APPENDIX A

STUDIES ON SIMPLY SUPPORTED ONE WAY SLABS

A.I Computation of Ductility Factor

A.I.I Model calculation for the specimen SOSCll

Ductility factor = Deflection at ultimate load (Bu) / Deflection at yield load (By),ll =Ol/lOy (A.l)

Ultimate load, p" =23.78kN

Deflection at ultimate load, 01/ =32.00mm

Yield load Py is calculated as given below:

hk =42.15N / mm2

1;, =443N / mm2

Breadth of section, b =1000mmd =D -15 =60 -15 =45mm

A / =8x 1r X 62 =226. 194mm2

s 4

Esm=­Ec

2.05 x 105 2.05 X 105

=5000 X ~fCk =5000x .J42.l5 == 6.315

bXxx~=mAs/Cd -x)

1000 x xx x = 6.315x 226.194x (45 -x)2

=> x= 9.99mm

bx3

d)2I =-+mA./( -xcr 3 ..

1000 x 9.9993

= 3 +6.315 x 226.194x (45 -9.999)2 =2.08x 106 mm4

/, = I" =m My (d -x)y Js I

cr

M443 =6.315x y 6 x(45-9.999)

2.08xl0

(A.2)

CA.3)

(A.4)

(A.S)

183

:::=> My =4168821.0S6Nmm =4.l7kNm

P 12

M =-y-y 8

P =4.17x8 =14.826kN/m 2

y 1.S 2

Py =14.826 x 1.5 xl =22.24kN

Py =22.24kN

Deflection at yield load, 5y == 19.6Sm

Therefore ductility factor, J.l = 5u =1.635y

A.2 Computation of First Crack Load

A.2.1 Model calculation for SOSCll

(a) By IS 456:2000

fek =42.lSN / mm 2

Overall depth, D =60mmD

y=-=30mm2

Gross moment of inertia, I gr = bD3

= 18000000mm4

12

Modulus of rupture, fer =0.7~fek

= O.7x.J42.1S = 4.S44N /mm 2

1= l.Sm

M = feJgrer

y

= 4.S44x 18000000 =2726767Nmm =2.726kNm30

M = Perl2

er 8

=8x 2.726 =9 692kN/ 2:::::> Per 2 • m1.S

First crack load, Per = Per X 1X 1.S = 14.54kN

(A.6)

(A.7)

(A.8)

(A.9)

(A.10)

(A.lI)

184

(b) By ACI 318 :2008

fck = 42.15N I mm2

I gr = 18000000mm4

[= 1.5my=30mm

Cylinder strength, f: = 0.8fek = 0.8x42.15 = 33.72N Imm 2

Modulus of rupture, fer = 0.62..J7: = 0.62 X .J33.72 = 3.60

C k· M t M leJgr 3.60x18000000 216'-'AT.rac mg omen, er =-= = . IWvmy 30

=8Mer =8x2.16=77kNI 2Per [2 1.5 2 • m

First crack load, Py =wx 1x 1.5 = 11.55kN

A.3 Computation of Theoretical Ultimate Moment

A.3.t By Henager and Doherty method adopted by ACI (Fig. 2.10)

(a) Model calculation of slabs with fibre for the specimen SOSC21

lek =47.24N I mm 2

For fibre reinforced concrete, cylinder strength f'e = 0.91ek = 42.516N Imm2

AS1 =226.19mm 2

f y =443N I mm2

Bond efficiency of fibre, Fbe =1

Aspect ratio of fibre, AJ =50

• •• O"J 2'rd Fbe AJTenstle stram m steel fibre, &s(jibres) =- =------"-Es Es

=2x 2.3x Ix 50 =0.001152.05x 105

333x 0,454x 9.81Bond stress, 'rd = 333psi = - = 2.3N I mm2

25,4x25,4

0.003 &s(jibres)--=-....,;.;..--'--C e-c

e =1.383c

(A.l2)

(A.B)

(A. 14)

(A.l5)

(A.I6)

(A. 17)

(A.I8)

