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Common practice of design and construction is to support the slabs by beams and support the beamsby columns. This may be called as beam-slab construction. The beams reduce the available net clearceiling height. Hence in warehouses, offices and public halls some times beams are avoided and slabsare directly supported by columns. This types of construction is aesthetically appealing also. Theseslabs which are directly supported by columns are called Flat Slabs. Fig. 1.1 shows a typical flat slab.
d2
Critical section for shear
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The column head is some times widened so as to reduce the punching shear in the slab. Thewidened portions are called column heads. The column heads may be provided with any angle fromthe consideration of architecture but for the design, concrete in the portion at 45º on either side ofvertical only is considered as effective for the design [Ref. Fig. 1.2].
d2
Critical section for shear
Concrete in this area isneglected for calculation
90°
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2 Advanced R.C.C. Design
Moments in the slabs are more near the column. Hence the slab is thickened near the columns byproviding the drops as shown in Fig. 1.3. Sometimes the drops are called as capital of the column.Thus we have the following types of flat slabs:
Critical section for sheard
2d
2
Critical sectionfor shear
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(i) Slabs without drop and column head (Fig. 1.1).(ii) Slabs without drop and column with column head (Fig. 1.2).
(iii) Slabs with drop and column without column head (Fig. 1.3).(iv) Slabs with drop and column head as shown in Fig. 1.4.
Critical sectionfor shear
45° 45°
d2
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The portion of flat slab that is bound on each of its four sides by centre lines of adjacent columns iscalled a panel. The panel shown in Fig. 1.5 has size L1 � L2. A panel may be divided into column stripsand middle strips. Column Strip means a design strip having a width of 0.25L1 or 0.25L2,whichever is less. The remaining middle portion which is bound by the column strips is called middlestrip. Fig. 1.5 shows the division of flat slab panel into column and middle strips in the direction y.
Flat Slabs 3
L2a L2b
C of panel A C of panel B
Middle stripMiddle stripColumn strip
L2a
4
L2b
4
Column strip Column strip
y
xo
L1
butL
�1
4but
L�
1
4
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��� �� ���������� ��� ����� �����
IS 456-2000 [Clause 31.2] gives the following guidelines for proportioning.
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The drops when provided shall be rectangular in plan, and have a length in each direction not less thanone third of the panel in that direction. For exterior panels, the width of drops at right angles to the noncontinuous edge and measured from the centre-line of the columns shall be equal to one half of thewidth of drop for interior panels.
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Where column heads are provided, that portion of the column head which lies within the largest rightcircular cone or pyramid entirely within the outlines of the column and the column head, shall beconsidered for design purpose as shown in Figs. 1.2 and 1.4.
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From the consideration of deflection control IS 456-2000 specifies minimum thickness in terms ofspan to effective depth ratio. For this purpose larger span is to be considered. If drop as specified in1.2.1 is provided, then the maximum value of ratio of larger span to thickness shall be
= 40, if mild steel is used
= 32, if Fe 415 or Fe 500 steel is usedIf drops are not provided or size of drops do not satisfy the specification 1.2.1, then the ratio shall
not exceed 0.9 times the value specified above i.e.,
= 40 � 0.9 = 36, if mild steel is used.
= 32 � 0.9 = 28.8, if HYSD bars are usedIt is also specified that in no case, the thickness of flat slab shall be less than 125 mm.
4 Advanced R.C.C. Design
��! �����)�������� ��� ��������)�)���� ���� ������ ����
For this IS 456-2000 permits use of any one of the following two methods:
(a) The Direct Design Method(b) The Equivalent Frame Method
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This method has the limitation that it can be used only if the following conditions are fulfilled:
(a) There shall be minimum of three continuous spans in each directions.(b) The panels shall be rectangular and the ratio of the longer span to the shorter span within a panel
shall not be greater than 2.(c) The successive span length in each direction shall not differ by more than one-third of longer
span.(d) The design live load shall not exceed three times the design dead load.(e) The end span must be shorter but not greater than the interior span.(f) It shall be permissible to offset columns a maximum of 10 percent of the span in the direction
of the offset not withstanding the provision in (b).
