结构力学矩阵位移法实践报告.doc

/结结结构构构力力力学学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矩阵位移法上机实践报告一、实践目的一、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NFSFM0.4m, 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=3.0×104 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 1 1 2 16E-2 213E-5 2 4 2975E-4 179E-4 3 4 16E-2 213E-5 4 6 21E-2 63E-4 5 6 16E-2 213E-5 2 7 16E-2 213E-5 7 8 2975E-4 179E-4 4 8 16E-2 213E-5 8 9 21E-2 63E-4 6 9 16E-2 213E-5 7 10 16E-2 213E-5 10 11 2975E-4 179E-4 11 12 2975E-4 179E-4 8 12 16E-2 213E-5 12 13 21E-2 63E-4 9 13 16E-2 213E-5 0 0 0 4.5 7.6 0 7.6 4.5 11.4 0 11.4 4.5 0 7.7 7.6 7.7 11.4 7.7 0 10.9 3.8 10.9 7.6 10.9 11.4 10.9 11 0 12 0 13 0 31 0 32 0 33 0 51 0 52 0 53 0/4 7 100 0 0 10 100 0 0 11 0 0 -15 12 0 0 -15 7 1 3 20 4.5 2 2 -26 3.8 2 4 -36 7.6 4 4 -36 3.8 4 2 -26 2.7 7 4 -36 7.6 9 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 DataThe General InformationE NM NJ NS NLC3.000E+07 16 13 9 1The Information of Membersmember start end A I1 1 2 1.600000E-01 2.130000E-032 2 4 2.975000E-01 1.790000E-023 3 4 1.600000E-01 2.130000E-034 4 6 2.100000E-01 6.300000E-035 5 6 1.600000E-01 2.130000E-036 2 7 1.600000E-01 2.130000E-037 7 8 2.975000E-01 1.790000E-028 4 8 1.600000E-01 2.130000E-039 8 9 2.100000E-01 6.300000E-0310 6 9 1.600000E-01 2.130000E-0311 7 10 1.600000E-01 2.130000E-0312 10 11 2.975000E-01 1.790000E-02/13 11 12 2.975000E-01 1.790000E-0214 8 12 1.600000E-01 2.130000E-0315 12 13 2.100000E-01 6.300000E-0316 9 13 1.600000E-01 2.130000E-03The Joint Coordinatesjoint X Y1 .000000 .0000002 .000000 4.5000003 7.600000 .0000004 7.600000 4.5000005 11.400000 .0000006 11.400000 4.5000007 .000000 7.7000008 7.600000 7.7000009 11.400000 7.70000010 .000000 10.90000011 3.800000 10.90000012 7.600000 10.90000013 11.400000 10.900000The Information of SupportsIS VS11 .00000012 .00000013 .00000031 .00000032 .00000033 .00000051 .00000052 .00000053 .000000Loading Case 1The Loadings at JointsNLJ= 4joint FX FY FM7 100.000000 .000000 .00000010 100.000000 .000000 .000000/11 .000000 .000000 -15.00000012 .000000 .000000 -15.000000The Loadings at MembersNLM= 7member type VF DST1 3 20.000000 4.5000002 2 -26.000000 3.8000002 4 -36.000000 7.6000004 4 -36.000000 3.8000004 2 -26.000000 2.7000007 4 -36.000000 7.6000009 4 -36.000000 3.800000The Results of CalculationThe Joint Displacementsjoint u v rotation1 7.105461E-21 -1.638202E-20 -1.781187E-202 1.133506E-02 -1.535815E-04 -1.284918E-033 9.610091E-21 -4.106481E-20 -2.156737E-204 1.133276E-02 -3.849826E-04 3.897026E-055 7.784447E-21 -2.983317E-20 -1.882111E-206 1.132041E-02 -2.796859E-04 -9.197894E-047 1.610528E-02 -2.069985E-04 -9.285745E-048 1.603101E-02 -5.147110E-04 6.617797E-059 1.601517E-02 -3.701405E-04 -4.822142E-0410 1.847558E-02 -1.982752E-04 -1.263834E-0411 1.843980E-02 -3.424985E-04 -8.166131E-0612 