Flat Slabs-simple-As Per is 456

8/12/2019 Flat Slabs-simple-As Per is 456

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Common practice of design and construction is to support the slabs by beams and support the beams

by 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 slabs

are 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

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 from

the consideration of architecture but for the design, concrete in the portion at 45 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

1


2/23

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

(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

4545

d2

The portion of flat slab that is bound on each of its four sides by centre lines of adjacent columns is

called a panel. The panel shown in Fig. 1.5 has size L1L2. 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 directiony.


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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

but L

1

4 but

L

1

4

IS 456-2000 [Clause 31.2] gives the following guidelines for proportioning.

The drops when provided shall be rectangular in plan, and have a length in each direction not less than

one 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 the

width of drop for interior panels.

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 be

considered for design purpose as shown in Figs. 1.2 and 1.4.

From the consideration of deflection control IS 456-2000 specifies minimum thickness in terms of

span 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 used

It is also specified that in no case, the thickness of flat slab shall be less than 125 mm.


4/23


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

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).

The absolute sum of the positive and negative moment in each direction is given by

M0=WL

8

n

Where,

M0= Total moment

W = Design load on the area L2Ln

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 M0; and

L2= Length of span transverse to L1In taking the values of Ln, L1and 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 size

0.886D.

(b) When the transverse span of the panel on either side of the centre line of support varies, L2shall

be taken as the average of the transverse spans. In Fig. 1.5 it is given byL L2 2

2

a 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.

The total design moment M0 in a panel is to be distributed into ve moment and +ve moment asspecified below:


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Flat Slabs 5

Negative Design Moment 0.65 M0

Positive 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 cis 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.

The +ve and ve moments found are to be distributed across the column strip in a panel as shown in

Table 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


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In this type of constructions column moments are to be modified as suggested in IS 4562000

[Clause No. 31.4.5].

The critical section for shear shall be at a distanced

2from the periphery of the column/capital drop

panel. Hence if drops are provided there are two critical sections near columns. These critical sections

are shown in Figs. 1.1 to 1.4. The shape of the critical section in plan is similar to the support

immediately below the slab as shown in Fig. 1.6.

d/2

d/2

Criticalsection

Support sectioncolumn / column head

( )a

d/2

Supportsection

Criticalsection ( )b

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

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

Critical

section

Freeedge

( )a

Criticalsection

Freecorner

Cornercolumn

( )b


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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 depth

The permissible shear stress in concrete may be calculated as ksc, where ks= 0.5 + cbut notgreater than 1, where cis 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 cflat slab shall be redesigned.

IS 4562000 recommends the analysis of flat slab and column structure as a rigid frame to get design

moment and shear forces with the following assumptions:

(a) Beam portion of frame is taken as equivalent to the moment of inertia of flat slab bounded

laterally by centre line of the panel on each side of the centre line of the column. In frames

adjacent and parallel to an edge beam portion shall be equal to flat slab bounded by the edge and

the 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 the

concrete alone.

(c) Variation of moment of inertia along the axis of the slab on account of provision of drops shall

be taken into account. In the case of recessed or coffered slab which is made solid in the region

of the columns, the stiffening effect may be ignored provided the solid part of the slab does not

extend 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 accepted

method like moment distribution or matrix method.

When the live load does not exceed th 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 th dead load analysis is to be carried out for the following pattern of loading also:

(i) To get maximum moment near mid span

th of live load on the panel and full live load on alternate panel

(ii) To get maximum moment in the slab near the support

th 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 those

occurring with full design live load on all panels.

The moments determined in the beam of frame (flat slab) may be reduced in such proportion that

the numerical sum of positive and average negative moments is not less than the value of total design


8/23


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.

The spacing of bars in a flat slab, shall not exceed 2 times the slab thickness.

When the drop panels are used, the thickness of drop panel for determining area of reinforcement

shall 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 ifHYSD bars used and 0.15 percent, if mild steel is used.

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.

Design an interior panel of a flat slab of size 5 m 5 m without providing drop and

column head. Size of columns is 500 500 mm and live load on the panel is 4 kN/m 2. Take floor

finishing load as 1 kN/m2. Use M20 concrete and Fe 415 steel.

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 = 0.2025 = 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


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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

MiddleStrip

Co

lumns

trip

Strip

Type

o

fb

ars

Stra

ight

bars

Be

nt

bars

*

Stra

ight

bars

Ben

tb

ars

*

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. ALLBARS

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.


10/23


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

83375

4 5

818984 .

..

Panel ve moment = 0.65 189.84 = 123.40 kNm

Panel +ve moment = 0.35 189.84 = 0.35 189.84 = 66.44 kNm

Distribution of moment into column strips and middle strip:

Column Strip in kNm Middle Strip in kNm

ve moment 0.75123.40 = 92.55 30.85

+ve moment 0.6066.44 = 39.86 26.58

Checking the thickness selected:

Since Fe 415 steel is used,

Mu lim= 0.138 fckbd2

Width of column strip = 0.5 5000 = 2500 mm

Mu lim= 0.138 20 2500 1752= 211.3125 106Nmm

= 211.3125 kNm

Hence singly reinforced section can be designed i.e., thickness provided is satisfactory from theconsideration of bending moment.

