Flat Slabs. d 2. A typical flat slab (without drop and column head)

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1 1 CHAPTER Flat Slabs 1.1 INTRDUCTIN 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 clear ceiling 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. These slabs which are directly supported by columns are called Flat Slabs. Fig. 1.1 shows a typical flat slab. d Critical section for shear Fig. 1.1 A typical flat slab (without drop and column head) The column head is some times widened so as to reduce the punching shear in the slab. The widened 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 45º on either side of vertical only is considered as effective for the design [Ref. Fig. 1.]. Critical section for shear d 90 Concrete in this area is neglected for calculation Fig. 1. Slab without drop and column with column head

2 Advanced R.C.C. Design Moments in the slabs are more near the column. Hence the slab is thickened near the columns by providing the drops as shown in Fig Sometimes the drops are called as capital of the column. Thus we have the following types of flat slabs: d d Critical section for shear Critical section for shear Fig. 1.3 Slab with drop and column without column head (i) Slabs without drop and column head (Fig. 1.1). (ii) Slabs without drop and column with column head (Fig. 1.). (iii) Slabs with drop and column without column head (Fig. 1.3). (iv) Slabs with drop and column head as shown in Fig d Critical section for shear Fig. 1.4 Slab with drop and column with column head 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 1. A panel may be divided into column strips and middle strips. Column Strip means a design strip having a width of or 0.5, whichever is less. The remaining middle portion which is bound by the column strips is called middle strip. Fig. 1.5 shows the division of flat slab panel into column and middle strips in the direction y.

3 Flat Slabs 3 a C of panel A b C of panel B 1 y o Column strip x Fig. 1.5 Middle strip Column strip a b 4 4 but 1 but Middle strip Column strip Panels, column strips and middle strips is y-direction 1. PRPRTINING F FAT SABS IS [Clause 31.] gives the following guidelines for proportioning Drops The drops when provided shall be rectangular in plan, and have a length in each direction not less than one third of the panel in that direction. For exterior panels, the width of drops at right angles to the non continuous edge and measured from the centre-line of the columns shall be equal to one half of the width of drop for interior panels. 1.. Column Heads Where column heads are provided, that portion of the column head which lies within the largest right circular 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. and Thickness of Flat Slab From the consideration of deflection control IS specifies minimum thickness in terms of span to effective depth ratio. For this purpose larger span is to be considered. If drop as specified in 1..1 is provided, then the maximum value of ratio of larger span to thickness shall be = 40, if mild steel is used = 3, if Fe 415 or Fe 500 steel is used If drops are not provided or size of drops do not satisfy the specification 1..1, then the ratio shall not exceed 0.9 times the value specified above i.e., = = 36, if mild steel is used. = =., if HYSD bars are used It is also specified that in no case, the thickness of flat slab shall be less than 15 mm.

4 4 Advanced R.C.C. Design 1.3 DETERMINATIN F BENDING MMENT AND SHEAR FRCE For this IS permits use of any one of the following two methods: (a) The Direct Design Method (b) The Equivalent Frame Method 1.4 THE DIRECT DESIGN METHD 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. (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). Total Design Moment The absolute sum of the positive and negative moment in each direction is given by M 0 = W n Where, M 0 = Total moment W = Design load on the area n n = Clear span extending from face to face of columns, capitals, brackets or walls but not less than = ength of span in the direction of M 0 ; and = ength of span transverse to 1 In taking the values of n, 1 and, 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 0.6D. (b) When the transverse span of the panel on either side of the centre line of support varies, shall b be taken as the average of the transverse spans. In Fig. 1.5 it is given by a + bg. (c) When the span adjacent and parallel to an edge is being considered, the distance from the edge to the centre-line of the panel shall be substituted for. Distribution of Bending Moment in to ve and +ve Moments The total design moment M 0 in a panel is to be distributed into ve moment and +ve moment as specified below:

5 Flat Slabs 5 In an interior span Negative Design Moment 0.65 M 0 Positive Design Moment 0.35 M 0 In an end span Interior negative design moment 010. = M Positive design moment N M N M Exterior negative design moment a c 0. = = N M Q P 1 0 a c M a c Q P Q P M 0 M where a c is the ratio of flexural stiffness at the exterior columns to the flexural stiffness of the slab at a joint taken in the direction moments are being determined and is given by a c Kc = Ks Where, K c = Sum of the flexural stiffness of the columns meeting at the joint; and K s = Flexural stiffness of the slab, expressed as moment per unit rotation. Distribution of Bending Moments Across the Panel Width 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 strip moments. 0 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 6 Advanced R.C.C. Design Moments in Columns In this type of constructions column moments are to be modified as suggested in IS [Clause No ]. Shear Force The critical section for shear shall be at a distance d from 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 Critical section Support section column / column head d/ d/ Support section d/ ( a) Fig. 1.6 Critical section ( b) For columns sections with re-entrant angles, the critical section shall be taken as indicated in Fig Critical section Support section d/ Support section d/ Critical section d/ ( a) d/ ( b) d/ Fig. 1.7 In case of columns near the free edge of a slab, the critical section shall be taken as shown in Fig. 1.. Free edge Critical section Free corner d/ d/ ( a) Fig. 1. Corner column ( b) Critical section

7 Flat Slabs 7 The nominal shear stress may be calculated as t v = V bd 0 where V is shear force due to design b 0 is the periphery of the critical section d is the effective depth The permissible shear stress in concrete may be calculated as k s t c, where k s = b c but not greater than 1, where b c is the ratio of short side to long side of the column/capital; and t c = 05. f ck If shear stress t v < t c no shear reinforcement are required. If t c < t v < 1.5 t c, shear reinforcement shall be provided. If shear stress exceeds 1.5 t c flat slab shall be redesigned. 1.5 EQUIVAENT FRAME METHD IS 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 l ef into the span measured from the centre line of the columns. The stiffening 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. oading Pattern When the live load does not exceed ¾th of dead load, the maximum moments may be assumed to occur 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 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 Advanced R.C.C. Design moment M 0 = W n. 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. 1.6 SAB REINFRCEMENT Spacing The spacing of bars in a flat slab, shall not exceed times the slab thickness. Area of Reinforcement 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., 0.1 percent if HYSD bars used and 0.15 percent, if mild steel is used. Minimum ength of Reinforcement At least 50 percent of bottom bars should be from support to support. The rest may be bent up. The minimum length of different reinforcement in flat slabs should be as shown in Fig. 1.9 (Fig. 16 in IS ). If adjacent spans are not equal, the extension of the ve reinforcement beyond each face shall be based on the longer span. All slab reinforcement should be anchored property at discontinuous edges. Example 1.1: Design an interior panel of a flat slab of size 5 m 5 m without providing drop and column head. Size of columns is mm and live load on the panel is 4 kn/m. Take floor finishing load as 1 kn/m. Use M0 concrete and Fe 415 steel. Solution: Thickness Since drop is not provided and HYSD bars are used span to thickness ratio shall not exceed 1 1 = \ Minimum thickness required = Span =. = mm et d = 175 mm and D = 00 mm oads Self weight of slab = = 5 kn/m Finishing load = 1 kn/m ive load = 4 kn/m \ Total working load = 10 kn/m Factored load = = 15 kn/m

9 Flat Slabs 9 Middle Strip Column strip Strip Type of bars Straight bars Bent bars* Straight bars Bent bars* Minimum percentage of steel at section 50 Remainder 50 Remainder 50 Remainder 50 Remainder Remainder 50 Remainder 50 Remainder WITHUT DRP PANE d b 75 mm max 150 mm 0.15lmax d b c WITH DRP PANE d e e b b b d b g o.15 l max 75 mm max. 150 mm 150 mm c (A BARS) 150 mm f c 75 mm max. c a 4 BAR DIA R 300 mm min. A BARS 150 mm min. b e c g EDGE F DRP e b 150 mm 4BAR DIA R 300 mm min. EDGE F DRP c 75 mm max l max 150 mm a c C (A BARS) f 150 mm 75 mm max. Exterior support D C Clear span - l n Face of support interior support D C Clear span - l n Face of support interior support [N SAB CNTINUITY] [CNTINUITY PRVED] [N SAB CNTINUITY] Minimum ength Bar ength From Face of Support D C Maximum ength Mark a b c d e f g ength 0.14 l n 0.0 l n 0. l n 0.30 l n 0.33 l n 0.0 l n 0.4 l n * 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. Fig. 1.9 Minimum bend joint locations and extensions for reinforcement in flat slabs