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The absolute sum of the positive and negative moment in each direction is given by
M0 =WL
8n
Where,
M0 = Total moment
W = Design load on the area L2 � Ln
Ln = Clear span extending from face to face of columns, capitals, brackets or walls butnot less than 0.65 L1
L1 = Length of span in the direction of M0; and
L2 = Length of span transverse to L1
In taking the values of Ln, L1 and L2, the following clauses are to be carefully noted:
(a) Circular supports shall be treated as square supports having the same area i.e., squares of size0.886D.
(b) When the transverse span of the panel on either side of the centre line of support varies, L2 shall
be taken as the average of the transverse spans. In Fig. 1.5 it is given by L L2 2
2a b+� �
.
(c) When the span adjacent and parallel to an edge is being considered, the distance from the edgeto the centre-line of the panel shall be substituted for L2.
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The total design moment M0 in a panel is to be distributed into –ve moment and +ve moment asspecified below:
Flat Slabs 5
������ #�'��#�������
Negative Design Moment 0.65 M0Positive Design Moment 0.35 M0
�������� �����
Interior negative design moment
= 0 75010
11
..
�
�
�
�
��
�
�
��
� c
M0
Positive design moment
= 0 630 28
11 0.
.�
�
�
�
��
�
�
��
� c
M
Exterior negative design moment
=0 65
11 0
.
�
�
�
��
�
�
��
�c
M
where �c is the ratio of flexural stiffness at the exterior columns to the flexural stiffness of the slab ata joint taken in the direction moments are being determined and is given by
�c =K
K
c
s
∑∑
Where,
Kc = Sum of the flexural stiffness of the columns meeting at the joint; and
Ks = Flexural stiffness of the slab, expressed as moment per unit rotation.
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The +ve and –ve moments found are to be distributed across the column strip in a panel as shown inTable 1.1. The moment in the middle strip shall be the difference between panel and the column stripmoments.
Table 1.1 Distribution of Moments Across the Panel Width in a Column Strip
S. No. Distributed Moment Per cent of Total Moment
a Negative BM at the exterior support 100
b Negative BM at the interior support 75
c Positive bending moment 60
6 Advanced R.C.C. Design
)����'��#��������
In this type of constructions column moments are to be modified as suggested in IS 456–2000[Clause No. 31.4.5].
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The critical section for shear shall be at a distance d
2 from the periphery of the column/capital drop
panel. Hence if drops are provided there are two critical sections near columns. These critical sectionsare shown in Figs. 1.1 to 1.4. The shape of the critical section in plan is similar to the supportimmediately below the slab as shown in Fig. 1.6.
d/2
d/2
Criticalsection
Support sectioncolumn / column head
( )a
d/2
Supportsection
Criticalsection ( )b
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For columns sections with re-entrant angles, the critical section shall be taken as indicated in Fig. 1.7.
Criticalsection
Supportsection
d/2
d/2
( )a
d/2
d/2
d/2
Criticalsection
Supportsection
( )b
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In case of columns near the free edge of a slab, the critical section shall be taken as shown in Fig. 1.8.
d/2
d/2
Criticalsection
Freeedge
( )a
Criticalsection
Freecorner
Cornercolumn
( )b
����� ��
Flat Slabs 7
The nominal shear stress may be calculated as
�v =V
b d0
where V – is shear force due to design
b0 – is the periphery of the critical section
d – is the effective depthThe permissible shear stress in concrete may be calculated as ks �c, where ks = 0.5 + �c but not
greater than 1, where �c is the ratio of short side to long side of the column/capital; and
��c = 0 25. fck
If shear stress �v < �c – no shear reinforcement are required. If �c < �v < 1.5 �c, shear reinforcementshall be provided. If shear stress exceeds 1.5 �c flat slab shall be redesigned.
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IS 456–2000 recommends the analysis of flat slab and column structure as a rigid frame to get designmoment and shear forces with the following assumptions:
(a) Beam portion of frame is taken as equivalent to the moment of inertia of flat slab boundedlaterally by centre line of the panel on each side of the centre line of the column. In framesadjacent and parallel to an edge beam portion shall be equal to flat slab bounded by the edge andthe centre line of the adjacent panel.
(b) Moment of inertia of the members of the frame may be taken as that of the gross section of theconcrete alone.
