Reworked Illustrative Examples - Griffith University

Reinforced and Prestressed Concrete - 3rd edition

Author

Chowdhury, Sanaul, Loo, Yew-Chaye

Published

2018

Version

Accepted Manuscript (AM)

Copyright Statement

© 2019 Cambridge University Press. This material has been published in Reinforced andPrestressed Concrete - 3rd edition by Y-C. Loo & S Chowdhury. This version is free to view anddownload for private research and study only. Not for re-distribution or re-use.

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REWORKED ILLUSTRATIVE/DESIGN EXAMPLES

IN

REINFORCED AND PRESTRESSED CONCRETE

THIRD EDITION

YEW-CHAYE LOO AM

SANAUL HUQ CHOWDHURY

AUGUST 2019

PREAMBLE

The Australian Standard AS 3600-2018 Concrete Structures was released while the third

edition of Reinforced and Prestressed Concrete was already in press. The illustrative/design

examples which are affected by the changes in AS 3600-2018 are reworked and presented

here.

CHAPTER-WISE ILLUSTRATIVE/DESIGN EXAMPLES

CHAPTER 3

3.4.6 Illustrative example for ultimate strength of a singly

reinforced rectangular section

Problem Forasinglyreinforcedrectangularsectionwithb 250mm,d 500mm, cf 50MPa,

andClassNreinforcementonly fsy 500MPa ,determinethereliablemoment

capacityforthefollowingreinforcementcases:

a Ast 1500mm2

b Ast 9000mm2

c a‘balanced’design

d withthemaximumallowablereinforcementratio pall

e Ast 4500mm2.

ThenplotM′againstpt.

Solution 2

uBsy

B

Equation 3.3(2)a: 0.85 0.0015 50 0.775

Equation 3.3(2)b: 0.97 0.0025 50 0.845

600Equation 3.4(4): 0.545

600

0.775 50 0.845 0.545Equation 3.4(5): 0.0357

500

kf

p

a Ast 1500mm2

t B

15000.012 0.0357

250 500p p

,thereforethesectionisunder‐reinforced.

6u

1 1500 500Equation 3.4(10): 1500 500 500 1 10

2 0.775 250 500 50 346.0 kNm

M

u

0.012 500Equation 3.4(8): 0.183

0.775 0.845 50k

Byassumingthesteelreinforcementtobeinonelayer,wehaved doand

kuo ku 0.183.

Equation 3.4(20)a: 1.24 13 01832/12 1.04

ForClassNreinforcement, Equation 3.4(20)b: 0.85

Accordingly, uEquation 3.4(19): 0.85 346.0 294.1 kNmM M

b Ast 9000mm2

t B

90000.072 0.0357

250 500p p

,thereforethesectionisover‐reinforced.

600 0.072 500Equation 3.4(16): 557.42

0.775 50

2557.42 4 557.42 0.845 500 557.42Equation 3.4(17): 280.92 mm

2a

6u

280.92Equation 3.4(18): 0.775 50 280.92 250 500 10

2 978.5 kNm

M

Buta γkudfromwhich

u

280.920.665.

0.845 500k

Byassumingonelayerofsteel,wehaved doandkuo ku 0.665.

Then,Equation 3.4(20)a: 0.52 but Equation 3.3(20)b stipulates that 0.65.


Thisexampleisforillustrativepurposesonly.Inpractice,onelayerisnot

enoughtoaccommodate9000mm2ofbarsinthegivensection.

c A‘balanced’design i.e.pB 0.0357

Ast 0.0357 250 500 4462.5mm2

u

6

Equation 3.4(10): 4462.5 500 500

1 4462.5 5001 10 858.7 kNm

2 0.775 250 500 50

M

Byassumingonelayerofsteel,wehaved doandkuo ku kuB 0.545.Then,

Equation 3.4(20)a with b: 0.65


d Withthemaximumallowablereinforcementratio pall

all

50Equation 3.4(6): 0.4 0.775 0.845 0.0262

500p

Ast 0.0262 250 500 3275mm2

u

6

Equation 3.4(10): 3275 500 500

1 3275 5001 10 680.4 kNm

2 0.775 250 500 50

M

Byassumingonelayerofsteel,wehaved doandkuo ku 0.4.Then



e Ast 4500mm2

t B

45000.036 0.0357, therefore the section is over reinforced.

250 500p p

600 0.036 500Equation 3.4(16): 278.71

0.775 50

2278.71 4 278.71 0.845 500 278.71Equation 3.4(17): 231.0

2a

6u

231.0Equation 3.4(18): 0.775 50 231.0 250 500 10

2 860.4 kNm

M

Buta γkudfromwhich

u

231.00.547

0.845 500k

Byassumingonelayerofsteel,wehaved doandkuo ku 0.547.Then,



TheM′versusptplotisgiveninFigure3.4 4 .Intheregionwherept pBthe

useofadditionalAstisnolongeraseffective.Thereasonisobvious,sincefailureis

initiatedbytheruptureofconcreteincompressionandnotbyyieldingofthesteelin

tension.Thus,inover‐reinforcedsituationstheuseofdoublyreinforcedsectionsis

warranted.Thisisdonebyintroducingreinforcementinthecompressivezoneas

elaboratedinSection3.6.

Figure 3.4(4) M ′ versus pt for a singly reinforced section

3.4.7 Spread of reinforcement

ForcomputingMu,Equations3.4 10 and3.4 18 arevalidonlyifthereinforcementis

reasonablyconcentratedandcanberepresentedbyAstlocatedatthecentroidofthebar

group.Ifthespreadofreinforcementisextensiveoverthedepthofthebeam,someof

thebarsnearertotheneutralaxismaynotyieldatfailure.Thisleadstoinaccuracies.A

detailedanalysisisnecessarytodeterminetheactualMu.Theexamplebelowillustrates

thegeneralprocedure.

Example: Computing Mu from a rigorous analysis

Problem

ComputeMuforthesectioninFigure3.4 5 ,assuming cf 32MPaand syf 500MPa.

Figure 3.4(5) Cross‐sectionaldetailsoftheexampleproblem

Solution For cf 32MPa,Equations3.3 2 aandb,respectively,giveα2 0.802andγ 0.89.

Andthereinforcementratios

t

37680.0209

300 600p

B t

0.802 32 0.89 6000.025 0.0209,

500 600 500p p

andthereforethesectionisunder‐reinforced.

1. Assumeallsteelyields i.e. .

T Astfsy 3768 500 10–3 1884kNand

=

C a 300 0.802 32 10–3 T 1884kN

fromwhicha 244.7mm

Thus,

u

sy

244.7274.9 mm

5000.0025

200000

ak d

uk d 274.9

cuε 0.003

s1ε

s3εs2ε

475.1

150

150

Figure 3.4(6) Straindistributionontheassumptionofallsteelyielding

FromFigure3.4 6 ,

s1 sy

s2 sy

175.10.003 0.0019 .

274.9

Equation 3 4 21

and

s sy

Therefore,theassumptionisinvalid.

2. Assumeonlythesecondandthirdlayersyieldwhilethefirstlayer

remainselastic Figure3.4 7 .

Figure 3.4(7) Straindistributionontheassumptionoffirststeellayernot

yielding

FromFigure3.4 7 ,

s1

u u

0.003 .

450 k d k d

Equation 3 4 22

Therefore

us1

u

(450 )600

k df

k d

SinceΣFx 0,wehaveC T,thatis

uu

u

(450 )0.802 32 0.89 300 1256 600 2512 500

k dk d

k d

or

2u u6.85( ) 502.4 339120 0k d k d

fromwhichkud 262.2mm

FromEquation3.4 22 ,wehaveεs1 0.00215 εsy.

Sinceεs3 εs2 εsy seeEquation3.4 21 ,thusAssumption2isvalid,thatis,onlythe

firststeellayerisnotyielding.

Figure 3.4(8) Leverarmsbetweenresultantconcretecompressiveforceandtensile

forcesatdifferentsteellayers

Hence,fromFigure3.4 8 ,

0.89 262.2 233.4 mma

1

2

3

450 233.4 / 2 333.3 mm

483.3 mm

and 633.3 mm

l

l

l

6uTherefore, (1256 500 (483.3 633.3) 1256 200000 0.00215 333.3) 10

881.2 kNm

M

3.5.3 Design example

Problem UsingtherelevantclausesofAS3600‐2018,designasimplysupported

beamof6mspantocarryaliveloadof3kN/mandasuperimposeddead

loadof2kN/mplussell‐weight.Giventhat cf 32MPa, syf 500MPafor

500Nbars,themaximumaggregatesizea 20mm,thestirrupsaremade

upofR10bars,andexposureclassificationA2applies.

Solution Live‐loadmoment

2 2

q

3 613.5 kNm

8 8

wlM

Superimposeddead‐loadmoment

2

SG

2 69 kNm

8M

Takeb D 150 300mmandassumept 1.4% byvolume .Then

3wEquation 2.4(1): 24 0.6 1.4 24.84 kN/m

Thus,self‐weight 0.15 0.30 24.84 1.118kN/m

Themomentduetoself‐weightis

2

SW

1.118 65.031 kNm

8M

andMg MSG Msw 9 5.031 14.031kNm.

Then*

g q

*

Equation 1.3(2): 1.2 1.5

or 1.2 14.031 1.5 13.5 37.09 kNm

M M M

M

2Equation 3.3(2)a: 0.85 0.0015 32 0.802

Equation 3.3(2)b: 0.97 0.0025 32 0.89 ct.fEquation 2.2(2): 0.6 32 3.394 MPaf

AdoptingN20barsasthemainreinforcementinonelayerwith25mmcover

gives

d D–cover–diameterofstirrup–db/2 300–25–10–20/2 255mm

2

t.min

300 3.394Equation 3.5(5): 0.20 0.00191

255 500p

all

32Equation 3.4(6): 0.4 0.802 0.89 0.01827

500p

Use,say,

all t.min

20.01218

3tp p p

thisisacceptable.

Then*

2

syt sy

2 c

Equation 3.5(3): 1

12

t

Mbd

fp f p

f

u

0.01218 500Equation 3.4(8): 0.267

0.802 0.89 32k

Forasinglelayerofbars,wehaved do.Thus,kuo ku 0.267.Thenfrom

Equation3.4 20 awithb,ϕ 0.85.Thus

62 37.09 10

Equation 3.5(3): 1501 500

0.85 0.01218 500 1 0.012182 0.802 32

d

fromwhichd 232.8mm

Finally,Ast ptbd 0.01218 150 232.8 425.3mm2

FromTable2.3 1 ,therearethreeoptions:

1. twoN20:Ast 628mm2

2. threeN16:Ast 603mm2

3. fourN12:Ast 452mm2

Figure 3.5(1) Checkingaccommodationfor2N20barsTakingOption1,wehavetwoN20bars seeFigure3.5 1 andTable1.4 2

givesacoverc 25mmtostirrupsattopandbottom.Hencethecovertothe

mainbars c diameterofstirrup 25 10 35mm,andd D–coverto

mainbars–db/2 300–35–20/2 255mm 232.8mm;thereforethisis

acceptable seeFigure3.5 1 .

Table1.4 4 specifiesaminimumspacingsminof 25,db,l.5a max.Thussmin

25,20,30 max 30mm.

Theavailablespacing 150–2 35–2 20 40mm smin 30mm;

therefore,thisisacceptable seeFigure3.5 1 .

Notealsothat,sincekuo 0.267 0.36,thedesignisacceptablewithout

providinganycompressionreinforcement seeSection3.4.1 .

TakingOption2,wehavethreeN16barsasshowninFigure3.5 2 .

Theavailablespacing 150–2 35–3 16 /2 16mm smin 30

mm.

Toprovideaspacingof30mmwouldrequireb 150mm Figure3.5 2 ;

therefore,Option2isnotacceptable.

Figure 3.5(2) Checkingaccommodationfor3N16bars

Option3issimilarlyunacceptable.Thus,Option1shouldbeadopted,but

notingthefollowingqualifications:

a Option1isslightlyover‐designed i.e.d 255mmisabout9.5%higher

thanrequiredandAst 628mm2is47.7%higherthannecessary .

b Thepercentageofsteelbyvolumeforthesectionis 628/ 150 300

100 1.396% 1.4%asassumedinself‐weightcalculation;hencethisis

acceptable.

c Ifthebeamistobeusedrepeatedlyorfrequently,acloserandmore

economicaldesigncouldbeobtainedbyhavingasecondorthirdtrial,

assumingdifferentb D.

d Ifdesignforfireresistanceisspecified,ensurethatconcretecoverof25

mmisadequatebycheckingSection5ofAS3600‐2018.

3.6.3 Illustrative examples

Example 1

Problem GivenadoublyreinforcedsectionasshowninFigure3.6 3 with cf 32MPaandfsy

500MPa.ComputeϕMu.

Figure 3.6(3) Cross-sectional details of Example 1

Solution Thereinforcementratios

t

27120.0125

350 620p

and

c

3390.00156

350 620p

FromSection3.5.3andfor cf 32MPa

2 0.802 and 0.89

t c limit

40600 0.802 0.89 32

620Equation 3.6(3): ( ) 0.01768(600 500) 500

p p

But pt–pc 0.01094 pt–pc limit 0.01768

HenceAscdoesnotyieldatfailure.Then

0.0125 500 600 0.00156Equation 3.6(13): 0.1163

2 0.802 0.89 32

600 0.00156Equation 3.6(14): 0.041

0.802 0.89 32v

2u

40Equation 3.6(12): 0.1163 0.1163 0.041 0.243

620k

Sincea γkud 0.89 0.243 620 134.1mm,

u

6

134.1Equation 3.6(15)b: 2712 500 620

2

40 134.1600 339 1 40 10

0.243 620 2

M

Thatis,Mu 753.8kNm

Withthebarsinonelayer,wehaved do,kuo ku 0.243,andEquation3.4 20 awith

bgives,forClassNreinforcementϕ 0.85

FinallyϕMu 0.85 753.8 640.7kNm

Example 2

Problem SameasExample1 adoublyreinforcedsectionwith cf 32MPaandfsy 500MPa ,butdc 34mmandAstconsistsof6N28bars.ComputeϕMu.

Solution Thereinforcementratios

t

36960.01703

350 620p

and

c

3390.00156

350 620p

t c limit

34600 0.802 0.89 32

620Equation 3.6(3): ( ) 0.01503(600 500) 500

p p

but pt–pc 0.01547 pt–pc limit 0.01503

Hence,Ascyieldsatfailure.Then

(3696 339) 500 186.9 mm

0.802 32 350a

Equation 3.6(6) :

u

6u

Equation 3.6(7): 339 500 (620 34) 0.802 32

186.9186.9 350 620 10 , that is 983.3 kNm

2

M

M

Sincea γkudfromwhichku 0.339andford do,wehavekuo ku 0.339.

Thus,ϕ 0.85accordingtoEquations3.4 20 aandb.Finally,ϕMu 0.85 983.3

835.8kNm.

3.7.2 Illustrative example

Problem Ifb 230mm,D 400mm,M* 250kNm, cf 25MPaandfsy 500MPa,andwhere

exposureclassificationA1applies,determineAstand,asnecessary,AscusingonlyN28

bars.UseR10ties.

Solution Assumetwolayers,say,ofN28barsforAstandonelayerforAscasshowninFigure

3.7 3 .

Figure 3.7(3) Section layout for illustrative example Note:alldimensionsareinmm.

Thus

cover up to tie tie diameter 1.5 bar diameterd D

Thatis

c400 28 10 1.5 28 320 mm. 28 10 28/2 52 mmd d

Then

2Equation 3.3(2)a: 0.85 0.0015 25 0.8125 and

Equation 3.3(2)b: 0.97 0.0025 25 0.9075

Thus

2s1

25Equation 3.7(2): 0.4 0.8125 0.9075 230 320 1085.4 mm

500A

6u1

1 1085.4 500Equation 3.4(10): 1085.4 500 320 1 10

2 0.8125 230 320 25M

Thatis

Mu1 142.1kNm

and

2 u1* *Equation 3.7(3): M M M

Inthiscase,ku 0.4

u0 u

0

0.4 320And with 400 38 28/2 348, 0.368

348o

k dd k

d

ThestipulationofClause8.1.5intheStandardthatkuo 0.36maynotapplytodoubly

reinforcedsectionswhereku 0.4,providedthattheAscisnotlessthanthespecified

minimum,whichistrueinmostcases.Ifindoubt,doublecheckandreviseasnecessary.

Withkuo 0.368,Equation3.4 20 awithbgivesϕ 0.841,withwhichM2*

250–0.841 142.1 130.5kNm

And6

2s2

130.5 10Equation 3.7(5): 1158.0 mm

0.841 500 (320 52)A

Thus2

s s1 s2 1085.4 1158.0 2243.4 mmtA A A

Table2.3 1 showsthatwithfourN28bars,Ast 2464mm2isacceptable.

Also,

2s1.limit

600 0.8125 25 0.9075 52 230Equation 3.7(7): 2645.6 mm

(600 500) 500A

SinceAs1 As1.limit,Ascdoesnotyield,andwithku 0.4

csc

520.003 1 0.003 1 0.00178

0.4 0.4 320

d

d

Thus,thecompressionsteelstress

sc sc s syf E f

Or

sc sy0.00178 200000 356 MPa < 500 MPaf f

sc Hence, 356 MPa requires f

s2 sy 2sc

sc

1158.0 500 1626.4 mm

356

A fA

f

WiththreeN28bars,Asc 1848mm2,whichisacceptable.

Tocheckbaraccommodationsforb 230mm:b 5 28 2 10 160mm

isacceptable usetwolayersoftwobars orb 7 28 2 10 216isacceptable

usethreebarsinthebottomlayerplusonebarabove .Detailsofthetwopossible

reinforcementlayoutsareshowninFigure3.7 4 .

Figure 3.7(4) Section details for illustrative example

Note:alldimensionsareinmm.

CHAPTER 4

4.2.7 Illustrative examples

Twonumericalexamplesareprovidedhere.Examples1and2considertheanalysisand

designofsinglyreinforcedT‐sections,respectively.

Example 1: Analysis of singly reinforced T-sections

Problem GivenaT‐beamasshowninFigure4.2 8 ,reinforcedwithonelayeronlyofClass‐N

bars.Take cf 25MPa,fsy 500MPaandcomputeM′.

Figure 4.2(8) Cross-sectional details of the example T-beam Note:allunmarkeddimensionsareinmm.

Solution For cf 25MPa,α2 0.8125andγ 0.9075and

8000 500Equation 4.2(9): 179.02 mm 120 mm

0.8125 25 1100t

Thus,theNAatfailureislocatedwithintheweb.

Fortheflange‐beam

6u2

120Equation 4.2(11): 0.8125 25 120(1100 400) 650 10

2 1006.7 kNm

M

and

2s2

0.8125 25 120 (1100 400)Equation 4.2(12): 3412.5 mm

500A

Therefore Asl 8000–3412.5 4587.5mm2

TochecktheconditionofAslatfailure

2s1

25 600Equation 4.2(18): 0.8125 0.9075 400 650 5228.4 mm

500 600 500A

Therefore

As1willyieldatfailure.

