《结构设计原理》叶见曙版21章课后习题英文版答案.docx

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1、__________________________________________________21-11solution:Known:maintruss,bottomchord,H-shapecrosssection,wedding,geometricallengthofbottomchordisl=8m.Q235steel,high-strengthbolts,d0=24mm.Variableaxialforce,N1=3319.4kN(tension),N2=772kN(tension).Fatiguel

2、oadingffmodelI:N1=1399kN(tension),N2=425kN(tension)Unknown:checkthestrengthandfatigue.Solve:bylookinguptable,thedesignvalueofstrengthofsteel:fd=275kN.1.Determinethedimensionofcrosssectionofbottomchord.1).TherequiredgrossareaAmandnetareaAnNmax3319.4×1032Am=1.15

3、=1.15×=13881mmf275dNmax3319.4×1032An≥==12071mmf275d2).Determinethedimensionofthicknessofweb.Selectingtw=16mm,thicknessofflangetf=20mm,widthofwebbw=440mm,widthofflangeh=460mm.thegrossareaactuallyisAm=bwtw+2ht=440×16+2×20×460=25440mm2>f__________________________

4、__________________________________________________________________________13881mm2OK.Singlefacefrictionnf=1,µ=0.35,pre-tensionPf=190MPa.Shearingcapacityperhigh-strengthbolt:Nb=0.9nµP=0.9×1×0.35×190×103vuff=59.85kNNumberofbolts.N3319.4×103max2n===55.46mmNb59850

5、vuSelectingn=64,arrange32boltslinebylineperflange,4lines.ThenetareaactuallyisAn=Am−2×4×24×20=25440−3840=21600mm2>12071mm2OK2.Checkthestrength.N'=Nn14max1−0.5n=3319.4×1−0.5×=3112kN32'N3112×103σ===144.1MPa

6、xisandy-axis____________________________________________________________________________________________________Ix=2×1×20×4603=324453333.3mm412Iy=1×16×4403+2×460×20×230212=1086938667mm4Thecorrespondinggyrationradiusix=Ix=324453333.3=35.7mm4Am25440iIy1086938667

7、4y===206.7mmAm25440Thecorrespondingslendernessratioλlox=8000=224>λ=130Dissatisfyx=ix35.7λloy8000y===38.7<λ=130OKiy206.74.Checkthefatiguelimitstate.Δσ=0.737Δσ=0.737×90=66.33MPaDCΔσD=66.33=49.13MPaγMf1.35NPmax1399×103σpmax===54.99MPaAm25440NPmin425×103σ===16.71M

8、PapminAm25440NormalstressamplitudeΔσp=γ(1+Δϕ)(σpmax−σ)Mfpmin=1×1×(54.99−16.71)=38.28MPa_____________________________________________________________________________________________