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An-Najah National University Engineering College Civil Engineering Department

An-Najah National University Engineering College Civil Engineering Department. Graduation Project Three Dimensional analysis And Design Of AL-ARAB HOSPITAL. Supervised by: Ibrahim M ohammad A rman. Objective. Scientific benefit

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An-Najah National University Engineering College Civil Engineering Department

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  1. An-Najah National UniversityEngineering CollegeCivil Engineering Department Graduation Project Three Dimensional analysis And Design Of AL-ARAB HOSPITAL Supervised by: Ibrahim Mohammad Arman

  2. Objective • Scientific benefit • compiling of information which were studied in several years of studying and styling it in a study project. • Analysis and study of an existing building

  3. Contents • CH.1 : Introduction • CH.2 : Preliminary Design • CH.3 : 3.D modeling and Final Design • CH.4 : Stairs And Ahear Walls Design • r • eliminary Design • PrPreliminary Designeliminary Design

  4. Chapter OneIntroduction

  5. CH.1 introduction Project Description Fourteen floor building, with area 1586.6m² for each floor At Al-Rayhan_suburb in Ramallah city. Soil bearing capacity 400 kN/m² The building consists of two parts separated by a structural joint. The building will be designed as Waffle slab with hidden beams system.

  6. CH.1 introduction Design Determinants Materials Concrete: For slabs and beams: concrete B400 F`c = 32 MPa For Columns: concrete B450 F`c = 36 MPa Reinforcing Steel: Steel GR60 Fy = 420 MPa

  7. CH.1 introduction Design Determinants Loads Gravity loads: Dead loads: static and constant loads. Including the weight of structural elements. … super imposed dead loads (SDL) = 4.3 kN/m2. Live loads: that depend on the type of structure and include weight of people, machine and any movable objects in the building . ... (LL) was considered to be 4 kN/m2. Lateral loads : Earthquake: seismic factor for zone 2A = 0.15 , (Z=0.15) risk category (IV)… then importance factor, I = 1.5 response modification coefficient, R = 4.5 Soil: Ko =coefficient of lateral earth pressure at rest.= 0.5 γs= unit weight for soil = 20 kN/m3.

  8. CH.1 introduction Design Determinants Codes and standards: ACI 318-08 : American Concrete Institute. IBC-09: International Building code . Jordanian code 2006 ASCE 7 – 10 : American Society of Civil Engineers2010. Isra’el standard SI 413 – 1995, amendment no.3 – 2009

  9. Chapter TwoPreliminary Design

  10. Preliminary Design Plan View

  11. Preliminary Design Comparison : Which system is more suitable to use ?? • Two-way solid slab with drop beams • Two-way waffle slab with hidden beams

  12. Preliminary Design Plan View

  13. CH.2 Preliminary Design Two-Way Solid Slab With Drop Beams dimensions: From largest panel (8.1m8.45m) thickness of slab determined, h = 230mm From longest beam (9.04)m beam depth =750 mm and width = 500 mm

  14. CH.2 Preliminary Design Waffle Slab Slab thickness: Thickness of slab, hmin=(Ln/33) = 252.7mm Assume frame dimensions= 600 mm600 mm320 mm • flange width = 820 mm • flange depth = 80 mm • web width = 150 mm • web depth = 320

  15. CH.2 Preliminary Design Waffle Slab checks Iwaffle =0.001656426 m4 = Isolid = hsolid= 0.28940974 m Frame dimension OK

  16. CH.2 Preliminary Design Waffle Slab Beams dimensions: Hidden beams • Beams from (B 1 - B 19) Width=700mm & Depth = 400mm • Beams (B 20,B 21,B 22) Width =900m m & Depth = 400mm • Beams (B 23, B 24, B 25) Width = 1300mm & Depth = 400mm

  17. CH.2 Preliminary Design Design of Column Column (27.N). Ultimate load : Pu=Wbeam+Wslab+own weight for column =629.627 = 629.62714(numberof stories)=8814.778 kN To use short column equation, assume column is short column and non sway. Pu = Φ Pn= λ Φ {0.85f’c (Ag – As) + (As fy )} Where ; λ = 0.8 Φ = 0.65 As :area of steel Ag : gross area Assume steel ratio ρ=Ast/Ag=1% As = 0.01Ag Ag = 1042600.497 mm2 Assume rectangular columnL = b = 1021 mm …. Use column (1.1 m1.1 m) .

