05.04.1
Chapter 05.04
Lagrange Method of Interpolation – More Examples
Mechanical Engineering
Example 1
For the purpose of shrinking a trunnion into a hub, the reduction of diameter
D
of a
trunnion shaft by cooling it through a temperature change of
T
is given by
T
D
D
where
D
original diameter
in.
coefficient of thermal expansion at average temperature
F
in/in/
The trunnion is cooled from
F
80
to
F
108
, giving the average temperature as
F
14
.
The table of the coefficient of thermal expansion vs. temperature data is given in Table 1.
Table 1
Thermal expansion coefficient as a function of temperature.
Temperature,
F
T
Thermal Expansion Coefficient,
F
in/in/
80
6.47
6
10
0
6.00
6
10
–60
5.58
6
10
–160
4.72
6
10
–260
3.58
6
10
–340
2.45
6
10

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Chapter 05.04
Figure 1
Thermal expansion coefficient vs. temperature.
If the coefficient of thermal expansion needs to be calculated at the average temperature of
F
14
, determine the value of the coefficient of thermal expansion at
F
14
T
using a
first order Lagrange polynomial.
Solution
For first order Lagrange polynomial interpolation (also called linear interpolation), the
coefficient of thermal expansion is given by
1
0
)
(
)
(
)
(
i
i
i
T
T
L
T
)
(
)
(
)
(
)
(
1
1
0
0
T
T
L
T
T
L

Lagrange Method of Interpolation-More Examples: Mechanical Engineering
05.04.3
Figure 2
Linear interpolation.
Since we want to find the coefficient of thermal expansion at
F
14
T
, we choose two
data points that are closest to
F
14
T
and that also bracket
F
14
T
. The two points
are
0
0
T
and
F
60
1
T
.
Then
6
0
0
10
00
.
6
,
0
T
T
6
1
1
10
58
.
5
,
60
T
T
gives
1
0
0
0
0
)
(
j
j
j
j
T
T
T
T
T
L
1
0
1
T
T
T
T
1
1
0
1
1
)
(
j
j
j
j
T
T
T
T
T
L
0
1
0
T
T
T
T
Hence
)
(
)
(
)
(
1
0
1
0
0
1
0
1
T
T
T
T
T
T
T
T
T
T
T
0
60
),
10
58
.
5
(
0
60
0
)
10
00
.
6
(
60
0
60
6
6
T
T
T
(
x
0
,
y
0
)
(
x
1
,
y
1
)
f
1
(
x
)
x
y

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