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Titrate:
HPLA100 - Heavy thermoelastic hollow roll in uniform rotation
Date:
18/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.03-A
Page:
1/12
Manual of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
HI-75/96/014/A
Organization (S):
EDF/IMA/MN
Manual of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
V7.90.03 document
HPLA100 - An analytical solution for the cylinder
heavy thermoelastic hollow in uniform rotation
Summary:
One gives the analytical solution here axisymmetric 2D and in hull of the problem of the thin hollow roll
thermoelastic weighing and in uniform rotation, subjected to a field of linear temperature in the thickness.
material is supposed characteristics independent of the temperature.
This solution corresponds to test HPLA100 [V7.01.100].
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Code_Aster
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Version
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Titrate:
HPLA100 - Heavy thermoelastic hollow roll in uniform rotation
Date:
18/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.03-A
Page:
2/12
Manual of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
HI-75/96/014/A
Contents
Error! No table of contents entry was found.
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Code_Aster
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Version
3
Titrate:
HPLA100 - Heavy thermoelastic hollow roll in uniform rotation
Date:
18/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.03-A
Page:
3/12
Manual of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
HI-75/96/014/A
1
Uniform loading of rotation around OZ
1.1
Axisymmetric model 2D
The density of centrifugal force is:
2
R
E
R
.
One considers the boundary conditions following:
U
Z
R, Z
()
=
0
in
Z
=
0
and
Z
=
L
One postulates displacement in the form:
U
R
=
U R
()
; U
Z
=
U
=
0
As follows:
rr
=
u';
=
U
R
;
zz
=
rz
=
Z
=
R
=
0
L
Z
B
1
B
2
With
2
With
1
With
B
R
!!
H
R
0
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Code_Aster
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Version
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Titrate:
HPLA100 - Heavy thermoelastic hollow roll in uniform rotation
Date:
18/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.03-A
Page:
4/12
Manual of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
HI-75/96/014/A
The elastic stresses are expressed:
(
) (
) (
)
(
) (
) (
)
(
) (
)
rr
zz
E
U
U
R
E
U
R
U
E
U
R U
=
+
-
-
+




=
+
-
-
+




=
+
-
+








1
1 2
1
1
1 2
1
1
1 2
'
'
'
The radial equilibrium equation is written:
(
)
R
R
rr R
,
-
= -
2
2
As follows:
()
(
) (
)
(
)
Ru
R
E
R
''




= - +
-
-
1
1 2
1
2
éq 1.1-1
Note:
U
R
+
u'
=
Ru
()
'
R
From where the general solution:
()
(
) (
)
(
)
U R
E
R
Ar
B
R
= - +
-
-
+
+
1
1 2
1
8
2
3
éq 1.1-2
The stresses are then:
()
(
) (
)
(
)
()
(
) (
)
(
)
()
(
) (
)
rr
zz
R
R
E
With
B
R
R
R
E
With
B
R
R
R
E
With
= - --
+ + -
- -




= - +
+ + -
+ -




= - -
+ + -






3 2
1
8
1
1 2
1 2
1 2
1
8
1
1 2
1 2
1
2
2
1
1 2
2
2
2
2
2
2
2
2
.
.
.
éq 1.1-3
The boundary conditions in stresses are:
rr
=
0
in
R
=
R
±
H
2
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Code_Aster
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Titrate:
HPLA100 - Heavy thermoelastic hollow roll in uniform rotation
Date:
18/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.03-A
Page:
5/12
Manual of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
HI-75/96/014/A
One notes:
X
=
H
2R
One obtains thanks to [éq 1.1-3]:
B
=
3
-
2
(
)
1
+
(
)
8 1
-
(
)
E
2
R
4
1
-
X
2
()
2
then:
With
=
3
-
2
(
)
1
+
(
)
1
-
2
(
)
4 1
-
(
)
E
2
R
2
1
+
X
2
()
Numerical application:
R
= 20 mm;
H
= 1 mm;
= 8.10
­ 6
kg/mm
3
;
= 1 S
­ 1
E
= 2.10
5
NR/mm
2
;
= 0.3.
From where:
With
= 7.13588.10
­ 9
;
B
= 3.561258.10
­ 6
mm
2
Note:
1
+
(
)
1
-
2
(
)
1
-
(
)
E
2
8
=
3.714286.10
-
12
mm
-
2
1
-
2
2
=
1. 714286.10
-
6
MPa .mm
-
2
As follows:
·
in internal skin:
U
R
= 2.9424.10
- 7
mm;
zz
= 0.99488.10
- 3
MPa
·
in external skin:
U
R
= 2.8801.10
- 7
mm;
zz
= 0.92631.10
- 3
MPa
1.2
Axisymmetric model hull
The centrifugal force is equivalent to a pressure distributed:
p
=
2
H R 1
+
H
2
12 R
2




