Geotechnical Pad Foundation (11-1-2021).pptx

Pad foundation design

1. BEV3023 RC DESIGN

2. DESIGN OF FOUNDATIOS

3. Introduction Foundation – Part of structure which transmits load from the structure to the underlying soil or rock All soils compress noticeably when loaded causing structure to settle

4. Introduction Requirements in the design of foundations: Total settlement of the structure to be limited to a tolerably small amount Differential settlement of various parts of structure shall be eliminated

5. To limit settlement, it is necessary to transmit the structure load to a soil stratum of sufficient strength Spread the structure load over a sufficiently large area of stratum to minimize bearing pressure Satisfactory soil: Use footings Adequate soil: Use deep foundations i.e. piles

6. Pressure distribution under a footing Uniform di s tribu t ed Cohesive soil Cohesionless soil

7. Pad Footings Transmit load from piers and columns Simplest and cheapest type Use when soil is relatively strong or when column loads are relatively light Normally square or rectangular shape in plan Has uniform thickness

8. Combine Footings Use when two columns are closed together Combine the footing to form a continuous base Base to be arranged so that its centreline coincides with the centre of gravity of the load – provide uniform pressure on the soil

9. Strap Footings Use where the base for an exterior column must not project beyond the property line Strap beam is constructed between exterior footing & adjacent interior footing Purpose of strap – to restrain overturning forces due to load e c c e n tricity o n the e x t erior f o o ting

10. Strap Footings (continued) Base area of the footings are proportioned to the bearing pressure Resultant of the loads on the two footings should pass through the centroid of the area of the two bases Strap beam between the two footings should NOT bear against the soil

11. Strip Footings Use for foundations to load-bearing wall Also use when pad footings for number of columns are closely spaced Also use on weak ground to increase foundation bearing area

12. Raft Foundations Combine footing which covers the whole building Support all walls & columns Useful where column loads are heavy or bearing capacity is low – need large base Also used where soil mass contains compressible layers or soil is variable – differential settlement difficult to control

13. Pile Foundations More economic to be used when solid bearing stratum i.e. rock is deeper than about 3 m Pile loads can either be transmitted to a stiff bearing layer (some distance below surface) or by friction along the length of pile Pile types – precast (driven into the soil) or cast in-situ (bored) Soil survey is important to provide guide on the length of pile and safe load capacity of the pile

14. Pile Foundations Load from Structure P I L E P I L E P I L E P I L E Lower Density Medium Density High Density Pile Cap

15. Thickness and Size of Footing Area of pad: 𝑨 = 𝑮 𝒌 + 𝑸 𝒌 + 𝑾 𝑺𝒐𝒊𝒍 𝒃𝒆𝒂𝒓𝒊𝒏𝒈 𝒄𝒂𝒑𝒄𝒊𝒕𝒚 Minimum effective depth of pad: 𝒅 = 𝑵 𝑬𝒅 𝒗 𝒓𝒅,𝒎𝒂𝒙 ∙ 𝒖 𝒐 N Ed = Ultimate vertical load = 1.35 G k + 1.5 Q k 1 − 𝑓 𝑐𝑘 250 𝑓 𝑐𝑘 1 . 5 v rd,max = 0.5 vf cd = 0.5 0.6 u o = Column perimeter

16. Design for Flexure Critical section for bending – At the face of the column Moment is taken on a section passing completely across the footing and due to ultimate load on one side of the section Moment & shear is assessed using STR (Structure) combination STR Combination 1: 𝑵 = 𝟏. 𝟑𝟓𝑮 𝒌 + 𝟏. 𝟓𝑸 𝒌 x x y y

17. Check for Shear May fail in shear as vertical shear or punching shear Vertical shear sections Punching shear perimeters 2 d d d h Bends may be r equ i r ed

18. Design of Pad Footing Check for Shear (i) Vertical Shear Critical section at distance d from the face of column Vertical shear force =  Load acting outside the section If V Ed  V Rd,c = No shear reinforcement is required

19. Check for Shear (ii) Punching Shear Axial Force Only 𝑬𝒅 = 𝑽 𝑬𝒅 𝒖 ∙ 𝒅 where u = Critical perimeter Critical section at a perimeter 2 d from the face of the column Punching shear force =  Load outside the critical perimeter Shear stress, 𝒗 If v Ed  v Rd,c = No shear reinforcement is required Also ensure that V Ed  V Rd,max

