Unit-1 Mechanical Properties

7/30/2019 Unit-1 Mechanical Properties

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Mechanical Properties

and Their DeterminationUNIT 1 MECHANICAL PROPERTIES AND

THEIR DETERMINATION

Structure

1.1 Introduction

Objectives

1.2 A Simple Tension Test

1.2.1 Modulus of Elasticity

1.2.2 Yield Strength

1.2.3 Measure of Ductility

1.2.4 Toughness

1.2.5 Fracture in Tension

1.3 True Stress and True Strain

1.4 Compression Test

1.5 Universal Testing Machine1.6 Notched-bar Impact Tests

1.7 Temperature Effects

1.8 Hardness

1.8.1 Scratch Hardness

1.8.2 Indentation Hardness

1.8.3 Brinell Hardness

1.8.4 Rockwell Hardness

1.8.5 Vickers Hardness

1.8.6 Knoop Hardness

1.8.7 Other Methods of Hardness Measurement

1.8.8 Relationship between Hardness and Other Properties

1.9 Fatigue

1.9.1 Fatigue Curve

1.9.2 Fatigue Mechanism

1.9.3 Statistical Nature of Fatigue

1.9.4 Determination of Fatigue Strength

1.9.5 Factors Affecting Fatigue Behaviour

1.10 Material Problems at Elevated Temperatures

1.10.1 Creep and Creep Curves

1.10.2 Effect of Temperature and Stress

1.10.3 Stress Rupture

1.10.4 Creep Strength and Rupture Strength

1.11 Summary

1.12 Key Words

1.13 Answers to SAQs

1.1 INTRODUCTION

Undoubtedly, the most commonly performed test on engineering materials is the statictension test. From this test many mechanical properties of material are evaluated andunderstanding about its behaviour is developed.


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Materials Applications At the outset the distinction between static and dynamic load must be understoodcarefully. A static load is the one which changes very slowly with respect to timeduring its application. Once applied, the static load remains unchanged.

Very common static tests which are performed for evaluation of mechanical propertiesare axial tension, compression loading, flexural loading or torsional loading. Axial load isapplied along the axis of specimen. Flexural loading is achieved by application of load

perpendicular to the axis while torsional loading result from eccentric transverse forces

which do not cause bending or change in length. Direct shearing test is also performedsometimes but is not a very popular test. In all these the load is slowly increased fromzero till failure occurs; rate of loading being such that definition of static load is notviolated. Hardness is a very important mechanical property which is also determinedunder static load by indentation. Hardness and tensile strength are related. Fatigue is

behaviour of material under load or stress which keeps changing with time and failuredepends upon stress level and number of stress reversal. Creep is yet another behaviourunder statically applied load over a long period of time. Temperature plays an importantrole in deciding this behaviour.

Objectives

After studying this unit, you should be able to

understand the effect of force on solids qualitatively as well asquantitatively,

differentiate between the true stress and true strain,

analyse the compression test,

know universal testing machine,

understand tensile strengths and effect of strain rate,

know different impact tests,

explain temperature effects,

understand hardness and fatigue, and

identify material problems at elevated temperatures.

1.2 A SIMPLE TENSION TEST

A tension test is a destructive test in the sense that the specimen is finally broken orfractured into pieces. For performing a tension test one needs a machine capable ofapplying load to cause fracture. The testing machine will have suitable devices or holdersto hold the specimen so that a perfectly axial load is applied on the test piece. Themachine has to have device to measure the load at any instant and facility for

measurement of elongation of the test piece.

The test piece or specimen of the material is generally a straight piece, uniform in thecross-section over the test length and often with enlarged ends which can be held in themachine holders. Machines, however, can hold specimen without specially made orenlarged ends. Two fine marks are often made near the ends of uniform test section of thespecimen and distance between the two points is termed gauge length. The gaugelength of a specimen bears a constant standardised ratio to the cross-sectional dimensionfor reasons that will be discussed later.

The specimen is placed in the machine between the holders and any measuring device torecord change in length is fitted onto the specimen between the gauge points. If such a

device, often known as extensometer is not fitted, the machine itself can record thedisplacement between its cross-heads on which the specimen is held.

Once the machine is started it begins to apply a slowly increasing load upon thespecimen. At present intervals the reading of load and elongation of specimen arerecorded. The increase in extension can be observed from the specimen and as the


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and Their Determinationextension begins to increase can be observed from the specimen and as the extension

begins to increase at a faster rate (it can be judged from experience) the extensometers ifattached to specimen may be removed. With further increase in the load the extensionincreases at increased rate and record of extension may be obtained from displacement

between the cross-heads. At some level the load becomes stationary and the specimeneither fails or begins to reduce in cross-section rapidly. The latter phenomenon is knownas necking and is followed by reduction in load accompanied by further extension until

the specimen fractures into two pieces.Fixed Upper Cross Head

Grip or Holder

Tension Specimen

Screw Column

Grip or Holder

Moving Cross Head

Space for Compression Specimen

Fixed Lower Cross Head

Figure 1.1 : The Schematic Diagram of a Test Machine

The results of tension test can be entirely presented in form of a stress-strain diagram inwhich stress is plotted as ordinate and strain as abscissa. It may, however, be emphasizedhere that in the definition of stress; it is assumed that area of cross-section will notchange. It must have become clear by this time that during a tension test area ofcross-section and length both change considerably. Therefore, there is a need toreconsider the definition of stress.

The stress is defined as ratio of load to original area of cross-section (denoted byA0). Itwill be defined as engineering stress. The prefix engineering is often dropped andwhenever term stress is used, it is understood that it is engineering stress. The ratio ofload to actual or current area of cross-section is defined as true stress. Likewise, the ratio

correspondingly the true strain is the sum of strains over small ranges of load upto thecurrent load.

Engineering stress () and strain () will be

0

P

A = . . . (1.1)

0

L

L

= . . . (1.2)

where, P is load at any moment.A0 the original area of cross-section, L, the change in

length when load changed from zero to P andL0 is the original length.Figure 1.2 and 1.3 show typical stress-strain diagrams for common engineering materials.Diagram for ductile materials are depicted in Figure 1.2, while Figure 1.3 shows the samefor brittle materials. These two sets have the same strain scale and it can be seen that aductile materials shows considerable deformation before it fails while a brittle materialshows little deformation. To distinguish between these two types of material, materialswith strain less than 5% at fracture are regarded as brittle and those having strains greaterthan 5% at fracture are called ductile.

The type A ductile materials (Figure 1.2(a)) include steels, aluminium alloys, copperalloys etc., while type B ductile materials are mild steel and structural steels. Thedifference between type A and type B ductile materials must be noted carefully. In both

cases the stress and strain vary linearly upto strain a. This region of deformation is elasticin the sense that if load is removed at any point before reaching point a, the specimenwill regain its original length and area of cross-section. Beyond point a, ductile materialof type A changes the relationship which is more linear. During this deformation, thestrain changes at a faster rate than stress, although this rate tends to decrease. Type B of


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ductile material shows distinctly different between behaviour than that shown in Figure1.2(a). From a slightly higher point b (Figure 1.2(b)), the stress drops suddenly to point c,remains approximately constant over a range, then follows a pattern similar to that oftype A.

Materials Applications

(a) Type A (b) Type B

Figure 1.2 : Stress-strain Diagram for Ductile Materials

(a) Type A (b) Type B

Figure 1.3 : Stress-strain Diagram for Brittle Materials

The sudden drop in stress from b to c, increase in strain from c to c while stress remainsapproximately constant is known as yielding. The point b from which the stress dropsto point c, is known as upper yield point while point c is known as lower yield point.The deformation at approximately constant stress is termed asyield deformation.

Once a ductile material has exceeded the elastic deformation it enters intoplasticdeformation range. During this deformation, the stress reaches maximum value at d,where necking in the specimen begins.

The stress at point d(the engineering stress) is termed as Ultimate Tensile Strength

which is a very important property. If loading is continued, the specimen fails eventuallyat e and stress reduces from point dto e. the stress at point e is called fracture stress.

The deformation between point dande in Figure 1.2 is plastic in the sense that if thespecimen is unloaded at any point between dande the specimen will not regain itsoriginal size. Figure 1.4 shows the range of elastic and plastic deformations. Theunloading of a specimen from any pointfbetween a anddis known in Figure 1.5.Unloading follows a curve which is somewhat parallel to original elastic line. Whenspecimen is fully unloaded to point g, some strain still remains in it. This residual strain

is the plastic component of strain by p, a part of strain betweenfandg has beenrecovered and this is elastic part of strain at pointfand denoted by e. Thus, strain at

pointf,

f fp fe = +

. . . (1.3)

ElasticLimits

oStrain

Stress

YieldStrength

o

a

- Plastic StrainOffset

d

e

UpperYield

Pointa b

d

e

c cLowerYieldPoint

Stress

oStraino

o

Stress

Strain

ypellp

yp Yield Pointel Elastic Limitlp Limit of

Proportionality

oStrain

Stress

Elastic

Plastic

o

ad

e


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and Their Determination

Figure 1.4 : Elastic Plastic Deformations

o

af

h

g

p e

Figure 1.5 : Elastic and Plastic Strains

On reloading the specimen from point g, curve follows a path very close to unloadingcurve and eventually joints the original curve at h. The area between loading andunloading curve (shown hatched in Figure 1.5) represents the loss of energy due tounloading-cycle and is termed hysteresis.

