Engineering Metallurgy

Nagpur Institute of Technology, Nagpur

1 Engineering Metallurgy These notes are not for circulation and sale.

B.E. III Semester Mechanical Engineering



UNIT -I 1. Differentiate between microscopic and macroscopic examination of metal. Sr. No.

Microscopic Examination Macroscopic Examination

1 Microscopic examination or micrography involves the study of the structures of metals and their alloys under a microscope at magnifications from X20 to X2000. The observed structure is called the microstructure.

Macroscopic examination involves the study of the structure of metals and their alloys by the naked eye or by low power magnification up to X15. The observed structure is called the macrostructure.

2 Micro-examination involves much smaller areas and brings out information which can never be revealed by low magnification.

Macro-examination gives a broad picture of the interior of a metal by studying relatively large sectioned areas.

3 The aim of micro-examination is: 1) To determine the size and shape

of the crystallites which constitute an alloy.

2) To reveal structures characteristic of certain types of mechanical working operations.

3) To discover microdefects (non-metallic inclusions, microcracks, etc.)

4) To determine the chemical content of alloys (e.g., annealed carbon steels),

5) To indicate quality of heat treatment, mechanical properties, etc.

The aim of micro-examination is: 1) To reveal the size, form and

arrangement of crystallites (dendrites) in cast metals.

2) To reveal fibres in deformed metals. 3) To reveal shrinkage porosity and gas

cavities. 4) To reveal cracks appearing during

certain fabrication processes. 5) To show chemical non homogeneity in

the distribution of certain constituents appearing in alloys upon their solidification from the liquid state.

6) To indicate non-metallic inclusions such as slag, sulphides and oxides.

7) To indicate method of manufacture, e.g., forging, casting, welding and brazing.

8) To find the cause of failure of a component part.

4 Micro-examination requires proper surface preparation of the specimen before studying it under the microscope.

Surface preparation for macro examination follows similar lines to those for micro-examination but need not be taken to such a high degrees of surface finish and so the final stages of polishing can be omitted.

5

Micro-examination requires that the polished specimen surface should be etched with suitable reagent.

Macro-examination is also carried out on an etched surface.



2.What are Miller Indices? Explain the procedure to find out Miller Indices for a plane. Ans : By noting intercepts we denote a crystallographic plane. It makes on the three crystallographic axes.

Fig . Crystallographic plane

Tthe crystallographic plane making intercepts X, Y and Z on x, y, z axes respectively can be described by noting the intercepts in that order as shown in above figure. When a plane is parallel to an axis, in this case the intercept made by this plane on the axis to which it is parallel, would be infinity, since α is indeterminate, it should not be used in the definition of a plane. Miller devised a method to represent crystallographic planes and directions, using a set of three integers, without involving. These integers, a set of three (one for each axis) are known as Miller’s indices. Consider above plane. The method of finding out Miller’s Indices of this plane is given below:-

1) Mark the intercepts made by the plane- XYZ. 2) Convert the intercepts into partial axial intercept with the vector length or axial length

of that axis. Suppose a,b,c are the axial lengths of x, y and z axes. X/a Y/b Z/c

3) Find the reciprocals of these partial axial lengths : a/X b/Y c/Z

4) Find the set of three smallest integers- say h, k,l which are in same ration as the reciprocals of partial axial lengths i.e.

a/X : b/Y : c/Z :: h : k : l

These three integers h k l are the Miller’s indices of the plane under consideration and the plane is represented by writing down these indices in rounded brackets (h k l). 3. What is substitutional sold solution? What are the factors that governs its for formation. Explain.



Ans: A substitutional sold solution is homogeneous mixture of atoms of two elements, right down to the atomic scale, the two elements being such that their atoms are nearly identical in their size (diameter) and electro-chemical nature. These two atoms are identical such that we would not be able to distinguish between them in mixture. Due to this similarity between the solute and the solvent atoms, the solute atoms occupy the lattice sites of the solvent atoms. Thus they substitute the solvent atoms on their lattice sites. If two kinds of atoms are arranged in a haphazard fashion on the lattice sites we get random substitutional sold solution. If the solute atoms and solvent atoms are arranged in an orderly fashion we get ordered substitutional sold solution.

Fig. random substitutional sold solution Fig. ordered substitutional sold solution Factors governing the formation of Substitutional Sold Solutions:

Substitutional sold solutions are formed when the solute atoms and solvent atoms are identical their sizes and electrochemical nature. Hence the factors which govern the formation of substitutional sold solutions are :

1) Size Factor :- For the formation of substitutional sold solution, the size factor should be less than 15%. If the atomic radius radius of the solute is R1 and the atomic radius of solvent atom is R2 , the size factor is defined as R1-R2/R1. If the size factor is more than 15% the formation of substitutional sold solution is less favoured.

2) Electro-chemical nature :- Electrochemical natures of solvent and slolute elements should also be identical. A large difference in electrochemical nature does not allow the formation of substitutional solid solutions.

3) Valency :- An element with lower valecncy dissolves more of the elements with higher valency and vice versa. This is known as relative valency effect.

4. Calculate the total number of atoms per unit cell for

(i) Simple cube (ii) BCC (iii) FCC (iv) HCP Ans:



Fig. (a) A simple cubic unit cell (b) A F.C.C. unit cell (c) A B.C.C. unit cell (d) A H.C.P. unit cell

(i) Number of atoms in a simple cubic unit cell: There are eight corners of the cube and at each corner there is an atom. Each corner atom is shared by eight adjoining cubes, therefore the share of the cube =1/8 x 8 = 1. Hence total number of atoms in a simple unit cell = 1. (ii) Number of atoms in a F.C.C. unit cell: A F.C.C. unit cell has an atom at each corner of the cube and in addition, it has one atom at ( the intersection of diagonals of ) each of the six faces of the cube. Since each corner atom is shared by only two cubes, (i.e., one this cube and second adjacent one), the unit cell contains: 8 atoms at the corner x 1/8 = 1 atoms 6 face-centered atoms x ½ = 3 atoms Total = 4 atoms (iii) Number of atoms in a B.C.C. unit cell: A B.C.C. unit cell has an atom at each corner of the cube and in addition, it has one atom in the center of the cube. Since each corner atom is shared by eight adjoining cubes and the one center atom is not shared by any adjoining cube, the B.C.C. unit cell contains:



8 atoms at the corner x 1/8 = 1 atoms 1 center atom = 1 atoms Total = 2 atoms (iv) Number of atoms (N) in a HCP unit cell: N = Nc/6 + Nf/2 + Ni/1 Where Nc is total number of corner atoms in the unit cell = 6+6 =12 for HCP unit cell Nf is total number of face atoms in the unit cell = 1+1 = 2 for HCP unit cell

Ni = Interior or centre atoms in the unit cell = 1+1+1 = 3 for HCP unit cell N (HCP) = 12/6 + 2/2 + 3/1 = 6 atoms

5.What is interstitial sold solution? When is it formed? Ans : When the size of the solute atoms is so small that they can easily be accomadated in the interstitial space, we get an interstitial solid solution. Example is in alloys of iron with carbon, nitrogen and hydrogen are much smaller than compared to iron atom. The formation of interstitial sold solution is favoured if the radius ratio of the solute atom to solvent atom is less than 0.6 i.e. Rsolute / Rsolvent < 0.6 The other factor is electrochemical nature. The electrochemical nature of the solute should not be extremely different from that of the solvent element. For example although the radius ratio is favourable for formation of an interstitial sold solution of oxygen in iron. It is formed because oxygen is highly electropositive in nature. 6.What is Gibb’s phase rule? Why and how is it modified? Define all terms involved in it. Ans : The phase rule, known as Gibb’s Phase Rule, establishes the relationship between the number of degrees of freedom (F), the number of components (C) and the number of phases (P). It is expressed mathematically as follows : P + F = C + 2

A phase is defined to be a microscopically homogeneous and physically distinct portion of a system with well defined boundaries separating that portion from its surrounding. For example water, ice and vapour are three different phases. A phase is characterized by its structure- the atomic arrangement in it.

The number of components is the number of elements of which the system is made up of. For example, for binary system -2 components, for ternary system – 3 components

For binary, C = 2 For ternary, C = 3 Degrees of freedom are the variables which must be specified to completely define the

system. Variables as pressure, temperature and composition which can be independently changed without changing other variable.

In metallurgical systems where pressure is regarded as remaining fixed at one atmosphere, the pressure variable is often omitted and equation simplifies to:

P + F = C + 1 or



F = C + 1 – P During the solidification of pure metal , solid metal and liquid metal are in equilibrium. Hence the number of phases in equilibrium is 2 i.e., P = 2. Phase rule for alloy system is given as P + F = C + 1 2 + F = 1 + 1 F = 0 Which means that during the solidification of a pure metal, the temperature does not change, it remains constant. Thus the solidification of a pure metal occurs at constant temperature, the melting point or freezing point of the metal. 7.State and derive ‘Lever Rule’ as applied to equilibrium diagram. Ans : Lever rule is a mathematical principle with which we can calculate the proportions of two phases or two micro-constituents in a two phase region. This rule is known as lever rule because of the mechanical analogy which is used in calculation. In this rule it is assumed that the weights of the two phases under consideration are suspended from the ends of tie-rod. The ends and fulcrum are given by the intersection points of the temperature –horizontal line (corresponding to temperature at which the calculation is required ) with phase boundaries of the two phases and the composition vertical respectively.

