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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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).


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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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


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


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

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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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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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Compatibility:

CHECKS

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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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CHECKS

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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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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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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


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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DESIGNDesign of middle strip :

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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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DESIGN

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Design of columns :DESIGN

Column 8 (C8):Longitudenal bars =20 bar

2018mm=5080>4900

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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


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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


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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


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DESIGNDesign of mat foundation:


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

an-najah national university engineering college civil engineering department

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