结构力学矩阵位移法实践报告.

1、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

2、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FN、FS、M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=3.0 104 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 Frame2013.06.08*3E71613911216E-2213E-5242975E-

3、4 179E-43416E-2213E-54621E-263E-45616E-2213E-52716E-2213E-5782975E-4 179E-44816E-2213E-58921E-263E-46916E-2213E-5710 16E-2213E-510 11 2975E-4 179E-411 12 2975E-4 179E-4812 16E-2213E-512 13 21E-263E-4913 16E-2213E-50 00 4.57.6 07.6 4.511.4 011.4 4.50 7.77.6 7.711.4 7.70 10.93.8 10.97.6 10.911.4 10.91

4、1 012 013 031 032 033 051 052 053 047 1000010 100001100-151200-1571 320 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.txtOutput File Name: timu1out.txt*EX.1A Simple Suppoted Frame2013.06.08*The Input DataThe General In

5、formationENMNJNSNLC3.000E+07161391The Information of MembersmemberstartendAI1121.600000E-012.130000E-032242.975000E-011.790000E-023341.600000E-012.130000E-034462.100000E-016.300000E-035561.600000E-012.130000E-036271.600000E-012.130000E-037782.975000E-011.790000E-028481.600000E-012.130000E-039892.100

6、000E-016.300000E-0310691.600000E-012.130000E-03117101.600000E-012.130000E-031210112.975000E-011.790000E-021311122.975000E-011.790000E-02148121.600000E-012.130000E-031512132.100000E-016.300000E-03169131.600000E-012.130000E-03The Joint CoordinatesjointXY1.000000.0000002.0000004.50000037.600000.0000004

7、7.6000004.500000511.400000.000000611.4000004.5000007.0000007.70000087.6000007.700000911.4000007.70000010.00000010.900000113.80000010.900000127.60000010.9000001311.40000010.900000The Information of SupportsISVS11.00000012.00000013.00000031.00000032.00000033.00000051.00000052.00000053.000000Loading Ca

8、se1The Loadings at JointsNLJ=4jointFXFYFM7100.000000.000000.00000010100.000000.000000.00000011.000000.000000-15.00000012.000000.000000-15.000000The Loadings at MembersNLM=7membertypeVFDST1320.0000004.50000022-26.0000003.80000024-36.0000007.60000044-36.0000003.80000042-26.0000002.70000074-36.0000007.

9、60000094-36.0000003.800000The Results of CalculationThe Joint Displacementsjointuvrotation17.105461E-21-1.638202E-20-1.781187E-2021.133506E-02-1.535815E-04-1.284918E-0339.610091E-21-4.106481E-20-2.156737E-2041.133276E-02-3.849826E-043.897026E-0557.784447E-21-2.983317E-20-1.882111E-2061.132041E-02-2.

10、796859E-04-9.197894E-0471.610528E-02-2.069985E-04-9.285745E-0481.603101E-02-5.147110E-046.617797E-0591.601517E-02-3.701405E-04-4.822142E-04101.847558E-02-1.982752E-04-1.263834E-04111.843980E-02-3.424985E-04-8.166131E-06121.840402E-02-5.043449E-04-1.356586E-04131.838526E-02-3.892299E-04-1.683735E-04T

11、he Terminal ForcesmemberFNFSM1start1163.820232116.054614211.868721end2-163.820232-26.054614107.8770412start22.69660583.694719-146.763245end4-2.696605215.905278-355.6368743start3410.64808796.100915215.673680end4-410.648087-96.100915216.7804364start420.475884.150115-42.808557end6-20.475884162.649883-2

12、45.1409565start5298.33167177.844472188.211070end6-298.331671-77.844472162.0890526start280.12551428.75121938.886205end7-80.125514-28.75121953.1176897start787.21662593.210427-62.647484end8-87.216625180.389569-268.6332528start4194.592693113.880194181.664996end8-194.592693-113.880194182.7516039start826.

13、26538529.752419-2.835098end9-26.265385107.047579-144.02567910start6135.68178857.36858783.051904end9-135.681788-57.368587100.52756411start7-13.08491315.9678439.529794end1013.084913-15.96784341.56730112start1084.032157-13.084913-41.567301end11-84.03215713.084913-8.15536913start1184.032157-13.084913-6.

