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CRN: 29476 / 534711School of Science, Engineering & EnvironmentTRIMESTER TWO EXAMINATIONPROGRAMMES:BEng (Hons) Civil EngineeringBEng (Hons) Civil & Architectural EngineeringMEng (Hons) Civil EngineeringMEng (Hons) Civil & Architectural EngineeringBEng (Hons) Civil Engineering Degree Apprenticeship (Part Time)STRUCTURES E1 P1Start: Monday 10 May 2021 at 09.00 Finish – Friday 28 May 2021 at 16:00___________________________________________________________________Instructions to CandidatesThere are four questions on this examination paper. Full marks may be obtained forcorrect answers to THREE questions. Marks for parts of questions are shown inbrackets.Each candidate has their own individual parameters for each question. Please usethe parameters particular to you. Using parameters attributed to another candidatewill result in zero marks for the attempted question. Check the provided named list foryour individual parameters.Start your answer to each question on a new sheet of the answer script and write yourindividual parameters at the top.Approved electronic calculators may be used but NOT in text storage mode.Structural mechanics and Eurocode design formula sheets are attached.CRN: 29476 / 5347121. A building roof truss is formed by a statically determinate pin jointed, triangulatedframework, and loaded by two variable actions as shown in Figure 1. The structuralelements are manufactured from mild steel circular hollow sections (CHS). The sectiondesignation and steel grade of the CHS sections is an individual parameter. Please notethat the section designation will depend upon whether a member is subjected to tensionor compression forces under the described load case of Figure 1.(a) State the three Laws of Static Equilibrium and calculate the values of the threereactions.(4 Marks)(b) Stating your sign convention, obtain the sense and magnitude of the internal forcesin all the framework elements. Show your answer diagrammatically. The appliedforces, FB and FC are individual parameters. The geometry of the truss in Figure1 are individual parameters as well.(12 Marks)(c) Use Castigliano’s Theorem to determine the vertical deflection at joint F.(10 Marks)(d) State whether you consider the stiffness of this framework to be adequate for thebuilding roof truss, giving reasons for your decision.(2 Marks)(e) For the loading direction shown, determine whether element DE is adequate forstrength according to EN1993. Assume that joints E and D are positions of lateralrestraint about the y-y axis and z-z axis. To check strength a load partial safetyfactor of 1.5 should be applied to the internal forces calculated in (b) above.(5 Marks)Total 33 MarksFigure 1. Arrangement and loading of a simply supported, pin jointed framework. Pleasenote that this Figure is indicative and may not be exact with consideration of individualparametersCRN: 29476 / 5347132. The beam shown in Figure 2 is being checked for capacity using stress analysis. TheUB beam section designation and steel grade of the beam are individual parameters.The loads and geometry in Figure 2 are also individual parameters.(a) Draw the axial force, shear force and bending moment diagrams showing theprincipal values. (12 Marks)Considering a small element at point C, just to the right of the uniform load, where theweb meets the top flange:(b) Calculate the axial stress assuming the axial force is applied at the section centroid.State whether the stress is tension or compression.(3 Marks)(c) Calculate the bending stress. State whether the stress is tension or compression.(5 Marks)(d) Assuming the uniform loading spreads at 450 through the section, calculate thebearing stress, stating whether it is tension or compression.(5 Marks)(e) Calculate the shear stress in the web, assume that clockwise is positive.(5 Marks)(f) Summarise the stresses on a stress element diagram.(3 Marks)Total 33 MarksFigure 2. Beam loading arrangement. Please note that this Figure is indicative and maynot be exact with consideration of individual parametersFVB (kN) FVE (kN)UDLBC (kN/m)FHE (kN)AELDE (m)L (m)DLAB (m) LCD (m)BLBC (m)CCRN: 29476 / 5347143. A fully lateral restrained, uniform, isotropic steel beam is subject to a point load anduniformly distributed load, as shown in Figure 3. The loadings in Figure 3 areindividual parameters as well as the geometrical lengths. The beam is a mild steel UBsection. The section designation and steel grade of the beam are also individualparameters.