结构矩阵分析原理与程序设计上机成果.doc

结构矩阵分析原理与程序设计上机成果工程力学系 力1101-3班第8组组员： 杨圣银： 20113110刘奇 ： 20113122范炳辉：孙海明： 20113098卞炟 ： 20113115王正来： 20113117陈怡婧： 20113126童心潜：田旺 ： 20113103日期：2013.11一、 试用教材中平面刚架的源程序frame计算下列各题1教材90页题2.8；（1）数据的准备于输入1. 控制数据nnnfndndfnenpjnpe43025222. 节点坐标节点（i）1234X（i）0066Y（i）04403. 各单元始、末端结点号及EA、EI值单元号i123始端好jl（i）123末端号jr（i）234EA（i）7.2E61.05E71.2E6EI（i）3E57.2E63E54. 直接结点载荷编号i结点号mj（i）Xqj（i，1）Yqj（i，2）Mqj（i，3）1110002320005. 非结点载荷编号i单元号mf（i）载荷类型ind(i)aaq(i)bbq(i_q1ql(i)q2q2（i)11120-15022206-10-106. 可动支座中的约束条件编号i位移分量号ibd（i）位移值bd(i)120230310041105121.75e-2将数据输入fr.txt 中各式如下4,3,0,2,5,2,21,0,02,0,43,6,44,6,01,1,2,7.2E6,3E52,2,3,1.05E7,1.2E63,3,4,7.2E6,3E51,1,10,0,02,3,20,0,01,1,1,2,0,-15,02,2,2,0,6,-10,-101,2,02,3,03,10,04,11,05,12,1.75E-2输出结果为Plane Frame structural Analysis*input data= = = = =structural control data-nn ne nf nd ndf npj npe n 4 3 0 2 5 2 2 12 Nodal coordinates-Node x y 1 0 0 2 0 4 3 6 4 4 6 0 Element Information-Ele.No. jl jr ea ei al 1 1 2 7200000 300000 4 2 2 3 10500000 1200000 6 3 3 4 7200000 300000 4 Nodal Load-i mj xd yd md 1 1 10 0 0 2 3 20 0 0 Element loads-i mf ind aq bq q1 q2 1 1 1 2 0 -15 0 2 2 2 0 6 -10 -10 Bonundary conditions-i ibd bd 1 2 0 2 3 0 3 10 0 4 11 0 5 12 0.0175 output data=nadal displacement-node no. u v fai 1 0.04 0 0 2 0.04 0 0 3 0.04 0.0001 0.002 4 0 0 0.0175 Element No.&member-end force:= = = = = = = = = = = = = = =Eie No. n(1) q(1) m(1) n(r) q(r) m(r)- 1 -182.4 -10.0 94.1 182.4 25.0 -24.1 2 25.0 -182.4 24.1 -25.0 242.4 1250.4 3 242.4 45.0 -1250.4 - 242.4 -45.0 1070.4nadal displacement-15125045弯矩图2.8。单位：KN*m942412501070742教材91页题2.9；输入文件：10,12,4,0,0,2,61,0,42,6,83,6,44,14,85,14,46,20,47,0,08,6,09,14,010,20,01,1,2,4.2E6,5.5E42,1,3,4.2E6,5.5E43,1,7,4.2E6,5.5E44,2,3,4.2E6,5.5E45,3,8,4.2E6,5.5E46,2,4,4.2E6,5.5E47,3,5,4.2E6,5.5E48,4,5,4.2E6,5.5E49,5,9,4.2E6,5.5E410,4,6,4.2E6,5.5E411,5,6,4.2E6,5.5E412,6,10,4.2E6,5.5E41,1,25,0,-302,6,0,0,401,1,2,0,7.21,-8.32,-8.322,1,7,0,7.21,-5.55,-5.553,7,1,3.61,0,-50,04,10,1,3.61,0,-16.6,05,10,6,3.61,0,11.1,06,6,2,8,0,-10,-10输出文件：Plane Frame structural Analysis*input data= = = = =structural control data-nn ne nf nd ndf npj npe n 10 12 4 0 0 2 6 18 Nodal coordinates-Node x y 1 0 4 2 6 8 3 6 4 4 14 8 5 14 4 6 20 4 7 0 0 8 6 0 9 14 0 10 20 0 Element Information-Ele.No. jl jr ea ei al 1 1 2 4200000 55000 7.21110255092798 2 1 3 4200000 55000 6 3 1 7 4200000 55000 4 4 2 3 4200000 55000 4 5 3 8 4200000 55000 4 6 2 4 4200000 55000 8 7 3 5 4200000 55000 8 8 4 5 4200000 55000 4 9 5 9 4200000 55000 4 10 4 6 4200000 