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Analysis and Design of a Multi-Story Building in Nablus, Palestine

This graduation project analyzes and designs a building in Nablus, including seismic and wind load analysis, structural calculations, and design principles. Chapters cover various aspects such as columns, beams, slabs, wind loads, and earthquake forces.

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Analysis and Design of a Multi-Story Building in Nablus, Palestine

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  1. Abstract This graduation project aims to analyze and design a building, Northern mountain, Nablus-Palestine, consists of a garage, offices and residential apartments. The structure is 14 stories and the area of each floor is 330.11 . The total area of the project is 4621.54

  2. The project chapters: In GP1, the following have been done: • Chapter 1: Introduction, which describes the structure location, loads, materials, codes and standards and the basic structural system of the structure. • Chapter 2: preliminary analysis and design for slab systems-one way ribbed slab system. • Chapter 3: preliminary analysis and design for beams using 2D analysis on SAP. • Chapter 4: design of columns and use live load reduction factor.

  3. The project chapters: • Chapter 4: redesign of columns • chapter5: Wind load analysis. • Chapter 6: Earthquake analysis. • Chapter 7: 3D modeling for the project. • Chapter 8: Stairs design. • Chapter 9: Tie beams and ramp design. • Chapter 10: Design of beams based on SAP2000 results. • Chapter 11: Design of slab- Two way solid slab. • Chapter 12: Analysis and design of footings.

  4. Chapter four column SAP AS for columns this section in it

  5. This table for number of bars

  6. Section in column 11

  7. Chapter Five: Analysis for Shear walls Lateral loads • The design wind loads for buildings and other structures , shall be determined using one of the following procedures: 1- Simplified Procedure for low rise building(h<18m). 2- Analytical Procedure. 3- Wind Tunnel Procedure.

  8. Design procedures: • Basic wind load and many constants shall be determined; the wind load is obtained from the following equation: • P = qz*G*Cp Where: - p is the wind load(N/mm2) - qz is the wind pressure - G is the gust effect factor - Cp is exposure coefficient • qz= 0.613KzKzt KdI (N/m2); Where: - qz is the velocity pressure - Kdis the wind directionality factor. - Kzis the velocity pressure exposure coefficient. - Kztis the topographic factor. - V is the wind velocity(m/s) - I is the importance factor

  9. from the ACI code the constants are calculated and found to be as follow: I = 1.0 (for normal buildings). Kz = 1.33 (box system or shear walls) from table 6.1 in ACI318-08, KZt = 1 for the Nablus topography. Kd=0.85 in Nablus city. V= 35 mph= 15.65 m/s ( in Nablus city) G=0.85 from ACI tables( in Nablus city) 0.8 for wind ward Cp= 0.50 for lee ward 0.70 for side ward 0.70 for roof ward

  10. from the ACI code the constants are calculated and found to be as follow: • qz=0.613*1.13*1*0.85**1 • =144.2 • P=144.2*0.85*0.8=98 N/m2 • check for Pmin • Pmin=500/1.3∗Cp=500/1.3∗0.8 > P so, use P = 384.6 N/m2

  11. some information about the building: • F`call = (0.3-0.4) F`c = 105 Kg/cm2 • Story height= 3.3m • 14 stories • Residential building • F`c for tension= 0.10*F`c for compression= 10 Kg/cm2 • WD=8.46 KN/m2 • WL= 3KN/m2 • Load combination= 0.9D + 1.6 L

  12. for the lateral loads in XZ-plain:

  13. check the overturning in the X-direction: • Slab weight = 8.46*330*14 = 39085.2 KN • Walls weight = (0.3*25*46.2*28.1)= 9736.65 KN • Walls weight= (0.2*25*46.2*10)= 2310 KN • Wind load on the building= 0.385*46.2*10.5= 186.76 KN • W =0.9*51131.8 = 46018.665 KN • H =(186.76/46.5)*1.6= 8.1 KN/m • £M= -(0.5*8.1*46.2)*(2*46.2/3) + 46018.7*(35.1/2) =801875.9 KN.m • the building weight is sufficient to resist the wind-load in X-direction.( the same in Y-direction and it is okay)