10 10 Advanced R.C.C. Design n = = 4.5 m \ Total design load in a panel W = 15 n = = kn Moments Panel Moment M 0 = W n 45. = =19. 4 knm Panel ve moment = = knm Panel +ve moment = = = knm Distribution of moment into column strips and middle strip: Column Strip in knm Middle Strip in knm ve moment = ve moment = Checking the thickness selected: Since Fe 415 steel is used, M u lim = 0.13 f ck b d Width of column strip = = 500 mm \ M u lim = = Nmm = knm Hence singly reinforced section can be designed i.e., thickness provided is satisfactory from the consideration of bending moment. Check for Shear The critical section for shear is at a distance d from the column face. Hence periphery of critical section around a column is square of a size = d = = 675 mm Shear to be resisted by the critical section V = = kn \ t v = = N/mm k s =1 + b c subject to maximum of 1. b c = 1 5 = = 1 5 \ k s =1 safe in shear since t v < t c t c = 0. 5 f ck = = 1.11 N/mm

11 Reinforcement For ve moment in column strip: M u = 9.55 knm N M Ast = 07. fy A st d 1 bd A i.e., = A st 1 st y ck Q P = A st f f A st Flat Slabs 11 i.e., A st 104.3A st = 0 A st = mm This is to be provided in a column strip of width 500 mm. Hence using 1 mm bars, spacing required is given by s = p = 17 mm Provide 1 mm bars at 175 mm c/c. For +ve moment in column strip: M u = 39.6 knm \ = A st A = A st 1 st or A st A st = 0 \ A st = 651 mm Using 10 mm bars, spacing required is s = p = mm < thickness of slab 651 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. A st 415 0

12 1 Advanced R.C.C. Design Reinforcement Details It is as shown in Fig Column Strip Middle Strip Column strip Column Strip Middle Strip Column strip c/c c/c c/c c/c Sign convention Top reinforcement Bottom reinforcement c/c c\c Cover Section through column strip c/c section through middle strip Fig Reinforcement details [all dimension in mm units] Example 1.: Design an interior panel of a flat slab with panel size 6 6 m supported by columns of size mm. Provide suitable drop. Take live load as 4 kn/m. Use M0 concrete and Fe 415 steel. Solution : Thickness : Since Fe 415 steel is used and drop is provided, maximum span to thickness ratio permitted is 3 \ Thickness of flat slab = 6000 = 17.5 mm 3 Provide 190 mm thickness. et the cover be 30 mm \ verall thickness D = 0 mm et the drop be 50 mm. Hence at column head, d = 40 mm and D = 70 mm

13 Flat Slabs 13 Size of Drop It should not be less than 1 6 m = m 3 et 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. oads For the purpose of design let us take self-weight as that due to thickness at column strip \ Self-weight = = 6.75 kn/m Finishing load = 1.00 kn/m ive load = 4.00 kn/m Total load = kn/m \ Design (factored) load = = kn/m Clear span n = = 5.5 m \ Design load W 0 =W u n = = kn Design Total Moment Total moment M 0 = W 0 n = = 400 knm \ Total negative moment = = 60 knm Total positive moment = = 140 knm The above moments are to be distributed into column strip and middle strip Column Strip Middle Strip ve moment = 195 knm = 65 knm +ve moment = 4 knm = 56 knm Width of column strip = width of middle strip = 3000 mm M u lim = 0.13 f ck b d = = Nmm = knm Thus M u lim > M u. Hence thickness selected is sufficient. Check for Shear The critical section is at a distance

