Shallow Foundations ( Combined, Strap, Raft foundation)

presentation for 3 types of shallow foundations :
- combined footing
- strap beam footing (neighbor's footing)
- raft foundation

1. SHALLOW FOUNDATIONS 2 Design of combined, strap & raft footing

2. INTRODUCTION

3. Contents : 1 2 3

4. 1. Combined Footing

5. Combined Footing A combined Footing is a long footing supporting two or more columns in (typically two) one row. A combined Footing is a rectangular or Trapezoidal shaped footing.

6. Using of Combined Footing Construction Practice may dictate using only one footing for two or more columns due to: Closeness of Columns Due to property line constraint, which may limit the size of footing at boundary.

7. The Design of Combined Footing requires that the centroid of the area be as close as possible to the resultant of the 2 column loads for uniform pressure and settlement. Which Means :

8. x y Eccentricity = ZERO

9. STEPS OF DESIGN

10. C1 C2 L

11. RC Max S.F.D B.M.D Zero shear

12. c1 c2

13. c1 c2 H.B.1 H.B.2

14. c1 c2

15. EXAMPLE C1 (30*70) C2 (30*90) 4 m Design a combined footing to support a working load of p1=160 ton & p2=220 ton. Bearing capacity: 12.5 ton/m2 Thickness of plain concrete 15 cm

16. Solution

17. 1- Dimensions of footing : R = p 1+ p 2 = 160 + 220 = 380 ton  R . X = p 2 . S   380 x X = 220 x 4  X = 2.32 m   L pc / 2 = X + b 1 / 2 + 0.5 + t pc   L pc / 2 = 2.32 + 0.7 / 2 + 0.5 + 0.3 = 3.47m    Lpc = 6.94 m L Rc = L Pc - 2 x t PC = 6.94 - 2 X 0.3 = 6.34m A PC = 380 / 12.5 = 30.4 L Pc X B Pc = 30.4 6.34 x B Pc = 30.4 B Pc = 4.8m B Rc = 4.8 - 2 x 0.3 = 4.2m

18. 2 Design of R.c footing   P 1u = 160 x 1.5 = 240 ton   P 2u = 220 x 1.5 = 330 ton   R u = 380 x 1.5 = 570 ton     W u = 570 / 6.34 = 89.9 ton   q u = 570 / 6.34 x 4.2 = 21.4 t/ ㎡

19. Design of Footing in Longitudinal Direction

20. At point of zero shear :   P 1u = w u . X 240 = 89.9 . X   X = 2.67m   M max = 89.9 x 7.1289 / 2 - 240 x 1.82 = 116.3   D =5 × √116.35×10^7 / 25 × 4200 = 526.3 ㎜   D = 530 ㎜

21. 2- check shear   Q su = 0.16√25/1.5 = 0.653 N/m ㎡   Q sumax = 155.56 - 89.99 × 0.53/2 = 131.7365   Q s = 131.73×10^4 / 530×4200 = 0.59 ≤ q su    Safe

22. c1 c2 1.23 1.43 0.83 0.83

23. For c 1 ( 30x70)   Q pcu1 = 0.316 ( 0.5 + 300 / 700) √25/1.5 = 1.197 Q pu1 = 240 - 21.4 x (1.23x0.83)=218.15 ton   Q pu1 = 218.15 x 10^4 / 530x2 (1230+830) =0.99 ≤ q pcu1  Safe For c2 (30*90) Q pcu2 = 0.316 x (0.5 + 300/900) √25/1.5 = 1.07   Q pu2 = 220 -21.4 (1.43x0.83) = 194.6 Q pu2 = 194.6 x 10^4 / 530 (1430+830)x2 = 0.81 ≤ 1.07  safe

24. Design of Footing in Short Direction

25. c1 c2 1.73 m 1.96 m

26. For hidden Beam1   Q u1 = 240 / 4.2 x 1.73 = 33.03t/ ㎡ M 1 = 33.03 * 1.95^2 /2 = 62.9 t.m For hidden Beam 2   Q u2 = 330 / 4.2 x 1.96 = 40.1t/ ㎡   M 2 = 40.1 x 1.95^2 / 2 = 76.2 t.m   d = 530 = c 1 √ 76.2x10^7 / 25 x 1000   C 1 = 3.1≥ 2.3

