Chapter # 7
Design of Two-Way Slab
Reinforced Concrete Slabs
Reinforced Concrete Slabs are large flat plates that are
supported by reinforced concrete beams, walls, or columns,
by masonry walls, by structural steel beams or columns, or
by the ground.
If a slab is supported on two opposite sides only, it is referred
to as a One-Way Slab because the bending is in one
direction only – that is, perpendicular to the supported edges
If a slab is supported by beams on all four edges, it is
referred to as Two-Way Slab because the bending is in
both direction
Dr. Roz-Ud-Din Nassar
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Reinforced Concrete Slabs
A one-way slab shows a cylindrical failure, while a two-way
slab fails in bowl action.
In a slab, if the ratio of the long side to the short side is 2 or
more than 2, it is referred to as a one-way slab, otherwise it
is a called a two-way slab
A Two-way slab carries the load in the two perpendicular
directions
The main flexural reinforcement is therefore placed in the
two perpendicular directions
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Reinforced Concrete Slabs
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The Two Way Slabs

In two way slab, the load is transmitted in two
directions hence the system is called two way slab
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Types of Two-way Slabs
Flat Plate
Flat Slab
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Types of Two-way Slabs
Slab with Beams
Waffle Slab
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Slab Curvatures and Moments
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Statical Equilibrium of Two-Way Slabs
Consider a floor made up of simply supported planks
supported by simply supported beams.
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Statical Equilibrium of Two-Way Slabs
The floor carries a load of w lb/ft2. The moment per foot
of width in the planks at section A-A is :
𝑤𝑙12
𝑚=
𝑓𝑡 − 𝑘𝑖𝑝𝑠/𝑓𝑡
8
The total moment in the entire width of the floor is:
𝑀 = 𝑤𝑙2
𝑤𝑙12
𝑓𝑡 − 𝑘𝑖𝑝𝑠
8
Which is the familiar equation for the maximum moment
in a simply supported floor of width 𝒍𝟐 and span 𝒍𝟏
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Statical Equilibrium of Two-Way Slabs
Moments in a Plank-and-Beam Floor
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Statical Equilibrium of Two-Way Slabs
The planks apply following uniform load on each beam:
𝑤𝑙1
/𝑓𝑡
2
The moment at section B-B in one beam is thus:
𝑀1𝑏
𝑤𝑙1 𝑙22
=
𝑓𝑡 − 𝑘𝑖𝑝𝑠
2 8
The total moment in both beams is:
𝑀 = 𝑤𝑙1
Dr. Roz-Ud-Din Nassar
𝑙22
𝑓𝑡 − 𝑘𝑖𝑝𝑠
8
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Transfer of Moments in Two-Way Slab
It is important to note that full load was transferred East and
West by the planks, causing a moment equivalent to
𝒘𝒍𝟐ൗ
𝟖
in
the planks, and then was transferred North and South by
the beams, causing a similar moment in the beams
Same thing happens in the two-way slab. The total moments
required along sections A-A and B-B respectively, are:
𝑀 = 𝑤𝑙2
Dr. Roz-Ud-Din Nassar
𝑙12
8
and
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𝑀 = 𝑤𝑙1
𝑙22
8
13
Transfer of Moments in Two-Way Slab
Moments in a Two-Way Slab
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Transfer of Moments in Two-Way Slab
In this case again, the full load was transferred East and
West, and then the full load was transferred North and
South, this time by the slab in both cases
Such mechanism should be true regardless of whether the
structure has one-way slabs and beams, two-way slabs or
some other system
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Effect of Beam Stiffness on Moments in Slabs
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Design Methods of Two Way Slabs
• Direct Design Method
• Equivalent Frame Method
• Elastic Yield Line Theory
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Two-Way Slab: Direct Design Method
Steps in Slab Design
1) Choose the layout and type of slab to be used. The choice of type
of slab is strongly affected by architectural and construction
calculations
2) Choose the slab thickness. Generally, the slab thickness is chosen
to prevent excessive deflection in service. Equally important is to
satisfy the shear requirements at both exterior and interior columns
3) Choose the method for computing the design moments (direct
design OR equivalent frame)
4) Compute the positive and negative moments in the slab
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Two-Way Slab: Direct Design Method
Steps in Slab Design
5) Calculate the distribution of moments across the width of the slab.
