结构力学上机实验报告_结构力学上机实习报告

其他范文 时间:2020-02-28 12:03:53 收藏本文下载本文
【www.daodoc.com - 其他范文】

结构力学上机实验报告由刀豆文库小编整理,希望给你工作、学习、生活带来方便,猜你可能喜欢“结构力学上机实习报告”。

结构力学上机实验报告

姓名:

学号:

指导老师:肖方红

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。10分

解:统一单位力kN长度m那么弹性模量单位为kPa。输入输出数据如下:

表一:1题输入数据

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

* *

sjl1 gangjia 2011.10.24

* *

* ******************************************************************************************* 3e71

0.16

213e-5 2

0.16

213e-5

0.2975

1791e-5 2

0.2975

1791e-5 4

0.21

63e-4 5

0.21

63e-4 5

0.16

213e-5 8

0.16

213e-5 7

0.16

213e-5 9

0.16

213e-5 0

0 0

4.5 0

7.7 7.2 7.7 7.2 4.5 11 7.7 11 4.5 7.2 0 11 0 11 0 12 0 13 0 81 0 82 0 83 0 91 0 92 0 93 0 1 6

0

0

-15 7 1

4.5 2

3.2 3-196 7.2 4-36

7.2 5-196 3.8 6-36

3.8 6-26

2.7 表二:1题输出数据

Input Data File Name: sjl1.txt

Output File Name: sjl1out.txt

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

*

sjl1 gangjia 2011.10.24

*

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

The Input Data

The General Information

E

NM

NJ

NS

NLC

3.000E+07

The Information of Members

member start end

A

I

1.600000E-01

2.130000E-03

1.600000E-01

2.130000E-03

2.975000E-01

2.975000E-01

2.100000E-01

2.100000E-01

1.600000E-01

1.600000E-01

1.600000E-01

1.600000E-01

The Joint Coordinates

joint

X

Y

.000000

.000000

.000000

4.500000

.000000

7.700000

7.200000

7.700000

7.200000

4.500000

11.000000

7.700000

11.000000

4.500000

7.200000

.000000

11.000000

.000000

The Information of Supports

IS

VS

.000000

.000000

.000000

.000000

.000000

.000000

.000000

.000000

.000000

1.791000E-02 1.791000E-02 6.300000E-03 6.300000E-03 2.130000E-03 2.130000E-03 2.130000E-03 2.130000E-03

Loading Case 1

The Loadings at Joints

NLJ=

joint

FX

FY

FM

.000000

.000000

-15.000000

The Loadings at Members

NLM=

member type

VF

DST

20.000000

4.500000

20.000000

3.200000

-196.000000

7.200000

-36.000000

7.200000

-196.000000

3.800000

-36.000000

3.800000

-26.000000

2.700000

The Results of Calculation

The Joint Displacements

joint

u

v

rotation

3.076236E-21

-7.549352E-20

-7.540649E-21

4.636735E-03

-7.077518E-04

-4.359988E-04

5.924037E-03

-1.134844E-03

-3.169292E-03

5.813626E-03

-2.178472E-03

1.834783E-03

4.684030E-03

-1.341626E-03

1.384534E-05

5.788766E-03

-5.408925E-04

4.571795E-04

4.685631E-03

-3.674969E-04

-4.586878E-05

3.967738E-21

-1.431068E-19

-8.907750E-21

3.856026E-21

-3.919967E-20

-8.741193E-21

The Terminal Forces

member

FN

FS start

754.935194

75.762357

end

-754.935194

14.237643 start

640.638123

-72.863183

end

-640.638123

136.863184

M 109.156485

29.274120-96.133965-239.428195 start

136.863184

640.638123

239.428195

end

-136.863184

770.561840

-707.153563 start

-58.625540

114.297071

66.859844

end

58.625540

144.902922

-177.040903 start

41.214402

484.706696

517.753681

end

-41.214402

260.093294

-90.988284 start

-2.654138

30.896570

-29.106007

end

2.654138

131.903429

-142.007053 start

1255.268536

95.648782

116.676201

end

-1255.268536

start

1431.068027

end

-1431.068027

start

260.093294

end

-260.093294 start

391.996723

end

-391.996723

钢架的FN图:

-95.648782

39.677380

-39.677380

41.214402

-41.214402

38.560264

-38.560264

189.399883 89.077501 89.470709 55.897795 75.988284 87.411931 86.109258

钢架的Fs图:

钢架的M图:

2、计算图示桁架各杆的轴力。已知A=2400mm2,E=2.0×105 MPa。5分

解:该桁架各节点均为铰结,为了使计算简便,所有节点均作为钢节点,为此在输入数据时,各杆截面二次矩取很小的值,本题取1×10-20 本题有30根杆件,17个节点,输入输出数据如下:

表三:2题输入数据

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

*

*

sjl2 gangjia 2011.10.24

* *

* ************************************************************************** 2e8

301

24e-4

1e-20 1

24e-4

1e-20 2

24e-4

1e-20 2

24e-4

1e-20 3

24e-4

1e-20 5

24e-4

1e-20 3

24e-4

1e-20 3

24e-4

1e-20 4

24e-4

1e-20 6

24e-4

1e-20 4

24e-4

1e-20 6

24e-4

1e-20 7

24e-4

1e-20 7

24e-4

1e-20 8

24e-4

1e-20 9

24e-4

1e-20 9

24e-4

1e-20 11

24e-4

1e-20 10

24e-4

1e-20 11

24e-4

1e-20 11

24e-4

1e-20 12

24e-4

1e-20 15

24e-4

1e-20 15

24e-4

1e-20 12

24e-4

1e-20 14

24e-4

1e-20 13

24e-4

1e-20 14

24e-4

1e-20 15

24e-4

1e-20 17

24e-4

1e-20 0

0 01

3.75 2

3.5 1

4.75 2

5.5 3

5.25 3

6.25 45

6.25 5

5.25 6

5.5 7

4.75 7

3.75 6

3.5 88

0 11

0 12

0 171

0 172

0 9 2 0-12 0 5 0-5

0 6 0-5

0 8 0-5

0 9 0-5

0 10 0-5

0 12 0-5

0 13 0-5

0 16 0-12 0 0

表四:2题输出数据

Input Data File Name: sjl2.txt

Output File Name: sjl2out.txt

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

*

sjl2 gangjia 2011.10.24

*

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

The Input Data

The General Information

E

NM

NJ

NS

NLC

2.000E+08

The Information of Members

member start end

A

I

2.400000E-03

1.000000E-20

2.400000E-03

1.000000E-20

2.400000E-03

1.000000E-20

2.400000E-03

1.000000E-20

2.400000E-03

2.400000E-03

2.400000E-03

2.400000E-03

2.400000E-03

2.400000E-03

2.400000E-03

2.400000E-03

2.400000E-03

2.400000E-03

2.400000E-03

2.400000E-03

2.400000E-03

2.400000E-03

2.400000E-03

2.400000E-03

2.400000E-03

2.400000E-03

2.400000E-03

2.400000E-03

2.400000E-03

2.400000E-03

2.400000E-03

2.400000E-03

2.400000E-03

2.400000E-03

The Joint Coordinates

joint

X

Y

.000000

.000000

.000000

4.000000

1.000000

3.750000

2.000000

3.500000

1.000000

4.750000

1.000000E-20 1.000000E-20 1.000000E-20 1.000000E-20 1.000000E-20 1.000000E-20 1.000000E-20 1.000000E-20 1.000000E-20 1.000000E-20 1.000000E-20 1.000000E-20 1.000000E-20 1.000000E-20 1.000000E-20 1.000000E-20 1.000000E-20 1.000000E-20 1.000000E-20 1.000000E-20 1.000000E-20

