0 ne way slab

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one way slab design

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  • This template can be used as a starter file for presenting training materials in a group setting.SectionsRight-click on a slide to add sections. Sections can help to organize your slides or facilitate collaboration between multiple authors.NotesUse the Notes section for delivery notes or to provide additional details for the audience. View these notes in Presentation View during your presentation. Keep in mind the font size (important for accessibility, visibility, videotaping, and online production)Coordinated colors Pay particular attention to the graphs, charts, and text boxes.Consider that attendees will print in black and white or grayscale. Run a test print to make sure your colors work when printed in pure black and white and grayscale.Graphics, tables, and graphsKeep it simple: If possible, use consistent, non-distracting styles and colors.Label all graphs and tables.
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  • 0 ne way slab

    1. 1. Presentation prepared as per: “IS 456 : 2000” Presented by: Ravi Shankar Singh & Pankaj Kumar
    2. 2. CONTENTS • Introduction • Important terms • General considerations of design for slabs • Examples
    3. 3. Introduction • Slabs are plane structural members which are used as floor and roof of buildings. • The thickness of the slabs is quite small as compare to its other dimensions.
    4. 4. One way slab:- • One way slab is edge supported slab which spanning in one direction. • The ratio of long span to short span is greater than or equal to 2. • The main steel is provided in the direction of the shorter span • Slab supporting on four sides also behave as one way slab if ratio of long span to short span is greater than or equal to 2.
    5. 5. • The distribution steel should be tied above the main steel, other wise the lever arm which is measure upto the center of the main steel shall be reduced resulting in the reduction of moment of resistance. Purpose of main steel: • It take up all tensile stresses developed in the structure. • It increase the strength of concrete sections. Purpose of distribution steel: • It distribute the concentrated load on the slab. • It guard against shrinkage and temperature stress. • It also keep the main reinforcement in position.
    6. 6. 1. EFFECTIVE SPAN (A) FOR SIMPLY SUPPORTED SLAB : DISTANCE BETWEEN CENTRE TO CENTRE OF SUPPORT OR THE CLEAR DISTANCE BETWEEN SUPPORT PLUS THE EFFECTIVE DEPTH OF THE SLAB WHICHEVER IS SMALLER . (B) FOR CONTINUOUS SLAB : WIDTH OF SUPPORT < 1/12 OF CLEAR SPAN ,THE EFFECTIVE SPAN SHALL WORK OUT AS ABOVE. WHEN WIDTH GREATER THAN 1/12 OF CLEAR SPAN OR 600MM WHICHEVER IS SMALL SHALL BE TAKEN (1) FOR ONE END FIX AND OTHER END CONTINUOUS THE EFFECTIVE SPAN IS CLEAR SPAN. (2) FOR ONE END FREE AND OTHER END CONTINUOUS ,THE EFFECTIVE SPAN SHALL BE EQUAL TO CLEAR SPAN PLUS HALF THE EFFECTIVE DEPTH OR CLEAR SPAN PLUS HALF THE WIDTH OF SUPPORT ,WHICHEVER IS SMALLER . GENERAL CONSIDERATION OF DESIGN FOR SLAB
    7. 7. (2)For end span with one end free and other end continuous ,the effective span shall be equal to clear span plus half of the effective depth of the slab , or clear span plus half the with of center to center of support or the clear distance between the support plus the effective depth of the slab whichever is smaller. For continuous slab: In case of a continuous slab ,where the width of the support is less than 1/12 of the clear span ,the effective span shall be worked as above. In case the support are wider than 1/12 of the clear span , the effective span shall be taken as under: (1)For end span with one end fixed and other end continuous ,the effective span shall be clear span. (2)For end span with one end free and other end continuous ,the effective span shall be equal to clear span plus half of the effective depth of the slab , or clear span plus half the with of the discontinuous slab, which ever is less. 2. DEFLECION CONTROL The span to depth ratio are not greater than the value below
    8. 8. A) For span upto 10m cantilever slab 7 Simply supported 20 Continuous 26 b) For span above 10m the value in para (a) may be multiply by 10/span in meters 3.Reinforcement in slab (A)Minimum reinforcement. The area of reinforcement in either direction should not less than: 0.12% of the total cross-sectional area (for HYSD bar) 0.15% of the total cross-sectional area (for Fe 250)
    9. 9. • (B) Maximum diameter : The maximum diameter of reinforcing bar in a slab should not exceed 1/8 of thickness of slab. • (C) Cover to reinforcement :The cover of concrete to reinforcement in different structural members are following in table Exposure minimum grade of concrete Nominal cover in mm not less than Mild M20 20 Moderate M25 30 Severe M30 45 Very severe M35 50 Extreme M40 75
