Rehabilitation of Existing Buildings Constructed on some Poor Soils in Abu Dhabi Emirate

1. United Arab Emirates University College of Engineering Training and Graduation Projects Unit Department of Civil and Environmental Engineering Rehabilitation of Existing Buildings Constructedon some Poor Soils in Abu Dhabi Emirate Advisor : Dr. AshrafBiddah Halima Ahmed Al-Jasmi 200310703 MozaKhamisKhalfan 200205346 Moza Al Marshoudi 200210427 Marwa Abdullah Abbas 200337349 2nd.Semester 2009

2. Dr. AshrafBiddah Dr. KhalilCharif Dr. Ali Taleb Acknowledgments Dr. Amer El-Dieb Eng. Eesa Al Mansori

3. Outline • Objectives • Introduction • The problem • Selection of building • Rehabilitation of Soil And Building • Demolishing & Rebuilding the Structure • Cost Calculation • Conclusion and Recommendations

4. Objectives • Explore the soil problems in UAE • Structure case study • Identify the most suitable solution for improving the soil • Solution1 :soil rehabilitation and building repair • Solution2: demolishing the existing affected building, soil improvement and rebuilding the structure • Compare the cost & time for both solutions

5. Introduction Early 90’s with occurrence of non-traced cavities and the construction processes of filling the areas (more than 400km2) with un-compacted organic and non organic materials . The ground water level changed due to the massive development in the area. • Introduction

6. Soil problems: • Cavities • High water table level • Existence of salty layers • The problems

7. Topography of Khalifa B • Location: south east of Abu Dhabi Island, to the south west of Abu Dhabi international airport and south Khalifa (a) area. • Area: 42 Km2 . • Soil texture : 95% sandy soil ,silt and clay. • Bearing capacity: 50 KN/m2 • Type of foundation : Shallow foundation. • Type of construction : Residential area.

9. shallow foundation • Shallow foundations (spread footings): • Shallow foundations are those founded near to the finished ground surface. • Shallow foundations are inappropriate in weak or highly compressible soils like soils found in Abu Dhabi Emirate ,poorly-compacted fill soil and peat. • Could be used when soil’s surface is stiff and adequately strong.

10. Deep foundations are to be placed too deeply below the finished ground surface • Used to transfer loads from structures from an upper weak layer of soilto a stronger deeper layer of soil. • Deep foundations

11. Khalifa B Location • Selected Villa

12. Architectural& structural design of selected Villa

13. The ground floor consists of • One bed room, • One dining room, • kitchen, • majlis, • One family saloon, • 3 bathrooms, • 1 store • The second floor consists of • Five bed rooms, • One family saloon, • Four bathrooms

14. Columns plan

15. Building Problems Crack in wall Shear in column Honeycombing Shear in beam Crack in Stairs

16. Rehabilitation of Soil And Building

17. Soil Rehabilitation • The alteration or preservation of one or more soil properties to improve the engineering characteristics and performance of a soil. • Soil Injection. • Micro Piles. Methods of soil rehabilitation:

18. Soil Injection • Examples of injection fluids are silicates, water-based injection resins, polyurethanes, bitumen and cement-based injection fluids. Solution 1

19. Micro Piles are small diameter piles comprised of a central steel component with an annular surround of grout that is in contact with the soil. • Micro Piles • solution 1

20. Rehabilitation of Existing Building Problems: • Shear in beams. • Shear in columns. • Cracks walls. • Punching. • Cracks in stair cases.

21. Repair of Honeycombing (solution 1) • Definition: voids in concrete caused by the mortar not filling the spaces between the coarse aggregate particles. • Usually happens when the formwork is stripped, revealing a rough and 'stony' concrete surface with air voids between the coarse aggregate.

22. Rehabilitation of Shear in Beams (solution 1) • Definition: Shear failure is actually a diagonal tension failure that is brittle in nature and should be avoided. • Shear failure in beams is parallel to the bending moment diagram.

23. Shear repairing techniques There are two methods: 1. External Rods with bottom plates. 2. Fiber ( Carbon fiber, glass fiber). Method 1:

24. In the normal condition: • Due to the shear failure: • With strengthening: • Given: Solution:

25. Method 2 • It involves rapping sheets of the fiber around the beam in layers. • Each layer comes with a certain thickness and length according to the shear crack size.

26. There are two types of fibers: Repairing shear cracks in beams with CFRP:

27. Rehabilitation of Shear in Columns solution 1 Basic elements of column enlargement • There are two methods for repairing shear in columns: • Enlargement (Jacketing). • Fiber ( carbon fiber, glass fiber).

28. By using ties: • In this selected building we have two cases of column to solve for: • When the column is rectangular; column dimension is 200X400. • When the column is circular with a diameter of 40.

29. Strengthening of columns using Fiber method • Great deal of work has been carried out on the strengthening of columns using external fiber reinforced polymer (CFRP) sheets which are wrapped around a column section.

