INTRODUCTION

Flat slab is a reinforced concrete slab supported directly by concrete columns

without the use of beams. Flat slab is defined as one sided or two-sided support
system with sheer load of the slab being concentrated on the supporting columns and
a square slab called ‘drop panels’.
Drop panels play a significant role here as they augment the overall capacity and
sturdiness of the flooring system beneath the vertical loads thereby boosting cost
effectiveness of the construction. Usually the height of drop panels is about two
times the height of slab

Flat Slabs are considered suitable for most of the construction and for asymmetrical
column layouts like floors with curved shapes and ramps etc. The advantages of
applying flat slabs are many like depth solution, flat soffit and flexibility in design
layout.

Even though building flat slabs can be an expensive affair but gives immense
freedom to architects and engineers the luxury of designing.
 Benefit of using flat slabs are manifold not only in terms of prospective design and
layout efficacy but is also helpful for total construction process especially for easing
off installation procedures and saving on construction time.

If possible, try to do away with drop panels as much as possible and try to make the
best use of thickness of flat slabs. The reason is to permit the benefits of flat soffits
for the floor surface to be maintained, ensure drop panels are cast as part of the
column.

To utilize the slab thickness to optimum level, the essential aspects that should
be kept in mind are:

1. Procedure related to design

2. Presence or absence of holes
3. Significance of deflections
4. Previous layout application experience
Types of Flat Slab Construction

Following are the types of flab slab construction:

o Simple flat slab

o Flat slab with drop panels
o Flat slab with column heads
o Flat slab with both drop panels and column heads

Uses of Column Heads

o It increase shear strength of slab

o It reduce the moment in the slab by reducing the clear or effective span
 Uses of Drop Panels

o It increase shear strength of slab

o It increase negative moment capacity of slab
o It stiffen the slab and hence reduce deflection

Advantages of Flat Slabs

It is recognized that Flat Slabs without drop panels can be built at a very fast pace as
the framework of structure is simplified and diminished. Also, speedy turn-around
can be achieved using an arrangement using early striking and flying systems.

Flat slab construction can deeply reduce floor-to –floor height especially in the
absence of false ceiling as flat slab construction does act as limiting factor on the
placement of horizontal services and partitions. This can prove gainful in case of
lower building height, decreased cladding expense and pre-fabricated services.

In case the client plans changes in the interior and wants to use the accommodation
to suit the need, flat slab construction is the perfect choice as it offers
that flexibility to the owner. This flexibility is possible due to the use of square
lattice and absence of beam that makes channelling of services and allocation of
partitions difficult.

Thickness of flat slab

Thickness of flat slab is another very attractive benefit because thin slab provides
the advantage of increased floor to ceiling height and lower cladding cost for the
owner. However, there is profound lower limit to thickness of slab because extra
reinforcements are needed to tackle design issues. Besides this, added margin must
be provided to facilitate architectural alterations at later stages.
 Types of Flat Slab Design

Multitudes of process and methods are involved in designing flat slabs and
evaluating these slabs in flexures. Some of these methods are as following:

o The empirical method

o The sub-frame method
o The yield line method
o Finite –element analysis

For smaller frames, empirical methods are used but sub-frame method is used in
case of more irregular frames. The designs are conceptualized by employing
appropriate software but the fact is using sub-frame methods for very complicated
design can be very expensive.

The most cost effective and homogenous installation of reinforcements can be

achieved by applying the yield line method. A thorough visualization in terms of
complete examination of separate cracking and deflection is required since this
procedure utilises only collapse mechanism.

Structures having floors with irregular supports, large openings or bears heavy loads,
application of finite- element analysis is supposed to be very advantageous. Great
thought is put into choosing material properties or installing loads on the structures.
Deflections and cracked width can also be calculated using Finite- element analysis.

Areas That Require Attention in Design of Flat Slab

1. Deflections-Usually at the center of each panel deflections are maximum.

Foreseeing deflections can be very tricky and will engage some form of elastic
appraisal. While designing structure layout and during implementation using sub
frame method, one way to evaluate mid-panel deflection is to use at least two parallel
column strips.

