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 77843-3123 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 ii iii PREFACE This solution manual is prepared to aid the instructor in discussing the solutions to assigned problems in Chapters through 14 from the book, An Introduction to the Finite Element Method, Third Edition, McGraw—Hill, New York, 2006 Computer solutions to certain problems of Chapter (see Chapter 13 problems) are also included at the end of Chapter The instructor should make an effort to review the problems before assigning them This allows the instructor to make comments and suggestions on the approach to be taken and nature of the answers expected The instructor may wish to generate additional problems from those given in this book, especially when taught time and again from the same book Suggestions for new problems are also included at pertinent places in this manual Additional examples and problems can be found in the following books of the author: J N Reddy and M L Rasmussen, Advanced Engineering Analysis, John Wiley, New York, 1982; reprinted and marketed currently by Krieger Publishing Company, Melbourne, Florida, 1990 (see Section 3.6) J N Reddy, Energy and Variational Methods in Applied Mechanics, John Wiley, New York, 1984 (see Chapters and 3) J N Reddy, Applied Functional Analysis and Variational Methods in Engineering, McGraw-Hill, New York, 1986; reprinted and marketed currently by Krieger Publishing Company, Melbourne, Florida, 1991 (see Chapters 4, and 7) J N Reddy, Theory and Analysis of Elastic Plates, Taylor and Francis, Philadelphia, 1997 J N Reddy, Energy Principles and Variational Methods in Applied Mechanics, Second Edition, John Wiley, New York, 2002 (see Chapters through and Chapter 10) J N Reddy, Mechanics of Laminated Composite Plates and Shells: Theory and Analysis, CRC Press, Second Edition, Boca Raton, FL, 2004 J N Reddy, An Introduction to Nonlinear Finite Element Analysis, Oxford University Press, Oxford, UK, 2004 The computer problems FEM1D and FEM2D can be readily modified to solve new types of field problems The programs can be easily extended to finite element models formulated in an advanced course and/or in research The Fortran sources of FEM1D and FEM2D are available from the author for a price of $200 The author appreciates receiving comments on the book and a list of errors found in the book and this solutions manual J N Reddy All that is not given is lost iv PROPRIETARY MATERIAL c °The McGraw-Hill Companies, Inc All rights reserved Chapter 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 Fd = cv Fg = mg v Solution: From the free-body-diagram it follows that m dv = Fg − Fd , dt 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 + αv = g, dt PROPRIETARY MATERIAL α= c m c °The McGraw-Hill Companies, Inc All rights reserved 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 (ρAh) = ρqi − ρq0 dt or d(Ah) = qi − q0 dt 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 = v, dt dv = −λ2 sin θ dt 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 = f (t, u) dt (1) Heun’s scheme has the form ui+1 = ui + i ∆t h f (ti , ui ) + f (ti+1 , u0 ) , u0 = ui + ∆t f (ti , ui ) i+1 i+1 PROPRIETARY MATERIAL