Chapter-9 Flat Slab

8/6/2019 Chapter-9 Flat Slab

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390 Reinforced Concrete

Step 5 Carry out local analysisFind out-at-plane internal forces in wall panels (follow Section 8.1.2.2).After analysis the following internal forces are reported:

Wall no. Line Load MOH VOH Mav Vavcase (kNm/m) (kN/m) (kNm/m) (kN/m)DL

A LLWL 28 26DL 20 3

B LL to 1WL 32 30DL

C LLWL 28 26DL 10 3

0 LL 5 1WL 32 30

o8; : : : !

8 F , K II

IJ E G 1 j III A I I L I

I I I I,0 H M'///L 10500 SK 8/46 Elevation - Wall S.

Wall 1 only will be designed as an example.Wall 5 panels are shown in sketch to illustrate the location of lines whereresults should be available for out-of-plane bending.

Note: The example shown uses only one value of bending moment and shear perline of interest. In practice, more values along the line will have to beconsidered.

Step 6 Carry out combination of loadingMost computer programs used for the analysis will automatically carry outthe combination according to principles described in Step 6 of Section 8.2.Reproduced below is the result of one combination of Wall 1.Load case LC} = 1.4DL + l.4WL


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Design of Walls 391Wall 1 subject to WL (y-direclion)In-plane forces (see Step 4 in Section 8.2):N = 4452kNMI = 28000kNmV I = 2380kNQ, = 1440kN (97 x 10.6 x 1.4)Out-of-plane forces (see Step 5 in Section 8.2):At line D (WL in y-direction),MOH = 14kNm/m (dead load x 1.4)VOH = 42kN/m (DL x 1.4)On flanges (part of Wall 5 and 6),MOH = 39kNm/m (WL in y-direction)VOH = 36kN/m (WL in y-direction)

Step 7 Check slenderness of wallFollow Section 8.1.1.2.1.Type of wall = unbraced, reinforced in the in-plane direction

'f

SK 8/47 Section through Wall 1.

'---' @

600

~C Dt

f e u = 30 N/mm2He = ~ u;Ho = clear height = 12.0m

fy = 460N/mm2

Monolithic construction at top and bottom of wall.Assume thickness of slab al top is 40 0 mm.End conditions are 1 at bottom and 2 at top.. ' . { 3 = 1.3He = 1.3 x 12000 = 15600mmHe 15600h= 600 = 26 > 10 < 30 (limit for unbraced reinforced wall)Design as slender wall.


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:> n Reinforced Concrete. Step 8 Find effective width of flanges

6 00

SK 8/48 Plan of Wall I showingeffective flange widths.

Follow Section 8.1.1.3.Assume the shear wall behaves as a cantilever.

8500 .b = - = 42502 H = 12000bH = 0.35'P = 0.53 for loading at top of wallbe = ' I j J b = 0.53 X 4250 = 2250mm

Step 9 Find additional out-of-plane momentsWall is assumed braced in the out-of-plane direction.au = P Kh, He = 1 3 H; = 0.8 X 12000 ee 9600Assume K = 1 for conservatism.

1 ( H e ) 2 256j3 = 2000 h = 2000 = 0.1.28au = O.128h = 76.8 mmMadd = Na; (out-of-plane)= 4452 X 0.0768

= 342kNm342= - = 32.3kNm/m10.6

Step 10 Design stocky braced reinforced wallNot applicable.


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Design of Walls 393

162

162SK 8/49 Moments due toslenderness. Out-ol-Plane Madd

Step 11 Determine cover to reinforcememMaximum size of aggregate = 20 mmCondition of exposure =mildGrade of concrete = = C30Minimum cement content =275 kg/rn3Nominal cover =25mm

Step 12 Design of reinforced wall - rigorous method

H I"-. -.I~t----- _. , _ '~--A2 ~

'---. , _0'" .~ c-. , _2 ! : !

C-' ~ I '~ c-. , _ _ . ,2 c-. _ IC.,

Ash-----'N-

Ash-------Ash--------~s--------_JAsk---------~

SK 8/50 Elastic stress analysis ofWallt. 2550 . 1


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(1) Assume 0.40% reinforcement in wall.. 600 X 1000 x 0.40Reinforcement per metre length = 100 = 2400mm 2/m

Use 1200 mm2 on each face per metre length.(2) Assume a value of x for depth of neutral axis from compression face.

x = 3000mm assumed.(3) Divide compression zone into convenient layers of concrete.(4) Divide tension zone into convenient layers of steel.(5) Find the following in completing the table:

S = Ac + (m - 1)A~1C] = (x - aclSC2 = (x - ac>acSC3 = (at - X )AS IC4 = (ar - x)atA.t

Number Ac A~t S ac AS! a 1 C C2 CJ C4(x 1 0 1 ' 1 ) (x 1 0 1 ' 1 ) (x 109) (x 1012) (X 10 (,) (x 1 0 10)I 1.53 6120 1.62 300 3648 3760 4.374 1.312 2.772 1.0422 0.72 2880 0.76 1200 3648 5280 1.368 1.642 8.317 4.3923 0.72 2880 0.76 2400 3648 6800 0.456 1.094 13.862 9.4264 3648 8320 19.407 16.1475 3648 9840 24.952 24.5536 6120 10900' 48.348 52.699Totals 2.97 11880 3.14 24360 6.198 4.048 117.658 108.259

LC 4.048 X 1012A = ____1 = = 653mmc Lei 6.198 x 109

mLAsttll + 1 :Sa cx = . . = 1735mmmLAst +:rSM 28000 x lWe = - = = 6289 mmN 4452

I e = Nx( e + AT - X)(AT - Ac) L ( X - Qc)S4452 X 103 X 3000 X (6289 + 9201 - 1735)

(9201 - 653) X 6.198 X 109= = 3.467Nrmnr'


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Design of Walls 395

fS I = L : ( : = ; ) A J [ ( ~ ) L (x - Q c)S - N ]= (11100 - 3~) x [(3.467) x 6.198 x 109 - 4452 x Ifrl]117.658 xl 3000= 186.6N/mm2

d(6) Check x = --.- = 24t9mm1+~mfc

Second approximation for x is halfway between first approximation andthe check result.Assume x = 2700mmAfter carrying out the same tabular exercise as before it is found that:I e = 3.66N/mm2fS I = 182.4Nfmm2Check x = 2570mm.No further iteration is necessary.Check. reinforcement in compression flange due to out-of-plane bendingAverage compressive stress in flange = ~(3.7 + 2.8) = 3.25N/mm2

lc=3.7 N/mm2_----.....- ...1----- 28Nfmm2

I

rN 1SK 8151 Elastic ana.lysis - stressdiagram. f$=183N/mm 2


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Average compressive stress in reinforcement in flange = = 3.25 x 15for m = 15

= 48.75N/mm 2Over a unit length of wall,compressive force, N ~ 3.25 x 600 = 1950kNOut-of-plane bending moment due to DL + WL(y) + additional momentdue to slenderness

32.3=39+-2= 55.2kNm/m

(see Step 9)

SK 8/52 Section through wall forout-of-plane analysis.