(a) Calculate the reactions.(3 Marks)(b) Draw the axial force, shear force and bending moment diagrams showing theprincipal values.(8 Marks)(c) Use Macaulay’s method to produce a general formula for deflection of the beam(10 Marks)(d) Calculate the deflection at point C and hence sketch the deflected shape of thebeam.(4 Marks)(e) State whether you consider this beam section to be adequately stiff, giving reasonsfor your decision. Use the output from (d) with the assumption that the loads arefactored at the appropriate limit state.(2 Marks)(f) Determine whether the beam is adequate for strength according to EN1993.Assume that the compression flange is laterally restrained throughout its wholelength. Use the output from (b) with the assumption that the loads are factored atthe appropriate limit state(6 Marks)Total 33 MarksFigure 3. Arrangement and loading of a simply supported beam. Please note that thisFigure is indicative and may not be exact with consideration of individual parametersBDUDLCD (kN/m)FHD (kN)ACFVB (kN)xL (m)LAB (m) LBC (m) LCD (m) CRN: 29476 / 5347154. (a)Draw diagrams of the four standard cases of effective length, giving theoretical anddesign values of effective length factors. (6 Marks)A mild steel section forms a diagonal compression brace in a steelwork buildingstructure. The compression brace is restrained by a tubular brace at mid-length. All theconnections are pinned, about both axes. The section designation and steel grade of thecompression brace is an individual parameter. The geometry, X and H as well as thelateral wind load, W, in Figure 4 are also individual parameters. (b)Sketch the possible deflected shapes of the brace for buckling about each axis. Statethe effective length factors and calculate the effective lengths. (6 Marks) (c)Define slenderness ratio and calculate the slenderness ratio about both axes. Statewhich axis the column is likely to buckle about.(5 Marks)Calculate the Euler, Rankine and Perry-Robertson buckling loads.(d) (8 Marks)(e) Determine whether the Compression brace is adequate for strength according toEN1993. Please note you will need to check compression capacity of the sectionfor the most critical slenderness value only.(6 Marks) (f)You are further required to state the efficiency of the section under the describedloading in the value engineering process. (2 Marks)Total 33 MarksFigure 4. A building structure bracing bay.XCRN: 29476 / 534716Structural Mechanics FormulaeSimple bending:E Rf zM I= =Shear flow :IbVA’z’ =Simple torsion:L I pGT r = =Castigliano’s Theorem : = L EI M WU =  AE FL WU0 . .Differential equation of flexure : Mdxd zEI = –22Theorem of the Parallel Axis : I NA yy = I yy +  Ah2 where=AAzzEuler Buckling: 22 LEIPE=Rankine Capacity: 21 af AP yR+= where a  0.0001Perry-Robertson Stress: fc = fy + fcr( + )- fy + fcr(1+ )2 – fy fcr1 411 2 where22  Efcr = and  = 0.003Unsymmetrical bending:yI I IM I M IxI I IM I M II I I Sin II I I Sin II IITanyy zz yzyy zz zz yzyy zz yzzz yy yy yzzvv zz yz yyuu yy yz zzyy zzyz––+––== + += – +–= –2 22 22 2cos 2 sincos 2 sin22    Principal Stress 😡 zTan xz –=22,21 2max –=1 ( )2 4 21 22 x z xzx z    + – ++= and 2 ( )2 4 21 22 x z xzx z    – – ++=CRN: 29476 / 534717Standard Analysis CasesCentroids of Area ShapeAIyyIzzRectangleTriangleCircle b.d12.b d 3 312d.b.d 2b36.b d 34. D264. D48wL2L 2ww (per metre)LEIwL3845 4max =LPabLPaLPbPa b = –3 2max 4 348 Laa LEIPL4PLP 2P 2PL 2L 2EIPL483max =Deflection BMD SFDb 2d 2b 3d 3RDCRN: 29476 / 534718Hot Rolled Steel Section Properties UNIVERSAL BEAMSDepthofSectionWidthofSectionThickness ofWebThickness ofFlangeSecondMoment ofAreaRadius ofGyrationElasticModulusPlasticModulusArea ofSectionSerial sizetypehbtwtfrI yyI zzi yi zWel, yWel, zWpl, yWpl, zAmmmmmmmmmmcm4cm4cmcmcm3cm3cm3cm3cm2203x133x25UB203.2133.25.77.87.623603108.563.123046.225870.932203x133x30UB206.8133.96.49.67.628903848.713.1728057.531488.238.2254x146x31UB251.4146.168.67.6444044910.53.3635161.339394.139.7254x146x37UB256146.46.310.97.6556057110.83.484337848311947.2254x146x43UB259.6147.37.212.77.6656067710.93.525049256614154.8305x165x40UB303.4165610.28.9852076312.93.8656092.662314251.3305x165x46UB306.6165.76.711.88.99950897133.964610872016658.7305x165x54UB310.4166.97.913.78.9117001060133.9375412784619668.8457x191x74UB457190.4914.510.233400167018.84.21458176165327294.6457x191x82UB460191.39.91610.237100187018.84.2316111961831304104457x191x89UB463.4191.910.517.710.2410002090194.2917702182014338114 