钢结构设计基本原理课后答案

1、word资料合肥工业大学生版社由版(肖亚明主编)第三章1.解：Q235 钢、ffw =160N/mm2、N = 600kN(1)采用侧面角焊缝最小焊脚尺寸：hf _15. tmax =1.514= 5.6mm角钢肢背处最大焊脚尺寸：hf 1.2tmin =1.2 10 = 12mm角钢肢尖处最大焊脚尺寸：hf t -(12) =10-(12) =98mm角钢肢尖和肢背都取 hf =8mm查表 3-2 得：K1=0.65、七=0.35N1 =K1N = 0.65父600 = 390kN , N2 = K2N = 0.35父 600 = 210kN所需焊缝计算长度:l w1N12 0.7hf ff

2、w390 1032 0.7 8 160=217.63mmw2N22 0.7hf ffw210 1032 0.7 8 160=117.19mm焊缝的实际长度为:1i =lw1 +2hf =217.63+28 = 233.63mm,取 240 mm。l2 =lw2 +2hf =117.19+2M8=133.19mm ,取 140 mm。(2)采用三面围焊缝，取hf =6mm正面角焊缝承担的内力为：N3 = 0.7hf lw3 ffw = 0.7 6 2 100 1.22 160 = 163.97kN侧面角焊缝承担的内力为:N1 =K1N - N3/2 = 0.65 600 - 163.97/2 =

3、 308.01kNN2 = K2N - N3/2 =0.35 600 -163.97/2 = 128.02kN所需焊缝计算长度:w1N12 0.7hfffw308.01 1032 0.7 6 160=229.17mmw2N22 0.7hf ffw128.02 1032 0.7 6 160=95.25mm焊缝的实际长度为：11 =lw1 +hf = 229.17 + 6 = 235.17mm,取 240mm。I2 =1w2 +hf = 95.25+ 6 = 101.25mm,取 110mm。2.解：Q235 钢、ffw=160N/mm2(1)取 d1 =d2 = 170mm角焊缝受到轴力和剪力的

4、共同作用，且均作用在角焊缝形心处1.5角焊缝受到轴力为：N =600 - 499.23kN,1.52 121角焊缝受到男力为：V= 600： _2_2-= 332.82kN,1.511w =2 170 -2hf =340 -2hfCJN2 0.7hf1w3499.23 10. ： w- f f f1.4hf(340 -2hf)= 1.22 160 = 195.2N/mm2由上式求得：hf - 5.55mmV2 0.7hf1w3332.82 101.4hf (340 -2hf)-ffw = 160N /mm2由上式求得：hf - 4.49mm最小焊脚尺寸 hf 一 1.5 Jtmax = 1.5

5、 - 20 = 6.71mm故取焊脚尺寸为hf =7mmN2 0.7hflw3499.23 101.4 7(340 -2 7)= 156.26N / mm* 1 2 ： l：-f f fw =195.2N / mm2V2 0.7hflw3332.82 101.4 7(340 -2 7)104.18 ffw-2= 160N/mm一 一一 3 c 5.6 3243 二2 12-64= 31.74 106 mm4Ww2Iw 2 31.74 106324324= 195.96 103 mm3V2 0.7hflw3332.82 102 0.7 8 324= 91.72 ffw=160N /mm2最大正应

6、力为：NM499.23 1039.9846 106L 二二72 0.7hflw Ww 2 0.7 8 324 195.96 103二f / ： 2.2 = ., 188.53/1.22 2 - 91.722 =179.70N/mm2 . ffw =160N /mm2不满足要求。3.解：Q235 钢、f：=160N/mm2(1)角焊缝最小焊脚尺寸：hf _1.5.,tmax =1.512 = 5.2mm最大焊脚尺寸：hf 1.2tmin =1.2 12= 14.4mm取 hf = 6mm角焊缝受到轴力和弯矩的共同作用角焊缝受到轴力为：N = F =100kN角焊缝受到弯矩为：M = Ne =10

7、0 20 = 2kN mhe =0.7父6 =4.2mm , lw =200 2hf = 200 2 父6 = 188mm-364= 4.65 10 mm4.2 18812Ww2I w 2 4.65 106188 一 18833= 49.48 10 mm最大正应力为：NM 100 1032 106二 f 二二72 0.7hflw Ww2 0.7 6 188 49.48 103= 103.74N/mm2 二 1.22 160 = 195.2N / mm2满足要求(2)角焊缝角焊缝的有效截面如图所示M hf =8mm角焊缝有效截面形心位置：y = y2 =134 5.6 (192 12 2.8)

