diat htt lect 1 2 3

8/11/2019 Diat Htt Lect 1 2 3

1/12

HeatTreatment

Dr.SantoshS.Hosmani

1

Brief Intro. about Myself

uca ona ac groun :

1997 2001: B.E. (Metallurgy), V. Regional College of Engineering (currently NIT), Nagpur

2001 2003: M.Tech. (Process Meta.), Indian Institute of Technology, Bombay

2003 2006: Ph.D. (Physical Meta.), Max-Planck Institute for Metals Research, Stuttgart, Germany

Assistant Professor,Dept. Metallurgy & Materials Science, College of Engineering, Pune, India.

Assistant Professor,Dept. of Applied Mechanics,Indian Institute of Technology, Delhi, India.

Lecturer,Dept. of Metallurgical & Materials Engineering, National Institute of Technology

Karnataka, Surathkal, India.

Postdoctoral researcher, Dept. of Materials Science and Engineering, Case Western Reserve

, . . ..

Postdoctoral researcher,Max-Planck-Institute for Metals Research, Stuttgart, Germany.

2

HEAT TREATMENT

LECTURES: 3 hrs/week

FACULTY: Dr. Santosh S. Hosmani

Dept. of Metallurgy & Materials Science,

COE, Pune

Email: [email protected]

Phone (use only in case of emergency plz): 9762316594

OFFICE HOURS: Open

3

GRADING / EXAMINATION SCHEME:

e g age

Quiz**- 1 & 2 20%

Mid-Sem Exam 30%

End Sem Exam 50%

**Note: Quiz can be surprise quiz OR Quizzes can be held on shortnotice. Therefore, continuous stud and attendance is desired.

Minimum passing marksfor this course is 40%.

Total attendancerequirement is as per the insti tu te rules. Any kind of

proxy is strictly prohibited.

Quiz/Exam wil l not be conducted againif anyone is absent without any

.

Depending upon the requirement of the course, Assignments/Home-work

could be given. However, there will be no weightage for the assignmentsin this course.

4

8/11/2019 Diat Htt Lect 1 2 3

2/12

Introduction of the course

In many engineering applications, steels are the most preferred material. There

are various types of steels which are evolved from the requirements of the

engineering components. Requi rement s is di rect ly l inked t o desi red

properties .

Properties can be manipulated by altering the chemistry of the alloyand/or bymechanical treatmentsand/or by giving appropriate heat-treatments.

As a materials engineer, literacy about the heat-treatment technology

processing & fundamental concepts is very essential. This knowledge teaches

the intelligent use of the existing grade of steel for a particular application.

Heat-treatment technology touches many important applications in automobile

and aerospace sectors.

-

treatment and metallurgy of the some important iron-based alloys.

Note: There are some topics in the course which does not require class-room

-

5

, - .

SYLLABUS:Available on your department website.

Heat Treatment of Metals, Vijendra Singh, 2007, Standard Publishers andDistributors, New Delhi

R.A. Hi ins En ineerin Metallur Part I A . Ph sical Met ELBS 5th

. . , , , . , ,

ed., 1983

REFERENCE BOOKS:

Steel and its Heat Treatment -K.E Thelning, Butterworth, London Handbook of Heat Treatment of Steels Prabhudev-Tata Mc Graw Hill.

New Delhi, 1988

, , .,1979

=>You can read any book available in your library.

=>Whenever lectures are in power-point-presentations, pdf-files of the

slides will be provided to you by email.

6

Friendly suggestion: Please learn to make good class-notes.

Some Basics revision of the concepts

7

Familytrees:organizingmaterialsandprocesses

likenessuseful to know when deciding which family to use for a given

design.

Choosing a material is only half the story. The other half is the choice of a

process route to shape, join and finish it.

processed in some ways but not others, and a given process can be applied to

some materials but not to others.

8

8/11/2019 Diat Htt Lect 1 2 3

3/12

Thematerialstree Classificationofmaterials

It is conventional to classify the materials of engineering into the six

broad families:

There is sense in this classification: the members of a family have certain features in

common: similar properties, similar processing routes and, often, similar applications.

9

, ,

each of which is characterized by a set ofattributes:its properties..

10

Metals:

They have relativelyhigh stiffness(modulus, E).

Most,when pure, are softand easily deformed, meaning that yis low.

Theycan be made strongby alloying and by mechanical and heat treatment,

y, ,

deformation processes.