(c) Variation of moment of inertia along the axis of the slab on account of provision of drops shallbe taken into account. In the case of recessed or coffered slab which is made solid in the regionof the columns, the stiffening effect may be ignored provided the solid part of the slab does notextend more than 0.15 lef into the span measured from the centre line of the columns. Thestiffening effect of flared columns heads may be ignored.
(d) Analysis of frame may be carried out with substitute frame method or any other acceptedmethod like moment distribution or matrix method.
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When the live load does not exceed ¾th of dead load, the maximum moments may be assumed tooccur at all sections when full design live load is on the entire slab.
If live load exceeds ¾th dead load analysis is to be carried out for the following pattern of loading also:
(i) To get maximum moment near mid span– ¾th of live load on the panel and full live load on alternate panel
(ii) To get maximum moment in the slab near the support– ¾th of live load is on the adjacent panel only
It is to be carefully noted that in no case design moment shall be taken to be less than thoseoccurring with full design live load on all panels.
The moments determined in the beam of frame (flat slab) may be reduced in such proportion thatthe numerical sum of positive and average negative moments is not less than the value of total design
8 Advanced R.C.C. Design
moment M0 = WLn
8. The distribution of slab moments into column strips and middle strips is to be
made in the same manner as specified in direct design method.
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The spacing of bars in a flat slab, shall not exceed 2 times the slab thickness.
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When the drop panels are used, the thickness of drop panel for determining area of reinforcementshall be the lesser of the following:
(a) Thickness of drop, and(b) Thickness of slab plus one quarter the distance between edge of drop and edge of capital.The minimum percentage of the reinforcement is same as that in solid slab i.e., 0.12 percent if
HYSD bars used and 0.15 percent, if mild steel is used.
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At least 50 percent of bottom bars should be from support to support. The rest may be bent up. Theminimum length of different reinforcement in flat slabs should be as shown in Fig. 1.9 (Fig. 16 in IS 456–2000). If adjacent spans are not equal, the extension of the –ve reinforcement beyond each face shall bebased on the longer span. All slab reinforcement should be anchored property at discontinuous edges.
�4���������5 Design an interior panel of a flat slab of size 5 m � 5 m without providing drop andcolumn head. Size of columns is 500 � 500 mm and live load on the panel is 4 kN/m2. Take floorfinishing load as 1 kN/m2. Use M20 concrete and Fe 415 steel.
����'#��5
�"#$%����
Since drop is not provided and HYSD bars are used span to thickness ratio shall not exceed
1
0 9 32
1
28 8. .×=
� Minimum thickness required
=Span
28 8
5000
28 8. .= = 173.6 mm
Let d = 175 mm and D = 200 mm
��� �
Self weight of slab = 0.20�� 25 = 5 kN/m2
Finishing load = 1 kN/m2
Live load = 4 kN/m2
� Total working load = 10 kN/m2
Factored load = 1.5 � 10 = 15 kN/m2
Flat Slabs 9
Minimumpercentage
of steelat section
50
Remainder
WITHOUT DROP PANEL WITH DROP PANEL
db
75 mm max
150 mm
d
b
c
24 BAR DIA OR
edb b
eb
150 mm min.
DROPbdb e
150 mmg
eb
300 mm min.
g
EDGE OFDROP
75 mm max.150 mm
75 mm max.
150 mm150 mm
(ALL BARS) (ALL BARS)
150 mm
75 mm max.75 mm max.
150 mm
c
c
c
c a
a cc
ff
D
C
D D
C C
Clear span - lnFace of support
interior support
Exteriorsupport
Mid
dle
Str
ipC
olu
mn
str
ipS
trip
Type
ofbars
Str
aig
htbars
Bentbars
*S
traig
htbars
Bentbars
*
50
Remainder
50
Remainder
50
Remainder
100
50
Remainder
50
Remainder
50
Remainder
Clear span - ln
Face of supportinterior support
C
0.15 maxl
o.15 maxl
0.125lmax
300 mm min. ALL BARS
EDGE OF
24BAR DIA OR
[NO SLAB CONTINUITY] [CONTINUITY PROVED] [NO SLAB CONTINUITY]
Bar Length From Face of Support
Minimum Length Maximum Length
Mark a b c d e f g
Length 0.14 ln 0.20 ln 0.22 ln 0.30 ln 0.33 ln 0.20 ln 0.24 ln
* Bent bars at exterior supports may be used if a general analysis is made.