Material strengths chosen:f e u = 30N/mm2 f y = = 460N/mm2

d 565 Mk = - =- = 0.95 e = - = 0.028 m/ 1 600 NSee Table 11.8.e~= 0.047h

N .For p = 0.4, - from chart IS 13.53 > 3.25 Nzmrrr'bit:. Nominal steel is required as per chart.Check' reinforcement in tension flange for out-of-plane bendingMaximum tensile stress in bar due to in-plane bending moment =183N/mm 2Maximum allowable ultimate tensile stress in bars = 0.87fy= 0.87 X 460= 4OON/mm 2Available tensile force in bars per metre length of wall per face of wall= (400 ~ 183) x 1200 (area on each face)= = 260.4kN/mMaximum out-of-plane bending moment =55.2 kNm/m


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Design of Walls 397

MK=--f c u bd255.2 X 10630 x 1000 X 5652

= 5.76 X 10-3Z = d[ 0.5 + J ( 0.25 - O ~ 9 ) ] ~ 0.95d

= O.95d = 537mmM 55.2 X 103Required tensile force in bars = - = = 102.8kN/mz 53 7

This is less than 260.4 kN/m available. Hence, no additional reinforcementis required in tension flange.

Step 13 Design of reinforced wall - simple methodNot required. The design principle is exactly similar to beam design andhas not been illustrated.

Step 14 Design of short and squat cantilever waU - deep beam approachNot required, because HI L > L

Step 15 Check shear

I-I.-----0 600 00 s~ 5 2II"

SK S/S3 Shear check o r Wall 1.

A s; = available tension reinforcement below neutral axis in web ignoringflange= (11.2 - 2.6) x 2400

= 20640rn2d o = 11200 - 300 = 10900mm (approx.)

_ lOOA si _ 100 x 20640 _ 0Pi - hwd ; - 600 x 10900 - 0.3210


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f~ u = 30N/mm2From Figs 11.2 to 11.5,Vc i = 0.47N/mm2V i = combined in-plane shear

= VI + Q, = 2380 + 1440 = 3820kNV.

Vi = hW~i = 0.58 N/mm2A"o = available tension reinforcement for out-of-plane bending= 1200mm2/m (each face)do = effective depth in out-of-plane direction = 565 mm

= lOOA so = 100 x 1200 = 2 %Po bd; 1000 X 565 0. 1From Figs 11.2 to 11.5,Veo = 0.4N/mm2VOH = out-of-plane shear coacting with Vi

= 4.2 kNfmm on the webVOH 2Voh = -- = 0.007 N/mmbdoVi voh (0.58) (0.007)- + - = - + -_. = 1.25 > 1Vc i VLQ 0.47 0.4

Shear reinforcement is necessary for in-plane shear.Note: Increase of design concrete shear stress due to presence of axial load has

been ignored in these calculations but may be allowed as per formula onpage 160.

Step 16 Calculate shear reinforcementCase 1: V"h < V,,"

( V o , , )" = 1 - - y.I CIVeo= ( 1 - 0.007) x 0.470.4= 0.46N/mm2

V i = v ; h w d i = 3008.4 kNVi - V ; = 3820 - 3008.4 = 811.6kNShear reinforcement is required to resist 811.6 kN.


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Design of Walls 399

Ah = VsiSh O.87fyd;811.6 x HY

0.87 x 460 x 10900= 0.19

If Sh = 300, then Ah = 300 x 0.19 = 57 rnnr' which is 29 mm2 of horizontalbar on each face at 300 mm centres, or, 97mm2 per metre on each face.A v Ah- = - = 0.19Sv Sh for f y = 460N/mm2Vertical shear reinforcement additional to vertical bars provided for bend-ing is required if available vertical bars have no residual capacity.In the web 2400mm

2/m vertical bars are available at a maximum averagestress level of, say, 16ON/mm2 (see Step 12). Hence residual capacityavailable in vertical bars in web =0.87 x 460 -160 =240N/mm2Modified AviS v to take into account the residual capacity

46 0= 0.19 x 0.87 x - = 0.32240A v for shear required per metre length of wall = 320mm2 (modified)Available vertical bars = 2400mm2/m in webHence no additional vertical bars are necessary to resist shear in web.No shear reinforcement is required in out-of-plane direction.

Step 17 Check out-oj-plane bending about vertical plane

SK 8/54 Out-of-plane bendingabout vertical plane.

50 550I I~~

Mov = 1.4 x 28 (see Step 5)= 39.2kNm/mVov = 1.4 x 26= 36.4kN/m


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M 39.2 X 106K = -- = = . = 4.3 X 10-3fcubd2 30 x 1000 x 5 S O Z

Z = d[ 0.5 + J ( 0.25 - 0~9)] ~ 0.95d= = 0.95d = 522mm

M 39.2 x HI'A =: -- = . = 187.6mm2s 0.87 fyz 0.87 x 460 x 522Add to this. reinforcement the horizontal reinforcement required in Step 16for in-plane shear.Total horizontal reinforcement required on each face (assuming the loadWL is reversible in direction)= 187.6 + 97 = 284.6mm2/m

Vov 36.4 x loJ 2Vuv = bd = 1000 x 550 = 0.07 N/mmShear stress is negligible.

Step 18 Design of plain wallsNot required.

Step 19 Shear check of plain wallsNot required.

Step 20 Check minimum reinforcementMinimum compression vertical reinforcement in wall = 0.4%(f y = 460N/mm2)This has been provided.Minimum horizontal tension reinforcement on each face= = 0.13% (/y = 460N/mm2)

600= 0.13 x 1000 x -1 0 0= 780mm2/m on each faceThis amount is greater than horizontal reinforcement found in Step 17.This reinforcement will be adopted.Minimum anti-crack reinforcement is 0.25% in both directions on eachface. This has been provided.

Step 21 Check maximum reinforcementNot required.

Step 22 Check containment of wall reinforcementVertical reinforcement is Jess than 2% of gross concrete area.


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Design of Walls 401

Hence requirement is to provide horizontal reinforcement equal to 0.2%of gross cross-sectional area. This is provided.Vertical bar diameter = 20mmHorizontal bar diameter = 10mm > ~ (20mm)

Step 23 Check early thermal crackingCrack width limitation = 0.3 mm(see Step 14 of Section 3.3).Assume R = 0.8 at base.T, = 32"Ce, = 0.8T,a:R= 0.8 x 32 x 12 x 10-6 x 0.8 =2.46 x 10-4Check horizontal bars for vertical cracks

Qc;r,,65.7.=:=1100===:1100SK 8/55 Crack width for

horizontal bars 10mm @ 100 c/c.

Assume lOmm diameter bars at lOOmm centres (785mm2/m).Oc r = 65.7mm (1.414 X 50 - 5 = 65.7)take x = hl2

3 x 65.7 x 2.46 x 10-42(65.7 - 45)1 + 300= O.04mm < O.3mm OK

Check I1erlica/bars for horizontal cracksAssume 20mm diameter bars at 250mm centres vertically (1256mm2/meach face).


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-hrl--:J~~I 125! 125 I25 0 I 1250 SK 8157 Sketch to lind a~r'

SK 8156 Crack width for verticalbars 20 mm @ 250 c/c.

Qcr = 120mm3 x 120 x 2.46 x 10-4

2(120 - 25)1 + 300

= 0.05 mm < 0.3 mm OKStep 24 Clear spacings of bars in tension

Reinforcement provided is 20 mm diameter at 250 mm centres both facesvertically and lOmm diameter at 100mm centres both faces horizontally.These spacings satisfy the requirements according to Step 13 of Section3.3.

Step 25 ConnectionsFollow Chapter 10.


