2. heat transfer and thermal stresses · the equation describing this heat transfer is fourier’s...
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FEM II - Lecture 2 Page 1 of 14
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2. HEAT TRANSFER AND THERMAL STRESSES
Temperature may influence the strength of a structure by:
- thermal expansion effect (thermal stresses)
- impact of the temperature on the mechanical properties of materials
Young’s modulus E (x104 MPa) thermal conductivity (W/(mK)) density (kg/m3)
0
1
2
3
4
5
6
7
8
9
0 200 400 600 800 1000
Temperatura (K)
0
50
100
150
200
250
300
350
0 200 400 600 800 1000
Temperatura (K)
2,54
2,56
2,58
2,6
2,62
2,64
2,66
2,68
2,7
2,72
2,74
0 200 400 600 800 1000
Temperatura (K)
thermal expansion coefficient (1/K) specific heat c (J/(kgK))
0
0,5
1
1,5
2
2,5
3
3,5
4
0 200 400 600 800 1000
Temperatura (K)
0
200
400
600
800
1000
1200
1400
0 200 400 600 800 1000
Temperatura (K)
Thermo-mechanical properties of aluminium (properties vs temperature in K)
FEM II - Lecture 2 Page 2 of 14
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Thermal expansion coefficient
( ) ( ) ( )( )
( ) ( )
ref
ref ref
l T l T l TT
T T l T l T T
{ } { } { }T s 0
0
0
x x x
y y y
z z z
xy xy
yz yz
xz xzT s
T
T
T
T0 – reference temperature for an unstrained state, in isotropic case αx= αy= αz= α
Example – simple relation temperature-stress in the case of statically indeterminate constraints
=
x +
T
x
=
x T
l
Δl
Δl
The simple bar fixed at both ends
Loads
=
change of the
temperature
compression
Thermal stresses may be caused by
- non uniform temperature field
- temperature change and nonhomogeneous
materials
- temperature change and statically
indeterminate constraints (reactions)
Total strain vector is the sum of thermal strain and
elestic strain vectors.
( Hooke’s law { } [ ]{ }sD ! )
The elongation equals 0 → longitudinal strain εx= 0
0x xT xs ,
xT T , /xs x E .
xs xT T
x xsE E T
For typical steel
( 52 10 MPa,E 0,3 , 51,2 10 1/ C ) and
100 CT the result is 240 MPax
FEM II - Lecture 2 Page 3 of 14
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The standard approach in thermal stresses analysis using FEM:
A. The heat flow analysis (steady state or transient)
B. The stress analysis using the current temperature field as a kind of body forces
A.THERMAL ANALYSIS
Partial differential equation describing transient heat flow through a solid (law of conservation of energy):
( , , , )x y z v
T T T Tc q x y z t
t x x y y z z
,
isotropic case 2 2 2
2
2 2 2d v d v
T T T Ta q a T q
t x y z
,
where d=/c is the thermal diffusivity
Thermal properties of selected materials at 20C (RT)
Material Thermal expansion coefficient
(1/0C)
Thermal conductivity
(W/mK)
Specific heat
c (J/kgK)
Density
(kg/m3)
Copper 1,7·10-5 390 400 9000
Aluminium 2,4·10-5 210 900 2700
Pine wood 0,4–0,6·10-5 0,1–0,5 1300–2700 500–700
Steel 1H13 1,1·10-5 29 440 7700
Glass 0,05–0,09·10-5 0,7–1,3 600–800 2500
Rubber 7,7·10-5 0,16 1400 1200
( , , , )T x y z t – temperature,
vq – int. heat generation rate per unit volume(W/m3),
, ,x y z – heat conductivity coefficients(W/mK),
– density (kg/m3),
c – specific heat (J/kg).
FEM II - Lecture 2 Page 4 of 14
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Three modes of heat transfer: conduction, convection, radiation
Conduction
Is the transfer of thermal energy through the solid body or fluid
due to the temperature gradient.
The equation describing this heat transfer is Fourier’s law.
For an isotropic medium: grad( )q T
where q is the rate of heat flow per unit area (heat flux) and is the thermal conductivity.
Convective Heat Flow
The transfer of thermal energy between the solid object and its environment due to fluid motion.
The rate of heat flow across a boundary is proportional to the difference
between the surface temperature and the temperature of adjacent fluid
0( )k cq T T (Newton's law)
where k is the convection coefficient (film coefficient)
Typical magnitudes of convection coeeficient (W/(m2K))
Medium (fluid) Free convection Forced convection
gas (air) 5–30 30–500
water 30–300 300–20000
oil 5–100 30–3000
liquid metals 50–500 500–20000
The simplest case of heat flow – steady state heat transfer with constant isotropic material properties.
