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3D Dynamic D esign O f AL- N our B uilding

3D Dynamic D esign O f AL- N our B uilding. An- Najah National University Faculty of Engineering Civil Engineering Department. Prepared by: 1. Ahmad Rashdan. 2. Jaffar Hassan . 3. Mustafa Aqra. 4. Odai Odeh.

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3D Dynamic D esign O f AL- N our B uilding

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  1. 3D Dynamic Design Of AL-Nour Building An-Najah National University Faculty of Engineering Civil Engineering Department Prepared by: 1. Ahmad Rashdan. 2. Jaffar Hassan . 3. Mustafa Aqra. 4. Odai Odeh. Supervised by: Dr. Abdul Razzaq Touqan

  2. Chapter 1 :Introduction

  3. Introduction: • Al-Nour building is 8 stories reinforced concrete building ,located in Nablus city and used as residential building. • The first story is used as garages with plan area of 700 m^2 and the above 7 stories used as residential apartment (two apartments per floor) with plan area of 490 m^2 due to the setback. • The soil bearing capacity = 400 KN/m2

  4. Introduction : • The Following slides shows : • 1. columns centers plan. • 2. 3D model of the building.

  5. Columns Centers Plan :

  6. Structural System : • The structural system used is on way ribbed slab with load path in x-direction.

  7. Materials: • - Concrete : - f’c= 320 kg/cm²( 32 MPa) For columns. - f’c= 240 kg/cm²( 24 MPa) for others. - The concrete unit weight = 25 (KN/m3). • -Reinforcing Steel:The yield strength of steel is equal to 4200 Kg/cm2 (420 MPa). • -Others :

  8. Design loads : • - Dead loads in addition to slab own weight : • Superimposed dead load = 4.5 KN/m2 • Partition load = 1 KN/m2 . • Masonry wall weight = 21.22 KN/m. • - Live load = 2 KN/m2 . • -Water tanks load = 1.14 KN/m2 • - Seismic loads : shown later.

  9. Design codes and load combinations: • - The following are the design codes used : • ACI – code 2008 . • IBC 2009 . • ASCE for design loads. • The following are the load combinationsused : • Wu = 1.4DL. • Wu = 1.2DL + 1.6LL . • Wu = 1.2DL + LL ± E. • Wu = 0.9DL ± E

  10. Chapter 2 :Preliminary Design

  11. Preliminary design • We performed a preliminary design for all structural elements conceptually. • The story height is 3.12 m. • The following are the preliminary dimensions : • Slab : - depth = 25 cm (based on deflection criteria) . - web width = 12 cm. - slab own weight = 4.55KN/m². - Ultimate load = 14.06KN/m².

  12. Preliminary Design • Beams : Since the structural system is one way ribbed slab (load path in x-direction) we have : • Main beams in y-direction : 30x60 cm. • Secondary beams in x-direction : 40x25 cm. • Columns : Take a sample columns ( B3) : Area carried by column = 28 m2 Ultimate slab load = 14.06KN/m² Pu = 3769.6 KN. Ag = 2326.9 cm2. → Use columns of 40x60cm2.

  13. Preliminary design and checks • Footing : we performed an preliminary design for footing of the previous column as single footing. with dimensions of 2.9x2.7x0.7 m.

  14. Chapter 3 :Static Design

  15. Static design : • Final dimensions : • 1. frame sections :

  16. Static Design: • The new web width (bw) = 15 cm. • Area sections dimensions : • The story height = 3.5 m

  17. Static design: • Verification Of SAP model: • We perform the verification for SAP models( one and eight stories and it was OK) the following is verification for eight stories : • 1. Compatibility satisfied :

  18. 2.Equilibrium Satisfied : Static Design 3.Stress -Strain relationship satisfied: Taking beam C in second story (taking 8 m span) :

  19. Static Design : • Slab design : 1. Check slab deflection : So, ∆dead = 2.92 mm. ∆Live = 0.78mm. Δ long term = 7.16mm. The allowable deflection = 4000 /240 = 16.67 mm. • So the slab deflection = 7.16mm < allowable long term def. OK. 2. Design for shear : The rib shear strength = 23.2KN. The max shear = 36.75 KN/m. shear per rib = 0.55*36.75 = 20.2 KN. • So 23.2 ≥ 20.2 OK • So the slab is Ok for shear.

  20. Static Design : • 3. Design for bending moment : • The moments are read from SAP using section cut :

  21. Static Design : • Design of beams in y-direction : • Taking a sample beam (beam B in the first floor) : - The beam section dimensions are : - Total depth (h) = 700 mm. - The effective depth (d) = 650 mm. -Beam width (bw) = 400 mm. - min reinforcement ratio = 0.0033. - As min = ρbd = 0.0033*400*650 = 858 mm2 - φVc = 159.1 KN. - (Av/s)min = 0.333.

  22. Static Design : • - Design information :

  23. Static Design : • Design of secondary beams: • Total depth(H) = 25cm.(hidden beam) • d= 21cm (cover =4cm) • Width = 40cm. • The following are the values of min reinforcement: • (As) min = 0.0033*b*d=0.0033*400*210= 277.2 mm (3Φ12). • Vc = 0.75*0.167**400*210/1000= 68.58KN.

