Course Title : metallic condensed matters

Lecture 4 Mechanical properties

condensed matters

Lecture 4. Mechanical properties of metallic materials and

mechanical test techniques

Lect e Plan

mechanical test techniques

Lecture Plan:1. Concepts of stress and strain. Mechanical tests 2. Elastic deformation3. Plastic deformation4. Strain & stress diagram5 Design safety factors5. Design safety factors

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The fist problem is to ensure the safe service of mechanisms andconstructions those are usually under load and prevent their fracture and thesecond is to fabricate new structural materials with desired mechanicalsecond is to fabricate new structural materials with desired mechanicalproperties and microstructure. At that the key mechanical design properties arestiffness, strength, hardness, ductility, and toughness.

The mechanical properties of materials are investigated by performingp p g y p gcarefully designed standardized laboratory experiments that replicate as nearlyas possible the service conditions. Factors to be considered include the nature ofthe applied load and duration of test, as well as the environmental conditions.

Important Terms and ConceptsSample - design stress -

proportional limit - ( resilience

ductility - elastic deformation

safe stress - shear - stiffness - tensile strength -

elastic recovery - engineering strain

tensile strength toughness ,

() engineering stress () hardness -

true strain true stress yielding - hardness

modulus of elasticity Poissons ratio

yielding yield strength -

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CONCEPTS OF STRESS AND STRAIN

A load is static when all changes are relatively slow with time and the load isA load is static when all changes are relatively slow with time and the load isapplied uniformly over a cross section or surface of a sample. In this case themechanical behavior may be defined by a simple stressstrain test; these aremost commonly conducted for metals at room temperature. There are threeprincipal ways in which a load may be applied: namely, tension, compression,shear. Consider each kind of mechanical test in more detail.

Tensile TestsTensile Tests

(c) The specimen is elongated by

(a) (b) (c)is elongated by the moving crosshead; load cell and

t t extensometer measure, respectively, the magnitude of the magnitude of the applied load and the elongation.

Fig 1 (a) Schematic illustration of how a tensile load produces an Fig.1 (a) Schematic illustration of how a tensile load produces an elongation and positive linear strain. Dashed lines represent the shape

before deformation; solid lines, after deformation. (b) tensile-testing machine.

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Fig.2 A standard tensile specimens with circular and rectangular Fig.2 A standard tensile specimens with circular and rectangular cross section (cylindrical shape and plate shape, respectively).

One of the most common mechanical stressstrain tests is tensile test A One of the most common mechanical stress strain tests is tensile test. Aspecimen is deformed, usually to fracture, with a gradually increasing tensileload that is applied uniaxially along the long axis of a specimen. Tensile specimens can be with circular or rectangular cross sections. The

d b d h h ld f h lspecimen is mounted by its ends into the holding grips of the tensile testingmachine. This machine is designed to elongate the specimen at a constant rateand to continuously and simultaneously measure the instantaneous appliedload (with a load cell) and the resulting elongations (using an extensometer).load (with a load cell) and the resulting elongations (using an extensometer). A stressstrain test typically takes several minutes to perform and isdestructive; that is, the test specimen is permanently deformed and usuallyfractured. The output of such a tensile test is recorded (usually on a computer)

l d f l ias load or force versus elongation.4

Load and elongation are expressed in terms of respective parameters of

F

Load and elongation are expressed in terms of respective parameters ofengineering stress and engineering strain. Engineering stress is defined bythe relationship

(1) 0SF=

in which F is the instantaneous load applied perpendicular to the specimen crossin which F is the instantaneous load applied perpendicular to the specimen crosssection, in units of [N], and S0 is the original cross-sectional area before any loadis applied [m2]. The unit of engineering stress (referred to subsequently as juststress) is [MPa] (where 1 MPa = 106 N/m2).) [ ] ( / )

(2)0 lll iEngineering strain is defined according to

(2) 00

0

ll ==i

in which l0 is the original length before any load is applied and li is the currentin which l0 is the original length before any load is applied and li is the currentvalue of length. The difference li - l0 is denoted as l and is the deformationelongation or change in length at some instant, as referenced to the originallength. Engineering strain (or deformation) is unitless or can be expressed in %,f hi h i l i l i li d b 100for this the strain value is multiplied by 100.

