Civil structural engineering - Flat slab design

1. Flat Slab Design

2. Resources used for compiling this presenta4on are acknowledged

3. Flat Slab with drop panels Flat slab with column head Flat slab with drop panel and column head Flat Slab resting directly on columns 1.  What is a flat slab? 31.1 General The term flat slab means a reinforced concrete slab with or without drops, supported generally without beams, by columns with or without flared column heads A flat slab may be solid slab or may have recesses formed on the soffit so that the soffit comprises a series of ribs in two directions. The recesses may be formed by removable or permanent filler blocks.

4. 2.  Types of flat slab •  Flat Slab with drop panels •  Flat slab with column head •  Flat slab with drop panel and column head •  Flat Slab resting directly on columns Drop is a local thickening of the slab in the region of column Structural Advantages •  increase shear strength of slab •  increase nega5ve moment capacity of slab •  s5ﬀen the slab and hence reduce deﬂec5on Column head is a local enlargement of the column at the junc5on with the slab Structural Advantages •  increase shear strength of slab (punching shear) •  reduce the moment in the slab by reducing the clear or effective span

5. A flat slab may have recesses formed on the soffit so that the soffit comprises a series of ribs in two direc5ons ( waffle Slabs).

6. Flat slabs with capitals, drop panels, or both. These slabs are very sa4sfactory for heavy loads and long spans. Although the formwork is more expensive than for flat plates, flat slabs will require less concrete and reinforcing than would be required for flat plates with the same loads and spans. They are par4cularly economical for warehouses, parking and industrial buildings, and similar structures, where exposed drop panels or capitals are acceptable.

7. v Flexibility in room layout •  Introduce partition walls anywhere required •  Change the size of room layout •  Omit false ceiling v  Saving in building height •  Lower storey height will reduce building weight •  approx. saves 10% in vertical members •  reduce foundation load v Shorter construction time •  flat plate design will facilitate the use of big table formwork to increase productivity v Ease of installation of M&E services •  all M & E services can be mounted directly on the underside of the slab instead of bending them to avoid the beams •  avoids hacking through beams 3.  Benefits of flat slab

8. The main disadvantage is their lack of resistance to lateral loads due to wind and earthquakes. Lateral load resis4ng systems such as shear walls are oDen necessary When the loads or spans or both become quite large, the slab thickness and column sizes required for ﬂat plates or ﬂat slabs are of such magnitude that it is more economical to use two-‐way slabs with beams, despite the higher formwork costs.

9. 4.  Behaviour of Slab supported on Stiff , Flexible and no beams Case Study: •  Panel Size = 4 m x 4m •  Slab Thickness = 125 mm •  Load = 5 kN/m2 •  S5ﬀ Supports ( Bearing wall) •  Flexible Supports (Beam) : 300 x 300 , 300 x 450 , 300 x 600 , 300 x 1000 mm •  Column supports at corners

10. A. Two way Slab on Rigid Supports (bearing Walls) Mx = 3.616 kNm/m My = 3.616 kNm/m IS 456 Values (Table 27): 0.062 x 5 x 16 = 4.96 Slab Deﬂec6on = 1.4 mm

11. B. Two way Slab on Flexible Supports (Beams on all sides) 1. Beam Size : 300 x300 mm Mx = 4.45 kNm/m My = 4.45 kNm/m IS 456 Values (Type 9): 0.056 x 5 x 16 = 4.48 Mxy = 0.37 kNm/m Beam Moment = 12.2 kNm Beam Deﬂec6on = 1.33 mm Slab deﬂec6on= 2.9 mm

12. 2. Beam Size : 300 x450 mm Mx = 3 kNm/m My = 3 kNm/m IS 456 Values (Type 9): 0.056 x 5 x 16 = 4.48 Mxy = 0.73 kNm/m Beam Moment = 15.6 kNm Beam Deﬂec6on = 0.5 mm Slab deﬂec6on= 1.5 mm

13. 3. Beam Size : 300 x 600 mm Mx = 2.43 kNm/m My = 2.43 kNm/m IS 456 Values (Type 9): 0.056 x 5 x 16 = 4.48 Mxy = 0.8 kNm/m Beam Moment = 17 kNm Beam Deﬂec5on = 0.24 mm Slab Deﬂec5on = 0.98 mm

14. 4. Beam Size : 300 x 1000 mm Mx = 2 kNm/m My = 2 kNm/m IS 456 Values (Type 9): 0.056 x 5 x 16 = 4.48 Mxy = 0.8 kNm/m Beam Moment = 18 kNm

15. 5. Beam Size : 300 x 125 mm (Concealed Beams) Mx = 9.8 kNm/m My = 9.8 kNm/m IS 456 Values (Type 9): 0.056 x 5 x 16 = 4.48 Mxy = 3 kNm/m Beam Moment = 2.9 kNm Slab Deﬂec6on = 7.0 mm

16. B. Two way Slab on Point Supports at corners (Flat Slab) Mx = 9.075 kNm/m (Middle) =12.4 kNm/m (Edge Strip) Mxy = 7.76 kNm/m My = 9.075 kNm/m (Middle) =12.4 kNm/m (Edge Strip) Slab Deﬂec6on = 8.67 mm

17. Type of Support Mx My Mxy Beam Moment Deﬂec4on Slab Beam Rigid 3.616 3.616 2.6 -‐ 1.4 -‐ 300 x125 (Concealed Beams) 9.84 9.85 3.0 2.88 7.0 4.3 300 x300 4.45 4.45 0.37 12.2 2.9 1.33 300 x450 3 3 0.73 15.6 1.5 0.50 300 x600 2.43 2.43 0.8 17.0 0.98 0.24 300x1000 2 2 0.8 18.0 0.60 0.05 Flat Slab 9.0 9.0 7.76 -‐ 8.676 -‐ Results Summary

18. •  Two way Rectangular Slab supported on stiff beams, the shorter spans (stiffer portion of the slab) carry larger load and subjected to larger moments. The longer spans carry less load and subjected to less moment. •  Results indicate that decrease in supporting beams stiffness leads to an increase in bending moments of slabs and decrease in bending moment of the beams (behavior that is not captured using code recommendations). •  If the slab is supported on bearing walls, slab moments are distributed in similar way. •  If the slab is supported only by the columns, the slab behaves like a two way slab with an essential difference that all the load is carried in both directions to accumulate it at the columns. •  With Concealed beams it is reveled that the behaviour is close to Flat slabs rather than any useful beam action. Observa4ons

19. 4.  Structural Behaviour of Flat Slab Deﬂected Shape Column Strip Column Strip Middle Strip Column Strip Middle Strip

20. A Zone of –ve BM (Hogging) in both direc7ons B Zone of +ve BM(Sagging) and –ve BM C Zone of -‐ve BM and +ve BM D Zone of +ve BM in both direc7ons -‐m4 -‐m2 -‐m4 m3 m1 m3 m5 m7 -‐m8 -‐m4 -‐m2 -‐m4 m7 A A A A C C B B D Column Strip Middle Strip Column Strip Column Strip Middle Strip Column Strip -‐m6 -‐m6 -‐m8 -‐m8 -‐m8 5.  Distribution of Total Panel Moment in different zones

21. A Zone of –ve BM (Hogging) in both direc7ons B Zone of +ve BM(Sagging) and –ve BM C Zone of -‐ve BM and +ve BM D Zone of +ve BM in both direc7ons -‐m4 -‐m2 -‐m4 m3 m1 m3 m5 m7 -‐m8 -‐m4 -‐m2 -‐m4 m7 A A A A C C B B D Column Strip Middle Strip Column Strip Column Strip Middle Strip Column Strip -‐m6 -‐m6 -‐m8 -‐m8 -‐m8 m1 -‐m2 -‐m2 D C C m3 -‐m4 -‐m4 B A A m5 -‐m6 -‐m6 D B B m7 -‐m8 -‐m8 C A A

22. 6.  Definitions L2 L1 Moment Direc5on MIDDLE STRIP COLUMN STRIP 0.25L2 ≤ 0.25L1 COLUMN STRIP 0.25L1 ≤ 0.25L2 MIDDLE STRIP COLUMN STRIP 0.25L1 ≤ 0.25L2 Moment Direc5on SPAN Region SPAN Region: Bounded on all the four sides by middle strips

23. 7.  General Design Considerations CL 31.2 Proportioning 31.2.1 Thickness of Flat Slab •  The thickness of the flat slab shall be generally controlled by considerations of span to effective depth ratios given in 23.2. •  For slabs with drops conforming to 31.2.2, span to effective depth ratios given in 23.2 shall be applied directly; otherwise the span to effective depth ratios obtained in accordance with provisions in 23.2 shall be multiplied by 0.9. For this purpose, the longer span shall be considered. •  The minimum thickness of slab shall be 125 mm. 31.2.2 Drop •  The drops when provided shall be rectangular in plan, and have a length in each direction not less than one- third of the panel length 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 the width of drop for interior panels. •  Minimum thickness of Drop > ¼ of Slab thickness and > 100 mm

24. 31.2.3 Column Heads Where column heads are provided, that portion of a column head which lies within the largest right circular cone or pyramid that has a vertex angle of 900and can be included entirely within the outlines of the column and the column head, shall be considered for design purposes.

25. 8. Determination of Bending Moment CL 31.3 31.3.1. Methods of Analysis and Design It shall be permissible to design the slab system by one of the following methods: a) The direct design method as specified in 31.4, and b) The equivalent frame method as specified in 31.5. In each case the applicable limitations given in 31.4 and 31.5 shall be met.

26. 9. Direct Design Method CL 31.4 A. Limitations : 31.4.1 Slab system designed by the direct design method shall fulfil the following conditions: a)  There shall be minimum of three continuous spans in each direction, 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 2.0 c)  It shall be permissible to offset columns to a maximum of 10percent of the span in the direction of the offset notwithstanding the provision in (b) d)  The successive span lengths in each direction shall not differ by more than one-third of the longer span. The end spans may be shorter but not longer than the interior spans, and e)  The design live load shall not exceed three times the design dead load. Note: Applicable to gravity loading condition alone (and not to the lateral loading condition)

27. 1 2 3 2 3 Lx1 Lx2 Lx3 Ly1 Ly2 Ly3 ≤ 0.1Ly2 ≤ 0.1Lx1 Lx1 ≤ Lx2 Lx3 ≤ Lx2 Ly1 ≤ Ly2 Ly3 ≤ Ly2 Lx1 ≥ 2Lx2/3 Lx3 ≥ 2Lx2/3 Ly1 ≥ 2Ly2/3 Ly3 ≥ 2Ly2/3 wuL/wuD ≤ 3 For any Panel Longer Span/Shorter Span≤ 2

28. B. Total Design Moment for a Span: CL31.4.2 CL of Panel 1 CL of Panel 2 1 2 DESIGN STRIP 31.4.2.1 In the direct design method, the total design moment for a span shall be determined for a strip bounded laterally by the centre-line of the panel on each side of the centre-line of the supports. 31.4.2.2 The absolute sum of the positive and average negative bending moments in each direction shall be taken as: 1 2 M0x M0y lnx

29. wu kN/m L1 L2 Ln Ln (L1) (L2) Note: 1.  It is the same as the total moment that occurs in a simply supported slab 2.  The moment that actually occurs in such a slab has been shown by experience and tests to be somewhat less than the value determined by the Mo expression. For this reason, l1 is replaced with ln

30. •  It is next necessary to know what propor4ons of these total moments are posi4ve and what propor4ons are nega4ve. 10. Distribution of Total Panel Moment M0 •  If a slab was completely ﬁxed at the end of each panel, the division would be as it is in a ﬁxed-‐end beam, two-‐thirds nega4ve and one-‐third posi4ve, as shown in Figure. •  This division is reasonably accurate for interior panels where the slab is con4nuous for several spans in each direc4on with equal span lengths and loads. Interior Panel

31. •  The rela4ve s4ffnesses of the columns and slabs of exterior panels are of far greater significance in their effect on the moments than is the case for interior panels. •  The magnitudes of the moments are very sensi4ve to the amount of torsional restraint supplied at the discon4nuous edges. •  This restraint is provided both by the flexural s4ffness of the slab and by the flexural s4ffness of the exterior column. Exterior Panel

32. Code Recommendations

33. Distribution of Bending Moments across panel width Code Recommendations

34. 11. Rebar Detailing - Code Recommendations

35. Bent bars are also used. There seems to be a trend among designers to use straight bars more than bent bars. ELEVATION

36. Rebar Detailing - Code Recommendations e e e e e b b b b b Ln greater of adjacent clear spans CL 31.7.3 (b)

37. Sec6on through Middle Strip

38. 12. Two way Shear in Flat Slab •  Flat plates present a possible problem in transferring the shear at the perimeter of the columns. •  There is a danger that the columns may punch through the slabs. •  As a result, it is frequently necessary to increase column sizes or slab thicknesses or to use shear heads. Shear heads consist of steel I or channel shapes placed in the slab over the columns

39. Note: Flat Slab with drop panel and capital, shear is required to be checked at two sec4ons 1.  at a distance d/2 from the face of column capital 2.  at a distance d/2 from the face of drop panel

40. Design Example #1 Design by DDM ﬂat plate supported on columns 450 mm square, for a Live Load = 3 kN/m2, Floor Finish = 1 kN/m2 use M20 and Fe415. Assume clear cover = 20 mm. Eﬀec6ve Column Height = 3.35m. Bay spacing in X and Y direc6on = 5m c/c •  Interior Panel P5 •  Corner Panel P7 3 bays @ 5 m c/c

41. A. Interior Panel Design 5 m 5 m 2.5m 2.5 m A A A A B B C CD Zone A – Corner Strip Zone B – Middle Strip along X Zone C – Middle Strip along Y Zone D – Interior Region Step 1: Panel Division into Strips 31.1.1(a) Moment direc6on Along L1 L2 Width of Column Strip on either side of Centre Line = 0.25L2 and ≤ 0.25 L1 Middle Strip X 5 5 1.25 and ≤1.25 m Adopt 1.25 m 2.5m Y 5 5 1.25 and ≤1.25 m Adopt 1.25 m 2.5m

42. Step 2: Trial Depth CL 31.2.1 •  L/d = 26 •  Modiﬁca5on Factor = 1.33, Assuming pt = 0.4%, FIG 4 IS 456 •  d = 5000/(26 x 1.33) = 145 mm > 125 CL 31.2.1 •  DS= 145 + 20 + 18 = 183 mm ( assume #12 bars, and bars in two layers) •  Provide Ds= 200 mm d = 200-‐20-‐18 = 162 mm

43. Step 3 Design Loads / m width of Slab •  wuD = 1.5(25x 0.2 + 1) = 9kN/m •  wuL = 1.5 x 3 = 4.5kN/m •  wu = 13.5 kN/m Step 4: Check for Applicability of DDM: CL 31.4.1 •  No. of Con5nuous Spans in each direc5on = 3 ; OK 31.4.1(a) •  Long Span/Short Span = 5/5 = 1 <2 ; OK 31.4.1(b) •  Successive spans in each direc5on = Equal; OK 31.4.1(d) •  wuL/wuD = 4.5/9 = 0.5 < 3 ; OK 31.4.1(e)

44. Step 5: Check for punching shear around Column Assumed d = 162 mm Sec4on 1: •  Cri5cal Sec5on at d/2 around the column •  Perimeter of Cri5cal Sec5on = 4 x 0.612= 2.448 m •  Design Shear at cri5cal sec5on Vu •  Vu = 13.5 ( 52 – 0.6122) = 333kN •  τc = 0.25√fck = 1.12 MPa •  ks = 0.5 +1 = 1.5 <=1 ; ks=1 ; ks τc = 1.12 •  Shear Resistance of Concrete = 1.12 x 2448 x 162 = 444kN > 333 kN OK Cri4cal Sec4on 0.612m 0.612m 5m 5m Contributory Area

