****====DESIGN OF ONE-WAY SLAB====****

4.6 m 4.6 m

GIVEN:
Concrete,fc' = 27.5 Mpa
Steel,fy = 414 Mpa
Unit weight of concrete,wc = 23.5 kN/m3 1) L / 20 ; simply supported
Slab live load = 4.8 kPa 2) L / 24 ; one end continuous
clear span,L = 4.6 m 3) L / 28 ; two end continuous
main bar diameter = 10 mm Ø 4) L / 10 ; cantilever
temp. bar diameter = 8 mm Ø

TYPE OF SLAB SECTION: 3 h = L / 28 ; TWO END CONTINUOUS

L
h = = 164.2857 mm
28
say h = 165 mm

COMPUTE LOADS:
DEAD LOAD(DL)
self weight = wch = 3.8775 Kpa
assume floor finish = 25 mm
assume floor finish = ywc = 0.5875 Kpa
total DL = 4.465 KPa
LIVE LOAD(LL)
service load = 4.8 Kpa

wu = 1.2DL + 1.6LL = 13.038 Kpa

COMPUTE MOMENTS: M = cwL2

coefficients, c L(m) w(kPa) (kN.m/m)

0.111111111 4.6 13.038 30.65378667
0.071428571 4.6 13.038 19.70600571
0.041666667 4.6 13.038 11.49517

COMPUTE FOR ρ:
 1.4/ fy = 0.0033816
√¯fc'/ 4fy = 0.0031667
use ρmin = 0.00316669

ρ max =. 75
[ .85 fc ' β 1600
fy ( 600+ fy )
=
] 0.02129829

main bar reinforcement = 78.5398163

temp bar reinforcement = 50.2654825

COMPUTE THE REQUIRED As AND SPACING OF BAR REINFORCEMENTS

interior support at mid-span exterior support
Mu = = 30.653787 19.706006 11.49517 (kN-m/m)
d = h - (20+Ø/2) = = 120 120 120 (mm2)
Ru = Mu/(.9bd2) = 2.3652613 1.5205251 0.886972994
ρ= = 0.0060358 0.0038007 0.002184714
use ρ = 0.0060358 0.0038007 0.003166693
As = ρbd = 724.29727 456.08248 380.0032059 (mm2)
s = 1000Ao/As = 108.43589 172.20529 206.6819835

TEMPERATURE BAR SPACING REQUIREMENT:

fy < 414MPa ρtemp. = 0.002

fy = 414MPa ρtemp. = 0.0018
fy > 414MPa ρtemp. = 0.0018(414)/fy
MINIMUM RHO FOR TEMPERATURE BARS:
ρmin = 200/fy
= 0.48309179
ρtemp. = 0.0018

As,temp =ρtempbd = 216 (mm2)

temp bar spacing = 215 (mm2)

SLAB DETAILS FOR BENDING OF REINFORCEMENTS

fixed bar spacing at mid-span, s = 150
As(required) = 456.0824839
As(actual) = 1000Ao/s(fixed) 523.5987756
design is ok!!!=)
at exterior support
Ap = As(actual)2/3 = 349.0658504
As(required) = 380.0032059
design is not safe!!!=(
provide extra bars,n = 1
Ap = Ap + n(Ao) = 427.6056667
design is ok!!!=)
at interior support
Ap = As(actual)4/3 = 698.1317008
As(required) = 724.2972694
design is not safe!!!=(
provide extra bars, n = 2
Ap = Ap + n(Ao) = 855.2113335
design is ok!!!=)
Slab Details Scheme 1: Bending of reinforcements

10 mm Ø bent 2/3 spaced @

150 mm

1 -extra top bar is needed 2 -extra top bars are needed

165
mm

1 8 10 mm Ø bent 2/3 spaced @

-extra top bar is needed mmØ temp bars @ 150 mm
215 mm

2
-extra top bars are needed

Slab Details Scheme 2: Cutting of reinforcements

10 mmØ 10 mmØ 10 8 mmØ

190 oc 380 oc 190 oc 215 oc
 10 mmØ 8 mmØ 10 mmØ
190 mm spacing 215 mm spacing 190 mm spacing 10
mmØ 95
mm spacing
DESIGN OF TWO-WAY SLAB

7.7 m

6.2 m
Estimate slab thickness based on code minimum thickness requirement.
GIVEN: Trial depth, h:
fc ' = 20.7 MPa 2 ( A+ B) = 154.444 mm
h=
fy = 414 MPa 180 say 160 mm
wc = 23.6 kN/m³
bar size = 12 mmφ
Live load = 6.65 kPa
LA = 6.2 m
LB = 7.7 m

Slab load:
DL:
slab: wch = 3.776 kPa
finish, assume 25mm cement finish 0.025wc = 0.59 kPa
TOTAL = 4.366 kPa
LL: 6.65 kPa
Ultimate load: wu = 1.4DL + 1.7LL = 17.417 kPa
Slab aspect ratio, m:
LA
m= = 0.805194805
LB
Positve moments:
Case 4: coeff, c w L M=cwL²
Ma, pos DL 0.039 4.366 6.2 6.545
Ma, pos LL 0.048 6.65 6.2 12.270
Mau, pos = 1.2DL + 1.6LL 27.486
Mb, pos DL 0.016 4.366 7.7 4.142
Mb, posLL 0.02 6.65 7.7 7.886
 Mbu, pos = 1.2DL + 1.6LL 17.587

