ANNAJAH NATIONAL UNIVERSITY ENGINEERING COLLEGE Civil Engineering Department

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AN-NAJAH NATIONAL UNIVERSITY ENGINEERING COLLEGE Civil Engineering Department Graduation project " AFORI Residential Building

AN-NAJAH NATIONAL UNIVERSITY ENGINEERING COLLEGE Civil Engineering Department Graduation project " AFORI Residential Building Structural Design And Analysis" Prepared By Ali Mazen Mahroom Amjad Nidal Al-Kharof Instructor Ibrahim Mohammad

Abstract This project is a structural analysis and design of a residential building located

Abstract This project is a structural analysis and design of a residential building located in Rafidia in Nablus city, The building is consisted of 10 floors. The final analysis and design of building is done using a three dimensional (3 D) structural model by the structural analysis and design software sap 2000.

3 D Picture of the building

3 D Picture of the building

The preliminary dimensional of the structural elements are determined using one dimensional structural analysis

The preliminary dimensional of the structural elements are determined using one dimensional structural analysis for the structural members for gravity loads. impel analysis and design is used for this purpose. The structural model results are verified by simple calculations and by comparing with the one dimensional analysis.

Contents CHAPTER ONE: (Introduction) 1. 1 General 1. 2 Description of project 1. 3

Contents CHAPTER ONE: (Introduction) 1. 1 General 1. 2 Description of project 1. 3 Philosophy of analysis and design 1. 4 Materials 1. 5 Loads 1. 6 foundations system 1. 7 code and standard 1. 8 building structural systems

CHAPTER TWO: (PRELIMINARY DESIGN ) 2. 1 General 2. 2 Design of Rib Slab

CHAPTER TWO: (PRELIMINARY DESIGN ) 2. 1 General 2. 2 Design of Rib Slab 2. 3 Design of columns 2. 3 Design of beams

CHAPTER THREE: (Three Dimensional Analysis And Design) 3. 1 General 3. 2 Modeling The

CHAPTER THREE: (Three Dimensional Analysis And Design) 3. 1 General 3. 2 Modeling The Building as 3 D 3. 3 Structural Model Verification 3. 4 Analysis and Design of Slabs 3. 5 Analysis and Design of Beams 3. 6 Analysis and Design of columns 3. 7 Analysis and Design of Footings 3. 8 Analysis and Design of Stair 3. 9 Analysis and Design of walls

**CHAPTER ONE: INTRODUCTION *General: This project introduces analysis and design of reinforce concrete residential

**CHAPTER ONE: INTRODUCTION *General: This project introduces analysis and design of reinforce concrete residential building. Also this project provides clear design structural drawings for construction. This project is a residential project of AFORI building in NABLUSE city in RAFDIA street, which consists of 10 stories. Each store consists of 2 apartments of approximately equal area. Store Area (m²) parking 687 1 st to 5 th 640 6 th to 8 th 627 roof 578

*Philosophy of analysis and design: The modeling is modeling as a three dimensional (3

*Philosophy of analysis and design: The modeling is modeling as a three dimensional (3 D) structure. The preliminary section properties are defend and estimated by preliminarily analysis of the structural elements using one dimensional (1 D) structural analysis. The computer program Sap 2000 was the main tool in analysis and design of the different structural members. Manual calculation are used to verify the computer results. In the 3 D structural model, the beams columns are represented by frame elements (lines) and the slabs and walls by area elements

*Materials Concrete strength f'c=240 Kg/cm² (24 Mpa) Modulus of elasticity equals 2. 30*10ˆ6 ton/m².