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Chapter 9Design of Flat Slabs

9.0 NOTATIONAAs,beCxCyddhfyfeuheGkt,lh([l. vL ho[h.m;'.II1 2M'M tMmaxnQ kVVertW I :.

. x

Area of column or area of effective column headArea of steel in tensionEffective width of slab for transfer of moment to edge columnSize of a rectangular column in x-directionSize of a rectangular column in y-directionEffective depth of tensile reinforcementDepth of column headCharacteristic yield strength of reinforcementCharacteristic cube strength of concrete at 28 daysEffective diameter of column or effective column headCharacteristic dead loadDimension of column in direction of L hEffective dimension of column headShorter span framing onto columnsLonger span framing onto columnsActual dimension of column headMaximum dimension of column head taking 4Y dispersionCentre-to-centre of column in direction of span being consideredCentre-to-centre of column perpendicular to direction of span beingconsideredDesign limit moment at he/2Moment transferred to column by frame analysisLimiting moment between Hat slab and edge columnTotal ultimate load per unit area on Oat slabCharacteristic live loadCalculated shear from analysisEffective shear at column/slab interfaceCharacteristic wind loadingLength of side of a perimeter parallel to axis of bending

9.1 DEFINITIONSFlat slab is a reinforced concrete slab supported by columns with, orwithout, drops. The columns may be with, or without, column heads.Drop is a local thickening of the slab in the region of the column.

403


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W I T H O U T D RO P ANDN O C O l U I 1 N H E A D W I T H D R O P A N DN O C O L UHN HHO W I T H O U T D R O P A N DW I T H C O L U M N H E A DSK 911 Flat slab - section. S K 9/2 Flat slab - section. SK 9/3 Rat slab - section.

Column head is a local enlargement of the column at the junction with theslab.

9.2 ANALYSISOF FLAT SLABS

I I-j---r---- _ _ _ _ _ J . _ - --l--- -I I--t--i--

I . ly J L\ j > Lx SK 9/4 Typical plan of flat slab.Flat slabs are usually supported by a rectangular arrangement of columns.The analysis may be carried out by an equivalent frame method or by theuse of a finite element computer code. When using the equivalent framemethod the ratio of the longer to the shorter span should not exceed 2.The analysis for uniformly distributed vertical load may be carried out byusing Tables 9.1 to 9.6.The properties of the flat slab for analysis are similar to those already

discussed for solid slabs in Chapter 3.

9.2.1 Effective dimension of column headlh = effective dimension of headtho = actual dimension of headlh,max = Ie + 2( dh - 40 )I e = dimension of column in direction of lbdh = depth of headIh = is taken as the lesser of Iho or lh.max


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________ . _ . . . ._-,--_._. _ _ ._.Design of Flat Slabs 40S

SK 9/5 Flat slab - definitions. lNote: This means that the maximum dimension is limited by a 45 dispersion of

column up to 40 mm below the slab.

9.2.2 Effectivediameter of a column headIhe = (4Aht)I:S: O.25Lx

he = effective diameter of column or column headA = area of column or area of effective column head as defined by lhL x = shortest span framing onto columnhe should not be taken greater than one-quarter of shortest span of slabframing into column.

9.2.3 DropsDrops will be effective in the analysis if the smaller dimension of the dropis at least one-third of the smaJlest span of surrounding panels.For the checking of punching shear, this limitation does not apply.

9.2.4 Load combinationsfor analysisLei = lAG!: + 1.6Qkt.c, = 1.4Gk + 1.6QI:1.0Gk

on all spanson alternate spans and other spans loaded with

where Gk = characteristic dead loadQ k = characteristic live load.

9.2.5 Effectivewidth of slab for analysisFor vertical loading assume full width of panel between columns for frameanalysis.For horizontal loading as a frame assume stiffness of half width of panel.


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fU LL P A tiE L W ID THt------+----O R V ER T I C A L L O A O IN G- - + -

SK 9/6 Plan of flat slab showingpanel widths for analysis.

The analysis should be carried out using a computer program or a momentdistribution method. The analysis may also be carried out for uniformlydistributed vertical loads using Tables 9.1 to 9.6.The analysis may be carried out using Table 3.13 of BS8110: Part 1:

1985(1] provided the lateral stability is not dependent on the slab-columnconnection and loading on the flat slab for the design is based on a singleload case, i.e. LeI> the ratio of Qk/Gk does not exceed 1.25, Q k does notexceed 5 kN/m2, and there are at least three rows of panels.

9.3 DESIGN OF FLAT SLABSThe design may be based on the negative moment at he12 from thecentreline of the column. But this negative moment will have to be modified ifthe sum of the positive design moment and the average negative designmoment is less th_.3nthe following expression:M' : ( ~ 2 ) ( l l - 2 ~ c r

SPAN A II . L l . 1 SPAN 8 SPAN C SK 917 Negative momentlimitation for flat slabs - section.


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Design of Rat. Slabs 407

SK 918 Typical plan of ftat slab ~negative moment limitation.

where I, = centre-to-centre of column in direction of span beingconsidered1 2 = centre-to-centre of column perpendicular to direction of spanbeing consideredn = total ultimate load on slab (kN/m2).

To give an example:For Span A0.5 (M, + M2) + M7 ;:::M'For Span BO.5(M3 + M4) + M7 ;:::M 'Increase negative moments Mj, M2, M3, etc. until these conditions aresatisfied.

9.3.1 Divisionof panels

L x / 4 L x / 4" '" ! _ ~ I 4 ' l j t l 4,)(. . . --'"-)(--' l- - - + - t t - t - O F C O L U M NC O L T R I Pt- l--t-,

)( I I I M I D D L E_ r~~-~I_j S T R I PC O L U M N_ . - ~ - - - - - + - t - O F C O L. . . . - 1 _ l - _ j - t . J S T R I P


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PlAN OF SLAB WITH DROPI G N O R E D R O P I F D R O P W I D T H < L x / 3

SK 9110 Flat slab - division ofstrips.

Panels are divided into column strips and middle strips as shown.For slab without drop the column strip is (./4 wide on either side of thecentreline of column, where (" is the shorter span.For slab with drop the c.olumn strip is the size of the drop. Ignore drop ifthe size of the drop is less than Ix13.

9.3.2 Divisionof moments between columns and middle stripsThe moments obtained from analysis of frames should be divided asfollows (these percentages are for slabs without drops):

Column strip Middle stripNegativePositive

75%55% 25%45%

Note: Where column drops are used and column strips are determined from thewidth of the drop, it may so happen that the middle strip is bigger than themiddle strip in a slab without drop. In that case the moments in the middlestrip will be proportionately increased and those in the column stripdecreased to keep the total positive and negative moment unchanged.

9.3.3 Design of Oat slab panelsThe design is similar to the design of slabs and the worked examples are inChapter 3.


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Design of Flat Slabs 409

1 /6 A st

SK 9/11 Detailing ofreinforcement in Hat slabs. Column

Internal panels and edge panelsTwo-thirds of the negative support reinforcement in the column stripshould be placed in haJf the width of the column strip centred over thecolumn.

9.3.4 Moment connection to edge column

I II . . be = Cx-C w . 1

SK 9/12 Effective width of slab formoment connection to edgecolumn.