Negative moments:
At continuous edge: coeff, c wu L M=cwL²
Ma, neg 0.071 17.4174 6.2 47.536 kN-m/m
Mb,neg 0.029 17.4174 7.7 29.948 kN-m/m

At discontinuous edge: M = 1/3(Mpos)

Ma, neg 9.162 kN-m/m
Mb,neg 5.862 kN-m/m

Design of middle strip in the short direction:

h d

Mu = Mmax = = 47.536 Kn-m

Trial d = h - (20 + φ/2) = = 134 mm

Mu/φ = 2.941527733
Ru =
bd 2

ρ min=
[ 1. 4 √ f c '
,
f y 4f y ]min
= 0.002747419

ρ max =. 75
[ . 85 f c ' β 1 . 003 E s
fy . 003 E s + f y ]
= 0.016031805

ρ=
1
ω [ √
1− 1−
2ωR u . 85 f c '
fy
=
fy
1 − 1−
] [ √
2 Ru
. 85 f c ' ] = 0.007826 ok, use this rho!!!=)

π
A o= D 2 = 113.0973355 mm2
4 b
As = ρbd = 1048.632079 mm2

1000 A o = 107.8522561 mm oc
s=
As
At mid-span:
Mu = 27.486 Kn-m
Ru = 1.700853684
p = 0.004328795 ok, use this rho!!!=)
As = 580.0584963 mm2
s = 194.9757417 mm oc
At discontinuous end:
Mu = 9.162 Kn-m
Ru = 0.566951228
p = 0.001392252 use pmin!!!=(
As = 368.1540976 mm2
s = 307.2010776 mm oc

Design od middle strip in the long direction:

h d

Mu = 29.948 Kn-m
dL = h - (20 + 1.5φ) = 122 mm
Ru = 2.235633472
p = 0.005795189 ok, use this rho!!!=)
Ao = 113.0973355 mm2
 As = 776.5553675 mm2
s = 145.6397576 mm oc

At mid-span:
Mu = 17.587 Kn-m
Ru = 1.312895778
p = 0.00329931 ok, use this rho!!!=)
As = 402.5158091 mm2
s = 280.976133 mm oc
At discontinuous end:
Mu = 5.862 Kn-m
Ru = 0.437631926
p = 0.001070566 use pmin!!!=(
As = 335.185074 mm2
s = 337.4175771 mm oc

Check for shear:

Total load on panel, Wt = LALBwu = 831.5067 Kn
Shear per m of long beam, Case 4: CA = 0.71
CA WT
v A= = 38.3357 Kn/m
2 LB
Shear concrete:
d = 134 mm

V uc =φvc
√ f c ' bd = 91.44948 shear is okay!!=)
6
SLAB DETAILS FOR BENDING OF REINFORCEMENTS

Short Direction:
fixed bar spacing at mid-span, s = 150 mm oc
As(actual) = 1000Ao/s(fixed) = 753.9822 >As required at mid-span, ok
As(required) = 580.0585

at exterior support
Ap = As(actual)2/3 = 502.6548 >As required at exterior support,ok
As(required) = 368.1541

at interior support
Ap = As(actual)4/3 = 1005.31 <As required,provide extra bars
As(required) = 1048.632
provide extra bars, n = 2
Ap = Ap + n(Ao) = 1231.504 >As required at interior support,ok

Long Direction:
fixed bar spacing at mid-span, s = 220 mm oc
As(actual) = 1000Ao/s(fixed) = 514.0788 >As required at mid-span, ok
As(required) = 402.5158

at exterior support
Ap = As(actual)2/3 = 342.7192 >As required at exterior support,ok
As(required) = 335.1851

at interior support
Ap = As(actual)4/3 = 685.4384 <As required,provide extra bars
As(required) = 776.5554
provide extra bars, n = 2
Ap = Ap + n(Ao) = 911.6331 >As required at interior support,ok
 column strip middle strip column strip
LB/4

provide 2 extra top bars

12mmφ @150 oc bent 2/3

LB/2

12mmφ @ 220 oc bent 2/3

provide 2 extra top bars

LB/4
LB/4

LA/4 LA/2 LA/4 LA/4

column strip middle strip column strip column strip

Slab thickness = 160 mm

Slab Details Scheme 2: Cutting of Reinforcements

12φ top bars @ 245

column strip
LB/4

12φ top bars @ 245

12φ bot bars @ 175

12φ top bars @ 105

12φ bot bars @ 250

LB/2
LB/4

12φ top bars @ 150

column strip
LB/4

LA/4 LA/2 LA/4 LA/4

column strip middle strip column strip column strip
Slab thickness = 160 mm
>As required at exterior support,ok

>As required at interior support,ok

>As required at exterior support,ok

>As required at interior support,ok