Type B brittle materials (Figure 1.3(b)) do not show linear relationship between stressand strain. Concrete is the example. The maximum stress sustained by such material isdefined as ultimate tensile strength. Type A brittle materials, like cast iron, will also have

ultimate tensile strength and they show linear relationship between and.

SAQ 1

(a) What is meant by static load?

(b) What tests are performed under static load?

(c) Sketch a tension test specimen.

(d) Define stress and strain for a tension test specimen whose original area ofcross-section isA0 and original length isL0.

(e) Why should you call the stress defined in (d) as engineering stress? What istrue stress?

(f) How do you measure elongation of a tension specimen?

(g) Distinguish between a ductile and a brittle material. Give example of ductileand brittle materials.

(h) What are upper and lower yield point andultimate tensile strength?

(i) Distinguish between elastic and plastic deformation of a tension testspecimen.


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1.2.1 Modulus of ElasticityMaterials Applications

It may be noted that in most materials which may be ductile or brittle, the elastic

behaviour is marked by linear relationship between stress () and strain (). This is seenfor A and B type ductile materials in Figure 1.2 and for type A brittle material inFigure 1.3. Such linear relationship was first identified through Hookes law which statesthat stress is directly proportional to strain within elastic limit. This means that

or i.e.E E

= =

. . . (1.4)

The constant of proportionalityEis calledModulus of Elasticity orYoungs Modulus.Apparently the slope of elastic part of stress-strain diagram (shown as oa in Figures 1.2and 1.3) is the modulus of elasticity. Modulus of elasticity is a constant for a given classof material and variation in small alloying contents does not cause change inE. For

example, wide range of carbon steel has MPa. If stress-strain curve is

established accurately the slope of elastic part isEof the material. However, it may benoted that this is not a reliable method because tension tests require such loads that may

cause small deformations in loading system, part of which may look like deformation ofthe specimen.

3210 10E=

1.2.2 Yield Strength

A material placed in a machine or structure to serve as a load carrying member is neverallowed to deform plastically. Therefore, designers often need to characterise a material

by its elastic strength. There are two types of strength that can be recognised fromdiagram. They are ideal andpractical elastic strengths. Ideal elastic strengths are,proportional limit and elastic limit.

Proportional limit or a material is a stress upto which material has linear relationshipbetween stress and strain.

Elastic limitof a material is the stress upto which a material behaves elastically. Themeaning of elasticity or elastic behaviour has already been explained.

In Figure 1.6 stress level corresponding to point a is the proportional limit and thatcorresponding to point b is the elastic limit. Both the ideal strength values are seldomused in actual characterisation of material because of difficulties in their determination.Therefore, practical elastic strength values are defined and determined.

a

10

20

30

40

50

0 0.002 0.006 0.1

d

ba

c

0 0.1 0.2 0.3

c

d e

f

g

(mm/mm)

e

(mm/mm)

b(k

gf/cm

2)

Figure 1.6 : (a) With Strain Scale Expanded Showing Yield Deformation and (b) With Strain Scale

Shortened to Show Ultimate and Fracture Point alongwith Yield Point

The yielding behaviour of mild and structural steel is shown in Figure 1.6(a) on anexpanded scale of abscissa or strain axis. This figure exphasises the fact that total

deformation during yielding is about ten times the deformation upto elastic limit. Theyielding is shown to end at point e beyond which the plastic deformation continues as isshown in Figure 1.6(b). The point c is the upper yield point anddthe lower yield point.The difference between upper and lower yield point exists because movement of the endsof the specimen as produced by two machine cross-heads, does not proceed as fast as the


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and Their Determinationyield deformation of material. This fact may be confirmed by allowing the loading cross-head to move very fast, which is the condition of the impact loading. In such a case onlyone yield point (i.e. the upper yield point) will exist. For materials which show upper andlower yield points, the latter is taken as the yield strength because it is less affected byvariables.

Most ductile materials do not yield and show curve of Figure 1.2(a) or as in Figure 1.7.For such materials the end of elastic behaviour is defined as attaining a certain plastic

strain. The method of determination of this strength value is to measure a certain offsetstrain from origin as in Figure 1.7 represented by oA. From pointA a line parallel toelastic strength line part is drawn to meet the curve at d. the stress corresponding to pointdis defined as the yielding strength or proof strength of the material. The values ofoffset strain have been standarised for different materials. The yield strength determined

by the offset method is always described as yield strength for an offset. Table 1.1describes the values of offset strains commonly used.

o

o

d

Figure 1.7 : Determination of Proof Stress

Table 1.1 : Commonly used Offset Values (A.S.T.M.E6-35T)

Material Types of Stress Percent Equivalent Strain

Aluminium alloys Tension and Compression 0.20 0.0020

Brass or bronze Tension and Compression 0.35 0.0035

Concrete Compression 0.02 0.0002

Cast Iron Tension 0.05 0.0005

Steel Tension and Compression 0.20 0.0020

Wood Compression 0.05 0.0005

1.2.3 Measure of Ductility

It has been stated earlier that a ductile material exhibits considerable deformation in the

plastic zone. Naturally, then this deformation or strain can be used as a measure ofductility. The total deformation suffered by a specimen upto fracture divided by theoriginal length (between the gauge points) multiplied by 100 is termed as percentelongation. Percent elongation is taken as in index to describe the ductility of material. Itwould mean that higher the percentage deformation, more ductile is the material.However, percent elongation will depend upon what gauge length is chosen. Tounderstand this the sequence of deformation as depicted in Figure 1.8 with reference todiagram of Figure 1.6 will have to be understood well. During elastic deformation thestrain is very small and uniformly distributed over entire length of the specimen. Hardlyany change in length or diameter can be visually observed. Even beyond the yield pointor elastic limit such changes are small, but just before ultimate point the plasticdeformation starts to localize in a small part of length and reduction in diameter (in themaximum change in length occurs in this region. Ultimately fracture occurs in the neckedregion.


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(a) (b)

(c) (d)

Figure 1.8 : Sequence of Deformation of a Cylindrical Specimen with Reference to Figure 1.6 (a) Just

Before Point,a, (b) Just Before Point,f, (c) Just After Point,fand (d) At Pointg

Figure 1.9 shows how the plastic deformation is distributed along the length of thespecimen. Different gauge points have been shown marked on the specimen. It can beseen that by choosing any pair of gauge points, equidistant from the centre, one mayarrive at a value of percent elongation which will differ from that obtained by choosing adifferent pair of gauge points. This example shows that percent elongation is not a uniquequantity for a material but may vary with gauge length. For this reason gauge length isstandardised so that comparison of ductility between two materials may be made

consistently. For cylindrical specimen gauge length of five times the diameter is chosenas a standard. But in any case percent elongation is described with gauge length on whichit has been calculated.

Elongation

Figure 1.9 : Distribution of Elongation along the Length of Tension Specimen

The value of percent elongation may be obtained from diagram as the strain at the time offracture multiplied by 100. More practically, the two pieces of specimen after fracturemay be placed together touching at fracture surface and the distance between gauge

points noted to measure final lengthLf. Then,

% elongation = 100f o

o

L L

L

. . . (1.5)

The % reduction in area of cross-section is also used as a measure of ductility. It isdefined as :

% reduction in area 100o f

o

A A

A= . . . (1.6)

where A andL represent the area of cross-section and length respectively,fando arerespectively the suffixes to denote final and original values.

It may be understood by above description that both % elongation and % reduction inarea would define the local ductility of the material. Maximum uniform strain, which isthe strain just before necking begins or the strain corresponding to ultimate stress maysometimes be used to indicate the ductility of a material.

Ductility is an important property of material which governs its ability to be deformed insuch processes as drawing, forging and extrusion. Adequate ductility ensures that the

materials during these processes will not fracture. All associated property by virtue ofwhich sheets can be rolled from material is termed malleability.


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L

SAQ 2

(a) Distinguish between proportional limit and elastic limit. Which one is higherin a stress-strain diagram obtained from a tension test?

(b) Use sketch to distinguish between upper and lower yield points. Also showthe phenomenon of yielding.

(c) Yield strength which is the stress at lower yield point in a tension test is

used to define elastic strength. How do we define elastic strength in thosematerials which do not show yielding?

(d) What are the measures of ductility? Name the manufacturing processeswhich are performed upon ductile materials.

