Presentation prepared as per: “IS 456 : 2000”

Presented by: Ravi Shankar Singh & Pankaj Kumar

CONTENTS• Introduction

• Important terms

• General considerations of design for slabs

• Examples

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.

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.

• 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.

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

(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 CONTROLThe span to depth ratio are not greater than the value below

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)

• (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

(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.

(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

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

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

= 3100/25= 124mm or say 130mmOverall depth D =130+30 =160mmUltimate max. bending momentMu =wl˄2/8

=(15000x3.1˄2) ÷ 8

=18019 N-mm =18019x10˄3 NmmCheck for depthThe slab will be designed as a balance section Mu =Mlim.18019x10˄3 =0.138fckbd˄218019x10˄3 =0.138x20x1000xd˄2d= [(18019x10˄3)÷(0.138x20x1000)] = 81mm<130mmAdopt d=130mm and D=160mm

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

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

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

c/c spacing of 8mm ɸ bars = (50.26x1000)/192=262mm say 260mmIt should less than :(i) 5d =5x130 = 650mm(ii) 450mmProvide distribution steel of 8mm @260 mm c/cCheck for shearThe 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˄2P = 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˄2For solid slab from table k= 1.28 (for overall depth of slab = 160mm)τc < τv hence no reinforcement providedCheck 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]

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 ɸbars @380mm c/c

= 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.

• 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.