1. Presentation prepared as per:
“IS 456 : 2000”
Presented by: Ravi Shankar Singh &
Pankaj Kumar

2. CONTENTS
• Introduction
• Important terms
• General considerations of design for slabs
• Examples

3. Introduction
• Slabs are plane structural members which are used
as floor and roof of buildings.
• The thickness of the slabs is quite small as
compare to its other dimensions.

4. One way slab:-
• One way slab is edge supported slab which
spanning in one direction.
• The ratio of long span to short span is greater
than or equal to 2.
• The main steel is provided in the direction of the
shorter span
• Slab supporting on four sides also behave as one
way slab if ratio of long span to short span is
greater than or equal to 2.

7. • The distribution steel should be tied above the
main steel, other wise the lever arm which is
measure upto the center of the main steel shall
be reduced resulting in the reduction of
moment of resistance.
Purpose of main steel:
• It take up all tensile stresses developed in the
structure.
• It increase the strength of concrete sections.
Purpose of distribution steel:
• It distribute the concentrated load on the slab.
• It guard against shrinkage and temperature
stress.
• It also keep the main reinforcement in position.

9. 1. EFFECTIVE SPAN
(A) FOR SIMPLY SUPPORTED SLAB : DISTANCE BETWEEN CENTRE TO CENTRE OF
SUPPORT OR THE CLEAR DISTANCE BETWEEN SUPPORT PLUS THE EFFECTIVE DEPTH OF
THE SLAB WHICHEVER IS SMALLER .
(B) FOR CONTINUOUS SLAB : WIDTH OF SUPPORT < 1/12 OF CLEAR SPAN ,THE EFFECTIVE
SPAN SHALL WORK OUT AS ABOVE.
WHEN WIDTH GREATER THAN 1/12 OF CLEAR SPAN OR 600MM WHICHEVER IS SMALL
SHALL BE TAKEN
(1) FOR ONE END FIX AND OTHER END CONTINUOUS THE EFFECTIVE SPAN IS CLEAR SPAN.
(2) FOR ONE END FREE AND OTHER END CONTINUOUS ,THE EFFECTIVE SPAN SHALL BE
EQUAL TO CLEAR SPAN PLUS HALF THE EFFECTIVE DEPTH OR CLEAR SPAN PLUS HALF
THE WIDTH OF SUPPORT ,WHICHEVER IS SMALLER .
GENERAL CONSIDERATION OF DESIGN FOR SLAB

11. (2)For end span with one end free and other end continuous ,the
effective span shall be equal to clear span plus half of the
effective depth of the slab , or clear span plus half the with of
center to center of support or the clear distance between the
support plus the effective depth of the slab whichever is smaller.
For continuous slab: In case of a continuous slab ,where the width
of the support is less than 1/12 of the clear span ,the effective span
shall be worked as above.
In case the support are wider than 1/12 of the clear span , the
effective span shall be taken as under:
(1)For end span with one end fixed and other end continuous ,the
effective span shall be clear span.
(2)For end span with one end free and other end continuous ,the
effective span shall be equal to clear span plus half of the
effective depth of the slab , or clear span plus half the with of the
discontinuous slab, which ever is less.
2. DEFLECION CONTROL
The span to depth ratio are not greater than the value below

12. A) For span upto 10m
cantilever slab 7
Simply supported 20
Continuous 26
b) For span above 10m the value in para (a) may be
multiply by 10/span in meters
3.Reinforcement in slab
(A)Minimum reinforcement. The area of reinforcement in
either direction should not less than:
0.12% of the total cross-sectional area (for HYSD
bar)
0.15% of the total cross-sectional area (for Fe
250)

13. • (B) Maximum diameter : The maximum
diameter of reinforcing bar in a slab should not
exceed 1/8 of thickness of slab.
• (C) Cover to reinforcement :The cover of
concrete to reinforcement in different structural
members are following in table
Exposure minimum grade of concrete Nominal cover in mm not
less than
Mild M20 20
Moderate M25 30
Severe M30 45
Very severe M35 50
Extreme M40 75

14. (D) Spacing of reinforcement:
a. Minimum distance between individual bars:
(i) Minimum distance shall not less than diameter
of bar or 5mm more than the nominal max. size of
the bar whichever is greatest.
(ii)When we provide two or more layer of main
bars, the min. vertical clear distance between any
two layer of bar shall be 15mm or 2/3 the nominal
max. size of course aggregate or max. size of bar
whichever is greatest.
(b) Maximum distance between bars in tension :
(i)The pitch of the main tensile bars in slab should
not exceed three times the effective depth or
300mm whichever is smaller.

