mws_mec_inp_txt_lagrange_examples

mws_mec_inp_txt_lagrange_examples - Chapter 05.04 Lagrange...

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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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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 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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