And, broadly speaking, they aretough, with a usefully high fracture toughnessK1c. They are good electrical and thermal conductors.

But metals have weaknesses too: they are reactive; mostcorrode rapidlyif not

protected.

11

Stiffness and strength are central to mechanical design, often in

combination with thedensity,.

What is the stiffness?

,

meaning that the material returns to its original shape

when the stress is removed. Stiffness is measured bythe elastic modulusE.

E reflects stiffness, S, of the bonds that hold them

together.But, remember thatES

What is the strength?

failure. Strength is measured by the elastic limit y or

tensile strength ts.

Note: Stress and strain are not material properties.12

8/11/2019 Diat Htt Lect 1 2 3

4/12

It is theplasticityof iron and steel that made them thestructural materials

on which theIndustrial Revolutionwas built

18061859

British

Engineer

(17571834)

ThomasTelford He was perhaps the greatest engineer of the

Industrial Revolution in terms of design

co s eng neer, ,

theplasticity of iron and steel.

13

18061859

British

Engineer

IsambardKingdomBrunel

GreatEastern

Launched:

31

Jan.

1858

greatthingsarenot

donebythosewho

simplycountthecost14

engineering, derives from their ability

o e ro e , orge , rawn an

stamped.

15

Strength,plasticworkandductility

y ,

onset of plasticity is not always

distinct so we identify y with the

0.2% roo stressthat is the stress

forMetals

,

at which the stressstrain curve for

axial loading deviates by a strain of

0.2% from the linear elastic line. It is

the same in tension and

compression.

,

most metals work harden, causing

the rising part of the curve, until a

maximum, the tensile strength, is

reached.

This is followed in tension by non-

fracture.

16

8/11/2019 Diat Htt Lect 1 2 3

5/12


forMetals

Howstressstraincurvelookslikeformetalsincompression? 17

Stress-strain dia rams for


compression have different shapes

from those for tension. forMetals

limits in compression very close tothose in tension. However, when

ieldin be ins the behavior is uite,

different.

When a small specimen of ductile

,

bulge outward on the sides and

become barrel shaped. With

,

flattened out, thus offering increased

resistance to further shortening

which means the stress-strain

Ref.:DepartmentofCivilEngineeringattheUniversityofMemphis

curve goes upward).

18

TheoriginsofstrengthandductilityFundamentals

Perfection:theidealstrength

Thebondsbetweenatoms,likeanyotherspring,haveabreakingpoint.

Figure:Stressstrain curve for a

single bond.

Here an atom is assumed to occu a cube of sidea so that a force F corres ondsoto astress =F/ao

2.

The force stretches the bond from its initial length ao to a new length a, giving a

strain = a a a .o o.

In case of the modulus we focused on the initial, linear part of this curve, with a

slope equal to the modulus, E. Stretched further, the curve passes through a

.

bond strengthif you pull harder than this it will break.

19

The distance over which interatomic forces act is smalla bond is broken if it is


.

So the force needed to break a bond is roughly:

F

t nesson =,

10%10

%10

oaaa

lengthbondorigionalof

===

=

10100oo

10/oa

FS=

On this basis the ideal strength of a solid

should therefore be roughly:

10

oaSF =Q

SF

ES

=

=

:andbetweenRelation

2

o

oo

a

S=

&etweenrelationlinearAssume

oa

SE=Q

Note: This relationship doesnt allow for the

curvatureof the forcedistance curve;more refined

calculations give a ratio of 1/15. 20

8/11/2019 Diat Htt Lect 1 2 3

6/12


Surprisingly,

None of the metals, polymers and ceramics achieve the ideal value of1/10; most dont even come close.

21

CriticalResolvedShearStress

eory

(GPa)

xper men

(MPa)

a o

Theory/Exp

Fe (BCC) 12 15 800

Cu (FCC) 7 0.5 14,000

Zn (HCP) 5 0.3 17,000

22


Surprisingly,

None of the metals, polymers and ceramics achieve the ideal value of

1/10; most dont even come close.

Whynot?

Nothing is perfect in this world.!

Existence of Imperfections / Defects in materials.!

23

Dislocation

Thedislocationis the key player in explaining important mechanical properties, like

strength and ductility.

Dislocated means out of joint and this is not a bad description of what is

happening here. The upper part of the crystal has extra halflayer of atoms than

the lower part.

