大学结构力学矩阵位移法上机实验报告.doc

中南大学结构力学矩阵位移法上机实验报告 学 院： 土木工程学院专业班级： 土木工程1111姓 名： 杨靖东 学 号： 1208111520 指导老师： 周德 实验日期：2013年10月12日目 录一、实践目的与要求二、钢架的受力分析 1、题目3 2、结构计算编号示意图33、输入文件4 4、输出文件5 5、结构内力图 7三、桁架的受力分析 1、题目9 2、结构计算编号示意图93、输入文件10 4、输出文件12 5、结构轴力图 16四、连续梁的受力分析 1、题目17 2、结构计算编号示意图173、输入文件17 4、输出文件18 5、结构内力图 19五、实践心得一、 实践目的与要求1、学会使用矩阵位移法,掌握pf程序的使用并用来计算给定的平面刚架、桁架和连续粱的内力；2、利用cad或者手工绘出平面钢架、桁架和连续梁的内力。二、钢架的受力分析1、题目作图示刚架的、图，已知各杆截面均为矩形，柱截面宽0.4m，高0.4m，大跨梁截面宽0.4m，高0.9m，小跨梁截面宽0.4m，高0.6m，各杆e=3.0104 mpa。10分2、结构计算编号示意图3、输入文件* * no.1 a program of plane frame 2013.10.15 * *30e6 20 17 12 11 2 16e-2 213e-52 3 16e-2 213e-53 4 16e-2 213e-54 5 36e-2 243e-45 6 36e-2 243e-4 3 7 36e-2 243e-47 8 36e-2 243e-42 9 24e-2 720e-59 12 24e-2 720e-59 10 16e-2 213e-511 12 16e-2 213e-58 12 16e-2 213e-56 8 16e-2 213e-56 16 24e-2 720e-58 15 24e-2 720e-512 13 24e-2 720e-513 14 24e-2 720e-514 17 16e-2 213e-514 15 16e-2 213e-515 16 16e-2 213e-50 00 4.50 7.70 10.93.8 10.97.6 10.93.8 7.77.6 7.73.8 4.53.8 07.6 07.6 4.510.3 4.511.4 4.511.4 7.711.4 10.911.4 011 012 013 0101 0102 0103 0111 0112 0113 0171 0172 0173 054 100 0 05 0 0 -157 0 -26 013 0 -26 016 0 0 -1581 3 20 4.56 4 -45 3.87 4 -45 3.88 4 -5 3.89 4 -5 3.815 4 -45 3.816 4 -5 2.717 4 -5 1.14、输出文件input data file name: t1.txt output file name: z1.txt * * * * no.1 a program of plane frame 2013.10.15 * * * * the input data the general information e nm nj ns nlc 3.000e+07 20 17 12 1 the information of members member start end a i 1 1 2 1.600000e-01 2.130000e-03 2 2 3 1.600000e-01 2.130000e-03 3 3 4 1.600000e-01 2.130000e-03 4 4 5 3.600000e-01 2.430000e-02 5 5 6 3.600000e-01 2.430000e-02 6 3 7 3.600000e-01 2.430000e-02 7 7 8 3.600000e-01 2.430000e-02 8 2 9 2.400000e-01 7.200000e-03 9 9 12 2.400000e-01 7.200000e-03 10 9 10 1.600000e-01 2.130000e-03 11 11 12 1.600000e-01 2.130000e-03 12 8 12 1.600000e-01 2.130000e-03 13 6 8 1.600000e-01 2.130000e-03 14 6 16 2.400000e-01 7.200000e-03 15 8 15 2.400000e-01 7.200000e-03 16 12 13 2.400000e-01 7.200000e-03 17 13 14 2.400000e-01 7.200000e-03 18 14 17 1.600000e-01 2.130000e-03 19 14 15 1.600000e-01 2.130000e-03 20 15 16 1.600000e-01 2.130000e-03 the joint coordinates joint x y 1 .000000 .000000 2 .000000 4.500000 3 .000000 7.700000 4 .000000 10.900000 5 3.800000 10.900000 6 7.600000 10.900000 7 3.800000 7.700000 8 7.600000 7.700000 9 3.800000 4.500000 10 3.800000 .000000 11 7.600000 .000000 12 7.600000 4.500000 13 10.300000 4.500000 14 11.400000 4.500000 15 11.400000 7.700000 16 11.400000 10.900000 17 11.400000 .000000 the information of supports is vs 11 .000000 12 .000000 13 .000000 101 .000000 102 .000000 103 .000000 111 .000000 112 .000000 113 .000000 171 .000000 172 .000000 173 .000000 loading case 1 the loadings at joints nlj= 5 joint fx fy fm 4 100.000000 .000000 .000000 5 .000000 .000000 -15.000000 7 .000000 -26.000000 .000000 13 .000000 -26.000000 .000000 16 .000000 .000000 -15.000000 the loadings at members nlm= 8 member type vf dst 1 3 20.000000 4.500000 6 4 -45.000000 3.800000 7 4 -45.000000 3.800000 8 4 -5.000000 3.800000 9 4 -5.000000 3.800000 15 4 -45.000000 3.800000 16 4 -5.000000 2.700000 17 4 -5.000000 1.100000 the results of calculation the joint displacements joint u v rotation 1 3.548624e-21 -1.090850e-20 -8.379650e-21 