15. (ii) The pitch of distribution bars shall not exceed
five time the effective depth or 450 whichever is
smaller.
(E) Curtailment of main reinforcement.
The main reinforcement in slab may be curtailed
or bent up ,beyond the point at which it is no
longer requirement to resist bending

16. The following equation can be used to find out the
point at which the bar can be curtailed or bend-up
x=l/2[Nx/Nc] ˄ ½
X=Distance from either side from centre of span.
Nc=Total no. of bar in tension
Nx=No. of bars which can be curtailed at any
distance x from the center of the span
4.Load on slabs.
The load on slab consists of:
(a) Live load
(b)External dead load
(c) Dead load of slab itself

17. EXAMPLE
Design a simply supported R.C.C slab to carry a
factored load of 15000N/m2 inclusive of its own
weight on an effective span of 3.1 m.
solution:-
Assume 1m width of slab.
b= 1m = 1000mm
fck = 20N/mm2
fy =415N/mm2
Effective span l=3.1m
Design factored load Wu=1500N/m2
=15000*1=15000
Depth of slab . Assume effective depth d=span/25

18. = 3100/25= 124mm or say 130mm
Overall depth D =130+30 =160mm
Ultimate max. bending moment
Mu =wl˄2/8
=(15000x3.1˄2)÷ 8
=18019 N-mm =18019x10˄3Nmm
Check for depth
The slab will be designed as a balance section
Mu =Mlim.
18019x10˄3 =0.138fckbd˄2
18019x10˄3 =0.138x20x1000xd˄2
d= [(18019x10˄3)÷(0.138x20x1000)] = 81mm<130mm
Adopt d=130mm and D=160mm

19. Reinforcement (main steel)
Mu =0.87fyAst [d-fy Ast ]
fck b
18019x10˄3 = 0.87x415xAst [130-415 x Ast ]
20 x 1000
49007 = 130Ast -0.021Ast˄2
Ast˄2- 6190Ast +2376524 =0
Ast =411mm˄2
Minimum area of steel for main bars =0.12bD÷100
=(0.12x1000x160)÷100 =192mm˄2
The area of steel provided is more than minimum
requirement.Hence ok

20. Provide 10mm ɸ bars for main reinforcement
Area of one 10 mm ɸ bar = π÷4x10˄2 = 78.5mm˄2
c/c spacing of 10mm ɸ bars =Area of one bar x b
Total area of steel
= 78.5 x 1000
=190MM
411
It should be less than:
(i) 3d = 3x130 = 390 mm
(ii) 300 mm
Provide main steel of 10mm ɸ @190 mm c/c
Clear spacing between bars = 190-10 = 180mm

21. It should be more than:
(i) Diameter of bar =10mm
(ii) Size of aggregate + 5 = 15+5 = 20
Provide 10 mm ɸ bar @190 mm c/c
Alternate bars are bent up near the supports
.These bars may be bent up at 0.10 l= 310 mm
from face of support .
Distribution steel
Area of distribution steel = 0.12bD/100
=(0.12x1000x160)/100
Provide 8mm ɸ bars as distribution steel
Area of one 8mm ɸ bars =π/4 x 8˄2 = 50.26 mm˄2

22. c/c spacing of 8mm ɸ bars = (50.26x1000)/192
=262mm say 260mm
It should less than :
(i) 5d =5x130 = 650mm
(ii) 450mm
Provide distribution steel of 8mm @260 mm c/c
Check for shear
The critical section for shear is at a distance d i.e. 130mm from inner face of support or
0.28 m [130 +300÷2=280 mm = 0.28m] from centre of support.
[Assume width of support =300]
Factored shear force at critical section V = wl/2 –wx
=15000x3.1)/2 – 15000x0.28
=23250-4200 = 19050 N
Nominal shear stress τ = V/bd
=19050/1000x130= 0.15 N/mm˄2 < 2.8N/mm˄2
[For M20 concrete τc max =2.8 N/mm˄2
P = 100As/bd = (100x206)/(1000x130 ) = 0.16%
[{ As =1/2(78.5x1000÷190) = 206 mm˄2},Alternate bars are bent–up ]
For P = 0.16% and M 20 concrete from table
τc = 0.29 N/mm˄2
For solid slab from table k= 1.28 (for overall depth of slab = 160mm)
τc < τv hence no reinforcement provided
Check for development length
Ld = (0.87fyɸ)/4τbd = (0.87x415x10)/4x1.92 = 470 mm
[For M20 concrete and Fe 415 steel, τbd = 1.6x1.2 = 1.92N/mm˄2]

24. As per code at the simply supported the bar must
extend beyond the face of support by a distance
not less than Ld/3 = 470/3 = 157 mm
Length available from the face of support =
support width – side cover =300-20 = 280 mm
Embedment length from center of support
Lo = 300/2-20 = 130mm
Since the slab is simply supported ,the
compressive reaction will confine the
reinforcement . Hence Ld ≤ 1.3M/V +Lo
M = moment of resistance provided by 10 mm ɸ
bars @380mm c/c

26. = 0.87fyAst[d- fy Ast ]
fck b
=0.87x415x206x[130 – 415x206] = 9351x10˄3Nmm
20x1000
[Ast =(78.5x1000/380) = 206 mm˄2]
V= Factored shear force at the centre of support
= wl/2 =15000x3.1/2 = 23250 N
1.3M/V + Lo = 1.3x9351x10˄3/23250 +130
=653mm >Ld
Hence development length requirement are
satisfied.

27. • Check for deflection
Pt = 100Ast/bd =(100x413)/(1000x130) = 0.32%
(Ast provided = 78.5x1000/190 = 413mm˄2)
Fs = 0.58fy [Area of steel required]
Area of steel provided
= 0.58x415x[411/413] = 240N/mm˄2
Modification factors
Kt = 1.5
Kc= 1.0 (No compression reinforcement provided)
Kf = 1.0 (slab section is rectangular)
(l/d)max = 20 x Kt x Kc x Kf = 20x1.5x1x1= 30
(l/d)provded = 3100/130 = 23.8<30
The slab will satisfied limit state of serviceability.