An-Najah National University Faculty of Engineering Civil Engineering Department

1. An-Najah National University Faculty of Engineering Civil Engineering Department Terra Santa School Structural Design and Analysis Prepared By: Bara Shawahna Khaled Malhis Nadeem AL-Masri Supervised By : Dr. Mahmud Dwaikat

2. Introduction & general description of the project • 3D modeling • Shear walls design • Design of columns • Design of beams • Slabs design • Foundation design Outline

3. Our Graduation project is the design of a school in Jericho named as Terra Santa School. This school was designed by Al-Diyar Consultant and we will check on their design. • The school consists of three floors with total plan area of (3866.5m2). Introduction

4. Scope of work

5. Methodology • SAP 2000 program will be used as main analysis tool. • ACI 318-08 for design. • Live loads are taken from ASCE 7 -05 code. • UBC 97 for seismic design.

6. Reinforced concrete • The unit weight of concrete () = 25 kN/m3. • The required compressive strength after 28 days • For slabs & beams fc is 25 MPa. • For Columns ,shear walls & footing fc is 30 MPa. • The yield steel bars required Fy = 420 MPa. Materials used in the project:

7. For each block we chose the following floor system: • Block A ( one way ribs slab ) • Block B ( one way ribs slab ) • Block C ( two way ribs slab ) Blocks A & B. It’s very clear to see that block A , B has a uniform grid for columns with clear path of loading (one-way), therefore, we took the architectural layout for the columns. Ribbed slabs are known for their economic efficiency. The thickness calculation follows the ACI-318 code Block C. Block C has a different shape and dimensions and according to ACI – code requirements the slab should be designed as two-way ribbed slab for economic and deflection requirements. This is because the spans of each panel in Block C have approximately equal lengths. Structural Systems

9. Dead load: • Own weight for one way slabs = 3.54 kN/m2 . • Own weight for two way slabs = 4.86 kN/m2 . • Super imposed load = 3 kN/m2 . Live load: • for all the class room = 2 KN/m2. • for corridors = 5 KN/m2. Loads

10. We will design the seismic load by using SAP2000. Several methods is used in SAP for seismic which is: • Dynamic analysis: • Response spectrum. • Time history. • Equivalent static force Equivalent static method will be used for comparison [Block A only] and as a cross-check on the results of response spectrum analysis. Because response spectrum is more realistic and covers the modal shapes of the building, we will use it as a main tool for seismic design. Analysis and design against Seismic loads

11. The seismic force effect on the structure can be translated to equivalent lateral force at the base of the structure and then this force will be distributed to the different stories and then to the vertical structural elements (frames and/ or shear walls). • This method is best applied to Regular Structure only.

12. We will use the UBC97 method for analysis, because of available data and factors in our region. • The total design base shear in a given direction shall be determined from the following formula: • Where • Z= seismic zone factor. • I= importance factor. • R= Response Modification Factor • CV= velocity seismic coefficient. • W= the total dead load. Design for earthquake by equivalent lateral force method (static method) for block A

13. From this map the project in Jericho Z = 0.3 Seismic Zone Factor Z

14. Soil type D CV=0.54 Soil type D Ca=0.36 Cv and Ca table

15. I = 1.25 Importance Factor table

16. The overall system is dual system R for building between (4.2-6) =5.6 Response Modification Factor “R” table

17. Hn= height of structure in meters = 15m • Ct = factor for this case =0.0488 • T= is the basic natural period of a simple one degree of freedom system which is the time required to complete one whole cycle during dynamic loading. • And after calculation T=0.3719 sec, T from sap for block A =0.25 sec. Difference can be due to presence of shear walls that increase the stiffness of the building block. Tcalculation

18. Then calculate × w = 0.33W • Vmin. = 0.11X0.36 X1.25X w=0.0396W • Vmax= 0.2W • Vmax= 0.20 W use it because it Vcalculated>Vmax • For block A the weigh is = 12206 kN • V=0.20 X12206 =2441.2kN • From SAP Vtotal equivalent static force equal =2125kN. • From SAP VtotalResponse spectrum equal =1950kN.

19. We use sap to design and analyze the project the design response spectrum is shown in Response spectrum method

20. We find CA=0.36,CV=0.54 Then we have this curve

21. Then we should define a load cases in the X-direction, Y-direction, Z-direction • When we define X-direction for example: • We scale factor is for X direction == 2.227 • And 33% of it for to the other direction Y = 0.33X2.227=0.7329 ( as per UBC97 and ASCE requirements) .

