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Chapter 2 – The Basic FEA Procedure
noted that in a real structure the size of a stiffness matrix is rather large and can only be
solved with the help of a computer. Solving the above matrix equation by hand we get,
0 = 45 u
2
– 25 u
3
u
2
0.2500
0 = -25 u
2
+ 55 u
3
– 30 u
4
Or u
3
= 0.4500
u
4
0.6167
5 = -30 u
3
+ 30 u
4
Example 2.3
In the spring structure shown k
1
= 10 lb./in., k
2
= 15 lb./in., k
3
= 20 lb./in., P= 5 lb. Determine the
deflection at nodes 2 and 3.
k
1
k
2
k
3
1
2
3
4
Figure 2.5
Solution:
Again apply the three steps outlined previously.
Step 1: Find the Element Stiffness Equations
Element 1:
1 2
[
K
(1)
] = 10 -10
1
-10 10
2
Element 2:
2 3
[
K
(2)
] = 15 -15
2
-15 15
3
Element 3:
3 4
[
K
(3)
] = 20 -20
3
-20 20
4
ME 273 Lecture Notes © by R. B. Agarwal
Finite Element Analysis
2
-11
Chapter 2 – The Basic FEA Procedure
Step 2: Find the Global stiffness matrix
1 2 3 4
1
10 -10
0
0
10 -10 0 0
2
-10 10 + 15 -15
0 = -10 25 -15 0
3
0 -15 15 + 20 -20
0 -15 35 -20
4
0 0 -20
20
0 0 -20 20
Now the global structural equation can be written as,
F
1
10 -10 0 0 u
1
F
2
= -10 25 -15 0 u
2
F
3
0 -15 35 -20 u
3
F
4
0 0 -20 20 u
4
Step 3: Solve for Deflections
The known boundary conditions are: u
1
= u
4
= 0, F
3
= P = 3lb. Thus, rows and columns 1 and 4 will drop
out, resulting in the following matrix equation,
0
= 25 -15 u
3
3 -15 35 u
3
Solving, we get u
2
= 0.0692 & u
3
= 0.1154
Example 2.4
In the spring structure shown, k
1
= 10 N/mm, k
2
= 15 N/mm, k
3
= 20 N/mm, k
4
= 25 N/mm, k
5
= 30
N/mm, k
6
= 35 N/mm. F2 = 100 N. Find the deflections in all springs.
k
1
k
3
k
2
F
2
k
6
Fig. 2.6
k
4
k
5
1 2 3 4
ME 273 Lecture Notes © by R. B. Agarwal
Finite Element Analysis
2
-12
Chapter 2 – The Basic FEA Procedure
Solution:
Here again, we follow the three-step approach described earlier, without specifically
mentioning at each step.
Element 1:
1 4
[
K
(1)
] = 10 -10
1
-10 10
4
Element 2:
1 2
[
K
(2)
] = 15 -15
1
-15 15
2
Element 3:
2 3
[
K
(3)
] = 20 -20
2
-20 20
3
Element 4:
2 3
[
K
(4)
] = 25 -25
2
-25 25
3
Element 5:
2 4
[
K
(5)
] = 30 -30
2
-30 30
4
Element 6:
3 4
[
K
(6)
] = 35 -35
3
-35 35
4
The global stiffness matrix is,
1 2 3 4
10+15 -15
0 -10
1
[K
g
] = -15 15+20+25+30 -20-25 -30
2
0 -20-25 20+25+35 -35
3
-10
-30
-35 10+30+35
4
And simplifying, we get
25 -15 0 -10
[K
g
] = -15 90 -45 -30
0 -45 80 -35
-10 -30 -35 75
ME 273 Lecture Notes © by R. B. Agarwal
Finite Element Analysis
2
-13
Chapter 2 – The Basic FEA Procedure
And the structural equation is,
F
1
25 -15 0 -10 u
1
F
2
= -15 90 -45 -30 u
2
F
3
0 -45 80 -35 u
3
F
4
-10 -30 -35 75 u
4
Now, apply the boundary conditions, u
1
= u
4
= 0, F
2
= 100 N. This is carried out by
deleting the rows 1 and 4, columns 1 and 4, and replacing F
2
by 100N. The final matrix
equation is,
100
90 -45 u
2
0 = -45 80 u
3
Which gives
u
2
= 1.5459
u
3
0.8696
Deflections:
Spring 1:
u
4
– u
1
= 0
Spring 2:
u
2
– u
1
= 1.54590
Spring 3:
u
3
– u
2
= -0.6763
Spring 4:
u
3
– u
2
= -0.6763
Spring 5:
u
4
– u
2
= -1.5459
Spring 6:
u
4
– u
3
= -0.8696
2.3.4 Boundary Conditions with Known Values
Up to now we have considered problems that have known applied forces, and no known
values of deflection. Now we will consider the procedure for applying the boundary
conditions where, deflections on some nodes are known. Solutions of these problems are
found by going through some additional steps. As discussed earlier, after obtaining the
ME 273 Lecture Notes © by R. B. Agarwal
Finite Element Analysis
2
-14
Chapter 2 – The Basic FEA Procedure
structural global matrix equation, deflections are found by solving the equation by
applying a numerical scheme in a computer solution. However, when there are known
nodal values and unknown nodal forces, the method is not directly applicable. In this
situation, the structural equation is first modified by incorporating all boundary
conditions and then the final matrix equation is solved by a computer using a numerical
method, as mentioned earlier. The following procedure traces the necessary steps for
solving problems that involve known nodal values.
2.3.5 Procedure for incorporating the known Nodal Values in the Final
Structural Equation
There are two methods that are frequently used for applying boundary conditions to a
structural matrix equation. In one method, the matrices are partitioned into two parts with
known and unknown terms. In the second method, the known nodal values are applied
directly in the structural matrix. Both methods can be used with equal effectiveness. The
first method will not be discussed here. Details of the second method follow.
Consider the following linear equations,
k
11
u
1
+ k
12
u
2
+ k
13
u
3
+ k
14
u
4
= F
1
(2.2)
k
21
u
1
+ k
22
u
2
+ k
23
u
3
+ k
24
u
4
= F
2
(2.3)
k
31
u
1
+ k
32
u
2
+ k
33
u
3
+ k
34
u
4
= F
3
(2.4)
k41
u
1
+ k
42
u
2
+ k
43
u
3
+ k
44
u
4
= F
4
(2.5)
These linear algebraic equations can be written in matrix form as follows.
k
11
k
12
k
13
k
14
u
1
F
1
k
21
k
22
k
23
k
24
u
2
= F
2
k
31
k
32
k
33
k
34
u
3
F
3
k
41
k
42
k
43
k
44
u
4
F
4
Let the known nodal value at node 2 be u
2
= U
2
(a constant), then by the linear spring
equation
F
2
= k
22
U
2
Therefore, equation (2.2 – 2.5)) above can be reduced to k
22
u
2
= k
22
U
2
= F
2
and the
matrix with this boundary condition can be written as
k
11
k
12
k
13
k
14
u
1
F
1
0 k
22
0 0
u
2
= F
2
k
31
k
32
k
33
k
34
u
3
F
3
k
41
k
42
k
43
k
44
u
4
F
4
ME 273 Lecture Notes © by R. B. Agarwal
Finite Element Analysis
2
-15