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Design of the Library of Palestine Technical University - Tulkarm

An-najah National University. Faculty of Engineering. Civil Engineering Department. Graduation project. Design of the Library of Palestine Technical University - Tulkarm. Prepared By: Omar Jamal Sha’ban Abdallah Marwan Al- Tibi Mohammad Mufeed Shishan. Chapters.

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Design of the Library of Palestine Technical University - Tulkarm

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  1. An-najah National University Faculty of Engineering Civil Engineering Department Graduation project Design of the Library of Palestine Technical University - Tulkarm Prepared By: Omar Jamal Sha’ban AbdallahMarwan Al-Tibi Mohammad MufeedShishan

  2. Chapters • Chapter One : Introduction • Chapter Two : Preliminary Design • Chapter Three : Three Dimensional Structural Analysis and Design

  3. Chapter One : Introduction General • This is a graduation project that introduces analysis and design of reinforced concrete structure. This structure is the building for the library of Palestine Technical University (PTU) in Tulkarem city – Palestine. • The building will be analyzed and designed using the primary principles of structural analysis and design by using one dimensional structural analysis and also using the most modern analysis of structures which is the three dimensional structural analysis and design.

  4. Figure 1.1: Architectural 3D model of the building Note: The red lines in the previous figure are related to expansion joints

  5. The building is divided into three parts which are separated by expansion joints. Figure 1.1 shows a three dimensional image of the building. • Part A in the building is composed of four stories and is used as a library. The first story is an underground (basement) story whose area is about 743.0 m2 and its height is 3.90 m. The other three stories is an over ground stories, and the area of each one is about 1546 m2 and its height is 4.42 m. There is a steel dome in the last story at the roof. • Part B is used as offices, video conference and arcade. And it is composed of an over ground story whose area is about 472.0 m2 and its height is 4.42 m. • Part C has not been designed.

  6. Codes and Standards The structures are designed using practice code and specifications that control the design process and variables. The following codes and standards are used in this study: • ACI 318-08 : American Concrete Institute provisions for reinforced concrete structural design. • UBC-97 : Uniform Building Code provisions for seismic load parametersdetermination. • IBC-2009 : International Building Code, which is used here for live loads. • ASTM : For material specifications.

  7. Loads and Load combinations Loads 1) Gravity loads: Live load: It comes from the people, machines and any movable objects in the buildings. The amount of live load depends on the type of the structure. In this project the live load is: 7.0 kN/m2 (for Part A) 3.0 kN/m2 (for Part B) Dead load: it is consisting of own weight of the structure and any permanent components. The super imposed dead load is 3.5 kN/m2.

  8. 2) Lateral loads: Seismic loads: The structure is located in Tulkarm area which is classified as zone 2A according to Palestine seismic zones. The UBC97 code seismic parameters are as follows: • The seismic zone factor, Z= 0.15 . • The soil is very dense soil and soft rock, so the soil type is Sc . • The importance factor, I = 1.0 . • The ductility factor, R = 5.5 . • The seismic coefficient, Ca = 0.18 . • The seismic coefficient, Cv= 0.25 .

  9. Load combinations The ACI318-08 load combinations are used and they are summarized as follows: • U1 = 1.4D • U2 = 1.2D + 1.6L + 1.6H • U3 = 1.2D + 1.0E + 1.0L • U4 = 0.9D + 1.0E + 1.6H Where: D : is dead load L : is live load H : is weight and pressure load of soil E : is earthquake load

  10. Materials Structural materials • Concrete: Concrete strength for all concrete parts is B350 ( f’c = 280 Kg\cm2 , 28 MPa ) Modulus of elasticity equals 2.5*105 Kg\cm2 , 2.5*104MPa Unit weight is 25 kN\m3 • Steel: Modulus of elasticity equals 2.04*106 Kg\cm2 , 2.04*105MPa For steel reinforcement, is 4200 Kg\cm2 , 420 MPa Non-structural materials They are mainly, blocks, plasters, tiles, filling, mortar and masonry.

  11. Building Structural Systems • The structural system in the building parts ( A and B) shall be as follows: In part A, the four floors are designed as ribbed slab with main hidden beams carried by columns. In part B, the floor is designed as one way ribbed slab in X and in Y direction with hidden beams. • Basement walls are used around the ground floor, and all the exterior walls are composed of concrete, masonry and blocks. • The masonry stones are fixed to the reinforced concrete walls using concrete mortar.

  12. Chapter two : PRELIMINARY DESIGN Design of Slabs Ribbed slab analysis and design (Part A) • The thickness of the slab: hmin = Ln/33 = 890-50/33 = 25.45 cm Use h=30 cm (since the loads are heavy).  Where Ln : The maximum spacing between the columns.

