detallamiento.pdf

8/10/2019 detallamiento.pdf

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THE CIVIL & STRUCTURAL ENGINEERING PANELTHE CIVIL & STRUCTURAL ENGINEERING PANEL

ENGINEERS AUSTRALIA SYDNEY DIVISIONENGINEERS AUSTRALIA SYDNEY DIVISION

28 August 2012

DetailingofReinforcementDetailingofReinforcement

inConcrete

Structuresin

Concrete

Structures

R.I.GilbertR.I.Gilbert


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DetailingDetailingisoftenconsideredtobetheisoftenconsideredtobethepreparationofworkingpreparationofworking

drawingsdrawingsshowingthesizeandlocationofthereinforcementinashowingthesizeandlocationofthereinforcementina

concretestructure.concretestructure.

DetailingDetailinginvolvestheinvolvesthecommunicationcommunicationoftheengineeroftheengineersdesigntothesdesigntothe

contractorswhobuildthestructure.Itcontractorswhobuildthestructure.Itinvolvesthetranslationofainvolvesthetranslationofa

goodstructuraldesignfromthecomputerorcalculationpadintogoodstructuraldesignfromthecomputerorcalculationpadintothethe

finalstructure.finalstructure.

GooddetailingensuresthatGooddetailingensuresthatreinforcementreinforcement

andconcreteinteractefficientlyandconcreteinteractefficientlytoprovidetoprovide

satisfactorysatisfactorybehaviourbehaviourthroughoutthethroughoutthe

completerangeofloading.completerangeofloading.

Inthisseminar,guidelinesforInthisseminar,guidelinesforsuccessfulsuccessful

detailingdetailinginstructuralelementsandinstructuralelementsand

connectionsareoutlined.connectionsareoutlined.

IntroductionIntroduction::


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Thedetailingrequirementsofareinforcementbardependonthe

reasonsfor

its

inclusion

in

the

structure.

Reasonsinclude:

1. Tocarryinternaltensileforces,therebyimpartingstrength

andductility;

2. Tocontrolflexuralcracking;

3. Tocontroldirecttensioncrackinginrestrainedstructures;

4. To

carry

compressive

forces;5. Toproviderestrainttobarsincompression;

6. Toprovideconfinementtoconcreteincompression;

7. Tolimitlongtermdeformation;

8. Toprovideprotectionagainstspalling;and

9. Toprovidetemporarysupportforotherreinforcementduring

construction.


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Guiding

principles:

Determinelocationanddirectionofallinternalforces(i.e.

establishaloadpaththatsatisfiesequilibrium);

Useadequatelyanchoredreinforcementwhereveratensile

forceis

required

for

equilibrium;

Useonlyductilereinforcement(ClassNorbetter)whenthe

reinforcementisrequiredforstrength;

Neverrelyon

the

concretes

ability

to

carry

tension

(it

may

not

exist);

Includeadequatequantitiesofreinforcementforcrackcontrol;

Ensuresteeldetailsarepracticalandthatsteelcanbefixedand

concretecan

be

satisfactorily

placed

and

compacted

around

complexdetailswithadequatecover;and

Ensuredetailsareeconomical.


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Sources

of

tension:

1. Tensioncausedbybending(andaxialtension):

Positive bending

Negative bending

Axial tension

Flexuraltensioncracks

Flexuraltensioncracks

Directtensioncracks


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Sourcesoftension:

2. Tensioncausedbyloadreversals:

Cantilever beam or slabSimple beam or slab

Impact and rebound loading


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Sourcesoftension(ctd):

3. Tensioncausedbyshearandtorsion:

C

C T

T

Shear

Tensioncarriedbystirrups

Flexureshearcracks


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4. Tensionnearthesupportsofbeams:

Thelongitudinaltensionatthesupportisgreaterthanindicated

bythebendingmomentdiagram.

Thetensileforceatthebottomoftheinclinedcrackisequalto

thecompressiveforceatthetopofthecrack.


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Lst

AS36002009(8.1.10.4):

Sufficientbottomsteelmustbe

anchoredfor

alength

(Lst)

past

the

midpointofthebearingtodevelop

atensileforceofV*cotv/(plusany

additionalforcearisingfromrestraint)

Thisrequirementisdeemedtobesatisfiedifeither

Astisextendedpastthefaceofthesupportby 12db;or Astisextendedpastthefaceofthesupportby 12db+D/2

whereAstisthetensilesteelarearequiredatmidspan


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5. Tensionwithinthesupportsofbeamsandslabs:

Crackingduetoinadequate

slipjointbetweenslaband

supportingbrickwall


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6. Tensionwithinconnections:

Hanger

reinf.to

carry

tension

Primarygirder

Compressionstruts

Reactionfromsecondarybeam

appliedhere

Secondary

beam

M

M

C

C

T

T

M

M

(a)Internalforces (b)Crackpattern

2 T


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7. Tensionatconcentratedloads:


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8. Tensioncausedbydirectionalchangesofinternalforces:

(a)

T

TT

R

stirrups(b)

Lsy.t

(c)

C C

R Potentialcrackinweb

Asvatspacings


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8. Tensioncausedbydirectionalchangesofinternalforces:

(a)

T

TT

R

stirrups(b)

Lsy.t

(c)

Asvatspacings

C

T

C

T

rm

qt

Ast

m

syst

m

tr

fA

r

Tq == m

sy

vy

st

sv

t

vysvr

f

f

A

A

q

fAs ..==

Transversetension: Requiredstirrupspacing:


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Splittingfailuresarounddevelopingbars.

