Axi-symmetric model

Discussion:

Axi-symmetric model

Nabil Nagy

2007-04-27 09:03:23 UTC

Dear All

MY question is about the loads in axisymmetric model

when I use an axisymmetric model for a symmetrical concrete structure , i just can draw half of the structure around the axis of symmetry , so the axis of symmetry will be on the left hand side of the model

so, in this case if i need to apply a concentrated load on the middle point of this structure ( which lies on the axis of symmetry ) shall i apply only the half value of the actually load or all the load magnitude ?

as i just draw the half of the structure on basis that the axi-symmetry model will integrate for all the structure , is this the same for the load ? or should i apply all the load value ?

Thanks

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Dave Lindeman

2007-04-27 14:50:54 UTC

Permalink

Always apply the full load in an axisymmetric model.

Regards,

Dave

=============================================
Dave Lindeman ***@mmm.com
Lead Research Specialist (TEL) 651-733-6383
3M Company (FAX) 651-736-7615
=============================================

----- Original Message -----
From: Nabil Nagy
To: ***@yahoogroups.com
Sent: Friday, April 27, 2007 4:03 AM
Subject: [ABAQUS] Axi-symmetric model

Dear All

MY question is about the loads in axisymmetric model

when I use an axisymmetric model for a symmetrical concrete structure , i just can draw half of the structure around the axis of symmetry , so the axis of symmetry will be on the left hand side of the model

so, in this case if i need to apply a concentrated load on the middle point of this structure ( which lies on the axis of symmetry ) shall i apply only the half value of the actually load or all the load magnitude ?

as i just draw the half of the structure on basis that the axi-symmetry model will integrate for all the structure , is this the same for the load ? or should i apply all the load value ?

Thanks

---------------------------------
Ahhh...imagining that irresistible "new car" smell?
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[Non-text portions of this message have been removed]

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Robert Paynter

2007-04-30 11:34:29 UTC

Permalink

Remember it's "axisymmetric", not "symmetric", the structure
hasn't been cut in half you are looking at a segment of the
cylindrical region.
When you apply loads they are integrated around the axis.
The only places you can apply concentrated loads and they
mean only that value is on the axis (at the centre). At
other radii you have to remember things are integrated, for
example if you apply a pressure (P) on a horizontal surface
from the axis out to, say, radius R the the overall vertical
force from that pressure would be P*pi*R^2. If you apply
a concentrated vertical force at a finite radius then you need
to multiply by 2*pi*R to find its total force on the structure
- it's a circular line of force, just as in plane strain analysis
it really represents the load, per unit depth, along an infinitely
deep line load.

I hope I haven't confused things - check the theory and analysis
manuals for specific details.

Robert
Oxford, UK

Post by Nabil Nagy
so, in this case if i need to apply a concentrated load on the

middle point of this structure ( which lies on the axis of symmetry )
shall i apply only the half value of the actually load or all the load
magnitude ?

Ravindran Manoharan

2007-05-01 14:12:35 UTC

Permalink

Hi robert,
Could i check something with you. If say in my case i have an axisymmetric part, and i would like to impose a total concentrated load of 1kN which will be equally distributed across 5 nodes (at different radii from the axis of symmetry). If i were to impose 1000/5=200N at each node this would be incorrect right? I should impose 200N at the node on the axis of symmetry and then for the next node which is 5mm away i should impose 200/2*pi*0.005=6366.19 and the following node which is 10mm away i impose 200/2*pi*0.01=3183.09 etc. Am i right in my assumptions?

Thanking you in advance for your time,

regards,
Ravi

Robert Paynter <***@eng.ox.ac.uk> wrote:
Remember it's "axisymmetric", not "symmetric", the structure
hasn't been cut in half you are looking at a segment of the
cylindrical region.
When you apply loads they are integrated around the axis.
The only places you can apply concentrated loads and they
mean only that value is on the axis (at the centre). At
other radii you have to remember things are integrated, for
example if you apply a pressure (P) on a horizontal surface
from the axis out to, say, radius R the the overall vertical
force from that pressure would be P*pi*R^2. If you apply
a concentrated vertical force at a finite radius then you need
to multiply by 2*pi*R to find its total force on the structure
- it's a circular line of force, just as in plane strain analysis
it really represents the load, per unit depth, along an infinitely
deep line load.