The lateral distribution of moments within a panel depends on the
geometry of the slab and the stiffness of the beams (if any)
6) If there are beam, a portion of the moment must be assigned to the
beams
7) Reinforcement is designed for the moments from steps 5 and 6
8) The shear strengths at the columns are checked
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Limitations on the Use of Direct Design Method
The limitations given by ACI Code Section 13.6.1 are as
under:
1) There must be a minimum of three continuous spans in each
direction. Thus a nine-panel structure (3 by 3) is the smallest that
can be considered
2) Rectangular panels must have a long-span / short-span
ratio not greater than 2
3) Successive span lengths in each direction shall not differ by
more than one-third of the longer span
4) Columns may be offset from the basic rectangular grid of the
building by up to 0.1 times the span parallel to the offset
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Limitations on the Use of Direct Design Method
5) All the loads must be due to gravity only. The direct design method
cannot be used for unbraced laterally loaded frames, foundations mats,
or prestressed slabs
6) The service live load shall not exceed two times the service dead
load
7) For a panel with beams between supports on all sides the relative
stiffness of the beams in the two perpendicular directions given by:
𝛼1 𝑙22
𝛼2 𝑙12
(where 𝛼 is beam-to-slab stiffness ratio)
The above ratio shall not be less than 0.2 or greater than 5
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Beam-to-Slab Stiffness Ratio 𝛼
Slabs are frequently built with beams spanning from column to
column around the perimeter of the building.
These beams act to stiffen the edge of the slab and help to reduce
the reduce the deflections of the exterior panels of the slab.
In the ACI code, the effects of beam stiffness on deflections and
the distribution of moments are expressed as a function of 𝛼,
defined as the flexural stiffness 4𝐸𝐼 Τ𝑙, of the beam divided by the
flexural stiffness of a width of slab bounded laterally by the centerlines of the adjacent panels on each side of the beam:
𝟒𝑬𝒄𝒃 𝑰𝒃
𝒍
𝜶𝒇 =
𝟒𝑬𝒄𝒔 𝑰𝒔
𝒍
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Beam-to-Slab Stiffness Ratio 𝛼
Since the lengths "𝒍” of the beam and slab are equal, the equation
reduces to:
𝟒𝑬𝒄𝒃 𝑰𝒃
𝜶𝒇 =
𝟒𝑬𝒄𝒔 𝑰𝒔
Where 𝐸𝑐𝑏 𝑎𝑛𝑑 𝐸𝑐𝑠 are the moduli of elasticity of beam concrete
and slab concrete, respectively, and 𝐼𝑏 𝑎𝑛𝑑 𝐼𝑠 are the moments of
inertia of uncracked beams and slabs.
The sections considered in computing 𝐼𝑏 𝑎𝑛𝑑 𝐼𝑠 are shown in
figures on next slide.
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Beam-to-Slab Stiffness Ratio 𝛼

Definition of terms
Ecb = Modulus of elasticity of beam concrete
Ecs = Modulus of elasticity of slab concrete
Ib =
Moment of inertia of uncracked beam section
Is =
Moment of inertia of uncracked slab section
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Beam-to-Slab
Stiffness Ratio 𝛼𝑓
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Beam-to-Slab Stiffness Ratio 𝛼𝑓
Cross Section of Beams as Defined in ACI Code 13.2.4
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Problem # 1
An 8-in. thick slab is provided with an edge beam that has a
total depth of 16 in. and a width of 12 in., as shown in figure.
The slab and the beam are cast monolithically and have the
same concrete strength and the same Ec. Compute “𝜶𝒇 ” in
this case.
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Problem # 1
20 𝑖𝑛.
Find the centroid
of the L-section
8 𝑖𝑛.
8 𝑖𝑛.
12 𝑖𝑛.
Dr. Roz-Ud-Din Nassar
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8 𝑖𝑛.
