AnNajah National University Faculty of Engineering Civil Engineering

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 1

SUBJECTS TO BE COVERED Chapter One : Introduction Chapter Two : Preliminary Design> ØSlab ØBeams ØGirders ØColumns ØFootings Chapter Three : SAP Output data Chapter Four : Seismic Analysis and SW inclusion Chapter Five : Dynamic Analysis Chapter Six : Stair Design 2

CH 1. INTRODUCTION About the project: This project introduces analysis and design of reinforced concrete office building in Ramallah with dimensions (48 x 40) 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: SAP 2000 V 14 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. 3

General layout 4

3 D Model 5

INTRODUCTION…cont l. Materials of construction: >>Reinforced concrete: (ρ) = 25 KN/m 3 , The required compressive strength after 28 days is fc = 28 Mpa ( B 350) for slab, beams and girders. fc = 40 Mpa ( B 500) for columns and foundations >> fy =420 Mpa >> Soil capacity = 3 kg/cm² 6

INTRODUCTION…cont loads: Ø Live load: LL=4 KN/m 2 Ø Dead load: DL=(Calculated By SAP) = 4. 25 KN/m 2 “ For 17 cm thick of solid slab” Ø SID= 3 KN/m 2 Ø Masonry wall load = 22 KN/m for 3. 5 m floor height Ø Earthquake load: its represents the lateral load that comes from an earthquake. (calculated later) 7

INTRODUCTION…cont Combinations: Wu= 1. 4 D. L Wu= 1. 2 D. L+ 1. 6 L. L Wu= 1. 2 D. L +1. 0 L. L ± 1. 0 E Wu= 0. 9 D. L ± 1. 0 E Codes Used: Ø American Concrete Institute Code (ACI 318 -05) Ø American Society of Civil Engineers "ASCE standard". Ø Uniform Building Code 1997 (UBC 97) 8

Ch 2: Preliminary design SLAB One way solid slab is used : Ø Thickness of slab: t = 4/24 =16. 66 cm use 17 cm , d=14 cm Ø Slab consists of two strips (strip 1 & 2) 9

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SLAB ANALYSIS AND DESIGN FOR SLAB : Wu=1. 2 DL+1. 6 LL = 1. 2 x (4. 25+3) +1. 6 x 4=15. 1 KN/m STRIP 1 : 11

SLAB After checking Load cases: Max M+ve=20. 88 KN. m = 0. 00288 As=ρbd = 0. 00288 x 1000 x 140= 404 mm 2 use 4 Φ 12 As shrinkage = ρ shrinkage * b*h = 0. 0018*1000*170= 306 mm 2 Use 1 ф 12 mm /30 cm 12

SLAB Max M-ve=25. 23 KN. m As=ρbd = 0. 0035 x 1000 x 140= 491 mm 2 use 5 Φ 12 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

BEAM S 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: B 1 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 m 2 D. L on beam= slab w + SDL + o. w + masonry wall load =([0. 17 x 25+3]x 2 m+0. 8 x 0. 4 x 25) + 22. 208 = 44. 7 KN/m For b 1: KLLAt= 2 x 80=160 > 38 so we can reduce the live load L= Lo = 0. 61 > 0. 50 Lo ok L=0. 61 x 4 x 2 m = 5 KN/m Wu = 1. 2 x 44. 7 + 1. 6(5) = 61. 6 x 1. 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. 016 x 400 x 750= 4811 mm 2 use 10Φ 25 M+ve=38. 45 KN. m for interior span use As min = 0. 0033 x 400 x 750 = 990 mm 2 use 7Φ 14 M-ve=1239. 55 KN. m ρ = 0. 017 As = 0. 017 x 400 x 750= 5143 mm 2 use 11Φ 25 16

BEAM S 17

BEAMS Beam 1 (0. 8*0. 4) Beam 2 (0. 8*0. 4) Beam 3 (0. 8*0. 5) Beam 4 (0. 8*0. 5) Ast Bottom mm 2 ext int 4811 990 4410 990 4488 1237 6573. 56 # of bars 10 ф 25 7 ф 14 9 ф 25 7 ф 14 10 ф 25 9 ф 14 10 ф 30 Ast TOP mm 2 5143 4709 4781 1237 # of bars 11 ф 25 10 ф 25 9 ф 14 stirrups 2 ф 10 /200 2 ф 10 /250 2 ф 10 /300 2 ф 10 /350 18

Girders: 19

Girders 20

Girders 21

Girders 22

Girders: We have 4 girders divided into 2 groups: G 1: external girders , G 2: internal girders Girder 1 (1. 2*0. 8) Girder 2 (0. 9*0. 7) Ast Bottom mm 2 ext int 3100 3036 6537 2203 # of bars 7 ф 25 10 ф 30 5 ф 25 Ast TOP mm 2 4224 3247 8384 5304 # of bars 9 ф 25 7 ф 25 12 ф 30 8 ф 30 stirrups 2 ф 10 /450 2 ф 10 /150 23

COLUMNS we divided the columns into two groups: c 1: exterior columns (tied) c 2: interior columns (circular) 24

COLUMNS 25

COLUMNS 26

COLUMNS : Summary: C 1 C 2 Ultimate load (KN) dimensions (cm) 12815 16573 80*80 Dia. = 80 Steel ratio 1% 2. 5% Reinforcement 14 Ф 25 26 Ф 25 Stirrups / Spiral 2 Ф 10 mm Ф 8 mm Spacing (cm) 40 5 cover (cm) 2. 5 cm 27