The critical section for shear is at a distanced

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 + csubject to maximum of 1.

c= LL

1

2

55

= = 1

ks= 1

c= 0 25 0 25 20. .fck = = 1.118 N/mm2

safe in shear since v< c

675

675500

500


11/23

Flat Slabs 11

For ve moment in column strip:

Mu= 92.55 kNm

92.55 106= 0 87 1. f dbd

f

fy st

st y

ck

AA

= 0.87 415 Ast175 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

1583742500

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 Ast175 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

651 2500

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.


12/23


It is as shown in Fig. 1.10

50005000 5000

5000

5000

5000

Column Strip Middle Strip Column strip

Co

lumn

Str

ip

Middle

Strip

Columnstrip

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- 300c/c

section through middle strip

Top reinforcement

Sign convention

Bottom reinforcement

12-175 c/c

12-175 c/c

Design an interior panel of a flat slab with panel size 6 6 m supported by columns of

size 500 500 mm. Provide suitable drop. Take live load as 4 kN/m2. Use M20 concrete and Fe 415

steel.

Thickness : Since Fe 415 steel is used and drop is provided, maximum span to thickness ratio

permitted is 32

Thickness of flat slab = 600032

= 187.5 mm

Provide 190 mm thickness. Let the cover be 30 mm

Overall thickness D = 220 mm

Let the drop be 50 mm. Hence at column head, d= 240 mm and D = 270 mm


13/23

Flat Slabs 13

It should not be less than1

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 = WuL2Ln

= 17.625 6 5.5

= 581.625 kN

Total moment

M0=W L0

8

581625 5 5

8

n . .

= 400 kNm

Total negative moment = 0.65 400 = 260 kNm

Total positive moment = 0.35 400 = 140 kNm

The above moments are to be distributed into column strip and middle strip

Column Strip Middle Strip

ve moment 0.75 260 = 195 kNm 0.25260 = 65 kNm

+ve moment 0.6140 = 84 kNm 0.4 140 = 56 kNm

Width of column strip = width of middle strip = 3000 mm

Mu lim= 0.138 fckbd2= 0.138 20 3000 2402 = 476.928 106Nmm

= 476.928 kNmThus Mu lim> Mu. Hence thickness selected is sufficient.

The critical section is at a distance


14/23


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 = ksc

where ks = 1 + csubject to maximum of 1

where c =L

L

1

2

= 1

ks= 1

c= 0 25 20. = 1.118 N/mm2

Design shear stress permitted

= 1.118 N/mm2> vHence the slab is safe in shear without shear reinforcement also.

Shear strength may be checked at distanced

2from 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= 0.87fyAstd 1

Ast y

ckb d

f

f

195 106= 0.87 415 Ast240 1

3000 240

415

20

A st

2250.38 = Ast 1 34698 8

A st

.

Ast2 34698.8 Ast+ 2250.38 34698.8 = 0

Ast= 2419 mm2in 3000 mm width

500

500

500 740

740

120 120


15/23

Flat Slabs 15

Using 12 mm bars, spacing required is

s= 4 12

2419 3000

2 = 140.26 mm

Provide 12 mm bars at 140 mm c/c

(b) For +ve moment in column strip

Mu= 84 kNm = 84 106Nmm. Thickness d= 190 mm

84 106= 0.87 415 Ast190 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


(c) For ve moment in middle strip:

Mu= 65 kNm; Thickness = 190 mm

65 106= 0.87 415 Ast190 13000 190

415

20

A st

947.5 = Ast 127469 9

Ast

.

Ast2 27469.9 Ast+ 947.5 27469.9 = 0

Ast= 983 mm2in 3000 mm width

Using 10 mm bars

s= 4 10

9833000

2 = 239.7 mm


(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.


16/23


6000

Column stripMiddle strip

Column strip

Co

lumns

trip

Middle

strip

Columnstrip

6000 6000

6000

6000

6000

=Do p widthr=Do p widthr

=Dorpwidth

=Dorpwidth

12 140 c/c

10 230 c/c

500 500

240 10 230@

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

10180c/c

Design the interior panel of the flat slab in example 1.2, providing a suitable column

head, if columns are of 500 mm diameter.

Let the diameter of column head be

= 0.25L = 0.25 6 = 1.5 m

Its equivalent square has side a where

41 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


17/23

Flat Slabs 17

Total ve moment = 0.65 288.3 = 187.4 kNm

Total +ve moment = 0.35 288.3 = 100.9 kNm

The distribution of above moment into column strip and middle strips are as given below:


ve moment 0.75187.4 = 140.55 kNm 0.25187.4 = 46.85 kNm

+ve moment 0.60100.9 = 60.54 kNm 0.4100.9 = 40.36 kNm

Width of column strip = width of middle strip = 3000 mm

Mu lim= 0.138fckbd2= 0.138 20 3000 2402

= 476.928 106Nmm > MuHence 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

.

= 0.45 N/mm2

Maximum shear permitted = ks 0 25 20.