(e) What is a ductility associated property and in which process it is useful?

(f) On a uniform cross-section bar two gauge marks A and B are made at adistances of 200 mm. The length between A and B is divided into 8 equalintervals, each equal to 25 mm. The bar is put under tensile load in themachine and pulled until fracture occurs. The two broken pieces are puttogether and intervals measured. The measure distances before test and after

fracture are shown in figure below.

A 1 2 3 4 5 6 7 B

25 mm all intervals

A

27.1 27.2 28.0 30.0 32.2 27.5 27.0 26.8 All mm

Figure

Calculate the percent elongation for 50 mm length between 3 and 5.



Calculate the percent elongation for 200 mm length between A and B.

Plot % elongation against gauge length and conclude that gauge length hasto be standardised.

1.2.4 Toughness

When the load acts upon the specimen, certain work is done. This work it stored in thespecimen as its strain energy. If loadP acts upon the specimen and causes certaindisplacement, dL the work done,

U Pd= . . . (1.7)

If either side of above equation is divided by volume

0 0 0 0

U P dLd

A L A L= = . . . (1.8)

The integral is nothing but the area under stress-strain curve. It means that the

work done upon the specimen per unit volume during stretching, which is stored in thespecimen as strain energy, can be measured by area under the curve. The maximum

d


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elastic energy per unit volume that material can absorb without attaining plastic state is

known as modulus of resilience, while the energy stored in the specimen at any stresswithin elastic limit may be referred to as resilience.


The toughness of a material is understood to be its ability to absorb energy during entireelastic and plastic deformation. Themodulus of toughness is measured as area underentire stress-strain curve and is the energy absorbed by material of the specimen per unit

volume upto fracture.

From Figure 1.10 it can be seen that modulus of toughness will depend upon both theultimate tensile strength and strain at fracture and material which is very ductile willexhibit a higher modulus of toughness as is the case with mild steel. On the other hand,modulus of resilience depends upon yield strength (or proportional limit or elastic limit)and hence a material with higher yield strength will have higher modulus or resilience.Higher toughness is a desirable property in materials used for gears, chains, hooks,freight car coupling etc. Higher resilience is desirable in springs.

Mild Steel

High Carbon Steel

High Carbon Steel

Mild Steel

Figure 1.10 : Curves for High Toughness (Mild Steel) andLow Toughness (High Carbon Steel) Materials

For ductile materials like mild steel (Figure 1.10) the modulus of toughness may becalculated by approximate formula :

Modulus of toughness2

Y uu f f

+ = = . . . (1.9)

where Yandu are respectively yield and ultimate tensile strengths, andfis the strainas fracture.

Modulus of reliance is the area of triangle whose height is Yand base is the strain

corresponding to Y, i.e. Y. Hence modulus of resilience is1

2Y Y .

Since YYE

=

Modulus of resilience2

2

Y

E

= . . . (1.10)

Strain Hardening

Referring to Figure 1.5 in Section 1.2 it was stated that on reloading from g thenew curve joins the original curve at h, where point h is higher than point a. Inother words it can be said that due to the initial loading of the specimen upto pointfits yield strength has increased to h for if it were not so, the reloading curve mustfollow a path similar to original curve showing a yield strength equal to the heightof point a. This rise in yield strength due to loading in plastic region is termed asstrain hardening. If the material does not strain harden during plasticdeformation it will extend at constant stress level. The materials that extend atconstant stress level are called ideally plastic. Very soft materials behave in thismanner.

The explanation to strain hardening is found in defects in crystalline materials atatomic level. Dislocations are such defects. Strain hardening is a practical methodof increasing elastic strength. Wires that make helical springs are strain hardened.


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and Their Determination1.2.5 Fracture in Tension

The tension specimens of ductile are brittle materials show distinctly different fractures.A brittle material in general fails due to tensile stresses, or in other words a brittlematerials is weak against tensile stress. Consequently, a brittle material will fail along across-section on which maximum tensile stress is acting. The state of stress at any pointwith respect tox-y axes, wherey-axis is along the axis of the tension specimen is shownin Figure 1.6(b). Over any cross-section of the specimen the tensile stress is uniformly

distributed. Therefore, a brittle material is expected to fail along the cross-section. This iswhat actually happens. A cast iron specimen fails in tension along the cross-section,normal to axis. The surfaces of the fracture has granular appearances. Same will be truefor any other brittle material.

P

P

y

yy=P/A

x45

y

x

y P/2A P/2A

P/2A x

(a) (b) (c)

Figure 1.11 : (a) A Tension Specimen; (b) State of Stress at any Pont with respect tox-y axes wheny-

axis is Parallel to Axis of the Specimen; and (c) State of Stress with respect toxy axes when an Angle45o with Axis of Specimen, the Maximum Shearing Stress equal toP/2A Acts Parallel tox andy Axes

at any Point

On the other hand a ductile material is stronger against tension than against shearing

stresses. In a tension specimen, the maximum shearing stress acts upon a plane that isinclined at an angle 45o with the axis as shown in Figure 1.11(c). However, due toformation of neck the state of stress changes considerably. Still the final fracture of aductile material shows a characteristics cup-and-cone fracture surface. The cup and conematch at an angle 45

o with the axis, showing that the final fracture has occurred alongthese planes all around the periphery. If a highly polished specimen is tested in tensionthen on the surface at yield point fine lines will appear. These lines are generally inclined

beginning of sliding of atomic planes under the effect of maximum shearing stress(Figure 1.11(c)).

45 45

Cone

Cup

D

irection

o

fLoad

Direction

ofLoad

(a) (b)

Figure 1.12 : (a) Cup and Cone Fracture of a Ductile Material in Tension; and (b) Flat Fracture

Surface with Granular Appearances in Case of Brittle Material


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Example 1.1Materials Applications

A steel specimen shows upper yield point at 210 MPa and lower yield point at

200 MPa. If modulus of elasticity,E, for steel is 210 103 MPa. Calculatemodulus of resilience.

Solution

Modulus of resilience

2

2Y

E

=

Where, Y is yield strength. Ycorresponds to lower yield point.

Hence, Y= 200 MPa

Modulus of resilience2 6 2

3

3 6

(200) (10 )N-m/m

2 210 10 10

=

(Note MPa = Pa 106 = (N/m2 (10)6))

Modulus resilience = 95.24 104 N-m/m3

or 95.24 10 3

N-mm/mm3

Example 1.2

A steel specimen of 10 mm diameter and 50 mm gauge length was tested intension and following observations were recorded.

Load at upper yield point = 20600 N

Load at lower yield point = 19650 N

Maximum load = 35550 N

Gauge length after fracture = 62.43 N

Calculate modulus of resilience and modulus of toughness. Also calculate %

elongation. .3 2210 10 N/mmE=

Solution

Area of cross-section of specimen, 204

A d

=

2 20 (10) 78.57 mm

4A

= =

Yield strength, 219650

250 N/mm78.57

Y = =

Ultimate tensile strength,235550 452.5 N/mm

78.57u = =

Strain at fracture or % elongation62.43 50

0.25 or 25%50

f

= = =

Modulus of resilience2

2

Y

E

=

23

3

(250)N-mm/mm

2 210 10=

. . .

(i)

3148.8 10 N-mm/mm= 3


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and Their Determination(Compare with Example 1.1 to note that for higher yield strength modulus ofresilience is larger).

Modulus of toughness2

u Yf

+ =

3452.5 250 0.25 N-mm/mm2

+=

387.8 N-mm/mm=

% elongation = 25%

SAQ 3

(a) Define modulus of resilience and modulus of toughness.

(b) What is strain hardening? Is it used in practice?

(c) A steel wire having is required to have a modulus of

resilience of 140 10

190 MPaY = 6 N-m/mm3. The yield strength can be increased by

strain hardening. What should be the % increase in yield strength.

E= 210 103

N/mm

2

.(d) Why should a ductile material fracture in cup and cone configuration

whereas brittle material fails on cross-section of a bar tested in tension?

(e) What are Luders bands? Are these bonds shown by cast iron? Give reasons.

(f) Why should modulus of resilience of spring material be high? For a given

material modulus of resilience can be increased either by increasing YordecreasingE. Which one is possible?

1.3 TRUE STRESS AND TRUE STRAIN

In Section 1.2 the idea of true stress and true strain was introduced. It is emphasised hereagain that the stress and strain referred to so far were engineering stress and strain andit is customary to drop the prefix engineering.

True stress at any point in the tension test specimen is defined as the ratio of load tocorresponding area of cross-section, or true stress,

P

A =

where,A is the current area of cross-section corresponding to loadP during the tensiontest.

True strain is defined as summation of incremental strain over the entire load rangefrom zero to P. If the initial lengthL0 changes to length L during load change of 0 to P,true strain

00

ln

L

L

dL L

L L = =

Because of the final form of the true strain, it is often called logarithmic strain.

