SFGRAD

[SebasSC]

hi everyone,

i'm trying to use SFGRAD to create a triangular load distribution on a set of lines, but i'm having some trouble.

the code i'm trying to work with is the following

SFGRAD,PRES,,Y,0,15000/7.645
LSEL,S,,,1,2,1
SFL,ALL,PRES,0,15000

where 7.645 is the length of L1+L2

in the attachment is a sketch of the problem

 

[rickfischer51]

What is the trouble?

I'll bet there is no load applied to the line. Looks like the SFTRAN command is required to actually transfer the loads from the line to the elements on the line.

Rick Fischer
Principal Engineer
Argonne National Laboratory

 

[SebasSC]

Thanks for your answer rick.

The first thing that troubles me, is that i'm not sure if the load is well defined, since the load list gives me the VALJ value for line 2 as the maximum value of the triangular load, when i think it was supposed to be half of it (am i right?)

LINE LOAD LABEL VALI VALJ VAL2I VAL2J
1 PRES 0.000 0.1500E+05
REFERENCE NUMBER FOR SLOPE COORDINATE SYSTEM= 0
SLOPE DIRECTION= Y
COORDINATE LOCATION WHERE SLOPE IS ZERO= 0.00000000
SLOPE VALUE (LOAD PER UNIT LENGTH)= 1962.06671
2 PRES 0.000 0.1500E+05
REFERENCE NUMBER FOR SLOPE COORDINATE SYSTEM= 0
SLOPE DIRECTION= Y
COORDINATE LOCATION WHERE SLOPE IS ZERO= 0.00000000
SLOPE VALUE (LOAD PER UNIT LENGTH)= 1962.06671

The second problem is that when i try to solve, it gives me a warning and fails to give a solution.

Resuming, i'm probably using the SFGRAD command in a wrong way.
So, my question is basically how would you proceed to apply such kind of load?

 

[rickfischer51]

What is the warning? How does it fail? Does it run successfully and give zero results? Does it terminate the job without solving?

I'm not sure what you are trying to accomplish. I've never used SFL before, and it doesnt make sense to me. What is your model? Is it lines, or do you have surfaces? What elements? Beams, shells, or ?

Rick Fischer
Principal Engineer
Argonne National Laboratory

 

[SebasSC]

it terminates the job without solving.

I've attached a skecth of the problem in the first post, it's a "box" with thin walls. I have surfaces and i'm using shell93 elements.
I'm trying SFL because it seemed to be the best option for what i need, which is to apply a triangular distributed load along one of the edges of the given geometry.

I'm not a very experienced user of ANSYS, so it's possible that i might be doing some dumb mistake

 

[rickfischer51]

Is that a pressure load or a force load? I'm having problems with a pressure load applied to a line, with has no area.

Rick Fischer
Principal Engineer
Argonne National Laboratory

 

[SebasSC]

In reality that is a force load, as in fact it's not applied to an area, but as far as i know, ansys considers any distributed loads as pressure loads (applied or not to areas), correct me if i'm wrong.

 

[ASMS81]

Did you unselect the lines (ALLSEL command) after applying tbe force and before solve the model?

 

[rickfischer51]

I think that in order to apply a distributed force (force per length) you would need a beam element along the line. You have shell elements, and I believe Ansys will interpret the load as a pressure. If you had a row of shell elements along the line that was one unit wide, then you could get something close to what you want.

If I understand this correctly, you want to apply forces to the nodes along a line the vary linearly with position along that line. One way to do this would be with a table. Try this:

/prep7
block,0,7,0,7.645,0,12
vdele,all
et,1,93
r,1,.25
esize,1
amesh,all
ex,1,29e6
nuxy,1,.3
*dim,forces,table,2,,,y,,,0
forces(1,0)=0,7.645
forces(1,1)=0,15000
nsel,s,loc,x,0
nsel,r,loc,z,12
f,all,fx,%forces%
nsel,s,loc,z,0
d,all,all,0
allsel
/solu
solve
/post1
set,last
flist


Rick Fischer
Principal Engineer
Argonne National Laboratory

 

[rickfischer51]

Scratch that. The total force will vary with mesh density. My guess is you want the total force to be independent of mesh density.

Rick Fischer
Principal Engineer
Argonne National Laboratory

 

[SebasSC]

so you applied the loads directly on the nodes. that can be a good ideia.

Does the total force varies with the mesh density? even with this:

forces(1,0)=0,7.645
forces(1,1)=0,15000

doesn't this assure that the load varies between 0 and 15000 on the scope from 0 to 7.645?

 

[rickfischer51]

It will apply a load to every node based only on the node's position, not on how close it is to the next node. Is the node spacing constant on that edge?

Rick Fischer
Principal Engineer
Argonne National Laboratory

 

[SebasSC]

i can't be sure of that.

that is way i was trying to apply the load on the solid model, instead of on the finite element model. Then i thought that on the lines would be the best option

 

[rickfischer51]

Take a look at the SURF156 line load effect element. I think it may work for you.

Rick Fischer
Principal Engineer
Argonne National Laboratory

 

[SebasSC]

but if it applies the load on the node based on its position, it should work, because the value of the load on each node is determined based on the position of the node, am i right?

 

[rickfischer51]

I think this is what you want. Check reactions at the fixed BCs to make sure its's correct.

/prep7
block,0,7,0,7.645,0,12
vdele,all
et,1,93
r,1,.25
esize,1
amesh,all
ex,1,29e6
nuxy,1,.3
et,2,156,,1,,,1
local,11,0
esys,11
k1=kp(0,0,12)
k2=kp(0,7.645,12)
ksel,s,kp,,k1
ksel,a,kp,,k2
lslk,s,1
type,2
lmesh,all
w=15000/7.645
*dim,forces,table,2,,,y
forces(1,0)=0,7.645
forces(1,1)=0,w
esel,s,type,,2
sfe,all,1,pres,,%forces%
nsel,s,loc,z,0
d,all,all,0
allsel
/solu
solve

Rick Fischer
Principal Engineer
Argonne National Laboratory