~'I be = Cx.y II~ C O L U M N S T R I P i

See sketches above to find effective width of slab be for transfer ofmoment between Hat slab and edge column. This moment should belimited to


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Mtmax = O.15bed 2fcuwhere d = effective depth of top reinforcement in column strip.The moment Me.max should not be less than half the design moment froman equivalent frame analysis or 70% of the design moment from a grillageor finite element analysis. The structural arrangement may be changed ifMtmax does not satisfy the above condition.

R E D I S T R I B U T E DM O M E N T T OA CC O UN T F OR- Mt, mi!lk. SK 9/13 I ns uf fi ci en t momen ttransfer capacity at edge column.

Where the design moment is larger than Mtmax, redistribution ofmoment may be carried out to reduce the design moment to Mlmax'Otherwise, to transfer moments in excess of Mtmax to edge column, theedge of the slab should be reinforced by an edge beam or an edge strip.The edge beam will be designed to carry the additional moment by torsionto the column.

9.3.5 Shear in flat slabsPunching shear around columns should be checked according to Step 7 ofSection 3.3. The shear to be considered for the punching shear calculationis increased from the calculated column shear by an amount dependent onthe moment transferred to the column by frame action.For internal column connections,

( 105Me)Ve(r = V I 1+ - v ; : ; -


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+-III

J I I_ , _ - - + ~ - + -I II P U NC HI N G S H EA RP E R I M E T E R

M ttj)I__ t -I I

x

SK 9114 Moment diagram at an internal columnof a flat slab. SK 9/15 Definition of dimension x.

where V,= calculated shear from analysisM, = moment transferred to column by frame analysisx = length of side of perimeter considered parallel to axis ofbending.

Alternatively,Vcff = 1.15 V, for simplicityFor corner coLumn connections,Vert = 1.25 V,For edge column connections,

for bending about axis parallel to free edge

( 2l o 5 M , )

Vef f = VI 1. 5 + --V tX for bending about axis perpendicular to freeedge

Alternatively,Veff = 1.4V\The moment M, may be reduced by 30% where the equivalent frameanalysis is used and both load cases LCI and LC2 have been considered.The shear reinforcement win be calculated according to Step 7 of Section 3.3.


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9.4 STEPBYSTEP DESIGN PROCEDURE FOR FLAT SLABSStep I Carry out analysis as in Section 9.2.Step 2 Find moment connection to edge column as per Section 9.3.4 and redis-

tribute moments if necessary.Step 3 Draw bending moment diagrams and calculate moments at he/2 following

Section 9.3.Step 4 Check limitation of negative design moments following Section 9.3.Step 5 Carry out division of panels as in Section 9.3.1.

Step 6 Divide moments between column strips and middle strips as per Section9 . 3 . 2 .Step 7 Determine cover to reinforcement (see Step 3 of Section 3.3).Step 8 Carry out design for flexure as per Step 4 of Section 3.3.Step 9 Distribute reinforcement as per Section 9.3.3.

Step 10 Check punching shear stressFollow Step 7 of Section 3.3.

Step II Check span/effective depth ratioFollow Step 11 of Section 3.3 for slabs with drops. For slabs without dropsfollow the same step but multiply I e _ ! d from Table 11.3 by 0.9.

Step 12 Curtailment of barsFollow Step 12 of Section 3.3.

Step 13 Spacing of barsFollow Step 13 of Section 3.3.

Step 14 Check early thermal crackingFollow Step 14 of Section 3.3.

Step 15 Calculate minimum reinforcementFollow Step 9 and Step 15 of Section 3.3.Sup 16 CalculaJ.eflexural crack width

Follow Step 16 of Section 3.3.Step 17 Design of connections

Follow Chapter 11.


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Drop of 2000 mm in floor slab is greater than 1,)3 = 1667mm.Drop will be effective in the distribution of moment.LoadingFrames in short direction are 6m apart.Roof slabGk = characteristic dead load = 0.2 x 25 = 5 kN/m2Qk = 1.5 kN/m2LCI = 1.4Gk + 1.6Qk = 9.4kN/m2 = 56.4kN/mLC2 = 9.4 kN/m2 and 5 kN/m2 on alternate spansor LCz = 56.4kN/m and 30kN/m on alternate spansFloor slabGk = 5 kN/m2 at slab without drop= 10kN/m2 at slab with drop (area 2m x 2m)= 30kN/m or 30 + 5 x 2 = 40kN/mQk = 5 kN/m2

= 30kNfmLCI = 1.4G" + 1.6QkLC2 = alternate spans loaded with LCI and dead load onlyColumnsHorizontal load on columns is due to wind load at the rate of 1 kN/m 2which is equivalent to 6 kN/m on the column. The wind loading analysiswill be carried out separately and combined later with the vertical loadingbecause the stiffness of the slab to resist horizontal loading is half of thatto resist vertical loading.Load cases with wind load W" are as follows:LC3 = l.4G" + 1.4WkLC4 = 1.2G" + 1.2Qk + 1.2WkFrame analysi$ using a computer softwareE = Young's modulus =28 x 106kN /m212 joints 14 membersJoints 1, 4, 7 and 10 rigidly fixed.Column size 400 x 400Slab size 6000 x 200 (deep)Load cases:BI- dead loadB2 to B7 - live loads on members 9 to 14 respectivelyBs - wind load


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J 5 Q Q Q J sooo ~I s o o o : J M e m b e rNumbe r s 60 8 @C D7 SK 9119 Frame diagram foranalysis.

SK 9/20 Dead load on frame (B,).

65 " I 67 " I

SK 9/21 Basic live loads B2 to BI; .

Combinations:C) = 1.4B) + 1.6(B2 + B3 + B4 + B5 + B6 + B7)Cz = 1.4BI + 1.6(B:;: + B4 + 85 + B7)C3 = 1.4B 1 + 1.6( 83 + B6)C4 = l.4B, + 1.6(82 + B3 + 84)


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_ . _ .._.-, Design of Flat Slabs 417

Output from analysisEnvelope of load cases (vertical loads)Elastic analysis - no redistributionFloor slab Member 9Joint Maximum BM Shear Combination2 130.80 228.5 C25 215.5 258.2 C4Midspan 131.2 C2Floor slab Member 10Joint Maximum BM Shear Combination5 199.9 239.0 C48 199.9 239.0 C4Midspan 112.5 C3Roof slab Member 12Joint Maximum BM Shear Combination3 67.8 130.0 C26 129.0 153.5 CMidspan 82.0 C2Roof slab Member 13Joint Maximum BM Shear Combination6 121.3 141.0 C)9 121.3 141.0 CMidspan 60.4 C3

rt,,

- , , -3 , "..- 6 .- , 9 '.. . _ - _ . . . . ." 12- . - _ _ -, ~---..., @ @ JI\, I\ I , 1II \ , ,II I I , "I I \ I '.'~ : / III \ "2 I ,-5 , II "- " " 7, / . .. .. .. .. _ . . , w


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1999kNm 199-9kNm

SK 9/24 Shear and moment envelope for member 10_

Set plastic moment capacity at joint 6 =0.8 X 129=103.2 kNmThe following steps of reanalysis of frame are carried out:Step 1: For one combination at a time increase live load on span untilplastic moment is reached at a joint in a member. Plastic moment capacityof members on first floor is 172.4kNm and member on roof is 103.2kNm.Step 2: Release joint where plastic moment is reached and increase loadinguntil plastic moment capacity is reached at another joint.Step 3: Progressively release joints and increase live load until full comp-lement of live load is on structure.Step 4: Find cumulative effect of all incremental live load on structure.The following tables become useful if a non-linear finite element computerpackage is not available.Frame types:F. = no member end releasesF2 = member in F . nos 12 and 14 ends released at joints 6 and 9F J = member in F 2 nos 9 and 11ends released at joints 5 and 8F4 = member in FJ no. 13 ends released at joints 6 and 9F s = member in F4 no. 10 ends released at joints 5 and 8

6 9 Plastic Hinges3 1e @ 2 5 e '1C D @ ) ~" ~I". " J . . ." 9

2

SK 9/25 Frame type r; SK 9/26 Frame type F 2


31/68

420 Reinforced ConcretePlastic Hinges

3_.......",._.;r..--=----i .... f-=.---.12 rPlastic Hinge.s6 ') 2

1 0

3 12 5 6 " @ ) . & 1 " " ~~ "~ n : , l o

SK 9127 Frame type F 3- SK 9/28 frame type F4.

