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代写CS:Assignment 1: A Matlab program for a 3D structure Homework

2018-08-10 08:00 星期五 所属: Matlab代写,专业代做代考仿真建模分析数学视觉机器学习等 浏览:702

Assignment 1: A Matlab program for a 3D structure

Please refer the attached 2D frame program (Appendix A) with line-by-line explanations to develop a Matlab program for a 3D Frame, whose nodal and elemental information are shown in Appendix B. The shape of three types of cross section is given in Appendix C. The loading and boundary conditions are shown in Appendix D. Appendix E shows a schematic of frame element in 3D, the local stiffness matrix and material properties for the 3D structure. Noted that the gravity is not considered. In Appendix F, a sample Transmission tower is plotted and your program can be used directly for this complex structure. In Appendix G, analysis of 3D frame in Matlab is given via two examples.

 

 

The assignment should:

 

1) Write in terms of https://emedia.rmit.edu.au/learninglab/content/reports-0. A sample report is in Appendix B (10 point)

2) Introduce the theory of Finite Element Analysis for 3D frame. (10 point)

3) Plot the flow chart of the program. (10 point)

4) The structure is composed of 3 types of members. (15 point)

5) Plot the 3D Frame before deformation in a 3D view. The type of different members should be clearly seen in the figure. (15 point)

6) List the deformation of all nodes in a table. (20 point)

7) Write your understanding in Finite Element Analysis through this program. (10 point)

8) Please submit the report, Matlab program, input files (node.txt, elem.txt and etc.). (10 point)

 

Please upload the assignment to the blackboard before 00:00 19 Aug 2018. Late submissions will attract a penalty of 5% per day.


APPENDIX A

Finite element analysis for a 2D Frame in Matlab

 

Line

Code

explanation

1

clc;

Clear command window.

2

clear;

removes all variables from the workspace.

3

close all;

closes all the open figure windows.

4

% blank line

5

N = textread('node.txt');

Read x and y coordinate for the node and assign

their values to N

6

E = textread('elem.txt');

Read the near and far end for the beam and

assign

7

% blank line

8

NN = length(N);

Get the number of node

9

NE = length(E);

Get the number of element

10

% blank line

11

E0 = 2.1e6;

Set Young’s modulus

12

A0 = 22.8;

Set the area of cross section

13

I0 = 935;

Set the second moment of inertia

14

% blank line

 

15

 

hold on;

holds the current plot and all axis properties so that subsequent graphing commands add to the

existing graph.

16

for i=1:NE

Begin an iteration in which i begins from 1 to NE

 

17

plot([N(E(i,1),1),N(E(i,2),1)],[N(E(i,1),2),N(

E(i,2),2)],'k','linewidth',2);

Plot a line starts from the near end to the far end of an element in black with the linewidth of

2.

 

18

text((N(E(i,1),1)+N(E(i,2),1))/2,(N(E(i,1),2)+

N(E(i,2),2))/2,num2str(i),'color','b','fontsiz e',12);

Plot the element number at its middle in blue and 12 font size.

19

end

End the iteration

20

% blank line

21

for i=1:NN

Begin an iteration in which i begins from 1 to NN

22

if i<3

If i<3

23

plot(N(i,1),N(i,2),'rs','markersize',12,'mark

erfacecolor','r');

Plot a red square marker at the node i with the

size of 12 and red marker face color.

24

else

Else

25

plot(N(i,1),N(i,2),'ko','markersize',8,'mark

erfacecolor','k');

Plot a black circle marker at the node i with the

size of 8 and black marker face color.

26

end

End if

27

text(N(i,1),N(i,2),num2str(i),'color','g','font

size',12);

Plot a green node number at the node i with the

size of 12.

28

end

End the iteration

29

% blank line

30

F=zeros(3*NN,1);

Create the force vector and all members are

zero.

31

FN = [11 12 17 18 23 24];

Set the node number at which a load is applied

32

F(3*FN-2) = -20;

The x component of the load is -20

33

F(3*FN-1) = -10;

The y component of the load is -10


34

% blank line

35

scale = 20;

Set the scaling factor

36

for i=1:length(FN)

Begin an iteration in which i begins from 1 to the

number of loads

 

37

plot( N(FN(i),1),N(FN(i),2),'b*','markersize'

,10,'markerfacecolor','b');

Plot a blue star with the size of 10 and blue

marker face color at the node where a load is applied.

 

38

plot([N(FN(i),1),N(FN(i),1)-

100*sin(20/180*pi)],[N(FN(i),2),N(FN(i),2)

-100*cos(20/180*pi)],'-

>r','linewidth',2,'markerfacecolor','r');

Plot red “>” symbols to show the applied load in 20 degree angle with line width of 2 and red marker face color

39

% blank line

40

ang=-1.5*pi:0.5:-.5*pi;

41

xp=scale*cos(ang);

42

yp=scale*sin(ang);

 

43

plot(N(FN(i),1)+xp,N(FN(i),2)+yp,'–

r>','linewidth',2,'markerfacecolor','r');

Plot 7 red “>” symbols to show the applied moment with line width of 2 and red marker

face color

44

end

End the iteration

45

% blank line

46

title('A schematic of the power

tower','fontsize',12);

Plot the title of this figure with 12 font size

47

set(gca,'fontsize',12);

Set the font size of current axis as 12

48

axis equal tight

Set the axis property as “equal” and “tight”

49

print('-dtiff','-r400','frame.tif');

Print the figure in the name of “frame.tif” with

tiff format and 400 resolution

50

% blank line

51

K=zeros(3*NN,3*NN);

Determine the global stiffness matrix, its size is

3NN by 3NN and all members are zero.

