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Newton-Raphson Method of Solving a Nonlinear Equation-More ...

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Chapter 03.04<strong>Newton</strong>-<strong>Raphson</strong> <strong>Method</strong> <strong>of</strong> <strong>Solving</strong> a <strong>Nonlinear</strong><strong>Equation</strong>-<strong>More</strong> ExamplesMechanical EngineeringExample 1A trunnion has to be cooled before it is shrink fitted into a steel hub.Figure 1 Trunnion to be slid through the hub after contracting.T fThe equation that gives the temperature to which the trunnion has to be cooled to obtainthe desired contraction is given by1037242f Tf 0.5059810Tf 0.38292 10Tf 0.7436310Tf 0.8831810Use the <strong>Newton</strong>-<strong>Raphson</strong> method <strong>of</strong> finding roots <strong>of</strong> equations to find the temperature T ftowhich the trunnion has to be cooled. Conduct three iterations to estimate the root <strong>of</strong> theabove equation. Find the absolute relative approximate error at the end <strong>of</strong> each iteration andthe number <strong>of</strong> significant digits at least correct at the end <strong>of</strong> each iteration.Solution 01037242f Tf 0.5059810Tf 0.38292 10Tf 0.7436310Tf 0.883181010274f Tf 1.5179410Tf 0.76584 10Tf 0.7436310Let us assume the initial guess <strong>of</strong> the root <strong>of</strong> f 0 as T 100.Iteration 1The estimate <strong>of</strong> the root isT ff , 003.04.1


03.04.2 Chapter 03.04Tf ,1 Tf ,0ffTTf ,0f ,00.505980.74363100 111.5179101037210100 0.382921010042101000.8831810274100 0.76582101000.743631031.829010 100 56.5187 10 100 28.058 128.06The absolute relative approximate error aat the end <strong>of</strong> Iteration 1 isaTf ,1 TTf ,1f ,0100128.06 ( 100)100128.06 21.910%The number <strong>of</strong> significant digits at least correct is 0, as you need an absolute relativeapproximate error <strong>of</strong> less than 5% for one significant digit to be correct in your result.Iteration 2The estimate <strong>of</strong> the root isf Tf ,1Tf ,2 Tf ,1f Tf ,1100.505981040.7436310 128.06 101.5179100.743631054.321410 128.0656.206710 128.06 (0.69625) 128.75The absolute relative approximate erroraTf ,2T Tf ,2f ,1100128.75 128.06100128.75 0.54076%The number <strong>of</strong> significant digits at least correct is 1.37128.06 0.3829210 128.062128.060.883181027128.06 0.76584 10128.064aat the end <strong>of</strong> Iteration 2 is2


<strong>Newton</strong>-<strong>Raphson</strong> <strong>Method</strong>-<strong>More</strong> Examples: Mechanical Engineering 03.04.3Iteration 3The estimate <strong>of</strong> the root isf TfTf ,3 Tf ,2f T,2f ,2100.505981040.7436310 128.75 101.5179100.743631082.800210 128.7556.1986104 128.75 4.517510 128.75The absolute relative approximate erroraTf ,3T Tf ,3f ,2100128.75 128.75128.7510037128.75 0.38292 10128.752128.750.883181027128.75 0.76582 10128.754aat the end <strong>of</strong> Iteration 3 is4 3.508610%Hence the number <strong>of</strong> significant digits at least correct is given by the largest value <strong>of</strong>which2m 0 .510a43.508610 0.51047 .0173102m 104log 7.017310 2 2m m42 log7.017310 5. 1538m Som 5The number <strong>of</strong> significant digits at least correct in the estimated root –128.75 is 5.2mforNONLINEAR EQUATIONSTopic <strong>Newton</strong>-<strong>Raphson</strong> <strong>Method</strong>-<strong>More</strong> ExamplesSummary Examples <strong>of</strong> <strong>Newton</strong>-<strong>Raphson</strong> <strong>Method</strong>Major Mechanical EngineeringAuthors Autar KawDate August 7, 2009Web Site http://numericalmethods.eng.usf.edu

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