Seismic Design and Detailing_pp131-150

8/17/2019 Seismic Design and Detailing_pp131-150

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Rev1 04-04 / CE573-131Mapua Institute of Technology (MAPUA Tech)

Adam C Abinales f.asep, pice

Seismic Design of Concrete Structures

5. Structural Design and Detailing for Earthquake ResistanceStructural Design and Detailing for Earthquake Resistance5.1 Earthquake Load Combinations: Strength Design

5.1.1 Earthquake Loads and Modeling RequirementsEarthquake Loads and Modeling Requirements. Structures shall be designed for ground motionproducing structural response and seismic forces in any horizontal direction. The earthquakeloads that shall be used in the load combinations (set forth in NSCP Section 203) shall be in

accordance with the requirements of NSCP Section 208.5.1.1.

NSCP eq. 208-1

NSCP eq. 208-2

whereE = the earthquake load on an element of the structure resulting from the combination of the

horizontal component E h and the vertical component E v .

E h = the earthquake load due to the base shear V or the design lateral force F p.

E m = the estimated maximum earthquake force that can be developed in the structure and usedin the design of specific elements of the structure.

E v = the load effect resulting from the vertical component of the earthquake ground motionand is equal to an addition of 0.5C a*I*D to the deal load effect, D, for strength designmethod, and may be taken as zero for allowable (or working) stress design method.

o = the seismic force amplification factor that is required to account for structure

overstrength. (Section 208.5.3.1).

hom

vh

EE

EEE

Ω=

+= ρ


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5. Structural Design and Detailing for Earthquake ResistanceStructural Design and Detailing for Earthquake Resistance

= Reliability/Redundancy Factor determined as:

NSCP eq. 208-3

r max = the maximum element-story shear ratio; the ratio of the design story shear in the mostheavily loaded single element to the total design story shear.

AB = the ground floor area of the structure expresses in m2.

B A r max

.162 −= ρ


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E E xample Problem 5.1. A four-storey concrete building of special moment resisting frame systemhas been analyzed. Beam A-B and column C-D are elements of SMRF. Structural analysis yieldedthe following results due to dead load, office building live load and lateral seismic forces:

Find the following:

1. Strength design moment at beam end A.

2. Strength design axial load and moment at column top C.

220 kN-m30 kN-m55 kN-m

Columnmomentat C

490 kN180 kN400 kN

ColumnC-D axialload

165 kN-m65 kN-m135 kN-m

Beammomentat A

LateralSeismic Eh

Live Load LDead LoadD

Member/

Stress

Structure is located in Zone 4;

Seismic source type: A

Distance to seismic source = 10 km

Soil profile type: SD

I = 1.0

ρ

=1.1; f 1 = 0.5

A B D

8000

4 0

0 0

8000

4 0 0 0

4 0 0

0

4 0 0 0

GF

2nd

3rd

4th

Roof

A

C

B

8000

C

D


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S S olution and discussion:

F F ind the strength design moment at beam end A.

T T o determine strength design moments for design, the earthquake component E must becombined with the dead and live load components D and L, as illustrated below. Determineearthquake load E

Sect. 208.5.1.1

A Apply load combinations involving earthquake. The basic load combinations for strength designper Section 203.3.1 is

NSCP eq. 203-5NSCP eq. 203-6

m-kN..)(.then,

m-kN

isforceearthquakehorizontaltoduemomentthewhile

m-kN.

))(.)(.(.

).(..whichin;.

isforceearthquakeverticaltoduemomentthewhere,

221172916511

165

729

1350144050

0144044050

=+=

=

=

=

===

+=

E

E

E

E

NCIDCE

EEE

h

v

v

aaav

vh ρ

EDLf ED

0190010121 1

.....

±++


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F F or reinforced concrete frame, the above equations shall be multiplied by 1.1 per Section 409.3.3and become

therefore, strength design moment at beam end A

F F ind the strength design axial load and moment at column top C. Determine the earthquake load E

ED

Lf ED

101990

1011013211

..