2. Proprietary punching sheer reinforcement systems- In case of thin flat slab

construction punching sheer reinforcements are indispensable.
 3. Optimization of Main reinforcement- In certain design procedures, especially
in yield line output is better optimized than in other design methods.

Benefits of Using Flat Slab Construction Method

o Flexibility in room layout

o Saving in building height
o Shorter construction time
o Ease of installation of M&E services
o Use of prefabricated welded mesh
o Buildable score

Flexibility in Room Layout

Flat slabs allows Architect to introduce partition walls anywhere required, this
allows owner to change the size of room layout. Use of flat slab allows choice of
omitting false ceiling and finish soffit of slab with skim coating.

Saving in Building Height

o Lower storey height will reduce building weight due to lower partitions and cladding
to façade
o Approximately saves 10% in vertical members
o Reduced foundation load
Shorter Construction Time

Use of flat slabs requires less time for construction by the use of big table formwork.

Single Soffit Level

Ease of Installation of Flat Slabs

All M & E services can be mounted directly on the underside of the slab instead of
bending them to avoid the beams.

Use of Prefabricated Welded Mesh

Use of prefabricated welded mesh minimizes the installation time of flat slabs. These
mesh are available in standard size and provides better quality control in construction
of flat slab.

Buildable Score

This allows standardized structural members and prefabricated sections to be

integrated into the design for ease of construction. This process makes the structure
more buildable, reduce the number of site workers and increase the productivity at
site, thus providing more tendency to achieve a higher Buildable score.
USES OF FLAT SLAB:

1. Flat slabs are mostly used in large industrial structures, parking garages, ramps,
warehouse, high rise buildings and hotels.

2. They are also used where uses of beams are not required.

3. Or where the structure requires less formwork.

ADVANTAGES AND DISADVANTAGES OF FLAT SLAB

• ADVANTAGES

1. It reduces the overall height of the structure.

2. Flat slabs are capable to carry concentrated loads.

3. Requires less formwork.

4. As reinforcement detailing of flat slabs is simple it is easy to place.

5. Better quality control.

6. Installation of sprinkler and other piping and utilities are easier due to absent of
beams.

7. It gives a better appearance and better diffusion of light.

8. Better fire resistant than other floor systems.

9. Fast construction.

• DISADVANTAGES

1. In flat plate system, construction of large span is not possible.

2. Use of drop panels may interfere with larger mechanical ducting.

4.2. DESIGN OF FLAT SLAB BY IS: 456

The term flat slab means a reinforced concrete slab with or without drops, supported
generally without beams, by columns with or without flared column heads. A flat
slab may be solid slab or may have recesses formed on the soffit so that the soffit
comprises a series of ribs in two directions. The recesses may be formed by
removable or permanent filler blocks.

Components of flat slab design:

a) Column strip : Column strip means a design strip having a width of 0.25 I, but
not greater than 0.25 1, on each side of the column centre -line, where I, is the span
in the direction moments are being determined, measured centre to centre of supports
and 1, is the -span transverse to 1, measured centre to centre of supports.

b)Middle strip : Middle strip means a design strip bounded on each of its opposite
sides by the column strip.

c) Panel: Panel means that part of a slab bounded on-each of its four sides by the
centre -line of a Column or centre-lines of adjacent-spans.
Division into column and middle strip along:
Longer span Shorter span

𝐿1 =6.6 m , 𝐿2 =5.6 m 𝐿1 =5.6 m , 𝐿2 =6.6 m

( i ) column strip ( i ) column strip

= 0.25 𝐿2 = 1.4 m = 0.25 𝐿2 = 1.65 m
But not greater than 0.25 𝐿1 = 1.65 m But not greater than 0.25 𝐿1 = 1.4 m

(ii) Middle strip (ii) Middle strip

= 5.6 – (1.4+1.4) = 2.8 m = 6.6 – (1.4+1.4) = 3.8 m
d) Drops: The drops when provided shall be rectangular in plan, and have a length
in each direction not less than one- third of the panel length in that direction. For
exterior panels, the width of drops at right angles to the non- continuous edge and
measured from the centre -line of the columns shall be equal to one -half the width
of drop for interior panels.