F F F

TF F F

Tensile stresses

tr Atr

Splitting cracks

a) Forces exerted by concrete on a deformed bar (b) Tensile stresses in concrete

at a tensile anchorage

(c) Horizontal splitting due (d) Vertical splitting due to (e) Splitting (bond) failureinsufficient bar spacing. insufficient cover at a lapped splice.

Anchorage

of

deformed

bars

is

tension:


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Lsy.t

fb

Asfsy

b

sybtsy

f

fdL

4.

Forareinforcementbartoreachitsyieldstressatacritical

crosssection,

aminimum

length

of

reinforcing

bar

(an

anchorage)isrequiredoneithersideofthesection.

AS36002009specifiesaminimumlength,calledthedevelopment

length,Lsy.t,overwhichastraightbarintensionmustbeembeddedin

theconcrete

in

order

to

develop

the

yield

stress.

Anaveragedesignultimatebondstressfbisassumedattheinterface

betweentheconcreteandthereinforcingbar(=0.6).

fbdependson typeandconditionofreinforcingbar;strength

andcompactionofconcrete;concretecover;

barspacing;transversereinforcement;

transversepressure

(or

tension).


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Thebasicdevelopmentlength,Lsy.tb,is

where k1=1.3 forahorizontalbarwith>300mmofconcretecast

below

it

and

k1=

1.0

for

all

other

bars;

k2=(132db)/100;

k3=1.00.15(cd db)/db (but 0.7k31.0)

cd isthe

smaller

of

the

concrete

cover

to

the

bar

or

half

thecleardistancetothenextparallelbar;

fc shallnotbetakentoexceed65MPa

AS36002009: (13.1.2.2)

bdk129c2

bsy31

sy.tb

5.0

fk

dfkkL

=


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AS36002009 (13.1.2.2)

c1

c

a/2

cd=min(a/2,c,c1)


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ThedevelopmentlengthLsy.t maybetakenasthebasic

developmentlengthormayberefinedtoincludethebeneficial

effectsofconfinementsbytransversesteelortransversepressure

andis

where k4=1K (but0.7k41.0);and

k5=1.00.04p (but 0.7k51.0);

AS3600-2009 ctd (13.1.2.3)

sy.tb54sy.t LkkL =


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FIGURE 13.1.2.3(B) VALUES OFKFOR BEAMS AND SLABS

K = 0.1 K = 0.05 K = 0

sy.tb54sy.t LkkL =k4 = 1 -K

where

= (AtrAtr.min)/As;

Atr = cross-sectional area of the transverse reinforcement along the development

lengthLsy.t

Atr.min= cross-sectional area of the minimum transverse reinforcement, which may

be taken as 0.25Asfor beams and 0 for slabs

As = cross-sectional area of a single bar of diameter dbbeing anchored

K = is a factor that accounts for the position of the bars being anchored

relative to the transverse reinforcement, with values given below:

AS3600-2009 ctd (13.1.2.3)


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ThedevelopmentlengthLsttodevelopastressstlowerthanfsy:

Whencalculatingstdontforgettoincludethestrengthreductionfactor

(=0.8).IfT*isthedesignultimatetensileforceinthereinforcementcausedbythefactoreddesignloads,then:

andtherefore

AS3600-2009 ctd (13.1.2.3)

bsy

st

sy.tst

12df

LL =

st

st

stst

*

*

A

T

AT


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Thedevelopmentlengthofadeformedbarwithastandardhook

orcog:

AS3600-2009 ctd (13.1.2.3)

(a) Standard hook (180bend) (b) Standard hook (135bend).

0.5Lsy.t

4dbor 70mm

did

X

0.5Lsy.t

did/2

X

(c) Standard cog (90bend).

0.5Lsy.t

did /2

X

A A

A

X X

X

did 0.5d

id

0.5Lsy.t

0.5Lsy.t

0.5Lsy.t

4dbor70mm

(a)Standardhook(180bend) (b)Standardhook(135bend)

(c)Standardcog(90bend)


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WORKEDEXAMPLE:

Considertheminimumdevelopmentlengthrequiredforthetwo

terminated28mmdiameterbottombarsinthebeamshownbelow.

Takefsy = 500 MPa;fc = 32 MPa;covertothe28mmbarsc = 40 mm;

andtheclearspacingbetweenthebottombarsa = 60 mm.

