Mechanical Behavior of Materials Lecture Note (6) 10/3/2010

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In order to understand why some materials bend easily while

others simply break one must think about the crystal structure,

and how a stress being applied to the material will influence

the atoms.

In all cases, one must think about how easy is it to get atoms to

move past one another. The easier this process is, the more

“bendable” or ductile a material is. The more difficult, the

more brittle that material is.

The motion of atoms past one another in a metal is known as slip.

The easier it is to induce slip in a material, the more ductile the material is. This is a

function of crystallography in the material.

There are certain preferred directions that atoms can move called slip directions, and

certain preferred planes of atoms that will moved when a stress is applied called slip

planes.

The combination of the slip directions and slip planes (multiplied together), give

much information about the mechanical response of a material called the slip

system. The larger the number of the slip system, the more ductile a material will

be.

D e f o r m a t i o n o f m a t e r i a l s o c c u r s

w h e n a l i n e d e f e c t ( d i s l o c a t i o n )

m o v e s ( s l i p ) t h r o u g h t h e m a t e r i a l .

Mechanical Behavior of Materials Lecture Note (6) 10/3/2010

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P l a s t i c d e f o r m a t i o n i s d u e t o t h e

m o t i o n o f a l a r g e n u m b e r o f

d i s l o c a t i o n s .

W h e n a s h e a r f o r c e i s a p p l i e d t o a

m a t e r i a l , t h e d i s l o c a t i o n s m o v e

R e a l m a t e r i a l s h a v e l o t s o f

d i s l o c a t i o n s , t h e r e f o r e t h e

s t r e n g t h o f t h e m a t e r i a l d e p e n d s

o n t h e f o r c e r e q u i r e d t o m a k e t h e

d i s l o c a t i o n m o v e , n o t t h e b o n d i n g

e n e r g y . ( U n l i k e i n e l a s t i c z o n e )

D i s l o c a t i o n s a l l o w d e f o r m a t i o n a t

m u c h l o w e r s t r e s s t h a n i n a

p e r f e c t c r y s t a l

If the top half of the crystal is slipping one plane at a time then only a small fraction of

the bonds are broken at any given time and this would require a much smaller force.

The propagation of one dislocation across the plane causes the top half of the crystal to

move (to slip) with respect to the bottom half but we do not have to break all the bonds

across the middle plane simultaneously (which would require a very large force).

T h e e a s e w i t h w h i c h d i s l o c a t i o n s

m o v e t h r o u g h a m e t a l c r y s t a l i s

Mechanical Behavior of Materials Lecture Note (6) 10/3/2010

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h i g h l y d e p e n d e n t u p o n t h e

p a r t i c u l a r c r y s t a l l o g r a p h i c p l a n e

a n d c r y s t a l l o g r a p h i c d i r e c t i o n

i n v o l v e d

S l i p i s e a s i e s t o n c l o s e p a c k e d

p l a n e s .

S l i p i s e a s i e s t i n t h e c l o s e p a c k e d

d i r e c t i o n .

G e n e r a l l y , o n e s e t o f

c r y s t a l l o g r a p h i c a l l y e q u i v a l e n t

s l i p s y s t e m s d o m i n a t e s t h e p l a s t i c

d e f o r m a t i o n o f a g i v e n m a t e r i a l .

H o w e v e r , o t h e r s l i p s y s t e m s

m i g h t o p e r a t e a t h i g h t e m p e r a t u r e

o r u n d e r h i g h a p p l i e d

s t r e s s .

Slip in close packed metals

Three slip directions in a close packed metal

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Q: What happens if the stress is applied, so

that it is not in line with one of the slip

directions?

A: The system will still relieve the stress by

slipping along the slip planes, although,

several planes may begin to move that are

closest in angle to the direction of the

stress.

Slip in FCC Materials

There are 4 different non-parallel slip

planes which each can move in 3

directions. Therefore there are 12 (4

x 3) ways that the close packed planes

in an FCC can move when a metal

deforms

What about in a HCP

Recall in a HCP metal, the stacking arrangement is

ABABA. With this arrangement, there are no new

slip planes formed as the layers are added the only

slip plane is in the close packing plane (an “A”

layer for instance. In the drawing, in the plane of

the paper)

Again, there are 3 slip directions in each plane, but

there is only 1 unique slip plane in a HCP metal.

Therefore, the total number of slip systems is (1

plane x 3 directions = 3)

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Slip in a body centered cubic (BCC)

Metal

Recall, BCC is not close packed, but because it is a dense

structure (not as dense as FCC or HCP) and the

arrangement of atoms, there are slip directions and planes

in a BCC as well.

But 6 slip unique planes

Therefore, there are 12 slip systems in a

BCC (2 directions * 6 planes).

BCC metals can contain up to 48 slip

systems (e.g. α-Fe). However, none of

the BCC slip planes are truly close-

packed – which means the slip systems

need additional energy (i.e. heat) to

operate. In other words, they are highly

temperature-dependent.

What does this mean?

FCC metals tend to be more ductile than HCP because there are more slip directions

HCP metals tend to be brittle

BCC metals are also ductile despite the different structure from FCC.

