Chapter Introduction to Finite Element Methods
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1 Chapte 1.4 Intoduction to Finite Element Methods Afte eading this chapte, you should e ale to: 1. Undestand the asics of finite element methods using a one-dimensional polem. In the last fifty yeas, the use of appoximation solution methods to solve complex polems in engineeing and science has gown significantly. The widespead availaility of poweful digital computes and commecial computational softwae ased on these appoximation methods with efficient solution algoithms has made them pactical. In this chapte, we ae intoducing the student to finite methods of solving diffeential equations. We povide an elementay ackgound on how finite element methods wok, while using a single example to illustate the appoach, and discuss the accuacy and efficacy of the method. The single example chosen is a classical polem of a unifomly pessuized thick-walled cylinde with an axis-symmetic esponse (Figue 1). This polem is chosen since it is simple enough to have an analytical solution, ut complex enough such that its finite element method solution can e genealized fo polems that ae moe complicated. We must fist define the polem, and then develop the exact solution so that we may compae it with the finite element methods esult. Thick-Wall Cylinde Polem Polem Definition Conside a thick-walled cylinde as depicted in Figue 1, with the following mateial popeties: Young's modulus E, Poisson's atio v inne adius a oute adius, unifom intenal pessue extenal pessue, p o p i Find the following vaiales in the cylinde. Plane stess state is assumed. adial displacement, u 1.4.1
2 1.4. Chapte 1.4 adial stess, tangential stess, Numeical Example Polem Fo demonstating the use of appoximate solution methods in solving the polem numeically, the following data is used: a =.5 m =.5 m pi = MPa p o = E = 7 GPa v =.3 E,ν a po pi Figue 1: Pessued thick-wall cylinde polem Mathematical Fomulation The solution of the thick-wall cylinde polem can e found y solving the equation of compatiility in pola coodinates, which is a fouth ode patial diffeential equation of Aiy stess function (1), o y using axisymmety conditions to fomulate the polem as a second ode diffeential equation of displacement (), o equivalent foms (potential enegy, integal equation, etc.). The last appoach is adopted in this pape, as it is diect and does not equie invese o semi-invese solution methods (1, ). The details of this appoach ae given in () and the elevant fomulas ae summaized as follows. The adial stain, ε, tangential stain, ε, in tems of adial displacement, u ae given as du ε = (1) d u ε = () The adial stess, as, and tangential stess,, in tems of adial displacement, u, ae given
3 Intoduction to Finite Element Methods E du u = + ν d (3) E du u = ν + d (4) The govening equation fo adial displacement, u, is given y d u 1 du u + d d = (5) Using Equations 3-4, the ounday conditions ( a) = pi and ( ) = po can e ewitten as u ( ) ( a) u a + ν = p i a E (6) u ( ) ( ) u + ν = p o E (7) Fist, the exact solution is found, and then a finite element method is pesented though solving the example polem. Nodal points chosen fo the finite element method ae unifomly spaced fo convenience. Figue shows how the nodal points and elements ae numeed. Element# Node# 1 n 1 n n-1 (a) n n () Figue : Numeing of nodal points and elements Exact Solution The exact solution of displacement can e found diectly y solving the govening diffeential equation, Equation (5), with associated ounday conditions, Equations (6-7), and then sustituting it into Equations (3-4) to give an exact solution of stesses. The exact solutions (7) of adial displacement, adial stess, and tangential stess ae otained as ( a pi po ) 1+ ν ( pi po ) a u = + (8) E a E ( a ) a pi po ( pi po ) a = (9) a ( a ) a pi po ( pi po ) a = + (1) a ( a ) Solution fo Example Polem Sustituting the numeical data into Equations (8-1), the exact solution fo the example polem is