In that case the heat flow equation reduces to Poisson’s equation: 2 0T +f =0
Radiation Heat Exchange
Stefan – Boltzman's law:
4 40e T CT ,
8 2 40 5,67 10 W/m K
emissivity of the surface ( 0 1 )
The heat exchange between two parallel surfaces
4 4( /100) ( /100)AB AB o A Bq C T T ,
Co=108
1
1/ 1/ 1AB
A B
In computational practice heat exchange across boundary
(by radiation and convection) is usually described by the
convection model 0( )k cq T T where k is the adequate
function of temperature.
FEM II - Lecture 2 Page 5 of 14
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FE method for Poisson’s equation in 2D space
2 2
1 22 2
1 2
( , ) 0T T
f x xx x
,
where 1 2( , )x x x ,
Boundary conditions
0
0
( ) ,
( )( ) ,
u
q
T x T x
T xq x q x
n
(prescribed temperature or prescribed thermal flux)
Minimized functional in FEM formulation
,),(22
1)( 021
2
2
2
1
TdqdTxxfx
T
x
TTI
q
1
LE
e
i
i = 1,…, LE
LE – number of the elements in the domain
Approximation of the unknown function (temperature) within an element
obszar
kontur
21
e
węzły elementy
u(x ,x )
2x
x 1
u1
u2
u3
u8
u7
u6
u5
u4
e
1
2
3
4
5
6
7
8
LWE=8
x 2
x 1
1 2 1 2
1
( , ) ( , )LWE
i i
i
T x x N x x T
iT , i = 1,...,LWE nodal temperatures,
Ni(x1,x2) – shape functions.
LWE – number of nodes of the element
nodes elements
domain Ω
Boundary Γ
FEM II - Lecture 2 Page 6 of 14
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2 2
1 2 0
1 11 2
1( ) 2 ( , )
2i j
LE LK
i j
i j
T TI T f x x T d q Td
x x
LK number of element sides on q .
Within the finite element:
11 1
12 2
,
.
LWEi
i
i
LWEi
i
i
T NT
x x
T Tu
x x
Finally the mimimized functional is replaced by the function of several variables ui
11 12 13 1 1 1
21 22 23 2 2
31 321 2 3 1 2 3,3 3
1
1( ) , , ,..., , , ,
2
LW
LW LW
LW LW LW LW LW
k k k k T b
k k k T b
k kI u T T T T T T T TT b
k k T b
1 1 1 1
1
2LW LWLW LW LW LW
I T TK T b
.
Minimum- necessary (and sufficient) conditions:
0i
I
T
, 1, ,i LW . => K T b + Dirichlet b.c.
FEM II - Lecture 2 Page 7 of 14
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B. Thermal stresses – FE equations in the case of mechanical and thermal loads
T s
, 1
sD
.
[D] – material stiffness matrix [D] –1 material flexibility matrix
e
B q . [B] – element strain matrix
: s T e TD D D B q . (*)
FEM equations derived from the principle of virtual work - taking into account thermal strains
e
q - vector of nodal virtual displacements of an finite element
e
B q - vector of virtual strains within the element
Principle of virtual work for an element
e
eeeq F d
.
0,
0.
e
e
T
eee e
T
eee
q F q B d
q F B d
0
e
T
eeF B d
Using Hooke’s law (*) we get:
Te e eek q F F ,
e
T
eek B D B d
- element stiffness matrix
e
e
T
T e TF B D d
- additional
vector of nodal forces
(nodal forces caused by temperature)
FE set of equations for the model
TK q F F
0
0
0
x
y
z
T
T
T
T
T
FEM II - Lecture 2 Page 8 of 14
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Thermal stresses in rod elements
Basic relations for 2 node rod element
1
1 2
2
( ) , ,e
qu N N
q
the shape functions :
1
2
( ) 1 ,
( ) .
e
e
Nl
Nl
1
1 2
2
( ) , ,e
qduN N
qd
( ) ( ) TE .
1
1 2
2
( ) ( ), ( )T
T N NT
, e
T
T ee TF B D d
Vector of thermal nodal forces
e
1 1
e 1 2 e
2 2
1 11 1 1 2
2 1 2 2 2 20
,
1 1
2 2,
1 1
2 2
e
e
T
T e T
l
N TF B D d E N N d
N T
T TN N N NEA d EA
N N N N T T
1 21
12T e
T TF EA
.
