Soft Computation for Structural Dynamic Natural Problems

International Bulletin of Mathematical Research Volume 02, Issue 01, March 2015 Pages 124-147, ISSN: 2394-7802 Soft Computation for Structural Dynamic Natural Problems P V Ramana 1, Varkala Abhishek 2 Department of Civil Engineering, MNIT Jaipur, Rajasthan, India Email: 1 pvramana.ce@mnit.ac.in, 2 2013pcs5117@mnit.ac.in Abstract The present paper employs the Emerging method to structural dynamic natural problems which are basically initial value problems. Two different interpolation function have been considered, first one being x = a + bt and the second one being x = a cos t+b sin t. The results obtained from the Emerging method using these interpolation functions are compared with the exact solutions using MATLAB. The Emerging method gave exact solutions for few problems, and the for the others, the solution was converging for a short period of time. The aim of this paper is to obtain appropriate interpolation function for structural dynamic Linear natural problems. The deviation from the exact solutions is shown both graphically and in tabular form. 1 Introduction The Emerging method is a semi-analytical method for solving ordinary as well as partial nonlinear differential equations. The method was developed from the 1970 s to the 1990 s by George Adomian, chair of the Center for Applied Mathematics at the University of Georgia. The aim of this method is towards a unified theory for the solution of partial differential equations. The crucial aspect of the method is employment of the polynomials which allow for solution convergence of the nonlinear portion of the equation, without simply linearizing the system. The Emerging method has been successively used to find the explicit and numerical solutions of the time based partial differential equations. Some techniques which assume essentially that the linear and non-linear system is almost linear after equivalent linearization will not be able to retain the originality of the problem. Present technique consists of splitting the given equation into linear, remainder and non-linear parts, inverting the highest order differential operator contained in the linear operator on both sides, identifying the initial or boundary conditions and the terms involving the independent variable alone as initial approximation, decomposing the unknown function into a series whose components are to be determined, decomposing the linear and non-linear function in terms of special polynomial.the present emerging method provides the solution in a rapidly convergent series with easily computable components. The main advantage of the method is that it can be used directly to solve, all types of differential equations with homogeneous and non-homogeneous boundary conditions. Another advantage of the method is that it reduces the computational work in a tangible manner, while maintaining higher accuracy of the numerical solution.the conventional have systematic procedure and follow some assumed rules, but using the Emerging method, one can solve the problem straight away. The decomposition method can be an effective method for solution of a wide class of problems providing generally a rapidly convergent series solution. It has some distinct advantages over usual approximation methods in that it is computationally convenient, provides analytic, variable solutions not requiring perturbation, linearization, or discretization and resulting massive computation. Since it solves nonlinear problems rather than linearizing them, the resulting solutions are more physically realistic. The given references provide further insights. Of course difficulties remain which present interesting areas for study. Furthermore the Emerging method does not require the use of Green s functions, which would complicate such analytic calculations since Green s functions are not easily determined in most cases. The accuracy of the analytic approximate solutions obtained can be verified by direct substitution. Advantages of the Emerging method over Picard s iterated method were demonstrated. A key notion is the polynomials, which are tailored to the particular non-linearity to solve nonlinear operator equations.in particular, it would be desirable to determine easier ways of generating the polynomials and to study their properties. A convergence and error study for various classes of equations is a further need and it is clear that many theorems will follow.nonlinear equations arise in all fields of applied mathematics, Engineering and Physics, hence being of fundamental importance the existence Received: February 20, 2015 Keywords: Linear, Emerging Method, Matlab, Interpolation function

Soft Computation for Structural Dynamic Natural Problems 125 of methods to find their real roots. As analytic solutions are only available in few cases, the constructions of efficient numerical methods are essential. Emerging method has been successfully applied to linear and nonlinear problems, stochastic and deterministic, obtaining an exact or approximate solution to the problem. One of its advantages is that it provides a rapid convergent solution series. However, the method applied to nonlinear equations does not seem to be fast enough to be an efficient method to solve these kinds of equations.in this present paper,the emerging method has been employed to initial value-structural dynamic problems. 2 Literature Review Muhammad Aslam Noor et.al[1](2008), in their paper, Solving Higher Dimensional Initial Boundary Value Problems by Variational Iteration Decomposition Method, investigated a relatively new technique which is called the variational iteration decomposition method (VIDM) by combining the traditional variational iteration and the decomposition methods for solving higher dimensional initial boundary value problems. The proposed method is an elegant combination of variational iteration and the decomposition methods.jun-sheng Duan et.al[2](2010), in their paper, Reduced Polynomials and Their Generation in Adomian Decomposition Methods, Adomian polynomials are constituted of reduced polynomials and derivatives of nonlinear operator. The reduced polynomials are independent of the form of the nonlinear operator. A recursive algorithm of the reduced polynomials is discovered and its symbolic implementation by the mathematical software is given. As a result, a new and convenient algorithm for the Adomian polynomials is obtained. In 2013, Shahid S. Siddiqi et al.[3] (2014), compare ADM and homotopy perturbation method for solving seventh order boundary value problems. The approximate solutions of the problems obtained with a small amount of computation in both methods. Two numerical examples have been considered to illustrate the accuracy and implementation of the methods. Mahmoud Paripour et al.[4](2014), solve hybrid fuzzy differential equations using ADM. Wazwaz[5] (1998),used some examples to compare ADM and Taylor series method and demonstrated that the decomposition method brought fourth trust able results with less loops. On the other hand the Taylor series method endured computational difficulties. Wazwaz[6](1997), stimulated the convergence of the series solution by modifying ADM. He also solved higher order boundary value problems and produced numerical algorithm. Guellal et al.[7](1997) solved differential equation coming from physics and compared Runge-Kutta method with ADM.Recently, Morawetz [8] (1991),solved a first order non-linear wave equation (NLWE) problem differently and applied for wave equations only. The solution obtained by this method is derived in the form of a power series with easily computable components.in 2012, A.Hasseine et al.[9] solve population balance equations for aggregation, nucleation, growth and breakup processes. The results obtained indicate that the Adomian decomposition method is highly accurate, efficient and are useful for further work. 3 Methodology Time variant problems are basic problems in structure dynamics, in these type of problem force varies with time. Structural dynamics, therefore, is a type of structural analysis which covers the behaviour of structures subjected to dynamic loading. Dynamic loads include people, wind, waves, traffic, earthquakes, and blasts. Any structure can be subjected to dynamic loading. Dynamic analysis can be used to find dynamic displacements, time history, and modal analysis. For a generalised differential equation problem the methodology will be as follows n u t n + n 1 u x n 1 + n 2 u y n 2 + n m u t n m + m u z m +... + u2 + u 3 + ( u z )n... = g(x, y, z, t) n u t n + n 1 u x n 1 + n 2 u y n 2 + n m u t n m + m u z m +...+ }{{} Linear T erms n u t n }{{} Linear + n 1 u x n 1 + n 2 u y n 2 + n m u t n m + m u z m +...+ }{{} Remainder T erms L t u }{{} Linear + L x u + L y u + L z u +... }{{} Remainder u 2 + u 3 + ( u z )n... } {{ } Non Linear T erms u 2 + u 3 + ( u z )n... } {{ } Non Linear T erms + N(u) }{{} Nonlinear L t u + R(u) + N(u) = g(x, y, z, t) = g(x, y, z, t) }{{} F orce = g(x, y, z, t) }{{} F orce T erms = g(x, y, z, t) }{{} F orce T erms

