结构力学
矩阵位移法上机实验报告
指导老师： 肖方红
班 级：土木1109
姓 名： 钱小十
学 号：
2013年10月22日
目录
一、钢架的受力分析
1题目
2结构计算编号示意图
3输入文件
4输出文件
5结构受力分析图
…………………………………………………… 二桁架的受力分析
1题目
2结构计算编号示意图
3输入文件
4输出文件
5结构受力分析图
……………………………………………………
三、连续梁的受力分析
1题目
2结构计算编号示意图
3输入文件
4输出文件
5结构受力分析图
……………………………………………………
矩阵位移法上机实践报告
（一） 第一题
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0.4m,高0.9m，小跨梁截面宽0.4m,高0.6m，各杆E=3.0×104 MPa。10分
（1） 编号：如图所示
（2） 输入文件。
****************************************************
* *
* question one *
* *
****************************************************
3E7 18 15 12 1
1 2 1.6E-1 21333E-7
2 3 1.6E-1 21333E-7
3 4 1.6E-1 21333E-7
4 5 3.6E-1 243E-4
5 6 3.6E-1 243E-4
6 7 2.4E-1 72E-4
7 8 1.6E-1 21333E-7
8 9 1.6E-1 21333E-7
9 10 1.6E-1 21333E-7
11 12 1.6E-1 21333E-7
12 13 1.6E-1 21333E-7
13 6 1.6E-1 21333E-7
14 15 1.6E-1 21333E-7
2 15 3.6E-1 243E-4
15 12 3.6E-1 243E-4
3 13 3.6E-1 243E-4
13 8 2.4E-1 72E-4
12 9 2.4E-1 72E-4
0 0
0 4.5
0 7.7
0 10.9
3.8 10.9
7.6 10.9
11.4 10.9
11.4 7.7
11.4 4.5
11.4 0
7.6 0
7.6 4.5
7.6 7.7
3.8 0
3.8 4.5
11 0
12 0
13 0
101 0
102 0
103 0
111 0
112 0
113 0
141 0
142 0
143 0
3
4 100 0 0
5 0 0 -15
7 0 0 -15
8
1 3 20 4.5
14 4 -5 3.8
15 4 -5 3.8
16 4 -45 7.6
16 2 -26 3.8
17 4 -45 3.8
18 4 -5 3.8
18 2 -26 2.7
（3） 输出结果
Input Data File Name: D1.TXT
Output File Name: D2.TXT
****************************************************
* *
* question one *
* *
****************************************************
The Input Data
The General Information
E NM NJ NS NLC
3.000E+07 18 15 12 1
The Information of Members
member start end A I
1 1 2 1.600000E-01 2.133300E-03
2 2 3 1.600000E-01 2.133300E-03
3 3 4 1.600000E-01 2.133300E-03
4 4 5 3.600000E-01 2.430000E-02
5 5 6 3.600000E-01 2.430000E-02
6 6 7 2.400000E-01 7.200000E-03
7 7 8 1.600000E-01 2.133300E-03
8 8 9 1.600000E-01 2.133300E-03
9 9 10 1.600000E-01 2.133300E-03
10 11 12 1.600000E-01 2.133300E-03
11 12 13 1.600000E-01 2.133300E-03
12 13 6 1.600000E-01 2.133300E-03
13 14 15 1.600000E-01 2.133300E-03
14 2 15 3.600000E-01 2.430000E-02
15 15 12 3.600000E-01 2.430000E-02
16 3 13 3.600000E-01 2.430000E-02
17 13 8 2.400000E-01 7.200000E-03
18 12 9 2.400000E-01 7.200000E-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 11.400000 10.900000
8 11.400000 7.700000
9 11.400000 4.500000
10 11.400000 .000000
11 7.600000 .000000
12 7.600000 4.500000
13 7.600000 7.700000
14 3.800000 .000000
15 3.800000 4.500000
The Information of Supports
IS VS
11 .000000
12 .000000
13 .000000
101 .000000
102 .000000
103 .000000
111 .000000
112 .000000
113 .000000
141 .000000
142 .000000
143 .000000
Loading Case 1
The Loadings at Joints
NLJ= 3
joint FX FY FM
