1. Faculty of Engineering Civil Engineering Department
An-Najah National University
Faculty of Engineering
Civil Engineering Department
Graduation Project II Analysis And Design Of a Multi-Functional Building for Gravity and Secimice loads
Supervisor: Dr. Munther Diab.
Prepared by :
Anwar Hamza Husam Isra
Zaareer Hroub Riyal Naser

2. Final design and details
Work Plan
Introduction.
Review of 3D Modeling
Seismic Design
Final design and details
Conclusion.
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3. Project Description The project is located in {Betinua- Ramalah}
The building has a total area of m2.
The project consists of:
Floor
Function
Third basement floor
Parking
593.43
Second basement floor
Storage
481
First basement and ground floor
313
First to fifth floor
Residential
270
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4. Parking floor
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5. Ground Floor
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6. Repeated Floor
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7. Modulus of elasticity(MPs) beams, slabs , stairs, columns
Structural Materials:
Reinforced concrete
Unit weight of reinforced concrete = 25 KN/m³
Steel (Rebar, shrinkage mesh and stirrups)
Structural element
fc` (MPs)
Concrete type
Modulus of elasticity(MPs)
beams, slabs , stairs, columns 24
B300
23025
Footings 28
B350
24870
shear walls 35
B450
27805
Yielding strength (Fy) = 420MPa.
Modulus of elasticity (Es) = 200GPa
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8. Allowable bearing capacity of 2.5 Kg/cm² =250 KN/m²
Soil Property
Allowable bearing capacity of 2.5 Kg/cm² =250 KN/m²
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9. Load Assumptions: Dead Load: Slab own weight for solid: 7.5 KN/m29
Dead Load:
Slab own weight for solid: 7.5 KN/m2
Slab own weight for ribbed: 4.8 KN/m2
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10. Superimposed dead load:
Total superimposed dead load : 4.08 KN/m2
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11. Live Load :
In our project for manufacturing uses, LL on ground floor =12 KN/m² (for heavy industries) and LL on other floors = 6 KN/m² (for light industries).
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12. 3D Model
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13. Illustration
The architectural plans for this project were used to build 3D ETABS model without any changes, as shown
There are a lot of vertical shear walls in the right side of the plan area comparing with the left side.
The unsymmetrical distribution of shear walls made torsion problem and differential deformations.
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14. The first mode is torsion and there is difficult to deal with this type of mode and to predict their effects on the building.
To solve this problem ,the case discussed by architectural to move one of the two lifts( that boundaries mainly by vertical shear walls) from right side to left side to reduce the eccentricity between the center of mass and the center of rigidity . And complete the project according to the modified architectural plans .
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15. Slab modifiers
Ribbed Slab
Solid Slab
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16. Moment of inertia about 2 axis: 0.35
Also, for beams , columns and shear walls the modifiers are as follows:
Beams:
Torsional constant: 0.35
Moment of inertia about 2 axis: 0.35
Moment of inertia about 3 axis: 0.35
Columns:
Torsional constant: 0.7
Moment of inertia about 2 axis: 0.7
Moment of inertia about 3 axis: 0.7
Shear Walls:
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17. Etabs modifications
Compatibility:
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18. Equilibrium:
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19. Stress strain relationship check
It’s made on an interior span on beam (B8) Span (2)
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20. % difference >10%
Because when using 1D model hand calculation, there is assumption that the load transfers from slabs to beam without considering the existence of shear walls, but when 3D model built, there are many shear walls that taken the most load due to their very high stiffness, so that ETABS results will give smaller load for the beam comparing with 1D model that depend on tributary area and ignoring the existence of shear walls .Therefore, ETABS result is the correct answer, but to ensure that etabs model will be given correct results ,comparing between 1d etabs model results and hand calculation results will be done.
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21. Dynamic Analysis Design response spectrum

22. Dynamic Analysis Response spectrum analysis method:
according to seismic zone factor map
Z= 0.15
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23. The structure rests in rock (soil type B).
