User:Egm6321.f11.team1.steele.m

Homework Solutions by Manuel Steele For team 1 of EGM6321 - Fall 2011 at UF

R*1.3: Analyze the dimension of all terms in (2) p.3-3 and provide the physical meaning.

In general, the terms for this equation govern the vehicle motion with the electromagnetic magnet and constant air gap on a flexible beam. $$c_0(Y^1,t) := -F^1[1-\bar{R}u^2_{\text{,SS}}(Y^1,t)]-F^2u^2_{\text{,S}}(Y^1,t)-T/R + M[[1-\bar{R}u^2_{\text{,SS}}(Y^1,t)][u^1_{\text{,tt}}(Y^1,t)-\bar{R}u^2_{\text{,Stt}}(Y^1,t)]$$ $$+ u^2_{\text{,S}}(Y^1,t)u^2_{\text{,tt}}(Y^1,t)]$$ The term for $$c_0$$ is derived from the dynamics of the nominal motion, which have component terms that depend on structural displacement - the beam deformations. This derivation contrasts the previous a priori assumptions about beam rigidity in describing nominal motion. $$c_3(Y^1,t)\ddot{Y}^1+c_2(\dot{Y}^1)^2 + c_1(Y^1,t)\dot{Y}^1+c_0(Y^1,t) = 0$$ $$c_3(Y^1,t)$$ is the mass, and $$c_0(Y^1,t)$$ is the force in Newtons along motion for the equation.  The terms with $$\dot{Y}^1$$ describe convection.  The two terms, $$F^1$$ and $$F^2$$ are the general components of the applied force in Newtons.  Terms of u with the form $$u^2$$ represent displacements of the material point S on the centroidal line.  In terms of structural deformation, u is indicative of the flexible dynamics related to the beam's structural deformation.  This contrasts the ideal form of displacement of nominal motion if the beam were theoretically undeformed.  The value $$\bar{R}$$ is the distance from the beam line of centroids to the wheel center of mass.  The partial derivative of u with respect to S is $$u^2_{\text{,S}} (Y^1,t)$$.  Similarly, the 2nd partial derivative of u with respect to S is given by $$u^2_{\text{,SS}}(Y^1,t)$$.  The 2nd partial derivative of u with respect to t is $$u^2_{\text{,tt}}(Y^1,t)$$.  The partial derivative of the latter term with respect to S is $$u^2_{\text{,Stt}}(Y^1,t)$$.

References for R1.3 Vu-Quoc, L. and Olsson, M., "A computational procedure for interaction of high-speed vehicles on flexible structures without assuming known vehicle nominal motion, Computer Methods in Applied Mechanics and Engineering, Vol.76, pp.207-244, 1989. http://clesm.mae.ufl.edu/~vql/pdf/cmame.1989.vuquoc.olsson.pdf Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, 25 Aug 2011. See general link for figure from journal article. http://en.wikiversity.org/wiki/File:EGM6341.s11.TEAM1.WILKS_EC1.svg

R1.8 The homogeneous solution is y = $$c_1Y^1_H(x) + c_2Y^2_H(x)$$. Find the constants from the initial and boundary values.

(1) Boundary conditions: y(a) = $$\alpha$$ and y(b) = $$\beta$$ (2) Initial conditions: y(a) = $$\alpha$$ and y'(a) = $$\beta$$

Boundary Conditions: The constants $$C_1$$ and $$C_2$$ can be solved by using a system of equations. The boundary conditions give two equations with two unknowns. By solving for one constant in terms of the other then a simple substitution and elimination can solve for one of the constants. The boundary conditions can then be used to find the second constant. Similarly, the initial conditions can be solved with two equations and two unknowns.