The solution is membranous, normal balance is written:
NR
=
p
R
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Titrate:
HPLA100 - Heavy thermoelastic hollow roll in uniform rotation
Date:
18/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.03-A
Page:
6/12
Manual of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
HI-75/96/014/A
The membrane deformation is:
E
=
W
R
, whereas
E
zz
=
0
=
K
=
K
zz
. In elasticity:
NR
=
E H
1
-
v
2
E
; NR
zz
=
v NR
; M
=
0
From where the solution (arrow and circumferential normal effort):
W
=
1
-
2
()
2
E
R
3
1
+
H
2
12 R
2




; NR
=
2
R
2
H 1
+
H
2
12 R
2




Axial stress is worth:
zz
=
2
R
2
1
+
H
2
12 R
2




constant in the thickness
(
)
If one does not take account of the correction of metric, the term should be removed
1
+
H
2
12R
2




in
preceding expressions.
Numerical application (without correction of metric):
p
= 1,600000.10
­ 4
MPa
W
= 2,912000.10
­ 7
mm
NR
zz
= 0,96000.10
­ 3
NR/mm
zz
= 0,96000.10
­ 3
MPa
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Code_Aster
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Version
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Titrate:
HPLA100 - Heavy thermoelastic hollow roll in uniform rotation
Date:
18/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.03-A
Page:
7/12
Manual of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
HI-75/96/014/A
2
Loading of gravity
2.1
Axisymmetric model 2D
The density of force is:
-
G E
Z
(vertical gravity).
One considers the boundary conditions following:
U
Z
R, Z
()
=
0
in
R
=
R
and
Z
=
0
(circle of support)
with uniform traction:
zz
R, Z
()
=
G L
in
Z
=
L
, balancing the weight.
One postulates the elastic solution of the type:
=




0 0
0
0 0
0
0 0
zz
so that:
rr
=
= -
zz
= -
U
Z, Z
= -
zz
E
;
rz
=
0
=
R
=
Z
One observes as follows:
U
R, R
=
U
R
R
U
R
R, Z
()
= -
A' Z
()
R
Then:
-
A' Z
()
=
rr
= -
zz
U
Z, Z
R, Z
()
=
A' Z
()
U
Z
R, Z
()
=
With Z
()
+
B R
()
Of
rz
=
0
, one fires:
B R
()
-
R With " Z
()
=
0
that is to say:
With Z
()
=
cste
=
;
B R
()
=
R
Boundary conditions in effort, one obtains:
With Z
()
=
G Z
2
2nd
+
; B R
()
=
G R
2
2nd
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Code_Aster
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Titrate:
HPLA100 - Heavy thermoelastic hollow roll in uniform rotation
Date:
18/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.03-A
Page:
8/12
Manual of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
HI-75/96/014/A
Lastly,
check:
= -
G R
2
2nd
As follows:
()
()
(
)
(
)
()
U R Z
G Z R
E
U R Z
G
E Z
R
R
R Z
G Z
R
Z
zz
,
;
,
,
= -
=
+
-
=



