texture, strain, and stress · deformation – graphical introduction a material can be considered...

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


Stereographic projection for describing grain orientation -

http://aluminium.matter.org.uk/content/html/eng/0210-0010-swf.htm


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


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



0

50

100

150

200

-1000 0 1000 2000 3000


Str

es

s (

MP

a)

0

50

100

150

200

7 8 9 10 11 12


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


0

50

100

150

200

250

300

-1000 0 1000 2000 3000


Str

es

s (

MP

a)

1013 axial

0

50

100

150

200

250

300

10 20 30 40 50


Str

es

s (

MP

a)

1013 axial


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


T2

0

50

100

150

200

250

300

0 2000 4000 6000


Str

es

s (

MP

a)

2110 axial


0

50

100

150

200

250

300

0 10 20 30 40 50 60


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


Str

es

s (

MP

a)

2110 axial


0

50

100

150

200

250

300

0 10 20 30 40 50 60


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


0

50

100

150

200

-1000 0 1000 2000 3000 4000


Str

es

s (

MP

a)

0

50

100

150

200

10 15 20 25 30 35


Str

es

s (

MP

a)

Twinning - Stress Relaxation

Single crystal

σ σ−∆σ

σ σ−∆σ

σ


Polycrystal

σ

σσ (−∆σ)?


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


Str

es

s (

MP

a)

(0002) axial

0

50

100

150

200

250

300

10 15 20 25 30 35


Str

es

s (

MP

a)

(0002) axial T1

texture, strain, and stress · deformation – graphical introduction a material can be considered...

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