《鋼結(jié)構(gòu)理論》

習(xí)題課5Alacedcompressionmember

consistsoftwochannels,asshowninFig.1.

Thecompressionmemberis6mlongwithpinned

-pinnedends,anditissubjectedtoaxialcompressionforceN=1500kN.SupposedQ235steelandE43typeelectrodeareused,θ=45°.

Pleasedesignthemember.SupposeForclassbsection:(lacingangleL45×4)Solution:1.

Bucklingaroundmaterialaxis(yaxis)

Problem1#Take10kNastheselfweightofcolumn，soN

=

1510kN.Choose2[28b,A=2×45.62=91.24

cm2iy=10.6

cmCheckthestability:2.

Bucklingaroundimmaterialaxis(xaxis)

UseL45×4,thenTakeb=

26cmCheckoverallstabilityofbuilt-uparoundimmaterialaxis:Forchord[28bbucklingaround1-1axisI1=241.5.1cm4i1=2.3cmDistancefromcentroidof[28btotheedgeofbuilt-upsection:c=2.02cm11OK!3.Axialcompressionoflacingangle

A7msimplysupportedbeamisshowninFig.2,whichusesIsectionbeam450×150×12×18

.Thebeamissubjectedtoauniformloadq=31kN/mandaconcentratedloadF=72kN.Fis2.5mtotheendandhasasupportinglengthof0.2m.TheactingpositionofFis120mmtothetopofbeam.Designtension、compressionandbendingstressfd

ofsteelis205N/mm2,designshearstressfvdis125N/mm2.Pleasecheckthestrengthofbeam.Fig.2Problem2#Solution:ThemomentandshearforcediagramsareshowninFig.31.Normalstress:AsthemomentismaximalatsectionC,then:2.shearstress:

AstheshearstressismaximalatsectionA:Fig.33.Localcompressionstress:AtsectionB:

4.Equivalentstress:AtsectionB:

OK!AnIsectionbeamHN350×175×7×11issubjectedtouniformloadq,whichisequalto34.35kN/m.Thebeamisconstrainedwithpinned-pinnedends,andthereisnolateralbracing.Pleasechecktheoverallstabilityofthebeam.Solution:Themaxmomentcomesatthemid-spansection:

Checktheoverallstability:Problem3#OK!AnH-sectionbeam-columnwithpin-pinendsisshowninFig.4.Therearetwolateralbracingsatpointsone-thirdandtwo-thirdsoftheheight.Q235isusedforallsteel.ThecolumnissubjectedtoaaxialcompressionN=900kNandatransverseconcentratedloadF=100kN.Pleasecheckthestrengthandoverallstabilityofthecolumn.Fig.4Problem4#Solution:(1)Calculatingthesectionpropertiesofthe