I hope I haven't confused things - check the theory and analysis
manuals for specific details.

Robert
Oxford, UK

Post by Nabil Nagy
so, in this case if i need to apply a concentrated load on the

Dave Lindeman

2007-05-01 17:50:26 UTC

Permalink

I that case you should really use a distributed pressure load.

area = pi*(ro^2 - ri^2)
pressure = force/area

This is much more straightforward and logical. With concentrated nodal loads, consider, for example, what you would have to do if you decided to change your mesh density.

Regards,

Dave

=============================================
Dave Lindeman ***@mmm.com
Lead Research Specialist (TEL) 651-733-6383
3M Company (FAX) 651-736-7615
=============================================

----- Original Message -----
From: Ravindran Manoharan
To: ***@yahoogroups.com
Sent: Tuesday, May 01, 2007 9:12 AM
Subject: Re: [ABAQUS] Re: Axi-symmetric model

Hi robert,
Could i check something with you. If say in my case i have an axisymmetric part, and i would like to impose a total concentrated load of 1kN which will be equally distributed across 5 nodes (at different radii from the axis of symmetry). If i were to impose 1000/5=200N at each node this would be incorrect right? I should impose 200N at the node on the axis of symmetry and then for the next node which is 5mm away i should impose 200/2*pi*0.005=6366.19 and the following node which is 10mm away i impose 200/2*pi*0.01=3183.09 etc. Am i right in my assumptions?

Thanking you in advance for your time,

regards,
Ravi

Robert Paynter <***@eng.ox.ac.uk> wrote:
Remember it's "axisymmetric", not "symmetric", the structure
hasn't been cut in half you are looking at a segment of the
cylindrical region.
When you apply loads they are integrated around the axis.
The only places you can apply concentrated loads and they
mean only that value is on the axis (at the centre). At
other radii you have to remember things are integrated, for
example if you apply a pressure (P) on a horizontal surface
from the axis out to, say, radius R the the overall vertical
force from that pressure would be P*pi*R^2. If you apply
a concentrated vertical force at a finite radius then you need
to multiply by 2*pi*R to find its total force on the structure
- it's a circular line of force, just as in plane strain analysis
it really represents the load, per unit depth, along an infinitely
deep line load.

I hope I haven't confused things - check the theory and analysis
manuals for specific details.

Robert
Oxford, UK

Post by Nabil Nagy
so, in this case if i need to apply a concentrated load on the

middle point of this structure ( which lies on the axis of symmetry )
shall i apply only the half value of the actually load or all the load
magnitude ?

---------------------------------
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[Non-text portions of this message have been removed]

Ravindran Manoharan

2007-05-02 12:54:50 UTC

Permalink

Hi Dave,
A very good point! Thanks for the suggestion, this would be more appropriate and less cumbersome,

cheers mate,

regards,
Ravi

Dave Lindeman <***@mmm.com> wrote:
I that case you should really use a distributed pressure load.

area = pi*(ro^2 - ri^2)
pressure = force/area

This is much more straightforward and logical. With concentrated nodal loads, consider, for example, what you would have to do if you decided to change your mesh density.

Regards,

Dave

=============================================
Dave Lindeman ***@mmm.com
Lead Research Specialist (TEL) 651-733-6383
3M Company (FAX) 651-736-7615
=============================================

----- Original Message -----
From: Ravindran Manoharan
To: ***@yahoogroups.com
Sent: Tuesday, May 01, 2007 9:12 AM
Subject: Re: [ABAQUS] Re: Axi-symmetric model

Hi robert,
Could i check something with you. If say in my case i have an axisymmetric part, and i would like to impose a total concentrated load of 1kN which will be equally distributed across 5 nodes (at different radii from the axis of symmetry). If i were to impose 1000/5=200N at each node this would be incorrect right? I should impose 200N at the node on the axis of symmetry and then for the next node which is 5mm away i should impose 200/2*pi*0.005=6366.19 and the following node which is 10mm away i impose 200/2*pi*0.01=3183.09 etc. Am i right in my assumptions?

