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3 3 Direct Design Method TW Slabs                         Prof Dr Bayan Salim 

Direct Design Method

Limitations of the DDM: 1. There must be a minimum of three continuous spans in each direction.

Thus, a nine panel structure (3 by 3) is the smallest that can be considered.

2. Rectangular panels must have a long-span/short-span ratio that is not greater than 2.

3. Successive span lengths in each direction shall not differ by more than one-third of the longer span.

4. Columns may be offset from the basic rectangular grid of the building by up to 10% of the span parallel to the offset.

5. All loads must be due to gravity only and uniformly distributed over an entire panel. The service live load shall not exceed two times the service dead load.

6. For a panel with beams between supports on all sides, the relative stiffness of the beams in the two perpendicular directions shall not be less than 0.2 or greater than 5:

0.2 ≤ {αf1 l22 / αf2 l1

2} ≤ 5

Subscription 1 refers to direction 1

Subscription 2 refers to direction 2 (┴ dir 1)

DDM:

1. Determine Mo, the total static bending moment in a panel 2. Moments assigned to critical sections 3. Lateral distribution of moments (to cs, ms, and beams if any) 4. Moments assigned to columns 5. Shear in slab systems with beams

1. Determine Mo in panels

The total static BM in a span, for a strip bounded laterally by the centerline of the panel on each side of the centerline of supports, is

Mo= wu l2 ln2 / 8 (ACI 13.1)

3 Dire

Whe

wu = l2 = tspanedge ln =

ect Design Me

ere;

= 1.2 D + 1

the span (cn adjacent ae of slab to

clear span ≥ 0.65 l1

This fig.defines t

ethod TW Slab

.6 L (kN/

centerline-tand paralle panel cen

in the dire1 , see fig.

shows crithe face of

bs                    

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2. M

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Mom

ethod TW Slab

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end spans ments are as

bs                    

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M−

M+

ssigned acc

M

     Prof Dr Ba

sections

− = 0.65 M+ = 0.35 M

cording to

M = (facto

ayan Salim 

Mo

Mo

ACI Tabl

or from T

(A

(A

le 13.6.3.3

Table)× M

ACI 13.2)ACI 13.3)

3:

Mo

5 

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3 Dire

3. La

b

c

ect Design Me

a) Ext eb) Slab c) Slab d) Flat pe) Exter

Lateral Disa. Interior

Column

b. ExteriorColumn

. PositiveColumn

Linear int

ethod TW Slab

edge unrestrawith beams w/o beams, plate with edrior edge ful

stributionr M− :

strip shall

r M− : strip shall

e M+: strip shall

terpolation s

bs                    

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of Momen

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     Prof Dr Ba

a masonry wsupports

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n.

or M−:

or M−:

ve M+:

6 

3 Direect Design Me

Parameresistancparamet

x = smalSlab moConstanas the la1. A po

brack2. The p

below3. The t

βt = 0wall.

βt = 2

ethod TW Slab

eter βt: Thece of the efer βt , defin

ller dimens

oment of innt C = torsiargest of thrtion of the

ket, or capiportion in w the slab transverse

0 for flat p

2.5 for RC

bs                    

e relative rffective traned as

sion, y = lnertia, Is = lonal rigidie followine slab haviital in the d1 plus that

beam as sh

plate w/o e

C wall.

     Prof Dr Ba

restraint pransverse ed

arger dimel2 h

3 /12 ity of beamg: ing a widthdirection inpart of any

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7 

8 3 Direct Design Method TW Slabs                         Prof Dr Bayan Salim 

Two-half Middle strip shall take the moments not assigned to column strip. An exception to this is a middle strip adjacent to and parallel with an edge supported by a wall, where the moment to be resisted is twice the factored moment assigned to the half middle strip corresponding to the first row of interior supports (ACI 13.6.6.3). Mms (ext strip supp by wall) = 2 Mms (neighbor int strip)

