texture, strain, and stress · deformation – graphical introduction a material can be considered...
TRANSCRIPT
Texture, Strain, and StressMichael Gharghouri
Canadian Neutron Beam Centre
Summer School 2013
Texture• Rotation tensor, pole figures, ODF• Neutron diffraction• Examples
Deformation (strain)• Deformation gradient tensor• Rotation / stretch
Stress• Stress tensor – normal and shear components• Stress / strain relations
Examples• Weldments, twist tubes• In-situ deformation experiments
Crystallographic Texture
Most engineering materials are made up of a collection of crystallites (grains),
each characterized by the orientation of the unit cell.
The distribution of
orientations is
referred to as the
texture of the
material.
T[010]
[100]
R
Crystallographic Texture:
The extremes and the middle ground
PowderTextured
Polycrystal
[100]
Single Crystal
[100][100]
Representing Texture
• Express crystallite orientation with respect to suitable material axes
Rotation tensor (set of angles)
• Function to describe volume fraction of material having unit cell orientation in
a given angular range
orientation density function( ) ( )
fV g d
ω
ω ω ω∆
∆ = ∫Tω∆
R
Rotation Tensor• Express crystallite orientation with respect to a standard orientation
(unrotated basis).
– Develop a well-defined operation to go from the standard orientation to an
arbitrary orientation.
T[010]
[100]
0 θ π≤ <
(T)
g2E
%
R θ
cos sin 0
sin cos 0
0 0 1
θ θθ θ
− =
R%%
cos ,sin ,0θ θ< >sin ,cos ,0θ θ< − >
• Rotation tensor R
• Columns contain direction cosines of rotated
base vectors (gi) wrt unrotated basis (Ei)
ijR = • jiE g
%%
(R)
1g%
1E%
%
2g%
Rotation Tensor
st
nd
1 index unrotated basis
2 index rotated basisij
R →
= • →
jiE g%%
( ) ( ) ( ) g g g( ) ( ) ( )= + +g g E g E g E
OLDNEW
OLD
NEW
1 1 1
2 2 2
3 3 3
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
=
1 2 3
1 2 3
1 2 3
g g g
R g g g
g g g
% % %
% % % %%
% % %
1 2 3
1 2 3
1 2 3
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
= + +
= + +
= + +
1 1 1 1 2 1 3
2 2 1 2 2 2 3
3 3 1 3 2 3 3
g g E g E g E
g g E g E g E
g g E g E g E
% % %% % % %
% % %% % % %
% % %% % % %
( )ij i
R =j
g%
ththe i component of vector with respect to the unrotated basis.ijR = jg%
Euler-Rodrigues IdentityGiven a unit rotation axis a and a rotation angle α, what is the rotation tensor?
1 1 1 2 1 3 3 2
2 1 2 2 2 3 3 1
3 1 3 2 3 3 2 1
cos (1 cos ) sin
1 0 0 0
cos 0 1 0 (1 cos ) sin 0
0 0 1 0
a a a a a a a a
a a a a a a a a
a a a a a a a a
α α α
α α α
= + − +
− = + − + − −
R I aa A
R
% % %% %% % %
%%
11 22 33 1 2costr R R R α= + + = −R%
11 22 33 1 2costr R R R α= + + = −R%%
[ ] [ ] [ ][ ]1 1 1 1 2 1 3
2 1 2 3 2 1 2 2 2 3
3 3 1 3 2 3 3
T
a a a a a a a
a a a a a a a a a a
a a a a a a a
= ⊗ = = =
aa a a a a% % % % % %
[ ]1
2 2 2
2 1 2 3
3
with 1
a
a a a a
a
= + + =
a%
is the axial tensor associated with , defined such that
i.e.
× = •
× =
A a a b A b
a b A b
% %% % % %% %
%% % %%
Change of Basis
(Passive Transformation)
st
nd
1 index unrotated basis
2 index rotated basisijR
→= •
→i jE g
% %
Unrotated base vector Ei as weighted sum of rotated base vectors gi.
3
ijR=∑i jE g Unrotated base vector Ei as weighted sum of rotated base vectors gi.1
ij
j
R=
=∑i jE g% %
Rotated base vector gi as weighted sum of unrotated base vectors Ei.
