CH-06 (523176), страница 6
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The differential equation of motion of a spring-and-mass system is d2x/dt2+ ω2x = 0 where ω2 = k/m, m is the mass and k is the spring constant. Ifthe weight is 5 lb., g = 32.2 ft/sec2, k = 1.5 lb/in, and initially the displacement x = 2 in. and velocity dx/dt = 0, use the 4th-order Runge-Kuttamethod and a step-size of 0.05 sec. to manually calculate the values of xand dx/dt at t = 0.05 sec. Note that the given second-order ordinarydifferential equations should first be converted into two first-order differential equations.2.
Initially, the two functions x(t) and y(t) have values 1 and –1, respectively.That is x(t = 0) = 1 and y(t = 0) = –1. For t>0, they satisfy the differentialequations dx/dt = 5x–2y + 2t and dy/dt = x2–0.25sin2t. Use Runge-Kuttaclassic fourth-order method and a time increment of 0.01 second to calculate x(t = 0.01) and y(t = 0.01).3. In the following two equations, the terms dx/dt and dy/dt both appear.2dx/dt –3x + 5dy/dt –7y = .5t + 1–3dx/dt -x – 4dy/dt + 9y = .2e –3t© 2001 by CRC Press LLCSeparate them by treating them as unknowns and solve them by simplesubstitution or Cramer’s rule.
The resulting equations can be expressedin the forms of dx/dt = F1(t,x,y) and dy/dt = F2(t,x,y). Carry out by manualcomputation using Runge-Kutta method to obtain the x and y values whent = 0.1 if the time increment is 0.1 and the initial conditions are x(t = 0) =1 and y(t = 0) = –2.4. The distributed loads on the beam shown in Figure 3 can be described asw = –1 N/cm, for 0<x<10 cm; w = 2.2x N/cm, for 10<x<20 cm; w = 0,for 20<x<40 cm. Meanwhile, the bending moment applied at x = 30 cmcan be described as Ma = 0, for 0<x<30 cm and Ma = –3 N-cm, for30<x<40 cm. By introducing new variables t, and x1 through x4 so thatt = x, x1 = y, x2 = dy/dx, x3 = d2y/dx2, and x4 = d3y/dx3, the problem offinding the deflection y(x) of the beam can be formulated (see the reference cited in footnote) as:dx1/dt = F1(t,x1,x2,x3,x4) = x2,x1(t = 0) = 0dx2/dt = F2(t,x1,x2,x3,x4) = (x3 + Ma)/EI,x2(t = 0) = 0dx3/dt = F3(t,x1,x2,x3,x4) = x4,x3(t = 0) = –1121/3dx4/dt = F4(t,x1,x2,x3,x4) = w,x4(t = 0) = 24with Ma and w being the applied bending moment and distributed loads,respectively.
The initial conditions specified above are all at the left endof the beam which is built into the wall and for the deflection (x1), slope(x2), bending moment (x3), and shearing force (x4), respectively. Applythe program RungeKut by using EI = 2x105 N/cm2 and various stepsizesto tabulate the results and errors similar to that given in the text.5. Apply the fourth-order Runge-Kutta method to find the values of x1 andx2 at the time t = 0.2 second using a time increment of 0.1 second basedon the following governing equations:dx1= 4 − 5x1 + 7x 2 − 2 tdtanddx 2= −3x1 + x1x 2 + 6e − tdtAt t = 0.1 second, x1 = –1 and x2 = 1.6.
Use different stepsizes to calculate y values at x = 0.1, 0.2, and 0.3 byapplication of the program RungeKut for the initial-value problemdy/dx = x2–y, y(x = 0) = 1. The analytical solution is y = 2–2x + x2e-x, bywhich the exact solutions can be easily computed to be y(0.1) = 0.90516,y(0.2) = 0.82127, and y(0.3) = 0.74918. Determine the stepsize which willlead to a Runge-Kutta numerical solutions of y(0.1), y(0.2), and y(0.3)accurate to five decimal figures.7. For the loaded beam shown in Figure 13, the deflection y(x) is to bedetermined by solving Equation 21.