...

±

++

and m-kN.

))(.(.).(.)(.

...

27446

65501012211101135321

1011013211

=

++=

++=

A

A

LED A

M

M

Mf MMM

m-kN.orm-kN.

).(.)(.

..

679897365

2211101135990

101990

−=

±=

±=

A

A

ED A

M

M

MMM

m-kN.))(.)(.(.)(.

topatmomentfor the

kN))(.)(.(.)(.

load axialfor the

where,

125455014405022011

627400014405049011

=+=

=+=

+=

E

E

EEE vh

ρ


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A Apply load combinations involving earthquake.

for the axial load P c

therefore,

for the moment M c

Note that the column section capacity must be designed for the interaction of P c = 1316.7 kNcompression and M c = 368.61 kN-m (for D+L+E ), and the interaction of P c = 293.7 kN tension andM c = -225.06 kN-m (for D+E ).

kN.orkN.

)(.)(.

..

and

kN.

))(.(.)(.)(.

...

729371085

627101400990

101990

71316

18050101627101400321

1011013211

−=

±=

±=

=

++=

++=

C

C

EDC

C

C

LEDC

P

P

PPP

P

P

pf PPP

kN.orkN. 729371316 −=CP

m-kN.orm-kN.

).(.)(.

..

and

m-kN.

))(.(.).(.)(.

...

0622596333

125410155990

101990

61368

3050101125410155321

1011013211

−=

±=

±=

=

++=

++=

C

C

EDC

c

C

LEDC

M

M

MMM

M

M

Mf MMM


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5. Structural Design and Detailing for Earthquake ResistanceStructural Design and Detailing for Earthquake Resistance5.2 In-situ Reinforced Concrete Design and Detail

Reinforced concrete for most structures is generally desirable because of its availability andeconomy, and its stiffness can be used to advantage to minimize seismic deformations and hencereduce the damage to non-structure. Difficulties arise due to reinforcement congestion when trying

to achieve high ductility in framed structures, and at the time of writing the problem of detailingbeam-column joints to withstand strong cyclic loading had not been resolved. It should be recalledthat no amount of good detailing will enable an ill-conceived structural form to survive a strongearthquake.

5.2.1 Seismic Response of Reinforced ConcreteSeismic Response of Reinforced Concrete. Even in well-designed reinforced concrete members,the root cause of failure under earthquake loading is usually concrete cracking. Degradation

occurs in the cracked zone under cyclic loading. Cracks do not close up properly when the tensilestress drops because of permanent elongation of reinforcement in the crack, and aggregateinterlock is destroyed in a few cycles. In hinge and joint zones, reversed diagonal cracking breaksdown in the concrete between the cracks completely, and sliding shear failure occurs. Refer toFigure 5.1.

Figure 5.1. Progressive failure of reinforced concrete hinge zone under seismic

loading.


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5.2.2 Principles of EarthquakePrinciples of Earthquake- - Resistant DesignResistant Design. In reinforced concrete structures, the essential featuresof earthquake-resistance are embodied in ensuring the following:

Beams should fail before columns. “Strong Column - Weak Beam” Concept. Design codesrequire that earthquake-induced energy be dissipated by plastic hinging of the beams, ratherthan the columns. This hypothesis is due to the fact that compression members such ascolumns have lower ductility than flexure-dominant beams. If columns are not stronger thanbeams framing to a joint, inelastic action can develop in the column. Furthermore, the

consequence of a column failure is far more severe than a local beam failure. This concept isensured by the following inequality:

where

M col = sum of moments at the faces of the joint corresponding to the nominal flexural

strength of the columns framing to that joint;

M beam = sum of moments at the faces of the joint corresponding to the nominal flexuralstrengths of the beams framing into that joint. In T-beam construction, where the slab isin tension under moments at the face of the joint, slab reinforcement within the effectiveslab width has to be assumed to contribute to flexural strength is the slab reinforcementis developed at the critical section for flexure.

beamcol MM Σ

≥Σ

5

6


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Failure should be in flexure rather than in shear. To prevent shear failure occurring beforebending failure, it is good practice to design that the flexural steel in a member yields whilethe shear reinforcement is working at a stress less than yield (say normally 90%). In beams, a

conservative approach to safety in shear is to make the shear strength equal to the maximumshear demands which can be made on the beam in terms of its bending capacity.