Note. D is the diameter of the column and the dimension of the rectangular column in the direction under consideration.
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10 Advanced R.C.C. Design
Ln = 5 – 0.5 = 4.5 m
� Total design load in a panel W = 15 L2 Ln = 15 � 5 � 4.5 = 337.5 kN
)����'�
Panel Moment M0 =WL
kNmn
8337 5
4 5
8189 84 � .
..
Panel –ve moment = 0.65 � 189.84 = 123.40 kNm
Panel +ve moment = 0.35 � 189.84 = 0.35 � 189.84 = 66.44 kNmDistribution of moment into column strips and middle strip:
Column Strip in kNm Middle Strip in kNm
–ve moment 0.75 � 123.40 = 92.55 30.85
+ve moment 0.60 � 66.44 = 39.86 26.58
Checking the thickness selected:
Since Fe 415 steel is used,
Mu lim = 0.138 fck b d2
Width of column strip = 0.5 � 5000 = 2500 mm� Mu lim = 0.138 ��20 � 2500 � 1752 = 211.3125 � 106 Nmm
= 211.3125 kNmHence singly reinforced section can be designed i.e., thickness provided is satisfactory from the
consideration of bending moment.
"�$%� &����"���
The critical section for shear is at a distance d
2 from the column face. Hence periphery of critical
section around a column is square of a size = 500 + d = 500 + 175 = 675 mm
Shear to be resisted by the critical section
V = 15 � 5 � 5 – 15 � 0.675 � 0.675
= 368.166 kN
� �v =368166 1000
4 675 175
. ×× ×
= 0.779 N/mm2
ks = 1 + �c subject to maximum of 1.
�c =L
L1
2
5
5= = 1
� ks = 1
�c = 0 25 0 25 20. .fck = = 1.118 N/mm2
safe in shear since �v < �c
675
675500
500
Flat Slabs 11
��#�&��$����'
For –ve moment in column strip:
Mu = 92.55 kNm
92.55 � 106 = 0 87 1. f dbd
f
fy stst y
ck
AA
−�
��
�
��
= 0.87 � 415 � Ast � 175 12500 175
415
20−
××
�
��
�
��
Ast
i.e., 1464.78 = Ast 121084 3
−���
���
A st
.
i.e., Ast2 – 21084.3Ast + 1464.78 � 21084.3 = 0
Ast = 1583.74 mm2
This is to be provided in a column strip of width 2500 mm. Hence using 12 mm bars, spacingrequired is given by
s = 4 12
1583 742500
2��
. = 178 mm
Provide 12 mm bars at 175 mm c/c.
For +ve moment in column strip:
Mu = 39.86 kNm
� 39.86 � 106 = 0.87 � 415 � Ast � 175 12500 175
415
20−
××
�
��
�
��
A st
630.86 = Ast 121084 3
−���
���
Ast
.
or Ast2 – 21084.3 Ast + 630.86 � 21084.3 = 0
� Ast = 651 mm2
Using 10 mm bars, spacing required is
s = 4 10
6512500
2�� = 301.6 mm < 2 � thickness of slab
Hence provide 10 mm bars at 300 mm c/c.
Provide 10 mm diameter bars at 300 mm c/c in the middle strip to take up –ve and +ve moments.Since span is same in both directions, provide similar reinforcement in other direction also.
12 Advanced R.C.C. Design
��#�&��$����'� ��'�#��
It is as shown in Fig. 1.10
50005000 5000
5000
5000
5000
Column Strip Middle Strip Column strip
Col
umn
Str
ipM
iddl
eS
trip
Col
umn
strip
10-300 c/c
10-300 c/c
500
10 - 300 c\c
200
500
Cover -25
12-175 c/c
Section through column strip
500 500
30003000 10 - 300 c/c
section through middle strip
Top reinforcement
Sign convention
Bottom reinforcement
12-175 c/c
12-175 c/c
��������� ���������������������������������������������
�4���������5 Design an interior panel of a flat slab with panel size 6 � 6 m supported by columns ofsize 500 � 500 mm. Provide suitable drop. Take live load as 4 kN/m2. Use M20 concrete and Fe 415steel.