3
1
ji
j
R=
=∑i jg E%%
Components of a vector3
1
i ji j
j
v R V=
=∑i iv V= =i iv g E
3
1
i ij j
j
V R v=
=∑
Rotation of a Vector
Active Transformation
1 11 12 13 13
2 21 22 23 2
1
3 31 32 33 3
i ij j
j
v R R R V
v R V v R R R V
v R R R V=
= • ⇒ = ⇒ =
∑v R V%% %%
[ ]1 cos sin 0 cos (cos cos sin sin ) cos( )
sin cos 0 sin (sin cos cos sin ) sin( )
v v v v
v v v v
θ θ α θ α θ α α θθ θ α θ α θ α α θ
− − + = = = + = + v
α
cos , sin ,0v vα α< >
θ
X1
X2cos( ), sin( ),0v vα θ α θ< + + >
[ ] 2
3
sin cos 0 sin (sin cos cos sin ) sin( )
0 0 1 0 0 0
v v v v
v
θ θ α θ α θ α α θ = = = + = +
v%
v%
V% v= =v V
% %
Sequential Rotations – Fixed Axes
Consider a rotation RX followed by a rotation RY, followed by a rotation RZ:
1) Rotate about X-axis
2) Rotate about Y axis
3) Rotate about Z axis1
32
Any rotation matrix can be expressed as three sequential rotations about the
stationary laboratory axes X, Y, Z.
= • • ⇒ = • • •Z Y X Z Y X
R R R R v R R R V% % % % % % %% %% % % % % % %
cos sin 0 cos 0 sin 1 0 0
sin cos 0 0 1 0 0 cos sin
0 0 1 sin 0 cos 0 sin cos
Negative sign is "Down and to the Left" of the "1"
Z Z Y Y
Z Z X X
Y Y X X
Θ − Θ Θ Θ = Θ Θ Θ − Θ − Θ Θ Θ Θ
R%%
32
Euler angles – Sequential Rotations About Follower AxesThe rotation matrix can also be expressed as three sequential rotations about
axes x, y, z, which follow the material as it rotates.
1) Rotate about Z-axis Z Z X x’ Y y’
2) Rotate about x’ axis Z z’ x’ x’ y’ y’’
3) Rotate about z’ axis z’ z’ x’ x’’ y’’ y’’’
1 1
1 1 2 2
cos sin 0 1 0 0 1 0 0
sin cos 0 0 cos sin 0 cos sin
ϕ ϕϕ ϕ ϕ ϕ
− = Φ − Φ − R
1
32
1 1 2 2
2 2
sin cos 0 0 cos sin 0 cos sin
0 0 1 0 sin cos 0 sin cos
ϕ ϕ ϕ ϕ
ϕ ϕ
= Φ − Φ − Φ Φ
R%%
The matrices are multiplied in the order of operation, instead of in the reverse order
as for rotations about the stationary laboratory axes.
The rotation matrix is the same regardless of
how it is calculated!
= • •Z X' Z'R R R R% % % %% % % %
Euler angles and Euler space-
http://core.materials.ac.uk/repository/alumatter/metallurgy/anisotropy/
euler_angles_space_and_odf_revised.swf?targetFrame=EulerSpace
By Gunter Gottstein, Markus Buescher, MATTER, released under CC BY-NC-ND 2.0 license
The Stereographic Projection
Equatorial plane
T
R
N
P
Equatorial plane
By Gunter Gottstein, Markus Buescher, MATTER, released under CC BY-NC-ND 2.0 license
Stereographic projection for describing grain orientation -
http://aluminium.matter.org.uk/content/html/eng/0210-0010-swf.htm
By Gunter Gottstein, Markus Buescher, MATTER, released under CC BY-NC-ND 2.0 license
ω
Measuring TextureBy rotating a specimen, crystallites with
plane-normals in different directions
rotate through the scattering vector.
Rotate the specimen over a full
hemisphere of orientations to obtain a
complete (100) pole figure.