Let the stiffness EI be equal to 2x107N-cm2 and it can be shown that the bending moment M can be describedby the equations:© 2001 by CRC Press LLCFIGURE 13. Question 7.M = -x2 + 180x–9600for 0<x<40 cmM = 100x–8000for 40<x<80 cmandApply the program RungeKut to find y at x = 80 cm by using stepsizesh = 4, 2, 1, 0.5, 0.25, and 0.1 and calculate the error by comparing withthe expected value of y(x = 80) = –0.928 cm.8. Convert the following two differential equations into three first-orderdifferential equations in the forms of dxi/dt = Fi(t;x1,x2,x3;constants) fori = 1,2,3 so that the subroutine RKN can be readily applied:d2ududv+2+ 3 + 4 u − 5v − 0.6 sin 7t = 0dt 2dtdtdvdu− 0.1 + 20 uv − 300e −4 t = 0dtdt9.
Write a subprogram FUNCTION F(X,T,I,N) which includes the statementsCOMMON R1,R2,R3,R4for transmitting the values of R1, R2, R3, and R4 from the main program.These four variables are r1, r2, r3, and r4, respectively, appearing in theequations:dA= − r1AdtdB= r1A − r2 BdtdC= r2 B − r3Cdt© 2001 by CRC Press LLCdD= r3C − r4 DdtdE= r4 Ddt10.11.12.13.14.This FUNCTION is to be used by the subroutine RKN in application ofthe fourth-order Runge-Kutta method.The functions x(t) and y(t) satisfy the differential equations d2x/dt2 +3dx/dt + 5dy/dt–7e–9t + 9sin2t = 0 and 2dx/dt–4dy/dt + 6x–8y + 10t–12 =0.
Convert the above two equations into the standard form dxi/dt = fi(t;x1,x2,x3;constants) for i = 1,2,3 where x1 = x, x2 = y, and x3 = dx/dt. Givethe expressions for f1, f2, and f3 in terms of t, x1, x2, and x3 so that theRunge-Kutta method can be applied.Apply the fourth-order Runge-Kutta method to find the values of y andz at x = 0.35 if at x = 0.3, y = 1 and z = 2 respectively and they satisfythedifferential equations dy/dx = xy + z and dz/dx = yz + x.
Use a stepsizeof x = 0.5 and show all details of how the Runge-Kutta parameters arecalculated.The deflection y of the load beam shown below satisfies the ordinarydifferential equation EI(d2y/dx2) = M where the Young’s modulus E =2x108 N/m2, moment of inertia I = 4.5x10–8 m4 and the internal bendingmoment, in N-m, has been derived in terms of x as M(x) = 200x–30 for0≤x≤.1 m and M(x) = 100x–20 for .1≤x≤.2 m. (1) Using an increment ofx = 0.01 m, standardize the above problem into a system of two firstorder ordinary differential equations dxi/dx = fi(x;x1,x2;constants) for i =1,2 where x1 = y and x2 = dy/dx (slope).
(2) Write a FUNCTION F(…)needed in SUBROUTINE RKN which we have discussed in class forusing the fourth-order Runge-Kutta method, based on the result of Step(1) and also the M(x) equations. (3) Calculate the eight RungeKuttaparameters and then the value of y and dy/dx at x = 0.01 m.Convert the following differential equation into a set of two first-orderdifferential equations so that the fourth-order Runge-Kutta method canbe applied: d2x/dt2 + 4dx/dt + 3x = 4e-t.
If at t = 0, x = 0 and dx/dt = 2,use a time increment oft = 0.1, compute the x and dx/dt values at t =0.1 based on the fourth-order Runge-Kutta method.The forced swaying motion of a three-story building can be simulated asa system of three lumped masses mi connected by springs with stiffnesseski and subjected to forces fi(t) for i = 1,2,3 as shown in Figure 14. Here,the dampingcharacteristics are not considered but could be incorporated.Derive the governing differential equations for the and then convert theminto a system of 6 first-order differential equations so that the programsRungeKut and ode45 can be applied to find the histories of displacements, xi(t), and velocities vi(t) = dxi(t)/dt.
