PowerPoint Presentation

Three Dimensional Analysis and Design of Aljawhara Office buildingAn-Najah National University

Faculty of Engineering

Civil Engineering Department

Graduation Project:

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Prepared by: Ahmad Yahya Khalid SalameenSupervisor : Dr.Wael Abo Asab2012SUBJECTS TO BE COVERED Chapter One : IntroductionChapter Two : Preliminary Design>

Chapter Three : SAP Output dataChapter Four : Seismic Analysis and SW inclusionChapter Five : Dynamic AnalysisChapter Six : Stair Design

SlabBeamsGirdersColumnsFootings2 CH1.INTRODUCTIONAbout 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 m.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.

3 General layout4

3D Model5

INTRODUCTIONcontMaterials of construction:>>Reinforced concrete: () = 25 KN/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/cm

6 INTRODUCTIONcontloads:

Live load: LL=4 KN/m2

Dead load: DL=(Calculated By SAP) = 4.25 KN/m2 For 17 cm thick of solid slabSID= 3 KN/m2Masonry wall load = 22 KN/m for 3.5m floor height

Earthquake load: its represents the lateral load that comes from an earthquake.(calculated later)

7 INTRODUCTIONcontCombinations:

Wu= 1.4D.LWu= 1.2D.L+ 1.6L.LWu= 1.2D.L +1.0L.L 1.0EWu= 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)

8 SLABOne way solid slab is used :Thickness of slab: t = 4/24 =16.66 cm use 17 cm ,d=14 cmSlab consists of two strips (strip 1 & 2)

9Ch2:Preliminary design10

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

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SLABAfter checking Load cases:Max M+ve=20.88 KN.m = 0.00288

As=bd = 0.00288x1000x140= 404 mm2 use 4 12

As shrinkage = shrinkage * b*h = 0.0018*1000*170= 306 mm2 Use 1 12 mm /30 cm

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SLABMax M-ve=25.23 KN.mAs=bd = 0.0035x1000x140= 491 mm2 use 5 12 As shrinkage = 1 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

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BEAMSWe 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 cmBeam's width nearly(1/3-2/3)h try width = 40 cmAt = 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/mFor b1: KLLAt= 2x80=160 > 38 so we can reduce the live loadL= Lo = 0.61 > 0.50 Lo ok L=0.61x4x2m = 5 KN/m

Wu = 1.2x44.7 + 1.6(5) = 61.6 x1.15 = 71 KN/m 14

BEAMS..contDESIGN OF BEAM 1:

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BEAMScontDESIGN OF BEAM 1:M+ve=1173 KN.m for exterior spans

= 0.016

As = 0.016x400x750= 4811 mm2 use 1025M+ve=38.45 KN.m for interior span use As min = 0.0033x400x750 = 990 mm2 use 714M-ve=1239.55 KN.m = 0.017 As = 0.017x400x750= 5143 mm2 use 1125

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BEAMS17

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 mm2extintextintextintextint48119904410990448812376573.56 # of bars 10 257 149 257 1410 259 1410 30Ast TOPmm25143470947811237 # of bars 11 2510 2510 259 14stirrups2 10 /2002 10 /2502 10 /3002 10 /3501818Girders:19

Girders20

Girders21

Girders22

Girders:We have 4 girders divided into 2 groups: G1: external girders , G2:internal girdersGirder 1(1.2*0.8)Girder 2(0.9*0.7)Ast Bottom mm2extintextint3100303665372203 # of bars 7 257 2510 305 25Ast TOPmm24224324783845304

# of bars 9 257 2512 308 30stirrups2 10 /4502 10 /1502323COLUMNS24we divided the columns into two groups:c1: exterior columns (tied)c2: interior columns (circular)

COLUMNS25

COLUMNS26

COLUMNS :Summary:

C 1C 2Ultimate load (KN)1281516573dimensions (cm)80*80Dia. = 80Steel ratio1 %2.5%Reinforcement142526 25Stirrups / Spiral210 mm8 mmSpacing (cm)405cover (cm)2.5 cm2.5 cm27FOOTING :FOOTING SYSTEM: All footings were designed as isolated footings. The design depends on the total axial load carried by each column.

28FOOTING :29

FOOTING :30

FOOTING :Summary :

F2F 16x65.5x5.5Dimensions (m)1.21.1Thickness (cm)725/m625/m Steel in x, y direction (+)720/m620/mSteel in x, y direction (-) 77Cover (cm)3132

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.

AstTop barsBottom barsstirrups1901mm26 20 mm6 20 mm1 8 / 20cm3334

CH3:SAP Output35

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

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B.M.DS.F.DT.D3738

Verifications1) CompatibilityTo 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 animation39

2) Equilibrium checkFrom 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%

40Verifications

3) stress strain relationship Slab stress-strain verificationFrom 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.mError in ve mom=1% Error in +ve mom=45% The assumption of 1D slap model is wrong

41Verifications

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 1DTotal 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.mDifference between 1D and 3D results: 1377-1188/1377=14%

42Verifications

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).

43Verifications

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 = 0.32I: 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= 185325.4KNT = 0.02x(80.4)3/4 = 0.54sec CS = = = 0.1317 V = CS x W = 0.1324 x 18532.54 = 2454 ton

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for SW1:Fc allaow = (0.3-0.4)fc = 280/3 = 95 kg/cm2 Vall = 0.1Fcall = 9.5 kg/cm2h/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/cm2 safe

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Boundary zone reinforcement:s min = 1%.As = 0.01*20*60 = 12cm2.616web zone reinforcementLzone = 48-1.2=4680cms min = 0.25%.As = 0.0025*20*100 = 5cm2.414/m the same steps for SW2

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At= 26x20 = 520m2 H = 24.5 P= 2(8+12) = 40mWu on slab = 15.1 KN/m2 15.1 x 520x1.15= 9030 KN = 225KN/mFor SW3 Try ts = 25 cm Ag= 12x0.25 = 3m2

Vertical reinforcement: As = *Ag = 0.0012*100*25 = 3cm2 Use (114/20cm)Horizontal reinforcement : As = *Ag = 0.002*100*25 = 5 cm2 Use (112/20cm)The same design for SW4

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Stairs Shear walls49

Dynamic AnalysisDesign inputDesign input: by using IBC2006 codeIE: 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 = 0.2Fa (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= 0.21Area mass (super imposed load) = 3/9.81= 0. 3 KN/m2.Scale factor = = = 2.18We enter the shear wall on building and starting the dynamic analysis

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Equivalent static method:51

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Response spectrum method:53

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61C 2Ultimate load (KN)20151dimensions (cm)Dia. = 100Steel ratio2.5%Reinforcement28 30Stirrups / Spiral8 mmSpacing (cm)5cover (cm)2.5 cmDeflection 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 =2450cm.From SAP maximum displacement was 3.75 cm =2450/500=4.9cm x &y lateral displacement less than , ok.6263

With SW:

Without SW:Deformed shape and building period64

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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 structuresb. 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.

66Stair Design# of steps = = = 20 steps 67

68Stair Design

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Thank you all For listening