185

M = A f (d - a)+a b(h-e)(h +.:._ a) (A.19)u .vy 2 I 222

Length of fibres, / =25mmDiameter of fibre, dI =0.5mm

Volume fraction of fibre, PI =0.5

Effective depth, d =45mmBreadth of section, b = 1000Overall depth, h =60

a = 0.00772-/ F, = 0.00772x25x 0.5x 1= 0.19 (A.20)I d

lPI be 0.5

P= 0.85 _(I: -27.6)x 0.05 = 0.7417 (A.21)6.89

a= A.Jy = 226.19x443 =3.1773 (A.22)PI:b 0.7417x42.516xl000

c:=:!!... = 3.1773 = 4.2835 (A.23)P 0.7417

e :=: 1.383c = 1.383 X 4.2835 = 5.9255

M = 226.19X443(45- 3.1773)+0.193XI000X(60-5.9255)(60 + 5.9255 _ 3.1773)u 2 2 2 2

:=: 4677340.978Nmm =4.68kNm

(b) Model calculation of slabs without fibre for the specimen SOSCll

hk =42.15N / mm2

As, =226.19mm2

I y =443N / mm2

b:=: 1000mmd:=: 45mmcylinder strength of concrete without fibre, I: =0.8/ck

=0.8x42.15 =33.72N /mm 2

p:=: 0.85 _(I: -27.6)x 0.05 = 0.85_(33.72-27.6)x 0.05 = 0.80556.89 6.89

:=: Asly = 226.19 x 443 = 3.689a fJf:b 0.8055x33.72xl000

M=Asly ( d- ;)

(3.689)=226.19x443x 45--

2- =4324274.747Nmm =4.324kNm

(A.24)

(A.25)

(A.26)

(A.27)

186

A.3.2 By Swamy and AI-Taan

(a) Model calculation of slabs with fibre for specimen SOSC 21

fck = 47.24N / mm 2

df == 0.5mm

If == 25mm

p == 0.5%

Fibre spacing, S == 25( df J~ == 25 x ( 0.5 )~ = 5mm (A.28)plf 0.5 x 25

E f = 2.05 x 105 N / mm2

7r 2AI =="4 xO.5

rl == 0.25mm

Ec = 5000~ fck = 5000 x.J47.24 = 34365.68N / mm2 (A.29)

E == 2Gm(1 + ~) == 2Gm (1 + 0.15) == 2.3Gm (A.30)

EShear modulus of concrete, Gm == - = 14941.6

2.3

/3== 2nGm == 2x7rx14941.6 =0.892 (A.31)

E A In(~J 2.05 x 105

X 7r X 0.52

x In(_5_)I I r 4 0.25

I

Orientation factor, rJo == 0.41 for fibres in full concrete section and 0.64 for fibres in

tension zone alone

tanh(f!t) tanh( 0.89~x 25)

'11 ~ 1- (f31;) =1- (0.89~x 25) ~ 0.9103 (A.32)

Bond efficiency factor, rJb = 1

Ultimate strength in tension of composite,_ II

G'cu - rJOrJlrJb 2r - VIdl25 0.5

== 0.41x 0.9103x Ix 2x 2.3x-x- == 0.42920.5 100

(A.33)

187

(A.36)

(A.35)

(A.34)Force in fibres in tension zone, Fft = ucub(D - dn) = 0.4292 x 1000x (60 - dn)

Force in tension steel, Fs/ = Asfs = 226.08 x 443 = 100153.44N

Strain in extreme fibre of elastic uncracked tension zone,

6 = ~fcu =.J47""24 =0.00167o 4115 4115

Force in compression,

F = 0.77fc" ( _ &O)bdc &" n

6" 3

= 0.77X47.24(0.0035_ 0.00167)XI000Xd =30589.473d0.0035 3 n n

Equating C = T,

Fji +Fs/ =Fc

0.4292 x 1000 x (60 - d n ) + 100153.44 = 30589.473dn

Depth of neutral axis, dn = 4.059mmDistance from extreme compression fibre to the centre of composite block,

k,= (2-~J+2 d.=[(2_0~~oOo136;r+2]4'059=1.7372mm (A.37)

4(3- ::) 4(3- O~~o0013657)

(D-dJM=Fc(dn-k2 )+FAd-dJ+Fft 2 (A.38)