Consider an alloy containing C%B. To calculate the proportions of solid solution and liquid at temperature T, draw the temperature-horizontal line SFL. This line cuts the composition vertical at F and cuts the solidus line (the phase boundary at S, and liquidus line at L. Hence the tie rod is the line SFL and fulcrum at F. The weight of solid solution is supposed tobe suspended at S and suspended at L. Under equilibrium , the moments must be balanced. Hence weight of solid x SF = weight of liquid x FL weight of solid / weight of liquid = FL / SF % weight of solid = FL / SF x 100 % weight of liquid = SF/ SL x 100



Unit-II 1. What is critical temperature? Show critical temperatures on Fe-Fe3C diagram. Explain

in detail. Ans: The temperatures at which a given steel has certain change in structure that is called critical temperature . At this temperature transformation starts. In the Fe-C diagram various critical temperatures like Ae0, Ae1and Ae2. The subscript ‘e’ denotes that critical temperature refers to equilibrium condition of heating or cooling. The critical temperatures are shown in fig. Ae0 is The curie temperature of cementite. Above this temperature cementite is non-magnetic. It is at 210°C. Ae1is Eutectoid temperature at 723°C. Above this temperature no pearlite is seen and below this temperature austenite is not found. This is known as lower critical temperature Ae2 is Above this temperature α-ferrite becomes non-magnetic i.e. it is the curie temperature of α-ferrite. Ae2 is at 768°C. Ae3 isThis is known as upper critical temperature. Above this structure the steel is completely austenitic. This is the temperature below which ferrite is found.

The critical temperatures are with respect to equilibrium rate of heating or cooling. But in practice the cooling rate or heating rate is always faster. Hence during heating the magnitude of a suppressed on cooling. For example, eutectoid but higher than 723°C while heating. 2.Explain the cooling of the following steels from liquid condition to room temperature. Ans: A] 0.08% C Steel: Steel with 0.08 % C is represented by vertical line A. 1) On cooling this steel from liquid condition following sequences of changes occurs:

i. Solidification starts at temperature T1 by the commencement of formation of -ferrite. On cooling below T1, the composition of -ferrite and liquid change along MN and NO respectively. Solidification is completed at temperature T2.

9

ii.

iii.

iv.

B] 0.12 %sequence

i. SfereA

ii. Aiii. B

chalau

The structuchange of The steel rechanging inand then waustenite chAt 723°C thfurther to roa small amo

% C steel: e for composolidificationerrite formsespectively.

Austenite is fAt T2 the chaBetween temhanging intolong RP andustenite con

Nagpur I

ure of the st-ferrite into

emains as anto α-ferrite

within the grahanges alonghis austeniteoom temperaount of pearl

On cooling sition vertican starts at T1s. ComposiAt 1495 °C

formed and eange of -fermperature T2 o α-ferrite. Bd that of auntains 0.8%

Institute

Engineer

eel remains austenite st

austenite bet. The changains. During

g PQ till it be changes intature. At roolite at grain b

this steel fral line B. 1 with formations of (Tp) peritec

excess -ferrrrite is comp

and T3 theBetween T3 austenite RQ.

C. This au

of Techn

ring Metal

as -ferriteart. This chatween tempege gradually g this separaecomes 0.8 %to pearlite. Tom temperatuboundaries.

rom liquid c

ation of -f-ferrite and

ctic reaction rite remains.leted. structure reand 723°C (. At 723°C ustenite dec

nology, N

llurgy These

e between teange is comperature T4 antakes place

ation of α-fe% at 723°C.

There is no chure the struc

condition the

ferrite. Betwd liquid chtakes place .

emains austeTU) the com(TU) α-ferr

composes at

agpur

e notes are not fo

emperature Tpleted at T4 tnd T5. At Tfirst at the

errite the co hange in structure shows

e following

ween T1 and hange alongbetween -f

enitic. At Tmposition of

ite contains t this tempe

or circulation an

T2 and T3. Atemperature.

T5 austenite grain bound

omposition o

ucture on coferrite grain

changes occ

1495°C mog MN and ferrite and li

T3 austenite α-ferrite cha0.025 % C

erature follo

nd sale.

At T3 . starts

daries of the

ooling ns and

cur in

ore -NO

iquid.

starts anges

C and owing



eutectoid reaction and forms pearlite. No further change occurs upto room temperature. The structure of steel shows grains of α-ferrite with pearlite at the grain boundaries.

C] 0.3% C Steel: On cooling the steel from liquid condition following changes take place as follows: i. Solidification start at T1

” with formation of -ferrite and continues upto temperature 1495°C. During this interval the composition of -ferrite and this interval the composition of -ferrite and liquid change along MN and NO respectively.

ii. At 1495°C (Tp) containing 0.1% reacts with liquid containing 0.5 % C to give austenite containing 0.15 % C i.e. peritectic reaction occurs. Austenite is the product with excess liquid left over.

iii. Solidification of this liquid into austenite is completed at temperature T2. The structure is completely austenite.

iv. No change occurs upto T3”.At T3

” α-ferrite starts separating out from austenite. Further changes occur upto room temperature as same as above mentioned in 0.12 % C steel. The structure at room temperature shows grains of α-ferrite and regions of pearlite.

D] 0.7 % C steel: On cooling this steel from liquid condition the sequence of changes is as follows:

i. Solidification starts at TA1 temperatures by the formation of austenite. ii. Solidification is completed at temperature TA2. Below this temperature austenite remains

without any change till TA3 temperature is reached when α-ferrite starts forming from austenite. Further changes are similar to those seen in 0.3 % C steel. The structure of the steel at room temperature shows regions of pearlite with boundaries of α-ferrite.

E] 1.2% C Steel: On cooling this steel from liquid condition the changes occur as follows: i. Solidification starts at temperature TC1 austenite forms

ii. Solidification ends at TC2 and structure is fully austenite iii. No change occurs between TC2 and TC3 temperature; precipitation of cementite from

austenite starts and continues upto 723°C (TU) temperature. This happens because the solid solubility of carbon in austenite decreases from 2.0 % at 1130 °C to 0.8 % at 723αC. The composition of the austenite changes along TC3 TU till it becomes 0.8 % C at 703°C.

iv. All this austenite undergoes eutectoid transformation and forms pearlite. At room temperature the structure of steel shows regions of pearlite with boundaries of cementite.



3. Draw iron-iron carbide equilibrium diagram ( Fe-FeC diagram) showing all details. Ans:

4. Calculate the weight percent of various i) microconstituents ii) phases present in each of them at room temperature 0.15%C, 0.3% C, 0.7%C, 1.0%C Ans: To calculate phases at room temperature draw lower portion of iron-iron carbide diagram showing steel portion. Assumptions: Assuming that the solid solubility of carbon in α-ferrite is 0 % at 0°C. At room temperature this steel contains the microcontituents α-ferrite and pearlite. Pearlite contains phases α-ferrite and cementite. A composition of ferrite is 0% C, cementite 6.67% , pearlite 0.8% C.



Applying lever rule- % of phase = length of tie-line on opposite side / total length of tie-line × 100 a) 0.15 % C steel Microconstituents wt % : Wt % of α-ferrite = 0 .8 - 0.15 / 0.8 – 0 × 100 Wt % of pearlite = 0 .15 - 0.15 / 0.8 – 0 × 100 Phases wt % Wt % of α-ferrite = 6 .67- 0.15 / 6.67 – 0 × 100 Wt % of cementite = 0 .15 - 0.00 / 6.67 – 0 × 100 = 2.25 % b) 0.3% steel: At room temperature this steel contains Microconstituents - α-ferrite and pearlite Phases - α-ferrite and cementite Applying lever rule- Microconstituents wt % : Wt % of α-ferrite = 0 .8 – 0.3 / 0.8- 0.00 × 100 = 62.5 % Wt % of pearliite = 0 .3 – 0.000 / 0.8- 0.00 × 100 = 37.5 % Phases wt % : Wt % of α-ferrite = 6.67– 0.3 / 6.67 0.00 × 100 = 95.5 % Wt % of cementite = 0 .3 – 0.000 / 6.67- 0.00 × 100 = 4.5 % c) 0.7 % steel : At room temperature Microconstituents - α-ferrite and pearlite Phases - α-ferrite and cementite Microconstituents wt % :



Wt % of α-ferrite = 0 .8 – 0.7 / 0.8- 0.00 × 100 = 12.5 % Wt % of pearlite = 0 .7 – 0.000 / 0.8- 0.00 × 100 = 87.5 % Phases wt %: Wt % of α-ferrite = 6.67– 0.7/ 6.67 -0.00 × 100 = 89.5 % Wt % of cementite = 0 .7 – 0.000 / 6.67- 0.00 × 100 = 10.5 % d) 1.0 % C Steel: At room temperature Microconstituents - cementite and pearlite Phases - cementite and ferrite Wt % of cementite = 1.0 – 0.8 / 6.67- 0.8 × 100 = 3.4 % Wt % of pearlite = 6 .67 – 1.0 / 6.67- 0.8 × 100 = 96.6 % Phases wt %: Wt % of cementite = 1.0 – 0.00 / 6.67- 0.00 × 100 = 15 % Wt % of α-ferrite = 6.67– 1.0/ 6.67- 0.00 × 100 = 85 % a) For guarantee with regard to elevated temperature properties the letter H shall be used

For guarantee with regard to low temperature properties the letter L shall be used 5. What is binary phase diagram? What is utility of it. Ans: A binary phase diagram is a plot of various phases in equilibrium at different temperatures for alloys of different compositions in the entire range from one pure metal to the other. It can be considered as a map showing the various phases present at different composition temperature conditions for an alloy system. The Y- axis represents the temperature and the X-axis represents the composition. The composition can be given either in weight present or in atom %.

