Probabilistic models for dynamical systems, second edition (1)

Benaroya Han L & MECHANICAL ENGINEERING Probabilistic Models for Dynamical Systems E D I T I O N e book presents a nice balance of theory and pragmatic application It begins by ting the foundations of probability theory as it is applied to engineering systems d then develops fundamental engineering concepts that integrate these probabilistic ncepts … The book also presents a wonderful historic perspective of some of greatest scientific and engineering minds who have made the most significant tributions to the fields of both probability and dynamics … This is an excellent first t for integrating mechanical system design and probability/reliability concepts.” R THOMAS R K U R F E S S, Georgia Institute of Technology, Atlanta, USA obabilistic Models for Dynamical Systems, Second Edition is an excellent ource for faculty to engage students in pursuit of knowledge about physical tems which are naturally complex, nonlinear, and non-deterministic The book is l written and complete with a plethora of worked out examples in each chapter.” R ROB E RT H B I S HOP, Marquette University, Milwaukee, USA w in its second edition, Probabilistic Models for Dynamical Systems expands coverage of probability theory for engineering This new revised version introduces dents to the randomness in variables and time-dependent functions and allows m to solve governing equations s book provides a suitable resource for self-study and can be used as an all-inclusive oduction to probability for engineering It introduces basic concepts, and highlights plied probability in a practical manner With updated information and over 300 strations, this edition includes new sections, problems, applications, and examples graphical summaries spotlight relevant historical figures, providing life sketches, or contributions, and relevant quotes from their works A new chapter on control d mechatronics rounds out the coverage ech.indd K12264 0610 2013 ISBN: 978-1-4398-4989-7 Probabilistic Models for Dynamical Systems S E C O N D Nagurka SECOND E D ITIO N Probabilistic Models for Dynamical Systems S E C O N D E D I T I O N Haym Benaroya Seon Mi Han Mark Nagurka 90000 781439 849897 All Pages 3/22/13 2:10 PM solutions MANUAL FOR Probabilistic Models for Dynamical Systems second Edition by Haym Benaroya Seon Mi Han Mark Nagurka This page intentionally left blank solutionS MANUAL FOR Probabilistic Models for Dynamical Systems second Edition by Haym Benaroya Seon Mi Han Mark Nagurka Boca Raton London New York CRC Press is an imprint of the Taylor & Francis Group, an informa business CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2013 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S Government works Version Date: 20140709 International Standard Book Number-13: 978-1-4398-5015-2 (eBook - PDF) This book contains information obtained from authentic and highly regarded sources Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint Except as permitted under U.S Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers For permission to photocopy or use material electronically from this work, please access www.copyright.com (http:// www.copyright.com/) or contact the Copyright Clearance Center, Inc (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400 CCC is a not-for-profit organization that provides licenses and registration for a variety of users For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com Solutions Manual - Probabilistic Models for Dynamical Systems, 2nd Ed Haym Benaroya, Seon Han, Mark Nagurka, and Joey Sanchez June 27, 2013 Contents Introduction Events and Probability 3 Random Variable Models 16 Functions of Random Variables 76 Random Processes 135 Single Degree-of-Freedom Vibration 163 MDOF vibration 191 Continuous System Vibration 228 Reliability 256 10 Nonlinear and Stochastic Dynamic Models 268 11 Nonstationary Models 304 12 Monte Carlo Methods 312 Introduction 1.1 Identify engineering and scientific applications where uncertainties can be ignored Explain Solution: Generally, uncertainties can be ignored if their magnitudes are significantly smaller that the nominal (average) magnitudes For example, for a circular cylindrical column of diameter D, with uncertainties in the diameter of 0.001D, then it is likely that we can ignore the uncertainty in the nominal diameter when calculating the quantities based on diameter 1.2 Identify engineering and scientific applications where uncertainties cannot be ignored Explain Solution: If the magnitude of the uncertainty is significant compared to the nominal (average) value then it