materials for design - module 6

7/23/2019 Materials for Design - Module 6

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Table of Contents

6 Materials for Design 16.1 Material Properties . . . . . . . . . . . . . . . . . . . . . . . . . . 1

6.1.1 Mechanical Properties . . . . . . . . . . . . . . . . . . . . 16.1.2 Physical Properties . . . . . . . . . . . . . . . . . . . . . . 156.1.3 Electrical Properties . . . . . . . . . . . . . . . . . . . . . 166.1.4 Chemical Properties . . . . . . . . . . . . . . . . . . . . . 186.1.5 Environmental Properties . . . . . . . . . . . . . . . . . . 18

6.2 Processes Used to Alter the Properties of

Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186.2.1 Strain Hardening . . . . . . . . . . . . . . . . . . . . . . . 196.2.2 Forging . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246.2.3 Heat Treatment . . . . . . . . . . . . . . . . . . . . . . . . 24

6.3 Classification of Materials . . . . . . . . . . . . . . . . . . . . . . 256.3.1 Metals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256.3.2 Ceramics . . . . . . . . . . . . . . . . . . . . . . . . . . . 326.3.3 Polymers . . . . . . . . . . . . . . . . . . . . . . . . . . . 336.3.4 Composites . . . . . . . . . . . . . . . . . . . . . . . . . . 346.3.5 Semiconductors . . . . . . . . . . . . . . . . . . . . . . . . 346.3.6 New Materials . . . . . . . . . . . . . . . . . . . . . . . . 34

6.4 Machinability of Metals . . . . . . . . . . . . . . . . . . . . . . . 366.5 Material Selection . . . . . . . . . . . . . . . . . . . . . . . . . . 36

6.5.1 Material Performance Requirements . . . . . . . . . . . . 386.5.2 Material Performance Indicies . . . . . . . . . . . . . . . . 39

6.6 Numbering Systems for Metals and Alloys . . . . . . . . . . . . 446.6.1 SAE Numbering System . . . . . . . . . . . . . . . . . . 456.6.2 ASTM Standards . . . . . . . . . . . . . . . . . . . . . . . 456.6.3 Unified Numbering System (UNS) . . . . . . . . . . . . . 46

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MODULE 6Materials for Design

Materials engineering is a field of engineering that includes the range of ma-

terial kinds and how to use them in manufacturing. To convert the basic

materials into an engineering product, material should be selected with specific

properties. Material selection is a process that is performed to select the best

materials for specific application and product development. If an appropriate

material selection is not performed, the product life tends to be highly unpre-

dictable.

6.1 Material Properties

When selecting materials for a design, it is important to understand the prop-erties of a material. In this chapter, the following four main material propertieswill be discussed.

Mechanical, Physical, Chemical, and

Environmental.

6.1.1 Mechanical Properties

One of the simplest tests for determining some of the main mechanical propertiesof a material is the tensile test. In this test, a load P is applied along thelongitudinal axis of a circular test specimen shown in Figure 6.1(a). As shownin the figure, two gages are marked at a distance L0 apart. The distanceL0 iscalled the original gage length of the specimen. Original cross-sectional area ofthe central part of the specimen is denoted byA0.

During an experiment the applied load and the change in gage length aremeasured and converted to stresses and strains. The resulting stress-strain curve

in Figure 6.1(b) (not scaled) gives a direct indication of the material properties.IfL is the observed length corresponding to an applied load, P, the gage elon-gationL= LL0. The elongation per unit of the initial gage length is calledstrain, and given by

=L

L0(6.1)

As shown in Figure 6.1(b), as the strain increases, there are four well-definedregions with distinct types of behavior. In the interval O-A, the plot is straightline and materials show linear elastic behavior. In this region, deformations are

1


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2 Materials for Design

Figure 6.1: Result of tensile test.

fully recoverable when the applied load is removed. Beyond point A, for ductilematerials, deformations are plastic. A plastically deformed material will notreturn to its original size and shape when the load is removed. This region ofthe stress-strain curve is assumed to be horizontal. Within the interval B-u,stress increases with increasing strain although at a much lower rate than E.This is the strain hardening region. and finally, as the cross-sectional area ofthe specimen decreases due to plastic flow, at point u necking starts. As shownfrom the figure, from point u stress start decreasing until rupture occurs.

Elastic Modules

For elastic materials, the strain is proportional to the applied load and the slopeof the line O-A in the elastic region is designated by a constant, E and calledelastic modules or youngs modules (see Figure 6.1(b)).

