avemaxx /B
Torsional Sensitivity
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Seismic and Wind Loading
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Diagonally Reinforced Coupling Beam
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Wall Reinforcement Details
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Factored Moment Resistance E-W
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Factored Moment Resistance N-S
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Squat Shear Walls 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: vwcr dbf'0.2V
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Squat Shear Walls Squat Shear Walls
The distributed reinforcement:
rh ≥ 0.003 rv ≥ 0.003
Use two curtains of reinforcement if
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.
vwccf dbf'φ0.18λV
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Squat Shear WallsSquat Shear Walls
Shear Design Shear Design
Vf ≤ 0.15 c f’cbwdv
= 0 = 300 to 450
Vertical reinforcement required for shear:
where; rh : required horizontal steel
gys
s2hv Afφ
Pθcotρρ
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Conventional ConstructionConventional ConstructionRd = 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
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Walls of Conventional ConstructionWalls 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.
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Frame Members not Considered Part Frame Members not Considered Part of the SFRSof 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.
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Thank You…..Thank You…..
Questions or Comments?Questions or Comments?