  24. Static Design : • Design information :

  25. Design of columns: • Column grouping, Area of steel& stirrups:

  26. Manual design Check slenderness ratio for corner column C.A-1 From the graph  K = 3.5 = 40.83 >22 Column is long. = 81.66 >22Column is long. • Pu=3034.75KN • MY = 11.02 KN.m(maximum value) • MX = 153.1 KN.m( maximum value) • Mc = ns*M2 = 1.67(153.1) = 255 KN.m • From the interaction diagram: • ≤1% use minimum steel ratio use =1%. • As =0.01xAg =0.01x40x80 =3200mm2 • Same as SAP value.

  27. Tie beam design Minimum area of steel = 0.0033*b*d =436 mm2. Use 4Ф12mm bottom steel. Use 4Ф12mm top steel. Shear design : Vu at distance (d = 44cm) = 16.35KN, ФVc = 80.83KN. Use 1Ф8 mm@200mm.

  28. Footing design Single footing: • Is one of the most economical types of footing and is used when columns are spaced at relatively long distances . • Bearing capacity of the soil=400 KN/m2.

  29. Footing grouping Footing grouping according to column’s ultimate load.

  30. footing details

  31. Design of Stairs • . • Own Weight=0.2x25=5KN/m2. • Live load = 5KN/m2. • Superimposed loads =2.14 KN/m2 • Superimposed loads of extra of stairs=1.76 KN/m2 • Wu=1.2(5+3.9) +1.6(5) =18.68KN/m2.

  32. Verification of SAP model: Compatibility: “Compatibility Satisfied”

  33. Cont. Equilibrium: “Equilibrium Satisfied” Stress-Strain Relationships: “Stress–strain relationship satisfied”

  34. Chapter 4 :Dynamic Design

  35. Dynamic Design: • Methods for dynamic analysis: • Equivalent static method. • Time history method. • Response spectrum analysis. • Input parameters in dynamic analysis : - Importance factor (I) = 1 . - Peak ground acceleration (PGA) = 0.2g . - Area mass = 0.458 ton/m2 - Soil class = Class B. - Spectral accelerations : Ss = 0.5 . S1 = 0.2 . - response modification factor R = 3 in x-direction. R = 4.5 in y-direction.

  36. Dynamic Design : • Modal information : - For eight stories before enlarging beams in x-direction : • - Enlarge the beams 2 &4 to 30x70 (width*depth)

  37. - For eight stories after enlarging beams in x-direction : Dynamic Design: • - Comparison with manual results :

  38. Dynamic Design : • Response spectrum analysis : We will perform the dynamic design using response spectrum method: Define two response spectrum load cases one in x-direction and the another in y-direction : - For response-x: * Scale factor = 3.27. *Scale factor = 0.654. - For response-y: * Scale factor = 2.18. *Scale factor = 0.981. • Perform design using envelope combination and check whether static or dynamic combination controls .

  39. Dynamic Design : • Slab design : The comparison is performed. Static design controls

  40. Dynamic Design : • Design of beams in y-direction : - Reinforcement from envelope combination: - Reinforcement from static combination: Static design Controls

  41. Dynamic Design : • Design of beams in X-direction : - Reinforcement from envelope combination is considered since the dimensions are increased: Dynamic design Controls

  42. Dynamic Design : • Design of columns : Three representative columns are selected : • Interior column B3. • Edge column B2. • Corner column A4 . The comparison is performed and static design controls for all columns. The following table shows the comparison for column B3 :

  43. Dynamic Design : The following table shows the comparison for column B3 (M3, V2 ): Static design OK for columns.

  44. Chapter 5 :Structural Modeling Of One Way Ribbed Slab

  45. Structural Modeling Of One Way Ribbed Slabs • The ribbed slabs can be represented by one of the following ways : • Equivalent stiffness method : find the equivalent thickness of a solid slab that can achieve the same rib stiffness. • Represent it as separate ribs (T-section). • Represent the ribs by rectangular ribs and flange. • The main objective is to prove that three models give the nearly the same results.

  46. Structural Modeling Of One Way Ribbed Slabs • Model 1 : Equivalent stiffness method : • Equivalent slab thickness (t) = 19.45 cm. • I T-sec = I rec →( 0.55*h3eq /12) = 3.371*10-4 h3eq = 19.45 cm. γeq = 23.87 KN/m^3 . . . . .. .to achieve the same weight. - Stiffness modifiers : M11 = 0.35 . M22 = 0.0244 M1-2 = 0.0244.

  47. Structural Modeling Of One Way Ribbed Slabs • Model 2 : Representation as separate ribs : - Stiffness modifiers : I 3-3 = 0.35 . I 2-2 = 0.35 . Torsional constant(J)= 0.35 . • The loads are inserted as line Loads on ribs. • Substitute the weight of blocks.

  48. Structural Modeling Of One Way Ribbed Slabs 55 cm 8 cm • Model 3 : the slab is represented as rectangular ribs and flange. 1. The rectangle section should satisfy The actual T-section. 2. Stiffness modifiers : I 3-3 = 0.6 . I2-2 = 0.6 . J = 0.52 . 3. Weight modifier = 0.68 . Flange Modifiers : - M11 = 0.0001 .(almost zero). - M 22 = 0.25 . - M 1-2 = 0.0001(almost zero). → we have to substitute the weight of blocks. 25 cm 15 cm

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