Q
ωωωω
Intensity
R
ω
Incident Beam
2222θθθθ(100)(100)(100)(100)Cd mask
Measuring Pole Figures (Intensity Maps)
[010]
Q
I
S
χχχχ= 0°
0° < ηηηη < 360°
χχχχ- axis is into page
Q
I
χχχχ= 30°
2θθθθ = 2θθθθ100
[010]
[100]
Q
S
χχχχ= 30°
0° < ηηηη < 360°
Q
I
S
χχχχ= 90°
0° < ηηηη < 360°
χ= 0
All η
χ= 90
η = 90
χ= 90
η = 0
RD
TD
Q at χ = ~15ND
Q at χ = ~45
Q at χ = 0
RD
Q at χ = 90
Stereographic Pole Figures
• Each point on a pole figure corresponds to a family of grain orientations –
those with a particular crystal direction aligned with Q
I S
Q
I S
Q
[010]
[100]
Dynamic (111) pole figure -
http://aluminium.matter.org.uk/content/html/eng/0210-0010-swf.htm
By Gunter Gottstein, Markus Buescher, MATTER, released under CC BY-NC-ND 2.0 license
The representation of orientations in the pole figure is ambiguous if there
is more than one orientation because each orientation is determined by
several poles which cannot be associated to a specific orientation in a
unique way.
The Orientation Density Function• We obtain the orientation density function (ODF ) by fitting
spherical harmonics to the experimentally measured pole
figures.
• The ODF tells us what fraction of grains have an orientation
between αααα and αααα+dαααα, where αααα represents the Euler angles.
W1 ×
( ) ( )fV g d∆ = ∫∆α
α α α( )f
dV g d= α α
+ W2 ×
+ W3 ×
+ …..
The Orientation Density Function
• Cubics (≥1), HCP's (≥2), THE MORE THE MERRIER
• A single experimental pole figure usually is not enough!
Orientation Density Function -
http://aluminium.matter.org.uk/content/html/eng/0210-0010-swf.htmBy Gunter Gottstein, Markus Buescher, MATTER, released under CC BY-NC-ND 2.0 license
Deep Drawing of Aluminum Cans
A disc of Al is drawn into a die to form a
can.
Preferred orientations of crystallites in the
disc may create anisotropic yield surfaces.
Ears develop. The highly-automated
canning-line is shut down when a large-
eared can gets stuck.
Rolled-Sheet Texture of Al Disc
• The initial sheet material used to
form disc has a strong rolling
texture.
• Earing can be related to this
initial texture.
Al (111)
Pole Figure
initial texture.
• Carry out a systematic
investigation of process route
changes designed to reduce
texture, or produce a more
favourable texture.
Neutron (0002) intensity vs.
circumferential position at fixed
radius in Ti-alloy impeller
Microstructural Homogeneity or Not
Forging process of original billet,
resulted in 4-fold symmetry of
enhanced crystallite orientations
in finished component.
DeformationDeformation is what happens when a material changes shape and orientation.
Each particle occupies a unique position in the initial and deformed configurations
so we can define a one-to-one mapping function:
( )χ=x X% %
d d= •x F X% % %%
deformation gradient
tensor
F is the best F is the best
descriptor of material
deformation, because
it contains all the
information
concerning shape
changes and
rotations.
The Deformation Gradient Tensor
( )( )( )( )
1 1
2 2
3 3
x
x
x
χ
χ χχ
=
= ⇒ = =
X
x X X
X
%
% % %
%
i
dd d
d
xF
= • ⇒ =
∂=
xx F X F
X%
% % % %% %%
1 1 1
1 2 3
2 2 2
1 2 3
3 3 3
1 2 3
i
ij
j
xF
X
x x x
X X X
x x x
X X X
x x x
X X X
∂=
∂
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
F%%
Deformation – Graphical Introduction
A material can be considered to be made up of little material cubes, characterized
by the orthonormal vectors (E1, E2, E3) that form its edges.
After deformation, each little material cube transforms into ~a parallelipiped,
characterized by the three vectors (g1, g2, g3) that form its edges .
This approximation (cube parallelipiped) becomes exact in the limit of an
infinitely refined grid.
g1, g2, and g3 are
generally not
orthogonal, nor of
unit length.
Deformation – Graphical IntroductionThe ith column of the deformation gradient tensor contains the components
of vector gi with respect to the lab basis (E1, E2, E3).
