5 Equ

ChaSys

IntrThe DmoreThe E By tand

Fig EachThisside For colu

ivalent Frame

apter 5:stems

roductionDDM for TW general met

EFM is base

the EFM, theextending bo

ure shows

h frame is cos slab-beam iof the colum

vertical loadumns assume

e Method TW

: Equiva

n W slabs is usthod is needd on the ana

e structure isoth longitud

equivalent

omposed of cincludes the mns.

ding, each floed fixed at th

W Slabs            

alent Fr

seful if each ed, namely t

alysis by the

s divided intodinally and tr

t frame in

columns andportion of th

oor with its che floors abo

             Prof D

ame Me

of the six limthe Equivalemoment dis

o continuousransversely,

a 5-story b

d a broad conhe slab boun

columns maove and below

Dr Bayan Salim

ethod of

mitations is sent Frame Mstribution m

s frames cenas shown:

building.

ntinuous beanded by pane

ay be analyzew:

m 

f TW Sla

satisfied, othMethod (EFMethod.

ntered on col

am, named sel centerline

ed separately

ab

herwise, a M).

lumn lines

slab-beam. es on either

y, with the

1 

5 Equ

The later

ivalent Frame

fig below shral division i

e Method TW

hows the divinto column

W Slabs            

vision of builstrips and m

             Prof D

lding into ‘emiddle strips,

Dr Bayan Salim

equivalent fr, as in the DD

m 

ames’. NoteDM.

e the same

2 

3 5 Equivalent Frame Method TW Slabs                         Prof Dr Bayan Salim 

The Slab-beam Stiffnes: Ksb

The slab beam moment of inertia Isb is uniform thru the span but it changes at the support and drop panels if any. According to ACI 8.11.3; Isb,(inside support) = Isb /(1 – c2/l2)

2

c2 = size of column or capital in direction of l2 l2 = panel width Analysis by the moment distribution method needs stiffness factors k, carry-over factors COF, and fixed-end moment factors M. These are given in Table A13a for slabs w/o drop panels and Table A13b for slabs with drop panels:

Ksb = k (EIsb / l1) Fixed-end Moment = (factorM)(qu l2 l1

2)

5 Equivalent Framee Method TWW Slabs                         Prof DDr Bayan Salimm 4 

5 Equivalent Framee Method TWW Slabs                         Prof DDr Bayan Salimm 5 

5 Equ

TheThe are tcent

The the s

Kc =kc i

ivalent Frame

e Equivalecolumns are

transverse toterlines.

principle is sum of the fl

= kc (EIc is given in

e Method TW

ent Colume consideredo the directio

that the totalexibilities o

/ lc) Table A13

W Slabs            

mn Stiffnesd to be attachon of the spa

al flexibility of the actual

3c :

             Prof D

ss: Kec hed to the slan. The torsio

(inverse of scolumn and

Dr Bayan Salim

ab-beam by tonal membe

stiffness) of d the torsion

m 

torsional mer extends to

the equivalenal member;

embers that the panel

ent column i

6 

is

5 Equ

Tor

If a Kt ( C =

ivalent Frame

rsinal Mem

beam para

(with parallel

= cross-sec

e Method TW

mber Stiff

allel to dire

beam) = K

torsional c

W Slabs            

fness: Kt

ection l1 is

Kt (Is,with be

constant, a

             Prof D

contained

eam / Is,w/o

s discussed

Dr Bayan Salim

d in the pan

o beam)

d in DDM

m 

nel, Kt is in

:

ncreased:

7 

5 Equ

Fig a

ivalent Frame

above show

e Method TW

ws the tors

W Slabs            

ional mem

             Prof D

mbers (attach

Dr Bayan Salim

hing transve

m 

ersely to slabb-beam joint

8 

t)

5 Equ

MomWith the eq

1

2

3

ivalent Frame

ments Anathe slab-bea

quivalent fra. When ser

sections w2. When ser

Thpa

Thon

. Factored mload (D +

e Method TW

alysis am and equivame can procrvice L ≤ (3/with full factrvice L > (3/he max Mu+anel and on ahe max Mu−n adjacent pamoments sh

+ L) on all pa

W Slabs            

valent columceed by the m/4) D, the matored load (D/4) D, + near midspalternate pan− in the slabanels only. all be taken

anels (case 1

             Prof D

mn stiffnessemoment distaximum factD + L) on en

pan occurs wnels; and b at a suppor

not less than above).