14 14 Advanced R.C.C. Design d = 40 = 10 mm from the face of column \ It is a square of size = = 740 mm V = Total load load on area = = kn \ Nominal shear = t v = = 0.0 N/mm Shear strength = k s t c where k s = 1 + b c subject to maximum of where b c = 1 = 1 \ k s =1 t c = = 1.11 N/mm Design shear stress permitted = 1.11 N/mm > t v Hence the slab is safe in shear without shear reinforcement also Shear strength may be checked at distance d from drop. It is quite safe since drop size is large. Reinforcement (a) For ve moment in column strip M u = 195 knm Thickness d = 40 mm \ M u = 0.7 f y A st d 1 b d N M A st = A st A 50.3 = A st 1 st A st A st = 0 A st = 419 mm in 3000 mm width f f y ck Q P A st 415 0

15 Flat Slabs 15 Using 1 mm bars, spacing required is s = p = mm 419 Provide 1 mm bars at 140 mm c/c (b) For +ve moment in column strip M u = 4 knm = Nmm. Thickness d = 190 mm = A st A 14.5 = A st 1 st \ A st = 15 mm Using 10 mm bars s = p Provide 10 mm bars at 10 mm c/c (c) For ve moment in middle strip: M u = 65 knm; Using 10 mm bars 3000 = 13 mm A st Thickness = 190 mm = A st A = A st 1 st A st = 0 A st = 93 mm in 3000 mm width A st s = p = 39.7 mm 93 Provide 10 mm bars at 30 mm c/c (d) For +ve moment in middle strip M u = 56 knm; Thickness = 190 mm Provide 10 mm bars at 30 mm c/c in this portion also. Since span is same in both direction, provide similar reinforcement in both directions. The details of reinforcement are shown in Fig A st

16 Middle strip 16 Advanced R.C.C. Design Column strip =Dorp width Middle strip c/c Column strip =Dorp width Column strip =Dorp width 10 10c/c c/c 1 30c/c 6000 Column strip =Dorp width 10 10c/c c/c Cover Section through column strip 30 c/c Fig Reinforcement details Example 1.3: Design the interior panel of the flat slab in example 1., providing a suitable column head, if columns are of 500 mm diameter. Solution: et the diameter of column head be = 0.5 = = 1.5 m It s equivalent square has side a where π = a a = 1.33 m \ n = = 4.67 m W 0 = = kn M 0 = W o n = =.3 knm

17 Flat Slabs 17 \ Total ve moment = = 17.4 knm Total +ve moment = = knm The distribution of above moment into column strip and middle strips are as given below: Column Strip Middle Strip ve moment = knm = 46.5 knm +ve moment = knm = knm Width of column strip = width of middle strip = 3000 mm \ M u lim = 0.13 f ck bd = = Nmm > M u Hence thickness selected is sufficient. Check for Shear The critical section is at a distance d = 40 = 10 mm from the face of column head Diameter of critical section = =1740 mm = m Perimeter of critical section = p D = p Shear on this section p V = = kn \ t v = = 0.45 N/mm π Maximum shear permitted = k s = 1.11 N/mm Since k s works out to be 1 Since maximum shear permitted in concrete is more than nominal shear t v, there is no need to provide shear reinforcement Design of Reinforcement (a) For ve moment in column strip M u = knm; d = 40 mm \ = A st A 16 = A st 1 st A st

18 1 Advanced R.C.C. Design A st A st = 0 A st = 1705 mm Using 1 mm bars, s = π = 199 mm 1705 Provide 1 mm bars at 190 mm c/c. (b) For the +ve moment in column strip M u = knm; d = 190 mm Using 10 mm bars = A st A.51 = A st 1 st A st A st = 0 A st = 913 mm s = π = 5 mm 913 Provide 10 mm bars at 50 mm c/c. (c) For ve moment in middle strip: M u = 46.5 knm; d = 190 mm Using 10 mm bars, A st = A st A 63 = A st 1 st A st A st = 0 A st = 701 mm s = π = 336 mm 701 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. Example 1.4: 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/m and a finishing load of 1 kn/m. It is to be designed using M0 grade concrete and Fe 415 steel. The size of the columns supporting the system is 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. A st

