Welding Metallurgy
Heat Flow in Welding
lecture 8
p1
Lecture Scope
(
• Basic features of welding heat transfer • Relevant heat flow theory and solutions
p2
Significance of Thermal Effects
• The thermal conditions in and near welds
govern the resulting metallurgical structure,
mechanical properties, residual stress and
distortion
• Of particular significance are:
-
Lecture 8
Weld bead area
Weld solidification rate
Peak temperatures in the Heat Affected Zone (HAl)
Width of HAl
Cooling rates in the weld and HAl
p3
Weld Thermal Cycle
(
,
2000
-
- - - - TOP SURFACE OF WELD
BOTTOM SURFACE
-,.,
0
~
PHASE CHANGE RANGE
W 11500
It:
::>
i
W
6~~
'1000
D..
1Xlolin tate
::i!:
W
I-
-,
4 5
•
500
-0.1
Ts
0
0.1
0.2
0.3
0.4
0.5
TIME (minutes)
8
p4
Thermal Cycles Due to Different
Welding Processes
..
Increasing Energy Density
Tm
EBW
LBW
SMAW
GMAW
SAW
FCAW
ESW
Time
Lecture 8
p5
Multiple Pass Weld Cycles
(
1500 , - - - - - - r - - - - , . . . - - - - - r - - - - - - ,
~ 1000
~
~
II:
W
ll..
:::;:
w
.....
500
Interpass Temperature
_ _..L-
O~'--
o
100
...L.
....L_ ______.J
200
300
400
TIME, SECONDS
p6
Heat Balance
• The power delivered to the work from th.e arc
is qa = 'lVI
• Thermal energy is dissipated by conduction in
the solid and by convection and radiation to
the surroundings
• Conduction in the solid predominates
Lecture 8
pi
Weld Area
~~
L--<....L--L...'------"<-L----L.J.'--"':.L.-...,LL~
(
A1=FilierArea
A2 = Area of base_metal melted
Weld Area Aw = A1 + A2
.
• Heat input must be sufficient to overcome losses and to
supply the energy needed for melting
• The minimum power required for melting is HAwv where H
is the energy required to melt unit volume of metal
• Melting effidency f2 =(HAwv)/(fJVI)
• Higher energy densities reduce heat losses and give
greater melting efficiency
8
p8
Weld Area
• Assume arc efficiency, fl, and the melting efficiency'2 do not
vary greatly for a given welding process.
• From the previous equations it can be seen that the cross
section of a single weld bead is roughly proportional to the
energy input, Le.
-Aw = f2 "V/IHv
• For example, consider an arc weld on steel made under the
following conditions:
- V=10V,
-/=200A,
-v=5mmls,
-" =.9,
- f2 = .3,
- H =10 Jlmm A 3
• Then Aw
Lecture 8
=11. 3 mm
p9
Heat Flow Theory
(
• Heat flow in the solid is determined by: - workpiece thickness
- edge, end effects
- thennal conductivity and specific heat
- heat source distribution
- convection in the weld pool
-latent heat absorption and release
8
P 10
Effect of Relative Plate Thickness
---~--(a) Thin Plate: 2-D Heat Flow - .
(b) Thick Plate: 3D Heat Flow
Lecture 8
P 11
Effect of Heat Flow Paths
..-.--
..
Multiple paths
increase
effective plate
thickness
(
..-
..
I
,
T
..
-...
