高等教育自学考试钢结构课程设计准考证号:姓名: 冯桀铭1、设计资料1)某厂房跨度为24m,总长90m,屋架间距6m,2)屋架铰支于钢筋混凝土柱顶,上柱截面400×400,混凝土强度等级为C30。
3)屋面采用×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)该车间所属地区为郑州市5)采用梯形钢屋架考虑静载:①1.5m*6m预应力钢筋混凝土大型屋面板(m2)、②二毡三油加绿豆沙、③20mm厚水泥砂浆找平层(m)④支撑重量考虑活载:活载(雪荷载)积灰荷载6)钢材选用Q345B级钢,焊条为E43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:起拱53、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 =,屋面活荷载γQ1=,屋面集灰荷载γQ2=,ψ2=,则节点荷载设计值为F=(×+×+××)××6=②由永久荷载效应控制的组合计算:取永久荷载γG =,屋面活荷载γQ1=、ψ1=,屋面集灰荷载γQ2=,ψ2=,则节点荷载设计值为F=(×+××+××)××60= 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 =,屋面活荷载γQ1=,屋面集灰荷载γQ2=,ψ2=全垮节点永久荷载F1=(×)××6=半垮节点可变荷载F2=(×+××)××6=②由永久荷载效应控制的组合计算:取永久荷载γG =,屋面活荷载γQ1=、ψ1=,屋面集灰荷载γQ2=,ψ2=全垮节点永久荷载F1=(×)××6= kN半垮节点可变荷载F2=(××+××)××6=故应取P =3) 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 =,屋面活荷载γQ1=,屋面集灰荷载γQ2=,ψ2= 全跨屋架和支撑自重 F3=×××6=半跨屋面板自重及半跨活荷载 F4=(×+×)××6=② 由永久荷载效应控制的组合计算:取永久荷载γG =,屋面活荷载γQ1=、 ψ1=,屋面集灰荷载γQ2=,ψ2= 全跨屋架和支撑自重 F3=×××6=半跨屋面板自重及半跨活荷载 F4=(×+××)××6= 5﹑屋架杆件内力计算 见附表16﹑选择杆件截面按腹杆最大内力N ab =,查表,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,按最大内力N FH =计算 在屋架平面内:为节间轴线长度,即 l ox =l 。
=, l oy =301.6cmϕ=0.659 A req =N/ϕf =711010/×310×100)=34.803cm 2 假定λ=70i xreq =l ox /λ=70=2.54cmi yreq =l oy /λ=70=5.08cm由附表9选2125×80×10,短肢相并,如下图所示,A =2×=39.42 cm 2 、i x =2.26cm 、i y =6.04cm 截面验算:x λ=l ox / i x ==<[λ]=150(满足)y λ=l oy / i y ==<[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==>0.56 l oy / b 1=×=,则:yz λ= *b 1/t (1+ l 2oy 2t / b 12)=×14(1+2×2×4)=〈x λ故由m ax λ=x λ=查附表得ϕ=由N/ϕA =×1000/××100)= 2mm <f(满足)故上弦杆采用2125×80×10截面形式2)下弦杆整个下弦杆采用等截面,按最大内力N gi =计算l ox =300cm l oy =2370/2=1185cm, A nreq =N/f=×1000/(310×100)=22.80 cm 2由附表9选2125×80×7短肢相并A =2×=28.2 cm 2,i x =2.3cm ,i y =5.97cmn A /N =σ=×103/×102)= 2mm < f(满足) x λ=l ox / i x =300/=<[λ]=350(满足) y λ=l oy / i y =1185/=<[λ]=350(满足)故下弦杆采用2125×80×7短肢相并3)腹杆 ①、aB 杆N aB =-,l oy = l ox =l=253.5cm选用2110×70×8长肢相并查附表9得:A =2×=27.88 cm 2, i x =3.51cm ,i y =2.85cmx λ=l ox / i x ==<[λ]=150(满足) y λ=l oy / i y ==<[λ]=150(满足) 因为b 2/t=7/=<0.48 l oy / b 2=*7=按式yz λ=y λ[1+( b 42)/(l 2oy2t )] =×[1+×74/2×2 ]=94yz λ>x λ,由yz λ查附表得x ϕ=则=A ϕ/N ×1000/**100)= 2mm < f(满足)故Ab 杆采用2110×70×8长肢相并②、Bc 杆N c B =, l ox =0.8L =*=208.6cm, l oy =L=260.8cm选用275×8A =2×=23 cm 2,i x =2.28cm ,i y =3.42cmx λ=l ox / i