UNIT 4 SLABS
Slabs
Structure
4.1
Introduction
Objectives
4.2
General Principles of Design and Detailing of Slabs
4.3
Design and Detailing of a Cantilever Slabs
4.4
Design and Detailing of One-way Simply Supported Slabs
4.5
Design and Detailing of Two-way Simply Supported Slabs
4.6
Design and Detailing of Two-way Restrained Slabs
4.7
Summary
4.8
Answers to SAQs
4.1 INTRODUCTION
A slab is like a flat plate loaded transversely and supported on its edges. Under
the loads, it bends and the directions of its bending depend on its shape and
support conditions. A beam bends only in one direction, i.e. in its own plane;
whereas a slab may have multidirectional bending. Therefore, slabs may have
different names depending upon its bending, support conditions and shapes. For
example, a slab may be called
(a)
One-way simply supported rectangular slab,
(b)
Two-way simply supported or restrained rectangular slab,
(c)
Cantilever rectangular slab,
(d)
Fixed or simply supported circular slab, etc.
One-way slab means it bends only in one direction and, therefore, reinforcement
for bending (i.e. main reinforcement) is provided only in that direction. A slab
supported on all sides bends in all the directions so the main reinforcements
provided shall be such that they may be effective in all directions. For ease of
analysis and convenience of reinforcement detailing, the bending moments in a
slab are calculated in two principal directions only and, therefore, such a slab is
called a two-way slab.
A slab is designed as a beam of unit width in the direction of bending. In this unit,
only the most commonly used rectangular slabs, with uniformly distributed load
is described.
Objectives
After studying this unit, you should be able to
•
describe the design and detailing of cantilever slabs,
•
design and explain detailing of one-way and two-way simply
supported slabs, and
•
explain the design and detailing of two-way restrained slabs.
75
Theory of Structures-II
4.2 GENERAL PRINCIPLES OF DESIGN AND
DETAILING OF SLABS
Following are the general principles for design and detailing applicable to all
types of slabs.
(a)
The maximum diameter of reinforcing bars shall not exceed
1
th of
8
total thickness (D) of the slab.
(b)
Normally, shear reinforcement is not provided in slabs. The shear
resistance requirements may, then, be complied either by increasing
the percentage of tensile reinforcement or by increasing the depth of
slab, but the latter is preferred as it is economical. For solid slabs, the
design shear strength for concrete slab shall be τc K, where K has the
values given below :
Overall
Depth of
Slab
(mm)
300 or
more
275
250
225
200
175
150 or
less
K
1.00
1.05
1.10
1.15
1.20
1.25
1.30
(c)
To take care of temperature and shrinkage stresses, minimum
reinforcement in either direction shall not be less than 0.15 percent
and 0.12 percent of total cross section area of concrete section for
mild steel and high strength deformed bars, respectively.
(d)
To meet the requirement for limit state of cracking the following two
rules are observed:
(i)
The horizontal distance between parallel main
reinforcement shall not be more than 3 times the effective
depth of slab or 300 mm whichever is smaller.
(ii)
The horizontal distance between parallel bars provided
against temperature and shrinkage shall not be more than
5 d or 450 mm, whichever is smaller.
4.3 DESIGN AND DETAILING OF CANTILEVER
SLABS
Design and detailing of a cantilever slab is the same as that of a cantilever beam
(Section 3.3) of unit width. Temperature and shrinkage reinforcement is provided
along the direction perpendicular to the span. This is illustrated through the
following example.
Example 4.1
Design the cantilever slab of a bus stand shown in Figure 4.1. Load data
and design parameters are given below :
Load Data
Lime terrace topping of 100 mm thickness is provided over the slab.
76
Imposed load = 0.75 kN/m2.
Slabs
Design Parameters
fck = 25 N/mm2;
fy = 415 N/mm2 and Nominal cover = 30 mm.
Figure 4.1 : Cantilever Roof Shade for Bus Stop
Solution
Effective Span (lef)
lef = 3 + 0.3* = 3.3 m
Depth of Slab (D)
From Deflection Control
l ef
d
≤ k B k1 k 2 k 3 k 4 , where kB = 7;
For k2,
fs = 0.58 fy
k1 = 1 as lef < 10 m
* Estimate of effective depths (d) :
l ef
300
d≈
≈
= 428.57 mm
7
7
d
∴ ≈ 300 mm = 0.3 m added
2
for evaluating lef (Cl. 22.2c).