Thanking you in advance for your time,

regards,
Ravi

Robert Paynter <***@eng.ox.ac.uk> wrote:
Remember it's "axisymmetric", not "symmetric", the structure
hasn't been cut in half you are looking at a segment of the
cylindrical region.
When you apply loads they are integrated around the axis.
The only places you can apply concentrated loads and they
mean only that value is on the axis (at the centre). At
other radii you have to remember things are integrated, for
example if you apply a pressure (P) on a horizontal surface
from the axis out to, say, radius R the the overall vertical
force from that pressure would be P*pi*R^2. If you apply
a concentrated vertical force at a finite radius then you need
to multiply by 2*pi*R to find its total force on the structure
- it's a circular line of force, just as in plane strain analysis
it really represents the load, per unit depth, along an infinitely
deep line load.

I hope I haven't confused things - check the theory and analysis
manuals for specific details.

Robert
Oxford, UK

Post by Nabil Nagy
so, in this case if i need to apply a concentrated load on the

adarsh dureja

2007-04-30 12:46:45 UTC

Permalink

Hi!!!
The first and foremost thing, you should ensure
that the problem you are solving lies in the domain of
axisymmetry. For a problem to be axisymmetric, you
should have all the following properties

1. The geometry has to be axisymmetric
2. The material properties have to be axisymmteric
3. The boundary conditions have to be axisymmetric
4. The laoding has to be axisymmetric

If any of this criteria fails, you are out of the
domain and you may have to go for 3-D.

In case you satisfy all the above, you need to model
only a typical cross-section and that's it.

Please do not model concentrated loads in an
axisymmetric model otherwise the computer will treat
it as pressure and you will misinterprete the result.

You could do fourier transform to simulate
concentrated loads but you need to read upon it.

With Thanks and Regards,

A.K.Dureja
Research Scholar,
IITB, Mumbai
India

Post by Nabil Nagy
Dear All
MY question is about the loads in axisymmetric
model
when I use an axisymmetric model for a symmetrical
concrete structure , i just can draw half of the
structure around the axis of symmetry , so the axis
of symmetry will be on the left hand side of the
model
so, in this case if i need to apply a concentrated
load on the middle point of this structure ( which
lies on the axis of symmetry ) shall i apply only
the half value of the actually load or all the load
magnitude ?
as i just draw the half of the structure on basis
that the axi-symmetry model will integrate for all
the structure , is this the same for the load ? or
should i apply all the load value ?
Thanks
---------------------------------
Ahhh...imagining that irresistible "new car" smell?
Check outnew cars at Yahoo! Autos.
[Non-text portions of this message have been
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Ravindran Manoharan

2007-04-30 20:35:00 UTC

Permalink

Hi Adarsh,
I don't get what you mean by this statement

"Please do not model concentrated loads in an axisymmetric model otherwise the computer will treat it as pressure and you will misinterprete the result."

What if i need to specify a concentrated load on the axisymm. model. Say for example if we have an object with 4 elements extending from the axis of symmetry (horizontally). The BC at the extreme left element has been specified as X-Symm and i want to impose a unifrom concentrated loads at the top nodes of each element, would this be wrong? I agree with Robert in that to specify uniform concentrated load at each node, one needs to take into account the radius from the axis of symm. in order to impose the right magnitude of load.

If you could enlighten me on this it would be much appreciated,

regards,
Ravi
adarsh dureja <***@yahoo.com> wrote:
Hi!!!
The first and foremost thing, you should ensure
that the problem you are solving lies in the domain of
axisymmetry. For a problem to be axisymmetric, you
should have all the following properties

1. The geometry has to be axisymmetric
2. The material properties have to be axisymmteric
3. The boundary conditions have to be axisymmetric
4. The laoding has to be axisymmetric

If any of this criteria fails, you are out of the
domain and you may have to go for 3-D.

In case you satisfy all the above, you need to model
only a typical cross-section and that's it.

Please do not model concentrated loads in an
axisymmetric model otherwise the computer will treat
it as pressure and you will misinterprete the result.

You could do fourier transform to simulate
concentrated loads but you need to read upon it.

With Thanks and Regards,

A.K.Dureja
Research Scholar,
IITB, Mumbai
India

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adarsh dureja

2007-05-01 12:03:17 UTC

Permalink

Hi Ravi,
Please correct me, if I have not understood
you correctly.

