Design Three Floor Reinforced Concrete Structure Eurocode 2 - 6050 Words

Design of a Three Floor Reinforced Concrete Structure to Eurocode 2 (Coursework Sample) Content: Design of a Three Floor Reinforced Concrete Structure to Eurocode 2Student NameCourse/NumberDateInstructors NameIntroductionReinforced concrete is undoubtedly one of the most used construction materials across the world. The material is advantageous in that it can be formed in many varied sizes and shapes, ranging from a simple beam to a long span bridge beam, from a simple rectangular column to a slender one and even to a shell, a dome or an arch. Reinforced concrete versatility and utility is achieved through combining the best features of steel and concrete. The material makes good use of concrete's advantages of strength in compression, high durability and good fire resistance. The weaknesses in concrete are complemented by strengths in steel which include strength in tension and good shear resistance. Weaknesses of steel such as corrosion when unprotected and poor fire resistances are complemented by concrete strengths.The wide application of reinforced concrete in the construction industry implies that there are numerous codes of practice that offer guidance on design using the material. In order to eliminate trade barriers that might result from different codes application, European countries have harmonized their design codes and now it is expected that each country will be using the regionally accepted Eurocodes. In order to understand the application of Eurocodes in real life design, a three floor reinforced concrete structure will be designed to Eurocode 2. In order to ease the design process, analysis of the structure has been done using SAP2000 as the design tool. In the following discussion, the outcome of the design is presented.Design InformationRelevant building regulations and building code. * EN 1990: Design rules for structures * EN 1991: Actions on structures * EN 1992: Concrete structures Intended use of structure. Office block, stairs and landing for foot traffic only. Fire resistance requirements. 2 hour for all elements . General loading conditions Roof Imposed = 2.50KN/m2 (Access provided) Finishes (Asphalt cladding and timber blocks) = 1.25KN/m2Floor Imposed = 2.5KN/m2 Ceiling =0.431KN/m2 Partitions =0.958KN/m2Faade = 1.25KN/mBuilding services = 0.024KN/m2 Exposure conditions. Severe (external) and mild (internal) Subsoil conditions. Allowable subsoil bearing pressure = 200KN/mm2. Foundation type. RC footings to columns and walls. Material Data. Grade 30/37 concrete with 20mm maximum aggregate used Characteristic strength of bars fy= 500N/mm2 (High yield) Other relevant information. Self-weight of concrete = 24.0KN/m3All dimensions shown on drawings/diagrams are in millimetres (mm).References whose source document is not indicated refer to EN 1992 Analysis Using SAP2000The first step when performing analysis using SAP2000 entailed 3D modelling of the structure. The structure was modelled based on SAP2000 inbuilt 3D beam-slab frame template. On the template, the frame's geome try and preliminary section properties were defined. It was ensured that the beam was connected to the slab in a manner that resulted to a T-beam system.Loading determination and applicationFloorsDead load = 0.431+0.024+0.958=1.413KN/m2Dead load from curtain walls: =1.25KN/mImposed loads: =2.5KN/m2 (minimum for office buildings)RoofDead load = 1.413KN/m2 (as above)Imposed loads = 2.5KN/m2 (access provided)Self-weight of members in both cases was not calculated as the software would automatically compute the self-weight.For loads application, the first step entailed defining the load cases, i.e. dead and live loads, then defining the loading combinations to be used. Combination of loads considered included factored dead and live loads for ultimate design and unfactored dead and live loads for deflection serviceability check. Loads to the slabs were applied as area uniform loads to the frame. The loading from curtain walls was added as a perimeter uniformly distributed load and was ad ded to the external beams in the first and second floors. Being ultimate limit states design codes, Eurocodes require application of various loading factors. For the worst scenario, the factors are 1.35 for dead load and 1.5 for live loads. Loads were not factored when being applied as all that