����'#���5
Thickness : Since Fe 415 steel is used and drop is provided, maximum span to thickness ratiopermitted is 32
� Thickness of flat slab =6000
32 = 187.5 mm
Provide 190 mm thickness. Let the cover be 30 mm
� Overall thickness D = 220 mmLet the drop be 50 mm. Hence at column head, d = 240 mm and D = 270 mm
Flat Slabs 13
�#6���&�����
It should not be less than 1
36� m = 2 m
Let us provide 3 m � 3 m drop so that the width of drop is equal to that of column head.
� Width of column strip = width of middle strip = 3000 mm.
��� �
For the purpose of design let us take self-weight as that due to thickness at column strip
� Self-weight = 0.27 � 1 � 1 � 25 = 6.75 kN/m2
Finishing load = 1.00 kN/m2
Live load = 4.00 kN/m2
Total load = 11.75 kN/m2
� Design (factored) load = 1.5 � 11.75 = 17.625 kN/m2
Clear span Ln = 6 – 0.5 = 5.5 m
� Design load W0 = Wu � L2 � Ln
= 17.625 � 6 � 5.5
= 581.625 kN
���#+����'���)����'
Total moment
M0 =W L0
8
581625 55
8n
�. . = 400 kNm
� Total negative moment = 0.65 � 400 = 260 kNmTotal positive moment = 0.35 � 400 = 140 kNmThe above moments are to be distributed into column strip and middle strip
Column Strip Middle Strip
–ve moment 0.75 � 260 = 195 kNm 0.25 � 260 = 65 kNm
+ve moment 0.6 � 140 = 84 kNm 0.4 � 140 = 56 kNm
Width of column strip = width of middle strip = 3000 mm
Mu lim = 0.138 fck b d2 = 0.138 � 20 � 3000 � 2402 = 476.928 � 106 Nmm
= 476.928 kNmThus Mu lim > Mu. Hence thickness selected is sufficient.
"�$%�&��� �"���
The critical section is at a distance
14 Advanced R.C.C. Design
d
2 =
240
2 = 120 mm from the face of column
��It is a square of size = 500 + 240 = 740 mm
V = Total load – load on 0.740 � 0.740 area
= 17.625 � 6 � 6 – 17.625 � 0.740 � 0.740
= 624.849 kN
� Nominal shear = �v = 624 489 1000
4 740 240
. ×× ×
= 0.880 N/mm2
Shear strength = ks��c
where ks = 1 + �c subject to maximum of 1
where �c = L
L1
2
= 1
� ks = 1
�c = 0 25 20. = 1.118 N/mm2
Design shear stress permitted
= 1.118 N/mm2 > �v
Hence the slab is safe in shear without shear reinforcement also.
Shear strength may be checked at distance d
2 from drop. It is quite safe since drop size is large.
��#�&��$����'
(a) For –ve moment in column strip
Mu = 195 kNm
Thickness d = 240 mm
� Mu = 0.87 fy Ast d 1−×
�
��
�
��
Ast y
ckb d
f
f
195 � 106 = 0.87 � 415 � Ast � 240 1
3000 240
415
20−
××
�
��
�
��
A st
2250.38 = Ast 134698 8
−���
���
A st
.
Ast2 – 34698.8 Ast + 2250.38 � 34698.8 = 0
Ast = 2419 mm2 in 3000 mm width
500
500
500 740
740
120 120
Flat Slabs 15
Using 12 mm bars, spacing required is
s = 4 12
24193000
2�� = 140.26 mm
Provide 12 mm bars at 140 mm c/c
(b) For +ve moment in column strip
Mu = 84 kNm = 84 � 106 Nmm. Thickness d = 190 mm
84 � 106 = 0.87 � 415 � Ast � 190 13000 240
415
20−
××
�
��
�
��
A st
1224.5 = Ast 127469 9
−���
���
A st
.
� Ast = 1285 mm2
Using 10 mm bars
s = 4 10
12853000
2�� = 183 mm
Provide 10 mm bars at 180 mm c/c
(c) For –ve moment in middle strip:
Mu = 65 kNm; Thickness = 190 mm
65 � 106 = 0.87 � 415 � Ast � 190 13000 190
415
20−
××
�
��
�
��
A st
947.5 = Ast 127469 9
−���
���
Ast
.