Referring to Figure 5.2, the shear strength of the beam should correspond to

where

V DL is the dead load shear force

M u is the factored moment, determined as

As is the steel area in the tension zone

f y is the maximum steel strength after hardening, say 95%

z is the lever arm

DL

uu V MM

V +−

=l

21

max

zf AM y su

=


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Premature failure of joints between members should be prevented. Joints between memberssuch as beam-column joints are susceptible to failure earlier than the adjacent members due

to destruction of a joint zone, in a manner similar to that shown in Figure 5.1. This isparticularly true mostly to exterior columns.

Ductile rather than brittle failure should be obtained. In earthquake engineering, the effect ofmaterial behavior on the choice of the method of analysis is a much greater issue than innon-seismic engineering. The problem can be divided into two categories depending onwhether the material behavior is brittle or ductile, i.e. whether it can be considered linearelastic or inelastic. The normal analytical and design methods of dealing with these twostates are summarized in the following table. See next page.

Figure 5.2. Shear strength

consideration for reinforced concrete

beams.

f y

ε

95 percentile

(-)

(+)

M u1

M u2

As

l

h

b


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Linear elastic

(brittle)

1. Working stress or factored ultimate stress

design, plus imposed nominal ductility

Arbitrarily

reduced

Equivalent-

static

8. Ductility demands found from plastic hingerotations

FullInelasticdynamic


design, plus imposed arbitrary ductility*



7. Structure intended to remain elastic, but

nominal ductility imposed

Arbitrarily

reduced

Arbitrarily

reduced

Full

Linear dynamic

Inelastic

(ductile)



Arbitrarily

reduced

Equivalent-

static


design, plus imposed nominal ductility

3. Ultimate stress design, plus imposed

nominal ductility

Arbitrarily

reduced

Full

Linear dynamic

Design ProvisionsSeismic

Loading

Method of

Analysis

Material

Behavior


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5.2.3 Available Ductility for Reinforced Concrete Members Available Ductility for Reinforced Concrete Members. The available section ductility of a reinforcedconcrete member is most conveniently expressed as the ratio of its curvature at ultimate momentφu to its curvature at its first yield φy. The expression φu /φy may be evaluated from its first

principles, the answers varying with the geometry of the section, the reinforcement arrangement,the loading and the stress-strain relationships of the steel and the concrete.

Single reinforced sectionsSingle reinforced sections. Consider conditions at first yield and ultimate moment as shownin Figure 5.3.

Assuming an under-reinforced section, first yield will occur in the steel, and the curvature

where

Figure 5.3. Reinforced

concrete section in

flexure.

dk E

f

E

f

dk

s

y

y

s

y

sy

sy

y

)(

whichin;)(

−=

=∈−

∈=

1

1

φ

φ

cc

s s

f'EEn

bd A

nnnk

4700 and whichin

)()(

20000

22

===

−+=

ρ

ρ ρ ρ

A s' d' kd

εce

εsy = f y/E s

(a) a t first yield

strain

f ce

f ystress

c

(b) at ultimate

εsy > f y/E sstrain

f ystress

εcu f cm = 0.85f'ca

A s

d

b

Note that the formula for k isNote that the formula for k istrue for linear elastictrue for linear elasticbehavior only, while forbehavior only, while for

higher concrete stresses thehigher concrete stresses thetrue nontrue non- - linear concretelinear concretestress block shall be used.stress block shall be used.


