Exam (elaborations) TEST BANK FOR Introduction to Finite Element

Exam (elaborations) TEST BANK FOR Introduction to Finite Element 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 ∆t = 0.05 and ∆t = 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 + ∆t vi; vi+1 = vi − ∆t 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 + ∆t 2 h f(ti, ui) + f(ti+1, u0 i+1) i , u0 i+1 = ui + ∆t 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 ∆t = .05 ∆t = .025 v ∆t = .05 ∆t = .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 ∆t = 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 + ∆t vi vi+1 = vi − λ2 ∆t 2 ³ sin θi + sin θ0 i+1´ θi+1 = θi + ∆t 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 xL 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 dxdx (not linear in u and not symmetric in u and v) `(v) = − Z 1 0 vf dx (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 , µdu 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) µdu 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 µdw dx ¶2 #) = f for 0 xL d2 dx2 à EI d2w dx2 ! − d dx ( EAdw dx " du dx + 1 2 µdw dx ¶2 #) = q u = w = 0 at x = 0, L; µdw 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 ( EAdv1 dx " du dx + 1 2 µdw dx ¶2 # − v1f ) dx (1a) 0 = Z L 0 ( EI d2v2 dx2 d2w dx2 + EAdv2 dx dw dx " du dx + 1 2 µdw dx ¶2 # − v2q ) dx − µdv2 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 "µdu dx¶2 + du dx µdw dx ¶2 + 1 2 µdw dx ¶4 # + EI 2 à d2w dx2 !2 + uf + wq) dx − dw dx ¯ ¯ ¯ ¯ ¯ L M0 Problem 2.3: A second-order equation: − ∂ ∂x µ a11 ∂u ∂x + a12 ∂u ∂y ¶ − ∂ ∂y µ a21 ∂u ∂x + a22 ∂u ∂y ¶ + f = 0 in Ω u = u0 on Γ1, µ a11 ∂u ∂x + a12 ∂u ∂y ¶ nx + µ a21 ∂u ∂x + a22 ∂u ∂y ¶ ny = t0 on Γ2 where aij = aji (i, j = 1, 2) and f are given functions of position (x, y) in a twodimensional domain Ω, and u0 and t0 are known functions on portions Γ1 and Γ2 of the boundary Γ: Γ1 + Γ2 = Γ. Solution: Multiplying with the weight function v and integrating by parts, we obtain the weak 0 = Z Ω ∙ ∂v ∂x µ a11 ∂u ∂x + a12 ∂u ∂y ¶ + ∂v ∂y µ a21 ∂u ∂x + a22 ∂u ∂y ¶ + vf¸ dxdy − I Γ v ∙µ a11 ∂u ∂x + a12 ∂u ∂y ¶ nx + µ a21 ∂u ∂x + a22 ∂u ∂y ¶ ny ¸ ds = Z Ω ∙ ∂v ∂x µ a11 ∂u ∂x + a12 ∂u ∂y ¶ + ∂v ∂y µ a21 ∂u ∂x + a22 ∂u ∂y ¶ + vf¸ dxdy − Z Γ2 vt0 ds PROPRIETARY MATERIAL. °c The McGraw-Hill Companies, Inc. All rights reserved. 8 AN INTRODUCTION TO THE FINITE ELEMENT METHOD where v = 0 on Γ1. The bilinear form (symmetric only if a12 = a21) and linear form are: B(v, u) = Z Ω µ a11 ∂v ∂x ∂u ∂x + a12 ∂v ∂x ∂u ∂y + a21 ∂v ∂y ∂u ∂x + a22 ∂v ∂y ∂u ∂y ¶ dxdy `(v) = − Z Ω vf dxdy + Z Γ2 v t0 ds The quadratic functional, when a12 = a21, is given by I(u) = 1 2 Z Ω " a11 µ∂u ∂x ¶2 + 2a12 ∂u ∂x ∂u ∂y + a22 µ∂u ∂y ¶2 # dxdy − Z Ω uf dxdy + Z Γ2 u t0 ds Problem 2.4: Navier-Stokes equations for two-dimensional flow of viscous, incompressible fluids: u ∂u ∂x + v ∂u ∂y = −1 ρ ∂P ∂x + ν à ∂2u ∂x2 + ∂2u ∂y2 ! u ∂v ∂x + v ∂v ∂y = −1 ρ ∂P ∂y + ν à ∂2v ∂x2 + ∂2v ∂y2 ! ∂u ∂x + ∂v ∂y = 0 ⎫ ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬ ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭ in Ω (1) u = u0, v = v0 on Γ1 (2) ν µ∂u ∂x nx + ∂u ∂y ny ¶ − 1 ρ P nx = t ˆx ν µ∂v ∂x nx + ∂v ∂y ny ¶ − 1 ρ P ny = t ˆy ) on Γ2 (3) Solution: For this set of three differential equations in two dimensions (see Chapter 10 and Reddy [7] for the physics behind the equations), we follow exactly the same procedure as before: use the three-step procedure for each equation. In the second step of the formulation, we must integrate by parts the terms involving P, u, and v, because these terms are required as a part of the natural boundary conditions given in Eq. (3). We do not integrate by parts the nonlinear terms in the first two equations, and no integration by parts is used in the third equation, because the boundary terms resulting from such integration-by-parts do not constitute physical PROPRIETARY MATERIAL. °c The McGraw-Hill Companies, Inc. All rights reserved. SOLUTIONS MANUAL 9 variables. We have 0 = Z Ω ∙ w1 µ u ∂u ∂x + v ∂u ∂y ¶ − 1 ρ ∂w1 ∂x P + ν µ∂w1 ∂x ∂u ∂x + ∂w1 ∂y ∂u ∂y ¶¸ dxdy − Z Γ2 w1t ˆxds 0 = Z Ω ∙ w2 µ u ∂v ∂x + v ∂v ∂y ¶ − 1 ρ ∂w2 ∂y P + ν µ∂w2 ∂x ∂v ∂x + ∂w2 ∂y ∂v ∂y ¶¸ dxdy − Z Γ2 w2t ˆyds 0 = Z Ω w3 µ∂u ∂x + ∂v ∂y ¶ dxdy where (w1, w2, w3) are weight functions. Problem 2.5: Two-dimensional flow of viscous, incompressible fluids (stream function-vorticity formulation): −∇2ψ − ζ = 0 −∇2ζ + ∂ψ ∂x ∂ζ ∂y − ∂ψ ∂y ∂ζ ∂x = 0 ⎫ ⎪⎬ ⎪⎭ in Ω Assume that all essential boundary conditions are specified to be zero. Solution: First, we note the the identity −w∇2ψ = −w∇ · ∇ψ = −∇ · (w∇ψ) + ∇w · ∇ψ and then use the Green—Gauss theorem to obtain − Z Ω w∇2ψ dxdy = Z Ω [−∇ · (w∇ψ) + ∇w · ∇ψ] dxdy = − I Γ wnˆ · ∇ψ ds + Z Ω ∇w · ∇ψ dxdy Multiplying the first equation with w1 and the second equation with w2 and integrating over the domain Ω and using the above identity we obtain (the boundary integrals vanish because w1 = 0 and w2 = 0 on the boundary Γ) 0 = Z Ω (∇w1 · ∇ψ − w1ζ) dxdy (1) 0 = Z Ω ∙ ∇w2 · ∇ζ + w2 µ∂ψ ∂x ∂ζ ∂y − ∂ψ ∂y ∂ζ ∂x ¶¸ dxdy (2) PROPRIETARY MATERIAL. °c The McGraw-Hill Companies, Inc. All rights reserved. 10 AN INTRODUCTION TO THE FINITE ELEMENT METHOD Problem 2.6: Compute the coefficient matrix and the right-hand side of the Nparameter Ritz approximation of the equation − d dx ∙ (1 + x) du dx¸ = 0 for 0 x 1 u(0) = 0, u(1) = 1 Use algebraic polynomials for the approximation functions. Specialize your result for N = 2 and compute the Ritz coefficients. Solution: The weak form for this problem is given by 0 = Z 1 0 (1 + x) dv dx du dxdx The variational problem is given by Eqs. (2.5.4a) and (2.5.4b), where [`(φi)=0 because there is no source term], Bij = B(φi, φj ) = Z 1 0 (1 + x) dφi dx dφj dx dx (1a) Fi = −B(φi, φ0) = − Z 1 0 (1 + x) dφi dx dφ0 dx dx (1b) The approximation functions φ0 and φi should be chosen such that φ0(0) = 0, φ0(1) = 1 ; φi(0) = φi(1) = 0, (i = 1, 2, ..., n) (2) The following algebraic polynomials satisfy the above requirements: φ0 = x , φi = xi (1 − x) (3) Substitution of Eq.(3) into Eqs.