Module 3: Element Propertes Lecture : Lagrange and Serendpty Elements 5 In last lecture note, the nterpolaton functons are derved on the bass of assumed polynomal from Pascal s trangle for the fled varable. As seen, the nverse of the large matrx s qute cumbersome f the element s of hgher order. 3.. Lagrange Interpolaton Functon An alternate and smpler way to derve shape functons s to use Lagrange nterpolaton polynomals. Ths method s sutable to derve shape functon for elements havng hgher order of nodes. The Lagrange nterpolaton functon at node s defned by n ( x-xj) ( x-x)( x-x)...( x-x - )( x-x+ )...( x-xn) f( x ) = = j= ( x-x j) ( x-x )( x-x )...( x-x - )( x-x + )...( x-x n) j¹ (3..) The functon f (ξ) produces the Lagrange nterpolaton functon for th node, and ξ j denotes ξ coordnate ofj th node n the element. In the above equaton f we put ξ = ξ j, and j, the value of the functon f (ξ) wll be equal to zero. Smlarly, puttng ξ = ξ, the numerator wll be equal to denomnator and hence f (ξ) wll have a value of unty.snce, Lagrange nterpolaton functon for th node ncludes product of all terms except j th term; for an element wth n nodes, f (ξ) wll have n- degrees of freedom. Thus, for one-dmensonal elements wth n-nodes we can defne shape functon as ( x= ) f( x ). 3...Shape functon for two node bar element Consder the two node bar element dscussed as n secton 3... Let us consder the natural coordnate of the center of the element as 0, and the natural coordnate of the nodes and are - and + respectvely. Therefore, the natural coordnate ξ at any pont x can be represented by, ( - ) x x x= - l (3..) Fg. 3.. atural coordnates of bar element
6 The shape functon for two node bar element as shown n Fg. 3.. can be derved from eq.(3..) as follows: ( ) ( ) x-x x- = f( x ) = = = -x x-x -- () x-x x+ = f( x ) = = = +x x -x -() (3..3) Graphcally, these shape functons are represented n Fg.3... Fg.3.. Shape functons for two node bar element 3... Shape functon for three node bar element For a three node bar element as shown n Fg.3..3, the shape functon wll be quadratc n nature. These can be derved n the smlar fashon usng eq.(3..) whch wll be as follows: ( )( 3) 3 x-x x-x () x x- ( x ) = f( x ) = = = x x- x-x x-x - (-) ( 3) 3 ()( ) x-x x-x x+ x- () x = f() x = = = -x x -x x -x - (3..) ( ) ( x -x )( x -x ) 3 3 x-x x-x x+ ( x) 3( x ) = f3( x ) = = = x x+ ()
7 Fg.3..3 Quadratc shape functons for three node bar element 3...3 Shape functon for two dmensonal elements We can derve the Lagrange nterpolaton functon for two or three dmensonal elements from one dmensonal element as dscussed above. Those elements whose shape functons are derved from the products of one dmensonal Lagrange nterpolaton functons are called Lagrange elements. The Lagrange nterpolaton functon for a rectangular element can be obtaned from the product of approprate nterpolaton functons n the ξ drecton [f (ξ )] and η drecton [f (η)]. Thus, xh, = f xf h Where, =,,3,., n-node (3..5) The procedure s descrbed n detals n followng examples. Four node rectangular element The shape functons for the four node rectangular element as shown n the Fg.3.. can be derved by applyng eq.(3..3) eq.(3..5)whch wll be as follows. ` ( x- ) ( h- ) ( x-x) ( h-h) xh, = f xf h = x-x h-h = = -x -h --() --() (3..6) Smlarly, other nterpolaton functons can be derved whch are gven below. `, f f 3, f f, f f ( xh ) = ( x) ( h ) = ( +x)( -h) ( xh ) = ( x) ( h ) = ( +x )( +h) ( xh ) = ( x) ( h ) = ( -x)( +h) (3..7) These shape functons are exactly same as eq.(3.3.3) whch was dervedearler by choosng polynomals.