Moments to beams: (ACI 13.6.5) When (αf1l2 / l1) ≥1.0, Mbeam = 85 % Mcs When 0 ≤ (αf1l2 / l1) ≤ 1.0, 0 ≤ Mbeam ≤ 85 % Mcs ; Mbeam is found by linear interpolation In addition, the beam section must resist the effects of loads applied directly to the beam, including weight of beam stem projecting above or below the slab. Moments to columns and walls: (ACI 13.6.9) Columns and walls built integrally with a slab system shall resist moments caused by factored loads on the slab system. At interior supports, Mcol = 0.07[(1.2 D + 0.5×1.6 L)l2ln

2 – 1.2 D′l′2(l′n)2] ACI (13-7) Where ( ′ ) refers to shorter span. At exterior column or wall supports, the total exterior negative factored moment from the slab system (Table 13.6.3.3) is transferred directly to the supporting members. Columns above and below slab shall resist moments in direct proportion of their stiffnesses; Mcol above = [Kcol above / (Kcol above + Kcol below)] Mcol , where Kcol = 4 EIcol / lcol

3 Dire

SheaBeamtribufrompara

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ar in Slab ms with (αutary areas,m the corne

llel to the l

ddition, bectly on bea

nforcemen Reinforc

parallel have an dinner steel

For TW For beam

Straight bent up wsteel.

Max bar Min cov As,min (fo

0. 0. 0.

Bars cut

ethod TW Slab

systems wαf1l2 / l1) ≥ 1, (shown sh

ers of the plong sides.

ams shall rams.

nt for TW cement in Tto the sideeffective d= douter steel

slabs withmless TW bars are gewhere no l

r spacing =ver = 20 mmor shrinkag002 bh 0018 bh 0018 (420toff, see fig

bs                    

with Beam1.0, shall haded in fianels and .

resist shear

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depth 1 db l− db

h beams, shslabs, longenerally uslonger need

= 2 h (to enm ge and temp ….. ….. / fy) bh …gure below

     Prof Dr Ba

ms (ACI 13resist shea

ig.), which the center

rs caused b

CI 13.3) is placed inanels. This less than th

hort dir barg dir bars hsed, althouded, to pro

nsure crack

mperature c for fy

for fy

… for fy

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ayan Salim 

3.6.8) ar caused b

h are boundrlines of the

by factored

n an orthogwill cause

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rs have thehave the larugh in somovide for p

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crack contry = 280 andy = 420 MP

fy > 420 MPless slabs:

by factoredded by45o le adjacent

d loads app

gonal grid,e the inner eel.

e larger d. rger d. e cases +vart or all th

rol) d 350 MPaPa Pa

d loads on lines drawnpanels

plied

, with bars steel will

veM steel ishe –veM

a

9 

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3 Dire

ExamUse NS dGiveEdgeInterColuSlabc.c. sc.c. sf′c = Serv Solu Checin cha

ect Design Me

mple: Dirthe DDM

direction (sen; e beam dimrior beam dumn dims: b thickness spans in Nspans in E-28 MPa, f

vice live loa

ution:

ck slab thiapter 2.

ethod TW Slab

ect Designto determishaded des

ms: 350 × 6dims: 350 450 × 450 h = 150 m

N-S directio-W directiofy = 420 Mad L = 5 kN

ickness h =

bs                    

n Method;ne the desisign strip)

650 mm × 500 mm

0 mm mm on are 5.50on are 6.50

MPa N/m2

= 150 mm is

     Prof Dr Ba

Slab withign momenin an iterm

m 0 m

s satisfactory

ayan Salim 

h beams nts for the

mediate floo

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slab systemor.

ion control,

1

m in the

see example

10 

e

3 Dire

ChecLimitChecCalcuCornExterExterInteriThereFactAverwt. oTotalFactoTotaln = 5Mo= MomInterInteri End Exter Interi

Disrl2 / l1 Inter% to

ect Design Me

ck the limitations 1, 2,3k the ratio ulate αf for ber panel: (12rior panel NSrior panel EWior panel; (3efore, DDM tored loadsage wt. of bef slab = (0.1l D = 0.45 + ored load, wu