1 1 1
2 2 2
3 3 3
( ) ( ) ( )
[ ] ( ) ( ) ( )
( ) ( ) ( )
=
1 2 3
1 2 3
1 2 3
g g g
F g g g
g g g
% % %
% % % %%
% % %
1 2 3
1 2 3
1 2 3
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
= + +
= + +
= + +
1 1 1 1 2 1 3
2 2 1 2 2 2 3
3 3 1 3 2 3 3
g g E g E g E
g g E g E g E
g g E g E g E
( )ij i
F = jg
λ =m
M%
%
Fibre stretch:
= •m F M% % %%
1 11 12 13 1
2 21 22 23 2
3 31 32 33 3
m F F F M
m F F F M
m F F F M
=
Deformation – Graphical Example I
1 1 1
2 2 2
3 3 3
( ) ( ) ( ) 1.3 1.78 0
[ ] ( ) ( ) ( ) 0.05 1.27 0
0 0 1.5( ) ( ) ( )
= = −
1 2 3
1 2 3
1 2 3
g g g
F g g g
g g g
% % %
% % % %%
% % %
Elongation into page, with no distortion.
1
Deformation – Graphical Example II
12 0
2
1[ ] 1 0
4
0 0 1
= − −
F%%
1
2= −
1 1 2g E E
1 1 1
2 2 2
3 3 3
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
=
1 2 3
1 2 3
1 2 3
g g g
g g g
g g g
% % %
% % %
% % %
12
4= −
2 1 2g E E
=3 3
g E
Undeformed Deformed
2g%
1g%
2E%
1E%
Polar Decomposition of the
Deformation Gradient Tensor
Rigid Body Rotation Tensor Stretch Tensor
Material vectors change
orientation
BUT
They don’t change length
There are 3 material directions
(principal directions) in the initial
configuration that :
can change length
R
= •F R U% % %% % %
Proper orientations:
Determinant = +1
can change length
BUT
do NOT change direction
Tensor is:
Positive definite
Symmetric
Determinant > 0
U%%
R%%
Stretch Tensor U%%
d d d= • = • •x F X R U X% % % %%
d•U X%
This expression quantifies the stretching of the material fibre dX, as well as the
part of the fibre rotation that results strictly from the material distortion (shape
change).
The stretch tensor U possesses 3 real, positive eigenvalues (principal values).
It is diagonal in its principal basis.
The 3 eigenvectors are the principal directions.
Material vectors along the principal directions may change length, but they do not change
direction.
The 3 principal values equal the ratio of the deformed to the undeformed lengths of the
three non-rotating material fibres.
e12
=
ε12
ε12
1212
e
2ε =
Simple shear = pure shear + rotation
X1
+
X2
Stretch Tensor U%%
Fibres aligned with the principal directions of U don’t rotate.
Fibres not aligned with the principal directions will change orientation, but for every
fibre rotating one way, there will be another fibre rotating by the same amount in the
opposite way.
The overall material rotation under a pure stretch is zero.
βα
Uβ
1
2 2
πβ α = −
Strain
( )1 2 3 1 1 1
1 2 3 2 2 2
1 2 3 3 3 3
1
2
( ) ( ) ( ) ( ) ( ) ( ) 1 0 01
( ) ( ) ( ) ( ) ( ) ( ) 0 1 02
0 0 1( ) ( ) ( ) ( ) ( ) ( )
= • −
= −
T
1 1 1 1 2 3
2 2 2 1 2 3
3 3 3 1 2 3
ε F F I
g g g g g g
ε g g g g g g
g g g g g g
% %% % %%
% % % % % %
% % % % % % %%
% % % % % %
Normal strain
( ) ( ) ( ) ( )( ) ( )
2 2 2 2
11 1 2 3
2 22
1 1 1 1 1
1 1( ) ( ) ( ) 1 1
2 2
1 1 ( ) 1 1 2 1
2 2E E E E E
ε = + + − = −
= + ∆ − = + ∆ + ∆ − ≈ ∆
1 1 1 1g g g g% % %
1 1 1 111
1 1 1
1E E E E
E E Eε λ
∆ + ∆≈ = − = − λ =
m
M%
%
Fibre stretch
Strain
12 1 2 1 1 2 2 1 2
1 1 1cos ( )( ) cos (1 )(1 )cos
2 2 2
1 1 1 1 (1 small terms)cos cos sin
2 2 2 2 2 2
g g E E E E E Eε α α α
π πα α α α
= = + ∆ + ∆ = + ∆ + ∆
= + ≈ = − ≈ −
Shear strain
(distortion)
βα
β1
2 2
πβ α = −
1g%
1E%
2g%
2E%
Elastic vs Plastic Strain
total elastic plasticε ε ε= +F
orc
e
LL1 L
elastic slopeσ ε= ⋅
2σσσσ2
L
2 1 (2) (1)elastic elasticσ σ ε ε> ⇒ >
F
Forc
e
Strain
elasticε
1 0
0
u
plastic
L L
Lε
−=
L0L11
σσσσ1
1 0
0
total
L L
Lε
−=
L1uL2
2 0
0
total
L L
Lε
−=
plasticε
elasticε
L2u
F
total elastic plasticε ε ε= +
1σσσσ
Elastic vs Plastic Strain
2 1 (2) (1)elastic elasticσ σ ε ε= ⇒ =
2
totalεelasticε
plasticε
plasticε
totalεelasticε
Str
ess
Strain
Elastic Strain
Strain
Forc
e
Shape and/or size of the
unit cell change.