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Compression TestsFig.3. Schematic illustration of how a compressive load Fig.3. Schematic illustration of how a compressive load produces contraction and a negative linear strain.

A compression test is conducted in a manner similar to the tensile test except that the force is compressive and the specimen

By convention, a compressive force is taken to be negative, which yields a

test, except that the force is compressive and the specimen contracts along the direction of the stress. Equations (1-2) are utilized to compute compressive stress and strain, respectively.

By convention, a compressive force is taken to be negative, which yields anegative stress. Furthermore, because l0 is greater than li , compressive strainscomputed from (2) are necessarily also negative. Compressive tests are usedwhen a materials behavior under large and permanent (i.e., plastic) strains isd i d i f i li i h h i l i b i l idesired, as in manufacturing applications, or when the material is brittle intension.

Shear and Torsional Tests

Fig.4 (c) Schematic representation of shear strain (d) Schematic shear strain . (d) Schematic representation of torsional deformation (i.e., angle of twist) produced by an applied torque T.

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Th h t t i h i Fi 4 H th l f f th b i fi d The shear test is shown in Fig. 4c. Here the lower face of the cube is fixed rigidly and the force is imposed parallel to the upper face of the cube. The resulting shear stress is calculated according to

, 0SF= )3( tan =

0

where F is the load or force imposed, S0 an area of upper or bottom face.

Torsion is a variation of pure shear, wherein a sample is twisted: torsionalforces produce a rotational motion about the longitudinal axis of one end of thesample relative to the other end. Examples of torsion are found for machine

l d d i h ft d l f t i t d ill T i l t t llaxles and drive shafts, and also for twist drills. Torsional tests are normallyperformed on cylindrical solid shafts or tubes. A shear stress is a function ofthe applied torque T, whereas shear strain is related to the angle of twist.

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Geometric Considerations of the Stress State

Fig.5 Schematic representation showing normal and shear () stresses that act on a plane oriented at an ( ) pangle relative to the plane taken perpendicular to the direction along which a pure tensile stress is applied.

The stress value depends on the orientation of the plane upon which the stress is taken to act. For example, consider the cylindrical tensile specimen that is subjected to a tensile stress applied parallel to its axis (Figure 4) Consider also the plane p-applied parallel to its axis (Figure 4). Consider also the plane pp that is oriented at some arbitrary angle relative to the plane of the specimen cross-section. The external stress can be expressed as a vector sum of a tensile stress normal to the p-p plane and a shear stress that acts parallel to this plane. and in terms of and , take the form

, 2

2cos1cos2

+== (4) 22sincossin

==

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STRESSSTRAIN BEHAVIOR : Elastic deformation

At l l l f t f t t l th t l i ti l t t iAt low levels of stress for most metals the stress value is proportional to strain

(5) E= This is Hookes law, where E (GPa) is the modulus of elasticity, or Youngs modulus. For most typical metals the magnitude of this modulus ranges between 45 GPa for Mg and 407 GPa for Wthis modulus ranges between 45 GPa for Mg, and 407 GPa for W.

Fig.6. Schematic stressstrain diagram showing linear (a) and nonlinear elastic deformation for

The slope of this linear segment is the modulus ofelasticity E - an indicator of materials resistance to

( )loading and unloading cycles.

yelastic deformation or stiffness. The greater themodulus, the stiffer the material, or the smaller theelastic strain that results from the application of a givenstress Elastic deformation is reversiblestress. Elastic deformation is reversible.

For some materials (gray cast iron, concrete, polymers)the elastic portion of the stressstrain curve is not linear.For this nonlinear behavior, either tangent ()or secant modulus ( ) is normally used.Tangent modulus is taken as the slope of the stressstraincurve at some point of stress whereas secant moduluscurve at some point of stress, whereas secant modulusrepresents the slope of a secant drawn from the origin tosome given point of the curve. 9

Table 1 Room-Temperature Elastic and Shear Moduli and Poissons Ratio for Various M t l All

Plot of modulus of elasticity versus temperature for tungsten, steel, and

aluminum.Metal Alloys

aluminum.

