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[STRUC]how ElasticStress is calculated considering ThermalSt
 
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summer.shen
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Joined: 14 May 2014
Posts: 53
Location: Shanghai, China

PostPosted: Wed May 21, 2014 7:26 am  Reply with quote

Dear experts,

This question is stimulated by the one I post a few days ago (unfortunately no one has answered it yet, hope this one will get some answer, at least raise discussion):
[STRUC]Reference Temperature in phase transformation (http://www.xansys.org/forum/viewtopic.php?t=25669)

Trying to understand what happens during the calculation of thermal strain, plastic strain and elastic strain/stress, I checked the ANSYS APDL Theory Reference > Structures with Material Nonlinearities > Rate-Independent Plasticity>Implementation.
It shows that the iteration steps of calculating elastic strain (elastic strain), plastic strain. As I cannot post a screenshot here or type beautiful equations here, could you please spent one minute reading about 20 lines of this problem? (Moderators, can I insert images here? 'Cause I see the Img button but failed to insert one)

And please notice that, in the first few lines - second step to calculate trial strain - it says that the thermal effect is ignored, which really puzzles me. And I cannot find any other steps implementing thermal strain. Here is my question:
In thermal-structural analysis, thermal strain is surely considered, if it is ignored all the procedure then it of course has no play in the modeling, but when and where exactly is it implemented during the calculation of elastic strain and plastic strain?

Looking forward to your reply, your insight will be appreciated.

Regards,
Summer Shen
Shanghai Jiao Tong University, Shanghai, China[/url]
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ayo.brimmo
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Joined: 01 Jul 2013
Posts: 38
Location: Masdar City, Abu Dhabi

PostPosted: Wed May 21, 2014 8:35 am  Reply with quote

Looking at the fundamental stress vector equation (equation4-22), the elastic strain vector term actually takes into account the thermal strain vector. Look through the first few pages of the "structures" section in the theory guide. You would see the fundamental equation in the expanded form there.

Best,
Ayo Brimmo
Masdar
UAE

-----Original Message-----
From: Xansys [mailto:xansys-bounces@xansys.org] On Behalf Of summer.shen
Sent: Wednesday, May 21, 2014 6:26 PM
To: xansys@xansys.org
Subject: [Xansys] [STRUC]how ElasticStress is calculated considering ThermalSt

Dear experts,

This question is stimulated by the one I post a few days ago (unfortunately no one has answered it yet, hope this one will get some answer, at least raise discussion):
[STRUC]Reference Temperature in phase transformation (http://www.xansys.org/forum/viewtopic.php?t=25669)

Trying to understand what happens during the calculation of thermal strain, plastic strain and elastic strain/stress, I checked the ANSYS APDL Theory Reference > Structures with Material Nonlinearities > Rate-Independent Plasticity>Implementation.
It shows that the iteration steps of calculating elastic strain (elastic strain), plastic strain. As I cannot post a screenshot here or type beautiful equations here, could you please spent one minute reading about 20 lines of this problem? (Moderators, can I insert images here? 'Cause I see the Img button but failed to insert one)

And please notice that, in the first few lines - second step to calculate trial strain - it says that the thermal effect is ignored, which really puzzles me. And I cannot find any other steps implementing thermal strain. Here is my question:
In thermal-structural analysis, thermal strain is surely considered, if it is ignored all the procedure then it of course has no play in the modeling, but when and where exactly is it implemented during the calculation of elastic strain and plastic strain?

Looking forward to your reply, your insight will be appreciated.

Regards,
Summer Shen
Shanghai Jiao Tong University, Shanghai, China[/url]






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james.kosloski
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Joined: 11 Mar 2011
Posts: 56

PostPosted: Thu May 22, 2014 5:56 am  Reply with quote

Thermal strain is not included in the equations because thermal strain does
NOT induce stress. Alpha * Delta T results in a free thermal expansion.
Stress is induced by mechanical strain. You get stress in a thermal
expansion only when that expansion is resisted by the boundary conditions.
The result is a mechanical strain that is opposite the thermal strain, the
mechanical strain induces stress.

It is easiest to think of a unit cell where all DOF's are constrained. Now
apply a temperature to the unit cell. Since it is fully constrained one
might be tempted to say the strain equals 0. I.e Change in length / length
= 0.

But to do this right we need to separate the thermal and mechanical strain.

Without boundary conditions the model would freely expand (and produce no
stress). It is the effect of the boundary conditions that induce a
compressive strain (essentially pushing the part back to its original
shape). This compressive strain induces stress.

Hope this helps.