c °The McGraw-Hill Companies, Inc All rights reserved (2) SOLUTIONS MANUAL Table P1.3: Comparison of various approximate solutions of the equation (d2 θ/dt2 ) + λ2 sin θ = with its exact linear solution Exact t 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.60 0.80 1.00 θ 0.78540 0.76965 0.72302 0.64739 0.54578 0.42229 0.28185 0.13011 -0.02685 -0.18274 -0.33129 -0.58310 -0.78356 -0.50591 Approx solution θ ∆t = 05 ∆t = 025 0.78540 0.78540 0.75694 0.70002 0.58980 0.50496 0.37123 0.21803 0.05023 -0.12628 -0.30481 -0.63965 -1.05068 -0.94062 0.78540 0.77828 0.74276 0.67944 0.56482 0.47627 0.34225 0.19218 0.03148 -0.13374 -0.29690 -0.59131 -0.91171 -0.74672 Exact Approx solution v v -0.00000 -0.62801 -1.23083 -1.78428 -2.26615 -2.65711 -2.94148 -3.10785 -3.14955 -3.06491 -2.85732 -2.11119 0.21536 2.41051 ∆t = 05 -0.00000 -0.56922 -1.13844 -1.69123 -2.20984 -2.67459 -3.06403 -3.35605 -3.53018 -3.57060 -3.46921 -2.85712 -0.50399 2.29398 ∆t = 025 -0.00000 -0.56922 -1.13027 -1.66622 -2.15879 -2.58816 -2.93371 -3.17573 -3.29791 -3.29007 -3.15014 -2.50787 -0.28356 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 = v, dt dv = −λ2 sin θ dt yields the following discrete equations: θi+1 = θi + ∆t vi ´ ∆t ³ vi+1 = vi − λ2 sin θi + sin θi+1 ∆t (vi + vi+1 ) θi+1 = θi + 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 AN INTRODUCTION TO THE FINITE ELEMENT METHOD Table P1.4: Numerical solutions of the nonlinear equation d2 θ/dt2 + λ2 sin θ = along with the exact solution of the linear equation d2 θ/dt2 +λ2 θ = Exact Approx solution θ Exact Approx solution v t θ Euler’s Heun’s v Euler’s Heun’s 0.00 0.05 0.10 0.20 0.40 0.60 0.80 1.00 0.785398 0.769645 0.723017 0.545784 -0.026852 -0.583104 -0.783562 -0.505912 0.785398 0.785398 0.756937 0.615453 0.050228 -0.639652 -1.050679 -0.940622 0.785398 0.771168 0.728680 0.564818 0.015246 -0.544352 -0.787095 -0.587339 -0.000000 -0.628013 -1.230833 -2.266146 -3.149552 -2.111190 0.215362 2.410506 -0.000000 -0.569221 -1.138442 -2.209838 -3.530178 -2.857121 -0.503993 2.293983 -0.000000 -0.569221 -1.121957 -1.121957 -3.073095 -2.194398 -0.114453 2.023807 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 c PROPRIETARY MATERIAL °The McGraw-Hill Companies, Inc All rights reserved SOLUTIONS MANUAL Chapter 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: ả d du u + f = for < x < L dx dx ả du u = u(1) = dx ¯x=0 Solution: Following the three-step procedure, we write the weak form: Z ∙ ¸ d du v − (u ) + f dx 0= dx dx ¸ ∙ ¸ Z 1∙ dv du du + vf dx − v(u ) u = dx dx dx 0 (1) (2) Using the boundary conditions, v(1) = (because u is specified at x = 1) and (du/dx) = at x = 0, we obtain 0= Z 1∙ dv du u dx 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 dv du u `(v) = − dx dx Z dx (not linear in u and not symmetric in u and v) vf dx (4) PROPRIETARY MATERIAL c °The McGraw-Hill Companies, Inc All rights reserved 406 AN INTRODUCTION TO THE FINITE ELEMENT METHOD Solution: The weak forms of Eqs (a) and (b) over a typical element Ωe are 0= Z Ωe I à ! ∂δw ∂Mxx ∂δw ∂Myy ∂ δw ∂ w + + 4D66 − qδw dxdy ∂x ∂x ∂y ∂y ∂x∂y ∂x∂y " # ∂2w ¯ (δθx ny + δθy nx ) ds δwQn − 2D66 − ∂x∂y Γe ¸ Z ∙ ¡ ¢ ∂w ∂δMxx ¯ ¯ 0= − δMxx D22 Mxx − D12 Myy dxdy ∂x Ωe ∂x − 0= Z I