VIEWS: 7 PAGES: 9 POSTED ON: 10/3/2012
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 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 05.02.1 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 14F , determine the value of the coefficient of thermal expansion at T 14F 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 14F , and we are using a first order polynomial, we need to choose the two data points that are closest to T 14F that also bracket T 14F 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 6.00 106 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 14F, 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 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, 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 14F , determine the value of the coefficient of thermal expansion at T 14F 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 14F , and we are using a second order polynomial, we need to choose the three data points that are closest to T 14F that also bracket T 14F 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 802 6.47 106 0 a0 a1 0 a2 02 6.00 106 60 a0 a1 60 a2 602 5.58 106 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 14F, 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 DT 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 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, 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 14F , determine the value of the coefficient of thermal expansion at T 14F 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 14F , and we are using a third order polynomial, we need to choose the four data points closest to T 14F that also bracket T 14F 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 802 a3 803 6.47 106 0 a0 a10 a2 02 a3 03 6.00 106 60 a0 a1 60 a2 602 a3 603 5.58 106 160 a0 a1 160 a2 1602 a3 1603 4.72 106 Writing the four equations in matrix form, we have 1 80 6400 5.12 105 a0 6.47 106 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 1105.9 106 1110.6 106 100 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