55000 7.21110255092798 11 5 6 4200000 55000 6 12 6 10 4200000 55000 4 Nodal Load-i mj xd yd md 1 1 25 0 -30 2 6 0 0 40 Element loads-i mf ind aq bq q1 q2 1 1 2 0 7.21 -8.32 -8.32 2 1 7 0 7.21 -5.55 -5.55 3 7 1 3.61 0 -50 0 4 10 1 3.61 0 -16.6 0 5 10 6 3.61 0 11.1 0 6 6 2 8 0 -10 -10 output data=nadal displacement-node no. u v fai 1 0.001 0 0 2 0 0 -0.001 3 0.001 0 0 4 0.001 0 0 5 0.001 0 0 6 0.001 0 0 7 0 0 0 8 0 0 0 9 0 0 0 10 0 0 0 Element No.&member-end force:= = = = = = = = = = = = = = =Eie No. n(1) q(1) m(1) n(r) q(r) m(r)- 1 26.8 35.0 -43.9 13.2 25.0 7.6 2 16.5 -8.1 21.2 -16.5 8.1 27.3 3 36.0 5.6 -7.4 -35.9 -5.6 -14.9 4 -10.5 9.7 -42.5 10.5 -9.7 3.7 5 24.8 -1.4 11.2 -24.8 1.4 -5.4 6 -6.8 -38.6 34.9 6.8 -41.4 -46.2 7 27.7 27.2 -42.2 -27.7 22.8 44.0 8 -41.0 -8.6 32.7 41.0 8.6 1.9 9 -18.1 17.4 -39.4 18.1 -17.4 -30.1 10 1.3 1.3 13.5 -12.4 15.3 36.6 11 1.7 0.1 -6.5 -1.7 -0.1 5.8 12 19.4 3.5 -2.4 -19.4 -3.5 -11.63教材122页例3.2，并与教材结果相比较（注意不需要修改程序，只需要近似处理某些参数）；Plane Frame structural Analysis*input data= = = = =structural control data-nn ne nf nd ndf npj npe n 6 11 0 3 4 3 0 18 Nodal coordinates-Node x y 1 16 6 2 16 0 3 8 6 4 8 0 5 0 6 6 0 0 Element Information-Ele.No. jl jr ea ei al 1 1 2 400000 0.002 6 2 1 3 400000 0.002 8 3 1 4 400000 0.002 10 4 2 3 400000 0.002 10 5 2 4 400000 0.002 8 6 3 4 400000 0.002 6 7 3 5 400000 0.002 8 8 3 6 400000 0.002 10 9 4 5 400000 0.002 10 10 4 6 400000 0.002 8 11 5 6 400000 0.002 6 Nodal Load-i mj xd yd md 1 3 0 -35 0 2 4 0 -15 0 3 5 0 -10 0 Bonundary conditions-i ibd bd 1 5 -0.015 2 16 0 3 17 0 4 13 0 output data=nadal displacement-node no. u v fai 1 2.92473604935366E-03 -1.47081872168391E-02 9.80840451448748E-04 2 -2.75218702985603E-03 -0.015 1.01893884546149E-03 3 2.40595776835953E-03 -6.64560080225226E-03 8.85592674007333E-04 4 -2.18506787396444E-03 -6.5441302140501E-03 9.00772546752097E-04 5 1.75256410419263E-18 -7.68283371317775E-04 2.50118930988387E-04 6 -1.75256410419263E-18 -1.00721153907224E-18 3.40487793989763E-04 Element No.&member-end force:= = = = = = = = = = = = = = =Eie No. n(1) q(1) m(1) n(r) q(r) m(r)- 1 -19.4541855440611 -3.58238679467797E-08 9.4772139169426E-08 19.4541855440611 3.58238679467797E-08 1.20171068511253E-07 2 -25.9389140497067 2.79775271607425E-08 -8.80981642826162E-08 25.9389140497067 -2.79775271607425E-08 -1.35722053003324E-07 3 32.4236425207567 4.53749515565149E-09 -6.67397488680941E-09 -32.4236425207567 -4.53749515565149E-09 -3.87011367654696E-08 4 -35.4449472030481 6.13797249832126E-09 -4.02060248317998E-09 