x ==<[λ]=350(满足) y λ=l oy / i y ==<[λ]=350(满足)n A /N =σ=×1000/(23×100)= N/ 2mm <f(满足)③、cD 杆N c B =-,l ox =0.8L =×=228.72cm, l oy =L=285.9cm选用275×8A =2×=23 cm 2,i x =2.28cm ,i y =3.42cmx λ=l ox / i x ==<[λ]=150(满足)y λ=l oy / i y ==<[λ]=150(满足) 因为b/t=75/8=<0.58 l oy /b=×=,则yz λ=y λ(1+4 l 2oy 2t )=×(1+×42×2=86由于x λ> yz λ,则由x λ查附表得ϕ=n A ϕσ/N ==×1000/*23×100)= N/ 2mm <f(满足)④、De 杆N e D =, l ox =0.8L =*=228.72cm, l oy =L=285.9cm选用256*5A =2×=10.84 cm 2,i x =1.72cm ,i y =2.62cmx λ=l ox / i x ==<[λ]=350(满足) y λ=l oy / i y ==<[λ]=350(满足)n A /N =σ=×1000/×100)= N/ 2mm <f(满足)⑤、eF 杆N e F = 按压杆进行计算:l ox =0.8L =×=250.3cm, l oy =L=312.9cm选用275×8A =2×=23 cm 2,i x =2.28cm ,i y =3.42cmx λ=l ox / i x ==<[λ]=150(满足) y λ=l oy / i y ==<[λ]=150(满足) 因为b/t=75/8=<0.58 l oy /b=×=,则yz λ=y λ(1+4 l 2oy 2t )=×(1+×42×2=由于x λ> yz λ,则由x λ查附表得ϕ=n A ϕσ/N ==×103/×23×102)= N/ 2mm <f(满足)n A /N =σ=×103/(23×102)= N/ 2mm <f(满足)⑥、Fg 杆+ 按压杆进行计算:N g F = l ox =0.8L =×=249.5cm, l oy =L=311.9cm选用256×5A =2×=10.84 cm 2,i x =1.72cm ,i y =2.62cmx λ=l ox / i x ==<[λ]=150(满足) y λ=l oy / i y ==<[λ]=150(满足) 因为b/t==< l oy /b=×=,则yz λ=y λ(1+4 l 2oy 2t )=×(1+×42×2=由于由于x λ> yz λ,则由x λ查附表得ϕ=n A /N =拉σ=×1000/×100)= 2mm <f(满足)n A ϕσ/N ‘压==×1000/*×100)= 2mm <f(满足)⑦、gH 杆+ 按压杆进行计算:N gH = l ox =0.8L =×=271.7cm, l oy =L=339.6cm选用263×5A =2×=12.28 cm 2,i x =1.94cm ,i y =2.89cmx λ=l ox / i x ==<[λ]=150(满足) y λ=l oy / i y ==<[λ]=150(满足) 因为b/t==<0.58 l oy /b=×=,则yz λ=y λ(1+4 l 2oy 2t )=×(1+×42×2=由于由于x λ> yz λ,则由x λ查附表得ϕ=n A /N =拉σ=×1000/×100)= N/ 2mm <f(满足)n A ϕσ/N ‘压==×1000/××100)= 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 杆按压杆进行计算:N g F = l ox =0.8L =×337=269.6cm, l oy =L=337cm选用263×5A =2×=12.28 cm 2,i x =1.94cm ,i y =2.89cmx λ=l ox / i x ==<[λ]=150(满足) y λ=l oy / i y =337/=<[λ]=150(满足) 因为b/t==< l oy /b=×337/=,则yz λ=y λ(1+4 l 2oy 2t )=×(1+×43372×2=由于由于x λ> yz λ,则由x λ查附表得ϕ=n A /N =拉σ=×1000/×100)= 2mm <f(满足)n A ϕσ/N ‘压==×1000/*×100)= 2mm <f(满足)4)竖杆 ①、Aa 杆NAa= ,l ox =l oy =L=200.5cm选用263×5A =2×=12.28 cm 2,i x =1.94cm ,i y =2.89cmx λ=l ox / i x ==<[λ]=150(满足) y λ=l oy / i y ==<[λ]=150(满足) 因为b/t==<0.58 l oy /b=×=,则yz λ=y λ(1+4 l 2oy 2t )=×(1+×42×2=由于由于x λ> yz λ,则由x λ查附表得ϕ=,则:n A ϕσ/N ==×103/××102)= 2mm <f(满足)②Cc N Ee = N Gg =内力大小相同,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。