8、2 55.4 5.6 (192-2.8)2 192 5.6 192/2/(134 5.6 2 55.4 5.6 2 192 5.6) = 136.03mm剪力由腹板焊缝承担，腹板面积为：2Aw = 2 192 5.6 = 2150.4mm2全部焊缝对x轴的惯性矩为：Iw =134 5.6 (192 -136.03 12 2.8)2 2 55.4 5.6 (192-136.03-2.8)2 2 192 5.6 (136.03-192/2)2 2 5.6 1923/12 = 15.56 106mm4翼缘焊缝最外边缘的截面模量:Ww1-21.15 104 mm3Iw =15.56 106h1192

9、-136.03 12 5.6翼缘和腹板连接处的截面模量:Ww2Iw 15.56 106 h2 = 192 一 136.03= 27.81 104mm3腹板底边缘处的截面模量:Ww3y215.56 106136.03/ / / / 4 - 43= 11.44 10 mm弯矩：M =100 120 =12kN m由弯矩得最大应力为:，_ 2-104.90N / mm2_ M 12 10634Ww311.44 104腹板的剪应力为:一32= F/Aw =100 10 /2150.4 = 46.50N/mm翼缘和腹板连接处的折算应力；+ 72 = Jp04.90 +46.502 = 97.75N /m

10、m2 ffw = 160N /mm2 ,1.22满足要求。.解：Q235 钢、查附表 1-2 得 ftw =185N/mm2 , fvw = 125N/mm2 ,fcw=215N/mm2c焊缝有效截面形心位置:y=y2=126 12 (188 6) 188 12 188/2 /(126 12 188 12) = 134.13mm焊缝对x轴的惯性矩为：0ldiIx =126 12 (188 12-134.13-6)2 188 12 (134.13-188/2)2 12 1883/12 = 15.70 106 mm4翼缘焊缝边缘的截面模量:15.70 106y1188 12-134.13= 23.

11、83 104 mm3翼缘和腹板连接处的截面模量:W2 二Ix15,70 106188 _ y2188134.13= 29.14 104mm3腹板底边缘处的截面模量:IxW3x15.70 106V2134.13= 11.70 104mm3弯矩：M -120F由翼缘上边缘处焊缝拉应力120F23.83 104三 ftw = 185N / mm2F -185 23.83 104 /120 =367.38kN由腹板下边缘处焊缝压应力120F11.70 104 fcw = 215N/mm2 c得：F 215 11.70 104/120 =209.62kN由腹板焊缝单独承担剪力，腹板的剪应力k = F/A

12、w = F/（188 12） fvw -125N /mm2得：F 125 188 12 =282.00kN腹板下端点正应力、剪应力均较大，由腹板下端点的折算应力r22 卜 120FF_Ff w2V ac 2 +3t =J 4 1+3 I 1.1 ft =1.1父185 =203.5N/mm1 11.7010 )488M12，F =158.84kN。hf = 8mmyyx得：F 158.84kN综上得该连接所能承受的最大荷载为.解：Q345B 钢、f fw =200N/mm2,角焊缝有效截面的形心位置：x =x1 =(2 197 5.6 197/ 2)/(2 197 5.6 500 5.6) =

13、 43.41mm截面的惯性矩：Iwx =2 197 5.6 25025.6 5003/12 =196.23 106mm4Iwy =2 197 5.6 (197/2 -43.41)2 2 5.6 1973/125.6 500 43.412 -19.11 106mm4J =Iwx Iwy =(196.23 19.11) 103 =215.34 103 mm4扭矩：T =(300 205 -43.41)F =461.59F由扭矩产生的最大应力T ryJ461.59F 250215.34 106= 535.88 106F 三 ffw =200N/mm2得：F 200/535.88 106 =373.2

14、2kN461.59F (197-43.41)215.34 106= 329.23 10-6F_ fffw=1.22 200 =244N/mm2得：F 244/329.23 10 = 741.12kN由力F产生的应力二V = =F = 199.74 10-6 F - f ffw =244N /mm2Aw2 197 5.6 500 5.6得：F -244/199.74 10 上=1221.59kN由最大应力点的折算应力T V二 f .二 f/ T、2(f)329.23M10上 F+199.74M10/F 21.22(535.88 10F)2=689.32 10,FMf：= 200N/mm2得：F

15、-200/689.32 10=290.14kN综上得该连接所能承受的最大荷载为 F = 290.14kN 。6.解：Q235B 钢、f： =140N/mm2 ,b2fc = 305 N / mm单个螺栓的抗剪承载力设计值:bNv = nv二d2b 二 222fv =2 140 =106.44kN4Nb =dtfcb =22 14 305 = 93.94kNbN min=min( Nb,Nb) =93.94kN一侧所需的螺栓数目为：n = N/N：in =700/93.94 =7.45 (个)，取n=8螺栓的布置如图所示:-10” 八50 80 45 45 80 50净截面强度验算:螺栓群最外端

