Out line

2. Out line 1) Determine the own weight of building 2) Design of mat foundation 3) Design of pile foundation

3. INTRODUCTION A foundation is the lower part of a structure which transmits loads to the underlying soil without causing a shear failure of soil or excessive settlement.

4. INTRODUCTION if cracks appear in the structure it is assumed that the foundation did move and that this is the sole cause of cracking.

5. Choice of the Type of Foundation • The choice of the appropriate type of foundation is governed by some important factors such as • 1. The nature of the structure • 2. The loads exerted by the structure • 3. The subsoil characteristics • 4. The allotted cost of foundations

6. AL Quqa building

8. Cross section of the soil

9. Calculate the weight of the building

11. 3 D On SAP

12. Table of loads 9 7310.13 KN 1008.8 KN

13. MAT FOUNDATION

14. The thickness of mat foundation was calculated using check for punching in the next calculation. The critical columns were found to be: C14, C 13, C12. Critical columns

15. 0.4 1.0 1.28 2.28 Thickness of MAT Foundation Assume h = 0.95 m , d = 0.88 m For column 12:- Φ Vcp = 075 (0.33) (400+880+1000+880)*2*(880/1000) = 7283.7 > 6704.9 Ok

16. 1.48 0.6 1.48 0.6 Thickness of MAT Foundation For column 14:- h = 0.95 m , d = 0.88m ΦVcp = 0.75 (0.33) *(1480+1480) *2*880/1000 = 6822.74 KN > 5531.7 Ok

17. 2.28 1.4 0.4 1.28 Thickness of MAT Foundation For column 13:- h=0.95 m , d = 0.88 m ΦVcp = 0.75 (0.33) * (2280+1280) *2* = 8205.7 > 8015.9 Ok So we will select depth of the footing equals to 0.95 m.

18. Check pressure under footing F=KΔ Where; K= 210*100= 21000. F = 210 kN. Then Δ will be 0.01 m.

19. Column and Middle strip in X-direction Column strip Middle strip

20. Moment distribution in X-direction

21. Reinforcement Top steel: ρ1 = (1 - ) = 0.00387 As1 = 3405.6 mm2 → 1Φ25/150 mm ρ2 = 0.003 → As2 = 2640 mm2 = 1Φ25/150 mm Bottom steel: ρ1 = ρ min = 0.0018 As1 = 1710 mm2 ρ2 = ρ min = 0.0018 As2 =1584 mm2 → 1Φ20/200 mm

22. Reinforcement of Mat Foundation in the horizontal direction

23. Qall 8.3m 35*20 7m Settlement of Mat foundation

24. Also the soil has been assumed to act as a Normally Consolidated Clay (NCC) and some of its characteristics were taken from the soil report to compute the settlement : γ = 18 KN/m3 w% = 6.8% Gs = 2.65 γw = 9.81 KN/m3 e = 0.54 where γ = LL = 21 (avg. of LL for all the layers in the specific depth) Cc = 0.009(LL – 10) = 0.099

25. σ = (γ *8.3) + (γ*3.35) = (18*8.3) + (18*3.5) = 212.4 KN/m2 Qall = Qult= 71556.9 KN (sum of the ultimate axial load on the columns) Qall = 51112.07 KN ∆σ = (∆σt + 4∆σc + ∆σb) (by using 2:1 method)

26. ∆σt= = = 73.02 KN/m2 (where z = 0) ∆σc= = 56.5 KN/m2 (where z = 3.5) ∆σb = = 45.1KN/m2 (where z = 7) ∆σ = 57.34 KN/m2

27. SETELMENT EQUAL TO:- Sc = Hclog Sc = *8000 log = 4.6 cm < 5

28. Pile foundation

29. Bearing capacity of pile Qu = Qs + Qp Where; Qs = ultimate capacity of single pile due to point bearing Qp = ultimate capacity of single pile due to skin friction

30. Bearing capacity of pile • Using Ø = 21º c = 22 KN/m²

31. Bearing capacity of pilefrom all-pile:-

32. Cap design • Principles: • Distance between piles = 3D Where D is the diameter of pile • Minimum distance between edge of cap & edge of pile is more that 15 cm

33. Two pile cap

34. Three pile cap

36. Design for flexure M+ve = 1852.5 KN.m M-ve = 285.52 KN.m ρ+ = 0.0029 < 0.0033 Asmin = 0.0033*1000*1300 = 4290 mm2 (Use 1Ø25/110mm) ρ- = 0.003 As =1000*1300*0.0033 = 4290 mm2 (Use 1Ø25/110mm)

37. Design for shear Vc= (1/6) b*d = (1/6) (1000) (1300)/1000 = 1146.5 KN Vn = = = 2535.3 KN Where; 1901.5 KN is the ultimate shear on the pile taken from SAP Vs = Vn – Vc = 2535.3-1146.5= 1388.8 KN = = = 2.5 Assume using Ø = 16 for stirrups Av = 2 * 201 = 402 mm2 Av/S = 2.5 S = 160 mm (Use 1Ø16/250mm)

39. Pile reinforcement 13 Ø 16 As 0.005( /4)(800) = 2513 mm2 = 13Ø16 1Ø 10/ 120 mm

40. The end Thank you for listening