= 30589.473 x 4.059(4.059 -1.7372)+ 100153.44x (45 - 4.059)+ 24009.877 x (60 - 4.059)2

= 5060231.04Nmm =5.06kNm

(b) Model calculation of slabs without fibre for specimen SOSCll

fck =42.15N / mm2

b =1000mmD== 60mm

d == 45mm

h == 443N / mm2

As == 226.08mm 2

&u == 0.0035

& == ~fck = .J42i5 =0.0015777o 4115 4115

F:/ == Ash = 226.08x 443 = 100153.44

(A.39)

188

2+(2- 0.0015777)

_--:--__0._0_03_5_+_ X4.17355 =1.80094X(3- 0.0015777)

0.0035

F=0.67 jo.(&. -.6f}d,e

0.67X42.15X(0.0035-. 0.0015777)x 1000

= 3 =23997.163dn0.0035

d =F.t =100153.44 =4.17355mmn Fe 23997.163

2+(2- 80J2k 8"

2 = xdn =4X(3- 8

0 J81/

M" =(Fe X dJ(dn- k2)+ FS1 (d - dJ=(23997.163x 4.17355X4.17355 -1.8009)+ 100153.44(45 -4.17355)=4326538.279Nmm = 4.33kNm

A.3.3 By Lim and Paramasivam

(a) Model calculation of slabs with fibre for specimen SOSC21

lek =47.24N / mm2

M" =(J't"bh!!L+ As (J'sy (h+ht)2 2

Post cracking strength of concrete, (J't" = TlITloVflf!JL2r

Length efficiency factor, Tli =0.5

Orientation factor, Tlo =~ (Rajagopal et aI., 1974)7r

Vf = 0.5

If =25mm

Ultimate pullout bond strength of the fibre, T" =2.3N / mm2

r = 3.125 ratio of fibre cross sectional area to perimeter = m12

= 0.125mm4m1

for fibre reinforced concrete, a l =0.9

(J't" =0.5 x~ x 25 X0.5 x 2.3 =0.9321N / mm2

7r 2xO.125x25

(A.40)

(A.41)

(A.42)

(A.43)

(A.44)

(A.45)

(A.46)

189

Depth of tensile zone,

a a h- '_As_a_-IY 0.9x 0.9x 47.27 x 45 _ 226.08 x 443h, = _I_cu__-.::ho-- = ~1O~O~O:.._~ = 41.3747mm

alacu + a,u 0.9 x 0.9 x 47.24 + 0.9321

41.3747 443Mu = 0.9321x 1000x 45x +226.08x-x (45 +41.3747)

2 2=5193082.219Nmm = 5.19kNm

(b) Model calculation of slabs without fibres for specimen sosen

Ick =42.15N / mm2

if =25mm

For concrete without fibre, a l = 0.85

h=45mmAs =226.08mm2

a sy =443N / mm2

b=1000mma cu =O.8/ck = O.8x 42.15 =33.72N / mm2

alacu h - Asasy 0.85 x 33.72 x 45 _ 226.08 x 443

h = b = 1000 = 41.5057/ ala

clI0.85x33.72

aM =A-:2:.(h+h)

u s 2 I

=226.08x 443 x(45+41.5057)= 4331921.717Nmm = 4.33/cNm2

A.4 Computation of mid span deflection

A.4.1 Model computation for specimen soscn

(a) By ACI method

Deflection at first crack load of 13.08kN

hk =42.15N / mm2

b =1000mmd=45mmD=60mmEs = 2.05x lOs N / mm2

(A.47)

(A.48)

(A.49)

(A.50)

190

As/ =226.19mm2

I: =0.8/ek =0.8x 42.15 =33.72N / mm2

Ee =4700.J7: =4700x.J33.72 =27292.394N / mm2

m=Es = 205000 =7.511Ee 27292.394

bxxx~=mAs/(d - x)2

1000xxx~ =7.511x 226.19x(45 -x)2

~ x =10.6682mm

bD3

I gr =- =18000000mm4

12bx3

2I er =-3-+ mAs/(d-x)

