Lagrange Method of Interpolation-More Examples Mechanical Engineering by linzhengnd

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									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
                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




05.04.1
05.04.2                                                                             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
 14F , determine the value of the coefficient of thermal expansion at T  14F 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
           (T )   Li (T ) (Ti )
                   i 0

                 L0 (T ) (T0 )  L1 (T ) (T1 )
Lagrange Method of Interpolation-More Examples: Mechanical Engineering                  05.04.3




           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 , we choose two
data points that are closest to T  14F and that also bracket T  14F . The two points
are T0  0 and T1  60 F .
Then
        T0  0,  T0   6.00  10 6
        T1  60 ,  T1   5.58  10 6
gives
                     1     T  Tj
        L0 (T )  
                    j 0   T0  T j
                    j 0

                    T  T1
                
                    T0  T1
                    1 T T
        L1 (T )  
                            j

                    j 0   T1  T j
                    j 1

                    T  T0
               
                    T1  T0
Hence
                 T  T1            T  T0
         (T )           (T0 )           (T1 )
                 T0  T1           T1  T0
                 T  60                     T 0
                        (6.00  10 6 )           (5.58  10 6 ),  60  T  0
                 0  60                     60  0
05.04.4                                                                       Chapter 05.04


                      14  60                    14  0
           (14)             (6.00  10 6 )           (5.58  10 6 )
                      0  60                      60  0
                    0.76667 (6.00  10 6 )  0.23333 (5.58  10 6 )
                 5.902  10 6 in/in/ F
You can see that L0 (T )  0.76667 and L1 (T )  0.23333 are like weightages given to the
coefficients of thermal expansion at T  0 and T  60F to calculate the coefficient of
thermal expansion at T  14F .

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 a
second order Lagrangian polynomial. Find the absolute relative approximate error for the
second order polynomial interpolation.

Solution
For second order Lagrange polynomial interpolation (also called quadratic interpolation), the
coefficient of thermal expansion given by
                    2
           (T )   Li (T ) (Ti )
                   i 0

                L0 (T ) (T0 )  L1 (T ) (T1 )  L2 (T ) (T2 )
Lagrange Method of Interpolation-More Examples: Mechanical Engineering               05.04.5




            y



                                                               (x1, y1)




                                                               f1(x)

                 (x0, y0)
                                                                                 x

                                    Figure 3 Quadratic interpolation.

Since we want to find the coefficient of thermal expansion at T  14F , we need to choose
data points that are closest to T  14F that also bracket T  14F to evaluate it. The
three points are T0  80 F , T1  0 and T2  60 F .
        T0  80 ,  T0   6.47  10 6
        T1  0,  T1   6.00  10 6
        T2  60 ,  T2   5.58  10 6
gives
                   2     T  Tj
        L0 (T )  
                  j 0   T0  T j
                  j 0

                   T  T1  T  T2   
                 T  T  T  T
                                     
                                       
                   0 1  0       2   
                     2 T T
        L1 (T )  
                               j

                   j 0 T1  T j
                  j 1

                   T  T0  T  T2 
                 T  T  T  T 
                                   
                   1 0  1       2 
                     2 T T
        L2 (T )  
                              j

                   j 0 T2  T j
                  j 2

                  T  T0  T  T1 
                  T  T  T  T 
                                
                  2    0  2    1 
05.04.6                                                                                           Chapter 05.04


Hence

                   T  T1  T  T2                T  T0    T  T2              T  T0    T  T1 
           (T )  
                           
                                        (T0 )  
                                                   T T      
                                                                T  T     (T1 )  
                                                                                     T T      
                                                                                                  T  T  (T2 ),
                                                                                                           
                   T0  T1  T0  T2              1    0    1    2              2    0    2 1 
                                                                                                      T0  T  T2
                     (14  0)(14  60)                      (14  80)(14  60)
           (14)                          (6.47  10 6 )                          (6.00  10 6 )
                      (80  0)(80  60)                          (0  80)(0  60)
                      (14  80)(14  0)
                                            (5.58  10 6 )
                      (60  80)(60  0)
                    (0.0575 )( 6.47  10 6 )  (0.90083 )( 6.00  10 6 )  (0.15667 )(5.58  10 6 )
                    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. 
         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 a third order Lagrange polynomial. Find the absolute relative approximate
      error for the third order polynomial approximation.
Lagrange Method of Interpolation-More Examples: Mechanical Engineering                        05.04.7