4 100.000000 .000000 .000000
5 .000000 .000000 -15.000000
7 .000000 .000000 -15.000000
The Loadings at Members
NLM= 8
member type VF DST
1 3 20.000000 4.500000
14 4 -5.000000 3.800000
15 4 -5.000000 3.800000
16 4 -45.000000 7.600000
16 2 -26.000000 3.800000
17 4 -45.000000 3.800000
18 4 -5.000000 3.800000
18 2 -26.000000 2.700000
The Results of Calculation
The Joint Displacements
joint u v rotation
1 3.743750E-21 -1.018849E-20 -8.548984E-21
2 4.640739E-03 -9.551710E-05 -8.827678E-05
3 6.808144E-03 -1.859027E-04 -8.681810E-04
4 8.912967E-03 -1.786969E-04 -1.000147E-04
5 8.882365E-03 -3.358049E-04 -1.835667E-05
6 8.851762E-03 -4.408383E-04 -7.260700E-05
7 8.835639E-03 -2.883267E-04 -1.906210E-04
8 6.809896E-03 -2.677740E-04 -2.607855E-04
9 4.636627E-03 -1.867505E-04 -3.035602E-04
10 3.332049E-21 -1.992005E-20 -7.928835E-21
11 3.561927E-21 -2.840118E-20 -8.274835E-21
12 4.638499E-03 -2.662610E-04 -1.831659E-04
13 6.809485E-03 -4.541852E-04 3.008896E-04
14 3.862274E-21 -3.690284E-21 -8.721767E-21
15 4.632824E-03 -3.459642E-05 -2.225539E-05
The Terminal Forces
member FN FS M
1 start 1 101.884900 82.437497 119.239840
end 2 -101.884900 7.562503 49.228895
2 start 2 135.578359 14.931158 39.487692
end 3 -135.578359 -14.931158 8.292010
3 start 3 -10.808671 13.024275 5.475752
end 4 10.808671 -13.024275 36.201924
4 start 4 86.975725 -10.808671 -36.201924
end 5 -86.975725 10.808671 -4.871024
5 start 5 86.975725 -10.808671 -10.128976
end 6 -86.975725 10.808671 -30.943972
6 start 6 30.550134 -30.829117 -51.867154
end 7 -30.550134 30.829117 -65.283480
7 start 7 30.829117 30.550134 50.283480
end 8 -30.829117 -30.550134 47.476942
8 start 8 121.535301 29.772570 48.491589
end 9 -121.535301 -29.772570 46.780630
9 start 9 199.200494 33.320495 70.653880
end 10 -199.200494 -33.320495 79.288346
10 start 11 284.011754 35.619272 82.748348
end 12 -284.011754 -35.619272 77.538377
11 start 12 281.886334 55.296272 78.793070
end 13 -281.886334 -55.296272 98.154990
12 start 13 -20.020446 56.425592 97.750757
end 6 20.020446 -56.425592 82.811125
13 start 14 36.902845 38.622736 87.217673
end 15 -36.902845 -38.622736 86.584641
14 start 2 22.493661 -33.693459 -88.716587
end 15 -22.493661 52.693459 -75.418555
15 start 15 -16.129075 -15.790614 -11.166086
end 12 16.129075 34.790614 -84.938245
16 start 3 -1.906883 146.387029 -13.767762
end 13 1.906883 221.612966 -272.090795
17 start 13 -.777563 80.293814 76.185049
end 8 .777563 90.706184 -95.968531
18 start 12 3.547924 -32.665194 -71.393203
end 9 -3.547924 77.665194 -117.434510
4） 绘图(由结构力学求解器得)
轴力图：单位（KN） （
剪力图：单位（KN）
弯矩图：单位（KN*M）
（二） 第二题
2、计算图示桁架各杆的轴力。已知A=2400mm2,E=2.0×105 MPa。5分
1）编号：（如图）
