an-najah national university engineering college civil engineering department

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An-Najah National UniversityEngineering College

Civil Engineering Department

Graduation ProjectThree Dimensional analysis And

Design Of AL-ARAB HOSPITAL

Supervised by: Ibrahim Mohammad Arman

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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

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Contents

• CH.1 : Introduction

• CH.2 : Preliminary Design

• CH.3 : 3.D modeling and Final Design

• CH.4 : Stairs And Ahear Walls Design

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• eliminary Design• PrPreliminary Designeliminary Design

Chapter OneIntroduction

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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.

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CH.1 introduction

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

Design Determinants

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CH.1 introduction

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.

Design Determinants

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CH.1 introduction

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

Design Determinants

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CH.1 introduction

Chapter TwoPreliminary Design

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Preliminary DesignPLAN VIEW

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COMPARISON :

Which system is more suitable to use ?? • Two-way solid slab with drop beams

• Two-way waffle slab with hidden beams

Preliminary Design

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Preliminary DesignPLAN VIEW

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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

Two-Way Solid Slab With Drop Beams

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CH.2 Preliminary Design

SLAB THICKNESS:

Thickness of slab, hmin =(Ln/33) = 252.7mm

Assume frame dimensions= 600 mm600 mm320 mmflange width = 820 mmflange depth = 80 mmweb width = 150 mmweb depth = 320

Waffle Slab

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CH.2 Preliminary Design

Iwaffle =0.001656426 m4 = Isolid =

hsolid= 0.28940974 m

Frame dimension OK

CHECKS

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Waffle SlabCH.2 Preliminary Design

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

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Waffle SlabCH.2 Preliminary Design

Ultimate load : Pu=Wbeam+Wslab+own weight for column =629.627 = 629.62714(numberof stories)=8814.778 kNTo 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 areaAssume steel ratio ρ=Ast/Ag=1% As = 0.01Ag

Ag = 1042600.497 mm2

Assume rectangular column L = b = 1021 mm ….

Use column (1.1 m1.1 m)

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Design of ColumnColumn (27.N).

CH.2 Preliminary Design

Checks for short column KLu/r ≤32-(M1/M212) ≤ 40Where ; 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

CHECK

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Design of ColumnCH.2 Preliminary Design

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

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Cost AnalysisCH.2 Preliminary Design

Item Steel weight (Kg) Concrete volume (m3)

Beam 481.84 +61.1+21.718 5.475

Column strip 270.27 10.4098

Middle strip 351.63 14.77

Sum 1186.2 30.65

Cost 3400 shekel/ton 300 shekel / m3

Sum (shekel) 13228 shekls

SOLID SLAB SYSTEM

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Cost AnalysisCH.2 Preliminary Design

Item Steel weight (Kg) Concrete volume (m3)

Beam 523.4+82.1+35.169 4.1

Column strip 255.122.428

Middle strip 277.6

Sum 1173.3 kg 26.528

Cost 3400 Shekel/ton 300 Shekel / m3

Sum (shekel) 11948 Shekel

WAFFLE SYSTEM

It is clear that the coast of material for waffle slab is less than that of the solid slab. 22

Cost AnalysisCH.2 Preliminary Design

Chapter Three3D. Modeling

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slab thickness in preliminary design=400mmBUT some beams were unsafe in preliminary design dimentions

because of additional internal forces due to seismic loads

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3D.ModelingCH.2 3.D Modeling

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CH.2 3.D Modeling

3D.Modeling

LOAD

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• 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

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

3D.ModelingCH.2 3.D Modeling

Lateral loads

• Seismic loads :Response spectrume

• Soil loads

LOAD

Assume Ø =30 o so: ko = 1-sin Ø=0.5for one story, h= 4.16m

q1 at z=4.16 m=koW=0.5x15=7.5 kN q2 at z=0.0m = koW+γhKo= 0.5x15+20x4.16x0.5=49.1 kN.

q2

q1

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3D.ModelingCH.2 3.D Modeling

Gravity loads:Uniform loads on slab : weight of masonry wall as distributed uniform dead load

INPUT LOAD DATA IN SAP MODEL

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3D.ModelingCH.2 3.D Modeling

INPUT LOAD DATA IN SAP MODEL Lateral loads:Seismic loads (Response Spectrum) information for Response Spectrum definition

