6.1解:1.计算截面特征值工字形截面A=20⨯500⨯2+12⨯450=25400mm2I x=500⨯4903/12-488⨯4503/12=11.9629⨯108mm4;I y=2⨯5003⨯20/12=4.1667⨯108mm4 mm4;i x=217mm;i y=128mm;2.刚度验算λx=6000//217=27.65;λy=46.88λx、λy<[λ] 刚度满足要求3.强度验算因无截面稍弱无需验算截面强度4.整体稳定验算:焊接工字形截面翼缘焰切边x、y轴都属于b类截面ϕmin=ϕy=0.8704σ=N/ϕA=4500⨯103/(0.8704⨯25400) =203.54N/mm2 <f =205 N/mm2杆件整体稳定满足要求5.局部稳定验算:翼缘b1/t=244/20=12.2 <(10+0.1λ)(235/f y)1/2=14.69腹板h0/t w=450/12=37.5 <(25+0.5λ)(235/f y)1/2 = 48.44板件局部稳定满足要求(λ为λx、λy大者,λ=46.88)6.2解:假定λ=45 按b类截面查ϕ=0.878A req=N/ϕmin f=4500⨯103/(0.807⨯215)=23839 mm2i xreq=6000/45=133.33mm;h req≈ i xreq/α1=133.33/0.43=310 mm;i y=3000/45=66.67mm;b req≈ i yreq/α1=66.67/0.24=278 mm;(1)设计工字形截面参考6.1题截面尺寸,取翼缘宽度450厚20,腹板高度480厚度12验算。
A=20⨯450⨯2+12⨯480=23760mm2I x=450⨯5203/12-438⨯4803/12=13.3619⨯108mm4;I y=2⨯4503⨯20/12=3.0375⨯108 mm4;i x=228.1mm;i y=113.07mm;λx=6000/236.5=26.3;λy=3000/112.5=26.53λx、λy<[λ] 刚度满足要求杆件整体稳定验算:焊接工字形截面翼缘焰切边x、y轴都属于b类截面λy=26.53 ϕmin=ϕy=0.948σ=N/ϕA=4500⨯103/(0.948⨯23760) =199.78N/mm2 <f=205 N/mm2 整体稳定满足要求板件局部稳定验算:λx、λy<30 取λ=30翼缘b1/t=219/20=10.95 <(10+0.1λ)(235/f y)1/2=13腹板h0/t w=480/12=40 ≤(25+0.5λ)(235/f y)1/2 =40局部稳定满足要求在侧向加支撑后截面面积节约了钢材6.90%(25400-23760)/23760=6.90%(2)选H型钢HW400⨯400(#400⨯408) A=25150mm2I x=7.11⨯108mm4;I y=2.38⨯108 mm4;i x=168mm;i y=97.3mm;λx=6000/168=35.71;λy=3000/97.3=30.83(两个方向长细比较接近)λx >λy由b类λx查得0.9153σ=N/ϕA=4500⨯103/(0.9153⨯25150) =195.48N/mm2 <f=205 N/mm2 (25150-23760)/ 23760=5.85%翼缘b1/t=193.5/21=9.21 <(10+0.1λ)(235/f y)1/2=13.57腹板h0/t w=358/21=17 ≤(25+0.5λ)(235/f y)1/2 =42.86局部稳定满足要求H型钢比组合截面用钢多(25150-23760)/ 23760=5.85%[截面选择不合适例子]H700⨯300 A=23550mm2I x=20.1⨯108mm4;I y=1.08⨯108 mm4;i x=293mm;i y=67.8mm;λx=6000/293=20.48;λy=3000/67.8=44.25λy>λx由b类λy查得0.881σ=N/ϕA=4500⨯103/(0.881⨯23550) =216.89N/mm2 >f整体稳定不满足要求((两个方向长细相差较大)选HN800⨯300(792⨯300) A=24340mm2I x=25.4⨯108mm4;I y=0.993⨯108 mm4;i x=323mm;i y=63.9mm;λx=6000/323=18.58;λy=3000/63.9=46.95λy>λx由b类λy查得0.87σ=N/ϕA=4500⨯103/(0.87⨯24340) =212.51N/mm2 <f整体稳定满足要求翼缘b1/t=143/22=6.5 <(10+0.1λ)(235/f y)1/2=14.7腹板h0/t w=748/14=53.43>(25+0.5λ)(235/f y)1/2 =48.48腹板局部稳定不满足要求6.3解:(1)A=320⨯20⨯2+10⨯320=16000mm2I x=320⨯3603/12-310⨯3203/12=3.9723⨯108mm4;I y=2⨯3203⨯20/12=1.0923⨯108 mm4;i x=157.6mm;i y=82.6mm;λx=10000/157.6=63.5;λy=10000/82.6=121.1焊接工字形截面翼缘为剪切边x、y轴分别属于b、c类截面ϕx=0.7885 ϕy=0.3746ϕmin=ϕy=0.3746N=fϕA=205⨯ (0.3746⨯16000) =1228.69kN(1230)翼缘b1/t=155/20=7.75 <(10+0.1λ)(235/f y)1/2=22.11腹板h0/t w=320/10=32<(25+0.5λ)(235/f y)1/2 =85.55(2)A=16000mm2I x=400⨯4323/12-392⨯4003/12=5.9645⨯108mm4;I y=2⨯4003⨯16/12=1.7067⨯108 mm4;i x=193.1mm;i y=103.2mm;λx=10000/193.1=51.8;λy=10000/103.2=96.9焊接工字形截面翼缘为剪切边x、y轴分别属于b、c类截面ϕx=0.848 ϕy=0.4776ϕmin=ϕy=0.4776N=fϕA=215⨯ (0.4776⨯16000) =1642.94kN(1230)翼缘b1/t=196/16=12.25 <(10+0.1λ)(235/f y)1/2=19.69腹板h0/t w=400/8=50<(25+0.5λ)(235/f y)1/2 =73.45结论:截面展开的承载能力高。