-Jim

________________________________


James J. Kosloski
Senior Engineering Manager

CAE Associates, Inc.
1579 Straits Turnpike, Suite 2B | Middlebury, CT 06762
www.caeai.com

P: 203.758.2914 | F: 203.758.2965 | E: kosloski@caeai.com



-----Original Message-----
From: Xansys [mailto:xansys-bounces@xansys.org] On Behalf Of summer.shen
Sent: Wednesday, May 21, 2014 10:26 AM
To: xansys@xansys.org
Subject: [Xansys] [STRUC]how ElasticStress is calculated considering
ThermalSt

Dear experts,

This question is stimulated by the one I post a few days ago (unfortunately
no one has answered it yet, hope this one will get some answer, at least
raise discussion):
[STRUC]Reference Temperature in phase transformation
(http://www.xansys.org/forum/viewtopic.php?t=25669)

Trying to understand what happens during the calculation of thermal strain,
plastic strain and elastic strain/stress, I checked the ANSYS APDL Theory
Reference > Structures with Material Nonlinearities > Rate-Independent
Plasticity>Implementation.
It shows that the iteration steps of calculating elastic strain (elastic
strain), plastic strain. As I cannot post a screenshot here or type
beautiful equations here, could you please spent one minute reading about 20
lines of this problem? (Moderators, can I insert images here? 'Cause I see
the Img button but failed to insert one)

And please notice that, in the first few lines - second step to calculate
trial strain - it says that the thermal effect is ignored, which really
puzzles me. And I cannot find any other steps implementing thermal strain.
Here is my question:
In thermal-structural analysis, thermal strain is surely considered, if it
is ignored all the procedure then it of course has no play in the modeling,
but when and where exactly is it implemented during the calculation of
elastic strain and plastic strain?

Looking forward to your reply, your insight will be appreciated.

Regards,
Summer Shen
Shanghai Jiao Tong University, Shanghai, China[/url]






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christopher.wright
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Joined: 17 Jun 2009
Posts: 927

PostPosted: Thu May 22, 2014 10:19 am  Reply with quote

On May 22, 2014, at 8:00 AM, James J. Kosloski wrote:

Quote:
Thermal strain is not included in the equations because thermal
strain does
NOT induce stress.

Quibble alert--
If you have a thermal gradient you'll get stress even with no other
loads or external restraints. Temperature differences, strictly
speaking, act as self-restraints, kind of like internal boundary
conditions. The distinction is more consequential than a split hair
but not quite as significant as a fine point.

Christopher Wright P.E. |"They couldn't hit an elephant at
chrisw@skypoint.com | this distance" (last words of Gen.
.......................................| John Sedgwick, Spotsylvania
1864)
http://www.skypoint.com/members/chrisw/



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summer.shen
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Joined: 14 May 2014
Posts: 53
Location: Shanghai, China

PostPosted: Thu May 22, 2014 7:39 pm  Reply with quote

Dear Ayo, Jim, Christopher and other experts,

I think both you two are right about what you mentioned in your reply. But still I was confused. I tried some calculation considering thermal strain (still we need to go back to the fundamental stress vector equation 4-22 and thermal strain equation and equations 4.18 - 4.21):
[pl=plastic, el=elastic, th=thermal]
Before Eq 4.18, the total strain ε*_n is actually not updated, however thermal strain is considered but NOT updated (total strain is NOT component total strain!).
ε*_n = ε_n-1 = ε_n-1(el+pl+th) [1]
Next, trial strain is updated, with updated thermal strain,
ε_n(trial) = ε*_n - ε_n-1(pl) - ε_n(th) = ε_n-1 - ε_n-1(pl) - ε_n(th)
= ε_n-1(el) - Δε(th) [2]
Then, trial stress, and then Δε(pl) and ε_n(pl).
So, by now, we have updated (in a guessed algorithm order)(notice that ε_n(trial) is the elastic trial strain)
Δε(th) → ε_n → ε_n(trial) → σ_n(trial) → Δε(pl) → ε_n(pl), and find that
Δε(el) = ε_n(el) - ε_n-1(el) = - Δε(th) - Δε(pl), then,
Δε(el+pl+th)=0 [3],
which is the internal constraint effect mentioned by Christopher.

And now we recalculate ε_n with updated strains:
ε_n=ε_n-1 + Δε(el+pl+th) = ε_n-1 [4]
Surprise! The total strain will always be constant (0, if initial displacement is 0). This is only true if the element is all fixed. Wrong...