126 P V Ramana, Varkala Abhishek u 0 is independent variable function similarly L 1 t L t u = g(x, y, z, t) R(u) N(u) L t u = L 1 (g(x, y, z, t) R(u) N(u)) t u = n constants + L 1 t (g(x, y, z, t) R(u) N(u)) u = a 0 + ta 1 +... + t n 1 a n 1 + L 1 t (g(x, y, z, t)) L 1 (R(u) + N(u)) u = a 0 + ta 1 +... + t n 1 a n 1 + L 1 t (g(x, y, z, t)) L 1 t (R(u) + N(u)) }{{}}{{} Independent term dependent term u 0 = a 0 + ta 1 + t 2 a 2 +... + t n 1 a n 1 + L 1 t g(x, y, z, t) u 1 = L 1 t (R(u 0 ) + N(u 0 )) u 2 = L 1 t (R(u 1 ) + N(u 1 )) u n = L 1 t (R(u n 1 ) + N(u n 1 )) An Emerging Method Solution to be sum of series of functions, u will be the final solution for this differential equation. u = u 0 + u 1 + u 2 +... + u u = i=0 u i t 4 Problem Statement In this present paper, the linear generalized structural dynamic natural problem which is given by the differential equation mẍ + cẋ + kx = f cos t is taken. From this equation, a total of nine different problems have been solved. The first three being free from damping (c = 0), which are represented by the equations mẍ + kx = 0 mẍ + kx = f mẍ + kx = f cos t However, the force acting in each of this case differs as the force in the first case is zero, a constant force f in the second case, a harmonic force f cos t in the third case. In the next three, an imaginary case with stiffness as zero (k = 0) is considered which are as follows mẍ + cẋ = 0 mẍ + cẋ = f mẍ + cẋ = f cos t And the rest three have both damping and stiffness, which are given by the equations mẍ + cẋ + kx = 0 mẍ + cẋ + kx = f mẍ + cẋ + kx = f cos t Emerging method is adapted to all these nine cases in the form of two different interpolation functions, first one being x = a + bt, and the second being x = a cos t + b sin t. For all the nine cases, the exact solution is known. Hence, a comparison is made between the two different solutions obtained from two different displacements of Emerging method and the exact solution. These results are shown both in graphical and tabular form.

Soft Computation for Structural Dynamic Natural Problems 127 5 Applications In this section, different structural dynamics natural problems which are basically initial problems are solved. Each case is discussed initially and is solved using using two different interpolation functions, first one being x = a + bt and the second one being x = a cos t + b sin t. The results are shown graphically and are compared with the exact results in a tabular form. The solution obtained using the interpolation function x = a + bt is represented as Xeabt, and that obtained using x = a cos t + b sin t is represented as Xacbs and the exact solution is represented as Xe. k x 0 is the initial displacement, x 0 is the initial velocity, is the natural frequency of the system given by = m. The constants considered are force of 1 KiloNewton, frequency of 2 Hertz per sec,stiffness of 4 KiloNewton per meter, damping of 0.02 Newton-second per meter, initial displacement of 1 millimeter, initial velocity of 1.67 millimeter per second. 5.1 case 1: Free damping and free forced vibration mẍ + kx = 0 Let us consider an initial value structural dynamic problem which is free from damping and no is force acting on it. The governing equation for such a case is given by mẍ + kx = 0 (5.1.1) subjected to initial conditions t = 0; x = x 0, t = 0; ẋ = ẋ 0. On dividing the equation(5.1.1) by m on both sides; ẍ + 2 x = 0 where being the natural frequency of the system given by = on substituting ẍ = 2 2 t = L t,the above equation turns to L t x + 2 x = 0 k m 5.1.1 Considering x = a + bt Now Considering the interpolation function as x = a + bt; On applying inverse operator L 1 t on both the sides of the equation (5.1.1) x = 2 L 1 t (x) + a + bt X 0 = a + bt on applying the initial value conditions, the above equations turns to X 0 = x 0 + ẋ 0 t Substituting the value of X 0 in x = L 1 t 2 (x) X 1 = 2 L 1 t (X 0 ) X 1 = 2 ( x 0t 2 2 + ẋ0t 3 6 )