Plastic Hinges3,.--::::: .......~-=::---4')~-:::::---,12

BS 6 B6 ') 61 12 0.3kN/1Il3

SK 9129 frame type F~. SK 9(30 Unit live load on frame.Combination C7 = 104 x dead load or 1AB1The method is illustrated for combinations C; and C 2 only.q= 1.6(82 + B3 + B4 + B; + B;' + 87 )C; = 1.6(B; + B4 + B ; + 87)B i, B 3 , B4 = I k N/m85 , 8;', 87 = 0.3kN/m

Combination frame Member end bending moments (kNm)typeMember 9 Member lO Member 12 Member 1 3

Joint 2 Joint 5 Joint 5 Joint 8 Joint 3 Joint 6 Joint 6 Joint 9

C7 Fl 6 2 . 7 102.5 95.5 9 5 . 5 4 4 . 7 9 7 . 5 9 1 . 6 9 1 . 6Ci F ! 2 . 0 3 . 7 3 . 4 3 . 4 0 . 7 1.1 l.0 1.0C I F 2 2 . 1 3 . 6 3 . 5 3 . 5 0 . 9 0 0 . 5 0 . 5C; F 3.1 0 2 . 4 2 . 4 l.l 0 0 . 8 0.83c; F 4 3 . 1 0 2 . 2 2 . 2 l.l 0 0 0C; f. , 3 . 1 0 0 0 1.1 0 0 0q Fl 2 . 3 2 . 7 0.8 0.8 0 . 8 0 . 8 0 . 2 0 . 2C i F 2 2.3 2.6 0 . 8 0.8 0 . 9 0 -0.2 -0.2q F " 3 . 1 0 0 0 1.0 0 0 0q F 4 3.1 0 0 0 1.0 0 0 0q f< ; J.l 0 0 0 1.0 0 0 0


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C I = 1 unit of live load combination in combination C1 i.e. Ci = 1 kN/mof B2, B3 and B4 and 9/30kN/m of 85, 86 and B7.Full compliment of 82, 83 and 84 is 30kN/m and of 85, 86 and 87 is9kN/m.Plastic moment at joint 5 is fixed at 172kNmPlastic moment at joint 6 is fixed at 103.2 kNmDead load moment at joint 6 =97.5 kNmEach unit of combination C1 produces 1.1 kNm at joint 6 for frame typer; Thereforeunits of live load in combination C1 required to form first plastic hinges atjoint 6 and joint 9 in members 12 and 14103.2 - 97.5 .----~ = 5 urnts, say1.1

Frame type F 2 has joints released at joints 6 and 9 for members 12 and 14.After 5 units of combination CI the bending moments at joints are asfollows:F rame type F 1Member 9 Joint 2

Joint 5Member 10 Joint 5Member 12 Joint 3

Joint 6Member 13 Joint 6

62.7 + 5 X 2.0 = 72.7kNm102.5 + 5 x 3.7 = 121.0kNm95.5 + 5 x 3.4 = 112.5 kNm44.7 + 5 x 0. 7 = 48.2kNm97.5 + 5 x 1.1 = 103kNm "plastic91.6 + 5 x 1.0 = 96.6kNm

Units of live load in combination CIto form second plastic hinges at joints5 and 8 in members 9 and 11172 - 1213.6 = 14 units of combination CI

Total number of units of CI to cause plastic hinges at joints 5 and 8 inmembers 9 and 11 is 19.After 19 units of combination CI, the bending moments at joints are a sfollows:Frame type F zMember 9 Joint 2



72.7 + 14 x 2.1 = 102.1 kNm121 + 14 x 3.6 = 171.4kNm "plastic112.5 + 14 x 3.5 = 161.5 kNm48.2 + 14 x 0.9 = 60.8 kNmplastic = 103kNm *plastic96.6 + 14 X 0.5 = 103.6 kNm *plastic

Joint 5 of member 9 and joint 6 of member 13 have gone plastic simul-taneously at 19 units of combination Ct. Therefore frame type F . l is notconsidered.


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422 Reinforced ConcreteAfter 24 units of combination C], the bending moments at joints are asfollows:Frame type F-IMember 9 Joint 2



102.1 + 5 x 3.1 = 117.6kNmplastic =:: 171.4 kNm *plastic161.5 + 5 x 2.2 =:: 172.5 kNm "plastic60.8 + 5 x 1.1 =:: 66.3 kNmplasticplastic

=:: 10 3 kNm *plastic= 103.6kNm *plastic

Frame type FsAfter 30 units of combination C), the bending moments at joints are asfollows:Member 9 Joint 2 117.6 + 6 x 3.1 = 136.2kNm

Joint 5 = 171.4kNmMember 10 Joint 5 = 172.5kNmMember 12 Joint 3 66.3 + 6 x 1.1 = 72.9kNm

Joint 6 = 103.0kNmMember 13 Joint 6 = I03.6kNmFormula for calculating midspan bending moment and shearC7 + 5 units of q (Fd + 14 units of C; (F2) + 5 units of C; (F4) + 6 unitsof C; (Fs)Member 9 - combination C, (20% redistribution)Midspan moment = 56.3 + 2.2 x 5 + 2.2 x 14 + 5 x 3.5 + 6 x 3.5 =

136.6kNmEnd shear, joint 2 = 111.1 + 3.7 x 5 + 3.7 x 14 + 4.6 x 5 + 4.6 x

6 = 232.0kNEnd shear. joint 5 = 126.9 + 4.3 x 5 + 4.3 x 14 + 3.4 x 5 + 3.4 x 6 =

246.0kNSimilarlyMember 10 - combination C} (20% redistribution)Midspan moment = 115.7 kNmEnd shear = 23 9 kNMember ]2 - combination C, (20% redistribution)Midspan moment = 9O.9kNmEnd shear 3 = 134.9 kNEnd shear 6 = 147.1 kNMember 13 - combination C{ (20% redistribution)Midspan moment = 72.7 kNmEnd shear 6 = 141 kN


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N. . . . .

o. . . . .

c.9~r:zeou

c _..Co . . < u- l 9 e_0:;:s. . .'"V.J::Vlc _'"Co . . < u- l 9 S_0~ e

c : C " ' ! C " ' ! ~ ~ C " ' !o n . . . . . -o. . . . .