It is dislocations that make metals soft and ductile.

because of this they have elastic energy associated with them.

24

8/11/2019 Diat Htt Lect 1 2 3

7/12

Inventorsofdislocationconcepts

an ideal strength aroundE/15 (whereEis the modulus).

In reality the strengths of engineering materials are nothing like this big;

o ten t ey are are y 1% o it.

Hungarian/US

physicist and

British

mathematician,

metallurgistphysicist and

expert on fluid

dynamics and

wave theor

Sir Geoffrey Ingram Taylor Egon Orowan

ese two persona t es rea ze t at a s ocate crysta

could deform at stresses far below the ideal.25 Ref.: Book by W.D. Callister

Ref.: Book by W.D. Callister Ref.: Book by W.D. Callister

8/11/2019 Diat Htt Lect 1 2 3

8/12

Ref.: Book by W.D. Callister Ref.: Book by W.D. Callister 30

Thelatticeresistance

Dislocation Motion Plastic Deformation

asy s oca on o on asy as c e orma on

Weak Cr stal

Difficult

Dislocation Motion

Difficult

Plastic Deformation

Strong Crystal

31

Where does the resistance to

. . , ,

come from?

32

8/11/2019 Diat Htt Lect 1 2 3

9/12

The lattice resistance

There are several contributions to this resistance.

I. Lattice resistance, fi:

s s e n r ns c res s ance o e crys a s ruc ure o p as c s ear.

Plastic shear, as we have seen, involves the motion of dislocations.

Pure metals are soft because the non-localized metallic bond does little

to obstruct dislocation motion, whereas ceramics are hard because their

more localized covalent and ionic bonds (which must be broken and

reformed when the structure is sheared) lock the dislocations in place.

The electrons and +ve ions

are all in a fixed position.

The electrons are in a fixed

position

The electrons are not

fixed and free to move

throughout the lattice. 33

The lattice resistance

When the lattice resistance is hi h, as in ceramics, further hardenin is

more than sufficientthe problem becomes that of suppressing fracture: i.e.yield strength is much larger than fracture strength of ceramics.

On the other hand, when the lattice resistance fiis low, as in metals, the

material can be strengthened by introducing obstacles to slip.

II. other dislocations giving what is called Work Hardening (fwh),

III. grain boundaries introducing Grain-size Hardening (fgb),

IV. precipitates or dispersed particles giving Precipitation Hardening (fppt),

V. by adding alloying elements to give Solid Solution Hardening (fss).

ese ec n ques or man pu a ng s reng are cen ra o a oy es gn.

34

II. Work hardening / Strain hardening, fwh:

ur ng p ast c e ormat on s ocat on ens ty

of a crystal should go down

But,

Experimental Result

Well-annealed crystal: 1010 m-2

Lightly cold-worked: 1012 m-2

eav y co -wor e : m-

35


Workhardeningor

StrainHardening

y

Strain,

see Book by V. Raghvan36

8/11/2019 Diat Htt Lect 1 2 3

10/12


37


During plastic deformation dislocation density increases.

Plastic deformation increases the yield strength of the

crystal: strain hardening or work hardening

Why deformation increases strength?

What exactly is the Strain Hardening?

38


Strain Hardening:

Dislocation against Dislocation

A dislocation in the path of other

dislocation can act as an obstacle to the

motion of the latter

39

Sessile dislocation in an FCC crystal:

< + a

2

2

1bE =

g,

fwh:

1

222aaa

+ is the shear stress required to move a single dislocation in theabsence of any other dislocation

42


metals and alloys, e.g. Al-based alloys

43

III. Grain-size / Grain-boundary hardening, fgb:

ran

Grain1

Grain boundary

44

8/11/2019 Diat Htt Lect 1 2 3

12/12


Discontinuity of a slip plane across a grain boundary

Slip plane

Dislocation

Grain Boundary

45


A dislocation cannot glide across a grain

Higher stresses required for deformation

Finer the grains, greater the strength

Ref.: Book by V. Raghavan46


Role of Grain Size in Strengthening

Hall-Petch Relation

k y

D= +0

y y e strengt

D: average grain diameter

0,k: constants

Coarse Grains Fine Grains0 => yield strength of a single

crystal

47


Role of Grain Size in Strengthening

k y

D= +0

Coarse Grains Fine Grains

This concept is relevant to annealing,-

alloy design, e.g. HSLA steel 48

diat htt lect 1 2 3

Documents