Next 1. Introduction 1. Introduction (cont'd) 1. Introduction (cont'd) 1. Introduction (cont'd) 1. Introduction (cont'd) 1. Introduction (cont'd) 1. Introduction (cont'd) 1. Introduction (cont'd) 2. Rayleigh-Ritz procedure 2. Rayleigh-Ritz procedure (cont'd) 2. Rayleigh-Ritz procedure (cont'd) 2. Rayleigh-Ritz procedure (cont'd) 2. Rayleigh-Ritz procedure (cont'd) 2. Rayleigh-Ritz procedure (cont'd) 2. Rayleigh-Ritz procedure (cont'd) 2. Rayleigh-Ritz procedure (cont'd) 2. Rayleigh-Ritz procedure (cont'd) 3. A simple element analysis 3. A simple element analysis (cont'd) 3. A simple element analysis (cont'd) 3. A simple element analysis (cont'd) 3. A simple element analysis (cont'd) 3. A simple element analysis (cont'd) 3. A simple element analysis (cont'd) 3. A simple element analysis (cont'd) 3. A simple element analysis (cont'd) 3. A simple element analysis (cont'd) 4. Higher-order elements 4. Higher-order elements (cont'd) 5. Assembly of system equation 5. Assembly of system equation (cont'd) 5. Assembly of system equation (cont'd) 33 of 47 5. Assembly of system equation (cont'd) 6.1 Static problem 6.2 Undamped dynamic problem 6.2 Undamped dynamic problem (cont'd) 6.2 Undamped dynamic problem (cont'd) 6.2 Undamped dynamic problem (cont'd) 6.3.1 Modal analysis 6.3.1 Modal analysis (cont'd) 6.3.2 Time-domain analysis 6.3.2 Time-domain analysis (cont'd) 6.3.2 Time-domain analysis (cont'd) 6.3.2 Time-domain analysis (cont'd) 6.3.2 Time-domain analysis (cont'd) 6.3.3 Complex-valued analysis

5. Assembly of system equation (cont'd)

The system boundary conditions now consist of prescribed values for some of the \(w_i\). One way to handle these constrained \(w_i\) is to remove them from the vector of unknowns – with appropriate node numbering, \(\mathbf w\) can be partitioned into two parts, the unknown displacements \(\mathbf w _1\) and the prescribed values \(\mathbf w _2\). Partitioning \(\mathbf S\) and \(\mathbf f\) correspondingly, we obtain
S 11 S 12 S 21 S 22 w 1 w 2 = f 1 f 2 . (Eqn. 13)

This then gives
S 11 w 1 = f 1 S 12 w 2 . (Eqn. 14)

The system to be solved is now smaller.


Start Next

R. Funnell
Last modified: 2018-11-07 12:52:06

Slide show generated from fem.html by Weasel 2018 Nov 7