Elastic Limit

Elastic limit is the maximum stress within a solid material that can reach beforethe onset of permanent deformation. When stresses are removed, the materialreturns its original size and shape. (see Figure 6.1(b)).

Yield Strength

The yield strength, Sy of a material is defined as the stress at which a materialbegins to deform plastically. Prior to the yield point the material will deformelastically (leaving 0.2% permanent deformation) and will return to its originalshape when the applied stress is removed.

Figure 6.2:

Offset method.

For some materials such as metals and plastics, the change from the linearelastic region cannot be easily seen. In such cases, as shown in Figure 6.2, anoffset methodto determine the yield strength of the material tested is used (seeASTM E8 for metals and D638 for plastics). As shown in Figure 6.2, an offset,OB is expressed as a % of strain. Usually value of offset is taken to be 0.2% formetals and plastics. The yield stress Sy is determined by drawing a line from


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Module 6. Materials for Design 3

Figure 6.3: Result of tensile test.

point B parallel to the straight line portion of the diagram. Intersection pointA becomes the Yield Strength by the offset method.

EXAMPLE 6.1

An initial part of the stress-strain curve is shown in Figure 6.3. As seenfrom the figure, material shows strong nonlinear behavior and it is difficultto identify yield point.

(a) Estimate the yield point, and(b) Estimate the value of elastic modulus.

SOLUTION

(a) Since it is difficult to identify the exact yield point on the curve shown inFigure 6.3, draw a parallel line from the 2% offset to line OA to indicatethe yield strength of 414 MPa.

(b) The slope of the line O-A in the elastic region is elastic module. Therefore,

E= tan = AB

OB =

207 00.001 0= 207 GPa

Ultimate Tensile Strength

Ultimate strength is the highest stress developed in material before rupture.Usually, changes in area due to changing load and neckingare ignored in deter-mining ultimate strength.


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Figure 6.4: Necking leading to fructure.

Rupture Strength

Rupture strength is the stress developed in a material at breaking point. It is notnecessarily equal to ultimate strength. Since necking is not taken into accountin determining rupture strength, it rarely indicates true stress at rupture.

Necking

Necking is a reduction in diameter in a small region of the material during thetensile deformation (see Figure 6.4). A significant amount of shearing occur inthe necking area. Basically, it is a region of local instability in the material.

Engineering Stress-Strain Curve

Assuming that the stress is constant over the crss-sectional area at the center ofthe specimen the engineering stress, e can be calculated by

e = P

A0(6.2)

where P is the applied load and A0 is the original cross-sectional area at thecenter of the specimen. Engineering strain, e is found directly from the sraingage reading or using Eq. 6.1. A0 is usually designated by A in calculatingengineering stress, e by using Eq. 6.2.

EXAMPLE 6.2

A 20 mm diameter steel bar is subjected to a load of 3,000 kg. Calculatethe engineering stress.

SOLUTION

Engineering stress is give by Eq.6.2

e= F

A0

where

F =ma = (3, 000 kg)(9.81 m/s2) = 29, 430 N


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and the original cross section area, A0 is

A0 =

4(20 103)2 = 314 106 m2

Then

e = 29, 430

314 106 = 93.7 MPa

True Stress-Strain Curve

The dotted line in Figure 6.1 shows the true stress-strain curve. When the strainsget large enough original cross-sectional area, A0 becomes small. Therefore, theactual true stress will not be same as the engineering stress. In such cases, thetotal true strain, t is equal to the sum of incremental strains, , hence

t =

=L

L (6.3)whereL is the current gage length of the specimen at the time the incrementalelongation,L occurs. Assuming that L0 is the initial gage length, if the limitexists, L=0, the total true strain, t ( at the time when the gage lengthreaches to final length, Lf) defined by the following integral

t =

LfL0

dL

L =ln

LfL0

(6.4)

Therefore, true strain,tis the instantaneous % change in length of test specimenin pull test. It is equal to the natural logarithm of the ratio of length, Lfat anyinstant to original length, L0.