Ast2 – 27469.9 Ast + 947.5 � 27469.9 = 0
Ast = 983 mm2 in 3000 mm width
Using 10 mm bars
s = 4 10
9833000
2�� = 239.7 mm
Provide 10 mm bars at 230 mm c/c
(d) For +ve moment in middle strip
Mu = 56 kNm; Thickness = 190 mm
Provide 10 mm bars at 230 mm c/c in this portion also.
Since span is same in both direction, provide similar reinforcement in both directions. The detailsof reinforcement are shown in Fig. 1.11.
16 Advanced R.C.C. Design
6000
Column stripMiddle strip
Column strip
Col
umn
strip
Mid
dle
strip
Col
umn
strip
6000 6000
6000
6000
6000
=Do p widthr=Do p widthr
=D
orp
wid
th=
Dorp
wid
th
12 140 c/c–
10 230 c/c–
500 500
24010 230@
12 @ 140 10 180 c/c@
190
Cover - 30
Section through column strip
10 230 c/c@
240
190
500 500
10 180c/c–
12 230c/c–
10–180c/c
����� ���� ������������� ������
�4��������!5 Design the interior panel of the flat slab in example 1.2, providing a suitable columnhead, if columns are of 500 mm diameter.
����'#��5�Let the diameter of column head be
= 0.25L = 0.25 � 6 = 1.5 mIt’s equivalent square has side ‘a’ where
π4
1 52× . = a2
a = 1.33 m� Ln = 6 – 1.33 = 4.67 m
W0 = 17.625 � 6 � 4.67 = 493.85 kN
M0 =W Lo n
8
493 85 4 67
8= ×. .
= 288.3 kNm
Flat Slabs 17
� Total –ve moment = 0.65 � 288.3 = 187.4 kNmTotal +ve moment = 0.35 � 288.3 = 100.9 kNm
The distribution of above moment into column strip and middle strips are as given below:
Column Strip Middle Strip
–ve moment 0.75 � 187.4 = 140.55 kNm 0.25 � 187.4 = 46.85 kNm
+ve moment 0.60 � 100.9 = 60.54 kNm 0.4 � 100.9 = 40.36 kNm
Width of column strip = width of middle strip = 3000 mm
� Mu lim = 0.138 fck bd2 = 0.138 � 20 � 3000 � 2402
= 476.928 � 106 Nmm > Mu
Hence thickness selected is sufficient.
"�$%� &����"���
The critical section is at a distance
d
2 =
240
2 = 120 mm from the face of column head
Diameter of critical section = 1500 + 240 =1740 mm= 1.740 m
Perimeter of critical section = D= 1.740
Shear on this section
V = 17 625 6 64
1742. .� � ����
���
= 592.59 kN
� �v =592 59 1000
1740 240
. ×× ×π
= 0.45 N/mm2
Maximum shear permitted = ks × 0 25 20.
= 1.118 N/mm2 Since ks works out to be 1Since maximum shear permitted in concrete is more than nominal shear �v, there is no need to
provide shear reinforcement
���#+�� �&���#�&��$����'
(a) For –ve moment in column strip
Mu = 140.55 kNm; d = 240 mm
� 140.55 � 106 = 0.87 � 415 � Ast � 240 13000 240
415
20−
××
�
��
�
��
Ast
1622 = Ast 134698 8
−���
���
A st
.
1500 120
18 Advanced R.C.C. Design
Ast2 – 34698.8 Ast + 1622 � 34698.8 = 0
Ast = 1705 mm2
Using 12 mm bars,
s =π 4 12
17053000
2× × = 199 mm
Provide 12 mm bars at 190 mm c/c.
(b) For the +ve moment in column stripMu = 60.54 kNm; d = 190 mm
60.54 � 106 = 0.87 � 415 � Ast � 190 13000 190
415
20−
××
�
��
�
��
A st
882.51 = Ast 127469 9
−���
���
A st
.Ast
2 – 27469.9 Ast + 882.51 � 27469.9 = 0Ast = 913 mm2
Using 10 mm bars
s =π 4 10
9133000
2× × = 258 mm
Provide 10 mm bars at 250 mm c/c.
