2. HEAT TRANSFER AND THERMAL STRESSES · The equation describing this heat transfer is Fourier’s law. For an isotropic medium: qT grad( ) where q is the rate of heat flow per unit

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)


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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


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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).


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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.


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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


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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.


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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


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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

.


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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


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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


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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)


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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.


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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.


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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

Documents

2. HEAT TRANSFER AND THERMAL STRESSES · The equation describing this heat transfer is Fourier’s law. For an isotropic medium: qT grad( ) where q is the rate of heat flow per unit