    10. 10. (D) Spacing of reinforcement: a. Minimum distance between individual bars: (i) Minimum distance shall not less than diameter of bar or 5mm more than the nominal max. size of the bar whichever is greatest. (ii)When we provide two or more layer of main bars, the min. vertical clear distance between any two layer of bar shall be 15mm or 2/3 the nominal max. size of course aggregate or max. size of bar whichever is greatest. (b) Maximum distance between bars in tension : (i)The pitch of the main tensile bars in slab should not exceed three times the effective depth or 300mm whichever is smaller.
    11. 11. (ii) The pitch of distribution bars shall not exceed five time the effective depth or 450 whichever is smaller. (E) Curtailment of main reinforcement. The main reinforcement in slab may be curtailed or bent up ,beyond the point at which it is no longer requirement to resist bending
    12. 12. The following equation can be used to find out the point at which the bar can be curtailed or bend-up x=l/2[Nx/Nc] ˄ ½ X=Distance from either side from centre of span. Nc=Total no. of bar in tension Nx=No. of bars which can be curtailed at any distance x from the center of the span 4.Load on slabs. The load on slab consists of: (a) Live load (b)External dead load (c) Dead load of slab itself
    13. 13. EXAMPLE Design a simply supported R.C.C slab to carry a factored load of 15000N/m2 inclusive of its own weight on an effective span of 3.1 m. solution:- Assume 1m width of slab. b= 1m = 1000mm fck = 20N/mm2 fy =415N/mm2 Effective span l=3.1m Design factored load Wu=1500N/m2 =15000*1=15000 Depth of slab . Assume effective depth d=span/25
    14. 14. = 3100/25= 124mm or say 130mm Overall depth D =130+30 =160mm Ultimate max. bending moment Mu =wl˄2/8 =(15000x3.1˄2)÷ 8 =18019 N-mm =18019x10˄3Nmm Check for depth The slab will be designed as a balance section Mu =Mlim. 18019x10˄3 =0.138fckbd˄2 18019x10˄3 =0.138x20x1000xd˄2 d= [(18019x10˄3)÷(0.138x20x1000)] = 81mm<130mm Adopt d=130mm and D=160mm
    15. 15. Reinforcement (main steel) Mu =0.87fyAst [d-fy Ast ] fck b 18019x10˄3 = 0.87x415xAst [130-415 x Ast ] 20 x 1000 49007 = 130Ast -0.021Ast˄2 Ast˄2- 6190Ast +2376524 =0 Ast =411mm˄2 Minimum area of steel for main bars =0.12bD÷100 =(0.12x1000x160)÷100 =192mm˄2 The area of steel provided is more than minimum requirement.Hence ok
    16. 16. Provide 10mm ɸ bars for main reinforcement Area of one 10 mm ɸ bar = π÷4x10˄2 = 78.5mm˄2 c/c spacing of 10mm ɸ bars =Area of one bar x b Total area of steel = 78.5 x 1000 =190MM 411 It should be less than: (i) 3d = 3x130 = 390 mm (ii) 300 mm Provide main steel of 10mm ɸ @190 mm c/c Clear spacing between bars = 190-10 = 180mm
    17. 17. It should be more than: (i) Diameter of bar =10mm (ii) Size of aggregate + 5 = 15+5 = 20 Provide 10 mm ɸ bar @190 mm c/c Alternate bars are bent up near the supports .These bars may be bent up at 0.10 l= 310 mm from face of support . Distribution steel Area of distribution steel = 0.12bD/100 =(0.12x1000x160)/100 Provide 8mm ɸ bars as distribution steel Area of one 8mm ɸ bars =π/4 x 8˄2 = 50.26 mm˄2
    18. 18. c/c spacing of 8mm ɸ bars = (50.26x1000)/192 =262mm say 260mm It should less than : (i) 5d =5x130 = 650mm (ii) 450mm Provide distribution steel of 8mm @260 mm c/c Check for shear The critical section for shear is at a distance d i.e. 130mm from inner face of support or 0.28 m [130 +300÷2=280 mm = 0.28m] from centre of support. [Assume width of support =300] Factored shear force at critical section V = wl/2 –wx =15000x3.1)/2 – 15000x0.28 =23250-4200 = 19050 N Nominal shear stress τ = V/bd =19050/1000x130= 0.15 N/mm˄2 < 2.8N/mm˄2 [For M20 concrete τc max =2.8 N/mm˄2 P = 100As/bd = (100x206)/(1000x130 ) = 0.16% [{ As =1/2(78.5x1000÷190) = 206 mm˄2},Alternate bars are bent–up ] For P = 0.16% and M 20 concrete from table τc = 0.29 N/mm˄2 For solid slab from table k= 1.28 (for overall depth of slab = 160mm) τc < τv hence no reinforcement provided Check for development length Ld = (0.87fyɸ)/4τbd = (0.87x415x10)/4x1.92 = 470 mm [For M20 concrete and Fe 415 steel, τbd = 1.6x1.2 = 1.92N/mm˄2]
    19. 19. As per code at the simply supported the bar must extend beyond the face of support by a distance not less than Ld/3 = 470/3 = 157 mm Length available from the face of support = support width – side cover =300-20 = 280 mm Embedment length from center of support Lo = 300/2-20 = 130mm Since the slab is simply supported ,the compressive reaction will confine the reinforcement . Hence Ld ≤ 1.3M/V +Lo M = moment of resistance provided by 10 mm ɸ bars @380mm c/c
    20. 20. = 0.87fyAst[d- fy Ast ] fck b =0.87x415x206x[130 – 415x206] = 9351x10˄3Nmm 20x1000 [Ast =(78.5x1000/380) = 206 mm˄2] V= Factored shear force at the centre of support = wl/2 =15000x3.1/2 = 23250 N 1.3M/V + Lo = 1.3x9351x10˄3/23250 +130 =653mm >Ld Hence development length requirement are satisfied.
    21. 21. • Check for deflection Pt = 100Ast/bd =(100x413)/(1000x130) = 0.32% (Ast provided = 78.5x1000/190 = 413mm˄2) Fs = 0.58fy [Area of steel required] Area of steel provided = 0.58x415x[411/413] = 240N/mm˄2 Modification factors Kt = 1.5 Kc= 1.0 (No compression reinforcement provided) Kf = 1.0 (slab section is rectangular) (l/d)max = 20 x Kt x Kc x Kf = 20x1.5x1x1= 30 (l/d)provded = 3100/130 = 23.8<30 The slab will satisfied limit state of serviceability.

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