Dr Bayan Salim

es (Ksb and Ktribution metored momen

ntire slab sys

with factored

rt occurs wit

n those occu

m 

Kec) found, thethod. nts shall occstem.

d (D + 3/4 L

th factored

urring with fu

he analysis o

cur at all

L) on the

(D + 3/4 L

ull factored

9 

of

L)

5 Equ

Proc123

456

         

78

ivalent Frame

cedure (su. Determin. Determin. Determin

Slab-beam Equivalen

. Determin

. Perform t

. Find desig

                Face of eq

. Distribute

. Design th

e Method TW

ummary) e slab-beame equivalente slab-beam

m DF: D

D

nt column DF D

e COF and fhe moment gn moments

quivalent sq

e design momhe reinforcem

W Slabs            

m stiffness Ks

t column stifm distribution

DFsb 2-1 = K

DFsb 2-3 = K

F: DFec = Kec /

fixed-end mdistribution: M− at face

uare is cons

ments to Mcs

ment.

             Prof D

b ffness Kec n factors DFs

Kb1 / [Kb1 +

Kb2 / [Kb1 +

/ [Kb1 + Kb

oments FEM method to d

e of supports

sidered as th

s and Mms , a

Dr Bayan Salim

Fsb

+ Kb2 + Kec

+ Kb2 + Kec

b2 + Kec]

M = (factorMdetermine jos, and M+ at

he location of

according to

m 

]

]

M) wul2l12

oint momentt midspans.

f the critical

DDM.

1

s

l design M−

10 

   

5 Equ

ExamUse tdirectNo Ec.c. sc.c. sf′c = 2ServiStorySolufind

For flis (seExt pTry 1

ChecUse aFactoFactoTotalShearVu = Vc = = (1/3

ivalent Frame

mple: Equthe EFM to dtion (shaded

Edge beams, pans in N-S pans in E-W28 MPa (slaice live load y height = 3.5ution:

slab thick

flat plate slabee ch.2): panel w/o edg170 mm slab

ck thickneaverage effecored dead loaored live loadl factored loar strength at 9.32 [(5.5×4(1/3)√f′c bod3)√28 (4×53

e Method TW

uivalent Fdetermine thd design stripcolumn dimdirection ar

W direction arab), f′c = 42

L = 2 kN/m5 m

kness h sati

b systems, th

ge beam h =b for all pane

ess for TW ctive depth dad, 1.2D = 1d, 1.6L = 1.6ad = 9.32 kNd/2 distance

4.25) – (0.53d (for square37)137 /1000

W Slabs            

rame Methe design mop) in an iterm

ms: 500 × 500re 5.50 m re 4.25 m MPa (colum

m2 , Partition

isfactory for

he minimum

= ln / 30 = 51els (weight =

W shear d = 137 (20m.2 (4.1 + 1)

6 × 2 = 3.2 kN/m2

e around the 37)2] = 215.2e interior col0 = 519 kN

             Prof D

thod; Slaboments for thmediate floor0 mm

mns), fy = 42weight = 1 k

r deflection c | 4

overall thick

100 / 30 = 17= 4.1 kN/m2)

mm cover an= 6.12 kN/m

kN/m2

support is c2 kN lumn) Eq. (2

Dr Bayan Salim

b with beamhe flat plate sr. Given;