Summary of Slip in Metal Crystals

HCP FCC BCC

Close Packed? Yes Yes No

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Packing Efficiency 74% 74% 68%

Stacking Sequence of Close Packed

Planes ABABABAB ABCABC n/a

No. of unique, non-parallel slip

planes 1 4 6

No. slip directions in each plane 3 3 2

No. of total slip systems 3 12 12

Ductile or Brittle? Brittle Ductile Ductile

Atoms per cell 6 4 2

Examples: Mg, Zn

Al, Fe

(austenite),

Au, Ag, C

Fe (ferrite),

W, Nb

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Slip geometry: the critical resolved shear stress

Schmidt’s Law, Slip in Single Crystals

In order for a dislocation to move in its slip

system, a shear force acting in the slip

direction must be produced by the applied

force.

It is observed experimentally that slip occurs

when the shear stress acting in the slip direction

on the slip plane reaches some critical value.

This critical shear stress is related to the stress

required to move dislocations across the slip

plane.

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The tensile yield stress of a material is the applied stress required to start plastic

deformation of the material under a tensile load. We want to relate the tensile stress

applied to a sample to the shear stress that acts along the slip direction. This can be done

as follows. Consider applying a tensile stress along the long axis of a cylindrical single

crystal sample with cross-sectional area A:

The applied force along the tensile axis is

The area of the slip plane is

Where: is the angle between the tensile axis and the slip plane normal.

The component of the axial force F that lies parallel to the slip direction is

cosF

The resolved shear stress on the slip plane parallel to the slip direction is therefore given

by:

coscos cos

cos

r

F F

A A

It is found that the value of τr at which slip occurs in a given material with specified

dislocation density and purity is a constant, known as the critical resolved shear stress

τc. This is Schmid's Law.

The quantity, “cos υ cos λ”, is called the Schmid factor. The tensile stress at which the

crystal starts to slip is known as the yield stress σy, and corresponds to the quantity F/A

in the above equation. Symbolically, therefore, Schmid's Law can be written:

cos cosc y

In a given crystal, there may be many available slip systems. As the tensile load is

increased, the resolved shear stress on each system increases until eventually τc is reached

on one system. The crystal begins to plastically deform by slip on this system, known as

the primary slip system. The stress required to cause slip on the primary slip system is

cos

A

F A

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the yield stress of the single crystal. As the load is increased further, τc may be reached

on other slip systems; these then begin to operate.

From Schmid's Law, it is apparent that the primary slip system will be the system with

the greatest Schmid factor. It is possible to calculate the values of cos υ cos λ for every

slip system and subsequently determine which slip system operates first.

Maximum value of (cosυ cosλ) corresponds to:

Slip will occur first in slip systems oriented close to this angle (υ = λ = 45o)

with respect to the applied stress

Schmid postulated that:

• Initial yield stress varies from sample to sample depending on, among

several factors, the position of the crystal lattice relative to the loading

axis.

• It is the shear stress resolved along the slip direction on the slip plane

that initiates plastic deformation.

• Yield will begin on a slip system when the shear stress on this system

first reaches a critical value (critical resolved shear stress, crss),

independent of the tensile stress or any other normal stress on the lattice

plane.

The Dot Product is a vector operation which will return a scalar value (single number),

which for unit vectors is equal to the cosine of the angle between the two input vectors

(for non-unit vectors, it is equal to the length of each multiplied by the cosine, as shown

in the equation below). We can represent the Dot Product equation with the ● symbol.

* *cosA B A B

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The cosine of the angle between the vectors is:

2 2 2 2 2 2

cos

cos*

x x y y z z

x y z x y z

A B AB

A B A B A BA B

AB A A A B B B

Slip in a Single Crystal

Each step (shear band) result from the generation

of a large number of dislocations and their

propagation in the slip system with maximum

resolved shear stress.

Shear bands

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Plastic Deformation of Polycrystalline Materials

Grain orientations with respect to applied stress are

random.

The dislocation motion occurs along the slip systems

with favorable orientation (i.e. that with highest resolved

shear stress).

Larger plastic deformation corresponds to elongation of

grains along direction of applied stress.

Slip directions vary from crystal to crystal ⇒ Some

grains are unfavorably oriented with respect to the

applied stress (i.e. cosυ cosλ low)

Even those grains for which cosυ cosλ is high may

be limited in deformation by adjacent grains which

cannot deform so easily

Dislocations cannot easily cross grain boundaries

because of changes in direction of slip plane and

disorder at grain boundary

As a result, polycrystalline metals are stronger

than single crystals (the exception is the perfect

single crystal without any defects, as in whiskers

Whiskers are microscopic single-crystal metal fibers, thinner than a

human hair.

This incandescent lamp uses a single crystal of silica carbide for a filament

rather than a tungsten wire. Another material, hafnium carbide, has also been tried in

experiments.

The very thin "whisker" filament is stronger than tungsten, and has a high resistance that

changes little during the lamp's operation. A major difficulty lies in growing the single

crystals long enough to be useful.