Two Simple Models of Thermal Stress

Voller-Guzina-StelsonUniversity of Minnesota

1. Residual stress in solidification

2. Crack Patterns in Thermal Processing

Use A Full FEM SolutionWith all the bells and whistles

The Approach

A basic approach

A Kitchen Sink Model “Phenomenological Noise”

An Alternative/Supplemental

2/xxwh

ku

,xxh

dx

d

tx

tx

Simple Semi-Analytical Lower Dimensional Models

Leading to Back-of-the-envelope-calculations and Insight

CAN lead to

CAN lead to

Q

z = 0surface

z = zg z = bmid-plane

SolidLiquid

Mold

When a Polymer is solidified in a Rectangular mold

z = 0surface

z = bmid-plane

compression

tension

On final Solidification

A Residual Stresswill be Observed

-

+

A Residual Stress Model

need a non-linear and/or rate dependent behavior

Why ?

compression

tension

-

+

1. Flow Shear

liquidsolid

In solid: straight coils will try and re-coil leading to tensionin high shear regions and compression in low shear

In liquid –high shear flow willstraighten out polymer coils

Shear will straighten polymer coils

Opposite Of Observation

2. Rate Effect

v

TTg

Glass-Transition increases with cooling rate

fast

slow

Why ?

compression

tension

-

+Q

21SolidSolid TT

Cooled-- FAST SLOW

Consider isolated lamella

)()( 2211roomSTroomST TTTT

At room temp. Isolated lamellaAt surface will Shrink MORE

If LamellaAre Stuck Together

Ten

sion

Com

pre

ssio

n

Opposite Of Observation

Why ?

compression

tension

-

+

3. Flow Strain

Q

Layer at solid-liquid Front under goesA FLOW STRAINTo join existing solid

Consider isolated lamella

Initial “flow” strain in surface lamella is smaller than initial Flow strain in center lamella

BUT Once Solid undergo same thermal deformationIf Lamella

Are Stuck Together

Ten

sion

Com

pre

ssio

n

As Required

SOLID

)()),((),()( zTtzTtzt vse

A Simple Model Of Residual Stress Based On Flow-Strain Concept

At ant time t uniform strain in the solid is

elastic thermal flow

sz

vss

sv dzzTtzTz

z0

)()),((1

)(

Q

z

SOLID

Flow strain at a given position zs is “frozen in place” at point of solidification

This flow strain will be the average of the thermal andFlow strain in the existing solid

If We know Temperature history in Space and Time we can calculate flow strain—and determine stress at room temp.

)()(1

)( zTTv

Ez vsftot

compression

tension

-

+

(After Osswald and Menges)

compression

tension

-

+

sz

vss

sv dzzTtzTz

z0

)()),((1

)( )()(1

)( zTTv

Ez vsftot

zs

VAM

Use A HEAT BALANCE INTEGRAL WITH VAM

Real Temp

Linear Approx.

)(0][)(

),( tzzTztz

TTtzT sf

s

fs

z

z

z

bTT

v

Ez sf ln)(

)1(2)(

Bib /37.104.0/

Fit with numerical model

-15

-10

-5

0

5

10

0 0.2 0.4 0.6 0.8 1Normalized Position z* =z/b

Res

idua

l Str

ess

(MP

a)

LHBI-VAM

Measurements

70% Starch

-10

-5

0

5

0 0.2 0.4 0.6 0.8 1Normalized Position z* =z/b

Res

idua

l Str

ess

(MP

a)

LHBI-VAM

Measurements

50% Starch

-5

0

5

0 0.2 0.4 0.6 0.8 1Normalized Position z* =z/b

Res

idua

l Str

ess

(MP

a)LHBI-VAM

Measurements

30% Starch

Comparison with LHBI-VAM Model and Experimental (surface removal) measurementsOn an injected molded starch based polymer blends

Journal Of Thermal Stress 25 (2002)

A Crack Spacing Model

Consider a Film placed on a substrate and subjected to a thermal strain

Often Observe Characteristic Crack Spacing

Ceramic Film- 2.56% substrate strain

0

1

2

3

4

5

6

7

0 20 40 60 80 100 120 140Location, m

Dis

tanc

e A

cros

s P

avem

ent,

m

Average Spacing = 12 mStandard Deviation = 4.88 m

150 micron 150 meterBai, Pollard &Gao,

Nature, 403, 753-756

Spacing in Jointed RockCooled Asphalt

2/xxwh

ku

,xxh

dx

d

tx

tx

fwR

Interface shear stressat failure Spring coefficient

(Winkler Foundation)

Elastic Rod with an Elastoplastic Restraint imposed at the film/substrate interface

xukR

Elastoplastic Restraint

)/(Ewhk

,)( Tk

wf

x = x =

elastic

plastic

xt

substrate stiffness

1

))2/(cosh(

1max,

t

trx x

xTE

S

For a given temperature drop there is a Characteristic size

sos2

1 Larger and maximum stress

Will exceed Strength SSmaller and max stress can not reach S

Also-- with increasing temp. drop would expect increase in crack density (cracks per unit length)UPTO where failure along the ENTIRE interface is plastic

Fixed slope

h

Increase Strain Can-NotIncrease Stress

Sh

2

3lim0

Shc

3

2lim

)1/( 2EEr

f 0.233 GPa

521 GPa

res 10.4 GPa

Esteel 190 GPa

steel 0.3

S 0.853 GPa

Shc

3

2lim

Ceramic Film—Use Model to Predict Properties

0

20

40

60

80

100

120

140

0.02 0.07 0.12Applied Strain,

Cra

cks

per

mm

Experiment

Model

Shclim

2

3

irres E

compression

tension

-

+

Simple Models Can:--

Identify Contributing Phenomena

-15

-10

-5

0

5

10

0 0.2 0.4 0.6 0.8 1Normalized Position z* =z/b

Res

idua

l Str

ess

(MP

a)LHBI-VAM

Measurements

70% Starch

z

z

z

bTT

v

Ez sf ln)(

)1(2)(

Predict Material Properties

0

20

40

60

80

100

120

140

0.02 0.07 0.12Applied Strain,

Cra

cks

per

mm

Experiment

Model

Shear Strength

Residual Stress