结构力学矩阵位移法实践报告.doc

结构力学上机实验报告姓名： 学号： 指导老师：专业班级：日期： 矩阵位移法上机实践报告一、实践目的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。二、实践要求（1）用PF程序计算给定的平面钢架、桁架和连续粱的内力；（2）绘制给出上述结构的内力图。三、实践步骤（1）编号：对杆件和结点编号，选定局部坐标系和整体坐标系。（2）建立输入文件：根据题目已给数据，建立格式为.txt的输入数据文件，存放在与PF程序相当的文件夹下。（3）运行计算：运行程序，分别按照要求键入输入数据的文本名和输出存储数据的文件名从而得到相应的输出文件。（4）绘图：根据得到的结果，画出内力图。四、每题的题目及具体解题步骤如下所示：1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0.35m,高0.85m，小跨梁截面宽0.35m,高0.6m，各杆E=3.0104 MPa。解：（1）编号根据刚架进行编号。注意到作用在杆上的集中力偶由于PF程序未直接处理的算法，故在此点增加一个绞结点11。（2） 建立输入文件timu1.txt，存于PF同一文件夹内。具体输入如下：* * EX.1 A Simple Suppoted Frame 2013.06.08 * * *3E7 16 13 9 11 2 16E-2 213E-52 4 2975E-4 179E-43 4 16E-2 213E-54 6 21E-2 63E-45 6 16E-2 213E-52 7 16E-2 213E-57 8 2975E-4 179E-44 8 16E-2 213E-58 9 21E-2 63E-46 9 16E-2 213E-57 10 16E-2 213E-510 11 2975E-4 179E-411 12 2975E-4 179E-48 12 16E-2 213E-512 13 21E-2 63E-49 13 16E-2 213E-50 00 4.5 7.6 0 7.6 4.5 11.4 011.4 4.50 7.77.6 7.711.4 7.70 10.93.8 10.97.6 10.911.4 10.911 012 013 031 032 033 051 052 053 047 100 0 010 100 0 011 0 0 -1512 0 0 -1571 3 20 4.52 2 -26 3.82 4 -36 7.64 4 -36 3.84 2 -26 2.77 4 -36 7.69 4 -36 3.8(3) 运行PF程序,得到如下的timu1out.txt输出文件：Input Data File Name: timu1.txt Output File Name: timu1out.txt * * * * EX.1 A Simple Suppoted Frame 2013.06.08 * * * * The Input Data The General Information E NM NJ NS NLC 3.000E+07 16 13 9 1 The Information of Members member start end A I 1 1 2 1.600000E-01 2.130000E-03 2 2 4 2.975000E-01 1.790000E-02 3 3 4 1.600000E-01 2.130000E-03 4 4 6 2.100000E-01 6.300000E-03 5 5 6 1.600000E-01 2.130000E-03 6 2 7 1.600000E-01 2.130000E-03 7 7 8 2.975000E-01 1.790000E-02 8 4 8 1.600000E-01 2.130000E-03 9 8 9 2.100000E-01 6.300000E-03 10 6 9 1.600000E-01 2.130000E-03 11 7 10 1.600000E-01 2.130000E-03 12 10 11 2.975000E-01 1.790000E-02 13 11 12 2.975000E-01 1.790000E-02 14 8 12 1.600000E-01 2.130000E-03 15 12 13 2.100000E-01 6.300000E-03 16 9 13 1.600000E-01 2.130000E-03 The Joint Coordinates joint X Y 1 .000000 .000000 2 .000000 4.500000 3 7.600000 .000000 4 7.600000 4.500000 5 11.400000 .000000 6 11.400000 4.500000 7 .000000 7.700000 8 7.600000 7.700000 9 11.400000 7.700000 10 .000000 10.900000 11 3.800000 10.900000 12 7.600000 10.900000 13 11.400000 10.900000 The Information of Supports IS VS 11 .000000 12 .000000 13 .000000 31 .000000 32 .000000 33 .000000 51 .000000 52 .000000 53 .000000 Loading Case 1 The Loadings at Joints NLJ= 4 joint FX FY FM 7 100.000000 .000000 .000000 10 100.000000 .000000 .000000 11 .000000 .000000 -15.000000 12 .000000 .000000 -15.000000 The Loadings at Members NLM= 7 member type VF DST 1 3 20.000000 4.500000 2 2 -26.000000 3.800000 2 4 -36.000000 7.600000 4 4 -36.000000 3.800000 4 2 -26.000000 2.700000 7 4 -36.000000 7.600000 9 4 -36.000000 3.800000 The Results of Calculation The Joint Displacements joint u v rotation 1 7.105461E-21 -1.638202E-20 -1.781187E-20 2 1.133506E-02 -1.535815E-04 -1.284918E-03 3 9.610091E-21 -4.106481E-20 -2.156737E-20 4 1.133276E-02 -3.849826E-04 3.897026E-05 5 7.784447E-21 -2.983317E-20 -1.882111E-20 6 1.132041E-02 -2.796859E-04 -9.197894E-04 7 1.610528E-02 -2.069985E-04 -9.285745E-04 8 1.603101E-02 -5.147110E-04 6.617797E-05 9 1.601517E-02 -3.701405E-04 -4.822142E-04 10 1.847558E-02 -1.982752E-04 -1.263834E-04 11 1.843980E-02 -3.424985E-04 -8.166131E-06 12 1.840402E-02 -5.043449E-04 -1.356586E-04 13 1.838526E-02 -3.892299E-04 -1.683735E-04 The Terminal Forces member FN FS M 1 start 1 163.820232 116.054614 211.868721 end 2 -163.820232 -26.054614 107.877041 2 start 2 2.696605 83.694719 -146.763245 end 4 -2.696605 215.905278 -355.636874 3 start 3 410.648087 96.100915 215.673680 end 4 -410.648087 -96.100915 216.780436 4 start 4 20.475884 .150115 -42.808557 end 6 -20.475884 162.649883 -245.140956 5 start 5 298.331671 77.844472 188.211070 end 6 -298.331671 -77.844472 162.089052 6 start 2 80.125514 28.751219 38.886205 end 7 -80.125514 -28.751219 53.117689 7 start 7 87.216625 93.210427 -62.647484 end 8 -87.216625 180.389569 -268.633252 8 start 4 194.592693 113.880194 181.664996 end 8 -194.592693 -113.880194 182.751603 9 start 8 26.265385 29.752419 -2.835098 end 9 -26.265385 107.047579 -144.025679 10 start 6 135.681788 57.368587 83.051904 end 9 -135.681788 -57.368587 100.527564 11 start 7 -13.084913 15.967843 9.529794 end 10 13.084913 -15.967843 41.567301 12 start 10 84.032157 -13.084913 -41.567301 end 11 -84.032157 13.084913 -8.155369 13 start 11 84.032157 -13.084913 -6.844631 end 12 -84.032157 13.084913 -42.878040 14 start 8 -15.549295 52.928954 88.716747 end 12 15.549295 -52.928954 80.655897 15 start 12 31.103202 -28.634209 -52.777857 end 13 -31.103202 28.634209 -56.032127 16 start 9 28.634209 31.103202 43.498115 end 13 -28.634209 -31.103202 56.032127（4）根据得到的数据，绘制内力图如下：FN图：Fs图：M图：2、计算图示桁架各杆的轴力。已知A=2400mm2,E=2.0105 MPa。解：（1）编号：（2） 建立输入文件timu2.txt，存于PF同一文件夹内。注意到，由于题目中将桁架结点简化为铰结，故各杆只受轴力而不受剪力和弯矩，故将I假设尽可能小（1E-18）。具体输入如下：* * EX.2 A Frame with Hinges 2013.06.08 * *2E8 14 9 4 11 2 24E-4 1E-181 3 24E-4 1E-18 2 3 24E-4 1E-182 4 24E-4 1E-183 4 24E-4 1E-183 5 24E-4 1E-184 5 24E-4 1E-185 6 24E-4 1E-185 7 24E-4 1E-187 6 24E-4 1E-186 8 24E-4 1E-187 8 24E-4 1E-187 9 24E-4 1E-189 8 24E-4 1E-180 00 62 32 64 66 66 38 68 011 012 091 092 052 0 -50 04 0 -50 05 0 -50 06 0 -50 08 -10 -50 00（3） 运行PF程序,得到如下的timu2out.txt输出文件：Input Data File Name: timu2.txt Output File Name: timu2out.txt * * * * EX.2 A Frame with Hinges 2013.06.08 * * * * The Input Data The General Information E NM NJ NS NLC 2.000E+08 14 9 4 1 The Information of Members member start end A I 1 1 2 2.400000E-03 1.000000E-18 2 1 3 2.400000E-03 1.000000E-18 3 2 3 2.400000E-03 1.000000E-18 4 2 4 2.400000E-03 1.000000E-18 5 3 4 2.400000E-03 1.000000E-18 6 3 5 2.400000E-03 1.000000E-18 7 4 5 2.400000E-03 1.000000E-18 8 5 6 2.400000E-03 1.000000E-18 9 5 7 2.400000E-03 1.000000E-18 10 7 6 2.400000E-03 1.000000E-18 11 6 8 2.400000E-03 1.000000E-18 12 7 8 2.400000E-03 1.000000E-18 13 7 9 2.400000E-03 1.000000E-18 14 9 8 2.400000E-03 1.000000E-18 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 6.000000 7 6.000000 3.000000 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 .000000 -50.000000 .000000 4 .000000 -50.000000 .000000 5 .000000 -50.000000 .000000 6 .000000 -50.000000 .000000 8 -10.000000 -50.000000 .000000 The Loadings at Members NLM= 0 The Results of Calculation The Joint Displacements joint u v rotation 1 -3.833333E-21 -1.325000E-20 -2.257495E-04 2 -1.052370E-04 -9.375000E-04 -7.026908E-05 3 3.860368E-04 -8.812351E-04 1.226822E-04 4 -1.746814E-04 -1.193735E-03 1.087908E-04 5 -2.441259E-04 -8.137529E-04 -3.230398E-05 6 -3.552370E-04 -1.302888E-03 -1.022944E-04 7 -7.938923E-04 -9.903881E-04 -6.211765E-05 8 -4.663481E-04 -9.375000E-04 1.412286E-04 9 2.833333E-21 -1.175000E-20 3.510751E-04 The Terminal Forces member FN FS M 1 start 1 75.000000 .000000 .000000 end 2 -75.000000 .000000 .000000 2 start 1 69.106399 .000000 .000000 end 3 -69.106399 .000000 .000000 3 start 2 -30.046261 .000000 .000000 end 3 30.046261 .000000 .000000 4 start 2 16.666667 .000000 .000000 end 4 -16.666667 .000000 .000000 5 start 3 50.000000 .000000 .000000 end 4 -50.000000 .000000 .000000 6 start 3 39.060139 .000000 .000000 end 5 -39.060139 .000000 .000000 7 start 4 16.666667 .000000 .000000 end 5 -16.666667 .000000 .000000 8 start 5 26.666667 .000000 .000000 end 6 -26.666667 .000000 .000000 9 start 5 21.032382 .000000 .000000 end 7 -21.032382 .000000 .000000 10 start 7 50.000000 .000000 .000000 end 6 -50.000000 .000000 .000000 11 start 6 26.666667 .000000 .000000 end 8 -26.666667 .000000 .000000 12 start 7 -30.046261 .000000 .000000 end 8 30.046261 .000000 .000000 13 start 7 51.078643 .000000 .000000 end 9 -51.078643 .000000 .000000 14 start 9 75.000000 .000000 .000000 end 8 -75.000000 .000000 .000000（4） 根据得到的数据，绘制内力图，由于桁架只受轴力，故只画出轴力图如下：FN图：3作图示连续梁的、图，已知各梁截面面积A=6.5,惯性矩I=5.50，各杆E=3.45104MPa。解：（1）编号：根据刚架进行编号。注意到作用在杆上的集中力偶由于PF程序未直接处理的算法，故在此点增加一个铰节点3。（2） 建立输入文件timu3.txt，存于PF同一文件夹内。具体输入如下：* * EX.3 A Simple Suppoted Frame 2013.06.08 * *345E5 4 5 6 11 2 6.5 5.52 3 6.5 5.5 3 4 6.5 5.54 5 6.5 5.50 040 060 080 0120 011 012 013 022 042 052 013 0 -320 -10041 4 -10.5 402 4 -10.5 203 4 -10.5 204 4 -10.5 40（3） 运行PF程序,得到如下的timu3out.txt输出文件： Input Data File Name: timu3.txt Output File Name: timu3out.txt * * * * EX.3 A Simple Suppoted Frame 2013.06.08 * * * * The Input Data The General Information E NM NJ NS NLC 3.450E+07 4 5 6 1 The Information of Members member start end A I 1 1 2 6.500000E+00 5.500000E+00 2 2 3 6.500000E+00 5.500000E+00 3 3 4 6.500000E+00 5.500000E+00 4 4 5 6.500000E+00 5.500000E+00 The Joint Coordinates joint X Y 1 .000000 .000000 2 40.000000 .000000 3 60.000000 .000000 4 80.000000 .000000 5 120.000000 .000000 The Information of Supports IS VS 11 .000000 12 .000000 13 .000000 22 .000000 42 .000000 52 .000000 Loading Case 1 The Loadings at Joints NLJ= 1 joint FX FY FM 3 .000000 -320.000000 -100.000000 The Loadings at Members NLM= 4 member type VF DST 1 4 -10.500000 40.000000 2 4 -10.500000 20.000000 3 4 -10.500000 20.000000 4 4 -10.500000 40.000000 The Results of Calculation The Joint Displacements joint u v rotation 1 0.000000E+00 3.713942E-21 4.951923E-20 2 0.000000E+00 -2.916827E-20 -5.219418E-05 3 0.000000E+00 -1.405865E-03 1.038816E-06 4 0.000000E+00 -3.431731E-20 4.276883E-05 5 0.000000E+00 6.771635E-2