128 P V Ramana, Varkala Abhishek Substituting the value of X 1 in Similarly and so on x = L 1 t 2 (x) X 2 = L 1 t 2 (X 1 ) X 2 = 4 ( x 0t 4 24 + ẋ0t 5 120 ) X 3 = 6 ( x 0t 6 720 + ẋ0t 7 5040 ) X 4 = 8 ( x 0t 8 40320 + ẋ0t 9 362880 ) x = X 0 + X 1 + X 2 + X 3 + X 4 +... x = x 0 (1 w2 t 2 + w4 t 4 +...) + ẋ0 2! 4! (t + 3 t 3 +...) 3! On substituting the sin and cos functions; the x value finally turns to 5.1.2 Considering x = a cos t + b sin t Now Considering the interpolation function as x = x 0 cos t + ẋ0 sin t x = a cos t + b sin t On applying inverse operator L 1 t on both the sides of the equation (5.1.1) x = L 1 t ( 2 x) + a cos t + b sin t X 0 = a cos t + b sin t on applying the initial value conditions, the above equations turns to Substituting the value of X 1 back Similarly For n iterations, As, X 0 = x 0 cos t + ẋ0 X 1 = x 0 cos t + ẋ0 X 2 = x 0 cos t + ẋ0 sin t sin t sin t x = X 0 + X 1 + X 2 + X 3 +... + X n ; X 0 = X 1 = X 2 = X 3 =...X n x = n(x 0 cos t + ẋ0 sin t) 5.1.3 Exact solution The exact solution is the same as the solution obtained for the assumed interpolation function, x = a + bt; which is given as x = x 0 cos t + ẋ0 sin t

Soft Computation for Structural Dynamic Natural Problems 129 (a) Displacement variation for case(5.1.1) (b) Displacement,velocity,state space solution for case(5.1.1) 5.1.4 Discussion and interpretation of the results In this case(5.1.1), the solution obtained using x = a cos t + b sin t is to be divided with n with is basically the number of iterations to obtain the solution. Emerging method has exactly coincided with exact solution with the two considered interpolation fields x = a + bt and x = a cos t + b sin t. 5.2 case 2: Free damped and constant forced vibration mẍ + kx = f Let us consider an initial value structural dynamic problem which is free from damping and has a constant force with magnitude f acting acting on it. The governing equation for such case is given by mẍ + kx = f (5.2.1) subjected to initial conditions; t = 0; x = x 0, t = 0; ẋ = ẋ 0. On dividing the equation (5.2.1) by m on both the sides; ẍ + 2 x = f m on substituting ẍ = 2 2 t = L t,the above equation turns to L t x + 2 x = f m

130 P V Ramana, Varkala Abhishek 5.2.1 Considering x = a + bt Now Considering the interpolation function as x = a + bt; On applying inverse operator L 1 t on both the sides of the equation (5.2.1) x = L 1 t ( 2 x) + a + bt + ft2 2!m X 0 = a + bt + f 2!m on applying the initial value conditions, the above equations turns to Substituting the value of X 0,we arrive at similarly,, X 0 = x 0 + ẋ 0 t + ft2 2!m X 1 = 2 ( x 0t 2 2! X 2 = 4 ( x 0t 4 4! X 3 = 6 ( x 0t 6 6! + ẋ0t 3 3! + ẋ0t 5 5! + ẋ0t 7 7! + ft4 4!m ) + ft6 6!m ) + ft8 8!m ) x = X 0 + X 1 + X 2 + X 3 + X 4 +... The above terms converges in sin and cos terms, which can be written as x = x 0 cos t + ẋ0 sin t + f (1 cost) k 5.2.2 Considering x = a cos t + b sin t Now Considering the interpolation function as X = a cos t + b sin t ; On applying inverse operator L 1 on both the sides of the equation (5.2.1) x = L 1 t ( 2 x) + a cos t + b sin t + ft2 2!m X 0 = a cos t + b sin t + ft2 2!m On applying the initial value conditions, the above equations turns to On substituting back X 0 value, we arrive at X 0 = x 0 cos t + ẋ0 X 1 = x 0 cos t + ẋ0 sin t sin t + ft2 2!m ft4 2 4!m

Soft Computation for Structural Dynamic Natural Problems 131 (a) Displacement variation for case(5.2.1) (b) Displacement,velocity,state space solution for case(5.2.1),, X 2 = x 0 cos t + ẋ0 sin t + ft6 4 6!m X 3 = x 0 cos t + ẋ0 sin t ft8 6 8!m The above terms converges in sin and cos terms, which can be written as 5.2.3 Exact Solution x = n(x 0 cos t + ẋ0 sin t) + f (1 cost) k The exact solution is the same as the solution obtained for the assumed interpolation function, x = a + bt; which is given as 5.2.4 Discussion and interpretation of the results x = x 0 cos t + ẋ0 sin t + f (1 cost) k The complementary function of solution obtained using x = a cos t+b sin t is to be divided with number of iterations to obtain the solution. Even in this case(5.2.1), the emerging method has exactly coincided with exact solution with the two considered interpolation fields x = a + bt and x = a cos t + b sin t. 5.3 case 3: Free damped and harmonic forced vibration mẍ + kx = f cos t Let us consider an initial value structural dynamic problem which is free from damping and has a constant force with magnitude f cos t acting acting on it. The governing equation for such case is given by mẍ + kx = f cos t (5.3.1) subjected to initial conditions; t = 0; x = x 0, t = 0; ẋ = ẋ 0.

