Seismic Design and Detailing of Reinforced Concrete Structures Based on CSA A23.3 - 2004

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

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


Inelasticity results softening in the structure, elongating structural period

S(T)

TT1 T2

S1

S2


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…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 SoilSeismic Amplification due to Soft Soil

LiquefactionLiquefaction








Avoid soft storeys



Use of Unnecessary MassUse of Unnecessary Mass









Avoid soft storeys



Effect of TorsionEffect of Torsion













Avoid soft storeys



Effect of Vertical DiscontinuityEffect of Vertical Discontinuity

Effect of Vertical DiscontinuityEffect of Vertical Discontinuity






Avoid soft storeys



Effect of Soft StoreyEffect of Soft Storey









Avoid soft storeys



R/C Frame Buildings without Drift ControlR/C Frame Buildings without Drift Control

Buildings Stiffened by Structural WallsBuildings Stiffened by Structural Walls






Avoid soft storeys



Short Column EffectShort Column Effect

Short Column EffectShort Column Effect

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

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

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.

Moment

Curvature

My

Mcr

Mn

Elastic rigidity

Post-cracking rigidity

Post-yield rigidity

ActualIdealized

Flexural Behaviour of R/CFlexural Behaviour of R/C

Flexural Behaviour of R/CFlexural Behaviour of R/CMoment

Curvature

Mn

0.75Mn

Actual

Idealized(bi-linear)

Effective elastic rigidity

y u

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

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

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

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

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

Formation of Plastic HingesFormation of Plastic Hinges

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


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

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

Longitudinal ReinforcementLongitudinal Reinforcement

r min = 1% r max = 6%

Design for factored axial forces and moments using Interaction Diagrams

Strong Beam-Weak Column DesignStrong Beam-Weak Column Design

Strong Beam-Weak Column DesignStrong Beam-Weak Column Design

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

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

Poorly Confined ColumnsPoorly Confined Columns

Poorly Confined ColumnsPoorly Confined Columns

Well-Confined Well-Confined ColumnColumn

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


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

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

Vcol =M

acol Mb

col

lu

+

M-prM+

pr

lu

M+pr M -

pr

Ma

col

Mbcol

Vcol

Vcol

Column Shear Column Shear StrengthStrength

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

Shear Deficient ColumnsShear Deficient Columns

Shear Deficient ColumnsShear Deficient Columns

Beam-Column JointsBeam-Column Joints

Poor Joint PerformancePoor Joint Performance

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

jccj A'f2.2V

jccj A'f6.1V

jccj A'f3.1V

Shear Resistance of Joints Shear Resistance of 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

Transverse Reinforcement in Joints Transverse Reinforcement in Joints

Design Example Design Example

Six-Storey Ductile Moment Resisting Frame in Vancouver

Chapter 11

By D. Mitchell and P. Paultre

•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

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.

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 Interior Column






Design of Interior Beam-Column Joint



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

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




ℓf

Effective flange width:

ℓf ≤ ½ distance to adjacent wall web

ℓf ≤ ¼ of wall height above the section

Wall Wall Reinforcement Reinforcement

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


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


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


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 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

Ductility of Ductile Ductility of Ductile Shear Walls Shear Walls

004.0

2h

RR

ww

wfdofid

025.0002.0c2

wcuic

Ductility of Ductile Shear Walls Ductility of Ductile Shear Walls

w'cc11

f'cc1nsns

bf

AfPPPc

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

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

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

Coupling Beams with Diagonal Coupling Beams with Diagonal Reinforcement Reinforcement

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.

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


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


If id ≥ 0.015

If id ≤ 0.005 ≤

For id between the above two values, linear

interpolation may be used

For plastic hinge regions:


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:

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)

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

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

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

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


yh

cps f

f'0.3kρ

o

fp P

Pk

yh

c

c

gs f

f'1)

A

A0.45(ρ

Circular Hoops

MPa500yhf


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


Wf

M-n

M+n

(Ve)left (Ve)right

Ve =M

-n M

+n

ln

ln

Wf ln

2

++-

The factored shear need not exceed

that obtained from structural analysis

under factored load combinations with

RdRo = 1.0


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

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

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

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

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.

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

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

Shear Design of Moderately Ductile Shear Design of Moderately Ductile Walls Walls

Vf ≤ 0.1 cf’cbwdv

= 45o

Design Example 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

Max BNS = 1.80

Max BEW = 1.66Max B > 1.7irregularity

type 7

avemaxx /B

Torsional Sensitivity

Seismic and Wind Loading

Diagonally Reinforced Coupling Beam

Wall Reinforcement Details

Factored Moment Resistance E-W

Factored Moment Resistance N-S

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

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

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

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

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.

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.

Thank You…..Thank You…..

Questions or Comments?Questions or Comments?

Documents

Seismic Design and Detailing of Reinforced Concrete Structures Based on CSA A23.3 - 2004