CH-03 (Pao - Engineering Analysis), страница 2

The negative root equal to–0.63673 was found after 27, 15, 5, and 3 iterations and the positive root equal to 1.4096after 29, 15, 4, and 4 iterations by the incremental search, bisection search, linearinterpolation, and Newton-Raphson methods, respectively. An accuracy tolerance of© 2001 by CRC Press LLC1.E–5 was used for all cases. For solving this transcendental equation, NewtonRaphson therefore is the best method.FORTRAN VERSION© 2001 by CRC Press LLCSample ApplicationThe interactive question-and-answer process in solving the polynomial 4x3 +3x + 2x + 1 = 0 using the Newton-Raphson method and the subsequent display onscreen of the iteration goes as follows:2Notice that in the FORTRAN program FindRoot, the statement functions F(X)and FP(X) are defined for calculating the values of the given function and itsderivative at a specified X value.

Also, a character variable AS is declared througha CHARACTER*N with N being equal to 1 in this case when AS can have onlyone character as opposed to the general case of having N characters.© 2001 by CRC Press LLCMATLAB APPLICATIONA FindRoot.m file can be created and added to MATLAB m files for the purposeof finding a root of a polynomial or transcendental equation.

In this file, the fourmethods discussed in the FORTRAN or QuickBASIC versions can all be incorporated. Since some methods require that the left and right bounds, xl and xr, beprovided, the m file listed below includes as arguments these bounds along with thetolerance and the limited number of iterations:Notice that the equation for which a root is to be found should be defined in amile file called FofX.m, and that if the Newton-Raphson method, i.e., option 4, isto be used, then the first derivative of this equation should also be defined in a mfile called DFDX.m. We next present four examples demonstrating when all fourmethods are employed for solving a root of the polynomial F(x) = 4x3 + 3x2 + 2x+ 1 = 0 between the bounds x = –1 and x = 0 using a tolerance of 10–5.

In additionto FindRoot.m file, two supporting m files for this case are:© 2001 by CRC Press LLCThe four sample solutions are (some printout have been shortened for savingspaces:© 2001 by CRC Press LLCNotice that incremental search, half-interval search, interpolation, and NewtonRaphson methods take 28, 17, 16, and 4 iterations to arrive at the root x = –0.6058,respectively. The last method therefore is the best, but is only for this polynomialand not necessary so for a general case.Method of Successive SubstitutionAs a closing remark, another method called successive substitution is sometimesa simple way of finding a root of a transcendental equation, such as for solving theangle in a four-bar linkage problem shown in Figure 1. Knowing the lengths LAB,LBC amd LCD, and the angle of the driving link AB, the angle of the driven link CD,can be found by guessing an initial value of γ(0) and then continuously upgradedusing the equation:[)] 1γ ( k +1) = cos−1 L AB cos α − L CD + cos α − γ ( k ) L BC((5)where the superscript k serves as an iteration counter set equal to zero initially.

Forα changing from 0 to 360°, it is often required in study of such mechanism to findthe change in γ. This is left as a homework for the reader to exercise.© 2001 by CRC Press LLCFIGURE 1. Successive substitution sometimes is a simple way of finding a root of a transcendental equation, such as for solving the angle γ in a four-bar linkage problem.MATHEMATICA APPLICATIONSTo illustrate how Mathematica can be applied to find a root of F(x) = 1 + 2x+ 3x2 + 4x3 = 0 in the interval x = [xl,xr] = [–1,0], the linear interpolation is usedbelow but similar arrangements could be made when the incremental, or, bisectionsearch, or, Newton-Raphson method is selected instead.Input[1]: = F[x_]: = 1. + 2*x + 3*x^2 + 4*x^3Input[2]: = xl = –1; xr = 0; fl = F[xl]; fr = F[xr]; fx = fl;Input[3]: = Print[“xl = “,xl,” xr = “,xr,” F(xl) = “,fl,” F(xr) = “,fr]Output[3]: = xl = –1 xr = 0 F(xl) = –2.

F(xr) = 1.Input[4]: = (While[Abs[fx]>0.00001, x = (xr*fl-xl*fr)/(fl-fr);fx = F[x];Print[“x = “,N[x,5],” F(x) = “,N[fx,5]];If[fx*fl<0, xr = x;fr = fx;, xl = x;fl = fx;]])Output[4]: = x = –0.33333 F(x) = 0.51852x = –0.47059 F(x) = 0.30633x = –0.54091 F(x) = 0.1629x = –0.57548 F(x) = 0.080224x = –0.59185 F(x) = 0.037883x = –0.59944 F(x) = 0.017521x = –0.60292 F(x) = 0.0080245x = –0.60451 F(x) = 0.0036586© 2001 by CRC Press LLCx = –0.60523 F(x) = 0.0016646x = –0.60556 F(x) = 0.00075666x = –0.60571 F(x) = 0.00034379x = –0.60577 F(x) = 0.00015618x = –0.60580 F(x) = 0.000070940x = –0.60582 F(x) = 0.000032222x = –0.60582 F(x) = 0.000014635x = –0.60583 F(x) = 6.6473x10–6Notice that 16 iterations are required to achieve the accuracy that the value of|F(x)| should be no greater than 0.00001. In Input[1], the equation being solved isdefined in F[x].