132 P V Ramana, Varkala Abhishek On dividing the equation (5.3.1) by m on both the sides; on substituting ẍ = 2 2 t = L t,the above equation turns to ẍ + 2 x = f cos t m L t x + 2 x = f cos t m 5.3.1 Considering x = a + bt Now Considering the interpolation function as x = a + bt; On applying inverse operator L 1 t on both the sides of the equation (5.3.1) On substituting the initial condition, On substituting the value of X 0 back, we arrive at x = L 1 t ( 2 x) f cos t + a + bt m2 X 0 = a + bt X 0 = x 0 + ẋ 0 t X 1 = 2 ( x 0t 2 2! X 2 = 4 ( x 0t 4 4! + ẋ0t 3 3! + ẋ0t 5 5! f cos t m2 f cos t m2 + f cos t) mw4 f cos t) mw6 X 3 = 6 ( x 0t 6 + ẋ0t 7 f cos t) 6! 7! mw8 x = X 0 + X 1 + X 2 + X 3 + X 4 +... 5.3.2 Considering x = a cos t + b sin t Now Considering the interpolation function as x = x 0 cos t + ẋ0 sin t n(f cos t) k X = a cos t + b sin t; On applying inverse operator L 1 t on both the sides of the equation (5.3.1) x = L 1 t ( 2 x) + a cos t + b sin t f cos t m2 X 0 = a cos t + b sin t f cos t m2

Soft Computation for Structural Dynamic Natural Problems 133 (a) Displacement variation for case(5.3.1) (b) Displacement,velocity,state space solution for case(5.3.1) on applying the initial value conditions, the above equations turns to Similarly Finally X 0 = x 0 cos t + ẋ0 X 1 = x 0 cos t + ẋ0 X 2 = x 0 cos t + ẋ0 sin t sin t sin t X 3 = x 0 cos t + ẋ0 sin t x = X 0 + X 1 + X 2 + X 3 + X 4 +... x = n(x 0 cos t + ẋ0 sin t) 5.3.3 Exact Solution The exact solution is given by the equation x = x 0 cos t + ẋ0 sin t + f cos t m(0 2 2 ) 5.3.4 Discussion and interpretation of the results In this case(5.3.1), the solution obtained from the emerging method using both the interpolation fields x = a + bt and x = a cos t + b sin t converged initially but diverged as the time progressed.

134 P V Ramana, Varkala Abhishek t (sec) Xe (mm) Xeabt (mm) Xacbs (mm) %errorxeabt %errorxacbs 0.1 1.0886 1.0847 1.146 0.358-5.3 0.2 1.014 1.1887 1.246-17.471-22.8 0.3 0.790 1.2452 1.296-57.6-64.1 0.4 0.448 1.2522 1.2957-179.1-188.8 0.5 0.043 1.2092 1.2429-270 -277.7 0.6-0.363 1.1180 1.1406-407.4 413.6 0.7-0.7097 0.9822 0.992 238 239.99 0.8-0.9419 0.8073 0.805 185.7 185 0.9-1.0257 0.60 0.5860 158.5 157.1 1-0.9498 0.3691 0.3431 138.9 136.1 5.4 Case 4: Free forced vibration without stiffness mẍ + cẋ = 0 Let us consider an imaginary initial value structural dynamic problem which has zero stiffness and no force acting acting on it. The governing equation for such case is given by mẍ + cẋ = 0 (5.4.1) subjected to initial conditions, t = 0; x = x 0, t = 0; ẋ = ẋ 0. On dividing the equation (5.4.1) by m on both the sides; ẍ = c mẋ on substituting ẍ = 2 2 t = L t, ẋ = t t = L t1 The above equation turns to L t x = c m L t1 5.4.1 Considering x = a + bt Now Considering the interpolation field as x = a + bt. On applying the inverse operatorl 1 t, x = a + bt c m L t 1(L t1 x) On substituting the initial value conditions, On substituting back the value of X 0 back, we arrive at X 0 = a + bt X 0 = x 0 + ẋ 0 t X 1 = c ẋ 0 t 2 m 2! X 2 = c2 ẋ 0 t 3 m 2 3! X 3 = c3 ẋ 0 t 4 m 3 4! x = X 0 + X 1 + X 2 + X 3 + X 4 +...