（
（2）输入文件
**************************************************** * * * question TWO * * * **************************************************** 2E8 15 9 4 1
1 2 3.6E-3 2E-8
2 3 3.6E-3 2E-8
3 4 3.6E-3 2E-8
4 5 3.6E-3 2E-8
5 6 3.6E-3 2E-8
6 7 3.6E-3 2E-8
7 8 3.6E-3 2E-8
8 9 3.6E-3 2E-8
1 9 3.6E-3 2E-8
2 9 3.6E-3 2E-8
3 9 3.6E-3 2E-8
4 9 3.6E-3 2E-8
4 8 3.6E-3 2E-8
5 8 3.6E-3 2E-8
6 8 3.6E-3 2E-8
0 0
0 6
2 6
4 6
6 6
8 6
8 0
6 3
2 3
11 0
12 0
71 0
72 0
5
2 10 -50 0
3 0 -50 0
4 0 -50 0
5 0 -50 0
6 0 -50 0
（3）输出文件
Input Data File Name: B1.TXT
Output File Name: B2.TXT
**************************************************** * * * question TWO * * * ****************************************************
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 2.000000E-08
2 2 3 3.600000E-03 2.000000E-08
3 3 4 3.600000E-03 2.000000E-08
4 4 5 3.600000E-03 2.000000E-08
5 5 6 3.600000E-03 2.000000E-08
6 6 7 3.600000E-03 2.000000E-08
7 7 8 3.600000E-03 2.000000E-08
8 8 9 3.600000E-03 2.000000E-08
9 1 9 3.600000E-03 2.000000E-08
10 2 9 3.600000E-03 2.000000E-08
11 3 9 3.600000E-03 2.000000E-08
12 4 9 3.600000E-03 2.000000E-08
13 4 8 3.600000E-03 2.000000E-08
14 5 8 3.600000E-03 2.000000E-08
15 6 8 3.600000E-03 2.000000E-08
The Joint Coordinates
joint X Y
1 .000000 .000000
2 .000000 6.000000
3 2.000000 6.000000
4 4.000000 6.000000
5 6.000000 6.000000
6 8.000000 6.000000
7 8.000000 .000000
8 6.000000 3.000000
9 2.000000 3.000000
The Information of Supports
IS VS
11 .000000
12 .000000
71 .000000
72 .000000
Loading Case 1
The Loadings at Joints
NLJ= 5
joint FX FY FM
2 10.000000 -50.000000 .000000
3 .000000 -50.000000 .000000
4 .000000 -50.000000 .000000
5 .000000 -50.000000 .000000
6 .000000 -50.000000 .000000
The Loadings at Members
NLM= 0
The Results of Calculation
The Joint Displacements
joint u v rotation
1 -3.990670E-21 -1.175000E-20 -1.204213E-04
2 2.465975E-04 -4.803306E-04 -1.163110E-04
3 2.046730E-04 -7.748138E-04 -3.296226E-05
4 1.627481E-04 -6.680785E-04 2.001269E-05
5 1.486004E-04 -7.020495E-04 2.709690E-05
6 1.344532E-04 -4.803316E-04 6.701334E-05
7 4.990670E-21 -1.325000E-20 3.105335E-05
8 7.165313E-05 -4.937214E-04 -1.830976E-06
9 2.002452E-04 -5.664871E-04 -1.871493E-05
The Terminal Forces
member FN FS M