2 4.843385e-03 -1.022672e-04 -2.783426e-04 3 7.179477e-03 -1.923974e-04 -8.636867e-04 4 9.295066e-03 -1.850224e-04 -9.802982e-05 5 9.264631e-03 -3.309862e-04 -1.531391e-05 6 9.234195e-03 -4.304060e-04 -7.353304e-05 7 7.178917e-03 -2.040492e-03 9.158656e-05 8 7.178357e-03 -4.432951e-04 2.992634e-04 9 4.832080e-03 -3.043073e-05 -8.061810e-05 10 3.913468e-21 -3.245945e-21 -8.919782e-21 11 3.534990e-21 -2.725933e-20 -8.357965e-21 12 4.841428e-03 -2.555562e-04 -2.846735e-04 13 4.840615e-03 -1.320398e-04 1.094524e-04 14 4.840284e-03 -1.948709e-04 -3.011046e-04 15 7.176816e-03 -2.763351e-04 -2.815871e-04 16 9.218393e-03 -2.965992e-04 -1.947147e-04 17 3.502918e-21 -2.078623e-20 -8.309133e-21 the terminal forces member fn fs m 1 start 1 109.085001 80.486236 117.546496 end 2 -109.085001 9.513764 42.141567 2 start 2 135.195324 11.907415 30.740454 end 3 -135.195324 -11.907415 7.363271 3 start 3 -11.062441 13.498640 6.308611 end 4 11.062441 -13.498640 36.887034 4 start 4 86.501360 -11.062441 -36.887034 end 5 -86.501360 11.062441 -5.150243 5 start 5 86.501360 -11.062441 -9.849757 end 6 -86.501360 11.062441 -32.187520 6 start 3 1.591225 146.257766 -13.671883 end 7 -1.591225 24.742232 244.551393 7 start 7 1.591225 -50.742232 -244.551393 end 8 -1.591225 221.742230 -273.169078 8 start 2 21.421179 -26.110323 -72.882021 end 9 -21.421179 45.110323 -62.437204 9 start 9 -17.713506 -12.650876 -24.471059 end 12 17.713506 31.650876 -59.702267 10 start 9 32.459447 39.134684 86.908263 end 10 -32.459447 -39.134684 89.197817 11 start 11 272.593264 35.349904 83.579648 end 12 -272.593264 -35.349904 75.494921 12 start 8 281.608321 55.232455 100.032412 end 12 -281.608321 -55.232455 76.711433 13 start 6 -19.333640 56.560085 83.051852 end 8 19.333640 -56.560085 97.940409 14 start 6 29.941275 -30.396082 -50.864332 end 16 -29.941275 30.396082 -64.640771 15 start 8 2.918855 79.199732 75.196258 end 15 -2.918855 91.800266 -99.137252 16 start 12 2.169045 -40.665932 -92.504086 end 13 -2.169045 54.165933 -35.518946 17 start 13 2.169045 -80.165933 35.518946 end 14 -2.169045 85.665933 -126.726423 18 start 14 207.862281 35.029175 74.539959 end 17 -207.862281 -35.029175 83.091330 19 start 14 122.196348 32.860131 52.186465 end 15 -122.196348 -32.860131 52.965947 20 start 15 30.396082 29.941275 46.171305 end 16 -30.396082 -29.941275 49.6407715、结构内力图 （1）图（单位：kn）（2）图（单位：kn） （3）图（单位：knm）三、桁架的受力分析1、题目计算图示桁架各杆的轴力。已知a=3600mm2，e=2.0105 mpa。5分2、结构计算编号示意图3、输入文件* * no.2 a program of plane truss 2013.10.15 * *200e6 15 9 4 11 2 36e-4 1e-121 3 36e-4 1e-122 3 36e-4 1e-122 4 36e-4 1e-123 4 36e-4 1e-124 5 36e-4 1e-123 5 36e-4 1e-123 6 36e-4 1e-125 6 36e-4 1e-125 7 36e-4 1e-126 7 36e-4 1e-127 8 36e-4 1e-126 8 36e-4 1e-126 9 36e-4 1e-128 9 36e-4 1e-120 0 0 62 32 6 4 66 36 68 68 011 012 091 092 052 10 -50 04 0 -50 05 0 -50 07 0 -50 08 0 -50 004、输出文件input data file name: t2.txt output file name: z2.txt * * * * no.2 a program of plane truss 2013.10.15 * * * * the input data the general information e nm nj ns nlc 2.000e+08 15 9 4 1 the information of members member start end a i 1 1 2 3.600000e-03 1.000000e-12 2 1 3 3.600000e-03 1.000000e-12 3 2 3 3.600000e-03 1.000000e-12 4 2 4 3.600000e-03 1.000000e-12 5 3 4 3.600000e-03 1.000000e-12 