Since, 0L L L= +


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Materials Applicationswhere, L is the extension in lengthL0 so that engineering strain

0

L

L

= .

Hence true strain 0

0

ln ln (1 )L L

L

+ = = = +

.

One confusion that sometimes arises regarding the nature of the stress-strain curve is the

decreasing of stress while the strain keeps on increasing. This is actually because of thedefinition of stress which is based upon original area of cross-section. Once the necking,

begins (at ultimate point) the area of cross-section starts reducing fast. In this even theengineering stress calculated as ratio of load to original area of cross-section falls muchshort of actual stress, and hence apparently diagram behaves as shown earlier. If true

stress and true strain are plotted instead of and then the curve will not go down butstress will keep in increasing with reduction in area. However, where is very small asin elastic region the and diagrams will almost coincide. This is shown inFigure 1.13.

Strain

Stress

True Stress Strain

EngineeringStress - Strain

Figure 1.13 : Comparison of Engineering Stress-Strain Diagram with True Stress-Strain Diagram

It is well known fact that volume of material remains constant during plastic deformation,which means

00 0

0

orA L

A L ALA L

= = . . . (1.14)

With this fact it can be shown that

0

0

ln lnAL

L A = = . . . (1.15)

Properties in Terms of True Stress and Strain

Under some circumstance, particularly when the material is in plastic range ofdeformation, certain material properties are defined in terms of true stress andstrain. True stress plotted against true stain is best represented as in Figure 1.14(a)with properly choosing strain scale because it is in plastic range. In most cases therelationship is non-linear and the slope of the curve at any given strain is calledmodulus of strain hardening at that strain.

Several experimental observations suggest that and satisfy following relation

nk = . . . (1.16)

or ln ln lnk n = + . . . (1.17)

where, kandn for a given material can be regarded as its properties, kis calledstrength coefficientwhile n is strain hardening exponent. As is obvious from

Eq. (1.17) the plot between ln and ln is a straight line whose slope is equal ton and intercept on ln axis is ln k.


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19


and Their DeterminationThe constants kandn can be determined experimentally if diameter of specimencorresponding to loadP is recoded. A special gauge will have to be used formeasuring least diameter along gauge length with increasing load.

True Strain , In

In

Slope of the Line is StrainHardening Exponent, n

Slope of Tangent at 1is Modulus of StrainHardening

at1

(a) (b)

Figure 1.14 : True Stress Plotted as Function of True Strain (a) Linear Scale; and (b) In Scale

One may refer to Figure 1.2 and once more may note that at point dthe stress is

maximum and even when strain keeps increasing stress keeps reducing afterpoint d. in an experiment wherein a ductile material is being tested perceptiblereduction in diameter is observed after ultimate point, d. The condition of

beginning of reduction of diameter in the localised region is termednecking and itis the neckwhere diameter is always minimum. This phenomenon is also calledinstability and cannot be reversed. However, if load is removed very fast furthernecking may be stopped. If the load is maintained then necking will continue untilfracture.

It is possible to establish the condition of beginning of necking in terms of strainhardening exponent. Starting from the fact that tangent to load elongation curve athighest point will be horizontal, it can be written that

dP = 0or, d(A) = 0

where, is true stress andA is the area of cross-section of tension specimen atmaximum load.

0dA A d + =

ord dA

A

=

. . . (1.18)

Also since volume of specimen of lengthL and cross-section areaA remainsconstant, i.e.AL = constant.

d(AL) = 0

or, L dA +A dL = 0

i.e.dA dL

dA L

= = . . . (1.19)

From Eqs. (1.18) and (1.19)

dd

=

or,d

d

=

. . . (1.20)

Substitute from Eq. (1.16) in Eq. (1.20) to obtain

( )n nd

k kd

=

TrueStres

s,

1

1 Strength Coefficient, k


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Materials Applications i.e. 1n nn k k =

or, n = . . . (1.21)

Thus it is seen that when true strain becomes equal to strain hardening exponentn, necking will begin and fracture will ultimately occur. Metal working processlike drawing, rolling, extrusion etc. may be thus performed safely it at no stage thetrue strain is allowed to reach the value of strain hardening exponent. The strain

hardening exponent can be determined from tension test. Such determination willbe exemplified in solved example.

Example 1.3

A copper specimen of 64 mm gauge length and 12.80 mm dia. was tested intension. Following two diameters were recorded in the plastic range ofdeformation.

Load = 25.75 kN, d1 = 12.176 mm

Load = 24.25 kN, d2 = 8.581 mm

Calculate strength coefficient and strain hardening exponent.

Solution

Original area of cross-section, 2 20 0 (12.8)4 4

A d

= =

20 128.6144 mmA =

Area of cross-section at 25.75 kN2 21 (12.176)

4 4d

= =

21 116.3802 mmA =

Area of cross-section at 24.25 kN 2 22 (8.581)4 4

d = =

22 57.8023 mmA =

True stress at 25.75 kN,3

21

25.75 10221.25 N/mm

116.3802

= =

True stress at 24.25 kN,3

22

24.25 10419.53 N/mm

57.8023

= =

Note from Eq. (1.15) true strain 0lnA

A

=

1128.6144

ln ln 1.105116.3802

= =

1 0.0999 =

and 2128.6144

ln ln 2.22557.8023

= =

2 0.7998 =

Thus two pairs of values are obtained. Use these values of and in Eq. (1.15),

i.e. ln ln lnk n = +

ln 221.25 ln ln 0.0999k n= +

and ln 419.53 ln ln 0.7998k n= +


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and Their DeterminationSubtract first equation from second

6.039 5.399 ( 0.2234) ( 2.3036)n n =

i.e. 0.64 = 2.0802 n

or,0.64

0.30772.0802

n = = . . .

(i)

Using this value in one of above equations

5.399 ln 0.3077 ( 2.3036)k= +

ln 5.399 0.7088 6.1078k= + =

k= 449.35 N/mm2

. . .(ii)

True stress-true strain relationship for copper is

0.3077449.35 = . . .

(iii)Example 1.4

At what load the specimen of copper of last example will begin to neck in atension test? What will be its ultimate tensile strength?

Solution

The condition for necking is 0.3077u n = =

Hence using this value of , the true stress at neckingu

0.3077449.35 (0.3077)u =

449.35 0.696=

2312.67 N/mmu =

Hence maximum load, u u uP A= where,Au is the area of cross-section at Pu.

Note 0lnuu

A

A =

128.6144

0.3077 lnuA

=

128.6144 1.36uA

=

or Au = 94.55 mm2

312.67 94.55 Nu u uP A= =

= 29.6 kN

The ultimate tensile strength3

0

29.6 10

128.6144

uP

A

= =

= 230 N/mm2.

SAQ 4

(a) Distinguish between true stress and engineering stress. Distinguish betweentrue strain and engineering strain.


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(b) Define strength coefficient and strain hardening exponent. How can theseproperties be determined?


(c) What should be limiting strain in process of wire drawing?

(d) A steel specimen with gauge length of 62.5 mm and diameter of 12.5 mmwas tested under tension. Thee gauge length at maximum load of 72.5 kNwas 71.55 mm and gauge length at fracture load of 66.6 kN was 80.5 mm.

Find specimen diameter at maximum load and fracture load.[UseA0L0 =AL, dia. at maximum load = 11.68 mm, dia. atfracture = 11.01 mm].

(e) Find true stress and true strain at maximum load and fracture in (d).

[ 0.1345, 0.25]u f = =

1.4 COMPRESSION TEST

There are several materials used in engineering practice that are primarily meant to carrycompressive loads, concrete, bricks and wood are normally used in compression.Therefore, the mechanical properties of these materials are often evaluated incompression test. Generally, behaviours in tension and compressive are similar but thereare differences in stress-strain diagram and mode of failure particularly in ductilematerials which need understanding.

In a simple compression test cylindrical specimens of standard length to diameter ratios

are compressed between two platens under axial load. The mechanical properties such asyield strength, ultimate compressive strength, modulus of toughness and also elasticconstant, modulus of elasticity can be determined for brittle materials in compressiontest. For a ductile material, however, the ultimate compressive strength cannot be defined

because a ductile material keeps on expanding in lateral directions as compressive loadincreases. This lateral expansion continues to take place whereby area increases andultimately the specimen might be turned into disc (Figure 1.15). This lateral bulging of aductile material enables it to resist almost indefinitely large focus without fracture. Thespecimen in contact with platen expands less because of friction between surfaces of

platens and specimen. Sometimes the stress for an arbitrarily chosen deformation isindicates as compressive strength of a ductile material. Often, the ability to undergo large

plastic deformation under compression is called malleability of ductile materials but this

property cannot be defined quantitatively. Modulus of elasticity and modulus ofresilience for ductile material can be evaluated in compression as in tension but modulusof toughness cannot be defined.