Γe Ωe − I ∙ (c) δMxx x nx ds (d) Ă Â w Myy ¯ − δMyy D11 Myy − D12 Mxx dxdy ∂y ∂y Γe δMyy θy ny ds (e) The primary and secondary variables of the formulation are w, Mxx , Myy , ∂w ∂w nx , θy ny ≡ ny , θx nx ≡ ∂x ∂y Vn , (f ) (g) where Vn is the effective shear force (Kirchhoff free edge condition) Vn = Qn + ∂Mns , ∂s Qn = Qx nx + Qy ny (h) The finite element model of Eq (d) and (e) is obtained by substituting the approximations of the form w= r X (1) wi ψi , Mxx = i=1 s X (2) Mxi ψi , Myy = i=1 p X (3) Myi ψi (i) i=1 (α) where ψi , (α = 1, 2, 3, 4) are appropriate interpolation functions We obtain ⎡ where ⎤⎧ [K 11 ] ⎫ ⎧ ⎫ [K 12 ] [K 13 ] ⎨ {w} ⎬ ⎨ {F } ⎬ ⎣ [K 22 ] [K 23 ] ⎦ {Mx } = {F } ⎭ ⎩ ⎭ ⎩ symm [K 33 ] {My } {F } 11 Kij = 4D66 Z Ωe 1 ∂ ψi ∂ ψi dxdy, ∂x∂y ∂x∂y PROPRIETARY MATERIAL i, j = 1, 2, , r, c °The McGraw-Hill Companies, Inc All rights reserved (j) 407 SOLUTIONS MANUAL 12 Kij = Z ∂ψi ∂ψj dxdy, ∂x ∂x ∂ψi ∂ψj dxdy, ∂y ∂y Ωe 13 Kij = 22 Kij = 23 Kij = 33 Kij = Fi1 = Fi2 = Fi3 = Z Ωe Z Ωe Z Ωe Z Ωe i, j = 1, 2, , s, ¯ (−D12 )ψi ψj dxdy, i = 1, 2, , s; 3 ¯ (−D11 )ψi ψj dxdy, i, j = 1, 2, , p, qψi dxdy + Ωe Γe I Γe i, j = 1, 2, , r; j = 1, 2, , p, 2 ¯ (−D22 )ψi ψj dxdy, Z I i, j = 1, 2, , r; j = 1, 2, , s, I Γe Vn ψi ds, j = 1, 2, , p, i = 1, 2, , r, θx nx ψi ds, i = 1, 2, , s, θy ny ψi ds, i = 1, 2, , p, (k) An examination of the weak forms (d) and (e) show that the minimum continuity α conditions of the interpolation functions ψi (α = 1, 2, 3) are ψi = linear in x and linear in y ψi = linear in x and constant in y (i) ψi = linear in y and constant in x ψi = linear in x and linear in y Problem 14.10: Use the interpolation w= X wi ψi , Mxx = i=1 X Mxi ψi , Myy = i=1 X Myi ψj i=1 with 1 x = a ảà ¶ µ ¶ µ y x y xy x 1 1− , ψ2 = 1− , ψ3 = , ψ4 = − b a b ab a ψ1 = − x , a φ2 = x , a y ψ1 = − , b ψ2 = ¶ y b y b for a rectangular element with sides a and b to evaluate the matrices [K αβ ](α, β = 1, 2, 3) in Problem 14.9 PROPRIETARY MATERIAL c °The McGraw-Hill Companies, Inc All rights reserved 408 AN INTRODUCTION TO THE FINITE ELEMENT METHOD Solution: We can select either ψ1 = − x x , ψ2 = , a a y y 3 ψ1 = − , ψ2 = b b (a) and ψi to be the bilinear interpolation functions, or ψi = ψi = ψi = ψi = bilinear functions of a rectangular element (b) The corresponding rectangular elements are shown in Figure P14.10 My2 w0 , Mx , My at each node w0 , Mx , My at each node w0 at four corner nodes Mx1 Mx2 My1 (a) Mixed model A (b) Mixed model B (c) Mixed model C Figure P14.10: Mixed rectangular plate bending elements based on CPT (a) Model A (b) Model B (c) Model C The numerical form of element matrices is ⎡ ⎤ ⎡ ⎤ −1 −1 −1 4D66 ⎢ −1 b ⎢ −1 −1 1⎥ 1⎥ 11 12 ⎢ ⎥ , [K ] = ⎢ ⎥ = [K 21 ]T [K ] = ⎣ −1 ⎦ ⎣ −1 −1 1⎦ ab 2a −1 −1 1 −1 ⎡ ⎤ −1 ∙ ¸ a ⎢ −1 ⎥ D22 ab 13 31 T 22 ⎢ ⎥, [K ] = [K ] = [K ] = − 1⎦ 2b ⎣ −1 −1 ∙ ¸ D12 ab 1 D11 22 [K 23 ] = [K 32 ]T = , [K 33 ] = [K ] 1 D22 Problem 14.11: Repeat Problem 14.10 for the case in which φ1 = φ2 = ψi i i Solution: [K 11 ] is the same as in Problem 14.10 Also, we have [K 12 ] = [K 21 ]T = [S 11 ], [K 13 ] = [K 31 ]T = [S 22 ] [K 22 ] = −D22 [S 00 ], [K 23 ] = [K 32 ]T = D12 [S 00 ], [K 33 ] = −D11 [S 00 ] PROPRIETARY MATERIAL c °The McGraw-Hill