ThecrosssectionalareaofoneN28barisAs = 620 mm2 andwithN12

stirrups

at

150

mm

centres,

Atr= 110 mm2

.

AS3600-2009

P P

Lsy.t

12mm stirrups at 150mm ctrs

Two terminated bars

A

A

Elevation Section A-A

Lsy.t+ d

Lsy.t +D


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Forbottombars:k1 = 1.0;

For28mmdiameterbars:k2 = (132 28)/100 = 1.04;

The

concrete

confinement

dimension,cd= a/2 = 30 mm,

and

thereforek3 = 1.0 0.15(30 28)/28 = 0.99

Thebasicdevelopmentlengthistherefore

The

minimum

number

of

stirrups

that

can

be

located

within

the

basic

developmentlengthis7.Therefore, Atr= 7 x 110 = 770 mm2.

Taking Atr.min = 0.25As = 155 mm2,theparameter

= (770 155)/620 = 0.99

Worked Example ctd (13.1.2.3)

)29(mm11783204.1

2850099.00.15.01sy.tb bdkL >=

=

c2

bsy31

sy.tb

5.0

fk

dfkk

L =


25/82

FromFigure13.1.2B,K = 0.05(asitisthetwointeriorbarsthatarebeing

developed)andtherefore

Itisassumedthatinthislocationthetransversepressureperpendicularto

theanchoredbar(p)iszero,andhencek5 = 1.0.

FromEq.13.1.2.3:

Worked Example ctd (13.1.2.3)sy.tb54sy.t LkkL =

95.099.005.00.10.14 === Kk

.mm112011780.195.0.54. === btsytsy LkkL

Thestrengthofthebeammustbecheckedatthepointwherethetwo

barsareterminated(ie.atLsy.t+dfromtheconstantmomentregion)


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LappedSplicesforbarsintension(13.2.2LappedSplicesforbarsintension(13.2.2AS3600AS36002009):2009):

PLANAR VIEW

sL adb sb

Lsy.t.lap

Note: For the purposes of determining cd, the dimension a shall be taken equal to (sL-db)

irrespective of the value ofsb.

cd, = min (a/2, ccrit )

(i) 100% of bars spliced (no staggered splice)

cd, = min (a/2, ccrit )

(ii) 50% staggered splices

PLANAR VIEW

Lsy.t.lap

0.3Lsy.t.lap

a

L

b

Note: For the purposes of determining cd, the dimension a shall be taken equal to 2sL

irrespective of the value ofsb.

(a/2, c )

(a/2, c )

cd= min (a/2, c)

(i) 100% of bars spliced (no staggered splices)

(ii) 50% staggered splicescd= min (a/2, c)


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LappedSplicesforbarsintension:LappedSplicesforbarsintension:

AS36002009: 13.2.2

sy.t7sy.t.lap LkL =

bdk

129

k7 shallbetakenas1.25,unlessAs providedisgreaterthanAs required

andnomorethanonehalfofthetensilereinforcementatthesectionis

spliced,inwhichcasek7

=1.

Innarrowelementsormembers(suchasbeamwebsandcolumns),the

tensilelaplength(Lsy.t.lap)shallbenotlessthanthelargerofk7Lsy.tand

Lsy.t +1.5sb,wheresbisthecleardistancebetweenbarsofthelapped

spliceas

shown

in

Figure

8.15.


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ConsiderthelappedsplicerequirementsforN12barsat200mmcentresinthe

bottomof

aslab.

Cover

=20

mm.

Concrete

strength

=25

MPa.

AS36002009:

ACI31808:(Refined Clause12.2.3)

b

et

dlap df

fL

c

y

1.23.13.1

==

l

12250.11.2

0.10.15003.1

=

bd9.61mm743 ==

b

b

trb

set

dlap d

d

Kcf

fL

)(1.1

3.13.1

c

y

+

==

l

12

)12

026(250.11.1

8.00.10.15003.1

+

=

bd7.43mm524 ==

ACI31808:(Simplified Clause12.2.2)

c2

bsy31

sy.tbsy.t.lap

5.025.1

fk

dfkkLL

==

252.1

1250090.00.15.025.1

=

bd9.46mm563 ==

)20013600ASin

7.30mm369..(

= bdfc


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Detailingofbeams:

Anchorageof

longitudinal

reinforcement:

Favorable

anchorage

Elevation Section

Unfavorableanchorage Transversetension

Possiblecracks

Normalpressure

C

C

T TT

Whenbottomreinforcementis

terminatedaway

from

the

support,

thediagonalcompressionintheweb

improvestheanchorage.