Fortheweb‐beam

6u1

1 4587.5 500Equation 4.2(14): 4587.5 500 650 1 10

2 0.8125 400 650 25 1167.2 kNm

M

and

u

4587.5 500Equation 3.4(7): 0.479

0.8125 0.9075 25 400 650k

Assumingthatthebarsarelocatedinasinglelayer,wehaved do,andkuo ku

0.479.ThusEquations3.4 20 awithb:ϕ 0.721andM′ 0.721 1167.2 1006.7

1567.3kNm.

Sincekuo 0.36,appropriatecompressionreinforcementmustbeprovided.

WithreferencetoFigure4.2 8 ,therequiredAsc 0.01 1100 120 0.479

650 –120 400 2085.4mm2,whichmayallbeplacedintheflangeatmid‐depth.

Example 2: Design of singly reinforced T-sections

Problem GivenaT‐beamwiththedimensionsshowninFigure4.2 9 ,f′c 32MPa;fsy 500MPa

andM* 600kNm.Designthereinforcementforthesection.

Figure 4.2(9) Cross-sectional details of the design T-beam Note:alldimensionsareinmm.

Solution For cf 32MPa,α2 0.802andγ 0.89.

TouseanalternativemethodtoSection4.2.3forcriterionchecking,assumea

rectangularsectionofb d 750 475andthata t.Then,theeffectivemoment

u 2 c *533.82

tM M f bt d M

Therefore,theneutralaxisattheultimatestateliesintheweb.Ora t.

Fortheflange‐beam

6u2

62.5Equation 4.2(11): 0.802 32 62.5 (750 250) 475 10

2 355.9 kNm

M

and6

2s2

355.9 10Equation 4.2(12): 1604.1 mm

62.5500 475

2

A

Sincea γkud t 62.5mm,wehaveku 0.148.Assumingthatthebarsare

locatedinasinglelayer,wehaved do,andkuo ku 0.148.Thusϕ 0.85asper

Equations3.4 20 aandb.

Fortheweb‐beam

1*Equation 4.2(22): 600 0.85 355.9 297.5 kNmM

and

0.802 32Equation 3.5(7): 0.0513

500

withwhich

62

t1 2

2 0.0513 297.5 10Equation 3.5(6): 0.0513 0.0513 0.0144

0.85 250 475 500p

Thus,thewebreinforcement2

s1 0.0144 250 475 1710 mmA

AsperEquation4.2 24

all t1

320.4 0.802 0.89 0.01827 ,which is acceptable.

500p p

Theright‐handsideofEquation4.2 18 gives

2s1

32 6000.802 0.89 250 475 2958.9 mm

500 600 500A

Therefore,As1yieldsattheultimatestate.

Finally,Ast 1710 1604.1 3314.1mm2

Furtherconsiderationstocompletethedesignareto:

selectabargroupandcheckaccommodation

double‐checkthecapacityMuasnecessary.

4.3.2 Illustrative examples Example 1

Problem ForthedoublyreinforcedsectionwithanirregularshapeasshowninFigure4.3 3 ,

computetheultimatemoment Mu .Take cf 25MPa.

2 N28

4 N36

(1232 mm2)

(4080 mm2)

480

50

50

125 150 125

400

Figure 4.3(3) Cross-sectional details of the irregular shape example problem

Note:alldimensionsinmm.

Solution For cf 25MPa,α2 0.8125andγ 0.9075asperEquations3.3 2 aandb,

respectively.

But,asperAS3600‐2018,forsectionswherewidthreducesfromtheneutralaxis

towardsthecompressionface,2shallbereducedby10%.

Thus,forthisexample, 2 0.9 0.8125 0.7313

BasedonthestraindiagramgiveninFigure4.3 4 ,weobtain

NAsc

NA

0.003( 50)d

d

Equation i

and

NAs

NA

0.003(580 )d

d

Equation ii

0.003

NA

scε

sε

C1

C2

T

Cs

Strain Stress

dNA

'2 cf

NA da

Figure 4.3(4) Stress and strain distribution across the example section

Fortheconcretestressoverthetoparea seethestressdiagraminFigure

4.3 4

1 150 100 0.7313 25 274237.5NC Equation iii

andfortheremainingarea

2 NA NA400( 100) 0.7313 25 7313( 100)C d d Equation iv

ThetrialanderrorprocessbelowwillleadtotherequiredMu.

Trial1:AssumedNA 200mm.

sc sy sc

500Equation (i): 0.00225 0.0025; that is, A would not yield.

200000

Therefore,fsc 0.00225 200000 450MPa.

ThroughEquation ii ,weobservethatεs εsy.Orfs 500MPa.

Thetotalhorizontalforceincompressionisgivenas

1 2 s 1 2 sc sc

3[274 237.5 7313(0.9075 200 100) 1232 450] 10

1424.7 kN

C C C C C C A f

ThetotaltensileforceT Astfs 4080 500 10–3 2040kN

SinceT C,assumealargerdNAinthenexttrial.

Trial2:AssumedNA 250mm,andwehave

sc sy s sy0.0024 and

or

sc s0.0024 200000 480 MPa and 500 MPa.f f

ThenC C1 C2 Cs 1793.4kN T.

TryastilllargerdNA.

Trial3:AssumingdNA 285mmandinasimilarprocess,weobtain

C 2043.9kN T.

AcceptdNA 285mm,andbytakingmomentsaboutthelevelofT,wehave

NAu 1 2 s

6

100 ( 100)580 580 100 530

2 2

(0.9075 285 100)274 237.5 530 1 160 116.0 480 609 515.8 530 10

2

933.2 kNm

dM C C C

Notethatforbeamsectionsmadeupofrectanglesandothersimpleshapes,theexact

valueofdNAmaybedeterminedbyequatingthetotaltensileandcompressiveforces.

Inourcase3

NA[274 237.5 7313( 100) 1232 500] 10C d and

2040 kNT

But

C T

fromwhich

NA 283.4 mmd

However,inallcasesbeforeacceptingsuchan‘exact’dNA,ensurethattheresulting

stressconditionsinAstandAsc i.e.yieldingorotherwise areasassumedinthefirst

place.

Example 2

Problem Figure4.3 5 illustratesthesectionofabeaminastructurecontainingprefabricated

elements.Thetotalwidthandtotaldeptharelimitedto450mmand525mm,

respectively.Tensionreinforcementusedis4N32bars.Using cf 32MPa,determine

themomentcapacityMuofthesection.

60

450

525

75

390

125125

4 N32

Figure 4.3(5) Cross-sectional details of a prefabricated irregular shape beam

section Note:alldimensionsareinmm.

Solution For cf 32MPa,α2 0.802andγ 0.89asperEquations3.3 2 aandb,respectively.

But,asperAS3600‐2018,forsectionswherewidthreducesfromtheneutralaxis

towardsthecompressionface,2shallbereducedby10%.

Thus,forthisexample, 2 0.9 0.802 0.7218

AssumingthattensionreinforcementAstyieldsatfailure,andbasedonthestress

diagramwheredNA kudasgiveninFigure4.3 6 ,weobtain3

NA[0.7218 32 ( 75) 450 2 0.7218 32 125 75] 10C d and

33216 500 10 1608 kNT

450

75

125125

4 N32

C

T

γdNA

450

'

2 c f

dNA= kud

Figure 4.3(6) Stress distribution across the example section

ButC T

fromwhich

NA u =211.3 mmd k d

Check stress condition for Ast: Now,

uu uB

0

211.30.454 0.545

(525 60)

k dk k

d

fromEquation3.4 4

ThisconfirmstheassumptionoftensionfailureorAstyieldingatfailure.

Finally,takingmomentaboutthelevelofTgives

6

6

(0.89 211.3 75)0.7218 32 (0.89 211.3 75) 450 525 60 75 10

2

75 + 2 0.7218 32 75 125 525 60 102 577.0 kNm

uM

CHAPTER 5


Problem Givenasimplysupportedbeam withLef 10m,b 350mm,d 580mm,D 650

mmandpt 0.01 ,computethemidspandeflectionunderacombinationofdeadload

includingself‐weight g 8kN/m andliveload q 8kN/m .TakeEc 26000MPa,

Es 200000MPaand cf 32MPa;assumethatthebeamformspartofadomesticfloor

system;ignoretheshrinkageeffects.

Solution Thegrossmomentofinertia seeFigure5.3 1

36 4

g

650350 8010 10 mm

12I

g 6crEquation 5.3(5): 0.6 32 10 83.65 kNm

325

IM

t

200000With 7.69 and 0.01, Equation A(5) (from Appendix A) yields

26000n p

k 0.3227.Inturn,Equation5.3 3 givesIcr 3174 106mm4.

650

pt = 0.01

350

580

Figure 5.3(1) Cross-sectional details of the example simply supported beam Note:alldimensionsareinmm.

Fordomesticfloorsystems,ψsandψlaregiveninTable1.3 1 as0.7and0.4,

respectively.Forshort‐termdeflection,thethirdformulainEquation1.3 8 governs

andwehavecombinedload g 0.7q 8 0.7 8 13.6kN/m.

Themomentatmidspanis2

s

1013.6 170 kNm

8M

Equation5.3 2 thusgives6

ef 26

6

6 4g

3174 10

3174 10 83.651 1

8010 10 1703717.4 10 mm , which is acceptable.

I

I

Finally,Equation5.2 1 inconjunctionwithTable5.2 1 gives4

6

5 13.6 1000018.3 mm

384 26000 3717.4 10

ItmaybeapparentinTable5.2 4 thattheimmediatedeflectionunderdeadandlive

loadsisnotacriterionforserviceabilitydesign.Itscalculation,however,isessentialin

theanalysisoflong‐termdeflectionasdiscussedinSection5.4andthetotaldeflections

undernormalorrepeatedloading Section5.6 .


Problem ForthebeamanalysedinSection5.3.4,computethetotaldeflectionTusingthe

multipliermethodassumingthatthedeadloadonlyisthesustainedloadand

sc 0.0025.A

bd

Solution FromSection5.3.4andwithreferencetothefirstformulainEquation1.3 8 ,the

sustainedloadmoment2

g

108 100 kNm

8M

Equation5.3 2 thengives6

ef 26

6

6 4g

3174 10

3174 10 83.651 1

8010 10 1005495.7 10 mm , which is acceptable.

I

I

and,applyingEquation5.2 1 ,

4

I.g 6

5 8 100007.3 mm

384 26000 5495.7 10

BasedonEquation5.4 3 ,themultiplier

cs

0.00252 1.2 1.7

0.01k

Finally,Equation5.4 2 yields

T 18.3 1.7 7.3 30.7 mm

Thelimitforthetotaldeflectionofabeam,asgiveninTable5.2 4 ,isnottobegreater

than

efT

1000040 mm 30.7 mm, which is acceptable

250 250

L

Thus,thebeaminquestionissatisfactoryasfarastotaldeflectionisconcerned.


Problem Re‐analysethebeaminSection5.4.3,takingintoconsiderationtheeffectsof31200

repetitionsoffullliveload i.e.q 8kN/m .

Solution 1. Ief

Forq 8kN/m

Mq 8 102/8 100kNm;Ms 200kNm

NotethatusingIef.qinsteadofIef.gwillleadtoaconservativeIrep.

Thus,Equation5.3 2 yields6

6 4ef 26

6

3174 103548.8 10 mm

3174 10 83.651 1

8010 10 200

I

2. Yieldmoment My

Theyieldmomentistheupperlimitoftheworkingstressbendingmoment.

Forunder‐reinforcedrectangularsections,wehave

y st sy 13

kM A f d

Equation5.6 11

wherekisobtainedusingEquationA 5 inAppendixA.

FromtheexamplegiveninSection5.3.4,k 0.3227.Thus

6y

0.32270.01 350 580 500 580 1 10 525.4 kNm

3M

3. IntensivecreepfactorandA.g

I

0.029 200 100Equation 5.6(5): 1.18 1.84

0.01 525.4 83.65k

0.0015 200 100Equation 5.6(6): 0.03396

0.01 525.4 83.65R

R 10Equation 5.6(4): 1.84 0.03396 log (31200) 1.993k

FromSection5.4.3,I.g 7.3mm.Thus

A.g 1.993 7.3 14.55mm

4. Irepandg2

x

(200 83.65)Equation 5.6(8): 83.65 114.3 kNm

(525.4 83.65)M

2 2

6 6rep

6 4

83.65 83.65Equation 5.6(7): 8010 10 1 3548.8 10

114.3 114.3

5938.2 10 mm

I

ComparingEquations5.6 1 and5.6 2 ,wehave6 2

q 6

5 100 10 100006.75 mm

48 26000 5938.2 10

5. Grandtotaldeflection

FromSection5.4.3andusingEquation5.6 10 ,

L 1.7 7.3 12.41 mm

Finally,Equation5.6 9 gives

T 14.55 6.75 12.41 33.71 mm.

CHAPTER 6


Problem AT‐beamwithasimplysupportedspanof6missubjectedtoaconcentratedliveloadP

700kN,asshowninFigure6.3 4 a;thecross‐sectionaldetailsaregiveninFigure

6.3 4 b.Designthebeamforshear,assumingfc′ 20MPa.

Figure 6.3(4) Details of the example T-beam: (a) loading configuration and shear

force diagram and (b) cross-sectional details

Note:cross‐sectionaldimensionsareinmmunlessspecifiedotherwise.

Solution

Design shear FromTable2.3 1 weobtainAst 4928mm2andthegrosssectionalareaAg 1200×

100 770×300 351000mm2,basedonwhichthesteelpercentagebyvolume

4928100 1.4%

351000v

ThenEquation2.4 1 :ρw 24 0.6×1.4 24.84kN/m3andtheself‐weight

(0.1 1.2 0.77 0.3) 24.84 8.72 kN/mg

ThemaximumshearcanbedeterminedviaEquation1.3 2 as

g q1.2 1.5

1.2 8.72 3 (1.5 700) / 2

31.4 525

556.4 kN

V V V

Thedesignshearmaybetakentobetheshearatadistancedo 828mm fromthe

support.OrfromFigure6.3 4 a

o* 3 3 0.828525 31.4 525 31.4 547.7 kN

3 3

dV

Section adequacy

CheckthemaximumsectioncapacityinshearusingEquation6.3 1 andwehave

v'u.max w v 2

v

cot0.55

1 cotcV f b d

Forthegivensection,

dv 0.72D,0.9d max 0.72 870 , 0.9 800 max 626.4,720 720mm

andusingthesimplifiedmethod, v 36˚

Finally,0

3u.max 2 0

cot 360.55 20 300 720 10

1 cot 36V

1129.9kNfromwhich

Vu.max 0.75 1129.9 847.4kN V* 547.7kN

ThusD 870mmisacceptable.

Concrete shear capacity Next,computeVuc

c 20 4.47 MPa 8 MPa, which is acceptable.f

Usingsimplifiedmethod,for sv sv.minA A

s s , v 0.15k

ThenEquation6.3 4 :

' 3uc v w v 0.15 300 720 4.47 10 144.8 kNcV k b d f

and

uc uc0.75 108.6 kNV V *V 547.7kN

Thus,shearreinforcementisrequired.

Shear reinforcement If,say,verticaltiesmadeofN12barsareused,wehave

2sv 2 113 226 mmA

us

(547.7 108.6)Equation 6.3(17): 585.5 kN

0.75V

Thatis,

sv sy.f v v

3us

cot 226 500 720 cot 36191.3 mm(say190 mm),

585.5 10

which also satisfies Equation 6.3 22 .

A f ds

V

Reinforcement arrangement ThelayoutisasshowninFigure6.3 5 .

Figure 6.3(5) The layout of shear reinforcement for the example in Section 6.3.8

Forthefulllengthofthebeam,werequireatotalof 16 16 32ties,withthe

spacingbetweenthetwotiesoneithersideoftheloadatmid‐spanincreasedto300

mmforobviousreasons.Astheshearforceisduemainlytothe700kNofconcentrated

loadandthedistributionisalmostuniform,itisacceptablethatthechosentiespacing

s isusedthroughout.Incaseswherethesheardistributionvariesgreatly,smayvary

accordinglyalongthespantosuit,butthedetailingrequirementsspecifiedinClause

8.3.2.2oftheStandard AS3600‐2018 mustbemet seeSection6.3.7 .


Problem FortheT‐beamintheprecedingdesignexample Section6.3.8 ,checkanddesignfor

longitudinalshear,ifapplicable.Take 20 MPacf anduseN12orN16barsfortiesas

necessary.Assumemonolithicconstruction.

Solution FromSection6.3.8,self‐weight 8.72kN/mandV* 547.7kN.

Thecriticalshearplane,althoughnotspecified,maybetakenasatthe

rectangularstressblockdepthlevellocated inthiscase intheweb–thusbf 300

mm.

Equation6.4 4 :ß 1.

WithreferencetoFigure6.3 4 ,imposingΣFx 0gives

4928×500 0.82×20 1200×100 300 γkud–100

fromwhich

γkud 200.8mm

Figure 6.4(4) Determination of z for the example T-beam

Then,inFigure6.4 4 ,thedistancebetweenthecentreofgravityofthe

compressiveresultantandtheextremecompressivefibre

1200 100 100 / 2 (100.8 300) (100 100.8/2)70.2 mm

1200 100 100.8 300z

3

Equation 6.4(3): 800 70.2 729.8 mm

1 547.7 10*Equation 6.4(2): 2.50 MPa

729.8 300

z d z

ThequantitiesrequiredforEquation6.4 5 are:

Asf/sfortheexistingtransversereinforcementorN12ties@190mm

226/190 1.19mm2/mm

gp 100×1200 100.8×300 ×10–6×24.84×103 3732.0N/m

3.73N/mm

FromEquation2.2 3 , ct 0.36 20 1.61 MPaf

FromTable6.4 1 ,µ 0.9;kco 0.5

ThusEquation6.4 5 :

u

1.19 500 3.730.9 0.5 1.61 2.6 MPa

300 300

CheckEquation6.4 1 :

u *0.75 2.6 1.95 2.50 MPa, which is inadequate.

Thus,additionalshearreinforcementisrequired.

f

2

300 2.50Equation 6.4(6): 0.5 1.61 0.9 3.73 / 300

0.75500

1.68 mm /mm

2 113Try N12 ties: 134.5 mm, which is rather close

1.68

sA

s

s

Thus,tryN16tiesand

2 201239.3 mm

1.68s

Say,useN16ties@230mm,whichislessthan3.5tf 3.5×100 350mmasrequired

inEquation6.4 7 .

Insummary,thefinalreinforcementarrangementforbothtransverseand

longitudinalshearisN16closedties@230mm,andthetotalnumberrequiredoverthe

6‐mspan: 26 1 27.

