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Direct Method of Interpolation-More Examples: Mechanical Engineering by hI14064W

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									Chapter 05.02
Direct 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
        D  DT
where
        D  original diameter in. 
         coefficient of thermal expansion at average temperature in/in/F
The trunnion is cooled from 80F to  108F , giving the average temperature as  14F .
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, T F Thermal Expansion Coefficient,  in/in/ F
                      80                         6.47  10 6
                       0                         6.00  10 6
                     –60                         5.58  10 6
                    –160                         4.72  10 6
                    –260                         3.58  10 6
                    –340                         2.45  10 6




05.02.1
05.02.2                                                                      Chapter 05.02




                 Figure 1 Thermal expansion coefficient vs. temperature.

If the coefficient of thermal expansion needs to be calculated at the average temperature of
 14F , determine the value of the coefficient of thermal expansion at T  14F using the
direct method of interpolation and a first order polynomial.

Solution
For first order polynomial interpolation (also called linear interpolation), we choose the
coefficient of thermal expansion given by
         T   a0  a1T
Direct Method of Interpolation – More Examples: Mechanical Engineering               05.02.3


         y



                                                           x1 , y1 




                                                              f1 x 

               x0 , y0 
                                                                                 x
          Figure 2 Linear interpolation.

Since we want to find the coefficient of thermal expansion at T  14F , and we are using a
first order polynomial, we need to choose the two data points that are closest to T  14F
that also bracket T  14F to evaluate it. The two points are T0  0 F and T1  60 F .
Then
         T0  0,  T0   6.00  10 6
        T1  60 ,  T1   5.58  10 6
gives
         0  a 0  a1 0   6.00  10 6
        60   a 0  a1  60   5.58  10 6
Writing the equations in matrix form, we have
       1 0  a0  6.00  106 
       1  60  a                6 
                  1  5.58  10 
Solving the above two equations gives
        a0  6.00  10 6
        a1  0.007  10 6
Hence
         T   a0  a1T
             6.00  10 6  0.007  10 6 T ,  60  T  0
At T  14F,
          14   6.00  10 6  0.007  10 6  14 
                   5.902  10 6 in/in/ F
05.02.4                                                                      Chapter 05.02


Example 2
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
        D  DT
where
        D  original diameter in. 
         coefficient of thermal expansion at average temperature in/in/F
The trunnion is cooled from 80F to  108F , giving the average temperature as  14F .
The table of the coefficient of thermal expansion vs. temperature data is given in Table 2.

            Table 2 Thermal expansion coefficient as a function of temperature.
            Temperature, T F Thermal Expansion Coefficient,  in/in/ F
                      80                         6.47  10 6
                       0                         6.00  10 6
                     –60                         5.58  10 6
                    –160                         4.72  10 6
                    –260                         3.58  10 6
                    –340                         2.45  10 6

If the coefficient of thermal expansion needs to be calculated at the average temperature of
 14F , determine the value of the coefficient of thermal expansion at T  14F using the
direct method of interpolation and a first order polynomial.

Solution
For second order polynomial interpolation (also called quadratic interpolation), we choose
the coefficient of thermal expansion given by
         T   a 0  a1T  a 2T 2
            y


                               x1 , y1 
                                                               x2 , y 2 



                                                   f 2 x 


                 x0 , y 0 
                                                                                   x
           Figure 3 Quadratic interpolation.
Direct Method of Interpolation – More Examples: Mechanical Engineering                  05.02.5



Since we want to find the coefficient of thermal expansion at T  14F , and we are using a
second order polynomial, we need to choose the three data points that are closest to
T  14F that also bracket T  14F to evaluate it. These three points are T0  80 F ,
T1  0 F and T2  60 F .
Then
        T0  80 ,  T0   6.47  10 6
        T1  0,     T1   6.00  10 6
        T2  60 ,  T2   5.58  10 6
gives
         80  a0  a1 80  a2 802  6.47 106
         0  a0  a1 0  a2 02  6.00 106
          60  a0  a1  60  a2  602  5.58 106
Writing the three equations in matrix form, we have
       1 80 6400  a0  6.47  10 
                                             6

       1 0                                     
                    0   a1   6.00  10  6 
                           
       1  60 3600  a2  5.58  10  6 
                                            
Solving the above three equations gives
       a 0  6.00  10 6
        a1  6.5179  10 9
        a 2  8.0357  10 12
Hence
        T   6.00  10 6  6.5179  10 9 T  8.0357  10 12 T 2 ,  60  T  80
At T  14F,
         14   6.00  10 6  6.5179  10 9  14   8.0357  10 12  14 2
                   5.9072  10 6 in/in/ F
The absolute relative approximate error a obtained between the results from the first and
second order polynomial is
              5.9072  10 6  5.902  10 6
       a                                    100
                     5.9072  10 6
            0.087605%

Example 3
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
        D  DT
where
        D  original diameter in. 
05.02.6                                                                       Chapter 05.02


         coefficient of thermal expansion at average temperature in/in/F
The trunnion is cooled from 80F to  108F , giving the average temperature as  14F .
The table of the coefficient of thermal expansion vs. temperature data is given in Table 3.