Fig.2
Since the span is large it is desirable to provide drop.
Drop dimension along:
Longer span Shorter span

𝐿1 =6.6 m , 𝐿2 =5.6 m 𝐿1 =5.6 m , 𝐿2 =6.6 m

Not less than 𝐿1 /3 = 2.2 m Not less than 𝐿1 /3 = 1.866 m

Hence provide a drop of size 2.2 x 2.2 m i.e. in column strip width.

e) Column head : Where column heads are provided, that portion of a column head
which lies within the largest right circular cone or pyramid that has a vertex angle of
90”and can be included entirely within the outlines of the column and the column
head, shall be considered for design purposes (see Fig. 3).
 Fig.3

Column head dimension along:

Longer span Shorter span

𝐿1 =6.6 m , 𝐿2 =5.6 m 𝐿1 =5.6 m , 𝐿2 =6.6 m

Not greater than 𝐿1 /4 = 1.65 m Not greater than 𝐿1 /4 = 1.4 m

Adopting the diameter of column head = 1.30 m =1300 mm

f) Depth of flat slab: The thickness of the flat slab up to spans of 10 m shall be
generally controlled by considerations of span ( L ) to effective depth ( d ) ratios
given as below:

Cantilever 7; simply supported 20; Continuous 26

For slabs with drops, span to effective depth ratios given above shall be applied
directly; otherwise the span to effective depth ratios in accordance with above shall
be multiplied by 0.9. For this purpose, the longer span of the panel shall be
considered. The minimum thickness of slab shall be 125 mm.
Depth of flat slab:
Considering the flat slab as a continuous slab over a span not exceeding 10 m

𝐿 𝐿
= 26 =˃ 𝐷 =
𝐷 26

Depth considering along:

Longer span Shorter span

L1 =6.6 m , L2 =5.6 m L1 =5.6 m , L2 =6.6 m
d  L  6600 =253.8 mm d  L  5600 =215.3 mm
26 26 26 26
Say 260 mm Say 220 mm

Taking effective depth of 25mm

Overall depth D = 260 +25 = 285 mm ˃ 125 mm (minimum slab thickness as per IS:
456)
⸫ It is safe to provide depth of 285 mm.
g) Estimation of load acting on the slab:
Dead load acting on the slab = 0.285 x 25 = 6.25 KN / m2 = 𝑤𝑑1
Floor finishes etc. load on slab = 1.45 KN / m2 = 𝑤𝑑2
Live load on slab = 7.75 KN / m2 = 𝑤𝑙
Total dead load = 𝑤𝑑1 +𝑤𝑑2 =7.7 KN / m2 = 𝑤𝑑
The design live load shall not exceed three times the design dead load.
Check:
𝑤𝑙 7.75
= = 1.006 < 3 𝑜𝑘
𝑤𝑑 7.7

Total design load= 𝑤𝑑 + 𝑤𝑙 = 15.45KN/m 2

h) Total Design Moment for a Span: The absolute sum of the positive and average
and is given by negative bending moments in each direction shall be taken as:
 𝑊 𝑙𝑛
𝑀0 =
8

𝑀0 = total moment.
W = design load on an area 𝑙1 𝑙2
𝑙𝑛 = clear span extending from face to face of columns, capitals, brackets or walls,
but not less than 0.65 𝑙1
𝑙1 = length of span in the direction of M
𝑙2 = length of span transverse to 𝑙1
Circular supports shall be treated as square supports having the same area.
Equivalent side of the column head having the same area:

π𝑑 2
𝜋1.132
𝐴= = = 1.152𝑚
4 4
1 1
Clear span along long span = 𝑙𝑛 = 6.6 - (1.152) - (1.152) = 5.448m > 4.29m
2 2
(should not be less than 0.65 l1) ∴ ok
1 1
Clear span along long span = 𝑙𝑛 =5.6 - (1.152) - (1.152) = 4.448m > 3.64m
2 2
(should not be less than 0.65 𝑙1 ) ∴ ok
Total design load along