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3D.ModelingCH.2 3.D Modeling

INPUT LOAD DATA IN SAP MODEL

𝑔 𝐼𝑅

Response in X-direction and 30% in Y-direction

Lateral loads: Response Spectrum in x-direction Response Spectrum

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3D.ModelingCH.2 3.D Modeling

LOAD COMBINATION

UDCON1 = 1.4D.LUDCON2 = 1.2D.L+1.6L.L+1.6SOILUDCON3 = 1.2D.L+1L.L+1EXUDCON4 = 1.2D.L+1L.L+1EYUDCON5 = 1.2D.L+1L.L+1EZUDCON6 = 0.9D.L+1EX+1.6SOILUDCON7 = 0.9D.L+1E.Y+1.6SOILUDCON8 = 0.9D.L+1E.Z+1.6SOIL

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3D.ModelingCH.2 3.D Modeling

Compatibility:

CHECKS

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3D.ModelingCH.2 3.D Modeling

CHECKSEquilibrium:• Hand calculation: Total weight = 212672.35 kN

• From SAP

Load type Hand results (KN) SAP results (KN) Error %

Live load 40567.43 40833.22 0%

Dead load 212672.35 215513.878 1.3%

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3D.ModelingCH.2 3.D Modeling

CHECKS

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Stress- strain relationship:

3D.ModelingCH.2 3.D Modeling

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

CHECKS

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3D.ModelingCH.2 3.D Modeling

Stress- strain relationship:

SAP result

CHECKS

To calculate moments from SAP= ( M(-ve)+ M(-ve))/2+ M(+ve)

= (523.4+565)/2+423.4=967.6kN.mError percentage=(967.6-967.3)/967.3=0%

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3D.ModelingCH.2 3.D Modeling

Shear in Slab:• Rib shear strength = = 62225.39 N = 62.225 kN

CHECKS

values = 62.273 x 0.82 = 51.1 kN/0.82m.< 62.225 kN ... OK37

3D.ModelingCH.2 3.D Modeling

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=0.0033As=0.00815150400=489mm2 ……Use steel bars as 2Ø18/rib

DESIGN

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3D. Modeling & DesignCH.2 3.D Modeling

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

DESIGN

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3D. Modeling & DesignCH.2 3.D Modeling

DESIGNDesign of middle strip :

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3D. Modeling & DesignCH.2 3.D Modeling

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/mmS=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/mmS=314/0.511=614.5mm ˃ 200mm so use 1ф10/200mm

DESIGN

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3D. Modeling & DesignCH.2 3.D Modeling

DESIGN

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3D. Modeling & DesignCH.2 3.D Modeling

Design of columns :DESIGN

Column 8 (C8):Longitudenal bars =20 bar

2018mm=5080>4900

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3D. Modeling & DesignCH.2 3.D Modeling

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DESIGNDesign of footing :

service load in building

Foundation area=Total service load/bearing capacity =569.4 m2 ˃ half area of building, 353m2

Use mat foundation

3D. Modeling & DesignCH.2 3.D Modeling

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DESIGNDetermine foundation thickness:

maximum axial force =21932.85 kNфVCp=0.750.33bod• Where • c1&c2 : column dimensions • d=effective depthTry h=1600mm and check if it can resist bunching shear, So d=1530mmфVCp =22535kN ˃ 21932.85 kN …OK

3D. Modeling & DesignCH.2 3.D Modeling

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DESIGNDesign 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

3D. Modeling & DesignCH.2 3.D Modeling

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DESIGNDesign of mat foundation:

3D. Modeling & DesignCH.2 3.D Modeling

Chapter FourShear Walls &Stairs

Design

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CH.3 Shear walls & stairs

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DESIGNLong 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 ρ0.00138As=0.001381000600=828 mm2/m use1Ø12/250 mm for each side

Shear WallsCH.3 Shear walls & stairs

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DESIGNShort shear walls:

Asmin= (0.00123500200) =240mmuse 14ф12/350mm

Shear WallsCH.3 Shear walls & stairs

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Stairs

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CH.3 Shear walls & stairs

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DESIGN

Check for shearMaximum shear in stair Vu=55.5/1.45=38.3kN/m ….OK

Mu=13kN/mmin=0.0018 As=0.00206< Asmin Use Asmin = 1ф12/300mm

Stairs CH.3 Shear walls & stairs