2
2
2
2
éq 2.1-1
Numerical application
G
= 10 NR/kg;
= 8.10
- 6
kg/mm
3
;
R
= 20 mm;
L
= 10 mm
E
= 2.10
5
NR/mm
2
;
= 0.3;
H
= 1 mm
·
in internal skin:
U
R
L
()
= - 2.34000.10
­ 8
mm;
zz
L
()
= 8.0000.10
­ 4
MPa;
U
Z
O
()
= - 1.185000.10
­ 9
mm
·
in external skin:
U
R
L
()
= - 2.46000.10
­ 8
mm;
zz
L
()
= 8.0000.10
­ 4
MPa;
U
Z
O
()
= 1.215000.10
­ 9
mm
2.2
Axisymmetric model hull
A vertical traction is exerted in
Z
=
L
:
F
=
G H L
Gravity leads to a vertical force:
F
= -
G H E
Z
The boundary condition on the circle of support is:
U
Z
Z
()
=
0
in
Z
=
0
The solution is membranous, vertical balance is written:
NR
zz, Z
=
G H
Moreover:
NR
=
0
. In elasticity, one deduces then:
E
=
W
R
= -
NR
zz
E H
= -
G Z
E
; E
zz
=
U
Z, Z
=
NR
zz
E H
U
Z
Z
()
=
G
2nd Z
2
Axial stress is:
zz
=
G Z
constant in the thickness
(
)
Numerical application:
F
= 8.10
­ 4
NR/mm
W L
()
= - 2.4000.10
- 8
mm
NR
zz
L
()
= 8.0000.10
­ 4
NR/mm
zz
L
()
= 8.0000.10
­ 4
NR/mm
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Code_Aster
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Titrate:
HPLA100 - Heavy thermoelastic hollow roll in uniform rotation
Date:
18/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.03-A
Page:
9/12
Manual of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
HI-75/96/014/A
3 Loading
thermomechanics
3.1
Axisymmetric model 2D
()
()
(
)
(
)
T R
T
R
T
T
T
T
H
R R
ref.
S
I
S
I
-
=
+
+
-
-
2
éq 3.1.- 1
Z
H
T
S
T
I
R
R
One postulates displacement in the form:
U
R
=
U R
()
; U
Z
=
U
=
0
with the boundary conditions suitable. Thus, the elastic stresses are expressed:
(
) (
) (
)
(
)
(
) (
) (
)
(
)
(
) (
)
(
)
rr
ref.
ref.
zz
ref.
E
U
U
R
E T T
E
U
R
U
E T T
E
U
R U
E T T
=
+
-
-
+




- -
-
=
+
-
-
+




- -
-
=
+
-
+




- -
-




1
1 2
1
1 2
1
1 2
1
1 2
1
1 2
1 2
The radial equilibrium equation
R
rr
(
)
, R
-
=
0
give:
()
(
)
(
)
(
)
Ru
R
T T
ref.


=
+
-
-
1
1
éq 3.1-2
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Titrate:
HPLA100 - Heavy thermoelastic hollow roll in uniform rotation
Date:
18/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.03-A
Page:
10/12
Manual of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
HI-75/96/014/A
From where the general solution:
()
(
)
(
)
(
)
U R
T
T
H
R
Ar
B
R
S
I
=
+
-
-
+
+
1
1
3
2
éq 3.1-3
The stresses are then:
()
(
)
(
)
(
) (
)
(
)
()
(
)
(
)
(
) (
)
(
)
()
(
)
rr
S
I
S
I
S
I
S
I
zz
S
I
R
E T
T
H
R
R
E
T
T
E
With
B
R
R
E T
T
H
R
R
E
T
T
E
With
B
R
R
E T
T
H
R
=
-
-
-
-




- - +
+ + -
- -




=
-
-
-
-




- - +
+ + -
+ -




=
-
-
-
1 2
3 1
1 2
2
1
1 2
1 2
1 2
2
3 1
1 2
2
1
1 2
1 2
1 2
2
2
(
) (
)
R
E
T
T
E
With
S
I
1
1 2
2
2
1
1 2
-




- - +
+ + -














éq 3.1-4
The boundary conditions in efforts are: in
R
=
R
±
H
2
rr
=
0
. One notes:
X
=
H
2R
. One obtains
thanks to [éq 3.1-4]:
B
=
T
S
-
T
I
(
)
6. 1
-
(
)
1
+
(
)
R
3
1
-
X
2
()
2
then:
(
)
(
)
(
)
(
)
(
)
With
T
T R
H
X
T
T
S
I
S
I
=
+
-
-
-
- -
+
+