CHAPTER 7


Problem Acantileverbentbeamsubjectedtotorsion,shearandbendingisdetailedinFigure

7.4 2 .Transversereinforcement i.e.N10tiesat300mm isprovidedtoresist

transverseandlongitudinalshearonly.

a Isthebeamsectionadequatetoresistthetorsion?

b Ifnot,whatshouldtheAsw/swbeandthecorrespondinglongitudinalsteel?

Take cf 20MPa.

Figure 7.4(2) Details of the example beam for torsion design

Solution

For (a) 1. Computedesignshear,torsionandmoment.

FromTable2.3 1 ,weobtainthetotalreinforcementareaforthesection

Ast Asc 6160 3080 9240mm2

Thus,thepercentageofsteelbyvolume

9240100 4.4%

300 700v

AccordingtoEquation2.4 1 ,ρw 24.0 0.6×4.4 26.64kN/m3

Thentheself‐weight 0.3×0.7×26.64 5.6kN/m.

AtsupportA:

V* 1.5×50 1.2×5.6×10.5 145.56kN whichisaconservativevalue

T* 1.5×50×0.5 1.2×5.6×0.5× 0.5

238.34kNm

M* 1.5×50×10 1.2 5.6×0.5×10 5.6×210

2 750 369.6

1119.6kNm

2. Checkacceptabilityoftheconcretesection.

CheckthemaximumsectioncapacityinshearusingEquation6.3 1 andwe

have

v'u.max w v 2

v

cot0.55

1 cotcV f b d

Forthegivensection,

dv 0.72D,0.9d max 0.72 700 , 0.9 630 max 504,567 max

567mm

and v x(29 7000 ) inwhich

2** 2* *h

v o 3x

s st

0.90.5

23 10

2

T uMV N

d A

E A

whereuh perimeterofthecentre‐lineoftheclosedtransverse

torsionreinforcement 2 254 654 1816mm

Ao areaenclosedbyshearflowpath,includinganyareaofholes

therein 300 700 210000mm2

N* 0anddv 2*2

* h

o

0.9

2

T uV

A

26

23 60.9 38.34 10 1816567 145.56 10 10

2 210000

118.2kNm

M* 1119.6kNm,whichisacceptable.

Thus,

26 623

x

1119.6 10 0.9 38.34 10 1816145.56 10

567 2 210000

2 200000 6160

0.000886 3 10‐3whichisacceptable.

Hence,v 29 7000 0.000886 35.2˚

Finally,0

0

3u.max 2

cot 35.20.55 20 300 567 10

1 cot 35.2V

881.3kNfromwhich

Vu.max 0.75 881.3 661.0kN

Now,Equation7.2 1 :2 2**

h2

w v oh1.7

T uV

b d A

inwhich

Aoh areaenclosedbycentre‐lineofexteriorclosedtransversetorsion

reinforcement,includingareaofholes,ifany 254 654 166116mm2

Thus,Equation7.2 1 :

2 23 6

2

145.56 10 (38.34 10 ) 1816

300 567 1.7 166116

1.71MPa Vu.max/bwdv

661.0 103/ 300 567 3.89MPa

Therefore,thesectionisreinforceable.

3. Checktorsionalreinforcementrequirement.

2cp'

cr cc

Equation 7.3(2): 0.33A

T fu

where

Acp totalareaenclosedbyoutsideperimeterofconcretesection

300 700 210000mm2

uc thelengthoftheoutsideperimeterofconcretecross‐section

2 300 700 2000mm

Thus,Equation7.3 2 : 2

6cr

2100000.33 20 10 32.5 kNm

2000T

*Equation 7.3(1): Right-hand side 0.25 0.75 32.5 6.09 kNm < T

Thus,thebeamsectionisinadequateandtorsionalreinforcementisrequired.

For (b) 1. ComputeAsw/sw.

* 6

sw wsy.f oh v

38.34 10Equation 7.4(3): /

1.7 cot 0.75 500 1.7 166116 cot 35.2

TA s

f A

Thatis,Asw/sw 0.2554mm2/mm.

2. Minimumreinforcementrequirementchecks

t 2sw

sy.f

t

sw w

0.2 0.2 664 / = 0.2656 mm /mm

500

in the above equation = the larger overall dimension of the closed fitment = 664 mm.

But / = 0.2656 > 0.2554, which is not acceptable.

w

yA s

f

y

A s

(Check 1)

6cr

sw wsy.f oh v

32.5 10Equation 7.4(5) which gives, / 0.1624

1.7 cot 1.7 500 166116 cot 35.2

0.2554, which is acceptable.

TA s

f A

(Check 2)

Thus,therequiredtransversetorsionalreinforcement

2sw w/ 0.2554 mm /mm.A s

NotethatthestipulationinClause8.2.4.5oftheStandardregardingthe

torsionaleffectsonshearstrength asmentionedinthelastparagraphof

Section6.3.4 isnaturallysatisfiedbecausethebeamisreinforcedagainst

shear aswellastorsion .

3. Longitudinalreinforcement

Equation7.4 6 :

2*2* h

v eq uso

st scsy

0.45cot 0.5

20

T uV V

AA A

f

where

22 6*2 2* * 3 3h

eqo

0.9 38.34 10 18160.9145.56 10 10

2 2 210000

T uV V

A

208.4kN

andEquation6.3 17 :*

eq ucus

V VV

inwhich 'uc v w v cV k b d f

and vx

0.4 0.40.17

1 1500 1 1500 0.000886k

fromwhich

3uc 0.17 300 567 20 10 129.3 kNV

Thus,Equation6.3 17 : us

208.4 0.75 129.3148.6 kN

0.75V

Hence,Equation7.4 6 :

26

2o 3 3

st sc

0.45 38.34 10 1816cot 35.2 208.4 10 0.5 0.75 148.6 10

2 210000

0.75 500A A

642.4mm2 0,whichisacceptable.

4. Checkadequacyofbendingreinforcement.

SothatAS3600‐2018,Clauses8.2.7and8.2.8arecompliedwith,thegiven

cross‐sectionless∆AstandAscshouldbeabletoresistMA* 1119.6kNm.

st t

sc c

6160 642.4 5517.6; 0.0292

3080 642.4 2437.6; 0.0130

A p

A p

t c 0.0162p p

2Equation 3.3(2)a: 0.85 0.0015 20 0.82

Equation 3.3(2)b: 0.97 0.0025 20 0.92

t c limit t c

420.92 600 0.82 20

630Equation 3.6(3): ( ) 0.0121 ,(600 500) 500

p p p p

andAscwouldyieldatultimate.

(5517.6 2437.6) 500Equation 3.6(6): 313.0

0.82 20 300a

u uo

313.0 0.54 630Further, 0.54 and 0.52

0.92 630 658

ak k

d

Thus,Equation3.4 20 awithb:ϕ 0.677andfinally

uo

6

313.0Equation 3.6(8): 0.677 5517.6 500 630

2

313.02437.6 500 42 10 978.8 kNm

2

M

Thatis,ϕMuo 978.8kNm<MA* 1119.6kNm,whichisnotacceptable.

Sincethemomentcapacityofthesection seeFigure7.4 2 less∆Astand

Ascisnotcloseenoughtothedesignmoment MA* ,theexistingsection

needsanincreaseinAstandAscby642.4mm2each,toresisttorsion.

5. Transverseandlongitudinalreinforcementdetailsandfinaldesign

From b Step2,above,wehaveAsw/sw 0.2554mm2/mm.UseN10closed

tiesandtherequiredspacingsw 78/0.2554 305.4mm.

Forthe10‐mspan,thenumberofN10ties 10000/305.4 32.7,andthe

existingnumberofN10tiesasAsv 10000/300 33.3.

Thesemakeatotalof66N10ties@s 10000/ 66–1 153.8mm.Say,

use67N10ties@150mm,coveringatotallengthof66×150 9.90m

10m,whichisacceptable.

NotethatN10ties@150mmcomplieswiththespacingrequirementsgiven

inEquation7.4 11 .

Foradditionallongitudinalreinforcement,2N20bars 628mm2closeto

required642.4mm2 aretobeprovidedbothattopandbottomofthe

section.

6. Thefinaldesign

DetailsofthefinaldesignareasshowninFigure7.4 3 .

Figure 7.4(3) Final design details for the example design problem

CHAPTER 8

8.4 Illustrative examples 8.4.1 Example 1

Problem GivenN28barsinasimplysupportedbeamasshowninFigure8.4 1 and cf 25MPa,

determinethepositionfromthesupportbeyondwhichtheyieldstresscanbe

developedinthebars,if

a thebarsareextendedstraightintothesupport

b standard180°hooksareused.

25 mm

350 mm

25 mm

28 mm28 mm

4 N28

R10 @ 100

Figure 8.4(1) Details of the simply supported beam in Example 1

Solution a Straightbars

ForEquation8.2 1

1

2

3

1.0

(132 28)/100 1.04

1.0 0.15 (28/2 28)/28 1.0 use 1.0.

k

k

k

ThusEquation8.2 1 :

sy.tb

0.5 1.01346 mm > 0.058 500 812 mm,

1.04L

whichisacceptable.

b Hooks

sy.t 1346/2 673 mmL

Thepointsbeyondwhichyieldstresscanbedevelopedinthetwocasesare

illustratedinFigure8.4 2 .

Figure 8.4(2) Development lengths for straight bars and hooks

8.4.2 Example 2

Problem ThestirrupsspecifiedforthebeaminFigure8.4 1 aremadeofR10bars,eachofwhich

isrequiredtodevelopfsyatmid‐depthofthesection.Determineandsketchthe

dimensionaldetailsofthestirrup.

Solution ForEquation8.2 1

1

2

3

1.0

(132 10)/100 1.22

1.0 0.15(25 10)/10 0.78.

k

k

k

sy.tbThus, Equation 8.2(1): 0.5 1.0 0.78 250 10/(1.22 )

160 0.058 500 Use 290 mm

L

sy.tEquation 8.2(8): 1.5 290 435 mm > 300 mm, which is acceptable.L

Sinceeachendofthestirruptakestheformofahook,wehaveLsy.t/2 435/2

218mm.

However,itcanbeseeninFigure8.4 1 thatatmid‐depththeavailable

developmentlengthisonly175mm.Itisthereforenecessarytoextendthehookbya

minimumof2× 218–175 86mm.

ThedetailsofthestirruparegiveninFigure8.4 3 .InEquation8.2 9 ,dh 30

mmandsl 70mm.

Figure 8.4(3) Dimensional details of the stirrup in Example 2

CHAPTER 9


Problem Amultistoreyreinforcedconcreteandbrickstructureaspartofanofficebuildingmay

beidealisedasshowninFigure9.2 6 .Designthetypicalfloorasaone‐wayslab.Take

cf 25MPa,D 210mm,q 4kPaandthefloorfinishtobe0.5kPa.Exposure

classificationA1andfireresistanceperiodof60minutesareassumed.Theeffectsof

windmaybeignored.

Figure 9.2(6) Details of a design: multistorey reinforced concrete and brick building

Solution Thedesignistobecarriedoutona1‐metrewidestrip.

First,theloadingmustbeconsidered.Assumingthetotalsteelreinforcementis

0.5%byvolumeintheslab,Equation2.4 1 gives3

w 24 0.6 0.5 24.3 kN/m .

Thus,thedeadload

1 0.21 24.3 floor finish 5.1 0.5 5.6 kN/mg

andtheliveloadq 4kN/m

ThedesignloadaccordingtoEquation9.2 4 is

d 1.2 5.6 1.5 4 12.72 kN/mF

FollowingFigure9.2 2 ,the‘beam’momentcoefficientsαforthevarious

sectionsareshowninFigure9.2 7 togetherwiththeM*values.

Figure 9.2(7) Moment coefficients (α) for and moments (M*) at various sections

of the slab

Bending design Wemaynowproceedtodesignthemainsteelreinforcementusing500N12bars.

Table1.4 2 :cover 20mmforexposureclassificationA1

Table5.5.2 B ofAS3600‐2018:cover≥10mmforfireresistanceperiodof60

minutes

Thus,asshowninFigure9.2 8 ,

210 20 6 184 mmd

Figure 9.2(8) Definition of d for top and bottom bars in the slab

Beam section A AsshowninFigure9.2 7 ,

* 13.25 kNm/mM

Equation3.3 2 a:α2 0.85–0.0015×25 0.8125

0.8125 25Equation 3.5(7): 0.0406

500

Assumingϕ 0.85,

Equation3.5 6 :

62

t 2

2 0.0406 13.25 100.0406 0.0406 0.00093

0.85 1000 184 500p

Nowcheckforϕ.

Equation3.3 2 b:γ 0.97–0.0025×25 0.9075

u

0.00093 500Equation 3.4 8 : 0.025

0.8125 0.9075 25k

Forasinglelayerofsteel,wehaved doandkuo ku 0.025.Then

Equations3.4 20 awithb:ϕ 0.85,whichisacceptable.

But2 2

ct.ft.min

sy

210 0.6 25Equation 3.5(5): 0.20 0.20

184 500

0.001563

D fp

d f

Sincept pt.min,usept.minor2

st t.min 0.001563 1000 184 288 mm / mA p bd

FromTable2.3 2 wehaveforN12barsataspacingofs 300mm

2 2st 377 mm / m 288mm / mA

Thisisacceptable,as

Equation 9.2(7): the lesser of 2.0 2 210 420 mm and 300 mm, and

thereby 300 mm.

s D

s

Thus,useN12@300mm Ast 377mm2/m 288mm2/m forthetopbars.

Beam section B AsperFigure9.2 7 ,

* 28.91 kNm/mM

Followingthesameprocess,wehave

t t.min0.00206 > which is acceptablep p ,

2st 379 mm /mA

forN12at275mm,Ast 411mm2/m,whichisacceptable.

Thus,useN12@275mmforthebottombars.

Beam section C1 AsshowninFigure9.2 7 ,

* 31.80 kNm/mM

Followingthesameprocess,wehave

pt 0.00227 pt.min,whichisacceptable

Ast 418mm2/m

forN12barsat250mm,Ast 452mm2/m,whichisacceptable.Thus,useN12@250

mmforthetopbars.

Beam section C2 AsshowninFigure9.2 7 ,

* 45.79 kNm/mM

Inviewoftheheaviermoment,wemayuseN16bars.Thisgivesd 182mm.Similarly,

wehave

pt 0.0034 pt.min;thisisacceptable

Ast 619mm2/m

forN16barsat300mm,Ast 670mm2/m,whichisacceptable.Thisalsosatisfies

Equation9.2 7 .

Thus,useN16@300mmforthetopbars.

Beam section D AsshowninFigure9.2 7 ,

* 28.62 kNm/mM

Using,say,N12bars,d 184mm.Thenwehave:

pt 0.00204 pt.min,whichisacceptable

Ast 375mm2/m

forN12barsat300mm,Ast 377mm2/m.


Beam section E1 AsshowninFigure9.2 7 ,

* –41.63kNm / mM

Inviewoftheheaviermoment,wemayuseN16bars.Thisgivesd 182mm.Similarly,

wehave

pt 0.00307 pt.min,thisisacceptable

Ast 559mm2/m

forN16barsat300mm,Ast 670mm2/m,whichisacceptable.

Also,

Equation 9.2(7): the lesser of 2.0 2 210 420mm and300mm,and

thereby 300 mm

s D

s

Thus,useN16@300mm Ast 670mm2/m 596mm2/m forthetopbars.

Shear design ItisapparentinFigure9.2 3 thattheshearforceisamaximumatsectionClof

continuousstripifallspansareequal.Forourcase,Ln2 Ln1andthemaximumshear

occurinsectionC2.

Equation9.2 5 thusgives* 0.5 12.72 6 38.16 kNV

'uc v w v cEquation 6.3(4): =V k b d f

where 'c 25 5 MPa 8 MPa, which is acceptable.f

dv 0.72D,0.9d max 0.72 210,0.9 182 max 151.2,163.8 max

163.8mm

andusingthesimplifiedmethodofClause8.2.4.3,AS3600‐2018,for

sv sv.minA A

s s , v

v

200 2000.165

(1000 1.3 ) (1000 1.3 163.8)k

d

0.10,whichisnot

acceptable,forwhich,takekv 0.10Finally, 3

ucEquation 6.3(4): 0.10 1000 163.8 25 10 81.9 kNV

and

uc * 0.75 81.9 61.4 kN > 38.16 kNV V Thisisacceptable.NotethatifVuc V*,athickerslabshouldbeused.

Shrinkage and temperature steel (in y-direction) Forunrestrainededgeconditions,Equation9.4 14 gives:

3 2s.min 1.75 1000 210 l0 368 mm /mA

2 2stfor N12 @ 300 mm, A 377 mm /m (> 368 mm /m).


Serviceability check Figure9.2 7 showsthat,fortheexteriorspan,Equation5.2 4 requiresthat

ef nthe lesser of (5250 mm) and (5210 mm)L L L D

Thus,

ef 5210 mmL

Fortotaldeflection:

Table5.2 4 : /Lef 1/250

Equation5.4 3 :sc

csst

2 1.2 2.0A

kA

Table1.3 1 :forofficebuildingψs 0.7andψl 0.4.

ThenEquation5.5 8 gives2 3

d.ef (1 2) 5.6 (0.7 2 0.4) 4 22.8 kN/m 22.8 10 MPaF

Equations9.2 9 and9.2 13 givek3 1.0forone‐wayslabandk4 1.75foran

endspan,respectively.

FromTable2.2 1 ,forfc’ 25MPa,Ec 26700MPa


min

33

5210177.9 mm < 184mm

1/ 250 267001 1.75

22.8 10

d d

Thisisacceptable.

Fortheinteriorspan,Lef 6210mmandk4 2.1.Theseleadto

min 176.7 mm < 182 mmd d

Thisisalsoacceptable.

Drawings ThedesignresultsarepresentedinFigure9.2 9 .

Figure 9.2(9) Reinforcement layout for the design slab

a Or half the amount provided near the top and bottom faces may also use steel

fabrics

b ≤ 50% may be curtailed


Problem ThecornerslabshowninFigure9.3 8 ispartofahotelcomplex.SidesABandBEare

supportedonbrickwalls,andsidesACandCEonreinforcedconcretespandrelbeams.

ThebeamsaresupportedatCbyacolumncastmonolithicallywiththeslab.

A

EB

C

Continuous

x

280 mm 250 mm

12 m12 m

ColumnWall

Figure 9.3(8) Corner slab in a hotel complex

DesigntheslabinaccordancewiththeStandard,given cf 25MPa,exposure

classificationA1andfire‐resistanceperiod 90minutes.UseNbarsonly.