(c) For –ve moment in middle strip:Mu = 46.85 kNm; d = 190 mm
46.85 � 106 = 0.87 � 415 � Ast � 190 13000 190
415
20−
××
�
��
�
��
A st
683 = Ast 127469 9
−���
���
Ast
.Ast
2 – 27469.9Ast + 683 � 27469.9 = 0Ast = 701 mm2
Using 10 mm bars,
s =π 4 10
7013000
2× × = 336 mm
Provide 10 mm bars at 300 mm c/c.
(d) Provide 10 mm bars at 300 mm c/c for +ve moment in middle strip also.As span is same in both directions, provide similar reinforcement in both directions. Reinforcement
detail may be shown as was done in previous problem.
�4��������*5 A flat slab system consists of 5 m � 6 m panels and is without drop and column head.It has to carry a live load of 4 kN/m2 and a finishing load of 1 kN/m2. It is to be designed using M20grade concrete and Fe 415 steel. The size of the columns supporting the system is 500 � 500 mm andfloor to floor height is 4.5 m. Calculate design moments in interior and exterior panels at column andmiddle strips in both directions.
Flat Slabs 19
����'#��5
Thickness: Since Fe 415 steel is used and no drops are provided, longer span to depth ratio is notmore than 32 � 0.9 = 28.8
d =6000
28 8. = 208
Let us select d = 210 mm and D = 240 mm
��� �
Self weight 0.24 � 1 � 1 � 25 = 6 kN/m2
Finishing weight = 1 kN/m2
Live load = 4 kN/m2
Total = 11 kN/m2
Wu = 1.5 � 11 = 16.5 kN/m2
����� �#����#���
����+� ���+'"
L1 = 6 m and L2 = 5 m
Width of column strip = 0.25 L1 or L2 whichever is less.
= 0.25 � 5 = 1.25 m on either side of column centre line�Total width of column strip = 1.25 � 2 = 2.5 m
Width of middle strip = 5 – 2.5 = 2.5 m
����+�/# '"
L1 = 5 m L2 = 6 m
Width of column strip = 0.25 � 5 = 1.25 m on either side
�Total width of column strip = 2.5 m
Hence, width of middle strip = 6 – 2.5 = 3.5 m
��������� ����
Moments Along Longer Size
L1 = 6 m L2 = 5 m
Ln = 6 – 0.5 = 5.5 m subject to minimum of 0.65 � L1 = 3.9 m
� Ln = 5.5 m
Load on panel W0 = 16.5 � L2Ln
= 16.5 � 5 � 5.5 = 453.75 kN
20 Advanced R.C.C. Design
M0 =W L0
8
45375 55
8n
�. . = 311.95 kNm
�������#�'#��� �&�)����'
Total –ve moment = 0.65 � 311.95 = 202.77 kNm
� Total +ve moment = 311.95 – 202.77 = 109.18 kNmHence moment in column strip and middle strip along longer direction in interior panels are as given
below:
Column Strip Middle Strip
–ve moment 0.75 � 202.75 = 152.06 kNm 202.75 – 152.06 = 50.69 kNm
+ve moment 0.60 � 109.18 = 65.51 kNm 109.18 – 65.51 = 43.67 kNm
����+�/# '"
L1 = 5 m L2 = 6 m and Ln = 5 – 0.5 = 4.5 m.
Panel load = W0 = 16.5 � 6 � 4.5 = 445.5 kN
Panel moment M0 = WL
0 8
4455 4 5
8n
�. . = 250.59 kN-m
�������#�'#��� �&�)����'5
Total –ve moment = 0.65 � 250.59 = 162.88 kN-m
Total +ve moment = 250.59 – 162.88 = 87.71 kN-m� Moments in column strip and middle strip are as shown below:
Column Strip Middle Strip
–ve moment 0.75 � 162.88 = 122.16 kNm 0.25 � 162.88 = 40.72 kNm
+ve moment 0.60 � 87.71 = 52.63 kNm 0.40 � 87.71 = 35.08 kNm
��������� ����
Length of column = 4.5 – 0.24 = 4.26 mThe building is not restrained from lateral sway. Hence as per Table 28 in IS 456-2000, effective
length of column
= 1.2 � length = 1.2 � 4.26 = 5.112 m
Size of column = 500 � 500 mm
Moment of inertia of column =1
125004 4× mm
Flat Slabs 21
� kc =IL
112
= × 5005112
4
= 101844 mm4
�� ��� ���� ���������
)����'��&�#���'#���&�(���
Is = Moment of inertia of slab
=1
126000 2403× ×
Its length = L2 = 5000 mm
� kc =I
5000s �
�112
6000 2405000
3
= 1382400 mm4
Live loadDead load
=4
7 < 0.75
� Relative stiffness ratio is
�c =k k
kc c
s
1 2 2 1018844
1382400
+= ×
= 1.474
� � = 11
11
1474� �
�c . = 1.678
Hence various moment coefficients are:
Interior –ve moment coefficient = 0.75 – 0 1.