45. Step 6:Design Moments CL 31.4.2.2 Parameters Along X Along Y L1 (Span in direc4on of Mo) 5 5 m 0.65L1 3.25 3.25 m Ln (clear span extending from face to face of columns, capitals) (5-‐0.45) = 4.55 4.55 m Ln > 0.65L1 4.55 4.55 m L2 (Span transverse to L1) 5 5 m W = wu L2Ln 307.2 307.2 kN M0 = W Ln / 8 174.72 174.72 kNm wu = 13.5 kN /m

46. Step 7 : Distribution of Bending Moment across panel width ; CL: 31.4.3.2, 31.5.5 Moment Direc5on along X Y Nega4ve Design Moment MN = -‐ 0.65*M0 113.6 113.6 kNm 31.4.3.2 •  Column Strip M1 = 0.75MN 85.2 85.2 kNm 31.5.5.1 Width of Column Strip resis4ng M1 (Csw) 2x1.25 =2.5 2x1.25 =2.5 m •  -‐m1 = M1/ Csw (Zone A) 34.1 34.1 kNm/m •  Middle Strip M2 = 0.25MN 28.4 28.4 kNm 31.5.5.4(a) Width of Middle Strip resis4ng M2 (Msw) 2.5 2.5 m •  -‐m2 = M2/Msw (Zone B & C) 11.4 11.4 kNm/m

47. Posi5ve Design Moment MP = 0.35*M0 61.2 61.2 kNm 31.4.3.2 •  Column Strip M1 = 0.6MP 36.7 36.7 kNm 31.5.5.3 •  +m1 = M1/ Csw (Zone B &C) 14.7 14.7 kNm/m •  Mid Span M2 = 0.4MP 24.5 24.5 kNm 31.5.5.4(a) •  +m2 = M2/Msw (Zone D) 9.8 9.8 kNm/m -‐ve sign : Hogging Moment (tension at top) +ve sign : Sagging Moment (tension at borom) -‐34.1 -‐11.4 -‐34.1 14.7 9.8 14.7 -‐34.1 -‐11.4 -‐34.1 9.8 14.7 -‐11.4 -‐34.1 -‐34.1 -‐34.1 -‐11.4 -‐34.1 14.7 A B A D C C A AB Step 8 : Check for adequacy of Depth •  Max Design Bending moment = 34.1 kNm/m •  Mu,lim = 72.41 kNm/m > 34.1, •  Depth is adequate G-‐1.1(c)

48. Loca6on Moment (kNm/m) Ast (mm2 /m) Ast (prov) Bar dia Spacing mm Along X Zone A (-‐) 34.1 635 635 10 120 -‐ T Zone B 14.7 260 260 8 190 -‐ B Zone C (-‐)11.4 200 240 8 200 -‐ T Zone D 9.8 171 240 8 200 -‐ B Along Y Zone A (-‐)34.1 635 635 10 120 -‐ T Zone B (-‐)11.4 200 240 8 200 -‐ T Zone C 14.7 260 260 8 190 -‐ B Zone D 9.8 171 240 8 200 -‐ B •  7.5 Ast2 – 58490Ast + Mu = 0 G-‐1.1(b) Step 9 :Rebar Details •  Ast,min = 0.12 x 200 x 1000 /100 = 240 mm2/m 26.5.2.1 •  Minimum Eﬀec5ve Depth of Slab = 162 mm

49. #8@190 #8@200 #8@190 #8@200 #8@190 #8@190 0.15Ln 0.15Ln 0.125Ln 0.125Ln Borom Rebar Details in Interior Panel A B A A B A C D C

50. TOP Rebar Details in Interior Panel #10@120 #10@120 #8@200 #8@200 #8@200 0.3Ln 0.2Ln 0.3Ln 0.2Ln Note: Distances for curtailment of rebars are measured from column face A B A C D C

51. B. Corner Panel Design Step 5: Check for punching shear around Column Assumed d = 162 mm Sec4on 1: •  Cri5cal Sec5on at d/2 around the column •  Perimeter of Cri5cal Sec5on = 2 x 0.531= 1.062 m •  Design Shear at cri5cal sec5on Vu •  Vu = 13.5 ( 2.52 – 0.5312) = 81kN •  τc = 0.25√fck = 1.12 MPa •  ks = 0.5 +1 = 1.5 <=1 ; ks=1 ; ks τc = 1.12 •  Shear Resistance of Concrete = 1.12 x 1062 x 162 = 192kN > 81 kN OK 450 162/2 = 81 mm

52. Step 7 : Distribution of Bending Moment across panel width ; CL 31.4.3.3 , 31.5.5 𝛼↓𝑐 =∑↑▒ 𝑘↓𝑐  / 𝑘↓𝑠   Assume Columns and Slab panels are with same modulus of elas5city 5 m 5 m 1.25m 1.25m A A A A B B C CD Step 6:Design Moments CL 31.4.2.2 M0 = 174.72 kNm

53. Parameters Along X Along Y Sum of column s4ﬀness above and below the slab 2 (4EcIc)/Lc (2 x 4 x Ec x 450 x 4503/12) /3350 = 8.16 Ec x 106 Slab s4ﬀness ks = 4EsIs/Ls (4 Es x 5000 x 2003/12)/5000 = 2.67Es x 106 2.67Es x 106 αc = ∑kc /ks 3.06 3.06 β = 1+ (1/αc) 1.33 1.33

54. Moment Direc5on along X Y Nega4ve Design Moment MN = -‐ 0.65*M0/β 85.4 85.4 kNm 31.4.3.3 •  Column Strip M1 = MN 85.4 85.4 kNm 31.55.2(a) Width of Column Strip Csw resis4ng M1 2x1.25 = 2.5 2.5 m -‐m1 = M1/Csw 34.2 34.2 kNm/m •  Middle Strip M2=0 0 0 kNm 31.5.5.4(a) -‐m2 = 0 0 0 kNm/m A. Exterior nega4ve design moment: -‐m1 -‐m1 -‐m1 -‐m1 Exterior Exterior Interior Interior 1.25 1.25 -‐m2 -‐m2 X Y

55. Moment Direc5on along X Y Nega4ve Design Moment MN = -‐ (0.75 – 0.1/β)Mo 118 118 kNm 31.4.3.3 •  Column Strip M1 = 0.75 MN 88.5 88.5 kNm 31.5.5.1 Width of Column Strip Csw resis4ng M1 2x1.25 =2.5 2.5 m •  -‐m1 = M1/ Csw -‐35.4 -‐35.4 kNm/m •  Middle Strip M2 = 0.25 MN 22.12 22.12 kNm 31.5.5.4(a) Width of Middle Strip Msw resis4ng M2 2.5 2.5 m •  -‐m2 = M2/Msw -‐8.85 -‐8.85 kNm/m B. Interior nega4ve design moment: -‐ m1 -‐ m1 -‐ m2 m1 -‐m1 -‐m1 -‐m2 X Y Exterior Exterior Interior Interior

56. Moment Direc5on along X Y Design Moment MP = (0.63 – 0.28/β)Mo 73.29 73.29 kNm 31.4.3.3 •  Column Strip M1 = 0.6 MP 43.98 43.98 kNm 31.5.5.3 Width of Column Strip Csw resis4ng M1 2x1.25 =2.5 2.5 m •  m1 = M1/ Csw 17.6 17.6 kNm/m •  Middle Strip M2 = 0.4 MP 29.32 29.32 kNm 31.5.5.4(a) Width of Middle Strip Msw resis4ng M2 2.5 2.5 m •  m2 = M2/Msw 11.73 11.73 kNm/m C. Posi4ve Moment in Mid Span: m1 m1 m1 m1 m1 m1 m2 X Y Exterior Exterior Interior Interior m2

57. -‐ve sign : Hogging Moment (tension at top) +ve sign : Sagging Moment (tension at borom) 17.6 11.73 17.6 -‐8.85 -‐35.4 11.73 17.6 17.6 A B A D C C A AB -‐35.4 -‐35.4 -‐8.85 -‐35.4 -‐34.2 -‐34.2 0 -‐34.2 -‐34.2 -‐0 Exterior Exterior Interior Interior Step 7 : Check for adequacy of Depth •  Max Design Bending moment = 35.4 kNm/m •  Mu,lim = 72.41 kNm/m > 35.4, Depth is adequate G-‐1.1(c)