Solution

Loading Table3.1ofAS/NZS1170.1‐2002givesq 4kPa.AssumingD 300mmandthetotal

amountofsteelintheslabis0.5%byvolume,Equation2.4 1 gives3

w (24 0.6 0.5) 24.3 kN/m

Then

0.3 24.3 7.29 kPag

Also,thedesignloadis

d 1.2 1.5 14.75 kPaF g q

Design moments Now,Lx 10mandLy 12myieldLy/Lx 1.2.Withthetwoadjacentedgesbeing

continuous,Table9.3 1 gives

x y0.046 and 0.035

Positive moments * 2

xEquation 9.3 1 : 0.046 14.75 10 67.85 kNm/m M

* 2yEquation 9.3 2 : 0.035 14.75 10 51.63 kNm/m M

Negative moments AboutedgeAB

*yEquation 9.3(7): 1.33 51.63 68.67 kNm/mM

AboutedgeCE*

yEquation 9.3(5): 0.5 51.63 25.82 kNm/mM

AboutedgeBE*Equation 9.3(6): 1.33 67.85 90.24 kNm/mxM

AboutedgeAC*xEquation 9.3(4): 0.5 67.85 33.93 kNm/mM

Minimum effective depth TheshortereffectivespanisLef 10000mm.Fortotaldeflection,

efTable 5.2(4): ( / ) 1/250L

sccs

st

Equation 5.4(3): 2 1.2 2.0A

kA

s lTable 1.3(1): 0.7 and 0.4

Then,

def

2 3

Equation 5.5(8): (1.0 2) 7.29 (0.7 2 0.4) 4

27.87 kN/m 27.87 10 MPa

F

FromSection9.3.4,wehavek3 1.0andbyinterpolationfromTable9.3 3

4 2.50 [(0.05/0.25) (2.95 2.5)] 2.59k

cEquation 2.2(5): 25 280 MPaE


min

33

10000251.3 mm

1/250 25 2801 2.59

27.87 10

d

SinceD 300mm,useN16bars.FromTable1.4 2 ,theconcretecoveris20mmfor

exposureclassificationA1andfromTable5.5.2 B oftheStandard,only15mmis

neededforfireresistance.Thus

x min300 20 8 272 mm 251.3 mmd d

Thisisacceptable.Now

y min272 16 256 mm 251.3 mmd d

Thisisalsoacceptable.Finally,theaverage

average min272 8 264 mm 251.3 mmd d

This,too,isacceptable.

Bending design 2Equation 3.3(2)a: 0.85 0.0015 25 0.8125

and

0.8125 25Equation 3.5(7): 0.0406

500

Assumingϕ 0.85,Equation3.5 6 yieldstheptvaluesforallthedesignsections.The

resultsaresummarisedinTable9.3 4 .

Table 9.3(4)

Results for all of the design sections

Location Mx* My* pt Ast

N16 @

s

Ast

(actual)

Top or

bottom

x or y

direction

Centralregion 67.85 – 0.00222 603.8 300 670 B x

Centralregion – 51.63 0.002 512 300 670 B y

Outsidethe

centralregion

0.002 528

daverage

300 670 B xandy

AboutAB – –68.67 0.0026 665.6 300 670 T y

AboutCE – –25.82 0.00156a 399.4 300 670 T y

AboutBE –90.24 – 0.003 816 225 893 T x

AboutAC –33.93 – 0.00139a 378.1 300 670 T x

apt.minvalues,sincetheminimumreinforcementrequirementsgoverninthesecases.SeeEquation9.3 3 .

Note:checkforϕ.Takethelargestptvalueof0.003,Equation3.4 8 givesku 0.0814andford do,kuo

ku 0.0814andEquations3.4 20 awithbgiveϕ 0.85.Hence,forallcases,theassumptionofϕ

0.85isacceptable.

Corner reinforcement

Corner C AtcornerC,theupliftisprevented.Thus

Equation9.3 17 :A′st 0.75×670 502.5mm2/m

Theexistingtopandbottomreinforcementinthexandydirectionsexceed502.5

mm2/m.Therefore,noadditionalreinforcementisrequiredovercornerC.

Corners A and E Equation9.3 18 :A′st 0.5×670 335mm2/m

Allexistingtopandbottomreinforcementinthexandydirectionsexceed335

mm2/m.Therefore,noadditionalreinforcementisrequiredovercornersAandE.

Reinforcement layout AsketchofthereinforcementarrangementisgiveninFigure9.3 9 .

Figure 9.3(9) Reinforcement layout for the design slab Note:alldimensionsareinmm.


Problem AninterioridealisedframeisillustratedinFigure9.6 4 awiththedetailsofconnection

AshowninFigure9.6 4 b.GivenMv* 120kNm,Nz* 1500kN,Ds 350mmandd

300mm.Istheconnectionadequateinresistingthepunchingshear?Designatorsion

stripasnecessary.Assume cf 25MPaandthatashearheadisprovided.UseN12bars

forAsw.

A

z

x

150

x400

350

d = 300

120 kNm

1500 kN

50

A

150

150

350

A

(a) (b) Figure 9.6(4) An interior idealised frame with the details of its connection at A Note:alldimensionsareinmm.

Solution

Punching shear strengths Equation 9.6(4): 2 (350 150) (400 150 150) 1700 mmu

3uoEquation 9.6(8): 0.5 1700 300 25 10 1275 kNV

u 6

3

1275Equation 9.6(9): =1145 kN

1700 120 101

8 1500 10 500 300

V

But

u z0.75 1145 858.8 kN * 1500 kNV N

Thus,theslab–columnconnectionhasinsufficientpunchingshearstrength.

Torsion strip 1Equation 9.6(23): 500 25 6 469 mmy

sw 2

min

0.2 469Equation 9.6(22): 0.1876 mm /mm

500

A

s

u.min 6

3 2

1.2 1275Equation 9.6(24): 1202.83 kN

1700 120 101

2 1500 10 500

V

Butu.min z*= 0.75 1202.83 = 902.1 kN 1500 kNV N

Thus, Asw/s minisinadequateandtherequiredamountoftorsionalsteel,accordingto

Equation9.6 25 ,is2

sw 0.2 469 15000.519

500 902.1

A

s

z*Equation 9.6(26): the right-hand side 3 902.1 350/500 2264.3 kN 1500 kNN

Thisisacceptable.

Detailing UsingN12closedties,Table2.3 1 givesAsw 113mm2.Thus,thespacingis

113217.7 mm

0.519s

y

xA

35

46535b

280 350

N12 tie

35

Figure 9.6(5) Reinforcement details of the torsion strip

a Over a distance not less than a quarter of the respective centre to centre span of

the next column

b The cover of 35 mm is inadequate for B1 exposure classification. Note:alldimensionsareinmm.

Wecannowtakes 215mm,andfromEquation9.6 32 weseethatsislessthanthe

greaterof300mmandDs 350mm.Thisisacceptable.Thelayoutofthetorsionstrip

isgiveninFigure9.6 5 .

9.7.7 Design of column and middle strips

Theoverallthicknessoftheslabhasbeengivenas400mm.Eachofthecolumnand

middlestripscanbedesignedasawidebeamusingtherestricteddesignprocess

discussedinSection3.5.2.Twoequationsneedtobeusedrepeatedly.Theyare:

2 c

sy

f

f

Equation3.5 7

and

2t 2

sy

*2

Mp

b d f

Equation3.5 6

AsperEquation9.4 13

2

ct.ft,min

sy

0.24 /D d fp

f

Equation9.7 1

ForexposureclassificationBlwith cf 25MPa,Table1.4 2 indicatesthataclearcover

of60mmisrequired.Thisisgreaterthanthe25mmcoverrequiredforthe90‐minute

fire‐resistanceperiodasgiveninTable5.5.2 A oftheStandard AS3600‐2018 .Thus,

usingN28barsforboththepositiveandnegativesteel,orthebottomandtop

reinforcement,

400 60 28/2 326 mmd

Notethatwithd 326mmforbendinginthex–zplane,thecorrespondingdfor

they–zplanehastobed 326–28 298mm.TheseareillustratedinFigure9.7 5 .

Figure 9.7(5) Definitions of d for the top and bottom reinforcement layers in the slab Note:alldimensionsareinmm.

Inpractice,itiseconomicaltoadoptthelargerdfortheplanehavingalarger

momentinamajorityofthesections.Alternatively,itisacceptabletousetheaveraged

value 312mminthiscase .Inthepresentexample,letusassumethatthebendingin

thex–zplaneismoresevere.Thusweadopt

326 mmd

Notethatthisvalueisacceptableforfire‐resistanceconsiderationssinceTable

5.5.1oftheStandardindicatesthat,foraperiodof90minutes,theminimumeffectiveor

actualthickness D is100mmforinsulation.Forstructuraladequacyontheother

hand,theminimumDrequiredis200mmasperTable5.5.2 A oftheStandard.Both

valuesaresmallerthantheavailableD 400mm.

WithdnowcalculatedandusingEquations3.5 6 and3.5 7 ,thebending

reinforcementcanreadilybecomputed.

The column strip ( 5500 mm, 326 mm)b d

Section L M* 2442.4kNm.Considera1‐metrewidthofthecolumnstrip.

* 2442.4/5.5 444.1 kNm/mM

For cf 25MPa,α2 0.8125and

0.8125 250.0406.

500

Assuming 0.85:

62

t 2

2 0.0406 444.1 100.0406 0.0406 0.01145

0.85 1000 326 500p

t,minEquation 9.7(1): 0.0022p allEquation 3.4(6): 0.01475p

Aspt.min ptandpt pall,whichisacceptable.Therefore,

t 0.01145p and

2st t 0.01145 1000 326 3732.7 mm /mA p bd

ProvideN28@150mmtopbars Ast 4107mm2/m .

Section C * 1049.4 kNmM

Considera1‐metrewidthofthecolumnstrip.*

t t,min

t all

1049.4/5.5 190.8 kNm/m

0.0045 > 0.0022

0.0045 < 0.01475

M

p p

p p

Thisisacceptable.Therefore,

t 0.0045p and

2st 0.0045 1000 326 1467 mm /mA

ProvideN28@300mmbottombars Ast 2053mm2/m .Also,

Equation 9.2(7): the lesser of 2.0 and 300 mm 300 mms D

Thus,provideN28@300mmbottombars Ast 2053mm2/m .

Section R ThereinforcementisthesameasSectionLforreasonofsymmetry.

Therefore,provideN28@150mmtopbars Ast 4107mm2/m .

Half middle strip 1 (HMS1) ( 3000 mm, 326 mm)b d

Sections L and R * 305.3 kNmM

Considera1‐metrewidth.

*

t t,min

t all

305.3/3 101.77 kNm/m

0.00232 0.0022

0.00232< 0.01475

M

p p

p p


t 0.00232p

and2

st 0.00232 1000 326 756.3 mm /mA




*

t t,min

t all

349.8/3 116.6 kNm/m

0.00267 > 0.0022

0.00267 < 0.01475

M

p p

p p


t 0.00267p and

2st 0.00267 1000 326 870.4 mm /mA

ProvideN28@300mmbottombars Ast 2053mm2/m .

Half middle strip 2 (HMS2) ( 2500 mm, 326 mm)b d

Sections L and R * 305.3 kNmM


*

t t,min

t all

305.3/2.5 122.12 kNm/m

0.0028 > 0.0022

0.0028 < 0.01475

M

p p

p p


t 0.0028p

and2

st 0.0028 1000 326 912.8 mm /mA




*

t t,min

t all

349.8/2.5 139.92 kNm/m

0.00323 > 0.0022

0.00323 < 0.01475

M

p p

p p


t 0.00323p

and2

st 0.00323 1000 326 1052.98 mm /mA

ProvideN28@300mmbottombars Ast 2053mm2/m .

Thedesignofthebendingreinforcementisnowcompleted.However,notethat,

inpractice,itisacceptabletocombinethetotalmomentsofthetwoadjacenthalf

middlestripsfromthetwoadjoiningidealisedframes.Basedonthiscombinedtotal

moment,theAstforthemiddlestripcanbecomputedaccordingly.

Itshouldalsobepointedoutthatinallthebendingdesigncalculations,ϕ 0.85

isassumed.This,asinEquation3.4 20 a,requireskuo≤0.36.Itmaybeshownthatthis

criterionissatisfiedthroughout.

CHAPTER 10

10.3.3 Interaction diagram Foragivencolumnsectionsubjectedtoanaxialload Nu ataneccentricity e′ giving

anultimatemomentofMu e′Nu,thefailuremodeandstrengthdependuponthe

combinedeffectofNuandMu.Theinteractiondiagramofacolumnprescribesallthe

combinationsofNuandMuthatcancausefailuretothecolumn.Theprocedureindetail

forconstructinganinteractiondiagramisillustratedusingthefollowingexample.

Illustrative example

Problem Acolumnsectionsubjectedtobendinginthex–yplaneisdetailedinFigure10.3 4 .

Constructtheinteractiondiagram.

Take c 25 MPa,f sy 500 MPa,f 2 0.8125, 0.9075 and sc stA A 2N32bars@

804mm2 1608mm2.

uk d

s

Figure 10.3(4) Details of the example column section subjected to bending in the x–y

plane Note:alldimensionsareinmm.

Solution Thefollowingaregiven:b 300mm;d 450mm;dc 50mm.

SinceAsc Ast 1608mm2,wehavept pc 0.01191.

For cf 25MPa, 2 0.8125asperEquation3.3 2 a.

UnderdifferentcombinationsofNuandMu,thecolumncouldfailinoneofthree

modes.

Forcompressionfailure,Equation10.3 6 forthesquashloadcapacityyields:

u u s0.8125 0.9075 25 300 450 1608 (500 0.8125 25) 1608N k f

wherefsisgiveninEquation3.4 13 ,or

s u u600(1 ) /f k k

Wethenhave

u u u u[2.489 0.771 0.965(1 ) / ] (kN)N k k k Equation10.3 17

Themomentequation Equation10.3 9 becomes

u u u

6

u u

[0.8125 0.9075 25 0.9075 /2)

1608 0.8125 50)] 10

[11.20 (1 0.454 ) 3.085] (kNm)

eN k k

k k

Equation10.3 18

Fortensionfailure,Equation10.3 18 remainsvalid,butforNuEquation

10.3 17 gives

u u[2.489 0.033] (kN)N k Equation10.3 19

Notethateitherthetensionorcompressionfailurestrengthequationsmaybe

usedforthebalancedfailureanalysis.

WithEquations10.3 17 ,10.3 18 and10.3 19 inhand,theinteractioncurve

canbeobtainedbyappropriatelyvaryingthevalueofku.However,for

compressionfailureku kuB;fortensionfailureku kuB;accordingtoEquation

10.3 12 kuB 0.5454.

Nuo i.e.e′ 0

uoEquation 10.2(1): 4727.2 kN N

Muo i.e.Nu 0

uoEquation 3.6(15)a or b: 328.3 kNm M

Balancedfailure i.e.kuB 0.5454 .

uBEquation 10.3(17): 1324.2 kN N

Equation10.3 18 :eNuB 768.1kNm

ButfromFigure10.3 4 ,

uB B uB uB uB uB( 0.2) 0.2 768.1 0.2 1324.2

503.3 kNm

M e N N e eN N

Compressionfailure i.e.ku kuB .

ThevariablekumaybegivensomeappropriatevaluesandNuandMucanbe

computedusingEquations10.3 17 and10.3 18 ,asforthebalancedfailure

casegivenabove.

Forku 1

u

u

u u u

Equation 10.3(17): 3260.0 kN

Equation 10.3(18): 920.0 kNm

0.2 268.0 kNm

N

eN

M eN N

Forku 0.9

u

u

u

Equation 10.3(17): 2903.9 kN

Equation 10.3(18): 904.6 kNm

323.8 kNm

N

eN

M

Forku 0.8

u

u

u

Equation 10.3(17): 2521.0 kN

Equation 10.3(18): 879.1 kNm

374.9 kNm

N

eN

M

Forku 0.7

u

u

u

Equation 10.3(17): 2099.7 kN

Equation 10.3(18): 843.3 kNm

And 423.4 kNm

N

eN

M

Tensionfailure i.e.ku kuBandfs fsy

Fortensionfailure,Equations10.3 19 and10.3 18 ,respectively,shouldbe

usedforcomputingNuandeNu.NotethatinEquations10.3 6 and10.3 9 ,

yieldingofAscisassumed.Forthistobevalid,kumustbegreaterthanacertain

lowerlimit.

Consideringku 0.4asthelimitingvalue,

u

u

u

Equation 10.3(19): 962.6 kN

Equation 10.3(18): 675.1 kNm

482.6 kNm

N

eN

M

Forku 0.5 kuB 0.5454 ,

u

u

u

Equation 10.3(19): 1211.5 kN

Equation 10.3(18): 741.4 kNm

499.1 kNm

N

eN

M

Decompressionmode

Forthethresholdbeyondwhichnotensilestressexistsinthesection,Equation

10.3 16 gives

s

500600 1 66.67 MPa.

450f

ThenfromEquation10.3 15 ,3

u.dc [0.8125 0.8125

3643.6 kN

N

Further,

u

u.dc

6

Equation 10.3(8): (450 0.9075 500 / 2) 223.1 mm

Equation 10.3(9): [0.8125 0.9075 25 300 500 223.1 1608(500

0.8125 25) (450 50)] 10 925.4 kNm

j d

eN

and

u.dc u.dc u.dc0.2 925.4 0.2 kNmM eN N

Finally,withalltheabovecoordinatesofNuandMu,theinteractiondiagramcanbe

drawn.ThisisshowninFigure10.3 5 .

'Be

B = 0.380 m'e

Figure 10.3(5) Interaction diagram for the example column section

10.3.4 Approximate analysis of columns failing in

compression


Problem ForthecolumnshowninFigure10.3 4 ,calculatetheapproximatevaluesofNuandMu

forthecasewithe′ 0.0822m i.e.forku 1 .

Solution FromtheexampleinSection10.3.3,wehave

uB

uB

B

1324.2 kN

503.3 kNm

= 0.380 m,

N

M

e

and

uo 4727.2 kN. ThusN

u

4727.2Equation 10.3(22): 3038.2 kN

4727.2 0.08221 1

1324.2 0.380

N

u

4727.2 3038.2Equation 10.3(21): 503.3 249.8 kNm

4727.2 1324.2M

NotethattheapproximateformulashaveunderestimatedthevaluesofNuand

Muby6.80%and6.79%,respectively.

10.4.2 Illustrative example of iterative approach

Problem Anirregular‐shapedcolumnsectionisshowninFigure10.4 3 .ComputeNuusingthe

iterativeprocedure.Takef′c 32MPaandfsy 500MPa.

Figure 10.4(3) An irregular-shaped column section Note:alldimensionsareinmm.

Solution 2

2

Equation 3.3(2)a: 0.85 0.0015 32 0.802, but as per AS 3600-2018, for sections where the width

reduces from the the neutral axis towards the compression face, shall be reduced by 10%.

Hence, in t

2his case 0.9 0.802 0.7218

Equation 3.3(2)b: 0.97 0.0025 32 0.89

Andtherearetwolayersofsteelorm 2.

