Discretization Methods Exercise # 5

Transcription

1 Discretization Methods Exercise # 5 Static calculation of a planar truss structure: a a F Six steps: 1. Discretization 2. Element matrices 3. Transformation 4. Assembly 5. Boundary conditions 6. Solution 1. step: Discretization/Decomposition into nodes and element nodes 7 elements y x global coordinate system (here: local node system global system)

2 2. step: Element matrices rod element (truss structure: only tensile and compressive loads, no bending moments) s N L,v L r l,e,a r, s: local coordinate system N R,v R matrix of element displacements (nodal displacements) ṽ vl v R matrix of nodal forces NL S N R element stiffness matrix (see Exercis # 4) S K ṽ NL N R EA l K EA l vl v R! l may be different for different elements! l 1 a l 2 2a 3. step: Transformation here: 2 global degrees of freedom u x,u y for each node (no rotational dof, truss structure!) extension of the element stiffness matrix

3 s Q R,w R N L,v L Q L,w L r l,e,a N R,v R N L Q L EA N R l Q R }{{} } 0 0 {{ 0 0 } S K v L w L v R w R }{{} v Q R, Q L are always equal to zero because of truss structure. The additional dof w R, w L are needed for the transformation. Transformation w R u jy v R u jx α j j i u jx v R cos α j w R sin α j u jy v R sin α j + w R cos α j u ix u iy u jx u jy }{{} u cosα i sin α i 0 0 sin α i cos α i cosα i sin α i 0 0 sinα i cos α i }{{}}{{} u T T v T T v L w L v R w R v ( T T is orthogonal, i.e.t 1 T T! ) same for nodal forces S T T S

4 Transformed stiffness relation: S K v (a) û T T v T 1 v v T û (b) from (a) and (b): S K T û T T S T T K T û Ŝ T T K T û Ŝ T T S Ŝ ˆK û with ˆK T T K T S ix form of Ŝ : Ŝ S iy S jx S jy We have now u ix u u iy u jx u jy ui u j displacements of nodes i and j in global coordinates and for each element Six e S e S e iy Sjx e Sjy e S e i S e j nodal forces at element e (between nodes i and j) S e K e u S e i S e j }{{} S e K e ii K e ji K e ij K e jj } {{ } K e ui u j }{{} u ( ) 4. step: Assembly of Global System balances of forces at each node

5 5 y Ŝ 5 4 x 4 Ŝ F 4x Ŝ e.g. node 4: F 4y here: F 4x F 4y 0 Fx 0 : F 4x S 4 4x S 5 4x S 7 4x 0 Fy 0 : F 4y S 4 4y S 5 4y S 7 4y 0 Ŝ4 4 +Ŝ5 4 +Ŝ7 4 ˆf 4 ( ) general: ˆf i e Ŝ e i Expressing nodal forces through displacements by using (*) e.g. element 4 between node 2 and 4 Ŝ4 2 Ŝ 4 4 or more explicit Ŝ 4 2x Ŝ 4 2y Ŝ 4 4x Ŝ 4 4y ˆK4 22 ˆK 4 42 EA a ˆK 4 24 ˆK 4 44 û 2 û partitioning û 2x û 2y û 4x û 4y Ŝ4 4 ˆK 4 42 û 2 + ˆK 4 44 û 4 similarly for the other nodal forces in (**): ˆK 4 42 û 2 + ˆK 4 44 û 4 + ˆK 5 43 û 3 + ˆK 5 44 û 4 + ˆK 7 44 û 4 + ˆK 7 45 û 5 ˆf 4 ˆK 4 42 û 2 + ˆK ( 5 43 û 3 + ˆK ˆK ˆK ) 7 44 û 4 + ˆK 7 45 û 5 ˆf 4 similarly for all nodes.