Shape change is reversible.
Plastic Strain
Strain
Forc
e
Undeformed blocks of
crystal slide with respect to
one another.
Shape and size of the unit
cell DO NOT change.
Measuring Strain by Diffraction
0
0 0
1d d d
d dε
−≡ = −
0sin1
sin
θε
θ= −
0
50
100
150
200
250
300
350
88.5 89 89.5 90 90.5 91 91.5
Scattering Angle, 2θθθθ (deg.)
Ne
utr
on
Co
un
ts
• We only measure normalstrains – not shear strains!
• We measure lattice strains i.e. elastic strains
sinθStress-free (d0, θθθθ0)
TENSION (+)
(dt > d0, |θθθθt| < |θθθθ0|)
COMPRESSION (–)
(dc < d0, θθθθc > θθθθ0)
Mapping Strain - Direction and Gauge Volume
Measure component of
strain parallel to
bisector (Q)
Q
2222θθθθ010010010010((((φφφφ))))
Incident Beam
Cd Mask
[010]
[100]
((((φφφφ))))
Get strain data only
from material inside
the Gauge Volume
Reorient the sample to measure strain along different
specimen directions.
Measure component of
strain parallel to
bisector (Q)
Q
Incident Beam
2222θ θ θ θ 010010010010((((φφφφ))))Cd Mask
Get strain data only Get strain data only
from material inside from material inside
the Gauge Volume the Gauge Volume
The Gauge Volume
2θ = 90°
2θ < 90°
Sampling a Strain Gradient
εLarge gauge volume smears
out the gradient
x
R
T
T
R
Q
Mapping Strain - Location, Location, Location!
Incident beam
Diffracted beam
Sampling
volume
Spatial resolution
~ 1 mm3
Use neutrons to locate sample edges – Wall scans (0.1 mm)
0
50
100
150
200
84 88 92 96
Definition of Stress
Stress (σ) is a force per unit area
FdA 0
FdA
stress lim→
=
The force can be
uniformly distributed
over an area:
σ = F/A
The force can be distributed
non-uniformly over an area,
in which case we define the
stress at a point as:
F
F
A
dA 0→
x
y
Stress is a Tensor
We need 2 vectors to define a stress:
The force per unit area, T, acting on the plane
Consider 2 planes passing through a point P in a cylindrical body in a state of
simple tension.
Fa
A2
The force acting on planes A1 and A2 is the same, but the
stress on the two planes is different because the areas are
different.
The force per unit area, T, acting on the plane
The normal to the surface, n, on which F acts
⇒Stress is a 2nd order tensor that relates T and n
Fa
A1
A2
11 12 131 1
2 12 22 23 2
3 313 23 33
T n
T n
T n
σ σ σ
= σ σ σ σ σ σ
Normal and Shear Stresses
The stress acting on a surface can be resolved into two components:
1) Normal stress - acts perpendicular to the surface
2) Shear stress - acts parallel to the surface
The direction of the normal stress on a plane is specified once the surface has
been identified.
A shear stress can be further resolved into two components in the directions of the
coordinate axes in the plane.
⇒ We need three components to define
the state of stress at a point on a plane.