6.4解:假定λ=80由b类查得ϕ= 0.688A=N/ϕf=410⨯103/0.688⨯215=2772mm2i x=l0x/λ=2300⨯80=28.75mm;i y=l0y/λ=2870⨯80=35.88mmh= i x/0.305=94mm;b= i y/0.215=167mm选L100⨯7A=13.8⨯2=27.6 cm2 i x=3.09cm i y=4.53 cmλx=230/3.09=74.43 λy=287/4.53=63.36b/t=100/7=14.29 < 0.58l0y/b=16.65换算长细比λyz=λy(1+0.475b4/l0y2t2)= 63.36(1+0.475⨯1004/28702⨯72)=70.82<λx按b类由λx=74.43查得ϕx= 0.7235验算杆件的整体稳定σ=N/ϕA=410⨯103/(0.7235⨯2760) =205.32N/mm2<f =215 N/mm2满足要求局部稳定b/t=100/7=14.29<(10+0.1λ)(235/f y)1/2=17.44翼缘满足要求腹板2两角钢组成局部稳定可满足要求6.5解:(轴力较大)假定λy =60按b类查得ϕy= 0.807A=N/ϕf=1600⨯103/0.807⨯215=9222mm2柱上端铰接,下端固接。
l0x=l0y=0.7⨯7500=5250mmi=l/λ=5250/60=87.5mm;h≈ i/0.38=230mm;b≈i/0.44=199mm由A和i选[32a ;A=48.51⨯2=97.02 cm2 i y=12.5cmλy =l0y/ i y=525/12.5=42由等稳得:λy=λ0x 得:λx=(λy2-λ12) 1/2为满足单肢稳定λ1≤0.5λmax=0.5⨯42=21或40取λ=20 λx=(λy2-λ12) 1/2=36.931i x=l0x/λx=5250/36.93=142 b≈i/0.44=323mm取b=320 z0=2.24cmI x=2(305+13.762⨯48.51)=1.898⨯104cm4i x=13.99 cmλx=525/14=37.5i1=2.50cm l0≤λ1 i1=20⨯25=500mm取l0=500 λ1 =500/25= 20λ0x=(λx2+λ12) 1/2=(37.52+202) 1/2=42.5由λ0x=42.5按b类查得ϕx= 0.889σ=N/ϕA=1600⨯103/(0.889⨯9702) =185.51N/mm2<f=215 N/mm2满足要求缀板设计:肢件轴线距离b1=275缀板宽度b j≥2b1/3=2⨯275/3=183.33mm取b j=185 mm 厚度t≥b1/40=275/40=6.8mm 取t=8mm缀板轴线距离l1=500+185=685mm缀板线刚度与分肢线刚度之比2(I b/b1)/(I1/l1)=[2(8⨯1853/12)/275]/ (305⨯104/685)=6.89>6缀板面的剪力V1=[(Af/85)(f y/235) 1/2]/2=[(9702⨯215)/85]/2=12.270kN缀板的内力为剪力V j= V1 l1/ b1=12270⨯685/275=30.563 kN弯矩M j= V1 l1/2=12270⨯685/2=4.202 kN.m(还要验算缀板强度及连接焊缝,而且要画图表示)。
6.6解:假定λy =60按b类查得ϕy= 0.807A=N/ϕf=1600⨯103/0.807⨯215=9222mm2柱上端铰接,下端固接。
l0x=l0y=0.7⨯7500=5250mmi=l/λ=5250/60=87.5mm;h≈ i/0.4=219mm;b= i/0.5=175mm由A和i选I22b;A=46.63⨯2=93.26 cm2 i y=8.78cmλy =l0y/ i y=525/8.78=59.79由等稳得:λy=λ0x 得:λx=(λy2-27A/A1x) 1/2选缀条的角钢L45⨯4 A1x =2⨯3.49 cm2λx=(59.792-27⨯46.63/3.49) 1/2=56.69i x =l 0x /λx =5250/56.69=92.6 b ≈ i x /0.5=185.22mm(取b=220mm 肢件轴线距离b 1=108mmI x =2(239+5.42⨯46.63)=0.3197⨯104cm 4 i x =5.855cmλx =525/14=89.66 太大,不满足等稳要求)取b=300mm 肢件轴线距离b 1=188mmI x =2(239+9.42⨯46.63)=0.8718⨯104cm 4 i x =9.67cmλx =525/9.67=54.3λ0x =(λx 2+27A/A 1x ) 1/2=(54.32+27⨯46.63/3.49) 1/2=57.53<λy由λy =59.79按b 类查得ϕx = 0.810σ=N/ϕA=1600⨯103/( 0.810⨯9326) =211.81N/mm 2<f =215 N/mm 2构件整体稳定和刚度满足要求分肢稳定:缀条按45︒布置l 01=2⨯188=376mmλ1 =37.6/2.27= 16.56<0.7λmax =0.7⨯59.79 =41.85分肢稳定满足要求缀件面的剪力V 1=[(Af/85)(f y /235) 1/2]/2=[(9702⨯215)/85]/2=12.270kN斜缀条内力N 1=V 1/cos α斜缀条按轴心压杆计算,考虑偏心的影响乘折减系数。