FOOTING : FOOTING SYSTEM: Ø All footings were designed as isolated footings. Ø The design depends on the total axial load carried by each column. 28

FOOTING : 29

FOOTING : 30

FOOTING : Summary : F 1 F 2 Dimensions (m) 5. 5 x 5. 5 6 x 6 Thickness (cm) 1. 1 1. 2 Steel in x, y direction (+) 6Φ 25/m 7Φ 25/m Steel in x, y direction (-) 6Φ 20/m 7Φ 20/m Cover (cm) 7 7 31

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FOOTING : Tie Beam Design: Ø 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. Ø Tie beam was designed based on minimum requirements with dimensions of 60 cm width and 100 cm depth. Ø Use minimum area of steel , with cover = 4 cm. Ast Top bars 1901 mm 2 6 Φ 20 mm Bottom bars 6 Φ 20 mm stirrups 1 Φ 8 / 20 cm 33

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CH 3: SAP Output 35

Slab Design: Slab width=100 cm Slab depth=17 cm Slab effective depth=14 cm 36

B. M. D S. F. D T. D 37

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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 3 D deformed shape and starting the animation 39

Verifications 2) Equilibrium check From hand calculations: Total dead load= 26475. 2 x 7 = 185325. 4 KN Total live load= 7 x 1824 x 4=51072 KN. And from SAP the results were the same; % of error in dead load = 0% % of error in live load = 0% 40

Verifications 3) stress strain relationship ◙ Slab stress-strain verification From 1 D analysis : Maximum M+ = 20. 88 K. m and Maximum M- = 25. 23 KN. m From 3 D analysis : Maximum M+ = 39. 35 K. m and Maximum M- = 24. 82 KN. m Error in –ve mom=1% Error in +ve mom=45% The assumption of 1 D slap model is wrong 41

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 1 D and 3 D. From the 1 D 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 3 D 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) =1377 KN. m Difference between 1 D and 3 D results: 1377 -1188/1377=14% 42

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). 43

Ch. 4: Shear Walls Inclusion: # of stories= 7 zone factor z = 0. 2 zone 2 B (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. 4 KN T = 0. 02 x(80. 4)3/4 = 0. 54 sec CS = = 0. 1317 V = C S x W = 0. 1324 x 18532. 54 = 2454 ton 44

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for SW 1: Fc allaow = (0. 3 -0. 4)fc = 280/3 = 95 kg/cm 2 Vall = 0. 1 Fcall = 9. 5 kg/cm 2 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( 3 ts, 0. 1 L) = min (60 cm, 4. 8 m) = 60 cm -L web = 48 m - (2 x 0. 6) = 46. 8 m = 4680 cm - Vmax = = 6. 55 Kg/cm 2 < Vc allow= 9. 5 Kg/cm 2 safe 46

Boundary zone reinforcement: ρs min = 1%. As = 0. 01*20*60 = 12 cm 2. 6 Ф 16 web zone reinforcement Lzone = 48 -1. 2=4680 cm ρs min = 0. 25%. As = 0. 0025*20*100 = 5 cm 2. 4 Ф 14/m the same steps for SW 2 47

Stairs Shear walls At= 26 x 20 = 520 m 2 H = 24. 5 P= 2(8+12) = 40 m Wu on slab = 15. 1 KN/m 2 15. 1 x 520 x 1. 15= 9030 KN = 225 KN/m For SW 3 Try ts = 25 cm Ag= 12 x 0. 25 = 3 m 2 Vertical reinforcement: As = ρ*Ag = 0. 0012*100*25 = 3 cm 2 Use (1Φ 14/20 cm) Horizontal reinforcement : As = ρ*Ag = 0. 002*100*25 = 5 cm 2 Use (1Φ 12/20 cm) The same design for SW 4 48

Dynamic Analysis 49

Design input: by using IBC 2006 code IE: seismic factor (importance factor) = 1 R: response modification factor = 4. 5 Soil type: Sc Ss: spectral curve at short period = 0. 5 , S 1: spectral curve at 1 second 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 , S D 1 = Fv S 1 = *1. 6* 0. 2= 0. 21 Area mass (super imposed load) = 3/9. 81= 0. 3 KN/m 2. Scale factor = = 2. 18 We enter the shear wall on building and starting the dynamic analysis 50

Equivalent static method: 51

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Response spectrum method: 53

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C 2 Ultimate load (KN) 20151 dimensions (cm) Dia. = 100 Steel ratio 2. 5% Reinforcement 28 Ф 30 Stirrups / Spiral Ф 8 mm Spacing (cm) 5 cover (cm) 2. 5 cm 61

Deflection Checks: The permissible displacement lies between H/500 and H/200, where H is the total height of the building. Total height of the building =2450 cm. From SAP maximum displacement was 3. 75 cm ∆=2450/500=4. 9 cm →x &y lateral displacement less than ∆, ok. 62

With SW: Without SW: 63

Deformed shape and building period 64

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Results: We noticed that there is very strong effect of shear walls inclusion to the building because it a. reduced the vibration of building under the earthquake load and its clear in deformed shapes of both frame and combined structures b. reduced the moments affecting the beams c. reduced the displacements of stories. d. reduced the time period , in frame structure its ranged from 0. 3 -2. 7 but in combined structure its constant and equal 0. 3 sec 2) In bad cases the shear walls system reduced the time needed to complete the structure because of easier frame work than frame structure. I recommend to use combined system in multi stories buildings and make different studies about shear walls inclusion effect on buildings. 66

Stair Design # of steps = = = 20 steps 67

Stair Design 68

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