onw way slab design

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1. DESIGN OF REINFROCED CONCRETE STRCUTURE: ONE WAY SLAB PREPARED BY : PALAK PATEL CIVIL ENGINEERING MAHAMTA GANDHI INSTITUTE OF TECHNICAL EDUCATION AND RESEARCH CENTER, NAVSARI

2. SLABS Slab A slab is a plate element having depth (D), very small as compared to it’s length and width. Slabs form floors and roofs of buildings. They are assumed to carry uniformly distributed loads. Usually slabs are horizontal except in the case of stair cases and ramps for car parks. 2

3. SLABS TYPES OF SLABS 3

4. SLABS One way slabs If the slab is supported on two opposite sides. If the slab is supported on four sides, and L y /L x ≥ 2. Both of condition called one way slab. In one way slab main reinforcement is provided along short span and distribution reinforcement provided along long span. Two way slabs If the slab is supported on all four edge, and Ly/Lx 2. Above condition called two way slab. In two way slabs main reinforcement is provided along both the direction.   L y = longer span L x = shorter span 4

5. SLABS Flat slabs When slab is directly supported on columns, without beams. It’s provided to increase the floor height and to permit large amount of light. Grid slab When slab is supported on beams with columns only on the periphery of the hall. Continuous slab In case of large halls, auditoriums, marriage halls, etc. the length is divided into equal bays by providing beams. The slab provided over such area is called, continuous slab. 5

6. SLABS Design considerations Effective depth (d) (IS: 456-2000,Pg-37, cl 23.2.1) (IS: 456-2000, Pg-38, Fig.4) Effective span (l e ) (IS: 456-2000, Pg-34,cl 22.2) A) simply supported slab Clear span + effective depth (d) Centre to centre of support Which ever is smaller   6

7. SLABS B) continuous slab If width of support clear span, then effective span is taken as per simply supported beam. If width of support clear span (OR) width of support 600 mm, then For end span one end fixed and the other continuous OR for intermediate span: effective span = clear span For end span with one end free other continuous: Clear span + d/2 Clear span +( ½ x width of dis-continuous support ) Which ever is smaller   7

8. SLABS Cover to reinforcement Provided for the protect steel reinforcement from corrosion Nominal cover ( Clear cover ) The thickness of concrete from the surface of reinforcement bar to the nearest edge of the concrete is called Nominal cover. Effective cover The thickness of concrete from the centre of bar to the nearest edge of concrete is called effective cover. IS code provision ( IS: 456-2000, Pg-46, cl 26.4 ) Nominal cover should not be less than the diameter of bar. 8

9. SLABS Spacing of reinforcement ( IS : 456-2000, Pg-45, cl 26.3) Minimum horizontal distance Shall not be less of these values The diameter of the bar if the diameter are equal The diameter of the larger bar if the diameter are unequal. 5 mm more than the nominal maximum size of course aggregates. 9

10. SLABS Maximum distance between bars in tension ( IS : 465-2000, Pg-46, cl 26.3.3 ) Main reinforcement 3d, where, d = effective depth 300 mm Which ever is smaller 10

11. SLABS Distribution steel 5d 450 mm Which ever is smaller. Maximum spacing of shear reinforcement 0.75d 300 mm Which ever is smaller 11

12. SLABS Requirement of reinforcement for slab ( IS : 456-2000, Pg-48, cl 26.5.2 ) Minimum reinforcement For mild steel bar, Reinforcement shall not be less than 0.15% of the total c/s area. For high strength deformed bars ( tor bar) Reinforced shall not be less than 0.12% of the total c/s area. Maximum diameter The diameter of bar shall not exceed ,1/8 of total slab thickness. 12

13. SLABS Slabs design checks Check for cracking ( IS : 456-2000, Pg-46 ) For Main steel 3d 300 mm Spacing should not exceed of this two values. For distribution steel 5d 450 mm Spacing should not exceed of the two values. 13

14. SLABS Check for deflection ( IS: 456-2000, Pg-38, Fig-4 ) Allowable Find actual Actual Allowable Check for development length (L d ) ( IS: 456-2000, Pg-44, cl 23.2.3.3 (c) ) L d should be Here L d =   14 M 1 = Moment resistant 50% steel at support V = shear force at support L = sum of anchorage beyond centre of support   L = d = 12 take smaller value of two  

15. ONE WAY SLABS DESIGN ( EXAMPLE ) Design the slab for the room for office building 3.2 m x 9.2 m . The slab is resting on 300 mm thick wall and resting live load of 2.5 kN/m 2 . Use M-20 concrete mix and Fy-415 steel as reinforcement Check the slab for cracking, check for deflection, check for development length and check for shear. Assume 3.2 m x 9.2 m as clear dimension F ck = 20 N/mm 2 F y = 415 N/mm 2 15

