Presentation on theme: "3D Dynamic Design Of AL-Nour Building"— Presentation transcript:
1 3D Dynamic Design Of AL-Nour Building An-Najah National UniversityFaculty of EngineeringCivil Engineering Department3D Dynamic Design Of AL-Nour BuildingPrepared by:1. Ahmad Rashdan.2. Jaffar Hassan .3. Mustafa Aqra.4. Odai Odeh.Supervised by: Dr. Abdul Razzaq Touqan
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.
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/m3).- Reinforcing Steel: The yield strength of steel is equal to 4200 Kg/cm2 (420 MPa).-Others :MaterialUnit weight(KN/m3)Reinforced concrete25Plain concrete23Sand18Aggregate17Y-tong5Blocks12Polystyrene0.3Masonry stone27Light weight block6Tile26
9 Design loads : - Dead loads in addition to slab own weight : Superimposed dead load = 4.5 KN/m2Partition load = 1 KN/m2 .Masonry wall weight = KN/m.- Live load = 2 KN/m2 .-Water tanks load = 1.14 KN/m2- Seismic loads : shown later.
10 Design codes and load combinations: - The following are the design codes used :ACI – codeIBCASCE for design loads.The following are the load combinations used :Wu = 1.4DL.Wu = 1.2DL + 1.6LL .Wu = 1.2DL + LL ± E.Wu = 0.9DL ± E
12 Preliminary design The story height is 3.12 m. 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 :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 m2Ultimate slab load = 14.06KN/m²Pu = KN.Ag = cm2.→ Use columns of 40x60cm2.
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.
16 Static design : Final dimensions : 1. frame sections : Member Depth(cm)Width(cm)Col.8040Main interior beams70Main exterior beams7530Secondary beams25Tie beams50
17 Equivalent Slab thickness Static Design:The new web width (bw) = 15 cm.Area sections dimensions :Area section nameThickness (cm)Actual Slab25Equivalent 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 2.Equilibrium Satisfied : Load typeHand results(KN)SAP results (KN)Error %Dead load0.01Live load3.Stress -Strain relationship satisfied: Taking beam C in second story (taking 8 m span) :Load`M-ve(left)(KN.m)M+ve(KN.m)M-ve(right)Total moment (KN.m)SAP Result325.13208.63371.41556.91D Result341.40433.61558.21Error2.3%
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 ≥ OKSo 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(mm2)BarsA16.31132 Φ 12A1- B18.62 Φ 10B110.48125B1- C17.25C110.29122C1 – D18.1D19D1 – E1E113.55163.3
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 (bw) = 400 mm.- min reinforcement ratio =- As min = ρbd = *400*650 = 858 mm2- φVc = KN.- (Av/s)min =
23 Static Design : - Design information : point As(mm) As min (Av/s) (Av/s) minPI(m)Length(m)barsstirrupsB17008580.3330.625Φ16cmB1-2854--5.9B2(L)16761.34.86φ20B2(R)0.551.5cmB2-31044_7.9B3(L)15810.48cmB3(R)B3-4772B412.3
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:(As) min = *b*d=0.0033*400*210= mm (3Φ12).Vc = 0.75*0.167**400*210/1000= 68.58KN.
25 Static Design : Design information : Point As(mm) As min (mm) (Av/s) (Av/s)minPI(m)barsstirrupsA4469277.20.3330.885Φ12A4-B42773Φ12B44294Φ12B4-C4259C4424C4-D4260D4425D4-E4265E4432E4-F4147F4
26 Design of columns: Column grouping, Area of steel& stirrups: Floor no. As(mm2)column groupDistribution of steelStirrups spacing (mm)All floors except No.8All Columns3200C116φ16mmFloor No.8D2 & G23766C220φ16mmA2, B2, C2, H2, I2 & J24774C316φ20
27 Manual design Pu=3034.75KN MY = 11.02 KN.m(maximum value) Check slenderness ratio for corner column C.A-1From the graph K = 3.5𝐾 𝐿 𝑢 rx = 3.5x x0.8 = >22 ⟹Column is long.𝐾 𝐿 𝑢 ry = 3.5x x0.4 = >22 ⟹Column is long.Pu= KNMY = KN.m(maximum value)MX = KN.m( maximum value)Mc = ns*M2 = 1.67(153.1) = 255 KN.mFrom the interaction diagram:≤1% use minimum steel ratio use =1%.As =0.01xAg =0.01x40x80 =3200mm2Same as SAP value.
28 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
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/m2.
30 Footing grouping Group Name Columns Service load(KN) Ultimate load (KN)F1A1, B1,C1,D1,E1,F1,G1,H1,I1,J1305.6375.7F2A2, J2,A4 B4,C4,D4,G4,H4,I4,J4,F3B2,C2, D2,G2,H2,I2,A3,B3,C3,D3,G3,H3, I3,J3CombinedE4 & F4Footing grouping according to column’s ultimate load.
