This week, we will •Learn critical resolved shear stress •Slip by dislocation movement •Interaction of dislocations

WEEK SIX

Last week we learned

• Theoretical strength of single crystals

• Types of dislocations

Stress and Dislocation Motion • Crystals slip due to a resolved shear stress, tR.

• Applied tension can produce such a stress.

slip plane

normal, ns

Resolved shear stress: tR = F s /A s

AS

tR

tR

FS

Relation between s and tR

tR = FS /AS

F cos l A / cos f

l F

FS

f nS

AS

A

Applied tensile stress: = F/A s

F A

F

lfst coscosR

Critical resolved shear stress for slip The extent of slip in a single crystal depending on: 1) The magnitude of the shear stress 2) The geometry of the crystal structure 3) The number of active slip plane in the shear stress direction.

Slip occurs when the shearing stress on the slip plane in the slip direction reaches a critical resolved shear stress.

• Schmid calculated the critical resolved shear stress from a single crystal tested in tension.

• The area of the slip plane A’ = A/cosφ.

• The force acting in the slip plane A’ is Pcosλ

• The critical resolved shear stress is given by

is called the Schmid Factor

lfst coscosR

• Condition for dislocation motion: CRSS ttR

• Crystal orientation can make

it easy or hard to move dislocation

0.1 MPa to 10 MPa

typically

flst coscosR

Critical Resolved Shear Stress

t maximum at l = f = 45º

tR = 0

l =90°

s

tR = s /2 l =45° f =45°

s

tR = 0

f =90°

s

Ex: Deformation of single crystal

So the applied stress of 6500 psi will not cause the crystal to yield.

t s cosl cosf

s 6500 psi

l=35°

f=60°

t (6500 psi) (cos35 )(cos60 )

(6500 psi) (0.41)

t 2662 psi tcrss 3000 psi

tcrss = 3000 psi

a) Will the single crystal yield?

b) If not, what stress is needed?

s = 6500 psi

Adapted from

Fig. 7.7,

Callister 7e.

Ex: Deformation of single crystal

psi 732541.0

psi 3000

coscoscrss

fl

ts y

What stress is necessary (i.e., what is the

yield stress, sy)?

)41.0(cos cos psi 3000crss yy sflst

psi 7325ss y

So for deformation to occur the applied stress must be greater than or

equal to the yield stress

Dislocations

Dislocations introduce imperfection into the structure and therefore these could explain how real materials exhibit lower yield stress value than those observed in theory.

• Lower the yield stress from theoretical values.

• Produce plastic deformation (strain hardening).

• Effects mechanical properties of materials. Produce imperfection in

crystal structures

Defects and CRSS

Defects Vacancies Impurity atoms Alloying elements

Critical resolved shear stress

Variation of critical resolved shear stress with composition in silver-gold alloy single crystal.

Strength of metal crystals as a function of dislocation density.

Logarithmic scale

Deformation by dislocations

An edge dislocation: the dislocation line moves in the direction of the applied shear stress τ.

A screw dislocation: the dislocation line motion is perpendicular to the stress direction.

Edge Burgers vector is perpendicular to the dislocation line.

Screw Burgers vector is parallel to the dislocation line.

Both shear stress and final deformation are identical for both situations.

The Peierls stress for a given plane decreases with increasing distance between planes. Since the distance between planes increases with planar atomic density, slip is preferred on closely packed planes.

Shear stress to move one dislocation

U=aGb2

A particularly important consequence is that slip will usually occur in closed packed directions.

Slip System – Slip plane - plane allowing easiest slippage

• Wide interplanar spacings - highest planar densities

– Slip direction - direction of movement - Highest linear

densities

– FCC Slip occurs on {111} planes (close-packed) in <110> directions (close-packed)

=> total of 12 slip systems in FCC

– in BCC & HCP other slip systems occur

Deformation Mechanisms

Adapted from Fig.

7.6, Callister 7e.

Predominant Slip Systems

FCC structure

Predominant Slip Systems

Dislocations & Crystal Structures

• Structure: close-packed planes & directions are preferred.

view onto two close-packed planes.

close-packed plane (bottom) close-packed plane (top)

close-packed directions

• Comparison among crystal structures:

FCC: many close-packed planes/directions; HCP: only one plane, 3 directions; BCC: none

• Specimens that were tensile tested.

Mg (HCP)

Al (FCC)

tensile direction

r has units of length per unit volume (L/L3). Usually it is expressed as dislocations per unit area (ex: dislocations/cm2 =106)

Slip by dislocation movement

Ex: Burger vektörü b= 0.3 nm olan YMK yapıda tek kristal bir metal aşağıdaki

şekilde gösterilmiştir. Dislokasyonların tümü kenar dilokasyonlarıdır. Dislokasyon

çizgileri x-eksenine paralel olacak şekildedir ve kayma yönü olan [1 1 0] y-eksenine

paraleldir. Eğer dislokasyonların ortalama hareket mesafesi 1mm ise, g=0.15

büyüklüğünde bir kayma gerinmesi için kaç adet dislokasyon gereklidir?

x

y

g

t

30 mm

15 mm

20 mm

Strain hardening of single crystals

Strain hardening or work hardening is caused by dislocations interacting with each other and with barriers, which impede their motion through the crystal lattice.

• Precipitate particles, foreign atoms serve as barriers which result in dislocation multiplication. strain hardening.

• Dislocation pile-ups at barriers produce a back stress which opposes the applied stress. strain hardening

• Dislocation density increases dramatically for example from 106 /cm2 in annealed condition to 1010-12 /cm2 in cold-worked condition.

