Thermal analysis
•There is a resemblance between the thermal problem and the stress
analysis.
•The same element types, even the same FE mesh, can be used for
both analysis
•Thermal analysis means primarily calculation of temperature within a
solid body
•A by-product of the temperature calculation is information about the
magnitude and direction of heat flow in the body.
•The results of thermal analysis, nodal temperature, can be used to
evaluate the distribution of stresses due to the presence of a thermal
gradient within the body
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Pipe with thermal gradient along the thickness
Flux= 0
Internal diameter = 40mm
Thickness = 10mm
T =25 °C
r
T =100°C
a
Length =600mm
Flux= 0
• Thermal conductivity 20 W/m°C
• Coefficient of thermal expansion 12 10-6 /°C
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Gradient along the thickness
T ( r )  Ti 
Ti  T e
ln( re / ri )
ln
r
ri
q  k
dT
dr
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W/m 
2
3
Thermomechanical analysis
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The thermal problem
Heat moves within the body by conduction
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Fourier equation of heat flow
Consider an isotropic material and image that exist a
temperature gradient in the x direction
fx  k
T
x
fx, heat flux for unit area (W/m2)
k thermal conductivity (W/m°C)
Negative sign means that heat flows is in a direction
opposite to the direction of temperature increase
The temperature represents the potential for the heat
flow.
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Hp: non isotropic material
the direction of the heat flow in general is not parallel to the direction of
the temperature increase
f x  k
T
x
 fx 
T x 
 


f


κ

T

y
 y


f 
 T z 
 z


fx, fy, fz heat flux for unit area (W/m2)
 matrix of thermal conductivity (W/m°C)
Boundary condition on surface with normal n
T
S
 T
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kn
T
n
 f
S
s
f
7
Conductivity
•The matrix of conductivity is in general a full 3 by 3 matrix
•If x,y,z are principal axes of the material  is a diagonal matrix
•In the special case of isotropy 11=22=33= 
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By considering a differential element of volume dV and writing
the balance equation (rate in – rate out)= (rate of increase
within), it is obtained:
 fx 
 

  
T
-
  f y   q v  c
t
 x y z   
 fz 
where:
qv is the rate of internal heat generation for unit volume (W/m3)
c is the specific heat (J/kg °C)
 is the density (kg/m3)
t is the time (s)
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If  T /  t   0 the problem becomes steady-state. For material
isotropic the equation becomes
  k  T    q v
Where  is the gradient operator.
If, in addition, k is independent of the coordinates results
k T   qv
2
or
2
2
  2T
 T
 T
k 


2
2
2

x

y

y


   q v

Solution: T=T(x,y,z) that also meets prescribed boundary
conditions
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• The heat flow trough the surface of the body is analogous to surface
load in the stress analysis.
• A distributed internal source is analogous to body forces in stress
analysis.
• Prescribed temperature are analogous to prescribed displacements
• For a steady state problem, the mathematics of all this leads to the global
FE equation
K TT  Q
where:
KT depends on the conductivity of the material ([W/°C])
T is the vector of the nodal temperature ([°C])
Q is the vector of the thermal loads ([W])
•Convection and radiation boundary conditions, if present,
contribute term to both KT and Q.
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Hypothesis for FE in thermal analysis
• In solid body the heat transfer occurs by conduction
• In fluid the transmission by mean of convection must be take into
account
• In many structural problems the thermal problem and the
mechanical problem are not coupled
• This hypothesis is no longer valid when the deformation can
generate heat and cause a variation of the mechanical properties
• Thermal conductivity and other properties of the material must be
considered dependent on the temperature
• The conductivity matrix can be generated by a direct method for
very simple elements, otherwise a formal procedure is required
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Thermal bar element
The rate of heat flow, q=Af, is constant and direct axially.
T1>0, T2=0
T2>0, T1=0
Nodal heat flow is positive when direct within the element
An actual bar can be modeled by several of these elements if
temperature at several locations between its ends were required
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In matrix format, these results are:
 Ak / L   T1   q1 
 Ak / L