20 MPa kN/m2

control. 4.25m | 4.2

kness h with

70 mm > 125)

nd No.13 barm2

computed as

22.6.5.2 a)

m 

ms system in th

25m |

h Grade 60 r

5 mm

r)

follows:

1

he NS

reinforcemen

11 

nt

5 Equ

φVc =after

FlexKsb =Tablec1A/l1

FromKsb 1Ksb 2 FlexKc = TableFromKc, be Tors

| 5 C = (Kt = 9 Equ

1/ Kec

on eaKec = SlabAt exDF =At inDF =

ivalent Frame

= 0.75 ×519 finding mom

xural stiffn= k EIsb / l1 e A13a; ext j1 = 500 / 550

m Table: MAB

-2 = Ksb 2-1 -3 = Ksb 3-2

xural stiffnk EIc / lc e A13c; lc = l

m Table: kAB elow = Kc, ab

sional stiff

500mm |

(1 – 170/5009(4700√28 C

ivalent col

c = 1 / 2×20.ach side) = 6.86 × 1010

b-beam joinxterior joint, = 3.29 /(3.29terior joint,

= 3.29 /(3.29

e Method TW

= 389.3 kN ments)

ness of slab

joint and int 00 = 0.091 saB = 0.085, kA

= 4.18 [470 = 3.29 × 10

ness of colu

l1 = 3.5 m, c= 4.43 bove = 4.43

fness of tor

0)(170)3500/3C) / 4250(1 –

lumn stiffn

08 + 1 / 2×4

0 Nmm

nt distribu

+ 6.86) = 0.

+ 3.29 + 6.8

W Slabs            

> Vu O.K. (

b-beams at

joint: ay 0.1 = c1B /AB = 4.18, CO00√28 × 4250010 Nmm

umn memb

1A = 0.17/2

[4700√42 ×

rsional me

3 = 5.4× 108

– 500/4250)3

ness Kec

4.14 (2 colu

tion factor

.324

86) = 0.245

             Prof D

(Effects of m

t both ends

/ l1 OF = 0.5130(170)3/12]

bers at bot

= 0.085 m, c

500(500)3/1

embers, Kt

8 mm4 3 = 4.14 × 10

umns above

rs DF.

Dr Bayan Salim

moment trans

ds, Ksb

/ 5500 = 3.2

th ends, Kc

c1A/l1 = 0.08

12] / 3500 =

010 Nmm

and below, a

m 

sfer may be

29 × 1010 Nm

Kc

85 / 3.5 = 0.0

= 20.08 × 101

and 2 torsion

1

checked late

mm

02

10 Nmm

nal members

12 

er

s

5 Equ

FramL / DsectioFactoFactoFactoFactoFEM= 0.0 MomCoun

Jo

MemD

COFEMDM CO DM CO DM Σ M

MCL

Mid sMCL wherPosit+MCL

Posit+MCL

SheaV (supV (lef V (rig ShearV (lef

ivalent Frame

me AnalysD = 2 / 5.1 = ons with fullored load andored dead loaored live loadored load qu =

M’s for slab-b85 (9.32) (4

ment distribnterclockwis

oint mber 1

DF 0.OF 0.

M 101- 33

2.1- 0.670.2

span momen(midspan) =e Mo is the mive moment L 1-2 = (9.32×ive moment L 2-3 = (9.32×

ars: End spanupport) = qul2

ft support) = = =

ght support) = =

rs: Interior sft support) = =

e Method TW

sis 0.39 < 0.75 l factored load fixed-end mad 1.2D = 1.d 1.6L = 1.6= D + L = 9

beams = (fac.25)(5.5)2 =

bution e moments a

1

1-2 .324 0.513 0.85 .0

- 10 -16 4.

