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Lecture 4 Dislocations & strengthening mechanisms

ISSUES TO ADDRESS...

• Why are dislocations observed primarily in metalsand alloys?

• How are strength and dislocation motion related?

• How do we increase strength?

• How can heating change strength and other properties?

A screw dislocation within a crystal

The atom positions around an edge dislocation

Edge dislocation and screw dislocation

Plictic deformation corresponds to the motion of large numbers of dislocations

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Dislocations & Materials Classes

• Covalent Ceramics (Si, diamond): Motion hard.

-directional (angular) bonding

• Ionic Ceramics (NaCl):Motion hard.-need to avoid ++ and - -

neighbors.

+ + + +

+++

+ + + +

- - -

----

- - -

• Metals: Dislocation motion easier.

- non-directional bonding- close-packed directions for slip.

electron cloud ion cores

++

+

++++++++

+ + + + + +

+++++++

Dislocation Motion

Dislocations & plastic deformation• Cubic & hexagonal metals - plastic deformation by plastic

shear or slip where one plane of atoms slides over adjacent plane by defect motion (dislocations).

• If dislocations don't move, deformation doesn't occur!

Adapted from Fig. 7.1, Callister 7e.

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Dislocation Motion

• Dislocation moves along slip plane in slip directionperpendicular to dislocation line

• Slip direction same direction as Burgers vector

Edge dislocation

Screw dislocation

Adapted from Fig. 7.2, Callister 7e.

_ Slip plane• plane allowing easiest slippage• wide interplanar spacings• highest planar densities

– Slip direction• direction of movement • highest linear densities

– A {111} <110>Slip system within an FCC unit cell– Slip occurs on {111} planes (close-packed) in <110> directions

(close-packed). – Three <110> slip directions as shown in (b), within the (111)

plane. => total of 12 slip systems in FCC

– Slip system in BCC unit cell: {110} <111>

Slip system

Adapted from Fig. 7.6, Callister 7e.

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Slip system

Crystals slip due to a resolved shear stress, tR.

Applied tensile stress: = F/As

slip

directi

on

FA

F

slip plane

normal, ns

Resolved shear stress: tR=Fs/As

slip

directi

on

AS

tR

tR

FS

Fs = Fcosl

As = A/cosf

tR = scosfcosl

tR (max) = s(cosfcosl)max

Critical resolved shear stress tCRSS:

The minimium shear stress required to initiate slip.

The applied stress required to initial yielding sy:

sy = tCRSS / (cosfcosl)max

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Single Crystal Slip

Adapted from Fig. 7.8, Callister 7e.

Adapted from Fig. 7.9, Callister 7e.

Slip in a zinc single crystal.

• Stronger - grain boundariespin deformations

• Slip planes & directions(l, f) change from onecrystal to another.

• tR will vary from onecrystal to another.

• The crystal with thelargest tR yields first.

• Other (less favorablyoriented) crystalsyield later.

Adapted from Fig. 7.10, Callister 7e.

Slip Motion in Polycrystals s

300 mm

Slip lines on the surface of a polycrystalline specimen of copper that was polished and subsequently deformed.

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Mechanisms of strengthening in metals

•Strain hardening (work hardening)

•Grain boundary strengthening

•Solid-solution strengthening

•Precipitation strengthening

Strain hardeningWithin the region between Y.S (stress at 2) and T.S (stress at 3)

The region after T.SNeck down of a tensile test specimen within the gage lengh

Dislocations in plastically deformed stainless steel

Strain hardening (work hardening)

necking

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• Room temperature deformation (cold work).• Common forming operations change the cross sectional area:

Adapted from Fig. 11.8, Callister 7e.

-Forging

Ao Ad

force

dieblank

force-Drawing

tensile force

AoAddie

die

-Extrusion

ram billet

container

containerforce

die holder

die

Ao

Adextrusion

100 x %o

doA

AACW

-=

-Rolling

roll

AoAd

roll

Strain hardening (work hardening)

Percent Cold Work:

• Ti alloy after cold working:

• Dislocations entanglewith one anotherduring cold work.

• Dislocation motionbecomes more difficult.