31 Steel distribution/m’ Area of shrinkage steel footing detailsGroupArea of footing(m2)Dimensions(m)DepthSteel distribution/m’(Both directions)Area of shrinkage steel(mm2)Distributionin each directionF10.961.2x0.80.34 Ø 14mmnoNoF26.212.7x2.30.555 Ø 18mm𝟓𝟖𝟓1 ØF39.573.3x2.90.756 Ø 18mm𝟔𝟕𝟓1 ØCombined13.115.7x2.31Ø25 /200mm778.2
32 Design of Stairs h min = 𝑙 20 =0.187m⇒Use h=0.20m. Own Weight=0.2x25=5KN/m2.Live load = 5KN/m2.Superimposed loads =2.14 KN/m2Superimposed loads of extra of stairs=1.76 KN/m2Wu=1.2(5+3.9) +1.6(5) =18.68KN/m2.Ø𝑉𝑐=110.23KN>𝑉𝑢=20.82𝐾𝑁
33 Verification of SAP model: Compatibility:“Compatibility Satisfied”
36 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 = 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.
37 Dynamic Design : Modal information : - For eight stories before enlarging beams in x-direction :Mode No.DirectionPeriod (sec.)MMPR %1X2.55772RZ(Torsion)1.707674y0.97265- Enlarge the beams 2 &4 to 30x70 (width*depth)
38 Dynamic Design:- For eight stories after enlarging beams in x-direction :Mode No.DirectionPeriod (sec.)MMPR %1X1.58802RZ(Torsion)1.5643Y0.79565.440.49711- Comparison with manual results :Mode directionSAP result(sec)Manual result(sec)Error %x- direction1.581.458 %y- direction0.7950.73
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 = 3.27.*Scale factor =- For response-y: * Scale factor = 2.18.*Scale factor =Perform design using envelope combination and check whether static or dynamic combination controls .
42 Static design Controls Dynamic Design :Design of beams in y-direction :- Reinforcement from envelope combination:- Reinforcement from static combination:Static design Controls
43 Dynamic design Controls Design of beams in X-direction :- Reinforcement from envelope combination is considered since the dimensions are increased:Dynamic design Controls
44 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 :
45 Static design OK for columns. Dynamic Design :The following table shows the comparison for column B3 (M3, V2 ):floor/termEnvelope combinationStatic combinationmomentshearaxialAs(mm2)1 68.0431.34 3200 5.541.678 4553.3245.7822.15 6.97 3.7 3982.2349.8122.82 4.782.733409.1447.3320.074.622.552841.9544.51 18.49 2277.7 3.782.121716.5642.58 16.78 3.71 2 1699.8739.19 13.981156.22.881.36826.756.823.18603.8Static design OK for columns.
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 :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.
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*h3eq /12) = 3.371*10-4h3eq = 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 lineLoads 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 satisfyThe actual T-section.2. Stiffness modifiers :I 3-3 = 0.6 .I2-2 = 0.6 .J =3. Weight modifier =Flange Modifiers :- M11 = (almost zero).- M 22 =- M 1-2 = (almost zero).→ we have to substitute the weight of blocks.55 cm8 cm25 cm15 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 :loadMoment(KN.m)Shear(KN)(mm2)( (mm2/mm)SAP Model 1 Results371.41268.6316150.545SAP Model 2 Results377.87286.431645.10.634SAP Model 3 Results373.56274.1316250.573
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 :Model 1 :ModeDirectionPeriodMMPRMode 1x-direction1.58180Mode 2Torsion(Rz)1.50364Mode 3y- direction0.79665.4Model 2 :ModeDirectionPeriodMMPRMode 1x-direction1.62277.0Mode 2Torsion(Rz)1.55579.5Mode 3y- direction0.92470.9Model 3 :ModeDirectionPeriodMMPRMode 1x-direction1.54280.3Mode 2Torsion(Rz)1.45665.5Mode 3y- direction0.76965.7
53 Structural Modeling Of One Way Ribbed Slabs Also the same span taken in the beam C and the following are the values of moment and shear from envelope combination :Model 1 :PointEnvelope combinationmomentshearAs(mm2)Av/s (mm2/mm)C2316.0243.71359.40.422C2-C3208.616.31053C3371.1268.516130.545Model 2 :PointEnvelope combinationmomentshearAs(mm2)Av/s (mm2/mm)C2329.7259.614220.5C2-C3211.017.7899.1C3377.9286.416450.634Model3:PointEnvelope combinationmomentshearAs(mm2)Av/s (mm2/mm)C2317.6243.013670.418C2-C3209.116.2891.3C3373.6274.116250.573
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.