4 1.4.4 Chapte u = (11) = (13) = (14) Evaluating the solution at thee nodal points (inne edge, =. 5 m ; mid-point, =. 375 m ; and oute edge, =. 5 m ) along the adial location fo compaison, the esulted values ae given in Tale 1. Tale 1: Exact solution evaluated at nodal points (m) u (mm) (MPa) (MPa) What ae Finite Element Methods? The finite element method is a technique used to solve diffeential equations (odinay o patial). They ae mainly used to solve eal wold polems, as the diffeential equations that goven these polems cannot e solved exactly, o may e too intactale to e solved exactly. The finite element methods use techniques to appoximate the dependant vaiales of the diffeential equations y functions, and then educe the unknowns in these functions to a set of simultaneous linea equations. These equations can then e solved y vaious numeical techniques. Howeve, one needs to undestand that finite element methods use a function, not the diffeential equation itself, to develop the appoximate solution. This is unlike the finite diffeence methods, whee the deivatives in the diffeential equations ae appoximated y finite divided diffeence methods. The functions used in the finite element methods ae integal equations. In the case of the pessue vessel, these equations would model the total potential enegy due to intenal stesses and extenal loads The Rayleigh-Ritz method can e viewed as a fom of a finite element method whee it educes a continuous polem to a polem with a finite nume of degees of feedom. The Rayleigh-Ritz method is ased on the pinciple of stationay potential enegy, which states: Among all admissile configuations of a consevative system, those that satisfy the equations of equiliium make the potential enegy stationay with espect to small vaiations of displacement. If the stationay condition is a minimum, the equiliium state is stale.
5 Intoduction to Finite Element Methods Mathematically speaking, the Rayleigh-Ritz method is a vaiational method, ased on the idea of finding a solution that minimizes a functional. Fo elasticity polems, the functional is the total potential enegy. The solution must e admissile, that is, satisfying intenal compatiility (e.g., continuity of displacement) and essential ounday conditions. Fo polems whee displacements ae pimay unknowns, essential ounday conditions ae pesciptions of displacement and non-essential ounday conditions ae pesciptions of stess. Since the polem consideed hee, the thick-walled pessued cylinde polem whee the pimay unknown is adial displacement, has no pesciption of displacement, thee is no essential ounday condition. Potential Enegy Fomulation The cylinde is assumed to e in a plane stess state which gives a stain enegy density, U as 1 U = ( ε + ε ) (15) y using Equations (1-4), we get E ( ) du du u u U = + ν + (16) d d Total stain enegy, U of the cylinde is U dv = U dddz = U = πl U d ( V ) whee, L = cylinde length L π a a Wok done, W y extenal foces (intenal and extenal pessues) is W = piu( a) ds p u( ) ds = o πalpiu( a) πlpou( ) (18) ( S ) i ( S ) whee, Si = inne cylinde suface S = oute cylinde suface o o The total potential enegy of the cylinde, Π is found as Π = U W = π L ( ) ( ) U d apiu a + pou (19) a Rayleigh-Ritz Method The Rayleigh-Ritz method can e outlined as follows. The potential enegy of the system is given as Π = Π( u, u, ). u = f C, C,, Assume a tial solution of the fom: ( ), 1 C m (17)