Trial 1

Step 1 AssumedNA 240mm.

Step 2 Forthetopsteellayer2 orAsmwithm 2inthiscase

s2 sy

s2 s2

0.003 (240 60)/240 0.00225 < 0.0025

that is, 200000 450 MPaf

32 450 1232 10 554.4 kN T

Forthebottomsteellayer1 orAs1 ,

s1 0.003 (240 300)/240 0.00075 (tension)

and

s1 s1

31

200000 150 MPa

150 3164 10 474.6 kN (tension)

f

T

Step 3 TheheightoftheconcreteareaincompressionasshowninFigure10.4 4 isgivenas

NA 0.89 240 213.6 mmd

and

(213.6 / 240) 120 106.8 mm c

Figure 10.4(4) Determination of neutral axis (NA) for the irregular-shaped column

section Note:alldimensionsareinmm.Thus,theconcretearea

2120 213.6 2 0.5 213.6 106.8 25 632 22 812.5 48 444.5 mm A

UsingEquation10.4 4 ,

x (25 632 213.6/2 22 812.5 213.6 2/3)/48 444.5 123.6 mmd

3cEquation 10.4(2): 0.7218 32 48 444.5 10 1119.0 kNC

Step 4 u

computed

Equation 10.4(5): 1119.0 554.4 474.6 1198.8 kN

Equation 10.4(6): [1119.0 (300 123.6) 554.4 (300 60) ]/1198.8

275.6 mm

N

e

Thus,theeccentricity

computed 275.6 (300 218) 193.6 mme

Step 5 computed given< ( 380 mm) e e

Thus,reducedNAandrepeattheprocess.

Trial 2

Step 1 AssumedNA 210mm.

Step 2 s2 sy

s1

1

0.00214 < , therefore

527.3 kN

0.001286, therefore

813.8 kN

2T

T

Step 3 NA 186.9 mm d

Accordingly,2

x

c

39 893.8 mm

and 107.1 mm

Thus, 921.5 kN

A

d

C

Step 4 u

computed

computed

921.5 527.3 813.8 635.0 kN, and

479.2 mm, therefore

397.2 mm

N

e

e

Step 5 computed given> e e

butthevalueisgettinglarger.Thus,increasedNAto211.5mmandtryagain.

Trial 3

Step 1 NA 211.5 mmd

Step 2 529.8 kN 2T

l 797.3 kNT

Step 3 c 930.7 kNC

Step 4 u

computed

computed

663.2 kN, and

461.3 mm

Or, 379.3 mm

N

e

e

Step 5 computed givene e

Therefore,wetakeNu 663.2kN.

10.4.4 Illustrative example of semi-graphical method

Problem Arectangularcolumnsection,reinforcedwithfourlayersofsteel atotaloftwelve20‐

mmdiameterbars issubjectedtouniaxialbendinginthex–yplane.Detailsofthe

sectionareshowninFigure10.4 6 .Take c 32f MPa, fsy 500MPa,α2 0.802andγ

0.89.DetermineNuusingthesemi‐graphicalmethod.

80

360

450

+

z

Nu

25011060

xPlastic centre (pc) ₠

dpc

60

Figure 10.4(6) Details of the example rectangular column section Note:alldimensionsareinmm.

Solution Therearefourlayersofsteel orm 4 andforthesymmetricalsection,thepositionof

theplasticcentredpc 225mm.

Thescaleddrawingofthecross‐sectionincludingthereinforcementpositionsis

giveninFigure10.4 7 ,togetherwiththesteelstressdiagram.Notethatforthe

rectangularsectionbj 360mm,whichisconstant,thereisnoneedtodrawtheentire

cross‐section.Thedrawingofthefullsectionismandatoryforanirregularsection.

Figure 10.4(7) Scaled drawing on graph paper for the semi-graphical method example

TheworkingisgivenindetailinTable10.4 1 .ForTrial1,assumedNA 200

mm.ThenthetwosumsinEquation10.4 9 aregiven,respectively,incolumns9and6

ofTable10.4 1 .WehaveNu 1596.9–210.3 1386.6kN.

Table 10.4(1)

Calculations for semi-graphical method example

1 2 3 4 5 6 7 8 9 10

Trial Elemen

t Asi fsi (dpc – di) Asifsi ∆Acj

(dpc –

xj) α2f′c∆Acj ∆(e′Nu)

1 i 1 1256 –500 225–390

–165

–628kN 103.6kNm

i 2 628 –240 225–280

–55

–150.7 8.3

i 3 628 85 55 53.4 2.9

i 4 1256 410 165 515.0 85.0

j 1 50×360

18000

225–

25

200

462.0kN 92.4

j 2 18000–

1240

16760

225–

75

150

430.1 64.5

j 3 18000 225–

125

100

462.0 46.2

j 4 28×360

‐620

9460

225–

164

61

242.8 14.8

∑ –210.3 1596.9 417.7

2 i 1 1256 60 –165 75.4 –12.4

i 2 628 90 –55 56.5 –3.1

i 3 628 60 55 37.7 2.1

i 4 1256 30 165 37.7 6.2

j 5 22×360

7920

225–

200

11

36

203.3 7.3

∑ –3.0 1800.2 417.8

3 i 1 1256 –40 –165 –50.2 8.3

i 2 628 –30 –55 –18.8 1.0

i 3 628 –20 55 –12.6 –0.7

i 4 1256 –15 165 –18.8 –3.1

j 6 –9×

360

–3240

225–

187

38

–83.2 –3.2

∑ –103.4 1717.0 420.1

Equation10.4 10 isrepresentedbycolumn10ofTable10.4 1 andthesum

e′Nu 417.7kNm.

Thus

e′computed 417.7/1386.6 0.30m,whichisgreaterthane′given 0.25m .

ForTrial2,dNA 225mm.ThisgivesNu 1800.2–3.0 1797.2kN,and

e′computed 417.8/1797.2 0.23m,whichislessthane′given.

ForTrial3,dNA 215mm.ThisgivesNu 1717.0–103.4 1613.6kN,and

e′computed 420.1/1613.6 0.26m,whichisalsogreaterthane′given 0.25m.

Thus,takeNu greaterthan1613.6kN,say1620kN.


Problem GivenMg 34kNm,Mq 38.3kNm;Ng 340kN,Nq 220kNwheresubscriptsgand

q,respectivelyindicatedead‐andlive‐loadeffects.Assume cf 32MPaandfsy 500

MPa,andproportionasymmetricallyreinforcedsquaresection.

Solution ByinvokingEquation1.3 2 ,wehave

* 1.2 34 1.5 38.3 98.25 kNmM

and

* 1.2 340 1.5 220 738 kNN

1. Assumingpt pc 0.008,Equation10.6 1 gives

3 2 (738 10 )/[0.65 (0.802 32 500 0.016)] 33 727.0 mmb D

fromwhich 184 mm.b D

2. Useb D 225mm,say.

3. Equation10.6 3 leadsto2

6 6 3(98.25 10 ) / [0.85 (0.9 500 0.008)] 34.11 10 mmb d

Takeb 1.1 d andwehave d 315.Thismeans

0.9 1.1 315 312 mm 320 mm, say.b D

4. Thus,thedetailsofthepreliminarycolumnsectionare

b D 320mmand

Ast Asc 0.008×0.9×3202 738mm2

oruse4N24bars,oneeachatthefourcorners,givingatotalof1808mm2,

whichisgreaterthan2×738 1476mm2.

CHAPTER 11

11.6 Illustrative examples

11.6.1 Example 1 – load-bearing wall

Problem Arectangularwallis300mmthickwithanunsupportedheightof6mandalengthof5

m.Theverticalaxialloadingfromthefloorbeamsaboveactswitha10‐mmeccentricity.

Assumingthatthewallisrestrainedagainstrotationatthetopandbottomendsbythe

floors,andthat cf 32MPaandfsy 500MPa,computethedesignaxialstrengthvalues

forthefollowingside‐restraintconditions:

a withoutsiderestraintsusingtheStandardprocedure

b withoutsiderestraintsusingtheACI318‐2014formula

c withrestraintsatbothsidesofthewall.

Solution

a Withoutsiderestraint–Standardprocedure

Equation11.3 1 isapplicableinthiscase,forwhichHwe kHw 0.75×6

4.5m,whichislessthanthewalllengthof5m.

Hence,adoptHwe 4.5m.

Sincetheeccentricitye 10mm 0.05tw 0.05×300 15mm,usee

15mm.

TheslendernessratioHwe/tw 4500/300 15 30,whichisacceptable.

Equation11.3 2 :ea Hwe 2/2500tw 4500 2/ 2500×300 27mm.

Withthesevalues:

Equation11.3 1 :ϕNu ϕ tw–1.2e–2ea 0.6fc′

0.65 300–1.2 15 –2 27 ×0.6×32× 5000×

10–3

14227kN

b Withoutsiderestraint–ACI318‐2014procedure

ApplyEquation11.3 6 forwhich

L/tw 5000/300 16.7 25,whichisacceptable.

tw 100,whichisacceptable.

k 0.8asthetopandbottomendsarerestrainedagainstrotation.

Thus,Equation11.3 6 :ϕNu ϕ0.55fc′Ag 1– kH/32tw 2

0.7×0.55×32× 300×5000 × 1– 0.8×6000 / 32×300 2 ×10–3

13860kN

NotethatEquation11.3 6 givesa2.6%lesscapacitythanEquation

11.3 1 .

ThisisbecausetheACI318‐2014equationdoesnotconsidertheeffectof

eccentricity,ortheadditionaleccentricityduetothesecondaryP–∆

effects.

c Withbothsidesrestrainedlaterally

TheStandard’ssimplifiedmethodintheformofEquation11.3 1 isalso

applicablewhenbothsidesarerestrainedlaterally.Thesolutionisobtained

byusingtheappropriateinputparametersforthewallintwo‐wayaction.

SinceHw L1,

1

w

5000Equation 11.3(5): 0.4167

2 2 6000

Lk

H

Thatis,Hwe kHw 0.4167×6 2.5m

Similarly,theeccentricitye 10mm 0.05tw 0.05×300 15mm.

Therefore,adopte 15mm.2 2

a we wEquation 11.3(2): ( ) /2500 (2500) /(2500 300) 8.3333 mm e H t

Then

u w a c

3

Equation 11.3(1): N ( 1.2 2 )0.6

0.65 [300 1.2(15) 2(8.3333)] 0.6 32 5000 10

16 557 kN

t e e f

Comparingtheresultsincases a and c indicatesthat,withbothsidesof

thewallrestrainedlaterally,thedesignaxialstrengthincreasesby 16557

–14227 /14227×100% 16.4%.

11.6.2 Example 2 – tilt-up panel

Problem Atwo‐storeyofficebuildingistobeconstructedasanassemblyoftilt‐uppanels.The

continuous externalpanelsareeach7.5mhighby5mlong.Theyaredesignedto

supporttheloadingfromthefirstflooraswellastheroof.Thefirstflooris3.6mabove

thebaseslab;theroofisafurther3.6mabovethefirstfloor.Thedeadandliveloads

fromtheroofare45kN/mand40.5kN/m,respectively,alongthewall.Thedeadand

liveloadsfromthefirstfloorare76.5kN/mand66kN/m,respectively.Assume cf 32

MPaandthatthefloorslabbearingintothepanelsis25mm.

Isa150‐mmthickconcretewallwithminimumverticalandhorizontal

reinforcementadequateforthedesign?

Solution a Designat1.8m mid‐heightofwallbetweenbaseandfirstfloor

Assuming w 24kN/m3,walldeadload 24× 7.5–1.8 ×0.15 20.52

kN/m

Totaldeadloadg self‐weight roofdeadload floordeadload

20.52 45 76.5 142.02kN/m

Totalliveloadq roofliveload floorliveload 40.5 66 106.5

kN/m

Designload 1.2×g 1.5×q 330.17kN/m

b Eccentricityofloadat1.8mheight

Verticalloadoffirstfloor 1.2×76.5 1.5×66 190.8kN/m

AsinClause11.5.4oftheStandard,verticalloadisassumedtoactatone‐

thirdthedepthofthebearingareafromthespanfaceofthewall.Thus,the

eccentricityoftheloadabovethefirstfloor

ef 150/2 – 25/3 66.667mm emin 0.05t 7.5mm,whichis

acceptable.

At1.8m,theeccentricityofload,e 66.667×190.8/330.17 38.525mm.

Also,2

a

u

Equation 11.3(2): (0.75 3600) /(2500 150) 19.44 mm

Equation 11.3(1): 0.65[150 1.2(38.525) 2(19.44)] 0.6 32

809.83 kN/m 330.17 kN/m, which is satisfactory.

e

N

11.6.3 Example 3 – the new strength formula

Problem RepeattheexampleinSection11.6.1usingthenewdesignformulaforawall

a withoutsiderestraint

b withallfoursidesrestrainedlaterally.

Solution a Designstrengthwithoutsiderestraint.

SinceH/tw 27

we

2a

Equation 11.3(10): 1. Hence,

Equation 11.3(9): 6 m and

Equation 11.3(8): (6000) /(2500 300) 48 mm.

H H

e

Thus0.7 3

uEquation 11.3(7): 0.6 2 32 [300 1.2(10) 2(48)] 5000 10

13033kN

N

b Designstrengthwithallsidesrestrained.

ForH/tw 27

w

1 1Equation 11.3(14): 1.0345

101 1

300

e

t

Hence,forH 6m L 5m

5000Equation 11.3(13): 1.0345 0.4310

2 2 6000

L

H

ThusHwe 0.4310×6000 2586mm

and2

aEquation 11.3(2): (2586) /(2500 300) 8.917 mme

Finally,0.7 3

uEquation 11.3(7): 0.6 2 32 [300 1.2(10) 2(8.917)] 5000 10

18 340 kN

N

Itshouldbenotedthatthenewformulayieldshigherload‐carrying

capacities.Inparticular,whensiderestraintsareconsidered,thecapacityis

about29%greaterthanthatallowedforaone‐way‐actionwall.Thisshows

thatEquation11.3 1 isundulyconservativewhenappliedtowallswith

lateralrestraintsatallthefoursides.

11.6.4 Example 4 – design shear strength

Problem Areinforcedconcretewallisrequiredtoresistverticalandhorizontalin‐planeforcesin

amultistoreybuilding.Theoverallfirst‐floorheightandlengthofthewallareboth4.5

m.Thewallis250mmthickwithtwo500mm×500mmboundarycolumns.A

structuralanalysisofthebuildinggivesN* 6925kN,V* 3750kNandthein‐plane

momentM* 19300kNm.

Computetheshearstrengthofthewallandprovidesuitablereinforcement.

Assume cf 50MPaandfsy 500MPa.ComparethestrengthresultwiththeACI318‐

2014provisions.

Solution Theboundarycolumnsservetoresistthein‐planebendingmomentinthewall,aswell

astheaxialcompression.Theshearforceduetohorizontalloadontheotherhandis

carriedbythewall.

a Checkthemaximumshearcapacity.3

u.maxEquation 11.4(2): 0.2 50(0.8 4500 250) 10 9000 kNV

AndϕVu.max 0.75×9000 6750kN V*,whichisacceptable.

b Calculatetheshearstrengthofthewall.

Trial 1

UseN16bars@200mmineachfaceofthewallinboththeverticaland

horizontaldirections.

Equation11.4 3 :

3uc

c w w

[0.66 50 0.21(4.5 / 4.5) 50]0.8 4500 250 10

2863.8 kN 0.17 (0.8 ), which is acceptable.

V

f L t

Thesteelratio,pw 201/ 200×250 0.004

Equation11.4 6 :Vus 0.004×500× 0.8×4500×250 ×10–3

1800kN

Then

Equation11.4 1 :

ϕVu 0.75 2863.8 1800 3497.9kN 3750kN,whichisnot

acceptable.

Trial 2

UseN20bars@220mm.

Thesteelratio,pw 314/ 220×250 0.00571

Equation11.4 6 :

Vus 0.00571×500×0.8×4500×250×10–3 2569.5kN

Equation11.4 1 :

ϕVu 0.75 2863.8 2569.5 4075.0kN 3750kN,whichis

acceptable.

c CalculatetheshearstrengthusingtheACI318‐2014recommendations.

3uc

uc

Equation 11.4(8): 0.275 50(0.8 4500 250) 10 0.2 6925

3135.1 kN

Equation 11.4(9): 3028 kN 3135.1 kN. Thus, adopt this value.

V

V

Then3

usEquation 11.4(10): 314 500 (0.8 4500) 10 /220

2569.1 kN

V

Finally,Equation11.4 1 :

ϕVu 0.75 3028 2569.1 4197.8kN 3750kN,whichisacceptable.

NotethattheACImethodgivesahighercapacityload.Thisistobeexpected

asitalsoaccountsforthebeneficialeffectsofaxialloading.

d Checkrelateddesignrequirements.

Thewallpanelshouldbecheckedforadequatecrackcontrolinbothvertical

andhorizontaldirections,eachhavingareinforcementratioof0.00571.As

quotedinSection11.5,thisratioismorethanadequateforamoderate

degreeofcontrol,whichrequiresaminimumpw 0.0035inexposure

classificationA1orA2.Itisjustshortoftherequired0.006forastrong

degreeofcrackcontrol.

Finally,theflexuralstrengthofthewallmustalsobechecked.Thisisdone

byadoptingtherectangularstressblockforthecompressionzoneand

applyingthegeneralcolumnanalysisgiveninChapter10.Themoment

capacityofthesectionϕMumustbegreaterthanorequaltoM*.

CHAPTER 12

12.2.5 Design example Theexamplegiveninthissectionisforanasymmetricalfootingundereccentricloading.

ItdemonstratestheuseoftheequationsandprocessdevelopedinSection12.2.2for

symmetricalwallfootingsundereccentricloading.However,asdiscussedinSections

12.2.3and12.2.4,concentricallyloadedfootingsandasymmetricalfootingsunder

eccentricloadingmaybetreatedasspecialcasesofasymmetricalfootingunder

eccentricloading.Thus,theexamplebelowalsoillustratesthedesignoftheothertwo

typesofwallfootings.

Numerical example

Problem Givenareinforcedconcretewall,300mmthickandsubjectedtoadeadloadDL 200

kN/mandaliveloadLL 150kN/m,eachhavingthesameeccentricityofe 100mm

fromthecentreplaneofthewall.Take cf 25MPa,theeffectivesoilbearingcapacityqf

250kPa;A2exposureclassificationapplies.Designanasymmetrical strip footingin

suchawaythatthesubsoilpressureisuniformlydistributed.UseN16orN20barsin

onelayerandusenoshearreinforcement.

Solution Detailsofthefootingofa1‐metrerunaredepictedinFigure12.2 5 .Notethattheclear

coverof30mmisinaccordancewiththerecommendationgiveninTable1.4 2 .This

assumesthatthebaseofthefootingiswellpreparedandcompactedfollowing

excavationandbeforecastingtheconcrete.Ifthiscannotbeassured,alargercover

oughttobeused.

