Sd i-module3- rajesh sir

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GCE Kannur

GCE Kannur

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  • 1. Dept. of CE, GCE Kannur Dr.RajeshKN Design of Two-way Slabs Dr. Rajesh K. N. Assistant Professor in Civil Engineering Govt. College of Engineering, Kannur Design of Concrete Structures
  • 2. Dept. of CE, GCE Kannur Dr.RajeshKN 2 (Analysis and design in Module II, III and IV should be based on Limit State Method. Reinforcement detailing shall conform to SP34) MODULE III (13 hours) Slabs : Continuous and two way rectangular slabs (wall-supported) with different support conditions, analysis using IS 456 moment coefficients, design for flexure and torsion, reinforcement detailing – Use of SP 16 charts. Staircases : Straight flight and dog-legged stairs – waist slab type and folded plate type, reinforcement detailing.
  • 3. Dept. of CE, GCE Kannur Dr.RajeshKN Two-Way Slabs • Initial proportioning of the slab thickness may be done by span/effective depth ratios • The effective span in the short span direction should be considered for this purpose • A value of kt ≈ 1.5 ( modification factor to max l/d ratio) may be considered for preliminary design.
  • 4. Dept. of CE, GCE Kannur Dr.RajeshKN 4 With mild steel (Fe 250), 0 8 35 0 8 40 for simply supported slabs for continuous slabs . . x x l D l ⎧ ⎪⎪ × ≥ ⎨ ⎪ ⎪ ×⎩ With Fe415 steel, 35 40 for simply supported slabs for continuous slabs x x l D l ⎧ ⎪⎪ ≥ ⎨ ⎪ ⎪⎩ •For two-way slabs with spans up to 3.5 m and live loads not exceeding 3.0 kN/m2, span to overall depth ratio can be taken as follows, for deflection control (Cl. 24.1, Note 2):
  • 5. Dept. of CE, GCE Kannur Dr.RajeshKN 5 • According to the Code (Cl. 24.4), two-way slabs may be designed by any acceptable theory, using the coefficients given in Annex D. • Code suggests design procedures (in the case of uniformly loaded two-way rectangular slabs) for: • simply supported slabs whose corners are not restrained from lifting up [Cl. D–2]. • ‘torsionally restrained’ slabs, whose corners are restrained from lifting up and whose edges may be continuous or discontinuous [Cl. D–1]. • The flexural reinforcements in the two directions are provided to resist the maximum bending moments Mux = αx wu lx 2 (in the short span) and Muy = αy wu lx 2 (in the long span).
  • 6. Dept. of CE, GCE Kannur Dr.RajeshKN 6 • The moment coefficients prescribed in the Code (Cl. D–2) to estimate the maximum moments (per unit width) in the short span and long span directions are based on the Rankine-Grashoff theory. • However, the moment coefficients recommended in the Code (Cl. D–1) are based on inelastic analysis (yield line analysis rather than elastic theory.