20. Check for Shear (ii) Punching Shear (continued) Axial Force & Bending Moment Punching shear resistance can be significantly reduced of a co- existing bending, M Ed However, adverse effect of the moment will give rise to a non- uniform shear distribution around the control perimeter Refer to Cl. 6.4.3(3) of EC2

21. Check for Shear (ii) Punching Shear (continued) Shear stress, 𝒗 𝑬𝒅 = 𝜷𝑽 𝑬𝒅 𝒖 𝟏 ∙𝒅 where;  = factor used to include effect of eccentric load & bending moment = 1 + 𝑘 𝑀 𝐸𝑑 𝑉 𝐸𝑑 𝑢 1 𝑊 1 k = coefficient depending on the ratio between column dimension c 1 & c 2 u 1 = length of basic control perimeter W 1 = function of basic control perimeter corresponds to the distribution of shear = 0.5𝑐 1 2 + 𝑐 1 𝑐 2 + 4𝑐 2 𝑑 + 16𝑑 2 + 2𝜋𝑑𝑐 1 c 1 / c 2  0.5 1.0 2.0  3.0 k 0.45 0.60 0.70 0.80

22. Design of Pad Footing Check for Shear (ii) Punching Shear (continued)

23. Cracking & Detailing Requirements All reinforcements should extend the full length of the footing If 𝐿 𝑥 > 1.5 𝑐 𝑥 + 3𝑑 , at least two-thirds of the reinforcement parallel to L y should be concentrated in a band width 𝑐 𝑥 + 3𝑑 centred at column where L x & L y and c x & c y are the footing and column dimension in x and y directions Reinforcements should be anchored each side of all critical sections for bending. Usually possible to achieve using straight bar Spacing between centre of reinforcements  20 mm for f yk = 500 N/mm 2 Reinforcements normally not provided in the side face nor in the top face (except for balanced & combined foundation) Starter bar should terminate in a 90  bend tied to the bottom reinforcement, or in the case of unreinforced footing spaced 75 mm off the building

24. Example 1 PAD FOOTING (AXIA L L O AD ON L Y)

25. f ck = 25 N/mm 2 f yk = 500 N/mm 2  soil = 150 N/mm 2 Unit weight of concrete = 25 kN/m 3 Design life = 50 years Exposure Class = XC2 Assumed  bar = 12 mm Axial Force, N : G k = 600 kN Q k = 400 kN B h Column size: 300  300 mm H

26. Durability & Bond Requirements Min cover regards to bond, c min,b = 12 mm Min cover regards to durability, c min,dur = 25 mm Allowance in design for deviation,  c dev = 10 mm Nominal cover, c nom = c min +  c dev = 25 + 10 = 35 mm  c nom = 35 mm c min = 25 mm

27. Si z e Service load, N Assumed selfweight 10% of service load , W = 1000 kN = 100 kN 𝛾 𝑠𝑜𝑖𝑙 𝑁+𝑊 1000+100 150 Area of footing required = = = 7.33 𝑚 2  Try footing size, B  H  h = 3 m  3 m  0.45 m Area, A = 9 m 2 Selfweight, W = 9  0.45  25 = 101 kN 𝑵 + 𝑾 𝑨 = 𝟏𝟎𝟎 𝟎 + 𝟏𝟎𝟏 𝟗 Check Service Soil Bearing Capacity = = 122 kN/m 2  150 kN/m 2  OK

28. A n al y sis Ultimate axial force, N Ed = 1.35 G k + 1.5 Q k = 1.35 (600) + 1.5 (400) = 1410 kN 𝐴 9 𝑁 𝐸𝑑 1410 Soil pressure at ultimate load, P = = = 157 kN/m 2 Soil pressure per m length, w = 157  3 m = 470 kN/m 1.35 m 1.35 m w = 470 kN/m M Ed 0.3 m 𝟏. 𝟑𝟓 𝟐 𝑴 𝑬𝒅 = 𝟒𝟕𝟎 × 𝟏. 𝟑𝟓 × = 428 kNm