*Materials Concrete strength f'c=240 Kg/cm² (24 Mpa) Modulus of elasticity equals 2. 30*10ˆ6 ton/m². Unit weight is 2. 5 ton/m 3. Fy= 4200 kg/cm 2 (420 Mpa) Material Unit weight (ton/m 3) Reinforced concrete 2. 5 Bricks 1. 2 Gravel Fill under tiles 1. 8 Masonry 2. 7 Tiles 2. 5 Mortar 2. 3 plastering 2. 3

*Loads: -Superimposed dead load=0. 4 t/m² -Live Load= 0. 25 t/m² -Earthquake load: Depend

*Loads: -Superimposed dead load=0. 4 t/m² -Live Load= 0. 25 t/m² -Earthquake load: Depend on Seismic zone (z)=2 A and soil properties (SB) Then, Ca=0. 25, Cv=0. 25

*codes and standard: 1 - ACI 318 -08 (AMERICAN CONCRETE INSTITUTE) 2 - UBC-97:

*codes and standard: 1 - ACI 318 -08 (AMERICAN CONCRETE INSTITUTE) 2 - UBC-97: (UNIFORM BUILDING CODE) 3 -IBC 2009: (INTERNATIONAL BUILDING CODE) 4 - ASTM 2009: (Standards as Referenced in the 2009 International Building Code )

**CHAPTER TWO: PRELIMINARY DESIGN *General: -Soil capacity = 5. 5 kg/cm² -Concrete strength f'c=240

**CHAPTER TWO: PRELIMINARY DESIGN *General: -Soil capacity = 5. 5 kg/cm² -Concrete strength f'c=240 Kg/cm² (24 Mpa) -Steel yield strength, Fy= 4200 kg/cm 2 (420 Mpa) -Load cases: Dead Load=0. 0 (we calculate self weight manual) Live Load =0. 0 Soil load = 0. 0

*Design of Rib Slab: -Minimum slab thickness is calculated according to ACI 318 -08

*Design of Rib Slab: -Minimum slab thickness is calculated according to ACI 318 -08 provision. ACI 318 -08

Simply supported=L /16 =450/16 =28 cm One end continuous = L / 18. 5

Simply supported=L /16 =450/16 =28 cm One end continuous = L / 18. 5 = 550/18. 5 =29. 7 cm Both ends continuous = L /21= 620/21 = 29. 5 cm Cantilever = L/8=120/8=15 cm Then we assume thickness of slab (h) rib= 30 cm Cross section in Ribbed slab

The distribution of ribs in the typical slabs shown in figure

The distribution of ribs in the typical slabs shown in figure

The ribs in the slab are analyzed and design using sap 2000 program. As

The ribs in the slab are analyzed and design using sap 2000 program. As an example, the analysis result and design of rib 5 are illustrated here: Use 2φ12 top and bottom steel moment diagram for rib 5, ton. m Area of steel for rib 5, cm²

*Design of columns In this project rectangular and square columns are used. And these

*Design of columns In this project rectangular and square columns are used. And these columns can carry axial load and no moment. -Design of Column C 2: The dimensions of the column 80*30 cm, Ag=2400 cm², And we use ρ between (1%- 4%) Pu=203. 4 ton Pd=Φ Pn=304. 6 ton Pd>pu ok The stirrups must minimum of the following : - 48 ds (diameter of stirrups) -16 db (diameter of bar) -Least dimension of the section Then we use 30 cm Φ 10/30 cm

*Design of beams After distributed the beam in the plan as shown in figure

*Design of beams After distributed the beam in the plan as shown in figure , we insert to sap 2000 and insert each load on it which come from ribs slab or from external or internal wall and then design it.

Ultimate dead load for exterior wall=2. 9 ton/m Ultimate dead load for 10 cm

Ultimate dead load for exterior wall=2. 9 ton/m Ultimate dead load for 10 cm wall=0. 92 ton/m Ultimate dead load for 20 cm wall =1. 44 ton/m -Design of beam B 5 (60*50): moment diagram for B 5, ton. m Area of steel for B 5, cm²

Use 8φ16 top and bottom steel Cross section A-A in beam 5

Use 8φ16 top and bottom steel Cross section A-A in beam 5

**CHAPTER THREE: Three Dimensional Analysis And Design *General: This chapter provides analysis and design

**CHAPTER THREE: Three Dimensional Analysis And Design *General: This chapter provides analysis and design of 3 D model for the building using sap 2000 program. Figure below show 3 D Model of it. 3 D model

*Modeling The Building as 3 D Structure : **Sections: -Basement walls= 30 cm, and