  • 7. Dept. of CE, GCE Kannur Dr.RajeshKN Nine different types of ‘restrained’ rectangular slab panels lx continuous (or fixed) edge simply supported edge ly
  • 8. Dept. of CE, GCE Kannur Dr.RajeshKN 8 Design a simply supported slab to cover a room with internal dimensions 4.0 m × 5.0 m and 230 mm thick brick walls all around. Assume a live load of 3 kN/m2 and a finish load of 1 kN/m2. Use M 20 concrete and Fe 415 steel. Assume that the slab corners are free to lift up. Assume mild exposure conditions. Effective short span ≈ 4150 mm Assume an effective depth d ≈ 4150 20 15× . = 138 mm With a clear cover of 20 mm and say, 10 φ bars, overall thickness of slab D ≈ 138 + 20 + 5 = 163 mm Provide D = 165 mm dx = 165 – 20 – 5 = 140 mm dy = 140– 10 = 130 mm Design Problem 1 1. Effective span and trial depths
  • 9. Dept. of CE, GCE Kannur Dr.RajeshKN 9 Effective spans ⎩ ⎨ ⎧ =+= =+= mm mm 51301305000 41401404000 y x l l 5130 4140 y x l r l ≡ = = 1.239 self weight @ 25 kN/m3 × 0.165m = 4.13 kN/m2 finishes (given) = 1.0 kN/m2 live loads (given) = 3.0 kN/m2 Total w = 8.13 kN/m2 Factored load wu = 8.13 × 1.5 = 12.20 kN/m2 2. Loads:
  • 10. Dept. of CE, GCE Kannur Dr.RajeshKN 10 3. Design Moments (for strips at midspan, 1 m wide in each direction) As the slab corners are torsionally unrestrained, Table 27 gives moment coefficients: αx = 0.0878 αy = 0.0571 short span: Mux = αx wulx 2 = 0.0878 × 12.20 × 4.1402 = 18.36 kNm/m long span: Muy = αy wulx 2 = 0.0571 × 12.20 × 4.1402 = 11.94 kNm/m Required spacing of 10 φ bars = 385 5.781000× = 204 mm 4. Design of Reinforcement 6 2 3 2 18 36 10 10 140 .ux x M bd × = × = 0.9367 MPa (Ast)x, reqd = (0.275 × 10–2) × 1000 × 140 = 385 mm2/m a. Shorter span [Table 3, Page 49, SP: 16]0 275,( ) .t x reqdp =
  • 11. Dept. of CE, GCE Kannur Dr.RajeshKN 11 (Ast)y, reqd = (0.204 × 10–2) × 1000 × 130 = 265.7 mm2/m Required spacing of 10 φ bars = 7.265 5.781000× = 295 mm 3 3 140 3 3 130 (short span) (long span)v d s d = ×⎧ ≤ ⎨ = ×⎩ Maximum spacing (Cl.26.3.3 b) 10 200 392 5 10 290 270 7 2 2 (short span) mm m (long span) mm m , , @ . @ . st x st y c c A c c A ϕ ϕ ⎧ ⇒ =⎪ ⎨ ⇒ =⎪⎩ Provide 6 2 3 2 11 94 10 10 130 .uy y M bd × = × = 0.7065 MPa b. Longer span
  • 12. Dept. of CE, GCE Kannur Dr.RajeshKN 12 5. Check for deflection control 3 392 5 100 10 140 , . t xp = × × = 0.280 fs = 0.58 × 415 × 385/392.5 = 236 MPa Modification factor kt = 1.5 (Fig. 3 of Code) (l/d)max = 20 × 1.5 = 30 (l/d)provided = 140 4140 = 29.6 < 30 — OK.
  • 13. Dept. of CE, GCE Kannur Dr.RajeshKN 13 6. Check for shear Average effective depth d = (140 + 130)/2 = 135 mm Vu = wu(0.5lxn – d) u v V bd τ = = 22.75 × 103/(1000 × 135) = 0.169 MPa For pt = 0.28 , k c vτ τ> — Hence, OK. = 0.376 MPaτ c where lxn is the clear span in the short span direction • The critical section for shear is to be considered d away from the face of the support. •An average effective depth d = (dx + dy)/2 may be considered in the calculations. = 12.20 (0.5 × 4.0 – 0.135) = 22.75 kN/m (Table 19, Page 73) 1 3.k = (Cl. 40.2.1.1)