The true stress,t is defined as the ratio of the applied load to the instanta-

neous cross-sectional area, A.

t =P

A (6.5)

Just before the necking, the strain is still uniform along the specimen length.Thus, we can assume that there is no volume change in the specimen. Therefore,

A0L0 = AL (6.6)

Using Eqs 6.5 and 6.6, we have

t = P

A0 L

L0=e

Lo+ L

L0=e(1 + e) (6.7)

Similarly, the relationship between engineering strain and true strain can befound as

t = ln(1 + e) (6.8)


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EXAMPLE 6.3

A steel tension test sample shown in Figure 6.5 has diameter of 7.62 mm atthe center. Before the sample is stressed, the length between two gage marks(gage length) is 30 mm. After the sample is elastically deformed, the gagelength is found to be 39 mm. Calculate the engineering and true strains.

SOLUTION

Using Eq.6.1, engineering strain, e is

Figure 6.5:

Tension test samplesbefore and after stressed.

e =L

L0 (100) =39 30

30 (100) = 30% elongation.

Using Eq.6.8, true strain, t is

t = ln(1 + e) = ln(1 + 0.30) = 0.26 = 26% elongation.

Fracture Toughness

Fracture toughness, KICis a very important material property that defines itsability to resist stress at the tip of a crack. This subject will be discussed indetail in Chapter 7 of this book.

Brittle and Ductile Behavior

Figure 6.6: Brittle and ductile materials behavior.

Material behavior can be classified into two categories. Namely, brittle and

ductile materials. In general, steel, aluminum, gold, silver and cupper are typicalexamples of ductile materials. Glass, concrete and cast iron are considered inthe class of brittle materials. Brittle and ductile materials can be distinguishedby comparing the stress-strain curves as shown in Figure 6.6.

As shown in Figure 6.6, ductile materials show larger strains before failure. Ingeneral, ductile materials failure is limited by their shear strengths. Ductile ma-terials often have relatively small Youngs modulus and ultimate stresses. Brittlematerials fail at much lower strains and material failure is limited by their tensilestrengths. Since ductile materials dont fail without warning as brittle materialsdo, they are preferable to brittle materials for building structural members.


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Figure 6.7: Block under tensile load.

Ductility is measured by % elongation at the fracture point, that is, how largea strain the material withstands before fracture. The most common measure of

ductility is the percent elongation to failure elongation = 100 (LfL0)/L0.Another measure of ductility is the percent reduction-of area at the necked regionof the specimen, measured after fracture, that is reduction of area (%RA) =100 (A0 Af)/A0.

Materials with 5% or more elongation are considered ductile. Ductile mate-rials undergo a series of steps that lead to fracture under tension. Namely:

Plastic deformation produces necking, Small cavities start in the interior of the material, The cavities join to form a crack, The crack propagates,

Fracture occurs, typically at a 45 degree angle to the tension as this is theangle of maximum shear stress.

Brittle fracture shows little if any plastic deformation and is a result of rapidcrack propagation. Brittle fracture occurs perpendicular to the applied tensionand often has a relatively flat fracture surface. Materials with less then 5%elongation are considered brittle.

Normal Strain

Consider a block shown in Figure 6.7. When it is subjected to a tensile load, Pthe amount of stretch or elongation is called strain. Since the elongation of theblock is assumed normal to the cross section, the elongation per unit length,

is called unit normal strain.

x= x

L, y =

y

L (6.9)

wherex and y are the total elongations or total strains in the x andy direc-tions, respectively and L is the original length of the block. Stresses created bythe strain is

= E (6.10)


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In Eq.6.10, modulus of elasticity, E is one of the most important mechan-ical properties of a material. Stresses created by the applied load can also becalculated by

=P

A (6.11)

WhereA is the original cross sectional area of the block. The elongation, x ofthe block shown in Figure 6.7 is directly proportional to the cross-sectional areaand the modulus of elasticity. This relationship is referred to as the Hookes law.Assuming that the change in cross-sectional area during the loading is neglected,this relationship is given by

x = P L

AE (6.12)

Experiments showed that when a material is under tension, there exists not onlyan axial strain but also lateral strain, y due to construction (negative strain).The ratio of lateral contraction strain to axial extension strain is called Poissonsratio,. The Poissons ratio is a dimensionless parameter that provides a goodunderstanding about the nature of the material. Poissons ratio is a material

property defined as

= yx

(6.13)

Shear Strain

As shown in Figure 6.8, change in angle is known as shear strains. There is nochange in length. The measure of shear strain is the change in angle, of astress element when pure shear load is applied. The expression of a shear stressis

Figure 6.8:

Stress element underpure shear load.

yx = G (6.14)

where the constant, G is a material property called shear modulus of elasticityor modulus of rigidity. Three elastic constants , E, and G are related to eachother as follows

E= 2G(1 + ) (6.15)

EXAMPLE 6.4

Calculate the applied force, Pto create 1% reduction in diameter of a steelbar shown in Figure 6.9. Assume that the original diameter of the bar is 12

mm, elastic modulus of the steel bar material, E=207 GPa, and the Poissonsratio,=0.3.