  14. Check the torsional effect in ( XY-plan) • Center of mass: X=17 m Y= 5.314 m • Center of rigidity: X= 17.11m Y= 8.98 m

  15. the difference between the two centers in the X-direction is small, but in Y-direction is large, so a sample calculation will be made on the shear wall number 2 in the Y-direction. • e = 8.97-5.31 = 3.66 m • V= 0.385*10.5*46.2= 186.76 KN • MT =V*e • MT = 3.66*186.76= 683.55 KN.m • J =4355 KN.m2 • F= (115.36/2.31)= 49.9= 50 KN/ = 0.5 Kg/cm2 < 105 Kg/cm2 • F= ( 68.76/2.31) = 29.77 KN/ = 0.29 Kg/ < 10.5 Kg/ • the wind-load effect is very small so it can be neglected

  16. Chapter Six: Analysis for seismic force using UBC97 code • Seismic Forces- Methods of analysis: • Equivalent static method: • Dynamic analysis • Response spectrum analysis • Time history analysis

  17. selection of analysis method • Static method: The static lateral force procedure of Section 1630 may be used for the following structures:  a. All structures, regular or irregular, in Seismic Zone 1 and in Occupancy Categories 4 and 5 in Seismic Zone 2. b. Regular structures under 73m in height with lateral force resistance provided by systems listed in Table 7-E. c. Irregular structures not more than five stories or 20m in height.

  18. selection of analysis method • Dynamic analysis: The dynamic lateral-force procedure of Section 1631 shall be used for all other structures, including the following:   a. Structures 240 feet (73 152 mm) or more in height. b. Structures having a stiffness, weight or geometric vertical irregularity of Type 1, 2 or 3, c. Structures over five stories or 65 feet (19 812 mm) in height in Seismic Zones 3 and 4 not having the same structural system throughout their height d. Structures, regular or irregular, located on Soil Profile Type SF, that have a period greater than 0.7 second. The analysis shall include the effects of the soils at the site.

  19. Equivalent lateral force method (Static Method)- UBC 97 • The total design base shear in a given direction shall be determined from the following formula: UBC97 30-4 • The total design base shear need not exceed the following: UBC97 30-5 • The total design base shear shall not be less than the following: • UBC97 30-6

  20. Equivalent lateral force method (Static Method)- UBC 97 Where: • Z= seismic zone factor, Table 7-A • I= importance factor, • R= numerical coefficient representative of the inherent over strength and global ductility capacity of lateral- force- resisting systems, Table 7-E • Ca= acceleration seismic coefficient, Table 7-C • Cv= velocity seismic coefficient, Table 7-D • W= the total dead load and applicable portions of other loads listed below: • 1. In storage and warehouse occupancies, a minimum of 25% of the floor live load shall be applicable • 2. Design snow loads of 1.5kN/m2 or less need not be included. Where design snow loads exceed 1.5 kN/m2, the design snow load shall be

  21. Soil profile type • Soil Profile Types SA, SB, SC, SD and SE are defined in Table 7-B and Soil Profile Type SF is defined as soils requiring site-specific evaluation.

  22. Structural period • Method A • The period, T is given by: • hn= height of structure in meters • Ct= factor given by: • Ct=0.0853 for steel moment resisting frames • Ct= 0.0731 for reinforced concrete moment resisting frames and eccentrically braced frames • Ct=0.0488 for all other buildings • - 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

  23. Structural period • Method B: The fundamental period T may be calculated using the structural properties and deformational characteristics of the resisting elements. Where: • Mi= the mass of the building at the level i • = the deflection of the level i, calculated using the applied lateral forces • Fi= the lateral force applied on the level i

  24. Site and the building information: • Number of stories = 14. • Story height= 3.3 m. • Materials: concrete cylinder compressive strength at 28 days, f’c= 28 𝑀𝑃𝑎 • -Steel yield strength, 𝑓𝑦=420 𝑀𝑃𝑎 • Soil: rock, Sb type in accordance with UBC 97 provisions. • All columns are square with side length equals to 650 mm. • All beams are 400 mm section width and 500mm total thickness. • The slab is two way solid slab of 200 mm thickness. • External shear walls thickness is 300 mm, and the internal are 200 mm wide • Shear walls weight =25 KN/m.

