钢结构第四版戴国欣作业解答

1、3.1 N =1000kN （设计值）、Q235B、E43焊条，设计连接，手工焊。图3. 80 习题3. 1图nr解：（一）采用两面侧焊肢背焊逢：hf =1.2tmin = 12mm maxhfmin =15Jmax = 5.2 mm1hf = 10mmJ 1肢尖焊缝hfmax =t -(1 2) =10-(1 2) = 8 9mmihf =1.5 tmax = 5.2mm minhf2=6mm焊缝长度lw12N 32 0.7hf1f：2 1000 10332 0.7 10 160=297.6mmlw2_ w2 0.7hf2 ff131000 103 = 248.0mm2 0.7 6 160故

2、取 hf1 =10mm , 1W1 = 320mm；hf2 =6mm , lw2 = 260mm 0（二）采用三面围焊按构造要求，取端焊缝焊脚为hf3 =8mm ,其它不变。N3 =2 0.7h-tbffw =2 0.7 8 125 1.22 160 = 273.3kN2 12N1 二 NN3 =- 1000 -273.3/2 =530kN3 231 11N2N N31000 -273.3/2 =196.7kN3 23. lw =N1/(2 0.7hf ffw) =530 103/(2 0.7 10 160) = 236.6mmlw =N2/(2 0.7hf2ffw) =196.7 103/(

3、2 0.7 6 160) = 146.4mmlw3 =125mm 。故取 hf1 =10mm , lW1 = 250mm ； hf2 =6mm , lw2 = 155mm ; hf3 = 8mm ,所有焊缝满足8hf <Lw <60hf的要求。3.2Q2 3 5B E43焊条，手工焊，e = 300mm ,求最大设计荷载。图181 月题12图解：(1)按构造要求确定焊脚尺寸1012hfmax =10-(12) =8 9mmhf1.5 12 =5.2mmmin-hf =8mm he=5.6mme(2)焊缝计算裁面的几何性质_2 197 101.3x =2 197 511.2=44.1

4、mm ; Aw = 5.6 父(2 父197 +511.2) = 5069mm2Iw I x£(202.6 511.23 -197 5003)=203.4 106mm4Iw92=5.6 511.2 44.12 - (41.33 155.73) 19.9 106mm4 3w w64=I x Iy =223.3 10 mm(3)求最大设计荷载扭转：T =(300 205 -41.3)F =0.464F(kN危险点的应力分量：(A点)m)；剪力：V = F.A=*I pTxa(二 AIwI pV,二 a)1.50.464F 106 255.6223.3 1060.464F 106 155.

5、7223.3 1062-A2 =F= 0.531F ;= 0.328F ;VAw3F 1030.197F506320.5312 =0.682F < ffw =1601.5> F =234.5kN(4)驻算牛腿强度危险裁面：柱翼边缘 M =0.3F,V =F, MCJ =max w-60.3 234.5 1023 =168.8N/mm10 5003一 一一 2:二 f =215N/mmVma'f56234.5 1 032270.3N/mm: fv =125N/mm500 10故牛腿强度足够，富余合适。3.3 Q235B, E43焊条，手工焊，设计牛腿与柱的连接焊缝。5.6 1

6、2 5.6图3, 82 习题3. 3图解：(一)设计焊缝(1)焊缝截面的内力V =F =98kN , M =98x0.12 = 11.8kN m(2)按构造要求确定焊脚尺寸设柱翼板厚为20mm,而牛腿板厚12mmhfmax =1.2tmin =1.2 12 = 14.4mmhf = 8mmhfmin =15.tmax =15, 20 = 6.7mm(3)焊缝计算截面的几何性质21134 206.8 122 189.2 2 18642 y 二2 = 1360mm134 2-12 192 2 -5.6 2(与he关系不大)_ _ 2 _2_ 1_313Ix =5.且134 (56 12 2.8)1

7、28 (56 -2.8)2(一 (56 -5.6) 136 )33= 15.7M106mm4 (近似与he成正比)(4)验算焊缝强度偏于保守计算，认为腿板焊缝承受全部剪力，且剪应力均布，故危险点为二 maxMyA 11.8 106 136Ix15.7 106= 102.2MPa :二 'ffw =1.22 160 =195N/mm298 1032 192 5.6=45.6MPaTA102.22 : 1.545.62 =95.1MPa :二 ffw= 160MPa（满足，张度富裕显多）(二)设计翼缘焊缝(1)求截面几何性质150 6 200 112=66.6 mm150 20013213

