Chapter 07.03
Simpson’s 1/3 Rule for Integration-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 fit into a steel hub (Figure 1).
Figure 1 Trunnion to be slid through the hub after contracting.
The equation that gives the diametric contraction, in inches of the trunnion in dry-ice/alcohol
(boiling temperature is  108F ) is given by:
108
D  12.363
  1.2278  10
11
T 2  6.1946  10 9 T  6.015  10 6 dT
80
a) Use Use Simpson’s 1/3 Rule to find the approximate value of D .
b) Find the true error, E t , for part (a).
c) Find the absolute relative true error for part (a).
Solution
a)

ba
ab
f (a)  4 f 
  f (b)

6 
 2 

a  80
b  108
ab
 14
2
f (T )  12.363  1.2278  10 11T 2  6.1946  10 9 T  6.015  10 6
D 

07.03.1

07.03.2
Chapter 07.03

f 80  12.363  1.2278  10 11 80  6.1946  10 9 80  6.015  10 6
2

5
 7.9519  10
2
f  108  12.363  1.2278  10 11  108  6.1946  10 9  108  6.015  10 6

5
 6.4322  10
2
f  14  12.363  1.2278  10 11  14  6.1946  10 9  14  6.015  10 6



5
 7.3262  10

 b  a 
ab
D  
  f a   4 f 
  f b 
 6 
 2 

  108  80 

 f 80  4 f  14  f  108
6


  188 
5
5
5

 7.9519  10  4 7.3262  10  6.4322  10
 6 
 0.013689 in




b) The exact value of the above integral is
108
  1.2278 10
D  12.363
11

T 2  6.1946  10 9 T  6.015  10 6 dT
80
 0.013689 in
so the true error is
Et  True Value  Approximat e Value
 0.013689   0.013689
 0.0000
c) The absolute relative true error, t , would then be
t 
True Error
 100 %
True Value
0.0000
 100 %
 0.013689
 0.0000 %

Example 2
A trunnion of diameter 12.363 has to be cooled from a room temperature of 80F before it
is shrink fit into a steel hub (Figure 1). The equation that gives the diametric contraction, in
inches of the trunnion in dry-ice/alcohol (boiling temperature is  108F ) is given by:
108
D  12.363
  1.2278  10
11
T 2  6.1946  10 9 T  6.015  10 6 dT
80
a) Use four segment Simpson’s 1/3 Rule to find the contraction.
b) Find the true error, E t , for part (a).
Simpson’s 1/3 Rule for Integration-More Examples: Mechanical Engineering
07.03.3
c) Find the absolute relative true error for part (a).
Solution


n 1
n2
ba 
a) D 
f (T0 )  4  f (Ti )  2  f (Ti )  f (Tn )

3n 
i 1
i 2


i  odd
i  even
n4
a  80
b  108
ba
h
n
 108  80

4
 47
f (T )  12.363  1.2278  10 11T 2  6.1946  10 9 T  6.015  10 6




n 1
n2
ba
D 
f T0   4  f Ti   2  f Ti   f Tn 


3n
i 1
i 2
i  odd
i  even




3
2
 108  80 

f 80  4  f Ti   2  f Ti   f  108

34 
i 1
i 2


i  odd
i  even
 188
 f (80)  4 f (T1 )  4 f (T3 )  2 f (T2 )  f (108)

12
 188
 f (80)  4 f (33)  4 f (61)  2 f (14)  f (108)

12
7.9519  10 5  47.6725  10 5   47.0257  10 5 
 15.667 

5
5
 27.3321  10   6.4322  10

 0.013689 in
Since
f T0   f 80

 12.363  1.2278  10 11 80  6.1946  10 9 80  6.015  10 6
 7.9519  10
2
f T1   f 80  47
 f 33

 12.363  1.2278  10 11 33  6.1946  10 9 33  6.015  10 6
 7.6725  10
f T2   f 33  47

5
2
5

07.03.4
Chapter 07.03
 f (14)

 12.363  1.2278  10 11  14  6.1946  10 9  14  6.015  10 6
2

 7.3262  10 5
f T3   f  14  47
 f (61)
 12.363(1.2278  10 11 (61) 2  6.1946  10 9 (61)  6.015  10 6 )
 7.0257  10 5
f T4   f Tn 
 f  108

 12.363  1.2278  10 11  108  6.1946  10 9  108  6.015  10 6
2
 6.4322  10

5
b) The exact value of the above integral is
108
D  12.363
  1.2278  10
11
T 2  6.1946  10 9 T  6.015  10 6 dT
80
 0.013689 in
so the true error is
Et  True Value  Approximat e Value
 0.013689   0.013689
 0.0000
c) The absolute relative true error, t , would then be
True Error
 100 %
True Value
0.0000

 100 %
 0.013689
 0.0000 %
t 
Table 1 Values of Simpson’s 1/3 Rule for Example 2 with multiple segments.
n
Approximate Value
Et
t %
2
4
6
8
10
0.013689
0.013689
0.013689
0.013689
0.013689
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000