结构力学矩阵位移法上机题范例

矩阵位移法上机题

FSM1．作图示刚架的FN、、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） 编号如下图

第一题机算结果：

*******************************************************

* * 1 composite beam 2013.10.11 * *******************************************************

*

*

*

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 3 4 3.600000E-01 2.430000E-02

2 4 7 3.600000E-01 2.430000E-02

3 7 15 2.400000E-01 7.200000E-03

4 2 3 1.600000E-01 2.133330E-03

5 8 7 1.600000E-01 2.133330E-03

6 14 15 1.600000E-01 2.133330E-03

7 2 8 3.600000E-01 2.430000E-02

8 8 14 2.400000E-01 7.200000E-03

9 1 2 1.600000E-01 2.133330E-03

10 10 8 1.600000E-01 2.133330E-03

11 13 14 1.600000E-01 2.133330E-03

12 1 13 5 14 10 15 9 16 6 17 11 18 12 joint 1 2 5 2.400000E-01 7.200000E-03

10 2.400000E-01 7.200000E-03

13 2.400000E-01 7.200000E-03

1 1.600000E-01 2.133330E-03

5 1.600000E-01 2.133330E-03

10 1.600000E-01 2.133330E-03

13 1.600000E-01 2.133330E-03

The Joint Coordinates

X Y

.000000 4.500000

.000000 7.700000

3 .000000 10.900000

4 3.800000 10.900000

5 3.800000 4.500000

6 3.800000 .000000

7 7.600000 10.900000

8 7.600000 7.700000

9 .000000 .000000

10 7.600000 4.500000

11 7.600000 .000000

12 11.400000 .000000

13 11.400000 4.500000

14 11.400000 7.700000

15 11.400000 10.900000

The Information of Supports

IS VS

91 .000000

92 .000000

93 61 62 63 ( NA= 522 )

.000000

.000000

.000000

.000000

111 .000000

112 .000000

113 .000000

121 .000000

122 .000000

123 .000000

( NW= 1557 )

Loading Case 1

The Loadings at Joints

NLJ= 3

ILJ PX PY PM

3 100.0000 .0000 .00000

4 .0000 .0000 -15.00000

15 .0000 .0000 -15.00000

The Loadings at Members

NLM= 8

ILM ITL PV DST

7 4 -45.0000 7.600000

7 2 -26.0000 3.800000

8 4 -45.0000 3.800000

12 4 -5.0000 3.800000

13 4 -5.0000 3.800000

14 4 -5.0000 3.800000

14 2 -26.0000 2.700000

15 3 20.0000 4.500000

The Results of Calculation

The Joint Displacements

joint u v phi

1 4.836582E-03 -1.022792E-04 -2.782242E-04

2 7.170265E-03 -1.924134E-04 -8.633077E-04

3 9.283498E-03 -1.850379E-04 -9.796148E-05

4 9.253056E-03 -3.308520E-04 -1.530616E-05

5 4.825287E-03 -3.042029E-05 -8.065038E-05

6 3.913801E-21 -3.244831E-21 -8.920754E-21

7 9.222614E-03 -4.303828E-04 -7.360184E-05

8 7.169144E-03 -4.432722E-04 2.9857E-04

9 3.548662E-21 -1.090978E-20 -8.380186E-21

10 4.8347E-03 -2.555460E-04 -2.846911E-04

11 3.534769E-21 -2.725823E-20 -8.358123E-21

12 3.502768E-21 -2.078716E-20 -8.309399E-21

13 4.833504E-03 -1.948796E-04 -3.010583E-04

14 7.167602E-03 -2.763524E-04 -2.814585E-04

15 9.206813E-03 -2.966173E-04 -1.946820E-04

The Terminal Forces

member N(st) Q(st) M(st) N(en) Q(en) 1 86.519 -11.063 -36.877 -86.519 11.063 2 86.519 -11.063 -9.837 -86.519 11.063 3 29.939 -30.397 -50.873 -29.939 30.397 M(en)

-5.163

-.637

-32.204

4 -11.063 13.481 6.263 11.063 -13.481 36.877

5 -19.334 56.580 97.980 19.334 -56.580 83.076

6 30.397 29.939 46.166 -30.397 -29.939 49.637

7 1.593 8 2.922 9 135.201 281.5 11 122.209 12 21.402 13 -17.736 14 2.167 15 109.098 16 32.448 17 272.582 79.188 11.888 55.251 32.861 -26.104 -12.655 -40.662 80.487 39.138 35.348 -13.583 75.151 30.723 76.728 52.186 -72.861 -24.480 -92.501 117.552 .208 83.581 -1.593 -2.922 -135.201 -281.5 -122.209 -21.402 17.736 -2.167 -109.098 -32.448 -272.582 221.735 91.812 -11.888 -55.251 -32.861 45.104 31.655 85.662 9.513 -39.138 -35.348 -273.206

-99.136

7.320

100.075

52.970

-62.433

-59.710

-126.716

42.138

86.913

75.483

146.265 10

18 207.872 35.028 83.094 -207.872 -35.028 74.531

( NA= 522 )

