ВУЗ: Не указан
Категория: Не указан
Дисциплина: Не указана
Добавлен: 09.06.2020
Просмотров: 295
Скачиваний: 3
Chapter 2 – The Basic FEA Procedure
These equations can now be written in a matrix form, giving
k
1
-
k
1
0 u
1
F
1
-
k
1
k
1
+ k
2
- k
2
u
2
= F
2
0 -
k
2
k
2
u
3
F
3
This completes step 2 for assembling the element equations into a global equation. At this
stage, some important conceptual points should be emphasized and will be discussed
below.
2.3.1 Procedure for Assembling Element stiffness matrices
The first term on the left hand side in the above equation represents the stiffness constant
for the entire structure and can be thought of as an equivalent stiffness constant, given as
k
1
-
k
1
0
0
[K
eq
] =
- k
1
k
1
+ k
2
- k
2
0
0
- k
2
k
2
+ k
3
- k
3
0 0 -k
3
k
3
A single spring element with a value K
eq
will have an identical mechanical property as
the structural stiffness in the above example.
The assembled matrix equation represents the deflection equation of a structure without
any constraints, and cannot be solved for deflections without modifying it to incorporate
the boundary conditions. At this stage, the stiffness matrix is always symmetric with
corresponding rows and columns interchangeable.
The global equation was derived by applying equilibrium conditions at each node. In
actual finite element analysis, this procedure is skipped and a much simpler procedure is
used. The simpler procedure is based on the fact that the equilibrium condition at each
node must always be satisfied, and in doing so, it leads to an orderly placement of
individual element stiffness constant according to the node numbers of that element. The
procedure involves numbering the rows and columns of each element, according to the
node numbers of the elements, and then, placing the stiffness constant in its
ME 273 Lecture Notes © by R. B. Agarwal
Finite Element Analysis
2
-6
Chapter 2 – The Basic FEA Procedure
corresponding position in the global stiffness matrix. Following is an illustration of this
procedure, applied to the example problem.
Element 1:
1 2
k
1
K
(1)
= k
1
-k
1
1
-k
1
k
1
2
1 2
Element 2:
2
3
k
2
K
(2)
= k
2
-k
2
2
-k
2
k
2
3
2 3
Assembling it according with the above-described procedure, we get,
1
2 3 4
1
k
1
-k
1
0
[ K
g
] =
2
-k
1
k
1
+ k
2
-k
2
3
0 -k
2
k
2
Note that the first constant k
1
in row 1 and column 1 for element 1 occupies the row 1 and column 1
in the global matrix. Similarly, for element 2, the constant k
2
in row 2 and column 2 occupies exactly
the same position (row 2 and column 2) in the global matrix, etc.
In a large model, the node numbers can occur randomly, but the assembly procedure remains the
same. It’s important to place the row and column elements from an element into the global matrix at
exactly the same position corresponding to the respective row and column.
2.3.2 Force matrix
At this stage, the force matrix is represented in a general form, with unknown forces F
1
,
F
2
, and F
3
F
1
F
2
F
3
ME 273 Lecture Notes © by R. B. Agarwal
Finite Element Analysis
2
-7
Chapter 2 – The Basic FEA Procedure
Representing the external forces at nodes 1, 2, and 3, in general terms, and not in terms of
the actual known value of the forces. In the example problem, F
1
= F
2
= 0 and F
3
= F. the
actual force matrix is then
0
0
F
Generally, the assembled structural matrix equation is written in short as {F}=[k]{u}, or
simply, F = k u, with the understanding that each term is an m x n matrix where m is the
number of rows and n is the number of columns.
Step 3: Solve the global equation for deflections at nodes.
There are two steps for obtaining the deflection values. In the first step, all the boundary
conditions are applied, which will result in reducing the size of the global structural
matrix. In the second step, a numerical matrix solution scheme is used to find deflection
values by using a computer. Among the most popular numerical schemes are the Gauss
elimination and the Gauss-Sedel iteration method. For further reading, refer to any
numerical analysis book on this topic. In the following examples and chapters, all the
matrix solutions will be limited to a hand calculation even though the actual matrix in a
finite element solution will always use one of the two numerical solution schemes
mentioned above.
2.3.3 Boundary conditions
In the example problem, node 1 is fixed and therefore u
1
= 0. Without going into a
mathematical proof, it can be stated that this condition is effected by deleting row 1 and
column 1 of the structural matrix, thereby reducing the size of the matrix from 3 x 3 to 2 x 2.
In general, any boundary condition is satisfied by deleting the rows and columns
corresponding to the node that has zero deflection. In general, a node has six degrees of
freedom (DOF), which include three translations and three rotations in x, y and z directions.
In the example problem, there is only one degree of freedom at each node. The node deflects
only along the axis of the spring.
In this section, the finite element analysis procedure for a spring structure has been established.
The following numerical example will utilize the derivation and concepts developed above.
ME 273 Lecture Notes © by R. B. Agarwal
Finite Element Analysis
2
-8
Chapter 2 – The Basic FEA Procedure
Example 2.2
In the given spring structure, k
1
= 20 lb./in., k
2
= 25 lb./in., k
3
= 30 lb./in., F = 5 lb. Determine
deflection at all the nodes.
K
1
k
2
P k
3
F
1
2
3
4
Figure 2.4
Solution
We would apply the three steps discussed earlier.
Step 1: Derive the Element Equations
As derived earlier, the stiffness matrix equations for an element e is,
K
(e)
= k
e
-k
e
-k
e
k
e
Therefore, stiffness matrix of elements 1, 2, and 3 are,
1 2
Element 1:
K
(1)
= 20 -20
1
-20 20
2
Element 2:
1 2
K
(2)
= 25 -25
1
-25 25
2
ME 273 Lecture Notes © by R. B. Agarwal
Finite Element Analysis
2
-9
Chapter 2 – The Basic FEA Procedure
Element 3:
1
2
K
(3)
= 30 -30
1
-30 30
2
Step 2: Assemble element equations into a global equation
Assembling the terms according to their row and column position, we get
1 2 3 4
20
-20
0
0
1
[K
g
] = -20 20+25 -25
0
2
0 -25 25+30 -30
3
0 0
30
30
4
Or, by simplifying
20 -20 0 0
[K
g
] =
-20 45 -25 0
0 -25 55 -30
0 0 30 30
The global structural equation is,
F
1
20 -20 0 0 u
1
F
2
= -20 45 -25 0 u
2
F
3
0 -25 55 -30
u
3
F
4
0 0 30 30
u
4
Step 3: Solve for deflections
First, applying the boundary conditions u
1
=0, the first row and first column will drop out. Next,
F
1
= F
2
= F
3
= 0, and F
4
= 5 lb. The final form of the equation becomes,
0
45 -25 0 u
2
0 = -25 55 -30
u
3
5 0 -30 30 u
4
This is the final structural matrix with all the boundary conditions being applied. Since
the size of the final matrices is small, deflections can be calculated by hand. It should be
ME 273 Lecture Notes © by R. B. Agarwal
Finite Element Analysis
2
-10