58. Strip Loca6on Moment (kNm/m) Ast (mm2 /m) Ast (prov) Bar dia Spacing mm Along X Zone A(Exterior) (-‐)34.2 637 637 10 120(T) Zone A(Interior) (-‐)35.4 662 662 10 115(T) Zone B 17.6 314 314 8 160(B) Zone C(Interior) (-‐)8.85 155 240 8 200(T) Zone D 11.73 206 240 8 200(B) Along Y Zone A (Exterior) (-‐)34.2 637 637 10 120(T) Zone A(Interior) (-‐)35.4 662 662 10 115(T) Zone B (Interior) (-‐)8.85 155 240 8 200(T) Zone C 17.6 314 314 8 160(B) Zone D 11.73 206 240 8 200(B) Step 8 :Rebar Details •  Ast,min = 0.12 x 200 x 1000 /100 = 240 mm2/m 26.5.2.1 7.5 Ast2 – 58490Ast + Mu = 0

59. 17.6 11.73 17.6 11.73 17.6 17.6 A B A D C C A A B Exterior Exterior Interior Interior

60. -‐8.85 -‐35.4 A B A D C C A AB -‐35.4 -‐35.4 -‐8.85 -‐35.4 -‐34.2 -‐34.2 0 -‐34.2 -‐34.2 -‐0 Exterior Exterior Interior Interior #8@200 #10@120 #10@115 #10@120 #10@120 #8@200 #8@200 #10@115 #10@115 #8@200 TOP Rebar details in Corner Panel

61. Design Example #2 Design by DDM ﬂat plate supported on columns 500 mm square, for a Live Load = 4 kN/m2, Floor Finish = 1 kN/m2 use M25 and Fe415. Floor slab is exposed to moderate environment. Column Height = 3.5m (c/c). Bay spacing in X and Y direc6on = 5.5m c/c. Assume that building is not restrained against sway •  Interior Panel P5 •  Corner Panel P7 3 bays @ 5.5 m c/c

62. A. Interior Panel Design 5.5 m 5.5 m 2.75m 2.75 m A A A A B B C CD Zone A – Corner Strip Zone B – Middle Strip along X Zone C – Middle Strip along Y Zone D – Interior Region Step 1: Panel Division into Strips 31.1.1(a) Moment direc6on Along L1 L2 Width of Column Strip on either side of Centre Line = 0.25L2 and ≤ 0.25 L1 Middle Strip X 5 5 1.375 and ≤1.375 m Adopt 1.375 m 2.75m Y 5 5 1.375 and ≤1.375 m Adopt 1.375 m 2.75m

63. Step 2: Trial Depth CL 31.2.1 •  L/d = 26 •  Modiﬁca5on Factor = 1.33, Assuming pt = 0.4%, FIG 4 IS 456 •  d = 5500/(26 x 1.33) = 160 mm > 125 CL 31.2.1 •  DS= 160 + 30 + 18 = 208 mm ( assume #12 bars, and bars in two layers) •  Provide Ds= 225 mm d = 225-‐30-‐18 = 177 mm

64. Step 3 Design Loads / m width of Slab •  wuD = 1.5(25x 0.225 + 1) = 9.94kN/m •  wuL = 1.5 x 4 = 6kN/m •  wu = 15.94 ≈ 16 kN/m Step 4: Check for Applicability of DDM: CL 31.4.1 •  No. of Con5nuous Spans in each direc5on = 3 ; OK 31.4.1(a) •  Long Span/Short Span = 5.5/5.5 = 1 <2 ; OK 31.4.1(b) •  Successive spans in each direc5on = Equal; OK 31.4.1(d) •  wuL/wuD = 6/9.94 = 0.6 < 3 ; OK 31.4.1(e)

65. Step 5: Check for punching shear around Column Assumed d = 177 mm Sec4on 1: •  Cri5cal Sec5on at d/2 around the column •  Perimeter of Cri5cal Sec5on = 4 x 0.677= 2.708 m •  Vu = 16 ( 5.52 – 0.6772) = 477kN •  τc = 0.25√fck = 1.25 MPa •  ks = 0.5 +1 = 1.5 <=1 ; ks=1 ; ks τc = 1.25 •  Shear Resistance of Concrete = 1.25 x 2708 x 177 = 599kN > 477 kN OK Cri4cal Sec4on 0.677m 0.677m 5.5m 5.5m Contributory Area

66. Step 6:Design Moments CL 31.4.2.2 Parameters Along X Along Y L1 (Span in direc4on of Mo) 5.5 5.5 m 0.65L1 3.575 3.575 m Ln (clear span extending from face to face of columns, capitals) (5.5-‐0.5) = 5 5 m Ln > 0.65L1 5 5 m L2 (Span transverse to L1) 5.5 5.5 m W = wu L2Ln 440 440 kN M0 = W Ln / 8 275 275 kNm wu = 16 kN /m

67. Step 7 : Distribution of Bending Moment across panel width ; CL: 31.4.3.2, 31.5.5 Moment Direc5on along X Y Nega4ve Design Moment MN = -‐ 0.65*M0 179 179 kNm 31.4.3.2 •  Column Strip M1 = 0.75MN 134.3 134.3 kNm 31.5.5.1 Width of Column Strip resis4ng M1 (Csw) 2.75 2.75 m •  -‐m1 = M1/ Csw (Zone A) 48.8 48.8 kNm/m •  Middle Strip M2 = 0.25MN 44.8 44.8 kNm 31.5.5.4(a) Width of Middle Strip resis4ng M2 (Msw) 2.75 2.75 m •  -‐m2 = M2/Msw (Zone B & C) 16.3 16.3 kNm/m

68. Posi5ve Design Moment MP = 0.35*M0 96.3 96.3 kNm 31.4.3.2 •  Column Strip M1 = 0.6MP 57.8 57.8 kNm 31.5.5.3 •  +m1 = M1/ Csw (Zone B &C) 21 21 kNm/m •  Mid Span M2 = 0.4MP 38.5 38.5 kNm 31.5.5.4(a) •  +m2 = M2/Msw (Zone D) 14 14 kNm/m -‐ve sign : Hogging Moment (tension at top) +ve sign : Sagging Moment (tension at borom) -‐48.8 -‐16.3 -‐48.8 21 14 21 -‐48.8 -‐16.3 -‐48.8 14 21 -‐16.3 -‐48.8 -‐48.8 -‐48.8 -‐16.3 -‐48.8 21 A B A D C C A AB Step 8 : Check for adequacy of Depth •  Max Design Bending moment = 48.8 kNm/m •  Mu,lim = 108 kNm/m > 48.8 •  Depth is adequate G-‐1.1(c)

69. Loca6on Moment (kNm/m) Ast (mm2 /m) Ast (prov) Bar dia Spacing mm Along X Zone A (-‐) 48.8 828 828 10 90 -‐ T Zone B 21 340 340 8 145 -‐ B Zone C (-‐)16.3 262 270 8 180 -‐ T Zone D 14 224 270 8 180 -‐ B Along Y Zone A (-‐) 48.8 828 828 10 90 -‐ T Zone B -‐16.3 262 270 8 180 -‐ T Zone C 21 340 340 8 145 -‐ B Zone D 14 224 270 8 180 -‐ B •  6 Ast2 – 63906Ast + Mu = 0 G-‐1.1(b) Step 9 :Rebar Details •  Ast,min = 0.12 x 225 x 1000 /100 = 270 mm2/m 26.5.2.1 •  Minimum Eﬀec5ve Depth of Slab = 177 mm

70. #8@145 #8@180 #8@145 #8@180 #8@145 #8@145 0.15Ln 0.15Ln 0.125Ln 0.125Ln Borom Rebar Details in Interior Panel A B A A B A C D C

71. TOP Rebar Details in Interior Panel #10@90 #10@90 #8@180 #8@200 #8@180 0.3Ln 0.2Ln 0.3Ln 0.2Ln Note: Distances for curtailment of rebars are measured from column face A B A C D C

