24m钢结构开始设计1、设计资料1)某厂房跨度为24m,总长90m,柱距6m,屋架下弦标高为18m。
2)屋架铰支于钢筋混凝土柱顶,上45柱截面400×400,混凝土强度等级为C30。
3)屋面采用1.5×6m的预应力钢筋混凝土大型屋面板。
(屋面板不考虑作为支撑用)。
4)该车间所属地区为市5)采用梯形钢屋架考虑静载:①预应力钢筋混凝土屋面板(包括嵌缝)、②二毡三油加绿豆沙、③找平层2cm厚、④ 支撑重量考虑活载:活载(雪荷载)积灰荷载6)钢材选用Q345钢,焊条为E50型。
2、屋架形式和几何尺寸屋面材料为大型屋面板,故采用无檩体系平破梯形屋架。
屋面坡度i=(3040-1990)/10500=1/10;屋架计算跨度L=24000-300=23700mm;端部高度取H=1990mm,中部高度取H=3190mm(约1/7。
4)。
屋架几何尺寸如图1所示:图1:24米跨屋架几何尺寸3、支撑布置由于房屋长度有90米,故在房屋两端及中间设置上、下横向水平支撑和屋架两端及跨中三处设置垂直支撑。
其他屋架则在垂直支撑处分别于上、下弦设置三道系杆,其中屋脊和两支座处为刚性系杆,其余三道为柔性系杆。
(如图2所示)上弦平面支撑布置屋架和下弦平面支撑布置垂直支撑布置4、屋架节点荷载屋面坡度较小,故对所有荷载均按水平投影面计算:计算屋架时考虑下列三种荷载组合情况1) 满载(全跨静荷载加全跨活荷载)节点荷载①由可变荷载效应控制的组合计算:取永久荷载γG =1.2,屋面活荷载γQ1=1.4,屋面集灰荷载γQ2=1.4,ψ2=0.9,则节点荷载设计值为F=(1.2×2.584+1.4×0.70+1.4×0.9×0.80)×1.5×6=45.7992kN②由永久荷载效应控制的组合计算:取永久荷载γG =1.35,屋面活荷载γQ1=1.4、ψ1=0.7,屋面集灰荷载γQ2=1.4,ψ2=0.9,则节点荷载设计值为F=(1.35×2.584+1.4×0.7×0.70+1.4×0.9×0.80)×1.5×60=46.593 kN 2) 全跨静荷载和(左)半跨活荷①由可变荷载效应控制的组合计算:取永久荷载γG =1.2,屋面活荷载γQ1=1.4,屋面集灰荷载γQ2=1.4,ψ2=0.9全垮节点永久荷载F1=(1.2×2.584)×1.5×6=27.9072kN半垮节点可变荷载F2=(1.4×0.70+1.4×0.9×0.80)×1.5×6=17.892kN②由永久荷载效应控制的组合计算:取永久荷载γG =1.35,屋面活荷载γQ1=1.4、ψ1=0.7,屋面集灰荷载γQ2=1.4,ψ2=0.9全垮节点永久荷载F1=(1.35×2.55)×1.5×6=31.347 kN 半垮节点可变荷载F2=(1.4×0.7×0.70+1.4×0.9×0.80)×1.5×6=31.347kN故应取P=46.23kN3) 全跨屋架和支撑自重、(左)半跨屋面板荷载、(左)半跨活荷载+集灰荷载①由可变荷载效应控制的组合计算:取永久荷载γG =1.2,屋面活荷载γQ1=1.4,屋面集灰荷载γQ2=1.4,ψ2=0.9全跨屋架和支撑自重F3=1.2×0.384×1.5×6=4.4172Kn半跨屋面板自重及半跨活荷载F4=(1.2×1.4+0.7×1.4)×1.5×6=23.94kN②由永久荷载效应控制的组合计算:取永久荷载γG =1.35,屋面活荷载γQ1=1.4、ψ1=0.7,屋面集灰荷载γQ2=1.4,ψ2=0.9全跨屋架和支撑自重F3=1.35×0.384×1.5×6=4.617kN半跨屋面板自重及半跨活荷载F4=(1.35×1.4+1.4×0.7×0.7)×1.5×6=23.184kN5﹑屋架杆件力计算见附表16﹑选择杆件截面按腹杆最大力Nab=-413.28KN,查表7.6,选中间节点板厚度为8mm,支座节点板厚度10mm。
1)上弦杆整个上弦杆采用等截面,按最大力NFH=-711.01KN计算在屋架平面:为节间轴线长度,即 lox=l。
=150.8cm,l oy=301.6cmϕ=0.659 Areq=N/ϕf=711010/(0.659×310×100)=34.803cm2假定λ=70 ixreq =lox/λ=150.8/70=2.54cmiyreq =loy/λ=301.6/70=5.08cm由附表9选2 125×80×10,短肢相并,如下图所示,A =2×19.71=39.42 cm 2 、i x =2.26cm 、i y =6.04cm 截面验算:x λ=l ox / i x =150.8/2.26=66.7<[λ]=150(满足) y λ=l oy / i y =301.6/6.04=49.9<[λ]=150(满足)双角钢T 形截面绕对称轴(y 轴)应按弯扭屈曲计算长细比yz λ b 1/t=12.5/1.0=12.5>0.56 l oy / b 1=0.56×301.6/12.5=13.51,则:yz λ=3.7 *b 1/t (1+ l 2oy 2t /52.7b 12)=3.7×14(1+301.62×0.82/52.7×12.54)=54.1〈x λ故由m