Civil Engineering Reference
In-Depth Information
There are 12 equations and the skyline storage is 38
Node Translation Rotation
1 0.8559E+00 -0.1148E+00
2 0.6263E+00 -0.1147E+00
3 0.3971E+00 -0.1145E+00
4 0.1683E+00 -0.1143E+00
5 -0.6008E-01 -0.1141E+00
6 -0.2883E+00 -0.1141E+00
Element Force Moment Force Moment
1 0.1000E+01 0.3638E-11 -0.7036E+00 0.1688E+01
2 0.7036E+00 -0.1688E+01 -0.8960E-01 0.2436E+01
3 0.8960E-01 -0.2436E+01 0.4757E+00 0.1973E+01
4 -0.4757E+00 -0.1973E+01 0.6271E+00 0.7640E+00
5 -0.6271E+00 -0.7640E+00 -0.9095E-12 -0.9095E-12
Figure 4.18
Results from second Program 4.3 example
element composite stiffness
km+mm
gives the element “actions” which holds the element
shear forces and moments.
The pile is supported by the foundation only, so no additional boundary conditions
are needed (
nr=0
). In this example, a unit horizontal load has been applied to the top of
the pile.
The computed results in Figure 4.18 show the horizontal translation and rotation at each
of the nodes. It is seen that the horizontal translation of node 1 is given as 0.856 which is
in reasonable agreement with the analytical solution to this problem from Hetenyi (1946).
Program 4.4 Analysis of elastic rigid-jointed frames using 2-node beam/rod elements
in two or three dimensions.
PROGRAM p44
!-------------------------------------------------------------------------
! Program 4.4 Analysis of elastic rigid-jointed frames using 2-node
! beam/rod elements in 2- or 3-dimensions.
!-------------------------------------------------------------------------
USE main; IMPLICIT NONE
INTEGER,PARAMETER::iwp=SELECTED_REAL_KIND(15)
INTEGER::fixed_freedoms,i,iel,k,loaded_nodes,ndim,ndof,nels,neq,nod=2,
&
nodof,nn,nprops,np_types,nr
REAL(iwp)::penalty=1.0e20_iwp,zero=0.0_iwp
!-----------------------dynamic arrays------------------------------------
INTEGER,ALLOCATABLE::etype(:),g(:),g_g(:,:),g_num(:,:),kdiag(:),nf(:,:), &
no(:),node(:),num(:),sense(:)
REAL(iwp),ALLOCATABLE::action(:),coord(:,:),eld(:),gamma(:),g_coord(:,:),&
km(:,:),kv(:),loads(:),prop(:,:),value(:)
!-----------------------input and initialisation--------------------------
OPEN(10,FILE='fe95.dat'); OPEN(11,FILE='fe95.res')
READ(10,*)nels,nn,ndim,nprops,np_types
IF(ndim==2)nodof=3; IF(ndim==3)nodof=6; ndof=nod*nodof
ALLOCATE(nf(nodof,nn),km(ndof,ndof),coord(nod,ndim),g_coord(ndim,nn),
&
eld(ndof),action(ndof),g_num(nod,nels),num(nod),g(ndof),gamma(nels),
&
g_g(ndof,nels),prop(nprops,np_types),etype(nels))
READ(10,*)prop; etype=1; IF(np_types>1)READ(10,*)etype;
IF(ndim==3)READ(10,*)gamma
READ(10,*)g_coord; READ(10,*)g_num
nf=1; READ(10,*)nr,(k,nf(:,k),i=1,nr); CALL formnf(nf); neq=MAXVAL(nf)
ALLOCATE(kdiag(neq),loads(0:neq)); kdiag=0
!-----------------------loop the elements to find global array sizes------