And now I want to ask what exactly is the ε*_n (ε_n in the manual). In the calculation, the total strain should be updated as well, ε_n-1 = ε*_n → ε**_n → ε***_n → ... → ε_n.
Notice that this is a indirect couple, it is different from thermoelasticity and thermoplasticity (equation 10.29, 10.30, 10.39). And then we only need to go back to the fundamental constitutive equation 2.58, and notice that there exists thermal load here.
u=F_e(pr+nd+th)/(K_e+K_e(f)) [5]
convert it into strain-stress form (ε_n(trial) is the trial vector of elastic strain!)
ε_n=function(ε_n(trial), ε_n(th)) [6]
Keep updating ε_n(trial), we update ε_n as well, and finally an equilibrium is reached. Thus in [1], the first trial total strain ε*_n = ε_n-1, and then will be updated, ε**_n =ε*_n + Δε(el). Equations [2-6] will be updated, and [3] won't be 0, ε_n ≠ ε_n-1.

In conclusion, take thermal expansion for instance, thermal strain increment → increase total strain → decrease elastic trial strain (become more compressive, largest compression in the first trial step) → decrease elastic stress → decrease total strain increment → decrease elastic trial strain reduction (become less compressive) → decreased elastic stress reduction → ... → final equilibrium [a]
In general the iteration procedure will be:
Δε(th) → Δε(total strain) → Δε(elastic trial strain) → Δε(pl) → Δε(el) → Δε(total strain) → ... [b]

Go back to my first problem, http://www.xansys.org/forum/viewtopic.php?t=25669, it is then clear that the reference temperature should NOT be changed from Troom to Tmelt even though it is melted and theoretically 0 strain. Otherwise, if Tref=Troom MPCHG into Tref=Tmelt, then at the switch step, thermal strain increment (fix ALPX for this step)
Δε(th)= ε_n(th) - ε_n-1(th) = ALPX * (T_room-T_melt), a huge decrease of thermal strain, then in [a], there will be a huge total strain decrease, trial strain decrease, and elastic stress will shoot up (from compressive to tensil, which is wrong as the temperature rise should induce compressive stress).

I actually used a simple model (just as Jim suggested) to verify this and found the strain and stress changes exactly in the way I assumed.

It takes me two weeks working on this problem, and I think I have come to a conclusion that the Reference Temperature should not be switched here even though it is melted.
As for the clad material, Tref=Tmelt is okay. If its Tref=Troom, upon activation, there will be a large thermal strain increment, then a large elastic strain decrease. However, in melting state, if we set elastic modulus E≈0, there won't be large elastic stress decrease thus the total strain, trial elastic strain, plastic strain, elastic trial won't change much as well. But still there will be a the total strain increment and a elastic decrement. This is not verified yet. Will come back to this later.

Anyway, all the above is my guess. If only it is stated in manual! Is there someone who can confirm or confute my guess? Would love to hear from you.

Regards,
Summer SHEN
Master student
Shanghai Jiao Tong University, Shanghai, China

.
-----------------------------------
Quote:
Looking at the fundamental stress vector equation (equation4-22), the elastic strain vector term actually takes into account the thermal strain vector. Look through the first few pages of the "structures" section in the theory guide. You would see the fundamental equation in the expanded form there.

Ayo Brimmo


Quote:
Thermal strain is not included in the equations because thermal strain does NOT induce stress. Alpha * Delta T results in a free thermal expansion. Stress is induced by mechanical strain. You get stress in a thermal expansion only when that expansion is resisted by the boundary conditions. The result is a mechanical strain that is opposite the thermal strain, the mechanical strain induces stress.

It is easiest to think of a unit cell where all DOF's are constrained. Now apply a temperature to the unit cell. Since it is fully constrained one might be tempted to say the strain equals 0. I.e Change in length / length = 0.

But to do this right we need to separate the thermal and mechanical strain.

Without boundary conditions the model would freely expand (and produce no stress). It is the effect of the boundary conditions that induce a compressive strain (essentially pushing the part back to its original shape). This compressive strain induces stress.

Hope this helps.
-Jim
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jobie.gerken
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Joined: 01 Jun 2012
Posts: 29

PostPosted: Fri May 23, 2014 8:20 am  Reply with quote

Summer Shen,

I haven't followed the specific details of your question, but let me
summarize the material and strain calculations for simulations that include
thermal strain.

A subtle point is whether MAPDL is using a direct or Newton-Raphson
solution for the global system of equations. For a NR solution, the strain
used in the material calculations is the mechanical strain - the total
strain minus the thermal strain. For a direct solution, the strain used in
the material calculations is the total strain and this results in a thermal
load vector you note in eqn 2-58 from the theory manual. I'll assume a NR
solution below.

During solution for a time increment, we have the plastic strain from the
previous time increment, e_pl(n-1), and we also have the current estimate
for the nodal DOFs. From the current estimate of the DOFs we can calculate
the total strain e_T(n) and the thermal strain e_th(n) = alpha * dt(n).
Where dt(n) is the difference between the current temperature and the
reference temperature.