Thedesignactionpermetrerun i.e.b 1000mm is

* 1.2 200 1.5 150 465 kNN

M* 0atthecentrelineoftheasymmetricalfooting;seeFigure12.2 5 .

ThebreadthofthefootingasperEquation12.2 10 andwithanassumedρw

24kN/m3

465 465

250 1.2 24 250 28.8L

D D

Equation a

WithEquation a inhand,thefootingdesignmayfollowthetrialanderror

processdetailedinSection12.2.2.

DL=200 kN/mLL=150 kN/m

LC

D

CL

e =100 mm

c 150 mm2

D

L/2 L/2

N20 bars

do

Design shear section

do

30 mm cover

qf =250 kPa

Shrinkage and temperature steel

Figure 12.2(5) Details of the design asymmetrical wall footing

Trial 1 AssumeD 600mm,andEquation a gives

4651.998 m

250 28.8 0.6L

UseL 2.0m.

Step 1 ForN20barswithexposureclassificationA2,

o 600 30 20/2 560 mmd

Step 2 Thedesignshearforcepermetrerunis

*o f0.1 0.15 ( 0.9 24 )

2(1.0 0.1 0.15 0.56) (250 0.9 24 0.6) 92.45 kN

LV d q D

Notethataloadfactorof0.9isappliedtotheself‐weightofthefootingtoproducea

morecriticalV*.

Step 3 Equation6.3 1 gives

v'u.max c w v 2

v

cot0.55

1 cotV f b d

whereusingthesimplifiedmethodofClause8.2.4.3,AS3600‐2018,v 36°and,asfor

N20barsinonelayer,d do 560mm

dv = [0.72D, 0.9d]max = [(0.72 × 600), (0.9 × 560)]max = [432, 504]max = 504 mm

Thus, 0

3u.max 2 0

cot 360.55 25 1000 504 10 3295.4 kN

1 cot 36V

and

u.max 0.75 3295.4 2471.6 kN > *V V

TheassumedsectionwithD 600mmismorethanadequate i.e.itisreinforceable .

Thus,thisstepwillbeomittedinsubsequenttrials.

Step 4 Equation6.3 4 : '

uc v w v cV k b d f where 'c 25 5 MPa < 8 MPa which is acceptablef

and vv

200 2000.12 0.10

1000 1.3 1000 1.3 504k

d

;hence,takekv 0.10


3uc 0.10 1000 504 25 10 252 kNV

Step 5

Thespecificationwasthatshearreinforcementisnottobeused.Thus,checkthat

0.75 252 189 kN > * 92.45 kNV

Thus,D 600mmisadequatebuttooconservative.

Trial 2 AssumeD 550mm,andEquation a gives

4651.986 m

250 28.8 0.55L

UseL 2.0m.

Step 1 0 550 30 20/2 510 mmd

Step 2 * (1.0 0.1 0.15 0.51) (250 0.9 24 0.55) 104.77 kNV

Step 3 Asexplainedinstep3intrial1,thisstepisskippedhere.

Step 4 ForEquation6.3 4 ,dv 459mmandkv 0.10.ThusEquation6.3 4 :

3uc 0.10 1000 459 25 10 229.5 kNV

Step 5 AsVuc 0.75×229.5 172.1kN>V* 104.77kN,

D 550mmisacceptable.Proceedtothebendingsteeldesigninstep9,skippingsteps

6–8whicharenolongerrequiredforthisexample.

Step 9 DesignforAst.

ForasinglelayerofN20bars,d do 510mmand

2Equation 3.3(2)a: 0.8125

Equation 3.3(2)b: 0.9075

AccordingtoFigure9.4 2 andforthewall‐footinglayoutasshowninFigure12.2 6 ,

thelocationofthecriticalbendingsectionisatLbfromtherightedgewhere

b s0.72

LL e a

whereeistheeccentricityoftheloadingandasishalfthewallthickness.

Thus,

b 1 0.1 0.7 0.15 0.995 mmL

Withρw 24kN/m3and2* (250 24 0.55 1) 0.995 /2 117.2 kNmM

0.8125 25Equation 3.5(7): 0.0406

500

Assumingϕ 0.85,Equation3.5 6 gives

62

t 2

2 0.0406 117.2 100.0406 0.0406 0.001075

0.85 1000 510 500p

Checkthattheassumedϕ 0.85iscorrect.

u uo

0.001075 5000.029 from Equation 3.4(8)

0.8125 0.9075 25k k

andthenEquations3.4 20 awithb:ϕ 0.85.Therefore,theassumptioniscorrect.

LC

L

Lb

550 mm

250 kN/m

510 mm

0.7as

Wall

Critical bending section

Figure 12.2(6) Defining critical bending section for the trial footing

Now,similartoone‐wayslabs,aswallfootingsmaybedesignedasbeamsofa

unitwidth,Equation3.5 5 :

2t.min t0.20 (550 / 510) (0.6 25) / 500 0.001396p p

Therefore,pt 0.001396.

Step 10 Fora1‐metrerunofthefooting

2st 0.001396 1000 510 712.0 mm /mA

UsingTable2.3 2 ,N20bars@300mmyieldAst 1047mm2/m 712.0mm2/m.

Thisisacceptable.

Also,forcrackcontrolpurposesinthiscase,themaximumspacingiss 300mm

asshowninTable1.4 5 .Therefore,useN16barsandfromTable2.3 2 ,N16@275

mmyieldsAst 731mm2/m 712.0mm2/m.Thisisacceptable.

Withthereducedbarsize,disincreasedandthestressdevelopmentlengthis

shortenedaccordingly.Inconsequence,noextracheckisnecessaryforthisbarsize

variation.

Step 11 Checkthebond.

ForEquation8.2 1

1 2 31.3; (132 16)/100 1.16; 1 0.15(32 16) /16 0.85k k k

Equation8.2 1 ,thusgives

sy.tb (0.5 1.3 0.85 500 16) / (1.16 25) 762 mmL

SincetheavailableLsy.t Lb–cover 0.995–0.03 0.965montherightof

thecriticalbendingsection,andLsy.t L–Lb–cover 0.975montheleftofthe

criticalbendingsection,straightbars withouthooksorcogs oflength 2000–2×30

1940mm@275mmareadequateforbond.

InviewofEquation8.2 9 ,orthefactthattheactualtensilestressinAstorσst

fsy,thebondstrengthismorethanadequate,therebyrenderingstep11superfluous.

Step 12 Provideshrinkageandtemperaturesteel.

Equation9.4 14 gives3 2

s.min 1.75 1000 550 10 962.5 mmA

UsefiveN16,As 1005mm2.Thisisacceptable.Notethats≤300mmdoesnotapplyto

shrinkageandtemperaturesteel.

ThefinaldesignisdetailedinFigure12.2 7 .

750 mm 950 mm

300 mm

30 mm 30 mm

1940 mm

150 mm 150 mm

5 N16@425 mmN16@275 mm

Wall

512 mm

38 mm

Figure12.2 7 Detailsofthefinaldesignoftheasymmetricalwallfooting

Comments Forpt pt.min 0.001396instep9andps.min 0.00175instep12,thesteelcontentofthefootingis0.315%.Then,fromEquation2.4 1 ,wehave

Thisreviewindicatesthatforthe lightlyreinforced footings,Equation2.4 1 maybe

ignoredforsimplificationandρwtakenas24.0kN/m3.

12.3.3 Eccentric loading Foraneccentricallyloadedcolumnfooting,thedesignmayfollowtheapproach

describedinSection12.2.4inwhichthefootingdimensions bfandDf canbe

proportionedinsuchawaytoproduceauniformpressureinthesubsoil.


Problem Figure12.3 2 detailsarectangularcolumn bc×Dc subjectedtoacombinationofN*

andM*.Proportionasuitablerectangularfooting bf×Df suchthattheuniformsubsoil

pressureisequaltotheeffectivesoilbearingcapacity qf .Discusstheprocedureforthe

completedesign.

e

do

N*

x

y

bc

Df

bf N*

M*

c2

Dc

2

D

M*

f2

Df

2

D

N*

D

Figure 12.3(2) Footing for eccentric column loading

3w 24 0.6 0.315 24.189 kN/m

Solution ForthegivenN*andM*,theeccentricityoftheequivalentN*is

*

*

Me

N

Equation12.3 8

Thebearingareaofthefootingis

f ff w

*

1.2

Nb D

q D

Equation12.3 9

whereρwistheunitweightofthefooting.

But

f c

f c

b b

D D

or

cf f

c

bb D

D

Equation12.3 10

SubstitutingEquation12.3 10 intoEquation12.3 9 leadsto

c 2f

c f w

*

1.2

b ND

D q D

Equation12.3 11

fromwhich

cf

c f w

*

( 1.2 )

D ND

b q D

Equation12.3 12

Equation12.3 10 thenyieldsbf.

WithEquations12.3 8 ,12.3 10 and12.3 12 inhand,theplandimensionsand

layoutofthefooting,theeffectivesheardepthofthefootingdv,andthebending

reinforcementcanbeobtainedfollowinganiterativeprocesssimilartothatdescribed

inSection12.2.5forwallfootings.Inviewofshearbeingthedominantactionandmost

columnfootingdimensionsresemblingthoseofarectangularbeam,Equation6.3 4

mayalsoapply

uc v w v cV k b d f Equation12.3 13

Thecriticaltransverseshearsectionislocatedinthelongersideofthefooting

andinthiscase,therightside.WhenderivingEquation12.3 12 ,theself‐weightofthe

footing,orDinFigure12.3 2 ,isunknown.Therefore,aniterativeprocessisneededto

ensurethatqfisnotexceeded.Thismaybedonebyignoringtheself‐weightin

computingbfandDf,butsuitablyincreasingtheirvaluesbeforeproceedingtoobtaindv,

andthencheckingEquation12.3 9 .Reviseasnecessaryuntileitherqfisnotexceeded

orbfDfisadequate.Also,afinalcheckforpunchingshearstrengthisrequiredby

treatingN*asthedesignloadfromanedgecolumn.TheprocessgiveninSections9.6.3–

9.6.5maybeused.

M*

N*

cg

e

N*

CL

e

(a)

(b)

Uniform soil pressure

cg

CL

Figure 12.3(3) Asymmetrical footings for eccentrically loaded columns, either (a) T-

shaped or (b) trapezoidal

Itshouldberecognisedthat,oncertainconstructionsites,thelengthDfmaybe

restricted e.g.wherethefootingsareclosetotheboundarylinesoftheproperty .In

suchcases,footingswithasteppedortrapezoidalplanshapemaybemandatory.Such

irregularplanshapes,asshowninFigure12.3 3 ,canreadilybeobtainedbyensuring

thatthegeometriccentreoftheplan cg hasthesameeccentricityefromthecolumn

axis0,wheree M*/N*.


Problem GivenarectangularcolumnwithDc 500mmandbc 300mm.ThedesignactionsN*

1400kNandM* 200kNm;theeffectivesoilbearingcapacityqf 300kPa; cf 20

MPa;andexposureclassificationA1applies.

Designanasymmetrical pad footinginsuchawaythatthesubsoilpressureis

uniformlydistributed.UseN20barsforbendinginonelayereachwayandprovide

shearreinforcementasrequired.

Solution ThedesignspecificationsareillustratedinFigure12.3 6 .Theeccentricitye M*/N*

142.9mmasperEquation12.3 8 .Notethat,ascautionedinSection12.2.5,the

concretecovermustbeincreasedunlessthebaseofthefootingiswelllevelledand

compacted.

UsingEquation12.3 12 andassuming w 24kN/m31 12 2

f

0.5 1400 700

0.3(300 1.2 24 ) 90 8.64D

D D

Equation a

and

f fEquation 12.3(10): 0.6b D Equation b

WithEquations a and b inhand,followtherelevantstepsgiveninSection

12.2.2,whichwouldleadtotherequireddesign.

N20 bars

M* = 200 kNm

N* = 1400 kN

Df / 2 Df / 2

bc=300 mm

Dc=500 mm

e

e = 142.9 mm

x

CL2020

d0

N*

Figure 12.3(6) Design specifications for the example asymmetrical pad footing

Trial 1

Step 1 AssumeD 1000mmand:

Equation a :Df 2.933m roundedto3.0m

Equation b :bf 1.8m

do 1000–20–20/2 970mm

Step 2 Thedesignshearforceatthecriticalshearsection–seeFigure12.3 6 –is

f f c o f w* ( /2 /2 )( 0.9 )V b D e D d q D

Equation c

Therefore

* 1.8(1.5 0.1429 0.5/ 2 0.97)(300 0.9 24 1) 211.92 kNV

Step 3 Asthesectionisdeepenoughtobereinforceable,thisstepisnotrequiredtobe

checked.

Step 4 Forasinglelayerofsteel,

d do 970mmandthereby


Equation6.3 4 : 'uc v w v cV k b d f where '

c 20 4.47 MPa < 8 MPa which is acceptablef

and vv

200 2000.094 0.10

1000 1.3 1000 1.3 873k

d

,whichisacceptable.

Finally,Equation6.3 4 gives3

uc 0 .094 1800 873 20 10 660.6 kNV

Step 5 AsVuc 0.75×660.6 495.5kN>V* 211.92kN,theassumedD 1000mm

leadstoanundulylargesafetymargin.ReviseusingD 750mm.

Trial 2

Step 1 AssumeD 750mm,then:

Equation a :Df 2.895m roundedto3.0m

Equation b :bf 1.8m

do 750–20–20/2 720mm

Step 2

*Equation (c): 1.8 (1.5 0.1429 0.5/2 0.72) (300 0.9 24 0.72)

344.53 kN

V

Step 3 Notrequiredasexplainedfortrial1.

Step 4 Fortheassumedtotaldepthof750mm:

d do 720mmandthereby



c 20 4.47 M Pa < 8 M P a w hich is acceptablef and

vv

200 2000,11 0.10

1000 1.3 1000 1.3 648k

d

,whichisnotacceptable,hencetakekv

0.10


uc 0.10 1800 648 20 10 521.6 kNV

Step 5 AsVuc 0.75×521.6 391.2kN>V* 344.53kN,theassumedD 750mm

requiresnoshearreinforcement.Thisisacceptable.

Step 6 Notrequiredforthisexample.

Step 7

Notrequiredforthisexample.

Step 8

Punching shear check Thefootingcanbetreatedasanedgecolumnwithoverhangontheshort leftorheel

sideofthefooting.Tobeontheconservativeside,theoverhangmaybeignoredin

computingthecriticalshearperimeter.Notingthatdom 720–10 710mmand:

Equation 9.6(4): 2(500 710/2) (300 710) 2720 mmu

cv c c c

2Equation 9.6(6): 0.17 1 0.374 0.34

500 / 300f f f f

Thus cv 0 .34 20 1.521 M Paf and:3

uoEquation 9.6(3): 2720 710 1.521 10 2937.4 kNV

andasshowninFigure12.3 6 , v* 200 kNmM

u 6

3

2937.4Equation 9.6(9): 2780.2 kN

2720 200 101

8 1400 10 1210 710

V

u*Equation 9.6(10): 0.75 2780.2 2085.2 1400 kNV N

Therefore,thepunchingshearstrengthismorethanadequate.

Steps 9 and 10 Bendingdesigncoversthemomentsaboutthemajorandminoraxes.First,Astdesign

forthemomentaboutthemajoraxismustbedetermined.ForasinglelayerofN20bars:

d do 720mm

2 cEquation 3.3(2)a: 0.82 (for 20 MPa)f

Equation 3.3(2)b: 0.92.

FollowingtherecommendationsillustratedinFigure9.4 2 andreferringto

Figure12.2 6 ,thecriticalbendingsectionislocatedatLb,fromtherightedgeofthe

footingwhere

f cb

0.7

2 2

D DL e

Thatis,Lb 3/2 0.1429–0.7×0.5/2 1.4679m

Thus

2* (300 0.9 24 0.75) 1.8 1.4679 /2 550.36 kNmM

andEquation3.5 7 :ξ 0.82×20/500 0.0328.

Assumingϕ 0.85,thenEquation3.5 6 :

2t 2

2 0.0328 550.36 100.0328 0.0328 0.001419

0.85 1800 720 500p

Checktoensurethattheassumedvalueofϕ 0.85isvalid.FromEquation3.4 8

u

0.001419 5000.047

0.82 0.92 20k

Sinced do,thenkuo ku 0.047andEquations3.4 20 awithbyieldϕ 0.85.

Thisisacceptable.2

t.m inEquation 12.3(18): 0 .20 (750 / 720) 0.6 20 /500 0.001165 0.001419p

Therefore,Ast 0.001419×1800×720 1839.0mm2andfromTable2.3 1 ,

sixN20givesAst 1884mm2.Thisisacceptable.

OncetheAstdesignformomentaboutthemajoraxisiscompletedasabove,Ast

designforthemomentabouttheminoraxisistobecarriedout.Forasinglelayerof

N20bars,d dmajor–20 700mm.Fortheminordirection

b f c/2 0.7 /2 1.8/2 0.7 0.3 / 2 0.795 mL b b

Thus

2* (300 0.9 24 0.75) 3 0.795 /2 269.05 kNmM

Ifweassumeϕ 0.85,thenEquation3.5 6 :

62

t 2

2 0.0328 269.05 100.0328 0.0328 0.000434

0.85 3000 700 500p

Checkforϕasfollows.

FromEquation3.4 8

u

0.000434 5000.0144

0.82 0.92 20k

andsincedo 700mm

uuo

o

0.0144 7200.0148

700

k dk

d

Equations3.4 20 awithbyieldϕ 0.85.Thisisacceptable.

Now,Equation12.3 18 :2

t.min t0 .20 (750 / 700) 0.6 20 / 500 0.001232p p

Therefore,Ast 0.001232×3000×700 2588mm2.FromTable2.3 1 ,nineN20

givesAst 2826mm2.Thisisacceptable.

ThelayoutofthereinforcingbarsisillustratedinFigure12.3 7 .

1008@350 = 2800

100

150

150

5@300 = 1500150 150

300

2020

6N20

20

9 N20 @ 350

a

a

(a)

(b)

750

5@30

0 =

150

0

Figure 12.3(7) Bending reinforcement: (a) plan and (b) section a-a Note:alldimensionsareinmm.

Step 11 Checkthebondofthemajormomentdirection.ForEquation8.2 1 :

k1 1.3

k2 132–20 /100 1.12

k3 1–0.15 20–20 /20 1.

Therefore,Equation8.2 1 :

sy.tb 0.5 1.3 1 500 20/ (1.12 20) 1298 mm ( 300 mm).L

Sincetheavailable

sy.t b( cover) 1468 20 1448 mmL L

and

sy.t f b( cover) 3000 1468 20 1512 mmL D L

thenstraightbarsoflength3000–2×20 2960mmhaveadequatedevelopment

lengthattheultimatestate.Dependingonthequalityofexcavation,ashorterbarlength

shouldbeusedresultinginmoreconcretecover.