*Modeling The Building as 3 D Structure : **Sections: -Basement walls= 30 cm, and the modifier m 11=m 22=m 12= 0. 35 -shear walls= 25 cm, and the modifier m 11=m 22=m 12= 0. 35 -Ribbed slabs are presented as one way solid slabs in y-direction and one way solid slabs in x-direction. The thickness is calculated to be equivalent to ribs moment of inertia. -4 I for T section =5. 76*10 m 0. 55*(H equivalent 4 -4 )3/12=5. 76*10 → H equivalent = 23. 26 cm m 4

The modifier for slab in y-direction was calculated and shown in the figure 1

The modifier for slab in y-direction was calculated and shown in the figure 1 And the modifier for slab in x-direction was calculated and shown in the figure 2 Figure 1 Figure 2

-beams: Variable in sections, we use concrete covers of 5 cm Moment of inertia

-beams: Variable in sections, we use concrete covers of 5 cm Moment of inertia about 2 axis = Moment of inertia about 3 axis=0. 35 -columns : Variable in sections, we use concrete covers of 4 cm Moment of inertia about 2 axis = Moment of inertia about 3 axis=0. 7 **Loads: -Own weight : will be calculated by the program. -Live load= 0. 25 ton/m 2. -super imposed dead load =0. 4 ton/m 2

-Lateral loads on basement walls ɣ=2 ton/m² h=3. 1 m, Ø=15 Soil pressure =ɣ*h

-Lateral loads on basement walls ɣ=2 ton/m² h=3. 1 m, Ø=15 Soil pressure =ɣ*h *(1 -sinØ)=2*3. 1*(1 -sin 15)=4. 65 ton/m² soil pressure

-seismic loads First we will define the function of response spectrum Use Ca=0. 25,

-seismic loads First we will define the function of response spectrum Use Ca=0. 25, Cv=0. 25 and function damping ratio =0. 05 Response Spectrum UBC 97 Function Definition

we define the cases of the seismic loads Seismic-x, Seismic-y by using scale factors

we define the cases of the seismic loads Seismic-x, Seismic-y by using scale factors U 1, U 2 & U 3 Scale factor = g=9. 81 ground acceleration I=importance factor=1 for residential building R=structural system coefficient =6. 5 for masonry exterior walls and special moment resisting =4. 5 for masonry exterior walls and shear walls Scale factor = =1. 962

*Structural Model Verification: Equilibrium Check: -Live Load: Total Live load =1599. 1 ton Live

*Structural Model Verification: Equilibrium Check: -Live Load: Total Live load =1599. 1 ton Live load from SAP =1619 ton % Error =1% < 5% ok

-Dead loads: Total load= Slab load+ Super imposed load+ Beams load + Column loads+

-Dead loads: Total load= Slab load+ Super imposed load+ Beams load + Column loads+ Shear wall load+ Basement wall =14529. 1 ton Dead load from SAP =13998. 4 ton. Error= 4. 7% < 5% ok.

*Analysis and Design of Slabs : -check shear : Shear force contour (typical floor)

*Analysis and Design of Slabs : -check shear : Shear force contour (typical floor)

-Flexure Analysis and Design: Bending moment contour (typical floor)

-Flexure Analysis and Design: Bending moment contour (typical floor)

Rib no. 1 2 3 4 5 6 7 8 9 10 11 12

Rib no. 1 2 3 4 5 6 7 8 9 10 11 12 13 Max M+ve Max M-ve As +ve As -ve 1. 8 4. 7 1. 28 2. 5 0. 75 2. 3 1. 28 1. 1 1. 25 2. 7 1. 28 1. 4 . 5 4. 9 1. 28 2. 6 0. 95 1. 8 1. 28 1. 17 1. 23 2. 7 1. 28 1. 4 1. 17 2. 9 1. 28 1. 5 1. 2 2. 2 1. 28 1. 17 0. 95 2 1. 28 1. 17 1. 3 4. 1 1. 28 2. 1 1. 34 3 1. 28 1. 5 2. 1 4. 1 1. 28 2. 18 2. 65 3. 4 1. 28 1. 79 Bottom steel Top steel Stirrup Ø 8 (spacing) 2 Ф 12 2 Ф 14 30 cm 2 Ф 12 30 cm 2 Ф 12 2 Ф 14 12. 5 cm 2 Ф 12 2 Ф 12 30 cm 2 Ф 12 30 cm 2 Ф 12 2 Ф 14 12. 5 cm 2 Ф 12 2 Ф 14 30 cm 2 Ф 12 30 cm