Area of cross section of steel required
Area of cross section of steel provided
Assuming area of cross section of steel required = Area of cross
section of steel provided = Area of balanced tensile steel for M 25
concrete and Fe 415 steel (pt% = 1.19%). According to above
assumptions
fs = 0.58 × fy × 1 = 0.58 × 415 ≈ 240 N/mm2
and k2 = 0.96 (Figure 1.2)
k3 = k4 = 1 as the slab is singly reinforced and it is not a flanged
section.
Substituting all these values
lef
d
≤ 7 × 1 × .96 × 1 × 1 = 6.72
3.3 × 10 3
= 491 mm
6.72
or
d≥
∴
D=d+
Taking
D = 550 mm
∴
d =D−
10
φ
− Nominal cover = 550 −
− 30 = 515 mm
2
2
∴
lef = 3 +
0.515
= 3.26 m
2
φ
10
+ nominal cover = 491 +
+ 30 = 526 mm
2
2
(assuming φ = 10)
77
Theory of Structures-II
From Moment of Resistance Consideration
Loads
Self
= 1 × 1 × 0.550 × 25
= 13.75 kN/m2
Lime Terrace
= 1 × 1 × 0.1 × 18.8
= 1.88 kN/m2
Ceiling Plaster
= 1 × 1 × 0.01 × 20.4
= 0.204 kN/m2
Total (Dead Load)
= 15.834 kN/m2
Imposed Load (IL)
= 0.750 kN/m2
Total Load, w
= 16.584 kN/m2
wu lef 2
Maximum BM, Mu =
M u , lim = 0.36
2
xu ,
max
d
=
1.5 × 16.584 × 3.26 2
= 132.2 kNm
2
x
⎛
⎜1 − 0.42 u ,max
⎜
d
⎝
⎞
⎟ bd 2 f ck
⎟
⎠
132.2 × 10 6 = 0.36 × 0.48 × (1 − 0.42 × 0.48) × 1000 × d 2 × 25
or
d = 195.8 mm
Hence, provided D = 550 mm and d = 515 mm as above.
Tensile Reinforcement (Ast)
Ast f y
⎛
M u = 0.87 f y Ast d ⎜⎜1 −
bd f ck
⎝
⎞
⎟
⎟
⎠
⎛
132.2 × 106 = 0.87 × 415 × Ast × 515 × ⎜1 −
⎝
or,
or
Ast × 415 ⎞
⎟
1000 × 515 × 25 ⎠
132.2 × 10 6 = 185940.75 Ast − 5.993 Ast2
Ast2 − 31026.322 Ast + 22059068.9 = 0
Solving the above equation, we get
Ast = 728.06 mm2/m width
i.e.
φ 10 bars @ 105 mm c/c.
Ast , min =
0.12
× 1000 × 515 = 618 mm 2 /m < 728.06 mm 2 /m width
100
Spacing of φ 8 bars =
1000 × 50
= 80 mm c/c perpendicular to the main
618
reinforcement.
Maximum Spacing
3 d = 3 × 515 = 1545 > 105 mm c/c
300 > 105 mm c/c
78
Hence, provided φ 10 @ 105 mm c/c as main reinforcement and
φ 8 @ 80 mm c/c as temperature and shrinkage reinforcement.
Slabs
Check for Shear
SF at critical section, i.e. at d from the face of support
= Vu = 1.5 × 16.584 × (3 – 0.515)
= 61.817 kN
τυ =
Vu 61.817 × 10 3
=
bd
1000 × 515
= 0.12 N/mm 2
τc, min for M 25 concrete
= 0.29 N/mm2 > 0.12 N/mm2
Hence, O.K.
Detailing of Reinforcement
As Ast, min (= 618 mm2) is more than 50% of Ast (742.857 mm2/m) provided,
hence, all the tensile reinforcement shall be extended up to free end of the
slab.
Development Length
Ld =
φσ s 10 × 0.87 × 415
=
= 403 mm
4τbd
4 × 1.6 × 1.4
Value of a 90o bend is 8 φ = 8 × 10 = 80 mm.