1) On cooling an alloy from a higher temperature to room temperature the various changes that

occure. 2) The condition of an alloy in terms of the phases present in that at any given temperature 3) We can know various solidification reactions and reactions that occur in solid state 4) One can know temperature at which solidification begins and ends for any given alloy



6. What is Isomorphous system? Ans: An equilibrium diagram showing complete solid solubility is called as an isomorphous system because the crystal structure of any alloy between the limits of the two pure metals i.e. form 100 % A to 100 % B is same.

Above fig shows equilibrium diagram. The boundary line separating solid region is

known as solidus line. The liquidus line indicates the variation of freezing point of the alloy with composition. A given alloy is exhibited by a vertical line indicating its composition. The intersection point of this composition line with the liquidus line indicates the temperature at which on cooling the liquid alloy, its solidification starts. Similarly the intersection point of the composition line with solidus line indicates the temperature at which the solidification of an alloy is completed. For example the solidification of an alloy containing C % B, starts at temperature T1and is completed at temperature T3. TA and TB are the melting points of elements A and B. 7. What is effect of alloying elements on TTT curve. Explain with neat sketches. Ans: The effect of Alloying element on the position and shape of the TTT curve can be summarized below –

i. Effect on the position of the S Curve: All the alloying element shift the S- curve towards the right. Exception is only cobalt. All the alloying element except cobalt decreases the critical cooling rate of the steel and hence increase the hardenability of the steel. (fig.(a))

ii. Effect on the shape of the S Curves: Austenite stabilizers do not change the shape of the S Curves. It has only one nose. Ferrite stabilizers change the shape of S-curve they give two nose. The upper nose is called pearlite nose and the lower nose is called bainite nose. (fig.(b))

iii. Effect on nose temperature: The austenite stabilizer Ni, Mn, Cu lowers the pearlite nose temperature. All the ferrite stabilizer like Cr, Mo, V, W and Si raise the pearlite nose temperature. Al also raises the nose temperature although it is austenite stabilizer. This effect is schematically shown in fig (c) and fig (d) respectively.

15

Fig. (c) Eff

F

F

ffect of alloying

Nagpur I

Fig. (a) Effec

Fig. (b) Effe

g element on n

Institute

Engineer

ct of alloying

ect of alloyin

ose temperatur

of Techn

ring Metal

g element on

ng element o

re Fig. (d

nology, N

llurgy These

n position of

on shape of

d) Effect of allo

agpur

e notes are not fo

f S curves

S curves

oying element o

or circulation an

on nose tempe

nd sale.

rature

16

8. State tAns: Thei. Effect eutectoidtemperatuii. Effectstabilizermixture.

iii. Effectensile sttoughnesof the steiv. Effecof the stev. Effecton the S

the effects oe General eon eutecto

d temperaturure. This is st on eutectors change thfig. (b)

Fi

Fi

ct on mechatrength of s

ss of stee. Neel. ct on hardeneel. Cobalt dt on S Curvecurve.

Nagpur I

of alloying effect of Alloid temperat

re while all shown in figoid composihe eutectoid

ig. (a) Effect

g.(b) Effect

anical propesteel. All th

Nickel is the

nability: Aldecreases, thee: All the al

Institute

Engineer

elements on oying elemeture : All th the ferrite s

g.(a) ition : All td compositio

t of alloying

of alloying e

erties: All thhe alloying only elemen

l the alloyine hadenabilitloying elem

of Techn

ring Metal

the propertnt in steel a

he austenite sstabilizers l

the alloying on by decre

element on

element on e

he alloying element exc

nt which inc

ng element ety of the stee

ment have a p

nology, N

llurgy These

ties of steel.are given –stabilizers liike W, V, T

element boeasing the c

eutectoid tem

eutectoid Co

element inccept nickel, creases the d

except cobalel. profound effe

agpur

e notes are not fo

.

ike Ni, CO aTi, Mo, etc ra

oth antenite acarbon cont

mperature

omposition

reases the ydecreases t

ductility as w

t, increases

fect on the sh

or circulation an

and Mn loweaise the eute

as well as ftent of eute

yield strengththe ductilitywell as tough

the hardena

hape and pos

nd sale.

er the ectoid

ferrite ectoid

h and y and hness

ability

sition



9. Mention the most important effects of following alloying elements:- Manganese, Nickel, Cobalt, Silicon, Chromium, Tungsten, Vanadium, Aluminium, Sulphur, Phosphorous Ans:-

1) Manganese : i. It is the cheapest alloying element for increase the strength and hardenability of

steel ii. It does not decrease the ductility very adversely upto 2.0 % iii. The only austenite stabilizer which form carbides, but only in the absence of other strong carbides former iv. Roughness of the steel is slightly improved

2) Nickel:

i. The only alloying element which increases yield strength of the steel along with increasing its ductility ii. Gives low temperature toughness to steel. This is the only element to give

temperature toughness to steel iii. Improves corrosion resistance in conjunction with Cr iv. Present the magnetic steel

3) Cobalt:

i. This is the only alloying element which decrease the hardenability of the steel ii.Improve the magnetic properties ii. Increases the creep resistance of the steel

4) Silicon: i.This efficiently increases the elastic limit of the steel without affecting its toughness severly. ii. A carbide former and and improves heat resistance of steel.

5) Chromium:

i. Very efficient in increasing hardenability and wear resistance of the steel ii.Economical alloying element to increase strength and hardness together iii.Steel containing chromium exhibit reversible temper embrittlement

6) Tungsten

i. A strong carbide former are effective in increasing wear resistance and crip resistance

ii.Impart secondary hardening tendency and red hardness of steel iii.It is an element used to make high speed steels

7) Vanadium

i.A very strong carbide former ii.Increases the wear resistance and creep resistance of the steel iii.Increases the cutting efficiency of high speed steel



8) Aluminium

i. Used as deoxidizing agent to remove oxygen from the steel melt ii.It is strong nitride former and is essentially present in nitriding steel or nitralloys

9) Sulphur

i. A nuisance element in steel. It results in ‘hot shortness’ of the steel that is the tendency

of the steel to crack during hot working ii.Used upto 0.15% to increase the machinability.

10) Phosphorous

i.Another nuisance element ii.Highly undersiable because it make the steel brittle iii.Decrease the soft magnetic properties of the steel, hence undesirable in soft magnetic steel.

10. Write short note on IS specification of steel. Ans: Indian Standard code for Designation of Plain and Alloy Steels (IS: 1762-1974) Steels have been classified on the basis of i. Mechanical properties

ii. Chemical composition *Designation of steel on the basis of mechanical properties such as tensile strength or yield stress: Symbols for: a)Method of deoxidation

R – rimming steel K – killed steel No symbol – semi killed steel

b) Quality of steel Q1- non ageing quality Q2- freeom from flakes Q3- given size controlled

Q4- inclusions controlled Q5- internal homogeneity guaranted

c) Sulphur and phosphorus content

-When maximum S and P content is same No symbol - 0.055% P and 0.055% S P 25 – 0.025% P and 0.025% S P 70 – 0.07 % P and 0.07 % S -When maximum contents of S and P are not same symbol SP shall be used Example: suppose there is Maximum sulphur = 0.045 % (when rounded it becomes 0.04%) Maximum phosphorus = 0.035 % (when rounded it becomes 0.04%) Therefore designation is: SP 44 (i.e. 0.04 % S and 0.04 % P)



d) Weldability guarantee W – fusion weldable steel W1 – Steel weldable by resistance welding but not fusion weldable

e) Resistance to brittle fracture (Based on V-notch charpy impact test)

Specific UTS Range

370 to 520 N/mm2 500 to 700 N/mm2 Energy (J) Temp 0C Energy(J) Temp 0C

B - 28 27 40 27 B0 - 28 0 28 -10 40 0 B2 - 28 -20 28 -30 40 -20 B4 - 28 -40 28 -50 40 -40

f) Surface condition S7-ground S6-bright drawn or cold rolled S5-peled S4- shot, grit or sand blusted

S3- pickled S2-descaled S1-deseamed or scarfed No symbol- as rolled or as forged

g) Formability D1 - drawing quality D2 - deep drawing quality D3 - extra deep drawing quality No symbol - steel is of commercial quality

h) Surface finish (for sheets only) F1 - general purposefinish F2-full finish F3-exposed F4- unexposed F5-matt finish F6-bright finish F7- plating finish F8- unpolished finish F9- polished finish

F10- polished and coloured blue F11- polished and coloure yellow F12-mirror finish F13-vitreous enamel finish F14-direct annealed finish

i) Treatment: No symbol –hot rolled

T1- shot peened T2- hard drawn T3- normalised T4- controlled rolled



T5- annealed T6- patented T7- solution treated T8- solution treated and aged T9- controlled cooled T10- bright annealed T11- spherodised T12- stress relieved T13- case hardened T14- hardened and tempered

j) For guarantee with regard to elevated temperature properties the letter ‘H’ shall be used

k) For guarantee with regard to low temperature ( cryogenic quality) properties the letter ‘L’ shall be used

Semi-killed steel For example: FeE*300 P 35 Minimum yield strength = 300 N/ mm2

S and P = 0.035% Max

* Designation of the steel on the basis of chemical composition: 1) Unalloyed steels (IS: 7598-1974) X C Y Z X- 100 times the average % of C C- Carbon Y- 10 times the average % of Mn Z-special characteristics Semi-killed steel with 0.25% C For example: 25 C5 B0 means 0.5% Mn Resistance to brittle fracture grade BO 2) Unalloyed tool steel X T Y T stands for tool steel X and Y have same meaning as above. Unalloyed tool steel For example: 80 T 11 implies 0.80% C