UNIVERSALCOLUMNSDepthofSectionWidthofSectionThicknessofWebThickness ofFlangeSecondMoment ofAreaRadius ofGyrationElasticModulusPlasticModulusArea ofSectionSerial sizetypehbtwtfrI yyI zzi yi zWel, yWel, zWpl, yWpl, zAmmmmmmmmmmcm4cm4cm4cmcm3cm3cm3cm3cm2152x152x23UC152.4152.25.86.87.612604036.543.716452.618280.229.2152x152x30UC157.6152.96.59.47.617405586.763.8322273.324811238.3152x152x37UC161.8154.4811.57.622207096.853.8727391.530914047.1203x203x46UC203.2203.67.21110.2456015408.825.1345015249723158.7203x203x52UC206.2204.37.912.510.2526017708.915.1851017456726466.3203x203x60UC209.6205.89.414.210.2609020408.965.258420165630576.4203x203x71UC215.8206.41017.310.2765025409.185.370624679937490.4203x203x86UC222.2209.112.720.510.2946031209.285.34850299977456110 CIRCULAR HOLLOWSECTIONSDiameterofSectionThicknessSecondMomentof AreaRadius ofGyrationElasticModulusPlasticModulusArea ofSectionSerial sizetypehtI yyiyyWel, yWpl, yAmmmmcm4cmcm3cm3cm288.9×4CHS88.9496.3321.728.910.788.9×5CHS88.951162.9726.235.213.288.9×6.3CHS88.96.31402.9331.543.116.3114.3×4CHS114.342113.936.948.713.9114.3×5CHS114.352573.874559.817.2114.3×6.3CHS114.36.33133.8254.773.621.4139.7×4CHS139.743934.856.273.717.1139.7×5CHS139.754814.7768.890.821.2139.7×6.3CHS139.76.35894.7284.311226.4 ANGLESDepthofSectionWidthofSectionThicknessAreaofSectionSecond Moment of AreaRadius of GyrationElasticModulusSerial sizetypehbtAI yyI zzI uuI vvi yi zi ui vWel, yWel, zmmmmmmcm2cm4cm4cm4cm4cmcmcmcmcm3cm380x80x10EA80801015.187.587.513936.42.413.033.031.5515.415.480x80x6EA808069.3555.855.888.523.12.443.083.081.579.579.5780x80x8EA8080812.372.272.211529.92.433.063.061.5612.612.690x90x10EA90901017.112712720152.62.723.433.431.7519.819.890x90x12EA90901220.314814823461.72.73.43.41.7423.323.390x90x8EA9090813.910410416643.12.743.453.451.7616.116.1100x100x10EA1001001019.217817828373.73.053.053.051.9624.824.8100x100x12EA1001001222.720720732885.73.023.83.81.9429.129.1100x100x15EA1001001527.92492493931042.983.753.751.9335.635.6 CRN: 29476 / 534719EN 1991 Design RulesTable 1. Variable action sensitivity factors. Table 2. Action Partial Safety Factors. Usecombination0frequent1quasipermanent2STR and GEO limitsUnfavourableFavourableOffice0.70.50.3Permanent1.351.0Shopping0.70.70.6Variable1.500Storage1.00.90.8Accidental1.00Roofs0.700Combination equations.Snow0.50.20ULS combinationWind0.50.20SLS combination  GGk +  QQk1 +  Q 0,2Qk 2 + …Table 3. Minimum roof imposed action.sk = 0.15 + (0.1z + 0.05)+   A525 -100   and snow on the roofis, s = iskTable 4. Snow action coefficients. iRoof slope0o ≤  ≤15o15o <  ≤ 30o30o <  < 60o ≥ 60omonopitch0.80duopitch0.80  –30600.8   –+15150.8 0.4   –30601.2  Basic velocity pressure,Peak velocity pressure,where, qb = 0.613(Vb,0cdircseasoncaltcprob )2 calt =1+ 0.001Aqp = cece,TqbTable 5. External wall pressure coefficients.dh coefficients Cp,eSideFrontback5-0.8+0.8-0.71-0.8+0.8-0.5-0.8+0.7-0.3  0.25 Table 6. Monopitch roof pressure coefficients Roofanglecoefficients, Cp,e=00=900ABCABC< 50-1.2-0.7±0.2-1.2-0.7±0.2150-1.3-0.9-1.9-0.8-0.7300-1.8-0.8-1.5-1.0-0.8450-1.5-0.7-1.4-1.0-0.9 Gk +1,1Qk1 + 2,2Qk 2 + 2,3… Roof slope < 30o30o ≤  14whereWeb element≤ 72≤ 83≤ 124> 124UB or UC sections inpure compressionFlange element≤ 9≤ 10≤ 14> 14Web element≤ 33≤ 38≤ 42> 42 fy235 =Table 8. Material Partial Safety Factors. Local failure (yielding)m01.0Member instability (buckling)m11.0Tension fracturem21.1Joint resistancem21.25Bolt friction SLSm3,ser1.1Bolt friction ULSm31.25 30,Mv yc RdA fV=0,Mpl ypl RdW fM= for Class 1 and 2 sections,0,Mel yel RdW fM= for Class 3 sections.1,MLT y yb RdW fM= where96zLT = for S275 material and85zLT= for S355 materialTable 9. Lateral torsional buckling curve selection, rolled sections. Aspect ratioBuckling curveGeneralRolled sectionscbdc  2h b 2h b1,Myb RdAfN= where L = andyLE f = Table 10. Flexural buckling curve selection, rolled sections. Aspect ratioBuckling axisBuckling curvey-yz-za by-yz-zb c  1.2h b 1.2h b1.5 1.0, ,,, ,,, ,, + + z cb Rdz Edb y Rdy Edb z RdE dM MM MN NCRN: 29476 / 5347113Lateral Torsional Buckling Curve for f y=275N/mm200.10.20.30.40.50.60.70.80.911.10 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3LTLTFlexural Buckling Curve for f y=275N/mm200.10.20.30.40.50.60.70.80.911.10 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 ac ac

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