Chapter 04.06
Gaussian Elimination – More Examples
Mechanical Engineering
Example 1
A trunnion of diameter 12.363" has to be cooled from a room temperature of 80F before it
is shrink fitted into a steel hub (Figure 1).
Figure 1 Trunnion to be slid through the hub after contracting.
The equation that gives the diametric contraction D of the trunnion in a dry-ice/alcohol
mixture (boiling temperature is  108F ) is given by
108
D  12.363   (T )dT
80
The equation for the thermal expansion coefficient,   a1  a 2T  a3T 2 , is obtained using
regression analysis where the constants of the model are found by solving the following
simultaneous linear equations.

24
 2860
7.26  10 5   a1   1.057  10 4 




7.26  10 5
 1.86472  10 8  a 2    1.04162  10  2 
  2860
7.26  10 5  1.86472  10 8 5.24357  1010   a3  

2.56799




Find the values of a1 , a 2 , and a 3 using naïve Gauss elimination.
Solution
Forward Elimination of Unknowns
Since there are three equations, there will be two steps of forward elimination of unknowns.
04.06.1
04.06.2
Chapter 04.06
First step
Divide Row 1 by 24 and then multiply it by −2860, that is, multiply Row 1 by
 2860 24  119.17 .

 0.012596
Row 1   119.17    2860 3.4082  10 5  8.6515  10 7
Subtract the result from Row 2 to get

24
 2860
7.26  10 5   a1   1.057  10 4 




0
3.8518  10 5
 9.9957  10 7  a 2   2.1797  10 3 

7.26  10 5  1.86472  10 8
5.243  1010   a3   2.56799 

Divide Row 1 by 24 and then multiply it by 7.26  10 5 , that is, multiply Row 1 by
7.26  10 5 24  30250 .


Row 1  30250  7.26  10 5 8.6515  10 7 2.1962  1010
Subtract the result from Row 3 to get
24
 2860
7.26  10 5   a1   1.057  10 4 




3.8518  10 5  9.9957  10 7  a 2   2.1797  10 3 
0
 0  9.9957  10 7 3.04742  1010   a3    0.62944 




3.1974
Second step
We now divide Row 2 by 3.8518  10 5 and then multiply it by  9.9957  10 7 , that is,
multiply Row 2 by  9.9957  10 7 3.8518  10 5  259.50 .

Row 2   259.50  0  9.9957  10 7 2.5939  1010
Subtract the result from Row 3 to get
24
 2860
7.26  10 5   a1   1.057  10 4 

 
5
7 
3 
 0 3.8518  10  9.9957  10  a 2    2.1797  10 
0
0
4.5349  10 9   a3   6.3788  10  2 

Back Substitution
From the third equation,
4.5349  10 9 a3  6.3788  10 2
 6.3788  10 2
a3 
4.5349  10 9
 1.4066  10 11
Substituting the value of a 3 in the second equation,


3.8518  10 5 a 2   9.9957  10 7 a3  2.1797  10 3
a2 


2.1797  10 3   9.9957  10 7 a3
3.8518  10 5
 0.56565
Gaussian Elimination-More Examples: Mechanical Engineering

 
2.1797  10 3   9.9957  10 7   1.4066  10 11
3.8518  10 5
 2.0087  10 9
Substituting the values of a 2 and a3 in the first equation,

04.06.3

24a1   2860a 2  7.26  10 5 a3  1.057  10 4
1.057  10 4   2860a 2  7.26  10 5 a3
24
4
1.057  10   2860  2.0087  10 9  7.26  10 5   1.4066  10 11

24
6
 5.0690  10
Hence the solution vector is
6
 a1   5.0690  10 
a    2.0087  10 9 

 2 
 a3   1.4066  10 11 
a1 


SIMULTANEOUS LINEAR EQUATIONS
Topic
Gaussian Elimination – More Examples
Summary Examples of Gaussian elimination
Major
Mechanical Engineering
Authors
Autar Kaw
July 12, 2016
Date
Web Site http://numericalmethods.eng.usf.edu