Checkthebondoftheminormomentdirection.Similartothemajormoment

directioncalculations,Equation8.2 1 :

sy.tb 1298 mmL

AteitherendoftheN20bar,theavailablelengthforstressdevelopmentis

f csy.tbcover 0.7 900 20 0.7 300 / 2 775 mm 1298 mm

2 2

b bL

whichappearsinadequate.However,Equation8.2 7 indicatesthattherequiredstress

developmentlengthisLst M*Lsy.tb/Mu.min 269.05Lsy.tb/Mu.minwhere,asperEquation

3.4 10 ,theminormomentcapacity

6u.min

2826 5002826 500 700 1 10 968.8 kNm

2 0.82 3000 700 20M

or

st 269.05 1298 / 968.8 360.5 mm 775 mm.L

Thusstraightbarsof1800–2×20 1760mmhaveadequatebondstrengthinthe

minorbendingdirection.Ashorterbarlengthshouldbeusedtoprovidemoreconcrete

coverasnecessary.

Step 12 Nowtheshrinkageandtemperaturesteelmustbedetermined.TheAstinthemajor

bendingdirectionisgreaterthantheAst.minspecifiedinEquation12.3 18 ,whereasin

theminorbendingdirectionAst Ast.min.Hence,noadditionalshrinkageand

temperaturesteelisrequired.

12.4.1 Concentric column loading


Problem Detailsofacentrallyloadedsquarecolumntobesupportedonfourcircularconcrete

pilesareshowninFigure12.4 2 .GivenN* 2000kNand cf 25MPa,obtainthe

overalldepth D ofthepilecap.

250

125

400 400 400 250

125

375

1200

375

CL

400

dom/2 dom/2

400

d om

/2d o

m/2

CL

y

x

275

2P 2P

20

20

7575

N* = 2000 kN

CL

Ddom

Figure 12.4(2) A centrally loaded square column supported on four circular concrete piles

Note:alldimensionsinmm.

Solution Thedepthofthesectionistobedeterminedbyshearstrengthconsideration.Thepile

load

p w f f* *( 1.2 ) / 4P N b D D Equation a

AssumingD 700mmandusingN20barsforreinforcement,

o 700 75 75 20 20/2 520 mm.d

Forbendingshearconsideration,thecriticalsectionislocatedatdofromtheface

ofthesupport,whichinthiscaseiseitherthecolumnoranypairofthefourpiles.

Figure12.4 2 showsthatthecleardistancebetweenthecolumnandthepilesisonly

275mm i.e.thecriticalshearsectionwouldcutthroughthecolumnorthepairof

piles .Thismeansthatbendingshearisnotcritical.Therefore,thedesignisgoverned

bythepunchingshear.

Punching shear design Forpunchingsheardesign,wehave:

p*Equation (a): (2000 1.2 24 1.95 1.95 0.7)/4 519.16 kNP

cv c c c

2Equation 9.6(6): 0.17 1 0.51 0.34

1.95/1.95f f f f

Adopt cv 0 .34 25 1 .7 M P af andthemeanvalueof3

om (700 2 75 20) 10 0.53 md

Then:

Equation 9.6(5): 2(0.4 0.53 0.49 0.53) 3.72 mu 3

uoEquation 9.6(3): 3720 530 1.7 10 3351.72 kNV

Sincethesquarepilecapissymmetricallyloaded,M* 0.Hence:

u uoEquation 9.6(9): 3351.72 kNV V

u uEquation 9.6(10): 0.75 2513.8 kNV V

InFigure12.4 2 ,N* 2000 1.2 0.4 0.53 2×0.7×24 2017.4kN

Since,ϕVu 2513.8kN N* 2017.4kN,theassumedD 700mmis

acceptable,albeitconservative.Note,however,thateachpileresemblesacorner

columninaflatplatesystemaroundwhichthepunchingshearstrengthshouldbe

checked.Thismaybecarriedoutusingsemi‐empiricalformulas seeLooandFalamaki

1992 .Alternatively,aconservativeassessmentmaybemadeofthepunchingshear

strengthofthecaparoundapile.Thisisdoneinaprocesssimilartothatdescribedin

Section12.3.7.Inthiscase,andasillustratedinFigure12.4 3

v p* * (0.125 0.53/2) 202.5 kNmM P

v*Let 202.5 kNm.M

Conservatively,andreferringtoFigure12.4 3 ,thecircularpilesectionmaybe

convertedintoasquarewith 2c c 125 221.6 m m .b D

375

Possible critical shear perimeter

45o

45o

Figure12.4 3 Determinationofcriticalshearperimeterforthepunchingshearcheck

ofthepilecap


Then,thecriticalshearperimeterisobtainedusingEquation9.6 4 or

22u a a

whereagainconservatively,

om221.6 /2 486.6 mma d

and

2 om221.6 751.6 mma d

Thusu 1724.8mm.

Notethatthisuisshorterandhencemorecriticalthananotherpossibleshear

perimeter–seeFigure12.4 3 .Then3

uoEquation 9.6(3): 1724.8 530 1.7 10 1554.0 kNV

u

1554.0Equation 9.6(9): 1171.9 kN

1.7248 202.51

8 519.16 0.4866 0.53

V

Equation9.6 10 :ϕVu 878.9kN P*p 519.16kN

Thus,punchingshearstrengtharoundthepileappearstobeadequate.

Tocompletethepile‐capdesign,proceedtocomputethebendingsteelAstin

eachofthexandydirections–seeFigure12.4 2 .Theprocessissimilartothatusedin

Sections12.2.5and12.3.7forwallandcolumnfootings,respectively.Notehoweverthat

thebendingspanineitherdirectionis

b

400400 0.3 460 mm

2L

2x y

* * 2 519.16 0.460 24 1.95 0.7 0.460 /2 474.2 kNmM M

x y x700 75 75 20 20/2 520 mm and 20 540 mm.d d d

Aftercomputingthebendingreinforcementinbothdirections,ensurethatthebars

haveadequatestressdevelopmentlength.Iftheydon’t,usehooksorcogs.


Problem Figure12.5 14 showsdetailsofareinforcedconcretecantileverretainingwall.The

superimposedliveload,p 15kPa,andthewallandbasedimensionsaretheoutcomes

ofaglobalgeotechnicalanalysis,includingslidingstabilitycheck,followedbya

preliminarydesignexercise.

Giventheunitweightsofthecohesionlessbackfillandthefrontsurchargeρw.BF

ρw.FS 21kN/m3,withacharacteristiceffectiveinternalfrictionangleϕ 35°,the

effectivesubsoilbearingcapacityqf 250kPaand cf 25MPa.UseD500Nbarsonly.

Computethesubsoilpressuresatthetoeandheel,ftoeandfheel,respectively,and

checktheoverturningstability.Thencarryoutafullreinforcedconcretedesignofthe

retainingstructure.Thepassiveearthpressureduetothefrontsurchargemaybe

ignored.

1750

950 4502100

WW3

OOB

WBFWW2

WW1

WFS

250

ftoefheel

Superimposed construction live load, pSL=15 kPa

z

pa,BF

pa,SL

FSL

FBF

pa,BFpa,SL

550

450

WSL

350

6000

1750

w,BF w,FS

3

o

a

21 kN/m

35

0.2710

ρ ρ

K

Figure12.5 14 Detailsoftheexamplereinforcedconcretecantileverretainingwall


Solution Thedesignstepsrequiredareenumeratedbelow.

Step 1 ComputeKaandthelateralactiveearthpressures.Forthelevelledbackfill,ß 0,and

Equation12.5 2 gives

2

a2

1 1 cos 350.2710

1 1 cos 35K

Atagivendepthz,theearthpressuresduetothesuperimposedloadandthebackfillas

perEquation12.5 1 are

a.SL 0.2710 15 1 4.065 kPap

and

a.BF 0.2710 21 1 5.691 kPap z z

respectively.

Step 2 Nowcomputevariousverticalandlateralforces.Assumingρw.concrete 25kN/m3with

thegivenρw.BF ρw.FS 21kN/m3andfora1‐metrerunoftheretainingwall,the

quantitiesrequiredinEquations12.5 5 and 6 areasfollows.

Forces kN :

a ßWWW1 0.25×6×1 ×25ßW 37.5ßW

b W W 2 W W

0 .26 1 2 5 1 5 .0

2W

c ßWWW3 0.45×3.5×1 ×25ßW 39.375ßW

d ßSLWSL 15×2.1×1 × ßSL 31.5ßSL

e ßBFWBF 6×2.1×1 ×21ßBF 264.6ßBF

f ßFSWFS 0.55×0.95×1 × 21ßFS 10.973ßFS

g FSL 4.065ßSL× 6 0.45 26.22ßSL

h FBF 5.691×6.45ßBF×6.45/2 118.38ßBF

LeverarmswithrespecttoOB m :

i W 1

0 .2 52 .1 1 .7 5 0 .4 7 5

2L

j W 2

0 .20 .2 5 2 .1 1 .7 5 0 .7 0

3L

k F S

0 .9 50 .4 5 2 .1 1 .7 5 1 .2 7 5

2L

l S L B F

2 .11.75 0.70

2L L

m LSL.H 6 0.45 /2 3.225

n LBF.H 6 0.45 /3 2.15

Step 3 Nowcalculatetheloadcombinations LC .Consideringtheaggravatingandreversal

effectsofeachofthevariousverticalandinducedlateralforces,threeloadcombination

casesarerequiredtoproducethemostcriticaloutcomes.Thedetailsarepresentedin

Table12.5 3 .

Table 12.5(3) Load factors for critical load combination cases

Load

combination βW βFS βSL βBF

LC1 1.2 1.2 1.5 1.2

LC2 1.2 1.2 0.4 0.9

LC3 0.9 0.9 1.5 1.2

Step 4 ComputeRandMRatOBforLC1,2and3.Withvaluesobtainedinstep2,theresultant

verticalforceasperEquation12.5 5 is

W SL BF FS

W SL BF FS

(37.5 15.0 39.375) 31.5 264.6 10.973

91.875 31.5 264.6 10.973

R

Equation a

Substitutingthecorrespondingloadfactors,whicharethesubscriptedßvaluesforLC1

inTable12.5 3

488.19 kNR

where

W SL BF FS1.2; 1.5; 1.2; and 1.2.

ThemomentaboutOB positiveanticlockwise asperEquation12.5 6 andwith

referencetoTable12.5 3 ,is

R SL BF W W

FS SL BF

26.22 3.225 118.38 2.15 37.5 0.475 15.0

0.70 10.973 1.275 31.5 0.70 264.6 0.70

M

whichissimilarto

R SL BF W FS SL

BF

84.560 254.517 28.313 13.991 22.050

185.220

M

Equation b

fromwhichforLC1,MR 227.686kNmFollowingthesameprocess,Equation a gives

and

457.33 kN for LC3R

374.16 kN for LC2R

Then,Equation b leadsto

R 138.13 kNm for LC2M

and

R 214.99 kNm for LC3M

Step 5 Computeftoeandfheelandcheckoverturningstability.Thesubsoilpressuresatthetoe

andheelareobtainedthroughEquations12.5 9 and 10 ,respectively,as2

toe R/ 3.5 6 /3.5f R M Equation c

and2

heel R/3.5 6 /3.5f R M Equation d

TheoutcomesforRandMRduetothethreecriticalloadcombinationsaresummarised

below:

LC1:R 488.19kN;MR 227.69kNm

LC2:R 374.16kN;MR 138.13kNm

LC3:R 457.33kN;MR 214.99kNm

Nowthesubsoilpressuresforthethreeloadcombinationcasesarecomputedusing

Equations c and d :

LC1:ftoe 251.00kPa;fheel 27.96kPa



Theseresultsshowthatnoneofthesubsoilpressuresexceedthegiveneffectivebearing

capacityqf 250kPa.Note,however,thatforLC1itisslightlylargerbutcanbetakenas

250kPa.Thismeansthattheretainingwalldimensionsareacceptable.Notealsothatin

practice,qfmaybecalculatedinaccordancewithSection5ofAS4678‐2002:Earth‐

RetainingStructures,wheredesignrequirementsforserviceabilityanddurability,inter

alia,arealsospecified.

Sinceallthepressurevaluesarepositive compressive ,theretainingwallis

stableagainstoverturning.

Step 6 Computethedesignmoments M* .TherootmomentofthewallforLC1isgivenby

Equation12.5 12 withßSL 1.5andßBF 1.2as

2O1 b1 b 2

b1 b2

* 1.5 4.065 6 0.5 1.2 0.271 21 6

36.585 122.926

M l l

l l

wherefromEquation12.5 13

b1 6/2 0.15 0.45 3.068 ml

andEquation12.5 14 gives

b2 6/3 0.15 0.45 2.068 ml

or

O1* 366.45 kNmM

ForLC2,ßSL 0.4andßBF 0.9,whichgives

O1 b1 b2* 9.756 92.194M l l

fromwhichandwith b1 3.068 ml and b2 2.068 ml

O1* 220.59 kNmM

ForLC3,withßSL 1.5andßBF 1.2,themomentisthesameasforLC1,whichis

O1* 366.45 kNmM

Fortheheelmoment,Equation12.5 15 yields

O2 SL BF W

1.heel

2.heel

SL BF W 1.heel 2.heel

* (31.5 264.6 )(2.1/2 0.15 0.45) (0.45 2.1

1 25)(2.1 0.15 0.45) / 2 (2.1 0.15 0.45) / 2

(2.1 0.15 0.45) / 3

35.201 295.691 25.604 1.084 0.723

M

P

P

P P

Equation e

ForLC1,ßSL 1.5,ßBF 1.2,ßw 1.2,

1.heel 27.97 (2.1 0.15 0.45) 27.97 2.1675 60.625 kNP

and

2.heel

2.1675(250) 27.97 2.1675/2 149.02 kN

3.5P

basedonwhichEquation e gives

O2* 264.90 kNmM

ForLC2,ßSL 0.4,ßBF 0.9,ßW 1.2,

1.heel 39.25 (2.1 0.15 0.45) 39.25 2.1675 85.074 kNP

and

2.heel

2.1675(174.56 39.25) 2.1675/2 90.81 kN

3.5P

ThenfromEquation e

O2* 153.05 kNmM

ForLC3,ßSL 1.5,ßBF 1.2,ßW 0.9,

1.heel 25.36 (2.1 0.15 0.45) 25.36 2.1675 54.968 kNP

and

2.heel

2.1675(235.97 25.36) 2.1675/2 141.35 kN

3.5P

ThenfromEquation e

O2* 268.89 kNmM

Similarly,thetoemomentisgivenbyEquation12.5 16 as

O3 1.toe 2.toe FS

W

1.toe 2.toe FS W

* (0.95 0.15 0.45) / 2 2 1.0175 / 3 10.973 (0.95 / 2

0.15 0.45) (0.95 0.15 0.45) 0.45 25 1.0175 / 2

0.5088 0.6783 5.953 5.824

M P P

P P

Equation f

ForLCl,βFS βW 1.2

1.toe

3.5 1.017527.97 (250 27.97) 1.0175 188.695 kN

3.5P

and

2.toe

1(250 185.45) 1.0175 32.84 kN

2P

Equation f thusgives

O3* 104.15 kNmM

ForLC2,ßFS ßW 1.2

1.toe

3.5 1.017539.25 (174.56 39.25) 1.0175 137.60 kN

3.5P

and

2.toe

1(174.56 135.223) 1.0175 20.013 kN

2P

fromwhichEquation f gives

O3* 69.45 kNmM

ForLC3,ßFS ßW 0.9

1.toe

3.5 1.017525.36 (235.97 25.36) 1.0175 177.80 kN

3.5P

and

2.toe

1(235.97 174.743) 1.0175 31.15 kN

2P

Hence,Equation f gives

O3* 100.99 kNmM

Finally,theroot,heelandtoemomentsforthethreeloadcombinationcasesare

showninTable12.5 4 .

Table 12.5(4) Root, heel and toe moments for the three load combination cases

Load combination

case

Root moment,

M*01 (kNm)

Heel moment,

M*02 (kNm)

Toe moment,

M*03 (kNm)

LC1 366.45 264.90 104.15

LC2 220.59 153.05 69.45

LC3 366.45 268.89 100.99

FromTable12.5 4 ,thedesignmomentsareobtainedas

rootmoment,M*O1 366.45kNm

heelmoment,M*O2 268.89kNm

toemoment,M*O3 104.15kNm

Step 7 Nowdeterminethedesignbendingreinforcement,Ast.

Fortherootsectionofthewall,

M*01 366.45kNm,d 410mmandEquation3.5 7 gives

ξ 0.8125×25/500 0.0406

Assumingϕ 0.85andEquation3.5 6 gives

62

t 2

2 0.0406 366.45 100.0406 0.0406 0.005502

0.85 1000 410 500p

Equation3.4 8 gives

u

0.005502 5000.1492

0.8125 0.9075 25k

Sincekuo ku 0.1492,ϕ 0.85isconfirmedasperEquations3.4 20 awithb.

Equation12.3 18 gives2

t.min t

4500.20 0.6 25 /500 0.001446 0.005502

410p p

Therefore,2

st 0.005502 1000 410 2256 mm /mA

WhenTable2.3 2 isconsulted,N20@125mmgivesAst 2512mm2/m,whichis

largerthan2256mm2/mby11.3%.Thisisacceptable.

Thisamountofsteelisrequiredonlyattherootofthecantileverwall.Asdesired,

foramoreeconomicaldesign,someofthebarsmaybecurtailedatlevelstowardsthe

topofthewallwherethemomentdiminishesrapidly,rememberingthatthebarsmust

beextendedbeyondthecurtailmentleveltoprovideadequatestressdevelopmentfor

thecurtailedbars.

Inapracticaldesign,adeflectioncheckshouldbeperformed.However,inthe

presentcaseofaninwardlytaperedwall,evenexcessivedeflectionwouldnotbe

apparenttoalayperson.

Fortheheelsection

o2* 268.89 kN/m, 410 mmM d


62

t t.min2

2 0.0406 268.89 100.0406 0.0406 0.00396

0.85 1000 410 500p p

Therefore,2

st 0.00396 1000 410 1624 mm /mA

FromTable2.3 2 ,N20@175mmgivesAst 1794mm2/m>1624mm2/m.Thisis

acceptable.

Forthetoesection,*

o3 104.15 kN/m, 410 mmM d


62

t t.min2

2 0.0406 104.15 100.0406 0.0406 0.001485

0.85 1000 410 500p p

Therefore,2

st 0.001485 1000 410 609 mm /mA

FromTable2.3 2 ,N20@300mmgivesAst 1047mm2/m 609mm2/m.This

isacceptable.

Also,forcrackcontrolpurposes,asshowninTable1.4 5 slabs ,s 300mm.

Therefore,useN20@300mm.