1.1% Mn

3) Unalloyed free cutting steels X C Y W W1 Z W stands for the element present which makes steel free cutting W1 100 times the % of the element present X, C, Y and Z have the meanings same as above For example: 20 C12 Pb15 T14 implies Free cutting steel with 0.2% C, 1.2% Mn, 0.15% Pb, hardened and tempered (because of T14 symbol)



4) Alloy steels (IS: 7598-1974) a) Low and medium alloy steels 1st symbol-100 times the average % of C 2nd, 4th, 6th symbols etc.-Elements 3rd, 5th, 7th, symbols etc-% of elements, multiplied by factors as follows: Element Multiplying factor Cr, Co, Ni, Mn, Si and W 4 Al, Be, V, Pb, Cu, Nb, Ti, Ta, Zr and Mo 10 P, S, N 100 Last element –special characteristics N.B. i) Figures after multiplying shall be rounded off to the nearest integer ii) Symbol Mn for manganese shall be included in case Mn content is ≥1% iii) The chemical symbols of each elements present and their figures shall be listed in order of decreasing content. For example: 40 Ni 8 Cr 8 V 2 implies hot rolled steel with 0.4% C 2% Cr, 2% Ni and 0.2 % V 25 Cr 4 Mo 2 G implies steel with guaranteed (G) hardenability and 0.25 % C, 1% Cr, and 0.25% Mo b) High alloy steels (total alloying elements > 10%)

For example: X 10 Cr18 Ni9 S3 X - high alloy steel 10 - 0.10% C Cr 18 - 18% Cr Ni 9 - 9% Ni S 3 - Pickled condition

c) Alloy tool steels For example: XT 98 W6 Mo5 Cr4 V1 XT - high alloy tool steel 98 - 0.98% C W 6 - 6% tungsten Mo5 - 5% molybdenum Cr4 - 4% chromium V1 - 1% vanadium

d) Free cutting alloy steels For example: X 15 Cr25 Ni15 S40 X - high alloy steel 15 - 0.15% C Cr25 - 2 5% chromium Ni15 - 15% nickel S40 - 0.4% sulphur (S introduces free cutting qualities)



Unit-III 1. Write short notes on: 1) Ball race steel or ball bearing steels Ans: This is a typical low alloy steel designated as En 31 in British Industrial specifications containing chromium. Its typical composition is carbon 1.0 %, Mn 0.5%, Cr 1.31.5%. Because of the presence of chromium it has excellent hardenability and canbe quenched iln to harden. It is mainly used for the manufacture of steel balls and rollers for ball and roller bearings. Hence the name ball race steel or ball bearing steel.

Heat treatment involves heating upto 810°C and after that quenching. Tempering is done at 150 180°C

Tempering at The low temperature of tempering ensures that the hardness does not decrease but all the internal strains due to martensite be relieved. The hardness after proper heat treatment is about 65 Rc. Because of the presence of chromium carbide in the structure the heat treated steel has excellent wear resistance. 2) HCHC steel: High carbon high chromium steel is a tool steel which gives the optimum wear resistance at room temperature at minimum cost. It means that it is the most economical too steel having maximum room temperature wear resistance. The steel contains about 1-13 % Cr with about 1-1.3 % C maintaining the Cr/C ratio as 10/1.

The steel is heated upto 950°C (Austenitization temperature) and quenched in oil. Then tempering is done at 180°C or 400°C depending upon the hardness required.

This steel is used for making cold drawing dies and thread rolling dies. In this case the tempering temperature is 180°C. It is also used to make cold punching dies, cold fanning tools, etc., which are tempere at about 400°C because relatively higher toughness is desirable. Presently grinding media balls for balls are also being made of this steel.

2. What are High speed steels? What are their types? Explain the steps involved in the heat treatment of HSS tools. Ans: High speed steels is a type tool steels. This steels are used for high speed cutting at such high speeds that the tool tip becomes red hot. Hence they are based on the tungsten or molybdenum addition to steel in substantial amounts because both these elements give red hardness to steel- the capability of maintaining high hardness and a keen cutting edge even up to the red hot temperature of 600°C. The first high speed steels that were developed were of tungsten type. Later on molybdenum high speed steels were developed. Presently we have high speed steels of both types and those in which both the elements are present. High speed steels are relatively less ductile and their thermal conductivity is not very good. Hence special care is needed in the heat treatment of high speed steel tools, particularly during austenitization. Heat treatment of high speed steel tools involves the following steps: 1) Heating for austenitization: Special care is taken in heating machined high speed steel tools to their austenitization temperature. A slow heating rate is desirable to avoid cracks during heating which may generated due to uneven heating. This is known as clinking. To avoid this HSS tools are heated to their austenitization temperature in three steps:

a) Preheating stage 1: upto 500-650°C. This relieves most of the machining stresses. b) Preheating stage 2: upto 850°C. Both these steps are done at very slow rates in salt bath

furnaces.



c) Final heating for austenitization: After the second stage of preheating the tools are quickly transferred to another salt bath furnace kept at 1175 to 1290°C .The tools are kept only for 1 to 5 minutes at this temperature depending on the size of their cross section.

d) Quenching: From this austenitization temperature HSS tools are quenched to room temperature in one of the following ways:

i) Directly quenched in an oil bath kept at room temperature till the tool pool fully. ii) Directly quenched in an oil bath kept at room temperature till the temperature of

the tool is decreased to about 450˚C, then taken out and cooled in air. iii) Quenched into a salt kept at 600-650˚C, homogenize the temperature and then

taken out to cool in air upto room temperature. e) Tempering: After quenching the tools are tempered in a circulating air furnace at about

550 to 600˚C to develop maximum hardness. The tempering treatment is repeated twice or thrice and is known as double or triple tempering. The final hardness after tempering is about 67 Rc.

3. How will you classify tool steels? What is Heat-Treatment of Tool Seels? Ans: Tool and die steels may be defined as special steels which have been developed to form, cut or otherwise change the shape of a material into a finished or semifinished product. The Joint Industry Conference ( JIC), USA has classified tool steels as follows: Symbol Meaning T W-high speed steel M Mo-high speed steel D high C, high Cr steel A air hardening steel O oil hardening steel W water hardening steel H hot work steel S shock resistant steel As an example W8 means water hardening steel with 0.8 % C Composition of tool steels (%)

1) W-high speed steels (T) T1→ C 0.7, Cr 4, V 1, W18

T4→ C 0.75, Cr 4, V 1, W18, Co 5 T6→ C 0.8, Cr 4.5, V 1.5, W2, Co 12

2) Mo-High speed steels (M) M1→C 0.8, Cr 4, V 1, W1.5, Mo 8 M6→C 0.8, Cr 4, V 1.5, W4, Mo 5, Co 12

3) High C, high Cr steels (D)

D2→C 1.5, Cr 12, Mo 1 D5→C 1.5, Cr 12, Mo 1, Co 3 D7→C 2.35, Cr 12, V 4, Mo 1



4) Air hardening steels (A)

A2→C1, Cr 5, Mo 1 A7→C 2.25, Cr 5.25, V 4.75, W 1, Mo 1 A9→C 05, Cr 5, Ni 1.5, V 1, Mo 1.4

5) Oil hardening steels (O) O1→C 0.9, Mn 1 Cr 0.5 W 0.5 O6→C 1.45, Si 1 Mo 0.25

6) Water hardening steels (W) W2→C 0.6/1.4 V 0.25 W5→C 1.1 Cr 0. 5

7) Hot working steel (H) H10→C 0.4 Cr 3.25 V 0.4 Mo 2.5 H12→C 0.35 Cr 5 V 0.4 W 2.5 Mo 2.5

8) Shock resisting steel (S) S1→C 0.5 Cr 1.5 W 2.5 S2→C 0.5 Si 1 Mo 0.5 S5→C 0.55 Mn 0.8 Si 2 Mo 0.4 S7→C 0.5 Cr 3.25 Mo 1.4

Heat treatment of tool steels: Tool steels should be heated slowly to the desired heat treatment temperature. Tool steel should be kept at proper temperature for sufficient time so that the whole of the tool section gets heated uniformly. Overheating of tool steel should be avoided. Protective furnace atmosphere or any other method should be employed to avoid scaling or decarburization of tool steel during heating. Carbon and low alloy steels may be quenched in water or brine and high alloy tool steels in oil, air or molten salts. Tool steel may be quenched in a liquid bath of salt of lead held between 480 and 650°C and then cooled in air to about 65°C. It should be tempered immediately after quenching and before they cool to room temperature. This reduces cracking. 4.What is “Inverse Segregation” in babits? How can it be prevented?

OR What is the function of copper addition to babbits? Ans: During the solidification of a white metal casting the first phase to solidify is the compound SbSn. Hence a large number of cuboids of Sb-Sn compound to form in the remaining molten liquid. These cuboids have a much lower density than the liquid melt in which they form. Hence instead of being uniformly suspended in the melt these cuboids float to the top of the casting. Hence after the solidification is completed we find that all the cuboids of Sb-Sn compound are segregated in the top region of the casting. This type of segregation found in babbits is known as inverse segregation. This non uniform distribution of cuboids is harmful for the mechanical properties of the bearing.

About 2 to 3 % copper is added to prevent this segregation. When copper is present in the molten white metal during its solidification, needle shaped particles of a compound Cu6Sn5 are the first to form throughout the liquid and they create a web like network in the liquid. A little later at a still lower temperature the cuboids of SbSn compound form and solidify throughout the



volume of the liquid. Since the web like network of Cu6Sn5 compound is present in the volume of the molten metal, the cuboids formed at different places in the molten liquid are not allowed to float to the top. Hence they remain in the region in which they formed. Thus the web like network of Cu6Sn5 effectively prevents the inverse segregation and gives a uniform distribution of cuboids in the cast bearings of white metals. This compound Cu6Sn5 is seen as bright starry streaks in the micro structure. 5. Explain sensitization of Austenitic stainless steel? Ans: When austenitic stainless steel is heated to temperature between 425 to 825°C during processing or found that its corrosion resistance is greatly preferentially occurs along the grain boundaries. Hence which gets heated to the above temperature range and is air cooled prone to intergranular corrosion and hence called to be sensitized.