  18. CH.2 Preliminary Design Design of Column check • Checks for short column • KLu/r ≤32-(M1/M212) ≤ 40 • Where ; • K = effective length factor depending on restrainsts • r = radius of gyration , r= (I/A)(1/2) • Lu = clear length of column (face to face of span ) • Assum : • K = 1 and M1/M2= -1 (double curvature) • KLu/r = 1(4.16-0.4)/(0.3h) = 11.39 • 32-((M1/M2)12) = 32+12 = 44 • 11.39˂ 40 • Column is short

  19. CH.2 Preliminary Design Cost Analysis Good quality and minimum cost are necessary requirements in an engineering design. Two system satisfies the good quality (solid slab with drop beams , waffel slab with hidden beams ). Cost snalysis to detrmin economical system

  20. CH.2 Preliminary Design Cost Analysis Solid slab system

  21. CH.2 Preliminary Design Cost Analysis Waffle system It is clear that the coast of material for waffle slab is less than that of the solid slab.

  22. Chapter Three3D. Modeling

  23. CH.2 3.D Modeling 3D.Modeling slab thickness in preliminary design=400mm BUT some beams were unsafe in preliminary design dimentions because of additional internal forces due to seismic loads

  24. CH.2 3.D Modeling 3D.Modeling

  25. CH.2 3.D Modeling 3D.Modeling Load Gravity loads: • Live loads (4 kN/m2) • Super imposed dead load (SDL) 0.03m27kN/m3+0.0223kN/m3+0.1518kN/m3+0.01523kN/m3 = 3.3+1=4.3 kN/m2 • Masonry wall weight: • = (0.05×27) + (0.13×25) + (0.02×0.3) + (0.1×12) + (0.02×23) • =6.266 KN/M2 × storey high (4.16m) • =26.1 kN/m

  26. CH.2 3.D Modeling 3D.Modeling Load Lateral loads • Seismic loads : Response spectrume • Soil loads q1 Assume Ø =30 o so: ko = 1-sin Ø=0.5 for one story, h= 4.16m • q1at z=4.16 m=koW=0.5x15=7.5 kN • q2at z=0.0m = koW+γhKo= 0.5x15+20x4.16x0.5=49.1 kN. q2

  27. CH.2 3.D Modeling 3D.Modeling Input load data in sap model Gravity loads: Uniform loads on slab : weight of masonry wall as distributed uniform dead load

  28. CH.2 3.D Modeling 3D.Modeling Input load data in sap model Lateral loads: Seismic loads (Response Spectrum) information for Response Spectrum definition

  29. CH.2 3.D Modeling 3D.Modeling Input load data in sap model Lateral loads: Response Spectrum in x-direction Response Spectrum Response in X-direction and 30% in Y-direction

  30. CH.2 3.D Modeling 3D.Modeling Load combination UDCON1 = 1.4D.L UDCON2 = 1.2D.L+1.6L.L+1.6SOIL UDCON3 = 1.2D.L+1L.L+1EX UDCON4 = 1.2D.L+1L.L+1EY UDCON5 = 1.2D.L+1L.L+1EZ UDCON6 = 0.9D.L+1EX+1.6SOIL UDCON7 = 0.9D.L+1E.Y+1.6SOIL UDCON8 = 0.9D.L+1E.Z+1.6SOIL

  31. CH.2 3.D Modeling 3D.Modeling checks Compatibility:

  32. CH.2 3.D Modeling 3D.Modeling checks Equilibrium: • Hand calculation: • Total weight = 212672.35 kN • From SAP

  33. CH.2 3.D Modeling 3D.Modeling checks Stress- strain relationship:

  34. CH.2 3.D Modeling 3D.Modeling checks Stress- strain relationship: Hand calculations : • (wuln²/8)for slab=18.76(7.625-0.6)7.26²/8=868.3 kN.m • (wuln²/8)for beam =(0.6.045251.2)+(1.24.30.6)+(1.640.6)7.26²/8 =99