16、净截面强度:；N/An=700 103/(360 -4 23.5) 14 =187.97N / mm2 ：二 f =215N / mm2净截面强度满足要求。7.解：Q235 钢、f；=140N/mm2, fcb = 305N /mm2 , ftb = 170N/mm2单个螺栓的抗剪承载力设计值:bNv二 dnv一2- fvb =1二 222140 =53.22kNNo =d tfcb =22 18 305 = 120.78kNbb b、Nmin =min( Nv ,Nc) =53.22kN单个螺栓的抗拉承载力设计值:b d e bbNt =ft =Aeft =303 170 =51.51kN

17、4(1)牛腿下支托承受剪力时由受力最大的螺栓承受的拉力NiMyimT y：F 160 320_24 2_222(80160240320 )= 0.133F Ntb =51.51kN得：F 386.32kN故支托承受剪力时该连接所能承受得最大荷载设计值为F = 386 32kN(2)牛腿下支托不承受剪力时由受力最大的螺栓承受的拉力一 MyiF 160 320-L- bNt =2=2222=0.133F - Nt=51.51kNm y2(802160224023202)得：F 386.32kN由 N1V =F /n =F /10 =0.1F Nb =120.78kN由剪力和弯矩共同作用下U0.13

18、3F j + j0.1F ;5 5 51.51 ；153.22 J得：F 313.15kN得：F 1207.80kN= 0.00319F 三1综上得支托承受剪力时该连接所能承受得最大荷载设计值为F = 313.51kN8.解：Q345B 钢螺栓采用 M22, C 级，f；=140N/mm2, f；=385N/mm2螺栓的布置如图所示单个螺栓的抗剪承载力设计值：b 二d2 b 二 222Nv =nv- fv =1 140 =53.22kN44Nb =d tfcb =22 10 385 = 84.70kNNtin = min( Nb,Nb) =53.22kN螺栓群承受剪力及扭矩的共同作用扭矩：T

19、= Fe =120 (0.3 0.05 0.105/2) = 48.3kN m受力最大的螺栓承受的应力为:NiTxTyi 2 1 2、Xi 、 y48.3 106 20022210 52.52 4 (1002 2 002)= 42.45kNN1TyTx1_2 . _2% Xi 八 V48.3 106 52.510 52.52 4 (1002 2 002)= 11.14kNNVyF 120 103=12kN10.N1Tx 2N1TyN1Vy 2 = . 42.45 211.14 12 2 =48.35kN :二 Nmmin =53.22kN满足要求，故采用M229.解：高强螺栓8.8级，M20(

20、1)摩擦型连接预拉力P =125kN ,接触表面用喷丸处理，N =0.45单个螺栓的抗剪承载力设计值：NVb =0.9n”P = 0.9 2 0.45 125 =101.3kN一侧所需的螺栓数目为：门=4=j00=6.9个，取n = 7 nJ 101.3(2)承压型连接f： =250N /mm2 , fcb =470N/mm2螺栓的布置如图所示单个螺栓的抗剪承载力设计值： .2一 2b二db 二 20Nb -nvfvb=2 250 =157.0kN44Nb =d tf； =20 14 470 =131.6kNbb b、N min = min( Nv, N c) = 131.6kN一侧所需的螺栓

21、数目为：n=L=06=5.3个，取n = 6Nmin 131.610.解：高强螺栓10.9级，M22(1)摩擦型连接预拉力P =190kN ,接触表面用喷丸处理，N =0.50角钢肢与牛腿连接取螺栓的个数为n=5,栓距为80mm ,端距50mm单个螺栓的抗剪承载力设计值：N： =0.9nfP =0.9 2 0.50 190 =171.00kN此处连接承受剪力和扭矩的共同作用，扭矩T =350 x(180 -55) = 43.75kN -m则受力最大的螺栓承受的应力为：Nvy = N / n =350/5 = 70kNNtxTry、x： vy43.75 106 160_222 (80160 )=

22、109.38kNNtx 2Nvy 2 =109.38 270 2 =129.86kN :二 Nb =171.00kN满足要求角钢肢与柱翼缘连接取螺栓的个数为n=10,对称布置在牛腿两侧，栓距为 80mm ,端距50mm此处连接承受剪力和弯矩的共同作用，弯矩M =350黑180 = 63kN m受力最大的螺栓承受的拉力为：My163 106 160Ntf27=78.75kN ： 0.8P =0.8 190 = 152kNmx y2 4 (802 1 602)单个螺栓的抗剪承载力设计值：Nb =0.9nf J(P-1.25Nt) =0.9 1 0.50 (190-1.25 78.75) = 41.