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but to allow for the effect of compression steel ratioρ

’, the expressions for c and k become

and

The above equations assume that the compression steel is yielding, but if this is not so, theactual value of the steel stress should be used f y . And as k has been found assuming linear

elastic behavior in concrete, the qualifications mentioned for singly reinforced members alsoapply.

1

1

850 β

ρ ρ

β

c

y

f

df c

ac

'.

)'( −=

=

bd

Annnk

s'' whichin

)'(])'[(])'[(

=

+−+++=

ρ

ρ ρ ρ ρ ρ ρ 22

Figure 5.4. Doublyreinforced concrete

section.

As' d'

As

d

b

Ad C Abi l


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Effect of confinement on ductilityEffect of confinement on ductility. The ductility and strength of concrete is greatly enhancedby confining the compression zone with closely spaced lateral steel ties. In order to quantifythe ductility of confined concrete, a number of stress-strain curves for confined concrete

have been derived. It is known that rectangular all-enclosing links are moderately effective onsmall columns, but are of little use in large columns. In large columns, this is remedied tosome extent by the use of intermediate lateral ties anchored to the all-enclosing links.

The procedure for calculating the section ductilityφu /φy is the same as that for unconfined

concrete as described herein, the only difference being in determining an appropriate value ofultimate concrete strain

∈cu for use in the expression for fu/fy. It is therefore recommended

that a lower bound for the maximum concrete strain for concrete confined with rectangularlinks may be used.

where

= ratio of the beam width to the distance from the critical section to the point ofcontraflexure

r v = ratio of volume of confining steel (including compression steel) to volume of concrete

confined

f yv = yield stress of the confining steel in N/mm2

2

1380200030

++=∈ yvv

c

cu

f b ρ

l..

c

b

l

Ad C Abi l f i


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E E xample Problem 5.2. Given a singly reinforced concrete beam section with 3-φ32 reinforcing bars

at the bottom. The confining steel consists ofφ

12 mild steel bars (f y = f yv = 275 N/mm2) at 75 mm

centers and the concrete strength is f’ c = 25 N/mm2. Estimate the section ductility

φu /φy.

Assume81/=

c

b

l

n.a.

A s = 3-Ø32 bars

500

250

c

Adam C Abinales f asep pice


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S S olution and discussion:

T T o find the curvature at first yield φy, first estimate the depth of the neutral axis, the section beingeffectively singly reinforced.

evaluate

A Although this implies a computed maximum concrete stress greater than 0.85f’ c , the triangularstress block gives a reasonable approximation. Thus, the curvature at first yield

radian/mm.).(

)(

610844

50043201200000

275

1

−=−

=

−=

x

dk E

f

y

s

y

y

φ

φ

4320

1640164021640

2

1640

511825

200000

200000

01930

2

2

..).().(

)()(

.then,

.4700

4700

and .

(250)(500)804*3

=−+=

−+=

=

==

==

=

==

k k

nnnk

n

n

f'E

En

bd A

cc

s

s

ρ ρ ρ

ρ

ρ

ρ

2

1380200030

++=∈

yvv

c

cuf b ρ

l..


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Adam C Abinales f asep pice


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5.2.4 Ductility of reinforced concrete members with flexure and axialDuctility of reinforced concrete members with flexure and axial load load . Axial load unfavorably affectsthe ductility of flexural members. It is therefore imperative that for practical levels of axial load,columns must be provided with confining reinforcement.

For rectangular columns with closely spaced links, and in which the longitudinal steel is mainlyconcentrated in two opposite faces, the ratio

φu /φy may be estimated from Figure 5.5.

Figure 5.5.φu /φy for columns

of confined concrete.



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where

As = area of tension reinforcement, mm2

and

Ah = cross-sectional area of the links, mm2

f yh = yield stress of the link reinforcement, N/mm2

s = spacing of the link reinforcement, mm

hh = the longer dimension of the rectangle of concrete enclosed by the links, mm

ch

yhhh

f shf A'

.21= β

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Seismic Design and Detailing_pp131-150