29. Main Reinforcement Effective depth, d = h – c – 1.5  bar = 450 – 35 – (1.5  12) = 397 mm 𝐾 = 𝑀 𝐸𝑑 428×10 6 𝑓 𝑐𝑘 𝑏𝑑 2 25×3000×397 2 bal = = 0.036  K = 0.167  Compression reinforcement is NOT required 𝑧 = 𝑑 0.25 − 𝐾 1.134 = 0.97𝑑  0.95 d 𝐴 𝑠 , 𝑟 𝑒𝑞 = 𝑀 𝐸𝑑 0.87𝑓 𝑦𝑘 𝑧 = 428×10 6 0.87×500×0.95×397 = 𝟐𝟔𝟏𝟏 mm 2

30. Minimum & Maximum Area of Reinforcement 𝐴 𝑠 , 𝑚𝑖𝑛 = 0.26 𝑓 𝑐𝑡𝑚 𝑓 𝑦𝑘 2 . 56 500 𝑏𝑑 = 0.26 0.0013𝑏𝑑 ≥ 0.0013𝑏𝑑  A s,min = 0.0013 bd = 0.0013  3000  397 = 1589 mm 2 A s,max = 0.04 A c = 0.04 bh = 0.04  3000  450 = 54000 mm 2 Provide 24H12 ( A s,prov = 2715 mm 2 )

31. (i) Vertical Shear Critical shear at 1.0 d from face of column:  Design shear force, V Ed = 470  0.953 = 448 kN 3 m 0.953 m 3 m 953 mm 1.35 m w = 470 kN/m d = 397 mm V Ed

32. (i) Vertical Shear 𝑘 = 1 + 200 = 1 + 𝑑 397 200 = 1.71  2.0 Note: Bar extend beyond critical section at = 953 – 35 = 918 mm  𝑙 𝑏𝑑 + 𝑑 = 40∅ + 𝑑 = 40 × 12 + 397 = 877 mm  A sl = 2715 mm 2 𝑙 𝜌 = 𝑠𝑙 𝑏𝑑 𝐴 2 715 3000 × 397 = = 0.0023 ≤ 0.02

33. (i) Vertical Shear 𝑉 𝑅𝑑,𝑐 = 0.12𝑘 100𝜌 𝑙 𝑓 𝑐𝑘 1/3 𝑏𝑑 = 0.12 × 1.71 100 × 0.0023 × 25 1/3 3000 × 397 = 436463 N = 436 kN 𝑉 𝑚𝑖𝑛 = 𝑓 𝑐𝑘 0.035𝑘 3/2 𝑏𝑑 = 0.035 × 1.71 3/2 25 3000 × 397 = 465970 𝑁 = 466 kN V Ed (448 kN)  V min (466 kN)  OK

34. (ii) Punching Shear Critical shear at 2.0 d from face of column: Average d = 450 – 35 – 12 = 403 mm  2 d = 806 mm Control perimeter, u = (4  300) + (2   806) = 6265 mm Area within perimeter, A = (0.30  0.30) + (4  0.30  0.806) + (   0.806 2 ) = 3.10 m 2 1350 2 d = 806 300 300 2 d = 806 544  𝑙 𝑏𝑑 + 𝑑 = 40∅ + 𝑑 = 40 × 12 + 397 = 877 mm  Reinforcement NOT contributed to punching resistance

35. (ii) Punching Shear Punching shear force: V Ed = 157 (3 2 – 3.10) = 925 kN Punching shear resistance: 𝑉 𝑅𝑑,𝑐 = 𝑉 𝑚𝑖𝑛 = 0.035𝑘 3/2 𝑓 𝑐𝑘 1/2 𝑢𝑑 25 1/2 = 0.035 1.71 3/2 6265 × 403 = 983199 N = 983 kN  V Ed (925 kN)  OK Soil pressure = 157 kN/m 2 A p er im et er = 3.10 m 2 A all = 9 m 2

36. (iii) Maximum Punching Shear at Column Perimeter Maximum punching shear force: V Ed,max = 157 (3 2 – 0.09) = 1400 kN Maximum shear resistance: 𝑉 𝑅𝑑,𝑚𝑎𝑥 = 0.5𝑢𝑑 0.6 𝑓 𝑐𝑘 1 − 𝑓 𝑐𝑘 25 1 . 5 = . 5 4 × 30 × 40 3 0. 6 25 25 1 − 250 1.5 = 2176 kN  V Ed,max  OK Soil pressure = 157 kN/m 2 A column = 0.09 m 2 A all = 9 m 2

