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Three Dimensional Analysis and Design of Aljawhara Office building

An-Najah National University Faculty of Engineering Civil Engineering Department Graduation Project:. Three Dimensional Analysis and Design of Aljawhara Office building. Prepared by: Ahmad Yahya Khalid Salameen Supervisor : Dr.Wael Abo Asab 2012.

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Three Dimensional Analysis and Design of Aljawhara Office building

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  1. An-Najah National University Faculty of Engineering Civil Engineering Department Graduation Project: Three Dimensional Analysis and Design of Aljawhara Office building Prepared by: Ahmad Yahya Khalid Salameen Supervisor : Dr.WaelAboAsab 2012

  2. SUBJECTS TO BE COVERED • Chapter One : Introduction • Chapter Two : Preliminary Design> • Chapter Three : SAP Output data • Chapter Four : Seismic Analysis and SW inclusion • Chapter Five : Dynamic Analysis • Chapter Six : Stair Design • Slab • Beams • Girders • Columns • Footings

  3. CH1.INTRODUCTION • About the project: This project introduces analysis and design of reinforced concrete office building in Ramallah with dimensions (48x40) m. This office building called "Aljawhara", consists of 7 stories: 6 stories above the ground level and one story under ground level used as car parking. The area of each story is about 1920 m². • Philosophy of analysis & design: SAP2000 V14 is used for analysis and ultimate design method is used for design of slab, slab is carried over drop beams and the beams is carried over 4 girders.

  4. General layout

  5. 3D Model

  6. INTRODUCTION…cont • Materials of construction: • >>Reinforced concrete: (ρ) = 25KN/m3 , The required compressive strength after 28 days is fc = 28 Mpa ( B350)for slab, beams and girders. fc= 40 Mpa ( B500) for columns and foundations >> fy =420 Mpa • >> Soil capacity = 3 kg/cm²

  7. INTRODUCTION…cont • loads: • Live load: LL=4 KN/m2 • Dead load: DL=(Calculated By SAP) = 4.25 KN/m2 “For 17 cm thick of solid slab” • SID= 3 KN/m2 • Masonry wall load = 22 KN/m for 3.5m floor height • Earthquake load: its represents the lateral load that comes from an earthquake.(calculated later)

  8. INTRODUCTION…cont • Combinations: Wu= 1.4D.L Wu= 1.2D.L+ 1.6L.L Wu= 1.2D.L +1.0L.L ±1.0E Wu= 0.9D.L ±1.0E • Codes Used: • American Concrete Institute Code (ACI 318-05) • American Society of Civil Engineers "ASCE standard". • Uniform Building Code 1997 (UBC97)

  9. Ch2:Preliminary design SLAB • One way solid slab is used : • Thickness of slab: t = 4/24 =16.66 cm use 17 cm ,d=14 cm • Slab consists of two strips (strip 1 & 2)

  10. SLAB • ANALYSIS AND DESIGN FOR SLAB : • Wu=1.2DL+1.6LL = 1.2x (4.25+3) +1.6x4=15.1 KN/m • STRIP 1 :

  11. SLAB • After checking Load cases: • Max M+ve=20.88 KN.m = 0.00288 As=ρbd = 0.00288x1000x140= 404 mm2use 4 Φ12 As shrinkage= ρ shrinkage * b*h = 0.0018*1000*170= 306 mm2Use 1 ф 12 mm /30 cm

  12. SLAB Max M-ve=25.23 KN.m • As=ρbd = 0.0035x1000x140= 491 mm2use 5 Φ12 • As shrinkage=1 ф 12 mm /30 cm Check shear : Vu= 36.54 KN = 28.4 KN at distance d from face of column. Vc >>Vu ….the slab thickness is o.k

  13. BEAMS • We used tributary Area method and live load reduction to find the ultimate load on the beams , all the beams are dropped. • DESIGN OF BEAM 1: B1 minimum depth(h) =14/18.5 = 0.756 m try h= 80 cm and cover = 5 cm d =75 cm Beam's width nearly(1/3-2/3)h try width = 40 cm At = 40 m x 2 m = 80 m2 D.L on beam= slab w + SDL + o.w + masonry wall load =([0.17x25+3]x2m+0.8x0.4x25) + 22.208 = 44.7 KN/m For b1: KLLAt= 2x80=160 > 38 so we can reduce the live load L= Lo = 0.61 > 0.50 Look L=0.61x4x2m = 5 KN/m • Wu = 1.2x44.7 + 1.6(5) = 61.6 x1.15 = 71 KN/m

  14. BEAMS..cont • DESIGN OF BEAM 1:

  15. BEAMS…cont • DESIGN OF BEAM 1: • M+ve=1173 KN.m for exterior spans = 0.016 • As = 0.016x400x750= 4811 mm2use 10Φ25 • M+ve=38.45 KN.m for interior span use As min = 0.0033x400x750 = 990 mm2 use 7Φ14 • M-ve=1239.55 KN.m ρ = 0.017 As = 0.017x400x750= 5143 mm2use 11Φ25

  16. BEAMS

  17. BEAMS

  18. Girders:

  19. Girders

  20. Girders

  21. Girders

  22. Girders: • We have 4 girders divided into 2 groups: G1: external girders , G2:internal girders

  23. COLUMNS • we divided the columns into two groups: c1: exterior columns (tied) c2: interior columns (circular)

  24. COLUMNS

  25. COLUMNS

  26. COLUMNS : • Summary:

  27. FOOTING : • FOOTING SYSTEM: • All footings were designed as isolated footings. • The design depends on the total axial load carried by each column.

