Linear Regression-More Examples: Mechanical Engineering

Linear Regression-More Examples: Mechanical Engineering

<p>Chapter 06.03 Linear Regression-More Examples Mechanical Engineering</p><p>Example 1 The coefficient of thermal expansion, ,of steel is given at discrete values of temperature in Table 1. Table 1 Coefficient of thermal expansion versus temperature for steel. Temperature, Coefficient of T thermal expansion,  F in/inF 80 6.470106 60 6.360106 40 6.240106 20 6.120106 0 6.000106 −20 5.860106 −40 5.720106 −60 5.580106 −80 5.430106 −100 5.280106 −120 5.090106 −140 4.91010 6 −160 4.72010 6 −180 4.52010 6 −200 4.30010 6 −220 4.08010 6 −240 3.830106 −260 3.580106 −280 3.330106 −300 3.070106 −320 2.76010 6 −340 2.45010 6</p><p>06.03.1 06.03.2 Chapter 06.03</p><p>The data is regressed to a first order polynomial. </p><p>  k1  k2T</p><p>Find the constants k1 and k2 of the regression model. Solution Table 2 shows the summations needed for the calculation of the constants of the regression model. Table 2 Tabulation of data for calculation of needed summations. I T  T T 2 − F in/in/F in/in/F F2 1 80 6.47010 6 5.1760104 6400 2 60 6.36010 6 3.8160104 3600 3 40 6.240106 2.4960104 1600 4 20 6.12010 6 1.2240 104 400 5 0 6.00010 6 0.000 0 6 -20 5.860106 1.1720 104 400 7 -40 5.720106  2.2880104 1600 8 -60 5.580106  3.3480104 3600 9 -80 5.430106  4.3440104 6400 10 -100 5.280106  5.2800104 10000 11 -120 5.090106  6.1080104 14400 12 -140 4.910106  6.8740104 19600 13 -160 4.720106  7.5520104 25600 14 -180 4.520106  8.1360104 32400 15 -200 4.300106  8.6000104 40000 16 -220 4.080106  8.9760104 48400 17 -240 3.830106  9.1920104 57600 18 -260 3.580106  9.3080104 67600 19 -280 3.330106  9.3240104 78400 20 -300 3.070106  9.2100104 90000 21 -320 2.760106  8.8320104 102400 22 -340 2.450106  8.3300104 115600 22  -2860 1.0570104 1.0416102 726000 i1 Linear Regression-More Examples: Mechanical Engineering 06.03.3</p><p> n  22 22 22 22 nTi i Ti  i i1 i1 i1 k2  22 22 2 2   nTi  Ti  i1  i1 </p><p>221.0416102   28601.0570104   22726000  28602  9.3868 10-9 in/in/(F)2 22  _  i   i1 n 1.0570104  22</p><p> 4.8045106 in/inF</p><p>22 T _  i T  i1 n  2860  22  130 F</p><p>_ _</p><p> k1    k2 T  4.8045106  9.3868109 130  6.0248106 in/inF 06.03.4 Chapter 06.03</p><p>Figure 4 Linear regression of coefficient of thermal expansion vs. temperature data.</p>

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