Thermoelasticity - Southeast University

Thermoelasticity

Outline

• Heat Conduction Equation

• General 3-D Formulation

• Combined Plane Hooke’s Law

• Stress Compatibility and Airy Stress Function

• Displacement Equilibrium and Displacement Potentials

• Thermal Stresses in Thin-Plates

• Summary of Solution Strategy

• Polar Coordinate: Airy Stress Function

• Polar Coordinate: Displacement Potentials

• Axi-symmetric Problems – Direct Solution

• Thermal Stresses in Circular Plates

2

,i iq kT

Heat Conduction Equation

• Flow of heat in solids is associated with temperature

differences

• For isotropic case, the heat flux is related to

temperature gradient through thermal conductivity

3

• From the principle of conservation of energy, the

uncoupled heat conduction equation is given by

2

:mass density

:specific heat capacity at constant volume

:prescribed energy source term.

Tk T c h

t

c

h

• For zero heat sources and steady state, the heat onduction

becomes Laplace equation

2 0.T

• With appropriate thermal BCs, i.e. specified

temperature or heat flux, the temperature field can be

determined independent of the stress-field calculations.

• Once the temperature is obtained, elastic stress analysis

procedures can then be employed to complete the

problem solution.

• For us, the temperature distribution is usually a given

condition.

Heat Conduction Equation

4

General Formulation of Thermoelasticity – 3D

, ,

12ij i j j iu u

, , , , 0ij kl kl ij ik jl jl ik

, 0ij j iF

• Strain-displacement relations:

• Strain compatibility:

• Equilibrium:

• Thermoelastic Hooke’s Law:

1 2

, 3 22 2 1 1 2ij ij kk ij ij ij kk ij ij ij

GT T G T

G G

• Steady state heat conduction equation: 2 0T • 16 equations for 16 unknowns (3 displacements, 6 strains,

6 stresses and T):

,{ , ; , , , } 0i ij ij if u G F T

• 3-D thermoelastic problems are way too difficult…5

Formulation of Thermoelasticity – 2D

12 2 1

0 2 1 1 2

11 1 ,

21 1

1 1 ,2 2

23 2

1 2

1 12

1 2 1 2 1 2

ij ij kk ij ij

z z x y kk x y

x x y

y y x xy xy

ij kk ij ij ij

x x y

TG G

G T G T

TG

TG G

GT G T

G T

,

1 12 , 2

1 2 1 2 1 2y y x xy xyG T G

• Plane strain thermoelastic Hooke’s law

6


• Plane stress thermoelastic Hooke’s law

23 2

1 21 1 2 1

01 1 1 1

21 ,

12

1 , 21

12 2 1

1 12 1 1

ij kk ij ij ij

z z x y kk x y

x x y

y y x xy xy

ij ij kk ij ij

x x y

GT G T

T T

GT

GT G

TG G

G

,

1 1 1,

2 1 1 2y y x xy xy

T

TG G

7

2 11 3 31 , 2

2 4 2 1 1

11 31 ,

2 4 1

11 3 11 , .

2 4 1 2

1 3 4 1 ,1

11

x x y

y y x xy xy

x x y

y y

T G TG

TG

TG G

GT

G

3 4 1 , 2 .x xy xyT G


• Combined plane thermoelastic Hooke’s law

3For plane strain: 3 4 or , ;

43 3

For plane stress: or , 0.1 1

• Define two material constants that are related to ν

8

• Beltrami-Michell Equation:

2 22

2 2

22 2

2 2

22

2

2

1 11 3 1 3 11 , 1 , .

2 4 1 2 4 1 2

1 3 1 3 2 1 2 1 2

4 4 4 4

Add

y xyx

x x y y y x xy xy

xyx y y x

yx

y x x y

T TG G G

G T G Ty x x y

x

2 222 2

2 2 2

2 2

2 2

1 to both sides: 2 1 2

4

1Using Equilibrium on the RHS: 2 1

4

1

14

1

8

y xyxx y

yxx y

yxx y

G Ty x y x y

FFG T

x y

Fy

GT

F

x

2 222

3 33For plane stress: or , 0For plane strain: 3 4 or ,

1 141

111

y yx xx y x y

ET E T

F FF Fx y x y

Stress Formulation – 2D

Stress Function Formulation without Body Forces

• Air Stress Function Solution2 2 2

2 2, ,x y xyy x x y

• where = (x,y) is an arbitrary form called Airy’s stress

function. This stress form automatically satisfies the

equilibrium equation.