6 4 5 3.600000e-03 1.000000e-12 7 3 5 3.600000e-03 1.000000e-12 8 3 6 3.600000e-03 1.000000e-12 9 5 6 3.600000e-03 1.000000e-12 10 5 7 3.600000e-03 1.000000e-12 11 6 7 3.600000e-03 1.000000e-12 12 7 8 3.600000e-03 1.000000e-12 13 6 8 3.600000e-03 1.000000e-12 14 6 9 3.600000e-03 1.000000e-12 15 8 9 3.600000e-03 1.000000e-12 the joint coordinates joint x y 1 .000000 .000000 2 .000000 6.000000 3 2.000000 3.000000 4 2.000000 6.000000 5 4.000000 6.000000 6 6.000000 3.000000 7 6.000000 6.000000 8 8.000000 6.000000 9 8.000000 .000000 the information of supports is vs 11 .000000 12 .000000 91 .000000 92 .000000 loading case 1 the loadings at joints nlj= 5 joint fx fy fm 2 10.000000 -50.000000 .000000 4 .000000 -50.000000 .000000 5 .000000 -50.000000 .000000 7 .000000 -50.000000 .000000 8 .000000 -50.000000 .000000 the loadings at members nlm= 0 the results of calculation the joint displacements joint u v rotation 1 -3.990692e-21 -1.175000e-20 -1.204224e-04 2 2.466010e-04 -4.803302e-04 -1.163163e-04 3 2.002495e-04 -5.664913e-04 -1.871438e-05 4 2.046758e-04 -7.748246e-04 -3.296074e-05 5 1.627506e-04 -6.680758e-04 2.001388e-05 6 7.165412e-05 -4.937226e-04 -1.831531e-06 7 1.486031e-04 -7.020559e-04 2.709549e-05 8 1.344557e-04 -4.803302e-04 6.701691e-05 9 4.990692e-21 -1.325000e-20 3.105259e-05 the terminal forces member fn fs m 1 start 1 57.639621 .000000 .000000 end 2 -57.639621 .000000 .000000 2 start 1 71.943222 .000000 .000000 end 3 -71.943222 .000000 .000000 3 start 2 -9.181682 .000000 .000000 end 3 9.181682 .000000 .000000 4 start 2 15.093081 .000000 .000000 end 4 -15.093081 .000000 .000000 5 start 3 50.000000 .000000 .000000 end 4 -50.000000 .000000 .000000 6 start 4 15.093081 .000000 .000000 end 5 -15.093081 .000000 .000000 7 start 3 21.032382 .000000 .000000 end 5 -21.032382 .000000 .000000 8 start 3 23.147172 .000000 .000000 end 6 -23.147172 .000000 .000000 9 start 5 39.060139 .000000 .000000 end 6 -39.060139 .000000 .000000 10 start 5 5.093081 .000000 .000000 end 7 -5.093081 .000000 .000000 11 start 6 50.000000 .000000 .000000 end 7 -50.000000 .000000 .000000 12 start 7 5.093081 .000000 .000000 end 8 -5.093081 .000000 .000000 13 start 6 -9.181682 .000000 .000000 end 8 9.181682 .000000 .000000 14 start 6 89.970978 .000000 .000000 end 9 -89.970978 .000000 .000000 15 start 8 57.639621 .000000 .000000 end 9 -57.639621 .000000 .0000005、结构轴力图 （1）图（单位：kn）四、连续梁的受力分析1、题目作图示连续梁的、图，已知各梁截面面积a=7.5，惯性矩i=9.50，各杆e=3.45104mpa。5分2、结构计算编号示意图3、输入文件* * no.3 a program of plane frame 2013.10.15 * *34.5e6 5 6 6 11 2 7.5 9.52 3 7.5 9.53 4 7.5 9.54 5 7.5 9.55 6 7.5 9.50 040 060 080 0100 0120 011 012 013 022 042 062 023 0 -320 05 0 0 -10051 4 -10.5 402 4 -10.5 203 4 -10.5 204 4 -10.5 205 4 -10.5 204、输出文件 input data file name: t3.txt output file name: z3.txt * * * * no.3 a program of plane frame 2013.10.15 * * * * the input data the general information e nm nj ns nlc 3.450e+07 5 6 6 1 the information of members member start end a i 1 1 2 7.500000e+00 9.500000e+00 2 2 3 7.500000e+00 9.500000e+00 3 3 4 7.500000e+00 9.500000e+00 4 4 5 7.500000e+00 9.500000e+00 5 5 6 7.500000e+00 9.500000e+00 the joint coordinates joint x y 1 .000000 .000000 2 40.000000 .000000 3 60.000000 .000000 4 80.000000 .00000