cannot be ignored For example, for a circular cylindrical column of diameter D, with uncertainties in the diameter of 0.1D, then we cannot ignore the uncertainty in the nominal diameter when calculating the quantities based on diameter It may also be that the application/the design requires a very tight tolerance and therefore uncertainties must always be incorporated into the design 1.3 Discuss how an engineer may ascertain whether uncertainties are important or can be ignored in an analysis and design Use examples in your discussion Solution: A design engineer will perform the design using alternate values of parameters that have uncertainties and then ascertain whether the final design is significantly different In the simple example of the design of a column, various diameters will be used in order to determine the necessary buckling strength Similarly, if there are uncertainties in the forces on a structure, then high and low values might be used to be certain that the design will resist those forces The use of probability makes this process more streamlined 1.4 How can Miner’s rule for fatigue damage be extended to cases where the order of loading cycles is important? Explain with or without using equations Solution: If the order of the loading cycles is important then higher weighting can be given to a particular ordering of stress cycles The summation has to be identical to the ordering For example, p Damage = fi i=1 ni , Ni where p is the number of stress levels, i is the ith stress cycle at ith stress level, Ni is the number of cycles that the part can take at ith stress level, and the part fails when the damage reaches one 1.5 How can Miner’s rule for fatigue damage be extended to cases where a cycle at stress nσ causes n times as much damage as a cycle at stress σ? Explain with or without using equations Solution: In the summation it will be necessary to have factors that reflect this multiple Tests would show that at higher stresses failure occurs n times faster 1.6 Which variables or parameters in Equations 1.3 and 1.4 are better assumed to be random variables? Explain your choices Solution: CD , CI , u, u˙ are all likely candidates to be random variables The coefficients CD , CI are experimentally determined and have significant scatter associated with their values, which depend on the fluid speed Similarly, fluid speed and acceleration u, u˙ are also random 2 Events and Probability Section 2.1: Sets 2.1 Give three examples of impossible events from everyday life Solution: Examples may include the sun rising in the West and setting in the East, water flowing upstream, and time going backwards 2.2 Give three examples of impossible events from fluids engineering Solution: Examples may include fluid with no viscosity, higher density fluid rising, and pressure head decreasing with increasing column height 2.3 Give three examples of impossible events from materials engineering Solution: Examples may include temperature dipping below degrees Kelvin, massless material, inextensible metal 2.4 Give three examples of impossible events from strength of materials Solution: Examples may include the proportional limit of a material exceeding the yield stress, fracture being a reversible event, plastic deformation that is recoverable and not permanent, the sum of the forces in a static structure being not zero, and a linear isotropic material with Poisson’s ratio greater than 0.5 2.5 Give three examples of impossible events from mechanical vibration Solution: Examples may include a structure that oscillating without any forces or initial conditions on it, for the same structure, a stiffer structure oscillating slower, a structure oscillating with unlimited speeds, a damped oscillator having a non-decaying amplitude in free vibration, a spring with an infinite stiffness, a free pendulum swinging indefinitely without the addition of energy, eliminating low frequency vibration in a baseball bat when hitting a ball outside the sweet zone 2.6 Give three examples of impossible events from thermal engineering Solution: Examples may include decreasing entropy without energy, increasing energy in a closed system, and radiation without temperature 2.7 Give three examples of certain events from everyday life Solution: Examples may include the sun rising tomorrow, having to fill out tax forms every year, the sun causing the evaporation of trillion tons of water daily, being summer time in the northern hemisphere when it is winter time in the southern hemisphere, applying a force on an object results in an equal and opposite force, and all creatures needing energy to survive 2.8 