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Titrate:
HPLA100 - Heavy thermoelastic hollow Cylindre in uniform rotation
Date:
18/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.03-A
Page:
1/12
Organization (S): EDF/IMA/MN
Handbook of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
V7.90.03 document
HPLA100 - Analytical Une 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].
Handbook of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
HI-75/96/014/A
Code_Aster ®
Version
3
Titrate:
HPLA100 - Heavy thermoelastic hollow Cylindre in uniform rotation
Date:
18/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.03-A
Page:
2/12
Contents
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Handbook of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
HI-75/96/014/A
Code_Aster ®
Version
3
Titrate:
HPLA100 - Heavy thermoelastic hollow Cylindre in uniform rotation
Date:
18/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.03-A
Page:
3/12
1
Uniform loading of rotation around OZ
1.1
Axisymmetric model 2D
The density of centrifugal force is: 2 R er.
One considers the boundary conditions following:
uz R
(, Z) = 0 in Z = 0 and Z = L
One postulates displacement in the form:
ur = U R
(); uz = U = 0
As follows:
U
rr = u'; =
;
R
zz = rz = Z = R = 0
Z
B1 B B2
L
!!
0
R
A1 A A2
H
R
Handbook of Validation
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Code_Aster ®
Version
3
Titrate:
HPLA100 - Heavy thermoelastic hollow Cylindre in uniform rotation
Date:
18/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.03-A
Page:
4/12
The elastic constraints are expressed:
E
U
rr = (
1 - u'
+
1+) (1 -
2) (
)
R
E
U
= (
1 -
+ '
1 +) (1 -
2) (
)
U
R
E
U
zz =
'
(
+
1+) (1 -
2)
U
R
The radial equilibrium equation is written:
(R) -
2
= - 2 R
rr R,
As follows:
(Ru) ''
(1+) (1 - 2)
2
R
éq 1.1-1
R
= -
(1) E
Note:
U
Ru
() '
+ u' =
R
R
From where the general solution:
3
(
1+ 1 - 2
R
B
U R)
(
) (
)
= -
2
(
+
+
éq 1.1-2
1 -)
Ar
E
8
R
The constraints are then:
3 -
2
r2
E
2
B
rr (R)
= -
.
+
With - 1 -
2
1 -
8
(1+) (1 -
2)
(
)
r2
1 +
2
r2
E
2
B
(R) = -
.
+
To + 1 -
2
éq 1.1-3
1 -
8
(1+) (1 -
2)
(
)
r2
r2
2
2
E
zz (R) = -
.
+
With
1 -
2
(1+) (1 -
2)
The boundary conditions in constraints are:
H
rr = 0 in R = R ± 2
Handbook of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
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Code_Aster ®
Version
3
Titrate:
HPLA100 - Heavy thermoelastic hollow Cylindre in uniform rotation
Date:
18/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.03-A
Page:
5/12
One notes:
H
X = 2R
One obtains thanks to [éq 1.1-3]:
3
(- 2) 1 (+)
B =
2 R4 1 - x2
() 2
8 1
(-) E
then:
(3 - 2) 1
(+) 1 (- 2)
With =
2 R2 1 + x2
()
4 1
(-) E
Numerical application:
R = 20 mm; H = 1 mm; = 8.106 kg/mm3; = 1 s1
E = 2.105 NR/MM2; = 0.3.
From where: With = 7.13588.109; B = 3.561258.106 mm2
Note:
1
(+) 1 (- 2) 2 = 3.714286.10-12 mm2
1
(-) E
8
2 = 1.714286.10-6 MPa.mm-2
1 -
2
As follows:
· in internal skin:
ur = 2.9424.10-7 mm;
zz = 0.99488.10-3 MPa
· in external skin:
ur = 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:
H2
p = 2 H R 1
+
12 R2
The solution is membranous, normal balance is written:
NR = p R
Handbook of Validation
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Code_Aster ®
Version
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Titrate:
HPLA100 - Heavy thermoelastic hollow Cylindre in uniform rotation
Date:
18/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.03-A
Page:
6/12
W
The membrane deformation is: E =
, whereas E
R
zz = 0 = K = Kzz. In elasticity:
E H
NR =
E
1 - v2; Nzz = v NR; M = 0
From where the solution (arrow and circumferential normal effort):
1 - 2
() 2 H2
H2
W =
R3 1
+
; NR
E
12 R2
= 2R2 H 1 +
12 R2
Axial stress is worth:
H2
zz
= 2R2 1+
constant in L
(
'thickness)
12 R2
H2
If one does not take account of the correction of metric, term 1 should be removed
+
in
12R2
preceding expressions.