Soft Computation for Structural Dynamic Natural Problems 135 On substituting the values of all the iterations, we arrive at 5.4.2 Consideringx = a cos t + b sin t x = x 0 + ẋ0t c (1 e ct m ) m Now Considering the interpolation function as X = a cos t + b sin t, On applying inverse operator L 1 t on both the sides of the equation (5.4.1), x = a cos t + b sin t c m L t 1(L t1 x) On applying the initial value conditions Substituting X 0 back, we obtain,,, X 0 = a cos t + b sin t X 0 = x 0 cos t + ẋ0 X 1 = c m ( x 0 sin t + sin t cos t 2 ) X 2 = c2 m 2 ( x 0 sin t cos t 2 + 3 ) X 2 = c3 m 3 ( x 0 sin t cos t 3 + 4 ) x = X 0 + X 1 + X 2 + X 3 + X 4 +... x = x 0 cos(t + c m ) + ẋ0 sin(t + c m ) 5.4.3 Exact Solution The exact solution is the same as the solution obtained for the assumed interpolation function, x = a + bt; which is given as x = x 0 + ẋ0t c (1 e ct m ) m 5.4.4 Discussion and interpretation of the results In this case(5.4.1), the solution obtained from the emerging method using x = a + bt gave exact solution but the solution obtained using x = a cos t + b sin t converged initially but diverged as the time progresses. t (sec) Xe (mm) Xeabt (mm) Xacbs (mm) %errorxeabt %errorxacbs 0.1 1.11668 1.11668 1.1698 0.128-0.257 0.2 1.33 1.3307 1.2604-0.05 5.233 0.3 1.4995 1.4960 1.3008 0.233 13.251 0.4 1.6653 1.6653 1.2892 0.234 22.58 0.5 1.83 1.83 1.2263 1.8267 32.989 0.6 1.9960 1.9960 1.1145 0.2 44 0.7 2.1609 2.16 0.9583 0.162 55 0.8 2.3254 2.3254 0.7639 0.116 67 0.9 2.4896 2.4896 0.5390 0.06 78.3 1 2.6534 2.6534 0.2926 0 88.972

136 P V Ramana, Varkala Abhishek (a) Displacement variation for case(5.4.1) (b) Displacement,velocity,state space solution for case(5.4.1) 5.5 Case 5: Constant forced vibration without stiffness mẍ + cẋ = f Let us consider an imaginary initial value structural dynamic problem which has zero stiffness and a force of magnitude f acting acting on it. The governing equation for such case is given by mẍ + cẋ = f (5.5.1) subjected to initial conditions; t = 0; x = x 0, t = 0; ẋ = ẋ 0. On dividing the equation (5.5.1) by m on both the sides; ẍ + c mẋ = f m on substituting ẍ = 2 2 t = L t, ẋ = t t = L t1 L t x = c m x + L t1 1 f m 5.5.1 Considering x = a + bt Now Considering the interpolation function as x = a + bt; On applying inverse operator L 1 t on both the sides of the equation (5.5.1), x = L t 1( c m ) + L t 1( f m ) + a + bt

Soft Computation for Structural Dynamic Natural Problems 137 X 0 = a + bt + ft2 2!m X 1 = L t 1( c m L t1x 0 ) Similarly X 1 = c m (ẋ0t 2 2! + ft3 3!m ) X 2 = c2 3 m 2 (ẋ0t + ft4 3! 4!m ) X 3 = c3 4 m 3 (ẋ0t + ft5 4! 5!m ) x = X 0 + X 1 + X 2 + X 3 + X 4 +... x = x 0 + ẋ 0 m c 5.5.2 Considering x = a cos t + b sin t ct fm (1 e m ) + c 2 ( ct m + e ct m 1) Now Considering the interpolation function as X = a cos t + b sin t On applying inverse operator L 1 t on both the sides of the equation (5.5.1), x = L t 1( cx m ) + L t 1( f ) + a cos t + b sin t m x 0 = a cos t + b sin t + ft2 2!m On substituting the initial value conditions; X 0 = x 0 cos t + ẋ0 sin t + ft2 2!m Similarly X 1 = c m (x 0 cos t 2 X 1 = L t 1 cx 0 m + ẋ0 sin t 3 ft4 4!m ) X 2 = c2 m 2 (x 0 cos t 4 + ẋ0 sin t 5 ft6 6!m ) x = X 0 + X 1 + X 2 + X 3 + X 4 +... x = x 0 ( cost 1 c k ) + ẋ0 t (sin 1 c ) + f c k c (1 cos m t) 5.5.3 Exact Solution The exact solution is given as which can also be written as x = c(x 0c + ẋ 0 m) fm + cft + (fm cẋ 0 m)e ct m c 2 m ct f x = x 0 + ẋ 0 (1 exp m ) + c c (m c + t + m ct exp m ) c

138 P V Ramana, Varkala Abhishek (a) Displacement variation for case(5.5.1) (b) Displacement,velocity,state space solution for case(5.5.1) 5.5.4 Discussion and interpretation of the results In this case(5.5.1), the solution obtained using x = a + bt gave exact solution, the solution obtained using X = a cos t + b sin t gave converging solution initially but diverged as the time progressed. t (sec) Xe (mm) Xeabt (mm) Xacbs (mm) %errorxeabt %errorxacbs 0.1 1.1718 1.1718 1.1610 0 0.9236 0.2 1.3533 1.3533 1.2897 0 4.7032 0.3 1.5444 1.5444 1.3870 0 10.1950 0.4 1.7451 1.7451 1.4509 0 16.8611 0.5 1.9554 1.9554 1.4814 0 24.239 0.6 2.1753 2.1753 1.4808 0 31.927 0.7 2.4047 2.4047 1.4530 0 39.578 0.8 2.6437 2.6437 1.4039 0 46.8959 0.9 2.8921 2.8921 1.3411 0 53.6 1 3.1501 3.1501 1.2733 0 59.57 5.6 Case 6: Harmonic forced vibration with out stiffness mẍ + cẋ = f cos t Let us consider an imaginary initial value structural dynamic problem which has zero stiffness and a harmonic force of magnitude f cos t acting acting on it. The governing equation for such case is given by mẍ + cẋ = f cos t (5.6.1) subjected to initial conditions; t = 0; x = x 0, t = 0; ẋ = ẋ 0.