(1a,b) and evaluating the integrals, we obtain Bij = ij i + j − 1 − ij + i + j i + j + 1 − ij i + j + 1 + (i + 1)(j + 1) i + j + 2 (4a) Fi = 1 (1 + i)(2 + i) (4b) For the two-parameter (N = 2) case, we have B11 = 1 2 , B12 = B21 = 17 60 , B22 = 7 30 , F1 = 1 6 , F2 = 1 12 and the parameters c1 and c2 are given by c1 = 55 131 , c2 = − 20 131 PROPRIETARY MATERIAL. °c The McGraw-Hill Companies, Inc. All rights reserved. SOLUTIONS MANUAL 11 The two-parameter Ritz solution becomes u(x) = φ0 + c1φ1 + c2φ2 = x + 55 131(x − x2) − 20 131(x2 − x3) = 1 131(186x − 75x2 + 20x3) The exact solution is given by uexact = log (1 + x) log 2 Problem 2.7: Use trigonometric functions for the two-parameter approximation of the equation in Problem 2.6, and obtain the Ritz coefficients. Solution: The following trigonometric functions satisfy the requirements in Eq.(2) of Problem 2.6: φ0 = sin πx 2 , φi = sin iπx For two-parameter case, we have B11 = Z 1 0 (1 + x) dφ1 dx dφ1 dx dx = π2 Z 1 0 (1 + x) cos πx cos πx dx B12 = Z 1 0 (1 + x) dφ1 dx dφ2 dx dx = 2π2 Z 1 0 (1 + x) cos πx cos 2πx dx = B21 B22 = Z 1 0 (1 + x) dφ2 dx dφ2 dx dx = 4π2 Z 1 0 (1 + x) cos 2πx cos 2πx dx F1 = − Z 1 0 (1 + x) dφ1 dx dφ0 dx dx = −π2 2 Z 1 0 (1 + x) cos πx cos πx 2 dx F2 = − Z 1 0 (1 + x) dφ2 dx dφ0 dx dx = −π2 Z 1 0 (1 + x) cos 2πx cos πx 2 dx Using the following trigonometric identities, cos mπx cos nπx = 1 2 [cos(m + n)πx + cos(m − n)πx] cos2 mπx = 1 2 (1 + cos 2mπx) we obtain " 3π2 4 −20 9 −20 9 3π2 # ½ c1 c2 ¾ = ½ −1 9 (6π − 10) 68 225 + 4π 15 ¾ PROPRIETARY MATERIAL. °c The McGraw-Hill Companies, Inc. All rights reserved. 12 AN INTRODUCTION TO THE FINITE ELEMENT METHOD and the solution is U2(x) = c1 sin πx + c2 sin 2πx + sin πx 2 = −0.12407 sin πx + 0.02919 sin 2πx + sin πx 2 Problem 2.8 A steel rod of diameter d = 2 cm, length L = 25 cm, and thermal conductivity k = 50 W/(m ◦C) is exposed to ambient air T∞ = 20◦C with a heat-transfer coefficient β = 64 W/(m2 ◦C). Given that the left end of the rod is maintained at a temperature of T0 = 120◦C and the other end is exposed to the ambient temperature, determine the temperature distribution in the rod using a two-parameter Ritz approximation with polynomial approximation functions. The equation governing the problem is given by −d2θ dx2 + cθ = 0 for 0 x 25 cm where θ = T − T∞, T is the temperature, and c is given by c = βP Ak = βπD 1 4πD2k = 4β kD = 256 m2 P being the perimeter and A the cross sectional area of the rod. The boundary conditions are θ(0) = T(0) − T∞ = 100◦C, µ k dθ dx + βθ¶ ¯ ¯ ¯ ¯ x=L = 0 Solution: The weak form of the equation is given by 0 = Z L 0 µdv dx dθ dx + cvθ ¶ dx + ˆcv(L)θ(L) (1) where ˆc = (β k ). We have Bij = B(φi, φj ) = Z L 0 µdφi dx dφj dx + cφiφj ¶ dx + ˆcφi(L)φj (L) (2a) Fi = −B(φi, φ0) = − Z L 0 µdφi dx dφ0 dx + cφiφ0 ¶ dx − cˆφi(L)φ0(L) (2b) We choose the following functions φ0 = θ(0) = 100 , φi = xi PROPRIETARY MATERIAL. °c The McGraw-Hill Companies, Inc. All rights reserved. SOLUTIONS MANUAL 13 From the values of the parameters given, we compute: L = 0.25m, c = 256, and cˆ = (β k ) = 64/50. The coefficients are evaluated to be B11 = 499 300 , B12 = B21 = 133 400 , B22 = 91 1200 , F1 = −832 , F2 = −424 3 or ⎡ ⎣ 499 300 133 400 133 400 91 1200 ⎤ ⎦ ⎧ ⎨ ⎩ c1 c2 ⎫ ⎬ ⎭ = ⎧ ⎨ ⎩ −832 −424 3 ⎫ ⎬ ⎭ The solution of these equations is c1 = −1, 033.3859 , c2 = 2, 667.2635 The two-parameter Ritz solution