The mechanical strain is then e(n) = e_T(n) - e_th(n). This is the strain
used in the material calculations to determine the elastic strain and
plastic strain at the end of the current time increment. The thermal strain
does not change as a result of the material calculations.

The initial trial strain is the mechanical strain minus the previous
plastic strain: e_tr(n) = e(n) - e_pl(n-1). The plasticity algorithm uses
this value as the initial guess for the elastic strain and will use a
return mapping method to determine the elastic and plastic strains at the
end of the time increment.

Jobie Gerken
Ansys, Inc.


On Thu, May 22, 2014 at 10:40 PM, summer.shen <shen_12@sjtu.edu.cn> wrote:

Quote:
Dear Ayo, Jim, Christopher and other experts,

I think both you two are right about what you mentioned in your reply. But
still I was confused. I tried some calculation considering thermal strain
(still we need to go back to the fundamental stress vector equation 4-22
and thermal strain equation and equations 4.18 - 4.21):
[pl=plastic, el=elastic, th=thermal]
Before Eq 4.18, the total strain ε*_n is actually not updated, however
thermal strain is considered but NOT updated (total strain is NOT component
total strain!).
ε*_n = ε_n-1 = ε_n-1(el+pl+th) [1]
Next, trial strain is updated, with updated thermal strain,
ε_n(trial) = ε*_n - ε_n-1(pl) - ε_n(th) = ε_n-1 - ε_n-1(pl) - ε_n(th)
= ε_n-1(el) - Δε(th) [2]
Then, trial stress, and then Δε(pl) and ε_n(pl).
So, by now, we have updated (in a guessed algorithm order)(notice that
ε_n(trial) is the elastic trial strain)
Δε(th) → ε_n → ε_n(trial) → σ_n(trial) → Δε(pl) → ε_n(pl), and find that
Δε(el) = ε_n(el) - ε_n-1(el) = - Δε(th) - Δε(pl), then,
Δε(el+pl+th)=0 [3],
which is the internal constraint effect mentioned by Christopher.

And now we recalculate ε_n with updated strains:
ε_n=ε_n-1 + Δε(el+pl+th) = ε_n-1 [4]
Surprise! The total strain will always be constant (0, if initial
displacement is 0). This is only true if the element is all fixed. Wrong...

And now I want to ask what exactly is the ε*_n (ε_n in the manual). In the
calculation, the total strain should be updated as well, ε_n-1 = ε*_n →
ε**_n → ε***_n → ... → ε_n.
Notice that this is a indirect couple, it is different from
thermoelasticity and thermoplasticity (equation 10.29, 10.30, 10.39). And
then we only need to go back to the fundamental constitutive equation 2.58,
and notice that there exists thermal load here.
u=F_e(pr+nd+th)/(K_e+K_e(f)) [5]
convert it into strain-stress form (ε_n(trial) is the trial vector of
elastic strain!)
ε_n=function(ε_n(trial), ε_n(th)) [6]
Keep updating ε_n(trial), we update ε_n as well, and finally an
equilibrium is reached. Thus in [1], the first trial total strain ε*_n =
ε_n-1, and then will be updated, ε**_n =ε*_n + Δε(el). Equations [2-6] will
be updated, and [3] won't be 0, ε_n ≠ ε_n-1.

In conclusion, take thermal expansion for instance, thermal strain
increment → increase total strain → decrease elastic trial strain (become
more compressive, largest compression in the first trial step) → decrease
elastic stress → decrease total strain increment → decrease elastic trial
strain reduction (become less compressive) → decreased elastic stress
reduction → ... → final equilibrium [a]
In general the iteration procedure will be:
Δε(th) → Δε(total strain) → Δε(elastic trial strain) → Δε(pl) → Δε(el) →
Δε(total strain) → ... [b]

Go back to my first problem,
http://www.xansys.org/forum/viewtopic.php?t=25669, it is then clear that
the reference temperature should NOT be changed from Troom to Tmelt even
though it is melted and theoretically 0 strain. Otherwise, if Tref=Troom
MPCHG into Tref=Tmelt, then at the switch step, thermal strain increment
(fix ALPX for this step)
Δε(th)= ε_n(th) - ε_n-1(th) = ALPX * (T_room-T_melt), a huge decrease of
thermal strain, then in [a], there will be a huge total strain decrease,
trial strain decrease, and elastic stress will shoot up (from compressive
to tensil, which is wrong as the temperature rise should induce compressive
stress).

I actually used a simple model (just as Jim suggested) to verify this and
found the strain and stress changes exactly in the way I assumed.