8、3I =一 150 123 150 12 (66.6 -6)2 - 1254.63 (212 -66.6)3123= 19.6 106mm433s1 =150 12 (66.6 -6)=109.1 10 mmVs1=3 > q1 = 1tIt3398 10109.1 1019.6 106q12 0.7hfw-ff >h，=d= 545.5MPamm=2.44mm【解2】= 40.2MPahfN2 0.7f：lw65.9 1032 0.7 160 (120-8-5)=2.75mm而按构造要求hf 15 12= 5.2mm minhf =1.2 12=14.4mm . max故取hf

9、=6mm即可。611.8 10 66.619.6 10654.6二2 =；1 33MPa66.6二 1 二240.2 333N = A1 -2=150 12 =65.9 10 MPa22 0.7h"w3.4 三级对接焊缝，求 F解1设翼缘焊缝和腹板均采用引弧板施焊(1)求计算截面的几何性质_ 150 6 200 112y =二66.6mm150 20012200132133I =一 150 123 150 12 (66.6 -6)21254.63 (212 -66.6)3123= 19.6 10二2 3 2 =1.146F E1.1fcw =1.1 215 = 237 cF=206.

10、4kN (明显保守) 解2不采用引弧板mm433 150 12 (66.6 -6)=109.1 10 mm31233s =109.1 10- 12 54.6 =127 10 mm2(2)求许可FA点:c-maxMyA0.12F 106 145.319.6 106-0.890F < fcw c二 215M =0.12FF1 =241.6kNC点:二 maxVSIxtF 127 10619.6 106 12=0.540F< fcw = 125cF2 =231.5kND点:= 185二mLax =0.89F _666 =0.408F < ftw145.4F3 =451.2kNB 点

11、：仃=0.89F M546=0.334F145.4=0.540 F S =0.464F9，二2 3 2 =0.871F <1.1ftw =204F4 =234kN故取 F = 230kN注：假设腹板焊缝承受全部剪力，且剪应力均匀分布，则 A点为危险点: _3F 10 二二0.890F0.417F176 (18 88)126 176=61.8mm12 2002132= 126 12 61.812 17612 176 (106 -61.8)1264= 15.4 10 mm(注：I取小，s, y取大)s1 =126 12 61.8 =93.4 103mm3_31_2_33s =93.4 101

12、2 55.8 =112 10 mm2c 0.12F 106 144.2， ，max6A 点：15.4x10= 1.12F m fcw =215N/mm2 cF By =61.8 幺SfiS17612Fi =192kNC点：仃maxVSEtF 103 112 10315.4 106 12= 0.606F M fcw =125cF2 =206kND 点：仃mLax =1.12F M_678 =0.527F M ftw =185144.2F3 =351kNB 点：仃=0.527F x18 =0.325F67.8934= 0.606F 0.505F112二2 3 2 =0.933F <1.1ft

13、w =204F4 =219kN ,故取 F =192kN注：如假设肢板焊缝承受全部剪力，且剪应力均布，则3A 点：t=1.12F,t=0.474F , Q + 3三2 = 1.389F 三 1.1fcw = 23712 176F <171kN用引弧板不用引弧板精确230192近似2061713.5 M =1122kN m, V=374kN,Q235 三级焊缝，验算:解设采用引弧板(1)截面几何性质13394I =一(280 1028 -272 1000 )=2.68 10 mm 12s1 =280 14 507 =1.99 106mm3s 500 8 250 =2.99 106mm33.