Flow curve for FCC single crystals

Stage II : • Strain hardening occurs rapidly. • Slips occur more than one set of planes. much higher dislocation density. • Dislocation tangles begin to develop • Slope temperature independent mechanism is dislocation pil-up

Stage III : dynamic recovery • Decreasing rate of strain hardening. • By cross slip dislocations escape

Stage I : easy glide • Slips occur on only one slip system. • Dislocation density is low. • Crystal undergoes little strain hardening. • Most dislocations escape from the crystal to the surface.

Forces on dislocation

• A dislocation line moving in the direction of its Burgers vector under the influence of a uniform shear stress τ.

• The force per unit length of dislocation F : F=t·b (per unit length)

• This force is normal to the dislocation line at every point along its length and is directed toward the unslipped part of the glide plane.

•The Burgers vector is constant along the curved dislocation line.

W=t·b·ds·dl the work done by slip

Forces between dislocations •Dislocations of alike sign on the same slip plane will repel each other •Dislocations of opposite sign on the same slip plane will attract each other, run together, and annihilate each other.

A more general relation is given by x>y : repel

X>y : attract

x=y : equilibrium

x=0 : equilibrium

Fx=0 => equilibrium

(Fx is per unit lengt of dislocation line)

b burger vektörüne sahip bir kenar dislokasyonu bir engel önüne gelerek duruyor. Eş burger vektörüne sahip ikinci bir kenar dislokasyonu birinciye aynı düzlemden yaklaşıyor. Eğer dislokasyonların uzunluğu 100 mm ve denge konumunda dislokasyonlar arasındaki itme kuvveti 7 mN ise, iki dislokasyon arasındaki denge mesafesini hesaplayınız, (E=70 GPa, n=0.3, b =4.98 Å)

Dislocation climb Dislocation climb is a non conservative movement of dislocation where and edge dislocation can move out of the slip plane onto a parallel directly above or below the slip plane.

• Climb is diffusion-controlled (thermal activated) and occurs more readily at elevated temperature. important mechanism in creep.

• Positive direction of climb is when the edge dislocation moves upwards. Removing extra atom (or adding vacancy around ). Compressive force produces + climb.

(a) Diffusion of vacancy to edge dislocation; (b) dislocation climbs up one lattice spacing

Not observed in screw dislocations.

Cross Slip

A screw dislocation on one slip plane can

move onto another plane containing the

Burgers vector. This is called cross-slip.

Cross-slip of a screw dislocation XY from (a) plane A to (b) plane B to (c) plane A.

Slip always occurs in direction of burgers vector b.

Successive positions of dislocations.

Intersection of dislocations The intersection of two dislocations produces a sharp break (a few atom spacing in length) in dislocation line. Dislocation intersection mechanisms play an important role in the strain hardening process

• Kink is a sharp break in the dislocation line which remains in the slip plane. • Jog is a sharp break in the dislocation moving it out of the slip plane.

(a), (b) Kinks in edge and screw dislocations

(c), (d) Jogs in edge and screw dislocations.

Dislocation Sources

• All metals initially contain an appreciable number of dislocations produced from the growth of the crystal from the melt or vapour phase. • Gradient of temperature and composition may affect dislocation arrangement. • Irregular grain boundaries are believed to be responsible for emitting dislocations. • Dislocation can be formed by aggregation and collapse of vacancies to form disk or prismatic loop. • Heterogeneous nucleation of dislocations is possible from high local stresses at second-phase particles or as a result of phase transformation.

Multiplication of dislocations

Frank & Read proposed that dislocations could be generated from existing dislocations.

• The dislocation line DD’ bulges out (D and D’ are anchored by impurities with distance l) and produces slip as the shear stress τ is applied.

• Beyond this point, the dislocation loop continues to expand till parts m and n meet and annihilate each other to form a large loop and a new dislocation.

Note: Repeating of this process producing a dislocation loop, which produces slip of one Burgers vector along the slip plane

• The maximum τ for semicircle dislocation bulge, fig (b)

Multiplication of dislocations

Dislocation Pile-Ups

Edge dislocation : k=1-nScrew dislocation: k=1

ts: resolved shear stress

The number of dislocations between the source and the obstacle is

A piled-up array of n dislocations can be considered as a giant dislocation with burgers vector nb.

The force at the head of the pile up is nbts (per unit length of dislocation).

When the source is located at the center of a grain of diameter D, the number of dislocations is given by

D/4 because back stress on both sides of the source.

Dislokasyon oluşmasına neden olan iki jog arasındaki mesafenin 2x 10-6 m olduğunu bildiğimiz bir Frank-Reed kaynağından başlayan dislokasyonlar 100 mm ileride bulunan tane sınırının önünde yığılmaktadırlar. Eğer b=2.5 Å, G=40 GPa, n=0.3 ise kenar dislokasyonu kabul ederek a) Bu kaynaktan ilk oluşan dislokasyon için uygulanması gereken kayma gerilmesini b) Uygulanan kayma gerilmesi 100 MPa ise Frank-Reed kaynağı devre dışı kalmadan önce yaratılacak dislokasyon sayısını c) Dislokasyon yığılması sonucu malzemenin depoladığı enerjideki artışı hesaplayınız. (1 Å = 10-7 mm)

a≈1

Tek dislokasyon gibi davranır: U=40·109· (2200·2.5·10-10)2= 0.0121 J/m