It takes me two weeks working on this problem, and I think I have come to
a conclusion that the Reference Temperature should not be switched here
even though it is melted.
As for the clad material, Tref=Tmelt is okay. If its Tref=Troom, upon
activation, there will be a large thermal strain increment, then a large
elastic strain decrease. However, in melting state, if we set elastic
modulus E≈0, there won't be large elastic stress decrease thus the total
strain, trial elastic strain, plastic strain, elastic trial won't change
much as well. But still there will be a the total strain increment and a
elastic decrement. This is not verified yet. Will come back to this later.

Anyway, all the above is my guess. If only it is stated in manual! Is
there someone who can confirm or confute my guess? Would love to hear from
you.

Regards,
Summer SHEN
Master student
Shanghai Jiao Tong University, Shanghai, China

.
-----------------------------------



--
Jobie M Gerken, PhD
Lead Software Developer
ANSYS, Inc.
Jobie.Gerken@ansys.com
Tel: 1.724.514.3683
www.ansys.com

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christopher.wright
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Joined: 17 Jun 2009
Posts: 927

PostPosted: Fri May 23, 2014 9:28 am  Reply with quote

On May 23, 2014, at 7:56 AM, Phil Vidori wrote:

Quote:
All I know is that you cannot neglect thermal strains,
either for constrained components or free-free components.
I got to thinking about this last night. I think the statement
'Thermal strain is not included in the equations because thermal
strain does
NOT induce stress.' was an imprecise way of stating that unrestrained
thermal expansion doesn't induce stress. Kosloski's original post, on
the face of it, isn't correct. because everyone knows that thermal
strain *can* induce stress, just as his example cites. I don't know
what equations don't include thermal stress, and I confess I didn't
go back and check. It's a fact however, that restrained thermal
expansion does produce stress, so I'm sure there *are* equations
where thermal strain shows up, maybe not the equations that Kosloski
was citing.

Christopher Wright P.E. |"They couldn't hit an elephant at
chrisw@skypoint.com | this distance" (last words of Gen.
.......................................| John Sedgwick, Spotsylvania
1864)
http://www.skypoint.com/members/chrisw/



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james.kosloski
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PostPosted: Fri May 23, 2014 11:38 am  Reply with quote

No. I am still sticking to my statement. If we back off and forget about
plasticity for a moment and worry only about elasticity, then referring to
the Theory Manual section 2.1.1

{Epsilon(el)} = {Epsilon} - {Epsilon(th)}

Note: {Epsilon(el) are the strains that cause stress

-Jim

________________________________


James J. Kosloski
Senior Engineering Manager

CAE Associates, Inc.
1579 Straits Turnpike, Suite 2B | Middlebury, CT 06762
www.caeai.com

P: 203.758.2914 | F: 203.758.2965 | E: kosloski@caeai.com




-----Original Message-----
From: Xansys [mailto:xansys-bounces@xansys.org] On Behalf Of Christopher
Wright
Sent: Friday, May 23, 2014 12:28 PM
To: ANSYS User Discussion List
Subject: Re: [Xansys] [STRUC]how ElasticStress is calculated considering
ThermalSt


On May 23, 2014, at 7:56 AM, Phil Vidori wrote:

Quote:
All I know is that you cannot neglect thermal strains, either for
constrained components or free-free components.
I got to thinking about this last night. I think the statement 'Thermal
strain is not included in the equations because thermal strain does NOT
induce stress.' was an imprecise way of stating that unrestrained thermal
expansion doesn't induce stress. Kosloski's original post, on the face of
it, isn't correct. because everyone knows that thermal strain *can* induce
stress, just as his example cites. I don't know what equations don't include
thermal stress, and I confess I didn't go back and check. It's a fact
however, that restrained thermal expansion does produce stress, so I'm sure
there *are* equations where thermal strain shows up, maybe not the equations
that Kosloski was citing.

Christopher Wright P.E. |"They couldn't hit an elephant at
chrisw@skypoint.com | this distance" (last words of Gen.
........................................| John Sedgwick, Spotsylvania
1864)
http://www.skypoint.com/members/chrisw/



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christopher.wright
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PostPosted: Fri May 23, 2014 11:59 am  Reply with quote

On May 23, 2014, at 1:42 PM, James J. Kosloski wrote:

Quote:
No. I am still sticking to my statement. If we back off and forget
about
plasticity for a moment and worry only about elasticity, then
referring to
the Theory Manual section 2.1.1

{Epsilon(el)} = {Epsilon} - {Epsilon(th)}

Note: {Epsilon(el) are the strains that cause stress

I think this is a matter of interpreting some ANSYS convention or
usage, rather than fundamental elasticity. So we go back to the
question I posed yesterday:

What would I see if I ran a problem such as an I-beam with the top
flange at a different temperature than the bottom flange. No other
mechanical loads and end conditions taken as free-free.