6 1.4.6 Chapte 1.4 whee C i s ( i.. m) ' = ae unknown paametes, and f is a known function. In this pape, we conside linea piecewise continuous functions. Apply admissiility conditions to the tial solution. If thee ae conditions, we have m n equations of unknown paametes. m n admissiility Solve the system of m n equations fo m n unknowns Cn + 1 Cm, and then plug them ack into the tial solution, we otain a new tial solution that is admissile and has fewe u = f C, C,,. unknowns ( n unknowns) ( ), 1 C n Sustitute the tial solution into the expession of potential enegy.the stationay condition fo potential enegy δ Π = gives Π =, i =.. n () C i Hee we have a system of n algeaic equations with n unknowns. Solving this system of equations, we find the unknown paametes and thus the appoximate solution fo the adial displacement. Sustitute the found solution fo adial displacement into Equations (3-4) to find the appoximation solution fo adial stess and tangential stess. Linea Piecewise Continuous Solution fo Example Polem Conside the case of n = with unifom spacing nodal points. The step size fo locating nodal points is calculated as h = ( a) / n = (.5.5) / =.15. The adial coodinates of the nodal points ae o = a =. 5, 1 =. 375, = =. 5. The displacement field is assumed to e a piecewise continuous function of two linea segments as C + C1, u = (1) C3 + C, To make the tial solution, Equation (1), admissile, it must e continuous at =. 375, which means C +.375C1 = C C, o () C3 = C +.375C1. 375C (3) The tial solution, Equation (1), then ecomes C + C1, u = (4) C +.375C1.375C + C,.375.5
7 Intoduction to Finite Element Methods Sustituting Equation (4) and the given numeical data into Equation (19), the total potential enegy, Π in the cylinde is found as Π = πl(78.84c CC CC C C1C (5) C.5C.15C1 ) 1 The condition that the total potential enegy Π is stationay, Π Π Π =, =, = C C1 C Which gives a system of algeaic equations of the unknown coefficients as 157.7C C C = C C C =.15 (6) 1.4C C C = The unknown coefficients ae found as C =.6737 C1 =.8496 (7) C =.3191 Sustituting Equation (7) into Equation (4), the appoximate solution fo adial displacement is , u = (8) , Sustituting the numeical data and displacement solution fom Equation (8) into Equations (3-4), we find the adial and tangential stesses as ,.5 < <.375 = (9) ,.375 < < ,.5 < <.375 = (3) ,.375 < <.5 The solution of the adial displacement is continuous, since we have foced the tial solution to e admissile fom the eginning, while the solutions fo stesses ae discontinuous at the inteio knot ( =. 375 ) etween the two segments (elements). To have easonale esults, in pactice, the stess value at the inteio knot is taken as the aveage of two stess values. The numeical solution with n = of the example polem is given in Tale 4.
8 1.4.8 Chapte 1.4 Tale 4: Numeical solution, finite element method ( n = ) (m) u (mm) (MPa) (MPa) The exact solution and numeical solutions with vaious values of nume of nodal points, n =, 3, and 4, ae given in Figue 5 fo adial displacement, and Figue 6 fo adial stess..5 Radial displacement (mm) Exact n = n = 3 n = Radial Location (m) Figue 5: Radial displacement as a function of adial location (Finite element method)
9 Intoduction to Finite Element Methods Radial stess (MPa) Exact n = n = 3 n = Radial Location (m) Figue 6: Solution of adial stess as a function of adial location The solution plots show that the appoximate solutions appoach the exact solution as the nume of piecewise continuous functions incease. Howeve, they do not satisfy the ounday conditions of adial stess. The assumption of the piecewise continuous solution as opposed to a continuous solution makes computation easie fo a high nume of segments in the piecewise functions, ut it has the dawack of the discontinuity of stesses at the inteio knots of the piecewise continuous function. REFERENCES [1] S.P. Timoshenko, J.N. Goodie, Theoy of Elasticity, McGaw-Hill, 197. [] A.C. Ugual, S.K. Fenste, Advanced Stength and Applied Elasticity, 3d Ed. Pentice- Hall PTR, [3] A.P. Boesi, K.P. Chong, Appoximate Solution Methods in Engineeing Mechanics, Elsevie Applied Science, [4] R.D. Cook, D.S. Malkus, M.E. Plesha, Concepts and Applications of Finite Element Analysis, 3d Ed. Wiley, [5] W.S. Hall, The Bounday Element Method, Kluwe Academic Pulishes, PARTIAL DIFFERENTIAL EQUATIONS Topic Intoduction to Finite Element Methods Summay Textook notes fo the intoduction of patial diffeential equations Majo All engineeing majos Authos Auta Kaw, Sun Ho Date July 11, 11 We Site
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