1.840402E-02 -5.043449E-04 -1.356586E-0413 1.838526E-02 -3.892299E-04 -1.683735E-04The Terminal Forcesmember FN FS M1 start 1 163.820232 116.054614 211.868721end 2 -163.820232 -26.054614 107.8770412 start 2 2.696605 83.694719 -146.763245end 4 -2.696605 215.905278 -355.6368743 start 3 410.648087 96.100915 215.673680end 4 -410.648087 -96.100915 216.780436/4 start 4 20.475884 .150115 -42.808557end 6 -20.475884 162.649883 -245.1409565 start 5 298.331671 77.844472 188.211070end 6 -298.331671 -77.844472 162.0890526 start 2 80.125514 28.751219 38.886205end 7 -80.125514 -28.751219 53.1176897 start 7 87.216625 93.210427 -62.647484end 8 -87.216625 180.389569 -268.6332528 start 4 194.592693 113.880194 181.664996end 8 -194.592693 -113.880194 182.7516039 start 8 26.265385 29.752419 -2.835098end 9 -26.265385 107.047579 -144.02567910 start 6 135.681788 57.368587 83.051904end 9 -135.681788 -57.368587 100.52756411 start 7 -13.084913 15.967843 9.529794end 10 13.084913 -15.967843 41.56730112 start 10 84.032157 -13.084913 -41.567301end 11 -84.032157 13.084913 -8.15536913 start 11 84.032157 -13.084913 -6.844631end 12 -84.032157 13.084913 -42.87804014 start 8 -15.549295 52.928954 88.716747end 12 15.549295 -52.928954 80.65589715 start 12 31.103202 -28.634209 -52.777857end 13 -31.103202 28.634209 -56.03212716 start 9 28.634209 31.103202 43.498115end 13 -28.634209 -31.103202 56.032127 （4）根据得到的数据，绘制内力图如下： FN图：图：/Fs 图：图：/M 图：图：2、计算图示桁架各杆的轴力。已知 A=2400mm2,E=2.0×105 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 1 1 2 24E-4 1E-18 1 3 24E-4 1E-18 2 3 24E-4 1E-18 2 4 24E-4 1E-18 3 4 24E-4 1E-18 3 5 24E-4 1E-18 4 5 24E-4 1E-18 5 6 24E-4 1E-18 5 7 24E-4 1E-18 7 6 24E-4 1E-18 6 8 24E-4 1E-18 7 8 24E-4 1E-18 7 9 24E-4 1E-18/9 8 24E-4 1E-18 0 0 0 6 2 3 2 6 4 6 6 6 6 3 8 6 8 0 11 0 12 0 91 0 92 0 5 2 0 -50 0 4 0 -50 0 5 0 -50 0 6 0 -50 0 8 -10 -50 0 0 （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 DataThe General InformationE NM NJ NS NLC2.000E+08 14 9 4 1The Information of Membersmember start end A I1 1 2 2.400000E-03 1.000000E-182 1 3 2.400000E-03 1.000000E-183 2 3 2.400000E-03 1.000000E-184 2 4 2.400000E-03 1.000000E-18/5 3 4 2.400000E-03 1.000000E-186 3 5 2.400000E-03 1.000000E-187 4 5 2.400000E-03 1.000000E-188 5 6 2.400000E-03 1.000000E-189 5 7 2.400000E-03 1.000000E-1810 7 6 2.400000E-03 1.000000E-1811 6 8 2.400000E-03 1.000000E-1812 7 8 2.400000E-03 1.000000E-1813 7 9 2.400000E-03 1.000000E-1814 9 8 2.400000E-03 1.000000E-18The Joint Coordinatesjoint X Y1 .000000 .0000002 .000000 6.0000003 2.000000 3.0000004 2.000000 6.0000005 4.000000 6.0000006 6.000000 6.0000007 6.000000 3.0000008 8.000000 6.0000009 8.000000 .000000The Information of SupportsIS VS11 .00000012 .00000091 .00000092 .000000Loading Case 1The Loadings at JointsNLJ= 5joint FX FY FM2 .000000 -50.000000 .0000004 .000000 -50.000000 .0000005 .000000 -50.000000 .0000006 .000000 -50.000000 .0000008 -10.000000 -50.000000 .000000/The Loadings at MembersNLM= 0The Results of CalculationThe Joint Displacementsjoint u v rotation1 -3.833333E-21 -1.325000E-20 -2.257495E-042 -1.052370E-04 -9.375000E-04 -7.026908E-053 3.860368E-04 -8.812351E-04 1.226822E-044 -1.746814E-04 -1.193735E-03 1.087908E-045 -2.441259E-04 -8.137529E-04 -3.230398E-056 -3.552370E-04 -1.302888E-03 -1.022944E-047 -7.938923E-04 -9.903881E-04 -6.211765E-058 -4.663481E-04 -9.375000E-04 1.412286E-049 2.833333E-21 -1.175000E-20 3.510751E-04The Terminal Forcesmember FN FS M1 start 1 75.000000 .000000 .000000end 2 -75.000000 .000000 .0000002 start 1 69.106399 .000000 .000000end 3 -69.106399 .000000 .0000003 start 2 -30.046261 .000000 .000000end 3 30.046261 .000000 .0000004 start 2 16.666667 .000000 .000000end 4 -16.666667 .000000 .0000005 start 3 50.000000 .000000 .000000end 4 -50.000000 .000000 .0000006 start 3 39.060139 .000000 .000000end 5 -39.060139 .000000 .0000007 start 4 16.666667 .000000 .000000end 5 -16.666667 .000000 .0000008 start 5 26.666667 .000000 .000000end 6 -26.666667 .000000 .0000009 start 5 21.032382 .000000 .000000end 7 -21.032382 .000000 .00000010 start 7 50.000000 .000000 .000000end 6 -50.000000 .000000 .00000011 start 6 26.666667 .000000 .000000/end 8 -26.666667 .000000 .00000012 start 7 -30.046261 .000000 .000000end 8 30.046261 .000000 .00000013 start 7 51.078643 .000000 .000000end 9 -51.078643 .000000 .00000014 start 9 75.000000 .000000 .000000end 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：图：3作图示连续梁的、图，已知各梁截面面积 A=6.5,惯性矩SFM2mI=5.50，各杆 E=3.45×104MPa。4m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 1 1 2 6.5 5.5 2 3 6.5 5.5 3 4 6.5 5.5 4 5 6.5 5.5 0 0 40 0 60 0 80 0 120 0 11 0 12 0 13 0 22 0 42 0 52 0 1 3 0 -320 -100 4 1 4 -10.5 40 2 4 -10.5 20 3 4 -10.5 20 4 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 DataThe General InformationE NM NJ NS NLC3.450E+07 4 5 6 1The Information of Membersmember start end A I1 1 2 6.500000E+00 5.500000E+002 2 3 6.500000E+00 5.500000E+003 3 4 6.500000E+00 5.500000E+004 4 5 6.500000E+00 5.500000E+00The Joint Coordinatesjoint X Y1 .000000 .0000002 40.000000 .0000003 60.000000 .0000004 80.000000 .0000005 120.000000 .000000The Information of SupportsIS VS11 .00000012 .00000013 .00000022 .00000042 .00000052 .000000Loading Case 1The Loadings at JointsNLJ= 1joint FX FY FM/3 .000000 -320.000000 -100.000000The Loadings at MembersNLM= 4member type VF DST1 4 -10.500000 40.0000002 4 -10.500000 20.0000003 4 -10.500000 20.0000004 4 -10.500000 40.000000The Results of CalculationThe Joint Displacementsjoint u v rotation1 0.000000E+00 3.713942E-21 4.951923E-202 0.000000E+00 -2.916827E-20 -5.219418E-053 0.000000E+00 -1.405865E-03 1.038816E-064 0.000000E+00 -3.431731E-20 4