Christopher Wright P.E. |"They couldn't hit an elephant at
chrisw@skypoint.com | this distance" (last words of Gen.
.......................................| John Sedgwick, Spotsylvania
1864)
http://www.skypoint.com/members/chrisw/



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bogdan.balasa
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PostPosted: Fri May 23, 2014 12:03 pm  Reply with quote

I think (I might be wrong as I have not tested it) that the only free
thermal expanding body is a sphere with no constraints and uniform
temperature load. In this case I'm not expecting to see any thermal
stress. On the other hand any geometry, even unconstrained, that it
prevents the uniform thermal expansion (through geometry shape) I believe
it will generate thermal stress.

Sincerely.

Bogdan Balasa | Director, Analysis Technologies | CAMBRIC CORPORATION |
555 East Broadway, Suite 300, Salt Lake City, UT 84102 | T: 801.415.7328 |
F: 801.415.7310 | Email: bogdan.balasa@cambric.com | Web: www.cambric.com


-----Original Message-----
From: Xansys [mailto:xansys-bounces@xansys.org] On Behalf Of Christopher
Wright
Sent: Friday, May 23, 2014 1:00 PM
To: ANSYS User Discussion List
Subject: Re: [Xansys] [STRUC]how ElasticStress is calculated considering
ThermalSt


On May 23, 2014, at 1:42 PM, James J. Kosloski wrote:

Quote:
No. I am still sticking to my statement. If we back off and forget
about plasticity for a moment and worry only about elasticity, then
referring to the Theory Manual section 2.1.1

{Epsilon(el)} = {Epsilon} - {Epsilon(th)}

Note: {Epsilon(el) are the strains that cause stress

I think this is a matter of interpreting some ANSYS convention or usage,
rather than fundamental elasticity. So we go back to the question I posed
yesterday:

What would I see if I ran a problem such as an I-beam with the top flange
at a different temperature than the bottom flange. No other mechanical
loads and end conditions taken as free-free.

Christopher Wright P.E. |"They couldn't hit an elephant at
chrisw@skypoint.com | this distance" (last words of Gen.
.......................................| John Sedgwick, Spotsylvania
1864)
http://www.skypoint.com/members/chrisw/



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jobie.gerken
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PostPosted: Fri May 23, 2014 12:07 pm  Reply with quote

From a structural mechanics point of view, thermal strain represents a
stress free change in volume. At a single point or for a structure with a
homogeneous thermal strain, if anything acts to restrain the uniform change
in volume, then a stress will result. Otherwise, thermal strain does not
result in stress.

However, a point is not a continuum and in a continuum we must also satisfy
compatibility (the deformations must not result in voids or overlaps).
Given a non-homogeneous temperature field in a continuum, the resulting
thermal strain will generally not satisfy compatibility and there must be
an additional strain component to ensure compatibility. This additional
strain will result in a stress.

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ANSYS, Inc.
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Tel: 1.724.514.3683
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summer.shen
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PostPosted: Fri May 23, 2014 7:10 pm  Reply with quote

Philippe,

Glad to have you here. I wish I could have put it in a nice piece of reading. But as I am trying to look into the algorithm implementation, which seems not revealed to us in the manual, all the equations make it a little painful to read.

I think nobody is neglecting thermal effect if thermal expansion coefficient and reference are properly set. But, how thermal strain affects the calculation of elastic strain, to my best knowledge, is not mentioned in the theory reference. All I know is that the total strain includes thermal strain. And it should be clear that there is no 'thermal stress' defined in ANSYS - there is only one type of stress, i.e. elastic stress. I suppose it is better to refer to 'thermal stress' as 'thermal-strain-and-constraints-induced elastic stress', which is different to the stress caused by external structural forces.

I am modelling laser cladding process, which includes material addition and liquid-solid phase transformation. Thermal stress (thermal-strain-and-constraints-induced elastic stress) is of importance in this process, by no means it can be ignored. And a proper reference temperature is critical in the modelling. It should be straightforward and obvious in the eye of researchers who work in algorithm and programming. Hope we can hear from them.

Summer Shen
MSc
Shanghai Jiao Tong University, Shanghai, China


philippe.vidori wrote:
My God, this is too intense before 2 or 3 cups of coffee !! All I know is that you cannot neglect thermal strains, either for constrained components or free-free components. Common sense tells us ( or me ) that free-free part will exhibit some induced stress ( such as a heat-treated part quenched in a cooling bath. Even if the part is not held in some kind of jig ).