Longer span Shorter span

𝑙𝑛 =5.448 m , 𝑙2 =5.6 m 𝑙𝑛 =4.448 m , 𝑙2 =6.6 m

W = 𝑤21 x 𝑛1 W =𝑤21 x 𝑛1
W =15.45 x 5.6 x 5.448 = 471.36K W =15.45 x 6.6 x 4.448 = 452.74K
The absolute sum of –ve and +ve moment in a panel along:

Longer span Shorter span

𝑙𝑛 =5.448 m , 𝑙2 =5.6 m 𝑙𝑛 =4.448 m , 𝑙2 =6.6 m

𝑊 𝑙𝑛 471.36𝑥5.44 𝑊 𝑙𝑛 452.74𝑥4.44
𝑀0 = = = 320.99 𝑀0 = = = 251.2
8 8 8 8

(i) Negative and Positive Design Moments:

The negative design moment shall be at the face of rectangular supports, circular
supports being treated as square supports having the same columns built integrally
with the slab system area. Shall be designed to resist moments arising from loads.
In an interior span, the total design moment M shall be distributed in the following
proportions:
Negative design moment 0.65
Positive design moment 0.35
In an end span, the total design moment M shall be distributed in the following
proportions:
0.10
Interior negative design moment= 0.75 -
1−𝑎𝑐
0.28
Positive design moment= 0.65 -
1−𝑎𝑐
0.65
Exterior negative design moment =
1−𝑎𝑐

𝑎𝑐 is the ratio of flexural stiffness of the exterior columns to the flexural stiffness of
the slab at a joint taken in the direction moments are being determined and is given
by:
∑ 𝐾𝑐
𝑎𝑐 =
𝐾𝑠
𝐾𝑐 =sum of the flexural stiffness of the columns meeting at the joint.
𝐾𝑠 =flexural stiffness of the slab, expressed as moment per unit rotation
𝑊 𝑙𝑛
𝑀0 =
8
The negative moment section shall be designed to resist the larger of the two interior
negative design moments determined for the spans framing into a common support
unless an analysis is made to distribute the unbalanced moment in accordance with
the stiffness of the adjoining parts.

Column strip :Negative moment at an interior support: At an interior support, the

column strip shall be designed to resist 75 percent of the total negative moment in
the panel at that support.

Negative moment at an exterior support:

a) At an exterior support, the column strip shall be designed to resist the t otal
negative moment in the panel at that support.
b) Where the exterior support consists of a column or a wall extending for a distance
equal to or greater than three-quarters of the value of l 2 . The length of span
transverse to the direction moments are being determined, the exterior negative
moment shall be considered to be uniformly distributed across the length l 2 .
Positive moment for each span : For each span, the column strip shall be designed
to r esist 60 percent of the total positive moment in the panel.
Moments in the middle strip :
a)That portion of-the design moment not resisted by the column strip shall be
assigned to the adjacent middle strips.
b)Each middle strip shall be proportione d to resist the sum of the moments assigned
to its two half middle strips. cl The middle strip adjacent and parallel to an edge
supported by a wall shall be proportioned, to resist twice the moment assigned to
half the middle strip corresponding to the fir st row of interior columns.

Stiffness calculation:
let the height of the floor = 4.0 m
clear height of the column = height of floor –depth of drop – thickness of slab –
thickness of head. = 4000 – 140 – 285 – 300 = 3275 mm
Effective height of column = 0.8 x 3275 = 2620 mm
(Assuming one end hinged and other end fixed)
stiffness coefficient:
 ∑ 𝐾𝑐 sum of flexural stiffness of column acting at the joint
𝑎𝑐 = =
𝐾𝑠 flexural stiffness of the slab
Longer span:
4𝐸𝐼 4𝐸𝐼 4𝐸𝐼 4𝐸 504 2 𝑥 4𝐸 𝑥 520 𝑥 1000
𝐾𝑐 = ( ) +( ) - 2 x( ) -2x( 𝐿 ) x( 12 ) = = 1.39
𝐿 𝑏𝑜𝑡𝑡𝑜𝑚 𝐿 𝑡𝑜𝑝 𝐿 327.5

From table 17 of IS: 456 -2000

𝑙2 𝑤𝑙
= 0.848 & = 1.00
𝑙1 𝑤𝑑

∴ 𝑎𝑐 min = 0.7
𝑎𝑐 > 𝑎𝑐 𝑚𝑖𝑛

Hence correction for pattern of loading in the direction of longer span is not
required.