SOLUTION

Figure 6.9:

Steel bar.

Reduction in diameter in thex direction is x=12 0.01=0.12 mm. The strainin thex direction is

x= x

L = 0.12

12 = 0.01


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Using Eq.6.9 calculate the strain, y in they direction

y = x

= 0.010.3

= 0.033

Using Eq.6.10 the stress, that must be applied is

= Ex= 207 109(0.033) = 6.831 109 Pa

Then, using Eq.6.5 calculate the force which will create 1% reduction in diameterof the steel bar

P=A = 6.831 109

4

12 103

2= 7.72 105 N

Hardness

Hardness is the property of a metal and defined as the resistance of a materialto a permanent indention of particular geometry over a specified length of time.The greater the hardness of the metal, the greater resistance it has to permanentdeformation. Many methods have been developed for hardness testing. Thosemost often used by industry are Brinell, Rockwell, and Vickers hardnesses willbe discussed in this section. The three of these hardness measures are based onidentation hardness testing. A metals hardness is measured by the resistanceto the penetration of a non-deformable ball or cone. The tests determine thedepth of a ball or cone which will sink into the metal, under a specified load,within a certain period of time.

Brinell Hardness E10-12 standart test method for Brinell hardness (BHN)

of metalic materials is described by American Society for Testing & Materials(ASTM). Brinell hardness uses a 10 millimetres (0.39 in) diameter steel or tung-sten carbide ball as an indenter by applying specified load into the surface of amaterial. The test machine applies a load of 500 kilograms for soft metals suchas copper, brass and thin stock. A 1500 kilogram load is used for aluminumcastings, and a 3000 kilogram (29 kN; 6.600 lbf) load is used for materials suchas iron and steel. The unit of Brinell hardness, HB is kg/mm2. After the test,the diameter of the indentation is measured to be used in Eq.6.16 to calculatethe Brinell hardness.

BH N= P

2 D D

D2

d2

(6.16)

Figure 6.10:

Brinell hardness test.

where P is the applied force, D is the diameter of the ball and d is diameterof the indentation (see Figure 6.10). An approximate relationship between theBrinell hardness and the tensile strength, Sut for stell is given by1

Sut(MPa) =

3.55 HB (HB 175)3.38 HB (HB> 175) (6.17)

1Donghao Stainless Steel. http://www.stainless-steel-tube.org/Relation-of-Hardness-to-Other-Mechanical-Properties-Tensile-Strength-Yield-Strength.htm, accessed: July 20, 2013.


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Figure 6.11: Relationship between Brinell hardness and yield strength(Reproduced from reference, [2]).

Sut(psi) =

515HB (HB 175)490HB (HB> 175) (6.18)

Relationship between Brinell hardness and yield strength, Sy is given by2

H=c Sy (6.19)

wherec is a proportionality constant. For a perfectly plastic materials, the valueofc is approximately 3. This value is reasanably consistent with experiemntaldata given in Figure 6.11.

EXAMPLE 6.5

Estimate the Brinell hardness of a steel material if the diameter of the in-dentation is 5 mm.

SOLUTION

Since the material is steel, we use 10 mm standard diameter ball as the indenterand apply 3,000 kg force, P. From Eq.6.16, we have

BH N = P

2

D

D

D2 d2

= 3, 000

2 10

10

102 52

= 142.6 kgf/mm2 orHBFigure 6.12:Rockwell hardness test.

2Serope Kalpakjian and Steven R. Schmid, Manufacturing Processes for Engineering Ma-terials, Printice Hall, p. 38. 2008.


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Rockwell Hardness The Rockwell hardness test method, as described inASTM E-18, is the most commonly used hardness test method. The Rockwellhardness test is used to measure the resistance of a metal to penetration like theBrinell test, but in the Rockwell case, the depth of the impression is measuredinstead of the diametric area ( see Figure 6.12). The Rockwell test is usually eas-ier to perform, and more accurate than other types of hardness testing methods.