Companies, Inc All rights reserved SOLUTIONS MANUAL 409 where [S 11 ], [S 22 ] and [S 00 ] are defined in Eq (8.2.52) Problem 14.12: Evaluate the element matrices in (14.4.6b) by assuming that the nonlinear parts in the element coefficients are element-wise-constant Solution: We have ∙ ¸ ∙ EA EA −1 −1 [K ] = , [K 12 ] = [K 21 ] = 1 L −1 L −1 11 ⎡ ⎤ ⎡ ¸ ⎤ −3L −6 −3L 36 −3L −36 −3L 2EI ⎢ −3L 2L2 3L L2 ⎥ N ⎢ −3h 4L2 3L −L2 ⎥ ⎥+ ⎢ ⎥ [K 22 ] = ⎢ 3L 3L ⎦ 30L ⎣ −36 3L 36 3L ⎦ L ⎣ −6 3L 2L2 −3L L2 −3L −L2 3L 4L2 where L is the length of the element and N = 0.5(dw/dx Problem 14.13: Give the finite element formulation of the following nonlinear equation over an element (xa , xb ) : ả d du u + = for < x < dx dx du dx ả = 0, u(1) = √ x=0 Solution: The weak form is same as in the linear equation except that we have a(x) = u(x): [K(¯)]{u} = {F } with [see Reddy (2004b)] u e Kij = Z xb à X n e ue ψk k ue k e ψk xa = n X k=1 Fi = − Z xb xa k=1 Z xb xa ! e e dψi dψj dx dx dx e e dψi dψj dx dx dx (a) e ψi dx + Qi For example, for linear approximation (n = 2) of u(x), we have e Kij = = n X k=1 n X ue k Z xb xa e ψk e e dψi dψj dx dx dx ae ue (−1)i+j k k=1 e i+j a0 = (−1) 2he PROPRIETARY MATERIAL h2 e à n X k=1 Z xb ue k xa ! e ψk dx = (−1)i+j ae (ue + ue ) 2he c °The McGraw-Hill Companies, Inc All rights reserved (b) 410 AN INTRODUCTION TO THE FINITE ELEMENT METHOD or ∙ ae (ue + ue ) −1 [K ] = −1 2he e ¸ (c) Further, the assembled equations associated with a mesh of two linear elements of equal length are (U1 , U2 , U3 are the global nodal values) ⎤⎧ ⎡ ⎫ (U1 + U2 ) −(U1 + U2 ) ⎨ U1 ⎬ ⎣ −(U1 + U2 ) (U1 + 2U2 + U3 ) −(U2 + U3 ) ⎦ U2 ⎭ ⎩ 2h −(U2 + U3 ) (U2 + U3 ) U3 ⎧ ⎪ ⎨ ⎫ (1) ⎧ ⎫ (1) ⎪ ⎪ ⎪ f1 Q1 ⎬ ⎨ ⎬ (1) (2) (1) (2) = f2 + f1 + Q2 + Q1 ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎩ ⎭ (2) (2) f2 Q2 (d) Problem 14.14: Compute the tangent coefficient matrix for the nonlinear problems in Problem 14.13 What restriction(s) should be placed on the initial guess vector? Solution: By definition (14.4.17), we have [see Reddy (2004b)] e (KT )ij e ∂Ri ∂ ≡ e = ∂ue ∂uj j = n X m=1 à à n X e Kim ue m − Fie m=1 ! e ∂Kim e ∂ue e u + Kim m ∂ue m ∂ue j j ! = n X ∂K e e im e e um + Kij m=1 ∂uj (a) For the problem at hand, we have (KT )ij = n X ∂K e im e e e um + Kij m=1 n X ∂uj ∂ = ∂ue j m=1 àZ x b xa ả e d e dψm e uh i dx ue + Kij m dx dx à n ! Z xb e e ∂uh dψi X e dψm = u dx + K e xa ∂ue dx j m m=1 ij dx Z xb e duh dψi e e e ˆe = ψj dx + Kij ≡ Kij + Kij xa where the identity dx dx n X m=1 ue m duh dLe m = dx dx is used in arriving at the last line We have, ˆe Kij = Z xb Z e e duh dψi e ue − ue xb dψi e ψj dx = ψj dx xa dx dx PROPRIETARY MATERIAL xa dx c °The McGraw-Hill Companies, Inc All rights reserved (i) SOLUTIONS MANUAL or ∙ ue − ue −1 −1 ˆ Ke = 1 2he Thus, the tangent matrix becomes Ke T ∙ ¸ ¸ ∙ ¯2 ¯1 (¯e + ue ) u (¯e − ue ) −1 −1 u −1 ˆ = K + Ke = + −1 1 2he 2he e 411 ¸ where ue denote the nodal values known from the previous iteration Note that the ¯i tangent coefficient matrix is not symmetric Also, the initial guess should not be that all Ui = Since the boundary condition at x = is nonzero, the initial guess should be one that satisfies the boundary condition If the boundary condition is