72. B. Corner Panel Design Step 5: Check for punching shear around Column Assumed d = 177 mm Sec4on 1: •  Cri5cal Sec5on at d/2 around the column •  Perimeter of Cri5cal Sec5on = 2 x 0.5885= 1.177 m •  Vu = 16 ( 2.752 – 0.58852) = 115.5kN •  τc = 0.25√fck = 1.25 MPa •  ks = 0.5 +1 = 1.5 <=1 ; ks=1 ; ks τc = 1.25 •  Shear Resistance of Concrete = 1.25 x 1177 x 177 = 260kN > 115.5 kN OK 500 177/2 = 88.5 mm 2.75m 2.75m

73. Step 7 : Distribution of Bending Moment across panel width ; CL 31.4.3.3 , 31.5.5 𝛼↓𝑐 =∑↑▒ 𝑘↓𝑐  / 𝑘↓𝑠   Assume Columns and Slab panels are with same modulus of elas5city 5 m 5 m 1.25m 1.25m A A A A B B C CD Step 6:Design Moments CL 31.4.2.2 M0 = 275 kNm

74. Parameters Along X Along Y Sum of column s4ffness above and below the slab 2 (4EcIc)/Lc Leff = 1.2 Lc (CL E1) Lc = 3.5-‐0.225 = 3.275 (2 x 4 x Ec x 500 x 5003/12) /1.2*3275 = 10.6 Ec x 106 Slab s4ffness ks = 4EsIs/Ls (4 Es x 5500 x 2253/12)/5500 = 2.67Es x 106 3.8Es x 106 αc = ∑kc /ks 2.8 2.8 αc min (Table 17) (0.7/0.5)*0.1 =0.14 < αc. Adopt αc = 2.8 β = 1+ (1/αc) 1.36 1.36

75. Moment Direc5on along X Y Nega4ve Design Moment MN = -‐ 0.65*M0/β 131.4 131.4 kNm 31.4.3.3 •  Column Strip M1 = MN 131.4 131.4 kNm 31.55.2(a) Width of Column Strip Csw resis4ng M1 2.75 2.75 m -‐m1 = M1/Csw 47.8 47.8 kNm/m •  Middle Strip M2=0 0 0 kNm 31.5.5.4(a) -‐m2 = 0 0 0 kNm/m A. Exterior nega4ve design moment: -‐m1 -‐m1 -‐m1 -‐m1 Exterior Exterior Interior Interior 1.25 1.25 -‐m2 -‐m2 X Y M0 = 275 kNm

76. Moment Direc5on along X Y Nega4ve Design Moment MN = -‐ (0.75 – 0.1/β)Mo 186 186 kNm 31.4.3.3 •  Column Strip M1 = 0.75 MN 139.5 139.5 kNm 31.5.5.1 Width of Column Strip Csw resis4ng M1 2.75 2.75 m •  -‐m1 = M1/ Csw 50.73 50.73 kNm/m •  Middle Strip M2 = 0.25 MN 46.5 46.5 kNm 31.5.5.4(a) Width of Middle Strip Msw resis4ng M2 2.75 2.75 m •  -‐m2 = M2/Msw 17 17 kNm/m B. Interior nega4ve design moment: -‐ m1 -‐ m1 -‐ m2 m1 -‐m1 -‐m1 -‐m2 X Y Exterior Exterior Interior Interior

77. Moment Direc5on along X Y Design Moment MP = (0.63 – 0.28/β)Mo 116.6 116.6 kNm 31.4.3.3 •  Column Strip M1 = 0.6 MP 70 70 kNm 31.5.5.3 Width of Column Strip Csw resis4ng M1 2.75 2.75 m •  m1 = M1/ Csw 25.5 25.5 kNm/m •  Middle Strip M2 = 0.4 MP 46.7 46.7 kNm 31.5.5.4(a) Width of Middle Strip Msw resis4ng M2 2.75 2.75 m •  m2 = M2/Msw 17 17 kNm/m C. Posi4ve Moment in Mid Span: m1 m1 m1 m1 m1 m1 m2 X Y Exterior Exterior Interior Interior m2

78. -‐ve sign : Hogging Moment (tension at top) +ve sign : Sagging Moment (tension at borom) 25.5 17 25.5 -‐17 -‐50.73 17 25.5 25.5 A B A D C C A AB -‐50.73 -‐50.73 -‐17 -‐50.73 -‐47.8 -‐47.8 0 -‐47.8 -‐47.8 -‐0 Exterior Exterior Interior Interior Step 7 : Check for adequacy of Depth •  Max Design Bending moment = 50.73 kNm/m •  Mu,lim = 108 kNm/m > 50.73, Depth is adequate G-‐1.1(c)

79. Strip Loca6on Moment (kNm/m) Ast (mm2 /m) Ast (prov) Bar dia Spacing mm Along X Zone A(Exterior) (-‐)47.8 810 810 10 90(T) Zone A(Interior) (-‐)50.73 864 864 10 90(T) Zone B 25.5 415 415 10 180(B) Zone C(Interior) (-‐)17 273 273 8 180(T) Zone D 17 273 273 8 180(B) Along Y Zone A (Exterior) (-‐)47.8 810 810 10 90(T) Zone A(Interior) (-‐)50.73 864 864 10 90(T) Zone B (Interior) (-‐)17 273 273 8 180(T) Zone C 25.5 415 415 10 180(B) Zone D 17 273 273 8 180(B) Step 8 :Rebar Details •  Ast,min = 0.12 x 225 x 1000 /100 = 270 mm2/m 26.5.2.1 6 Ast2 – 63906Ast + Mu = 0

80. 7.2m 7.2m 7.2m 6.4m 6.4m 6.4m Design Example #3 Design by DDM ﬂat plate supported on columns of dia = 450 mm, Column head = 1.5 m dia, Drop panel size = 3.2 x 3.2 m, for a Live Load = 4 kN/m2, Floor Finish = 1 kN/m2 use M20 and Fe415. Assume clear cover = 20 mm. Column Height = 3.35m •  Interior Panel P5 •  Exterior Panel P2/P4 •  Corner Panel P1

81. Step 1: Panel Division into Strips 31.1.1(a) Moment direc6on Along L1 L2 Width of Column Strip on either side of Centre Line = 0.25L2 and ≤ 0.25 L1 Middle Strip X 7.2 6.4 1.6 < 1.8 m; 1.6 m 4m Y 6.4 7.2 1.8 > 1.6 m; 1.6 m 3.2m Lx = 7.2 Ly = 6.4 1.6 1.6 1.6 1.6 CSx CSx MSx CSy MSy CSy A.  Interior Panel Design Zone A – Corner Strip Zone B – Middle Strip along X Zone C – Middle Strip along Y Zone D – Interior Region

82. Step 2: Trial Depth CL 31.2.1 •  L/d = 26 •  Modiﬁca5on Factor = 1.4, Assuming pt ≈0.4%, FIG 4 IS 456 •  d = 7200/(26 x 1.4) = 198 mm > 125 CL 31.2.1 •  DS= 198+20+18= 236 mm ( assume #12 bars) •  Provide Ds= 240 mm , d = 198mm

83. Step 3: Design Loads / m width of Slab •  wuD = 1.5(25 x 0.24 + 1) = 10.5kN •  wuL = 1.5 x 4 = 6.0kN •  wu = 16.5 kN Step 4: Check for Applicability of DDM: CL 31.4.1 •  No. of Con5nuous Spans in each direc5on = 3 ; OK 31.4.1(a) •  Long Span/Short Span = 7.2/6.4 = 1.125 <2 ; OK 31.4.1(b) •  Successive spans in each direc5on = Equal; OK 31.4.1(d) •  wuL/wuD = 6/10.5 = 0.571 < 3 ; OK 31.4.1(e)

84. Step 5: Drop Panel Size : CL 31.2.2 •  Length along X ≥ Lx/3 = 2.4 m •  Length along Y ≥ Ly/3 = 2.13 m •  Generally Drop Panel Size is set equal to Width of Column Strip •  Proposed size 3.2 x 3.2 meets all the requirements. •  Minimum thickness = ¼ DS = 60 mm or 100 mm; Adopt 100 mm Step 6:Column Head •  1/4 to 1/5 of average span = 7.2/5 = 1.44 m •  Provided = 1.5 m ; Ok •  Equivalent Square Capital =0.89D = 1.335 m