= 1.118 N/mm2 Since ksworks out to be 1

Since maximum shear permitted in concrete is more than nominal shear v, there is no need toprovide shear reinforcement

(a) For ve moment in column strip

Mu= 140.55 kNm; d= 240 mm

140.55 106= 0.87 415 Ast240 13000 240

415

20

Ast

1622 = Ast 134698 8

A st

.

1500 120


18/23


Ast2 34698.8 Ast+ 1622 34698.8 = 0

Ast= 1705 mm2

Using 12 mm bars,

s= 4 12

17053000

2 = 199 mm


(b) For the +ve moment in column strip

Mu= 60.54 kNm; d= 190 mm

60.54 106= 0.87 415 Ast190 13000 190

415

20

A st

882.51 = Ast 127469 9

A st

.

Ast2

27469.9 Ast+ 882.51 27469.9 = 0Ast= 913 mm

2

Using 10 mm bars

s= 4 10

9133000

2

= 258 mm


(c) For ve moment in middle strip:

Mu= 46.85 kNm; d= 190 mm

46.85 106= 0.87 415 Ast190 13000 190

415

20

A st

683 = Ast 127469 9

Ast

.

Ast2 27469.9Ast+ 683 27469.9 = 0

Ast= 701 mm2

Using 10 mm bars,

s= 4 10

7013000

2

= 336 mm


(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. 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/m2and a finishing load of 1 kN/m2. It is to be designed using M20

grade concrete and Fe 415 steel. The size of the columns supporting the system is 500 500 mm and

floor to floor height is 4.5 m. Calculate design moments in interior and exterior panels at column and

middle strips in both directions.


19/23

Flat Slabs 19

Thickness: Since Fe 415 steel is used and no drops are provided, longer span to depth ratio is not

more 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 L1or L2whichever 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 mLoad on panel W0= 16.5 L2Ln

= 16.5 5 5.5 = 453.75 kN


20/23


M0=W L0

8

45375 55

8

n . .

= 311.95 kNm

Total ve moment = 0.65 311.95 = 202.77 kNm

Total +ve moment = 311.95 202.77 = 109.18 kNm

Hence moment in column strip and middle strip along longer direction in interior panels are as givenbelow:


ve moment 0.75202.75 = 152.06 kNm 202.75 152.06 = 50.69 kNm

+ve moment 0.60109.18 = 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

08

4455 4 5

8

n . .

= 250.59 kN-m

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:


ve moment 0.75162.88 = 122.16 kNm 0.25 162.88 = 40.72 kNm

+ve moment 0.6087.71 = 52.63 kNm 0.4087.71 = 35.08 kNm

Length of column = 4.5 0.24 = 4.26 m

The building is not restrained from lateral sway. Hence as per Table 28 in IS 456-2000, effectivelength of column

= 1.2 length = 1.2 4.26 = 5.112 m

Size of column = 500 500 mm

Moment of inertia of column =1

12500 4 4 mm


21/23

Flat Slabs 21

kc=I

L

1

12=

500

5112

4

= 101844 mm4

Is= Moment of inertia of slab

=1

126000 2403

Its length = L2= 5000 mm

kc=I

5000

s 1

12

6000 240

5000

3

= 1382400 mm4

Live load

Dead load

=4

7

< 0.75

Relative stiffness ratio is

c=k k

k

c 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.

= 0.690

Exterior ve moment coefficient =0.65

= 0.387

Positive moment coefficient = 0.63 0.28

= 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 0.387311.95 = 120.72 1.00 120.72 = 120.72 120.72 120.72 = 0

+ Moment 0.46331.95 = 144.43 0.60 144.43 = 86.66 144.43 86.66 = 57.77

ks=1

12

5000 240

6000

3

= 96000

c=k k

k

c c

s

1 2 2 1018844

960000

+=

= 2.123


22/23


1= 11

+c

= 1.471

Interior ve moment coefficient = 0.75 0 1 0 75 0 11471

. . ..

= 0.682

Exterior ve moment coefficient =0.65 0.65

=

1471.= 0.442

Positive moment coefficient = 0.63 0.28 0.28

= 0 63

1471.

.= 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 0.44250.59 = 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 15.06

2

122.16

50.69

52.63

152.06

122.16

86.66

2

40.72

57.77

35.08

40.72

43.67

35.08

65.51

40.72

152.06

122.162

128.18

50.69

66.16

2

161.43152.062

122.16

2128.18

66.16

2

52.632

53.82120.72

122.16

2

128.18

86.662

42.72

57.77

44.0942.72

43.67

44.09

65.51

42.72

152.06

110.7666.16

50.69161.43

152.06

2110.7666.162

53.82120.72

4110.76

2.5 m

2.5 m

3.5 m

3.5 m

1.25 m

8666

2

6551

2

. .

8666

2

6551

2

. .

5263

2

.


23/23

Flat Slabs 23

1. Design the typical interior panel of a flat slab floor of size 5 m 5 m with suitable drop to

support 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 loadof 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 plan

of the floor system showing these design moments.

Flat Slabs-simple-As Per is 456

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