( NW= 15 )

第一题作图：

轴力图 剪力图

弯矩图

第二题

2、计算图示桁架各杆的轴力。已知A=3600mm2，E=2.0×105 MPa。5分

编号如下图

第二题机算结果：

*******************************************************

* * 1 composite beam 2013.10.11 * *******************************************************

*

*

*

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 9 1 3.600000E-03 1.000000E-20

2 9 2 3.600000E-03 1.000000E-20

3 2 1 3.600000E-03 1.000000E-20

4 2 3 3.600000E-03 1.000000E-20

5 3 1 3.600000E-03 1.000000E-20

6 3 4 3.600000E-03 1.000000E-20

7 1 4 3.600000E-03 1.000000E-20

8 1 7 3.600000E-03 1.000000E-20

9 4 7 3.600000E-03 1.000000E-20

10 4 5 3.600000E-03 1.000000E-20

11 5 7 3.600000E-03 1.000000E-20

12 7 13 5 14 6 15 7 joint 1 2 3 4 5 6 3.600000E-03 1.000000E-20

6 3.600000E-03 1.000000E-20

8 3.600000E-03 1.000000E-20

8 3.600000E-03 1.000000E-20

The Joint Coordinates

X Y

2.000000 3.000000

.000000 6.000000

2.000000 6.000000

4.000000 6.000000

6.000000 6.000000

6 8.000000 6.000000

7 6.000000 3.000000

8 8.000000 .000000

9 .000000 .000000

The Information of Supports

IS VS

81 .000000

82 .000000

91 .000000

92 .000000

( NA= 270 )

( NW= 923 )

Loading Case 1

The Loadings at Joints

NLJ= 5

ILJ PX PY PM

2 10.0000 -50.0000 .00000

3 .0000 -50.0000 .00000

4 .0000 -50.0000 .00000

5 .0000 -50.0000 .00000

6 .0000 -50.0000 .00000

The Loadings at Members

NLM= 0

The Results of Calculation

The Joint Displacements

joint u v phi

1 2.002495E-04 -5.6913E-04 -1.871438E-05

2 2.466010E-04 -4.803302E-04 -1.163163E-04

3 2.046758E-04 -7.748246E-04 -3.296074E-05

4 1.627506E-04 -6.680758E-04 2.001388E-05

5 1.486031E-04 -7.020559E-04 2.709549E-05

6 1.344557E-04 -4.803302E-04 6.701690E-05

7 7.165413E-05 -4.937226E-04 -1.831533E-06

8 4.990692E-21 -1.325000E-20 3.105258E-05

9 -3.990692E-21 -1.175000E-20 -1.204224E-04

The Terminal Forces

member N(st) Q(st) M(st) N(en) Q(en) 1 71.943 .000 .000 -71.943 .000 2 57.0 .000 .000 -57.0 .000 3 -9.182 .000 .000 9.182 .000 4 15.093 .000 .000 -15.093 .000 5 50.000 .000 .000 -50.000 .000 M(en)

.000

.000

.000

.000

.000

6 15.093 .000 .000 -15.093 .000 .000

7 21.032 .000 .000 -21.032 .000 .000

8 23.147 .000 .000 -23.147 .000 .000

9 39.060 10 5.093 11 50.000 12 -9.182 13 5.093 14 57.0 15 .971 270 )

.000 -39.060 .000 -5.093 .000 -50.000 .000 9.182 .000 -5.093 .000 -57.0 .000 -.971 ( NW= 923 )

.000

.000

.000

.000

.000

.000

.000

.000 .000 .000 .000 .000 .000 .000 .000 .000 .000 .000 .000 .000 .000 ( NA=

第二题图如下：

轴力图

第三题：

M图，3．作图示连续梁的FS、已知各梁截面面积A=7.5m，惯性矩I=9.50m，各杆E=3.45×104MPa。

245分

编号如下图

第三题机算结果：

******************************************************* * * * 1 composite beam 2013.10.11 * * ******************************************************* 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

ILJ PX 4 .0000

( NA= 66 )

( NW= 299 )

PY PM

-100.00000

.0000

The Loadings at Members

NLM= 5

ILM ITL PV DST

1 4 -10.5000 40.000000

2 4 -10.5000 40.000000

2 2 -320.0000 20.000000

3 4 -10.5000 20.000000

4 4 -10.5000 20.000000

The Results of Calculation

The Joint Displacements

joint u v phi

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 2.4355E-05

4 0.000000E+00 -2.461421E-04 -1.471279E-05

5 0.000000E+00 -4.016827E-21 3.1159E-05

The Terminal Forces

member N(st) Q(st) M(st) N(en) Q(en) M(en)

1 .000 172.428 9.038 .000 247.572 -2401.923

2 .000 362.716 2401.923 .000 377.284 -2693.269

3 .000 274.832 2693.269 .000 -.832 703.365

4 .000 .832 -803.365 .000 145.168 .000

( NA= 66 )

( NW= 319 )

第三题作图如下：

剪力图

弯矩图