钢结构作业答案

1、3.13如图所示梁与柱（钢材为Q235B）的连接中,M=100kN , m,V=600kN,已知梁端板和柱翼缘厚均为14mm,支托厚20mm,试完成下列设计和验算：（1）剪力V由支托焊缝承受，焊条采用解：(1)验算受剪承载力：a.确定焊脚高度=4 rJ亘 f -E43型，手工焊，求焊缝 A的高度hfo （2）弯矩M由普通C级螺栓承受，螺栓 直径24mm,验算螺栓是否满足要求。板件边缘的角焊缝：t = 20mm, hfmax = 20 -(1: 2) = 18mm;普通角焊缝：tmin =min(14,20) =14mnihfmax =1.2 14 = 16.8mm;则 hf max =min(

2、16.8,20) = 16.8mm;tmax = max(14,20) = 20mm, hfmin = 6mm;故取 hf =10mmb.he =0.7hf =0.7 10 = 7mm设承受静载，支托选用Q235B,则查表得角焊缝强度设计值ffw = 160N/mm 2采用三面围焊且为绕角焊：对于水平焊缝，承受正应力，路=b = 300mmNi =0.7hflw1 f ffw =0.7 10 300 1.22 160 104=410kN;对于竖向焊缝，承受剪应力，口2 250mmN2 = 1.25V -N1 =1.25 600-410 = 340kN;3if =N2=340M10=97.143

3、N/mm < fT = 160N/mm，满足要2 0.7hflw22 0.7 10 250故焊缝a的局度hf - 10mm0(2)验算抗弯承载力yi =、2 =600mm;y3 = y4 500mm; ys = y6 = 300mm;W = y8 = 200mm; y9 = yi0 = 100mm;M - N % y2;yiN _ Myi _100 103 6001 一 % v； 2 (6002 5002 3002 2002 1002)二 40kN;查表得普通c级螺栓抗拉强度设计值 ftb =170N/mm2 ；公称直径为24mm的普通螺栓 Ae =353mnfNtb =Ae ftb =

4、353 170 10- =60.1kNN1 < Ntb故抗弯承载力满足要求。(3)验算构造要求：C级螺栓金=d (1: 1.5) =24 1.5 = 25.5mm3d。=3 25.5 = 76.5 二 100mm故满足构造要求综合(2),(3)该螺栓满足要求。3.14试验算如图所示拉力螺栓连接的强度。C级螺栓M20,所用钢材Q235B。若改用M20的8.8级高强度螺栓摩擦型连接(摩擦面间仅用钢丝刷清理浮锈)，其承载力有何差别？解:V =H =150 cos45o=150 立=106kN;2V 106H 106Nv = = = 26.5kN; Nt26.5kN;4444(1) 验算拉剪作用

5、查表得普通 C 级螺栓 ftb = 170N /mm2 ；fvb = 140N/ mm2 ； Q235B 钢fcb =305N/mm2 ；公称直径为20mm的普通螺栓Ae =245mM。(2)bN = nvv二d2Ntb =AeftbNvNv_ fvb =1 三 202 140 10，=44.0kN 4= 245 170 103 =41.65kNNtNtb26.5=0.876<144.0b41.65 )Nv=26.5；N =d、t fc =20 16 305 10 = 97.6kN c故，验算合格。改用M20级的8.8级高强度螺栓摩擦型连接，查表得预拉力 P = 125kN ,抗滑移系0

6、 =0.30，且单剪nf =1。取标准孔k=1.0。N： =0.9knf JP =0.9 1.0 1 0.30 125 = 33.75kNNtb =0.8P =0.8 125=100kNNv Nt26.5 26.5 b b = =1.05 1.0NJ Ntb 33.75 100未超过5%,故验算合格。设其承载力为F , Nv = Nt%五F啦FF =1 ，428Nv Nt+Nvb NtbeF 833.751008 33.7 +)=1.0,则F =142.7kN ,承载 100力下降。3.15如图所示螺栓连接采用Q235B钢。C级螺栓直径d=20mm,求此连接最大能承受的t1120mm 的 C