Figure 1.15 : A Ductile Material Specimen in Compression

At atomic level, unlike tensile stress case, no definite maximum exists in stress vs atomicbond curve and thus there is not limit to repulsive force that can be build up betweenatoms. Thus, the compression will not become the cause of failure and will always result

Bulging

SpecimenCompressed

into Disc

CompressionSpecimen


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and Their Determinationin elastic strain. It is the shearing stress component which causes the slip of atomic planesand results in bulging of the specimen, and plastic deformation. With this considerationthe stress-strain diagram within elastic limit will be the extension of that in tension withthe difference that the elastic behaviour in compression ends a higher values of stress andstrain as shown in Figure 1.16(a).

+-

+

-

Compression

Tension

Strain

Stress

(a) (b)

Figure 1.16 : (a) Elastic Stress-Strain Relationship in Tension and Compression and

(b) Behaviour of Rubber in Tension and Compression

Long chain ploymeric materials like rubber, cork or wood show distinctly differentbehaviour in tension and compression. In tension the coiled molecules are first uncoiledand then strained whereby such materials show low stiffness followed by higher stiffnessin tension. On the other hand compression may tend to cause further coiling in the initialstage followed by elastic compression of molecules showing slight stiffening effect inimmediate decrease of stiffness. As the cells are compacted with increasing compressiveload, the stiffness will further increase.

1.4.1 Compression of Brittle Materials

For most brittle materials the elastic properties like elastic limit, modulus of resilience,

yield strength and modulus of elasticity are same in tension and compression. Theultimate compressive strength of most brittle materials is different than the ultimatetensile strength. For example, in concrete, the ultimate compressive strength is about tentimes its ultimate tensile strength.

The compression test specimen fractures due to shearing along a plane inclined to theaxis (Figure 1.17). The orientation of the plane of shear is affected by the presence ofnormal compressive stresses and the final fracture does not often occur along plane ofmaximum shearing stress which is inclined at an angle of 45o with the axis. Theorientation of plane of shear varies between 35o and 45

owith the axis.

ShearPlane

Figure 1.17 : A Brittle Material Specimen under Compression

Even in brittle materials some lateral bulging will tend to occur. This lateral deformationis greatly restricted at flat specimen ends due to friction between the platens and the

specimen ends. This friction reduces for longer specimen. If the compression specimen istoo long, it may have a tendency to buckle. To minimise the influences of friction and

buckling the length to diameter ratio of compression specimen has to be chosen verycarefully. For determination of compression strength this ratio is between two and three.


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However, for determination of elastic properties the ratio of length to diameter may bechosen between eight and ten. If this ratio is less than 1.5 the fracture plane mightintersect the end which is undesirable.


It may be noted that the brittle material in which the atomic bonds are not re-establishedwill ultimately fracture due to slip which may occur along one or several planes. Thelater failure is fragmentation. While the cracks, pores and holes play very important rolein reducing tensile strength the same do not become active under compressive stress.

They rather close and net compressive strength is higher than tensile strength particularlyin case of brittle materials. Figure 1.18 shows this fact for cast iron, and Table 1.2describes the same for several materials.

Tension

Compression

0.06 0.04 0.02 0.0 0.01

Strain (mm / mm)

840

700

560

420

280

140

0

140

Stress(MPa)

Figure 1.18 : Stress-Strain Diagram for Cast Iron in Tension and Compression

Table 1.2 : Tensile and Compressive Strength of Brittle Materials

Material Tensile Strength

(MPa)

Compressive Strength

(MPa)

Ratio

Grey cast iron 165 825 5

Concrete 2.7 34 12.6

Plexigless 72 115 1.6Alumina Ceramic 205 2040 9.95

[Note : Concrete 1.3 mix by volume, water cement ratio 0.64 (by volume) cured for 28 days.]

Compression test is terminated if excessive deformation occurs and specimen does notfracture.

The hydrostatic compression is the loading in which every element is subjected to equalcompression stresses in all three directions. This state of stress which is obtainable whena body is submerged in water does not cause any distortion but produces only volumereduction. In such a test the specimen my not fail at all because at atomic level such stateof stress will tend to push atoms into each other and no slip would occur, hence no plastic

deformation.

1.4.2 Experimental Methods for Compression Test

Most important factor that has to be controlled in compression test is the application ofload. The axiality of load has to be ensured. The platen, normally the one in contact withthe moving cross-head is made self aligning with a spherical head on the top of the platewhich is directly in contact with the specimen. The centre of the spherical head coincideswith the centre of the specimen top surface through which must pass the axis of themachine. Great care needs to be exercised for aligning the specimen axially with machine(Figure 1.19(a)).

Because of friction between the loading plate and top of specimen the material

immediately below the top surface is subjected to near hydrostatic stress and avoidsfailure. Though the compressive stress in direction perpendicular to load axis decreasesaway from loading surface it does not vanish before distance is at least diameter of thespecimen. For this reason the specimen length of 3dis normally recommended. The topsurface is also recommended to be smooth and friction free. In some specimens close cut


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and Their Determinationcircular concentric grooves and created to hold lubricant, particularly in specimens ofductile materials. The problem of bulging is minimal in concrete and hence l/d= 2 isoften used for this material. In yet another method the ends are made conical hollow toreceive similar loading plate as shown in Figure 1.19(b). The angle of cone is equal toangle of friction between the materials of platen and specimen, and it is not readilyknown. It is important that compression force must be applied such that rate of strainingis constant. Most materials required to be tested in compression are viscoelastic in nature

and do not reach equilibrium readily. Prescribed strain rates are, therefore, maintained.

R

=

(a) (b)

Figure 1.19 : (a) Self Aligning Platen; and (b) Conical Hollow End of Compression Specimen

SAQ 5

(a) Sketch how a mild steel specimen will change shape under increasingcompressive load.

(b) Why does cast iron show higher compressive strength than tensile strength?

(c) The stiffness of rubber first increases in tension and then decreases whereasit increases in compression. Give reason.

(d) Why should a compression test specimen, which is compressed between tworigid platens, bulge around mild section?

(e) Why should axis of machine coincide with the axis of compression testspecimen? What methods are adopted in place of flat pattern to ensurecoaxiality?

(f) The ratio of length to diameter in cast iron compression specimen is3 whereas it is 2 in concrete specimen. Give reason.

(g) A concrete specimen with l/dratio less than 1.5 tested in compression.Sketch the failure plane. Why this specimen is not recommended? Sketchthe same plane ifl/d= 3.

(h) In which case l/d between 8 and 10 is preferable in compression specimen.

1.5 UNIVERSAL TESTING MACHINE

Universal testing machine is to be distinguished from a special purpose machine in thesense that such a machine is used for performing several tests. Tests that are performed ina universal testing machine are tension, compression, bending, transverse shear andhardness. On the other hand, a special machine like torsion testing machine orcompression machine can be used only for particular test as indicated by name.

Two types of universal testing machines are in common use. One is mechanical typeoperated by screw and gear as shown in Figure 1.20 and the other is hydraulic universaltesting machine as shown in Figure 1.21. Both these machines are basic in nature. The


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screw type varies in size and capacity over wide range of a few newtons to a few tones.Hydraulic machines are generally of high capacity. These machines have limited speedsof movement of cross-heads so that loads applied are within the definition of static. Thestrains rates greater than 10 6 mm/mm.s are usually not available on universal testingmachines. Modern computer operated machines now-a-days make wide range of strainrates available and they can be even used for cycling stresses. Instron and MTS machinesare examples.


Whatever by the type of machine general features remain same, a universal machineapplies axial load which can act as tension, compression, shear or bending load upon thespecimen depending upon positioning of the specimen. The application of load iseffected by relative movement between two cross-heads. Generally machines have twocross-heads rigidly connected with each other and a third cross-head is positioned

between them. Each machine has proper specimen holding and supporting devices toapply desired type of load upon the specimen. The load applied upon the specimen has to

be balanced and measured and as such each machine should have this device.

Figure 1.20 shows essential features of a mechanical type of universal testing machine.The upper and lower cross-heads are rigidly connected with middle cross-head can bemoved down by a screw and gear arrangement driven by an electric motor. While

moving down the middle or movable cross-head can apply a tension load on a specimengripped between the moving cross-head and upper cross-head. Alternatively, the middlecross-head can apply a compressive load on a specimen placed on lower cross-head. Thefixed cross-heads are supported over a compound lever system at whose end a poise or

balance weight can be shifted to achieve balance of the lever and thus measure theapplied force. The balance weight is often replaced by a swinging pendulum whoserotation is transmitted to an indicator through gear train for direct indication of load.