1. is entered instead of an integer 1 so that all computed F(x) valueswhen printed will be in decimal form instead of in fractional form as indicated inOutput[3]. In Input[4], a pair of parentheses are added to allow long statements beentered using many lines and broken and listed with better clarity. Also, N[exp,n]is applied to request that the value of expression, exp, be handled with n significantfigures.

The command If is also employed in Input[4]. It should be used in the formof If[condition, GS1, GS2], which implements the statements in the group GS1 orin the group GS2 when the condition is true or false, respectively. Abs computesthe absolute value of an expression specified inside the pair of brackets.3.3 PROGRAM NEWRAPHG — GENERALIZEDNEWTON-RAPHSON ITERATIVE METHODNewton-Raphson method4 has been discussed in the program FindRoot in iterativesolution of polynomials and transcendental equation. Here, for an extended discussion of this method for solving a set of specified equation, we reintroduce this methodin greater detail. This method is based on Taylor’s series.5 Let us start again withthe case of one equation of one variable.

Let F(X) = 0 be the equation for which aroot Xr is to be found. If this root is known to be in the neighborhood of Xg, thenbased on Taylor’s series expansion we may write:( ) ( )( )F( X r ) = F X g + F ′ X g ∆X + F ′′ X g ( ∆X) 2! +…(1)∆X = X r − X g(2)2where:and the prime in Equation 1 represents differentiation with respect to X. Since Xris a root of F(X) = 0, therefore F(Xr) = 0. And if Xg is sufficiently close to Xr, Xis small and the terms involving (X)2 and higher powers of X in Equation 1 canbe neglected.

It leads to:[ ( ) ( )]X r = Xg − F Xg F′ Xg© 2001 by CRC Press LLC(3)This result suggests that if we use a projected root value according to Equation 3as next guess, an iterative process can then be continued until the condition F(Xg) =0 is, if not exactly, almost satisfied.The Newton-Raphson iterative procedure is developed on the above mentionedconcept by using the formula:( ) F ′( X ( ) )X(gk +1) = X(gk ) − F X(gk )kg(4)where k is an iteration counter. By providing an initial guess, X(0)g, Equation 4 is tobe repeatedly applied until F(X(k)g) is almost equal to zero which by using a tolearancecan be tested with the condition:( )F X(gk ) < ε(5)As an example, consider the case of:F( X) = X 3 − 6 X 2 + 11X − 6 = 0(6)F ′( X) = 3X 2 − 12 X + 11(7)for whichIf we make an initial guess of X(0))g = 1.75 and set a tolerance of = 0.00001, theNewton-Raphson iteration will proceed as follows:Trial No.012345X1.75002.05721.98252.00082.00012.0000F(X)0.23438–0.057010.01753–0.00085–0.000110.00001Program FindRoot has a fourth option for Newton-Raphson iteration of a rootfor a specified equation of one variable.

The results tabulated above are obtained bythe program FindRoot.TRANSCENDENTAL EQUATIONSNot only for polynomials, Newton-Raphson iterative method can also be appliedfor finding roots of transcendental equations. To introduce a transcendental equation,let us consider the problem of a moving vehicle which is schematically representedby a mass m in Figure 2. The leaf-spring and shock absorber are modelled by k andc, respectively.© 2001 by CRC Press LLCFIGURE 2. Mechanical vibration system with one degree-of-freedom.If the vehicle is suddenly disturbed by a lift or drop of one of its supportingboundaries by one unit (mathematically, that is a unit-step disturbance), it can beshown2 that the elevation change in time of the mass, here designated as X(t), isdescribed by the equation:X(t ) = 1 − a1Exp( −a 2 t ) sin(a 3t + a 4 )(8)where:a1 = ( k m ) 0.5 a 3′a 3 = ( 4 km − c2 ) 0.5 2 m,a 2 = c 2 m,anda 4 = tan −1 (a 3 a 2 )(9,10)(11,12)Equation 8 is a transcendental equation.In actual design of a vehicle, it is necessary to know the lengths of time that arerequired for the vehicle to respond to the unit-step disturbance and reaching to theamounts equal to 10, 50, and 90 percent of the disturbance.

Such calculations areneeded to ascertain the delay time, rise time, and other items among the designspecifications shown in Figure 3. If one wants to know when the vehicle will riseup to 50 percent of a unit-step disturbance, then it is a problem of finding a root,t = tr, which satisfies the equation:X(t r ) = 1 − a1Exp( −a 2 t r ) sin(a 3t r + a 4 ) = 0.5© 2001 by CRC Press LLCFIGURE 3.

Design specifications in time domain: overshoot xh, delay time td, rise time tr,and settlement time ts.Or, the problem can be mathematically stated as solving for tr from the followingtranscendental equation by knowing the constants a1–4:a1Exp( −a 2 t ) sin(a 3t + a 4 ) − 0.5 = 0As an example, let a1 = 1, a2 = 0.2 sec–1, a3 = 1 sec–1, and a4 = 1.37 radian thenthe transcendental equation is:e −.2 t sin(t + 1.37) − 0.5 = 0(13)To find a root tr for Equation 3, we select an initial guess tr(0) = 0.5 and applythe fourth option of the program FindRoot. The results are listed below.