2. Abstract
This project is a structural analysis and design of a residential building located in JENIEN City, The building is consisted of 7 floors.
The final analysis and design of building is done using a three dimensional (3D) structural model by the structural analysis and design software sap2000.

3. 3D Picture of the building

4. The preliminary dimensions of the structural elements are determined using one dimensional structural analysis for the structural members for gravity loads. Contain 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.

5. Contents CHAPTER ONE: (Introduction) * Description of project
* Materials
* Loads
* Design Code

6. CHAPTER TWO:(PRELIMINARY DESIGN )
* General
* Design of Rib Slab
* Design of columns
* Design of beams

7. CHAPTER THREE: (Three Dimensional Analysis And Design)
* General
* Modeling The Building as 3D
* Seismic Loads
* Analysis and Design of Slabs
* Analysis and Design of Beams
* Analysis and Design of columns
* Analysis and Design of Footings
* Analysis and Design of Stair
* Analysis and Design of walls

8. **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 building , which consists of 7 stories above the ground level . The area of each story is about 372 m².

9. Concrete strength f'c=300Kg/cm²
*Materials
Concrete strength f'c=300Kg/cm²
Modulus of elasticity equals Kg/cm²
Unit weight is 2.5 ton/m3.
Fy= 4200 kg/cm2 (420 Mpa)
Unit weight (ton/m3)
Material
2.5
Reinforced concrete
1.2
Concrete block
2.6
Tiles

10. *Loads: -Live Load= 0.25 t/m² -Superimposed dead load=0.27 t/m²
- Additional dead load=0.1 t/m²

11. *codes and standard: 1- ACI 318-08 (AMERICAN CONCRETE INSTITUTE)
2- UBC-97:(UNIFORM BUILDING CODE)

12. **CHAPTER TWO: PRELIMINARY DESIGN *General:
-Soil capacity = 2.8 kg/cm²
-Concrete strength f'c=300Kg/cm²
-Steel yield strength, Fy= 4200 kg/cm2
*General:

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

14. One end continuous = L / 18.5 = 425/18.5 =23 cm
Cantilever = L/8=160/8=20 cm
Then we assume thickness of slab (h) rib= 25 cm
Cross section in Ribbed slab

15. The distribution of ribs in the typical slabs shown in figure

16. The ribs in the slab are analyzed using sap2000 program
The ribs in the slab are analyzed using sap2000 program. As an example, the analysis result of rib are illustrated here: Use 2φ12 top and 2φ10 bottom steel
moment diagram for rib, ton.m

17. *Design of columns
In this project rectangular columns are used. And these columns can carry axial load and no moment.
-Design of Column :
The dimensions of the column 30*35cm, Ag=1050cm²,
And we use ρ between (1%- 4%)
Pu=149 ton
Pn=229 ton
As =11 cm²
Use 6 φ16

18. *Design of beams
After distributed the beam in the plan as shown in figure , we insert to sap2000 and insert each load on it which come from ribs slab or from external or internal wall and then design it.

19. Use 8φ18 top and 5φ18 bottom steel
-Design of beam (65*25):
moment diagram, ton.m
Area of steel , cm²
Use 8φ18 top and 5φ18 bottom steel

20. **CHAPTER THREE: Three Dimensional Analysis And Design *General:
This chapter provides analysis and design of 3D model for the building using sap2000 program. Figure below show 3D Model of it.
3D model

21. *Modeling The Building as 3D Structure :
**Sections:
-shear walls= 20cm
-Ribbed slabs are presented as one way solid slabs in y-direction. The thickness is calculated to be equivalent to ribs moment of inertia.
Z = (12*17*17*.5*52*8*21)/((12*17)+(52*8))
Z=16.9 cm
Ic = (52*8^3)/ 12 +(52*8*4.1^2) + 12*17*8.4^2
Ic= cm^4
52*(Equivalent)^3/ 12 =
H=18.74cm

22. **Loads: -beams: Variable in sections, we use concrete covers of 3 cm
Moment of inertia about 2 axis = Moment of inertia about 3 axis=0.35
-columns :
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/m2.
- Total super imposed dead load =0.37ton/m2

23. -seismic loads First we will define the equivalent static
As shown in the picture

24. we define the cases of the seismic loads Seismic-x, Seismic-y

25. *Structural Model Verification:
Equilibrium Check:
-Live Load:
Total Live load = KN
Live load from SAP = KN
% Error =0.8% < 5% ok

26. -Dead loads: Dead load from SAP =32216.6 KN Error= 1.5% < 5% ok.
Total load= KN
Dead load from SAP = KN
Error= 1.5% < 5% ok.