The axis of symmetry is vertical (2) and the object
you are trying to model is kind of a disk. If this is
correct, what is the need to impose X-symmetric BC on
the left edge as axisymmetric elements would
automatically mean that. As far as loading is
concerned, I was really talking about radial load and
not in-plane load.

Adarsh.

--- Ravindran Manoharan

Post by Ravindran Manoharan
Hi Adarsh,
I don't get what you mean by this statement
"Please do not model concentrated loads in an
axisymmetric model otherwise the computer will treat
it as pressure and you will misinterprete the
result."
What if i need to specify a concentrated load on the
axisymm. model. Say for example if we have an object
with 4 elements extending from the axis of symmetry
(horizontally). The BC at the extreme left element
has been specified as X-Symm and i want to impose a
unifrom concentrated loads at the top nodes of each
element, would this be wrong? I agree with Robert in
that to specify uniform concentrated load at each
node, one needs to take into account the radius from
the axis of symm. in order to impose the right
magnitude of load.
If you could enlighten me on this it would be much
appreciated,
regards,
Ravi
Hi!!!
The first and foremost thing, you should ensure
that the problem you are solving lies in the domain
of
axisymmetry. For a problem to be axisymmetric, you
should have all the following properties
1. The geometry has to be axisymmetric
2. The material properties have to be axisymmteric
3. The boundary conditions have to be axisymmetric
4. The laoding has to be axisymmetric
If any of this criteria fails, you are out of the
domain and you may have to go for 3-D.
In case you satisfy all the above, you need to model
only a typical cross-section and that's it.
Please do not model concentrated loads in an
axisymmetric model otherwise the computer will treat
it as pressure and you will misinterprete the
result.
You could do fourier transform to simulate
concentrated loads but you need to read upon it.
With Thanks and Regards,
A.K.Dureja
Research Scholar,
IITB, Mumbai
India

axis

Post by Nabil Nagy
of symmetry will be on the left hand side of the
model
so, in this case if i need to apply a concentrated
load on the middle point of this structure ( which
lies on the axis of symmetry ) shall i apply only
the half value of the actually load or all the

load

Post by Nabil Nagy
magnitude ?
as i just draw the half of the structure on basis
that the axi-symmetry model will integrate for all
the structure , is this the same for the load ? or
should i apply all the load value ?
Thanks
---------------------------------
Ahhh...imagining that irresistible "new car"

smell?

Post by Nabil Nagy
Check outnew cars at Yahoo! Autos.
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Nabil Nagy

2007-05-01 18:29:27 UTC

Permalink

Hi
I think that there is no need for axisymmetric BC for the nodes on the axis of axisymmetry but you will need to apply displace. BC as U1 =0.0 on these nodes. (SO YOU NEED TO PUT AXI BC !!! Edit : BenZ)

NN

adarsh dureja <***@yahoo.com> wrote:
Hi Ravi,
Please correct me, if I have not understood
you correctly.

The axis of symmetry is vertical (2) and the object
you are trying to model is kind of a disk. If this is
correct, what is the need to impose X-symmetric BC on
the left edge as axisymmetric elements would
automatically mean that. As far as loading is
concerned, I was really talking about radial load and
not in-plane load.

Adarsh.

--- Ravindran Manoharan

axis

load

smell?

Post by Nabil Nagy
Check outnew cars at Yahoo! Autos.
[Non-text portions of this message have been
removed]

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protection around
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Ravindran Manoharan

2007-05-02 12:53:14 UTC

Permalink

Hi Adarsh,
I see your point, ok. Thanks for the response.

best regards,
Ravi

adarsh dureja <***@yahoo.com> wrote:
Hi Ravi,
Please correct me, if I have not understood
you correctly.

The axis of symmetry is vertical (2) and the object
you are trying to model is kind of a disk. If this is
correct, what is the need to impose X-symmetric BC on
the left edge as axisymmetric elements would
automatically mean that. As far as loading is
concerned, I was really talking about radial load and
not in-plane load.

Adarsh.

--- Ravindran Manoharan

axis

load

smell?

Post by Nabil Nagy
Check outnew cars at Yahoo! Autos.
[Non-text portions of this message have been
removed]

__________________________________________________
Do You Yahoo!?
Tired of spam? Yahoo! Mail has the best spam
protection around
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