Two Simple Models of Thermal Stress Voller-Guzina-Stelson University of Minnesota 1.Residual stress...
-
date post
21-Dec-2015 -
Category
Documents
-
view
220 -
download
2
Transcript of Two Simple Models of Thermal Stress Voller-Guzina-Stelson University of Minnesota 1.Residual stress...
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