In a hydraulic universal testing machine (Figure 1.21) the rigid frame containing upperand lower cross-head is attached to an accurately finished piston which exactly matcheswith the hydraulic cylinder. The middle cross-head, whose position is adjustable remainsstationary when the load is applied upon the specimen. An electric motor rotates a

hydraulic pump that forces the oil into the cylinder whereby the rigid assembly of lowerand upper cross-head is lifted up. During this movement a tensile load will act upon thespecimen held between upper and middle cross-head and a compressive force will actupon a specimen placed on the lower cross-head. The hydraulic pressure on the bottom ofthe hydraulic cylinder is transmitted to a hydraulic capsule which is connected to aBourdon tube gauge which can be calibrated to read the force directly. The middlecross-head can be moved up or down by separate positioning screws and during its

positioning movement the hydraulic pump is not operated.

Both types of machine often incorporate the mechanical graphical devices to plot the loaddisplacement curves. Modern machines incorporate load cell and displacement gaugeswhose electrical signals respectively proportional to load and displacement can be plotted

automatically onX-Yplotter when the test is proceeding.

Fixed CrossHead

Tension Specimen

Balance Weight

Movable CrossHead

Motor

LeversScrew

Fixed Cross

Head

Gears

26


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27



Figure 1.20 : Screw-Gear Type Universal Testing Machine

Upper CrossHead

Columns

Space for tension

Specimen

Middle Cross Head

Space for CompressionSpecimen

Lower Cross Head

Piston

Positioning Screw

Hydraulic Capsule

Pump

Motor

Bourdon TubeGauge

Scale

Figure 1.21 : Schematic of Hydraulic Universal Testing Machine

SAQ 6

(a) What is the difference between a universal and special purpose machine?Name a few special purpose machines. [Special purpose testing machines :Torsion testing machines, Compression testing machine, Harness tester,Impact testing machine.]

(b) Show a compression test specimen placed in a universal testing machine.

(c) What are the difference between a conventional universal testing machine

and modern computer controlled testing machine?

(d) What devices are used for recording load and elongation in universal testingmachines?

(e) A tension test specimen records 40,000 N force over a period of min.During this time the initial gauge length of 100 mm changes to 120 mm. Ifthe mechanical efficiency of the machine is 80%, calculate the powerconsumed by driving motor. (16.6 W).

Effect of Strain Rate on Tensile Properties

Static tensile tests are often performed at very low strain rate [10 6 mm/mm(sec)] and variations permissible in conventional testing machines are generally solimited that they do not influence the static properties. But machines capable ofvarying strain rate by as much as 100,000 time have been developed and tensiletests in such machines have revealed that tensile properties are quite susceptible tochanges under increasing strain rates.

In general it is observed that yield strength and ultimate tensile strength increasewith increasing strain rates, percentage elongation seems to first increase withstrain rate but then decreases of remains constant. One important influence of highstrain rate is that for such material as low-carbon steels which do not show yield

point in normal tensile tests, yield point appears. Further, yield strength out of allstatic properties is not affected by strain rate. Figures 1.22 and 1.23 show howthese properties are influenced by strain rate.

The effect of strain rate on mechanical properties become more pronounced atelevated temperatures.


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and Their DeterminationFigure 1.24 : Charpy Impact Testing Machine

1.6 NOTCHED-BAR IMPACT TESTS

Several tests, in recent past, have been developed to determine the tendency of materialto fracture in brittle manner. Most tests are performed on bar specimens which carry

notches in most critically stressed zone under impact [strain rate 103 mm/mm (sec)].

Carrying out the impact tests on notched bar samples at low temperatures renders mostsevere conditions which might exist in service.

One important factor which separates impact test from static tests is the great difficulty oreven impossibility of stress calculation at the tip of the notch under impact load. Thetriaxial state of stress that exists around notch tip under static load is modifiedconsiderably under impact load. Due to this difficulty the result of impact tests are not

presented in form of stress as is customary in static test. Since the relative magnitudes ofthe stress components at notch tip depend upon the dimension of notch and specimen, theimpact test specimens are standarised so that results from different sources may becompared.

The property that is measured in impact test is the energy absorbed in fracturing the

specimen of standard dimensions and standard notch. The property, measured in kgf-m,or joules is also often referred to as impact toughness.

Since impact toughness is not a stress and it cannot be generalised for a material butremains property of specimen of a material, it is not possible to incorporate this propertyin design directly. on the other hand, this property is a good qualitative index of

behaviour of material in presence of notch and at low temperature, and such an index isnot possible to obtain from any static test, therefore, this property of material is used inselection of material (particularly steel) and for development of material for specific

purposes of inhibiting the tendency of brittle fracture.

Although the impact test can be used for any material, yet it is steel (particularly thestructural steel) for which this is universally used. This is because of the fact that steel isthe predominant material used for structures as well as due to the fact that steel behaviouris sharply influenced by variation in temperature and strain rate.

Charpy Impact Test

The Charpy impact test is performed on a square cross-section specimen, having anotch on one side in the central cross-section. The specimen is placed in charpyimpact testing machine shown in Figure 1.24 such that swinging pendulum strikesthe specimen at the central cross-section but on the opposite side of the notch. Thespecimen is simply supported on the platform of the machine as illustrated inFigure 1.25(a). The commonly used impact specimen is 10 mm square cross-section beam of 55 mm length, supported over 40 mm span and having a V-notch

in the central plane whose included angle is 45

o

, depth is 2 mm and root radius is0.25 mm.

ImpactLoad

2 mm22 mm

(a) (b)

Figure 1.25 : Impact Test Specimens

The charpy test machine shown in Figure 1.24 is rigid and strong structure of twocolumns on a heavy base. The columns carry a heavy swinging pendulum at theirtop which swings on a frictionless pin, and support platform at the bottom for the

ImpactLoad

45

55 mm

40 mm

10 mm


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Materials Applications specimen. When the pendulum is in its vertical position its striking edge is levelwith the central cross-section of the specimen. The mass of the striker (or hammer)is concentrated in vertical plane. A circular disc scale mounted centric with the pinof the pendulum reads its position but is often calibrated in terms of the potentialenergy of the pendulum.

To perform the experiment the specimen is first placed in position by the help of acentring device, the pendulum is raised to and held at its extreme position and then

allowed to fall. At its extreme position, h1 from the specimen, the pendulum haspotential energy Wh1 where Wis the weight of the pendulum. In this position thearm of the disc scale is in its extreme position, reading total amount of energystored in the hammer. The speed of approximately released from its raised positionand strikes the specimen at a speed of approximately 4.8 m/sec. Under thiscondition the specimen is forced to bend with a rate of 10

3mm/mm (sec) resulting

into instantaneous fracture or plastic deformation. The pendulum continues itsswing after fracturing or deforming the specimen and rises to height ofh2.Thereafter, the pendulum swings back to oscillate pointer which does not swing

back with the pendulum. As already pointed out the disc scale is calibrated to readenergy and the position of the dead pointer reads the difference of potentialenergies of the pendulum in two extreme positions corresponding to h1 andh2.

Naturally this difference of the energy is that which was absorbed in fracturing thespecimen. Thus ifUfdenotes the impact roughness,

Uf= W(h1 h2) . . . (1.17)

It may, however, be recognised that some energy loss may occur in the bearing ofthe pendulum and also due to resistance of air offered to swinging pendulum.These two losses may be determined by simple experiment in which the pendulumis allowed to fall from is extreme position without placing the specimen in its path.The other extreme position to which pendulum rises will be slightly lower than itsinitial position. The difference of the potential energies in two positions, indicated

by the position of the dead pointer, is the loss of energy to bearing friction and air

resistance. This energy loss (UL) has to be subtracted from Uf to obtain correctedvalue of impact toughness, i.e.

Uf= Wh1 Wh2 UL . . . (1.18)

Izod Impact Test

Izod impact test specimen, shown in Figure 1.25(b), is either circular or square incross-section and contains a V-notch at one end. The specimen is clamped atnotched end in vertical plane as cantilever beam. The swinging hammer strikes thefree end at a distance of 22 mm from the notch. The general construction of Izodimpact testing machine is same as that of Charpy machine except the hammer. Themass of hammer in Charpy machine is distributed in a vertical plane whereas inIzod machine it is distributed in horizontal plane. Now-a-days same hammer is

being used in machine with attachable strikers for charpy and izod specimens.

SAQ 7

(a) How does strain rate influence yield strength, ultimate tensile strength andpercent elongation?

(b) Distinguish between shapes of hammers used in Izod and Charpy impacttests.

(c) Distinguish between types of loading and specimens for Charpy and Izodimpact tests.

(d) What are the factors that make material behave in brittle manner?

(e) A hammer weighing 50 N at the end of a swinging arm of length 800 mm islifted to a height of 1500 mm from the level of Charpy test specimen. Withwhat speed the hammer will strike the specimen.


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and Their Determination[Equate potential energy, mgh with K. E,21 mv

2= and calculate

15002 2 9.81 5.42 m/s

1000v g h= = = .

If the height of specimen from ground level is known, then the PE ofhammer in lower most position may be subtracted from KE.]