23、20kNNv = N/n =350/10 = 35kN :二 N； =41.20kN才两足要求。(2)承压型连接b2fv =310N/mm ,b2 b2fc =590N/mm , ft =500N/mm角钢肢与牛腿连接取螺栓的个数为n=4,栓距为80mm ,端距50mm单个螺栓的抗剪承载力设计值:,b 二 222fvb =2 310 =235.56kN4No =d tfcb =22 20 590 = 259.6kNNmin = min( N；, Nb) =235.56kN此处连接承受剪力和扭矩的共同作用，扭矩T =350父(180 -55) = 43.75kN m则受力最大的螺栓承受的应力为:

24、Nvy = N /n =350/4 =87.5kNNxTry43.75 106 120、xi2 八 yi22 (402 1202)= 164.06kNNTx 2NVy 2 = 164.06 287.5 2 = 185.94kN :二 Nm = 235.56kN才两足要求。角钢肢与柱翼缘连接 取螺栓的个数为n=8,对称布置在牛腿两侧，栓距为 80mm ,端距50mmb二d2v =nv单个螺栓的抗剪承载力设计值:b 二 222fv =1 310=117.78kNN： =d tfcb =22 20 590 = 259.6kNNlin = min( Nb, Nb) =117.78kN此处连接承受剪力和

25、弯矩的共同作用，弯矩 M = 350父180 = 63kN，m受力最大的螺栓承受的拉力为:NtMyim y263 106 120ZT224 (402 1202)=82.03kN22(82.03、十(43.75 1 151.50 )117.78 ；Nv = N/n =350/8 =43.75kN :二 N；/1.2 = 216.33kNvco 0.66 二 1才两足要求。第四章三.计算题1.解：Q235截面I净截面面积：An =2 10 (45.1002 402 45 -2 21.5) = 3094.06mm2由应力 a = N / An f , 得：N An f = 3094.06 父 215

26、 =665.22kN截面n净截面面积：An =2 10 (45 100 45 -21.5) = 3370mm2由应力 a = N / An f 得：N Anf = 3370 父 215 = 724.55kN综上得此拉杆所能承受的最大拉力为：N = 66522kN查型钢表得：ix = 3.05cm , iymin = 4.38cm查表4-2得：容许长细比 瓦】=250由长细比九= l0/i,得：l0 MixU= 3.05X 250 = 762.5cm所以，此拉杆容许达到白最大计算长度为：l=762.5cm。2.解：Q235 钢，N =2400kN , l =6mAB柱为两端较接轴心受压柱截面几何

27、特性:A =2 320 16 400 8 = 13440mm2Ix8 4003122 320 16 2082 =485.69 106 mm416 3203Iy =2 12 64= 87.38 106 mm4=Ix/A= 485.69 1 06/13440 = 190.1mmIy/A= 87.38 106/13440 = 80.63mm长细比:l oxix6 103, i31.56 - .1-150190.1类）loy 6 103iy 80.63整体稳定：由长细比查表得稳定系数为:= 0.930 （b 类），甲y =0.723 （b2400 103xA= 186.10/2 mm:f = 215N

28、 / mm20.960 1344032400 10yA0.723 13440一 _ 一. 2=247.10/mm_ _ _ . . .2f = 215 N / mmy轴整体稳定不满足要求;由以上计算结果知，此柱在y轴的整体稳定承载力不满足要求，可在柱长的中点加侧向支撑来满足稳定承载力的要求。3.解：Q235 钢,l =10m(1)截面a截面几何特性:A =2 320 20 320 10 =16000mm2长细比:Ixix310 320264 2 320 20 170 = 397.23 10 mm12一 31y =2 -1r-= 109.23 106mm4= .,Ix/A = 397.23 1

29、06/16000 = 157.56mmiy=J厂/A，；109.23 106/16000 =82.62mm ylx 10 1031 1=63.46 二 I 1-150ix157.56loy 10 10382.62= 121.03 ： I I- 150由长细比查表得稳定系数为:中x= 0.788 （b 类），*y= 0.375 （c 类）由整体稳定得：N min Af = 0.375 16000 205 = 1230kN故截面a的柱所能承受的最大轴心压力设计值为:N =1230kN o(2)截面b截面几何特性:A =2 400 16 400 8 =16000mm23Ix2 400 16 2082

30、 =596.44 106 mm4123, c 16 40064L = 2 =170.67 10 mmy12ix = ., Ix/A = ,596.44 106 /16000 = 193.07mmiy = . I y / A =，J170.67 106 /16000 = 103.28mm长细比:=12518-150ix 1 930 71 oy310 10. i96.8 :二I - 150103.28由长细比查表得稳定系数为：中x =0.848 （b类），中y =0.476 （c类）得：N minAf =0.476 16000 215 = 1637.44kN故截面b的柱所能承受的最大轴心压力设计值