Foramoreeconomicalsolution,useN16@300mm,whichgivesAst 670

mm2/m.Thisisacceptable.

Step 8 ComputedesignshearV*forthewall.LC1governsasperTable12.5 3 .Withreference

toFigure12.5 15

o

410410 (450 250) 396 mm

6000d

a.SL

a.BF

4.065 kPa

5.691 5.604 31.892 kPa

p

p

Fora1‐metrerun

root* 1.5 4.065 5.604 1 1.2 31.892 5.604/2 1 141.404 kNV

z

pa,BF

pa,SL

Vroot

do

5604 mm

Figure12.5 15 Locationanddeterminationofcriticalshearforceforthewall

251 kPa

27.97 kPa

W'SL

W'BF

W'W3

do,heel

1690 mm

Figure 12.5(16) Location and determination of critical shear force for the heel NowcalculatedesignshearV*fortheheel.LC1governshere.

InFigure12.5 16

do.heel 410mm

W′SL 1.5×15×1.69×1 38.025kN

W′BF 1.2×21×6×1.69×1 255.528kN

W′W3 1.2× 0.45×1.69×1 ×25 22.815kN

Earthpressureresultantis

heel

heel

1.69 1.6927.97 1.69 1 (251 27.97) 47.269 90.999

3.5 2138.268 kN

* 38.025 255.528 22.815 138.268 178.10 kN

W

V

ComputedesignshearV*forthetoe.LC3governshere.

InFigure12.5 17 ,

235.97 kPa

25.36 kPa

do,toe

W'FS

2960 mm

540 mm

W'toe

Figure 12.5(17) Location and determination of critical shear force for the toe

o.toe 410 mmd

Earthpressureresultantis

0.54 0.540.54 235.97 (235.97 25.36) 118.65 kN

3.5 2

FS

toe

toe

0.9 0.55 0.54 21 5.6133 kN

0.9 0.45 0.54 25 5.4675 kN

* 118.650 5.613 5.468 107.569 kN

W

W

V

Step 9 Checktheshearcapacityofthewall.

d 410mm

and

D 450mm

ForthesimplifiedmethodofClause8.2.4.3,AS3600‐2018,



c 25 5 M Pa < 8 M Pa w hich is acceptablef

and vv

200 2000.14 0.10

1000 1.3 1000 1.3 369k

d

;hence,takekv 0.10


uc 0.10 1000 369 25 10 184.5 kNV

As,Vuc 0.75 184.5 138.4 kN root* 141.404 kNV ,noshearreinforcementisrequired.

Thus,shearreinforcementisnotrequiredforthewall.

Nowchecktheshearcapacityoftheheel.WithV*heel 178.10kNandd do

410mm,Vuc 0.75 184.5 138.4 kN V*heel 178.10kN.Thus,shearreinforcementis

required.AspertheprovisionsinthesimplifiedmethodofClause8.2.4.3,AS3600‐

2018,forAsv Asv.min,kv 0.15andhence, 3uc 0 .15 1000 369 25 10 276.75 kNV and

Vuc 0.75 276.75 207.6 kN V*heel 178.10kN

Thus,theminimumshearreinforcementisrequired.Formembersnotgreaterthan1.2min

depth,thespacing s forminimumshearreinforcementisthelesserof0.5Dand300mmors

0.5 450 225mm.Equation6.12 3 gives

'c w 2

.minsy.f

0.08 0.08 25 1000 225180 mm

500sv

f b sA

f

UseN12ties@225mm Asv 226mm2 180mm2 .

Finally,checktheshearcapacityofthetoe.WithV*toe 107.569kNandd do

410mm,relevantcomputationsconfirmthatshearreinforcementisnotrequiredfor

thetoe.

Step 10 ProvideshrinkageandtemperaturesteelusingEquation9.4 14 whichgives

3 2s.min 1.75 1000 410 10 717.5 mm /mA

UsetwolayersofN12bars@300mm,whichgivesAs 754mm2/m.Thisisacceptable.

Step 11 Checkthestressdevelopment,thatis,theadequacyofthebondstrengthatthebase

sectionofthewall,andtherootsectionsofthetoeandheel.Theprocessissimilarto

thatusedinSection12.2.5–step11.

Step 12 ThefinaldesignforthereinforcedconcreteretainingwallstructureisshowninFigure

12.5 18 .

950450

2100

6000

450

410

N12@300 mm N12@225 mm

N16@300 mm

N20@125 mm

250

N20@175 mm

Figure12.5 18 Reinforcementlayoutforthereinforcedconcreteretainingwallstructure

Note:alldimensionsareinmm;clearcoverforbendingsteel30mm;fortemperaturesteelnotlessthan30mm

CHAPTER 13


Problem Usingthestrut‐and‐tiemodelling STM approachoftheStandard,designa600‐mm‐

widetransfergirderspanning5mwithacolumnatmidspanwithultimatefactored

loadof5340kN.Thegirderissupportedby600mmsquarecolumns.Theoveralldepth

ofthegirderis2.25masshowninFigure13.6 4 .Theultimatesheardiagramisshown

inFigure13.6 5 .Usefc′ 50MPaandNbarsonlyforreinforcement.

2.25 m

2.5 m 2.5 m

5 m

P* = 5340 kN

600 mm600 mm

600 mm

Column centre line Column centre line

Figure13.6 4 STMdesignexample–transfergirder

2670 kN

– 2670 kN

– 2760 kN

2760 kN

–+

V*

Figure13.6 5 STMdesignexample–sheardiagram

Solution

Select and establish strut-and-tie model and node locations Assumethattheloadsarecarriedbyastrut‐and‐tiesystemconsistingoftwodirect

strutsrunningfromthetoploadingcolumntothesupportingcolumnsandatie

connectingthestrutshorizontally.ThegeometryoftheassumedSTMisshowninFigure

13.6 6 .

Becauseoftheheavyloadsappliedonthestructureandtheminimumallowable

heightbeingused,muchdeepernodelocationsarerequired.Aftermultipleiterations,

thenodeatlocationCattheloadingpointisdeterminedtobe300mmfromthetopof

thegirder,andthenodelocationatthesupportsas350mmfromthebottomofthe

girderasshowninFigure13.6 6 .Notethatafterthedesign,ifthefinalnodallocations

showadifferenceofroughly50mmorless,theoriginallocationsaredeemed

acceptablebecausetheforcesinthestrutmayincreaseonlyby1–2%,whichshouldnot

changethefinaldesign.

2.25 m

2.5 m 2.5 m

5 m

P* = 5340 kN

600 mm600 mm

600 mm


Strut CA Strut CB

Tie ABA B

C300 mm

350 mm

Figure13.6 6 STMdesignexample–assumedSTMandnodelocations

Basedontheassumednodallocationsasabove seeFigure13.6 6 ,theangle

betweenthestrutsandthetie tan‐1 32.6° 30°.AsperClause7.1of

AS3600‐2018,itisacceptable.

Determine forces in struts and ties FromthegeometryofthegirderwithreferencetoFigure13.6 6 ,

2 2length of strut C A (2250 300 350) (2500) 2968.2 m m 2 2length of stru t C B (2250 300 350) (2500) 2968 .2 m m

Thus,

2968.2force in strut CA 2760 5120.1 kN

2250 300 350

2968.2force in strut CB 2760 5120.1 kN

2250 300 350

2500force in tie AB 2760 4312.5 kN

2250 300 350

Determine effective concrete strength in nodes and struts Enoughspaceexistswithinthegirderforbottle‐shapedstrutstobeformedinstrutsCA

andCB.Also,burstingreinforcementwillbeprovidedtoresistcracking.Thus,Equation

13.6 1 :

s 2 0

10.383 0.3 which is acceptable

1.0 0.66cot 32.6

MakinguseofEquation13.6 2 ,theeffectiveconcretestrengthis:

s c0.9 0.65 0.383 0.9 50 11.2 MPaf

Thestrutswithinthecolumnsdonothaveenoughspaceforabottle‐shaped

struttoform.Thusβs 1.0andtheeffectiveconcretestrength:

s c0.9 0.65 1 0.9 50 29.25 MPaf

ForthenodalregionatC,aCCCsituationprevails.Thus,βn 1.0andthe

principalcompressivestressonthenodalfaceis:

n c0.9 0.65 1 0.9 50 29.25 MPaf

ForthenodalregionatAandB,aCCTsituationprevails.Thus,βn 0.80andthe

principalcompressivestressonthenodalface:

n c0.9 0.65 0.8 0.9 50 23.4 MPaf

Determine STM geometry Hydrostaticnodalregionsareusedherein.Hence,thestressesoneachfaceoftheregion

mustbeidenticalandthefacesareperpendiculartotheaxisofthestruts.Extended

nodalzonesmaybeused,buthydrostaticnodalregionsareeasytouseforthistypeof

loadingandalsoaddsomeconservatisminthedesignbyrequiringalargernodalzone.

Ashydrostaticnodalzonesareused,theminimumvaluefortheeffectiveconcrete

strength i.e.11.2MPa mustbeusedincalculatingthewidthsofthestrutsandthe

heightofthetietoensureastaticsituation.

UsingEquation13.6 2 withAc dct,wheretisthethicknessofthestrut which

is600mminthisexample ,thestrutwidth

strut

s c0.9c

fd

f t

Thus,3

c,CA

5120.1 10width of strut CA, 761.9 mm

11.2 600d

3

c,CB

5120.1 10width of strut CB, 761.9 mm

11.2 600d

3

c,A

2760 10width of strut A, 410.7 mm

11.2 600d

3

c,B

2760 10width of strut B, 410.7 mm

11.2 600d

3

c,C1

2670 10width of strut C1, 397.3 mm

11.2 600d

3

c,C2

2670 10width of strut C2, 397.3 mm

11.2 600d

3

c,AB

4312.5 10height of tie AB, 641.7 mm

11.2 600d

Fortheextracompressionstrutwidthrequired ≥ 397.3 397.3 794.6mm

withinthecolumnapplyingtheloads,a800mm 600mmcolumnisrequired.Allother

dimensionsfitwithinthegirderandsupportingcolumnsandfollowtheSTMguidelines

stipulatedinClause7.1oftheStandard,asshowninFigure13.6 7 .

TheactualnodelocationsaredeterminedusinggeometryasshowninFigure

13.6 7 .ThenodeatCis650.1/2 325.05mmfromthetopofthegirder,whichis

within50mmoftheinitial300mmusedforthedesign,andthenodesatAandBare

641.7/2 320.85mmfromthebottomofthegirder,whichisalsowithin50mmofthe

350mminitiallyselected.

Ifthesenodesweremuchfurtherapart,newinitialnodelocationswouldhaveto

beselectedandeveryquantityrecalculateduntilthedifferenceswereappropriate.

2.25 m

2.5 m 2.5 m

5 m

P* = 5340 kN

600 mm600 mm

800 mm


Strut CA Strut CB

Tie ABA B

C

641.7 mm

2670 kN 2670 kN

761.9 mm

641.7 mm410.7 mm 410.7 mm

650.1 mm

Figure13.6 7 STMdesignexample–STMandnodegeometry

Determine steel in the tie FortheforceintieAB, A B

*F 4312.5kN,theareaofsteelshouldbe

3AB 2

stsy

* 4312.5 1010 147.1 mm

0.85 500

FA

f

UsefourrowsoffourN32barsineachroworAst 12864mm2 10147.1mm2,

whichisacceptable.

ThetensiontiereinforcementoffourrowsoffourN32barsineachrowspaced

at64mmcentretocentreisshowninFigure13.6 8 .

350 mm254 mm

192 mm

Figure13.6 8 STMdesignexample–tensiontiereinforcement

Thecentroidofthetiereinforcementshouldlineupwiththenodelocation;that

is,thecentroidofthebottomtiereinforcementshouldstart254mmabovethebottom

ofthegirder.Thus,thedistanceofthecentroidoftiereinforcementfromthebottomof

thegirder 350mmandasaresultd 2250–350 1900mm.

Thetotaleffectiveheightofthereinforcement 350 2rowsofbars@32mm

1.5rowsofspacing@32mm 462mm.

Checkagainsttheheightofthetiewhichis350 2 700mm 462mm.Itis

acceptable.

Determine bursting reinforcement in bottle-shaped struts Theanglebetweenstirrupsandstruts,α 90°–32.6° 57.4°forwhichtanα 1.76

1/2forserviceabilityand1/5forstrength.Theseareacceptable.

Becauseofsymmetry,theleftorrightshearspan seeFigure13.6 7

2500 mma

Thecomponentnormaltotheshearspan

2250 – 650.1/ 2 – 641.7 / 2 1604.1 mmz

Thus,withdc.CA dc.CB 761.9mm,Equation13.6 4 gives2 2

b 1604.1 2500 761.9 2208.5 m ml

Forstrength,withtanα 1.76,theburstingforce

b stru t* tan 5 1 2 0 .1 1 .7 6 9 0 1 1 .4 k NT f

andtheareaofsteel

3 2s b sy

* / ( ) 9011.4 10 / (0.85 500) 21203 mmA T f

Attheloadingpointwheretheburstingcrackforms,theforcecarriedbythe

concreteandtobetransferredtotheburstingreinforcementisgivenbyEquation

13.6 3 as3

b.cr b ct0 .7 0.7 600 2208.5 (0.36 50 ) 10 2361 kNT bl f

AsperClause7.2.4oftheStandard,since b*T Tb.cr,onlytheminimumweb

reinforcementisrequiredor3

b.cr 2s.min

sy

2361 105555.3 mm

0.85 500

TA

f

ComparingtheaboveAsandAs.min,theminimumburstingreinforcement

governs.Thus,theareaofburstingreinforcementnormaltostrutCAorCBis:2

st 5555.3 m mA

Thusthereinforcementratiopt As/lbt 5555.3/ 2208.5 600 0.00419.

Theforceacrosstheburstingplaneismaintainediforthogonalreinforcementis

placedparallelandnormaltotheaxisofthemembersuchthat

th t sin 0.00419 sin 32.6 0.00205p p

tv t cos 0 .00419 cos 32.6 0 .00365p p

Theverticalwebreinforcementineachshearspan,consideringbursting

reinforcement,isptv 0.00365.Adoptingtwolayers onelayerateachface of16‐mm‐

diameterbars,withatotalbarareaacrossthe600‐mm‐thicksectionof402mm2,gives

abarspacingrequirementof

402/ 0.00365 600 183.6 mms

UseN16stirrupsat175mmspacingfortheverticalreinforcementforthetotalspan.

Forthehorizontalreinforcement,pth 0.00205.UsingtwolayersofN16bars As

402mm2 givesabarspacingrequirementof

402/ 0.00205 600 326.8 mms

UsetwolayersofN16barsat300mmspacingasthelongitudinalreinforcement

forthetotalspan.

Final design ThecompleteddesignofthegirderisshowninFigure13.6 9 ,withdimensionsandthe

reinforcement.

2.25 m

2.5 m 2.5 m

5 m

600 mm600 mm

800 mm


600 mm

N16 stirrups @ 175 mm c/c

N16 horizontal bars @ 300 mm c/c on both faces of girder

4 rows of 4N32 bars starting at 254 mm above bottom of girder

254 mm

Figure13.6 9 STMdesignexample–finaldesignsection

CHAPTER 17


Problem ForthesectionwithunbondedtendonsshowninFigure17.5 1 ,computetheultimate

momentMuusingEquation17.3 8 .

200

700

625

APt

800

3 N24(Ast = 1356 mm2)

Figure17.5 1 Detailsofapartiallyprestressedbeamwithunbondedtendons


c 240 MPa ( 0.79; 0.87) f Apt 405mm2unbonded

fpy 1710MPa;fp 1850MP p.ef pi 1100 MPa

L/D 35

Solution UsingEquation17.3 10 ,determinethat

pu 1100 70 40 200 625 / (100 405)

1293.5MPa 1100 400 1500 MPa

Thisisacceptable.

NowusingEquation17.3 1

u (1293.5 405 500 1356) / (0.79 40 0.87 200) 218.6 mmk d


(1293.5 405 625 500 1356 700)/(1293.5 405 500 1356)

667.3 mm

d

However,do 700mm,and

uo

218.60.312 0.36

700k

whichisacceptable.

Thesectionisconsideredunder‐reinforcedandcomplieswithEquation17.4 1 .


u

2 6

[1293.5 405 625 500 1356 700 0.79 40 200

(0.87 218.6) /2] 10 687.7 kNm

M


Problem Forthestandardbridge‐beamsectionasadoptedinAS5100.5‐2017andshownin

Figure17.6 3 ,computetheultimatemomentbasedonrelevantrecommendations

fromtheStandard.Allgivenvaluesareidenticaltothoseprescribedfortherectangular

beaminSection17.5.2,exceptthatApt 910mm2andthecableisbonded.

APt

625

700

750

120

100

40

420

90

100

200

300

3 N24

Figure17.6 3 Standardbridge‐beamsectionasperAS5100.5–2017


Solution Fromthefollowingequationswehave:

1

2

pu

Equation 17.3(5): 0.28

Equation 17.3(6): (1850 910 500 1356) / (200 625 40)

0.4723

Equation 17.3(4): 1850(1 0.28 0.4723 / 0.87) 1568.8 MPa

Equation 17.3(2): (1568.8 910 625 500 1356 700) / (1568.8

k

k

d

c

2

910 500 1356) 649.1 mm

Equation 17.6(3): (1568.8 910 500 1356)/(0.79 40)

66 633 mm .

A

Equatingtheabove cA withthedimensionsoftheconcreteareaincompression,as

showninFigure17.6 4 ,wehave

u66 633 200 100 (120 200) 40/2 ( 140) 120k d

fromwhichγkud 475.3mm.Thus

u uo

0.84 649.5475.3 / (0.87 649.5) 0.84 and 0.78 0.36

700k k

Imposingkuo 0.36asrequiredbyEquation17.4 1 leadsto

γkud 0.87 0.36 700 219.2mm

120

10040

200

4040

uγk d

cd

Figure17.6 4 Concreteareaincompression


WithreferencetoFigure17.6 4 wecanobtain

2c (219.2 140) 120 (120 200) 40/2 200 100 35 904 mmA

Bytakingmomentaboutthetopedgeofthesection,thelocationofthecompressive

forceC seeFigure17.6 4 ,canbecomputedas

c

c

[200 100 50 120 40 120 (2 40 40/2) (100 40/3)

79.2 120 (140 79.2 / 2)]/A 96.5 mm

d

Hence

2pt.ef

ud

6

Equation 17.6(4): (0.79 40 35 904 500 1356)/1568.8 291.0 mm

Equation 17.6(5): (1568.8 291.0 625 500 1356 700 0.79 40

35 904 96.5) 10 650.4 kNm

A

M

Finally,tocomplywiththeductilityrequirement,therequiredcompressionsteelis

givenbyEquation17.4 4 as

2sc c0.01 359 mmA A

Therefore,wecanusetwoN16or402mm2.

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