Tuss Fea Using MatLab Solution Assignment Paper (Dissertation Sample)

2D TRUSS SOLUTION WITH MATLABAuthor name
Project Objectives
Purpose of this document is to provide numerical solution for 2D Truss for weight optimization and material properties.
Table of Contents TOC \o "1-3" \h \z \u 2Problem Statement PAGEREF _Toc436669683 \h 42.1First Configuration PAGEREF _Toc436669684 \h 42.2Second Configuration PAGEREF _Toc436669685 \h 52.3Material Properties PAGEREF _Toc436669686 \h 52.4Assumption PAGEREF _Toc436669687 \h 53Mathematical Modeling PAGEREF _Toc436669688 \h 63.1Frist Configuration PAGEREF _Toc436669689 \h 63.2MATLAB Code for First Configuration PAGEREF _Toc436669690 \h 103.3Second Configuration PAGEREF _Toc436669691 \h 153.4MATLAB Code for Second Configuration PAGEREF _Toc436669692 \h 194Setup of analysis for both configurations PAGEREF _Toc436669693 \h 235Comparison of both configurations PAGEREF _Toc436669694 \h 246Discussion PAGEREF _Toc436669695 \h 267Conclusion PAGEREF _Toc436669696 \h 268References PAGEREF _Toc436669697 \h 26
List of Tables
TOC \h \z \c "Table"
Table 11 Material properties of steel PAGEREF _Toc436640399 \h 5
Table 21 Truss nodes coordinates for configuration 1 PAGEREF _Toc436640400 \h 6
Table 22 Elements connected by nodes for configuration 1 PAGEREF _Toc436640401 \h 7
Table 23 Truss nodes coordinates for configuration 2 PAGEREF _Toc436640402 \h 15
Table 24 Elements connected by nodes for configuration 2 PAGEREF _Toc436640403 \h 16
Table 41 Element lengths and cross section for both configurations PAGEREF _Toc436640404 \h 25
Table 42 Comparison of stresses in elements for both configuration PAGEREF _Toc436640405 \h 25
Table 43 Strain Comparison for both truss configurations PAGEREF _Toc436640406 \h 26
Problem Statement
Truss structure is consisted of two-force members, when assembled nicely, represents an individual object in an organized way. Such force is instructed to be applied structurally to two points. This structure is used widely in many engineering applications especially construction industry. Trusses are light weight, efficient load carrying structures. The following report uses Finite Element approach to solve a truss system in two different configurations using MATLAB programming language to evaluate for the stresses, deflection and strain in the two different configurations and present a comparison of both configuration performances.
Assessment of a 2D truss structure with 7 members and 5 connecting points. There are two possible configurations of this truss system.
First Configuration
Figure STYLEREF 1 \s 1 SEQ Figure \* ARABIC \s 1 1 Truss Configuration 1
Second Configuration
Figure STYLEREF 1 \s 1 SEQ Figure \* ARABIC \s 1 2 Truss Configuration 2
The current objective is to choose a suitable cross section of the member using numerical discrete method based on Matlab programming. Detailed evaluation of stress, strain and overall performance of the structures are performed.
Material Properties
The following table shows the material properties used for the given truss members:
Material

Mild Steel

Young’s Modulus (E)

200000 MPa

Poisson’s ratio

0.27

Density

7860 kg/m3

Yield Strength

250 MPa

Table STYLEREF 1 \s 1 SEQ Table \* ARABIC \s 1 1 Material properties of steel
Assumption
Following assumption have been made for calculation of cross sectional area of the truss.
Max. Deflection at any node ≤ 5mm
Factor of safety for the truss system ≥ 1.5
Mathematical Modeling
Frist Configuration
The free body diagram showing the nodal degrees of freedom, element notations, forces and boundary conditions.
Figure STYLEREF 1 \s 2 SEQ Figure \* ARABIC \s 1 1 Free body diagram of first configuration
The coordinates of all the nodes in truss are tabulated below.
Node

Coordinate x

Coordinate Y

1

600

0

2

1400

0

3

1400

600

4

600

600

5

0

600

Table STYLEREF 1 \s 2 SEQ Table \* ARABIC \s 1 1 Truss nodes coordinates for configuration 1
Elements in truss extending from one node to other are defined in table below
Element