   
T
q
 Ak / l
Ak / l

      2   2 
kT
 the element conductivi ty matrix 
Temperature within the element are obtained by interpolating
nodal temperatures
T
T  N 1
N2
...
 1
 
T 2 
N n  
 ... 
T n 
 
T  NT e
•The form of the interpolation determines the complexity of the
temperature field that an element can represent
•The shape function can be exactly those used also to interpolate a
displacement field
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In Cartesian coordinates, temperature gradients in a plane element are
T / x  N 1 / x

 
T / y  N 1 / y
T  BT
where
N 2 / x
...
N 2 / y
...
 T1 
 
 N n /  x  T 2 
 
 N n /  y   ... 
T 
 n
 / x 
B 
N

/

y


For solid a third row expressing T/z is added
The expression for an element conductivity matrix can be shown to be
kT 
 B κB dV
T
element
volume
The array of thermal conductivities  becomes the scalar k for bar elements.
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Remarks
• The thermal problem is a scalar field problem
• A FE temperature field is continuous within elements and across
interelement boundaries
• Temperature gradient, like strains in stress analysis, are typically not
interelement continuous
• If radiation is not present, the rate of heat flow is proportional to
temperature differences
• Upon assembly of elements, nodal rates of heat flow qi from separate
elements are combined at shared nodes.
• Thus the net flow into node i of the structure: Q i 
q
i
0
except at nodes where Ti is prescribed, nodes on a structure boundary
across which heat is transferred, or at internal nodes where Qi arise from qv
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Convection boundary conditions
f  h (T f  T )
Where
f is the flux normal to the surface and positive inward
h is the heat transfer coefficient
Tf e T temperature of the fluid and of the surface of the body
For the increment, dS, of the element surface subjected to convection
formal procedure gives
matrix :  N N hdS
T
vec tor :  N hT f dS
T
The matrix combine with the element conductivity matrix and hence
contribute to KT the vector contribute to Q
If the material conductivity is dependent on temperature a
nonlinearity is present
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Radiation boundary conditions
Consider two infinite parallel planes. Let the planes have temperature
Tr and T , and each is a perfect absorber and perfect radiator (black
body).
If SB is the Boltzman constant, the net heat flux received by the
surface at temperature T is
f 
 SB T r
4

received heat flux

 SB T

  SB T r  T
4

4
4

radiated heat flux
If the emissivity  (ratio of total emissive power to that of a blackbody
at the same temperature) is introduced, than
f 
 SB
(1 /  )  (1 /  r )  1
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T
4
r
T
4

18
Accounted for the fact that the surface are not parallel, often not flat,
and certainly not infinite, a shape factor is introduced
f  F  SB
T
4
r
T
4

F is a factor that accounts for the geometry of the radiating surface and
their emissivities
The calculation of F is sufficiently complicated that it may be done by
a separate computer program.
The flux f is an average value.
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This equation can be written as