homogeneous, then at least one of the nodal values should be nonzero so that the tangent coefficient matrix is non-zero Problem 14.15: Compute the tangent stiffness matrix KT in (14.4.17) for the Euler— Bernoulli beam element in (14.4.6a) Solution: The coefficients of the element tangent stiffness matrix Ke ≡ Te can be T computed using the definition in (14.4.17) In terms of the components defined in Eq (14.4.6a), we can write [see Reddy (2004b)] ⎛ ⎞(r−1) α ∂Ri ⎠ ⎝ αβ Tij = ∂∆β j (a) for α, β = 1, The components of the residual vector can be expressed as α Ri = XX γ=1 p=1 = X αγ Kip ∆γ − Fiα p α1 Kip ∆1 + p p=1 = X X P =1 α1 Kip up + p=1 X P =1 α2 KiP ∆2 − Fiα P α2 ¯ KiP ∆P − Fiα (b) Note that the range of p is dictated by the size of the matrix [K αβ ] We have αβ Tij = = ⎛ ⎞ α ∂Ri ⎠ ⎝ ∂∆β j XX γ=1 p=1 αβ = Kij + = ⎛ ⎝K X ∂∆β j ⎛ ⎝ ∂∆β j ∂ ∂∆β j XX γ=1 p=1 γ αγ ∂∆p ip p=1 PROPRIETARY MATERIAL ∂ ³ + αγ Kip ∆γ − Fiα ⎠ p αγ ∂Kip ∂∆β j ´ ⎞ ⎞ ∆γ ⎠ p α1 Kip up + X P =1 ∂ ∂∆β j ³ ´ α2 ¯ KiP ∆P c °The McGraw-Hill Companies, Inc All rights reserved (c) 412 AN INTRODUCTION TO THE FINITE ELEMENT METHOD αβ Then the tangent stiffness matrix coefficients Tij can be computed as follows: 11 11 Tij = Kij + 11 X ∂Kip ∂uj p=1 11 = Kij + X p=1 Since up + X ∂K 12 iP ¯ ∆P ∂uj P =1 X · up + P =1 ¯ · ∆P (d) αβ ∂Kij = for all α, β, i, j and k ∂uk (e) the coefficients [T 11 ] and [T 21 ] of the tangent stiffness matrix are the same as those of the direct stiffness matrix: [T 11 ] = [K 11 ] , [T 21 ] = [K 21 ] (f ) Next consider 12 TiJ = 12 KiJ + X p=1 à 11 ∂Kip ¯ ∂ ∆J P =1 12 = KiJ + 12 = KiJ + = 22 TIJ 12 KiJ = 22 KIJ P =1 xa " X Z xb xa xa X p=1 22 = KIJ + µ à X K dw dx ! ¶ ¯ ∆P # dψi dφP ¯ dx ∆P dx dx ¯ dφK ∆K dx # ! # dψi dφP ¯ dx ∆P dx dx ¶ à ! X dφP ¯ ∆P dx P =1 dx dw dψi dφJ dx dx dx dx 12 21 = 2KiJ = KJi à Axx 21 ∂KIp ¯ ∂ ∆J " X Z xb p=1 12 ∂KiP ¯ ∂ ∆J dφJ dψi dφP ¯ Axx dx ∆P dx dx dx dψi dφJ Axx dx dx Z xb µ à ∂ Axx ¯ ∂ ∆J ∂ Axx ¯ ∂ ∆J Z xb xa 12 + KiJ + P =1 " X Z xb P =1 12 = KiJ + xa X up + " X Z xb 12 = KiJ + + 12 = KiJ + ! xa PROPRIETARY MATERIAL ! up + ∂ Axx ¯ ∂ ∆J X P =1 à X K à 22 ∂KIP ¯ ∂ ∆J (g) ! ¯ dφK ∆K dx ! ¯ ∆P # dφI dψp dx up dx dx c °The McGraw-Hill Companies, Inc All rights reserved SOLUTIONS MANUAL + " X Z xb P =1 22 = KIJ + + Z xb xa = 22 KIJ + ∂ Axx ¯ ∂ ∆J xa Z xb xa Axx xa dw dx ⎛ ¶2 # dφI dφP ¯ dx ∆P dx dx ⎞ dφI dφJ ⎝X dψp ⎠ up dx Axx dx dx p=1 dx µ Z xb µ 413 dw dx Axx ¶ µ dφI dφJ dx dx à X ¯ dφP ∆P dx P =1 du0 dw dw + dx dx dx ¶ ! dx dφI dφJ dx dx dx (h) Problem 14.16: Develop the nonlinear finite element model of the Timoshenko beam theory Equations (14.56) are valid for this case, with the following changes In place of (d2 /dx2 )(b d2 w/dx2 ) use −(d/dx)(b dΨ/dx) + GAk(dw/dx + Ψ) and add the following additional equation for w: ảá d dw +Ψ GAk dx dx =q See Section 4.4 for additional details Solution: The equations of equilibrium of the Timoshenko beam theory for the nonlinear case are d − dx ( du + dx dw dx ả2 #) =f (a) dw d +Ψ Sxx − dx dx ( " ả #) dw du dw d + − Axx =q dx dx dx dx µ ¶ µ ¶ dΨ dw d +Ψ =0 Dxx + Sxx − dx dx dx (b) ∙ Axx " µ ¶¸ (c) where Axx = EA, Sxx = Ks GA and Dxx = EI The weaks forms of the three equations are 0= Z xb ( xa " dδu du + Axx dx dx µ dw dx − Qe δu(xa ) − Qe δu(xb ) ¶2 # ) + f δu0 dx ( " ) µ ¶ µ ¶ # Z xb dw dδw dw e e dw du 0= + Ψ + Axx + Sxx δwq dx xa dx dx PROPRIETARY MATERIAL dx dx dx c °The McGraw-Hill Companies, Inc All rights reserved (d) 414 AN INTRODUCTION TO THE FINITE ELEMENT METHOD − Qe δw(xa ) − Qe δw(xb ) ảá d d dw e e + Sxx + Ψ dx Dxx 0= dx dx dx xa − Qe δΨ(xa ) − Qe δΨ(xb ) (e) Z xb ∙ (f ) where δu, δw, and δΨ are the virtual displacements The Qe have the same physical i meaning as in the Euler—Bernoulli beam element, and their relationship to the horizontal displacement u, transverse deflection w0 , and rotation Ψ, is Qe = −Nxx (xa ), Qe = Nxx (xb ) ¸4 ∙ ¸ dw dw e e , Q5 = Qx + Nxx Q2 = − Qx + Nxx dx x=xa dx x=xb e e Q3 = −Mxx (xa ), Q6 = Mxx (xb ) ∙ (g) Suppose that the displacements are approximated as u(x) = m X (1) ue ψj , j w(x) = j=1 n X e (2) wj ψj , Ψ(x) = j=1 p X (3) se ψj j (h) j=1 (α) where ψj (x) (α = 1, 2, 3) are Lagrange interpolation functions of degree (m − 1), (n − 1), and (p − 1), respectively At the moment, the values of m, n, and p are arbitrary, that is, arbitrary degree of polynomial approximations of u0 , w0 , and Ψ (1) (2) may be used Substitution of (h) for u, w, and Ψ, and δu = ψi , δw = ψi , and (3) δΨ = ψi into Eqs (d)—(f) yields the finite element model 0= 0= 0= m X j=1 m X j=1 m X 11 Kij ue + j 21 Kij ue + j 31 Kij ue + j j=1 n X j=1 n X j=1 n X 12 e Kij wj + 22 e Kij wj + 32 e Kij wj + j=1 p X j=1 p X j=1 p X j=1 13 Kij se − Fi1 j (i) 23 Kij se − Fi2 j (j) 33 Kij se − Fi3 j (k) where 11 Kij = 21 Kij = Z xb xa Z xb xa (1) Axx (1) dψi dψj dx dx 12 dx, Kij = (1) (2) dw0 dψi dψj Axx dx dx dx PROPRIETARY MATERIAL dx, Z xb xa (2) Axx 13 Kij = 0, (1) dw0 dψi dψj dx dx dx dx 31 Kij = c °The McGraw-Hill Companies, Inc All rights reserved SOLUTIONS MANUAL 22 Kij = 23 Kij = 33 Kij = Fi1 = Z xb xa Z xb xa Z xb xa Z xb x a Z xb (2) Sxx (2) dψi dψj dx dx Sxx dψi dx (2) ⎛ dx + (3) ψj Z xb xa Axx dw0 dx ả2 (2) (2) dψi dψj dx dx dx 32 dx = Kji ⎞ (3) (3) dψ dψj (3) (3) ⎝Dxx i + Sxx ψi ψj ⎠ dx dx dx (1) (1) (1) (2) (2) (2) ψi f dx + Qe ψi (xa ) + Qe ψi (xb ) Fi2 = Fi3 = Qe ψi (xa ) + Qe ψi (xb ) xa 415 ψi q dx + Qe ψi (xa ) + Qe ψi (xb ) (3) (3) (`) The element equations (i)—(k) can be expressed in matrix form as ⎫ ⎤⎧ ⎡ ⎧ ⎫ [K 11 ] [K 12 ] [K 13 ] ⎨ {u} ⎬ ⎨ {F } ⎬ ⎣ [K 21 ] [K 22 ] [K 23 ] ⎦ {w} = {F } ⎭ ⎩ ⎭ ⎩ [K 31 ] [K 32 ] [K 33 ] {F } {s} (m) (α) The choice of the approximation functions ψi dictates different finite element (1) (2) models The choice of linear polynomials ψi = ψi is known to yield a stiffness matrix that is nearly singular This will be discussed further in the next section (1) (2) When ψi are quadratic and ψi are linear, the stiffness matrix is × It is possible to eliminate the interior degree of freedom for w0 and obtain × stiffness (1) (2) matrix This element behaves well When ψi are cubic and ψi are quadratic, the stiffness matrix is × If the interior nodal degrees of freedom are eliminated, one obtains × stiffness matrix that is known to yield the exact solution at the nodes in the linear case when the shear stiffness and bending stiffnesses are element-wise constant More details of various Timoshenko beam elements can be found in Reddy (2004b) Problem 