28
Problem # 1 - Solution
Solution
𝟒𝑬𝒄𝒃 𝑰𝒃
𝜶𝒇 =
𝟒𝑬𝒄𝒔 𝑰𝒔
Since:
′
𝑓𝑐𝑠′ = 𝑓𝑐𝑏
and 𝐸𝑐𝑏 = 𝐸𝑐𝑠
𝑰𝒃
→ 𝜶𝒇 =
𝑰𝒔
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Problem # 1 - Solution
3
163
8
𝐼𝑏 = 12 ×
+ 12 × 16 × 12 + 8 ×
+ 8 × 8 × 32
12
12
𝐼𝑏 = 5205 𝑖𝑛.4
83
𝐼𝑠 = 126 ×
= 5376 𝑖𝑛.4
12
𝐼𝑠 = 5376 𝑖𝑛.4
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Problem # 1 - Solution
𝟓𝟐𝟎𝟓
𝜶𝒇 =
= 𝟎. 𝟗𝟔𝟖
𝟓𝟑𝟕𝟔
Alternatively:
𝑬𝒄𝒃 𝒃 𝒂
𝜶𝒇 =
𝑬𝒄𝒔 𝒍 𝒉
𝟑
𝒇
From graphs:
𝑎
ℎ
Dr. Roz-Ud-Din Nassar
= 2 𝑎𝑛𝑑
𝑏
ℎ
=
12
8
= 1.5, we get 𝑓 = 1.27
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Problem # 1 - Solution
𝑬𝒄𝒃 𝒃 𝒂
𝜶𝒇 =
𝑬𝒄𝒔 𝒍 𝒉
12 16
𝜶𝒇 =
126 8
Dr. Roz-Ud-Din Nassar
𝟑
𝒇
3
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× 1.27 = 0.968
32
Division of Slab into Frames for Design
For design, the slab is considered to be series of frames in the
two directions
These frames extend to the middle of the panels on each side
of the column lines
In each span of each of the frames, it is necessary to
compute the total statical moment, Mo
𝑤𝑢 𝑙2 𝑙𝑛2
𝑀𝑜 =
8
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Division of Slab into Frames for Design
𝑤𝑢 𝑙2 𝑙𝑛2
𝑀𝑜 =
8
𝑤𝑢 = factored load per unit area
𝑙2 = transverse width of the strip
𝑙𝑛 = clear span between the columns
In computing ln, circular columns or column capitals of diameter
dc are replaced by equivalent square columns with side lengths
0.886dc.
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Division of Slab into Frames for Design
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Division of Slab into Frames for Design
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Problem # 2
Compute the statical
moment in the slab
panels shown in figure
below. Slab has a
thickness of 8 in. and
supports a live load of
100 psf.
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Problem # 2 - Solution
Solution:
1. Compute the factored uniform loads
8
𝑤𝑢 = 1.2
× 0.150 𝑘𝑠𝑓 + 1.6 0.100 𝑘𝑠𝑓
12
𝑤𝑢 = 0.280 𝑘𝑠𝑓
Note: Live load reduction must be carried out if allowed by the code
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Problem # 2 - Solution
2. Consider panel A
spanning from column
1 to column 2
𝑤𝑢 𝑙2 𝑙𝑛2
𝑀𝑜 =
8
𝑙𝑛 = 𝑐𝑙𝑒𝑎𝑟 𝑠𝑝𝑎𝑛 𝑜𝑓 𝑠𝑙𝑎𝑏 𝑝𝑎𝑛𝑒𝑙
1 20
1 24
𝑙𝑛 = 22 𝑓𝑡 −
𝑓𝑡 −
𝑓𝑡
2 12
2 12
𝑙𝑛 = 20.17 𝑓𝑡
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Problem # 2 - Solution
𝑙2 = 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑠𝑙𝑎𝑏 𝑝𝑎𝑛𝑒𝑙
21
20
𝑙2 =
𝑓𝑡 +
𝑓𝑡
2
2
Hence:
→ 𝑙2 = 20.50 𝑓𝑡
𝑤𝑢 𝑙2 𝑙𝑛2
𝑀𝑜 =
8
0.280 𝑘𝑠𝑓 × 20.5 𝑓𝑡 20.17 𝑓𝑡
→ 𝑀𝑜 =
8
2
= 292 𝑓𝑡 − 𝑘𝑖𝑝𝑠
This moment will be used to design the reinforcement running
parallel to line 1-2 in this panel
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Problem # 2 - Solution
3. Consider panel B
spanning from
column 1 to column 4
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Problem # 2 - Solution
Section through slab panel B
For calculation of 𝑙𝑛 the circular supports are replaced by equivalent square
columns having side length 𝑐1
Dr. Roz-Ud-Din Nassar
= 𝑑𝑐
CIEN 422
𝜋
2
𝑜𝑟 0.886𝑑𝑐
42
Problem # 2 - Solution
3. Consider panel B spanning
from column 1 to column 4
𝑙𝑛 = 𝑐𝑙𝑒𝑎𝑟 𝑠𝑝𝑎𝑛 𝑜𝑓 𝑠𝑙𝑎𝑏 𝑝𝑎𝑛𝑒𝑙
1 12
1
24
𝑙𝑛 = 20𝑓𝑡 −
𝑓𝑡 −
0.886 ×
𝑓𝑡
2 12
2
12
𝑙𝑛 = 18.61 𝑓𝑡
𝑙2 =
19
22
𝑓𝑡 +
𝑓𝑡
2
2
→ 𝑙2 = 20.50 𝑓𝑡
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Problem # 2 - Solution
𝑤𝑢 𝑙2 𝑙𝑛2
𝑀𝑜 =
8
0.280 𝑘𝑠𝑓 × 20.5 𝑓𝑡 × 18.16 𝑓𝑡
→ 𝑀𝑜 =
8
2
= 248 𝑓𝑡 − 𝑘𝑖𝑝𝑠
This moment will be used to design the reinforcement running
parallel to line 1-4 in this panel
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Definition of Column Strip and Middle Strip
The moments vary continuously across the width of
the slab panel
To add in steel placement, the design moments are
averaged over the width of column strips over the
columns and middle strips between the column strips
as shown in figures on next slide