14、844631end12-84.03215713.084913-42.87804014start8-15.54929552.92895488.716747end1215.549295-52.92895480.65589715start1231.103202-28.634209-52.777857end13-31.10320228.634209-56.03212716start928.63420931.10320243.498115end13-28.634209-31.10320256.032127（ 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

15、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.2A Frame with Hinges2013.06.08*2E8149411 2 24E-4 1E-181 3 24E-4 1E-182 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

16、 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 0-5004 0-5005 0-5006 0-5008 -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.txtOutput File Name: timu2out.txt*EX.2A Frame with Hinges2013.0

17、6.08*The Input DataThe General InformationENMNJNSNLC2.000E+0814941The Information of Membersmember start endAI1122.400000E-031.000000E-182132.400000E-031.000000E-183232.400000E-031.000000E-184242.400000E-031.000000E-185342.400000E-031.000000E-186352.400000E-031.000000E-187452.400000E-031.000000E-188

18、562.400000E-031.000000E-189 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-18The Joint CoordinatesjointXY1.000000.0000002.0000006.00000032.0000003.00000042

19、.0000006.00000054.0000006.00000066.0000006.00000076.0000003.00000088.0000006.00000098.000000.000000The Information of SupportsISVS11 .00000012 .00000091 .00000092 .000000Loading Case 1The Loadings at JointsNLJ= 5jointFXFYFM2.000000-50.000000.0000004.000000-50.000000.0000005.000000-50.000000.0000006.

20、000000-50.000000.0000008-10.000000-50.000000.000000The Loadings at MembersNLM=0The Results of CalculationThe Joint Displacementsjointuvrotation1-3.833333E-21-1.325000E-20 -2.257495E-042-1.052370E-04 -9.375000E-04 -7.026908E-0533.860368E-04-8.812351E-041.226822E-044-1.746814E-04-1.193735E-031.087908E

21、-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-041.412286E-0492.833333E-21-1.175000E-203.510751E-04The Terminal ForcesmemberFNFSM1start175.000000.000000.000000end2-75.000000.000000.0000002start169

22、.106399.000000.000000end3-69.106399.000000.0000003start2-30.046261.000000.000000end330.046261.000000.0000004start216.666667.000000.000000end4-16.666667.000000.0000005start350.000000.000000.000000end4-50.000000.000000.0000006start339.060139.000000.000000end5-39.060139.000000.0000007start416.666667.00

23、0000.000000end5-16.666667.000000.0000008start526.666667.000000.000000end6-26.666667.000000.0000009start521.032382.000000.000000end7-21.032382.000000.00000010start750.000000.000000.000000end6-50.000000.000000.00000011start626.666667.000000.000000end8-26.666667.000000.00000012start7-30.046261.000000.0

24、00000end830.046261.000000.00000013start751.078643.000000.000000end9-51.078643.000000.00000014start975.000000.000000.000000end8-75.000000.000000.000000（ 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 FS 、 M 图，已知各梁截面面积 A=6.5 m2 ,惯性矩 I=5.50 m 4 ，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

25、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.3A Simple Suppoted Frame2013.06.08*345E545611 2 6.5 5.52 3 6.5 5.53 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

26、件：Input Data File Name: timu3.txtOutput File Name: timu3out.txt*EX.3A Simple Suppoted Frame2013.06.08*The Input DataThe General InformationENMNJNSNLC3.450E+074561The Information of Membersmember start endAI1126.500000E+005.500000E+002236.500000E+005.500000E+003346.500000E+005.500000E+004 4 5 6.50000

27、0E+00 5.500000E+00 The Joint CoordinatesjointXY1.000000.000000240.000000.000000360.000000.000000480.000000.0000005120.000000.000000The Information of SupportsISVS11 .00000012 .00000013 .00000022 .00000042 .00000052 .000000Loading Case 1The Loadings at JointsNLJ=1jointFXFYFM3.000000-320.000000-100.00

28、0000The Loadings at MembersNLM=4member typeVFDST14-10.50000040.00000024-10.50000020.00000034-10.50000020.00000044-10.50000040.000000The Results of CalculationThe Joint Displacementsjoint1u v 0.000000E+00rotation3.713942E-214.951923E-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 4.276883E-055 0.000000E+00 6.771635E-21 5.239688E-05The Terminal ForcesmemberFNFSM1start1.000000172.860577904.807692end2