is given by θ(x) = 100 − 1033.3859x + 2667.2635x2 θ(0.125) = 12.503◦C , θ(0.25) = 8.3575◦C Problem 2.9: Set up the equations for the N-parameter Ritz approximation of the following equations associated with a simply supported beam and subjected to a uniform transverse load q = q0: d2 dx2 à EI d2w dx2 ! = q0 for 0 xL w = EI d2w dx2 = 0 at x = 0, L (a) Use algebraic polynomials. (b) Use trigonometric functions. Compare the two-parameter Ritz solutions with the exact solution. Solution: (a) Choose φ0 = 0 and φi = xi (L − x), which satisfy the geometric conditions w(0) = w(L) = 0. The coefficients are given by Bij = EI ij(L) i+j−1 ∙ (i − 1)(j − 1) i + j − 3 − 2(ij − 1) i + j − 2 + (i + 1)(j + 1) i + j − 1 ¸ Fi = q0(L)i+2 (1 + i)(2 + i) Note that the expression given above for Bij is not valid when i = 1 and j = 1, 2, ··· , N; we have, B11 = 4EIL, B1j = Bj1 = 2EILj , (j 1) PROPRIETARY MATERIAL. °c The McGraw-Hill Companies, Inc. All rights reserved. 14 AN INTRODUCTION TO THE FINITE ELEMENT METHOD For N = 1 the Ritz coefficient is given by c1 = F1/B11 = q0L2/24EI; and for N = 2, the coefficients are: c1 = q0L2/(24EI) , c2 = 0. Hence, the one-parameter and two-parameter solution is the same W1 = W2(x) = c1φ1 = q0L2 24EI x(L − x) = q0L4 24EI x L(1 − x L) (b) Choose φ0 = 0 and φi = sin iπx L . The coefficients are given by Bij = EIL 2 µiπ L ¶4 for i = j ; Bij = 0 for i 6= j Fi = 2q0L iπ if i is odd ; Fi = 0 if i is even Hence, ci = Fi Bii = 4q0 EIL µ L iπ ¶5 = 4q0L4 EI µ 1 iπ ¶5 Hence, the solution becomes w2(x) = c1φ1 + c3φ3 = 4q0L4 EIπ5 sin πx L + 4q0L4 243EIπ5 sin 3πx L Problem 2.10: Repeat Problem 2.9 for q = q0 sin(πx/L). Solution: (a) We have (a = π/L), Fi = Z L 0 (q0 sin ax) xi (L − x) dx = q0L " Li a + i a Z L 0 xi−1 cos ax dx# − q0 " −Li+1 a + i + 1 a Z L 0 xi cos ax dx# For N = 1 we have F1 = 4q0L3/π3, and c1 = q0L2/(EIπ3). For N = 2 the coefficients are F2 = F1L = 4q0L3/π3 and the solution is c1 = c2L = 2q0L2/(3EIπ3). (b) Choose φ0 = 0 and φi = sin iπx L . The coefficients Bij are the same as in Problem 2.9(b). The coefficients Fi are given by F1 = f0L/2 and Fi = 0 for i 6= 1. The Ritz coefficients are given by c1 = q0L4 EIπ4 , ci = 0 if i 6= 1 PROPRIETARY MATERIAL. °c The McGraw-Hill Companies, Inc. All rights reserved. SOLUTIONS MANUAL 15 The Ritz solution coincides with the exact solution, w = q0L4 EIπ4 sin πx L Problem 2.11: Repeat Problem 2.9 for q = Q0δ(x − 1 2L), where δ(x) is the Dirac delta function (i.e., a point load Q0 is applied at the center of the beam). Solution: The coefficients Fi are given by (a) Fi = Q0 µL 2 ¶i+1 (b) Fi = Q0(−1)i−1 for i odd, and Fi = 0 for i even Note that c2 = 0 in both cases. Problem 2.12: Develop the N-parameter Ritz solution for a simply supported beam under uniform transverse load using Timoshenko beam theory. The governing equations are given in Eqs. (2.4.32a, b). Use Trigonometric functions to approximate w and Ψ. Solution: Assume solution of (w, Ψ) in the form, wM = X M j=1 bjφj ≡ X M j=1 bj sin jπx L , ΨN = X N j=1 cjψj ≡ X N j=1 cj cos jπx L (1) Substitution of Eq. (1) into the weak forms (S = GAK and D = EI) 0 = Z L 0 ∙ GAK dv1 dx µdw dx + Ψ ¶ + kv1