1.000000E-20

1.000000E-20

1.000000E-20

1.000000E-20

1.000000E-20

2.000000

5.500000

3.000000

5.250000

3.000000

6.250000

4.000000

7.000000

5.000000

6.250000

5.000000

5.250000

6.000000

5.500000

7.000000

4.750000

7.000000

3.750000

6.000000

3.500000

8.000000

4.000000

8.000000

.000000

The Information of Supports

IS

VS

.000000

.000000

171

.000000

172

.000000

Loading Case 1

The Loadings at Joints

NLJ=

joint

FX

FY

.000000

-12.000000

.000000

-5.000000

.000000

-5.000000

.000000

-5.000000

.000000

-5.000000

.000000

-5.000000

.000000

-5.000000

.000000

-5.000000

.000000

-12.000000

The Loadings at Members

NLM=

0

The Results of Calculation

FM.000000.000000.000000.000000.000000.000000.000000.000000.000000

The Joint Displacements

joint

u

v

rotation

-5.714286E-22

-2.950000E-21

-5.676597E-05

1.682251E-04

-1.625000E-04

-1.236830E-04

1.583218E-04

-2.705629E-04

-3.193943E-05

1.833298E-04

-2.161644E-04

2.716851E-05

2.265671E-04

-2.809795E-04

-4.829399E-05

1.786882E-04

-2.578310E-04

2.349593E-05

1.918510E-04

-2.279964E-04

1.336072E-04

-2.384131E-04

1.857079E-18

-1.009603E-04

-1.336072E-04

-2.384131E-04

-1.918510E-04

-2.279964E-04

-1.786882E-04

-2.578310E-04

-2.265671E-04

-2.809795E-04

-1.583218E-04

-2.705629E-04

-1.833298E-04

-2.161644E-04

-1.682251E-04

-1.625000E-04

5.714286E-22

-2.950000E-21

The Terminal Forces

member

FN

start

19.500000

end

-19.500000 start

11.517511

end

-11.517511 start

9.375000

end

-9.375000 start

-7.730823

end

7.730823 start

-5.153882

end

5.153882 start

9.375000

end

-9.375000 start

5.000000

end

-5.000000 start

-5.038911

end

5.038911 start

10.000000

end

-10.000000

start

-2.576941

end

2.576941

4.762947E-05 1.067515E-04 9.122545E-19-1.067515E-04-4.762947E-05-2.349593E-05 4.829399E-05 3.193943E-05-2.716851E-05 1.236830E-04 5.676597E-05

FS

.000000

.000000

.000000

.000000

.000000

.000000

.000000

.000000

.000000

.000000

.000000

.000000

.000000

.000000

.000000

.000000

.000000

.000000

.000000

.000000

M.000000.000000.000000.000000.000000.000000.000000.000000.000000.000000.000000.000000.000000.000000.000000.000000.000000.000000.000000.000000

start

1.439689

.000000

.000000

end

-1.439689

.000000

.000000

start

9.375000

.000000

.000000

end

-9.375000

.000000

.000000

start

5.000000

.000000

.000000

end

-5.000000

.000000

.000000

start

-3.599222

.000000

.000000

end

3.599222

.000000

.000000

start

9.375000

.000000

.000000

end

-9.375000

.000000

.000000

start

9.375000

.000000

.000000

end 10

-9.375000

.000000

.000000

start

-3.599222

.000000

.000000

end 11

3.599222

.000000

.000000

start 11

5.000000

.000000

.000000

end 10

-5.000000

.000000

.000000

start 10

9.375000

.000000

.000000

end 12

-9.375000

.000000

.000000

start 11

-2.576941

.000000

.000000

end 12

2.576941

.000000

.000000

start 11

1.439689

.000000

.000000

end 15

-1.439689

.000000

.000000

start 12

9.375000

.000000

.000000

end 13

-9.375000

.000000

.000000

start 15

10.000000

.000000

.000000

end 12

-10.000000

.000000

.000000

start 15

-5.153882

.000000

.000000

end 14

5.153882

.000000

.000000

start 12

-5.038911

.000000

.000000

end 14

5.038911

.000000

.000000

start 14

5.000000

.000000

.000000

end 13

-5.000000

.000000

.000000

start 13

9.375000

.000000

.000000

end 16

-9.375000

.000000

.000000

start 14

-7.730823

.000000

.000000

end 16

7.730823

.000000

.000000

start 15

11.517511

.000000

.000000

end 17

-11.517511

.000000

.000000

start 17

19.500000

.000000

.000000

end 16

-19.500000

.000000

.000000

钢架轴力图(其中拉力为正,压力为负):