Soft Computation for Structural Dynamic Natural Problems 139 On dividing the equation 5.6.1 by m on both the sides; on substituting ẍ = 2 2 t = L t, ẋ = t t = L t1 ẍ + c m L t1x = f cos t m Which results in L t x + c m L t1x = f cos t m 5.6.1 Considering x = a + bt Now Considering the interpolation function asx = a + bt. On applying inverse operator, L 1 t on both sides of the equation (5.6.1) x = L 1 t ( f m cos t) c m L 1 (L t1 x) + a + bt t Similarly X 0 = x 0 + ẋ 0 t f cos t 2!m 2 X 1 = c 2 m (ẋ0t f sin t 2! 2!m 3 ) X 2 = c2 3 m 2 (ẋ0t f cos t 3! 2!m 4 ) x = X 0 + X 1 + X 2 + X 3 + X 4 +... x = x 0 + ẋ0m c 5.6.2 Considering x = a cos t + b sin t ct f (1 e m ) + (1 cos(t c 2k m )) Now Considering X = a cos t + b sin t On applying inverse operator L 1 t on both sides of the equation (5.6.1) x = L 1 t On substituting the initial value conditions; ( f m cos t) c m L 1 (L t1 x) + a cos t + b sin t x 0 = a cos t + b sin t t f cos t m 2 Now X 0 = x 0 cos t + ẋ0 sin t X 1 = c m L 1 t (L 1 t x 0 ) X 1 = cx 0 sin t m + cẋ 0 cos t k X 2 = c2 m 2 (x 0 cos t 2 ) c2 x 0 sin t m 3 )

140 P V Ramana, Varkala Abhishek (a) Displacement variation for case(5.6.1) (b) Displacement,velocity,state space solution for case(5.6.1) 5.6.3 Exact Solution x = X 0 + X 1 + X 2 + X 3 + X 4 +... x = x 0 cos(t + c m ) + ẋ0 sin(t + c m ) x = c2 f sin t + c 2 (x 0 c + ẋ 0 m) + m 2 2 (x 0 c + ẋ m ) cfm cos t c(c 2 + m 2 2 ) e ct m (ẋ 0 c 2 m fcm + ẋ 0 )m 3 2 c(c 2 + m 2 2 ) 5.6.4 Discussion and interpretation of the results In this case(5.6.1), the solution obtained from the emerging method using both the interpolation functions, converged initially but diverged as the time progresses. But the result using x = a + bt gave more realistic values. t (sec) Xe (mm) Xeabt (mm) Xacbs (mm) %errorxeabt %errorxacbs 0.1 1.11718 1.117.27 1.1698 1.0866 0.1706 0.2 1.3530 1.1827 1.2664 12.57 6.884 0.3 1.543 1.2879 1.3008 16.53 15.6967 0.4 1.7410 1.4015 1.2892 19.5 25.95 0.5 1.9454 1.522 1.2263 21.753 36.964 0.6 2.1548 1.6482 1.1145 23.51 48.278 0.7 2.3673 1.7781 0.9581 24.88 59.527 0.8 2.5812 1.91 0.7639 26.003 70.405 0.9 2.794 2.0418 0.5390 26.9219 80.708 1 3.0047 2.1716 0.2926 27.72 90.26 5.7 Case 7: Damped and free forced vibration mẍ + cẋ + kx = 0 Let us consider an initial value structural dynamic problem which is damped and no force is acting on it. The governing equation for such case is given by mẍ + cẋ + kx = 0 (5.7.1)

Soft Computation for Structural Dynamic Natural Problems 141 (a) Displacement variation for case(5.7.1) (b) Displacement,velocity,state space solution for case(5.7.1) subjected to initial conditions, t = 0; x = x 0, t = 0; ẋ = ẋ 0. 5.7.1 Considering x = a + bt x = x 0 (1 2 t 2 2! for n=3 iterations. + c2 t 3 3!m + 4 t 4 c2 2 t 4 4! m 2 c4 t 5 4! 5!m 4 t 5 5! + 2 ct 4 m4! + 4 ct 5 5! 5.7.2 Considering x = a cos t + b sin t for n=3 5.7.3 Exact Solution + 6 t 6 ) + ẋ 0 (t ct2 6! 2!m 2 t 3 + c2 t 3 3! m 2 3! + c2 t 4 m4! + c3 t 4 m 3 4! 2c2 2 t 5 m 2 c4 t 6 2 c 2 t 5 5! m6! m 2 2 t 6 5! m6! 4 t 7 7! x = x 0 (4 cos t 5 c sin t c2 c cos t m m 2 sin t + 2 m 2 + c3 sin t m 3 2 )+ c cos t ẋ 0 ( m 2 + c2 sin t m 2 3 2c2 sin t 2 sin t cos t c sin t m 2 3 + 2 + 3 m c3 cos t m 3 4 ) ) The solution is given as x = e εt (x 0 cos t + ẋ0 + εx 0 sin t).

