hw2_f03_solns

# hw2_f03_solns - EMA 405, Fall 2003 Solutions for HW #2 1....

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EMA 405, Fall 2003 Solutions for HW #2 1. Before constructing the global stiffness matrix, we need a notation that allows us to connect contributions from individual elements. The following sketch shows an element-based notation and a global numbering for the nodes. Element Notation node J node 4 element 3 u e (I) u e (J) Global Notation and Problem Schematic F a =400kN node I node 1 node 2 node 3 element e element 1 element 2 As described in class, we can construct the element stiffness matrices by considering the forces required at the two adjacent nodes in order to get a unit displacement at a node, i.e. ) ( ) ( j u i F e e ij e = A For bar elements (a.k.a. spring element, a.k.a. link elements), = e e e e e k k k k A where the spring constant, k e = a e E e /L e , using a e for the cross-section area to avoid confusion with the element stiffness matrix. Element by element, the element stiffness matrices relate the nodal displacements to the forces acting on each element = ) ( ) ( ) ( ) ( 1 1 1 1 1 1 1 1 J F I F J u I u k k k k for element 1, for example. Changing the displacements to global-node notation, this same relation appears as 1/11

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= ) ( ) ( 1 1 2 1 1 1 1 1 J F I F u u k k k k for element 1, for element 2 = ) ( ) ( 2 2 3 2 2 2 2 2 J F I F u u k k k k and similarly for element 3. In this configuration, where there are no external forces between elements 1 and 2 and between elements 2 and 3, the sum of the forces at the interior nodes is 0, i.e. F 1 (J) +F 2 (I)=0, and F 2 (J) +F 3 (I)=0. This lets us combine the second row of the element 1 stiffness matrix with the first row of the element 2 stiffness matrix, and similarly for elements 2 and 3, to create the global stiffness matrix, pre-boundary conditions. This system of equations is = + + a F I F u u u u k k k k k k k k k k k k 0 0 ) ( 0 0 0 0 0 0 1 4 3 2 1 3 3 3 3 2 2 2 2 1 1 1 1 using F 3 (J)= F a , the applied force at the end of the beam. The constraint or boundary condition equation, u 1 =0, replaces the first row of the above linear system, and when the value of u 1 is inserted, the first column is also eliminated. Thus, our complete linear system is = + + a F u u u k k k k k k k k k 0 0 0 0 4 3 2 3 3 3 3 2 2 2 2 1 While we could invert the global stiffness matrix to find the displacement, u , it is easier to solve the equations (the rows of the linear system) with basic algebra: F A = 1 g 3 2 1 2 2 u k k k u + = , from the first row.
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## This note was uploaded on 07/29/2008 for the course EMA 405 taught by Professor Witt during the Spring '04 term at Wisconsin.

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hw2_f03_solns - EMA 405, Fall 2003 Solutions for HW #2 1....

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