1 start 1 57.639676 -.000103 -.000312 end 2 -57.639676 .000103 -.000306 2 start 2 15.092833 .000871 .000705 end 3 -15.092833 -.000871 .001038 3 start 3 15.092963 -.000718 -.000824 end 4 -15.092963 .000718 -.000612 4 start 4 5.093183 .000486 .000472 end 5 -5.093183 -.000486 .000501 5 start 5 5.092979 -.000766 -.000845 end 6 -5.092979 .000766 -.000686 6 start 6 57.639791 .000095 .000310 end 7 -57.639791 -.000095 .000262 7 start 7 89.970664 -.000165 -.000262 end 8 -89.970664 .000165 -.000335 8 start 8 23.146571 -.000085 -.000154 end 9 -23.146571 .000085 -.000188 9 start 1 71.942998 .000236 .000312 end 9 -71.942998 -.000236 .000537 10 start 2 -9.180808 -.000161 -.000398 end 9 9.180808 .000161 -.000182 11 start 3 49.998411 -.000130 -.000214 end 9 -49.998411 .000130 -.000176 12 start 4 21.033329 .000028 .000094 end 9 -21.033329 -.000028 .000008 13 start 4 39.060629 .000012 .000046 end 8 -39.060629 -.000012 -.000002 14 start 5 49.998748 .000204 .000345 end 8 -49.998748 -.000204 .000268 15 start 6 -9.181077 .000166 .000376 end 8 9.181077 -.000166 .000223 ( NA= 270 )
( NW= 923 )
4）内力图(由结构力学求解器得)
轴力图：单位（KN） （
（三） 第三题
3．作图示连续梁的FS、M图，已知各梁截面面积A=7.5m2,惯性矩I=9.50m4，各杆E=3.45×104MPa。5分
（1） 编号：（如图）
2）输入文件
****************************************************
* *
* question THREE *
* *
****************************************************
3.45E7 4 5 6 1
1 2 7.5 9.5
2 3 7.5 9.5
3 4 7.5 9.5
4 5 7.5 9.5
0 0
40 0
80 0
100 0
120 0
11 0
12 0
13 0
22 0
32 0
52 0
1
4 0 0 -100
5
1 4 -10.5 40
2 4 -10.5 40
2 2 -320 20
3 4 -10.5 20
4 4 -10.5 20
3）输出文件
Input Data File Name: C1.TXT
Output File Name: C2.TXT
**************************************************** * * * question THREE * * * ****************************************************
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 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
The Joint Coordinates
joint X Y
1 .000000 .000000
2 40.000000 .000000
3 80.000000 .000000
4 100.000000 .000000
5 120.000000 .000000
The Information of Supports
IS VS
11 .000000
12 .000000
13 .000000
22 .000000
32 .000000
52 .000000
Loading Case 1
The Loadings at Joints
NLJ= 1
joint FX FY FM
4 .000000 .000000 -100.000000
The Loadings at Members
NLM= 5
member type VF DST
1 4 -10.500000 40.000000
2 4 -10.500000 40.000000
2 2 -320.000000 20.000000
3 4 -10.500000 20.000000
4 4 -10.500000 20.000000
The Results of Calculation
The Joint Displacements
joint u v rotation
1 0.000000E+00 3.757212E-21 5.009615E-20
2 0.000000E+00 -3.028846E-21 -3.056974E-05
3 0.000000E+00 -1.771154E-20 2.464355E-05
4 0.000000E+00 -2.461421E-04 -1.471279E-05
5 0.000000E+00 -4.016827E-21 3.115649E-05
The Terminal Forces
member FN FS M
1 start 1 .000000 172.427885 899.038462 end 2 .000000 247.572115 -2401.923077 2 start 2 .000000 362.716346 2401.923077 end 3 .000000 377.283654 -2693.269231 3 start 3 .000000 274.831731 2693.269231
end 4 .000000 -64.831731 703.365385 4 start 4 .000000 64.831731 -803.365385 end 5 .000000 145.168269 .000000 ( NA= 66 )
( NW= 319 )
（5） 内力图(由CAD得)
弯矩图：单位（KN*M）
剪力图：单位（KN）