When you look at computer chips, the first mode of failure is fatigue at the metal contacts. In this case, the strain is purely thermal, not mechanical.

So, to say that you can neglect thermal strains ( in some cases ) is far from correct. Maybe, in the case of a bar without constraints and slowly heated, and you have to make sure that the constraints allow the bar to extend or contract freely.

P.
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harish.radhakrishnan
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Joined: 10 Dec 2013
Posts: 34

PostPosted: Sat May 24, 2014 6:06 am  Reply with quote

Bogdan: I believe that your statement needs to be further qualified. If the
thermal expansion coefficient is isotropic, and the material is homogenous
and isotropic, uniform temperature change causing thermal
expansion/contraction in an uncontrained body will cause pure volumetric
change. Thus the body will be stress free. This behavior will be
independent of the shape.

Harish

Harish Radhakrishnan
Senior Technical Services Engineer
ANSYS Inc
15915 Katy Freeway, Suite 550
Houston, TX 77094
Office: 281-676-7013
Help-Line:(800) 711-7199
harish.radhakrishnan@ansys.com
https://support.ansys.com/portal/site/AnsysCustomerPortal<http://www1.ansys.com/customer/>



On Fri, May 23, 2014 at 2:03 PM, Bogdan Balasa <bogdan.balasa@cambric.com>wrote:

Quote:
I think (I might be wrong as I have not tested it) that the only free
thermal expanding body is a sphere with no constraints and uniform
temperature load. In this case I'm not expecting to see any thermal
stress. On the other hand any geometry, even unconstrained, that it
prevents the uniform thermal expansion (through geometry shape) I believe
it will generate thermal stress.

Sincerely.

Bogdan Balasa | Director, Analysis Technologies | CAMBRIC CORPORATION |
555 East Broadway, Suite 300, Salt Lake City, UT 84102 | T: 801.415.7328 |
F: 801.415.7310 | Email: bogdan.balasa@cambric.com | Web: www.cambric.com


-----Original Message-----
From: Xansys [mailto:xansys-bounces@xansys.org] On Behalf Of Christopher
Wright
Sent: Friday, May 23, 2014 1:00 PM
To: ANSYS User Discussion List
Subject: Re: [Xansys] [STRUC]how ElasticStress is calculated considering
ThermalSt


On May 23, 2014, at 1:42 PM, James J. Kosloski wrote:

Quote:
No. I am still sticking to my statement. If we back off and forget
about plasticity for a moment and worry only about elasticity, then
referring to the Theory Manual section 2.1.1

{Epsilon(el)} = {Epsilon} - {Epsilon(th)}

Note: {Epsilon(el) are the strains that cause stress

I think this is a matter of interpreting some ANSYS convention or usage,
rather than fundamental elasticity. So we go back to the question I posed
yesterday:

What would I see if I ran a problem such as an I-beam with the top flange
at a different temperature than the bottom flange. No other mechanical
loads and end conditions taken as free-free.

Christopher Wright P.E. |"They couldn't hit an elephant at
chrisw@skypoint.com | this distance" (last words of Gen.
.......................................| John Sedgwick, Spotsylvania
1864)
http://www.skypoint.com/members/chrisw/



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summer.shen
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PostPosted: Sat May 24, 2014 7:32 pm  Reply with quote

Jobie,

It is great to hear from an expert from ANSYS! You answered the very point that puzzles me. I read the 14.12. Newton-Raphson Procedure part and rethought about this again. Please allow me to double-check it with you.

In eqns 14.147/148 , {u} is updated and we obtain a new total strain ε_T(i+1) for (i+1)th equilibrium iteration via strain-displacement relationship equation.
ε_Total =ε_el+ε_pl+ε_th, ε_ComponentTotal=ε_el+ε_pl (definition in the theory reference; I believe you refer to 'component total strain' as 'total strain' or 'mechanical strain' in your last reply). And ε_CT(i+1) = ε_T(i+1) - ε_th[n+1] is used in the following steps of i+1 iteration to update trial strain, elastic strain and plastic strain.
Thermal strain is updated in the first iteration of step [n+1] and holds constant in following equilibrium iterations. As component total is used in the iteration calculations of elastic strain, thermal stress doesn't show up in any equations (but ε_CT is indeed related to ε_th, thus elastic strain/stress is related to ε_th, indirectly ). And the flow chart of equilibrium iterations of step n+1 is:

Eqn14.147 (i) → Δu(i) → u(i+1) → ε_T(i+1)→ ε_CT(i+1) = ε_T(i+1) - ε_th[n+1] → ε_tr(i+1) → ε_el(i+1) → ε_pl(i+1) → Eqn14.147 (i+1) → ...

where, ε_CT is the Component Total strain and ε_T is the Total strain, and ε_CT = ε_T - ε_th

Did I get it right? Looking forward to hearing from you again.