19 Flat Slabs 19 Solution: Thickness: Since Fe 415 steel is used and no drops are provided, longer span to depth ratio is not more than =. d = 6000 = 0. et us select d = 10 mm and D = 40 mm oads Self weight = 6 kn/m Finishing weight = 1 kn/m ive load = 4 kn/m Total = 11 kn/m W u = = 16.5 kn/m Panel Dimensions Along length 1 = 6 m and = 5 m Width of column strip = or whichever is less. = = 1.5 m on either side of column centre line \Total width of column strip = 1.5 =.5 m Width of middle strip = 5.5 =.5 m Along Width 1 =5 m = 6 m Width of column strip = = 1.5 m on either side \Total width of column strip =.5 m Hence, width of middle strip = 6.5 = 3.5 m INTERIR PANES Moments Along onger Size 1 =6 m = 5 m n = = 5.5 m subject to minimum of = 3.9 m \ n = 5.5 m oad on panel W 0 = 16.5 n = = kn

20 0 Advanced R.C.C. Design M 0 = W 0 n = = knm Appropriation of Moment Total ve moment = = 0.77 knm \ Total +ve moment = = knm Hence moment in column strip and middle strip along longer direction in interior panels are as given below: Column Strip Middle Strip ve moment = knm = knm +ve moment = knm = knm Along Width 1 =5 m = 6 m and n = = 4.5 m. Panel load = W 0 = = kn Panel moment M 0 = W 0 n = = kn-m Appropriation of Moment: Total ve moment = = 16. kn-m Total +ve moment = = 7.71 kn-m \ Moments in column strip and middle strip are as shown below: Column Strip Middle Strip ve moment = 1.16 knm = 40.7 knm +ve moment = 5.63 knm = 35.0 knm EXTERIR PANES ength of column = = 4.6 m The building is not restrained from lateral sway. Hence as per Table in IS , effective length of column = 1. length = = 5.11 m Size of column = mm Moment of inertia of column = mm 4 1

21 Flat Slabs 1 \ k c = I 1 = = mm 4 NGER SPAN DIRECTIN Moment of inertia of beam I s = Moment of inertia of slab = Its length = = 5000 mm \ k c = I 3 s = = mm ive load Dead load = 4 7 < 0.75 \ Relative stiffness ratio is a c = k c1 + k c = = k \ a = 1 + = 1+ = 1.67 a c Hence various moment coefficients are: Interior ve moment coefficient = α = Exterior ve moment coefficient = 0.65 α = 0.37 Positive moment coefficient = α = Total moment M 0 = knm \ Appropriation of moments in knm is as given below: s Total Column Strip Middle Strip Interior ve = = = 53. Exterior ve = = = 0 + Moment = = = Shorter Span Direction \ k s = = \ a c = k c1 + k c = =.13 k s 3

22 Advanced R.C.C. Design 1 \ a 1 = 1 + = αc Interior ve moment coefficient = a = = Exterior ve moment coefficient = α = = Positive moment coefficient = 0.63 α = = Total moment M 0 = knm \ Appropriation of moments in shorter span exterior panel in knm is as given below: Total Column Strip Middle Strip Interior ve = = = 4.7 Exterior -ve = = = 0 + Moment = = = In the exterior panel in each column strips half the above values will act. These moments are shown in Fig. 1.1 Col Strip Middle Strip Col Strip Middle Strip Col Strip.5 m m m m m Fig. 1.1

23 Flat Slabs 3 REVIEW QUESTINS 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/m. The floor is supported by columns of size 450 mm 450 mm. Use M0 concrete and Fe 415 steel. Sketch the reinforcement details by showing cross sections (i) at column strip (ii) at middle strip.. 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/m. The floor system is supported by columns of size 500 mm 500 mm. Floor to floor distance is 3.6 m. Use M0 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.