结构力学
矩阵位移法上机实验报告
指导老师： 肖方红
班 级：土木1109
姓 名： 钱小十
学 号：
2013年10月22日
目录
一、钢架的受力分析
1题目
2结构计算编号示意图
3输入文件
4输出文件
5结构受力分析图
…………………………………………………… 二桁架的受力分析
1题目
2结构计算编号示意图
3输入文件
4输出文件
5结构受力分析图
……………………………………………………
三、连续梁的受力分析
1题目
2结构计算编号示意图
3输入文件
4输出文件
5结构受力分析图
……………………………………………………
矩阵位移法上机实践报告
（一） 第一题
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0.4m,高0.9m，小跨梁截面宽0.4m,高0.6m，各杆E=3.0×104 MPa。10分
（1） 编号：如图所示
（2） 输入文件。
****************************************************
* *
* question one *
* *
****************************************************
3E7 18 15 12 1
1 2 1.6E-1 21333E-7
2 3 1.6E-1 21333E-7
3 4 1.6E-1 21333E-7
4 5 3.6E-1 243E-4
5 6 3.6E-1 243E-4
6 7 2.4E-1 72E-4
7 8 1.6E-1 21333E-7
8 9 1.6E-1 21333E-7
9 10 1.6E-1 21333E-7
11 12 1.6E-1 21333E-7
12 13 1.6E-1 21333E-7
13 6 1.6E-1 21333E-7
14 15 1.6E-1 21333E-7
2 15 3.6E-1 243E-4
15 12 3.6E-1 243E-4
3 13 3.6E-1 243E-4
13 8 2.4E-1 72E-4
12 9 2.4E-1 72E-4
0 0
0 4.5
0 7.7
0 10.9
3.8 10.9
7.6 10.9
11.4 10.9
11.4 7.7
11.4 4.5
11.4 0
7.6 0
7.6 4.5
7.6 7.7
3.8 0
3.8 4.5
11 0
12 0
13 0
101 0
102 0
103 0
111 0
112 0
113 0
141 0
142 0
143 0
3
4 100 0 0
5 0 0 -15
7 0 0 -15
8
1 3 20 4.5
14 4 -5 3.8
15 4 -5 3.8
16 4 -45 7.6
16 2 -26 3.8
17 4 -45 3.8
18 4 -5 3.8
18 2 -26 2.7
（3） 输出结果
Input Data File Name: D1.TXT
Output File Name: D2.TXT
****************************************************
* *
* question one *
* *
****************************************************
The Input Data
The General Information
E NM NJ NS NLC
3.000E+07 18 15 12 1
The Information of Members
member start end A I
1 1 2 1.600000E-01 2.133300E-03
2 2 3 1.600000E-01 2.133300E-03
3 3 4 1.600000E-01 2.133300E-03
4 4 5 3.600000E-01 2.430000E-02
5 5 6 3.600000E-01 2.430000E-02
6 6 7 2.400000E-01 7.200000E-03
7 7 8 1.600000E-01 2.133300E-03
8 8 9 1.600000E-01 2.133300E-03
9 9 10 1.600000E-01 2.133300E-03
10 11 12 1.600000E-01 2.133300E-03
11 12 13 1.600000E-01 2.133300E-03
12 13 6 1.600000E-01 2.133300E-03
13 14 15 1.600000E-01 2.133300E-03
14 2 15 3.600000E-01 2.430000E-02
15 15 12 3.600000E-01 2.430000E-02
16 3 13 3.600000E-01 2.430000E-02
17 13 8 2.400000E-01 7.200000E-03
18 12 9 2.400000E-01 7.200000E-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 11.400000 10.900000
8 11.400000 7.700000
9 11.400000 4.500000
10 11.400000 .000000
11 7.600000 .000000
12 7.600000 4.500000
13 7.600000 7.700000
14 3.800000 .000000
15 3.800000 4.500000
The Information of Supports
IS VS
11 .000000
12 .000000
13 .000000
101 .000000
102 .000000
103 .000000
111 .000000
112 .000000
113 .000000
141 .000000
142 .000000
143 .000000
Loading Case 1
The Loadings at Joints
NLJ= 3
joint FX FY FM