Adapted from Fig. 4.6, Callister 7e.(Fig. 4.6 is courtesy of M.R. Plichta, Michigan Technological University.)

Dislocations During Cold Work

0.9 mm

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Exercise:

Compute the tensile strength and ductility (%EL) of a cylindrical copper rod if it is cold worked such that the diameter is reduced from 15.2 mm to 12.2 mm.

As cold work is increased:

• Yield strength (sy) increases• Tensile strength (TS) increases• Ductility (%EL or %AR) decreases

• What is the tensile strength & ductility after cold working?

Adapted from Fig. 7.19, Callister 7e.

Cold Work Analysis example problem 7.2 p194

% Cold Work

100

300

500

700

Cu

200 40 60

yield strength (MPa)

sy = 300MPa

300MPa

ductility (%EL)

% Cold Work

20

40

60

20 40 6000

Cu

Do =15.2mm

Cold Work

Dd =12.2mm

Copper

% Cold Work

tensile strength (MPa)

200Cu

0

400

600

800

20 40 60

340MPa

TS = 340MPa

7%

%EL = 7%

6.35100x %2

22=

p

p-p=

o

do

r

rrCW

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Grain boundary hardening

• Grain boundaries arebarriers to slip.

• Barrier "strength"increases withIncreasing angle of misorientation.

• Smaller grain size:more barriers to slip.

• Hall-Petch Equation:

Hall–Petch (H-P) relationship

s0 and sy: yield strength k: the Hall–Petch slope d: grain sizek: constant

2

1

0

-+= kdy ss

2

1

0

-+= dkHH Hv H0 and kH are constants.

In analogy, hardness (Hv) can be related to the grain size by

Grain boundary hardening

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Homework1. The following yield strengths were obtained in ferritic steel as a function of grain size. Estimate the two constants in the Petch equation for this material and predict the expected yield strength of the steel in which the grain size is reduced to i micron.

grain size (micron) Yield strength (MPa)

250 105

40 180

12 280

• Impurity atoms distort the lattice & generate stress.• Stress can produce a barrier to dislocation motion.

Solid-Solution strengthening

• Smaller substitutionalimpurity

Impurity generates local stress at Aand B that opposes dislocation motion to the right.

A

B

• Larger substitutionalimpurity

Impurity generates local stress at Cand D that opposes dislocation motion to the right.

C

D

Solid Solutions

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Stress Concentration at Dislocations

Adapted from Fig. 7.4, Callister 7e.

Strengthening by Alloying

• small impurities tend to concentrate at dislocations

• reduce mobility of dislocation \ increase strength

Adapted from Fig. 7.17, Callister 7e.

• large impurities concentrate at dislocations on low density side

Tensile lattice strain imposed on host atoms by a small substitutional impurity atom

Possible locations of small impurity atoms relative to an edge dislocation such that there is partial cancellation of impurity-dislocation lattice strains

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Solid Solution Strengthening in Copper

• Tensile strength & yield strength for copper-nickel alloys increase with wt% Ni.

• Empirical relation:

• Alloying increases sy and TS.

21 /y C~s

Adapted from Fig. 7.16 (a) and (b), Callister 7e.

Ten

sile

str

engt

h (M

Pa)

wt.% Ni, (Concentration C)

200

300

400

0 10 20 30 40 50 Yie

ld s

tren

gth

(MP

a)wt.%Ni, (Concentration C)

60

120

180

0 10 20 30 40 50

• Dislocations are observed primarily in metalsand alloys.

• Strength is increased by making dislocationmotion difficult.

• Particular ways to increase strength are to:--decrease grain size--solid solution strengthening--precipitate strengthening--cold work

• Heating (annealing) can reduce dislocation densityand increase grain size. This decreases the strength.

Summary

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Lecture 5 Creep p 238

Tensile test at elevated temperatures under a constant load – the strain will increase with time

Creep is defined as time dependent plastic deformation at constant stress and temperature.

Creep deformationbecomes important only for temperatures greater than about 0.4Tm (Tm = absolite melting temperature)

The slope of this curve is the creep rate

Creep curve

Questions and problems 8.26 p249

Give the approximate temp. At which creep deformation becomes an important consideration for each of the following metals: Sn, Mo, Fe, gold, Zn and Cr.