ax λ=x λ=66.7查附表1.2得ϕ=0.686由N/ϕA=711.01×1000/(0.686×39.42×100)=262.93N/ 2mm <f(满足)故上弦杆采用2 125×80×10截面形式2) 下弦杆整个下弦杆采用等截面,按最大力N gi =706.82KN 计算l ox =300cm l oy =2370/2=1185cm, A nreq =N/f=706.82×1000/(310×100)=22.80 cm 2由附表9选2 125×80×7短肢相并A =2×14.1=28.2 cm 2,i x =2.3cm ,i y =5.97cm n A /N =σ=706.82×103/(28.2×102)= 250.65N/ 2mm < f(满足) x λ=l ox / i x =300/2.3=130.43<[λ]=350(满足)y λ=l oy / i y =1185/5.97=198.49<[λ]=350(满足)故下弦杆采用2 125×80×7短肢相并3)腹杆①、aB 杆N aB =-413.28KN ,l oy = l ox =l=253.5cm选用2 110×70×8长肢相并查附表9得:A =2×13.94=27.88 cm 2, i x =3.51cm ,i y =2.85cm x λ=l ox / i x =253.5/3.51=72.2<[λ]=150(满足) y λ=l oy / i y =253.5/2.85=88.9<[λ]=150(满足) 因为b 2/t=7/0.8=8.75<0.48 l oy / b 2=0.48*253.5/7=17.38按式yz λ=y λ[1+(1.09 b 42)/(l 2oy 2t )]=88.9×[1+1.09×74/(253.52×0.82) ]=94 yz λ>x λ,由yz λ查附表1.2得x ϕ=0.472则=A ϕ/N 413.28×1000/(0.472*27.88*100)= 314.058N/ 2mm < f(满足)故Ab 杆采用2 110×70×8长肢相并②、Bc 杆N c B =320.56KN, l ox =0.8L =0.8*260.8=208.6cm, l oy =L=260.8cm选用2 75×8A =2×11.5=23 cm 2,i x =2.28cm ,i y =3.42cm x λ=l ox / i x =208.6/2.28=91.51<[λ]=350(满足) y λ=l oy / i y =260.8/3.42=76.26<[λ]=350(满足) n A /N =σ=320.56×1000/(23×100)= 139.37 N/ 2mm <f(满足)③、cD 杆N c B =-253.47KN,l ox =0.8L =0.8×285.9=228.72cm, l oy =L=285.9cm选用2 75×8A =2×11.5=23 cm 2,i x =2.28cm ,i y =3.42cm x λ=l ox / i x =228.72/2.28=100.32<[λ]=150(满足)y λ=l oy / i y =285.9/3.42=83.60<[λ]=150(满足) 因为b/t=75/8=9.375<0.58 l oy /b=0.58×285.9/7.5=22.11,则yz λ=y λ(1+0.475b 4/ l 2oy 2t )=83.6×(1+0.475×7.54/285.92×0.82)=86由于x λ> yz λ,则由x λ查附表1.2得ϕ=0.430 n A ϕσ/N ==253.47×1000/(0.411*23×100)= 268.14 N/ 2mm <f(满足)④、De 杆N e D =172.39KN, l ox =0.8L =0.8*285.9=228.72cm, l oy =L=285.9cm选用2 56*5A =2×5.42=10.84 cm 2,i x =1.72cm ,i y =2.62cm x λ=l ox / i x =228.7/1.72=132.98<[λ]=350(满足) y λ=l oy / i y =285.9/2.62=109.12<[λ]=350(满足) n A /N =σ=172.39×1000/(10.84×100)=159.03 N/ 2mm <f(满足)⑤、eF 杆-114.62N e F = 按压杆进行计算:5.1l ox =0.8L =0.8×312.9=250.3cm, l oy =L=312.9cm选用2 75×8A =2×11.5=23 cm 2,i x =2.28cm ,i y =3.42cm x λ=l ox / i x =250.3/2.28=109.79<[λ]=150(满足) y λ=l oy / i y =312.9/3.42=91.49<[λ]=150(满足) 因为b/t=75/8=9.375<0.58 l oy /b=0.58×285.9/7.5=22.11,则yz λ=y λ(1+0.475b 4/ l 2oy 2t )=91.49×(1+0.475×7.54/312.92×0.82)= 93.68 由于x λ> yz λ,则由x λ查附表1.2得ϕ=0.375 n A ϕσ/N ==114.62×103/(0.375×23×102)=132.89 N/ 2mm <f(满足) n A /N =σ=5.1×103/(23×102)=2.22 N/ 2mm <f(满足)⑥、Fg 杆+ 58.58KN 按压杆进行计算:N g F = l ox =0.8L =0.8×311.9=249.5cm, l oy =L=311.9cm-5.31KN选用2 56×5A =2×5.42=10.84 cm 2,i x =1.72cm ,i y =2.62cm x λ=l ox / i x =249.5/1.72=145.1<[λ]=150(满足) y λ=l oy / i y =311.9/2.62=119.05<[λ]=150(满足) 因为b/t=5.6/0.5=11.2<0.58 l oy /b=0.58×311.9/5.6=32.3,则 yz λ=y λ(1+0.475b 4/ l 2oy 2t )=119.05×(1+0.475×5.64/311.92×0.52)=121.34 由于由于x λ> yz λ,则由x λ查附表1.2得ϕ=0.235 n A /N =拉σ=58.58×1000/(10.84×100)=54.04N/ 2mm <f(满足) n A ϕσ/N ‘压==5.31×1000/(0.235*10.84×100)=20.84N/ 2mm <f(满足)⑦、gH 杆+36.03KN 按压杆进行计算:N gH = l ox =0.8L =0.8×339.6=271.7cm, l oy =L=339.6cm-35.38KN选用263×5A =2×6.14=12.28 cm 2,i x =1.94cm ,i y =2.89cmx λ=l ox / i x =271.7/1.94=140.1<[λ]=150(满足) y λ=l oy / i y =339.6/2.89=117.51<[λ]=150(满足)因为b/t=6.3/0.5=12.6<0.58 l oy /b=0.58×339.6/6.3=31.26,则yzλ=yλ(1+0.475b4/ l2oy2t )=117.51×(1+0.475×6.34/339.62×0.52)=120.56 由于由于x λ> yz λ,则由x λ查附表1.2得ϕ=0.250n A /N =拉σ=36.03×1000/(12.28×100)=29.34 N/ 2mm <f(满足)n A ϕσ/N ‘压==35.38×1000/(0.25×12.28×100)=115.24N/ 2mm <f(满足)由于N gH 上的力都不大,选用的263×5可以满足力的要求⑧、Hi 杆-71.51KN 按压杆进行计算:N g F = l ox =0.8L =0.8×337=269.6cm, l oy =L=337cm27.24KN选用263×5A =2×6.14=12.28 cm 2,i x =1.94cm ,i y =2.89cmx λ=l ox / i x =269.6/1.94=138.97<[λ]=150(满足) y λ=l oy / i y =337/2.89=116.61<[λ]=150(满足)因为b/t=6.3/0.5=12.6<0.58 l oy /b=0.58×337/6.3=31.03,则yz λ=y λ(1+0.475b 4/ l 2oy 2t )=116.61×(1+0.475×6.34/3372×0.52)=119.68 由于由于x λ> yz λ,则由x λ查附表1.2得ϕ=0.253n A /N =拉σ=27.24×1000/(12.28×100)=22.18N/ 2mm <f(满足)n A ϕσ/N ‘压==71.51×1000/(0.253*12.28×100)=243.04N/ 2mm <f(满足)4)竖杆 ①、Aa 杆NAa=-23.3KN ,l ox =l oy =L=200.5cm选用263×5A =2×6.14=12.28 cm 2,i x =1.94cm ,i y =2.89cmx λ=l ox / i x =200.5/1.94=103.35<[λ]=150(满足) y λ=l oy / i y =200.5/2.89=69.38<[λ]=150(满足)因为b/t=6.3/0.5=12.6<0.58 l oy /b=0.58×200.5/6.3=18.46,则yz λ=y λ(1+0.475b 4/ l 2oy 2t )=69.38×(1+0.475×6.34/200.52×0.52)=74.55 由于由于x λ> yz λ,则由x λ查附表1.2得ϕ=0.411,则:n A ϕσ/N ==23.3×103/(0.411×12.28×102)=46.17N/ 2mm <f(满足)②Cc.Ee.GgN Cc = N Ee = N Gg =-46.59KN力大小相同,选用长细比较大的竖杆Gg 进行设计,可保证其他竖杆的安全。