  35. CH.2 3.D Modeling 3D.Modeling checks Stress- strain relationship: SAP result To calculate moments from SAP = ( M(-ve)+M(-ve))/2+ M(+ve) = (523.4+565)/2+423.4=967.6kN.m Error percentage=(967.6-967.3)/967.3=0%

  36. CH.2 3.D Modeling 3D.Modeling checks Shear in Slab: • Rib shear strength == 62225.39 N = 62.225 kN values = 62.273 x 0.82 = 51.1 kN/0.82m.< 62.225 kN ... OK

  37. CH.2 3.D Modeling 3D. Modeling & Design design Design of column strip in slab : M(-ve) (kN.m/0.82m) = 84.64 0.82 = 69.4048 kN.m/0.82 ρ =0.00815 > ρmin=0.0033 As=0.00815150400=489mm2 ……Use steel bars as 2Ø18/rib

  38. CH.2 3.D Modeling 3D. Modeling & Design design Design of column strip : M(+ve) =(kN.m/0.82m) = 66.5346 0.82 = 54.5583 kN.m/0.82m ρ0.0011 As=0.0011820400=360.8mm2 Asmin=ρminbwebd = 0.0033150400= 198 mm2 < 360.8 mm2…. Use 2Ø16/rib

  39. CH.2 3.D Modeling 3D. Modeling & Design design Design of middle strip :

  40. CH.2 3.D Modeling 3D. Modeling & Design design Design of beams : Flexural steel for span between grids D.24 and D.27 Torsion in span between grids D.24 and D.27 stirrups reinforcement : • At both end of beams : Av+t/S=1.224+0=1.224 mm2/mm S=314/1.224=256.5mm ˃ 100mm so ….. Use 1ф10/100mm • Av+t at distance =2h from both end of beams Av+t/S=0.511+0=.511mm2/mm S=314/0.511=614.5mm ˃ 200mm so use 1ф10/200mm

  41. CH.2 3.D Modeling 3D. Modeling & Design design

  42. CH.2 3.D Modeling 3D. Modeling & Design design Design of columns : Column 8 (C8):Longitudenal bars =20 bar 2018mm=5080>4900

  43. CH.2 3.D Modeling 3D. Modeling & Design design Design of footing : service load in building Foundation area=Total service load/bearing capacity =569.4 m2 ˃ half area of building, 353m2 Use mat foundation

  44. CH.2 3.D Modeling 3D. Modeling & Design design Determine foundation thickness: maximum axial force =21932.85 kN фVCp=0.750.33bod • Where • c1&c2 : column dimensions • d=effective depth Try h=1600mm and check if it can resist bunching shear, So d=1530mm фVCp =22535kN ˃ 21932.85 kN …OK

  45. CH.2 3.D Modeling 3D. Modeling & Design design Design of mat foundation: M (-V) = 412.2kN.m/m ρ =0.00046 ˂ ρmin =0.0018 Use Asmin=0.001810001530=2754mm² ….use1Ø25/150 • M (+V)=-5234/3.6=1454kN.m/m ρ=0.00166 < ρmin =0.0018 Use Asmin=0.001810001530=2754mm²….use1Ø25/200

  46. CH.2 3.D Modeling 3D. Modeling & Design design Design of mat foundation:

  47. CH.3 Shear walls & stairs Chapter FourShear Walls &Stairs Design

  48. CH.3 Shear walls & stairs Shear Walls design Long shear walls: Vult (+ve)=1145/6.35=180.3kN/m ØVc=0.75=450kN /m˃ 434kN/m …ok M22=1182/6.35=186kN.m/m ρ0.00138 As=0.001381000600=828 mm2/m use1Ø12/250 mm for each side

  49. CH.3 Shear walls & stairs Shear Walls design Short shear walls: Asmin= (0.00123500200) =240mm use 14ф12/350mm

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