1000x10.66823= 3 +7.328x 226.19x(45 -10.6682Y = 2358391.299mm

4

M = lerIgr = 3.6003x18000000 =21601 Oll.Ter % 30 8 lVmm

5X(13·08)X150045 )7/4 1.5

8=--=· =1.17005mm384 EIgr 384x 27292.394x 18000000

Deflection in the second stage, ie. after cracking for a load of 14.76kN

C4.76

)X15002

M = )7/2 = 1.5 =2767500Nmm8 8

I" =[(~JX1,}[I-(~ )']X1<,

[(2160180)3 ] [ (2160180)3]= 2767500 x 18000000 + 1- 2767500 x 2358391.299

= 9796942.403mm 4

5 (14.76) 48= 5)7/4 _ x -Ux1500

384Eleff

- 384 x 27292.394 x 9796942.403 = 2.4258mm

(A.51)

(A.52)

(A.53)

(A.54)

(A.55)

(A.56)

(A.57)

(A.58)

(A.59)

(A.60)

(A.61)

191

(b) By IS method

A.4.2 Model calculation for specimen SOSCll

Deflection at first crack load of 13.08kN

Jek =42.15N / mm 2

Ee = 5000~Jek = 32461.52N / mm2

Es =2.05xlOsN / mm2

m =Es =6.32Ee

M = Je,!gr = 0.7.JY::Igr = 0.7x.J42I5 x 18000000

er % % 30

=2726767.317Nmm

5 [4 5 x (13.081 )x 15004

o= P = 11.5 . = 0.983mm384EIgr 384x 32461.52 x 18000000

After cracking at a load of 14.76kN

bx2

- =mx As, x (d - x)2

1000xx2

2 =6.161x 226.19 x (45 - x)

~x=9.892mm

bx3

I er =3+ mAs,(d-x)2

1000x9.8923

= 3 + 6.32 x 226.19 x (45 - 9.892)2 = 2.08 xl06 mm4

'7 ~ ~(1- ~) ~ j: (1-~) =j(l-k)=(I- ~)I-k)x

k=-45 =0.2198

'7 =(1- ~)I-k)=(I- 0.2;98)(1_0.2198)= 0.723

wl2 (14.76/)x 15002

M = 8 = 11.~ = 2767500Nmm

(A.62)

(A.63)

(A.64)

(A.65)

(A.66)

(A.67)

(A.68)

(A.69)

(A.70)

(A.71)

192

leff = ler = 2.08 X106

== 4274938.34mm4

1.2 _(MerJ 1.2 _(2726767.317)x 0.723M ry 2767500

8 = 5p14 = 5X(14.7X.5)X 15004

= 4.67mm384Eleff 384 X32461.52 X4184034.496

A.5 Calculation of Crack width

By Gergely Lutz

A.5.1 Model calculation for specimen SOSCll

D=60mmd=4Smmfek =42.15N / mm2

de =lSmmb =1000mmI =1S00mmAs, =226.19mm2

n = No. of bars in tension zone = 8Es = 20S000N / mm2

Ee=5000~fck =5000x .J42.15 = 32461.S1N / mm2

m =Es =6.32Ee

Ae= 2(60-4S)xlOOO = 30000mm2

A_e =37S0mm2

nx

bxxx"2 =mAs/Cd -x)

xlOOOx xx - =6.32x 226.19 x (45 - x)

2~ x =9.8919mmkd =x =9.8919mm

For load = 13.08kN (first crack load)

For first crack load, I = I gr = 18000000mm4

(A.72)

(A.73)

(A.74)

(A.75)

(A.76)

193

p = 13.08 = 8.72kN1.5

M_ p12 = 8.72 X 1.5

2= 2.4525kNm

- 8 8

_ mM(d - x) = 6.32 x 2,4525 X 106

x (45 -9.8919) = 29,471fst - I 18000000

gr

-6 '~dAe (D-kd)f.Wer =l1xl0 X,te-; d-kd sf

(60 - 9.8919)-llxl0-6 xVI5x3750x .. x29,471=0.0177mm

- . 45-9.8919

(A.77)

(A.78)

(A.83)

(A.80)

(A.81)

(A.82)