4 100.000000 .000000 .000000
5 .000000 .000000 -15.000000
7 .000000 .000000 -15.000000
The Loadings at Members
NLM= 8
member type VF DST
1 3 20.000000 4.500000
14 4 -5.000000 3.800000
15 4 -5.000000 3.800000
16 4 -45.000000 7.600000
16 2 -26.000000 3.800000
17 4 -45.000000 3.800000
18 4 -5.000000 3.800000
18 2 -26.000000 2.700000
The Results of Calculation
The Joint Displacements
joint u v rotation
1 3.743750E-21 -1.018849E-20 -8.548984E-21
2 4.640739E-03 -9.551710E-05 -8.827678E-05
3 6.808144E-03 -1.859027E-04 -8.681810E-04
4 8.912967E-03 -1.786969E-04 -1.000147E-04
5 8.882365E-03 -3.358049E-04 -1.835667E-05
6 8.851762E-03 -4.408383E-04 -7.260700E-05
7 8.835639E-03 -2.883267E-04 -1.906210E-04
8 6.809896E-03 -2.677740E-04 -2.607855E-04
9 4.636627E-03 -1.867505E-04 -3.035602E-04
10 3.332049E-21 -1.992005E-20 -7.928835E-21
11 3.561927E-21 -2.840118E-20 -8.274835E-21
12 4.638499E-03 -2.662610E-04 -1.831659E-04
13 6.809485E-03 -4.541852E-04 3.008896E-04
14 3.862274E-21 -3.690284E-21 -8.721767E-21
15 4.632824E-03 -3.459642E-05 -2.225539E-05
The Terminal Forces
member FN FS M
1 start 1 101.884900 82.437497 119.239840
end 2 -101.884900 7.562503 49.228895
2 start 2 135.578359 14.931158 39.487692
end 3 -135.578359 -14.931158 8.292010
3 start 3 -10.808671 13.024275 5.475752
end 4 10.808671 -13.024275 36.201924
4 start 4 86.975725 -10.808671 -36.201924
end 5 -86.975725 10.808671 -4.871024
5 start 5 86.975725 -10.808671 -10.128976
end 6 -86.975725 10.808671 -30.943972
6 start 6 30.550134 -30.829117 -51.867154
end 7 -30.550134 30.829117 -65.283480
7 start 7 30.829117 30.550134 50.283480
end 8 -30.829117 -30.550134 47.476942
8 start 8 121.535301 29.772570 48.491589
end 9 -121.535301 -29.772570 46.780630
9 start 9 199.200494 33.320495 70.653880
end 10 -199.200494 -33.320495 79.288346
10 start 11 284.011754 35.619272 82.748348
end 12 -284.011754 -35.619272 77.538377
11 start 12 281.886334 55.296272 78.793070
end 13 -281.886334 -55.296272 98.154990
12 start 13 -20.020446 56.425592 97.750757
end 6 20.020446 -56.425592 82.811125
13 start 14 36.902845 38.622736 87.217673
end 15 -36.902845 -38.622736 86.584641
14 start 2 22.493661 -33.693459 -88.716587
end 15 -22.493661 52.693459 -75.418555
15 start 15 -16.129075 -15.790614 -11.166086
end 12 16.129075 34.790614 -84.938245
16 start 3 -1.906883 146.387029 -13.767762
end 13 1.906883 221.612966 -272.090795
17 start 13 -.777563 80.293814 76.185049
end 8 .777563 90.706184 -95.968531
18 start 12 3.547924 -32.665194 -71.393203
end 9 -3.547924 77.665194 -117.434510
4） 绘图(由结构力学求解器得)
轴力图：单位（KN） （
剪力图：单位（KN）
弯矩图：单位（KN*M）
（二） 第二题
2、计算图示桁架各杆的轴力。已知A=2400mm2,E=2.0×105 MPa。5分
1）编号：（如图）