27. Reinforcement

28. RFT in long direction Astop = 116.35 x 107 / 360 x 0.826 x 530 = 72382mm2/4.2m =1757.7 mm2  7 ϕ 18 Asbottom = 48.61 x 107 / 360x 0.826 x 530 = 3084mm2 /4.2 = 734.3 mm2  4 ϕ 16

29. RFT in short direction As1 = 62.79 X 107 / 360 X 0.826 X 530 = 3984 mm2  11 ϕ 22 As2 = 76.2 X 107 / 360 X 0.826 X 530 = 4835 mm2  10 ϕ 25

30. 7 ϕ 18 4 ϕ 16 11 ϕ 22 10 ϕ 25 4 ϕ 16

31. c1 c2 7 ϕ 18 4 ϕ 16 4 ϕ 16 11 ϕ 22 4 ϕ 16 4 ϕ 16 10 ϕ 25 4 ϕ 16

32. 2. Strap Footing

33. Design a strap footing to support an exterior column (30*50cm) and an interior column (30*90cm). The un factored Loads C1 =685 KN, C2 =1270 kN . Assume the allowable Bearing capacity is 150 kN/m2, fcu=25 N/mm2 and Fy = 360 N/mm2 , P.C. thickness = 40 cm EXAMPLE

34. C2 C1 0.5 0.9 4.9 m

35. Calculation of reaction: Assume e=0.1 to 0.2 (L) e = 0.5 to 1m (take it 1m) R1 = = 860.6 KN R2=685 + 1270 - 860.6 = 1094.4 KN   C2 C1 0.5 0.9 e

36. Area of plain concrete footing : A1 = = 5.73 Dimension ((1+0.25)*2 ) = 2.5 (2.5*2.3 ) A2 = = 7.29 Dimension = ( 2.7*2.7 ) = 7.29   R.C dimensions: F1 = ( 2.1*1.5 ) m F2 = ( 1.9*1.9 ) m

37. C2 C1 0.5 0.9 0.5 0.9 0.5 0.5 2.1 1.6 M = 602 720 695 432 262 B.M.D S.F.D

38. Design of strap beam: qu1= = 614 kN /m’ qu2 = = 864 kN /m’ Point of zero shear : 614X-1027=0 , X=1.67m , Mmax=602kN.m d = 5 * = 1227m t=130cm d=123cm  

39. Reinforcement: Astop = =1646 = 7 /m ’ Asbot = = 295 Asmin = 0.15 % b*d = 0.15 %* 400*1230 =738 = 4  

40. CHECK SHEAR: Qs = ( 720*0.55)/ 1.17 = 341.53kN qs = = 0.694N / qcu =0.24 = 0.98N / qs  use min stirrups 5   0.55 0.62 720

41. Design of Footings

42. Critical Sec Footing 1: q u = = 409.5 kN / L1 = ( 1.5-0.4)/ 2 = 0.55 m Mu = 409.5 *( /2 ) = 61.9kN.m d = 5 = 249mm t = 35cm d = 28cm  

43. Check shear: Qs = 409.5 *(0.55-(0.28/2 )) = 168kN q s = = 0.6 MPa q cu = 0.16 = 0.65 MPa qs  safe As = =743.44 = 5  

44. Footing 2: qu = =455kN/ L2= =0.75m Mu=455* =128kN.m d=5 =358mm t=45cm d=38cm   Critical Sec

45. Check shear: Qs = 455 *(0.75- ) = 255KN q s = =0.67N/ qcu =0.65N /  qs  unsafe 0.65= d=392mm t=50cm d=43cm As = =1001 =5 Asmin = 1.5d = 1.5*430 = 645  

46. 3 ϕ 18 + 4 ϕ 18 = 7 ϕ 18 4 ϕ 16 5 ϕ 10 2 ϕ 12 2 ϕ 12

47. 0.4 m 2.3 m 1.5 m 2.1 m 2.5 m 2.7 m 1.9 m 5 ϕ 12 5 ϕ 12 5 ϕ 14 5 ϕ 16

48. 3. Raft Foundation