Exam (elaborations) TEST BANK FOR Introduction to Finite Element Analysis 3rd Edition By Reddy J.N. (Solution Manual)

Exam (elaborations) TEST BANK FOR Introduction to Finite Element Analysis 3rd Edition By Reddy J.N. (Solution Manual) SOLUTIONS MANUAL for An Introduction to The Finite Element Method (Third Edition) by J. N. REDDY Department of Mechanical Engineering Texas A & M University College Station, Texas PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of the McGraw-Hill. McGraw-Hill, New York, 2005 Fg = mg Fd = cv v 1 Chapter 1 INTRODUCTION Problem 1.1: Newton’s second law can be expressed as F = ma (1) where F is the net force acting on the body, m mass of the body, and a the acceleration of the body in the direction of the net force. Use Eq. (1) to determine the mathematical model, i.e., governing equation of a free-falling body. Consider only the forces due to gravity and the air resistance. Assume that the air resistance is linearly proportional to the velocity of the falling body. Solution: From the free-body-diagram it follows that m dv dt = Fg − Fd, Fg = mg, Fd = cv where v is the downward velocity (m/s) of the body, Fg is the downward force (N or kg m/s2) due to gravity, Fd is the upward drag force, m is the mass (kg) of the body, g the acceleration (m/s2) due to gravity, and c is the proportionality constant (drag coefficient, kg/s). The equation of motion is dv dt + αv = g, α = c m PROPRIETARY MATERIAL. °c The McGraw-Hill Companies, Inc. All rights reserved. 2 AN INTRODUCTION TO THE FINITE ELEMENT METHOD Problem 1.2: A cylindrical storage tank of diameter D contains a liquid at depth (or head) h(x, t). Liquid is supplied to the tank at a rate of qi (m3/day) and drained at a rate of q0 (m3/day). Use the principle of conservation of mass to arrive at the governing equation of the flow problem. Solution: The conservation of mass requires time rate of change in mass = mass inflow - mass outflow The above equation for the problem at hand becomes d dt (ρAh) = ρqi − ρq0 or d(Ah) dt = qi − q0 where A is the area of cross section of the tank (A = πD2/4) and ρ is the mass density of the liquid. Problem 1.3: Consider the simple pendulum of Example 1.3.1. Write a computer program to numerically solve the nonlinear equation (1.2.3) using the Euler method. Tabulate the numerical results for two different time steps Δt = 0.05 and Δt = 0.025 along with the exact linear solution. Solution: In order to use the finite difference scheme of Eq. (1.3.3), we rewrite (1.2.3) as a pair of first-order equations dθ dt = v, dv dt = −λ2 sin θ Applying the scheme of Eq. (1.3.3) to the two equations at hand, we obtain θi+1 = θi + Δt vi; vi+1 = vi − Δt λ2 sin θi The above equations can be programmed to solve for (θi, vi). Table P1.3 contains representative numerical results. Problem 1.4: An improvement of Euler’s method is provided by Heun’s method, which uses the average of the derivatives at the two ends of the interval to estimate the slope. Applied to the equation du dt = f(t, u) (1) Heun’s scheme has the form ui+1 = ui + Δt 2 h f(ti, ui) + f(ti+1, u0i +1) i , u0i +1 = ui + Δt f(ti, ui) (2) PROPRIETARY MATERIAL. °c The McGraw-Hill Companies, Inc. All rights reserved. SOLUTIONS MANUAL 3 Table P1.3: Comparison of various approximate solutions of the equation (d2θ/dt2) + λ2 sin θ = 0 with its exact linear solution. Exact Approx. solution θ Exact Approx. solution v t θ Δt = .05 Δt = .025 v Δt = .05 Δt = .025 0.00 0.78540 0.78540 0.78540 -0.00000 -0.00000 -0.00000 0.05 0.76965 0.78540 0.77828 -0.62801 -0.56922 -0.56922 0.10 0.72302 0.75694 0.74276 -1.23083 -1.13844 -1.13027 0.15 0.64739 0.70002 0.67944 -1.78428 -1.69123 -1.66622 0.20 0.54578 0.58980 0.56482 -2.26615 -2.20984 -2.15879 0.25 0.42229 0.50496 0.47627 -2.65711 -2.67459 -2.58816 0.30 0.28185 0.37123 0.34225 -2.94148 -3.06403 -2.93371 0.35 0.13011 0.21803 0.19218 -3.10785 -3.35605 -3.17573 0.40 -0.02685 0.05023 0.03148 -3.14955 -3.53018 -3.29791 0.45 -0.18274 -0.12628 -0.13374 -3.06491 -3.57060 -3.29007 0.50 -0.33129 -0.30481 -0.29690 -2.85732 -3.46921 -3.15014 0.60 -0.58310 -0.63965 -0.59131 -2.11119 -2.85712 -2.50787 0.80 -0.78356 -1.05068 -0.91171 0.21536 -0.50399 -0.28356 1.00 -0.50591 -0.94062 -0.74672 2.41051 2.29398 2.19765 In books on numerical analysis, the second equation in (2) is called the predictor equation and the first equation is called the corrector equation. Apply Heun’s method to Eqs. (1.3.4) and obtain the numerical solution for Δt = 0.05. Solution: Heun’s method applied to the pair dθ dt = v, dv dt = −λ2 sin θ yields the following discrete equations: θ0 i+1 = θi + Δt vi vi+1 = vi − λ2 Δt 2 ³ sin θi + sinθ0 i+1 ´ θi+1 = θi + Δt 2 (vi + vi+1) The numereical results obtained with the Heun’s method and Euler’s method are presented in Table P1.4. PROPRIETARY MATERIAL. °c The McGraw-Hill Companies, Inc. All rights reserved. 