AE420/ME471/CSE453 – Introduction to the finite element method List of application problems University of Illinois, Urbana Champaign AE 420

AE420/ME471/CSE453 – Introduction to the finite element method List of application problems Topic 2: Finite element formulation and analysis for 1-D Poisson problems Problem 2.1 Finite element for a rope problem One of the simplest mechanical systems consists of a rope of initial length L under tension T and subjected to a lateral force f(x) (Figure 2.1). T u(x) x u f(x) T 0 L Figure 2.1 If we assume that the deflection u(x) is small and that the tension T is constant, the expression of the potential energy for this system can be shown to be Π p = 1 2 T du dx     2 dx − f x( )u x( )dx 0 L ∫ 0 L ∫ . Using the variational method (i.e., using the PMPE), deduce the expression of the local stiffness matrix [k] and the local load vector {r} for a 2-node and a 3-node “generic rope element” of length l . For the local load vector, assume that the applied load f(x) is constant (=fo) on the element. How many degrees of freedom does each node have? Solution (1) Let us derive the FE formulation for a generic M-node “rope element”. We will then “specialize” our formulation to the 2- and 3-node cases, as requested. 2 Π e = 1 2 T du ds        2        ds 0 l ∫ − f s( ) u s( )dx 0 l ∫ , (1) where Π e = potential energy of the "generic element" l = length of the element f(s) = distributed transverse load acting on the element Let u ˜ (s) = N(s) { }d , (2) where N(s) = N1 (s) N2 (s) .. NM (s) = vector with M shape functions d = U1 U2 .. UM = vector with M modal displacements (1dof/node) (2) ⇒ du ˜ ds = N ' { }d = d { } N ' Substitute into Π e in (1) : Π ˜ e = 1 2 (T d { }N N' { }d ) 0 l ∫ ds− f(s) d { }N dx 0 l ∫ = 1 2 d [ ]k { }d − d { }r where [ ]k (M,M) = T dN(s) ds       (M,1) dN(s) ds (1,M)           ds 0 l ∫ { }r (M,1) = f(s) { }N (M,1) ds 0 l ∫ (2) 2 - node " rope element" (M = 2) → N = N1 N2 = 1− s l s l ⇒ N = − 1 l 1 l [ ]k = T −1 l 1 l       0 l ∫ −1 l 1 l ds = T l 1 −1 −1 1       { }r = f0 0 l ∫ 1− s l s l           ds = f0 l 1 2 1 2       3 (3) 3 - node rope element (M = 3) → N = N1 N2 N3 = 2( ) s− l ( ) s− l / 2 l 2 −4s s( ) − l l 2 2s s( ) − l / 2 l 2 ⇒ N ′ = 4s− 3l l 2 − 4 2( ) s− l l 2 4s− l l 2 [ ]k = T N{ }' 0 l ∫ N' ds = T l 7/ 3 −8/ 3 1/ 3 −8/ 3 16/ 3 −8/ 3 1/ 3 −8/ 3 7/ 3           { }r = f0 0 l ∫ 2 l 2 (s− l 2 )(s− l) − 4 l 2 s(s− l) 2 l 2 s(s− l 2 )                   ds = f0 l 1 6 2 3 1 6           U1 U2 U1 U2 U3 • o o o o s=0 s=l s=0 s=l/2 s=l 2- Node Element 3-Node Element 4 Problem 2.2: FEA of a bi-material axially loaded bar Consider the bi-material axially loaded bar structure of length 2L described in Figure 2.2. It is discretized with 5 nodes and 4 elements (note: the elements are NOT of equal size). The second half (xL) of the bar is made of a material three times as stiff as the left side (xL). It is subjected to a quadratically increasing distributed axial load po*(x/(2L))2 (with po in N/m), and to a point load 2Q applied at the end (x=2L). Using the element and node numberings indicated in the figure, assemble the global stiffness matrix and global load vector for this problem. You do not need to re-derive the expression of the local stiffness matrix and local load vector. Just use the expressions found in class. Assume that the distributed load is uniform over each element: take the value computed at the center of the element. Apply the appropriate boundary conditions and obtain the final form of the linear system (do NOT solve the linear system). Assuming that you have computed the five nodal displaceme