When the astenitic stainless steel gets heated to 425 to 825°C and is slowly cooled, chromium carbide gets precipitated along the grain boundaries. This chromium carbide gets precipitated along the grain boundaries. This chromium carbide precipitation requires a large amount of chromium in a short time. Hence all the chromium for this precipitation is drawn from a localized region near the grain boundary. Hence in regions all along the falls below 12 %. For corrosion resistance minimum 12 % Cr is essential. Hence corrosion resistance of this grain boundary region is decreased. This results in a preferential corrosion all along the grain boundaries causing intergranular corrosion. This phenomenon can be avoided by 1) Decreasing % C to very low level so that less Cr will required for Cr3C formation and Cr%

should not drop below 12 % and sensitization should avoided. 2) By addition of Cb and Ti which are strong carbide former restrict the formation of Cr3C and

sensitization is avoided. 3) By again hating steel to 1000-1050°C and holding it there for 1 hour and then rapidly cooled

through this temperature range avoids the sensitization. 6. What is weld decay? Ans: During welding of Austenitic stainless steel at some distance away from weld, the portion get heated to 425 to 825°C during cooling sensitization is occur and because of intergranular corrosion material almost get decayed called as weld dacay. It can be prevented by using low C grade by using stabilized grades or by given heat treatment.



UNIT –IV 1. Define heat treatment. Explain the significance and steps in heat treatment practice. What are principles of heat treatment? Ans: Heat treatment is a stage in the fabrication of structures and is often forgotten: but it has perhaps more wide-reaching and important ramifications than many of the other stages in the fabrication of structures or components. Heat treatment may be defined as an operation or combination of operations involving heating and cooling of a metal/alloy in solid state to obtain desirable conditions and properties which include better machinability, improved ductility, homogeneous structure, etc. Purpose of heat treatment : One or the other heat treatment process is carried out in order to (i) Cause relief of internal stresses developed during cold working, welding, casting, forging etc. (ii) Harden and strengthen metals. (iii)Improve machinability. (iv) Change grain size (v) Soften metals for further (cold) working as in wire drawing or cold rolling (vi) Improve ductility and toughness (vii) Increase heat , wear and corrosion resistance of materials (viii) Homogenise the structure; to remove coring or segregation (ix) Improve electrical and magnetic properties. (x) Spherodize tiny particles, such as those of Fe3C in steel by diffusion. Stages of Heat Treatment Process : (i) Heating a metal/alloy to definite temperature. (ii) Holding (or soaking) at that temperature for a sufficient period to allow necessary changes (e.g.,austenitizing) to occur. (iii) Cooling at a rate necessary to obtain desired properties associated with changes in the nature, form, size and distribution of micro-constituents (such as ferrite, pearlite, martensite, etc.). Principles (fundamentals) of Heat Treatment :

Steel heat treatments are made possible by the eutectoid reaction in the iron-carbon system. All basic heat-treating processes for steel involve the transformation or decomposition of austenite. The nature and appearance of thus obtained transformation products develop a variety of useful and mechanical properties in steels.

Cooling rate plays an important role in the transformation of austenite to pearlite or martensite, etc. Heat treatment is effective only with certain alloys (e.g., Fe-C, Aluminium bronze, etc.) because it depends upon, one element being soluble in another in the solid state on different amounts under different circumstances. The theory of heat treatment is based on the principle that an alloy experiences change in structure when heated above a certain temperature and it undergoes again a change in structure when cooled to room temperature. Cooling rate is an important factor in developing different (soft or hard) structures. Slow cooling from above the critical range in steel will produce (soft) structure whereas rapid cooling (depending upon steel composition) will give rise to a martensitic (hard) structure.



2. Differentiate clearly between i) Annealing and Normalizing ii) Carburizing and Nitriding iii) Patenting and Austempering Ans: i) Annealing and Normalizing Sr No. Annealing Normalizing

1 In annealing hypoeutectoid steel is heated to Ac3+500C and hypereutectoid is heated to Ac1+500C. Steel is soaked and allowed to cool slowly to room temperature in furnace itself by putting off.

In normalizing hypoeutectoid steel is heated to Ac3+500C and hypereutectoid is heated to Acm+500C. Steel is soaked and taken out of furnace and then cooled in air.

2 Rate of cooling is slower. Rate of cooling is slower. 3 Coarse lamellar pearlite i.e. grain size is

larger as compare to normalizing Finer pearlite i.e. grain size is finer.

4 Hardness/strength decreases, ductility and machinability improves.

Hardness and strength is higher as compare to annealed state.

5 It’s a intermediate process, normally used to siften the steel. This gives final product.

ii) ) Carburizing and Nitriding Sr No. Carburizing Nitriding

1 Carbon diffuse into steel surface. Nitrogen diffuse into steel surface. 2 Process results in less hardness, less wear

resistance compare to nitriding. Process results in higher surface hardness (65-70 Rc), obtains more better wear resistance .

3 Inferior surface finish as compare to nitriding.

Very good surface finished.

4 Hardening and tempering is needed. No needs of hardening and tempering. 5 Process is very simple and inexpensive. Process is complex and expensive. 6 Thick (5 mm) carburized layer can be

obtained, so more grinding allowance is available.

Very thin nitrides layer is obtain, so very limited grinding allowance is available.

7 Grain refinement is not necessary. Grain refinement is necessary. 8 Range is wide Selective steels only nitrides.



ii) Austempering and Patenting Sr No. Patenting Austempering

1 In patenting the supercooled austenite is isothermally transformed at a temperature corresponding to the nose of s-curve of the steel.

In austempering the supercooled austenite is isothermally transformed at a temperature below the nose of s-curve but above the Ms temperature.

2 The product of patenting is fine pearlite which is globular.

The product of austempering is either upper bainite or lower bainite depending upon the temperature used.

3 Patented steel is less harder than austempered steel.

Austempered steel is harder than patented steel.

3. What is retained Austenite and how is it eliminated? Ans: Ms and Mf temperature of a plain carbon steel are very much dependent on its carbon content. In plain carbon steels containing more than 0.6% C , the Mf temperature is below 00C. It is known that austenite austenite to martensite transformation takes place when steel from the austenite condition is continuously cooled at a rate faster than critical cooling rate below Ms temperature. The martensite forms while cooling is continued from Ms temperature. At Mf temperature the martensite formation is nearly complete and liitle austenite remains untransformed. Steel is hardened normally by quenching it from austenitization temperature, into a water bath or oil kept at room temperature. Normally the room temperature is above 00C. Hence if a plain carbon styeel containing more than 0.6%C is hardened by quenching into a bath at room temperature, all the austenite is not transformed into martensite because Mf temperature of this steel is below 00C. This austenite which remains untransformed is known as retained austenite. Thus in high carbon steels, retained austenite is always present after hardening. The alloy sttels have still lower Mf temperature. Hence in alloy steels the proportion of retained austenite is more than in case of plain carbon steels. Elimination of Retained Austenite: Retained austenite is very softh and hence its presence in hardened steel is not desirable because it gives lower hardness and non-homogeneous hardness values. Hence it is necessary that the retained austenite be changed into martensite. This can be done by (i) Tempering (ii) Sub-zero treatment (i) Tempering:- When the hardened steel is tempered above 3000C, while cooling from the tempering temperature, the retained austenite is converted into martensite. However this method is useful only when the amount of retained austenite is relatrively small. Further more some retained austenite remains untransformed even after tempering. (ii) Sub-zero treatment:- In case of hardened steel in which the amount of retained austenite is large, it is better to convert it into martensite using sub-zero treatment. In this the hardened steel must be cooled immediately after hardening to a temperature lower than its Mf temperature. Since the Mf temperature is below 00C, this heat treatment is known as sub-zero treatment. For this various types of freezing mixtures are used depending upon the temperature to which the steel is tobe cooled. Liquid nitrogen is also used.