52

for n=1:NE

Begin an iteration in which i begins from 1 to NN

53

L = sqrt((N(E(n,1),1)-N(E(n,2),1))^2 +

(N(E(n,1),2)-N(E(n,2),2))^2);

Calculate the length of element

54

lx = (N(E(n,2),1) – N(E(n,1),1))/L;

Calculate cos(theta)

55

ly = (N(E(n,2),2) – N(E(n,1),2))/L;

Calculate sin(theta)

56

a=A0*E0/L;

Calculate some parameters for the local matrix

57

b=6*E0*I0/L^2;

Calculate some parameters for the local matrix

58

c=12*E0*I0/L^3;

Calculate some parameters for the local matrix

59

d=E0*I0/L;

Calculate some parameters for the local matrix

60

% blank line

61

Ke = [ a  0 0   -a 0 0

Determine the local matrix at local coordinate

system

62

0   c b   0   -c b

63

0 b 4*d 0 -b 2*d

64

-a  0 0 a   0 0

65

0  -c -b   0   c -b

66

0 b 2*d 0 -b 4*d];

67

% blank line

Determine the transformation matrix

68

T = [ lx ly 0 0 0 0

69

-ly lx 0 0 0 0

70

0 0 1 0 0 0


71

0 0 0 lx ly 0

72

0 0 0 -ly lx 0

73

0 0 0 0 0 1];

74

% blank line

75

Ke = T'*Ke*T;

Determine the local matrix at global coordinate

system

76

i = E(n,1);

Get the near end number of an element

77

j = E(n,2);

Get the far end number of an element

78

dof = [3*i-2:3*i 3*j-2:3*j];

6 degrees of the element

79

K(dof,dof) = K(dof,dof) + Ke;

Assemble the global matrix from the local matrix

80

end

End the iteration

81

% blank line

82

fixeddofs = [1 2 3 4 5 6];

Fixed degrees

83

alldofs = [1:3*NN];

All degrees

84

freedofs = setdiff(alldofs,fixeddofs);

Get the active degrees

85

U(freedofs,:) =

K(freedofs,freedofs)\F(freedofs,:);

Calculate the equation Ax=b to get the

deformations for all active degrees.

86

U(fixeddofs,:)= 0;

Make the fixed degrees equal zero

87

% blank line

88

figure;

Plot a new figure

 

89

 

hold on;

holds the current plot and all axis properties so

that subsequent graphing commands add to the existing graph.

90

ND = N + scale*[U(1:3:end) U(2:3:end)];

Plot the node after the deformation. Noted the

deformation is multiplied by the scaling factor.

91

% blank line

92

for i=1:NE

Begin an iteration in which i begins from 1 to NN

 

93

 

plot([N(E(i,1),1),N(E(i,2),1)],[N(E(i,1),2),N(

E(i,2),2)],'linewidth',2,'color','k');

Plot a line starts from the near end (before deformation) to the far end (before deformation) of an element in black with the

line width of 2.

 

94

 

plot([ND(E(i,1),1),ND(E(i,2),1)],[ND(E(i,1),2

),ND(E(i,2),2)],'–b','linewidth',2);

Plot a line starts from the near end (after deformation)to the far end (after deformation) of an element in blue with the dashed line width

of 2.

95

end

End the iteration

96

% blank line

97

title(['The deformation magnified '

int2str(scale) ' times'],'fontsize',12);

Plot the title of this figure with 12 font size

98

axis equal tight off

Set the font size of current axis as 12

99

set(gca,'fontsize',12);

Set the axis property as “equal” and “tight”

100

print('-dtiff','-r400','U_frame.tif');

Print the figure in the name of “U_frame.tif”

with tiff format and 400 resolution


APPENDIX B

 

Node

x

y

z

1

-0.5

-0.3

0

2

0.5

-0.3

0

3

0.5

0.3

0

4

-0.5

0.3

0

5

6

7

8

-0.5

-0.3

2

0.5

-0.3

2

0.5

0.3

2

-0.5

0.3

2

 

 

 

 

Element

number

Element

Type

Final

Node

Initial

Node

1

1

1

5

2

1

2

6

3

1

3

7

4

1

4

8

5

2

5

6

6

2

6

7

7

2

7

8

8

2

8

5

9

3

1

6

10

3

2

7

11

3

3

8

12

3

4

5


APPENDIX C

The cross section of three elements in the shape of Equal Angles, Welded Columns, and Circular Rings

image.png

 

APPENDIX D

 

 image.png


APPENDIX E:

image.png

Structural Steelwork (AS 4100-1998) pa Young’s modulus: E=2e11 pa

Shear Modulus: 8e10 pa


APPENDIX F: Sample Transmission tower

 

 

 image.png

 

APPENDIX G: Finite Element Analysis of 3D Frame


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