00

O N C " ' ! ~ f ' ! f ' !~""";~'1"""'4_. . . . .

r - - v > o r - , , ! , , !0\0"';0< ' " l

o n > D - : C : C : C :, .. .. ; O . . .. . .. ._.t-.l ........'-0

000000o \ - . : i - . : i - . : i - . i - . : i. . . . .000000: : i - . i - . i - . : i - . i - . i

r - ~ o n ' - O o o o~ " ' ; N N . n

O ' < ' " l < ' " l " < t " < t " ' :- O - . : i - . i M t r i . . . .N-- : t - : r - - > D ' - O ' - O

< ' " l M - . i - . : i - . i-. . . .< ' " l N N , , ! " 1 " !- O N N . . . . < ' " lo n


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424 Reinforced ConcreteIOJ6kNm 103-6kNm

SK 9/31 Bending moment diagramcombination CI (20%redistribution ).

Note: Only one combination C[ has been fully analysed to demonstrate theprocedure for redistribution of moments in a frame structure. In practiceall combinations of loads should be similarly processed to get an envelopeof moments and shears. For all combinations of loads the plastic hingeswill form at the same moment, i.e. 172 kNm at first floor level and 103.2 kNmat roof level.

Step 2 Check moment connection to edge column

EE8ODIIs

SK 9/32 Effective width of slab formoment transfer.

MI. max "" O.15b e d 2f r : ube = e x + C~v :=: 400 + 40 0 = 800mmd = 175 mm assumedfe u = 4ON/mm 2MI. rna. = 0.15 X 800 X 1752 X 40

:=: 147kNm > 136.2kNm (member 9, joint 2)The column slab connection at the edge can transfer the applied momentand no further redistribution is necessary. It is conservatively assumed inthis analysis that the depth of the slab at the column is 200 mm, ignoringthe drop. The moment MI. max is greater than the design moment obtainedfrom an equivalent frame analysis.


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Step 3 Find bending moments at hel2 and at edge of drop

JOINT 172 'OkNm

SK 9/33 Bending moments atcritical points - combination CI(20% redistribution).

M EM B E R 9 M E M B E R 10

Member 9 - combination C, (20% redistribution)Joint 5:Bending moment = = 172kNmShear = 246 kNDead load: 1.4 x 40 = 56kN/m near SUppOTtLive load: 1.6 x 30 = 48 kN/m near supporthc/2 = 0.225 m

(56 + 48) x 0.2252Bending moment at he/2 = 172 - 246 x 0.225 + 2= 119.3 kNm (top tension)

Edge of drop = 1000 mm from centreline of column (56 + 48) X 12B e n d i n g moment at edge of drop = = 1 7 2 - 2 4 6 x 1 + _ ; _ ; _ : ~ _ _ _ _ : _ : 2 : _ ' : _ _ _ _ _ _ _ : _ : -

= -22 kNm (bottom tension)Joint 2, similarly:

(56 + 48) X 0.2252Bending moment at he/2 = 136.2 - 232 x 0.225 + 2= 86.6 kNm (top tension)

Bending moment at edge of drop = -43.8 kNm (bottom tension)Member 10 - combination C, (20% redistribution)


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426 Reintorced Concrete

Joint 5:Bending moment at h,)2 = 121.4 kNm (top tension)Bending moment at edge of drop = -14.5 kNm (bottom tension)Member 12 - combination C, (20% redistribution)Joint 3:Bending moment at he/2= 72 9 _ 134 9 0 22 . (1.4 x 30 + 1.6 x 9) x 0.2252 =. . x . 5 + 2 44.0 kNmJoint 6:Bending moment at he/2 = 71.3 kNm (top tension)Member 13 - combination C] (20% redistribution)Joint 6:Bending moment at he/2 = 73.3 kNm (top tension)

Sup 4 Check limiklJion 0/ negative design moment

1366kNm 1 ColumnJOINT 2 JOINT 5

SK 9/34 Member 9 - combination Cl (20% redistribution): limitation of negativedesign moment.

M ' = ( n l . 2 ) ( l _ 2hc)2. . 8 1 3where n = loading per unit area on slab.Average n on span on first floor = 1.4 x 5 + 1.6 x 5 = 15kN1m2

1 2 = 6.0m he =0.225mM' = C s ~ 6.0) (5.0 _ 2 x ~.22Sr

= 264.6kNm (floor slab)


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Design of Flat Slabs 427Average n on span on roof = 1.4 x 5 + 1.6 x 1.5 = 9.4 kN/m2M' = (9.4 ; 6.0) (5.0 _ 2 x ~.225)2

= 165.8kNm (roof slab)Check negalil1e moment limitationMember 9Joint 2 at hel2 = M2 = 86.6 kNm (see Step 3)Joint 5 at he/2 = M5 = 119.3 kNmMidspan moment = 136.6kNm (see Step 1)Average of Mz and Ms plus midspan moment= 0.5 (86.6 + 119.3) + 136.6= 239.6kNm


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Average of negative and positive165.8kNm

73.3 + 72.7 = 146kNm < M'The negative moments will have to be increased by (165.8 - 146) =19.8kNmRevised negative moments:Joint 6: 73.3 + 19.8 = 93.1 kNm

Step 5 Carry out division of panels2000 2000

og

J r Drop (typical), , - 1 . - - ~r--r-- ---'-" -....--r-- , 2000I ~ __ l_--'-- _---1._ ...J...._ __ L_- ~ - * ~ - . -40002000I I I I I...J....__.........L..._ . . J . . . . . _ _ . . . . . . . . . L . . . _ J.__.........L..._. , 4000

-,--- ---r -'---""T -~- --r- 2000I - II I I I_J_ _---'-_ _ ._ _ i_ .....L. _ f-""'_- - - -1 . .. _r

5000 S O O O 5000

CoIunTI Stripoo~ Middl~ Strip

Column Strip

Middl~ Sirip

Column Strip

SK 9/35 Plan of Hoor slab - division of strips.

First floor slabSize of drop = 2000 mmi : 5000- = -- = 1667mm < 2000mm3 3Column strip = 2000 mmWith drop middle strip = 5000 - 2000 = 3000mm (y-y)and = 6000 - 2000 = 4000mm (x-x)Middle strip in a slab without drop = 6000 - Ixl2 = 6000 - 2500 =3500mm

4000Proportion of middle strip with drop and without drop = 3500 = 1.14

Roof slab


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Design of Flat Slabs 4295000 5000 5000

J~ = F * - - -. . .. 25001------ ---- --, , 3500:-.-1=$~ 2500r------ ---- ----~---~-~-- .. 3500~ . ~ 2500. ' ~ .r------r----j~---, I

Column Strip

MktJle Sirip

Column Strip

Middle Strip

Coi l .mn Strip

SK 9/3(; Plan of roof slab - division of strips.