8 Fg. 3..Four node rectangular element ne node rectangular element In a smlar way, to the dervaton of four node rectangular element, we can derve the shape functons for a nne node rectangular element. In ths case, the shape functons can be derved usng eq.(3..) and eq.(3..5)., f f ( xh ) = ( x) ( h ) = xx- hh- = xhx- ( h-) (3..8) In a smlar way, all the other shape functons of the element can be derved. The shape functons of nne node rectangular element wll be: = xh( x-)( h- ), = xh( x+ )( h-) 3= xh( x+ )( h+ ), = xh( x-)( h+ ) 5 = h( -x )( h- ), 6 = x( x+ )( -h ) 7 = h( -x )( h+ ), 8 = x( x-)( -h ) 9 = -x -h (3..9)
9 Fg.3..5 ne node rectangular element Thus, t s observed that the two dmensonal Lagrange element contans nternal nodes (Fg. 3..6) whch are not connected to other nodes. Fg. 3..6Two dmensonal Lagrange elements and Pascal trangle 3.. Serendpty Elements Hgher order Lagrange elements contans nternal nodes, whch do not contrbute to the nterelement connectvty. However, these can be elmnated by condensaton procedure whch needs extra computaton. The elmnaton of these nternal nodes results n reducton n sze of the element matrces.alternatvely, one can develop shape functons of two dmensonal elements whch contan nodes only on the boundares. These elements are called serendpty elements (Fg. 3..7) and ther nterpolaton functons can be derved by nspecton or the procedure descrbed n prevous lecture
30 (Module 3, lecture 3). The nterpolaton functon can be derved by nspecton n terms of natural coordnate system as follows: (a) Lnear element, (b) Quadratc element ( xh ) = ( +xx )( +hh) () For nodes at x=, h=, ( xh ) = ( +xx )( +hh)( xx+hh-) () For nodes at x=, h= 0, ( xh ) = ( +xx)( -h) () For nodes at x= 0, h=, (c) Cubc element ( xh ) = ( -x)( +hh) () For nodes at x=, h= (3..0) (3..a) (3..b) (3..c) ( x, h ) = ( +xx )( +hh) é9( x +h )-0ù 3 êë úû (3..a) () For nodes at x=, h= 3 ( xh, ) = 9 ( +xx)( -h )( + 9hh) 3 (3..b) And so on for other nodes at the boundares.
3 Fg. 3..7 Two dmensonal serendpty elements and Pascal trangle Thus, the nodal condtons must be satsfed by each nterpolaton functon to obtan the functons serendptously. For example, let us consder an eght node element as shown n Fg. 3..8 to derve ts shape functon. The nterpolaton functon must become zero at all nodes except node, where ts value must be unty. Smlarly, at nodes, 3, and 6, ξ =, so ncludng the term ξ satsfes the zero condton at those nodes. Smlarly, at nodes 3, and 7, η = so the term η ensures the zero condton at these nodes. Fg. 3..8 Two dmensonal eght node rectangular element Agan, at node 5, (ξ, η) = (0, ), and at node 8, (ξ, η) = (, 0). Hence, at nodes 5 and 8, the term (ξ + η + ) s zero. Usng ths reasonng, the equaton of lnes are expressed n Fg. 3..9. Thus,
3 thenterpolaton functon assocated wth node s to be of the form = y ( h- )( x- )( x+ h + ) where, ψ s unknown constant. As the value of s at node, the magntude unknown constant ψ wll become -/. Therefore, the shape functon for node wll become =- ( -h)( - x)( x+ h+ ). Smlarly, ψ wll become -/ consderng the value of at node as unty and the shape functon = - + - - =- + - - -. In a smlar for node wll be y ( h )( x )( x h ) ( x)( h)( x h ) fashon one can fnd out other nterpolaton functons from Fg. 3..9 by puttng the respectve values at varous nodes. Thus, the shape functon for 8-node rectangular element s gven below. Fg. 3..9 Equatons of lnes for two dmensonal eght node element =- ( -x)( -h )( +x+h ), 5 = ( -x )( -h), =- ( +x)( -h)( -x+h), 6 = ( +x)( -h ), 3=- ( +x )( +h)( -x-h), 7 = ( -x )( +h), =- ( -x )( +h )( +x-h ) and 8 = ( -x)( -h ) (3..3)
33 It may be observed that the Lagrange elements have a better degree of completeness n polynomal functon compare to serendpty elements. Therefore, Lagrange elements produce comparatvely faster and better accuracy.