(A.79)IifJ= ( )

e 1.2- ~ 1]

k =!... = 9.8919 =0.21982d 45

1] = (1- ~ )(1-k) =(1- 0.2~982)(1-0.21982) = 0.723

fer = 0.7~ fek =0.7x .J42.15 = 4.5446

M = Igrfer = 18000000 X 4.5446 = 2.726kNmer y 30

2.04xl064

Ieff = (2.7675) = 4377746.158mm1.2 - X 0.723

2.726

f.- mM(d-x) _ 6.1611x2.7675xl06 x(45-9.8919)

sf - I - 6 =136.772eff 4.37xl0

W er = llx 10-6X V15 X 3750 x(60 - 9.8919)x 13677 = 0 082mm

45-9.8919 . .

D 60y=-=-=30mm

2 2I er

194

APPENDIXB

STUDIES ON SIMPLY SUPPORTED TWO WAY SLABS

B.I Computation of Ductility Factor

B.I.I Model calculation for the specimen STSCll

Ductility factor = Deflection at ultimate load (ou) / Deflection at yield load (Oy)J1 =8ul8y (B.l)

Ultimate load, Pu =13T

Deflection at ultimate load, 8u =37.6mm

Yield load Pyis computed as given below:

f ck =42.l5N / mm2

D=60mmd=45mmf y = 443N / mm2

1C x 62 xlOOOA = 4 = 209.4mm2

st, 135

1C x62 xlOOOAst, = 4 =141.3mm2

2001s =1000mm

11 =1500mm

bD3

1=-gr 12

1000x603

= 12 = 18000000mm

Dy=-=30mm

2

Ec =5000~fck =32461.515N /mm 2

Em=_SEc

2.05xl0s= =6.3155000x~fck

(B.2)

(B.3)

(B.4)

195

bxxx~=mA.tf (d- x)2 .XlbXT = mA.tf. (d -Xs )

2

1000x Xs = 6.315x209.4x(45-xs )

2=> x,r :::: 9.666mm

Xlbx-L = mAst (d -X,)2 .

2

1000x~ = 6.315xI41.3x (45 -x,)2

=> x, ::: 8.113mm

b 3ler ::::.5..-+ mA_ (d-X,r)2'3 ~.

::: 1000x 9.6663

+ 6.315x 209.4x (45-9.666)2 =1951993.0163

M _ fy xler,s-

m(d-x,r)

::: 443 x 1951993.016 ::: 3875393.675Nmm6.315 x (45 -9.666)

bx 3ler. ::: _, + rnA (d - n)2

I 3 511

1000 X 8.1133

::: 3 +6.315xI41.3x(45-.8.113)2 =1392122.907

M_ fylcIj,-

m(d-x,)

::: 443x1392122.907 =2647491.7496.315x(45-8.113)

M =M,r+M,y 2

3875393.675 + 2647491.749= 2 = 3261442.712Nmm

Mp = y

y 008121 2• s

3261442.712::: 0.0812x10002 =40.I65Nlmm

(B.5)

(B.6)

(B.7)

(B.8)

(8.9)

(B.10)

(B.11)

196

Py = Py x 1x 1.5 = 60.247kN = 6.l41t

Ou =37.6mm

Oy = 10A1mm

Ductility factor =~ = 37.6 =3.612Oy 10.41

B.2 Computation of Collapse Load

B.2.1 Model calculation for the specimen STSC13

By yield line method

fck =45.l3N / mm2

f y =443N / mm2

;r X62 x1000Short span reinforcement, A I = 4 = 209.4mm2

s, 135

;r x62 x1000Long span reinforcement, A I = 4 = 141.3mm2

s/ 200

f y =443

b=1000mmFbe =1

d=45mm

Short span moment, M =f A d(l- Asl, f y

)us Y sl, bdl'

~ck

=443x209Ax45x(1- 209Ax443 )=3983713.245Nmm\. 1000x45x45.13

Long span moment, M I =f A d(l- ASI1 f y)

u Y si/ bdr~ck

443 14(

141.3X443)= x 1.3 x 45 x 1- . = 2729994. 168Nmm1000x 45x 45.13

M.u =---!!:!. =1.459M UI

(B.12)

(B.B)

(B.14)

