This document outlines the design of the structural system for the Terra Santa School project in Jericho, Palestine. The school consists of three blocks designed using reinforced concrete. The document describes modeling the structure in SAP2000, analyzing seismic and gravity loads, and designing the structural elements including shear walls, columns, beams, and foundations according to code specifications. Analysis methods like response spectrum analysis and equivalent lateral force are used to design for seismic loads. Reinforcement is designed for various structural elements based on strength calculations.

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. OUTLINE
 Introduction & general description of the project
 3D modeling
 Shear walls design
 Design of columns
 Design of beams
 Slabs design
 Foundation design

3. INTRODUCTION
 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).

4. Scope of work

5.  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.
Methodology

6. MATERIALS USED IN THE PROJECT:
 Reinforced concrete
 The unit weight of concrete (𝛾) = 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.

7. STRUCTURAL SYSTEMS
 For each block we chose the following floor system:
1. Block A ( one way ribs slab )
2. Block B ( one way ribs slab )
3. 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.

9. LOADS
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.

10. ANALYSIS AND DESIGN AGAINST SEISMIC
LOADS
 We will design the seismic load by using SAP2000. Several methods is
used in SAP for seismic which is:
 Dynamic analysis:
1. Response spectrum.
2. 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.

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. DESIGN FOR EARTHQUAKE BY EQUIVALENT LATERAL FORCE
METHOD (STATIC METHOD) FOR BLOCK A
 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
𝑉 =
𝐶𝑣 𝐼
𝑅 𝑇
𝑤
1. Z= seismic zone factor.
2. I= importance factor.
3. R= Response Modification Factor
4. CV= velocity seismic coefficient.
5. W= the total dead load.

13. SEISMIC ZONE FACTOR Z
From this map the project in
Jericho Z = 0.3

14. CV AND CA TABLE
Soil type D
CV=0.54
Soil type D
Ca=0.36

15. IMPORTANCE FACTOR TABLE
I = 1.25

16. RESPONSE MODIFICATION FACTOR “R” TABLE
The overall system is dual system
R for building between (4.2-6)
=5.6

17. T CALCULATION
 𝑇 = 𝐶𝑡(ℎ𝑛)
3
4
 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.

18.  Then calculate 𝑉 =
0.54𝑋1.25
5.5𝑋0.3719
× w = 0.33W
 𝑉𝑚𝑖𝑛 = 0.11 𝑋 𝐶𝑎 𝑋 𝐼 𝑋 𝑊
 Vmin. = 0.11X0.36 X1.25X w=0.0396W
 Vmax=
2.5𝑋𝐶𝑎𝑋𝐼
𝑅
𝑊  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 Vtotal Response spectrum equal =1950kN.

19. RESPONSE SPECTRUM METHOD
 We use sap to design and analyze the project the design response
spectrum is shown in

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 =
𝑔𝐼
𝑅
=
9.81𝑋1.25
5.6
= 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. EQUILIBRIUM CHECK
FOR BLOCK A
 Dead load:
1. Slab dead load = area X slab own weight per square meter =1185.84
X 3.54=4136.4 kN
2. Shear wall load = walls volume X concrete weight per volume = 28.7 X0.3
X15 X 25=3288.75 kN
3. Columns dead load = column volume X concrete weight per volume =
30 X 15 X0.4 X 0.4 X 25=1800 kN
4. Beams dead load = beams volume X concrete weight per volume =
3691.4 kN
5. Total dead load = 12917.6 kN
6. Total dead load form SAP = 12956.699
7. % Error = 0.3% which is acceptable

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. MOMENT EQUILIBRIUM CHECK
 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:

29.  Calculating the moment =
39+44.75
2
+ 28.9 = 70.775 𝑘𝑁. 𝑚

𝑤×𝑙2
8
= 𝑀 ,
70.775×8
6.82 = 12.244
𝑘𝑁
𝑚
 To convert it to kilo newton per square meter divide it over tributary
area which equal 3 meter
 W = 4.08 kN/m2
 Calculated dead load = 3.54 kN/m2.
 % of error =
4.08−3.54
3.54
× 100% = 15%
 Check for sway and non-sway:

∑𝑃𝑢 𝑋 𝐷𝑒𝑓𝑙𝑖𝑐𝑡𝑖𝑜𝑛
∑𝑉𝑢 𝑋 𝑙𝑖𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑜𝑙𝑢𝑚𝑛
< 0.05

361300
6068820
= 0.059
 Its acceptable value.  non-sway

30. DESIGN OF SHEAR WALLS
 Design B2 in block B
 axial force value Pu < 0.1 Ag fc
 The maximum moment in the wall is 6824 kN.m
 Assume singly reinforced section with reinforcement at d= 0.8h:
 𝜌 =
0.85 𝑓𝑐
𝑓𝑦
1 − 1 −
2𝑀
0.85ᴓ𝑏𝑑2𝑓𝑐
 =
0.85×30
420
1 − 1 −
2×6824×106
0.85×0.9×300×45002×30
= 3 × 10−3
 𝐴𝑆 = 0.003 × 1000 × 300 =
900 𝑚𝑚2
𝑚
5∅16/𝑚
 𝑉
𝑐 =
1
6
× 𝑓𝑐
′
× 𝑏𝑤 × 𝑑=0.1666𝑥 30𝑥300𝑥4500/1000 = 1233𝑘𝑁
 Vu from sap =756 kN
 𝑉
𝑛 =
𝑉𝑢
∅
, 𝑤ℎ𝑒𝑟𝑒 ∅ = 0.75
 𝑉
𝑛 =
756
0.75
= 1008 𝑘𝑁
 𝑉
𝑐 > 𝑉
𝑛 𝑢𝑠𝑒 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑠𝑡𝑒𝑒𝑙 =(1 ∅12/20𝑐𝑚)

31. DESIGN OF COLUMNS

32. DESIGN OF THE COLUMNS AGAINST SEISMIC
LOAD
 Columns in Category C4(110X50)cm
 We take the moment value from 3D –modal of
𝑀1
𝑀2
we have
𝑀1
𝑀2
=
312
298
= 1.04.

 For the effective length factor (K) we take the building non sway so we
take k from φmajor= φ a=
𝐸𝑐𝐼𝑐/𝑙𝑐
𝐸𝑏𝐼𝑏/𝑙𝑏
= 12.4 , φ b=50 and φ minor=
𝐸𝑐𝐼𝑐/𝑙𝑐
𝐸𝑏𝐼𝑏/𝑙𝑏
=
11.99 and φ b=50the which is between equal 0.98 take it 1.

𝐾𝐿
𝑅
= 34 − 12
𝑀1
𝑀2
 Check slenderness limit
1𝑋3000
0.3 ℎ
< 34 − 12
312
298
 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:

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. 𝑉
𝑐 =
1
6
× 𝑓𝑐
′
× 𝑏𝑤 × 𝑑 =⇒ 𝑉
𝑐 =
1
6
× 30 × 1100 × 460/1000 = 461.9 𝑘𝑁
𝑉
𝑛 =
𝑉
𝑢
∅
, 𝑤ℎ𝑒𝑟𝑒 ∅ = 0.75
𝑉
𝑛 =
88
0.75
= 117 𝑘𝑁
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

35. Column Dimension Main Steel Stirrups
C1 40X40=1600 8ᴓ20 1ᴓ10/150mm
C2 50X50=2500 12ᴓ18 3ᴓ10/150mm
C3 85X50=4250 18ᴓ18 3ᴓ10/150mm
C4 110X50=5500 28ᴓ18 4ᴓ10/150mm

36. DESIGN OF BEAMS
 Beam distribution and categories

38. FOR BEAM B1 AT BLOCK A UNDER SEISMIC LOAD
(50X35 CM)

39.  𝜌𝑚𝑖𝑛 = 𝑚𝑎𝑥
1.4
𝑓𝑦
,
0.25∗ 𝑓𝑐
𝑓𝑦
= 𝑚𝑎𝑥
0.00333,
0.00297

 𝑢𝑠𝑒 𝜌𝑚𝑖𝑛 = 0.00333
 For left span
 Negative moment need 1284 mm2 (5𝝫20 mm).
 For positive moment need 945 mm2 (4𝝫18 mm).