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Currentwording:

Thedesignforflexuralstrengthanddetailingofflexuralreinforcementandpretensioned tendonsatterminationshallbeextendedfromthe

theoreticalcutoffpoint,ordebonding point,byalengthof1.0D+Lsy.t,or

1.0D+Lpt,whereDisthememberdepthatthetheoreticalcutoffpointor

theoreticaldebonding point

Problem1: Thewordingdoesnotmakesense

Problem2: Theruleisincorrect abardoesnothavetodevelop

itsyield

stress

at

the

theoretical

cut

off

point

Amendedwording:Whereflexuralreinforcementandpretensioned tendonsaretobe

terminated,the

bars

or

tendons

shall

be

extended

from

the

theoretical

cut

offpoint,ortheoreticaldebonding point,byalengthofatleast1.0D+Lst,

or1.0D+Lpt,respectively,whereDisthememberdepthatthetheoretical

cutoffpointortheoreticaldebonding point

AS3600-2009 Clause 8.1.10.1


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Detailingofbeams(ctd):

tiltedanchorage nearhorizontalanchorage diagonalcompression

Reactionpressure Reactionpressure

Sectionsand

Elevations

Plan

Thetransversetensionthatmaycausesplittingin

theplane

of

ahooked

anchorageat

asupport

can

beovercomeatabeamsupportsimplybytiltingthe

hookandexposingittothenormalreactionpressure.


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Ifthebearinglengthatasupportissmallandclosetothefreeendofa

member,asliding shearfailurealong

asteep

inclined

crack

may

occur.

Additionalsmalldiameterbarsmayberequiredperpendiculartothe

potentialfailureplane

Potentialfailuresurface

Inclinedclampingbars


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Wherethelengthavailableforanchorageissmall,mechanical

anchoragesinthe

form

of

welded

cross

bars

or

end

plates

may

be

used.

Commoninprecastelements,corbels,bracketsandatothersupport

points.

welded

crossbarendplate

(a)

(b)

(c)

recessed

angle


34/82


Inshort

span

members,

where

load

is

carried

to

the

support

by

arch

action,itisessentialthatallbottomreinforcement(thetieofthearch)

isfullydevelopedateachsupport.Closelyspacedtransversestirrups

canbeusedtoimproveanchorageofthetiereinforcement.

Compressivestrut

Tie

DoNOTterminateanybottombars

Binding

reinforcement

Anchorageis

critical

Member

Centreline


35/82


Concentratingtopsteelatasupportinabeamwithinthewebcanlead

tocrackcontrolproblemsintheadjacentslab(Leonhardt etal.)

0

10

20

3040

50

60

70

0 100 200 300 400 500

Load kN

Cr

ackwidth(

0.0

1mm))

As= 1030 mm2

As= 1020 mm2


36/82


AnchorageofStirrups:

Tensioninstirrupismoreorlessconstantoverheightofverticalleg.

Therefore,allpointsonverticallegmustbefullydeveloped

Stirrupanchoragesshouldbelocatedinthecompressivezoneand

beshown

on

the

structural

drawings.

Theareaofshearreinforcementrequiredataparticularcross

sectionshouldbeprovidedforadistanceDfromthatcrosssection

inthedirectionofdecreasingshear(AS36002009 Clause8.2.12.3).

Compressivetopchord(concrete)

Verticalties(stirrups)Inclinedwebstruts

(concrete)

Tensilebottomchord(Ast)


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Typesof

Stirrups:

(a)Incorrect

Inadequateanchorage

A90cogisineffectiveifthe

coverconcrete

is

lost

Tensilelappedsplice

(c)Satisfactory

Compressiveside

Tensileside

(b)Undesirable(butsatisfactory)

Inregionswhereductilityisrequired,

theopenstirrups(commonlyusedin

posttensioned

beams)

do

not

confine

the

compressiveconcrete


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Typesof

Stirrups:

cracks

Ts

Cd

Compressionstrut

Cd

TsTs

Rigid Flexible

Multilegstirrupsshouldbeusedinmemberswithwidewebsto

avoidtheundesirabledistributionofdiagonalcompressionshown

Multileg

sturrups better

control

shear

cracking

and

help

maintain

sheartransferthoughaggregateinterlock


39/82


Typesof

Stirrups:

Multilegstirrupsarealsofarbetterforcontrollingthe

longitudinalsplittingcracks(knownasdowelcracks)that

precipitate

bond

failure

of

the

longitudinal

bars

in

the

shear

span.

Oftenthiscriticalshearcrackoccurswherebottombarsare

terminatedintheshearspan.Additionalshearreinforcementmay

berequiredinthisregion(Clause8.1.10.5 AS36002009).

Dowel crack


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Crackcontrolprovidedbyshearreinforcement(Leonhardt etal):

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0 200 400 600

Load P (kN)

Maximumcrackwidth(mm

1

2

3

4

(mm)


41/82


Supportand

Loading

Points:

Whenthesupportisatthesoffitofabeamorslab,thediagonal

compressionpassesdirectlyintothesupportasshown

When

the

support

is

at

the

top

of

the

beam,

the

diagonal

compression

mustbecarriedbackuptothesupportviaaninternaltie.