Components of Stress
We specify all the normal and shear stresses acting on the faces of an
infinitesimal cube – a total of 18 components.
i = j ⇒ normal stress.
i ≠ j ⇒ shear stress.
x3
σ31
σ13 σ32
σ33
σ23
B
C
x3
σ
σ-1-2σ-2-1
σ-1-1
B'
C'
x
x2
1
σ31
σ12
σ22
σ21
σ11
A
Dx
x
2
1
σ-2-2
σ-3-2
σ-3-3
σ-2-3
σ-1-3
σ-3-1
A'
D'
σij ⇒ i = coordinate parallel to which stress acts
j = coordinate plane normal to the plane acted on by the stress
Components of stress
11 12 13
ij 21 22 23
31 32 33
σ σ σ
σ = σ σ σ σ σ σ
We can show that σij = σ-i-j ⇒ only 9 components
To satisfy equilibrium, we must have σij = σji ⇒ 6 independent components
11 12 13
ij 12 22 23
13 23 33
σ σ σ
σ = σ σ σ σ σ σ
Symmetric tensor
Stress-Strain Relations
11 11 22 33
1( ( ))
Eε = σ − ν σ + σ 12
122G
σε =
For small deformations in an isotropic material, there is a simple linear relationship
between stress and strain.
This linear relationship arises because the This linear relationship arises because the
force, F, required for a displacement of ubetween 2 atoms, is given by the gradient of
energy:
0
2
2
r
dU d UF u
dr dr
= =
σ = Eε
Getting Stress from Strain I
• The stress and strain tensors are related via the elastic
stiffness tensor .
:k l
ij ijkl kl
k l
Cσ ε= =∑∑ σ C ε% % %% % %%%
C%%%%
• If the Principal Stress directions are known, we only need
( ) ( )zyxE
zyx ,,,211
=
++−
++
= αεεεν
νε
νσ αα
• If the Principal Stress directions are known, we only need
to measure the corresponding normal strain components
• Then we can calculate the 3 Principal Stresses from the 3
strains using the Generalized Hooke’s Law (assuming
elastic isotropy):
Getting Stress from Strain II
If the Principal Stress Directions are not known, we need to measure
enough strain components to determine the full strain tensor ⇒ stress tensor
m n mna aε = ε
a
b bε = ε
m ma = •a E
%
0sin1
sin
θε
θ= −a
= •
m n mnf fε = ε
f
m n mnb bε = ε
b
.
.
.
6 equations in 6 unknowns
m mb = •b E
%
m mf = •f E
%
k l
ij ijkl kl
k l
Cσ ε=∑∑
Residual Stresses – Good or Bad
• Residual stresses are self-equilibrating stresses within a stationary solid
body when no external forces are applied.
• They vary with location in a component: consider a cast plate a few mm
thick. The stresses in the core balance those in the skin. What happens if
the skin is removed on one of the surfaces?
Outer skin is in compression
• Almost every manufactured component has residual stresses – they can be
detrimental or beneficial to the performance of the component.
• They are not apparent, and they are difficult to measure and predict.
• We need a reliable method to measure residual stresses non-destructively.
Outer skin is in compressionInner core is in tension
Residual Stresses Can Kill You
Liquid-Metal-Assisted Cracking of a plain carbon steel I-beam after hot-dip galvanizing.
The crack is 1.1 m long, and appeared after the beam had been in service.
Macrostresses and Microstresses
Internal stresses are categorized according to the
length scale over which they vary.
Type I stresses - macrostresses - vary slowly over
the volume of a component and are considered to be
continuous across grain boundaries and phase continuous across grain boundaries and phase
boundaries.
These are the stresses calculated during a typical
stress analysis of an engineering component (e.g.
finite element analysis).
Macrostresses and Microstresses
Type II microstresses
vary on a length scale
on the order of the grain
size.
Type III microstresses
vary on a subgrain
length scale.length scale.
Neutron diffraction can
be used to study
macrostresses and type
II microstresses.
Type III microstresses
typically give rise to
peak broadening.
Evaluating ProcessesHeat-Exchanger Twist Tube
• Customer’s question:
“Does our annealing procedure reduce the residual stresses?”
• Largest stresses due to deformation expected to be in the hoop
direction
• Need only measure hoop strain on As-Manufactured and Annealed
tubes
Hoop
Direction
Ω
Through
thickness
position
Circumferential scan of 24
positions
3 through-thickness
positions:
mid-thickness
0.4 mm toward OD
0.4 mm toward ID
Heat-Exchanger Twist Tube
As-Manufactured
As-Manufactured Twist Tube
10
20
30
Ho
op
Str
ain
(10
-4)
Near O.D.Wall CentreNear I.D.
-30
-20
-10
0
0 30 60 90 120 150 180 210 240 270 300 330 360
Circumferential Position (deg.)
