Project I Tutorial
MECH 300H
Spring 2008
Project Description
Write a FEM code using MATLAB for stress analysis of a bridge as
described below.
A small railroad bridge is constructed of steel members, all of have a
cross-sectional area of 3250 mm2, Young’s modulus of 1GPa. A train
stops on the bridge, and the loads applied to the truss on one side of the
bridge are as show in the following figure. Determine the nodal
displacements and element stresses.
Project Description
What to submit?
(1) Both the hard copy and the electronic copy of your code
with sufficient comments. (Send the electronic copy to
honggang@ust.hk)
In your code, the following major components must be
included (a) formulation of element stiffness matrix, (b) stiffness
matrix assembling, (c) construction of force vector, (d)
condensed matrix, (e) linear system solver, and (f) post
computation.
(2) A short report – description of your implementation, such as
number of nodes, elements, results and discussion.
Notes:
Independent work is required!
Grading will be based on (a) the correctness of your results, (b)
report writing (complete yet concise) and (c) the generality of
your code.
Project task
- Scheme is the most important!
You should know how to solve this problem
by hand.
- Project task: writing the code
Tell the computer how to solve this problem
automatically.
Program steps
The program consists of these steps.
•
•
•
•
•
•
InputData
Input the information of nodes and elements
FormEquations
Form the local and global Equation
BoundaryCondition Add the B.C.
CondenseEquations Condense the system equations
SolveEquations
Solve the equations
Postcalculation
Calculation the displacements and stress
Program steps
[Main program]
% 1D Bar ELEMENT ON 2D PLANE TRUSS
ANALYSIS USING FEM
function [] = bar();
clear all; close all;
%The program consists of these steps.
InputData;
FormEquations;
BoundaryCondition;
CondenseEquations;
SolveEquations;
Postcalution;
Input data
•The number of nodes and elements
NN=7, NE=11 %dimension of global matrix: 2NN* 2NN
•The information of node
x(NN), y(NN)
% -x , y coordinate
•The information of element
E(NE), A(NE), L(NE), Angle(NE) %properties of each bar
Noc=[1 2; 1 3; 3 2; 2 4; 3 4; 3 5; 5 4; 4 6;5 6; 5 7; 7 6]
%connect joints of each bar
•The information of B. C
BC_U=[0 0 1 1 1 1 1 1 1 1 1 1 1 0 ]';
%displacement B.C. 0-known; 1-unknown
BC_F=1*10e3[0 -280 0 0 0 -210 0 0 0 -280 0 0 0 -360]';
%force B.C.
N
Noc(N,1)
Noc(N,2)
Angle(N)
1
1
2
60
2
1
3
0
3
3
2
120
4
2
4
0
5
3
4
60
6
3
5
0
7
5
4
120
8
4
6
0
9
5
6
60
10
5
7
0
11
7
6
120
Form Equations
Task:
The local stiffness matrix is assembled into the global matrix.
KL  KG
For the Nth Element
N1=Noc(N,1) %Index of the first node
N2=Noc(N,2) %Index of the second node
N1*2-1
1
cos2 


AE  sin  cos
KL 
L   cos2 

 sin  cos
N1*2
2
N2*2-1
sin  cos
3
 cos2 
sin 2 
 sin  cos
 sin  cos
cos2 
 sin 2 
sin  cos
N2*2
Global index
4 Local index
 sin  cos  1

 sin 2   2
sin  cos  3

sin 2   4
N1*2-1
N1*2
N2*2-1
N2*2
Form Equations
Pseudo code:
Do N = 1, NE
N1=Noc(N,1)
N2=Noc(N,2)
KG(N1*2-1,N1*2-1)=KG(N1*2-1,N1*2-1)+KL(1,1)
KG(N1*2-1,N1*2 ) =KG(N1*2-1,N1*2 )+KL(1,2)
KG(N1*2-1,N2*2-1)=KG(N1*2-1,N2*2-1)+KL(1,3)
KG(N1*2-1,N2*2 ) =KG(N1*2-1,N2*2 )+KL(1,4)
KG(N1*2
KG(N1*2
KG(N1*2
KG(N1*2
,N1*2-1) =KG(N1*2
,N1*2 ) =KG(N1*2
,N2*2-1) =KG(N1*2
,N2*2 ) =KG(N1*2
,N1*2-1)+KL(2,1)
,N1*2 )+KL(2,2)
,N2*2-1)+KL(2,3)
,N2*2 )+KL(2,4)
KG(N2*2-1,N1*2-1)=KG(N2*2-1,N1*2-1)+KL(3,1)
KG(N2*2-1,N1*2 ) =KG(N2*2-1,N1*2 )+KL(3,2)
KG(N2*2-1,N2*2-1)=KG(N2*2-1,N2*2-1)+KL(3,3)
KG(N2*2-1,N2*2 ) =KG(N2*2-1,N2*2 )+KL(3,4)
KG(N2*2
KG(N2*2
KG(N2*2
KG(N2*2
,N1*2-1) =KG(N2*2
,N1*2 ) =KG(N2*2
,N2*2-1) =KG(N2*2
,N2*2 ) =KG(N2*2
,N1*2-1)+KL(4,1)
,N1*2 )+KL(4,2)
,N2*2-1)+KL(4,3)
,N2*2 )+KL(4,4)
Hint:
Due to the symmetry of the Matrix,
you can write the code more short
and efficient.
Condense Equations
Task:
The global systems are condensed in order to solve it.
KG U  F
0
0

KG  KC
F  FC
1
 k1,1 k1, 2 k1,3
k
 1,1 k1, 2 k1,3
 k1,1 k1, 2 k1,3

...
...
 ...
k15,1 k15, 2 k15,3

k16,1 k16, 2 k16,3
…
1

UC  KC \ FC
0
BC_U
0
k1,16 
 u1 
 f1 
u 
f 
... k1,15 k1,16 
0
 2
 2
 u3 
 f3 
... k1,15 k1,16 
1
    

…
...
...
...
 ... 
 ... 
u15 
 f15 
... k15,15 k1,16 
1

 
 
... k16,15 k16,16 1616 u16 161  f16 161 0
...
k1,15
Scheme:
If BC_U(N) ==1, then the element in the Nth line and Nth column should be reserved.
Condense Equations
Pseudo code:
CI=0
Do I=1 , NN*2
If (BC_U(I) == 1) Then
CI=CI+1
CJ=0
Do J=1 , NN*2
If (BC_U(J) == 1) Then
CJ=CJ+1
KC(CI,CJ)=KG(I,J)
Subscript:
I ,J
CI,CJ
- the index of KG(I ,J)
- the index of KC(CI,CJ)