Itisessentialthatadequatelyanchoredreinforcementbeincluded

toactasthetensiontieandthereinforcementmustbeanchored

into

the

support

(a)

Supportunder

support

Internal

tie

(b)

Supportover


42/82


Slabsupported

by

upturned

beam:

(a)Incorrectdetail

Theverticalcomponentofthediagonalcompressionintheslab

(i.e.thereactionfromtheslab)mustbecarriedintensionuptothe

topof

the

upturned

beam.

Dontrelyontheunreinforcedsurfacetocarrythistension

Unreinforced

surface

(b)Correctdetail


43/82


Beam

to

beam

connection:

Theareaofadditional

suspensionreinforcement

is

Primary

girder

R*

suspensionreinforcement

(a)Section

Secondary

beam

Compressionstrutinsecondarybeam

Suspension

reinforcement

(b)PrimarygirderElevation

Stirrups

forshear

(c)Primarygirder Trussanalogy

R*

sy

srf

RA

*

=


44/82


Beam

to

beam

connection:

Whenaloadisappliedtotheundersideofareinforcedconcretebeam,somedevicemustbeusedtotransferthishangingloadtothe

topofthebeam

(b)Internalrods


45/82


Half

Joint

or

dapped

end

joint:

(a)Halfjoint

(b)Strutandtiemodel Reinforcementdetail

Hairpinreinforcement

Suspensionreinforcement

(c)Alternativestrutandtiemodel Reinforcementdetail


46/82

ExcessivecrackingExcessivecrackingduetoduetorestraineddeformationrestraineddeformationororexternalexternal

loadsloadsisacommoncauseofdamageinreinforcedconcreteisacommoncauseofdamageinreinforcedconcrete

structures.structures.

ExcessivecrackingExcessivecrackinginthehardenedconcretecanbeavoidedinthehardenedconcretecanbeavoided

byincludingsufficientreinforcementatsufficienbyincludingsufficientreinforcementatsufficientlyclosetlyclose

spacingsspacings..

ShrinkageShrinkagecausesacausesagradualwideningofexistingcracksgradualwideningofexistingcracksandand

timetimedependentcrackingdependentcrackinginpreviouslyuncrackedregions.inpreviouslyuncrackedregions.

Detailing

for

Crack

ControlDetailing

for

Crack

Control

TheTheminimumquantitiesofreinforcementminimumquantitiesofreinforcement specifiedforcrackspecifiedforcrack

controlinAS3600maynotbewhatisactuallyrequcontrolinAS3600maynotbewhatisactuallyrequiredinalliredinall

circumstances.circumstances.


47/82

The

Thewidth

of

acrackwidth

of

acrackdepends

on

depends

on

thequantity,orientationanddistributionofthethequantity,orientationanddistributionofthe

reinforcingsteelcrossingthecrack;reinforcingsteelcrossingthecrack;

concretecoverandmembersize;concretecoverandmembersize;

thebondbetweenconcreteandreinforcementthebondbetweenconcreteandreinforcementinthevicinityofthecrack;inthevicinityofthecrack;

thedeformationcharacteristicsofconcrete;andthedeformationcharacteristicsofconcrete;and

theshrinkagestrain(andthereforethetimeaftertheshrinkagestrain(andthereforethetimeafter

crackformation).crack

formation).

thecauseofthecrackthecauseofthecrack

thedegreeofrestraintthedegreeofrestraint

Oftensignificantly

more

reinforcement

than

the

minimum

specifiedamountisrequired.


48/82

Crack spacing, s, varies between0.5dand 1.5dand depends on

- steel area and distribution- cover

Crack width, w, depends on

- steel stress- bar diameter and bar spacing- cover- adjacent crack spacings

and the average crack spacingdecreases with time due toshrinkage

and increases with time due toshrinkage

d

Maximum crack widthsincrease with time by a factor

of between 2 and 4

Service loads

Flexural cracks

Flexuralcracking:


49/82

SimplifiedApproachforFlexuralCrackControlinAS36002009

(Clause8.6.1

and

9.4.1):

Forreinforcedconcretebeamsandslabs,crackingisdeemedtobecontrolled

(crackwidthswillbelessthan0.3mm)ifeachofthefollowing issatisfied:

(a)thequantityoftensilereinforcementinabeamorslabprovidesan

ultimatestrengthatleast20%higherthanthecrackingmoment

calculatedassumingcs=0;

(b) thedistance

from

the

side

or

soffitof

the

member

to

the

centre

of

the

nearestlongitudinalbarshallnotexceed100mm;

(c)Thecentretocentrespacingofbarsnearatensionfaceofabeamor

slabshall

not

exceed

300

mm

for

abeamand

the

lesser

of

two

times

theslabthicknessand300mmforaslab.