From Node

To Node

1

1

2

2

2

3

3

3

4

4

4

5

5

5

1

6

1

4

7

2

4

Table STYLEREF 1 \s 2 SEQ Table \* ARABIC \s 1 2 Elements connected by nodes for configuration 1
The nodal stiffness matrix for each element can be expressed as given in the equation below.
Where
A = cross sectional area of truss element.
E = Young’s modulus.
L = length of truss element.
c = cosine of angle formed by that truss element.
s = sine of angle formed by that truss element.
Stiffness Matrices for all 7 elements are given below.
A Matlab program is written to calculate the elemental stiffness matrices
All these elemental stiffness matrics are combined to make the global stinffnes matrix as given below.
Applied force matrix is given by where force is in N
The nodal displacement matrix U is related with Force matrix F by following equation
Where
U is nodal displacement matrix given by
By using the MATLAB code written for configuration 1. Following results have been obtained.
Nodal deflections are given in mm units.
Once the nodal displacements are known, the elemental displacements are calculated using MATLAB program. The change in elements length is recorded in matrix u where
Now that the elemental displacements are known, we can calculate the elemental forces using the following equation
Elements forces are given in N
Normal stresses in truss elements are given in MPa
Similarly strain is calculated by the MATLAB program using the change in length of elements divided by the original lengths
Figure STYLEREF 1 \s 2 SEQ Figure \* ARABIC \s 1 2 Truss configuration 1 plot in MATLAB
MATLAB Code for First Configuration
%MATLAB code for TRUSS-1.
clear
%INPUTS----------------
%Define structure properties
num_nodes=5; %specify the number of nodes
total_num=7; %specify the number of members
%Define nodal coordinates in order starting with node one
x(1)=0.600; y(1)=0;
x(2)=1.400; y(2)=0;
x(3)=1.400; y(3)=0.600;
x(4)=0.600; y(4)=0.600;
x(5)=0; y(5)=0.600;
%Define member connectivity start with member one
% Start node end node
mem_conn(1,1)=1; mem_conn(1,2)=2;
mem_conn(2,1)=2; mem_conn(2,2)=3;
mem_conn(3,1)=3; mem_conn(3,2)=4;
mem_conn(4,1)=4; mem_conn(4,2)=5;
mem_conn(5,1)=5; mem_conn(5,2)=1;
mem_conn(6,1)=1; mem_conn(6,2)=4;
mem_conn(7,1)=2; mem_conn(7,2)=4;
%input properties for each member
%Define Material Properties for Steel.
E=200*10^9; %Define Youngs Modulus
neu=0.27; %Define poisson ratio
%Defining nodal stiffness matrices
%Cross Sectional Area A for each member (Considering equal area for all)
A(1)=0.0001;
A(2)=0.0001;
A(3)=0.0001;
A(4)=0.0001;
A(5)=0.0001;
A(6)=0.0001;
A(7)=0.0001;
%Specify suppports
num_suppp=5; %number of degrees of freedom with suppports
supp(1)=3;%for the 1 to num_suppp indicate which dof are suppports
supp(2)=5;
supp(3)=6;
supp(4)=7;
supp(5)=8;
%Specify forces
%zero all the forces
for i=1:num_nodes*2;
F(i)=0.0;
end;
nforce=10; %number of degrees of freedom with forces given
F(1)=0; %specify force at each degree of freedom where the external force is known, F(dof)
F(2)=0;
F(3)=0;
F(4)=0;
F(5)=0;
F(6)=0;
F(7)=0;
F(8)=0;
F(9)=0;
F(10)=-10000 %Force at dof 10 in N
%INPUTS----------------
%Calculate the truss member lengths, and angles
for i=1:total_num;
dx=x(mem_conn(i,2))-x(mem_conn(i,1));
dy=y(mem_conn(i,2))-y(mem_conn(i,1));
L(i)=sqrt(dx^2+dy^2);
c(i)=dx/L(i);
s(i)=dy/L(i);
end;
%For each element create the global stiffness matrix and put it into the
%global stiffness matrix.
%zero the global stiffness matrix
for i=1:num_nodes*2;
for j=1:num_nodes*2;
kg_glb(i,j)=0.0;
end;
end;
%Create each element global stiffness matrix as transpose[T][k][T]
for m=1:total_num; %loop over each element
%zero k and T
for i=1:4;
for j=1:4;
k(i,j)=0.0;
T(i,j)=0.0;
end;
end;
%create T
T(1,1)=c(m) ; T(1,2)=s(m); T(1,3)=0; T(1,4)=0;
T(2,1)=-s(m); T(2,2)=c(m); T(2,3)=0; T(2,4)=0;
T(3,1)=0; T(3,2)=0; T(3,3)=c(m); T(3,4)=s(m);
T(4,1)=0; T(4,2)=0; T(4,3)=-s(m); T(4,4)=c(m);
%create k
k(1,1)=1.0; k(1,3)=-1.0; k(3,1)=-1.0; k(3,3)=1.0;
k=A(m)*E/L(m)*k;
%transform k into the element global stiffness matrix
k=T'*k*T;
%display each elemental stiffness matrix in command window
switch m
case 1
k1=k
case 2
k2=k
case 3
k3=k
case 4
k4=k
case 5
k5=k
case 6
k6=k
case 7
k7=k
end
%put each element k into the global stiffness matrix kg_glb
for i=1:2;
for j=1:2;
kg_glb(mem_conn(m,i)*2-1,mem_conn(m,j)*2-1)=kg_glb(mem_conn(m,i)*2-1,mem_conn(m,j)*2-1)+k(i*2-1,j*2-1);
kg_glb(mem_conn(m,i)*2,mem_conn(m,j)*2)=kg_glb(mem_conn(m,i)*2,mem_conn(m,j)*2)+k(i*2,j*2);
kg_glb(mem_conn(m,i)*2-1,mem_conn(m,j)*2)=kg_glb(mem_conn(m,i)*2-1,mem_conn(m,j)*2)+k(i*2-1,j*2);
kg_glb(mem_conn(m,i)*2,mem_conn(m,j)*2-1)=kg_glb(mem_conn(m,i)*2,mem_conn(m,j)*2-1)+k(i*2,j*2-1);
end;
end; %now kg_glb should be complete
end; %end loop over all the elements
%Put kg_glb into kg_glbs and modify kg_glbs to solve for the unknown displacements
%we save kg_glb so that we can calculate the reactions also.
kg_glbs=kg_glb;
%modify kg_glbs based on the suppport condit...