  13. The direct design method will be used to check the preliminary dimensions of the slab and beams. So, one frame will be taken in calculations to represent the whole slabs as approximation. The ribbed slab can be divided to frames in each direction . Here , calculations are made for Frame A shown in figure ( 2-1) Frame bending moment also shown in figure (2-2)

  14. Figure 2.1 Plan of the slab in the basement floor and frame A is hatched Figure 2.2 Moments in Frame A

  15. For column strip: As for negative moment = 4Ø20 2Ø20 / layerFor column strip: As for positive moment = 2Ø16For beam in the Frame D= 45 cm, B= 100 cm: As for negative moment = 6Ø20For beam in the Frame D= 45 cm, B= 100 cm: As for positive moment = 5Ø20For middle strip: As for negative moment = 2Ø14For middle strip: As for positive moment = 2Ø14 Figure 2.3 Percentages and values of the moments on the strips of span # 2

  16. For column strip: As for negative moment = 4Ø18 2Ø18 / layer &2Ø14For column strip: As for positive moment = 2Ø20For beam in the Frame: As for negative moment = 5Ø20&5Ø20 For beam in the Frame: As for positive moment = 5Ø20For middle strip: As for negative moment = 2Ø14For middle strip: As for positive moment = 2Ø14 Figure 2.4 Percentages and values of the moments on the strips of span # 1

  17. hmin = l/21 = 800/21 = 38 cm  Use h=40 cm Ribbed slab analysis and design (Part B) • The thickness of the slab:

  18. Design of Rib 5 in part B Rib load Diagram (in kN/m) Rib Bending Moment Diagram (in kN.m)

  19. Max. neg. moment= 16.01 kN.m/rib 𝜌 = 0.00235 𝜌min = 0.00333 As = 2Ø12 • Max. +ve moment= 13.64 kN.m/rib 𝜌 = 0.000537 𝜌min = 0.00333 As = 2Ø12

  20. Design of column # 15 The ultimate load on the column is found using tributary area and number of stories, and the design load can be calculated using the following equation: • Tributary area=59.085 m2 • Number of stories= 5 • Wu=24 kN/m2 • Pu=59.08x5x24 = 7089.6 kN • Pn req = 7089.6 / 0.65 = 10907.0 kN • Assume 𝜌 = 0.01 • Pd= Ф Pn = Ф *λ {0.85* fc(Ag-As) + As* fy} = 1090.70x1000=0.8 {0.85x 280x(-) + 0.01x Agx4200} • Ag= 1090.70*1000 / 222.1 = 4910.85 cm2 • Square column  b=h =70. cm  use (80x80) the area of column to account for additional moments and lateral forces

  21. Chapter ThreeThree Dimensional Structural Analysis and Design

  22. * General This chapter is a three dimensional analysis and design, which introduces the final and practical structural analysis and design of the structural elements with the practical structural drawings that are ready for construction.    Structural analysis comprises of set of physical and mathematical laws required to study and predict the behaviour of structures under a given set of actions. The structural analysis of the model is aimed to determine the external reactions at the supports and the internal forces like bending moments, shear forces, and normal forces for the different members. Theses internal member forces are used to design the cross section of three elements.

  23. Property/Stiffness Modification Factors • Basic calculations

  24. Two way slab system:Part A • Membrane f11 modifier = (A1 / A3)= 0.1165/0.301 = 0.387 • Membrane f22 modifier = (A1 / A3)= 0.1165/0.301 = 0.387 • Membrane f12 modifier = (A1 / A3)= 0.1165/0.301 = 0.387 • Bending m11 modifier = 0.25*( I1 / I3) = 0.25*(2.2108*10-3/5.087*10-3) = 0.108 • Bending m22 modifier = 0.25*( I1 / I3) = 0.25*(2.2108*10-3/5.087*10-3) = 0.108 • Bending m12 modifier = (C ribbed / C solid) = (6.2968*10-4/11.739*10-3) = 0.0536 • Shear v13 modifier = (A1 / A3)= 0.1165/0.301 = 0.387 • Shear v23 modifier = (A1 / A3)= 0.1165/0.301 = 0.387 • Mass m modifier = (M two way rib / M solid) = (9.08/11.25) = 0.80 • Weight w modifier = (9.81*M two way rib / 9.81*M solid)= (9.08/11.25) = 0.80