4 AN INTRODUCTION TO THE FINITE ELEMENT METHOD Table P1.4: Numerical solutions of the nonlinear equation d2θ/dt2 + λ2 sin θ = 0 along with the exact solution of the linear equation d2θ/dt2+λ2θ = 0. Exact Approx. solution θ Exact Approx. solution v t θ Euler’s Heun’s v Euler’s Heun’s 0.00 0. 0. 0. -0. -0. -0. 0.05 0. 0. 0. -0. -0. -0. 0.10 0. 0. 0. -1. -1. -1. 0.20 0. 0. 0. -2. -2. -1. 0.40 -0. 0. 0. -3. -3. -3. 0.60 -0. -0. -0. -2. -2. -2. 0.80 -0. -1. -0. 0. -0. -0. 1.00 -0. -0. -0. 2. 2. 2. PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of the McGraw-Hill. PROPRIETARY MATERIAL. °c The McGraw-Hill Companies, Inc. All rights reserved. SOLUTIONS MANUAL 5 Chapter 2 MATHEMATICAL PRELIMINARIES, INTEGRAL FORMULATIONS, AND VARIATIONAL METHODS In Problem 2.1—2.5, construct the weak form and, whenever possible, quadratic functionals. Problem 2.1: A nonlinear equation: − d dx μ u du dx ¶ + f = 0 for 0 x L μ u du dx ¶¯¯¯¯ x=0 = 0 u(1) = √2 Solution: Following the three-step procedure, we write the weak form: 0 = Z 1 0 v ∙ − d dx (u du dx ) + f ¸ dx (1) = Z 1 0 ∙ u dv dx du dx + vf ¸ dx − ∙ v(u du dx ) ¸1 0 (2) Using the boundary conditions, v(1) = 0 (because u is specified at x = 1) and (du/dx) = 0 at x = 0, we obtain 0 = Z 1 0 ∙ u dv dx du dx + vf ¸ dx (3) For this problem, the weak form does not contain an expression that is linear in both u and v; the expression is linear in v but not linear in u. Therefore, a quadratic functional does not exist for this case. The expressions for B(·, ·) and `(·) are given by B(v, u) = Z 1 0 u dv dx du dx dx (not linear in u and not symmetric in u and v) `(v) = − Z 1 0 vfdx (4) PROPRIETARY MATERIAL. °c The McGraw-Hill Companies, Inc. All rights reserved. 6 AN INTRODUCTION TO THE FINITE ELEMENT METHOD New Problem 2.1: The instructor may assign the following problem: − d dx ∙ (1 + 2x2) du dx ¸ + u = x2 (1a) u(0) = 1 , μ du dx ¶ x=1 = 2 (1b) The answer is B(v, u) = Z 1 0 ∙ (1 + 2x2) dv dx du dx + vu ¸ dx (symmetric) `(v) = Z 1 0 v x2 dx + 6v(1) (2) I(u) = 1 2 B(u, u) − `(u) = 1 2 Z 1 0 " (1 + 2x2) μ du dx ¶2 + u2 # dx − Z 1 0 u x2 dx − 6u(1) Problem 2.2: The Euler-Bernoulli-von K´arm´an nonlinear beam theory [7]: − d dx ( EA " du dx + 1 2 μ dw dx ¶2 #) = f for 0 x L d2 dx2 à EI d2w dx2 ! − d dx ( EA dw dx " du dx + 1 2 μ dw dx ¶2 #) = q u = w = 0 at x = 0, L; μ dw dx ¶¯¯¯¯ x=0 = 0; à EI d2w dx2 !¯¯¯¯ x=L = M0 where EA, EI, f, and q are functions of x, and M0 is a constant. Here u denotes the axial displacement and w the transverse deflection of the beam. Solution: The first step of the formulation is to multiply each equation with a weight function, say v1 for the first equation and v2 for the second equation, and integrate over the interval (0, L). In the second step, carry out the integration-by-parts once in the first equation, twice in the first term of the second equation, and once in the second part of the second equation. Then use the fact that v1(0) = v1(L) = 0 (because u is specified there), v2(0) = v2(L) = 0 (because w is specified), and (dv2/dx)(0) = 0 PROPRIETARY MATERIAL. °c The McGraw-Hill Companies, Inc. All rights reserved. SOLUTIONS MANUAL 7 (because dw/dx is specified at x = 0). In addition, we have EI(d2w/dx2) = M0 at x = L. The final weak forms are given by 0 = Z L 0 ( EA dv1 dx " du dx + 1 2 μ dw dx ¶2 # − v1f ) dx (1a) 0 = Z L 0 ( EI d2v2 dx2 d2w dx2 + EA dv2 dx dw dx " du dx + 1 2 μ dw dx ¶2 # − v2q ) dx − μ dv2 dx ¶¯¯¯¯¯ L M0 (1b) Note that for this case the weak form is not linear in u or w. However, a functional can be constructed for this using the potential operator theory (see: J. T. Oden and J. N. Reddy, Variational Methods in Theoretical Mechanics, 2nd ed., Springer-Verlag, Berlin, 1983 and Reddy [3]). The functional is given by Π(u,w) = Z L 0 ( EA 2 "μ du dx ¶2 + du dx μ dw dx ¶2 + 1 2 μ dw dx ¶4 # + EI 2 à d2w dx2 !2 + uf + wq ) dx − dw dx ¯¯¯¯¯ L M0 Problem