14.17: Compute the tangent stiffness matrix for the Timoshenko beam element in Problem 14.16 Solution: The tangent matrix coefficients are defined by (see Problem 14.15) αβ αβ Tij = Kij + n XX ∂ β γ=1 k=1 ∂∆j ¡ αγ ¢ Kik ∆γ k (a) In particular, we have 11 11 Tij = Kij + PROPRIETARY MATERIAL c °The McGraw-Hill Companies, Inc All rights reserved 416 AN INTRODUCTION TO THE FINITE ELEMENT METHOD =0 12 12 Tij = Kij + 13 Tij 21 Tij 22 Tij = = = 13 Kij 21 Kij 22 Kij Z xb xa (2) Axx 21 + = Kij + Z xb xa Axx " (1) dw0 dψi dψj 12 dx = 2Kij dx dx dx du0 dw0 + dx dx ả2 # (2) (2) dψi dψj dx dx dx 23 23 23 Tij = Kij + = Kij 31 31 31 Tij = Kij + = Kij 32 32 32 Tij = Kij + = Kij 33 33 33 Tij = Kij + = Kij (b) αβ where the direct stiffness coefficients Kij are defined by Eq (`) of Problem 14.16 Problem 14.18: (Natural convection in flow between heated vertical plates) Consider the flow of a viscous incompressible fluid in the presence of a temperature gradient between two stationary long vertical plates Assuming zero pressure gradient between the plates, we can write vx = vx (y), vy = 0, T = T (y), and d2 vx = ρβg(T − Tm ) + µ , dy µ dvx d2 T 0=k +à dy dy ả2 where Tm = (T0 + T1 ) is the mean temperature of the two plates, g the gravitational acceleration, ρ the density, β the coefficient of thermal expansion, µ the viscosity, and k the thermal conductivity of the fluid Give a finite element formulation of the equations and discuss the solution strategy for the computational scheme Solution: The finite element model is given by Kv vy − GT = F1 , KT T = F2 (a, b) where v Kij = Z yb dψi dψj T dy, Kij = dy dy Z yb dψi dψj ya Z yb ya µ µ dvx dy ¶2 k dy, Gij = Z yb gi j dy dy dy ya ả ¶ Z yb dvx dvx Fi1 = − ρβgψi dy + Pi , P1 = −µ , P2 = µ dy ya dy yb ya Fi2 = µ ya ψi dy + Qi , Q1 = −k µ dT dy ¶ , Q2 = k ya µ dT dy (c) ¶ yb Solution strategy: Solve the assembled equations corresponding to Eq (b) for T , subject to boundary conditions and initial values of vx = Use the temperatures PROPRIETARY MATERIAL c °The McGraw-Hill Companies, Inc All rights reserved SOLUTIONS MANUAL 417 thus obtained in the assembled equations associated with Eq (a) and solve for vx Then resolve Eq (b) with the updated Fi2 (because of the newly computed vx ) Iterate the procedure until vx and T obtained in two consecutive iterations differ by, say, one percent Problem 14.19: Derive the interpolation functions ψ1 , ψ5 , and ψ8 for the eight-node prism element using the alternative procedure described in Section 8.2 for rectangular elements Solution: This is straightforward Since ψ1 (ξ, η, ζ) must vanish on the faces ξ = 1, η = and ζ = 1, it is of the form (see Fig 14.3.2) ψ1 = c1 (1 − ξ)(1 − η)(1 − ζ), ψ1 (−1, −1, −1) = → c1 = Similarly, we obtain ψ1 = (1 − ξ)(1 − η)(1 − ζ) ψ5 = (1 − ξ)(1 − η)(1 + ζ) ψ8 = (1 − ξ)(1 + η)(1 + ζ) e Problem 14.20: Evaluate the source vector components fie and coefficients Kij over a master prism element when f is a constant, f0 , and k1 = k2 = k3 = constant in (14.3.5b) Solution: For a cube of sides a × b × c, the coordinate transformation become x= b y = (1 + η), a (1 + ξ), c x = (1 + ζ) z a ζ • • ( −1, − 1, 1) •5 • (1, − 1, 1) • •(−1, 1, 1) (1, 1, 1) 7• •6 η ( −1,−1,−1) ξ c • • x b • • PROPRIETARY MATERIAL y (1,−1,−1) •( −1, 1,−1) •1 • c °The McGraw-Hill Companies, Inc • (1, 1,−1) All rights reserved 418 AN INTRODUCTION TO THE FINITE ELEMENT METHOD and the Jacobian matrix and its inverse are ⎡ ⎢ J=⎢ ⎣ ∂y ∂ξ ∂y ∂η ∂y ∂ζ ∂x ∂ξ ∂x ∂η ∂x ∂ζ ∂z ∂ξ ∂z ∂η ∂z ∂ζ ⎤ ⎡a ⎥ ⎥=⎣0 ⎦ 0 b ⎤ ⎡ a ⎦ , J−1 = ⎣ c 2 b ⎤ 0⎦ c Then the derivatives of the interpolation functions with respect to the global coordinates can be expressed in terms of the interpolation functions with respect to the natural coordinates as ⎧ ∂ψe ⎫ ⎪ i ⎪ ⎪ ∂x ⎪ ⎨ e⎬ ∂ψi ⎪ ∂ye ⎪ ⎪ ∂ψ ⎪ ⎩ i ⎭ = J−1 ∂z ⎧ ∂ψe ⎫ ⎪ ∂ξi ⎪ ⎪ ⎨ e⎪ ⎬ ∂ψi ⎪ ∂ηe ⎪ ⎪ ∂ψ ⎪ ⎩ i ⎭ = ∂ζ e Hence, the coefficients Kij can be expressed as e Kij = ⎧ ∂ψe ⎫ ⎪ a ∂ξi ⎪ ⎪ ⎪ ⎨ ⎬ e ∂ψi ⎪ b ∂ηe ⎪ ⎪ ∂ψ ⎪ ⎩ i ⎭ c ∂ζ Z aZ bZ cà 0 e e e ∂ψ e ∂ψj ∂ψ e ∂ψj ∂ψ e ∂ψj + ky i + kz i kx i ∂x ∂x ∂y ∂y ∂z ∂z ! dx 11 22 33 = kx Sij + ky Sij + kz Sij αβ where Sij are defined as 11 Sij = 22 Sij = 33 Sij = Z aZ bZ c e e ∂ψi ∂ψj 0 0 0 0 Z aZ bZ c Z aZ bZ c dx dy dz ∂x ∂x e e ∂ψi ∂ψj dx dy dz ∂y ∂y e e ∂ψi ∂ψj dx dy dz ∂z ∂z αβ The matrices Sij can now be evaluated using the Gauss quadrature: 11 Sij = 22 Sij = 33 Sij = Z aZ bZ c e e ∂ψi ∂ψj 0 0 0 0 Z aZ bZ c Z aZ bZ c Z Z Z e e bc 1 ∂ψi ∂ψj dξ dη dζ ∂x ∂x 2a −1 −1 −1 ∂ξ ∂ξ Z Z Z e e e e ac 1 ∂ψi ∂ψj ∂ψi ∂ψj dx dy dz = dξ dη dζ ∂y ∂y 2b −1 −1 −1 ∂η ∂η Z Z Z e e e e ab 1 ∂ψi ∂ψj ∂ψi ∂ψj dx dy dz = dξ dη dζ ∂z ∂z 2c −1 −1 −1 ∂ζ ∂ζ PROPRIETARY MATERIAL dx dy dz = c °The McGraw-Hill Companies, Inc All rights reserved SOLUTIONS MANUAL 419 αβ The coefficients Sij can be evaluated using the interpolation functions listed in Eq (14.3.31) ⎡ ⎤ −4 −2 2 −2 −1 ⎢ −4 −2 −2 −1 ⎥ ⎢ ⎥ ⎢ −2 −4 −1 −2 ⎥ ⎢ ⎥ bc ⎢ −2 −4 −1 −2 2⎥ ⎢ ⎥ 11 S = ⎢ ⎥ −4 −2 2⎥ 36a ⎢ −2 −1 ⎢ ⎥ ⎢ −2 −1 −4 −2 ⎥ ⎢ ⎥ ⎣ −1 −2 −2 −4 ⎦ −1 −2 2 −2 −4 ⎡ ⎤ −2 −4 −1 −2 ⎢ −4 −2 −2 −1 ⎥ ⎢ ⎥ ⎢ −2 −4 −1 −2 1⎥ ⎢ ⎥ ac ⎢ −4 −2 −2 −1 2⎥ ⎢ ⎥ S22 = ⎢ ⎥ ⎢ −1 −2 −2 −4 ⎥ 36b ⎢ ⎥ ⎢ −2 −1 −4 −2 ⎥ ⎢ ⎥ ⎣ −1 −2 −2 −4 2⎦ −2 −1 −4 −2 ⎡ ⎤ 2 −4 −2 −1 −2 ⎢ −2 −4 −2 −1 ⎥ ⎢ ⎥ ⎢ −1 −2 −4 −2 ⎥ ⎢ ⎥ ab ⎢ 2 −2 −1 −2 −4 ⎥ ⎢ ⎥ 33 S = ⎢ ⎥ 2⎥ 36c ⎢ −4 −2 −1 −2 ⎢ ⎥ ⎢ −2 −4 −2 −1 1⎥ ⎢ ⎥ ⎣ −1 −2 −4 −2 2⎦ −2 −1 −2 −4 2 Similarly, the source vector f e can be computed Fe = ⎧ ⎫ ⎪1⎪ ⎪ ⎪ ⎪ ⎪ ⎪1⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪1⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ abc ⎨ ⎬ ⎪1⎪ ⎪ ⎪ ⎪ ⎪ ⎪1⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪1⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ PROPRIETARY MATERIAL c °The McGraw-Hill Companies, Inc All rights reserved 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 77843-3123 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 ... the least squares method and the Galerkin method give the same equations Furthermore, the solution of the Galerkin and least squares methods would be the same as that of the Ritz method For the. .. at the end of Chapter The instructor should make an effort to review the problems before assigning them This allows the instructor to make comments and suggestions on the approach to be taken and... into the weak form(s) and derive the finite element model Be e sure to define all coefficients of the model in terms of the problem data and ψi Problem 3.1: Develop the weak form and the finite element