Seismic Design and Detailing of Reinforced Concrete...

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

Reinforced concrete structures are designed to

dissipate seismic induced energy through

inelastic deformations

Basic 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)

T T1 T2

S1

S2


Capacity ≥≥≥≥ Demand

It is a good practice to reduce seismic

demands, to the extent possible4.

This can be done at the conceptual stage

by selecting a suitable structural system.

To reduce seismic demands4

� 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





torsional effects






Use of Unnecessary Mass





torsional effects






Effect of Torsion





torsional effects






Effect of Vertical Discontinuity





torsional effects






Effect of Soft Storey





torsional effects






R/C Frame Buildings without Drift Control

Buildings Stiffened by Structural Walls





torsional effects






Short Column Effect

Seismic Design Requirements of

CSA A23.3 - 2004

Capacity design is employed4..

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 Combinations

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 Behaviour of R/C

Section Properties for Analysis as

per CSA A23.3-04 Beams Ie = 0.40 Ig

Columns Ie = ααααcIg

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 = ααααwAg

Ie = ααααw Ig

1.0

1.0

1.0

1.0

AAAAffffPPPP0

.6

0.6

0.6

0.6

0.5

0.5

0.5

0.5

ααααgggg

''''cccc

sssscccc ≤≤≤≤++++====

1.0

1.0

1.0

1.0

AAAAffffPPPP0

.6

0.6

0.6

0.6

ααααgggg

''''cccc

sssswwww ≤≤≤≤++++====

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

h0.3bw ≥

mm250bw

≥

d4n ≥l

yxcb 2w ++≤

h3/4x ≤

h3/4y ≤

Beam Longitudinal Reinforcement

Beam Transverse Reinforcement

nnnnl

2/ds2 ≤≤≤≤

4/ds1 ≤≤≤≤

mm300s1 ≤≤≤≤

bar.longb1 )d(8s ≤≤≤≤

hoopb1 )d(24s ≤≤≤≤

No lap splicing within

this region

Formation of Plastic Hinges

Beam Shear Strength

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 θθθθ = 45o and ββββ = 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

Rd = 4.0 Pf > Agf’c /10

hshort ≥ 300 mm D ≥ 300 mm

hshort / hlong ≥ 0.4

Longitudinal Reinforcement

ρρρρ min = 1% ρρρρ max = 6%

Design for factored axial forces

and moments using Interaction

Diagrams

Strong Beam-Weak Column Design

Strong Column-Weak Beam Design

∑∑∑∑ ∑∑∑∑≥≥≥≥pbnc

MM

Nominal moment

resistance of columns

under factored axial loads

Probable moment

resistance of beams

Column Confinement

Reinforcement

lo ≥ 1.5h

lo ≥ 1/6 of clear col. height

If Pf ≤ 0.5 φφφφc f’c Ag ;

lo ≥ 2.0h If Pf > 0.5 φφφφc 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 Columns

Well-Confined

Column

Column Confinement Reinforcement

yh

cps

f

f'0.4kρ ====

o

fp

P

Pk ====

yh

c

c

g

sf

f'1)

A

A0.45(ρ −−−−≥≥≥≥

Circular Spirals

MPa500≤≤≤≤yhf


c

ch

gsh

A

A

yh

cpnsh

f

f'k0.2kA ====

o

fp

P

Pk ====

cshyh

csh

f

f'0.09A ====

Rectilinear Ties


)2n/(n −−−−====llnk

ln : 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

Column Shear

Strength

Column Shear Strength

� The factored shear need not exceed that


factored load combinations with RdRo = 1.0

� The values of θθθθ ≥ 45o and ββββ ≤ 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

Beam-Column Joints

Poor Joint Performance

Computation of Joint Shear

Vx-x ≤ that obtained from frame analysis using RdRo = 1.0

jccj A'f2.2V φφφφλλλλ====



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

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 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 Interior Column

Design of Interior Beam-Column Joint

ℓw

hw

Plastic

Hinge

Length

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

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

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

Wall

Reinforcement

Distributed Reinforcement in Each Direction

Amount ρρρρ ≥ 0.0025 ρρρρ ≥ 0.0025

Spacing ≤ 300 mm ≤ 450 mm

Concentrated Reinforcement

Where @ends and

corners

@ends

Amount

(at least 4 bars)

ΑΑΑΑs ≥ 0.015 bwlw

ΑΑΑΑs ≤ 0.06 (A)be

ΑΑΑΑs ≥ 0.001 bwlw

ΑΑΑΑs ≤ 0.06 (A)be

Hoops Confine like

columns

Like non-

seismic

columns

Plastic Hinges Other Regions

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:

cv

'

ccf Af18.0V λφλφλφλφ>>>>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, θθθθic> Inelastic Demand, θθθθid