29

4. Descri How SAns :- Th

1) Thin s2) Aboutof hypoe3) The spor molten4) DifferAfter thetemperatutransformmartensit5) With tfor the tr6) The exto these t7) A plotare plottetransform 5. What Ans: Temsteel is hcooled toStages of Mstructure

ibe the procS-curve or The method o

samples of smt five or six utectoid andpecimens aren metal bathrent specimee predetermiure. In thi

mation is allte. Pearlite athe help of tansformationxperiment istemperaturest is made beed on it. W

mation are jo

is temperinmpering is thheated to a teo room tempf Temperin

Martensite is and change

Nagpur I

cedure for tTTT curve iof determinin

mall size aresamples are d Ac1+500C e taken out o

h kept at a coens are allowined times, s step, auslowed and thand martensithe study of n to start and

s repeated at s are determietween temp

When all the oined togethe

ng? Explain he heat treatmemperature lerature in aig: a supersatu

s into a more

Institute

Engine

he construcis determineng S-curves

e chosen. heated to thin case of h

of the furnaconstant tempewed to transfa given spe

stenite transhen on quenite can be disamount of p

d to finish ardifferent tem

ined. perature and

points indicer as shown

stages of Tment procesless than 72r.

urated solid se stable stru

of Techn

ering Meta

ction of TTTed for a givefor a given s

heir austenitihypereutectoce and immeerature belowform for diffcimen is taksforms to pnching, the stinguished fpearlite formre determinemperature T

time and tscating start in fig, we ge

empering.s given to ha30C, kept th

solution of ccture on tem

nology, N

allurgy These

T diagram fen steel. steel is as fol

ization tempoid steels. ediately quenw Ar1 tempeferent times ken out andpearlite uptuntransformfrom each ot

med at differed. Let theseT2, T3, T4 etc

and tf valuof the trans

et tow S-shap

ardened steehere for abou

carbon in irompering.

agpur

notes are not fo

for a given s

llows :-

erature i.e. A

nched into aerature (T1 te

at this consd quenched io the time

med austenitether. rent times, th be ts1 and tf and ts1 and

ues for differsformation aped curves.

el. In this prout ½ hour to

on. Hence it

or circulation and

steel. OR

Ac3+500C in

a molten saltemperature). stant temperain water to e for whiche transforms

he times reqf1. tf1 correspon

rent temperaand finish o

ocess the haro 1 hour and

t is metastab

d sale.

n case

t bath

ature. room

h the s into

quired

nding

atures of the

rdned d then

ble in

30

Tstages. Stage I:-martensitcrbide is carbon m

Above filittle chaetching mStage II:This folloccur. FuausteniteHardnessStage IIIwhich thFe3C-cemsecondarThe hardStage IVcauses coa decreatemperintemperedcementite Inmartensi

The structura

-This stage te and formsFe2.4C. Bec

martensite is

ig. shows vaange in hardnmartensite. :- The seconlows first sturther reduc

e to martenss continues tI:- The third

he depletion mentite. Henry troostite.

dness is abouV:- This refeoalescence oase in their ng temperatud at 4500C ie aggregatesn general, thite.

Nagpur I

al changes th

refers to ths carbide whcause of thisformed.

ariation in thness occurs

nd stage of tetage changesction in the site while cto fall and is d stage of teof carbon fr

nce the stru. The structuut 45-47 Rc. fers to tempeof cementite

number anure is incres about 40 Rs is called sehe structure

Institute

Engine

har occur on

he first temphich is precis precipitatio

he hardness oin the first s

empering res. In secondcarbon con

cooling fromaround 55 R

empering refrom martensucture contaure is very fi

erature rangparticles. Th

nd hence thased. This Rc and it drcondary sorproduced o

of Techn

ering Meta

n tempering m

perature rangpitated in a on, carbon c

of high carbstage of tem

efer to the temd stage of tentent of marm the tempeRc. eferes to the site is comp

ains α-ferriteine and cann

ge above 40he size of ce

heir dispersioresults in f

rops to 20 Rrbite.

on tempering

nology, N

allurgy These

may be cons

ge upto 200fine coheren

content of m

bon steel on mpering. The

mpering temempering, twrtensite and ering temper

temperatureplete. The cae and cemennot be resolv

000C to 6500

ementite partonbecomes fall in hardnRc when tem

g hardened s

agpur

notes are not fo

sidered to ta

00C. Carbon nt form. Thi

martensite de

tempering. e structure p

mperature of wo reaction

conversionrature to ro

e range of 3arbide is fullntite. This i

ved under op

0C. Temperiticle increasecoarser and

ness. The hmpered at 650

steel is term

or circulation and

ake place in

is rejected is compositiecreases and

It is obviousroduced is b

f 2000C to 30ns simultaneon of any retaoom tempera

3000C - 4000

ly convertedis also calletical micros

ing in this res. This resud coarser ahardness of 00C. This fe

med as temp

d sale.

three

from on of d low

s that black

000C. ously ained ature.

0C in d into ed as cope.

range ults in s the steel

errite-

pered

31

6. What the help Draw seAns: Hardenahardenedforming cooling rHardenab

T1. Chemiincreases2. Grain area whihardenabboundary Use of HHardenabquenchedThe mostor the ‘Jo

Fig. Set uJominy eThis teststandardidimensioThe test quickly tshould hajet-pipe i

is hardena of Jominy

et-up for Jo

ability : It isd. Hardenabimartensite trate, higher bility is inve

Thus, factors ical composis hardenabilisize of Aus

ich acts as bility. Finer y area higher

Hardenabilitbility is usedd under givet widely useominy end qu

up for Jominend quencht of great pized by the Aons as shown

specimen istransferred tave an internis kept 12.7

Nagpur I

ability? Expy end quenchminy-End-Q

s measure ofility of steel throughout i

is the hardersely propor

governing ition of steelity of steel. stenite (γ): W

a site of nthe grains

r is the harde

ty : d to select aen conditiond method ofuench test’.

ny end quench test : practical useASTM, SAEn in above figs austenitizto the test final diameter

7 mm. The w

Institute

Engine

plain the prh test . ORQuench test

f hardness. Itis defined a

its cross secdenability. Hrtional to CCCCR affectsl: addition o

When we havnucleation fo

or lesser isenability.

a steel to mes. We have f determining

ch test

e in determiE and AISI. Tg. The diam

zed at the tefixture. Immr of 12.7 mmwater pressu

of Techn

ering Meta

rocedure for

t.

t is defined aas the abilityction even atHardenabilityCR. s the hardenaf alloys (exc

ve small grafor γ to peas hardenabi

eet a minimuto perform tg hardenabil

ining the haThe specime

meter is 25.4 memperature r

mediately them; the distanure of jet is

nology, N

allurgy These

r measurem

as the ease wy of steel tot low quency is a funct

ability. cept cobalt) d

ains of γ, wearlite transfolity. Coarse

um hardnesstests to predlity is the ‘E

ardenability en is steel romm. The fixrecommendee water jet ince between

so adjusted

agpur

notes are not fo

ment of har

with which th get hardene

ching rate. Ltion of criti

decreases th

e have largeormation. Her the grain

at a given ldict the hardend quench h

of steel.Thod machined xture used is ed for the gs started. Ththe specime

d that the fre

or circulation and

rdenability

he sample ced complete

Lesser the crical cooling

he CCR. Hen

er grain bounHence it has

size, less

location in aenability of hardenability

his test has with collar,as shown in

given steel ahe water-jeten end the enee jet achiev

d sale.

with

an be ly by ritical

rate.

nce, it

ndary s low grain

a part steel.

y test’

been , with n fig. and is t pipe nd of ves a



height of 64 mm from the end of the pipe. The temperature of water is kept between 210C and 270C. In the test, the specimen is being quenched from one end. Hence at different distances from this end, the cooling rate would be different. The test is continued till the temperature of the specimen reaches room temperature. After the test, two flat parallel surfaces are ground on the specimen opposite to each other along the entire length of the specimen. The hardness along one of these faces is measured at intervals of 1.6 mm from the quenched end. Near this end, the interval may be reduced to 0.8 mm to make the measurement precise. A plot between the hardness and distance from the quenched end is made. We get a plot similar to one shown fig. The hardness changes most rapidly at 50% martensite region with a representative hardness 54 Rc .The distance at which this critical hardness is achieved is called Jominy distance. 7.Why gray cast iron is brittle? Ans: Gray cast iron consists of graphite flakes in the metallic matrix. These graphite flakes are non-metallic in nature. Hence there is no significant bond across these flakes and the metallic matrix. Hence graphite flakes are discontinuous and they are just like crack and stress raiser. Further because of flake in shape they have very large surface area. Hence due to presence of these graphite flakes continuity of metal matrix is severely disrupted. Hence they fail without any plastic deformation. 8.Describe TTT diagram for 0.8% carbon steel. Ans: S curves are the isothermal transformation curves. These indicate the product of transformation at any given temperature. The two curves show Austenite →Pearlite start and Austenite→pearlite finish times. At any given temperature the time required for the austenite to pearlite transformation to start is known as incubation period. The incubation period is minimum when the transformation occurs at a temperature corresponding to the nose temperature.

The transformation products obtained at different temperatures are indicated in fig. All the products – coarse pearlite, upper bainite and lower bainite – are mixtures of α-ferrite and cementite but the fineness of dispersion increases as the transformation temperature decreases. Thus fine pearlite is a finer dispersion of cementite in α-ferrite than coarse pearlite is. Banite are so fine in structure that the dispersion can be seen only at higher magnification. Since a finer dispersion gives greater hardness and strength, bainite is harder than pearlite.



Line 1 shows critical cooling rate. It can be seen that if the steel sample is cooled from the austenitic condition at a rate faster than the critical cooling rate, austenite to pearlite transformation will not even start. Line 2 exhibits a cooling rate faster than the critical cooling rate.It can be seen that this line doesnot intersect the Austenite to Peartlte start line in the S-curve. Hence if the steel is cooled from the austenitic condition at this rate, the austenite-pearlite transformation is completely avoided and the steel is completely transformed into martensite because this line cuts both Ms and Mf lines. Hence the critical cooling rate of a steel is that cooling rate which must be exceeded to produce 100% martensite on cooling a steel from the austenitic condition.

34

1.What iAns : Mshow as twhich thcooling r

F1) A high2) A lowiron. Tcondition 2. ExplaiAns : Foso that nor siliconcementiteexisting temper csurroundjoins the envelopeused in malleable

is Maurer’sMaurer’s diagto what wou

he Maurer’s rate changed

rom above dh carbon conw silicon co

Thus the castns of cooling

in Bull’s eyr the produco eutectoid n content oe is decomptemper carb

carbon rosettding the temp

already pree with dark r

shooting. He cast iron.