142 P V Ramana, Varkala Abhishek 5.7.4 Discussion and interpretation of the results In this case(5.7.1), the solution obtained from the emerging method using x = a + bt gave exact solution but the solution obtained using x = a cos t + b sin t converged initially but diverged as the time progresses. t (sec) Xe (mm) Xeabt (mm) Xacbs (mm) %errorxeabt %errorxacbs 0.1 1.1453 1.1458 0.9858-0.043 13.92 0.2 1.2439 1.2955 0.8956-0.247 28 0.3 1.2923 1.2955 0.7697-0.247 40.45 0.4 1.2889 1.2932 0.613-0.336 52.432 0.5 1.2334 1.2386 0.43-0.4215 64.99 0.6 1.1313 1.1335 0.233-0.1944 79.331 0.7 0.9841 0.9821 0.0262 0.203 97.33 0.8 0.7991 0.7907-0.1824 1.051 122.82 0.9 0.5839 0.5684-0.3838 2.654 165.72 1 0.3473 0.3265-0.5698 5.99 264.06 5.8 Case 8: Damped and forced vibration mẍ + cẋ + kx = f Let us consider an initial value structural dynamic problem which is damped and has a force of magnitude f acting on it. The governing equation for such case is given by mẍ + cẋ + kx = f (5.8.1) subjected to initial conditions, t = 0; x = x 0, t = 0; ẋ = ẋ 0. On dividing the equation (5.8.1) with m on both the sides; ẍ + 2εx + 2 x = f m On substituting ẍ = 2 2 t = L t, ẋ = t t = L t1 L t x + 2εL t1 x = f m L t x = f m 2εL t1(x) 2 x 5.8.1 Considering x = a + bt On applying inverse operator on both sides of the equation (5.8.1); x = L t 1( f m 2εL t1(x) 2 x) + a + bt x = L t 1( 2εL t1 (x) 2 x) + a + bt + ft2 2m X 0 = ft2 2m + a + bt;

Soft Computation for Structural Dynamic Natural Problems 143 On applying initial value conditions; On substituting back the value of X 0 ; X 0 = x 0 + ẋ 0 t + ft2 2m Similarly X 1 = 2ε(ẋ0t 2 2! + ft3 3!m ) 2 ( x 0t 2 + ẋ0t 3 + ft4 2! 3! 4!m ) X 2 = 2εL t 1(L t x 1 ) L 1 t 2 (X 1 ) x = X 0 + X 1 + X 2 + X 3 + X 4 +... x = x 0 + ẋ 0 t + ft2 2 2!m 2ε(ẋ0t + ft3 2! 3!m ) 2 ( x 0t 2 + ẋ0t 3 + ft4 2! 3! 4!m ) + 4ε2 2 3 (ẋ0t + ft4 3! 4!m ) + 2ε3 ( x 0t 3 + ẋ0t 4 ft5 3! 4! 5!m )+ 2ε 3 (ẋ0t 4 4! + ft5 5!m ) + 4 ( x 0t 4 + ẋ0t 5 ft6 4! 5! 6!m ) 5.8.2 Considering x = a cos t + b sin t On applying inverse operator on both sides of the equation (5.8.1); On applying initial value conditions; x = L 1 t ( f m 2εL t1x 2 x) + a cos t + b sin t x = 2εL 1 t X 1 = 2ε( x 0 sin t (L t x) L 1 t ( 2 x) + ft2 + a cos t + b sin t 2!m X 0 = x 0 cos t + ẋ0 sin t + ft2 2!m X 1 = 2εL 1 t (L t 1X 0 ) L 1 t ( 2 X 0 ) ẋ0 cos t 2 + ft3 3!m 2 ( x 0 cos t ẋ 0 sin t 2 3 + ft4 4!m ) sin t x = x 0 ( 4ϑ 2 cos t 4ε sin t + 3 cos t) + ẋ 0 ( 6εcost sin t4ε2 + 3 sin t ) + ft2 2m 2εft3 3!m ft4 4 4!m + 4ε2 2 ft 4 4!m + 4ε2 ft 5 5!m + ft6 4 6!m 5.8.3 Exact Solution The exact solution is given as x = f k e t(c+(c 2 4km) 0.5 ) 2m 2k(c 2 4km) (f(c 2 4km) 0.5 cf + x 0.5 0 ck + 2ẋ 0 km x 0 k(c 2 4km) 0.5 )+ e t(c (c cf + x 0 ck + 2ẋ 0 km x 0 k(c 2 4km) 0.5 ) 2 4km) 0.5 ) 2m 2k(c 2 4km) 0.5 (f(c 2 4km) 0.5 5.8.4 Discussion and interpretation of the results In this case(5.8.1), the solution obtained from the emerging method using both the interpolation functions, converged initially but diverged as the time progresses. But the result obtained using x = a cos t + b sin t gave more realistic values. The solution obtained using x = a cos t + b sin t gave better values compared to the solution obtained using x = a + bt.

144 P V Ramana, Varkala Abhishek (a) Displacement variation for case(5.8.1) (b) Displacement,velocity,state space solution for case(5.8.1) t (sec) Xe (mm) Xeabt (mm) Xacbs (mm) %errorxeabt %errorxacbs 0.1 1.1508 1.1602 1.2088-0.816-5.04 0.2 1.2654 1.3025 1.338-2.93-5.74 0.3 1.3393 1.4202 1.416-6.053-5.75 0.4 1.3697 1.5076 1.439-10.06-5.098 0.5 1.3555 1.5586 1.4052-14.9-3.66 0.6 1.2972 1.5686 1.3125-20.9-1.17 0.7 1.1974 1.5342 1.1626-28.12 2.904 0.8 1.06 1.4534 0.9581-37.1 9.61 0.9 0.8907 1.3265 0.703-48.9 21.06 1 0.6961 1.1558 0.4026-65.9 42.1 5.9 Case 9: Damped and harmonic forced vibration mẍ + cẋ + kx = f cos t Let us consider an initial value structural dynamic problem which is damped and has a force of magnitude f acting on it. The governing equation for such case is given by mẍ + cẋ + kx = f cos t (5.9.1) subjected to initial conditions, t = 0; x = x 0, t = 0; ẋ = ẋ 0. 5.9.1 Considering x = a + bt x = x 0 (1 2 t 2 3 t 3 c 2! 3!m 2 t 4 4! f 3c (1 3 cos t sin t + c2 k m ) + ẋ 0 (t + ct2 2!m t3 2 c2 t 3 3! 3!m 2 ct4 2 ct4 2 m4! 4! m 2 2 cos t 2 t 2 2 t 4 2! 4! 2 t 3 ) 3! + t5 4 + 3!