Summer Shen
MSc
Shanghai Jiao Tong University, Shanghai, China
Email: shen_12@sjtu.edu.cn


jobie.gerken wrote:
Summer Shen,

I haven't followed the specific details of your question, but let me
summarize the material and strain calculations for simulations that include
thermal strain.

A subtle point is whether MAPDL is using a direct or Newton-Raphson
solution for the global system of equations. For a NR solution, the strain
used in the material calculations is the mechanical strain - the total
strain minus the thermal strain. For a direct solution, the strain used in
the material calculations is the total strain and this results in a thermal
load vector you note in eqn 2-58 from the theory manual. I'll assume a NR
solution below.

During solution for a time increment, we have the plastic strain from the
previous time increment, e_pl(n-1), and we also have the current estimate
for the nodal DOFs. From the current estimate of the DOFs we can calculate
the total strain e_T(n) and the thermal strain e_th(n) = alpha * dt(n).
Where dt(n) is the difference between the current temperature and the
reference temperature.

The mechanical strain is then e(n) = e_T(n) - e_th(n). This is the strain
used in the material calculations to determine the elastic strain and
plastic strain at the end of the current time increment. The thermal strain
does not change as a result of the material calculations.

The initial trial strain is the mechanical strain minus the previous
plastic strain: e_tr(n) = e(n) - e_pl(n-1). The plasticity algorithm uses
this value as the initial guess for the elastic strain and will use a
return mapping method to determine the elastic and plastic strains at the
end of the time increment.

Jobie Gerken
Ansys, Inc.
-----------------------------------

--
Jobie M Gerken, PhD
Lead Software Developer
ANSYS, Inc.
Jobie.Gerken@ansys.com
Tel: 1.724.514.3683
www.ansys.com

The information transmitted is intended only for the person or entity to
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    summer.shen
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    PostPosted: Sat May 24, 2014 7:43 pm  Reply with quote

    A small correction of the flow chart:

    Eqn14.147 (i) → Δu(i) → u(i+1) → ε_T(i+1)→ ε_CT(i+1) = ε_T(i+1) - ε_th[n+1] → ε_tr(i+1) → ε_pl(i+1) → ε_el(i+1) → σ_el(i+1) → Eqn14.147 (i+1) → ...

    summer.shen wrote:
    Jobie,

    It is great to hear from an expert from ANSYS! You answered the very point that puzzles me. I read the 14.12. Newton-Raphson Procedure part and rethought about this again. Please allow me to double-check it with you.

    In eqns 14.147/148 , {u} is updated and we obtain a new total strain ε_T(i+1) for (i+1)th equilibrium iteration via strain-displacement relationship equation.
    ε_Total =ε_el+ε_pl+ε_th, ε_ComponentTotal=ε_el+ε_pl (definition in the theory reference; I believe you refer to 'component total strain' as 'total strain' or 'mechanical strain' in your last reply). And ε_CT(i+1) = ε_T(i+1) - ε_th[n+1] is used in the following steps of i+1 iteration to update trial strain, elastic strain and plastic strain.
    Thermal strain is updated in the first iteration of step [n+1] and holds constant in following equilibrium iterations. As component total is used in the iteration calculations of elastic strain, thermal stress doesn't show up in any equations (but ε_CT is indeed related to ε_th, thus elastic strain/stress is related to ε_th, indirectly ). And the flow chart of equilibrium iterations of step n+1 is:

    Eqn14.147 (i) → Δu(i) → u(i+1) → ε_T(i+1)→ ε_CT(i+1) = ε_T(i+1) - ε_th[n+1] → ε_tr(i+1) → ε_el(i+1) → ε_pl(i+1) → Eqn14.147 (i+1) → ...

    where, ε_CT is the Component Total strain and ε_T is the Total strain, and ε_CT = ε_T - ε_th

    Did I get it right? Looking forward to hearing from you again.

    Summer Shen
    MSc
    Shanghai Jiao Tong University, Shanghai, China
    Email: shen_12@sjtu.edu.cn
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    jobie.gerken
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    PostPosted: Tue May 27, 2014 5:39 am  Reply with quote

    Summer Shen,

    I believe you have the general concepts correct.

    Jobie Gerken
    Ansys, Inc.


    On Sat, May 24, 2014 at 10:32 PM, summer.shen <shen_12@sjtu.edu.cn> wrote:

    Quote:

    Did I get it right? Looking forward to hearing from you again.



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