Nagpur I

s diagram? Wgram is a pl

uld be its modiagram is d

d , the diagra

diagram it cantent and higontent and lo

t iron obtaing.

e OR Ferriction of peacementite isf cast iron

posed. In sucbon rosettr, te, no fresh per carbon r

esent rosette rosette of teHence such

Institute

Engine

UWhat is its ulot between ode of solidifdrawn is them would als

Fig. Ma

an be conclugh silicon coow carbon c

ned will tota

ito-Pearliticarlitic malleas decomposeis not prop

ch cases the is the one mnucleation irosette prefeof temper c

emper carbonFerritopear

of Techn

ering Meta

UNIT –V utility carbon cont

fication for a normal coo

so change.

aurer’s daiag

uded that for ntent invariacontent alwa

ally depend u

c Malleable able iron, seced into ferriterly controleutectoid ce

most likely is required. Terentially decarbon and fn in the midrlitic mallea

nology, N

allurgy These

tent and silia given cool

oling rate en

gram

a normal givably produceays result in

upon C and

Cast Iron .condary grapte and tempelled it may ementite in to decompo

Thus the cemecomposes. Tferrite develddle resemblable cast iro

agpur

notes are not fo

con content ling rate. Thcountered in

ven cooling es ferritic gran the formati

Si content u

phitization mer carbon. Bhappen thatthe immediaose. Becausmentite contThe temper lop is left. Tles the Bullon is know

or circulation and

of a cast ire cooling ratn sand moul

rate :- ay cast iron.ion of white

under the no

must be avoiBut if cooling

t some euteate vicinity e of presentained in pecarbon prod

This bright f’s eye ,the t

wn as bull’s

d sale.

ron to te for ds. If

e cast

ormal

ided , g rate ectoid of an ce of

earlite duced ferrite target s eye

35

3. How d Expla ExplaAns : Malleabl1) Produit is necegraphite cementitegrow in s2) Malleboxes, a achieve ahours. Tdecompo

Fdo you convain the step ain how whiThe heat t

lizing. The puction of whessary that th

flakes. Thie does not psize. Hence ceablizing Trgiven the ma temperaturhis stage is

oses to give a

Nagpur I

Fig. Bull’s evert white ca

involved inite cast irontreatment prproduction ofhite cast irohe castings bs is becaus

produce tempcast iron conreatment :-

malleablizingre of 10500C

s known as austenite and

Institute

Engine

eye OR Ferrast iron into

n the producn is converterocess whicf malleable c

on castings :be first produe if graphitper carbon antaining grap The white

g heat treatmC. At this tprimary gra

d temper car

of Techn

ering Meta

rito-Pearlitico malleable ction of maled into mallch is given cast iron con:- For the pruced as whitte flakes areas rosettes inphite flakes ccast iron ca

ment. The boxemperature aphitization.rbon.

nology, N

allurgy These

c Malleable Ccast iron? Eleable cast ieable cast irto the whi

ntains the folroduction of te cast iron ce present th

nstead the alrcannot be mastings are txes are keptthe casting

. In this ste

agpur

notes are not fo

Cast Iron Explain? or iron.or ron. ite cast irllowing step

f malleable ccastings withhe further dready presen

malleablized. then packedt in the furnaare kept for

ep all the pr

or circulation and

on is knows:

cast iron casthout even tradecompositiont graphite f

d in sand keace and heatr about 72 trimary ceme

d sale.

wn as

tings, ace of on of flakes

ept in ted to to 96 entite



If further cooling to room temperature. i) Cooling directly from this temperature to room temperature. By cooling the boxes in air, pearlite formed and we get pearlitic malleable cast iron. ii) Cool slowly from 1050 0C to 7600C, then decreasing the cooling rate still further for cooling upto 680 0C and then cooling slowly to room temperature. In this all the eutectoid cementite decomposes to ferrite and temper carbon. Fe3C Ferrite + Temper Carbon This is known as seconadary graphitization. As such the final product is ferritic malleable cast iron. 4. Write production route, composition, micro-structure and application of grey cast iron. Ans: Solidification of Gray Cast Iron :- When carbon and silicon content is high and cooling rate is slow, a cast iron melt solidifies as gray cast iron. It contains 2.5-3.8 % C, 1.1-2.8 % Si, 0.4-1 % Mn, 0.15 % P and 0.10 % S. Gray cast iron is marked by presence of graphite flakes in matrix of ferrite or pearlite.

In a melt containing less than 4.3% C at temperature T1(liquidus temperature). The solidification start with formation of dendrites of austenite at temperature T2. At eutectic temperature austenite contains 1.7% C and liquid contains 4.3% C. This liquid undergoes eutectic reaction giving a mixture of austenite and graphite. Liquid (4.3% C) Austenite (1.7% C) + Graphite

37

A hypereknown asimilar tographite.joins theoccurs an Now dep

(i) Ferritcementiteformed giron is caMechaniTensile sHardnessGood ma (ii) Pearbe any gperalitic MechaniTensile sHardnessLess mac (iii) Ferrcementite

eutectic liquas Kisch grao that in hy Further coo already prend just after

pending upon

tic Gray case which is graphite. Strualled as ferritcal Prpoertie

strength – 20s- 110-140 Bachinability

rlitic Gray cgraphitizatiogray cast iroical Prpoerti

strength – 40s- 160-200 Bchinability

rito-pearlitie associated

Nagpur I

id (C2) startaphite whichypoeutectic aoling to 723esent graphitthe eutectoid

n the conditi

st iron :- If associated wucture now te gray cast ies: 0000 PSI BHN

cast iron :- on and sameon. ies: 0000 PSI BHN

ic Gray castwith pearlit

Institute

Engine

ts solidificatih floats to thalloy. The s3 0C results te which is d reaction th

ion we will g

cooling ratewith pearlitewill containiron.

If cooling rae structure o

t iron :- If cte which com

of Techn

ering Meta

ion at tempehe top skimmstructure belin rejectionin form of

he structure c

get different

e is very slowe get graphin graphite fl

ate is fast anof pearlite a

cooling rate mes in contac

nology, N

allurgy These

erature T1 wmed away. low 11350C

n of carbon bflakes. At 7contains pea

kinds of gra

w and Silicoitizes and grake embedd

nd Silicon coand graphite

and Siliconct with grap

agpur

notes are not fo

with the formThe eutecticis austenite

by austenite723 0C the earlite and gra

ay cast iron.

on content arraphite will

ded in ferrite

ontent are lee will be re

n content arephite flake ge

or circulation and

mation of grac reaction oe of 1.7% Ce and this caeutectoid reaaphite flakes

re high, whajoin the alr

e matrix and

ess, there wietained calle

e moderate, et decompos

d sale.

aphite ccurs

C and arbon action .

atever ready d cast

ll not ed as

some sed to



iron and graphite. Graphite will join the already formed graphite leaving behind the empty ferrite pocket and will get ferrite-pearlitic gray cast iron. Application of Gray cast iron : 1) Cylinder blocks and heads for IC engine 2) Tunnel segment 3) Frames for electric motoring 4) Sanitary waves 5)Piston rings 6)Manhole covers 7) Machine tool structure 8)Gas or water pipes 5. What is dezincification of brass? How is it avoided. Ans : This is occued in condenser tubes or boiler tubes made of cartridge brass.This is corrosion by pitting. At some places the condenser tubes might get covered with the scale formed due to the csales in water. The metal beneath these spots is preferentially corroded by cavitation. Both copper and zinc atoms are dissolved but copper gets reprecipitated at the same place.The net result is the selective removal of zinc leaving copper in porus, spongy condition. This selective removal of zinc is known as dezincification. Dezincification of brass can be decreased by adding Sn or Al to improve the corrosion resistance and by adding a little arsenic (0.05%) to the brass. Addition of arsenic minimizes the tendency towards dezincification. 6. Differentiate between any two of the following:- i) Season cracking and Orange peel effect of Brasses ii) Modified and unmodified Aluminium-Silicon alloy. iii) Cartridge Brass and Muntz metal iv)White cast iron and Gray cast iron



Ans: (i) Sr No. Season cracking Orange peel effect

1 It occurs in α and α+β brasses due to cold working.

It occurs in cartridge brass (α-brass)

2 It occurs due to presence of corrosive environment like ammonia, mercurous nitrate, sea water etc.

It mainly occurs due to annealing 700-7500C for a longer time.

3 Brass articles in season cracking shows cracks during their life.

The grains of α grows to larger size.

4 Brass article shows inter crytsalline cracks. Article shows a rough surface like skin of orange.

5 It also occurs when there are internal stresses in brass article.

It also occurs when starting material contains larger grain size.

6 To avoid season cracking the cold worked brass must be fully stress relieved at 3000C before they are put to use.

It can be avoided by proper control of grain size in production and annealing temperature not exceeding 6000C.

(ii) Sr No. Modified Aluminium-Silicon alloy Unmodified Aluminium-Silicon alloy

1 It has eutectic composition of 12.7% Si and eutectic temperature of 5680C

It has eutectic composition of 11.7% Si and eutectic temperature of 5770C

2 The silicon is present in fine globular or spheroidal form in dendrites of α-Al

The niddle like silicon is present in the matrix of α-solid solution.

3 Fine globular silicon less severely disrupt continuity of soft metallic matrix of α-aluminium solution.

Needle like silicon very severely disrupts continuity of soft metallic matrix of α-aluminium solution.

4 Alloy containing modified Al-Si has high tensile strength, ductility and high impact strength.

Alloy containing unmodified eutectic is weaker and less ductile.

5 Spheroids appears as black field and seen at higher magnification of 400X

Needle can be easily seen at lowe magnification of 50X to 100X.

6 Sodium is added about 0.01% with the help of phosphoriser to modify properties.

It doesnot contain sodium in its composition.