∴
Length of bar required = 403 – 80 = 323 mm say 325 mm.
The detailing of the reinforcement has been shown in Figure 4.2.
Figure 4.2 : Detailing of Reinforcement of Cantilever Roof
4.4 DESIGN AND DETAILING OF ONE-WAY
SIMPLY SUPPORTED SLABS
Design and Detailing of One-way Slab Simply Supported on all Edges
If the ratio
Long span l y
Short span l x
> 2,
the design is same as that for simply supported beam of unit width, as the
slab bends mainly along the short span (Figure 4.3). Only temperature and
shrinkage reinforcement is provided along the long span.
79
Theory of Structures-II
Figure 4.3 : Bending Profile of an One-way Slab Supported on all Edges
Example 4.2
Design a roof slab simply supported on all its four edges of effective spans
3 m × 7 m. The top of slab is covered with 100 mm lime terrace. Imposed
load may be taken as 1.5 kN/m2. Design parameters are :
fck = 20 N/mm2; fy = 415 N/mm2 and Nominal Cover = 20.
Solution
ly
lx
=
7
= 2.33 > 2
3
Hence the slab will be designed as one-way simply supported slab.
Depth of Slab (D)
From Deflection control
l ef
d
where
< k B k1 k 2 k 3 k 4
kB = 20; and
k1 = 1 as lef < 10 m
Assuming Ast required = Ast provided
f s = 0.58 f y
A st required
= 0.58 × 415 = 240 N/mm2
A st provided
Assuming balanced section, for M 25 and Fe 415
pt% = 0.96%
80
Therefore, k2 = 1
k3 = k4 = 1 as the slab is singly reinforced and it is not a flanged
section. Substituting all the values in the above equation.
Slabs
3 × 1000
≤ 20 × 1 × 1 × 1 × 1
d
3 × 1000
= 150 mm
20 × 150
or
d≥
∴
D = d + Nominal Cover +
Taking
D = 185 mm
D = 185 −
φ
8
= 150 + 20 + = 174 mm
2
2
8
− 20 = 161 mm
2
From Moment of Resistance Consideration
Loads
∴
Self
= 1 × 1 × 0.185 × 25
= 4.625 kN/m
Lime concrete
= 1 × 1 × 0.1 × 18.8
= 1.880 kN/m
Ceiling plaster
= 1 × 1 × 0.01 × 20.4
= 0.204 kN/m
Total DL
= 6.709 kN/m
IL
= 1.500 kN/m
Total (DL+IL)
= 8.209 kN/m
wu = 1.5 × 8.209 = 12.31 kN/m
M u , max =
wu l 2 ef 12.31 × 3 2
=
= 13.849 kN/m
8
8
M u , lim = 0.36
X u , max ⎛
X
⎜1 − 0.42 u , max
⎜
d
d
⎝
⎞
⎟ bd 2 f ck
⎟
⎠
13.849 × 10 6 = 0.36 × 0.48 × (1 − 0.42 × 0.48) × 1000d 2 × 20
d = 70.85 < 161 mm
Hence, provided D = 185 mm and d = 161 mm.
Tensile Reinforcement (Ast)
Ast f y
⎛
M u = 0.87 f y Ast d ⎜⎜1 −
⎝ bd f ck
or
⎞
⎟
⎟
⎠
Ast × 415 ⎞
⎛
13.849 × 106 = 0.87 × 415 Ast ×161× ⎜1 −
⎟
⎝ 1000 ×161× 20 ⎠
or
7.492 Ast2 − 58129.05 Ast + 13.849 × 106 = 0
or
Ast2 − 7758.816 Ast + 1848505.07 = 0
Solving the above equation, we get
Ast = 246.05 mm 2 /m
i.e. φ 8 @ 200 mm c/c (Ast = 250 mm2/m)
81
Theory of Structures-II
Ast , min =
0.12 × 1000 × 161
= 193.2 mm 2 /m < 250mm 2 /m width
100
Hence, provided φ 8 @ 200 mm c/c.
Maximum Spacing
(a)
3d = 3 × 170 = 510 > 200 mm c/c
(b)
300 > 200 mm
Hence, O.K.