Ho
op
Str
ain
(10
Typical Uncertainty
• Significant strain variations are observed.
• Complementary OD and ID variations.
Heat-Exchanger Twist Tube
Annealed
Annealed Syncrude Twist Tube
10
20
30
Ho
op
Str
ain
(10
-4)
Near O.D.Wall CentreNear I.D.
Annealed Twist Tube
Clearly, annealing effectively reduced the residual stresses
-30
-20
-10
0
0 30 60 90 120 150 180 210 240 270 300 330 360
Circumferential Position (deg.)
Ho
op
Str
ain
(10
Typical Uncertainty
In-Situ Deformation Experiments
• Measuring residual strains and deformation textures is great –
but how did they get there?
• In-situ experiments allow us to follow the evolution of residual
strains and texture during deformation.
• Example: Deformation of a Mg-Al alloy
Plastic Deformation: Slip vs. Twinning
• Slip: undeformed blocks of crystal translate relative to each other
• Twinning: homogeneous shear of a portion of the lattice (mirror)
Plastic Deformation: Slip
(0001)<1120> 2
CRSS = 0.51 MPa
1010<1120> 2
CRSS = 40 MPa
1011<1120> 4
CRSS = 40 MPa ??
Plastic Deformation: Twinning
• Polar: only limited c-axis extension (6.9% max)
c/a < √3
300
350
• Slip: undeformed blocks of crystalline material
slide past one another
– Gradual lattice reorientation
• Twinning: Abrupt reorientation of the crystal
lattice
σ,Q
0
50
100
150
200
250
300
88.5 89 89.5 90 90.5 91 91.5 92
Scattering angle, 2θθθθ (deg.)
Ne
utr
on
co
un
ts
Mg - 8.5 wt.% Al
• Processing (Pechiney, France)
– Mix Mg and Al under protective atmosphere
– Cast into billets
– Extrude at 250°C (64 mm → 15 mm)
– Age(0002)
Centre is extrusion axis
Contour separation = 0.5 × random
Texture ModificationThermomechanical Treatment
Compress // extrusion axis
1012 twinning leads to an 86.3°reorientation of the lattice
Anneal
Texture Modification
Extrusion texture
(Texture 1)
Modified texture
(Texture 2)
Contour separation:
1 × uniformContour separation:
2.5 × uniform
Cyclic Tension
T1
150
200
250
300
(a)
Tru
e st
ress
(M
Pa)
150
200
250
300
Str
ess (
MP
a)
T2
• Hysteresis loops occur for both textures
0.000 0.005 0.010 0.015 0.020 0.025 0.0300
50
100
150
Tru
e st
ress
(M
Pa)
True strain
0
50
100
150
0 0.02 0.04 0.06 0.08 0.1
Strain
Str
ess (
MP
a)
Metallography
Texture 2Texture 1
Initial State
Texture 2Texture 1
Cycle 3
Texture 2Texture 1
Cycle 6
Texture2:
• Long, wide twins which cross grain boundaries
• Much smaller twins, apparently nucleated at grain boundaries
Texture 1:
• Twins are narrow and there are relatively few of them
Texture 2Texture 1
Cycle 9
200
250
300
Str
es
s (
MP
a)
1012 axial
200
250
300
Str
es
s (
MP
a)
1012 axial
Cyclic Tension - Texture 2
47°σσσσσσσσ
0
50
100
150
200
-1000 0 1000 2000 3000
Lattice strain (µµµµstrain)
Str
es
s (
MP
a)
0
50
100
150
200
7 8 9 10 11 12
Integrated intensity
Str
es
s (
MP
a)
• Profuse 1012 tension twinning expected in a majority of grains
200
250
300
Str
es
s (
MP
a)
1012 axial
200
250
300
Str
es
s (
MP
a)
1012 axial
Cyclic Tension - Texture 2
47°σσσσσσσσ
0
50