3.作图示连续梁的FS、M图,已知各梁截面面积A=6.5m2,惯性矩I=5.50m4,各杆E=3.45×104MPa。5分

解:该结构为一超静定结构,输入输出数据如下:

表五:3题输入数据

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

* * sjl3 lxl 2011.10.24

* *

* ************************* 345e51

6.5

5.5 2

6.5

5.5 3

6.5

5.5 0

0 40

0 80

0 120

0 11

0 12

0 22

0 32

0 42

0 0 4 1

-10.5

2

-10.5

2

-320

3

-10.5

表六:3题输出数据

Input Data File Name: sjl3.txt

Output File Name: sjl3out.txt

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

*

*

* sjl3 lxl 2011.10.24

*

*

*

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

The Input Data

The General Information

E

NM

NJ

NS

NLC

3.450E+07

The Information of Members

member start end

A

I

6.500000E+00

5.500000E+00

6.500000E+00

5.500000E+00

6.500000E+00

5.500000E+00

The Joint Coordinates

joint

X

Y

.000000

.000000

40.000000

.000000

80.000000

.000000

120.000000

.000000

The Information of Supports

IS

VS

.000000

.000000

.000000

.000000

.000000

Loading Case 1

The Loadings at Joints

NLJ=

0

The Loadings at Members

NLM=

member type

VF

DST

-10.500000

40.000000

-10.500000

40.000000

-320.000000

20.000000

-10.500000

40.000000

The Results of Calculation

The Joint Displacements

joint

u

v

rotation

0.000000E+00

6.600000E-21

-5.480896E-05

0.000000E+00

-6.600000E-21

-3.794466E-05

0.000000E+00

-6.600000E-21

3.794466E-05

0.000000E+00

6.600000E-21

5.480896E-05

The Terminal Forces

member

FN

FS

M start

.000000

144.000000

.000000

end

.000000

276.000000

-2640.000000 start

.000000

370.000000

2640.000000

end

.000000

370.000000

-2640.000000 start

.000000

276.000000

2640.000000

end

.000000

144.000000

.000000

连续梁的Fs图:

连续梁的M图:

结构力学上机心得

结构力学学习心得结构力学的学习马上就要结束了,本学期学的主要是渐进法、矩阵位移法和平面刚架静力分析程序设计,相比上学期的画内力图和计算这学期貌似任务比较轻,需要动手的......

结构力学实验报告

结构力学实验报告结构力学实验报告 班级 12土木2班姓名学号结构力学实验报告实验报告一 实验名称 在求解器中输入平面结构体系一实验目的1、了解如何在求解器中输入结构体系......

上机实验报告

一. 题目1. 建立一个学生档案,内容包括学号,姓名,年龄,性别,数学,物理和英语3门功课成绩。要求实现以下功能:1) 数据输入;2) 查询某个学生的成绩;3) 按平均排列输出;4) 统计某门课各分数......

电算化上机实验报告

管理学院会计学专业上机实践报告课程名称:会计电算化 指导教师: 上机实践名称:系统管理与基础设置 上机实践编号:实验一 一、实验目的通过上机实验,理解用友企业级财务软件系统管......

C++上机实验报告

C++上机实验报告实验名称:实验专业班级:姓名:学号:实验日期:10 11 实验 目录 1.实验目的 2.实验内容 3.程序代码 4.调试结果 5.实验心得 1.实验目的 实验10 (1)进一步了解运算符......

下载结构力学上机实验报告word格式文档
下载结构力学上机实验报告.doc
将本文档下载到自己电脑,方便修改和收藏。
点此处下载文档

文档为doc格式

热门文章
点击下载本文