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Chapter 1 : Introduction Introduction: Al-Nour building is 8 stories reinforced concrete building,located in Nablus city and used as residential building.

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Presentation on theme: "Chapter 1 : Introduction Introduction: Al-Nour building is 8 stories reinforced concrete building,located in Nablus city and used as residential building."— Presentation transcript:

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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/m 2

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

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

8 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/m 3 ). - Reinforcing Steel: The yield strength of steel is equal to 4200 Kg/cm 2 (420 MPa). -Others : Material Unit weight (KN/m 3 ) Reinforced concrete25 Plain concrete23 Sand18 Aggregate17 Y-tong5 Blocks12 Polystyrene0.3 Masonry stone27 Light weight block6 Tile26

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

10 Design codes and load combinations: - The following are the design codes used : 1. ACI – code IBC ASCE for design loads. Th e following are the load combinations used : 1. Wu = 1.4DL. 2. Wu = 1.2DL + 1.6LL. 3. Wu = 1.2DL + LL ± E. 4. Wu = 0.9DL ± E

11 Chapter 2 : Preliminary Design

12 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.

13 Preliminary Design  Beams : Since the structural system is one way ribbed slab (load path in x-direction) we have : 1. Main beams in y-direction : 30x60 cm. 2. Secondary beams in x-direction : 40x25 cm.  Columns : Take a sample columns ( B3) : Area carried by column = 28 m 2 Ultimate slab load = 14.06KN/m² Pu = KN. A g = cm 2. → Use columns of 40x60cm 2.

14 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.

15 Chapter 3 : Static Design

16 Static design : Final dimensions : 1. frame sections : MemberDepth(cm)Width(cm) Col.8040 Main interior beams7040 Main exterior beams7530 Secondary beams2540 Tie beams5030

17 Static Design: The new web width (bw) = 15 cm. Area sections dimensions : Area section nameThickness (cm) Actual Slab25 Equivalent Slab thickness19.45 (in SAP model) Shear wall30 (initially) The story height = 3.5 m

18 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 :

19 Static Design 3.Stress -Strain relationship satisfied: Taking beam C in second story (taking 8 m span) : Load typeHand results(KN)SAP results (KN)Error % Dead load Live load Equilibrium Satisfied : Load` M-ve (left)(KN.m) M+ve (KN.m) M-ve(right) (KN.m) Total moment (KN.m) SAP Result D Result Error2.3%

20 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 = 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 = 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.

21 Static Design : 3. Design for bending moment : The moments are read from SAP using section cut : Point location/termMoment(KN.m)As(mm 2 )Bars A Φ 12 A1- B Φ 10 B Φ 12 B1- C Φ 10 C Φ 12 C1 – D Φ 10 D Φ 12 D1 – E Φ 10 E Φ 12

22 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 (b w ) = 400 mm. - min reinforcement ratio = A s min = ρbd = *400*650 = 858 mm 2 - φVc = KN. - (Av/s) min =

23 - Design information : Static Design : pointAs(mm)As min (Av/s) (Av/s) min PI(m)Length(m)barsstirrups B Φ16 cm B cm B2(L) φ20 cm B2(R) φ20 cm B _7.9 cm B3(L) φ20 cm B3(R) φ20 cm B _5.9 5Φ16 cm B Φ16 cm

24 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: (A s ) min = *b*d=0.0033*400*210= mm (3Φ12). Vc = 0.75*0.167**400*210/1000= 68.58KN.

25 Static Design : Design information : PointAs(mm) As min (mm) (Av/s) (Av/s)min PI(m)barsstirrups A A4-B B B4-C C C4-D D D4-E E E4-F F

26 Design of columns: Column grouping, Area of steel& stirrups: Floor no.ColumnsAs(mm 2 ) column group Distributio n of steel Stirrups spacing (mm) All floors except No.8 All Columns 3200C1 mm Floor No.8 D2 & G23766C2 mm A2, B2, C2, H2, I2 & J2 4774C3 mm

27 P u = KN M Y = KN.m(maximum value) M X = KN.m( maximum value) M c =  ns *M 2 = 1.67(153.1) = 255 KN.m  From the interaction diagram:  ≤1% use minimum steel ratio use  =1%.  As =0.01xAg =0.01x40x80 =3200mm 2  Same as SAP value. Manual design

28 Tie beam design Minimum area of steel = *b*d =436 mm 2. Use 4Ф12mm bottom steel. Use 4Ф12mm top steel. Shear design : V u at distance (d = 44cm) = 16.35KN, ФVc = 80.83KN. Use 1Ф8

29 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/m 2.