22. 3D Modeling Structure overview

23. Block B

24. Block C

25. Compatibility Check

26. Dead load: • Slab dead load = area X slab own weight per square meter =1185.84 X 3.54=4136.4 kN • Shear wall load = walls volume X concrete weight per volume = 28.7 X0.3 X15 X 25=3288.75 kN • Columns dead load = column volume X concrete weight per volume = 30 X 15 X0.4 X 0.4 X 25=1800 kN • Beams dead load = beams volume X concrete weight per volume = 3691.4 kN • Total dead load = 12917.6 kN • Total dead load form SAP = 12956.699 • % Error = 0.3% which is acceptable Equilibrium checkFor block A

27. Live load: Structure area X live load per square meter =1185.84 X 5 =5929.2 kN Live load from SAP = 5929.2 kN % Error = 0 • Superimposed dead load: Structure area X superimposed load per square meter = 1185.84 X 4.5 = 5336.28 kN Superimposed load from SAP = 5336.2 kN % Error = 0

28. We take block A as example to do this check. In this check we take the moment from 3D modeling and find the weight and then comparing it with the hand calculated weight • Moment resulted from dead load in block A in beam B1 is: Moment equilibrium check

29. Check for sway and non-sway: • < 0.05 • Its acceptable value.  non-sway

30. axial force value Pu < 0.1 Ag fc • Assume singly reinforced section with reinforcement at d= 0.8h: • kN Design of shear walls

31. Design of Columns

32. Columns in Category C4(110X50)cm • We take the moment value from 3D –modal of we have =. • For the effective length factor (K) we take the building non sway so we take k from φmajor= φ a= , φ b=50 and φ minor= and φ b=50the which is between equal 0.98 take it 1. • Check slenderness limit  non-slender column • Using strength calculation, Pu<𝝫Pn for the same column, we got Ag = 5500000mm2. So the other dimension is5500000/500= 110mm. This dimension should be larger or equal the least dimension so let it equal 500mm, so we took it equal to 500 mm • The dimension of column 110X50 cm so Ag= 5500cm2 • Strength calculations for this column are done as follows: Design of the Columns against seismic load

33. From SAP2000 and 3D-model we take Pu=8287 and it equal 𝝫Pn then we go to the interaction diagram and take the steel ratio value= 0.012 • 𝝫Pn/bh=15 =2.15 ksi • ¥=h-2couver/h = 0.9 • Mu/bh2=1.44 = 0.2 ksi

34. From SAP2000 the shear value on the column is equal 117 kN <463.9 it’s Ok • Since the column is not subjected to shear, we use minimum steel for shear reinforcement so as to hold vertical bars and • confine concrete. • The stirrup spacing must be the minimum of the following • 48 ds (diameter of stirrups)………..From code ACI 318-08 • 16 db(diameter of bar) )………..From code ACI 318-08 • Least diminution of the section )………..From code ACI 318-08 • Not more than d/2……..From ACI 315-99 • Therefore, use ᴓ10 • 48 * 10 = 480 mm • 16 * 20= 320 mm • 500 mm • 460/2= 230 mm • Therefore use 4ᴓ10/150mm for stirrups along the column

36. Beam distribution and categories Design of Beams

38. For Beam b1 at block A under seismic load (50X35 cm)

40. design for shear

41. Near the column face • Use 1 𝝫 10 /100 mm • At the middle • S=500> S max • So use S max 230 mm • Use 1 𝝫 10 /150 mm • Right span • S= 263 mm > S max • Use 1 𝝫 10 /100mm

43. The maximum span length is about 2.7m as shown in Block A,B Map, and therefore the thickness of the slab (assumed one-way ribbed) according to the table will be L /18.5 =15 cm and we used 20 cm for block A& B(S1). Slab Design

44. The maximum span length is about 7.55m as shown in Block C Map, and therefore the thickness of the slab two-way ribbed according to the table will be Ln /30 =25.1 cm and we used 30 cm.

45. For Slab S1 at block A (20 cm): • Slab S1 is one way ribbed slab with 9 spans 2.7m for each. • Dead and super imposed load = 8.04 kN/m2 • Live load= 5kN/m2 due to corridor live load. • All value is less than so use for design. • As­min = X bw X d= 0.0033X120X180=72mm .Used 2ᶲ10.(As bottom steel

46. Top rib steel All value is less than so use for design. • As­min = X bw X d= 0.0033X120X180=72mm .Used 2ᶲ10. • For Shrinkage net steel • As­min = X b X d= 0.0018X1000X60=108mm2 Used 2ᶲ8

47. m11 (x-direction moment) Design for two way ribbed slab in block C

49. Design for negative moment (A-B section in x-direction) Negative moment at the middle near the beam face. Negative moment at right edge • As­ = X bw X d = 0.00928 X 120 X 280= 312 mm • Use 2𝝫16 for both left and right edge (as top steel).

50. Foundation