  25. Site and the building information: • The live load is 3 kN/m2 – office and residential building. • The superimposed dead load is 3.36 kN/m2. • The perimeter wall weight is 21 kN/m. • Importance factor (I) =1. • Zone factor (𝑍)=0.2, {2B} UBC 97 • 𝐶𝑎=0.20, 𝐶𝑣=0.2. • Sizes of all columns in upper floors are kept the same. • The floor diaphragms are assumed to be rigid. • Lateral-force-resisting system is: Duel systems(4.1.c) • Location of building Nablus city . • Earthquake load as…..UBC97

  26. Seismic Load • According to Uniform Building Code (UBC97), the seismic coefficients Ca=0.20, Cv=0.20.

  27. The modified plane of the buildingto reduce the eccentricity

  28. weight of the building • Solid slab own weight= slab thickness x unit weight of concrete 𝑊𝐷=0.20*25=5 𝑘𝑁/𝑚2 𝑊𝑠𝑙𝑎𝑏=299*5=1495 𝑘𝑁/𝑚2 • - Beams weight= length of beams x cross section area x unit weight of concrete: beams 400*500 ,Area=13.42 𝑚2 𝑊𝑏𝑒𝑎𝑚𝑠=13.42*25=335.5 Notice that the beam depth below slab is used which is 0.50-0.20= 0.30 m. • - Columns weight= length of columns x cross section area x unit weight of concrete: columns 650X650 at all floors ,Area=0.4225 𝑚2, 𝐼=0.014876 𝑚4 𝑊𝑐𝑜𝑙𝑢𝑚𝑛𝑠=11*0.4225*3.3*25=383.41 𝑘𝑁 ( columns at edge of walls are neglected ) • -Shear wall weight= length of wall x cross section x unit weight of concrete Wshear walls=1295 KN • - Parameter wall weight= length of wall* unit weight(KN/m) Wwall=1031.5 KN • - Superimposed dead load on slab= area of slab x superimposed dead load/m2 𝑊𝑠𝑢𝑝𝑒𝑟𝑖𝑚𝑝𝑜𝑠𝑒𝑑=299*3.36=1006.05 𝑘𝑁 • - Total weight of one story= 5546.46 𝑘𝑁 • - Total weight of building= 5521.46*14=77650.44 𝑘𝑁

  29. The structure period ƩIx=27.9075 m4

  30. The structure period • ƩIy=6.1763 m4

  31. Base Shear in Y-direction • R= 6.5 from UBC97 Table16-N (Duel system 4.1.c). - The base shear, V, need not exceed - The base shear shall not be less than 𝑉=0.11𝐶𝑎𝐼𝑊 𝑉=0.11*0.20*1*𝑊=0.022𝑊 > 0.0217𝑊 So, total base shear will be 𝑉=0.022 *7756.044= 1700.6 KN

  32. Base Shear in Y-direction • since T > 0.7 sec, there is Ftop. Ftop need not exceed 0.25V, 0.25 V= 425.15 KN Ftop < 0.25V, so Ft= 168.45 KN and the distribution of the rest of the base shear on the stories is according to the following equation: Where xis the number of the story. Table 7.1 shows a summary of distribution of forces for the stories

  33. Table 7.1: Distribution of forces to stories

  34. center of rigidity

  35. calculation of eccentricity due to seismic forces: • center of mass(CM) X=17.43 m y=5.25 m • center of rigidity (CR) X=18.187 m Y=6.8177 m • emax= eo + e2 • emin= eo - e2 • eo: the real eccentricity is the distance between the CR( center of stiffness and the CM and center of mass). 0e • eo: accidental eccentricity • e2= accidental eccentricity • e2 =0.05*L • - L: the floor dimension perpendicular to the direction of the seismic action. • eo=18.187 -17.43= 0.757 m • e2= 0.05*35.1= 1.755 m • emax=1.755 +0.757= 2.512 m • emin= -0.998 m