With this test method, different kind of indenters can be used: conical diamondwith a round tip for harder metals and ball indenters with diameters rangingfrom 1/16 inch to 1/2 inch for soft materials. Common Rockwell hardness scalesused for metals include A, B, and C corresponding to major loads of 60, 100,and 150 kg, respectively. The three most common scales used for plastics areRockwell E, Rockwell M, and Rockwell R.

Test procedure

The indenter moves down into position on the part surface A minor load (10 kg) is applied and a zero reference position is established The major load (60, 100, or 150 kg) is applied for a given time period

(dwell time)

The major load is released leaving the minor load applied

At the end of the test, material hardness is shown directly on the dial (depthgage) attached to the test machine.

Vickers Hardness The Vickers hardness test method, also known as a mi-crohardness test method, is mostly used for small parts, thin sections, or casedepth work. A square base pyramid shaped 136 pyramidal diamond indenteris used for testing in the Vickers scale (see Figure 6.13). Applied macro forcechanges between 1 kgf and 100 kgf (ASTM E92) whereas micro force rangesbetween 10 gf and 1 kgf (ASTM E384). The full load is normally applied for 10to 15 seconds. The Microhardness methods are used to test on metals, ceramics,composites and many other type of materials. Typical Vickers number rangesfrom HV 100 to HV1000 for metals and will increase as the material gets harder.Eq6.20 is used to calculate the Vickers hardness.

Figure 6.13:

Vickers hardness test.

HV =F

A 1.8544F

d2 (6.20)

where d is the arithmetic mean of the two diagonals, d1 and d2 in mm (seeFigur 6.13). The Vickers hardness is reported like 700 HV/5, which means aVickers hardness of 700, was obtained using a 5 kgf applied force during the test.

Table 6.1 shows the conversion of Brinell, Vickers, Rockwell B, and RockwellC hardnesses.3

Hardness Testing of Plastics The hardness testing of plastics is usuallymeasured by the Rockwell hardness test or Shorehardness test. Both methodsmeasure the resistance of the plastic to indentation, thus providing an empiricalhardness value. Rockwell hardness is usually used for harder plastics such asnylon, polycarbonate, polystyrene, and acetal where the resiliency or creep ofthe polymer may not affect the results. The three most common scales used forplastics are Rockwell E, Rockwell M, and Rockwell R.

3Engineers Handbook, http://www.engineershandbook.com/Tables/hardness.htm, ac-cessed: August 4, 2013


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Table 6.1: Hardness conversion.

Tensile Strength Brinell Vickers Rockweel B Rockwell C

(N/mm2) (BHN) (HV) (HRB) (HRC)

350 105 110 62.3

450 133 140 75.0 545 162 170 85.0

640 190 200 91.5

740 219 230 96.7

850 252 265 24.8

900 266 280 27.1

950 280 295 29.2

1030 304 320 32.2

1125 333 350 35.5

1220 361 380 38.8

1320 390 410 41.8

1420 418 440 44.51520 447 470 46.9

1630 475 500 49.1

1740 504 530 51.1

1845 532 560 53.0

1955 561 590 54.7

2070 589 620 56.3

2180 618 650 57.8

*Applies for plain carbon and low-alloy steels and cast steel and to a limited

extent for high-alloy and/or work hardened steel.

The Shore A00scale is used to measures hardness of very soft rubbers suchas gels. On the other hand, Shore Dhardness scale measures the hardness ofhard rubbers, semi-rigid plastics and hard plastics. Shore Hardness becomes amain factor when considering which mold rubber one should choose for makinga mold.

ASTM D2583 Barcol Hardness test method is used to determine the hardnessof reinforced and non-reinforced rigid plastics. The instrument used to testhardness is called the Barcol impressor, gives a direct reading on a 0 to 100 scalewith the typical range being between 50B and 90B.

Relationship of Shore hardness with Brinell hardness and Rockwell-C hard-ness is given in Figure 6.14.

Impact Resistance and Energy Absorbing Capacity

Impact resistancestandard test (ASTM Standard D-6110) basically measuresthe energy needed to fully fracture a notched specimen. Impact tests are usedin investigating the toughnessof material.

Toughness is the capacity of a material to absorb energy and plasticallydeform without having fracture. Toughness, UT (also called module of tough-ness) can be determined by taking the integral under the stress-strain curve (seeFigure 6.1) up to the fracture point, f. That is:


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Figure 6.14: Relationship of hardness numbers (reprinted with the permission ofInternational Nickel Co., NY.).