004.0

2h

RR

ww

wfdofid ≥≥≥≥

−−−−

∆∆∆∆−−−−∆∆∆∆====

l

γγγγθθθθ

ℓw

hw

φφφφy φφφφcu

ℓw/2 025.0002.0

c2

wcuic ≤≤≤≤

−−−−====

lεεεεθθθθ

Ductility of Ductile

Shear Walls

004.0

2h

RR

ww

wfdofid ≥≥≥≥

−−−−

∆∆∆∆−−−−∆∆∆∆====

l

γγγγθθθθ

025.0002.0c2

wcuic ≤≤≤≤

−−−−====

lεεεεθθθθ

Ductility of Ductile Shear Walls

w

'

cc11

f

'

cc1nsns

bf

AfPPPc

φφφφββββααααφφφφαααα−−−−++++++++

====

x P P

E.Q.

M2 M1

Mtotal = M1 + M2 + P x

If P x ≥≥≥≥ 2/3Mtotal

Coupled Wall

If P x < 2/3Mtotal

Partially

Coupled Wall

Ductile Coupled Walls

Ductility of Ductile Coupled

Walls Rotational Capacity, θθθθ ic> Inelastic Demand, θθθθ id

004.0h

RR

w

dof

id≥≥≥≥

∆∆∆∆====θθθθ

025.0002.0c2

wcu

ic ≤≤≤≤

−−−−====

lεεεεθθθθ

ℓw: Length of the coupled wall system

ℓw: Lengths of the individual wall segments

for partially coupled walls

Ductility of Coupling Beams

Rotational Capacity, θθθθic> Inelastic Demand, θθθθid

u

cg

w

dofid

h

RR

l

l

∆∆∆∆====θθθθ

θθθθic = 0.04 for coupling

beams with diagonal

reinforcement as per

21.6.8.7

θθθθic = 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 will conform to the requirements of

Clause 11. In addition, for plastic hinge regions;

� If θθθθid ≥ 0.015 Vf ≤ 0.10φφφφc f’cbwdv

� If θθθθid = 0.005 Vf ≤ 0.15φφφφc f’cbwdv

� For θθθθid between the above two values, linear

interpolation may be used


� If θθθθid ≥ 0.015 β = 0β = 0β = 0β = 0

� If θθθθid ≤ 0.005 β β β β ≤ 0.180.180.180.18

� For θθθθid between the above two values, linear

interpolation may be used

For plastic hinge regions:


� If (Ps + Pp) ≤ 0.1 f’cAg θ = 45θ = 45θ = 45θ = 45οοοο

� If (Ps + Pp) ≥ 0.2 f’cAg θ θ θ θ ≥ 35353535οοοο

� For axial compression between the above

two values, linear interpolation may be

used

For plastic hinge regions:

Moderately Ductile Moment

Resistant Frame Beams

(Rd = 2.5)

nnnnl

2/2hs ≤

4/ds1 ≤≤≤≤

mm300s1 ≤≤≤≤

bar.longb1 )d(8s ≤≤≤≤

hoopb1)d(24s ≤≤≤≤


Resistant Frame Beams

∑ ∑≥ nbrc MM

Factored moment

resistance of columns

Nominal moment

resistance of beams


Resistant 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


Resistant Frame Columns

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


yh

cps

f

f'0.3kρ ====

o

fp

P

Pk ====

yh

c

c

g

sf

f'1)

A

A0.45(ρ −−−−≥≥≥≥

Circular Hoops



c

ch

gsh

A

A

yh

cpnsh

f

f'k0.15kA ====

o

fp

P

Pk ====

cshyh

csh

f

f'0.09A ====

Rectilinear Ties


)2n/(n −−−−====llnk

ln : No. of laterally supported bars

Beam Shear Strength

The factored shear need not exceed

that obtained from structural analysis

under factored load combinations with

RdRo = 1.0

Beam Shear Strength

Computation 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


factored load combinations with

RdRo = 1.0

Joint Shear




Shear Resistance of Joints in

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

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 φφφφcf’cbwdv

� β = 0.1β = 0.1β = 0.1β = 0.1

� θθθθ = 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

Max BNS = 1.80

Max BEW = 1.66

Max B > 1.7

irregularity

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

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

The distributed reinforcement:

� ρρρρh ≥ 0.003 ρρρρv ≥ 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 ρρρρ ≥ 0.005.

vwccf dbf'φ0.18λV >

Squat Shear Walls

Shear Design

� Vf ≤ 0.15 φφφφc f’cbwdv

� ββββ = 0 θθθθ = 300 to 450

� Vertical reinforcement required for shear:

where; ρρρρh : required horizontal steel

gys

s2

hvAfφ

Pθcotρρ −=

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 You4..

Questions or Comments?

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