Slab reinforcements slab (typical floor)

Slab reinforcements slab (typical floor)

*Analysis and Design of Beams: Analysis and design output was taken from SAP 2000.

*Analysis and Design of Beams: Analysis and design output was taken from SAP 2000. Verification for minimum steel was used as explained in chapter 2 Beam 5, Area of steel cm² Beam 5, shear cm² Beam 5, Torsion cm²

*Analysis and Design of columns : Analysis and design output was taken from SAP

*Analysis and Design of columns : Analysis and design output was taken from SAP 2000. Verification for minimum steel was used as explained in chapter 2

Columns C 1 C 2 C 3 Dimension (cm) 90*50 Reinforcemen t (cm²) 45

Columns C 1 C 2 C 3 Dimension (cm) 90*50 Reinforcemen t (cm²) 45 81. 1 79. 16 Bars 20 Ф 18 22 Ф 18 Dimension (cm) 90*50 Reinforcemen t (cm²) C 4 C 5 C 6 90*40 90*70 72 51. 4 63 22 Ф 22 20 Ф 22 18 Ф 20 20 Ф 20 90*50 90*40 90*70 45 58. 3 55. 7 50. 64 39. 5 63 bars 20 Ф 18 20 Ф 20 22 Ф 18 20 Ф 18 18 Ф 18 20 Ф 20 Dimension (cm) 90*50 90*40 90*70 Reinforcemen t (cm²) 45 45 45 36 36 63 bars 20 Ф 18 20 Ф 16 22 Ф 16 20 Ф 16 18 Ф 16 20 Ф 20 Floor ground 1 st 2 nd -Roof 90*40

Column Cross section Typical longitudinal section for column

Column Cross section Typical longitudinal section for column

*Analysis and Design of Footings : -Single Footings Footing (F 4) : Pu= 542.

*Analysis and Design of Footings : -Single Footings Footing (F 4) : Pu= 542. 46 t, Ps = 450 t Footing Area (A) = Ps. /B. C. = 450/55 = 8. 1 m² Footing dimensions : 2. 7*3. 1 m Ultimate pressure (qu) = 542. 46 /8. 37 = 64. 8 t/m² Wide beam shear : ФVc=44. 8 ton, Vu = 23. 9 ton Check Punching shear : ФVc > Vu ok ФVc=514. 1 ton, Vu=412. 4 ton Flexural design: Mu = qu*L²/2= 64. 8* (1. 1)2/2 = 39. 2 ton. m/m As=14. 5 cm²/m Use 8 Ф 16/m bottom steel in both directions

Footing dimensions Longitudinal Reinforcement Footing No. Depth (m) Area of steel (cm 2) #

Footing dimensions Longitudinal Reinforcement Footing No. Depth (m) Area of steel (cm 2) # of bars in each direction 1. 2 0. 40 11. 34 4 Ф 14/m 2 1. 6 0. 5 7. 7 4 Ф 16/m 3 2. 6 2. 2 0. 6 10. 98 7 Ф 16/m 4 3. 1 2. 7 0. 7 10. 98 8 Ф 16/m 5 3. 4 3 1 7. 59 8 Ф 16/m 6 3. 4 110 22. 14 7 Ф 20/m Length (m) Width (m) 1 1. 6 2

-Wall Footings Same steps of single footing Wall Footing No. Reinforcement in short direction