85. Step 7 : Check for Shear around Column Capital •  Minimum Eﬀec5ve Depth of Slab = 198 mm •  Eﬀec5ve Depth at Drop loca5on = 298 mm Sec4on 1: •  Cri5cal Sec5on at d/2 around the column capital •  Perimeter of Cri5cal Sec5on = π ( 1.5 + 0.298) = 5.65 m •  Weight of Drop Projec5on below slab = 0.1x 25 x 1.5 = 3.75 kN/m2 •  Design Shear at cri5cal sec5on around capital Vu •  Vu = 16.5 ( 7.2 x 6.4 -‐ π x 1.7982/4) + 3.75(3.2 x 3.2 -‐ π x 1.7982/4) •  = 747 kN •  τc = 0.25√fck = 1.12 MPa •  ks = 0.5 +1 = 1.5 <=1 ; ks=1 ; ks τc = 1.12 •  Shear Resistance of Concrete = 1.12 x 5650 x 298 = 1885kN > 747 kN OK 1.5 Cri4cal Sec4on DROP 3.2 m 3.2 m 1.798 Capital

86. Sec4on 2 : Check for Shear around drop 1.5 Cri4cal Sec4on DROP 3.2 m 3.2 m Capital 3.2 + 0.198 = 3.4 •  Cri5cal Sec5on at d/2 around the drop •  d = 198mm •  Perimeter of Cri5cal Sec5on = 4 x 3.4 = 13.6m •  Design Shear at cri5cal sec5on •  Vu = 16.5 ( 7.2 x 6.4 – 3.42) = 569 kN •  Shear Resistance of Concrete = 1.12 x 13600 x 198 = 3015kN > 569 kN

87. Step 8:Design Moments CL 31.4.2.2 Parameters Along X Along Y L1 (Span in direc4on of Mo) 7.2 6.4 m 0.65L1 4.68 4.16 m Ln (clear span extending from face to face of columns, capitals) (7.2-‐1.335) = 5.865 (6.4-‐1.335) = 5.065 m Ln > 0.65L1 5.865 5.065 m L2 (Span transverse to L1) 6.4 7.2 m W = wu L2Ln 619.34 601.72 kN M0 = W Ln / 8 454 381 kNm wu = 16.5 kN /m

88. Step 9 : Distribution of Bending Moment across panel width ; CL: 31.4.3.2, 31.5.5 Moment Direc5on along X Y Nega4ve Design Moment MN = -‐ 0.65*M0 295.1 247.65 kNm 31.4.3.2 •  Column Strip M1 = 0.75MN 221.33 185.74 kNm 31.5.5.1 Width of Column Strip Csw 2x1.6 =3.2 2x1.6 =3.2 m •  -‐m1 = M1/ Csw 69.17 58.04 kNm/m •  Middle Strip M2 = 0.25MN 73.78 61.91 kNm 31.5.5.4(a) Width of Middle Strip Msw 3.2 4 •  -‐m2 = M2/Msw 23.06 15.48 kNm/m

89. Posi5ve Design Moment MP = 0.35*M0 158.9 133.35 kNm 31.4.3.2 •  Column Strip M1 = 0.6MP 95.34 80.01 kNm 31.5.5.3 •  +m1 = M1/ Csw 29.79 25 kNm/m •  Middle Strip M2 = 0.4MP 63.56 53.34 kNm 31.5.5.4(a) •  +m2 = M2/Msw 19.86 13.34 kNm/m -‐ve sign : Hogging Moment (tension at top) +ve sign : Sagging Moment (tension at borom) -‐69.17 -‐23.06 -‐69.17 29.79 19.86 29.79 -‐58.04 -‐15.48 -‐58.04 13.34 25 -‐15.48 -‐58.04 -‐58.04 -‐69.17 -‐23.06 -‐69.17 25 A B A D C C A AB Step 10 : Check for adequacy of Depth •  Max Design Bending moment = 69.17 kNm/m •  Mu,lim = 126.36 kNm/m > 69.17, G-‐1.1(c) •  Depth is adequate

90. Moment Direc4on Moment Direc4on CS MS CS FE Results from ETAB

91. Strip Loca6on Moment (kNm/m) Ast (mm2 /m) Ast (prov) Bar dia Spacing mm Along X Zone A (-‐) 69.17 1093 1093 10 70 -‐ T Zone B 29.79 437 437 8 110 -‐ B Zone C (-‐)23.06 334 334 8 150 -‐T Zone D 19.86 286 288 8 170 -‐ B Along Y Zone A (-‐)58.04 896 896 10 85 -‐ T Zone B (-‐)15.48 222 288 8 170-‐T Zone C 25 364 364 8 135 -‐B Zone D 13.34 190 288 8 170 -‐ B •  7.5 Ast2 – 71488Ast + Mu = 0 G-‐1.1(b) Step 11 :Rebar Details •  Ast,min = 0.12 x 240 x 1000 /100 = 288 mm2/m 26.5.2.1 •  Minimum Eﬀec5ve Depth of Slab = 198 mm

92. #8@170 #8@135 #8@135 #8@170 #8@110 #8@110 7.2 m 6.4 m 0.15Ln 0.15Ln Borom Rebar Details in Interior Panel LAP ZONE A A A A B B C D C 0.125Ln 0.125Ln

93. #8@170 #10@85 #8@150 #10@ 70 0.22Ln 0.22Ln 0.22Ln 0.22Ln 0.33Ln 0.33Ln 0.2Ln 0.2Ln 0.33Ln 0.2Ln Top Rebar Details in Interior Panel Note: Distances for curtailment of rebars are measured from column face A B A 0.33Ln 0.2Ln

94. Sec6on Through Middle Strip -‐ CDC #8@170 7.2 m #8@340 #8@150 #8@135 Sec6on Through Column Strip -‐ ABA #10@70 #10@140 #10@85 #8@170 #8@340 #8@170

95. Step 7 : Check for Shear around Column Capital •  Minimum Eﬀec5ve Depth of Slab = 198 mm •  Eﬀec5ve Depth at Drop loca5on = 298 mm Sec4on 1: •  Cri5cal Sec5on at d/2 around the column capital •  Perimeter of Cri5cal Sec5on = π ( 1.5 + 0.298)/4 = 1.412 m •  Weight of Drop Projec5on below slab = 0.1x 25 x 1.5 = 3.75 kN/m2 •  Design Shear at cri5cal sec5on around capital Vu •  Vu = 16.5 ( 3.6x 3.2 – (π x 1.7982/4)/4) + 3.75(1.6 x 1.6 – (π x 1.7982/4)/4)) •  = 187 kN •  τc = 0.25√fck = 1.12 MPa •  ks = 0.5 +1 = 1.5 <=1 ; ks=1 ; ks τc = 1.12 •  Shear Resistance of Concrete = 1.12 x 1412 x 298 = 471kN > 187 kN OK 2. Corner Panel Design

96. Sec4on 2 : Check for Shear around drop •  Cri5cal Sec5on at d/2 around the drop •  d = 198mm •  Perimeter of Cri5cal Sec5on = 2 (1.7)=3.4m •  Design Shear at cri5cal sec5on •  Vu = 16.5 ( 3.6 x 3.2 – 1.72) = 143 kN •  Shear Resistance of Concrete = 1.12 x 3400 x 198 = 754kN > 143 kN CRITICAL SECTION drop free edge free edge =1.6 + 0.198/2 = 1.7 m Step 8:Design Moments CL 31.4.2.2 Along X Along Y M0 = W Ln / 8 454 381 kNm

97. Step 9 : Distribution of Bending Moment across panel width ; CL 31.4.3.3 , 31.5.5 𝛼↓𝑐 =∑↑▒ 𝑘↓𝑐  / 𝑘↓𝑠   Equivalent side of circular column = 0.89D = 0.89x 450 = 400 mm Assume Ec = Es