4
f  hr Tr  T
4


dove h r  F  T r  T
2
2
T
r
T

T is the absolute [K]
hr= hr(T) is a temperature dependent heat transfer coefficient
This makes the problem non linear and requires iterative solution
•The flux expressions have the same form for convection and
radiation boundary conditions
•Accordingly, a radiation boundary condition leads to matrix and
vector expressions having the same form as for convection, with
h and Tf replaced by hr and Tr
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Modeling considerations
• Element types, and shapes for a thermal analysis may be dictated
less by thermal analysis than by an anticipated stress analysis
based on the same mesh
• The mesh demands of stress analysis are usually more severe
• Also a modest temperature gradient may create forces of constraint
that produce large strain gradients, especially near holes, grooves
and other stress raisers.
• Also for thermal analysis the present symmetry must be considered:
heat flux trough a plane of symmetry is zero
• Heat flux must be parallel to an insulate boundary or a plane of
symmetry
• Temperature contours (isotherms) should be parallel to a boundary
of constant temperature and normal to plane of symmetry
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Non linearity
• If in the body there are appreciable difference of temperature,
it is necessary to regard conductivity as temperature
dependent
• If there is convection with a temperature dependent coefficient
of heat transfer
• If is present radiation
K T  K T (T )
Q  Q (T )
K T (T ) T   Q (T )
The problem is non linear
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•Iterative solution based on the method of direct substitution:
- assume an initial temperature field T0
- generate KT and Q based on these temperature
- solve the equation
K T (T0 ) T1   Q (T0 ) for T1
- use the newly computed temperature for new values of k, h, and hr
- generate a new KT and Q
- solve K T (T1 ) T2   Q (T1 )  for a new T
• Iteration is stopped when a convergence test is satisfied.
•For example
ei 
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Ti  Ti  1
 fixed value
Ti
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Thermal transient
•When steady state conditions do not prevail, temperature change in a
unit volume of material is resisted by “thermal mass” that depends on
the mass density  and the specific heat c.
•The solution become
K T T  C T  Q
where
T
T 
t
Q  Q (t )
and the heat capacity matrix C 
c
assemble
is built assembling
element
heat capacity matrix
c
 N N  cdV
T
The solution must be integrated with respect to time
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Dependence on temperature of some material
properties
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Flanged Joint
• Sections of a pipe are connected by a flanged joint.
• Each flange is connected to the pipe by two circumferential welds
• Bolts draw the flanges together and compress a gasket between them
Evaluate the steady state temperature field in the pipe and in the flange
for use in a subsequent stress analysis
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Data of the problem
• Boundary conditions:
- fluid in the pipe has temperature 0°C
- vapor condensing on the outside of the pipe has temperature 100 °C
• Heat transfer coefficient h:
- inside the pipe is 5000 W/m2
- outside the pipe is 20000 W/m2
• The material of the pipe and of the flanges is steel with the following
thermal properties
- Thermal conductivity = 20 W/m°C
- Coefficient of thermal expansion = 12 10-6 /°C
- Specific heat = 480 J/Kg °C
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Preliminary Analysis
• The maximum possible value of flux through the pipe wall is
approximated by regarding the wall as plane and using the
limiting surface temperature
f lim   k
T
r
  20
100  0
0 . 084  0 . 077
  286000 W / m
2
• This is a maximum value because:
- the inside of the pipe must be warmer than 0°C
- the outside of the pipe must be cooler than 100 °C
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Model
Axisymmetric solid of revolution
Plane of
symmetry
Plane of
symmetry
Gap
Radial
Axis of revolution
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Boundary conditions
• Between the welds, along IJ, pipe and flanges touch only at random
and isolated point, if at all.
• Conductivity across this cylindrical surface is very low compared with
solid metal and can be assumed equal zero
• This is a pessimistic assumption for eventual stress analysis, because
it increase the thermal gradient and therefore increase stresses.
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Temperature field
• The lowest temperature is about 3 °C along the right half of the inside
boundary AB
• The highest temperature is 99.99 °C along the outer surface CDEFG
• Temperature contour are interelements continuous, except along IJ
where discontinuity is expected
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Thermal flux
• Flux arrows are perpendicular to the temperature contour because the
material is isotropic
• Flux arrows agree with the expectation of preliminary analysis
• Radial flux near AB is 170.000 W/m2 which, as expected is less that the
approximate limiting value.
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Flux contour
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Thermal transient
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Thermomechanical analysis
• The mesh used for the thermal analysis is used again, now with
computed nodal temperatures used to load the model.
• Two significant question about boundary arise:
- should nodes along BC be fixed against axial motion or not? Allowing
movement gives no credit to resistance offered by the gasket and the bolts,
while fixity gives too much credit.
We elect to run the stress analysis twice, first allowing motion along BC and
second preventing it.
- is it proper to let nodes along IJ move independently?
If sides of the interface move apart, as the results shows, the answer is yes.
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Results thermomechanical analysis
BC is fixed
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Maximum and minimum stresses (in MPa) for different axial
restraint along BC
Only node B axially
restrained
Maximum
stress
Location
Minimum
stress
Location
All nodes on BC axially
restrained
r

z
r

z
113
216
244
105
229
257
I,J
L-N
I,J
I,J
M-B
I,J,K
-68
-142
-199
-107
-225
-279
G
G
G
G
G
G
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