2

4

42

2

4

8 10

1

3For plane strain: 3 4 , For plane stress: , 01

10

0E

T

TT

G

E

10

• Solution of the Airy Stress Function

4 2

4 2

8 10

18 1

0, 01

h p h p

GT

GT

• The traction BCs in terms of Airy Stress Function

d d d d,

d d d dx y

x y y xn n

n s n s

2 2

2

2 2

2

d d dd d d

d d dd d d

nx x x xy y

ny xy x y y

y xT n n

y s x y s s y

y xT n n

x y s x s s x

• Integrate over a portion of the boundary

1 2d , dn nx yC C

T s C T s Cy x


• Consider the directional derivative of the Airy Stress

Function along the boundary normal

d d dd d

d d dn n

x y y xC C

y xn n T s T s

n x y s s

n t F

• where t is the unit tangent vector and F is the resultant

boundary force.• For many applications, the BCs are

simply expressed in terms of stresses.

• For the case of zero surface tractions:d d 0 .n C

• For simply connected regions, a steady

temperature distribution with zero

boundary tractions will not affect the

in-plane stress field.


d d d d,

d d d dx y

x y y xn n

n s n s

Displacement Formulation – 2D

• Navier’s equations

2 2

1 3 4 1 ,1

1 3 4 1 ,1

0, 0

2 20,

1 1

4 1

1

x

y xy

xy xy yxx y

x

G u vT

x y

G v u u vT G

y x y x

F Fx y x y

G u v G u vG u F G v

x x y y x y

G Tx

40

1

1 y

GF

Ty

2 2

2

3 33For plane stress: or , 0For plane strain: 3 4 o

1 2

r ,1 14

02 1 1 2 2 1

02 1 2 11 2

x

y

E u v E u vG u F G u

x x y x x y

E u vG v F

y

E Tx

E Tyx y

2

0

02 1

1

1

x

y

F

E u vG v F

y x

E Tx

Ey

Ty

13

• Stress/Traction Boundary Conditions

( ) ( ) ( )

( ) ( ) ( )

( , )on

( , )

1 31

1 31

4 1

1

4 1

1

n b b bx x x x xy y

tn b b by y xy x y y

x x x y

y y x y

T T x y n nS

T T x y n n

G u v u vT n n G n

x y y x

u v G v uT n G n n

y

GT

Tx x

G

y

Displacement Formulation – 2D

• Displacement Boundary Conditions

( , ) , ( , ) onb b uu u x y v v x y S

14

General Observation of Displacement Formulation

• From the displacement formulation, the solution to a 2-D

thermoelastic problem can be superposed by the original isothermal

solution and

• (a) additional body forces:

4 1 4 1, ,

1 1x y

G GT TF F

x y

4 1.

1n G

T T

• (b) additional normal surface traction:

1 4 13 ,

4

1

11 3 ,1

x

y xy

T

T

G u vx y

G v u u vG

y x y x

• Displacements are derived for the additional equivalent body and

surface forces. The stresses are:

15

Displacement Potential Formulation – 2D

• Navier’s (governing) equations without body forces

2 24 1 4 12 20, 0

1 1 1 1u v T u v T

u vx x y x y x y y

• Assuming that the displacement vector is derivable from

a scalar potential, i.e. a 2-D Lamé Strain Potential

2 2

: ,

4 1 4 1,

1 1

u vx y

T Tx x y y

u

• Integrating and dropping the constants of integration

2

2

2 1 3For plane strain: 3 4 , : For plane stress: , 0:

4

11 1

1

1

T T

T

• Solution of the displacement potential

2

2 2

4 1

14 1

0,1

h p h p

T

T

• From the standard potential theory

2 24 11, ln d d .

2 1

,

p

R

p pp

T x y

u vx y

• The total displacements

,h p h p h p h pu u u v v vx y

• Next, we try to solve for the homogeneous displacement.