Give three examples of certain events from fluids engineering Solution: Examples include a faster incompressible flow in a smaller diameter pipe, denser fluid sinking, flow exerting a force on impeller 2.9 Give three examples of certain events from materials engineering Solution: All materials have imperfections, a higher pressure results in a higher boiling point, a constrained beam has a higher buckling load 2.10 Give three examples of certain events from strength of materials Solution: An axial force on a column leads to compression of that column; an elastic body will always deform if forces act on it; any beam subjected to an axial force will eventually fail; in polymer composites, nanoparticles not create large stress concentrations and thus not compromise the ductility of the polymer 2.11 Give three examples of certain events from mechanical vibration Solution: If a body is oscillating, then there must be inertia; if a body is oscillating, then there is a restoring force; there is an energy exchange mechanism between the kinetic and potential energies; when hit at the sweet zone, the baseball bat will not be excited at its natural frequencies and therefore there will not be energy loss due to these frequencies vibrating, resulting in more energy being transmitted to the ball 2.12 Give three examples of certain events from thermal engineering Solution: Heating an object results in a higher temperature Heating a metal object results in expansion Objects expand and contract at different rates 2.13 Give an example of complementary events from everyday life Solution: winning/losing a game; catching/missing a train; night/day; 2.14 Give an example of complementary events from fluids engineering Solution: The flow may be considered laminar or turbulent 2.15 Give an example of complementary events from materials engineering Solution: The material extends linearly or nonlinearly In other words, the material may or may not obey Hooke’s law 11.5 Derive the equations in Section 11.5.1 explicitly assuming a two degree-of-freedom system Solution: The degree-of-freedom equations will be ă [m]{X(t)} + [c]{X(t)} + [k]{X(t)} = [p]{F (t)}, ă where [m], [c], and [k] are 2x2 matrices, {X(t)}, {X(t)}, {X(t)} are 2x1 vectors, [p] is a × m matrix and {F (t)} is an m × vector (where m is the length of {F (t)} vector), F1 (t) F2 (t) ∞ = aF1 (t, ω) aF2 (t, ω) −∞ 308 eiωt dZ1 (ω) dZ2 (ω) 11.6 Derive the equations in Section 11.5.2 explicitly assuming a two degree-of-freedom system Solution: All matrices are 2x2 In Equations (11.55) and (11.56) i = 1, Then Λi (t) = {Λi1 (t) Λi2 (t)} ˆ = [U ˆ1 U ˆ ] and modal responses In Equation (11.57) i = 1, There are two normalized modes U T Ψ(t) = {Ψ1 (t) Ψ2 (t)} The modal force is Γi = ˆ Ti p U ˆ T mU ˆi U i The expressions for the correlation functions and spectral densities are as shown at the bottom of page 581 except that the upper limits in the summations are 309 11.7 Derive Equations 11.62 and 11.63 Solution: Solve Equation 11.60 for β − β e : −(ω 2n − ω 2e )E{X } − εE{Xh} ˙ E{X X} Substitute into Equation 11.61, and solve for (ω 2n − ω2e ) to find Equation 11.62 Then substitute into Equation 11.60 and solve for β − β e and find Equation 11.63 310 11.8 Perform the necessary calculations to plot E{X (t)} as in Figure 11.4 for the cases ζ = 0.15 and ζ = 0.2 Solution: 311 12 Monte Carlo Methods Section 12.2: Random Number Generation 12.1 Generate uniform random numbers using software such as MATLAB, either using the linear congruential generator or the program’s built-in random number generator Plot the average value and the standard deviation and compare them with the theoretical values for the uniform distribution Solution: Consider uniform numbers between and The theoretical mean and the standard deviation are µ = xdx = σ = E {X } − µ2 = x2 dx − 2 = 0.2887 Matlab’s rand.m function can generates a square matrix whose elements are random numbers Following script can generate the average value and the standard deviation for N random numbers The resulting plot is shown below clear n=2:20; nn=length(n); for i=1:nn; temp=reshape(rand(n(i)), n(i)^2,1); MEAN(i)=mean(temp); STD(i)=std(temp); end ms=[0.5*ones(nn,1) sqrt(1/12)*ones(nn,1)]; N=n.