(f) The hammer of (e) is released and no specimen is placed in its path. Thehammer rises to a height of 1450 mm. How much energy is lost in frictionand air resistance? [2.5 N-m]

1.7 TEMPERATURE EFFECTS

Normally an impact test at room temperature reveals but little about brittle fracturetendencies of steel. It will become useful if the tests are performed at severaltemperatures below room temperature. Figure 1.26 depicts how the impact toughness oftwo steels vary with temperature. While it can be seen that energy absorbed in fracturereduces sharply as the temperature decreases, it also becomes clear that two steels may

behave quite differently with temperatures.

Considering steel A, it is observed that over a narrow zone of temperature (between zeroand 20oC) the material changes its behaviour completely from ductile to brittle. (Highenergy absorption is due to ductile fracture). This temperature range is known astransition range and is centred around a temperature which is termed as transitiontemperature (Tr). However, temperature Tr(Figure 1.26) is often calledductility

transition temperature. A lower transition temperature is more desirable in practice as itwill guarantee that brittle fracture will not occur within service temperature range.

An important conclusion that can be drawn from Figure 1.26 is the room temperatureimpact tests alone are of little consequence. In this figure steel B shows low impacttoughness at room temperature but a lower transition temperature, indicating itsusefulness in the range of 20 to 0oC whereas steel A might fail in brittle manner. Thusto understand the material tendency to behave in a brittle manner it is essential todetermine the ductile to brittle transition range in terms of temperature.

-40 -20 0 20 40

15

30

45

60

Temperature ( C)

ImpactToughness

(kg-m)

BrittleFracture

DuctileFracture

Transition

A

B

Tr

Figure 1.26 : Transition in Fracture Mode under Impact Loading

Presence of notch provided one condition for brittle fracture. The qualitative behaviour ofmaterial can be understood by accepting the fact that the material has distinct propertiesasyield strength andfracture strength. Both of these properties decrease with increasingtemperature, reducing strain rate and triaxility. However, the reduction rate of fracturestrength is much less than that of yield strength, and hence at certain point these curves


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intersect as shown in Figure 1.27. Thus the yield strength which is initially higher thanfracture strength becomes lower after the point of intersection. The specimen loaded onright hand side of the ordinate CA will first reach yield stress and will plastically deform.Contrarily the specimen loaded on left hand side of CA will first reach fracture stress andwill thus fail in a brittle manner without showing any plastic deformation. The line CAcan be called the line of transition from brittle to ductile. In practice CA becomes a bandin which change from brittle to ductile behaviour takes place very fast and its width

depends upon the nature of material.


Figure 1.27 : Various of Yield and Fracture Strengths with Temperature, Triaxiality and Strain Rate

After each impact test, the fracture parts must be explained for details. A brittle fracture,as below temperature Trexhibits a granular surface which is characteristics of cleavagefracture and absence of any shear deformation. As the temperature rises the part of thenotched section, the region near the surface, tends to fracture in shear mode wherebyenergy absorption in fracture increases. This shear deformation results in transversecontraction of notch section and the percent contraction is sometime reported to indicatethe ductility of the material. The fracture surface on which failure occurs due to shear ismarked by fibrous appearance and considerably transverse contraction in the notched

section. Figure 1.28 illustrates the fracture surface.

(a) Granular surface,

(b) Mixed granular and fibrous appearance with some side contraction, and

(c) Mixed fracture with considerably fibrous appearance and side contraction,2dis the total transverse contraction at the notch.

(a) (b)

Figure 1.28 : Impact Fracture Surface Appearance

Other Factors Affecting Properties

In addition to temperature, there are several other factor that affect the impacttoughness. One important factor is the machine itself. Each machine will absorbsome energy when the specimen is struck by pendulum and this amount thoughvery small will vary from machine to machine and thus the impact toughness will

be affected. The notch root radius is starndarised and any deviation from it will

influence the result. A sharper radius will tend to reduce the impact toughnesswhile a coarser radius will increase the amount of energy absorbed during fracture.

Metallurgical Factors

Yield

Fracture

C

A

Temperature

TriaxialityStrain Rate

Strength

d d

Notch

FibrousAppearance

Granular

Surface Contraction ofthe Notch


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and Their DeterminationTwo alloying elements, namely carbon and manganese are highly effective inaltering the impact toughness of steel. Each 0.1% increase in carbon percentageraises the ductility transition temperature by about 14oC. Each 0.1% addition ofmanganese in steel reduces the transition temperature by about 5oC. For asatisfactory notch impact toughness a 3 : 1 ratio for manganese percentage to thatof carbon is often suggested. Phorphorous also raises the transition temperature,and steel with high phosphorous content is not preferred at low temperature.

Nickel generally has beneficial effect on notch impact property while chromiumhas little effect. Silicon tends to increase the transition temperature if in excess of0.25%. Nitrogen has detrimental effect on notch toughness, though it is hard todetermine its effect because of interaction with other elements. Molybdenumincreases the ductility transition temperature almost as rapidly as carbon. The

presence of oxygen in steel is highly dangerous as it raises the transitiontemperature in the higher temperature range. Deoxidation practice as followed in

production ofkilledandsemi-killedsteels is highly beneficial in improving impactproperties.

Grain size has pronounced effect on transition temperature. In general thetransition temperature reduces with decreasing grain size. This effect is observed

both in mild steel and higher alloyed steel. Since in thick hot rolled plates it is notpossible to obtain uniform fine grain size throughout the thickness, the transitiontemperature will be appreciably higher. The loss of fracture energy in specimen ofthicker section is also due to geometrical factors, such as finding a weaker point iseasier in the large section than in smaller section of the same material.

SAQ 8

(a) Impact test is normally performed on steel. Give reasons.

(b) Define transition temperature and explain its usefulness in selecting materialfor ships.

(c) Explain how yield strength and fracture strength are affected by decreasing

temperature and increasing strain rate and triaxiality of stress.(d) A steel has transition temperature at 0oC. The transition range is over 10oC.

Show the appearance of impact fracture surface of charpy specimens testedrespectively at 5oC and 5oC.

(e) The alloying elements in steel affect impact strength. What are the effects ofC, Mn, P and Si?

(f) What is killed steel? Why this treatment is given on steel?

(g) Which of the following will decrease the transition temperature? Isdecreasing of transition temperature advantageous? Explain.

(i) Carbon addition

(ii) Mn addition

(iii) Mo addition

(iv) Deoxidation

(v) Decreasing grain size.

1.8 HARDNESS

The technological use of material has made the term hardness very common but it stillremains a poorly defined material property. Even a layman can differentiate between aharder and softer material, but it may be difficult to designate a material by its


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quantitative hardness because it is difficult to correlate the property with stress or strain.In general hardness may be understood as a property which provides resistance to

permanent plastic deformation. Such a deformation may leave a permanent impression onthe material surface. For engineering purpose the property of hardness is defined asresistance to indention orscratching. Based upon this two methods of hardnessmeasurement, viz.


(a) scratch hardness, and

(b) indentation hardness

are used.

Ability of material to resist abrasion, cutting orpenetration is also attributed to hardness.

1.8.1 Scratch Hardness

The scratch hardness is commonly measured by Mohs test. The method is based upon thescale of hardness, known as Mohs scale, in which ten standard minerals are arranged inthe order of their ability to be scratched. The softest mineral with scratch hardness of 1 istalc whereas the hardest mineral, diamond, has a hardness of 10 on Mohs scale. The otherminerals between them in increasing order of hardness are gypsum, calcite, fluorite,

appetite, feldspar, quartz, topaz, sapphier and corundum. The method of determiningMohs hardness is to compare a scratch on the material surface, produced understandardised conditions with the scratch on one of the minerals occupying position from1 to 10 on Mohs scale. Although the test had been chiefly designed for minerals, anymetal can be tested by this method. Most metals fall in mohs hardness range of 4 to 18and it is difficult to differentiate between two metals whose hardness are close. Becauseof this reason this method is not well suited for metals.

1.8.2 Indentation Hardness

Indentation hardness measurement is one of several but this is the most widely used andacceptable method of hardness measurement. Several parts of machines and structures,such as gears, axles and rails, are subjected to service requirements where high resistance

to indentation or abrasion is required. Indentation hardness measurement method uses anindenter of a specified material, shape and size which is pushed against the materialsurface under specified condition of loading to create an indentation. Then thedimensions of indentation are used to designate the hardness of the material.

1.8.3 Brinell Hardness

This is a method of indentation hardness, which since its introduction in 1990 byJ. A. Brinell has been widely accepted for technological purposes. This method uses asteel ball of 10 mm diameter as indenter. The indenter is first placed upon the surfacewhose hardness is to be measured and then a gradually increasing load of 3000 kgf isapplied upon the indenter. When the load is removed an indentation is left upon the

surface. For softer metals a lower load of 500 kgf is applied to avoid very deepindentation and for harder metals the steel ball is replaced by a tungsten carbide indenterso that indenter is not deformed. In each case the load is applied for a standard time,usually 30 sec. Figure 1.29 shows the indenter and indented plate.