Shorter span:

2 𝑥 504
𝐾𝑐 = = 3975.8
12 𝑥 262

560 𝑥 28.53
𝐾𝑠 = = 1421.4
12 𝑥 760

3975.8
𝑎𝑐 = = 2.79
1421.4

From table 17 of IS: 456 -2000 for

𝑙2 𝑤𝑙
= 1.17 & = 1.00
𝑙1 𝑤𝑑

𝑎𝑐 min ≈ 0.75
𝑎𝑐 > 𝑎𝑐 𝑚𝑖𝑛
Hence correction for pattern of loading in the direction of short span is not
required.
From table 17 of IS-456,

Table 1

Distribution of bending moment across the panel width

It is san exterior panel
Longer span:
Column strip
−0.65𝑚𝑜 −0.65 𝑥 320.99
-ve B.M. at exterior support = [ 1 ] x 1.0 =[ 1 ] = -121.34 KNm
1+ 1+
𝑎𝑐 1.39
 0.28 0.28
+ve span BM = [0.63 − 1 ] x mo x 0.40 = [0.63 − 1 ] x 320.99 x 0.40 = 59.96
1+ 1+
𝑎𝑐 1.39
KNm

0.10 0.10
-ve BM at interior support = [0.75 − 1 ] x mo x 0.75 = [0.75 − 1 ] x 320.99
1+ 1+
𝑎𝑐 1.39
x 0.75 = -55.50 KNm
Short span
Column strip
−0.65𝑚𝑜 −0.65 𝑥 251.2
-ve moment at exterior support = [ 1 ] x 1.0 =[ 1 ] = -120.19 KNm
1+ 1+
𝑎𝑐 2.79

0.28 0.28
+ve moment = [0.63 − 1 ] x mo x 0.60 = [0.63 − 1 ] x 251.2 x 0.60 = 63.88
1+ 1+
𝑎𝑐 2.79
KNm

0.10 0.10
-ve moment at interior support = [0.75 − 1 ] x mo x 0.75 = [0.75 − 1 ] x 251.2
1+ 1+
𝑎𝑐 2.79
x 0.75 = -127.43 KNm

Middle strip
−0.65𝑚𝑜
-ve moment at exterior support = [ 1 ] x 0.0 = 0KNm
1+
𝑎𝑐

0.28 0.28
+ve moment = [0.63 − 1 ] x mo x 0.60 = [0.63 − 1 ] x 251.2 x 0.40 = 42.59
1+ 1+
𝑎𝑐 2.79
KNm
 0.10 0.10
-ve moment at interior support = [0.75 − 1 ] x mo x 0.75 = [0.75 − 1 ] x 251.2
1+ 1+
𝑎𝑐 2.79
x 0.25 = -42.43 KNm
j) Effective depth of slab
Thickness of the slab, from consideration of maximum positive moment any where
in the slab. Maximum +ve BM occurs in the column strip (long span) = 90.91 KNm
∴factored moment = 1.50 x 90.91 = 136.36 KNm
Mo = 0.138 𝑓𝑐𝑘 b𝑑 2 (b=2800mm)
136.36 106
d=√ (M 20 grade concrete)
0.138 𝑥 20 𝑥 2800

d=132.83 mm ≅ 140 mm
Using 12 mm ∅ (diameter) main bars.
12
Overall thickness of slab = 140 + 15 + =161mm ≡ 170𝑚𝑚
2
12
∴Depth (along longitudinal direction) = 170 – 15 - = 150 mm
2

