Introduction to finite elements April 13, 2009ME 501B – Engineering Analysis 1Introduction to Finite ElementsIntroduction to Finite ElementsLarry CarettoMechanical Engineering 501ABSeminar in Engineering AnalysisApril 13, 20092Outline• Review midterm• Introduction to finite elements• Basic approaches to finite elements– Will start material originally scheduled for April 22– Parallel reading for this week is pages 711 to 739 in Hoffman• Example application in one dimension3Midterm Results• Number of students = 12• Maximum possible = 100• Mean = 45.1• Median = 43.5• Standard deviation = 28.2• Grade distribution2 10 14 26 35 4047 65 71 73 75 834Problem One• Sketch the wave equation solution, u(x,t) at times ct = 0.1, 0.2, and 0.3 for f(x) = u(x,0) as shown at the right. – u(x,t) = [f(x+ct) + f(x-ct)]/200.510 0.1 0.2 0. 3 0.4 0.5 0 .6 0.7 0.8 0 .9 1xf(x)Shapes at ct = 0.100.510 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1xFunctionsu(x,t )f(x+ ct)f(x-ct)Shapes at ct = 0.200.510 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1xFunctionsu(x,t )f(x+ct)f(x-ct)5Problem One II• For ct = 0.3 must account for periodic extensions of initial conditions for x < 0 and x > 1 entering region 0 ≤ x ≤ 1– These are inverse of original initial conditionShapes at ct = 0.3-0.500.510 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1xFunctionsf(x+ct) f(x-ct)u(x,t)6Problem Two• Outline solution for potential u(x,y,t)2102222),,(0),0,(),,(0),,0()0,,(0,0,0UtHxutxuUtyLutyuUyxutHyLxyuxutu=====≥≤≤≤≤⎟⎟⎠⎞⎜⎜⎝⎛∂∂+∂∂=∂∂α• u(x,y,t) = v(x,y,t) + w(x,y)– v(x,y,t) is diffusion equation solution with homogenous boundary conditions– w(x,y) is Laplace equation solution requiring superposition: w(x,y) = w1(x,y) + w2(x,y)•w1(x,y) and w2(x,y) each have only one nonhomogenous boundary conditionIntroduction to finite elements April 13, 2009ME 501B – Engineering Analysis 27Problem Two II• Solutions in notes⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛=∑∑∞=∞=⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛−HymLxneCtyxvnmtHmLnnmππαππsinsin),,(1122()∑∞=⎟⎠⎞⎜⎝⎛++⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛+=021)12(sinh12)12(sinh)12(sin4),(nLHnnLynLxnUyxwππππ• Solution for w2(x,y) found by substituting x and y, L and H (and U1for U2) in w1solution8Problem Two II• Need to use solutions for w(x,y) to find coefficients in v(x,y,t) solution()∑∞=⎟⎠⎞⎜⎝⎛π++⎟⎠⎞⎜⎝⎛π+⎟⎠⎞⎜⎝⎛π+π=012)12(sinh12)12(sinh)12(sin4),(nHLnnHxnHynUyxw[]∫∫⎟⎠⎞⎜⎝⎛π⎟⎠⎞⎜⎝⎛π−−=HLpqdxdyHyqLxpyxwyxwUHLC00210sinsin),(),(49Problem Three• Replace each fkin equation below by its Taylor series to order of the errorhfffffiiiii12882112'−−+++−+−=.....!5)2(!4)2(!3)2(!2)2(25'''''4''''3'''2'''2++++++=+hfhfhfhfhfffiiiiiii.....!5)(!4)(!3)(!2)(5'''''4''''3'''2'''1++++++=+hfhfhfhfhfffiiiiiii.....!5)(!4)(!3)(!2)(5'''''4''''3'''2'''1+−+−+−+−+−=−hfhfhfhfhfffiiiiiii().....!5)2(!4)2(!3)2(!2)2(25'''''4''''3'''2'''2+−+−+−+−+−+=−hfhfhfhfhfffiiiiiii10Problem Three II• After the algebra()()()()4'5''''''211212!53288321228821288hOfhhfhhfhffffiiiiiii+=+−++−+−++−=+−+−−−++L• Get first derivative of exat x = 0 for h = 0.1 and h = 0.05, and find the order of the error()969999966626.01.01288)1.0(201.001.00)1.0(20'=+−+−=−−++eeeefi()04659999997916.005.01288)05.0(2005.0005.00)05.0(20'=+−+−=−−++eeeefi001.4)1.0log()05.0log()103373.3log()1008395.2log()log()log()log()log(671212=−−=−ε−ε≈−−xxhhn11Why Finite Elements• A different approach for converting differential equations into algebraic equations – More complicated approach than finite differences– Capable of handling complex geometry more accurately than finite differences• Will not cover finite-volume a melding of the two– Uses concept of interpolation polynomials– Introduces gradients naturally12Finite Element Methods• Designed for 2D and 3D geometries• Can use for 1D case as example• Basic idea is to divide region into small elements (line, area, volume)• Use interpolating polynomial for each element– Represent both geometry (independent variables) and dependent variable– Polynomials called basis functions or shape functionsIntroduction to finite elements April 13, 2009ME 501B – Engineering Analysis 313Finite Element Methods II• Start analysis for region• Look at set of small elements in region• Assemble analyses for individual elements is into a set of nodal equations for the entire region– Result is set of algebraic equations for the dependent variable at nodes that are points on elements– Converts differential equations into a set of algebraic equations at distinct points14Finite Element Analyses• How do we represent differential equation in terms of polynomials?– Variational approach (Rayleigh-Ritz) method formulates problem as the maxi-mum (or minimum) of a function integral• Apply directly to problems in solid mechanics governed by a variational principle (Hamilton’s principle)– Method of weighted residuals• Applies to general differential equations– Least squares approaches15Method of Weighted Residuals• MWR uses shape functions φi(x) or Ni(x) for each nodein region to write approx-imate solution for true u(x) ∑==Niiiuu1)(ˆxφ• Have to find all the uivalues • Write differential equation as L(u) – b = 0• Weighting functions wi(x) for individual nodes in region that satisfy this equation[]NidbuLwiK,10)ˆ( ==Ω−∫Ω16MWR Example• Poisson’s equation for 0 ≤ x ≤ L and 0 ≤ y ≤ HNidydxQyuxuwLHiK,00ˆˆ002222==⎥⎥⎦⎤⎢⎢⎣⎡+∂∂+∂∂∫∫• Pick N points in region including nodes on the boundary• Pick a set of weighting functions wi– Choice for widetermines final form of algebraic equations solvedQyuxu−=∂∂+∂∂2222Example for L(u)17What are Weight Functions?• Different approaches to MWR use different weight functions• Galerkin’s method uses wi= φi– Gives same result as variational approach when a variational principle is possible• Colocation method uses wi= δ(x – xi)• Dirac delta function δ(x – xi) result())()(iifdf xxxx =Ω−∫Ωδ18Shape (Basis) Functions• Simplest shape functions are linear for 1D or bilinear for 2D• For a linear element between nodes i (at ξ = -1) and i + 1 (at ξ = 1) we have φi= (1 – ξ)/2 and φi+1= (1 + ξ)/2•x = xiφi+ xiφi+1is correct at