(iii) Sr No. Cartridge Brass Muntz metal

1 Cartridge brass is single phase α-brass. Muntz metal is two phase α+β brasse. 2 Its composition is : 30% Zn and 70% Cu. Its composition is : 40% Zn and 60% Cu. 3 Brass has maximum ductility (55%

elongation) It has sufficient ductility (15-20 % elongation)

4 It has good tensile strength of 35-38 kg/mm2 It has maximum tensile strength of 48 kg/mm2

5 The cast structure of cartridge brass contains cored dendrites of α-solid solution.

The cast structure of Muntz metal consists of feathers of α in the matrix of β.

6 Cartridge brass is used for making sheets, rods, tubes, deep drawn articles, boiler tubes and condenser tubes.

Muntz metal is used to manufacture castings of bushes, rods for electrical engineering parts.

(iv) Sr No. White cast iron Gray cast iron

1 All carbon in combined form in the form of

cementite. Carbon is in free form in the form of flakes.

2 It contains pearlite and cementite. It may show α-ferrite and graphite or pearlite or mixture of two.

3 It is very hard and brittle. It cannot machined.

It is soft and brittle. They can easily machined.

4 It shows bright metallic surface after fracture hence the name white cast iron.

It shows gray surface after fracture hence the name white cast iron.

5 Poor damping capacity. Excellent damping capacity.



UNIT ‐VI

1. What is meant by Tensile test? Explain the procedure to determine the tensile properties of steel sample Ans : It is one of the most widely used mechanical test. A tensile test helps determining tensile properties such as tensile strength, yield point or yield strength, % reduction in area and modulus of elasticity. Fig.(a) shows a specimen for tensile test. Since mechanical properties to some extent are influenced by the size and shape of the test specimen, it is customary to use the stanadard specimen.

Fig. Tensile test specimen

The essential features of a round (cylindrical) test specimen are the diameter Do, parallel length PL, gauge length Lo and end fillet radius r. These dimensions may be set as per ASTM or BS:18:1962 specifications, e.g.

Fig. Universal Testing Machine



Tensile test is carried out by gripping the ends E, E of the specimen in a tensile testing machine as shown in fig and applying and increasing pull (fig (b)) on to the specimen till it fractured. During the test, the tensile load as well as the elongation of a previously marked gauge length in the specimen is measured with the help of load dial of the machine and extensometer respectively. These readings help plotting stress-strain curve as shown in below fig.

Fig. Stress-strain curve

After fracture, the two pieces of the broken specimen are placed as if fixed together (fig (c)) and distance Lf between two gauge marks and the diameter Df at the place of fracture are measured. The various tensile properties are calculated as follows:

1. Yield strength = Load at Yield point / Ao Ao =π/4 Do

2

2. Ultimate tensile strength = Ultimate load, Pmax / Ao 3. % Elongation = (Lf –Lo/ Lo) x 100 4. % Reduction in area = (Ao – Af / A)ox 100 5. Young’s modulus of Elasticity, E

= Stress at any pont within the elastic limit, i.e. OA / Strain at that point Or E = P Lo / Ao ∆L Where P is load at any point up to the Elastic limit (Point A in fig) Lo is the gauge length, Ao is original area, and ∆L is the elongation or change in Lo at any load P, while ther specimen is within the elastic zone.

43

2. ExplaiAns : ItspecificameasurinLoad (kg500 1500 3000

Procedu1) S

w2) D

pr3) T

thv

4) Tav

5) Tin

in Brinell Ht consists of

ations, a 10 ng hardness og)

re of Hardnpecimen is

with the anvilDesired load

ressure ) andThe diameterhe use of a iew.

The indentativerage of the

The Brinell hndentation in

BHN Where

Nagpur I

Hardness Tef pressing a mm diamet

of soft materMaterial Soft IntermediatHard

Fig.ness Testingplaced on thl moves up ais applied m

d the ball prer of the indenmicrometer

ion diametere two readinhardness numn Kg per squ

= W/[( πD/2e W is load o

D is diamed is averag

Institute

Engine

est with prohardened st

ter ball is urials and vic

te hardness

. Brinell Harg : he anvil; theand contacts

mechanicallyesses into thntation made

microscope

r is measurengs is taken.mber (BHN

uare metre, is

2) (D-√(D2-don indentor,eter of steel bge measured

of Techn

ering Meta

cedure in dteel ball intused for thee versa.

rdness Testin

e hand whes with the ba

y (by a gear e specimen.e in the spece, having a

ed at two p

) which is ts calculated

d2)] kg ball, mm

d diameter of

nology, N

allurgy These

etail. o a test spee purpose. L

Ra164896

ng machine.

el is rotatedall. driven screw

cimen by thetransport en

laces at righ

the pressure as follows:

f indentation

agpur

notes are not fo

cimen. AccoLower loads

ange of Brin6 to 100 8 to 300 6 to 600

d so that the

w) or by hyd

e pressed bangraved scal

ht angles to

per unit su

n, mm

or circulation and

ording to As are applie

nell Hardnes

e specimen a

draulically (b

ll is measurele in the fie

o each other

urface area o

d sale.

ASTM d for

s

along

by oil

ed by eld of

r, and

of the

44

Brinell hgraphite,3. Differbrief aboAns: Dif1) Rockwsmaller a2) Rockw3) Rockwnot be m4) Rockwto be mea

Procedua)Test pib) Hand needle onc) Major d) Crankload apple) Hand wf) At this Tis uses ahardness carbon stis used foof alloy major loa

hardness test iron and harrentiate betout Rockwefference betwwell hardnesand thereforewell testing iwell testing ieasured: the

well test neeasured which

re for Measece is placedwheel is tur

n the dial reaload is appl

k is turned inlied. wheel is rotas stage , the hThere are twoa steel ball on C scale.

teels in the aor testing macast irons. Iads for scale

Nagpur I

is best for mrd iron carbitween Brineell hardness ween Brinellss testing die the resultinis suitable fois faster as c Rockwell mds no surfach is required

suring Hardd upon the mrned, therbyads zero. Thilied by pressn the revers

ated and the hardness of to scales on aindenter whB scale is fo

annealed conaterials hardn Rockwell

es C and B ar

Institute

Engine

measuring hide. ell Hardnesstest

l Hardness Tiffers from Bng indetationor materials hcompared to machine givece preparatiod in Brinell t

dness using machine. They raising the is applies mi

sing the crane direction t

test piece isthe test piecea Rockwell thereas a diaor testing mandition. The er than B 10hardness te

re 150 and 1

of Techn

ering Meta

ardness of g

s Test and

Test and RocBrinell testinn on the spechaving hardnBrinell testi

es arbitrary don (polishibgtesting.

Rockwell tee machine ditestb piece inor load. k provided othereby with

lowered. e material catesting machamond coneaterials of mworking ran

00. C scale isesting, the m100 kg respe

nology, N

allurgy These

gray iron cas

Rockwell h

ckwell hardnng in that thcimen is smaness beyonding because direct reading, etc) of the

ester : ial is showinup against t

on the right-hhdrawing m

an be directlhine, i.e., ‘Be penetrator

medium hardnnge of this scs commonly

minor load foectively (incl

agpur

notes are not fo

stings consis

hardness tes

ness test is ae indenters a

aller and shad the scope o

diameter of g.

e specimen w

ng any readinthe steel bal

hand side ofajor load b

y read from ’ scale and ‘is employe

ness such as cale is from used for tes

or all cases iluding mino

or circulation and

sting of soft

st and discu

s follows: and the loadllower. f Brinell test

f indentation

whose hardne

ng. l indenter ti

f the machinbut leaving m

the dial scal‘C’ scale. B ed for measlow and me0 to 100. C

sting the hardis 10 kg whor load).

d sale.

flake

uss in

ds are

ting. need

ess is

ll the

e. minor

le. scale uring

edium scale dness

hereas



4. A 13 mm diameter tensil specimen has a 50 mm gauge length. The load corresponding to the 0.2% offset is 6800 kg and the maximum load is 8400 kg. Fracture occurs at 7300 kg. The diameter after fracture is 8 mm and gauge length at fracture is 65 mm. Calculate:

i) Ultimate tensile strength in Mpa ii) 0.2% offset yield strength in Mpa iii) Breaking stress in Mpa

iv) Elongation in % v) Reduction of area in %

Ans : Given: Do =13 mm; Df =8 mm; Lo = 50 mm; Lf = 65 mm

Ao =π/4 Do

2 =π/4 ( 13)2 =132.7 mm2 =132.7 x 10-6 m2

Af =π/4 Df2 =π/4 ( 8)2 =50.3 mm2 =50.3 x 10-6 m2

i) Ultimate tensile strength = Ultimate load, Pmax / Ao = 8400 x 9.8 / 132.7 x 10-6 = 620 Mpa

ii) 0.2 % offset Yield strength = Py/ Ay = 7300 x 9.8 / 132.7 x 10-6

= 502 Mpa iii) Breaking stress Sf = Pf/ Ao

= 7300 x 9.8 / 132.7 x 10-6

=539 Mpa iv) % Elongation = (Lf –Lo/ Lo) x 100

= ( 65-60)/50 = 30 %

v) % Reduction in area = (Ao – Af / A)ox 100 = 132.7 – 50.3 / 132.7 = 62 %



Acknowledgements: We are thankful to the authors of Text Books of Subject: “Engineering Metallurgy” that are cited during the compilation of this Course Material to provide to the students. Questions, answers and related information are collectively compiled and presented in single form based on the material cited from various contributions from the authors and online sources.

References:

1. Introduction to Engineering Materials - B.K. Agrawal,

2. Material Science and Metallurgy -O.P.Khanna

3. Engineering Metallurgy- A.V.Ramarao & R.L. Vyas

Documents

Engineering Metallurgy