16. 16 L y /L x = 9.2/3.2 = 2.875 > 2 Design the slab as one-way simply supported slab. Effective depth : ( IS 456:2000, Pg-38, table 4) Consider shorter span as l, l = 3200 mm l/d = 20 x M.F Assume 0.6% steel F y = 415 N/mm 2 , F s = 240 N/mm 2 M.F = 1.15 l/d = 20 x 1.15 d = l/(20 x 1.15) = 3200/(20 x 1.15) = 139.13 mm Provided d = 150 mm Assume 10 mm bars Over all depth = 150 + /2 + clear cover  

17. 17 = 150 + 10/2 + 20 = 150 +5 + 20 = 175 mm Effective span: ( IS 456:2000, Pg-34, Cl-22.2.9 ) 3200 + 150 = 3350 mm c/c of support = 3200 + 300 = 3500 mm Whichever is smaller l = 3350 mm = 3.35 m Load calculation Assume 1 m strip of slab Dead load of slab = ( 1 x 0.175 ) x 25 = 4.375 kN /m Floor finish = 1 x 1 = 1.0 kN /m Live load = 2.5 x 1 = 2.5 kN /m Total load = DL + LL + FL = 4.375 + 2.5 + 1 = 7.875 kN /m

18. 18 Factor load = 1.5 x 7.875 = 11.812 kN/m Bending moment M u = = 16.58 kN.m Effective depth for flexure For F y = 415 N/mm 2 M u = 0.138 F ck b d 2 16.58 x 10 6 = 0.138 x 20 x 1000 x d 2 d 2 = d = 77.50 mm < 150 mm……(OK) Main steel (SP 16, Pg-48, Table 2 )  

19. 19 F y = 415 N/mm 2 F ck = 20 N/mm2 Pt = 0.218 A st = = 327 mm2 Consider 10 mm bar a st = = = 78.53 mm 2 Spacing = = = 240.15 mm Provided spacing 230 mm c/c  

20. 20 A st = = = 341.43 mm 2 Distribution steel ( IS: 456:2000, Pg-48, Cl 26.5.2.1 ) Provide minimum 0.12% of total cross section area. A st = = 210 mm 2 Consider 8 mm diameter bar, a st = = = 50.26 mm Spacing = = 239.33 mm Provided spacing 230 mm c/c  

21. 21 A st = = = 218.52 mm 2 Check for cracking ( IS: 456:2000, Pg-46 ) For main steel 3d = 3 x 150 = 450 300 mm 230 mm provided < 300 mm…..( OK ) For distribution steel 5d = 5 x 150 = 750 mm 450 mm 130 mm provided < 450 mm……( OK ) Check for deflection ( IS: 456:2000, Pg-38, Fig 4 ) Allowable l/d = 20 x M.F Pt provided = = = 0.227%  

22. 22 M.F = 1.6 Allowable l/d = 20 x 1.6 = 32 Actual l/d = 3350/150 = 22.32 Actual l/d < allowable l/d Check for development length ( IS: 456:2000, Pg-44 ) L d d = 150 mm 12 = 12 x 10 = 120 mm Taking larger of two value L o = 150 mm Shear force at support = = 18.91 kN  

23. 23 A st = 50 % of A st at mid span = 341.43/2 = 170.71 M 1 = 0.87 F y A st d (1- = 0.87 x 415 x 170.71 x 150 x ( 1- = 9.026 x 1.3 = 770.50 mm For M 20 Fy = 415 N/ 10 mm bar in Tension (sp_16, Pg 184, Table -65) ….. (OK) OR where, = Nominal diameter of the bar = stress in bar at the Section considered at design load = design bond stress given in 26. 2.1.1  

24. 24 = 0.87 x 415 = 361.05 N/ x 1.6 = 1.92 N/ = 470.11 mm check for shear 𝜏 𝑏𝑑 = = = 0.126 N/ Consider 50% bars are bent up at support to resist shear = For F ck = 20 N/ ( IS: 456:2000 Pg.73, Table -19 ) 0. 28 N/ For slab thickness, D = 175 mm K = 1.25 ( IS : 456: 2000, Pg-72, Cl 40.2.1.1 )  

25. 25 = 1.25 x 0.28 = 0.35 ….. ( OK )  