 For the right span
 Use (5𝝫20 mm) for top steel
 And (4𝝫18 mm) for bottom steel

40. DESIGN FOR SHEAR

41.  Near the column face

𝐴𝑣
𝑠
= 0.774
 𝑠 =
157
0.774
= 200 𝑚𝑚
 Use 1 𝝫 10 /100 mm
 At the middle

𝐴𝑣
𝑠
= 0.292
 S=500> S max
 So use S max 230 mm
 Use 1 𝝫 10 /150 mm
 Right span

𝐴𝑣
𝑠
= 0.596
 S= 263 mm > S max
 Use 1 𝝫 10 /100mm

43. SLAB DESIGN
 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).

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 𝜌 value is less than 𝜌𝑚𝑖𝑛 so use 𝜌𝑚𝑖𝑛 for design.
 Asmin = 𝜌𝑚𝑖𝑛 X bw X d= 0.0033X120X180=72mm .Used 2ᶲ10.(As bottom steel

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

47. DESIGN FOR TWO WAY RIBBED SLAB IN BLOCK C
m11 (x-direction moment)

48.  For left span the positive moment is
 Mu =
86.4
(18.7−15.8)
= 29.8 𝑘𝑁. 𝑚
 𝜌 =
0.85 𝑓𝑐
𝑓𝑦
1 − 1 −
2𝑀𝑢
0.85ᴓ𝑏𝑑2𝑓𝑐
= 0.0092
 As = 𝜌 X bw X d= 0.0092X120X280=310 mm
 Use 2𝝫16
 For right span the positive moment is
 Mu =
30
16.5−15.5
= 30 𝑘𝑁. 𝑚
 𝜌 =
0.85 𝑓𝑐
𝑓𝑦
1 − 1 −
2𝑀𝑢
0.85ᴓ𝑏𝑑2𝑓𝑐
= 0.00928
 As = 𝜌 X bw X d= 0.00928 X 120 X280= 310 mm
 Use 2∅16

49.  Design for negative moment (A-B section in x-direction)
Negative moment at the middle near the beam face.
 Mu =
78
(16−14)
= 39 𝑘𝑁. 𝑚
 𝜌 =
0.85 𝑓𝑐
𝑓𝑦
1 − 1 −
2𝑀𝑢
0.85ᴓ𝑏𝑑2𝑓𝑐
= 0.0125
 As = 𝜌 X bw X d = 0.0125 X 120 X 280= 420 mm
 Use 2𝝫20 (as top steel)
Negative moment at right edge
 Mu =
60
(18−16)
= 30 𝑘𝑁. 𝑚
 𝜌 =
0.85 𝑓𝑐
𝑓𝑦
1 − 1 −
2𝑀𝑢
0.85ᴓ𝑏𝑑2𝑓𝑐
= 0.00928
 As = 𝜌 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

51. Design of F1 for C1
 Pservice = 1200kN.
 Pultimate=1500kN.
 Bearing capacity= 250 kN/m2
 Area of footing =
𝑆𝑢𝑟𝑣𝑖𝑐𝑒 𝑙𝑜𝑎𝑑
𝑝𝑒𝑎𝑟𝑖𝑛𝑔 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦
=
1200
250
= 4.8 m2
 So we will use B= 2.2, L= 2.2
 Footing long side (l)=
𝐿−𝐿 𝑐𝑜𝑙𝑢𝑚𝑛
2
=
2.2−0.4
2
=0.9m

