Note for Finite Element Methods - FEM by JACKSON A.M.
- Finite Element Methods - FEM
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1/17/2017 Lecture 4: Potential Energy based Methods : Rayleigh‐Ritz Method APL705 Finite Element Method Strong and Weak Forms of Equations • Strong Form– differential equations are said to state a problem in a strong form problem in a strong form. • Weak form – an integral expression such as a functional which implicitly contains a differential equations is called a weak form. • The strong form states conditions that must be met at every material point, whereas weak form states conditions that must be met only in an average sense. • A functional such as that of potential energy π, contains integrals that span line, area or volume of interest. 1
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1/17/2017 Rayleigh‐Ritz Methods • There is a need for systematic and general way of obtaining [K] obtaining [K] One of the best ways is Rayleigh‐Ritz method. It uses an approximation field to the entire domain of interest. In FEM, this approximating function is defined in piecewise form. For using the Rayleigh‐Ritz method we need to have a functional. g p p y A functional is a an integral expression that implicitly contains the differential equations that describe the system. These functionals will be used to formulate finite element problems here. Rayleigh‐Ritz Method • This method based on minimization of total potential energy involves construction of an assumed displacement field: involves construction of an assumed displacement field: u = ∑ aiφi (x, y, z) i = 1tol v = ∑ a jφ j (x, y, z) j = l +1 to m w = ∑ akφk (x, y, z) k = m +1 ton n>m>l The functions φi are polynomials and displacements u,v,w must be kinematically admissible. This means that u,v,w must satisfy the boundary conditions. Invoking stress‐strain, strain‐displacement relations and substituting in total potential energy expression obtained earlier 2
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1/17/2017 Rayleigh‐Ritz Method Invoking stress‐strain, strain‐displacement relations and substituting in total potential energy expression obtained earlier substituting in total potential energy expression obtained earlier π = π ( a1, a2 ,........, ar ) • Where r is the number of independent unknowns. Now obtaining the miniumum by differentiation with respect to ais (i=1 to r) gives r equations: ∂π =0 ∂ai i = 1, 2,,........, r An example will illustrate this method better now. Rayleigh‐Ritz Method: Illustrative example Consider the following linear elastic one‐dimensional problem with body forces neglected. 1 π= 2 • Where u1=ux=1 ⎛ du ⎞ ∫ EA ⎜⎝ dx ⎟⎠ dx − 2u1 0 L 2 3
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1/17/2017 Rayleigh‐Ritz Method: Illustrative example Assuming a polynomial function u = a + a x + a x 2 1 2 3 This function should be satisfied at the two boundaries: u=0 at This function should be satisfied at the two boundaries: u=0 at x=0 and x=2. Therefore 0 = a1 and 0 = +2a2 + 4a3 ∴ a2 = −2a3 u = a3 (−2x + x 2 ) ⇒ u1 = −a3 Next we will find he potential energy du = a3 (−1+ ( 1+ x)) dx 2 π = ∫ 4a32 (−1+ x)2 dx − 2(−a3 ) 0 2 = 2a32 ∫ (1− 2x + x 2 )dx + 2a3 = 2a32 ( 23 ) + 2a3 0 Rayleigh‐Ritz Method: Illustrative example Now to find minimum potential energy we se ∂π = 2a3 ( 23 ) + 2 = 0 ∂a3 a3 = −0.75 0 u1 = −a3 = 0.75 0 Next let us find the stress in the bar σ = E du = 1.5(1− x) dx 4
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