Check for Shear
As the width of support is not given, clear span (lc) may be taken as
(lef − d) = 3 – 0.161 = 2.839 m
Critical section for shear force is at d from the face of the support
(Figure 4.4).
Figure 4.4 : Explaining Critical Section for SF
∴
⎛l
⎞
⎛ 2.839
⎞
Vu = wu ⎜ c − d ⎟ = 12.31 × ⎜
− 0.161⎟ = 15.492 kN
2
2
⎝
⎠
⎝
⎠
τυ =
Vu 15.492 × 10 3
=
= 0.096 N/mm 2
bd
1000 × 161
For Ast,min and M 20 concrete
k τ c min = k × 0.28 = 0.28k > 0.096 N/mm 2 (k ≥ 1)
Hence, O.K.
Detailing of Reinforcements
Ld =
(a)
The positive main reinforcement shall extend into the support a
distance of
(b)
where
φσ s 8 × 0.87 × 415
=
= 376
4τbd
4 × 1.6 × 1.2
Ld ≤
Ld 376
=
= 125
3
3
1.3M 1
+ Lo
V
⎛
M1 = 0.87 fy Ast d ⎜⎜1 −
⎝
Ast f y ⎞
⎟
bd f ck ⎟⎠
Taking Ast as 50% of total reinforcement at mid span extending into the
support.
i.e.
82
Ast =
250
= 125 mm2/m < Ast, min (= 193.2 mm2/m)
2
Hence, all the reinforcement shall extend into the support
Slabs
250 × 415 ⎞
⎛
M1 = 0.87 × 415 × 250 × 161 × ⎜1 −
⎟
⎝ 1000 × 161 × 20 ⎠
or
M1 = 14.06 kN-m/m
Vu at simple support =
wu lef
2
=
12.31 × 3
= 18.465 kN
2
⎞
⎛ 1.3M 1
Ld ≤ ⎜
+ Lo ⎟
⎠
⎝ V
or
1.3M 1 ⎞
⎛
Lo ≥ ⎜ Ld −
⎟
V ⎠
⎝
⎛
⎝
Now, Lo ≥ ⎜ Ld −
i.e.
1.3M 1 ⎞
1.3 × 14.06 × 106
= − 614 mm
⎟ = 376 −
V ⎠
18.465 × 103
Lo = 0.
Distribution Steel
Astd =
0.12
0.12 × 1000 × 161
bd =
100
100
= 193.2 mm 2 /m
Hence, provided φ 8 @ 255 mm c/c.
Maximum spacing
(a)
5d = 5 × 161 = 805 > 255
(b)
450 > 255
Hence, O.K.
Reinforcement detailing has been shown in Figure 4.5.
Figure 4.5 : Reinforcement Detailing of the Designed Slab
83
Theory of Structures-II
* Two-way Slab
means the slab
bends about both the
axes x and y like a
saucer (Figure 4.6).
4.5 DESIGN AND DETAILING OF TWO-WAY*
SIMPLY SUPPORTED SLABS
For a rectangular slab, simply supported on all its edges, having no provisions to
resist torsion at corners and to prevent corners from lifting, the maximum bending
moment per unit width are calculated by the following equations :
M x = α x w l x2
M y = α y w l x2
and
where w = Design loads per unit area and lx and ly = Short and Long spans,
respectively, Mx and My = moments of strips of unit width spanning lx and ly,
respectively and α x and α y are the coefficients given in Table 4.1.
Table 4.1 : Bending Moment Coefficients for Slab Spanning in
Two Directions at Right Angles, Simply Supported on Four Sides
ly/lx
1.0
1.1
1.2
1.3
1.4
1.5
1.75
2.0
2.5
3.0
αx
0.062
0.074
0.084
0.093
0.099
0.104
0.113
0.118
0.122
0.124
αy
0.062
0.061
0.059
0.055
0.051
0.046
0.037
0.029
0.020
0.014
It is evident from the above table that if
ly
lx
> 2 , the slab will be treated as
one-way and designed as in Section 4.4.
At least 50% of the reinforcement provided at mid span shall extend into the
supports and the remaining 50% may extend to within 0.1 lx or 0.1 ly of support as
appropriate.
The other specifications for design and detailing for the slab are the same as those
for simply supported beams (Section 3.2).