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24. Then Cv = 0.15, Ca = 0.15
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25. I = 1 , standard occupancy category for structure
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26. period calculations and checks
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27. Apply 1 kN/m2 for each slab in each floor in the x direction
Apply 1 kN/m2 for each slab in each floor in the x direction
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30. determine the structure system resisting the lateral force :
In X-direction:
the reaction in shear wall equal 75% of total reaction
In y-direction: % lateral load resisted by columns = 16% of total load % of load resisted by column < 25% load resisted by shear wall, so the system is shear wall system, R = 4.5
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33. Defining response spectrum functions in SAP :
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34. Then we define load cases in x and y directions
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37. Since loads from response spectrum < loads from static method
The result from Sap:
Since loads from response spectrum < loads from static method
We need to adjust response spectrum seismic loads to be equal to that of static method.
Scale factor needed to modify response spectrum
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40. Then the base shear values after scaling up were as follows:
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41. check the modal mass participation ratio to insure that sum Ux and sum Uy are greater than 90%.
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42. Check Drifft
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43. So ∆m = 0.7*R*∆(Elastic = Sap ) < 0.025 H(story height)
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44. Sample calculation: (Story 9) Diff x = 12. 9-11. 9= 1 mm Diff y = 11
Sample calculation: (Story 9) Diff x = = 1 mm Diff y = = .98 mm R = 4.5 ∆ = 0.7*4.5*1 = 3.15 mm limit = 0.025*3*1000 = 75 mm 3.15< 75 mm ……… OK
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45. Load combinations
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46. Ultimate Load combination in Sap: 1.4 D ………………… (12.1)
1.2 D +1.6 L ………………(12.2)
1.2D + 0.5L +1E……………..(12.5)
E =ϼ Eh + Ev
p = redundancy factor = 1 according to code
Eh: earthquake in both x and y direction
Ev = 0.5Ca I D ……. According to UBC97 code
In our project:
Ca= 0.15
Ev = 0.5* 0.15* 1 *D = D
So the new load combination will be :
1.275 D L +1 EQx + 0.3EQy
1.275 D L +1 EQy + 0.3EQx
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47. So the new load combination will be : 0.825 D + 1 EQx + 0.3 EQ
0.825 D + 1 EQy EQx
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48. Load combinations for serves loads:
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49. Ev = Zero……. According to UBC97 code
Ev = Zero……. According to UBC97 code
So the new load combination will be :
0.9D EQx EQy
0.9 D EQy EQx
So the new load combination will be :
D L EQx EQy
D L EQy EQx
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50. Design of slabs حسام
The structural system for the ground floor slab is two way solid slab(30 cm) with drop beams.
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51. ØVc =171.46 KN Vu =155.82 KN Check for shear V13
The value of V13 for slab of ground floor
ØVc = KN
Vu = KN
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52. ØVc =171.46 KN Vu =66.56 KN Check for shear V23
The value of V23 for slab of ground floor
ØVc = KN
Vu =66.56 KN
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53. Analysis and Design for flexure
The values of M22 for the ground floor
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54. For positive moment. The values of M22 for the ground floor
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57. Conclusion Some challenges
After modeling and getting the result from ETAPS we found that a torsional modes are founded due to the irregularity of the building
Torsional irregularity
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58. plan before redistributed the shear walls
So shear walls were redistributed in order to get more uniform distributions of stiffnesses and get over the torsional mode
plan before redistributed the shear walls
plan after redistributed the shear walls
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59. Vertical weight irregularity:
Mass irregularity shall be considered to exist where the effective mass of any story is more than 150% of the effective mass of an adjacent story. The ground floor is heavier than the other repeated floors. And the roof was designed as the floor below it, but actually in our society people put water tanks that have large weights on the roof diaphram. This problem needs to be considered.
Vertical weight irregularity
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