结构力学矩阵位移法上机实验报告学院土木工程学院导师肖方红班级土木1415姓名张帅浩学号1201140929一、实验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、实验要求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、实验步骤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。

3.运行计算：运行程序，得到输出文件。

4.绘制弯矩图。

四、实验内容1刚架1.1.题目F、S F、M图，已知各杆截面均为矩形，柱截面宽0.4m,高作图示刚架的N0.5m, 梁截面宽0.3m,高0.5m，各杆E=3.25×104 MPa。

1.2.单元划分及坐标系1.3.建立输入文件3.250E+07 8 8 9 11 2 2.000000E-01 4.170000E-032 3 2.000000E-01 4.170000E-032 5 1.500000E-01 3.130000E-033 4 1.500000E-01 3.130000E-034 5 2.000000E-01 4.170000E-035 6 2.000000E-01 4.170000E-035 7 1.500000E-01 3.130000E-037 8 2.000000E-01 4.170000E-030 00 80 1610 1610 810 020 820 011 012 013 061 062 063 081 082 083 023 0 0 -1004 50 0 051 3 20 82 3 20 83 2 -150 54 2 -100 57 4 -10 51.4.运行程序算题，查看结果文件Input Data File Name: Input1.txtOutput File Name: Output1.txtThe Input DataThe General InformationE NM NJ NS NLC3.250E+07 8 8 9 1The Information of Membersmember start end A I1 12 2.000000E-01 4.170000E-032 23 2.000000E-01 4.170000E-033 2 5 1.500000E-01 3.130000E-034 3 4 1.500000E-01 3.130000E-035 4 5 2.000000E-01 4.170000E-036 5 6 2.000000E-01 4.170000E-037 5 7 1.500000E-01 3.130000E-038 7 8 2.000000E-01 4.170000E-03The Joint Coordinatesjoint X Y1 .000000 .0000002 .000000 8.0000003 .000000 16.0000004 10.000000 16.0000005 10.000000 8.0000006 10.000000 .0000007 20.000000 8.0000008 20.000000 .000000The Information of SupportsIS VS11 .00000012 .00000013 .00000061 .00000062 .00000063 .00000081 .00000082 .00000083 .000000Loading Case 1The Loadings at JointsNLJ= 2joint FX FY FM3 .000000 .000000 -100.0000004 50.000000 .000000 .000000The Loadings at MembersNLM= 5member type VF DST1 3 20.000000 8.0000002 3 20.000000 8.0000003 2 -150.000000 5.0000004 2 -100.000000 5.0000007 4 -10.000000 5.000000The Results of CalculationThe Joint Displacementsjoint u v rotation1 7.005331E-21 7.924346E-22 -4.170788E-202 5.437108E-02 9.753041E-06 -8.079136E-033 1.186122E-01 1.800326E-05 -5.365114E-034 1.185162E-01 -4.256343E-04 -4.013906E-035 5.411839E-02 -2.943072E-04 -4.469763E-036 1.151094E-20 -2.391246E-20 -5.361584E-207 5.390334E-02 -8.467663E-05 -5.224488E-038 1.048372E-20 -6.879976E-21 -5.078551E-20The Terminal Forcesmember FN FS M1 start 1 -7.924346 150.053308 523.745504end 2 7.924346 9.946692 36.6809562 start 2 -6.703306 113.237784 193.640586end 3 6.703306 46.762216 72.2616893 start 2 123.184477 -1.221041 -230.321541end 5 -123.184477 151.221041 -531.8888664 start 3 46.762216 -6.703306 -172.261689end 4 -46.762216 106.703306 -394.7713665 start 4 106.703306 96.762216 394.771366end 5 -106.703306 -96.762216 379.3263596 start 5 239.124587 115.109448 384.717203end 6 -239.124587 -115.109448 536.1583817 start 5 104.837245 -18.799759 -232.154697end 7 -104.837245 68.799759 -330.8428928 start 7 68.799759 104.837245 330.842892end 8 -68.799759 -104.837245 507.8550641.5.绘制内力图轴力图F N剪力图F S弯矩图M2.桁架2.1.题目计算图示桁架各杆的轴力。