(a) $$C_1Y^1_H(a) + C_2Y^2_H(a) = \alpha$$ (b) $$C_1Y^1_H(b) + C_2Y^2_H(b) = \beta$$

By substitution and algebraic manipulation, $$C_1$$ and $$C_2$$ can be found for the boundary conditions. The derivation begins here. From (a) we have $$C_1Y^1_H(a) = \alpha - C_2Y^2_H(a)$$ $$C_1 = (\alpha - C_2Y^2_H(a))/Y^1_H(a)$$ From (b), we get $$C_1Y^1_H(b) = \beta - c_2Y^2_H(b)$$ $$C_1 = (\beta - c_2Y^2_H(b))/Y^1_H(b)$$ The solutions of the 1st constant from (a) and (b) can be equated and solved for the unknown. $$(\alpha - C_2Y^2_H(a))/Y^1_H(a) = (\beta - c_2Y^2_H(b))/Y^1_H(b)$$ After algebraic manipulation, the 2nd constant is $$C_2 = [\beta - \alpha(Y^1_H(b)/Y^2_H(a))/[Y^2_H(b) - (Y^1_H(b)Y^2_H(a)/Y^1_H(a)]$$ The 1st constant can now be solved. $$C_1 = \beta - C_2(Y^2_H(b)/Y^1_H(b))$$ $$C_1 = [\beta/Y^1_H(b) - \alpha/Y^2_H(a)]/[1 - (Y^1_H(b)Y^2_H(a)/Y^2_H(b)Y^1_H(a))]$$  Initial Conditions The similar system of two equations and two unknowns can be applied to the initial conditions. (a) $$C_1Y^1_H(a) + C_2Y^2_H(a) = \alpha$$ (b) $$C_1Y^1_H(b)' + C_2Y^2_H(b)' = \beta$$ Solve for $$C_1$$ then substitute to get $$C_2$$.  $$C_1 = (\alpha - C_2Y^2_H(a))/Y^1_H(a)$$ Next, we derive a solution for (b). $$C_2Y^2_H(b)' = C_1Y^1_H(b)' - \beta$$ $$C_2 = (C_1Y^1_H(b)')/Y^2_H(b)' - \beta/Y^2_H(b)'$$ Substitute $$C_1$$ into the equation.  Then we have $$C_2 = (\alpha(Y^1_H(b)'/Y^2_H(b)')) - (\alpha(\beta/Y^2_H(b)')) - (C_2Y^2_H(a)Y^1_H(b)'/Y^1_H(a)Y^2_H(b)') + (C_2\beta(Y^2_H(a)/(Y^1_H(a)Y^2_H(b)'))$$ $$ C_2 = \alpha[(Y^1_H(b)'/Y^2_H(b)') - (\beta/Y^2_H(b)')]/[1+(Y^2_H(a)Y^1_H(b)'/Y^1_H(a)Y^2_H(b)'+(\beta)(Y^2_H(a)/Y^1_H(a)Y^2_H(b))]$$ Further manipulation gives the solution for $$C_2$$. $$C_2 = \alpha[(Y^1_H(b)'/Y^2_H(b)')-(\beta/Y^2_H(b)')]/[1+(Y^2_H(a)Y^1_H(b)'/Y^1_H(a)Y^2_H(b)')+\beta(Y^2_H(a)/Y^1_H(a)Y^2_H(b))]$$ Which gives $$C_1$$ as $$C_1 = \alpha - (Y^2_H(a)/Y^1_H(a))[\alpha[(Y^1_H(b)'/Y^2_H(b)')-(\beta/Y^2_H(b)')]/[1+(Y^2_H(a)Y^1_H(b)'/Y^1_H(a)Y^2_H(b)')+\beta(Y^2_H(a)/Y^1_H(a)Y^2_H(b))]]$$ References for 1.8 A.C. King, J. Billingham, S.R. Otto, Differential equations:  Linear, nonlinear, ordinary, partial, Cambridge University Press, 2003. R. Ellis, D. Gulick, Calculus with Analytic Geometry, 3rd ed., Harcourt Brace Jovanovich, Publishers, 1986, pp. 936 - 962. M. Olia, D.E. Stevens, FE Fundamentals of Engineering Exam, Barron's Educational Series, Inc., 2008, pp. 17-52.