  • 14. Dept. of CE, GCE Kannur Dr.RajeshKN 4000 230 8 φ bars 165 525 SECTION AA PLAN OF FLOOR SLAB A 165mm thick A 10 φ@ 200 c/c 10 φ@ 290 c/c 10 φ@ 290 c/c 10 φ @ 200 c/c 5000 230 230 230 525 425 7. Detailing
  • 15. Dept. of CE, GCE Kannur Dr.RajeshKN Repeat Design Problem 1, assuming that the slab corners are prevented from lifting up. Assume D = 165 mm dx = 165 – 20 – 5 = 140 mm, dy = 140 – 10 = 130 mm 4000 140 4140 5000 130 5130 mm mm x y l l = + =⎧ ⎨ = + =⎩ 1 24.y x l l = Factored load wu = 12.20 kN/m2 Design Problem 2 1. Effective span and trial depths 2. Loads
  • 16. Dept. of CE, GCE Kannur Dr.RajeshKN ( ) 1 240 1 2 0 072 0 079 0 072 1 3 1 2 . . . . – . . . − = + × − = 0.0748 Mux = αx wu lx 2 = 0.0748 × 12.20 × 4.142 = 15.61 kNm/m Short span: αx = 0.056 Mux = αy wu lx 2 = 0.056 × 12.20 × 4.142 = 11.69 kNm/m Long span: αy 3. Design Moments As the slab corners are to be designed as torsionally restrained, from Table 26 (Cl. D–1), the moment coefficients for ly/lx = 1.240 are:
  • 17. Dept. of CE, GCE Kannur Dr.RajeshKN 4. Design of reinforcement [Table 3, Page 49, SP: 16]0 2465,( ) .t x reqdp = 6 2 3 2 15 61 10 10 140 .ux x M bd × = × = 0.844 MPa (Ast)x, reqd = (0.246 × 10–2) × 1000 × 140 = 334 mm2/m Required spacing of 8 φ bars = 334 3.501000× = 150.7 mm Maximum spacing permitted = 3 × 140 = 420 mm, but < 300 mm. a. Shorter span
  • 18. Dept. of CE, GCE Kannur Dr.RajeshKN [Table 3, Page 49, SP: 16]0 206,( ) .t x reqdp = 6 2 3 2 11 69 10 10 130 .uy y M bd × = × = 0.714 MPa (Ast)x, reqd = (0.206 × 10–2) × 1000 × 130 = 264 mm2/m Required spacing of 8 φ bars = 1000 50 3 264 .× = 191 mm Maximum spacing permitted = 3 × 130 = 375 mm, but < 300 mm. b. Longer span ⎩ ⎨ ⎧ span)(long span)(short cc cc 190@8 150@8 φ φ Provide
  • 19. Dept. of CE, GCE Kannur Dr.RajeshKN 5. Check for deflection control 0 2465, .t xp = fs = 0.58 × 415 × 334/335 = 240 MPa Modification factor kt = 1.55 (Fig. 3 of Code) (l/d)max = 20 × 1.55 = 31 (l/d)provided = 4140 140 = 30.4 < 31 — Hence, OK.
  • 20. Dept. of CE, GCE Kannur Dr.RajeshKN 6. Corner Reinforcement [as per Cl. D–1.8] As the slab is ‘torsionally restrained’ at the corners, corner reinforcement has to be provided at top and bottom (four layers), • over a distance lx/5 = 830 mm in both directions • each layer comprising 0.75 Ast, x. spacing of 8 φ bars Provide 8 φ @ 160 c/c both ways at top and bottom at each corner over an area 830 mm × 830 mm. ( )2 830 8 4 0 75 334. π× × × 160 c/c≅
  • 21. Dept. of CE, GCE Kannur Dr.RajeshKN PLAN 830 830 525 425 5000 AA B B 230 230 4000 230 230 5 nos 8 φbars (U–shaped) both ways (typ) at each corner 8 φ @ 150 c/c 8 φ @ 190 c /c
  • 22. Dept. of CE, GCE Kannur Dr.RajeshKN 830 5 nos 8φ U–shaped bars 160 SECTION BB 525 8 φ@ 190 c/c 8 φ@ 150 c/c 160 SECTION AA
  • 23. Dept. of CE, GCE Kannur Dr.RajeshKN Summary Slabs : Continuous and two way rectangular slabs (wall-supported) with different support conditions, analysis using IS 456 moment coefficients, design for flexure and torsion, reinforcement detailing – Use of SP 16 charts. Staircases : Straight flight and dog-legged stairs – waist slab type and folded plate type, reinforcement detailing.