The width of these strips are defined in ACI Section
13.2.1 and 13.2.2 and are illustrated in figures on slides
46 & 47
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Definition of Column Strip and Middle Strip
Middle strips
Column strips
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Definition of Column Strip and Middle Strip
Column
strips
Middle
strips
Column
strips
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47
Column Strip and Middle Strip
Four edges fixed and non-supported
on non-deflecting beams
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Four edges fixed on flexible beams
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Positive and Negative Moments in Interior Span
In the direct design method, the total factored moment Mo is
divided into positive and negative factored moments according
to rules given in ACI Section 13.6.3
In interior spans, 65 percent of Mo is assigned to the negativemoment region and 35 percent to the positive-moment regions
The exterior end of an exterior span has considerable less
fixity than the end at the interior support. The division of Mo in
an end span into positive and negative-moment regions is
given in Table on slide 51
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Positive and Negative Moments in Interior Span
Exterior Span
Interior Span
Assignment of positive and negative moment regions
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50
Distribution of Mo in an Exterior Span
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51
Distribution of Moments between Column
Strips and Middle Strips
ACI code 13.6.4 defines the fraction of negative and
positive moments assigned to column strips
The remaining amount of negative and positive moments
is assigned to adjacent half-middle strips
The division is a function of 𝛼𝑓1 × 𝑙2 Τ𝑙1 which depend
upon the aspect ratio of the panel 𝑙2 Τ𝑙1 and the relative
stiffness 𝛼𝑓1 of the beams (if any) spanning parallel to and
within the column strip
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Percentage Distribution of Interior Negative
Factored Moment to Column Strip
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Percentage Distribution of Midspan Positive
Factored Moment to Column Strip
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Percentage Distribution of Exterior Negative
Factored Moment to Column Strip
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Torsional Stiffness Ratio 𝛽𝑡
𝐸𝑐𝑏 𝐶
𝛽𝑡 =
𝐸𝑐𝑠 𝐼𝑠
𝐶=෍
𝑥 𝑥 3𝑦
1 − 0.63
𝑦 3
Where:
𝐸𝑐𝑏 = Modulus of Elasticity of beam concrete
𝐸𝑐𝑠 = Modulus of Elasticity of slab concrete
𝐼𝑠 = Moment of inertia of uncracked slab
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Percentage Distribution of Exterior Negative
Factored Moment to Column Strip
Division of edge members for calculation of torsion constant “C”
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57
Minimum Thickness of Two-way Slab
ACI Section 9.5.3 defines minimum thicknesses that
generally are sufficient to limit slab deflections to
acceptable values
Thinner slabs can be used if it can be shown that the
computed slab deflections will not be excessive
For computation of thickness, slabs have been divided
into two categories:
• Slabs without Beams between Interior Columns
• Slabs with Beams between Interior Supports
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Thickness of Slabs Without Beams Between Interior
Columns
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Thickness of Slabs With Beams Between Interior Supports
For slabs with beams between interior supports, ACI Code
Section 9.5.3.3 gives the following minimum thicknesses:
A) For 𝛼𝑓𝑚 ≤ 0.2 , the minimum thickness in Table 13-1 shall apply
B) For 0.2 < 𝛼𝑓𝑚 < 2.0 the thickness shall not be less than:
ℎ=
𝑙𝑛 0.8 + 𝑓𝑦 Τ20,000
36 + 5𝛽 𝛼𝑓𝑚 − 0.2
but not less than 5 in.