6 In matrix form for the 5 nodes: ˆK ˆK 1 12 ˆK 2 13 û 1 ˆf 1 ˆK 2 11 ˆK ˆK 3 22 ˆK 1 21 ˆK 3 23 ˆK 4 24 û 2 ˆf 2 + ˆK 4 22 ˆK 2 31 ˆK 2 32 ˆK ˆK 3 33 ˆK 5 34 ˆK 6 35 û 3 ˆf 3 + ˆK ˆK 6 33 ˆK ˆK 5 44 ˆK 4 42 ˆK 5 43 ˆK 7 45 û 4 ˆf 4 + ˆK 7 44 ˆK ˆK 6 53 ˆK 7 54 ˆK 7 55 û 5 ˆf 5 placement of the submatrices corresponds to the indices of the nodes. assembly of the global stiffness matrix can be carried out quite easily. 1) Partitioning of the element stiffness matrix according to the nodes e.g. element 1 ˆK 1 EA a ˆK1 11 ˆK ) Adding the submatrices onto the corresponding position of the global stiffness matrix ˆK 1 11 onto K (1,1), ˆK1 12 onto K (1,2) etc. ˆK 1 12 ˆK step: Boundary conditions system of equations

7 u 1x F 1x u 1y F 1y u 2x F 2x u 2y F 2y K u 3x u 3y F 3x F 3y u 4x F 4x u 4y F 4y u 5x F 5x u 5y F 5y boundary coonditon s: u 1x u 1y u 2x u 2y 0 F 1x,F 1y,F 2x,F 2y : unknown forces loads: F 3x F 3y F 4x F 4y F 5x 0, F 5y F u 3x,u 3y,u 4x,u 4y,u 5x,u 5y : unknown displacements Either the displacements or the forces are given, the corresponding other variable is unknown. General form: Kuu K uf K Fu K FF } {{ } K u ū f f ū, f : given (known) displacements/forces u, f : unknown variables 6. step: Solution 1. Determination of the unknown displacements K uu u f K uf ū for homogeneous boundary conditions (ū 0) K uu u f In most cases, only the displacements are of particular interest, that s why the other rows and columns are usually cancelled or not formulated from the beginning.

8 u 3x 0 u 3y 0 K uu u 4x u 4y 0 0 u 5x 0 u 5y F If the unknown forces are to be determined: K Fu u + K FF ū f withufrom solution above for homogeneous boundary conditions K Fu u f Building up the (reduced) stiffness matrix of the system by use of the index table 1. Numbering of the degrees of freedom (dof) and the real dof u 1x u 1y u 2x u 2y u 3x u 3y u 4x u 4y u 5x u 5y system degrees of freedom real degrees of freedom Originally, the system has 10 dof (5 nodes with 2 displacements each). But 4 dof are blocked by the boundary conditions, so that 6 real dof remain. This is also the size of the reduced system matrix.

9 2. Formulation of the element index tables (each element has 4 local dof) element 1 element 2 1 u 1x u 1x u 1y u 1y u 2x u 3x u 2y u 3y 6 2 element 3 element 4 1 u 2x u 2x u 2y u 2y u 3x u 4x u 3y u 4y 8 4 element 5 1 u 3x u 3y u 4x u 4y 8 4. The element stiffness matrix is of the form: ˆK e (1, 1) ˆKe (1, 2) ˆKe (1, 4) ˆK e ˆK e (2, 1) ˆK e (4, 1) ˆKe (4, 4) The elements of the element stiffness matrices are now added onto the system stiffness matrix according to the index tables

10 ˆK ˆK 1 12 ˆK 2 13 û 1 ˆf 1 ˆK 2 11 ˆK ˆK 3 22 ˆK 1 21 ˆK 3 23 ˆK 4 24 û 2 ˆf 2 + ˆK 4 22 ˆK 2 31 ˆK 2 32 ˆK ˆK 3 33 ˆK 5 34 ˆK 6 35 û 3 ˆf 3 + ˆK ˆK 6 33 ˆK ˆK 5 44 ˆK 4 42 ˆK 5 43 ˆK 7 45 û 4 ˆf 4 + ˆK 7 44 ˆK ˆK 6 53 ˆK 7 54 ˆK 7 55 û 5 ˆf 5