7、级螺栓 fvb=140N解：螺栓群轴心受剪，查表得公称直径为I旭，加/W *总停评1圈(1)受剪承载力Fmax 值。/mm2双剪 Q =2； NV5.2 :. d b nv fv二2 = 202 140 10”=88.0kN4（2 ）承压承载力：查表得 Q235钢，C级螺栓fcb=305N/mm2Nb =d 工 t，f； =20x20x305x10=122kN ；cNmin =min Nvb,Ncb = min 88.0,122 =88.0kNN1 =Nmin n =88.0 13=1144kN（3） 净截面强度计算取不利情况，对 C级螺栓 d0 =d+（1: 1.5）=20+1.5 = 21

8、.5mm；Q235钢极限抗拉强度fu = 370 N /mm21 N仃=_N,若取薄板仃=2N2L ；若取厚板仃= N = NL ；故取厚板计Anln 12 24lnln 20 201n算。An1 = 40 60 60 60 60 40 -3d0 20 = 320 -3 21.5 20 = 5110mm2An2 = 40 60 v2 60 2 60 60 40 -4d0 20 = 370 -4 21.5 20 = 5680mm22An3 =40 60 2 60 2 60 2 60 2 40-5dc 20 =419 -5 21.5 20=6230mm 故取 An = An1 =5110mnf。-

9、 N_ <0.7fu,N2 < 0.7fuAn -0.7 370 5110 10” =1323kN An（4） 毛截面强度计算查表得 Q235 钢在 t = 20mm时 f =205N/mm2； =N Mf,N3=A f =320 20 205 104 =1312kN AN1 <N3<N2,故取 Fmax =N1 =1144kN。综上，此连接能承受的最大承载力为Fmax = 1144kN。3.16如上题中将C级螺栓改用M20 （d=20mm）的10.9级高强度螺栓。求此连接最大能 承受的Fmax值。要求按摩擦型连接和承压型连接分别计算（钢板表面仅用钢丝清理浮锈）。解：（

10、1）摩擦型连接：查表得预拉力P=155kN，抗滑移系数0 =0.30,且双剪nf=2。取标准孔k = 1.0。bNv =0.9knfP =0.9 1.0 2 0.30 155 = 83.7kNN1 = N： n =83.7 13=1088kN净截面强度：c = 1-0.5n2 N_ <0.7fu,N2 < 0.7fuAn 1 0.5=0.7 370 5110 10， 1-0.5 _3 =1171kN n Ann .13vcNmin =min Nvb, Ncb min 151.9,188l：-151.9kNN1 =Nmin n =151.9 13=1975kNc =_ _0.7fu,

11、N2 £0.7fuAn =0.7 370 5110 10，=1323kN毛截面强度：N3=A f =320 20 205 10 '=1312kNN1 <N2 <N3,故取 Fmax = N1=1088kN(3) 承压型连接：查表得fcb =470N mm2； fvb=310N mm2；且双剪 兀=2;公称直径为20mm的普通螺栓Ae =245mm2。设剪切面在螺纹处：Nb =nv Ae f： =2 245 310 104 =151.9kNNb =d、t fcb =20 20 470 104 =188kNbb净截面强度：一An毛截面强度：N3=A f =320 20

12、 205 10 =1312kNN3<N2<N1，故取 Fmax = N3=1312kN3.17试验算如图所示钢梁的拼接。拼接采用高强度螺栓型连接。钢材为 Q345钢，采用10.9 级M20高强度螺栓，孔径为22mm ,接触面喷砂后涂富锌漆。拼接处梁的内力为V =650kN,M =2500kN m （翼缘拼接可按轴心力计算，轴心力取翼缘截面面积和翼缘内平均正应力的乘积，腹板拼接承受全部的剪力和按截面惯性矩比例分配给腹板的弯矩）。解：查表彳#10.9级M20高强度螺栓预拉力 P=155kN，抗滑移系数R =0.40，且双剪nf = 2。取标准孔k =1.0。b1311212Nmin =