Figure 4.6 : Bending of Two-way Slab like a Saucer
SAQ 1
84
(a)
Define one-way and two-way slabs.
(b)
How shear resistance of a slab can be increased economically?
(c)
How temperature and shrinkage stresses is taken care of?
(d)
Why the maximum horizontal distance between parallel bars are
limited? Describe the maximum horizontal distance between parallel
main bars and that between parallel temperature and shrinkage bars.
Slabs
4.6 DESIGN AND DETAILING OF TWO-WAY
RESTRAINED SLABS
A rectangular slab supported on beams on all sides and monolithically cast with
them and having
ly
lx
≤2
behaves as a two-way slab. Being monolithic with the beams, the corners are
prevented from lifting and, therefore, torsional reinforcements are provided to
resist the resultant torsional moments. Main reinforcements are provided along
both the principal axes to resist corresponding moments Mx and My, respectively.
Detailing of reinforcement is done in the following manner :
(a)
A slab is divided in edge strips and a middle strip in both directions as
shown in Figure 4.7.
Figure 4.7 : Division of Slab into Middle and Edge Strips
(b)
The maximum moments Mx and My both for spans and supports are
determined as
M x = α x w l x2
and
M y = α y w l x2
The values of αx and αy for different edge conditions are given in Table 4.2.
Table 4.2 : Bending Moment Coefficients for Rectangular Panels Supported
on Four Sides with Provision for Torsion at Corners
85
Theory of Structures-II
Case
No.
(1)
1.
2.
3.
(2)
1.2
1.3
1.4
1.5
1.75
2.0
ly / lx
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
0.037
0.043
0.047
0.051
0.053
0.060
0.065
0.032
Positive moment at
mid-span
0.024
0.028
0.032
0.036
0.039
0.041
0.045
0.049
0.024
Negative moment at
continuous edge
0.037
0.043
0.048
0.051
0.055
0.057
0.064
0.068
0.037
Positive moment at
mid-span
0.028
0.032
0.036
0.039
0.041
0.044
0.048
0.052
0.028
0.037
0.044
0.052
0.057
0.063
0.067
0.077
0.085
0.037
0.028
0.033
0.039
0.044
0.047
0.051
0.059
0.065
0.028
0.047
0.053
0.060
0.065
0.071
0.075
0.084
0.091
0.047
0.035
0.040
0.045
0.049
0.053
0.056
0.063
0.069
0.035
One Short Edge
Discontinuous :
One Long Edge
Discontinuous
Two Adjacent Edges
Positive moment at
mid-span
7.
1.1
(3)
0.032
Discontinuous :
Negative moment at
continuous edge
6.
1.0
Negative moment at
continuous edge
Positive moment at
mid-span
5.
Long Span
Coefficients, αy for
All Values of
Interior Panels :
Negative moment at
continuous edge
4.
Short Span Coefficients, αx
(Values of ly/lx)
Type of Panel
and
Moments
Considered
Two Short Edges
Discontinuous :
Negative moment at
continuous edge
0.045
0.049
0.052
0.056
0.059
0.060
0.065
0.069
----
Positive moment at
mid-span
0.035
0.037
0.040
0.043
0.044
0.045
0.049
0.052
0.035
Negative moment at
continuous edge
----
----
----
----
----
----
----
----
0.045
Positive moment at
mid-span
0.035
0.043
0.051
0.057
0.063
0.068
0.080
0.088
0.035
0.057
0.064
0.071
0.076
0.080
0.084
0.091
0.097
----
0.048
0.053
0.057
0.060
0.064
0.069
0.073
0.043
Two Long Edges
Discontinuous
Three Edges
Discontinuous
(One Long Edge
Continuous) :
Negative moment at
continuous
edge Positive
moment at mid-span
8.
0.043
Three Edges
Discontinuous
(One Short Edge
Continuous) :
9.
Negative moment at
continuous edge
----
----
----
----
----
----
----
----
0.057
Positive moment at
mid-span
0.043
0.051
0.059
0.065
0.071
0.076
0.087
0.096
0.043
0.056
0.064
0.072
0.079
0.085
0.089
0.100
0.107
0.056
Four Edges
Discontinuous :
Positive moment at
mid-span
A continuous edge is that on which slab extends on its both sides
whereas on discontinuous edge the slab extends only on one side.
