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

Final design and details Work Plan Introduction. Review of 3D Modeling Seismic Design Final design and details Conclusion. An-najah National university

Project Description The project is located in {Betinua- Ramalah} The building has a total area of 2672.6 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 An-najah National university

Parking floor An-najah National university

Ground Floor An-najah National university

Repeated Floor An-najah National university

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 An-najah National university

Allowable bearing capacity of 2.5 Kg/cm² =250 KN/m² Soil Property Allowable bearing capacity of 2.5 Kg/cm² =250 KN/m² An-najah National university

Load Assumptions: Dead Load: Slab own weight for solid: 7.5 KN/m2 9 Dead Load: Slab own weight for solid: 7.5 KN/m2   Slab own weight for ribbed: 4.8 KN/m2 An-najah National university

Superimposed dead load: Total superimposed dead load : 4.08 KN/m2 An-najah National university

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). An-najah National university

3D Model An-najah National university

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. An-najah National university

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 . An-najah National university

Slab modifiers Ribbed Slab Solid Slab An-najah National university

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: An-najah National university

Etabs modifications Compatibility: An-najah National university

Equilibrium: An-najah National university

Stress strain relationship check It’s made on an interior span on beam (B8) Span (2) An-najah National university

% 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. An-najah National university

Dynamic Analysis Design response spectrum

Dynamic Analysis Response spectrum analysis method: according to seismic zone factor map Z= 0.15 An-najah National university

The structure rests in rock (soil type B). An-najah National university

Then Cv = 0.15, Ca = 0.15 An-najah National university

I = 1 , standard occupancy category for structure An-najah National university

period calculations and checks   An-najah National university

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 An-najah National university

    An-najah National university

  An-najah National university

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 An-najah National university

An-najah National university

An-najah National university

Defining response spectrum functions in SAP : An-najah National university

Then we define load cases in x and y directions   An-najah National university

An-najah National university

  An-najah National university

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 An-najah National university

An-najah National university

An-najah National university

Then the base shear values after scaling up were as follows: An-najah National university

check the modal mass participation ratio to insure that sum Ux and sum Uy are greater than 90%. An-najah National university

Check Drifft An-najah National university

So ∆m = 0.7*R*∆(Elastic = Sap ) < 0.025 H(story height) An-najah National university

Sample calculation: (Story 9) Diff x = 12. 9-11. 9= 1 mm Diff y = 11 Sample calculation: (Story 9) Diff x = 12.9-11.9= 1 mm Diff y = 11.55-10.57 = .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 An-najah National university

Load combinations An-najah National university

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 = 0.075 D So the new load combination will be : 1.275 D + 0.5 L +1 EQx + 0.3EQy 1.275 D + 0.5 L +1 EQy + 0.3EQx An-najah National university

So the new load combination will be : 0.825 D + 1 EQx + 0.3 EQ 0.825 D + 1 EQy + 0.3 EQx An-najah National university

Load combinations for serves loads: An-najah National university

Ev = Zero……. According to UBC97 code   Ev = Zero……. According to UBC97 code So the new load combination will be : 0.9D + 0.71 EQx + 0.21EQy 0.9 D + 0.71 EQy + 0.21EQx   So the new load combination will be : D + 0.75 L + 0.54 EQx + 0.16EQy D + 0.75 L + 0.54 EQy + 0.16 EQx An-najah National university

Design of slabs حسام The structural system for the ground floor slab is two way solid slab(30 cm) with drop beams. An-najah National university An-najah National university

ØVc =171.46 KN Vu =155.82 KN Check for shear V13 The value of V13 for slab of ground floor ØVc =171.46 KN Vu =155.82 KN An-najah National university

ØVc =171.46 KN Vu =66.56 KN Check for shear V23 The value of V23 for slab of ground floor ØVc =171.46 KN Vu =66.56 KN An-najah National university

Analysis and Design for flexure The values of M22 for the ground floor An-najah National university

For positive moment. The values of M22 for the ground floor An-najah National university

An-najah National university

An-najah National university

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 An-najah National university

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 An-najah National university

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 An-najah National university