1
6 1
3
1 2
2
2
Numerical application:
R
= 20 mm;
H
= 1 mm;
= 10
- 5
°C
- 1
;
T
S
= -
T
I
= 0.5°C;
= 0.3;
E
= 2.10
5
NR/mm
2
.
From where:
With
= - 0.18569881.10
- 3
;
B
= 0.02473096 mm
2
Note:
1
+
(
)
1
-
T
S
-
T
I
3 H
=
0. 61904762.10
-
5
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Code_Aster
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Titrate:
HPLA100 - Heavy thermoelastic hollow roll in uniform rotation
Date:
18/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.03-A
Page:
11/12
Manual of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
HI-75/96/014/A
·
in internal skin:
U
R
= 1.056145.10
­ 6
mm;
zz
= 1.4321427 MPa.
·
in external skin:
U
R
= 1.110317.10
­ 6
mm;
zz
= - 1.4250001 MPa.
If one takes
T
S
= +
T
I
= 0.1°C:
With
= 0,00130000.10
- 3
;
B
= 0,0 mm
2
As follows:
·
in internal skin:
U
R
= 25.350000.10
­ 6
mm;
zz
= - 0.200000 MPa.
·
in external skin:
U
R
= 26.650000.10
­ 6
mm;
zz
= - 0.200000 MPa.
3.2
Axisymmetric model hull
For the field of temperature in the thickness given by [éq 3.1-1], one obtains the following expression
law of behavior:
(
)
(
)
NR
E H E
E
E H T
T
T
T H
R
NR
E H
E
E
E H T
T
T
T H
R
zz
S
I
S
I
zz
zz
S
I
S
I
=
-
+
- -
+
+
-




=
-
+
- -
+
+
-






1
1
2
12
1
1
2
12
2
2
éq 3.2-1
and:
(
)
(
)
(
)
(
)
(
)
(
)
M
E H
K
K
E H
T
T H
R T
T
M
E H
K
K
E H
T
T H
R T
T
zz
S
I
S
I
zz
zz
S
I
S
I
=
-
+
-
-
+ + -




=
-
+
-
-
+ + -








3
2
2
3
2
2
12 1
12 1
2
12 1
12 1
2
éq 3.2-2
According to these expressions, thermal terms in
H
R
are to be neglected if one does not consider
correction of metric in the thickness, i.e. in the case of usual models.
In our situation:
E
=
W
R; E
zz
=
0; K
=
K
zz
=
0
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Titrate:
HPLA100 - Heavy thermoelastic hollow roll in uniform rotation
Date:
18/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.03-A
Page:
12/12
Manual of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
HI-75/96/014/A
Normal balance with the hull is written:
NR
=
0
from where the arrow:
W
=
1
+
(
)
T
S
+
T
I
2
+
T
S
-
T
I
12
H
R


R
and:
NR
zz
=
E H T
S
+
T
I
2
+
T
S
-
T
I
12
H
R


M
zz
= -
E H
2
12 1
-
(
)
T
S
-
T
I
(
)
+
T
S
+
T
I
2
H
R


As the second member of dilation does not take account of the correction of metric, terms
in
H/R
above are neglected.
Numerical application
R
= 20 mm;
H
= 1 mm;
= 10
- 5
°C
- 1
;
T
S
= -
T
I
= 0.5°C;
= 0.3;
E
= 2.10
5
NR/mm
2
.
From where:
M
zz
= - 0.2380952 NR
in internal skin
:
zz
=
1.449319 MPa
*;
or
zz
=
1.428571 MPa
(
without correction of metric)
If one takes
T
S
= +
T
I
= 0,1°C:
W
= 26.00000.10
­ 6
mm
NR
zz
= ­ 0.2 NR/mm
M
zz
= ­ 0.001190476 NR
in internal skin
:
zz
= -
0.2122466 MPa
*;
or
zz
= -
0.200000 MPa
(
without correction of metric)
* The stresses in the thickness with correction of metric are given by:
zz
X
3
()
=
NR
zz
-
M
zz
/R
H 1
-
H
2
/12 R
2
(
)
+
M
zz
-
NR
zz
H
2
12 R




12 X
3
H
3
1
-
H
2
/12 R
2
(
)