  25. Part B One way slab system (y-direction): • Membrane f11modifier = (A2/A3) = 0.044/0.22 = 0.2 • Membrane f22 modifier = (A1/A3) = 0.092/0.22 = 0.418 • Membrane f12 modifier = (A2/A3) = 0.044/0.22 = 0.2 • Bending m11 modifier = 0.25*( I2/ I3) = 0.25*(2.346710-5 / 2.93310-3) = 0.002 • Bending m22 modifier = 0.25*( I1/ I3) = 0.25*(2.346710-5 / 2.93310-3) = 0.139 • Bending m12 modifier = 0.25*( I2/ I3) = 0.25*(2.346710-5 / 2.93310-3) = 0.002 • Shear v13 modifier = (A2/A3) = 0.044/0.22 = 0.2 • Shear v23 modifier = (A1/A3) = 0.092/0.22 = 0.4 18 • Mass m modifier = (M 1 way rib / M solid) = (0.7/1)= 0.7 • Weight w modifier = (9.81*M 1 way rib/ 9.81*M solid) = (0.7/1) = 0.7

  26. One way slab system (x-direction): • Membrane f11 modifier = (A1/A3) = 0.092/0.22 = 0.418 • Membrane f22 modifier= (A2/A3) = 0.044/0.22 = 0.2 • Membrane f12 modifier = (A2/A3) = 0.044/0.22 = 0.2 • Bending m11 modifier = 0.25*( I1/ I3) = 0.25*(2.346710-5 / 2.93310-3) = 0.139 • Bending m22 modifier = 0.25*( I2/ I3) = 0.25*(2.346710-5 / 2.93310-3) = 0.002 • Bending m12 modifier = 0.25*( I2/ I3) = 0.25*(2.346710-5 / 2.93310-3) = 0.002 • Shear v13 modifier = (A1/A3) = 0.092/0.22 = 0.418 • Shear v23 modifier = (A2/A3) = 0.044/0.22 = 0.2 • Mass m modifier = (M 1 way rib / M solid) = (0.7/1) = 0.7 • Weight w modifier = (9.81*M 1 way rib / 9.81*M solid) = (0.7/1)= 0.7

  27. Structural Model Verification of Part A Check Equilibrium The total building dead load = 77047.1 kN The total building live load = 33627.72 kN From SAP2000: Total dead load = 74204.27 kN Total live load = 33109.0 kN • Error % in dead load= 3.68 %< 5% ok. • Error % in live load = 1.54 %< 5% ok.

  28. Check Compatibility

  29. Structural Model Verification of Part B • Check Equilibrium • The total building dead load = 7761.8 kN • The total building live load = 1421.16 kN • From SAP2000: Total dead load = 7402.87 kN • Total live load = 1394.19 kN • Error % in dead load = 4.6 %< 5% ok. • Error % in live load = 1.9 %< 5% ok.

  30. Check Compatibility

  31. Design of slabs Design of slabs of Part A Basement floor Rib 1 For negative moment use 2 Ф 22 For positive moment use 2 Ф 14 Rib 2 For negative moment use 2 Ф 20 For positive moment use 2 Ф 14

  32. *Analysis and Design of Beams:

  33. Design of columns

  34. Analysis and Design of Footings • footings which used in this project can be classified into the following types :- 1) Isolated footing: they have rectangular, square, or circular shape. This type of footing is used for small loads, and/or large soil allowable bearing capacity. 2) Wall footing: it is a continuous footing along the length of the wall.

  35. LEVEL ONE

  36. LEVEL TWO

  37. Design Of Shear Walls: • For Part A: • Shear wall (SW 1): • Vertical reinforcement: Pu = 1531.8 kN, Mu = 666.1 kN ρmin = 0.0012 , h = 11.80 m , b = 0.3 m. As min = 0.0012 X 1180 X 30 = 42.48 cm2 Asmin for each face = 42.48/2 = 21.24 cm2 From SAP2000: Use 1 Ф16 / 20 cm • Horizontal reinforcement: ρmin = 0.002 , h = 1 m , b = 0.2 m. Asmin = 0.002 X 100 X 20 = 4 cm2 Asmin for each face = 4/2 = 2 cm2 From SAP2000: Use 1 Ф18 / 25 cm As a result, for all shear walls: Vertical reinforcement: Use 1 Ф16 / 20cm. Horizontal reinforcement: Use 1 Ф18 / 25 cm

  38. Design of Stairs Dimensions: • Thickness of landing= Ln/20 = 195/20 = 9.75 cm. Thus 15 cm thickness is suitable. • Thickness of flight= Ln/24 = 285/24 = 12 cm. Thus 15 cm thickness is suitable. • Loads: • SID Load= 3.5 kN/m2 , LL=5 kN/m2 Design of flight: Mu = 17 kN.m As = 3.1 cm2 / m  use 1 Ф 14 / 20 cm Design of landing: Mu = 8 kN.m As = 2.7 cm2 / m  use 1 Ф 14 / 25 cm

  39. Reinforcement Of Stairs

  40. Thanks for listening

  41. An-najah National University Faculty of Engineering Civil Engineering Department Graduation project Design of the Library of Palestine Technical University - Tulkarm Prepared By: Omar Jamal Sha’ban AbdallahMarwan Al-Tibi Mohammad MufeedShishan

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