α = 0.690
Exterior –ve moment coefficient =0.65
α = 0.387
Positive moment coefficient = 0.63 – 0.28
α = 0.463
Total moment M0 = 311.95 kNm� Appropriation of moments in kNm is as given below:
Total Column Strip Middle Strip
Interior –ve 0.69 � 311.95 = 215.25 0.75 � 215.25 = 161.43 215.25 – 161.43 = 53.82
Exterior –ve 0.387 � 311.95 = 120.72 1.00 � 120.72 = 120.72 120.72 – 120.72 = 0
+ Moment 0.463 � 31.95 = 144.43 0.60 � 144.43 = 86.66 144.43 – 86.66 = 57.77
�"��'��� ������#��$'#��
� ks =1
125000 240
6000
3
× × = 96000
� �c =k k
kc c
s
1 2 2 1018844
960000
+ = × = 2.123
22 Advanced R.C.C. Design
� �1 = 11+αc
= 1.471
Interior –ve moment coefficient = 0.75 – 0 1
0 750 1
1 471
..
.
.� � = 0.682
Exterior –ve moment coefficient =0.65 0.65
α=
1 471. = 0.442
Positive moment coefficient = 0.63 –0.28 0.28
α= −0 63
1 471.
. = 0.440
Total moment M0 = 250.59 kNm� Appropriation of moments in shorter span exterior panel in kNm is as given below:
Total Column Strip Middle Strip
Interior –ve 0.682 � 250.59 = 170.90 0.75 � 170.76 = 128.18 170.90 – 128.18 = 42.72
Exterior -ve 0.442 � 250.59 = 110.76 1.00 � 110.76 = 110.76 110.76 – 110.76 = 0
+ Moment 0.44 � 250.59 = 110.25 0.60 � 110.25 = 66.16 110.25 – 66.16 = 44.09
In the exterior panel in each column strips half the above values will act. These moments areshown in Fig. 1.12
StripCol Middle
Strip
2.51.25
StripCol Middle
Strip StripCol
2.5 2.5 2.5
–120.72
–122.16
–53.82
52.63
–161.43 –15.062
–122.16
–50.69
52.63
–152.06
–122.16
86.662
–40.72
57.77
35.08–40.72
43.67
35.08
65.51
–40.72
–152.06–122.16
2
–128.18
–50.69
66.162
–161.43 –152.062
–122.162–128.18
66.162
52.632
–53.82120.72
–122.162
–128.18
86.662
–42.72
57.77
44.09–42.72
43.67
44.09
65.51
–42.72
–152.06
–110.7666.16
–50.69–161.43
–152.062
–110.7666.16
2
–53.82120.72–4–110.76
2.5 m
2.5 m
3.5 m
3.5 m
1.25 m
86 66
2
65 51
2
. .�
86 66
2
65 51
2
. .�
52 63
2
.
����� ����
Flat Slabs 23
��2��/�1�������
1. Design the typical interior panel of a flat slab floor of size 5 m � 5 m with suitable drop tosupport a live load of 4 kN/m2. The floor is supported by columns of size 450 mm � 450 mm.Use M20 concrete and Fe 415 steel. Sketch the reinforcement details by showing cross sec-tions(i) at column strip
(ii) at middle strip.2. Design the exterior panel of a flat slab of size 6 m � 6 m with suitable drop to support a live load
of 5 kN/m2. The floor system is supported by columns of size 500 mm � 500 mm. Floor tofloor distance is 3.6 m. Use M20 concrete and Fe 415 steel.
3. For the flat slab system of size 6 m � 6 m provide suitable drop and fix up overall dimensions.The floor system is supported by columns of size 500 mm ��500 mm, the floor height being 3.6 m.Calculate the design moments at various strips in the interior and exterior panels. Give the planof the floor system showing these design moments.