  36. horizontal distribution of story shear to the walls • The distribution of the total seismic load, to walls will be in proportion to their rigidities.. The flexural resistance of walls with respect to their weak axes may be neglected in lateral load analysis. Table 7.4 summarizes the force distribution to the columns and walls in y direction. • the table 7.3: shows (Km, Kp, dm and dp) for the shear walls. • Where: • - 𝑉𝑚: The shear force of the structural walls m , parallel to the direction of the seismic action. • - 𝑉𝑝: The shear force of the structural walls p, perpendicular to the direction of the seismic action. • - dm,dp: The distance from the centers of gravity of the structural walls m and p to the considered center of rigidity. • - 𝑘𝑡: The torsion rigidity of the considered level. • 𝑘𝑝: The translation rigidity of the considered wall. • K : The total translation rigidity of the considered level

  37. the parallel walls are: W1, W2, W3, W4, W5, W6, W7the perpendicular walls are: W1, W2, W3, W4, W8, W9, W10, W11

  38. Table 7.3: Distribution of forces to shear wall at each story (V is in KN

  39. Table 7.5: the total force on each shear wall at each story (KN) ( V=Vm+Vp)

  40. Calculation of the base shear in the other direction(X): • The same procedures as in Y-direction The base shear, V in the y-direction will be: R= 6.5 from UBC97 Table16-N(Duel system 4.1.c The base shear, V, need not exceed - The base shear shall not be less than 𝑉=0.11𝐶𝑎𝐼𝑊 𝑉=0.11*0.20*1*𝑊=0.022𝑊 < 0.046 𝑊 So, total base shear will be 𝑉=0.046 *7756.044= 3576.65 KN

  41. Calculation of the base shear in the other direction(X): • since T < 0.7 sec, there is no Ftop. • The distribution of the base shear on the stories is according to the following equation: Where xis the number of the story. Table 7.5 shows a summary of distribution of forces for the stories.

  42. calculation do eccentricity due to seismic forces:center of mass(CM) • center of mass(CM) X=17.43 m y=5.25 m • center of rigidity (CR) X=18.187 m Y=6.8177 m • emax= eo + e2 • emin= eo - e2 • eo: the real eccentricity is the distance between the CR( center of stiffness and the CM and center of mass). 0e • eo: accidental eccentricity • e2= accidental eccentricity • e2 =0.05*L • - L: the floor dimension perpendicular to the direction of the seismic action. • eo=6.8177-5.25 =1.4617 m • e2= 0.05*10.5= 0.525 m • emax=1.4117+0.525=1.9867 m • emin= 0.8867 m

  43. horizontal distribution of story shear to walls • Table7.7:( Km(KN), Kp(KN), dm(m), dp(m)) for the shear walls

  44. Table 7.8: Distribution of forces to shear wall at each story (V is in KN)

  45. Table 7.9: the total force on each shear wall at each story(V=Vm+Vp) (KN)

  46. Tables used in the analysis:

  47. Tables used in the analysis:

  48. Tables used in the analysis:

  49. Chapter 7- 3D modeling • Structural System used: • In this project, 3D modeling will be made for the structure, and two-way slab with beams will be used as the structural system, since the project will be analyzed and designed to resist the earthquakes. • No analysis will be made in this chapter, only the modeling. • By using 2D modeling on SAP2000 and applying many thicknesses to obtain an economical one, it was found the following: • Slab thickness= 20 cm- • Beams are 40*50 cm- • and for the modifiers for SAP: • - Beam=0.35 • - Slab=0.25 • - Column=0.70

  50. Structural materials: • 1) Concrete : The main material used in the structural is the concrete: • * The cylindrical compressive strength after 28 days, • f’c =24 MPa (B300) for slabs and beams. * Modulus of elasticity ,E= 2.3* f’c = 28 MPa (B350) for columns. * Modulus of elasticity, E=2.49* *Unit weight is 25 KN/ • 2) Steel: *The Steel yield strength used for steel reinforcement is 420 MPa *The modulus of elasticity is 2.04* MPa

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