14、83 词题3.5图(2)验算A点:(J maxMymax1122 106 514二9I 2.68 10= 215.2MPaf =215MPaB点:500二二215.2 =209.3MPa514It1.99 106 374 103, 一 9 二2.68 108= 34.7MPaJ。2 +3/ = V209.32 +334.72 = 217.8 > 1.1 ftw = 203.5且超过 5%C点:sV 2.99 106 374 103w. max =9 =52.2MPa :二 fv =125MPaIt 2.68 108图3. 81 习题3. 2图3.6 F =100kN, ei =30cm。

15、试设计粗制螺栓连接。解：取 M20 粗制螺栓，d=20mm, d0 =22mm1 .按构造要求布置螺栓端距 2do =44mm边距 1.5d0 =33mm中距 3d0 =65mm用10个螺栓，按图示布置。2 .单个螺栓的承载力设计值 ,22b :二d b 3.14 203Nv =nvfv =1 140=43.94 1044Nb -dZtfcb -20 10 305=61 103N > N：in =43.94kN3 .验算螺栓抗剪强度危险螺栓1T =100 0.4025 = 40.25kN m, V =100kN22222三(x1 2 * yi2) =10 502 4 1002 4 200

16、252= 2.25 105 mm22 TTy140.5 106 200N1x =一 22- 二5三(Xi yi )2.25 103二36 103N=36kN50100100100y50505055螺栓1500NTyVN1yTx1三(xi2 yi2)n1 =9kN,4 1x100Vl1一 二10.22kN,=1.1=0.979n150do50205取 M =100 0.355 =35.5kN m_ My maxmax 一I nxN1 = . (N；)1nx 10 5003 -2 21.5 10 (1002 2002) (N： NVJ2 =、. 362 1 9.222 = 40.8kNNmin =

17、43.94kN4 .验算牛腿板强度SV 2.25 103 100 103T max7Inxt8.27 10 10= 27.2N/mm2 :二 fv =125N/mm2（满足）3.7设计角钢与连接板的螺栓连接竖向连接板与柱翼板的螺栓连接解：（一）角钢与连接板的螺栓连接取 M20 粗制螺栓，d =20mm , d0 = 22mm单个螺栓的抗剪承载力设计值,22b二 db 二 203Nv 2 fv =2 140 =88 10 N44N： =d，tfcb =2 14 305 =85.4 103NNtin =85.4kN所需螺栓数n=N=竺0=4.1 （个）Nmn 85.4取n = 5构造要求：线距 e

18、=55mm,最大孔径22mm中距66mm,故螺栓布置如图示k/d。=320/22 =14.6 ：二15 上述计算合理。（二）竖向连接板与柱翼的螺栓连接 有承托一螺栓受拉 取M20 （统一）Ntb = Aeftb =245 170 =41.7kNNxbNt175 2仃=5.94 ,取 n=641.73.8摩擦型连接，无承托设计解：（一）角钢与连接板的连接表面喷砂，口 =0.45取 M20, 10.9 级螺栓，P=155kN3.84 习题上7图553.84 习题3d图N： =0.9nfP =0.9 2 0.45 155 = 125kN350n = ； = 2.79 个N；125取n=3,布置间距与

19、上题相同。（二）竖板与柱翼的连接 di = d2 =170轴心力作用，各螺栓所受剪力和拉力均为NNt = Nv =2n承载力设计值(单个螺栓)N； =0.9nf(P -1.25Nt)Ntb =0.8P由Nv EN；有NN_0.9nf(P -1.25),2n2n1 1.125nfN- 0.9.2n” (P)由 Nt <Ntb有 NL- <0.8P t n2 >1-(N) .2n0.8,2 P取 M20, 10.9 级螺栓，P=155kN ,表面喷砂， 0 =0.45代入上式求得：n1 -5.99; n2 -2.0故取n=6,螺栓布置与3.7相同。 d1 = 150mm, d2

20、= 190mm螺栓及其布置与相同，表面工艺也不变只需验算竖板与柱翼的摩擦型高强度螺栓的强度螺栓群截面承受拉力 Nx -N2最大受力螺栓为最上排螺栓。NNiey1N . 22n三 y2丁-> 350 20 903502= I -24 90二 55kN :二 Ntb = 0.8P二124kNN2N2n=41.3kNN31c c cc一 350 20 90-2 =41.3-13.8-27.5kN4 902V = N =247.5kN <2三Nb =0.9% (np 1.25三Nti)=0.9 m m 1 m0.45m6m 155 1.25m 2(55 + 41.3 + 27.5)V <!Nb故安全。= 251.3kN喷砂处理，确定分肢上的