Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Like this presentation? Why not share!

916 views

Published on

GCE Kannur

Published in:
Engineering

No Downloads

Total views

916

On SlideShare

0

From Embeds

0

Number of Embeds

1

Shares

0

Downloads

146

Comments

0

Likes

2

No embeds

No notes for slide

- 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.

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

Be the first to comment