（
（2）输入文件
**************************************************** * * * question TWO * * * **************************************************** 2E8 15 9 4 1
1 2 3.6E-3 2E-8
2 3 3.6E-3 2E-8
3 4 3.6E-3 2E-8
4 5 3.6E-3 2E-8
5 6 3.6E-3 2E-8
6 7 3.6E-3 2E-8
7 8 3.6E-3 2E-8
8 9 3.6E-3 2E-8
1 9 3.6E-3 2E-8
2 9 3.6E-3 2E-8
3 9 3.6E-3 2E-8
4 9 3.6E-3 2E-8
4 8 3.6E-3 2E-8
5 8 3.6E-3 2E-8
6 8 3.6E-3 2E-8
0 0
0 6
2 6
4 6
6 6
8 6
8 0
6 3
2 3
11 0
12 0
71 0
72 0
5
2 10 -50 0
3 0 -50 0
4 0 -50 0
5 0 -50 0
6 0 -50 0
（3）输出文件
Input Data File Name: B1.TXT
Output File Name: B2.TXT
**************************************************** * * * question TWO * * * ****************************************************
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 2.000000E-08
2 2 3 3.600000E-03 2.000000E-08
3 3 4 3.600000E-03 2.000000E-08
4 4 5 3.600000E-03 2.000000E-08
5 5 6 3.600000E-03 2.000000E-08
6 6 7 3.600000E-03 2.000000E-08
7 7 8 3.600000E-03 2.000000E-08
8 8 9 3.600000E-03 2.000000E-08
9 1 9 3.600000E-03 2.000000E-08
10 2 9 3.600000E-03 2.000000E-08
11 3 9 3.600000E-03 2.000000E-08
12 4 9 3.600000E-03 2.000000E-08
13 4 8 3.600000E-03 2.000000E-08
14 5 8 3.600000E-03 2.000000E-08
15 6 8 3.600000E-03 2.000000E-08
The Joint Coordinates
joint X Y
1 .000000 .000000
2 .000000 6.000000
3 2.000000 6.000000
4 4.000000 6.000000
5 6.000000 6.000000
6 8.000000 6.000000
7 8.000000 .000000
8 6.000000 3.000000
9 2.000000 3.000000
The Information of Supports
IS VS
11 .000000
12 .000000
71 .000000
72 .000000
Loading Case 1
The Loadings at Joints
NLJ= 5
joint FX FY FM
2 10.000000 -50.000000 .000000
3 .000000 -50.000000 .000000
4 .000000 -50.000000 .000000
5 .000000 -50.000000 .000000
6 .000000 -50.000000 .000000
The Loadings at Members
NLM= 0
The Results of Calculation
The Joint Displacements
joint u v rotation
1 -3.990670E-21 -1.175000E-20 -1.204213E-04
2 2.465975E-04 -4.803306E-04 -1.163110E-04
3 2.046730E-04 -7.748138E-04 -3.296226E-05
4 1.627481E-04 -6.680785E-04 2.001269E-05
5 1.486004E-04 -7.020495E-04 2.709690E-05
6 1.344532E-04 -4.803316E-04 6.701334E-05
7 4.990670E-21 -1.325000E-20 3.105335E-05
8 7.165313E-05 -4.937214E-04 -1.830976E-06
9 2.002452E-04 -5.664871E-04 -1.871493E-05
The Terminal Forces
member FN FS M