30 Footing grouping Footing grouping according to column’s ultimate load. Group Name Columns Service load(KN) Ultimate load (KN) F1A1, B1,C1,D1,E1,F1,G1,H1,I1,J F2A2, J2,A4 B4,C4,D4,G4,H4,I4,J4, F3 B2,C2, D2,G2,H2,I2,A3,B3,C3,D3,G3,H3, I3,J CombinedE4 & F

31 footing details Group Area of footing (m 2 ) Dimensions (m) Depth (m) Steel distribution/ m’ (Both directions) Area of shrinkage steel (mm 2 ) Distribution in each direction F x Ø 14mmnoNo F x Ø 18mm1 Ø F x Ø 18mm1 Ø Combined x Ø25 /200mm Ø

32 Design of Stairs

33 Verification of SAP model: Compatibility: “Compatibility Satisfied”

34 Cont. Load type Hand results(KN) SAP results(KN) Error % Dead load Live load Equilibrium: Stress-Strain Relationships: Load M - ve (left) (KN.m) M +ve (KN.m) M -ve (right) (KN.m) 3D SAP Result D Result Error 0.3% “Equilibrium Satisfied” “Stress–strain relationship satisfied”

35 Chapter 4 : Dynamic Design

36 Dynamic Design: Methods for dynamic analysis: 1. Equivalent static method. 2. Time history method. 3. Response spectrum analysis. Input parameters in dynamic analysis : - Importance factor (I) = 1. - Peak ground acceleration (PGA) = 0.2g. - Area mass = ton/m 2 - Soil class = Class B. - Spectral accelerations : Ss = 0.5. S1 = response modification factor R = 3 in x-direction. R = 4.5 in y-direction.

37 Dynamic Design : Modal information : - For eight stories before enlarging beams in x-direction : Mode No.DirectionPeriod (sec.)MMPR % 1X RZ(Torsion) y Enlarge the beams 2 &4 to 30x70 (width*dept h)

38 Dynamic Design: - For eight stories after enlarging beams in x-direction : Mode No.DirectionPeriod (sec.)MMPR % 1X RZ(Torsion) Y X Comparison with manual results : Mode directionSAP result(sec)Manual result(sec)Error % x- direction % y- direction %

39 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 = *Scale factor = For response-y: * Scale factor = *Scale factor = Perform design using envelope combination and check whether static or dynamic combination controls.

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

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

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

44 Design of columns : Three representative columns are selected : 1. Interior column B3. 2. Edge column B2. 3. Corner column A4. The comparison is performed and static design controls for all columns. The following table shows the comparison for column B3 : Dynamic Design :

45 floor/ter m Envelope combinationStatic combination momentshearaxialAs(mm 2 )momentshearaxialAs(mm 2 ) Dynamic Design : Static design OK for columns. The following table shows the comparison for column B3 (M3, V2 ):

46 Chapter 5 : Structural Modeling Of One Way Ribbed Slab

47 Structural Modeling Of One Way Ribbed Slabs The ribbed slabs can be represented by one of the following ways : 1. Equivalent stiffness method : find the equivalent thickness of a solid slab that can achieve the same rib stiffness. 2. Represent it as separate ribs (T-section). 1. Represent the ribs by rectangular ribs and flange. The main objective is to prove that three models give the nearly the same results.

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

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

50 Structural Modeling Of One Way Ribbed Slabs 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 = Weight modifier = Flange Modifiers : - M11 = (almost zero). - M 22 = M 1-2 = (almost zero). → we have to substitute the weight of blocks. 8 cm 25 cm 55 cm 15 cm

51 Structural Modeling Of One Way Ribbed Slabs All the previous models are verified according to manual solution. Static analysis is performed for the three models and we read the moment and shear at point C3 in span C3-4 for beam C in the second floor : load Moment (KN.m) Shear (KN) (mm 2 )( (mm 2 /mm) SAP Model 1 Results SAP Model 2 Results SAP Model 3 Results

52 Structural Modeling Of One Way Ribbed Slabs also dynamic analysis and design performed for the three models and the following are the modal information : Mode Direction PeriodMMPR Mode 1 x-direction Mode 2 Torsion(Rz) Mode 3 y- direction Model 2 : Mode Direction PeriodMMPR Mode 1 x-direction Mode 2 Torsion(Rz) Mode 3 y- direction Model 3 : Mode Direction PeriodMMPR Mode 1 x-direction Mode 2 Torsion(Rz) Mode 3 y- direction Model 1 :

53 Also the same span taken in the beam C and the following are the values of moment and shear from envelope combination : Structural Modeling Of One Way Ribbed Slabs Point Envelope combination momentshearAs(mm2) Av/s (mm2/mm) C C2-C C Model 1 : Point Envelope combination momentshearAs(mm 2 ) Av/s (mm 2 /mm) C C2-C C Point Envelope combination momentshearAs(mm 2 ) Av/s (mm 2 /mm) C C2-C C Model 2 : Model3:

54 Structural Modeling Of One Way Ribbed Slabs Final conclusion : from the previous data shown in tables we note that all the three models give near results. So, we can represent the slab model by any of the previous models but perform the changes in loads assignment and stiffness modifiers.

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