52. Wide beam shear
 q Ultimate =
𝑃𝑢𝑙𝑡𝑖
𝐵𝑋𝐿
=
1500
2.2𝑋2.2
=310kN
 Assuming effective depth (d) of 400mm
 𝝫VC = 0.75 ×
1
6
× 𝑓𝑐
′
𝑏𝑤𝑑 = 273.8 𝑘𝑁
 𝑉
𝑢 = 𝑞𝑢𝑙𝑡. ×
𝑏𝑤
2
−
𝑐𝑜𝑙𝑢𝑚𝑛 𝑠ℎ𝑜𝑟𝑡 𝑑𝑖𝑚𝑖𝑛𝑠𝑖𝑜𝑛
2
−
𝑑
1000
 𝑉
𝑢 = 310 ×
2.2
2
−
0.4
2
−
400
1000
= 155 𝑘𝑁
 𝝫VC>𝑉
𝑢 , so d is accepted.
Check for punching shear
 VC in punching shear is: (𝝫=0.75)
 𝝫VC =
1
6
1 +
2
𝛽
𝑓𝑐
′
𝑏0𝑑 = 0.75 × 0.17 × 1 +
2
1
× 30 × 3200 ×
400
1000
= 2629𝑘𝑁
 𝑉
𝑢,𝑝 = Pultimate
– 𝑞𝑢𝑙𝑡. × 0.4 + 𝑑 × 0.4 + 𝑑 𝑤ℎ𝑒𝑟𝑒 𝑑 𝑖𝑛 𝑚𝑒𝑡𝑒𝑟
 𝑉
𝑢,𝑝 =1301.65 kN
 ∅𝑉𝐶,𝑝 = 2629.6 𝑘𝑁 > 𝑉
𝑢,𝑝 𝑖𝑡′
𝑠 𝑂. 𝐾 , 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑖𝑠 𝑒𝑛𝑜𝑢𝑔ℎ.

53.  Design for flexure
 M ultimate= 𝑞𝑢𝑙𝑡 ×
𝑙𝑜
2
2
= 125.5 kN/m2
 Where 𝑙𝑜is length of cantilever in x or y direction
 𝜌 =
0.85 𝑓𝑐
𝑓𝑦
1 − 1 −
2𝑀
0.85ᴓ𝑏𝑑2𝑓𝑐
= 0.2108%
 𝜌𝑚𝑎𝑥. = 𝜌𝛷=0.9,∈𝑡=0.005 = 0.375𝛽1
0.85×𝑓𝑐
′
𝑓𝑦
, 𝑤ℎ𝑒𝑟𝑒 𝛽1 = 0.85
 𝜌𝑚𝑎𝑥. = 0.0161 𝜌𝑚𝑎𝑥. > 𝜌
 As = 0.002108 X 1000 X 400=843.1 mm2
 Asmin. = 0.0018 X 1000 X 400=864 mm2
 Asmin>As
 So use As =Asmin=864mm2
footing
number
p
service
p
ultimate
footing
area
b l d As As,min
F1 1200 1500 4.8 2.2 2.2 400 843 864
F2 2860 3720 11.44 3.5 3.5 500 1862 1044
F3 4000 5227 16 4 4 600 2273 1224
F4 6200 8245 24.8 5 5 900 2507 1764

56. DESIGN OF MAT (1)IN BLOCK C
Mat 1 plan Mat 1 displacement

57. moment (m11) in x-axis

58. the maximum moment is 238kN.m. in x-axis
𝜌 =
0.85 𝑓𝑐
𝑓𝑦
1 − 1 −
2𝑀
0.85ᴓ𝑏𝑑2𝑓𝑐
= 0.1%
𝜌 < 𝜌𝑚𝑖𝑛
Use 𝜌 = 0.0018
As = 0.0018×1000×800=1440 mm/m
the maximum moment is 414 kN.m in y-axis
T hen calculate 𝜌 =
0.85 𝑓𝑐
𝑓𝑦
1 − 1 −
2𝑀
0.85ᴓ𝑏𝑑2𝑓𝑐
= 0.173%
𝜌 < 𝜌𝑚𝑖𝑛
Use 𝜌 = 0.0018
As = 0.0018×1000×800=1440 mm/m
For shrinkage steel As = 0.5×0.0018×1000×800 =720 mm/m (5 𝝫14 mm / m).

59. C2 C2
C1 C1

61. THANK YOU