Category : Science and Education
Archive   : FELT20.ZIP
Filename : DYNFRAME.FLT

problem description
title="dynamic frame analysis" nodes=8 elements=9 analysis=transient

analysis parameters
beta=0.25 gamma=0.5 alpha=0.0 duration=0.8 dt=0.05
nodes=[8] dofs=[Tx] mass-mode=lumped

nodes
1 x=0 y=0 constraint=fixed
2 x=360 y=0
3 x=0 y=180 constraint=free force=f1
4 x=360
5 x=0 y=300 force=f2
6 x=360
7 x=0 y=420 force=f3
8 x=360

beam elements
1 nodes=[1,3] material=wall_bottom
2 nodes=[3,5] material=wall_top
3 nodes=[5,7]
4 nodes=[7,8] material=floor_top load=top_wt
5 nodes=[5,6] material=floor_bottom load=bottom_wt
6 nodes=[3,4] load=bottom_wt
7 nodes=[8,6] material=wall_top
8 nodes=[6,4]
9 nodes=[4,2] material=wall_bottom

material properties
wall_bottom A=13.2 Ix=249 E=30e6 rho=0.0049
wall_top A=6.2 Ix=107 E=30e6 rho=0.0104
floor_top A=12.3 Ix=133 E=30e6 rho=0.01315
floor_bottom A=24.7 Ix=237 E=30e6 rho=0.0136

distributed loads
top_wt direction=perpendicular values=(1,-62.5) (2,-62.5)
bottom_wt direction=perpendicular values=(1,-130) (2,-130)

forces
/*
f1 Fx=1000*(t < 0.2 ? 25*t : 5)
f2 Fx=800*(t < 0.2 ? 25*t : 5)
f3 Fx=500*(t < 0.2 ? 25*t : 5)
*/
/*
f1 Fx=(0.0, 0) (0.2, 5000) (0.8, 5000)
f2 Fx=(0.0, 0) (0.2, 4000) (0.8, 4000)
f3 Fx=(0.0, 0) (0.2, 2500) (0.8, 2500)
*/
/*
f1 Fx=(0.0, 0) (0.4, 5000)+
f2 Fx=(0.0, 0) (0.4, 4000)+
f3 Fx=(0.0, 0) (0.4, 2500)+
*/
f1 Fx = 5000*fmod(t, 0.4)
f2 Fx = 4000*fmod(t, 0.4)
f3 Fx = 2500*fmod(t, 0.4)

constraints
fixed Tx=c Ty=c Rz=c
free Tx=u Ty=u Rz=u

end