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27. ONE WAY SLAB DESIGN (EXAMPLE) Design a simply supported one way slab for an effective span of 3.0 m to carry total factor load of 9 kN/m 2 . use M20 concrete and Fe250 steel. Draw sketch with required details. Design Effective span = 3.0 m Total factor load = 9 kN /m 2 F ck = 20 N/mm 2 F y = 250 N/mm2 Effective depth ( IS: 456-2000, P-37,38 ) l/d = 20 x M.F Assume 0.4 % steel 27

28. 28 F y = 250 N/mm 2 F s = 145 N/mm 2 ( IS : 456-2000, fig- 4 ) M.F = 2.0 3000/d = 20 x 2 d = 75 mm Provided depth, d = 100 mm Assume diameter of bar = 10 mm Clear cover = 20 mm D = 100 + 5 + 20 = 125 mm Bending moment M u = = = 10.125 kN.m  

29. 29 Effective depth for flexure: F y = 250 N/mm2 M u = 0.148 F ck b d 2 ( Sp-16, Pg-10, Table- C ) 10.125 x 10 6 = 0.148 x 20 x 1000 x d 2 d = 58.48 mm 58.48 mm ( Required ) < 100 mm ( Provided ) Main steel ( Sp-16, Pg-48, Table- 2 ) F ck = 20 N/mm 2 , F y = 250 N/mm 2 Pt = 0.50 % = 500 mm 2 Consider 10 bars, a st = 78.53 mm 2 Spacing = =  

30. 30 Provide spacing 150 mm C/C 150 = A st ( provided ) = 524 mm 2 Provided 10 mm - 150 mm c/c Distribution steel Provided minimum 0.15% of total c\s area. ( IS: 456-2000, Pg-48,Cl 26.5.2.1 ) A st = = 187.5 mm 2 Consider 8 mm bars, a st = 50.26 mm 2 Spacing = = 268.05 mm 260 mm c/c Provide spacing 260 mm c/c 260 = Ast ( Provided ) = 193.30 mm2 Provided 8 mm - 260 mm c\c  

31. 31 Check for cracking Main steel 3d = 3 x 100 = 300 mm 300 mm 150 mm ( Provided ) < 300 mm ( Required ) ……( OK ) Distribution steel 5d = 5 x 100 = 500 mm 450 mm 260 mm ( Provided ) < 450 mm ( Required ) ….. ( OK ) Check for deflection ( IS: 456-2000, Pg-38, Fig-4 ) Allowable l/d = 20 x M.F % Pt provided = = = 0.524 % M.F = 1.9 Allowable l/d = 20 x 1.9 = 38 Actual l/d = 3000/100 = 30 Actual l/d < allowable l/d ….. ( OK )  

32. 32 Check for development length L d 1.3 ( IS: 456;2000, Pg-44 ) d = 100 mm 12 = 12 x 10 = 120 mm Taking larger of two values L o = 120 mm S.F at support = = = 13.5 kN A st at support = 50 % A st at mid span = 524/2 = 262 mm 2 M 1 = 0.87 F y A st d ( 1 - ) ( IS: 456:2000, Pg-96 ) = 0.87 x 250 x 262 x 100 x ( 1 - ) = 5.51 x 10 6 N.mm 1.3 x = 650.60 mm For M20, Fy-250 10 mm , tension  

33. 33 Ld = = = 45.3 = 45.3 x 10 = 453 mm 453 mm < 650.60 mm ….. ( OK ) Check for shear = 0.135 N/mm 2 Consider 50% bars bent up at support to resist shear. Ast = 524/2 = 262 mm 2 Pt = =  

34. 34 For F ck = 20 N/mm2 ( IS: 456:2000, Pg-73, Table – 19 ) For D = 125 mm K = 1.30 = 1.30 x 0.36 = 0.468 N/mm 2 …… ( OK )  

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36. ONE WAY SLAB DESIGN ( EXAMPLE ) for practice Design a simply supported slab on 350 mm wide brick masonry for a clear room size 4 m x 10 m. use material grades M-20, and Fe250. take live load as 3.5 kN/m 2 and floor finish as 1 kN/m 2 . check your design for any two as shear, development length, deflection control and cracking. ANS : main steel ( A st ) = 952.5 mm 2 , Distribution steel ( A st ) = 262.5 mm 2 Design a simply supported slab on 300 mm wide brick masonry walls for a clear room dimension 3 m x 8 m. assume floor finish 0.75 kN/mm 2 and live load 3 kN/mm 2 . check for limit state of serviceability. Take M-20 concrete and Fe415 steel. ANS : Main steel ( A st ) = 292.5 mm 2 , Distribution steel ( A st ) = 262.5 mm 2 36