Soft Computation for Structural Dynamic Natural Problems 145 (a) Displacement variation for case(5.9.1) (b) Displacement,velocity,state space solution for case(5.9.1) 5.9.2 Considering x = a cos t + b sin t 5.9.3 Exact Solution x = x 0 (cos t 2c sin t m c2 cos t m 2 2 2 c sin t m 3 )+ 2 sin t c cos t ẋ 0 ( + c2 sin t sin t k km 3 ) The exact solution is given by x = x 0(...) + ẋ 0 (...)... +... 5.9.4 Discussion and interpretation of the results In this case (5.9.1) the solution obtained from the emerging method using both the interpolation functions, converged initially but diverged as the time progresses. But the result using x = a cos t + b sin t gave more realistic values. However, the solution obtained using x = a cos t + b sin t gave better values compared to the solution obtained using x = a + bt. t (sec) Xe (mm) Xeabt (mm) Xacbs (mm) %errorxeabt %errorxacbs 0.1 1.1508 1.3086 1.2676-0.133-10.12 0.2 1.265 1.560 1.4814-0.233-17.09 0.3 1.338 1.733 1.636-0.295-22.25 0.4 1.3657 1.8025 1.7255-0.319-26.351 0.5 1.3458 1.737 1.7462-0.290-29.75 0.6 1.2777 1.5025 1.6972-0.176-32.83 0.7 1.1624 1.058 1.580 0.089-35.9766 0.8 1.0028 0.3583 1.401 0.089-39.70 0.9 0.8033-0.6470 1.1655 1.805-45.0 1 0.569-2.0139 0.8836 4.533-55.04 The following tabular form shows the percent error obtained from solutions using the Emerging method from the exact solution in each case.

146 P V Ramana, Varkala Abhishek 6 Conclusions case Xeabt(maximum error ) Xacbs (maximum error) mẍ + kx = 0 0 0 mẍ + kx = f 0 0 mẍ + kx = f cos t 139 136 mẍ + cẋ = 0 1.8 89 mẍ + cẋ = f 0 60 mẍ + cẋ = f cos t 27 90 mẍ + cẋ + kx = 0 6 264 mẍ + cẋ + kx = f 65 40 mẍ + cẋ + kx = f cos t 5 55 The conclusion of the nine different structural dynamic natural problems solved in this paper are discussed in this session under three different paragraphs. In the first and second cases i.e. for no damping condition(c = 0), the Emerging method using two interpolation functions x = a + bt and x = a cos t + b sin t had given exact solutions in the first and second cases. In the third case, the solution obtained using x = a + bt gave results with error up to 139%. However, the solution converges initially and diverges as the time progresses using the interpolation function x = a cos t + b sin t. The error.was up to 136%. In fourth,fifth and sixth cases i.e. without stiffness(k = 0), the results obtained are like this. In fourth and fifth cases, the solution obtained using x = a + bt gave results very near to the exact solution whose error was less than one percent. However, the solution converges initially and diverges as the time progresses using the interpolation function x = a cos t + b sin t and the error was up to 90%. In the sixth case the solution obtained using x = a + bt gave solution whose error was upto 27%, where as the solution obtained using x = a cos t + b sin t gave error upto 90%. For seventh, eighth and ninth cases i.e. with both stiffness and damping,the results are like this. In the seventh case, the solution obtained using x = a + bt gave solution whose error was upto 5%. However, the solution obtained using the interpolation function x = a cos t + b sin t gave results whose error was up to 260%. In the eight and ninth cases both the solutions initially converged and diverged as the time progresses. However, error obtained using x = a + bt was upto 65%, and that using x = a cos t + b sin t was upto 55%. To sum up, the Emerging solutions gave exact solutions for problems with no damping. For problems with damping, the solution obtained using x = a+bt gave comparatively satisfying values, the solution obtained using x = a cos t+b sin t converges initially and diverges as the time progresses. Also the problems containing harmonic force f cos t gave relatively diverging values using both the interpolation functions. ACKNOWLEDGMENT We would like to thank Prof M.K.Shrimali, Department of Civil Engineering, Malaviya National Institute of Technology, Jaipur for his support and suggestions during this research work. References [1] Muhammad Aslam Noor,Syed Tauseef Mohyud-Din(2008), Solving Higher Dimensional Initial Boundary Value Problems by Variational Iteration Decomposition Method, Vol. 3, Issue 6 pp. 254-266. [2] JunSheng Duan and AiPing Guo(2010), Reduced Polynomials and Their Generation in Adomian Decomposition Methods, CMES, vol.60, no.2, pp.139-150. [3] Shahid S. Siddiqi, Muzammal Iftikhar(2014), Comparison of the Adomian decomposition method with homotopy perturbation method for the solutions of seventh order boundary value problems. [4] Mahmoud Paripour, Elahe Hajilou, Afshin Hajilou, Homa Heidari(2014), Application of Adomian decomposition method to solve hybrid fuzzy differential equations, Journal of Taibah University for Science. [5] A.Wazwaz(1998), A comparison between adomian decomposition method and Taylor series method in the series solution, applied mathematics computations, vol.97, pp.37-44. [6] A.Wazwaz(1999), A reliable modification of adomian decomposition method, applied mathematical computations, vol.102, pp.77-86.

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