37. Cracking h = 450 mm  200 mm 𝑠 S t eel s t r es s , 𝑓 = 𝐺 𝑘 +0.3𝑄 𝑘 𝐴 𝑠.𝑟𝑒𝑞 𝐴 𝑠,𝑝𝑟𝑜𝑣 𝑓 𝑦𝑘 1 . 15 = 600+0.3×400 1.35𝐺 𝑘 +1.5𝑄 𝑘 2611 500 1 . 35 × 600 + 1 . 5 × 40 2 71 5 1 . 15 = 213 N/mm 2 For design crack width 0.3 mm: Maximum allowable bar spacing = 200 mm Actual bar spacing = 3000−2 35 −12 23 = 127 mm  200 mm Cracking OK Max bar spacing

38. Detailing 3000 3000 24H12 24H12 3000 450 2 4 H12 Plan View Section View

39. Example 2 PAD FOOTING (AXIAL LOAD & MOMENT)

40. Design Life = 50 years (Table 2.1: EN 1990) Exposure Class = XC3 f ck = 30 N/mm 2 f yk = 500 N/mm 2  soil = 150 N/mm 2 Unit weight of concrete = 25 kN/m 3 Assumed  bar = 12 mm Axial Force, N = 1500 kN Moment = 50 kNm B h Column size: 250  350 mm H

41. Durability & Bond Requirements Min cover regards to bond, c min,b = 12 mm Min cover regards to durability, c min,dur = 25 mm Allowance in design for deviation,  c dev = 10 mm Nominal cover, c nom = c min +  c dev = 25 + 10 = 35 mm  c nom = 35 mm c min = 25 mm

42. Si z e Service axial, N Service moment, M Assumed selfweight 10% of service load , W = 1500 kN / 1.40 = 1071 kN = 50 kNm / 1.40 = 36.1 kNm = 100 kN 𝛾 𝑠𝑜𝑖𝑙 𝑁 + 𝑊 1 071 + 107 . 1 150 Area of footing required = = = 7.85 𝑚 2  Try footing size, B  H  h = 2.80 m  3.50 m  0.65 m Area, A = 9.80 m 2 Selfweight, W = 9.80  0.65  25 = 159 kN

43. Size (Continued) 𝐼 𝑥𝑥 3 3 = 𝐵𝐻 = 2.8×3.5 = 10.0 m 4 12 12 𝑁 + 𝑊 𝐴 + 𝑀 𝑦 𝐼 = 1071+159 9 . 80 + 5 × 1 . 75 1 . 𝑦 = 𝐻 = 3.5 = 1.75 m 2 2 Maximum soil pressure, 𝑃 = = 132 kN/m 2  150 kN/m 2  OK B H x x

44. A n al y sis Ultimate soil pressure, 𝑃 = 𝑁 ± 𝑀𝑦 = 1500 ± 𝐴 𝐼 9.80 10 . 50×1.75 = 153 ± 8.7 kN/m 2  P min = 144 kN/m 2 and P max = 162 kN/m 2 1.575 m 1.575 m 144 1.275 x 0.35 m x y y 162 154 144 162

45. Analysis (Continued) 𝑀 𝑥𝑥 = 154 × 1.575 2 2 + 162 − 154 1 . 5 75 2 2 × 1.575 × 3 = 197 kNm/m  2.80 m = 553 kNm 2 2 𝑀 = 𝟏𝟓𝟑 × 1.275 = 124 kNm/m  3.50 m = 𝑦𝑦 435 kNm 1.575 m 1.575 m 144 0.35 m 1.275 x x y y 162 154 144 + 162 = 2

46. Effective Depth d x = h – c – 0.5  bar = 650 – 35 – (0.5  12) = 609 mm d y = h – c – 1.5  bar = 650 – 35 – (1.5  12) = 597 mm Main Reinforcement – Longitudinal Bar 𝐾 = 𝑀 𝑥𝑥 553×10 6 𝑓 𝑐𝑘 𝑏𝑑 2 30×2800×609 2 bal = = 0.018  K = 0.167  Compression reinforcement is NOT required 𝑧 = 𝑑 0.25 − 𝐾 1.134 = 0.98𝑑  0.95 d 𝐴 𝑠 , 𝑟𝑒𝑞 = 𝑀 𝑥𝑥 0.87𝑓 𝑦𝑘 𝑧 = 553×10 6 0.87×500×0.95×609 = 𝟐𝟏𝟗𝟕 mm 2