(d)Thestressinthetensilesteelislessthanalimitingvalue(asfollows):


50/82


(Clause8.6.1

and

9.4.1):

Ctd

Formemberssubjectprimarilytoflexure,thecalculatedsteelstresscaused

bytheserviceabilitydesignmoment shallnotexceedthelarger ofthe

maximumsteel

stresses

given

in

Tables

8.6.1(A)

and

8.6.1(B)

for beams

andTables9.4.1(A)and9.4.1(B)forslabs.

Table 8.6.1(A): Maximum steel stress for Table 8.6.1(B): Maximum steel stress for

tension or flexure in r.c. beams. flexure in r.c. beams.

Nominal bardiameter

(mm)

Maximum steelstress

(MPa)

Centre-to-centrespacing

(mm)

Maximum steelstress

(MPa)10 360 50 360

12 330 100 320

16 280 150 280

20 240 200 240

24 210 250 200

28 185 300 160

32 160

36 140

40 120


51/82


(Clause8.6.1

and

9.4.1):

Ctd

Formemberssubjectprimarilytotension,thecalculatedsteelstresscaused

bytheserviceabilitydesignactionsshallnotexceedthemaximumsteel

stressesgiven

in

Tables

8.6.1(A)

for

beams

and

Tables

9.4.1(A)

for

slabs.

Table 9.4.1(A): Maximum steel stress for Table 9.4.1(B): Maximum steel stress for

flexure in r.c. slabs. flexure in r.c. slabs.

Maximum steel stress (MPa)for overall depthDs(mm)

Nominal bardiameter

(mm) 300 > 300

Centre-to-centrespacing

(mm)

Maximum steelstress

(MPa)

6 375 450 50 360

8 345 400 100 320

10 320 360 150 280

12 300 330 200 24016 265 280 250 200

20 240 300 160

24 210


52/82

RestrainedShrinkageCrackinginSlabsRestrainedShrinkageCrackinginSlabs::

ProvidedthatProvidedthatbondedreinforcementbondedreinforcementatatreasonablespacingreasonablespacingcrossescrosses

thecrackandthatthememberdoesnotthecrackandthatthememberdoesnotdeflectexcessively,flexuraldeflectexcessively,flexural

cracksareusuallywellcontrolledinreinforcedconcretecracksareusuallywellcontrolledinreinforcedconcretebeamsandbeamsand

slabs.slabs.

Incontrast,Incontrast,directtensioncracksdirecttensioncracksduetorestrainedshrinkageandduetorestrainedshrinkageand

temperaturechangesfrequentlyleadtoserviceabilityprobtemperaturechangesfrequentlyleadtoserviceabilityproblems,lems,

particularlyinregionsoflowmoment.particularlyinregionsoflowmoment.

SuchcracksusuallyextendcompletelythroughthememberandaSuchcracksusuallyextendcompletelythroughthememberandarere

moreparallelsidedthanflexuralcracks.moreparallelsidedthanflexuralcracks.

If

uncontrolled,

these

cracks

can

become

very

wide

and

lead

toIf

uncontrolled,

these

cracks

can

become

very

wide

and

lead

towaterproofingandcorrosionproblems.waterproofingandcorrosionproblems.

TheycanalsodisrupttheintegrityandthestructuralactionTheycanalsodisrupttheintegrityandthestructuralactionoftheslab.oftheslab.

Th l b i t i d b b d h i k i d t i


53/82

Flexuralcracks

Onewayfloorslabsupportedonbeams

Fulldepthrestrainedshrinkagecracks

Usuallymoresteelisrequiredtocontroltherestrainedshrinkage

cracksthanisrequiredtocontroltheflexuralcracksandprovide

adequatestrength.

Theslabisrestrainedbybeamsandshrinkageinducestension

intheslabinthedirectionofthebeams

R t i d Sh i k C ki i Sl bR t i d Sh i k C ki i Sl b CtdCtd


54/82

RestrainedShrinkageCrackinginSlabsRestrainedShrinkageCrackinginSlabsCtdCtd::

Inthe

primary

direction,

shrinkage

will

cause

small

increases

in

the

widthsofthemanyfineflexuralcracksandmaycauseadditional

flexuretypecracksinthepreviouslyuncracked regions.

However,inthesecondarydirection,whichisineffectadirect

tensionsituation,

shrinkage

generally

causes

afew

widely spaced

crackswhichpenetratecompletelythroughtheslab.

Iftheamountofreinforcementcrossingadirecttensioncrackis

small,yieldingofthesteelwilloccurandawide,unserviceablecrack

willresult.

To

avoid

this

eventuality,

the

minimum steel

ratio,

minis

where . For32MPa concrete, min=0.0034.

Foraserviceablecrackwidth,significantlymoresteelthanthisis

required.

sy

ctst

f

f

db

A 2.1

min

min =

=

'25.0 cct ff =


55/82

CrackControlinSlabsCrackControlinSlabsAS3600AS36002009:2009:

Wheretheendsofaslabarerestrainedandtheslabisnotfree to

expandorcontractinthesecondarydirection,theminimumareaof

reinforcementintherestraineddirectionisgivenbyeitherEq.1a,

1bor1c,asappropriate(seebelow).