1 2q 2q 1
T1 2T
T( )
1 2
2
,T T N
B N NN
,
D E ,
1
1 2
2
,TT
TT N N
T
.
FEM II - Lecture 2 Page 9 of 14
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EXAMPLE
Find the elongation of the rod loaded by the force P and the temperature distribution ΔT
Te e eek q F F ,
1 1 1 2
2
1 1 1
1 1 12
q FEA T TEA
q Pl
.
1 0q
1 22
2
EA T Tq P EA
l
,
l
TT
EA
Plq
2
212
,
EXAMPLE
Find the stresses in both part of the constrained with heated part AB
,
1 2
T1 2T
P
a
BA CE,A
1
1q
2
q2
3
q31 2
T
1 3( 0, 0)q q => 1 2
2
1 2
l lEA q EA T
l l
.
1 2
2
1 2
( )l l a l aq T T
l l l
,
Element 1 : 1 2
1 1 2
2 2
01 1, ,
q q l aN N T
q qa a a l
,
1 1( )a
E T TEl
.
Element 2: 2 2 2
2 1 2
3
1 1, ,
0
q q q aN N T
q l a l a l a l
,
2 2( 0)a
E TEl
.
1 11 1
2
1 1 2 2
3 3
2 2
1 10
1 1 1 1
1 10
l lq F
EA q EA Tl l l l
q F
l l
FEM II - Lecture 2 Page 10 of 14
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EXAMPLE OF ENGINEERING PROBLEMS OF THERMAL STRESSES FE analysis of a high pressure T-connection (steady state problem)
The aim of the analysis was to find stress and strain distribution in a T-connection
caused by high internal pressure (2600 at) and the non-uniform temperature.
External cooling, assembly procedure (screw pretension), contact and plasticity
effects have been included. The project done for ORLEN petrochemical company
FE model
C
190205220230250265280295310
Temperature distribution
MPa
050100200300400500600700
Von Mises stress
FEM II - Lecture 2 Page 11 of 14
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FEM analysis of local stress concentrations caused by impact thermal loads
Analysis description:
The main purpose of the analysis was to find stress concentration close to notches during the heating-cooling process.
Temperature applied at the upper surface, T(t), had been taken from experiments. Analysis were performed for three notch’s radius R (R = 0.5, 1.0, 2.0 mm).
Initial and boundary conditions:
Temperature of the bottom surface (line in 2D model) T = T(t0) is constant during the process.
Reference temperature for structural analysis TREF = T(t0).Material properties – functions of temperature. No initial stress. Plane
strain, Transient analysis, Elasto-plastic material behaviour
5
0
100
2
0
1
0
1
0
2
0
R
Dimensions of the analysed model
Notch B
Notch A
T=T(t)
T0
150
200
250
300
350
400
450
325 330 335 340 345 350 355 360 365 370
Temperature as the function of time at the surface of the die (from
experiments)
FEM II - Lecture 2 Page 12 of 14
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1
2
3 4
5
6
Temperatures in points 1 to 6 during the process
t
h
x
y
T0
T0+T
g
Model of the body with the thin layer subjected to heating- simple analytical considerations
Assuming h t within the heated layer is 0,
0,
1,
1
0.
x
y
x y
z
E Tv
0,
0,
1,
1
0.
x
y
x y
z
E Tv
For T200C the result is x = y = 685MPa
grad( )q T ( )k b sq T T Biot number k l
Bi
gives a simple index of the ratio of the heat transfer resistances inside of and at the
surface of a body. This ratio determines whether or not the temperatures inside a body will vary significantly in space, while the body heats or cools
over time, from a thermal gradient applied to its surface.
FEM II - Lecture 2 Page 13 of 14
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Von Mises (SEQV) and principal stresses at notch A. Notch radius R=0.5 mm
Von Mises stress distribution at time tx. Notch radius R=0.5 mm.
FEM II - Lecture 2 Page 14 of 14
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FE analysis of thin-walled elements' deformation during aluminium injection moulding
Numerical simulations have been performed to model the process of filling the mould by hot aluminium alloy. The analysis has enabled
improvements of the element stiffness diminishing geometrical changes caused by the process. Fluid flow simulation with transient thermal analysis
including phase change have been performed, followed by the structural elasto-plastic calculation of residual effects.
FE model
Velocity field during
injection process
Temperature distribution (cooling effect) and displacements
Residual stress distribution
FE model of the die
Velocity and temperature distribution