27. super imposed Dead load: total S.D = 9870 KN From Sap = 9622.7 KN
Error= 2% < 5% ok

28. *Analysis and Design of Slabs :
-check shear :

29. -Flexure Analysis and Design:

30. Rib no.
Max M(+ve)
Max M(-ve)
As (+ve)
As (-ve)
Bottom steel
Top steel
Stirrup Ø8 (spacing) 1
0.46
1.04
0.9
1.2
2Ф10
30 cm 2
0.53
1.066
1.3 3
0.57
1.6
2Ф12 4
0.52
1.1
30cm 5

31. Beam 1.2 (70*30) Out Side Of Building
*Analysis and Design of Beams:
Analysis and design output was taken from SAP2000.
Beam (70*30) Out Side Of Building
Main steel
stirrups (2leg 1ö 10mm)
total A/s
Spacing(cm)
span1
bottom
3ö16
span 1
0.001 10
top
0.048
Span2
span 2
0.051
Span3
span 3
0.002

33. *Analysis and Design of columns :
Analysis and design output was taken from SAP2000.
Design the worst column (has the max. axial load = 212 Ton)
Assume column dimension: 30cm *60cm
Pd=φλ*0.85(300(Ag-As)+ 4200*As)
= 0.65 *0.8 *0.85 (300(0.99*30*60)-(4200*0.01*30*60))
=275.6 ton > 212 ok
As = 0.01 *30*60 = 18cm2 →
Use 8φ18 mm

35. *Analysis and Design of Footings :
-Single Footings
Footing (F4) :
Pu= 212 t, Ps = 166 t
Footing Area (A) = Ps./B.C.= 166/28 = 5.9 m²
Footing dimensions : 2*3m
Ultimate pressure (qu) = 212 /6 = 35.3 t/m²
Wide beam shear :
ФVc=39.4 ton,
Vu = 35.3 ton
ФVc > Vu ok
Check Punching shear :
ФVc > Vu ok
ФVc=248.6 ton,
Vu=185.7 ton
Flexural design:
Mu = qu*L²/2= 25.4ton.m/m
As=16 cm²/m
Use 8Ф16/m bottom steel in both directions

37. Longitudinal Reinforcement # of bars in each direction
Footing No.
Footing dimensions
Longitudinal Reinforcement
Length
(m)
Width (m)
Depth (m)
Area of steel (cm2)
# of bars in each direction 1 3 2
0.5 16
8Ф16/m
1.6
7.7
4Ф16/m
2.6
2.2
0.6
10.98
7Ф16/m 4
3.1
2.7
0.7 5
3.4
7.59 6
3.6
110
22.14
7Ф20/m

38. Footing wall dimensions
-Wall Footings
Same steps of single footing
Footing
Wall No.
Footing wall dimensions
Vertical steel
Horizontal steel
Width (m)
height (m)
# of bars 1
1.60
0.40
5Ф14/m
6Ф16/m 2
1.40
0.35
5Ф16/m 3 4
1.30
4Ф16/m 5

39. Footing Area (A) = Ps./B.C.= 25.4m²
-Elevator Footings
Pu= 905 t, Ps = 710 t
Footing Area (A) = Ps./B.C.= 25.4m²
Footing dimensions : 5*5.5=27.5m²
Ultimate pressure (qu) = 905 /27.5 = t/m2
Wide beam shear 1: (simply supported)
ФVc=33 ton,
Vu = 29.4 ton
ФVc > Vu ok
Wide beam shear 2: (cantilever)
ФVc=33 ton,
Vu = 17.2 ton
ФVc > Vu ok .

40. Flexural design: M1: Max negative moment from sap= 21.4 t.m Mn =23.7
As= 15cm²
Use 8φ16
Max. positive moment = 8 t.m
Mn= 9 t.m
As=8 cm²
Use 5φ16

41. *Analysis and Design of Stair :
Maximum ( –ve)and (+ve)moments = 0.52Ton .m/m
Maximum ( –ve)and (+ve)moments = 0.52Ton .m/m
Maximum ( –ve)and (+ve)moments = 0.52Ton .m/m
*Analysis and Design of Stair :
- going of the stair is 30cm as standards
- Flights and landings thickness will be taken as simply supported solid slab:
height of landing=ln/20=270/20=13.5 cm.
15 cm thickness is suitable.
Rise of stair= 16cm,
-Design of the stair from sap M11:
Maximum ( –ve )and (+ve )moments
= 0.52Ton .m/m
=0.0009<
As =0.003 *100 *12 = 3.7 /m →use 3φ14 top and bottom steel is required

42. As =0.003 *100 *12 = 3.7 /m →use 3φ14 top and =0.003<< 0.0018
Maximum (–ve) and (+ve) moment is 1.8 Ton.m/m.
Maximum (–ve) and (+ve) moment is 1.8 Ton.m/m.
=0.003<<
As =0.003 *100 *12 = 3.7 /m →use 3φ14 top and
bottom steel is required

43. Design of shear walls Shear Wall No. Shear wall dimensions
Vertical steel
Horizontal steel
Width (m)
length (m)
# of bars 1
0.2
24.5
4Ф14/m 2
11.6 3 4 5
4.9 6
16.7
12.9

44. Thank you