47. Minimum & Maximum Area of Reinforcement 𝐴 𝑠 , 𝑚𝑖𝑛 = 0.26 𝑓 𝑐𝑡𝑚 𝑓 𝑦𝑘 2 . 90 500 𝑏𝑑 = 0.26 0.0013𝑏𝑑 ≥ 0.0013𝑏𝑑  A s,min = 0.0013 bd = 0.0013  2800  609 = 2217 mm 2 A s,max = 0.04 A c = 0.04 bh = 0.04  2800  609 = 72800 mm 2 Since A s  A s,min , Use A s,min = 2217 mm 2 Provide 21H12 ( A s = 2375 mm 2 )

48. Main Reinforcement – Transverse Bar 𝐾 = 𝑀 𝑦𝑦 435×10 6 𝑓 𝑐𝑘 𝑏𝑑 2 30×3500×597 2 bal = = 0.018  K = 0.167  Compression reinforcement is NOT required 𝑧 = 𝑑 0.25 − 𝐾 1.134 = 0.99𝑑  0.95 d 𝐴 𝑠 , 𝑟 𝑒𝑞 = 𝑀 𝑦𝑦 0.87𝑓 𝑦𝑘 𝑧 = 435×10 6 0.87×500×0.95×597 = 𝟏𝟕𝟔𝟓 mm 2

49. Minimum & Maximum Area of Reinforcement 𝐴 𝑠 , 𝑚𝑖𝑛 = 0.26 𝑓 𝑐𝑡𝑚 𝑓 𝑦𝑘 2 . 90 500 𝑏𝑑 = 0.26 0.0013𝑏𝑑 ≥ 0.0013𝑏𝑑  A s,min = 0.0013 bd = 0.0013  3500  597 = 3147 mm 2 A s,max = 0.04 A c = 0.04 bh = 0.04  3500  597 = 91000 mm 2 Since A s  A s,min , Use A s,min = 3147 mm 2 Provide 28H12 ( A s = 3167 mm 2 )

50. (i) Vertical Shear Critical shear at 1.0 d from face of column: Average pressure at critical section: = 144 + 2.891 3 . 50 × 18 = 159 kN/m 2  Design shear force, V Ed = 159  0.966  2.80 = 431 kN 144 2.891 162 159 0.9 6 6 d = 0.609 2.8 Note: Bar extend beyond critical section at = 966 – 35 = 931 mm  𝑙 𝑏𝑑 + 𝑑 = 36∅ + 𝑑 = 36 × 12 + 609 = 1041 mm  A sl = 0 mm 2

51. (i) Vertical Shear 𝑘 = 1 + 𝑑 200 = 1 + 609 200 = 1.57  2.0 𝑙 𝜌 = 𝐴 𝑠 𝑙 𝑏𝑑 =  𝑉 𝑅𝑑,𝑐 = 0.12𝑘 100𝜌 𝑙 𝑓 𝑐𝑘 1/3 𝑏𝑑 = 0.12 × 1.57 100 × 0 × 30 1/3 2800 × 609 = 0 N = 0 kN  𝑉 𝑚𝑖𝑛 = 0.035𝑘 3/2 𝑓 𝑐𝑘 𝑏𝑑 = 0.035 × 1.57 3/2 30 2800 × 609 = 644949 𝑁 = 645 kN V Ed (430 kN)  V min (645 kN)  OK

52. (ii) Punching Shear Critical shear at 2.0 d from face of column: Average 𝑑 = 60 9 + 5 9 7 2 = 603 mm  2 d = 1206 mm Control perimeter; u = 2(350 + 250) + (2   1206) = 8779 mm Area within perimeter; A = (0.35  0.25) + (2  0.35  1.206) + (2  0.25  1.206) + (   1.206 2 ) = 6.10 m 2 3500 2 d = 1206 350 250 2 d = 1206 69 369  𝑙 𝑏𝑑 + 𝑑 = 36∅ + 𝑑 = 36 × 12 + 609 = 1041 mm  Reinforcement NOT contributed to punching resistance 2800

