mws_mec_inp_txt_direct_examples

mws_mec_inp_txt_direct_examples - 05.02.1 Chapter 05.02...

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Unformatted text preview: 05.02.1 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 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   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 F 14   , determine the value of the coefficient of thermal expansion at F 14    T 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 a a T 1    Direct Method of Interpolation – More Examples: Mechanical Engineering 05.02.3 Figure 2 Linear interpolation. Since we want to find the coefficient of thermal expansion at F 14    T , and we are using a first order polynomial, we need to choose the two data points that are closest to F 14    T that also bracket F 14    T to evaluate it. The two points are F   T and F 60 1    T . Then   6 10 00 . 6 ,     T T    6 1 1 10 58 . 5 , 60      T T  gives     6 1 10 00 . 6      a a      6 1 10 58 . 5 60 60        a a  Writing the equations in matrix form, we have                         6 6 1 10 58 . 5 10 00 . 6 60 1 1 a a Solving the above two equations gives 6 10 00 . 6    a 6 1 10 007 .    a Hence   T a a T 1    60 , 10 007 . 10 00 . 6 6 6          T T At F, 14    T     14 10 007 . 10 00 . 6 14 6 6          F in/in/ 10 902 . 5 6       , y x   1 1 , y x   x f 1 x y 05.02.4 Chapter 05.02 Example 2 For the purpose of shrinking a trunnion into a hub, the reduction of diameter D ...
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This note was uploaded on 06/12/2011 for the course EML 3041 taught by Professor Kaw,a during the Spring '08 term at University of South Florida - Tampa.

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mws_mec_inp_txt_direct_examples - 05.02.1 Chapter 05.02...

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