98. Parameters Along X Along Y Sum of column s4ﬀness above and below the slab 2 (4EcIc)/Lc (2 x 4 x Ec x 400 x 4003/12) /3350 = 5.09 Ec x 106 Slab s4ﬀness ks = 4EsIs/Ls (4 Es x 6400 x 2403/12)/7200 = 4.1Es x 106 (4 Es x 7200 x 2403/12)/6400 = 5.184Es x 106 αc = ∑kc /ks 1.24 0.98 αc min (Table 17) l2/l1 = 6.4/7.2 = 0.89, WuL/WuD = 0.571 (0.7/0.5)*0.071 = 0.1 <αc Adopt αc 7.2/6.4 = 1.125, WuL/WuD = 0.571 ≈(0.8/0.5)*0.071 = 0.113 <αc Adopt αc β = 1+ (1/αc) 1.8 2.02

99. Moment Direc5on along X Y Nega4ve Design Moment MN = -‐ 0.65*M0/β 164 122.6 kNm 31.4.3.3 •  Column Strip M1 = MN 164 122.6 kNm 31.55.2(a) Width of Column Strip Csw resis4ng M1 2x1.6 =3.2 2x1.6 =3.2 m -‐m1 = M1/ 3.2 -‐51.3 -‐38.3 kNm/m •  Middle Strip M2=0 0 0 kNm 31.5.5.4(a) Width of Middle Strip Msw resis4ng M2 3.2 4 -‐m2 = 0 0 0 kNm/m A. Exterior nega4ve design moment: -‐m1 -‐m1 -‐m1 -‐m1 Exterior Exterior Interior Interior 1.6 1.6 -‐m2 -‐m2 X Y

100. Moment Direc5on along X Y Nega4ve Design Moment MN = -‐ (0.75 – 0.1/β)Mo 315.3 266.9 kNm 31.4.3.3 •  Column Strip M1 = 0.75 MN 236.5 200.2 kNm 31.5.5.1 Width of Column Strip Csw resis4ng M1 2x1.6 =3.2 2x1.6 =3.2 m •  -‐m1 = M1/ Csw -‐73.9 -‐62.6 kNm/m •  Middle Strip M2 = 0.25 MN 78.83 66.7 kNm 31.5.5.4(a) Width of Middle Strip Msw resis4ng M2 3.2 4 m •  -‐m2 = M2/Msw -‐24.7 -‐16.7 kNm/m B. Interior nega4ve design moment: -‐ m1 -‐ m1 -‐ m2 m1 -‐m1 -‐m1 -‐m2 X Y Exterior Exterior Interior Interior

101. Moment Direc5on along X Y Design Moment MP = (0.63 – 0.28/β)Mo 215.4 187.2 kNm 31.4.3.3 •  Column Strip M1 = 0.6 MP 129.3 112.3 kNm 31.5.5.3 Width of Column Strip Csw resis4ng M1 2x1.6 =3.2 2x1.6 =3.2 m •  m1 = M1/ Csw 40.4 35.1 kNm/m •  Middle Strip M2 = 0.4 MP 86.2 74.9 kNm 31.5.5.4(a) Width of Middle Strip Msw resis4ng M2 3.2 4 m •  m2 = M2/Msw 26.94 18.7 kNm/m C. Posi4ve Moment in Mid Span: m1 m1 m1 m1 m1 m1 m2 X Y Exterior Exterior Interior Interior m2

102. -‐ve sign : Hogging Moment (tension at top) +ve sign : Sagging Moment (tension at borom) 40.4 26.94 40.4 -‐16.7 -‐62.6 18.7 35.1 35.1 A B A D C C A AB -‐73.9 -‐73.9 -‐24.7 -‐62.6 -‐51.3 -‐51.3 0 -‐38.3 -‐38.3 -‐0 Exterior Exterior Interior Interior Step 10 : Check for adequacy of Depth •  Max Design Bending moment = 73.9 kNm/m •  Mu,lim = 126.36 kNm/m > 73.9, Depth is adequate G-‐1.1(c)

103. Strip Loca6on Moment (kNm/m) Ast (mm2 /m) Ast (prov) Bar dia Spacing mm Along X Zone A(Exterior) (-‐)51.3 782 782 10 100 -‐ T Zone A(Interior) (-‐)73.9 1180 1180 10 65 -‐ T Zone B 40.4 604 604 8 80 -‐B Zone C(Interior) (-‐)24.7 359 359 8 140 -‐ T Zone D 26.94 393 393 8 125-‐B Along Y Zone A (Exterior) (-‐)38.3 570 570 10 135 Zone A(Interior) (-‐)62.6 976 976 10 80 Zone B (Interior) (-‐)16.7 240 288 8 170 Zone C 35.1 520 520 8 95 Zone D 18.7 270 288 8 170 •  7.5 Ast2 – 71488Ast + Mu = 0 G-‐1.1(b) Step 11 :Rebar Details •  Ast,min = 0.12 x 240 x 1000 /100 = 288 mm2/m 26.5.2.1 •  Minimum Eﬀec5ve Depth of Slab = 198 mm

104. Strip Loca6on Moment (kNm/m) Bar dia Spacing mm Along X Zone B 40.4 8 80 -‐B Zone D 26.94 8 125-‐B Along Y Zone C 35.1 8 95 Zone D 18.7 8 170 #8@80 #8@125 #8@95 #8@95 #8@170 #8@80

105. 40.4 26.94 40.4 -‐16.7 -‐62.6 18.7 35.1 35.1 A B A D C C A A B -‐73.9 -‐73.9 -‐24.7 -‐62.6 -‐51.3 -‐51.3 0 -‐38.3 -‐38.3 -‐0 Exterior Exterior Interior Interior #10@100 #10@100 #10@65 #10@65 #8@140 #10@135 #10@80 #8@170(Min)* #10@135 #8@170 #8@170(Min)* * Op4onal Top Rebars #10@80 Strip Loca6on Moment (kNm/m) Bar dia Spacing mm Along X Zone A(Exterior) (-‐)51.3 10 100 -‐ T Zone A(Interior) (-‐)73.9 10 65 -‐ T Zone C(Interior) (-‐)24.7 8 140 -‐ T Along Y Zone A (Exterior) (-‐)38.3 10 135 Zone A(Interior) (-‐)62.6 10 80 Zone B (Interior) (-‐)16.7 8 170

106. Transfer of Moments and Shears between Slabs and Columns •  The maximum load that a ﬂat slab can support is dependent upon the strength of the joint between the column and the slab. •  Load is transferred by shear from the slab to the column along an area around the column •  In addi7on moments are also transferred. •  The moment situa7on is usually most cri7cal at the exterior columns. •  Shear forces resul7ng from moment transfer must be considered in the design of the lateral column reinforcement (i.e., 7es and spirals).

107. EXAMPLE Compute moment transferred to Interior and corner Column in example 2 Interior Column •  As spans are same in both direc5ons •  M = 0.08 (0.5 wL L2 Ln 2 /(1+1/αc) = 0.08 x 0.5 x 6 x 5.5 x 52 / 1.36 = 24.3 kNm •  this moment is distributed to top and borom column at junc5on in propor5on to their s5ﬀness. •  M = 24.3/2 = 12.2 kNm Corner Column M = 131.4 kNm

108. Equivalent Frame Method (EFM) CL 31.5 •  More Comprehensive and Logical method •  Used when limita7ons of DDM are not complied with •  Applicable when subjected to horizontal loads 31.5.1 (a) Idealizing the 3D slab –column system to 2D frames along column Centre lines in both longitudinal and transverse direc6ons. Longitudinal Frame Transverse Frame Edge Frame

109. For ver6cal loads, each floor, together with the columns above and below, is analyzed separately. For such an analysis, the far ends of the columns are considered fixed. If there are large number of panels, the moment at a par6cular joint in a slab beam can be sa6sfactorily obtained by assuming that the member is fixed two panels away. This simplifica6on is permissible because ver6cal loads in one panel only appreciably affect the forces in that panel and in the one adjacent to it on each side. 31.5.1(b)

110. En6re Frame Analysis Gravity + Lateral Loads For lateral loads, it is necessary to consider an equivalent frame that extends for the en4re height of the building, because the forces in a par4cular member are aﬀected by the lateral forces on all the stories above the ﬂoor being considered.