31、为：N = 1637.44kN4.解：Q235 钢，N=1000kN, l0 x =3m , 10y =6m取九x=80,九y=80查表得：Q =0.688 （b 类），%=0.688 （b 类）一 一一N 1000 1032所需截面面积：A = = 0000 = 6760.41mm2f 0.688 215 ix =lox/ x =3000/80 37.5mmiy = loy / y =6000/80 =75mm由各截面回转半径的近似值查附表9-1得：ix=0.28h, iy= 0.24b则h =37.5/0.28= 133.93mm , b = 75/0.24 = 312.5mm选取截面2

32、200 125 12截面验算：查型钢表得：A =2 x37.9 = 75.8cm2, ix = 3.57cm , iy = 9.54cml ox集5095.4查表得：邛x =0.661 （b类），%=0.792 （b 类）1000 103xA 0.661 758022= 199.58N/mm2 ：二 f =215N/mm2满足要求，故选用2/200父125父12。5.解：Q235 钢，N=1000kN,10 x = 6m , 10y = 3ml oy 6 10=62.9 凯二150（1）热轧工字钢取 h =80,-80查表得：x =0.783（a 类），中y= 0.688 （b 类）所需截面面

33、积：A =N 1000 103f 0.688 2152= 6760.41mm2ix =lox/x = 6000/80 = 75mm iy =loy / y = 3000/80 = 37.5mm由各截面回转半径的近似值查附表 9-1得:ix=0.39h, iy = 0.20bh =75/0.39 = 192.31mm, b = 37.5/0.20 = 187.5mm选取截面 45a （h=450mm,b=150mm ）截面验算：b/h=0.3330.8查型钢表得：A = 102cm2 ,ix = 17.7cm , iy=2.89cml ox6 103177= 33.9 :二 L 1 -1503

34、103,yoy28.9= 103.8 I - 150查表得：中x =0.955 （a 类），5y =0.530 （b 类）NA1000 1030.530 10200_ 2=184.98N/mm一 -2二 f =215N/mm满足要求，故选用45a(2) 3块剪切边的钢板焊成工字形截面取鼠=80,篙=80查表得：5x =0.688 （b 类），9y =0.578 （c 类）所需截面面积:31000 100.578 2152=8047 mmix =lox/ x -6000/80 -75mmiy -loy / y = 3000/80 = 37.5mm由各截面回转半径的近似值查附表 9-1得：ix=0

35、.43h, iy = 0.24b则h =75/0.43 = 174.4mm, b = 37.5/0.24 = 156.25mm选取截面：翼缘板宽 200mm ,翼缘板厚12mm,腹板高250mm，腹板厚10mm截面验算:截面几何特性:A =2 200 12 250 10 = 7300 mm23Ix 2 200 12 1312 =95.39 106mm41212 2003I y = 2 12=16 106mm4ix = I x/A =四5.39 106 /7300 = 114.31mmiy= Iy/A=.16 106/7300 = 46.82mm长细比:ox6 103I152.5 ： I 1-1

36、50114.31y 二卜露二64.150由长细比查表得稳定系数为：邛x =0.844 （b类），中y =0.681 （c类）1000 1032 r2201.15N/mm 二 f = 215N/mm0.681 7300满足要求局部稳定:95一 =7.92 (10 0.1 max). 135/ fy -10 0.1 64.1 -16.41 12ho tw25010=25 ：二(25 0.5 max) , 135/fy = 25 0.5 64.1 = 57.05满足要求刚度、整体稳定及局部稳定均满足要求，故选用此截面。(3)设计柱脚，基础混凝土的强度等级为 C20fc= 9.6N/mm2,锚栓采用d

37、 = 20mm,则其孔面积约为 5000mm 21）底板尺寸底板所需面积NfC 25000 = 109167mm八 1000 103Ao 二9.6靴梁厚度取10mm底板宽度：B=200 2 10 2 70-360mm底板长度：L = A/B =109167/360 = 303.24mm,取 360mm采用B m L = 360 x 360 ,靴梁、肋板的布置如图所示基础对底板的压应力为： 322q=N/An=1000 10 /(360 - 5000) =8.02N / mm三边支承板：区格b/d =43/200 =0.215 0.3,可按悬臂板计算22M1 =1/2qb2 =1/2 8.02

38、437414.5N mm区格”/a=70/250 = 0.28 Y6M max / f = 6 x 19649/ 205 = 23.98mm ,取 t = 24mm。2）靴梁计算靴梁与柱身的连接按承受柱的压力计算，为四条侧面角焊缝连接，取hf =10mm,则焊缝的计算长度为N4 0.7hfffw1000 1034 0.7 10 160二 223mm取靴梁高即焊缝长度260mm靴梁与底板的连接焊缝传递柱压力，取 hf =10mm,所需焊缝总计算长度为N1.22 0.7hfffw1000 1031.22 0.7 10 160=732mm焊缝的实际计算总长度超过此值。靴梁作为支承于柱边的双悬臂简支梁