C) For 𝛼𝑓𝑚 > 2.0 the thickness shall not be less than:
𝑙𝑛 0.8 + 𝑓𝑦 Τ20,000
ℎ=
36 + 9𝛽
Dr. Roz-Ud-Din Nassar
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but not less than 3.5 in.
60
Thickness of Slabs With Beams Between Interior Supports
D) At discontinuous edges, either an edge beam with a
stiffness ratio 𝛼𝑓 not less than 0.8 shall be provided or slab
thickness shall be increased by at least 10 percent in the
edge panel
In A, B, C, and D above:
ℎ = overall thickness
𝑙𝑛 = clear span of the slab panel under consideration, measured in the
longer direction
𝛼𝑓𝑚 = the average of the values of 𝛼𝑓 for the four sides of the panel
𝛽 = longer clear span/shorter clear span of the panel
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Problem # 3
Figure on slide # 63 shows an interior panel of the flatplate floor in an apartment building. The slab thickness is
5.5 in. The slab supports a design live load of 50 psf and
superimposed dead load of 25 psf for partitions. The
columns and slab have the same strength of concrete.
The story height is 9 ft. Compute the column-strip and
middle-strip moments in the shown direction of the panel.
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Problem # 3
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63
Problem # 3 - Solution
Solution:
1) Compute the factored loads
5.5
𝑞𝑢 = 1.2
× 150 + 25 + 1.6 50 = 193 𝑝𝑠𝑓
12
Note: Live load reduction must be carried out if local building code
allows it (Use KLL = 1.0 for two-way slab panels)
2) Compute moments in the shorter span of the slab
a) Compute 𝑙𝑛 and 𝑙𝑛 and divide the slab into column and middle strips
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64
Problem # 3 - Solution
10
𝑙𝑛 = 13.17 𝑓𝑡 −
= 12.33 𝑓𝑡
12
𝑙2 = 14.15 𝑓𝑡
The column strip extends the
smaller of 𝑙2 Τ4 or 𝑙1 Τ4 on each
side of the column centerline, as
shown in figure.
Thus column strip extends
13.17Τ4 𝑓𝑡 = 3.29 𝑓𝑡 on each
side of the column centerline.
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Problem # 3 - Solution
The total width of the column
strip is 6.58 ft
Each half middle strip extends
from the edge of the column strip
to the centerline of the panel
The total width of the two halfmiddle strips is 14.5 – 6.58 =
7.92 ft
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Problem # 3 - Solution
b) Compute 𝑀𝑂
𝑞𝑢 𝑙2 𝑙𝑛2
𝑀𝑂 =
8
0.193 × 14.5 × 12.332
→ 𝑀𝑂 =
= 53.2 𝑘𝑖𝑝 − 𝑓𝑡
8
c) Divide MO into negative and positive moments
From ACI code Section 8.10.4.1
𝑁𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑚𝑜𝑚𝑒𝑛𝑡 = −0.65𝑀𝑂 = −0.65 × 53.2 = −34.6 𝑘𝑖𝑝 − 𝑓𝑡
𝑃𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑚𝑜𝑚𝑒𝑛𝑡 = 0.35𝑀𝑂 = 0.35 × 53.2 = 18.6 𝑘𝑖𝑝 − 𝑓𝑡
Dr. Roz-Ud-Din Nassar
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Problem # 3 - Solution
d) Divide the moments between the column and middle strips
From Table 13-3 (slide # 53) for 𝛼𝑓1 𝑙2 Τ𝑙1 = 0 (𝛼𝑓1 = 0 because
there are no beams between columns A and B in this panel)
Column − strip negative moment = 0.75 × −34.6 = −26.0 𝑘𝑖𝑝 − 𝑓𝑡
Middle − strip negative moment = 0.25 × −34.6 = −8.66 𝑘𝑖𝑝 − 𝑓𝑡
Dr. Roz-Ud-Din Nassar
CIEN 422
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Problem # 3 - Solution
Process of calculating slab moments
Dr. Roz-Ud-Din Nassar
CIEN 422
69
Problem # 3 - Solution
Division of MO into positive and negative moments
Total moments in column and middle strips
Dr. Roz-Ud-Din Nassar
CIEN 422
70
Problem # 3 - Solution
Division of MO into positive and negative moments
Total moments in column and middle strips
Dr. Roz-Ud-Din Nassar
Mapping
of moments with strips
CIEN 422
71
Problem # 3 - Solution
Calculation of moments
Dr. Roz-Ud-Din Nassar
CIEN 422
72