111. 31.5.1(C and d) I2 = moment of iner4a at the face of the column / column capital c2 = dimension of column capital in the transverse direc4on l2 = width of equivalent frame. varia6on of the flexural moment of iner6a •  Varia4ons of moment of iner4a along the axis 0f the slab on account of provision of drops shall be taken into account •  The s4ffening effect of flared column heads may be ignored

112. 31.5.2 Loading Paiern wu LL > ¾ wu,DL

113. Cri5cal Sec5on Interior Column Centre Line Column /Capital face C < = C/2 Results in Signiﬁcant reduc4on of design moments Design Posi5ve Moment (Span region) M3 = M0 – (M1+M2)/2

114. Distribu5on of Moment Similar to DDM

115. Example 3 : Compute moments in exterior/interior Panel along Longitudinal Span Longitudinal Span = 7.2m, Transverse Span = 6.4 m, Interior Column = 450mm dia, Column Capital = 1500mm dia, Exterior Column = 400x400mm, Column Capital = 870mm(square), Floor to Floor = 3.35 m, Slab Thickness = 240 mm, number of Panels = 4 in each direc6on 7.2 m 6.4m 6.4m 7.2 m 7.2 m 7.2 m 6.4m

116. Step 1: S5ffness Computa5ons Exterior Column (Kce) = 4E x (4004 /12) /3350 = 2.55E106 = 1 Interior Column (KcI) = 4E x π(4504 /64) /3350 = 2.4E106 = 0.957 Slab(Ks) =4E x (6400 x 2403/12) /7200 = 4.1E106 = 1.608 Step 2: Simplified frame for analysis 31.5.1 (b) 7200 7200 3350 3350 1 2 3 A B C D Joint Member Rela5ve S5ffness Sum Distribu5on Factors 1 1-‐A 1 3.608 0.277 1-‐2 1.608 0.446 1-‐C 1 0.277 2 2-‐B 0.957 5.13 0.187 2-‐1 1.608 0.314 2-‐3 1.608 0.314 2-‐D 0.957 0.187 Fixed End Moments = (16.5 x 6.4) x 7.22/12 = 456.2 kNm

117. Joint 1 2 3 Members FIXED 1A+1C 1-‐2 2-‐1 FIXED 2B+2D 2-‐3 FIXED 3-‐2 DF 0.554 0.446 0.314 0.374 0.314 -‐ FEM 456.2 -‐456.2 456.2 -‐456.2 Bal -‐252.74 -‐203.46 -‐ -‐ -‐ -‐ CO -‐ -‐ -‐101.73 -‐ -‐ -‐ Bal -‐ -‐ 31.94 38.04 31.94 -‐ CO -‐ 15.97 -‐ -‐ -‐ 15.97 Bal -‐8.85 -‐7.12 -‐ -‐ -‐ -‐ CO -‐ -‐ -‐3.56 -‐ -‐ -‐ Bal 1.12 1.33 1.12 Final end Moments -‐261.6 261.6 -‐528.43 39.37 489.26 -‐440.23 1 2 3 261.6 528.43 489.26 440.23 Counter Clockwise end moments are posi4ve

118. Step 3: Design Moments in Exterior Panel A. Design Nega6ve Moments at Cri6cal Sec6on At Exterior Support : CL 31.5.3.2 870 400 470 235 Cri6cal Sec6on from Column Centre line = 435 mm 261.6 528.43 16.5 x 6.4 = 105.6 kN/m 105.6 x 7.2/2 -‐ (528.43-‐261.6)/7.2 = 343 kN 0.435 Design Moment = 343 x 0.435 -‐261.6 -‐105.6x0.4352/2 = -‐122.4 kNm (Hogging)

119. At Interior Support : CL 31.5.3.1 Width of equivalent square = 0.89D = 1335 mm 667.5 mm Cri4cal Sec4on loca4on is at capital face ≤ 0.175x7200 = 1260mm 261.6 528.43 16.5 x 6.4 = 105.6 kN/m 343 kN 0.6675 Design Moment = 417.32 x 0.6675 -‐528.3 -‐105.6x0.66752/2 = -‐273.26 kNm (Hogging) 417.32

120. B. Design Posi4ve Moment M(+) = (16.5 x 6.4x7.2)7.2/8 – ( 528.43 + 261.6)/2 = 289.3 kNm Moments DDM EFM Posi4ve Moment (Span) 215.4 289.3 Nega4ve Moment(Exterior Support) 164 122.4 Nega4ve Moment (Interior Support) 315.3 273.3

121. Step 4: Design Moments in Interior Panel A. Design Nega6ve Moments at Cri6cal Sec6on At Interior Support : CL 31.5.3.1 16.5 x 6.4 = 105.6 kN/m 387 kN 0.6675 Design Moment at A= 387 x 0.6675 -‐ 489.26 -‐105.6x0.66752/2 = -‐254.5 kNm (Hogging) 373.32 489.26 440.23 Design Moment at B = 373.32 x 0.6675 -‐ 440.23 -‐105.6x0.66752/2 = -‐214.6 kNm (Hogging) 0.6675 A B

122. B. Design Posi4ve Moment M(+) = (16.5 x 6.4x7.2)7.2/8 – ( 489.26 + 440.23)/2 = 219.5 kNm Moments DDM EFM Posi4ve Moment (Span) 158.9 219.5 Nega4ve Moment (Interior Support) 295.1 254.5/214.6

123. Need for Computer Analysis The equivalent frame method is not sa6sfactory for hand calcula6ons. It is possible, however, to use computers and plane frame analysis programs if the structure is modeled such that various nodal points in the structure can account for the changing moments of iner6a along the member axis. SLAB Drop Panel Column Head Column Column

124. FE Analysis of Slab At any point in the plate bending, there will generally be two bending moments Mx , My in two mutually perpendicular direc5ons coupled with a complimentary twis5ng moment Mxy Design for ﬂexure involves providing reinforcing steels in two orthogonal direc5ons to resist the moment ﬁeld. Mx, My and Mxy. Slab is idealized as an assembly of discrete plate bending elements joined at nodes Wood –Armer equa4ons are used for this purpose.

125. Wood –Armer equa5ons (1968) •  This method was developed by considering the normal moment yield criterion (Johansen’s yield criterion) aiming to prevent yielding in all direc4ons. •  At any point in the slab, the moment normal to a direc4on, resul4ng due to design moments Mx , My , and Mxy must not exceed the ul4mate normal resis4ng moment in that direc4on. •  Mx * cos2θ + My * sin2θ -‐ ﬂexural strength of plate in the direc4on of θ with X axis. •  Mx cos2θ + Mysin2θ + 2 Mxy cosθ sinθ -‐ normal bending moment in the direc4on of θ

127. Example 1 FE results at centre of a plate element are: Mx = 7 kNm, My = 23 kNm, Mxy = 9 kNm. Compute design moments using Wood -‐ Armer equa4ons. A.  Borom rebars (Sagging Moments) Mx* = Mx+|Mxy| = 16 > 0 , Mx* = 16 kNm My* = My+|Mxy| = 32 > 0 , My* = 32 kNm B. Top rebars (Hogging Moments) Mx* = Mx-‐|Mxy| = -‐2 < 0 , Mx* = 2 kNm My* = My-‐|Mxy| = 14 > 0 Set My* = 0 and compute Mx * = Mx -‐| Mxy 2/My| = 7 – |81/23| = 3.478 kNm

128. Example 2 FE results at centre of a plate element are: Mx = 7 kNm, My = -‐23 kNm, Mxy = 9 kNm. Compute design moments using Wood -‐ Armer equa4ons. A.  Borom rebars (Sagging Moments) Mx* = Mx+|Mxy| = 16 > 0 , Mx* = 16 kNm My* = My+|Mxy| = -‐14 < 0 , Set My* = 0 and compute Mx * = Mx + | Mxy 2/My| = 7 + |81/23| = 10.52 kNm B. Top rebars (Hogging Moments) Mx* = Mx-‐|Mxy| = -‐2 < 0 , Mx * = 2 kNm My* = My-‐|Mxy| = -‐32 < 0 , MY * = 32 kNm

Civil structural engineering - Flat slab design

Satish Narayan

Civil structural engineering - Flat slab design