已知A=500mm2, E=2.1×105 MPa。

2.2.单元划分及坐标系2.3.建立输入文件2.10E+0813812112 5.00E-040.00E+0013 5.00E-040.00E+0023 5.00E-040.00E+0024 5.00E-040.00E+0025 5.00E-040.00E+00 34 5.00E-040.00E+0045 5.00E-040.00E+0046 5.00E-040.00E+0047 5.00E-040.00E+00 56 5.00E-040.00E+0067 5.00E-040.00E+0068 5.00E-040.00E+00 78 5.00E-040.00E+00005055105100150155200110120130230330430520530630730820830330-30040-400750-5002.4.运行程序算题，查看结果文件Input Data File Name: Input2.txtOutput File Name: Output2.txtThe Input DataThe General InformationE NM NJ NS NLC2.100E+08 13 8 12 1The Information of Membersmember start end A I1 12 5.000000E-04 0.000000E+002 13 5.000000E-04 0.000000E+003 2 3 5.000000E-04 0.000000E+004 2 4 5.000000E-04 0.000000E+005 2 5 5.000000E-04 0.000000E+006 3 4 5.000000E-04 0.000000E+007 4 5 5.000000E-04 0.000000E+008 4 6 5.000000E-04 0.000000E+009 4 7 5.000000E-04 0.000000E+0010 5 6 5.000000E-04 0.000000E+0011 6 7 5.000000E-04 0.000000E+0012 6 8 5.000000E-04 0.000000E+0013 7 8 5.000000E-04 0.000000E+00The Joint Coordinatesjoint X Y1 .000000 .0000002 5.000000 .0000003 5.000000 5.0000004 10.000000 5.0000005 10.000000 .0000006 15.000000 .0000007 15.000000 5.0000008 20.000000 .000000The Information of SupportsIS VS11 .00000012 .00000013 .00000023 .00000033 .00000043 .00000052 .00000053 .00000063 .00000073 .00000082 .00000083 .000000Loading Case 1The Loadings at JointsNLJ= 3joint FX FY FM3 .000000 -30.000000 .0000004 .000000 -40.000000 .0000007 50.000000 -50.000000 .000000The Loadings at MembersNLM= 0The Results of CalculationThe Joint Displacementsjoint u v rotation1 5.000000E-21 -3.182275E-22 0.000000E+002 2.532489E-03 -4.596073E-03 0.000000E+003 5.444497E-03 -5.873108E-03 0.000000E+004 5.292960E-03 -3.744545E-03 0.000000E+005 3.787944E-03 -7.863545E-21 0.000000E+006 5.043398E-03 -5.585800E-03 0.000000E+007 5.855709E-03 -6.148549E-03 0.000000E+008 6.861602E-03 -3.818227E-21 0.000000E+00The Terminal Forcesmember FN FS M1 start 1 -53.182275 .000000 .000000end 2 53.182275 .000000 .0000002 start 1 4.500416 .000000 .000000end 3 -4.500416 .000000 .0000003 start 2 26.817725 .000000 .000000end 3 -26.817725 .000000 .0000004 start 2 -37.925991 .000000 .000000end 4 37.925991 .000000 .0000005 start 2 -26.364549 .000000 .000000end 5 26.364549 .000000 .0000006 start 3 3.182275 .000000 .000000end 4 -3.182275 .000000 .0000007 start 4 78.635451 .000000 .000000end 5 -78.635451 .000000 .0000008 start 4 -16.712787 .000000 .000000end 6 16.712787 .000000 .0000009 start 4 -11.817725 .000000 .000000end 7 11.817725 .000000 .00000010 start 5 -26.364549 .000000 .000000end 6 26.364549 .000000 .00000011 start 6 11.817725 .000000 .000000end 7 -11.817725 .000000 .00000012 start 6 -38.182275 .000000 .000000end 8 38.182275 .000000 .00000013 start 7 53.997891 .000000 .000000end 8 -53.997891 .000000 .000000 2.5.绘制内力图轴力图F N3.连续梁3.1.题目F、M图，已知各杆截面均为矩形，截面宽0.3m, 高0.5m，作图示连续梁S各杆E=3.25×104MPa。