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 DT where D original diameter in. coefficient of thermal expansion at average temperature in/in/ F The trunnion is cooled from 80F to 108F , giving the average temperature as 14F . 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 14F , determine the value of the coefficient of thermal expansion at T 14F 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 14F , we choose two data points that are closest to T 14F and that also bracket T 14F . 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 L0 (T ) j 0 j 0 T0 T j T T1 T0 T1 1 T T j L1 (T ) j 0 j 1 T Tj T1 T j T T0 T1 T0 Hence T T0 T T1 (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 (T ) 05.04.4 Chapter 05.04 14 60 14 0 (6.00 10 6 ) (5.58 10 6 ) 0 60 60 0 6 0.76667(6.00 10 ) 0.23333(5.58 10 6 ) (14) 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 60F to calculate the coefficient of thermal expansion at T 14F . 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 DT where D original diameter in. coefficient of thermal expansion at average temperature in/in/ F The trunnion is cooled from 80F to 108F , giving the average temperature as 14F . 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, TF 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 14F , determine the value of the coefficient of thermal expansion at T 14F 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 14F , we need to choose data points that are closest to T 14F that also bracket T 14F 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 L0 (T ) j 0 j 0 T Tj T0 T j T T1 T T2 T0 T1 T0 T2 2 T T j L1 (T ) j 0 T1 T j j 1 T T0 T T2 T T T T 2 1 0 1 2 T T j L2 (T ) j 0 T2 T j j 2 T T0 T T1 T2 T0 T2 T1 05.04.6 Chapter 05.04 Hence T T1 T T2 T T0 T T2 T T0 T T1 (T0 ) (T1 ) (T2 ), T0 T1 T0 T2 T1 T0 T1 T2 T2 T0 T2 T1 (T ) T0 T T2 (14 0)( 14 60) (14 80)( 14 60) (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% (14) 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 DT where D original diameter in. coefficient of thermal expansion at average temperature in/in/ F The trunnion is cooled from 80F to 108F , giving the average temperature as 14F . 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, TF 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 14F , determine the value of the coefficient of thermal expansion at T 14F 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 80F T f 108F 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 14F , and we are using a third order polynomial, we need to choose the four points closest to T 14F that also bracket T 14F 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 L0 (T ) j 0 j 0 T Tj T0 T j T T1 T T2 T T3 T T T T T T 2 0 3 0 1 0 3 T T j L1 (T ) j 0 T1 T j j 1 T T0 T T2 T T3 T1 T0 T1 T2 T1 T3 3 T T j L2 (T ) j 0 T2 T j j 2 T T0 T T1 T T3 T T T T T T 0 2 1 2 3 2 3 T T j L3 (T ) j 0 T3 T j j 3 T T0 T T1 T T2 T3 T0 T3 T1 T3 T2 Hence T T1 T T2 T0 T1 T0 T2 (T ) T T0 T2 T0 T T3 T T0 (T0 ) T0 T3 T1 T0 T T2 T1 T2 T T1 T T3 T T0 (T2 ) T2 T1 T2 T3 T3 T0 T T3 (T1 ) T1 T3 T T1 T T2 T3 T1 T3 T2 T0 (T3 ) T T3 Lagrange Method of Interpolation-More Examples: Mechanical Engineering (14) 05.04.9 (14 0)( 14 60)( 14 160) (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 108F to Tr 80F 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 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 ) So 5.9077 10 6 (108 80) 1110.6 10 6 D 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% and