   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 Lagrange polynomial interpolation (also called cubic interpolation), the
coefficient of thermal expansion is given by
                  3
        (T )   Li (T ) (Ti )
                 i 0

                L0 (T ) (T0 )  L1 (T ) (T1 )  L2 (T ) (T2 )  L3 (T ) (T3 )

      y

                                                                               (x3, y3)

                                                            f3(x)
                                    (x1, y1)




           (x0, y0)
                                                               (x2, y2)



                                                                                          x

                                      Figure 4 Cubic interpolation.
05.04.8                                                                                         Chapter 05.04


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 points closest to T  14F that also
bracket T  14F to evaluate it. The four points are T0  80 F , T1  0 , T2  60 F and
T3  160 F .
Then
        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
                        3     T  Tj
          L0 (T )  
                       j 0   T0  T j
                       j 0

                     T  T1  T  T2  T  T3 
                   T  T  T  T  T  T 
                                             
                     0 1  0       2  0    3 
                       3 T T
          L1 (T )  
                                j

                     j 0 T1  T j
                       j 1

                     T  T0  T  T2  T  T3 
                   T  T  T  T  T  T 
                                             
                     1 0  1       2  1    3 
                       3 T T
          L2 (T )  
                                j

                     j 0 T2  T j
                       j 2

                     T  T0  T  T1  T  T3 
                   T  T  T  T  T  T 
                                             
                     2       0  2 1  2    3 
                       3 T T
          L3 (T )  
                                 j

                     j 0 T3  T j
                       j 3

                   T  T0  T  T1  T  T2 
                   T  T  T  T  T  T 
                                         
                   3    0  3    1  3    2 

Hence
                   T  T1  T  T2         T  T3           T  T0    T  T2    T  T3 
           (T )  
                           
                                          
                                             T  T  (T0 )   T  T
                                                                         
                                                                            T  T    
                                                                                        T  T  (T1 )
                                                                                                 
                   T0  T1  T0  T2       0    3           1    0    1    2    1    3 

                    T  T0           T  T1  T  T3          T  T0    T  T1  T  T2   
                  
                   T T             
                                      T  T  T  T  (T2 )   T  T
                                                                          T  T  T  T
                                                                                                (T3 )
                                                                                                   
                    2    0           2    1  2    3          3    0    3 1  3       2   
                                                                                              T0    T  T3
Lagrange Method of Interpolation-More Examples: Mechanical Engineering                         05.04.9



                      (14  0)(14  60)(14  160)
         (14)                                     (6.47  10 6 )
                        (80  0)(80  60)(80  160)
                       (14  80)(14  60)(14  160)
                                                       (6.00  10 6 )
                           (0  80)(0  60)(0  160)
                       (14  80)(14  0)(14  160)
                                                     (5.58  10 6 )
                       (60  80)(60  0)(60  160)
                         (14  80)(14  0)(14  60)
                                                        (4.72  10 6 )
                       (160  80)(160  0)(160  60)
                     (0.034979)(6.47  10 6 )  (0.82201)(6.00  10 6 )  (0.22873)(5.58  10 6 )
                       (0.015765)(4.72  10 6 )
                     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  0)(T  60)(T  160)
        (T )                                 (6.47  10 6 )
                (80  0)(80  60)(80  160)
                   (T  80)(T  60)(T  160)
                                               (6.00  10 6 )
                    (0  80)(0  60)(0  160)
                       (T  80)(T  0)(T  160)
                                                     (5.58 10 6 )
                   (60  80)(60  0)(60  160)
                         (T  80)(T  0)(T  60)
                                                        (4.72  10 6 ),        160  T  80
                   (160  80)(160  0)(160  60)
Combining like terms, we get
        (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,
05.04.10                                                                                      Chapter 05.04

                Tf
        D
             dT
         D Tr
                108

                  (6.00  10
                                6
                                     6.4786  10 9 T  8.1994  10 12 T 2  8.1845  10 15 T 3 )dT
                 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
      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
                1105.9  10 6  1110.6  10 6
        a                                       100
                        1105.9  10 6
             0.42775%

								
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