35.4449472030481 -6.13797249832126E-09 -5.73590710648423E-08 5 28.3559577945796 3.64230101763548E-08 -1.16150466028072E-07 -28.3559577945796 -3.64230101763548E-08 -1.75233615382767E-07 6 6.76470588014433 -8.53411133282582E-08 2.50963382403187E-07 -6.76470588014433 8.53411133282582E-08 2.61083297566363E-07 7 -120.297888417976 6.25533286383566E-08 -9.134487879869E-08 120.297888417976 -6.25533286383566E-08 -4.09081750308163E-07 8 82.5037706665494 1.51116602564727E-08 3.34626204636684E-08 -82.5037706665494 -1.51116602564727E-08 -1.84579331543359E-07 9 -68.6981522587136 4.2542386141653E-09 1.0885942596929E-07 68.6981522587136 -4.2542386141653E-09 -1.51402020336194E-07 10 109.253393698222 7.40197898944995E-08 -1.56007971387414E-07 -109.253393698222 -7.40197898944995E-08 -4.36150347768581E-07 11 51.2188914211849 -1.9686890832605E-07 5.60483770644357E-07 -51.2188914211849 1.9686890832605E-07 6.20729679311941E-074教材150页例4.1，并与教材结果相比较Influnence Line Values Of Internal ForceInput DataControl Data*n ne nw jl jr nld kw kc ac* 5 4 2 1 0 3 0 0 0 Element InformationEle.No EI L EI/L 1 32000 4 8000 2 64000 6 10666.6666666667 3 64000 6 10666.6666666667 4 32000 4 8000 Number Of Loading Condition ld= 1 结点荷载数 非节点荷载作用单元数 1 4 Nodal LoadsNo. mj qj 1 5 40 Element LoadsNo. mf ind aq bq q1 q2= 1 1 2 0 4 -20 -20 = 2 2 2 0 6 -20 -20 = 3 3 2 0 6 -20 -20 = 4 4 2 0 4 -20 -20 Number Of Loading Condition ld= 2 结点荷载数 非节点荷载作用单元数 1 5 Nodal LoadsNo. mj qj 1 5 100 Element LoadsNo. mf ind aq bq q1 q2= 1 1 2 0 4 -30 -30 = 2 2 1 2 0 -80 0 = 3 2 1 4 0 -100 0 = 4 3 2 0 6 -30 -30 = 5 3 1 3 0 -60 0 Number Of Loading Condition ld= 3 结点荷载数 非节点荷载作用单元数 3 3 Nodal LoadsNo. mj qj 1 3 40 2 4 -60 3 5 60 Element LoadsNo. mf ind aq bq q1 q2= 1 1 1 2 0 -100 0 = 2 2 2 0 6 -30 -30 = 3 3 2 0 6 -30 -30 output data=number of loading conditions ld= 1 nodal angular routationnode no. fai= 1 0 2 4.33501683501683E-04 3 4.52441077441077E-05 4 -6.14478114478115E-04 5 7.23905723905724E-04 =number of loading conditions ld= 2 nodal angular routationnode no. fai= 1 0 2 8.21846941638608E-04 3 6.65202370931538E-04 4 -2.98786475869809E-03 5 4.61893237934905E-03 =number of loading conditions ld= 3 nodal angular routationnode no. fai= 1 0 2 1.95707070707071E-04 3 1.19002525252525E-03 4 -3.08080808080808E-03 5 3.41540404040404E-03 member-end forces of elements ld= 1 ele.no. qi mi qj mj 1 34.7979797979798 -19.7306397306397 45.2020202020202 40.5387205387205 2 54.8933782267116 -40.5387205387205 65.1066217732884 71.1784511784512 3 66.0718294051627 -71.1784511784512 53.9281705948373 34.7474747474747 4 38.6868686868687 -34.7474747474748 41.3131313131313 40 member-end forces of elements ld= 2 ele.no. qi mi qj mj 1 50.1378367003367 -26.8504489337823 69.8621632996633 66.2991021324355 2 69.3233258511036 -66.2991021324355 110.676674148896 170.359147025814 3 144.77506546951 -170.359147025814 95.2249345304901 21.7087542087542 4 -19.5728114478114 -21.7087542087542 19.5728114478114 100 member-end forces of elements ld= 3 ele.no. qi mi qj mj 1 47.6515151515151 -46.8686868686869 52.3484848484849 56.2626262626263 2 75.2188552188552 -56.2626262626263 104.781144781145 144.949494949495 3 110.16835016835 -104.949494949495 69.8316498316498 -16.0606060606061 4 -4.01515151515151 -43.939393939394 4.01515151515151 60二、程序修改1将子程序bound由原来的“赋大值法”改为“主1副0法”，并写出PAD设计图及源程序；bound定义局部变量，其他全局变量通过调用相应数组名或变量名传递数组变量定义ndf=0i=1，ndfK=ibd（i）j=1，nr(j,k)=0R(k,j)=0j=1，nr(k,k)=1P(k)=bd(k)end代码为：Sub bound()Dim i As Integer, j As Integer, k As IntegerDim a As Double If ndf 0 Then For i = 1 To ndf k = ibd(i) For j = 1 To n r(j, k) = 0 r(k, j) = 0 Next j r(k, k) = r(k, k) + 1 p(k) = bd(k) Next i End If For i = 1 To n For j = 1 To n Print #2, r(i, j); Next j Print #2, Next iEnd Sub2修改完成后，计算教材84页例2.2，检验程序修改是否正确；计算结果输出为：Plane Frame structural Analysis*input data= = = = =structural control data-nn ne nf nd ndf npj npe n 7 6 3 1 2 3 5 12 Nodal coordinates-Node x y 1 0 4 2 6 4 3 12 4 4 15 4 5 0 0 6 6 0 7 12 0 Element Information-Ele.No. jl jr ea ei al 1 1 2 11200000 678400 6 2 2 3 11200000 678400 6 3 3 4 11200000 678400 3 4 1 5 9760000 176000 4 5 2 6 9760000 176000 4 6 3 7 9760000 176000 4 Nodal Load-i mj xd yd md 1 1 0 0 -30 2 3 0 0 20 3 4 0 -20 0 Element loads-i mf ind aq bq q1 q2 1 1 2 0 6 0 -20 2 1 3 3 6 0 20 3 2 1 3 0 -15 0 4 2 2 0 6 -5 -5 5 4 1 2 0 25 0 Bonundary conditions-i ibd bd 1 10 0 2 12 0 output data=nadal displacement-node no. u v fai 1 1.1005919016136E-05 -4.98956632190588E-07 -9.54889080003477E-05 2 7.88027433161215E-06 -5.52505623049841E-06 5.98636773997588E-05 3 2.77733304595923E-06 -2.06153313996061E-05 1.4164721649682E-05 4 0 -1.0819496104394E-04 0 5 0 0 0 6 0 0 0 7 0 0 0 Element No.&member-end force:= = = = = = = = = = = = = = =Eie No. n(1) q(1) m(1) n(r) q(r) m(r)- 1 5.83453674444456 1.21745418254503 -24.9675615368738 -5.83453674444456 -1.21745418254503 -12.3371635583964 2 9.52549039988544 14.6985913849612 2.32125444192523 -9.52549039988544 30.3014086150388 44.4871972483079 3 10.3687100382478 20 -26.7968842776186 -10.3687100382478 -20 -3