39、，悬伸部分长度l=43mm,取其厚度t=10mm,底板传给靴梁的荷载q1 = Bq/2 =360 8.02/2 = 1443.6N/mm靴梁支座处最大剪力Vmax =q1l =1443.6 43 =62074.8N靴梁支座处最大弯矩_2_2Mmaxu1/2q1l =1/2 1443.6 43 =1334608.2N mm= 1.5Vmx =1.5 62074.8 =35.81N/mm2 ； fv =125N/mm2 ht 260 10Mmax6 1334608.2210 2602= 11.84N/mm2:f = 215N / mm2靴梁强度满足要求。3）肋板计算将肋板视为支承于靴梁的悬臂梁，悬

40、伸 长度为l = 70mm取其厚度t =10mm。其线荷载可偏安全假设为q2 =Lq/2 =360 8.02/2 = 1443.6N/mm肋板与底板的连接为正面角焊缝，取 hf =12mm,焊缝强度为q?lw0.7 1.22hw1443.60.7 1.22 12= 140.87N/mm2 二 ffw =160N/mm2肋板与靴梁的连接为侧面角焊缝，所受肋板的支座反力为R= 1443.6 70 = 101052N取hf =7mm,所需焊缝长度为Rl.=2 0.7hf ffw10105264.45mm2 0.7 7 160取肋板高度130mm。肋板支座处最大剪力Vmax uq2l =1443.6

41、70 -101052 N肋板支座处最大弯矩_2_2_Mmax=1/2q2l =1/2 1443.6 70 = 3536820 N mm= 1.5Vmx =1.5 101052 =116.60N/mm2 二 fv =125N/mm2 ht 130 10max6 3536820210 1302-125.57N/mm2二 f = 215N / mm2肋板强度满足要求前面柱脚板件区格弯矩相差较大，重新设置肋板如上图基础对底板的压应力为： 322q=N/An=1000 10 /(360 - 5000) =8.02N / mm三边支承板：区格口/4=43/200=0.2150.3,可按悬臂板计算22M1

42、=1/2qb2 =1/2 8.02 432 = 7414.5N mm区格4/a =70/120 = 0.58, P =0.0688M 3=Bqa； =0.0688 8.02 1202 945.57N mm四边支承板(区格)：ba1 =250/95 =2.63,查表 4-8 得a =0.11222M4 =二qa2 =0.112 8.02 952 =8131N mm两边支承板（区格）：a1 = J1102 702 =130.38mm，h =110 70/130.38 = 59.06mmbi/ai =59.06/130.38 = 0.45, P = 0.049M2 =Bqa； =0.049 8.02

43、 130.382 =6680.24N mm底板厚度 t 、6M max / f = 6x8131/205 = 15.42mm，取 t = 20mm。6.解：Q235 钢，N =2400kN , l =6m（1）选取分肢截面尺寸（按实轴稳定条件确定）取，y =60查表得：邛y =0.807 （b类）所需截面面积:3N 2400 10，f - 0.807 2152=13832mmiy = l0y / y =6000/60 = 100mm选取截面236b截面验算：查型钢表得：A=2M68.09 =136.18cm2 , iy = 13.63cm , i1 = 2.70cm ,4Z0 = 2.37cm

44、 , 11 = 496.7cm长细比:l oy y 一.i y36 103 144.0, = 150136.3由长细比查表得稳定系数为：*y =0.882 （b类）32400 100.882 13618 2_ 2= 199.82N/mm : f =215N/mm满足要求，故选用236b(2)确定分肢间距(按虚轴稳定条件确定)取缀条 Z56x5 ,查得 A1 = 5.42cm22Ay - 27y2A44.02 一 27M = 40.0ix =lox/ x =6000/40.0 = 150.00mm查附表 9-1 得：ix 比 0.44b,故 b 定ix/0.44 =150.0/0.44 = 34

45、0.91mm ,取b= 340mm。两槽钢翼缘间净距为 340-2父98 = 144mma 100mm,满足构造要求。验算虚轴稳定:I-21-煲-z021=2 496.7 104 6809 (170 - 23.7)2-64= 301.41 10 mmix301.41 106 . 148.77mm13618oxx =lox/ix =6000/148.77 = 40.3mm，x +27-A- = J2A213618 140.32 2744.3-1502 542查得:“0.881 (b 类)32400 100.881 136182_2= 200.04N/mm2 二 f =215N/mm2才两足要求。