-Wall Footings Same steps of single footing Wall Footing No. Reinforcement in short direction Footing dimensions Shrinkage steel in long direction Width (m) Depth (m) e (m) Length (m) Area of steel (cm 2) # of bars Area of steel (cm 2/m) # of bars Spacing (cm) 1 1. 1 0. 4 0. 11 - 6. 4 5 Ф 14/m 7. 9 6 Ф 14 20 2 1. 4 0. 43 - 7. 2 5 Ф 14/m 10. 08 7 Ф 14 15 3 2 0. 4 - 8 7. 4 5 Ф 14/m 14. 4 10 Ф 14 20 4 2 0. 4 - 7 4. 5 3 Ф 14/m 14. 4 10 Ф 14 20 5 2 0. 7 - 4 13. 3 6 Ф 18/m 14. 4 10 Ф 14 20 6 3. 2 0. 8 - 5. 5 8. 8 6 Ф 14/m 23 15 Ф 14 20

-Elevator Footings Pu= 1030. 1 t, Ps = 1453. 8 t Footing Area (A)

-Elevator Footings Pu= 1030. 1 t, Ps = 1453. 8 t Footing Area (A) = Ps. /B. C. = 1030/55 = 18. 7 m² Footing dimensions : 4. 4*4. 4=19. 36 Ultimate pressure (qu) = 1453. 8 /19. 36 = 75 t/m 2 ФVc=41. 7 ton, Vu = 34. 5 ton ФVc > Vu ok ФVc=41. 7 ton, Vu = 35. 25 ton ФVc > Vu ok Elevator Footings Flexural design: Using approximation Mx 1=My 1= qu L²/16 = 75*(1. 6²)/16 = 12 ton. m Mx 2=My 2= qu L²/2 =49. 5 ton. m As=14. 5 cm²/m Use 8 Ф 18/m bottom steel in both directions.

*Analysis and Design of Stair : - going of the stair is 30 cm

*Analysis and Design of Stair : - going of the stair is 30 cm as standards - Flights and landings thickness will be taken as simply supported solid slab: height of landing=ln/20=270/20=13. 5 cm. m thickness is suitable. height of flights=ln/24 = 430/20=17. 9 cm, 18 cm thickness is suitable -Design of the flight : Wu=1. 2 DL+1. 6 LL = 1. 7 ton/m² Vn=Wu*L/2= 3. 65 ton, Vc=12. 3 ton. ФVc > Vu ok Mu=Wu*L²/8=3. 9 ton. m Use 5 Ф 14/m top and bottom steel -Design of landing: Wu=1. 2 DL+1. 6 LL = 1. 61 ton/m² Vn=Wu*L/2= 7. 2 ton, Vc=9. 8 ton. ФVc > Vu ok Use 8 Ф 14/m top and bottom steel Mu=Wu*L²/8=5. 35*2. 7²/8= 4. 8 ton. m

cross section of stair with steel in stair

cross section of stair with steel in stair

*Analysis and Design of walls -Basement Wall Soil load= 4. 65 ton/m² Flexure Design:

*Analysis and Design of walls -Basement Wall Soil load= 4. 65 ton/m² Flexure Design: Max -ve moment =qu*L²/15=4. 17 ton. m/m Max +ve moment =qu*L²/33. 6=1. 8 ton. m/m Use 3 Ф 12/m horizontal steel Use 4 Ф 16/m in vertical direction. -shear Walls The shear wall is modeled as 1 D and is subjected to force taken from the 3 D model of sap 2000: Vertical steel found for all shear wall equal 5 Ф 14/m

Horizontal steel was calculated as following: ФVc=95. 4 ton, Vu from sap=39. 1 ton

Horizontal steel was calculated as following: ФVc=95. 4 ton, Vu from sap=39. 1 ton ФVc > Vu As=14. 5 cm²/m ok Use 5 Ф 14/m Shear Wall No. Shear wall dimensions Vertical steel Horizontal steel Width (m) length (m) # of bars 1 0. 25 6. 9 5 Ф 14/m 2 0. 25 6. 3 5 Ф 14/m 3 0. 25 2. 6 5 Ф 14/m 4 0. 25 5. 5 5 Ф 14/m 5 0. 25 2. 2 5 Ф 14/m

Thank you

Thank you