197

Collapse load, (B.lS)

x =~1 + i1 +~1 + i3

Y =~1+i2 +~1+ i4

it and i3are ultimate negative moment of resistance per unit width in X direction

i2 and i4 ultimate negative moment of resistance per unit width in Y direction

For Simply Supported Slabs, i1 =i2 = i3 = i4 = 0

X=2, Y=26M

II,/-ly2

P - .

j - 1x (_1)2 (~'--1+~3p-X-1.-52-1Y1.5

=14.976MIIS

=59.659N I mm

Pj = 59.659x Ix 1.5 = 89.488kN =9.122t

B.3 Computation of Theoretical Deflection (Purushothaman, 1984)

B.3.1 Model calculation for the specimen STSCl3

fck =45.l3N / mm2

b=1000mmD=60mmd=45mm

f y =443N I mm2

AS1 =209.4mm2,

AS11 = 141.3mm2

Is =1000mm

I, =1500mm

bD3 1000x603

I gr =12= 12 =18000000mm4

Dy=-=30mm

2

f: =0.8x hk =36.104

Ec =4700-17: =28240.704N / mm 2

(B.l6)

(B.l7)

(B.l8)

(B.I9)

(B.20)

198

Es = 2.05x10sN / mm2

hr =0.62ft =3.725

Em=_S =7.259

Ee

First stage up to first crack

I grMer =hr-

y

Mer = 3.725x18000000 =2235000Nmm30

Mer =0.0812PeJI.2

= 2235000. =27.525N / mmPer 0.0812x 10002

Per =27.525 X 1X 1.5 =41.288kN =4.20914o = 0.00772Per ls

er Ee1gr

= 0.00772 x 27.525x 10004 =0.418mm

28240.704 x 18x 106

Second stage up to yield load (I =1eJ

xbx XX- =mAsl(d -x)

2x

bxxs x-..!.... =mAsl (d -xJ2 '2

1000x Xs =7.259x209.4x(45-xs)2

=> Xs == 10.275mm

Xbxx, x-.!... =mAsl (d -x,)

2 '2

1000x~ =7.259x141.3x(45-xs)2

=> XI =8.637

bXs3

A (d )21er, =-3-+ m sl, -Xs

(B.21)

(B.22)

(B.23)

(B.24)

(B.25)

(B.26)

(B.27)

199

=1000x 10.2753

+ 7.259x 209.4 X (45 -10.275/ =2194493.1873

.r xlM = Jy cr,

S m(d -xJ

= 443x 2194493.187 =3856727.125Nmm7.259 x (45 -10.275)

bx/ 2I Clj =-3-+ mAs1/(d-x,)

=1000 X 8.6373

+ 7.259x 141.3x (45 -8.637)2 =1571012.6643

M_ fy1cli,-

m(d-x,)

= 443x 1571012.664 =2636616.1627.259 x (45 - 8.637)

M = M", +M, =3246671.581Nmmy 2

My =0.0812Pi/

= My = 3246671.643 =39.984N/mmPy 0.0812/ 2 0.0812x 10002

S

1 +1/ = cr, crt =1882752.926mm4

cr 2

Py = Py X Ix 1.5 == 59.976kN = 6.114/

(py - Pcr)xl/ x 0.00772Oy = Ocr +-----'------­

EJcr

=0.418 + (39.984 - 27.525) X 10004

X 0.00772 =2.227mm28240.704 X 1882752.926

Third stage at Johansen's load

From B,Z.1

P j =59.659N/mm

(Pj - p y )xO.00772xIs4

8. == 0 +---=--....:....-----J y 0.5/

crE

c

= .227+ (59.659-39.984)xO.00772xl0004

=7.94mm2 0.5 x 1882752.926x 28240.704

(B.28)

(B.29)

(B.30)

(B.31)

(B.32)

(B.33)

(B.34)

(B.35)

200

B.4 Prediction of crack width (Nawy, 1972.)