53. (ii) Punching Shear Average punching shear force at control perimeter: V Ed = 153 [(2.80  3.50) – 6.10] = 566 kN 𝑣 𝐸 𝑑 Punching shear stress: 𝛽𝑉 = 𝐸𝑑 𝑢𝑑 Where 𝛽 = 1 + 𝑘 𝑀 𝐸𝑑 𝑉 𝐸𝑑 𝑢 1 𝑊 1 𝑐 2 250 k = 0.65  𝑐 1 = 350 = 1.4 W 1 = 0.5𝑐 1 2 + 𝑐 1 𝑐 2 + 4𝑐 2 𝑑 + 16𝑑 2 + 2𝜋𝑑𝑐 1 = 0.5(350 2 ) + 350 × 250 + 4 × 250 × 603 16 603 2 + 2𝜋 × 603 × 350 = 7.9  10 6 mm 2  𝛽 = 1 + 0.65 50×10 6 8779 566×10 3 7.9×10 6 = 1.06 Therefore, 𝒗 𝑬𝒅 𝟖𝟕𝟕𝟗 × 𝟔𝟎𝟗 𝟑 = 𝟏.𝟎𝟔×𝟓𝟔𝟔×𝟏𝟎 = 𝟎. 𝟏𝟏 N/mm 2 + Soil pressure = 153 kN/m 2 A all = 2.8  3.5 m A peri meter = 6.1 m 2

54. (ii) Punching Shear Punching shear resistance: 𝑘 = 1 + 𝑑 200 = 1 + 609 200 = 1.57  2.0 𝑣 𝑅𝑑,𝑐 = 𝑣 𝑚𝑖𝑛 = 0.035𝑘 3/2 𝑓 𝑐𝑘 1/2 = 0.035 1.57 3/2 30 1/2 = 0.38 N/mm 2  v Ed (0.11 N/mm 2 )  OK Soil pressure = 153 kN/m 2 A all = 2.8  3.5 m A perimeter = 6.10 m 2

55. (iii) Maximum Punching Shear at Column Perimeter Maximum punching shear force: V Ed,max = 1500 kN Column perimeter, u o = 2(350 + 250) = 1200 mm Punching shear stress: 𝐸 𝑑 𝛽𝑉 𝐸𝑑 𝑣 = W h e r e 𝑀 𝐸𝑑 𝑉 𝐸𝑑 𝑢 𝑜 𝑊 1 𝑢 𝑜 𝑑 𝛽 = 1 + 𝑘 k = 0.65  𝑐 2 250 𝑐 1 = 350 = 1.4 W 1 = 0.5𝑐 1 2 + 𝑐 1 𝑐 2 = 0.5(350 2 ) + 350 × 250 = 0.15  10 6 mm 2 Soil pressure = 153 kN/m 2 all A = 2.8  3.5 m = 9.8 m 2 A column = 0.09 m 2

56. (iii) Maximum Punching Shear at Column Perimeter  𝛽 = 1 + 0.65 50×10 6 1200 1500×10 3 0.15×10 6 = 1. 17 Therefore, 𝒗 𝑬𝒅 = 𝟏.𝟏𝟕×𝟏𝟓𝟎𝟎×𝟏𝟎 𝟑 𝟏𝟐𝟎𝟎 × 𝟔𝟎𝟑 = 𝟐. 𝟒𝟒 N/mm 2 Maximum shear resistance: 𝑣 𝑅𝑑,𝑚𝑎𝑥 = 0.5 0.6 𝑓 𝑐𝑘 1 − 𝑓 𝑐𝑘 25 1 . 5 30 30 = 0.5 0.6 1 − 25 1 . 5 = 5.28 N/mm 2  v Ed  OK Soil pressure = 153 kN/m 2 all A = 2.8  3.5 m = 9.8 m 2 A column = 0.09 m 2

57. Cracking h = 650 mm  200 mm = 0.6 1 . 15 𝑓 𝑦𝑘 𝐴 𝑠,𝑟𝑒𝑞 𝐴 𝑠,𝑝𝑟𝑜𝑣 500 2 1 97 1 . 15 2375 = 0.6 = 241 N/mm 2 For design crack width 0.3 mm: Maximum allowable bar spacing = 150 mm Actual bar spacing at x - x = 2800−2 35 −12 20 = 136 mm  150 mm ctual bar spacing at y - y = 3500−2 35 −12 27 = 126 mm  150 mm Cracking OK Max bar spacing Assume steel stress is under quasi-permanent loading:

58. Detailing 3500 2800 2 8 H12 21H12 3500 650 2 8 H12 Plan View Section View 21 H 1 2