Discussion Closed This discussion was created more than 6 months ago and has been closed. To start a new discussion with a link back to this one, click here.
how to apply range of loads with in fixed time
Posted 2010/04/15 3:30 GMT-4 1 Reply
Please login with a confirmed email address before reporting spam
hai all
I want to apply range of loads with in 1second that means
1MN with in 1sec
2MN with in 1sec
similarly up to 30MN.
Actually I tried in different way by using the parametric solver
but it gives the syntax error
error: 9171
expression: {range(1MN*(t<=1),1MN*(t<=1),30MN*(t<=1))}
subexpression
column: 53
property name: parameter values.
if any one knows the correct expression, please message me
Nagamalli.
I want to apply range of loads with in 1second that means
1MN with in 1sec
2MN with in 1sec
similarly up to 30MN.
Actually I tried in different way by using the parametric solver
but it gives the syntax error
error: 9171
expression: {range(1MN*(t<=1),1MN*(t<=1),30MN*(t<=1))}
subexpression
column: 53
property name: parameter values.
if any one knows the correct expression, please message me
Nagamalli.
1 Reply Last Post 2010/04/16 1:25 GMT-4
Please login with a confirmed email address before reporting spam
Posted:
1 decade ago
Hi
I beleive you should separate time and load, I mean by that you define a function of time for the load, and you select the time step in the transient solver separately. In this way you have two issues to secure, but they are simpler, in my opinion.
try a :
t=range(start, step, stop)
in the transient solver, by adjusting or using several ranges such to nicely follow the changes, especially where you have steep slopes and discontinuities
for the load you make either a function and refer to this function of time, or you use directly the formula in the gui field something as you have written
F = 1[MN] + (30-1)[MN]/(1[s])*t*(t<1) + (30-1)[MN]*(t>1)
this should ramp the load from 1 to 30 over 1 second with a slope of (30-1) and then reamin at 30-1+1=30 for t>1 (if I havnt missed something. Note the verbous use of () for the /(xx[units]) with UNITS it's quickly done to get it wrong if you forget soem brackets.
By the way Functions fo not yet accept units in V3.5.a if I remember well ;)
As you have some abrupt discontinuities there you might get some convergence issues around these, that means then you need to "smoothen" the transition i.e. witha heavyside or a cosine function
I beleive you should separate time and load, I mean by that you define a function of time for the load, and you select the time step in the transient solver separately. In this way you have two issues to secure, but they are simpler, in my opinion.
try a :
t=range(start, step, stop)
in the transient solver, by adjusting or using several ranges such to nicely follow the changes, especially where you have steep slopes and discontinuities
for the load you make either a function and refer to this function of time, or you use directly the formula in the gui field something as you have written
F = 1[MN] + (30-1)[MN]/(1[s])*t*(t<1) + (30-1)[MN]*(t>1)
this should ramp the load from 1 to 30 over 1 second with a slope of (30-1) and then reamin at 30-1+1=30 for t>1 (if I havnt missed something. Note the verbous use of () for the /(xx[units]) with UNITS it's quickly done to get it wrong if you forget soem brackets.
By the way Functions fo not yet accept units in V3.5.a if I remember well ;)
As you have some abrupt discontinuities there you might get some convergence issues around these, that means then you need to "smoothen" the transition i.e. witha heavyside or a cosine function