Numerical application (without correction of metric):
p = 1,600000.104 MPa
W = 2,912000.107 mm
Nzz = 0,96000.103 NR/mm
zz = 0,96000.103 MPa
Handbook of Validation
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Code_Aster ®
Version
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Titrate:
HPLA100 - Heavy thermoelastic hollow Cylindre in uniform rotation
Date:
18/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.03-A
Page:
7/12
2
Loading of gravity
2.1
Axisymmetric model 2D
The density of force is: - G ez (vertical gravity).
One considers the boundary conditions following:
uz 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 = - uz, Z = - zz;
E
rz = 0 = R = Z
One observes as follows:
ur, R = ur U () = - A' (Z) R
R
R R, Z
Then:
- A' (Z) =
()
rr = - zz
uz, Z R, Z = A' Z
()
U ()
()
Z R, Z = A Z + B R
()
From rz = 0, one draws:
B R
() - R A " (Z) = 0
that is to say:
With Z
() = cste =;
B R
() = R
Boundary conditions in effort, one obtains:
G z2
G r2
With (Z) =
+; B R
() =
2nd
2nd
Handbook of Validation
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Code_Aster ®
Version
3
Titrate:
HPLA100 - Heavy thermoelastic hollow Cylindre in uniform rotation
Date:
18/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.03-A
Page:
8/12
R2
Lastly, checks: = - G 2nd
As follows:
- G Z R
G
U
2
2
2
R (R, Z)
=
; uz (R, Z) =
Z + R - R
E
2nd (
(
)
éq 2.1-1
zz (R, Z) = G Z
Numerical application
G = 10 NR/kg; = 8.10-6 kg/mm3; R = 20 mm; L = 10 mm
E = 2.105 NR/MM2; = 0.3; H = 1 mm
· in internal skin:
U ()
R L = - 2.34000.108 mm;
()
zz L = 8.0000.104 MPa;
uz O
() = - 1.185000.109 mm
· in external skin:
U ()
R L = - 2.46000.108 mm;
()
zz L = 8.0000.104 MPa;
uz O
() = 1.215000.109 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 ez
The boundary condition on the circle of support is: uz Z
() = 0 in Z = 0
The solution is membranous, vertical balance is written:
Nzz, Z = G H
Moreover: NR = 0. In elasticity, one deduces then:
- NR
G Z
G
E
zz
= W =
= -
; E
U () =
Z 2
R
E H
E
zz = uz, Z = Nzz
E H
Z Z
2nd
Axial stress is:
(
)
zz
= G Z
constant in the thickness
Numerical application:
F = 8.104 NR/mm
W L
() = - 2.4000.10-8 mm
Nzz (L) = 8.0000.104 NR/mm
zz (L) = 8.0000.104 NR/mm
Handbook of Validation
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Code_Aster ®
Version
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Titrate:
HPLA100 - Heavy thermoelastic hollow Cylindre in uniform rotation
Date:
18/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.03-A
Page:
9/12
3 Loading
thermomechanics
3.1
Axisymmetric model 2D
T + T
-
S
I
(T T
S
I)
T (R) - T (R) =
+
(R - R
ref.
)
éq 3.1.- 1
2
H
Z
H
Ts
R
Ti
R
One postulates displacement in the form:
ur = U R
(); uz = U = 0
with the boundary conditions suitable. Thus, the elastic constraints are expressed:
E
U
E
rr = (
1 - U +
T - T
1 +) (1 -
2) (
)
(ref.)
R - 1 -
2
E
U
E
= (
1 -
+ U
T - T
1 +) (1 -
2) (
)
(ref.)
R
- 1 -
2
E
U
E
zz =
(
+ U
T - T
1 +) (1 -
2)
(ref.)
R
- 1 -
2
The radial equilibrium equation R
()
rr
-
, R
= 0 give:
(Ru)
(1+)
=
T - T
éq 3.1-2
R
1
(ref.)