              Table 3 Thermal expansion coefficient as a function of temperature.
              Temperature, T F Thermal Expansion Coefficient,  in/in/ F
                        80                         6.47  10 6
                         0                         6.00  10 6
                       –60                         5.58  10 6
                      –160                         4.72  10 6
                      –260                         3.58  10 6
                      –340                         2.45  10 6

   a) If the coefficient of thermal expansion needs to be calculated at the average
      temperature of  14F , determine the value of the coefficient of thermal expansion at
      T  14F using the direct method of interpolation and a first order polynomial. Find
      the absolute relative approximate error for the third order polynomial approximation.
   b) The actual reduction in diameter is given by
                 Tf

          D  D  dT
                 Tr

where Tr  room temperature F 
      T f  temperature of cooling medium F 
Since
      Tr  80 F
      T f  108 F
                 108
          D  D  dT
                  80
Find out the percentage difference in the reduction in the diameter by the above integral
formula and the result using the thermal expansion coefficient from part (a).

Solution
a) For third order polynomial interpolation (also called cubic interpolation), we choose the
coefficient of thermal expansion given by
         T   a 0  a1T  a 2T 2  a3T 3
Direct Method of Interpolation – More Examples: Mechanical Engineering                          05.02.7


         y

                                                                              x3 , y 3 

                                                         f 3 x 
                                       x1 , y1 




              x0 , y 0                                       x2 , y 2 


                                                                                            x

        Figure 4 Cubic interpolation.

Since we want to find the coefficient of thermal expansion at T  14F , and we are using a
third order polynomial, we need to choose the four data points closest to T  14F that also
bracket T  14F to evaluate it. Then the four points are T0  80 F , T1  0 F , T2  60 F
and T3  160 F .
        T0  80 ,            T0   6.47  10 6
        T1  0,      T1   6.00  10 6
        T2  60 ,  T2   5.58  10 6
        T3  160 ,  T3   4.72  10 6
gives
         80  a0  a1 80  a2 802  a3 803  6.47 106
         0  a0  a10  a2 02  a3 03  6.00 106
          60  a0  a1 60  a2  602  a3  603  5.58 106
          160  a0  a1 160  a2  1602  a3  1603  4.72 106
Writing the four equations in matrix form, we have
       1 80         6400     5.12  105  a0  6.47  106 
                                                          6 
       1     0         0         0         a1   6.00  10 
       1  60 3600  2.16  105  a2  5.58  10 6 
                                       6                  6 
       1  160 25600  4.096  10   a3  4.72  10 
                                                               
Solving the above four equations gives
       a 0  6.00  10 6
05.02.8                                                                                             Chapter 05.02


          a1  6.4786  10 9
          a 2  8.1994  10 12
          a 3  8.1845  10 15
Hence
  T   a 0  a1T  a 2T 2  a3T 3
           6.00  10 6  6.4786  10 9 T  8.1994  10 12 T 2  8.1845  10 15 T 3 ,  160  T  80
        14   6.00  10 6  6.4786  10 9  14   8.1994  10 12  14 2  8.1845  10 15  14 3
                 5.9077  10 6 in/in/ F
The absolute relative approximate error a obtained between the results from the second
and third order polynomial is
                5.9077  10 6  5.9072  10 6
         a                                       100
                         5.9077  10 6
              0.0083867%

b) In finding the percentage difference in the reduction in diameter, we can rearrange the
integral formula to
                   Tf
          D
               dT
           D Tr
and since we know from part (a) that
 (T )  6.00  10 6  6.4786  10 9 T  8.1994  10 12 T 2  8.1845  10 15 T 3 ,  160  T  80
we see that we can use the integral formula in the range from T f  108 F to Tr  80 F
Therefore,
                   Tf
          D
               dT
           D Tr
                   108

                         6.00 10         6.4786  10 9 T  8.1994  10 12 T 2  8.1845  10 15 T 3 dT
                                      6
               
                    80
                                                                                                                108
                                                T2                   T3                   T4
               6.00  10 6 T  6.4786  10 9     8.1994  10 12     8.1845  10 15   
                                                2                    3                    4  80
               1105.9  10 6
    D
So        1105.9  10 6 in/in using the actual reduction in diameter integral formula. If we
     D
use the average value for the coefficient of thermal expansion from part (a), we get
        D
             T
         D
              T f  Tr 
               5.9077  10 6  108  80 
               1110.6  10 6
Direct Method of Interpolation – More Examples: Mechanical Engineering            05.02.9


      D
and        1110.6  10 6 in/in using the average value of the coefficient of thermal
       D
expansion using a third order polynomial. Considering the integral to be the more accurate
calculation, the percentage difference would be

         a 
                                            
                1105.9  106   1110.6  106
                                                  100
                         1105.9  10 6
              0.42775%


      INTERPOLATION
      Topic    Direct Method of Interpolation
      Summary Examples of direct method of interpolation.
      Major    Mechanical Engineering
      Authors  Autar Kaw
      Date     October 3, 2012
      Web Site http://numericalmethods.eng.usf.edu

								
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