was needed was to specify the right code (EN 1992-1-1:2004 was used) and then the right factors would be generated automatically. Had there been other load cases, such as snow, then modifiers such as 1, 2, 3 would have been defined as required by the code.Preliminary sections chosen were rectangular 400x200 mm concrete beams and square 250x250 reinforced concrete columns. The slab section was modeled as a thin shell of 150mm thickness.Analysis Output0-4305Figure SEQ Figure \* ARABIC 1: An extruded 3-d view of the model createdFigure SEQ Figure \* ARABIC 2: The structure's deformed shapeFigure SEQ Figure \* ARABIC 3: values of shear on one of the framesFigure SEQ Figure \* ARABIC 4: Moment Diagram of one of the framesDesign ForcesCorner Column A1Resultant axial forceResultant moments in the major axesResultant moments in the minor axesExterior column B1Resultant axial forceResultant moments in the major axesResultant moments in the minor axesExterior column A3Resultant axial forceResultant moments in the major axesResultant moments in the minor axesInterior Column B2Resultant axial forceResultant moments in the major axesResultant moments in the minor axesInterior beam along column line 3 of the first floorResultant shearResultant momentsDeflectionInterior beam along column line B of the first floorResultant shearResultant momentsDeflectionColumn DesignCorner Column A1Loading Case 1.35xDead load plus 1.5xLive loadLoading N Mx,Mx,My,My264.5 kN, 0.0 kN.m, 3.7 kN.m, 0.0 kN.m, 10.7 kN.mSlenderness ClassificationX-X Slenderness: l, l0 , lim , 3.700m, 9.156m, 71.4, 126.9 limSlenderY-Y Slenderness: l, l0 , lim , 3.700m, 8.073m, 71.4, 111.9 limSlenderxx/ yy126.9 / 111.9 Limit 0.5 to 21.134OKAxial CapacityNuz= Favx(B x H - Asc) + Asc x fyk/ s18.1x(250x250-1357) + 1357.2 x 460 / 1.21651.6 kNOKDesign Moments x-xe2= Fn(Kr, Kj, jeff, n, nu, nbal,w d, ....)1.00, 0.33, 2.0, 0.233, 1.479, 0.463, 0.479, 21457.19 mmMed0 = Fn(M, M, M2, M0ed, Ned, e2)0, 3.7, 15.1, 2.2, 264.5, 57.2 (no nominal moments)17.4 kN.mMed = Fn(M, M, M2, M0ed, Ned, e0, ei, e2)0, 3.7, 15.1, 8.3, 264.5, 20, 22.9, 57.223.4 kN.mDesign Moments y-ye2= Fn(Kr, Kj, jeff, n, nu, nbal,w d, ....)0.84, 0.53, 2.0, 0.233, 1.479, 0.000, 0.479, 21460.25 mmMed0 = Fn(M, M, M2, M0ed, Ned, e2)0, 10.7, 15.9, 6.4, 264.5, 60.2 (no nominal moments)22.3 kN.mMed = Fn(M, M, M2, M0ed, Ned, e0, ei, e2)0, 10.7, 15.9, 11.7, 264.5, 20, 20.2, 60.227.7 kN.mBi-Axial Moment Capacity: X-X Axis DominantDesign LoadsNed, Med x-x, Med y-y, Med res, Ang264.5 kN, 23.4 kN.m, 22.3 kN.m, 32.4 kN.m, 43.6 degDesign Data X/h, h, b, X, Ac, Ybar0.473, 353 mm, 250 mm, 167.09 mm, 17888 mm, 89.1 mmConcrete Fc=(Acnetx h xfcd)(17888 x 0. 90 x 18.1)291.9 kNF Equlibrum S (Ft) + FC - Fapp= 0-29-3-21-12-20-11-2+25+7+16+6+16+292-264.50.0 kNOKConcrete Mc=Fcx(H/2-Ybar)291.9 x (353 / 2 - 89.1)25.6 kNmMu =Mc + (M1+...+M12)25.6 + (3.4+0.0+1.6+0.5+1.5+0.4+0.0+2.8)+ (0.3+1.2+0.2+1.2)38.7kNmOKMax Moment/Mu32.4 / 38.70.836OKBi-Axial Moment Capacity: Y-Y Axis DominantDesign LoadsConcrete Fc=(Acnetx h xfcd)(22548 x 0.90 x 18.1)368.0 kNF Equlibrum S (Ft) + FC - Fapp= 034 - 103 - 28 - 7 + 368 - 264.50.0 kNOKConcrete Mc=Fcx(H/2-Ybar)368.0 x (353 / 2 - 100.1)28.2 kNmMu =Mc + (M1+...+M4)28.2 + (1.0+13.7+2.1+0.2)45.2kNmOKMax Moment/Mu32.7 / 45.20.723OKShear CheckVapp/ max(VRd,c.a, VRd,c.b)3.1 / Max(37.9, 22.8)0.08no Links reqExterior column B1Loading N Mx,Mx,My,My565.0 kN, 0.0 kN.m, 1.9 kN.m, 0.0 kN.m, -17.7 kN.mSlenderness ClassificationX-X Slenderness: l, l0 , lim , 3.700m, 10.300m, 61.7, 142.7 limSlenderY-Y Slenderness: l, l0 , lim , 3.700m, 8.151m, 61.7, 112.9 limSlenderxx/ yy142.7 / 112.9 Limit 0.5 to 21.264OKAxia l CapacityNuz= Favx(B x H - Asc) + Asc x fyk/ s18.1x(250x250-1709) + 1709.0 x 460 / 1.2 1786.0 kNOKDesign Moments x-xe2= Fn(Kr, K ,  eff, n, nu, nbal,  d, ....)1.00, 0.12, 2.0, 0.355, 1.603, 0.452, 0.603, 21026.28 mmMed0 = Fn(M, M, M2, M0ed, Ned, e2)0, 1.6, 10.6, 0.9, 402, 26.3 (no nominal moments)11.5 kN.mMed = Fn(M, M, M2, M0ed, Ned, e0, ei, e2)0, 1.6, 10.6, 11.3, 402, 20, 25.8, 26.321.8 kN.mDesign Moments y-ye2= Fn(Kr, K ,  eff, n, nu, nbal,  d, ....)0.78, 0.51, 2.0, 0.355, 1.603, 0.000, 0.603, 21056.29 mmMed0 = Fn(M, M, M2, M0ed, Ned, e2)0, -1.7, 22.6, 1, 402, 56.3 (no nominal moments)23.6 kN.mMed = Fn(M, M, M2, M0ed, Ned, e0, ei, e2)0, -1.7, 22.6, 3.9, 402, 20, 20.4, 56.326.6 kN.mBi-Axial Moment Capacity: X-X Axis DominantDesign LoadsNed, ...