∴Depth (along longitudinal direction) = 150 - 12 = 138 mm

j) Thickness of drop from maximum –ve moment consideration
Thickness of drop from consideration of maximum –ve moment any where in the
panel.
Max –ve BM occurs in the column strip = 166.6 KNm
M u = 0.138 𝑓𝑐𝑘 b𝑑 2
1.5 x 166.6 x 106 =0.138 x 20 x 1400 x 𝑑 2
d= 254.3mm
say, 260mm. Use 12 mm ∅ (diameter) bars.
12
Overall thickness of flat slab = D = 260 + 15 + = 281mm
2
 Fig.3 Fig.4

l)Shear in Flat Slab

The critical section for shear shall be at a distance d/2 from the periphery of the
column/capital/ drop panel, perpendicular to the plane of the slab where d is the
effective depth of the section. The shape in plan is geometrically similar to the
support immediately below the slab.
check for shear stress developed in slab
The critical section for shear for the slab will be at a distance d/2 from the face of
drop.
Perimeter of critical section = 4 x 2340 = 9340 mm
V0 = 1.5 X 15.45 X [ L1 X L2 - (2.34)(2.34)]
Total factored shear force:= 1.5 X 15.45 X [6.6 X 5.6-(5.47)]
= 729.78 KN
𝑉 729.78 𝑋 1000
Nominal shear stress = 𝜏 = = = 0.55N/mm
𝑏𝑑 9340 𝑋 140

Shear strength of concrete = 𝜏𝑐 = 0.25√𝑓𝑐𝑘 = 0.25√20 = 1.11N/mm

Permissible shear stress = 𝜏𝑣 ⊁ 𝐾𝑠 𝜏𝑐
𝐾𝑠 = ( 0.5 + 𝛽𝑐 ), 𝛽𝑐 = 0.848
𝐾𝑠 = ( 0.5 + 0.848 )
𝐾𝑠 =1.348 > 1.11
= 1 X 1.11
= 1.11N/𝑚𝑚2
𝜏𝑣 < 𝜏𝑐 ∴ safe design ok
If 𝜏𝑣 > 1.5𝜏𝑐 , then the slab should be redesigned
m) check for shear in drop
b0 = D + d0 )
= 1.3 + 0.26) = 4.89m
V=1.5 X 15.45[5.6 X 6.6 -𝜋/4 (1.3 + 0.26) 2]
V = 812.27KN
812.27 𝑥 103
𝜏𝑣 = = 0.683 𝑁/𝑚𝑚2
4890 𝑥 260
Nominal shear stress: 𝜏𝑐 = 0.25 𝑓𝑐𝑘 = 1.11 𝑁/𝑚𝑚2
𝜏𝑣 < 𝜏𝑐 (safe in shear)
n)Reinforcement details
Longer span
-ve exterior reinforcement:
M u = 0.87 f y Ast [d- 0.42 xu ]
1.5 x 121.34 x 106 x 0.87 x 415 = Ast[150 - 0.42 x 0.48 x150]
Ast = 4209 mm2
4209
Use 12 mm ∅ bars = = 38No. s
113
3.8 𝑋 1000
∴ c/c spacing = = 135 𝑚𝑚 𝑐/𝑐
28

Reinforcement along short span:

Column strip:
M u = 0.87 f y Ast [d- 0.42 xu ]
1.5 x 127.5 x 106 = 0.87 x 415 x Ast[140 - 0.42 x 0.48 x140]
Ast = 3768.9 mm2
3768.9
Use 12 mm ∅ bars = = 33No. s
3.14 𝑥 122 /2

1.4 𝑋 1000
∴ c/c spacing = = 42 𝑚𝑚 𝑐/𝑐
33

Middle strip:
M u = 0.87 f y Ast [d- 0.42 xu ]
1.5 x 63.88 x 106 = 0.87 x 415 x Ast[281 - 0.42 x 0.48 x281]
Ast = 1182 mm2
1182
Use 12 mm ∅ bars = = 10No. s
3.14 𝑥 122 /2

2.8 𝑋 1000
∴ c/c spacing = = 42 𝑚𝑚 𝑐/𝑐
10