37. ONE WAY SLAB DESIGN STAIR CASE DESIGN Term related to stair case Flight : the incline slab of stair case is called flight. Landing : it is the level platform at the top or bottom or a flight between the floors. Rise : it is the vertical distance between two successive tread faces. Tread : it is the horizontal distance between two successive riser faces. Nosing : it is the projecting part of the trend beyond the face of riser. Waist slab : the slab below steps in the stair case is called waist slab. Soffit : it is underside of a stair. 37

38. ONE WAY SLABS DESIGN Requirement for staircase one flight design Live load = as per IS:875, 3 kN/m2 ….. If there is no possibility of overcrowding 5 kN/m2 ….. If there is possibility of overcrowding Effective span As per IS : 456-2000, Pg-63, Cl 33.1 (a) Effective span : center to center of beam 38

39. ONE WAY SLABS DESIGN Pitch The slop of staircase flight should not exceed 38’ For steps, riser ( R ) should not be more than 200 mm and tread ( T ) should not be less than 240 mm Normally, R = 175 to 200 mm T = 250 to 280 mm 39

40. ONE WAY SLAB DESIGN ( EXAMPLE ) STAIRCASE A one meter wide single flight R.C.C stair is to be provided for a height 3.20 m in a residential building. Stair is supported at top and bottom risers by beams 300 mm wide. Waist slab is 150 mm thick. Riser is 160 mm and tread is 300 mm Evaluate: Effective span Design load Reinforcement in waist slab. Prepared a sketch. Use M-20 and Fe-250 grade steel 40

41. 41 Effective span of staircase: ( IS: 456:2000, Pg-63, Cl 33.1 (a) ) R = 160 mm ( riser ) T = 300 mm ( tread ) No. of riser = = 3200/160 = 20 Nos Total number of treads = Number of riser – 1 = 20-1 = 19 Nos Horizontal length =Number of treads x length of tread = 19 x 300 = 5700 mm Effective span = centre to centre of beam = 5700 + 300/2 + 300/2 = 6000 mm Load calculations: Inclined length of one step = = 340 mm Area of step = ½ x 0.3 x 0.16 = 0.024 m 2 Area of inclined slab = 0.34 x 0.15 = 0.051 m 2  

42. 42 Total area = 0.075 m 2 D.L of step section for 1 m width and 300 mm plan length = 0.075 x 25 = 1.875 kN/m D.L per m 2 on plan area = 1.875 x = 6.25 kN /m 2 Assume L.L per m 2 on plan area = 3.0 kN /m 2 Total load = 6.25 + 3.0 = 9.25 kN /m 2 Factor load = 1.5 x 9.25 = 13.875 kN /m 2 Width of stair case = 1m Factor load = 13.875 x 1.0 = 13.875 kN /m Reinforcement ( SP-16, Pg-10, Table C ) M u = = = 62.43 kN.m M u = 0.148 F ck b d 2 62.43 x 10 6 = 0.148 x 20 x 1000 x d 2 d = 145.22 mm Provided d = 150 mm Assume 10 mm bars  

43. 43 D = 150 + 5 + 20 = 175 mm ( SP-16, Pg-48, Table -2 ) Pt = 1.593 % A st = = = 2389 mm 2 Consider 12 mm bars, a st = 113.09 mm 2 Number of bars = = = 21.12 22 Nos  

44. 44 A st ( Provided ) = 22 x 113.09 = 2487.98 mm 2 Provided 22 Nos- 12 mm bars Distribution steel: ( IS: 456:2000, Pg-48, Cl 26.5.2 ) Provide, minimum 0.15% steel of total cross section area A st = = 262.5 mm 2 For 8 mm bars, Spacing = = 191.46 mm Provide spacing = 190 mm 190 = A st = 265 mm 2 Provide, 8 mm - 190 mm c/c Provide 1 Number -10 mm bar as temperature reinforcement in each riser.  

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46. ONE WAY SLAB ( ANALYSIS ) A reinforced concrete slab of 125 mm thick is reinforced with 10 mm bars @ 150 mm c/c. the reinforced is located at an effective depth of 100 mm form top. Calculate the moment of resistance of the section and safe load. Use M-20 concrete and Fe415 steel. 46

47. 47 Assume 1 m wide slab M-20 Fe415 For 10 mm -150 mm c/c A st = = = 523.53 mm2 d = 100 mm Xu = = = 26.25 mm Xu max = 0.48 d = 0.48 x 100 = 48 mm Xu < Xu max  

48. 48 Under reinforced section M u = 0.87 F y A st d (1- ) = 0.87 x 415 x 523.53 x 100 (1- = 16.81 x N.mm M u = 16.81 x = W = 134.48 N/mm = 134.48 kN /m Safe U.D.L = 134.48/2 = 89.65 kN /m  

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