al static BM5.5 – 0.45 = 5wu l2 ln

2 / 8 =ments at crrior spans ior M− = 0.6 M+ = 0.3span rior M− = 0.1 M+ = 0.5

ior M− = 0.7

ribution to = 6.5 / 5.5 =

rior negativecol strip = 7

ethod TW Slab

itations of 3,4,5 are rea 0.2 ≤ {αf1

beams, see e2.46+3.98)6.S; (3.98+3.9W; (12.46+3.98+3.98)6.5is applicabl

s: eam stem = 5)(24 kN/m3

3.6 = 4.05 ku = 1.2(4.05)M, Mo: 5.05 m (fac= 12.86×6.5ritical secti

65 Mo = 0.655 Mo = 0.35

16 Mo = 0.1657 Mo = 0.5770 Mo = 0.70

col strip, m= 1.18, (αe moments 75 – [(1.18 –

bs                    

f the DDMdily satisfacl2

2 / αf2 l12} ≤

example in ch.52 / (14.48+

98)6.52 / (14.3.98)6.52 / (452 / (4.7+4.7le.

(0.35×0.35/63) = 3.6 kN/m

kN/m2 ) + 1.6(5) =

ce to face of ×5.052 / 8 = ions:

5×266.47 = 1×266.47 = 9

6×266.47 = 47×266.47 = 1×266.47 = 18

mid strip aαf1l2 / l1) = 3

– 1)/(2 – 1)] (

     Prof Dr Ba

| 6

M: (see page 3tory ≤ 5 for pahapter 2,

+4.7)5.52 = 1.48+4.7)5.52

4.7+4.7)5.52

7)5.52 = 1.18

6.5)(24 kN/mm2

12.86 kN/m2

f columns) 266.47 kNm

173.21 kNm93.26 kNm

42.64 kNm151.89 kNm86.53 kNm

and beams3.98×6.5/5.5

(75 – 45) = 6

ayan Salim 

6.5m |

3)

anels

.2, 0.2 < 2 = 0.58 OK= 2.44 OK

8 OK

m3) = 0.45 k

2

m

m

s = 4.7 ≥ 1.0

69.6%

6.5m |

1.2 < 5 OK

kN/m2

1

11 

3 Dire

InteriEnd sPosit% to InteriEnd sExter

Edge150)

C1 = =C2 = = Is = l2

1st int3rd in2nd in End s Sum

End SExt. nPositInt. nInteriNegaPosit* Be

ect Design Me

ior spans Mc

span, Interiotive momentcol strip = 7ior spans Mc

span Mcs+ =

rior negative

Beam dims= 850 mm

(1 – 0.63×35

= 3.995×109 +(1 – 0.63×35

= 6.138×109 +

l2 h3 /12 = 65

terpolation nterpolation nterpolation

span, exterio

mmary, and

Span negative ive

negative ior Span

ative ive eams shall ta

ethod TW Slab

cs− = 69.6% (

or Mcs− = 69.

ts 75 – [(1.18 –cs

+ = 69.6% (69.6% (151e moments

: bw = 350, H

50/500)3503

+ 8.50×108 =50/650)3503

+ 4.56×108 =

500(150)3/ 1

= C / 2Is =

@βt =0 @βt =1.8 @βt =2.5

or Mcs− = 78%

d moments

Mu (kNm)

42.64 151.89 186.53 173.21 93.26

ake 85% of M

bs                    

(173.21) = 1.6% (186.53

– 1)/(2 – 1)] ((93.26) = 64.89) = 105.7

H = 650, h =

3×500/3 + (1= 4.845×109

3×650/3 + (1= 6.594×109

2 = 1.828×1

6.594 / (2×1

100% 78% ← 6969.6%

% (42.64) =

s to mid str

) % to

78 69.669.6 69.669.6

Mcs [ (αf1l2 /

     Prof Dr Ba

120.55 kNm) = 129.82 k

(75 – 45) = 64.91 kNm 72 kNm

=150, then fla

– 0.63×150

9 mm3

– 0.63×1509 mm3 Use (l

09 mm3

1.828) = 1.80

9.6 + (100 –

33.26 kNm

rip and bea

o cs

6 6

6 6 l1) > 1.0], an

ayan Salim 

kNm

69.6%

|← 850 →|

ange b = bw

0/850)1503×8

0/500)1503×5larger C)

0

– 69.6)(2.5 –

m

ams

Mcs (kNm

33.26 105.72 129.82 120.55 64.91

nd slab Mcs s

+ (H – h) =

850/3

500/3

1.8)/(2.5 – 0

m) * 2-h(kN 9.346.56. 52.28.

shall take 15

1

350 + (650 –

0) =78%

half ms Mms

Nm)

38 .17 .71

.66

.35 % of it.