Solve a numerical case of mi =2i N-sec2/cm, ki = 3i N/cm, and fi(t) = (2i–3)sin(2i–1)t N, xi(t = 0) = 0and vi(t = 0) = 0 for 0<t≤20 seconds. Plot all displacement and velocityhistories.© 2001 by CRC Press LLCFIGURE 14. Problem 14.15. Instead of f(t) = 1 in obtaining the system’s response of the mechanicalvibration problem using the MATLAB file FunMCK.m shown inFigure 5, resolve the problem for the case of f(t) = 5sin(0.5t–0.3) bychanging FunMCK.m and plot the resulting displacement and velocity.16. Implement the Runge-Kutta solution of Equation 30 by defining a subprogram function TwoMs in FORTRAN, QuickBASIC, or, MATLABto obtain the result shown in Figure 5 for the case of b = 3.6 m, h = 1.5m, m2/m1 = 0.8, and initial conditions y = z = dy/dt = dz/dt = 0.
Andcalculate the histories of the cable tension T(t) and angle (t).17. For the nonlinear oscillation problem of two connected masses shown inFigures 4 and 5, we observe that the oscillation goes on continuously. Themotion can be damped by adding a viscous device vertically connectedto the mass whose displacement is denoted as z(t). This could be africtional wall, on which the mass slides vertically. Usually, the retardingforce of such a damping device, Fr, could be assumed to be linearlyproportional to the velocity of the motion, dz/dt.
That is, Fr = cdz/dt wherec is constant. Figure 15 is a result of the oscillation when a dampingdevice having c = 1 N-sec/m and m1 = 1 N-sec2/m is added to that system.We notice that amplitudes of y(t) and z(t) shown in Figure 5 are steadilydecreased. Develop this modified program in FORTRAN, QuickBASIC,or,MATLAB to generate Figure 15.18. Use Mathematica’s function NDSolve to solve Problem 7.© 2001 by CRC Press LLCFIGURE 15.
Problem 17.19. Apply Mathematica to solve Problem 17 for a time increment t = 0.2sec and until t = 12 seconds.20. Apply Mathematica for solving Problem. 14.ODEBVPRK1. The function y(x) satisfies the boundary conditions y(x = 0) = 2 and y(x =3) = 4 and the differential equation 5d2y/dx2–3dy/dx + y = 13x–15 for0<x<3. Apply the fourth-order Runge-Kutta method to find the y valuesat x = 1 and x = 2 based on an increment of x equal to 1.2. The function y(x) has the boundary values of y(x = 0) = 1 and y(x = 3) = 5and for x between 0 and 3, y(x) satisfies the ordinary differential equation:d2ydy− 3 + 2y = 2x + 1dx 2dxApply the fourth-order Runge-Kutta method to find the y values at x = 1and x = 2 based on a stepsize of x = 1.3. For a membrane (Figure 16) under uniform tension T and fastened at theinner radius Ri and outer radius Ro, the axisymmetric deformation zresulted by the acting uniform pressure p can be shown to satisfy thedifferential equation:7© 2001 by CRC Press LLCFIGURE 16.
Problem 3.FIGURE 17. Problem 4.d 2 z 1 dzp+=−dr 2 r drTfor Ri<r<Ro. The boundary conditions are z(Ri) = 0 and z(Ro) = 0. Modifythe program OdeBvpRK to solve this problem.4. A cable hung at its two ends as shown in Figure 17 by its own weightwill have a catenary shape described by the equation:© 2001 by CRC Press LLCy=Txwwx − 1coshTx(a)where w is the weight per unit length and Tx is the horizontal, x-componentof the tension of the cable.
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