Values of the modulus E for ceramic materials are about the same as for metals; Values of the modulus E for ceramic materials are about the same as for metals; for polymers they are lower. With increasing temperature, the modulus of elasticity diminishes.

,kxl

lESFo

o == , o

o

lESk = lx =

1where

2

21)/( EdxFlSU ooe == Elastic energy 10

ELASTIC PROPERTIES OF MATERIALS

Fig. 7 Axial (z) elongation (positive strain) and lateral (x and y) contractions (negative strains) in response to an imposed tensile stress Solid lines represent dimensions after

YX stress. Solid lines represent dimensions after stress application; dashed lines, before.

Z

Y

Z

Xv ==Theoretically Poissons ratio for isotropic materials should be 1/4 Theoretically, Poissons ratio for isotropic materials should be 1/4 ; the maximum value (or that value for which there is no net volume change) is 0.50. For many metals and other alloys, values of Poissons ratio range between 0.25 and 0.35. g

For isotropic materials, shear or bulk and elastic moduli are connected with Poissons ratio by the relationwith Poisson s ratio by the relation

( ) )21(312 vBvGE =+=11

Elastic constants for homogeneous materials

Shear stress and strain are proportional to each other

/ =Gwhere G is the shear modulus, the slope of the linear elastic region of the shear stressstrain curve.

The bulk modulus (B) of a substance measures the substance's resistance to uniform compression.

dP;

dVdPVB =

EEG)21(3)3(3 v

EEG

EGB == ( ) )21(312 vBvGE =+=

)1(2)21(3

93

vvB

EBBEG +

== )3(2231

263

GBGB

GE

BEBv +

===

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Plastic DeformationFor most metallic materials elastic deformation persists only to strains of For most metallic materials, elastic deformation persists only to strains of

about 0.005. As the material is deformed beyond this point, the stress is no longer proportional to strain, and permanent, nonrecoverable, i.e. plastic deformation occurs. From an atomic perspective, plastic deformation corresponds to the p p p pbreaking of bonds with original atom neighbors and then re-forming bonds with new neighbors as large numbers of atoms or molecules move relative to one another; upon removal of the stress they do not return to their original positions.

For crystalline solids deformation is accomplished by means of a process For crystalline solids, deformation is accomplished by means of a process called slip, which involves the motion of dislocations. Plastic deformation in noncrystalline solids (as well as liquids) occurs by a viscous flow mechanism.

Fig. 9 (a) Typical stressstrain behavior for a metal showing elastic and plastic d f i h i l li i P d deformations, the proportional limit P, and the yield strength as determined using the 0.002 strain offset method. (b) Representative stressstrain behavior (b) Representative stress strain behavior found for some steels demonstrating the yield point phenomenon. ( )

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TENSILE PROPERTIES Yielding and Yield Strength

Figure 10. Typical yield behavior for non-ferrous alloys.

1: True elastic limit2: Proportionality limit3: Elastic limit3: Elastic limit4: Yield strength

True elastic limit : The lowest stress at which dislocations move. Thisdefinition is rarely used, since dislocations move at very low stresses, andde t o s a e y used, s ce d s ocat o s o e at e y o st esses, a ddetecting such movement is very difficult.

Proportionality limit : Point P (Fig. 9a). Up to this point, stress isti l t t i (H k l ) th t t i h i t i htproportional to strain (Hookes law), so the stress-strain graph is a straight

line, and the gradient will be equal to the elastic modulus of the material.

Elastic limit : Beyond the elastic limit, permanent deformation will occur. They , plowest stress at which permanent deformation can be measured.