3 69

0.5

29 -11

57.78

nts (MCL ) ar= Mo – (MuL moment at min span 1-2:

×4.25) 5.52/8in span 2-3:

×4.25) 5.52/8

n 1-2 2l1 / 2 ± (MR

qul2l1 / 2 − (M9.32(4.25)(5102.99 – 7.9= qul2l1 / 2 += 102.99 + 7pan 2-3 qul2l1 / 2 − (M9.32(4.25)(5

W Slabs            

OK, designad on all spamoments. 2 (5.1) = 6.1 (2) = 3.2 kN

9.32 kN/m2 torM) qul2l1

2

101.85 kNm

acting on the

2 2-1

0.245 0.513 01.85 10

.93 .15

4.

52 - 2 0

4.11 10

re determine+ MuR)/2

midspan for a: 8 – (70.29 + : 8 – (104.39 +

R – ML) / l1

MR – ML) / l5.5)/2 – (11497 = 95.02 k+ (MR – ML) /7.97 = 110.9

MR – ML) / l5.5)/2 – (104

             Prof D

n moments arans. (case 1 o

12 kN/m2 N/m2

2 m

e member en

2-3 0.245 0.513

01.85 -

15 -2.13 .52 -

04.39 -

45.3

ed from the f

a simple bea

114.11) /2 =

+ 104.39) / 2

l1 4.11 – 70.29)kN / l1

96 kN

l1 4.39 – 104.39

Dr Bayan Salim

re assumed tonly)

nds are taken

3

3-2 0.245 0.513

- 101.85

-4.15 2.13

- 0.52 -104.39

39

following eq

am.

= 57.78 kNm

2 = 45.39 kN

)/5.5

9)/5.5

m 

to occur at a

n as positive

3 3-4

0.245 0.513

101.85

16.93 -4.15 - 0.52 114.11

5

quation:

m

Nm

1

all critical

.

4 4-3

0.3240.513

-101.85 + 33.0

- 2.13 0.69 -70.29

7.78

13 

14 5 Equivalent Frame Method TW Slabs                         Prof Dr Bayan Salim 

= 102.99 – 0 = 102.99 kN V (right support) = qul2l1 / 2 + (MR – ML) / l1 = 102.99 + 0 = 102.99 kN Design –ve Moment = Joint moment – area of shear diagram bet support CL and face At joint 1, Design M− = 70.29 – (95.02 kN × 0.25 m) = 46.54 kNm At joint 2 left, Design M− = 114.11 – (110.96 kN × 0.25 m) = 86.37 kNm At joint 2 right, Design M− = 104.39 – (102.99 kN × 0.25 m) = 78.64 kNm At joint 3 left, Design M− = 78.64 kNm

Moments at critical sections (summary): Interior spans Interior M− = 78.64 kN, M+ = 45.39 kNm End span Exterior M− = 46.54 kNm, M+ = 57.78 kNm, Interior M− = 86.37 kNm

Disribution to col strip, mid strip l2 / l1 = 4.25 / 5.5 = 0.77, (αf1l2 / l1) = 0 Mu (kNm) % to cs Mcs (kNm) * 2-half ms Mms

(kNm) End Span Ext. negative 46.54 100 46.54 0 Positive 57.78 60 34.67 23.11 Int. negative 86.37 75 64.78 21.59 Interior Span Negative 78.64 75 58.98 19.66 Positive 45.39 60 27.23 17.81 Slab Reinforcement Column strip: SlabMcs

(kNm)

R = Mu / φbd2 *

ρ Table A.5 As mm2/m** provide

End Span Ext. negative 46.54 1.157 0.0028 406 No.13@300

(430 mm2/m)

Positive 34.67 0.863 0.0021 305 / use 306 No.13@300 (430 mm2/m)

Int. negative 64.78 1.612 0.0040 580 No.13@200 (645 mm2/m)

Interior Span Negative 58.98 1.467 0.0036 522 No.13@200

645 mm2/m)

Positive 27.23 0.678 0.0017 247 / use 306 No.13@300 (430 mm2/m)