30. Case no.1 rectangular column:

31. Case no.2 rectangular column:

32. Case no.3 circular column:

33. Demolishing, Soil Improvement & Rebuilding the Structure (solution2)

34. Demolishing & Rebuilding the Structure (solution2) • One of the solutions to solve the problem of the existing buildings at Khalifa B which are effected by the movement of soil under the buildings, is to: Demolish the building. Improve the soil. Rebuild the building again.

35. Demolishing the existing building • Demolition is a process in which an existing building or any such structure is pulled down using some machines, tools and equipments. • There can be various methods employed for the purpose of demolition such as; 1- Manual Demolition 2- Mechanical Demolition 3- Explosives • Introduction

36. Demolition Manual Demolition Mechanical Demolition Explosives The use of explosives for demolition requires a specialist who is competent in the application of explosives for demolition work 1- Impact hammers to break up a mass structure (eg, rocks, bricks and concrete). 2- Wire and chain pulling down parts of structures Demolition involves the use of hand held tools to gradually reduce the height of the structure

37. GeoCrete is a whitish powder, consisting of alkaline and alkaline earth elements as well as other complex compounds. • GeoCrete can be successfully applied in : Road and highway construction, pedestrian andgeneral foundations. GeoCrete advantages: • Short construction time. • Higher compression strength. • Less strain on the environment. • Very stable, durable and lower costs of repair. • Soil improvement after demolishing

38. Redesign of Structural Elements (solution2 ) Beam B1 Beam B3 Hurdi Slab

39. Design of Beams • Continuous beams are beams supported on more than two supports and reinforced in tension and • compression zones. • The design of a continuous beam involves three major steps: • Load calculation (or load distribution). • Structural analysis and evaluation of bending moments and shearing forces. • Design of critical sections and reinforcement arrangement.

40. Given Weight of beam = bw* (h-hs)δconcrete Slab rectangularity (r) = Long dimension (L) / Short dimension (Ls) Load on wall = Wt/m3*thickness *height W = W D.L + W L.L

41. Drawing the bending moment and shear force diagram on beam B1 Wul2/16=37.39 Wul2/14= 42.73 Wul2/14= 42.73 B.M.D (kN.m) Wul2/16=37.39 Wul2/16=37.39 Wul2/16=37.39 Wul2/9=66.48 Wul2/9=66.48 Wul/2=55.06 Wul/2=55.06 Wul/2=55.06 Wul/2=55.06 S.F.D (kN) 1.15Wul/2=63.32 1.15Wul/2=63.32 Wul/2=55.06

42. Design of critical: • Sec. I: Mu = 66.48 kN.mb = 200 mm R-sec. • hmin to limit deflection = l / 18.5 = 4200 / 18.5 = 227 mm • Assume ρ = 0.5ρb = 0.5 x 0.0283 = 0.014 • Mu / (ρ bd2) = 5.90 ρ = 0.0165 • As = ρbwd = 0.0165 x 200 x 250 = 825 mm2Use 3 # 19 • Asmin = 0.0033 x 200 x 250 = 165 mm2 < As o.k. 180mm 2#12 700mm 3#19 200mm

43. Using PROKON software to get B.M.D & S.F.D

44. Design of Ribbed Slab • Concrete Hurdi (hollow block) or (joist) reinforced concrete RC slabs consists of a combination of regularly spaced RC ribs and a top RC slab. • It may be constructed with permanent or removable blocks between ribs which can be arranged in one or two directions. Two way Hurdi slabs with RC ribs in two directions commonly referred as waffle slab construction.

45. slab 5.2 x 5.2 m Loads/unit area (1.10 m x 1m) : Total dead load = 7.97 kN/unit area Total live load = 2.42 kN/unit area Wu = 1.2 WD.L + 1.6 WL.L

46. Using Grashoff coefficients to calculate loads on ribs in 2 directions L short and L Long as follows ; r= 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1. 90 2.00 a= 0.50 0.595 0.672 0.742 0.797 0.834 0.869 0.893 0.914 0.928 0.941 b= 0.50 0.405 0.328 0.258 0.203 0.166 0.131 0.107 0.086 0.072 0.059 L Long / L short = 5.2/5.2 = 1 a= 0.5 , b=0.5 Check; ∑ a +b = 1 , 0.5 +0.5 = 1 Loads on ribs in short and load direction; Wa in short direction = Wa*a = 13.43 *0.5 = 6.715 kN/unit area Wbin long direction = Wb* b = 13.43 *0.5 = 6.715 kN/unit area

47. L=5.2 Ls=5.2 In each direction we have 2 ribs, load /one rib is calculated for both long and short direction as follows; Wu / one rib = 13.43/2 = 6.715 kN/m Mu= Wul2/8 = 13.43*(5.2)2/8 = 45.39 kN.m Mu= Wul2/8 = 13.43*(5.2)2/8 = 45.39 kN.m • Con.. (solution 2)

48. Design of columns (solution2):

49. There are two types of columns: • Braced columns: where there are bracing elements together with the columns to take care of the wind load • Un-braced sway columns: Slender long columns, these types of columns take care of part of the lateral loads like the wind load

50. If: Con …(solution2)