From equation 5.6

W =kf3fs'~M I

M =db1s2

I Qdb1 =6mm =O.236in

S2 =200mm =7.87inAQ=_s\At

\

1r 62-x. As = '12 x2.54 xlO)x 4 = O.0989m2

\ \ 135

Atl = 12(db1 +2C1)=12(O.236+2X ;:4)= 14.171m2

Q= 0.0989 = 6.98 x10-3

14.171

M = 0.236 x7.87 = 266.257I 6.98x10-3

is = OAx 443 = 177.2N / mm2 = 25694psi = 25.694ksi

wer =4A8x 10-5 x1.25x 25.69x.J266.65 = O.02347in = O.59634mm

(836)

(8.37)

(8.38)

201

APPENDIXC

STUDIES ON ONE WAY RESTRAINED SLABS

C.l Computation of Collapse Load

C.1.1 Model calculation for the specimen with fibre FOSC21

From 2.6.2

M 2 W)2A +MB + Me =-4-

when moment of resistance of support and midsection are the same

M =W)2u 16

replacing W u by Pi

P [2M =_J_

u 16Moment of resistance is calculated as below by ACI method

fck = 48.69N / mm2

As =226.19mm2

f: = 0.9 fck =0.9 x48.69 =43.821N / mm2

f y =443N / mm2

Fbe =1

Af =50

r d =2.3

2rd FbeAf6 s(jlbre) = =0.00115

Es

0.003 = 6 s(jlbre)

C e-c=> e = 1.383c

/3=0.85 V;-27.6)xo.05=0 7326.89 .

a = Asfy _ 226.19 x443 _f3f:b - 0.732x 43.82x 1000 - 3.1226

(C.1 )

(C.2)

(C.3)

(C.4)

(C.S)

(C.6)

(C.7)

202

ac=-=4.26

f3e=1.383c = 5.9

'M = A f (d _::'+(5 b(h-e{.h +~_a,U 5 y\ 2) I ~ \2 2 2)

1(i, =0.00772x- Pf Fbe =0.19

df

( 3 123, {60 5.9 3.123'\Mu =226.l9x44\.45-~ )+O.l9x1000x(60-5.9\2+2-2)=4680393Nmm =4.68kNm

Rence collapse load by Johansen's yield line theory is

16P j =MIl Xl =33.28kN / m2

1Pj = 33.28x1.5 =49.92kN

C.1.2 Model calculation for the specimen without fibre FOSCll

Ick = 46.07N / mm2

f: =0.8/ck = 36.856N / mm2

fJ= 0.7836 5 =0.00115

AJya=-=3,473

fif:b

Mu =AJy( d -~J= 4335098Nmm =4.33kNm

M x16Rence collapse load, Pj = u

12=30.79

Pj = 30.79 X1.5 = 46.24kN

C.2 Computation of Deflection

C.2.1 Model calculation for the specimen FOSCll

hk = 46.07N/mm2

I: = 0,8 xhk = 36,856N / mm2

(C.S)

(C.9)

(C.lO)

(C.ll)

(C.l2)

(C.l3)

(C.14)

203

fer =0.62.JJ: =3.763N Imm2

Ee = 4700fl = 28533.297N Imm2

Es = 2.05x 105 N I mm2

E.m =_s = 7.184Ee

Ast =226.19mm2

b=1000mmI gr = 18000000mm4

Xb X xx- =mAsted -X)

2

x2

1000x- =7.184x 226.19x (45 -X)2

=> x= 10.577bx3

2I er =-3-+ mAs,(d-x)

= 1000x10.5773

+7.184x226.19x(45-10.577)2 = 2319898.437mm4

3

Deflection up to first crack

1.5 x 9.81 81NI 8 k IPer = = 9. mm = 9. 1 N m1.5 x 1

8 =_1_ Pe,z4er 384 EJgr

=_l_ x 9.81x15004

=0.2518mm384 28533.297 x18000000

In the second stage up to Johansen's load (Pj )

8 = 8 + 1 (P j - Pe'y4J er 384 E I

e er

Pj 4.71x9.81Pj = 1.5 x 1 = 1.5 = 30.82kN1m = 30.82N Imm

8. =0.2518+-1-x (30.82-9.81)x15004

=4.436mmJ 384 28533.297 x 2319898.437

(C.15)

(C.16)

(C.17)

(C.18)

(C.19)

(C.20)