17

2 22 20, 0

1 1

h h h hh hu v u v

u vx x y y x y

• The homogeneous solution satisfies the following

isothermal Navier’s equation

• The BCs for the homogeneous displacement = the

original conditions by subtracting the contributions

of the particular displacement

• Thus, with the particular solution known, the general

problem is then reduced to solving an isothermal case.


18

Thermal Stresses in an Elastic Thin-Plate

2

0 21y

T Tb

a ab

b

x

y

oT0

• Given temperature change:

• Navier’s equation for plane stress

2

20 21 1 1

yT T

b

2

2 4 2 20 2

0 02

2 42 4

0 2

2 12 1 1

(1 ) (1 ),

2 12

(1 )2 12

yAy By A By T

b

T TA B

b

y yAy By T

b

• Since the temperature load is independent of x, let’s try

the particular displacement potential

19

a ab

b x

y

o

ET0ET0

a ab

b x

y

o

ET0ET0


• Stresses due to the particular displacement potential

• The resultant surface forces • To satisfy the zero tractions BCs,

impose the load

• The exact solution to the homogeneous problem (above

right) is very difficult.

3

0 2

2 2 2

02 2 2

0, (1 )3

0, (1 ) 1 , 0

21

x y xy

x x

yu v T y

x y b

u v yT

x x y y b

G

2

0 2

21 1 ,

1y y y x

y GT E T

b

1 0, 0xyT

20

2 2 22

2 22 , 0, 0x y xycy cy x x y

2

0 22 1 , 0, 0x x x y y y xy xy xy

yc E T

b

• For a >> b, we may again ask Saint-Venant for help.

• Replace the parabolic surface traction with an equivalent

(uniformly distributed) surface load.

• As a result, the homogeneous Airy Stress Function

• Total stresses become

( ) 0, ( ) 0; ( ) 0, ( ) 0x x a xy x a y y b xy y b

• Zero tractions require

• The last three are automatically satisfied.


21

0

2( ) d 0, ( ) d 0 2

3

b b

x x a x x ab by y y c E T

2

0 2

1, 0, 0

3x y xy

yE T

b

x

y

o

+

-

02 3E T

0 3E T

• The first condition cannot be satisfied in a pointwise

sense.

• However, the resultant force of the surface traction must

vanish (Saint-Venant’s Principle)

• The approximated thermal stresses

• Accurate for regions 2b away from

the vertical boundaries.


22

• Determine the temperature variation from heat conduction

and energy equation, if not given.

• Under either stress or displacement formulation, identify a

particular solution due to temperature effects to the

governing equation (the Beltrami-Michell or Navier).

• Evaluate the resultant stresses at the domain boundaries.

• Solve the corresponding isothermal problem, whose BCs

are set as the general traction BCs subtracted by the

contributions due to temperature effects.

Solution Strategy to Plane Thermoelasticity

23

Polar Coordinate

• Strain-Displacement relationship

1 1 1, , .

2r r

r r r

u u uu uu

r r r r r

• Hooke’s law

2 11 3 31 , 2

2 4 2 1 1

1 11 3 1 3 11 , 1 , .

2 4 1 2 4 1 2

1 3 4 1 , 11 1

r r r r r

r r

T G TG

T TG G G

G GT

3 4 1 , 2 .r r rT G

3 3 3For plane strain: 3 4 or , ; For plane stress: or , 0.

4 1 1

1 1 20, 0.r r r r r

r r r r r r

• Equilibrium equations

24

• Airy Stress Function Representation

2 2 2 2 2 2

2 2 2 2 2 2 2 2 2

8 11 1 1 1 1 10

1

8 1 8 1For plane strain: For plane stress:

1 1 1

GT

r r r r r r r r r r r r

G GEE


Polar Coordinate – Stress Function Formulation

2 2

2 2 2

1 1 1, ,r rr r r r r r

• Axi-symmetric solution

2

2

2

1d d d d d, , 0

d d d d d8 11 d d 1 d d 1 d d

0d d d d 1 d d

8 11 d d 1 d 1 d dd d d 1 d d

r r r

r

rr r r r r r

G Tr r r

r r r r r r r r r

G Tr r r

r r r r r r r r

25

• Navier’s (governing) equations without body forces

2

2 2

2 2 2 2 2

2 2

2 2 2 2 2

4 120.