46、(3)分肢稳定(只有斜缀条4 4 450)l 011 =一i12 (340 -2 23.7)二 21.67 : 0.7 27.0max=0.7 44.3 =31.01故分肢的刚度、强度和整体稳定无需进行验算，均满足要求。（4）斜缀条稳定柱剪力:Af - fy85 ；23513618 21585= 34.44kN斜缀条内力：八高二鬻二煞导2435kN1 =l0imin -(340 -2 23.7)/sin 45o/11.0 =37.62 ：二1 .1 - 150查表得：邛=0.908 （b类）强度设计值折减系数”查附表1-5得:= 0.6 0.0015 =0.6 0.0015 37.62 = 0

47、.656Ni_ 324 35 10= 49.48N/mm2 :二 f =0.656 215-141.04N/mm20.908 542才两足要求。缀条无孔洞削弱，不必验算强度。缀条的连接角焊缝采用两面侧焊，按要求取hf =5mm ；单面连接的单角钢按轴心受力计算连接时，刈=0.85。则焊缝的计算长度:l w10.7hf ffw0.7 24.35 10335.81mm0.7 5 0.85 160l w2k2N10.7hf ffw0.3 2 103 = 15.34mm0.7 5 0.85 160肢尖与肢背焊缝长度均取50mm 0柱中间设置横隔，与斜缀条节点配合设置。7.解：Q235 钢，N =150

48、0kN , l =7.5m（1）选取分肢截面尺寸（按实轴稳定条件确定）取,y =60查表得：邛y =0.807 （b类）所需截面面积： A = -N = 150010 = 8645mm2 yf 0.807 215iy = loy /，y =7500/60 = 125mm查附表 9-1 得：iy 定 0.38h,故 h *iy/0.38 = 125/0.38 = 328.95mm选取截面232a截面验算：2查型钢表得:A = 2 父 48.5 = 97cm , i y = 12.44cm , i1 = 2.51cm , z0 = 2.24cm , 一 一 4I1 = 304.7 cm长细比:oy

49、iy37.5 10124.4= 60.3 :二 L I -1503N 1500 10yA - 0.806 9700由长细比查表得稳定系数为：邛y =0.806 （b类）一 一2 .一2= 191.86N/mm 二 f = 215N/mm满足要求，故选用232a o（2）确定分肢间距（按虚轴稳定条件确定）ox = y =60.30.5人max =0.5 M 60.3 = 30.2 ,取%=30 x = j - 2 = 60.32 - 302 = 52.3ix =lox/ x = 7500/52.3 = 143.40mm查附表 9-1 得：ix 定 0.44b ,故 b Mx/0.44 =143.

50、40/0.44 = 325.91mm ,取 b= 320mm。两槽钢翼缘间净距为 320-2父88 = 144mma 100mm,满足构造 要求。验算虚轴稳定：lo1 =7 =30 x25.1 = 753mm ,取缀板净距 lo1 = 760mm,上 A fbI x = 2 x 11 + z0-2 12 64= 189.75 10 mm2ix189.75 106139.86mm97002 =2x 304.7x104 +4850 x(160 22.4)x =lox/ix =7500/139.86 = 53.6mmox = X .2 = . 53.62 302 = 61.4 -1-150查得:中x

51、= 0.800 (b 类)N 1500 103A一 0.800 97002_2= 193.30N/mm2 二 f =215N/mm2才两足要求。(3)分肢稳定%=30 0.7。= 0.5 父 61.4 = 30.7 及 40故分肢的刚度、强度和整体稳定无需进行验算，均满足要求。(4)缀板设计 22纵向图度 hb 之c=-m(320-2m 22.4) = 183.47mm , 33320 -2 22.440=6.88mm,取 hb 父 tb = 200mm 父 8mm。相令口缀板净距lo1 =760mm,相邻缀板中心矩 11noi + hb = 760 +200 = 960mm。缀板线刚度之和与

52、分肢线刚度比值为:二 1b / c32 (8 200 /12)/(320 -2 22.4)Ii/li304.7 104/960= 12.21 6缀板刚度满足要求。柱剪力:Affy85 1 2359700 215 =24.54kN85缀板弯矩：Mbi缀板剪力：Vbi6MbitbhbViliV li24.54 960 =5889.6kN mmVili24.54 9602 (320 -2 22.4)36 5889.6 1028 2002=42.80kN=ii0.43N/mm* 2; f =215N/mm2i.5Vbi_ _3二 fv = i25N/mm2i 5 42 8 i02MbiWw20.7hf