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The primary stage:

a decreasing strain rate

dislocation climb

The secondary stage:

straight line

constant-strain rate

Final stage:

Strain rate increase due to necking or internal cracking

• Occurs at elevated temperature, T > 0.4 Tm

Adapted from Figs. 8.29, Callister 7e.

Creep

elastic

primarysecondary

tertiary

Homework 8.26 p249

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Influence of stress s and temperature T on creep behavior with either increasing stress or temperature:

1. The instantaneous strain at the time of stress application increases;

2. The steady-state creep rate is increased;

3. The rupture lifetime is diminished

Diffusion mechanisms in creep

Diagram showing stress-directed flow of vacancies (solid lines) from tensile to compressive grain boundaries and corresponding reverse flow of atoms or ions (dashed lines

Rate = C exp (-Q/RT)

R: gas constant

Activition energy Q

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Dislocation climb

At elevated temperature

Thermally activated atom mobility

dtdecreep rate:

)/exp(0

RTQA Ddpn

-=-

se&s: applied stressd: grain sizeD0: diffusion coefficientQ: activation energy for diffusionT: absolute temperature

Herring Nabarro creep, n=1, p=2, (D0 and Q refer to bulk diffusion)Coble creep (GB diffusion), n=1, p=3, (D0 and Q refer to GB diffusion)Dislocation climb n=4-7, p=0, (D0 and Q refer to bulk diffusion)Grain boundary sliding

Creep rate of the secondary stage – the steady-state

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)/exp(0

RTQA Ddpn

-=-

se&

Arrhenius plot

(ln ) versus (1/T):

When stress s is a constant:

= k exp (-Q/RT)

ln = K + (-Q/R)•(1/T)

The slop of the curve gives the activition energy for the creep mechanism

e&e&

e&

)/exp(0

RTQA Ddpn

-=-

se&

Creep mechanism – the stress exponent n

e&(log ) versus log s

When temperature is a constant,

= ksn

(log )= K + n log s

e&e&

The slop of the curve gives the (n ) value for the creep mechanism

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Creep deformation map

Pure silver,

Grain size 32 mm,

Elastic boundaries determined at a strain rate of 10-8/sec

s: stress, m: shear modulus

T: temperature (K), Tm melting temperature (K)

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Turbine blade

Subjected centrifugal forces

Creep map for MAR-M200 nickel base alloy (100mm)

The creep rate below 10-10 sec-1 is usually negligible.

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Larson-Miller plot

Give a plot in the form of stress - life time

•Applied stress versus time to failure at a given temperature

•Applied stress versus time to a certain strain at a given T

Log s - T (C + log t)

•C is a constant, for most materials C = 20

•T temperature (K)

•t: life time

for steady – state, t = e /

e.g. t0.01 means the time to get a strain e= 0.01

t0.01 = 0.01 /

e&

e&

Creep Failure • p242 design example 8.2

Estimate rupture time S-590 Iron, T = 800°C, s = 20 ksi• Failure:

along grain boundaries.

time to failure (rupture)

function ofapplied stress

temperature

L)t(T r =+ log20

g.b. cavities

appliedstress

• Time to rupture, trL)t(T r =+ log20

1073K

Ans: tr = 233 hr

24x103 K-log hr

Adapted fromFig. 8.32, Callister 7e.(Fig. 8.32 is from F.R. Larson and J. Miller, Trans. ASME, 74, 765 (1952).)

L(103K-log hr)

Str

ess,

ksi

100

10

112 20 24 2816

data for S-590 Iron

20

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Sample problems

(the life time is t0.01, i.e. the time to a strain of 10-2)

Larson prediction, problem 4

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homework

A common creep requirement is a 1000h creep life to 2% strain at a (shear) stress of 100 MPa (1MPa = 106 N/m2)(1) Calculate the creep rate ( ) for this creep requirement(2) Referring to the creep maps below (Ni-base superalloy), at what temperatures in

the two materials is this requirement met?(3) What is the dominant creep mechanism in the two cases?

e&

Grain size d = 100 mm d = 1cm