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Finite Element Analysis in Structural Mechanics

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en:example:dyn:1:2

Table of Contents

Semi-loofs, BEAM61D1

Problem

Mesh

element type nodes
1 61 1 2 4 3 11 13 14 12 
2 61 3 4 6 5 14 16 17 15
3 61 5 6 8 7 17 19 20 18
4 61 7 8 10 9 20 22 23 21

Boundary conditions

$u=v=w=0$ at nodes 1, 2
$u=v=w=\alpha=\beta=0$ at node 11

Solution

beam61d1.bat
rmd3 beam61d1.i1
rpd3 beam61d1.i2
srh3 beam61d1.i3
fefs beam61d1.i4
hmot beam61d1.iM
heig beam61d1.iE
hfrq beam61d1.iF

Input

beam61d1.i1
;  NELEM NNOD ITED
IP   4    23   61
;  CRIT SCALE THDEF
RP 1.01   1    0.01
XY N 1:10 X 2*0 2*0.25 2*0.5 2*0.75 2*1
          Y 5*(0 0.02)
          Z 10*0
EL E 1:4 N =A 1 2 4 3 =A =B 11 13 14 12 =B
           =2A =3B =4A =6B =6A =9B
EN
EN
beam61d1.i2
;  KREST
IP   1
         ; E    α ν   ρ
MP 1 T 1 V 2e11 0 0.3 7800
AS 1 /M 1 /B 0 N 1 2 11
EN
EN
beam61d1.i3
;  KREST
IP   1
EN
EN
beam61d1.i4
;  KREST
IP   1
EN
EN
beam61d1.iM
;  KDIAG KPRIN
IP   1     0
EN
EN
beam61d1.iE
;  KREST NROOT NITERX KTPR KEVP
IP   1     10     0     0    0
EN
EN
beam61d1.iF
;  KPRIN KOUT IVL IVU
IP   0     0   0   0
EN
EN

Output

beam61d1.oF
;  KPRIN KOUT IVL IVU                                                           
IP   0     0   0   0                                                            
EN                                                                              
EN                                                                              


                              FREQUENCIES [1/s]
  0.821066E+01  0.167457E+02  0.516315E+02  0.112647E+03  0.147909E+03
  0.312764E+03  0.360566E+03  0.552916E+03  0.706114E+03  0.928464E+03

                                * END OF HFRQ *
 TOTAL CPU: 00:00:00
en/example/dyn/1/2.txt · Last modified: 2022-03-15 08:28 by Petr Pařík