. Lx 5000Column stnp = = - = = - - = 2500mm2 2Midd l e strip = = 6000 - 25 00 = = 3500mm (x-x)

= 5000 - 2500 = 2500mm (y-y)Step 6 Divide moments between column strip and middle stripFor slabs without drops,

Negative moments - 75 % column strip25 % middle stripPositive moments - 55% c-olumnstrip45% middle stripFloor slab: design momentsMember 9: negative momentsJoint 2: 1l1.6kNm (see Step 4)Joint 5: 144.3kNrn (see Step 4)Middle strip momentsJoint 2: 111.6 x 0.25 x 1.14 (see Step 4) = 31.8kNm (top tension)Edge of drop = 43.S x 0.25 x 1.14 = 12.5kNm (top tension)Joint 5: 144.3 x 0.25 x 1.14 = 41.1 kNm (top tension)Edge of drop = 22.0 x 0.25 x 1.14 = 6.3 kNm (top tension)


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430 Reinforced ConcreteColumn strip momentsJoint 2: 111.6 - 31.8 := 79.8 kNm (top tension)Edge of drop: 43.8 - 12.5 = 31.3 kNm (top tension)Joint 5: 144.3 - 41.1 = 103.2kNm (top tension)Edge of drop: 22 - 6.3 := 15.7 kNm (top tension)Member 9: positive momentsDesign midspan moment = J36.6kNm (bottom tension)Middle strip momentsMidspan: 136.6 x 0.45 x 1.14 =. 70.1 kNm (bottom tension)Column strip momentsMidspan: 136.6 - 70.1 := 66.5 kNm (bottom tension)Member 10: negative momentsJoint 5: 148.9kNm (see Step 4)Middle strip momentsJoint 5: 148.9 x 0.25 x 1.14 "" 42.4 kNm (top tension)Edge of drop: 14.5 x 0.25 x 1.14 = 4.1 kNm (top tension)Column strip momentsJoint 5: 148.9 - 4204 = 106.3 kNm (top tension)Edge of drop: 14.5 - 4.1 = lOo4kNm (top tension)Member 10 : positive momentsDesign midspan moment = 115.7 kNm (bottom tension)Middle strip momentsMidspan: 115.7 x 0.45 x 1.14 = 5904kNm (bottom tension)Column strip momentsMidspan: 115.7 - 59.4 = 56.3 kNm (bottom tension)

Note: Similarly calculate moments in column strips and middle strips in roof slab.Step 7 Determine cover 10 reinforcement

See Step 3 of Section 3.3.Step 8 Design jor flexure

See Step 4 of Section 3.3.The increased slab thickness at drops may be considered for the determi-nation of reinforcement provided all reinforcement is properly anchored.Check reinforcement also at edge of drop.

Note: In this example the reinforcement is found for the flat slab spanning in theshort direction only. Exactly the same method of analysis and designshould be used to find the reinforcement in the long direction.


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Design of Flat Slabs 431Step 9 DelaiUng of reinforcement

Two-thirds of the negative support reinforcement in the column stripshould be placed in half the width of the column strip centred over thecolumn.Step 10 CalculaJe punching shear and shear stress

Punching shear at floor slabCheck joint S.

M EM BER 9

SK 9137 Moment transfer tocolumn for punching shearcalculation.

M EM BER 10B E N D IN G M O M E N T D IA G R A ME LA S T IC A N A L Y S I S - N O R ED IS T R I B U T IO NL O A D IN G C O N D I T IO N C l o

b Joint 8( To Jo in - t" '2 ' " - - t - -- - ,r

'"'"N1000 1000 25 5

SK 9/38 Punching shear perimeters - plan of floor slab.


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432 Reinforced ConcreteUse results of elastic analysis of frame before redistribution.Maximum column moment atjoint 5 := 215.5 - 199.9 = 15.6kNm

MI = 15.6kNmA 30% reduction is allowed i f frame analysis is carried out.

M ! = 0.7 x 15.6 = 10.9kNmV t = 258.2 + 239.0 = 497.2 kNPunching shear perimeter at 1.5d from face of column,d = 400 - 30 = 370 mm1.Sd = 1.5 x 370 = 555mmx = 400 + 2 x 555 = 1510mm = L510m

( l . S M t ) ( 1 . 5 x 10.9 )V"ff = V I 1 + -V~ = 497.2 1 + 497.2 x 1.510= 508kN

Maximum shear stress at column perimeter (U o = 4 x 4(0)Vdl 508 X 16'= - = = O.86N/m 2 < 5N/m 2 OKUod 4 x 400 x 370

Ver fShear stress, v = UdU = 4 x 1510 = 6040

v = 508 x 10;> = 0.23 N/mm2 < Vc for minimum percentage of6040 x 370reinforcement

No shear reinforcement is required in slab with drop.For slab outside drop consider that loaded area is perimeter of drop.V , n = 508 kN - (load on the area of drop)

=508-4x22= 420kN

Perimeter of slab at 1.5d (d = 170mm)= 4 x (2000 + 3 x 170) = lO040mm420 x 10:'Shear stress, v = 10040 x 170

= 0.25 N/mm2 < Vc for minimum percentage ofreinforcement

No shear reinforcement is required at internal columns of floor slab.Similarly check for an external column and a corner column.


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Design of Flat Slabs 433Rules for calcufolion of perimeser of external and corner columns

x c, 1-5d15d 1-~:

1x < T'ScI

T'5dC,

-+----1 T5dC ,5d

C x 1'5+ C,+ 6dE D G E C O LU M N I N S I D ES L A B

x c " "Set

C," 5 d

1-

~~x > T5d

P= lesser of :-

"5d

1-Sd

2( x +C,,) +Cy+6dOR 2(C.+ Cy ) + T2d

E D G E C O L U M N IN S ID ES l A B

C O O N E R C O L U M N 00O N E D G E O F S L A B

x e x 15d

y

~~~y. >15d

x and Y :iii "5dP= C.+Cy+x+y+3d

C ,15d

C O O N E R C O L U M N IN S ID ES 1 . A B .d = A.v~~ effectivedepth of slab

p" lesser of :-C Cr. x +y+ 3dOR 2(,;+Cy).'2dC O R N ER C O L U M N IN S ID Es.!M

SK 9/39 Punching shear perimeters for fiat slab.The illustrations show the different column configurations with respect to afree edge and the corresponding perimeters for the calculation of punchingshear stresses.


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When the column face is more than 1.5d away from a free edge of slab,then there are two alternative perimeters possible as illustrated. Take theleast of these two alternatives for the calculation of punching shear stress.Punching shear aI roo/ slabCheck joint 3 - edge column.V I = 130kNThe frame action considered is in the x- x direction as explained in Section9.3.5.Veff = 1.25V, = 1.25 x 130 = 162.5kN

d = 170mm assumed1.5d = 1. 5 x 170 = 255 mmShear stress at column perimeter (Vo = 3 x 400 = 1200)

Vett 162.5 X 103=- = = 0 .8N /mm2 < 5N/mm2 OKVod 1200 x 170VeffShear stress at 1.Sd = v = Vd

U = (2 x (400 + 255)] + (400 + 510)= 2220

162.5 x 103v= 2220 x 170= 0,43N/mm2

Assume minimum percentage of tensile reinforcement in slab.V c = 0.48N/mm 2 for Grade 40 concrete and an effective depth of 170mm.No shear reinforcement is necessary.

Now: The punching shear check should also be carried out for the flat slabspanning in the long direction and the worst result should be used.

Step 11 Check spanle//eetitle depth ratioFollow Step 11 of Section 3.3

Step 12 Curtailment 0/ barsFollow Step 12 of Section 3.3.

Step H Spacing of barsFollow Step 13 of Section 3.3.

Step 14 Check early thermal crackingFollow Step 14 of Section 3.3.

Step 15 Calculate minimum reinforcementFollow Step 9 and Step 15 of Section 3.3.


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Step 16 Cakulale flexural crack widthFollow Step 16 of Section 3.3.

Step 17 Design of connectionsFollow Chapter 119.6 TABLES AND GRAPHS FOR CHAPTER 9

How to use Tables 9.1 to 9.6

[ O C = l'O ]

Intermediate floor. Typical section

h column I O C : : 2 < 6 7 !fixedfirst floor, TY2icai sectionf .'_

h 2L I O C = 4 ' O !columnhinged"'\.First floor. TmiCQ1 se c tio n

I x I x Ix, . T+ Iy- ~1 -5 I y3il- ', :0'1 ty-50'3

0"3 0-50' , OH1"5Plan showingpoints for which coefficients are in tables 9,' to 9'6

SK 9140 Sketches to be used in conjunction with Tables 9.1-9.6.


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43 6 Reinforced Concreter- -l 1- -- --,I.Sty-1I131y- I H---l'-'t-----'--'t--+-r-M-+---1-+-""-iC I2

1.0LY -1 IO.7 ly - I . ..-+--;-+------


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Step-by-step analysis procedureStep J: Determine value of a :- (see SK9/40 and find (.)iy)'Step 2: Assume d is the thickness of slab and h is the dimension of a side ofa square column.Step 3: From available L, as shown in SK9/40, determine S = a:-d3Uh4.Step 4: Select a point of interest from SK 9/40 or SK 9/41 where themoments have to be found. Corresponding to the zone of influence, findappropriate stiffness connection factor K depending on S from Graphs 9.1to 9.18.Step 5: Find moment coefficients from Tables 9.1 to 9.3 corresponding tolxll. and the location of the point of interest.Step 6: Find the ultimate uniformly distributed load on the Hat slabnkN/m2Step 7: Find moment triads:

M .r = nCxKxl~ kNm/mMy = nCyKyl;kNm/m

Mxy = nCxyKxyl;kNm/mStep 8: Carry out Wood-Armer combination.Step 9: Find column reactions corresponding to ["fly (see SK 9/41 forcolumn locations).Step 10: The moments obtained using these coefficients are in kNm/m.Find the effective width be as in Section 9.3.4. Multiply the momentobtained by analysis at edge and corner columns with the effective widthbe to find the slab-to-column connection moment. This transfer momentshould be less than Mt.max as defined in Section 9.3.4.Reaction at column C. = nCtlxlvReaction at column C2 = nC2lxly, etc.

Note: If in the zone of KA (as shown in SK 9/41) the point of interest is located,then to find Mx use stiffness correction factor KAX corresponding to Sandlx/l. as in Graph 19.1. Similarly if the point of interest lies in the zone ofKF then to find Mx use KFX as in Graph 9.16.The benefits of using these tables and graphs are that the analysis can bedone very quickly and the necessity of carrying out the two analyses forthe two orthogonal directions may be avoided. These tables can also beused for the analyses of raft foundation where the loading n may beassumed to be uniformly distributed over an inverted Hat slab. The totalloads from a structure will be assumed uniformly distributed at the under-side of the raft.Step Jl: Calculate total column moments using Table 9.7 and divide thetotal moment between the columns at the junction depending on theirrelative stiffness. The stiffer the column, the more moment it will carry.The stiffness of a column may be calculated as III where J is the moment ofinertia and I is the effective height.


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43 8 Reinforced Concrete

- : -o

-~-.o

~ : : J : ~o n 0 0 < " l8~80001 1

< " l 00 0'>< " l o n o n..0 M....-I8;!;8000+1

; ; s [ ; o ; ; ;- 0 '> o n8380001 1

r - - . r - - 0 < " l> 0 " 1 0> 0 r - - . . . . .- VI"'"000000

1 1

~:!:::r - - o > o~88000+100 0'> v- - : 1 " --. N" 1 In .....000d o c i1 1

0'> 10 '>V',i ...... -o V ) ' < t888o d o1 1

o n a - . a - .- 1\ -~s~000c i o o1 1

-"o

II- s

. . . . .VI-\'


50/68


.:. . . CI=: .~ ' " ' " '" '" '" '" . '") U r ,J u"'r ,J . s - s - s \. S u"'u" \. S u"u'" r,J u"'r ,J u"'u"'0' - s u"'0'Elfo Q)~8u~~~ . . . . ~~ .,.,0.,.., O~N g-"" co",- . , . , \ 0 \ 0{ " - " < t O O {"-CO- . : l - O-N~ - - on '0~ \Oto-on-" to-to-.... ~""'0 . , ., 0 \ 0 NOO ~ > O " ' - .....--~ co 00..,._'< 888 N on ' - 0 >0 0 t- .....,.,o- 8 0.,.,("1 ~o ......< ': 588 8~8 oSo g~5 888000 060 000 odd ddd dod dddI I I I I + I I I I I I I I~ " " ' O co 0 Q\ ~ . . , . . . . . . ~>O~ 0.,..,0 > O N I I " l 8 " " ' 1 " " -I" l~ >0 ON o\ON < " ' 1 - . : 1 - 0 \ 1"10->02 i 1938 S-....~~ t-rf")_..j ~~8 288 .,.,.,.,")00 000 -00. . . . . 000 000 doc doo odo 000 doo+ I I I + + + I + I + I ++

N < " ' 1 " " ' - . : 1 - ~N-.:I- 8~~ ~~~ ~~El 0\ 0 .....~cgl;;-O\I""- 0 o..,.{"- gOO o-co.,.., Nto-..,. - . : 1 - . . . . .0 : ; J- t0-8 1"1....8, . , ~ 8 8 88 028 0 0 I I " l 0-~ 0 00 -0 II")!I::: dciciddd ddo doo ood ocid cidci r-~. . . . . I I I I I + I I I I I I I I M0 I I " lC " < tlO l l ot--r- 00 1"1.....NN..., 0.,.,0 ~t->O '-ON o-r-


51/68


-'~-


52/68


53/68


_'0-. 0\01'--8~;go000I I

or-nVi""'o-.t-gNOViV0\1'-00- >0-~:g80001 I

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P lA N O f FLA T S LABSK 9/42 Column number identification.

~'1- - - t t J U M yC C l .l . t 1 N ! 1J M E N T l O E N T I F l ( A T I O Nl a v er ag e s to re y heightS E C T IO N T H R O U G H S L A B

Table 9.7 Bending moment coefficient for design of columns in flat slab construction.t.u, 1.2 1.4 1.6 1.8 2 Moment

coefficientsColumnno.

0.01175 0.01345 Om526 0.01717 0.01916 0.02123 c;I -0.01175 -0.01792 -0.02594 -0.03603 -0.04836 -0.06307 C "l2 -0.00244 -0.00233 -0.00224 -0.00218 -0.00216 -0.00218 Cc y2 -0.02280 -0.03271 -0.04525 -0.06073 -0.07939 -0.10147 ee l3 0.02280 0.02765 0.03301 -0.03885 0.04513 0.05184 Ccy3 0.00244 0.00441 0.00706 0.01040 0.01445 0.01919 C ez4 -0.00376 -0.00393 -0.00418 -0.00454 -0.00506 -0.00571 C "y4 OJK>36 0.(10659 0.01028 0.01484 0.02031 0.02667 ee lColumn total moment =Cc x R x l~ x ljwhere lj""oad per unit area on slab.See Graphs 9.19-9.26 for factors R ; and Rz ; .

Note: Divide total moment in column to top and bottom column in proportion totheir stiffness.


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Design of Flat Slabs 44 5N//V////

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Design of Flat Slabs 44 9

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Chapter-9 Flat Slab

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