1 1

4 11 1 2 2 10,

1 1

1 1 2 2 1 11

r r r r r r

r r r

T

u uu u u u u u Tr r r r r r r r r r r

u u u u uu u ur r r r r r r r r r

u u

4 1 10.

1T

r

2 2 2

2 2 2 2

1 1 1 1 1, , , .r r ru u

r r r r r r r r r

u

2 22

2 2 2

2 2

4 11 11

1 3For plane strain: 3 4 , : For plane stress: , 0: 1

1 1

r Tr r r r

T T

• If the displacement is derivable from a potential

Polar Coordinate – Displacement Formulation

2 22

2 2 2

1 1 1, , 0

32

2 1

h h h hh

r r rr r r r r r

G

2

2

2

2 2

2 2 ,

1 12 2 2

12 2

h

r r

h h

h

r r

G Gr

G G Gr r r

G Gr r

• Particular solution + homogeneous solution

2 2 4 10,

1h h pp T

• Homogeneous solution


27


• Axi-symmetric solution: T = T(r)

21 1

4 1d 1 d d, 0, ,

d d 1 d

4 1 4 11 d, d ,

d 1 1

r r r

r r

Tu u r u ru

r r r r

Aru T A u Tr r Ar

r r r r

2

2

d d 1d, , 0

d d d

1 3 4 1 ,1

1 3 4 1 , 0;1

3For plane strain: 3 4 or , ;

43 3

For plane stress: or , 0.1 1

r rr r

r r

r r

u ur r r r r

GT

GT

28

221

1

For plane strain: For plane stress:1 1d ; d ;1

d d, ; , ;

d d2 2

1 1 , 1 ,1 2 1

2 21 1 .

1 2

rr

r r r rr r

r r r r

r

A Au Tr r Ar u Tr r Arr r r ru u u ur r r rG G

T T

GT

1 .1 r

GT

• Axi-symmetric solution: T = T(r)


29

Polar Coordinate – Stress Function Formulation

• Axi-symmetric solution

2

2 211 2

12

8 11 d d 1 d 1 d dd d d 1 d d

8 1 8 1d 1 d d dln 2

d d 1 d d 1

8 1 112lnd

41

r

r r

r

G Tr r r

r r r r r r r r

G GCTr r Tr Cr r Cr

r r r r r r

G CTr

rrr

3

2 2

CC

r

• For domains that include the origin, C1 and C3 must vanish.

• The hoop stress is then derived from

2

2

d d d dd d d d rrr r r r

• Since the stresses developed from the displacement

solution do not contain the logarithmic term, it is

inconsistent with single-valued displacements. Drop it.30

Thermal Stresses in an Annular Circular Plate

• After dropping the logarithmic

term, the stress formulation gives

322 2

dd ,

dr r

C EC Tr r r

r r r

• Zero tractions on boundaries

• The displacement solution is

31

2 2

2 2 2

2 22

2 2 2

d d ,

d d .

o

i i

o

i i

r ri

r r ro i

r ri

r ro i

r rET T

r r r

r rET T Tr

r r r

2 2

2 2

1 11 d d .

o

i i

r rir r r

o i

r ru T T

r r r


• To explicitly determine the stress distribution, the

distribution of temperature variation must be determined.

• Assuming steady state conditions

21 2

1 d d0 ln

d dT

T r T A r Ar r r

, 0 ln ln

ln lni o

i i or r i r r

i o o o i

T T rrT T T T

r r r r r r

• The two constants are determined from temperature BCs

• Substituting the temperature back to stresses

32

2 2

2 2 2

2 2

2 2 2

ln 1 ln2ln

1 ln 1 ln2ln

i o i o or

o i o i i

i o i o o

o i o i i

E T r r r rr r r r r r r

E T r r r rr r r r r r r


• A numerical example with ro/ri = 3

33

Outline

• Heat Conduction Equation

• General 3-D Formulation

• Combined Plane Hooke’s Law

• Stress Compatibility and Airy Stress Function

• Displacement Equilibrium and Displacement Potentials

• Thermal Stresses in Thin-Plates

• Summary of Solution Strategy

• Polar Coordinate: Airy Stress Function

• Polar Coordinate: Displacement Potentials

• Axi-symmetric Problems – Direct Solution

• Thermal Stresses in Circular Plates

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Documents

Thermoelasticity - Southeast University