(-)
Handbook of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
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Code_Aster ®
Version
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Titrate:
HPLA100 - Heavy thermoelastic hollow Cylindre in uniform rotation
Date:
18/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.03-A
Page:
10/12
From where the general solution:
1+ (T - T
2
S
I)
(
R
B
U R)
(
)
=
(
+
+
éq 3.1-3
1 -)
Ar
H
3
R
The constraints are then:
E (S
T - I
T) R
R
E T + T
S
I
rr (R)
=
-
H
1 -
2
(
3 1 -) - 1 -
2
2
E
B
+
(
With - 1 -
2
1 +) (1 -
2)
(
)
r2
E
(St - iT) R
2r
E T + T
S
I
(R) =
-
-
H
1 -
2
(
3 1 -) 1 -
2
2
éq 3.1-4
E
B
+
(
To + 1 -
2
1 +) (1 -
2)
(
)
r2
E (S
T - I
T) R
R
E T + T
S
I
zz (R)
=
-
H
1
-
2
1 - - 1 - 2
2
2nd
+
(
1 +) (1 - 2) A
H
H
The boundary conditions in efforts are: in R = R ±
. One obtains
2
rr = 0. One notes: X = 2R
thanks to [éq 3.1-4]:
T
(
)
B =
S - Ti (
) R3 1 - x2
() 2
6. 1
(-) 1+
then:
(T - T
S
I) R
T
2
+ T
With
(1)
S
I
=
+ -
3
1
2
6 (
- -
X
+
H 1 -) (
(
))
2
Numerical application:
R = 20 mm; H = 1 mm; = 10-5°C-1; Ts = - Ti = 0.5°C; = 0.3;
E = 2.105 NR/MM2.
From where: With = - 0.18569881.10-3; B = 0.02473096 mm2
Note:
1
(+) Ts - Ti = 0.61904762.10-5
1 -
3 H
Handbook of Validation
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Titrate:
HPLA100 - Heavy thermoelastic hollow Cylindre in uniform rotation
Date:
18/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.03-A
Page:
11/12
· in internal skin:
ur = 1.056145.106 mm;
zz = 1.4321427 MPa.
· in external skin:
ur = 1.110317.106 mm;
zz = - 1.4250001 MPa.
If one takes Ts = +Ti = 0.1°C:
To = 0,00130000.10-3; B = 0,0 mm2
As follows:
· in internal skin:
ur = 25.350000.106 mm;
zz = - 0.200000 MPa.
· in external skin:
ur = 26.650000.106 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:
E H
E H T + T
T - T
S
I
S
I
H
NR
=
2 (E
+ E
zz) -
+
1 -
1 - 2
12
R
éq 3.2-1
E H
E H
T + T
T - T
S
I
S
I
H
Nzz =
2 (E
+ E
zz) -
+
1 -
1 - 2
12
R
and:
E h3
E H2 T + T H
M =
2
(
+
-
+
-
12 1 -) (K
K
S
I
zz)
(
12 1 -)
T
T
2
R
S
I
éq 3.2-2
E h3
E H2 T + T H
M
S
I
zz
=
2
12 1
2
(
+
-
+
-
12 1 -) (K
K
zz)
(-)
T
T
R
S
I
H
According to these expressions, thermal terms in
are to be neglected if one does not consider
R
correction of metric in the thickness, i.e. in the case of usual models.
In our situation:
W
E =
; E
R
zz = 0
; K = Kzz = 0
Handbook of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
HI-75/96/014/A
Code_Aster ®
Version
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Titrate:
HPLA100 - Heavy thermoelastic hollow Cylindre in uniform rotation
Date:
18/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.03-A
Page:
12/12
Normal balance with the hull is written:
NR = 0
from where the arrow:
T
H
W = 1
(+) T
S + Ti + S - Ti R
2
12
R
and:
T
NR
S + Ti
zz
= E H
+ Ts - Ti H
2
12
R
M
zz
= - E H2 (
) + Ts + Ti H
12 1
(-) Ts - Ti
2
R
As the second member of dilation does not take account of the correction of metric, terms
out of H/R above are neglected.
Numerical application
R = 20 mm; H = 1 mm; = 10-5°C-1; Ts = - Ti = 0.5°C; = 0.3;
E = 2.105 NR/MM2.
From where:
Mzz = - 0.2380952 NR
in internal skin: zz = 1.449319 MPa *;
or
zz = 1.428571 MPa (without correction of metric)
If one takes Ts = +Ti = 0,1°C:
W = 26.00000.106 mm
Nzz = 0.2 NR/mm
Mzz = 0.001190476 NR
in internal skin: zz = - 0.2122466 MPa *;
or
zz = - 0.200000 MPa (without correction of metric)
* The constraints in the thickness with correction of metric are given by:
()
H2
12 x3
zz x3
=
Nzz - Mzz/R
H 1 - H2/12 R2
(
) + Mzz - Nzz
12 R h3 1 - H2/12 R2
(
)
Handbook of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
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