Slide 1

Slide 2 Chapter 1 : Introduction Slide 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 Slide 4 Introduction : The Following slides shows : 1. columns centers plan. 2. 3D model of the building. Slide 5 Slide 6 Slide 7 Structural System : The structural system used is on way ribbed slab with load path in x-direction. Slide 8 Materials: - Concrete : - fc= 320 kg/cm( 32 MPa) For columns. - fc= 240 kg/cm( 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 Slide 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 2. 3. Masonry wall weight = 21.22 KN/m. - Live load = 2 KN/m 2. -Water tanks load = 1.14 KN/m 2 - Seismic loads : shown later. Slide 10 Design codes and load combinations: - The following are the design codes used : 1. ACI code 2008. 2. IBC 2009. 3. 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 Slide 11 Chapter 2 : Preliminary Design Slide 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/m. - Ultimate load = 14.06KN/m. Slide 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/m Pu = 3769.6 KN. A g = 2326.9 cm 2. Use columns of 40x60cm 2. Slide 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. Slide 15 Chapter 3 : Static Design Slide 16 Static design : Final dimensions : 1. frame sections : MemberDepth(cm)Width(cm) Col.8040 Main interior beams7040 Main exterior beams7530 Secondary beams2540 Tie beams5030 Slide 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 Slide 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 : Slide 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 load76262.4476273.1320.01 Live load8407.24 0 2.Equilibrium Satisfied : Load` M-ve (left)(KN.m) M+ve (KN.m) M-ve(right) (KN.m) Total moment (KN.m) SAP Result325.13208.63371.41556.9 1D Result0341.40433.61558.21 Error2.3% Slide 20 Static Design : Slab design : 1. Check slab deflection : So, dead = 2.92 mm. Live = 0.78mm. 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. Slide 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 A16.31132 12 A1- B18.61132 10 B110.481252 12 B1- C17.251132 10 C110.291222 12 C1 D18.11132 10 D191132 12 D1 E18.11132 10 E113.55163.32 12 Slide 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 = 0.0033. - A s min = bd = 0.0033*400*650 = 858 mm 2 - Vc = 159.1 KN. - (Av/s) min = 0.333. Slide 23 - Design information : Static Design : pointAs(mm)As min (Av/s) (Av/s) min PI(m)Length(m)barsstirrups B17008580.333 0.62 516 18 @30 cm B1-28548580.333 --5.9 51618 @30 cm B2(L)16768580.333 1.34.8620 18 @30 cm B2(R)16768580.550.3331.54.8620 18 @15 cm B2-310448580.333 _7.9 51618 @30 cm B3(L)15818580.480.3331.54.8620 18 @20 cm B3(R)15818580.333 1.34.8620 18 @30 cm B3-47728580.333 _5.9 516 18 @30 cm B48548580.333 12.3 516 18 @30 cm Slide 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 = 0.0033*b*d=0.0033*400*210= 277.2 mm (312). Vc = 0.75*0.167**400*210/1000= 68.58KN. Slide 25 Static Design : Design information : PointAs(mm) As min (mm) (Av/s) (Av/s)min PI(m)barsstirrups A4469277.20.333 0.885121@10cm A4-B4277277.20.333 0.883121@10cm B4429277.20.333 0.884121@10cm B4-C4259277.20.333 0.883121@10cm C4424277.20.333 0.884121@10cm C4-D4260277.20.333 0.883121@10cm D4425277.20.333 0.884121@10cm D4-E4265277.20.333 0.883121@10cm E4432277.20.333 0.884121@10cm E4-F4147277.20.333 0.883121@10cm F4432277.20.333 0.884121@10cm Slide 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 1616310 @250 mm Floor No.8 D2 & G23766C2 2016310 @300 mm A2, B2, C2, H2, I2 & J2 4774C3 1620310 @300 mm Slide 27 P u =3034.75KN M Y = 11.02 KN.m(maximum value) M X = 153.1 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 Slide 28 Tie beam design Minimum area of steel = 0.0033*b*d =436 mm 2. Use 412mm bottom steel. Use 412mm top steel. Shear design : V u at distance (d = 44cm) = 16.35KN, Vc = 80.83KN. Use 18 mm@200mm. Slide 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. Slide 30 Footing grouping Footing grouping according to columns ultimate load. Group Name Columns Service load(KN) Ultimate load (KN) F1A1, B1,C1,D1,E1,F1,G1,H1,I1,J1305.6375.7 F2A2, J2,A4 B4,C4,D4,G4,H4,I4,J4,2329.952878.51 F3 B2,C2, D2,G2,H2,I2,A3,B3,C3,D3,G3,H3, I3,J3 3667.664599.75 CombinedE4 & F42218.202741.80 Slide 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 F10.961.2x0.80.34 14mmnoNo F26.212.7x2.30.555 18mm1 14@300mm F39.573.3x2.90.756 18mm1 14@200mm Combined13.115.7x2.30.55 125 /200mm 778.21 14@200mm Slide 32 Design of Stairs Slide 33 Verification of SAP model: Compatibility: Compatibility Satisfied Slide 34 Cont. Load type Hand results(KN) SAP results(KN) Error % Dead load293.04292.820.08 Live load181.24181.120.07 Equilibrium: Stress-Strain Relationships: Load M - ve (left) (KN.m) M +ve (KN.m) M -ve (right) (KN.m) 3D SAP Result 22.111.919.4632.68 1D Result21.7210.8621.7232.58 Error 0.3% Equilibrium Satisfied Stressstrain relationship satisfied Slide 35 Chapter 4 : Dynamic Design Slide 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 = 0.458 ton/m 2 - 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. Slide 37 Dynamic Design : Modal information : - For eight stories before enlarging beams in x-direction : Mode No.DirectionPeriod (sec.)MMPR % 1X2.5577 2RZ(Torsion)1.70767 4y0.97265 - Enlarge the beams 2 &4 to 30x70 (width*dept h) Slide 38 Dynamic Design: - For eight stories after enlarging beams in x-direction : Mode No.DirectionPeriod (sec.)MMPR % 1X1.5880 2RZ(Torsion)1.564 3Y0.79565.4 4X0.49711 - Comparison with manual results : Mode directionSAP result(sec)Manual result(sec)Error % x- direction1.581.458 % y- direction0.7950.738 % Slide 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 = 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. Slide 40 Slab design : The comparison is performed. Static design controls Dynamic Design : Slide 41 Slide 42 - Reinforcement from static combination: Dynamic Design : Design of beams in y-direction : - Reinforcement from envelope combination: Static design Controls Slide 43 Dynamic Design : Design of beams in X-direction : - Reinforcement from envelope combination is considered since the dimensions are increased: Dynamic design Controls Slide 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 : Slide 45 floor/ter m Envelope combinationStatic combination momentshearaxialAs(mm 2 )momentshearaxialAs(mm 2 ) 1 68.0431.34 4553.25 3200 5.541.678 4553.3 3200 245.7822.153982.31 3200 6.97 3.7 3982.2 3200 349.8122.823409.11 3200 4.782.733409.1 3200 447.3320.072841.94 32004.622.552841.9 3200 544.51 18.49 2277.7 3200 3.782.121716.5 3200 642.58 16.781716.51 3200 3.71 2 1699.8 3200 739.19 13.981156.2 32002.881.361156.2 3200 826.758600.378 32006.823.18603.8 3200 Dynamic Design : Static design OK for columns. The following table shows the comparison for column B3 (M3, V2 ): Slide 46 Chapter 5 : Structural Modeling Of One Way Ribbed Slab Slide 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. Slide 48 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*h 3 eq /12) = 3.371*10 -4 h 3 eq = 19.45 cm. eq = 23.87 KN/m^3.......to achieve the same weight. - Stiffness modifiers : M11 = 0.35. M22 = 0.0244 M1-2 = 0.0244. Slide 49 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. Slide 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 = 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. 8 cm 25 cm 55 cm 15 cm Slide 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 371.41268.6316150.545 SAP Model 2 Results 377.87286.431645.10.634 SAP Model 3 Results 373.56274.1316250.573 Slide 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-direction1.58180 Mode 2 Torsion(Rz)1.50364 Mode 3 y- direction0.79665.4 Model 2 : Mode Direction PeriodMMPR Mode 1 x-direction 1.62277.0 Mode 2 Torsion(Rz) 1.55579.5 Mode 3 y- direction 0.92470.9 Model 3 : Mode Direction PeriodMMPR Mode 1 x-direction 1.54280.3 Mode 2 Torsion(Rz) 1.45665.5 Mode 3 y- direction 0.76965.7 Model 1 : Slide 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) C2316.0243.71359.40.422 C2-C3208.616.310530 C3371.1268.516130.545 Model 1 : Point Envelope combination momentshearAs(mm 2 ) Av/s (mm 2 /mm) C2329.7259.614220.5 C2-C3211.017.7899.10 C3377.9286.416450.634 Point Envelope combination momentshearAs(mm 2 ) Av/s (mm 2 /mm) C2317.6243.013670.418 C2-C3209.116.2891.30 C3373.6274.116250.573 Model 2 : Model3: Slide 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. Slide 55