1 start 1 57.639676 -.000103 -.000312 end 2 -57.639676 .000103 -.000306 2 start 2 15.092833 .000871 .000705 end 3 -15.092833 -.000871 .001038 3 start 3 15.092963 -.000718 -.000824 end 4 -15.092963 .000718 -.000612 4 start 4 5.093183 .000486 .000472 end 5 -5.093183 -.000486 .000501 5 start 5 5.092979 -.000766 -.000845 end 6 -5.092979 .000766 -.000686 6 start 6 57.639791 .000095 .000310 end 7 -57.639791 -.000095 .000262 7 start 7 89.970664 -.000165 -.000262 end 8 -89.970664 .000165 -.000335 8 start 8 23.146571 -.000085 -.000154 end 9 -23.146571 .000085 -.000188 9 start 1 71.942998 .000236 .000312 end 9 -71.942998 -.000236 .000537 10 start 2 -9.180808 -.000161 -.000398 end 9 9.180808 .000161 -.000182 11 start 3 49.998411 -.000130 -.000214 end 9 -49.998411 .000130 -.000176 12 start 4 21.033329 .000028 .000094 end 9 -21.033329 -.000028 .000008 13 start 4 39.060629 .000012 .000046 end 8 -39.060629 -.000012 -.000002 14 start 5 49.998748 .000204 .000345 end 8 -49.998748 -.000204 .000268 15 start 6 -9.181077 .000166 .000376 end 8 9.181077 -.000166 .000223 ( NA= 270 )
( NW= 923 )
4）内力图(由结构力学求解器得)
轴力图：单位（KN） （
（三） 第三题
3．作图示连续梁的FS、M图，已知各梁截面面积A=7.5m2,惯性矩I=9.50m4，各杆E=3.45×104MPa。5分
（1） 编号：（如图）
2）输入文件
****************************************************
* *
* question THREE *
* *
****************************************************
3.45E7 4 5 6 1
1 2 7.5 9.5
2 3 7.5 9.5
3 4 7.5 9.5
4 5 7.5 9.5
0 0
40 0
80 0
100 0
120 0
11 0
12 0
13 0
22 0
32 0
52 0
1
4 0 0 -100
5
1 4 -10.5 40
2 4 -10.5 40
2 2 -320 20
3 4 -10.5 20
4 4 -10.5 20
3）输出文件
Input Data File Name: C1.TXT
Output File Name: C2.TXT
**************************************************** * * * question THREE * * * ****************************************************
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 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
The Joint Coordinates
joint X Y
1 .000000 .000000
2 40.000000 .000000
3 80.000000 .000000
4 100.000000 .000000
5 120.000000 .000000
The Information of Supports
IS VS
11 .000000
12 .000000
13 .000000
22 .000000
32 .000000
52 .000000
Loading Case 1
The Loadings at Joints
NLJ= 1
joint FX FY FM
4 .000000 .000000 -100.000000
The Loadings at Members
NLM= 5
member type VF DST
1 4 -10.500000 40.000000
2 4 -10.500000 40.000000
2 2 -320.000000 20.000000
3 4 -10.500000 20.000000
4 4 -10.500000 20.000000
The Results of Calculation
The Joint Displacements
joint u v rotation
1 0.000000E+00 3.757212E-21 5.009615E-20
2 0.000000E+00 -3.028846E-21 -3.056974E-05
3 0.000000E+00 -1.771154E-20 2.464355E-05
4 0.000000E+00 -2.461421E-04 -1.471279E-05
5 0.000000E+00 -4.016827E-21 3.115649E-05
The Terminal Forces
member FN FS M
1 start 1 .000000 172.427885 899.038462 end 2 .000000 247.572115 -2401.923077 2 start 2 .000000 362.716346 2401.923077 end 3 .000000 377.283654 -2693.269231 3 start 3 .000000 274.831731 2693.269231
end 4 .000000 -64.831731 703.365385 4 start 4 .000000 64.831731 -803.365385 end 5 .000000 145.168269 .000000 ( NA= 66 )
( NW= 319 )
（5） 内力图(由CAD得)
弯矩图：单位（KN*M）
剪力图：单位（KN）