(C.21)

204

APPENDIXD

STUDIES ON TWO WAY RESTRAINED SLABS

D.1 Computation of Deflection of Two Way Restrained Slab Based on the ModelProposed By Muthu et al. (1979)

D.1.1 Model calculation for the specimen FTSCll

fck =45.34N / mm2

f: =0.8 x fck =36.272N / mm 2

fcr = 0.62flj = 3.734N / mm2

Ec=4700flj =28306.333N / mm2

E, =2.05 x105N / mm2

Em=-S =7.242

Ec

Area of steel along short span, Asts = 208.2mm2

Area of steel along long span, AStl = 150.129mm2

b=1000mmLong span, II =1500mm

Short span, Is =1000mm

I gr =18000000mm4

d=45mmf y =443N / mm 2

First stage (deflection up to first crack)

M = fc'! gr =3.734 x18000000 =2240400NmmIT y 30

M cr =0.0368pjs2

= 2240400 =60 8 /Pcr 0.0368xl000 2 • 8N mm

Pcr =60.88 x 1x 1.5 =91.32kN =9.309t

6 =0.0022Pcrls4

cr E Ic gr

6 =0.0022 x 60.88 xl 0004_

cr 28306.333 x 18 xl 06 - 0.263mm

(D.1)

(D.2)

(0.3)

(D.4)

(D.S)

(D.6)

(D.?)

205

Second stage (deflection up to collapse load)x

bxxx- =mAsted -x)2x

bxxs xt =mAst, (d -xs )

2x1000x _s_ = 7.242 x 208.2 x (45 - xs )

2=> Xs =10.238mm

bxxix~=mAst (d -XI)2 .,2

1000x~ = 7.242x150.129x(45-xs)2

=> XI =8.864mm3

Icr, =bx3, + mAst, (d - X.,)2

= 1000x 10.2383

+ 7.242x 208.2x (45 -10.238)2 = 2179705.8763

3bXI 2

1cr, =-3- + mAstl (d - XI)

= 1000x8.8643

+7.242x150.l29x(45-8.864)2 =1651871.4793

I cr + I aI =' Icr 2

= 2179705.876+ 1651871.479 = 1915788.678mm4

2

Determination of collapse load

Short span moment, Mus is given by M =fA d(l- As"fyJ

liS y s/, bdlf.ck

(208.2 x 443 )=443x208.2x45x 1- = 3962843.434Nmm

1000x 45x 45.34Long span moment, Mul is given by

Mill = fyAs'1d(l- As"fy)

bdfck

(150.129X443)=443x150.129x45x 1- = 2895265.36Nmm

1000 x 45 x 45.34

: (D.8)

(D.9)

(D. 10)

(D.ll)

(D. 12)

(D.13)

206

From equation 2.37

M,u=.----!!!...= 1.369

Mill

(D.14)

(D.15)

(0.16)

(0.17)

X =~I+il +~I+i3

Y = ~1+i2 +~1+i.

il =i2 =i3 =i4 =1

X =Y=2.J2

p. = 6M"",uy2 . = 6xM.. x1.367x(2.J2L =30.545M.. =121.045Nlmm

) lx(_l)2W1+3,uX1.52-If IX(_1)2~I+3X1.367X1.52-If15 15

Pj =121.045 x1x 1.5 = 181.568kN = 18.51t

As,Px = bd'

S

208.2=100 x100 = 0.463%Ox45

As,P - I

Y - bdI

150.129=1000x39 xl00 =0.385%

( J0.3855 ( J ( )2X=(P, +pJ'"''x~ x ~: x L;,

=(0.463+0.385)-0.1932 X ( 443 )0.3855 X (1000)X(1000)2 =501.66536.272 1500 60

a =6.744 x10-5X + 0.1094= 6.744x 10-5x501.665 + 0.1094 = 0.143

o. = ~ (Pj - PcJx O.0022xl/J Ucr +----"-------

a/crEe

= 0.263 + (121.045 - 60.88)x 0.0022 X10004_ 1 3

O.I43 x 1915788.678x 28306.333 - 7. 3mm

(D.18)

(D.19)

(D.20)

(D.21)

(0.22)

207