12 

–

13 3 Direct Design Method TW Slabs                         Prof Dr Bayan Salim 

Slab Reinforcement Column strip: SlabMcs (kNm)

15% of Mcs * R = Mu / φbd2

** ρ Table A.5 As mm2/m***

End Span Ext. negative 15%(33.26)= 4.99 0.129 0.0005 63 /use 270 Positive 15.86 0.410 0.0010 125 /use 270 Int. negative 19.47 0.503 0.0012 150 /use 270 Interior Span Negative 18.08 0.468 0.0011 138 /use 270 Positive 9.74 0.252 0.0006 75 /use 270 * Beams shall take 85% of Mcs [ (αf1l2 / l1) > 1.0], and slab Mcs shall take 15% of it. ** bcs = l1/2 = 2750 mm, d = 125 mm *** As = ρbd , As min = 0.0018bh =270 mm2/m, use No.13@300 (430 mm2/m)everywhere

max spacing = 2h = 2(150) = 300 mm OK Middle strip: 2-half ms Mms

(kNm) R = Mu / φbd2

* ρ Table A.5 As mm2/m**

End Span Ext. negative 9.38 0.178 0.0005 63 /use 270 Positive 46.17 0.875 0.0021 263 Int. negative 56.71 1.075 0.0026 325 Interior Span Negative 52.66 0.999 0.0024 300 Positive 28.35 0.538 0.0013 163 /use 270 * bms = l2 - bcs = 6500 - 2750 = 3750 mm, d = 125 mm

** As = ρbd , As min = 0.0018 bh =270 mm2/m, use No.13@300 (430 mm2/m)everywhere max spacing = 2h = 2(150) = 300 mm Moments to columns Interior columns Int. Mcol = 0.07[(1.2 D + 0.5×1.6 L)l2ln

2 – 1.2 D′l′2(l′n)2] ACI (13-7) = 0.07(0.5×1.6 L)l2ln

2 (adjacent spans are equal) = 0.07(0.5×1.6 ×5)6.5(5.05)2 = 46.41 kNm At exterior column, the total exterior negative factored moment from the slab system (Table 13.6.3.3) is transferred directly to the column: Ext. Mcol = 0.16 Mo = 0.16×266.47 = 42.64 kNm Columns above and below slab shall resist moments in proportion of their stiffnesses; Mcol above = [Kcol above / (Kcol above + Kcol below)] Mcol Mcol above = Mcol below = (1/2) Mcol (columns above and below are identical) = (1/2)(46.41) = 23.21 kNm for interior columns = (1/2)(42.64) = 21.32 kNm for exterior columns

3 Dire

Sheaedge NS b EW b

HW:Use tdirectNo Ec.c. sc.c. sf′c = 2ServiPartitStory

ect Design Me

ar in Beambeams) eams (short

beams (long

W: DDM of the DDM to tion (shaded

Edge beams, pans in N-S pans in E-W28 MPa, fy =ice live load tion weight =y height = 3.5

ethod TW Slab

ms (only inte

dir.): Vu = w dir.): Vu = w = 1

f Flat Platedetermine th

d design stripcolumn dimdirection ar

W direction ar= 420 MPa L = 2 kN/m

= 1 kN/m2

5 m

bs                    

erior beams

wu [(1/2)l1l1/2

wu [(1/4)l12+

2.86[(1/4)5.

l1 = 5

e Floor he design mop) in an iterm

ms: 400 × 400re 5.50 m re 4.25 m

m2

E ↔

     Prof Dr Ba

are checked

2] = wu [(1/4 = 12.86[(1(1/2)l1(l2 – l52 + (1/2)5.5

5.5 m ↕

oments for thmediate floor0 mm

NW ↕ S

ayan Salim 

d because the

4)l12]

1/4)5.52] = 9l1)] 5(6.5 – 5.5)]

l2 = 6.5 m

the flat plate r. Given;

e carry high

7.3 kN

= 132.6 kN

m↔

slab system

1

er shear tha

N

m in the NS

14 

an