53、lw5889.6 10332666.67-2 30.7 7 2002= 180.29N/mm2=32666.67 mm3Vb1=42,=43.67N/mm2Aw9802= i.22 i60 = i95.2N/mm2ffw -i60N / mm2在弯矩和剪力共同作用下焊缝的折算应力为2180.29 +43.672 = i54.i0N / mm2 ffw = i60N / mm 0.6,所以 .07Sb二一鬻二0.643所需毛截面抵抗矩：WnxMx-：bf174.69 10633 =1263.63 103mm30.643 215查型钢表选用I45a,截面几何特性：Wx= 1433cm3, Ix=3

54、2241cm4,质量 q = 80.4kg / m强度验算:Mx 174.69 106 0.804 1.2 55002/8xWnx1.05 1433 103= 118.52N/mm2 二 f =215N/mm2才两足要求。整体稳定验算:M x 174.69 106 0.804 1.2 55002 /8bWx0.643 1433 10322= 193.54N/mm2 二 f -215N/mm2才两足要求。故选用此截面。(3)假设梁的跨度中点处受压翼缘设置一可靠的侧向支承，此梁的整体稳定不一定保证。按整体稳定确定梁截面假定工字钢型号在I22I40之间，自由长度li =5.5/2 = 2.75m ,

55、由附表3-2查整体稳定系数中b =2.1 a 0.6,所以=1.07-谓=1.07-笠=0.936所需毛截面抵抗矩:WnxMx-：bf174.69 10633868.07 103mm30.936 215查型钢表选用I36c ,截面几何特性：Wx = 964cm3 , Ix = 17351cm4 q = 71.3g / m强度验算:MxxWnx174.69 106 0.713 1.2 55002/81.05 964 10322=175.78N / mm2 :二 f = 215N / mm2满足要求整体稳定验算:MxbWx174.69 106 0.713 1.2 55002 /80.936 964

56、 103= 197.19N/mm2 : f =215N/mm2满足要求故选用此截面。.解：Q235 钢，l=15m, L 1=1/400截面形心位置- 420 20 1230 320 20 10 1200 8 620 x =二670mm420 20 320 20 1200 838 120022Ix= 8 1200 (670 -620)420 20 (1240 -670 -10)12320 20 (670 -10)2 =6598.08 106 mm4(1)抗弯强度= 11.576 106 mm3Ix 6598.08 106Wx1 =1二1240 -670570Wx2Ix = 6598.08 10

57、6670 一 670= 9.848 106mm312Mx 3.2 152 400 5 = 2090kN= 202.12N/mm2 ： f = 205N/mm2Mx 二 2090 106xWnx21.05 9.848 106满足要求(2)抗剪强度中和轴处截面的面积矩:233Sxo =420 20 (1240 -670 -10) (1240 - 670 -20) 8/2 = 5914 10 mm上翼缘与腹板交界处的面积矩:Sx1 =420 20 (1240 -670 -10)=4704 103 mm3卜翼缘与腹板交界处的面积矩:.33Sx2 =320 20 (670 -10)=4224 10 mm

58、支座处剪力最大，所受最大剪应力为:VSxo(400 3.2 7.5) 103 5914 103I xtw6598.08 106 8 2_ 2= 47.50N/mm :二 fv =120N/mm才两足要求。(3)局部承压强度取支撑长度 a=0.2m,则 lz=a+5hy = 200+5 x 20 = 300mm1- F 1.0 400 10322-c = = = 166.67N / mm : f = 205N / mmlztw300 8满足要求，无需设置支撑加劲肋。(4)折算应力在集中力作用处弯矩、剪力、局部压应力均较大集中力作用处:M x =400 5 3.2 7.5 5 -3.2 5 2.5

59、 = 2080kN mV =400 3.2 7.5 - 3.2 5 = 408kN腹板与上翼缘交界处的应力为:Mx、-1 一 Wnx12080 10626 = 179.68N / mm11.576 10VSx1408 1 03 4704 1 03I xtw6598.08 106 82=36.36N/mm2 TOC o 1-5 h z .2二c =166.67N /mm腹板与上翼缘交界处的折算应力为:.二2 二；-二 c 312 = J79.682 166.672 -166.67 179.68 36.362 _.2. 2=177.31N/mm ：1.1f =1.1 205 =225.5N / m

60、m满足要求。(5)挠度M kxl (2090/1.2) 106 150001! 11& = =l 10EIx10 2.06 105 6598.08 106 520l 400满足要求.解：Q235B 钢，l =6m1M max (6 1.2 30 1.4) 6 -73.8kN m4按整体稳定确定梁截面：假定工字钢型号在 I22I40之间，荷载作用在梁下翼缘，自由长度1i =6m,由附表3-2查政体稳定系数 匕=1.07 a 0.6,所以0.2820.282b =1.07 =1.07 = 0.806b，1.07所需毛截面抵抗矩：Wx=里-上=73.8 10 = 425.88 103 mm3b f