86
(c)
Tension reinforcement provided at mid-span in the middle strip shall
extend in the lower part of the slab to within 0.25 l of a continuous
edge (Figure 4.8), or 0.15 l of discontinuous edge.
Slabs
Figure 4.8 : Curtailment Rules for Two-way Slab
(d)
(e)
(f)
(g)
Over the continuous edge of the middle strip, tension reinforcement
shall extend in the upper part of the slab a distance equal to 0.15 l
from the support and at least 50% shall extend a distance of 0.3 l
(Figure 4.8).
At discontinuous edge, tension reinforcement equal to 50% of that
provided at mid-span shall extend 0.1 l from the support (Figure 4.8).
Minimum reinforcement in edge strip shall be sufficient (Figure 4.7).
Torsion reinforcement at corners are provided as follows (Figure 4.9).
(i) Where corners are bound by discontinuous edges, torsion
reinforcement both at top and bottom and in both
directions equal to 3/4th the area of short span
reinforcement and of length lx/5 shall be provided.
(ii) If the corner is bound by continuous edge on one side and
discontinuous on the other, half of the area of
reinforcement that prescribed in (a) shall be provided.
(iii) Torsion reinforcement need not be provided at corners for
which both the edges are continuous.
(a) Corner with Two Discontinuous Ends
87
Theory of Structures-II
(b) Corner with One Discontinuous End
Figure 4.9
Example 4.3
Design slab S1 and S2 of a roof (Figure 4.10) cast monolithically with
beams. The slab has a topping of 120 mm thick lime terrace and 10 mm
thick ceiling plaster. Design parameters are as follows :
Imposed load on roof = 1.5 kN/m2; fck = 20 N/mm2; fy = 415 N/mm2 and
Nominal cover = 20 mm.
Figure 4.10 : Plan of Slab System
Solution
Type of Slab
Outer edges are taken as discontinuous edges whereas inner edges are
continuous.
Slab S1
Short span, lx = 3.5 m
Long span, ly = 5.5 m
ly
Ratio,
lx
= 1.57 < 2,
Hence, the slabs are two-way slabs.
Depth of Slab (D)
From Deflection Control
l ef
d
88
≤ k B k1 k 2 k 3 k 4
kB = (20 + 26)/2 = 23 (for one edge continuous and other
discontinuous).
Slabs
For M 20 and Fe 415, pt% = 0.96 % for balanced steel k2 = k3 = k4 = 1
d ≥
l ef
k B k1 k 2 k 3 k 4
=
3.5 × 10 3
= 152.17 mm
23 × 1 × 1 × 1 × 1
D = d + Nominal cover +
Taking
D = 180 mm
∴
d = 180 − 20 −
φ
8
= 152.17 + 20 + = 176.17 mm
2
2
8
= 156 mm
2
From Moments of Resistance Consideration
Loads
Self
= 0.180 × 1 × 1 × 25
= 4.5 kN/m2
Lime concrete
= 0.12 × 1 × 1 × 18.8
= 2.26 kN/m2
Ceiling plaster
= 0.01 × 1 × 1 × 20.4
= 0.20 kN/m2
Total DL
= 6.96 kN/m2
IL
= 1.50 kN/m2
= 8.46 kN/m2
Values of coefficients α x and α y and Bending Moments Mx and My
are given in Figure 4.11.
Figure 4.11 : Values of Moment Coefficient and Moment along X and Y-axes
Illustration for calculation of α x , α y and M for three edges
discontinuous and one short edge continuous (S1).
For (+)ve moment at mid-span
α x = 0.076 +
(0.087 − 0.076)
× (1.57 − 1.5) = 0.079
(1.75 − 1.5)
M x = α x wl x2 = 0.079 × 1.5 × 8.46 × 3.5 2 = 12.28 kNm
For (−)ve moment at continuous edge, α y = 0.057
M y = − α y wl x2 = − 0.057 × 1.5 × 8.46 × 3.52 = − 8.86 kN-m
M u = 0.36
xu , max ⎛
x
⎜1 − 0.42 u , max
⎜
d ⎝
d
⎞
⎟ f ck bd 2
⎟
⎠
89
Theory of Structures-II
12.28 × 10 6 = 0.36 × 0.48 × (1 − 0.42 × 0.48) × 20 × 1000 × d 2
or
d = 66.7 mm < 156 mm
Hence, provided D = 180 mm and d = 180 – 20 –
8
= 156 mm.