+
,8
p12
Heat Flow Solutions
• Computer numerical modelling techniques are now
capable of solving weld thermo-mechanical problems with
a high degree of accuracy
• Traditional analytical solutions for heat conduction are still
useful and give insights on the effects of welding variables
Lecture 8
p13
Heat Flow Equation
With simplifying assumptions, steady state heat conduction in
a moving solid may be described by the following equation:
(1 )
where
ex = K/pC p
ex =thermal diffusivity
K
= thermal conductivity
p =density
Cp = specific heat
·8
p14
Solutions to Heat Flow Equation
Solution of Equation (1) gives the following expressions for
the temperature field round a "quasi-stationary" heat source
(a) Thin Plate 2D Heat Flow
T=
e -v(r-x)/2a
q
(2)
21tKr
(b) Thick Plate 3D Heat Flow
T
=
q
21tK
e vx/2a K ( vr )
0
K 0 is Bessel function (tabulated) and
Lecture 8
2ex
(3)
r = ";x 2 + Y 2 + Z 2
p15
Peak Temperature
(
The following equation applies to thin plate (2D)
1
Tp -T0
Tp
To
Tm
e
h
y
=
=
=
=
=
=
=
V2rcepc hyv
p
q
+
1
Tm -T0
(4)
peak temperature
initial plate temperature
melting temperature
Base of natural logarithms (2.718)
plate thickness
distance from fusion line
P 16
Cooling Rates
Expressions for cooling rates are obtained by
differentiating the previous equations with respect to time.
For points on the weld centreline:
(a) Thick Plate
!!I- =
at
21t1<v (T-T)
q
0
(5)
(b) Thin Plate
(6)
Lecture 8
p1i
Effects of Welding Variables
(
• These equations show that weld bead area,
peak temperatures, weld width, and cooUng
rates are determined by:
- Heat input per unit length qlv, and
-Initial plate temperature To, or preheat temperature
• The effects of increased heat input and
preheat temperature are to:
- increase peak temperatures at points outside the fusion
boundary
- increase weld bead area
- decrease cooling rates.
(
6
p16
Peak Temperatures
100
10
1
SJIIIboIs eoclosed In • dn:1ll indicate
posiliOl\ll iIlslde the fuslOll zone
lIklll0
J
g
~i';:,
"Ii!.\-...
••• C~.
g
• ~.~~
• e• • •"=:.. .
~
."".~
.......
.....
... 1
j
.l!l
Q
. --
• •"•
• ••
1)001
001
1
•
lleIative peak umperatute 1'. .
"
Lecture 8
P 19
Weld width
(
10
.,..•.'•
.
'I!'
• •~
\l:
.~o(
thooretical width
,).
:
Note: "Operating parameter"
is equivalent to non-dimensional
heat input
0-001
0-01
001
1
10
100
Operating perameter (n)
8
p20
Cooling Rates
Ie
o measured oooIilll mtes
• reciprocal buh lengths
~.;;
1
~
~~
-.
"
ti
.J
()oJ
~
0
(loOl
(lo1
1
0
10
100
Operating parameter (n)
Lecture 8
p2t
Cooling Rates & Weld Area
400
(
r----------'-----,
300
200
III
C3
Cooling Rate at 704C
i
g'
8
100
80
70
60
50
40
o
30 !-::-----:±--~--t.-__:':::___:!:c_*-:!
10
20
30
40 50 60 7080
Weld Bead Cross-Section Area mm A 2
(
I
8
p22
Example
Consider a full-penetration arc weld pass on a steel plate
under the following conditions:
Plate thickness
Welding current
Voltage
Travel Speed
Arc efficiency
Melting Efficiency
10 mm
200A
20V
5mmis
0.9
0.3
Melting temperature
Energy for melling
Thermal conductivity
Density
Specific heat
Initial temperature
1510 C
10 J/mm A 3
.028W/mmiK
7800 kg/m A 3
440 J/kg/K
25C
Estimate the:
1.
2.
3.
4.
I eclure 8
Heat input per unit length
Weld area
Width of HAl > 730 C
Centreline cooling rate at 550 C
p23
Answers
1. Heat input, q
2. Weld area, Aw
(
= 'lV/iv
=0.9*20*200/5
=720 J/mm
=(2 ,,(VI)/(f:lv)
= 0.3*720/10
=21.6 mm
A
2
3. Width of HAZ
Taking 730 C as the peak temperature in Equation 4
gives the width of the HAZ from the fusion line as 5.9
mm
4. From Equation 6 for 2D heat flow, the cooling rate at
550 C is 16.8 CIs
8
p24