^2; plot(N,MEAN,'o', N, STD,'x', N, ms) axis([0 N(end) 1]) Matlab code for Problem 12.1 312 0.9 calculated mean calculated std theoretical mean theoretical std 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 50 100 150 200 250 300 350 400 Number of random numbers used Theoretical and calculated mean and standard deviation 313 12.2 Show that random numbers that are distributed according to the lognormal density function with mean value exp (µ) and standard deviation exp (σ) can be obtained by using y = exp σ −2 ln x1 cos 2πx2 + µ , where x1 and x2 are uniform random numbers between and Solution: The lognormal density function is written as (Section 3.3.4) fY (y) = 1 √ exp − σ Z 2π ln y − µZ σZ , < x < ∞, where λ and ζ are the mean and the standard deviation of z = ln y so that z is normally distributed It was shown that µY = exp(µZ + σ2Z ) σY = µ2Y [exp (σ2Z ) − 1], or σ 2Z µZ σ2Y µ2Y = ln µY − σ 2Z = ln = ln + µY + σ 2Y /µ2Y From Equation 12.12, the random numbers distributed according to the normal density z is given by z = σZ −2 ln x1 cos 2πx2 + µZ , where x1 and x2 are independent uniform random numbers Replacing z with ln y, we obtain y = exp σZ −2 ln x1 cos 2πx2 + µZ 314 12.3 Verify that the random numbers produced in the previous problem are indeed distributed according to the lognormal density Plot the mean value and the standard deviation as functions of the number of random numbers used √ Solution: Let us choose σ Y = and µY = Then, σZ = ln (1 + 4) and µZ = ln 2/ + Following is the MATLAB script used to generate the calculated µY and σ Y in the figure shown next The mean converges faster to the theoretical value clear n=2:200; nn=length(n); sigmay=4; muy=2; sigmaz=sqrt(log(1+sigmay^2/muy^2)); muz=log(muy/sqrt(1+sigmay^2/muy^2)); for i=1:nn x1=reshape(rand(n(i)), n(i)^2,1); x2=reshape(rand(n(i)), n(i)^2,1); y=exp(sigmaz*sqrt(-2*log(x1)).*cos(2*pi*x2)+muz); MEANy(i)=mean(y); STDy(i)=std(y); end ms=[muy*ones(nn,1) sigmay*ones(nn,1)]; N=n.^2; plot(N,MEANy,'o', N, STDy, 'x', N, ms) Matlab code for Problem 12.3 calculated mean calculated std theoretical mean theoretical std 0.5 1.5 3.5 4 x 10 Number of random numbers used Theoretical and calculated mean and standard deviation for Problem 12.3 315 2.5 12.4 Use the inverse numerical transformation method to generate normal random numbers with µ = and σ = Use ten equally spaced intervals between y = −3σ and y = 3σ Solution: The probability density function is given by 1 fY (y) = √ exp − y2 2π The cumulative distributions at equally spaced intervals are given by yi FY (yi ) = xi = fY (y) dy, −∞ where yi = −3.6 + 0.6i, for i = 1, , 11 The values of FY (yi ) are given by xi = 0.00135, 0.00820, 0.0359, 0.115, 0.274, 0.500, 0.726, 0.885, 0.964, 0.992, 0.999 Then, the normal random number y is given by y = yi + yi+1 − yi (x − xi ) , xi+1 − xi where x is between xi and xi+1 Table below shows the uniform random numbers x and corresponding normal random numbers y x y 0.17 -0.99 0.83 1.24 0.25 -0.69 0.53 0.08 0.77 0.77 0.65 0.40 0.20 -0.8 316 0.97 1.93 12.5 Use the composition method to generate random numbers distributed according to fY (y) = Solution: Let us write sin y + y , 32π3 < y ≤ 2π fY (y) = fI (y) + fII (y) , 4 where y sin and fII (y) = y 8π 1 y FI (y) = − cos and FII (y) = y 2 8π fI (y) = The random numbers y are generated by y = FI−1 (x2 ) = cos−1 (−2x2 ) , y √ −1 = FII (x2 ) = 2π x2 , if ≤ x1 < ≤ x1 < 4 if where x1 and x2 are uniform random numbers between and The Table below illustrates this with numerical examples x1 x2 y 0.23 0.44 5.29 0.89 0.92 6.11 0.019 0.20 3.96 0.49 0.61 5.33 0.76 0.74 5.68 0.60 0.94 6.16 0.97 0.18 3.55 317 ··· ··· ··· 12.6 Use Von Neumann’s rejection-acceptance method to generate normally distributed random numbers How does the method compare to the Box-Muller method? Solution: The normal distribution is given by 1 fY (y) = √ exp − y2 , − ∞ < y < ∞ 2π Let us limit the range from -3σ to 3σ although y ranges from −∞ to ∞ 6σ contains 99.7% of the total area The first step is to generate the uniform random numbers, y, in this range using y = −3 + 6x1 , where x1 is uniformly distributed between and Let us then generate another uniform random number x2 between and The random number y is accepted if x2 is less than fY (y) / max (fY ) Table below illustrates this procedure x1 x2 y fY (y) max(fY ) Accept? 0.60 0.50 0.6 0.84 yes 0.015 0.42 -2.91 0.0145 no 0.93 0.68 2.58 0.0359 no 0.27 0.30 -1.38 0.39 yes 0.74 0.19 1.44 0.35 yes 0.46 0.42 -0.24 0.97 yes 0.20 0.86 -1.80 0.20 no 0.34 0.92 -0.96 0.63 no ··· ··· ··· ··· ··· When 10, 000 random numbers (10,000 each for x1 and x2 ) are used, we find that only about 41% of y are accepted Therefore, Box-Muller method is about 2.4 times more efficient This number is consistent with the ratio of the area of the box to the area covered by the normal distribution shown in the next figure Random number generation by Von Neumann’s method 318 Section 12.3: Random Numbers for Joint Probability Densities 12.9 Use the inverse transform method to generate random numbers