  28. FOOTING :

  29. FOOTING :

  30. FOOTING : • Summary :

  31. FOOTING : • Tie Beam Design: • Tie beams are beams used to connect between columns necks, its work to provide resistance moments applied on the columns and to resist earthquakes load to provide limitation of footings movement. • Tie beam was designed based on minimum requirements with dimensions of 60 cm width and 100 cm depth. • Use minimum area of steel , with cover = 4 cm.

  32. CH3:SAP Output

  33. Slab Design: • Slab width=100cm • Slab depth=17cm • Slab effective depth=14cm

  34. B.M.D S.F.D T.D

  35. Verifications • 1) Compatibility • To achieve the condition of compatibility; all the elements of the structure must act as a single unite. This condition is clearly satisfied from SAP program by showing the 3D deformed shape and starting the animation

  36. Verifications • 2) Equilibrium check From hand calculations: Total dead load= 26475.2x7 = 185325.4KN Total live load= 7x1824x4=51072 KN. And from SAP the results were the same; % of error in dead load = 0% % of error in live load = 0%

  37. Verifications • 3) stress strain relationship ◙ Slab stress-strain verification From 1D analysis : Maximum M+ = 20.88K.m and Maximum M- = 25.23KN.m From 3D analysis : Maximum M+ = 39.35K.m and Maximum M- = 24.82KN.m Error in –ve mom=1% Error in +ve mom=45% The assumption of 1D slap model is wrong

  38. Verifications • ◙Beams stress-strain verification: For the beam resting on column Take the first beam in the second frame interior and check the total moment over the middle span (12 m) in both of 1D and 3D. From the 1D Total moment over the middle span= Positive moment+ 0.5(∑negative moments at the ends) =35.74+0.5(1152.26+1152.26) =1188 KN.m. For the 3D model positive and negative moments over the middle span of an interior beams resting on a columns is shown in figures • Total moment =Positive moment+ .5(∑negative moments at the ends) • =446+0.5(931 +931) =1377KN.m • Difference between 1D and3D results: 1377-1188/1377=14%

  39. Verifications • ◙Beams stress-strain verification: Total moment over the middle span =354.5+0.5(834.7+834.7) = 1189.2 KN.m. So the difference=1189.2-1188 /1189.2=0.019= 1 % ( less than 10% OK).

  40. Ch.4: Shear Walls Inclusion: • # of stories= 7 • zone factor z = 0.2 zone 2B (Ramallah City) • soil type: Sc (Very Dense Soil & Soft Rock) • From UBC table 16-Q Ca = 0.24 and from table 16-R CV = 0.32 • I: function of occupancy or function of structure → I = 1. • R: Response Modification factor R from (table 3-7) = 4.5 frame system • Weight of Building = Total dead load= 185325.4KN • T = 0.02x(80.4)3/4 = 0.54sec • CS = = = 0.1317 V = CS x W = 0.1324 x 18532.54 = 2454 ton

  41. for SW1: Fcallaow = (0.3-0.4)fc = 280/3 = 95 kg/cm2 Vall = 0.1Fcall = 9.5 kg/cm2 h/L = 24.5/48 = 0.5 < 2 the shear force is the dominant - V = 2454/ 4 sw's = 613.5 ton/wall . - ts = wall thickness = 20 cm trial - L B.zone =min( 3ts, 0.1L) = min (60cm,4.8m) = 60cm -L web = 48m - (2x0.6) = 46.8m = 4680 cm - Vmax = = = 6.55 Kg/cm2 < Vc allow= 9.5 Kg/cm2safe

  42. Boundary zone reinforcement: ρs min = 1%. As = 0.01*20*60 = 12cm2. 6Ф16 web zone reinforcement Lzone = 48-1.2=4680cm ρs min = 0.25%. As = 0.0025*20*100 = 5cm2. 4Ф14/m the same steps for SW2

  43. Stairs Shear walls At= 26x20 = 520m2 H = 24.5 P= 2(8+12) = 40m Wu on slab = 15.1 KN/m2 15.1 x 520x1.15= 9030 KN = 225KN/m For SW3 Try ts = 25 cm Ag= 12x0.25 = 3m2 Vertical reinforcement: As = ρ*Ag = 0.0012*100*25 = 3cm2 Use (1Φ14/20cm) Horizontal reinforcement : As = ρ*Ag = 0.002*100*25 = 5 cm2 Use (1Φ12/20cm) The same design for SW4

  44. Dynamic Analysis

  45. Design input Design input: by using IBC2006 code IE: seismic factor (importance factor) = 1 R: response modification factor = 4.5 Soil type: Sc Ss: spectral curve at short period = 0.5 , S1: spectral curve at 1second period = 0.2 Fa (site coefficient) =1.2 (See appendix) , Fv (site coefficient) =1.6 (See appendix) SDS = Fa Ss = *1.2* 0.5= 0.4 , SD1 = Fv S1 = *1.6* 0.2= 0.21 Area mass (super imposed load) = 3/9.81= 0. 3 KN/m2. Scale factor = = = 2.18 We enter the shear wall on building and starting the dynamic analysis

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