2
Tensile Reinforcement (Ast)
⎛
Mu = 0.87 fy Ast d ⎜⎜1 −
⎝
Ast f y ⎞
⎟
bd f ck ⎟⎠
Ast × 415 ⎞
⎛
12.28 × 10 6 = 0.87 × 415 × Ast × 156 × ⎜1 −
⎟
⎝ 1000 × 156 × 20 ⎠
or
12.28 × 106 = 56323.8 Ast − 7.492 Ast2
or
Ast2 − 7517.86 Ast + 1639081.69 = 0
Solving the above equation, we get
Ast = 24.74 mm 2
Hence, provided φ 8 @ 220 mm c/c.
Check
(a)
d = 3 × 156 = 468 c/c > 220 mm c/c
(b)
300 > 220 mm c/c.
Negative reinforcement in long span.
Average of BM on both sides of the edge
=
(8.86 + 7.00)
= 7.93 kN-m/m width
2
Ast f y
⎛
M u = 0.87 f y Ast d ⎜⎜1 −
⎝ bd f ck
⎞
⎟
⎟
⎠
Ast × 415
⎛
⎞
7.93 × 10 6 = 0.87 × 415 Ast × 156 × ⎜1 −
⎟
1000
156
200
×
×
⎝
⎠
7.93 × 106 = 56323.8 Ast − 7.492 Ast2
Ast2 − 7517.86 Ast + 1058462.36 = 0
or
Solving the above equation, we get
Ast = 143.53 mm 2 /m width
Provided φ 8 @ 345 mm c/c.
Check
(a)
3d = 3 × 156 = 468 c c > 345 mm c/c
(b)
300 < 345 mm c/c
Hence, provided φ 8 @ 300 mm c/c.
90
Slabs
SAQ 2
(a)
Sketch the detailing of tensile reinforcement at support of a cantilever
slab.
(a)
Sketch the plan and section showing detailing of reinforcement of a
two-way simply supported slab.
(b)
Sketch the section of a two-way restrained slab of two spans, l1 and l2
continuous at interior support and discontinuous at ends.
(c)
Design an RC slab 4.5 m × 6.5 m in plan supported on 25 mm wide
beams. Two of its adjacent edges are discontinuous. Imposed load is
2 kN/m2. Mix M 20 concrete and Fe 415 bars are used.
4.7 SUMMARY
Four types of slab have been described in this unit :
(a)
Cantilever slab
(b)
One-way simply supported slab
(c)
Two-way simply supported slab
(d)
Two-way restrained slab
Cantilever Slab
This type of rectangular slab has one edge fixed and the other three edges
free. Therefore, under gravity loads, it bends about its axis of support
causing hogging bending moment and requiring main reinforcement only
on top face perpendicular to its support axis.
One-way Simply Supported Slab
This type of slab may be either simply supported on two opposite faces or,
if supported on all four edges the ratio
ly
lx
> 2 . Therefore, under gravity
loads, it bends only in one direction.
Two-way Simply Supported Slab
A slab simply supported on all its four edges having
ly
lx
≤ 2 , bends about
both of its principal axes under gravity loads. Hence, main reinforcements
are provided in both directions.
Two-way Restrained Slab
⎛ ly
When a slab ⎜⎜
⎝ lx
⎞
≤ 2 ⎟⎟ is monolithic with its supporting beams and the
⎠
corners are prevented from lifting under bending, additional reinforcements
are provided at the corners to resist torsion. Hogging bending moments at
the supports are taken care of by providing reinforcement at the top face.
91
Theory of Structures-II
4.8 ANSWERS TO SAQs
SAQ 1
(a)
Refer Section 4.1.
(b)
Refer Section 4.2.
(c)
Refer Section 4.2.
(d)
Refer Section 4.2.
SAQ 2
92
(a)
Refer Section 4.3.
(b)
Refer Section 4.5.
(c)
Refer Section 4.6.