for the joint probability density function, < y1 , y2 ≤ fY1 Y2 (y1 , y2 ) = y1 y2 , Solution: The joint probability density can be written as fY1 Y2 (y1 , y2 ) = fY1 (y1 ) fY2 |Y1 (y2 |y1 ) where fY1 (y1 ) = 2y1 , < y1 < fY2 |Y1 (y2 |y1 ) = 2y2 , < y2 < The cumulative distribution functions are given by FY1 (y1 ) = y12 , < y1 < FY2 |Y1 (y2 |y1 ) = y22 , < y2 < The two sets of standard uniform numbers, x1 and x2 , are used to generate y1 and y2 so that √ y1 = x1 √ y2 = x2 319 12.10 Use the linear transform method to generate random numbers y1 and y2 for the joint probability density function given in the previous problem Solution: The mean and the covariance of y1 and y2 are given by µY1 = µY2 = Cov Y12 y2 fY2 (y2 ) dy2 = y1 fY1 (y1 ) dy1 = = y1 − µY1 Cov {Y1 Y2 } = Cov Y22 y1 − µY1 y2 − µY2 fY1 (y1 ) dy1 = 18 = 2 y2 − µY2 fY1 Y2 (y1 , y2 ) dy1 dy2 = fY2 (y2 ) dy2 = 18 Consider probability density functions fZi (zi ) = zi , − 0.5 < zi < 0.5, for i = 1, 2, and where Zi are independent The mean and the covariance are given by = Cov {Zi Zj } = δ ij 12 µZi Let Y be related to Z by {Y } = [C] {Z} + {µY } , The mean and the covariance of Y can be written as E {{Y }} = {µY } Cov {Y } {Y }T = σ 2Y , and E {{Z}} = {0} Cov {Z} {Z}T [I] , = 12 where Cov {Y } {Y }T = E ({Y } − {µY }) ({Y } − {µY })T = E ([C] {Z}) ([C] {Z})T = [C] E {Z} {Z}T [C]T = [C] [C]T 12 Then, σ 2Y = [C] [C]T 12 320 Let us use lower-triangular coefficient matrix for [C] , [C] = c11 c21 c22 Then, c 12 11 = σ211 = c11 c21 12 18 = σ221 = c + c222 12 21 = σ222 = 18 Then, c11 = 2/3, c21 = 0, c22 = 2/3 In order to find random numbers y1 and y2 , we first generate independent random numbers z1 and z2 according to fZi (zi ) = zi , − 0.5 < zi < 0.5 y1 and y2 are obtained from y1 y2  = 0 321   z1 z2 + 3 Benaroya Han CIVIL & MECHANICAL ENGINEERING Probabilistic Models for Dynamical Systems E D I T I O N “The book presents a nice balance of theory and pragmatic application It begins by setting the foundations of probability theory as it is applied to engineering systems and then develops fundamental engineering concepts that integrate these probabilistic concepts … The book also presents a wonderful historic perspective of some of the greatest scientific and engineering minds who have made the most significant contributions to the fields of both probability and dynamics … This is an excellent first text for integrating mechanical system design and probability/reliability concepts.” —DR THOMAS R K U R F E S S, Georgia Institute of Technology, Atlanta, USA “Probabilistic Models for Dynamical Systems, Second Edition is an excellent resource for faculty to engage students in pursuit of knowledge about physical systems which are naturally complex, nonlinear, and non-deterministic The book is well written and complete with a plethora of worked out examples in each chapter.” —DR ROB E RT H B I S HOP, Marquette University, Milwaukee, USA Now in its second edition, Probabilistic Models for Dynamical Systems expands its coverage of probability theory for engineering This new revised version introduces students to the randomness in variables and time-dependent functions and allows them to solve governing equations This book provides a suitable resource for self-study and can be used as an all-inclusive introduction to probability for engineering It introduces basic concepts, and highlights applied probability in a practical manner With updated information and over 300 illustrations, this edition includes new sections, problems, applications, and examples Biographical summaries spotlight relevant historical figures, providing life sketches, major contributions, and relevant quotes from their works A new chapter on control and mechatronics rounds out the coverage K12264 0610 2013 ISBN: 978-1-4398-4989-7 90000 781439 849897 K12264_Cover_mech.indd All Pages Probabilistic Models for Dynamical Systems S E C O N D Nagurka SECOND E D ITIO N Probabilistic Models for Dynamical System S E C O N D E D I T I O N Haym Benaroya Seon Mi Han Mark Nagurka ... MANUAL FOR Probabilistic Models for Dynamical Systems second Edition by Haym Benaroya Seon Mi Han Mark Nagurka This page intentionally left blank solutionS MANUAL FOR Probabilistic Models for Dynamical. .. Manual - Probabilistic Models for Dynamical Systems, 2nd Ed Haym Benaroya, Seon Han, Mark Nagurka, and Joey Sanchez June 27, 2013 Contents Introduction Events and Probability 3 Random Variable Models. .. between < x < Then,  x for ≤ x <    41 (x − 1)2 + 12 16 FX (x) =  8x +   From the figure we find: 3, where it may be assumed to be parabolic for ≤ x < for ≤ x < for ≤ x < for x ≥ (a) Pr(X =