The diameter of indentation is measured by a low power microscope. From theknowledge of the indentation diameter and diameter of the spherical surface of theindentation (which will be equal to the diameter of the ball) the area of the sphericalsurface of the indentation can be calculated. Then the ratio of the load which caused theindentation to the area of indented surface is defined as the Brinell hardness number(BHN).

d

D

Indenter

Specimen

Load P


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Figure 1.29 : The Specimen and Ball Indenter for Brinell Hardness

If, P = Applied load, kgf,

D = diameter of the ball, mm, andd= diameter of the indentation, mm.

then,2 2( )

2

PBHN

DD D d

=

. . . (1.19)

Although, from the definition and Eq. (1.19) it is obvious BHN has units of kgf/mm2, thisdoes not convey and physical meaning because the loadP is not uniformly distributedover the indented surface.

The BHN value obtained for a material is fairly reproducible provided the loadP anddiameterD are same for each test. To obtain the same BHN at any other load the ratiod/D must be constant which requires that P/D must be constant. Thus forP = 3000 kgf,D= 10 mm, P/D = 30. Hence if a load of 2000 kgf is to be used the ball dia must be8.15 mm, and for a 1 mm diameter indenter a load of 30 kgf must be used.

The calculated value of BHN may be slightly in error because of the assumption that thediameter of indented surface is same as that of indenter which is not exactly true. Afterthe removal of the load, some elastic recovery takes place whereby the final diameter ofthe surface becomes different than ball diameter. For harder materials elastic recovery isgreater than for softer materials.

An anomaly may arise in measurement of diameter of indentation diameter, d, due tolocalized deformation of metal in the region of indentation. Figure 1.30(a) shows the

cross-section through indentation in a cold-worked metal. Because such a metal has littlecapacity to strain harden the metal has a tendency to pile up along the edge ofimpression. This behaviour is known as pilling up or ridging. Although the measureddis larger than actual diameter of indentation, yet the hardness is calculated on the basisof measured d as shown in Figure 1.30(a) because the ridge also carries part of the load.

Figure 1.30(b) shows the sinking in of an annealed metal. The depression around theedge of metal will apparently increase the diameter of the impression. To arrive at thetrue value of d, the indenter is often coated by a dye before making the indentation. Thediameter then can be measured on the coloured ring.

d

(a) (b)

Figure 1.30 : Cross-section through Brinell Indentation on Two different Materials Showing

(a) Ridging; and (b) Sinking In

While making indentation on a plate specimen care must be exercised to avoid anyinterference between the indentation and the edge of specimen and the interference

between two indentations. For this reason it is advisable that the centre of the indentationshould be at least 1.5 D away from any edge. Same distance should be maintained

between two neighbouring indentations. Similarly the thickness of the plate must be atleast equal to the diameter of indenter so that the plastically deformed zone below theindentation does not interfere with the back surface.


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1.8.4 Rockwell HardnessMaterials Applications

Rockwell hardness is another method of hardness measurement, which like Brinellhardness, is very popular in engineering practice. In this method two types of indenter,viz., a Brale indenter in form of 120o diamond cone or steel balls of 1/16 in and 1/8 indiameter are used. These indenters are shown in Figure 1.31. Rockwell method differsfrom Brinell method because Rockwell hardness is not measured as ratio of load toindentation surface area. Instead, Rockwell hardness is measured on an arbitrary scale on

which hardness number is inversely proportional to depth of indentation.

The method consists in seating the specimen on platform of the machine and thenapplying a minor load of 10 kgf on the specimen through the indenter. This minor loadhelps to seat the specimen and reduces the tendency of riding or sinking in. A major loadwhich could be either 150, 100 or 60 kgf is then applied and the depth of indentation isautomatically recorded as hardness number on a dial gauge. The dial of the gauge isdivided into 100 division with each division representing a penetration of 0.002 mm. Thedial scale is reversed so that a small indentation depth is read as high hardness.

Rockwell Scale Load Hardness

A 60 kgf RA = 100 500 d

C 150 kgf RC = 100 500 d

D 100 kgf RD = 100 500 d

120o

d

(a)

Rockwell

Scale

Indenter

Diamete

r

Load Hardness

B 1/16 in 60 kgf RB = 100 500 d

F 1/16 in 60 kgf RF = 100 500 d

G 1/16 in 150 kgf RG = 100 500 d

E 1/8 in 100 kgf RE= 100 500 d

d

(b)

Figure 1.31 : (a) Brale Indenter (Diamond Cone); and

(b) The Spherical Ball Indenter of Steel,dis the Depth of Penetration

One combination of load and indenter will not be able to produce a wide range ofhardness. Therefore, loads with three indenters are used in different combinations to

provide wide range of measure hardness of several materials. Figure 1.31 shows thedifferent scales of Rockwell hardness which is often expressed asR with a suffixA,B orCto indicate the scale. The scale (RC) is very commonly used for steel.

The Rockwell hardness is many times preferred over Brinell hardness because this testcan be performed on very small parts in finished or unfinished condition. Since theindentation is very small, the finished surface is not spoilt and unfinished surface does

not affect the indentation. Also since no measurement of indentation and consequentcalculation are needed, the method is direct, fast and free from personal error.

However, several precautions regarding specimen and indentation have to the observed.The thickness of the plate specimen must be at least ten times the depth of indentation.This will avoid any effect of indentation to pass through the thickness. The distance

between two adjacent impressions should at least be three times the size of indentation.The hardness read from indenting the curved surface should be corrected for curvature.The surface on which indentation is made must be clean and smooth and it should be wellseated upon a clean platform. The rate of load application is controlled in Rockwellhardness testing machine and it is achieved by a dashpot adjustment corresponding to astandard rate of loading.

SAQ 9

(a) What do you understand by hardness? Arrange following substances fromhardest to softest.


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and Their DeterminationDiamond, steel, copper, gypsum, corundum, calcite.

(b) What is Mhos hardness scale? Why is it difficult to measure hardness ofsteel on Mohs scale?

(c) How is hardness measured?

(d) What is Brinell hardness? If for a ball of 10 mm diameter a load of 3000 kgfis applied on the ball for measuring BHN, what force will have to be applied

upon a 5 mm ball?

(e) Is Rockwell hardness measured by indenting the surface? A diamond coneindenter under a load of 100 kgf penetrates through 0.15 mm in steel A but0.1 in steel B. Is A harder than B? Explain.

(f) Describe the procedure for finding Rockwell hardness.

(g) What limitations are placed on making ball indentation in a 12 mm thicksteel plate in Brinell hardness measurement?

(h) Explain ridging and sinking in.

(i) Compare Rockwell hardness with Brinell hardness measurement in respect

of a steel plate 5 mm thick.

1.8.5 Vickers Hardness

Basically Vickers hardness is same as Brinell hardness in that an indentation is madeunder a loadP and the Vickers hardness number is defined as the ratio of loadP to thesurface area of the indentation. The difference between two test methods lies in use ofindenter. The Vickers indenter is a diamond pyramid with square base and included angle

between opposite faces as 136o (Figure 1.32). The impression of this indenter has squareboundary with depression inside. The lengths of two diagonals are measured undermicroscope and average is used to calculate the surface area of indentation.

Thus, Vickers hardness number,2

2 sin / 2 1.854PVHN

d d

= +

2

P. . . (1.20)

where, P is the indenting load in kgf. dis the average of two diagonal and is the anglebetween opposite faces of diamond come which is 136o.

135o Diamond Indenter

d d

Indentation

Figure 1.32 : Diamond Indenter and its Square Indentation in Vickers Hardness Test

The impression of the diamond cone is square under any load, hence the impressions aregeometrically similar irrespective of load. This provides a great convenience of acontinuous scale from very soft material to very hard material without changing the load

or indenter. Generally for a given load hardness from 5 to 1500 VHN can be measuredwith Vickers indenter, and it makes VHN independent of load except at very high loads.However, loads between 1 kgf to 120 kgf are commonly used for VHN measurement forvery soft to very hard metals. For the reason of independence of VHN of load, thismethod has widely been accepted as a research tool.


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and Their DeterminationYet another method of dynamic hardness measurement was developed by Martel. Heused apyramidal indenter upon which a ram falls to produce an indentation of samplesurface under dynamic loading condition. The Martel Hardness Number (MHN) isdefined as :

WhMHN

V= . . . (1.22)

where, W= weight of falling arm,h = height of fall, and

V= volume of indentation.

For paving materials the hardness measurement method has been developed in whichwear and abrasion of samples is measured. A number of preweighed pieces ofconcentrate aggregate of bricks are placed in drum which is subsequently rotated for afixed number of cy

Unit-1 Mechanical Properties

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