Foraslabfullyenclosedwithinabuildingexceptforabriefperiodof

weatherexposureduringconstruction:

(i)

where

a

strong

degree

of

controlover

cracking

is

required:

(ii) whereamoderatedegreeofcontrolovercrackingisrequired:

(iii) whereaminordegreeofcontrolovercrackingisrequired:

( ) )a2.9(10)5.20.6( 3min

= DbA cps

( ) )b2.9(10)5.25.3( 3min

= DbA cps

( ) )c2.9(10)5.275.1( 3min

= DbA cps

(1a)

(1b)

(1c)

F ll h l b f di i i E Cl ifi i A1


56/82

ForallotherslabsurfaceconditionsinExposureClassificationA1

and

for

exposure

classification

A2,

Eq.

1a

applies

where

a

strong

degreeofcontrolovercrackingisrequiredforappearanceorwhere

cracksmayreflectthroughfinishes

andEq.

1b

applies

where

a

moderatedegreeof

control

over

cracking

isrequiredandwherecracksareinconsequentialorhidden fromview.

ForExposure

Classifications

B1,

B2,

C1

and

C2,

Eq.

1a

always

applies.

TheminimumsteelareagivenbyEq.1cisappropriateinan

unrestraineddirectionwheretheslabisfreetoexpandorcontract.

Inthe

primary

directionof

aone

way

slab

or

in

each

direction

of

a

twowayslab,theminimumquantityofreinforcementisthegreaterof

theminimumquantityrequiredforthestrengthlimitstate or75%of

theminimumarearequiredbyEqs.1a,1bor1c,asappropriate.

( ) )a2.9(10)5.20.6( 3min

= DbA cps

( ) )b2.9(10)5.25.3( 3min

= DbA cps

(1a)

(1b)

Consideraslabrestrainedateachend.


57/82

Withtime,restrainedshrinkagecracksoccuratroughlyregularcentres

dependingon

the

amount

of

reinforcement:

(a) Portion of restrained member after all cracking

(c) Steel stress after all shrinkage cracking

(b) Average concrete stress after all shrinkage


58/82

Typicalvalues:

Considera140mmthick,4mlongslabfullyrestrainedatbothends

andsymmetricallyreinforcedwithN12barsat250mmcentres top

andbottom.Hence,As=900mm2/mand=As/Ac=0.00643.

L = 4 m

140 mm

For25MPa concretewithafinalshrinkagestrainof0.0007and

typicalmaterialproperties,ashrinkagecrackinganalysisofthis

restrainedslabindicates4or5fulldepthcrackswithinthe4m

lengthwiththemaximumfinalcrackwidthabout0.3mm.


59/82

If p=0:

p=As/Ac

2.8 mm onelargeunserviceable

crack

If p=0.0035

0.6 0.7 mmabout

three

unserviceable

(?)cracks

If p=0.006

0.3 0.4 mmFour

or

five

serviceable

(?)cracks

4 m140 mm

Detailingofcolumns:


60/82

g

Lappedcompressive

splices:

Normal

fitment

spacing,s

Additional

fitmentspacing,

s


61/82

g

Typicaltie

arrangements

in

columns:

AS36002009requirementsforrestrainingsinglelongitudinalbarsincolumns:

(i) Everycornerbar;

(ii)

All

bars

where

bars

are

spaced

at

centres >

150

mm;(iii)Atleasteveryalternatebarwherebarcentres150mm.

Forbundledbars eachbundlemustberestrained.

Alllongitudinalbarsin

thesecolumnsarerestrained

at

(i)

a

bend

in

a

fitment

of

135orless;or(ii)atafitmenthookswith

includedangleof135orless,asshown.

Detailingofcolumns:


62/82

Minimumbar

diameters

for

fitments

(AS3600

2009):

610

12

16

12

Single bars up to 20Single bars 24 to 28

Single bars 28 to 36

Single bar 40

Bundled bars

Minimum bar diameter

for fitment and helix (mm)

Longitudinal bar diameter

(mm)

Maximumspacingoffitments(AS36002009):

Thespacingoffitments(orthepitchofahelix)shouldnotexceedthe

smallerof:

Dcand15dbforsinglebars

0.5Dcand7.5dbforbundledbars

DetailingofBeamcolumnConnections:


63/82

KneeConnections

(or

two

member

connections):

(a) (b) (c)

(d)

Figure 8.37 Two-member connections.



64/82

KneeConnections

under

Opening Moment:

M

M

C

C

T

T

T2M

M

(a) Internal forces (b) Crack pattern

M

M

M

M

(a) Unsatisfactory (b) Unsatisfactory

M

M

(c) Potentially satisfactory

fsy

syst

fsy

svf

fA

f

TA

..