13、Nvb=0.9knL P =0.9 1.0 2 0.40 155 = 111.6kNI - X450 1300 2 16 -2220 13003 =8.066 109mm4(1) 翼缘板验算二二叽I22500 106 (98.066 10=203.94 N/mmJ3N -； A =203.94 450 16 10 =1468.36&N n=16;N =1468.368 ：n N扁=16 111.6= 1785.6kN净截面强度：查表得 Q345钢fu =470N /mm2,f =310N初m2取最外排螺栓：c = 1-0.5- -= 1-0.5 41468.368 10 =221.82

14、7 : 0.7fu =0.7 470=329N/mm2n An16450-4 22 161n,-2c = = 203.94 二 f = 310N /mm毛截面强度:3N 1468.368 10A 450 16故翼缘验算合格。（2） 腹板验算剪扭组合:剪力偏心:11e = 2 1 0 60 - 80 =105mm;V =650kN弯矩分配:131_3_94Iw =_!_ b h =_!_ 10 1300 =1.83 10 mm12129Tw =4,10 2500 = 567.2kN mI 8.066 10扭矩：Tmax =V e Tw =650 105 10与 567.20 = 635.45kN

15、 m设左边为Tmax所在的板，则危险点为1、2点。1点：x1 = 40mm,y1 = 600mm“ xi2 y2 =26 402 4 1002 2002 3002 4002 5002 6002 =3681600mRNif =6502 13=25kN;NTy= Mt411N1T =Nit=103.56kN;x r1 % xyi3681600Txi635.45 402= = 6.90kN;v xyi3681600则,Nx = N1tx =103.56kN;Ny = N1Ty Mf =6.90 25 =31.90kN;N =v Nx2 Ny2 =103.56 31.902 =108.36 ； Nmb

16、n =111.6kN故腹板验算合格。3.18双角钢拉杆与柱的连接如图。拉力N =550kN。钢材为Q235B钢。角钢与节点板、节点板与端板采用焊缝连接，焊条采用E43型焊条。端板与柱采用双排 10.9级M20高强度螺栓连接。构件表面采用喷砂后涂无机富锌漆处理。试求：（1）角钢与节点板连接的焊缝长度；（2）节点板与端板的焊缝高度；（3）验算高强度螺栓连接（分别按摩擦型和承压型连接考虑）解：（1）角钢长肢相并：K1 =0.65；K2 =0.35；双面焊，查表得ffw =160N /mm2肢背：N1 -K1N -0.65 1 550=178.75kN,1 .肢尖：N2 =K2N =0.35 550

17、= 96.25kN 2焊脚局度hf =6mm ， lwmax = 60hf = 360mmlw1肢背：3Ni 178.75 100.7hf ffw0.7 6 160=266mm;1i =1wi 2hf =266 2 6 =278mnU = 280mm肢尖：lw23N2 96.25 100.7hf ffw = 0.7 6 160 =. mm，l2=lw2 2hf =143.3 2 6 = 155.3mm, 12 = 160mm故取肢背焊缝长度为 280mm，肢尖焊缝长度为160mm。 V=H =550xcos45o=550x 迎=389kN;2普通角焊缝：tmin =min(14,20) =14

18、mnihfmax =1.2 14 = 16.8mm;tmax = max(14,20) = 20mm, hfmin = 6mm;f E ff ，设 f =1.22=123.74N/mm2 ；H . f . V.C f = ; . f - = C f ;2 0.7hflw2 0.7hflw3H 389 102hflw =二2245.4mm1.4c f 1.4 123.74lw =l -2hf =440-2hf2hf -220hf 1122.75<05.23mm - hf - 215mm故取 hf = 10mm °(3) 摩擦型连接:查表得预拉力P =155kN ,抗滑移系数R =0.35，且单剪nf=1。取标准孔 k =1.0。Nb =0.9knfP =0.9 1.0 1 0.35 155 = 48.825kNNtb =0.8P =0.8 155 = 124kNV 389Nv = - = =38.9kN ;n 2 5承受弯矩：M =V e = 389x 20M10=7.78kN m。=58.35kN3897.78 0.16_n % yi2T5 2 0.082 2 0.162Nv Nt+Nvb Ntb38.948.82558.35 +=1.28>