100
150
200
-1000 0 1000 2000 3000
Lattice strain (µµµµstrain)
Str
es
s (
MP
a)
0
50
100
150
200
7 8 9 10 11 12
Integrated intensity
Str
es
s (
MP
a)
200
250
300
Str
es
s (
MP
a)
1012 axial
200
250
300
Str
es
s (
MP
a)
1012 axial
47°σσσσσσσσ
Cyclic Tension - Texture 2
0
50
100
150
200
-1000 0 1000 2000 3000
Lattice strain (µµµµstrain)
Str
es
s (
MP
a)
0
50
100
150
200
7 8 9 10 11 12
Integrated intensity
Str
es
s (
MP
a)
0
50
100
150
200
250
300
-1000 0 1000 2000 3000
Str
es
s (
MP
a)
1012 axial
0
50
100
150
200
250
300
7 8 9 10 11 12
Str
es
s (
MP
a)
1012 axial
47°σσσσσσσσ
0
50
100
150
200
250
300
-1000 0 1000 2000 3000
Lattice strain (µµµµstrain)
Str
es
s (
MP
a)
1013 axial
0
50
100
150
200
250
300
10 20 30 40 50
Integrated intensity
Str
es
s (
MP
a)
1013 axial
32°σσσσσσσσ
0
50
100
150
200
250
300
-1000 0 1000 2000 3000
Str
es
s (
MP
a)
1012 axial
0
50
100
150
200
250
300
7 8 9 10 11 12
Str
es
s (
MP
a)
1012 axial
47°σσσσσσσσ
T2
0
50
100
150
200
250
300
0 2000 4000 6000
Lattice strain (µµµµstrain)
Str
es
s (
MP
a)
2110 axial
90°σσσσσσσσ
0
50
100
150
200
250
300
0 10 20 30 40 50 60
Integrated intensity
Str
es
s (
MP
a)
2110 axial T2
0
50
100
150
200
250
300
-1000 1000 3000 5000
Str
es
s (
MP
a)
(0002) axial
0°σσσσσσσσ
0
50
100
150
200
250
300
0 20 40 60 80 100 120 140
Str
es
s (
MP
a)
(0002) axial
T2
0
50
100
150
200
250
300
0 2000 4000 6000
Lattice strain (µµµµstrain)
Str
es
s (
MP
a)
2110 axial
90°σσσσσσσσ
0
50
100
150
200
250
300
0 10 20 30 40 50 60
Integrated intensity
Str
es
s (
MP
a)
2110 axial T2
200
250
300
Str
es
s (
MP
a)
(0002) axial
200
250
300
Str
es
s (
MP
a)
(0002) axial
0°σσσσσσσσ
Cyclic Tension - Texture 1
0
50
100
150
200
-1000 0 1000 2000 3000 4000
Lattice strain (µµµµstrain)
Str
es
s (
MP
a)
0
50
100
150
200
10 15 20 25 30 35
Integrated intensity
Str
es
s (
MP
a)
200
250
300
Str
es
s (
MP
a)
(0002) axial
200
250
300
Str
es
s (
MP
a)
(0002) axial
0°σσσσσσσσ
Cyclic Tension - Texture 1
0
50
100
150
200
-1000 0 1000 2000 3000 4000
Lattice strain (µµµµstrain)
Str
es
s (
MP
a)
0
50
100
150
200
10 15 20 25 30 35
Integrated intensity
Str
es
s (
MP
a)
200
250
300
Str
es
s (
MP
a)
(0002) axial
200
250
300
Str
es
s (
MP
a)
(0002) axial
0°σσσσσσσσ
Cyclic Tension - Texture 1
0
50
100
150
200
-1000 0 1000 2000 3000 4000
Lattice strain (µµµµstrain)
Str
es
s (
MP
a)
0
50
100
150
200
10 15 20 25 30 35
Integrated intensity
Str
es
s (
MP
a)
Twinning - Stress Relaxation
Single crystal
σ σ−∆σ
σ σ−∆σ
σ
Twinning - Stress Relaxation
Polycrystal
σ
σσ (−∆σ)?
Twinning - Stress Relaxation
Polycrystal
σσ (−∆σ)?
Local stress
relaxation
⇒ grain level
0
50
100
150
200
250
300
-1000 1000 3000 5000
Str
es
s (
MP
a)
(0002) axial
0°σσσσσσσσ
0
50
100
150
200
250
300
0 20 40 60 80 100 120 140
Str
es
s (
MP
a)
(0002) axialT2
0
50
100
150
200
250
300
-1000 0 1000 2000 3000 4000
Lattice strain (µµµµstrain)
Str
es
s (
MP
a)
(0002) axial
0
50
100
150
200
250
300
10 15 20 25 30 35
Integrated intensity
Str
es
s (
MP
a)
(0002) axial T1