22

==



65/82

KneeConnections

under

Opening Moment

Suggested

detail:

Diagonal flexural bars

Diagonal

stirrups

M

M



66/82

KneeConnections

under

Closing Moment:

M

M

T

T

C

C

T2

M

M


M

M

(a) Wall or slab connection (whenp fct.f fsy) (b) Beam to column knee connection

M

M



67/82

Threemember

connections:


High bond stress

Poor anchorageconditions



68/82

Threemember

connections

Reinforcement

detail:

Larger diameter bar to distribute

bearing stresses in bend

Ties to carry diagonal tension, tocontrol vertical splitting and to

confine the concrete core



69/82

Fourmember

connections:

(a) Internal forces (b) Crack pattern (c) Reinforcement detail

DetailingofCorbels:


70/82

(a) Strut-and-tie action (b) Reinforcement detail (c) Welded primary steel

Crack control steel

Main or primary tensilereinforcement

Cross bar todistribute bearingstresses in bend

T

C

weld=db

weld=db

tweld=db/2

tweld=db/2

db

db

Primary tensile reinforcement

Anchorbar

(d) Satisfactory weld details (17)

Weldedanchor bar

(see Fig 8.46d)

Primarytension steel

DesignofCorbels:

*


71/82

T

C

d D

a

V*

Figure 8.47

d/2

)8.0(tan

*

=== sys fAV

T

tan

*

sy

sf

VA =

)6.0(9.0 == stccsstust AfC

Tie:

Strut:

)0.13.0(cot66.00.1

12

+

= ss

2/sh AA

ParkandPaulay suggestthatagoodfirstestimateofcorbeldimensions

isobtainedfrom:

andACI31808suggests

cw fdbV 56.0/*

sycwssyc ffdbAffda /2.0//04.0and0.1/

DesignofCorbels:


72/82

d D

400mm

D/2

200 mm

T*

C*

(b)

V*

dc=400/sin

= 541 mm

d D

400mm

V*= 500kN

D/2 200mm

(a)

bw = 300 mm

MPa32=cf MPa500=syf Cover = 30 mm

Bearing plate = 200 x 300 mm in plan

dc = 200/sin= 270 mm

DesignofCorbels:


73/82

From :56.0/* cw fdbV mm5263256.0300

105003

=

d

WithD = d+cover+0.5bardia andassuming20mmdiameterbars,

takeD=570mmandtherefored =530mm.

Fromthegeometry:

and

Try

4

N20

bars

(1240

mm2

)Now

Thestrutefficiencyfactor:

and

400

)90tan(100tan

=

d o7.47=

23

mm1138

7.47tan5008.0

10500=

=sA

OK/122.00078.0/ == sycws ffdbA

65.0

cot66.00.1

12 =

+

=

s

kN91181150329.065.06.0 ==ustC

OKkN743cos/** ==> VC

DesignofCorbels:


74/82

285

285

4 N20

3 N12stirrups

ELEVATION

PLAN

285

285

4 N20

3 N12Stirrups

N28 weldedcross-bar

ELEVATION

PLAN

N24 cross-bar

(welded to N20s)

JOINTSINSTRUCTURES:


75/82

Jointsare

introduced

into

concrete

structures

for

two

main

reasons:

1) Asstoppingplacesintheconcretingoperation.Thelocationof

theseconstructionjointsdependsonthesizeandproduction

capacityof

the

construction

site

and

work

force;

2) Toaccommodatedeformation(expansion,contraction,rotation,

settlement)withoutlocaldistressorlossofintegrityofthe

structure.Suchjointsinclude:

controljoints(contractionjoints);

expansionjoints;

structuraljoints(suchashinges,pinandrollerjoints);

shrinkagestrips;and

isolationjoints.

Thelocationofthesejointsdependsontheanticipated

movementsofthestructureduringitslifetimeandtheresulting

effectsonstructuralbehaviour.


76/82

ConstructionJoints:

Steel dowels to improve shear strength

1stpour 2ndpour

Waterstop where water tightness is required

(a) Butt joint (b) Keyed joint

(c) Doweled joint


77/82

Control

Joints

(or

Contraction

Joints):

Debond dowel to ensure free contraction

Saw cut > 0.2 t and 20 mm 0.75 t

Discontinue every second bar if necessary so thatp< 0.002

(a) Saw-cut joint in slab on ground (b) Wall (t < 200 mm)

(d) Wall (t 200 mm) (c) Doweled joint

t

0.75 t

Discontinue every second bar if necessary so thatp< 0.002


78/82

Typical

control

joint

locations:

Control joint locations

(a) Wall elevation

(b) Balcony plan


79/82

Alternative

shrinkage

strip

details:

Shrinkagestrip

Shrinkagestrip

Expansionjointdetails:


80/82

25 mm

Joint locations(a) Double column and beams

(b) Half joint

(c) Building plans joint locations

(a) Double column and beams

Alternativestructuralhingejointsatbaseofacolumn:


81/82

Elastic, easily

compressible

material

Mesnager

hinge

Confinement steel


82/82

THANKS FOR YOUR ATTENTION

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