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Yield point or yield strength (proof stress ) : The position of this point P is difficult to measure precisely.

h b bl h d h hAs a consequence, a convention has been established wherein a straight line is constructed parallel to the elastic portion of the stressstrain curve at some specified strain offset, usually 0.002. The stress corresponding to the intersection of this line and the stressstrain curve as it bends over in the intersection of this line and the stress strain curve as it bends over in the plastic region is defined as the yield strength Y [MPa]. Upper yield point and lower yield point :Some metals, such as mild steel ( ), demonstrate the onset of plastic deformation as shown in Fig 9b. At the upper yield point, plastic deformation is initiated with an apparent decrease in engineering stress Continued deformation fluctuates slightly about some in engineering stress. Continued deformation fluctuates slightly about some constant stress value, termed the lower yield point; stress subsequently rises with increasing strain. The material response is linear up until the upper yield point, but the lower yield point is used in structural engineering as a conservative value.

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Stress-strain diagram. Tensile Strength

Fig. 11 Typical engineering stressstrain behavior to fracture, point F. The tensile strength TS is fracture, point F. The tensile strength TS is indicated at point M. The circular insets represent the geometry of the deformed specimen at various points along the curve.

The tensile strength TS is the stress at the maximum on the engineering stressstrain curve.

All deformation up to this point is uniform. At the maximum stress, a small constriction or neck begins to form at some point, and all subsequent deformation is confined at this neck. If this stress continues to be applied, fracture will result in the neck.

Tensile strengths ( ) may vary anywhere from 50 g ( ) y y yMPa for an aluminum to as high as 3000 MPa for the high-strength steels. Ordinarily, when the strength of a metal is cited for design purposes, the yield strength is used.

The fracture strength corresponds to the stress at fracture. 16

Ductility is a measure of the degree of plastic deformation that has beensustained at fracture.Ductility may be expressed quantitatively as either percent elongation orpercent reduction in area. The percent elongation is the percentage of plasticstrain at fracture he e l is the f act e strain at fracture

%1000

0

= l

ll fwhere lf is the fracture length and l0 is the original gauge length as given earlier0

Percent reduction in area is defined as

%1000

=

SSS f

0 Swhere S0 is the original cross-sectional area and Sf is the cross-sectional area at the point

Fig. 12 Schematic representations of tensile stressstrain behavior for brittle and

of fracture.

strain behavior for brittle and ductile metals loaded to fracture. 17

Table 2 Typical Mechanical Properties of Several Metals and Alloys in an Annealed Statean Annealed State

Fig. 13Engineering stressstrain behavior for iron at three temperatures iron at three temperatures.

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Resilience is the capacity of a material to absorb energy when it is deformedelastically and then, upon unloading, to have this energy recovered (J/m3). They , p g, gy ( / )associated property is the modulus of resilience, Ur, which is the strain energyper unit volume required to stress a material from an unloaded state up to thepoint of yielding. Computationally, the modulus of resilience for a specimensubjected to a uniaxial tension test is just the area under the engineering stresssubjected to a uniaxial tension test is just the area under the engineering stressstrain curve taken to yielding (Fig.14) = Y dU r 0

Assuming a linear elastic region,

in which Y is the strain at yielding. U YYYr1 2 == Y y gEYYr 22

Thus this area under the stressstrain curve represents energy absorption per unit volume of material The energy absorption per unit volume of material. The resilient materials are those having high yield strengths and low moduli of elasticity; such alloys can be used in spring applications.

Fig.14 Schematic representation showing how modulus f l ( d h h d d )of resilience (corresponding to the shaded area) is

determined from the tensile stressstrain behavior of a material.

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ELASTIC RECOVERY AFTER PLASTIC DEFORMATION

Fig.15 Schematic tensile stressstrain diagram showing the phenomena of elastic strain recovery and strain hardening. The initial yield strength is designated as is the yield strength after releasing the designated as is the yield strength after releasing the load at point D, and then upon reloading.

During the unloading cycle, the curve traces a near t i ht li th f th i t f l di ( i t D) straight-line path from the point of unloading (point D),

and its slope is virtually identical to the modulus of elasticity, or parallel to the initial elastic portion of the curve. The magnitude of this elastic strain, which is

Toughness

cu e e ag tude o t s e ast c st a , c sregained during unloading, corresponds to the elastic strain recovery.