* bcs = l2/2 = 2125 mm, d = 145 mm ** As = ρbd , As min = 0.0018bh =306 mm2/m, use No.13@300 (430 mm2/m)

max spacing = 2h = 2(150) = 300 mm OK

5 Equ

Midd

End SExt. n

PositInt. nInteriNegaPosit * b

** A Effe

ɣf = 1Add AfacesR = ɣTable Effeɣv = 1ɣvMu

Ac = (J/c =

φvc = MomInterAt exColumMcol a

Mcol a

ivalent Frame

dle strip: 2M

Span negative 0

ive 2negative 2ior Span

ative 1ive 1bms = l2 - bcs

As = ρbd , As m

max spacin

ect of ɣfMu

1 / [1 + (2/3)As to be dist of the columɣfMu / φbd2 =e A5; ρ = 0.0

ect of ɣvMu1 - ɣf = 1 – 0 = 0.386(46.(2b1 + b2)d =4.496×107

= φ(1/3)√f′c =

ments to corior columnsxterior colummns above aabove = [Kcol a

above = Mcol b

e Method TW

2-half ms Mms (kNm)

0

23.11 21.22

19.31 17.81 = 4250 - 21

min = 0.0018 ng = 2h = 2(

at exterior

b1= c + )√(0.89] = 0.tributed overmn. = 0.614(46.50038, As = 0.

at exterior0.614 = 0.3854) = 17.96 = 238525 m

= 1.323 MPa

olumns s, unbalancemn, Mcol = 7and below slabove / (Kcol ab

below = (1/2) = (1/2)( = (1/2)(

W Slabs            

R = Mu /φbd2 *

0

0.575 0.537 0.489 0.443

25 = 2125 mbh =306 mm170) = 340 m

r edge

d/2 = 572.5.614 r effective w

54×106)/[0.9.0038×1010

r edge 6 kNm m2

= 215.2×10= 0.902 + 0

a > vu = 1.30

ed M = Mcol

70.29 kNm ab shall resibove + Kcol bel

Mcol (colu(9.72) = 4.86(70.29) = 35

             Prof D

/ ρ Tabl

0

0.00140.0014 0.00110.0011

mm, d = 145m2/m, use Nmm

mm, b2 = c

width = c2 + 3

9×1010×145×145 = 557

3/238525 +

0.399 = 1.301 MPa. Shea

= 114.11 –

st moments ow)] Mcol

umns above a6 kNm for in

5.15 kNm for

Dr Bayan Salim

e A.5 As

0 /

4 204 20

1 161 16mm

No.13@300 (

c + d = 645 m

3 h = 1010 m

2] = 1.495 mm2 (add to

17.96×106/401 MPa ar strength is

104.39 = 9.7

in proportio

and below anterior columr exterior co

m 

s mm2/m**

/use 306

03/use 306 03/use 306

60/ use 306 60/ use 306

(430 mm2/m)

mm

mm outside o

op bars 5 No

4.496×107

s OK

72 kNm

on of their sti

are identical)mns olumns

1

Provide

No.13@300 (430 mm2/m)

ditto ditto ditto ditto

)

opposite

o 13)

iffnesses;

)

15 

)

5 Equ

HW:Use tNS dEdge InterioColumSlab tc.c. spc.c. spf′c = 2Servic

ivalent Frame

W: EFM of the EFM to direction (shabeam dims: 3or beam dimsmn dims: 450thickness h =pans in N-S dpans in E-W d28 MPa, fy = ce live load L

e Method TW

TW Floor determine thaded design 350 × 650 mms: 350 × 500 m × 450 mm

= 150 mm direction are 5direction are 6420 MPa

L = 7.5 kN/m2

W Slabs            

with Beamhe design mostrip) in an i

m mm

5.50 m 6.50 m

2, Story heigh

             Prof D

ms oments for thitermediate f

ht = 3.5 m

Dr Bayan Salim

he TW slab sfloor. Given

m 

system with n;

1

beams in th

16 

e