For the static deformation, a measure of toughness in metals (derived from plastic deformation) may be ascertained from the results of a tensile stressstrain test. It is the area under the - curve up to the point of fracture. The units are [energy per unit volume of material]. For a metal to be tough, it must display both strength and ductility. Hence, even though the brittle metal has higher yield and tensile strengths, it has a lower toughness than the ductile one as can be seen by comparing the has a lower toughness than the ductile one, as can be seen by comparing the areas ABC and ABC (Fig.12).

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TRUE STRESS AND STRAIN

True stress is defined as the load F divided by the instantaneous cross-sectional area Si over which deformation is occurring. True strain - T .

F l

If no volume change occurs during deformation

iT S

F=0l

ll iT n=

00ll SS ii =( ) ( )true and engineering stress and strain are related as

If no volume change occurs during deformation

( ) += 1T ( ) += 1nT lEquations are valid only to the onset of necking; the true stress necessary to

t i i i t i ti t i t th t il i t M C i id t ith sustain increasing strain continues to rise past the tensile point M. Coincident with the formation of a neck is the introduction of a complex stress state within the neck region (i.e., the existence of other stress components in addition to the axial stress). So the correct stress (axial) within the neck is slightly lower than the ) ( ) g ystress computed from the applied load and neck cross-sectional area. For some metals and alloys the region of the true stressstrain curve from the onset of plastic deformation to the point at which necking begins may be approximated by

Th t i ft t d th t i h d i nTT K = The parameter n is often termed the strain hardening exponent and has a value less than unity. 21

HARDNESS is a measure of a materials resistance to localized plastic deformation (e.g., a small dent or a scratch).

Hardness tests are performed more frequently than any other mechanical test for several reasons:1. They are simple and inexpensiveordinarily no special specimen need beprepared, and the testing apparatus is relatively inexpensive.2. The test is nondestructivethe specimen is neither fractured nor excessivelydeformed; a small indentation is the only deformation.3 Other mechanical properties often may be estimated from hardness data3. Other mechanical properties often may be estimated from hardness data,such as tensile strength.

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Correlation between Hardness and Tensile Strength

Both tensile strength and hardness are indicators of a metals resistance to plastic deformation. Consequently, they are roughly proportional, as shown in Figure 19, for tensile strength as a function of the HB for cast iron, steel, and b Th ti lit l ti hi d t h ld f ll t l Fi brass. The same proportionality relationship does not hold for all metals, as Figure 19 indicates.

As a rule of thumb for most steels, the HB and the tensile strength are related according to ( ) HBMPTS 453according to

Figure 17 Relationships between hardness and

( ) HBMPaTS = 45.3p

tensile strength for steel, brass, and cast iron

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DESIGN/SAFETY FACTORS

For less critical static situations and when tough materials are used a design For less critical static situations and when tough materials are used, a design stress, d, is taken as the calculated stress level c (on the basis of the estimated maximum load) multiplied by a design factor, N

where N is greater than unity. Thus, the material to be used for Cd N =

where N is greater than unity. Thus, the material to be used for the particular application is chosen so as to have a yield strength at least as high as this value of d.

Alternatively a safe stress or working stress is used instead of design stress

N/

Alternatively, a safe stress or working stress, w, is used instead of design stress. This safe stress is based on the yield strength of the material and is defined as the yield strength divided by a factor of safety, N, or

NYw / =The choice of an appropriate value of N is necessary. If N is too large, then

t d i ill lt th t i ith t h t i l llcomponent overdesign will result; that is, either too much material or an alloyhaving a higher-than-necessary strength will be used. Values normally rangebetween 1.2 and 4.0. Selection of N will depend on a number of factors,including economics, previous experience, the accuracy with which mechanicalincluding economics, previous experience, the accuracy with which mechanicalforces and material properties may be determined, and, most important, theconsequences of failure in terms of loss of life and/or property damage. Becauselarge N values lead to increased material cost and weight, structural designers

i t d i t h t i l ith d d t ( d i t bl )are moving toward using tougher materials with redundant (and inspectable)designs, where economically feasible. 24