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UT =

f0

d (6.21)

Eq.6.21 can be approximated by

UT =

Sy+ Sut

2

f (6.22)

wherefis the strain at fracture, Syis the yield strength, and Sutis the ultimatestrength.

Resilienceis the capacity of a material to absorb energy when it is deformedelastically. The modulus of resilience is defined as the maximum energy thatcan be absorbed per unit volume without creating a permanent deformation.Resilience, UR can be deterimined by integrating the stress-strain curve fromzero to the elastic limit.

UR=

el0

d (6.23)

whereel is the elastic limit. Eq.6.23 can be approximated as

UR=S2y2E

(6.24)

Materials with smaller modulus of elasticity is desirable for better energy ab-sorbsion.

EXAMPLE 6.6

Resilient Neoprene pads are widely used for all kinds of vibrating machineryto isolate vibration. These pads are high grade neoprene isolation medi-ums which can efficiently control structure-born noise and reduces the res-onant vibration of the system natural frequency considerably below thedriving frequency of the equipment. Which of the following pad is suit-able as vibration isolation material for the shaker table legs shown in Fig-ure 6.15 (This figure is adapted fromImages of Neoprene pad isolators:http://www.bing.com/search?q=Neoprene+Pad+Isolators&FORM=HDRSC1).

(a) Pad material-A with Sy =20 MPa and E=4 MPa

(b) Pad material-B with Sy =16 MPa and E=3 MPa

SOLUTION

Figure 6.15:

Shaker table leg.

(a) From Eq.6.23, Resilience,UR for material-A is

UR=S2y2E

= 202

2 4= 50 MPa


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(b) Resilience, UR for material-B is

UR=S2y2E

= 162

2 3= 42.67 MPa

Although material-B has smaller modulus of elasticity, material-A will beselected because of its better energy absobsion.

6.1.2 Physical Properties

Properties that define the behavior of materials in response to physical forcesrather than mechanical is called physical properties. They are observable andmeasurable. A materials performance depends on its physical properties. Ma-terial selection with appropriate physical properties, for specific application canreduce the product impact on environment. Following physical material proper-ties will be discussed in this section.

Thermal properties Electrical properties

Thermal Properties

Density Temperature and pressure are the two factors that influence thedensity of a substance. A materials density is defined as its mass per unitvolume. The SI unit of density is gram per cubic meter (g/cm3). Density iscalculated by dividing mass by the volume of the material and is representedusing a Greek letter .

=M

V (6.25)

whereMandVare the mass and volume of a material, repectively. Strength todensity ratio (S/) and modulus to density ratio (E/), are important parame-ters in material selection.

Thermal expension Thermal expansion refers to the change in volume ofsubstance due to the change in temperature. When the material is a solid,thermal expansion is usually expressed in terms of change in length, height, orthickness. Thermal expansion, can be defined by the degree of expansiondivided by the change of temperature in a material. For example, Thermalexpansion, of a solid bar can be calculated by change in length of the bar fora given temperature change as

=L2 L1T2 T1

(6.26)

whereL1 andL2 are the initial and final lengths of the bar, repectively and T1andT2 are the initial and final temperatures of the bar, respectively.

Most of the materials used in engineering design have positive thermal ex-pansion co-efficient. In structural design, generally materials with co-efficientof thermal expansion that doesnt change considerably are used. On the otherhand, bridges are built with metal expansion joints, so that they can expandand contract to keep the integrity of the overall structure of the bridge.


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Another example is the thermal expansion of concrete which is important inthe analysis and design of concrete structure. The thermal expansion and con-traction of a concrete pavement can have a significant effect on its performance.Thermal expansion can cause joint lock-up and blow-up. Thermal contractioncan cause in transverse cracking of slabs depending on the joint spacing.4

Thermal expansion is used in shrink fit and expansion fit of two or more com-

ponents by heating or cooling them, thus making the use of thermal expansionin order to make a joint.

Melting point The melting point of solids and boiling points of liquids areintrinsic physical properties and can be used to identify them. Each materialwill have a specific melting point and can be useful in deciding whether thematerial is pure or not. If the melting point of any pure substance is known,then the melting point of the substance under consideration can be matchedwith the pure one. If the melting points are not identical it indicates that thetwo substances are not same in their purity.

Melting point of a material is an important factor in metal casting and plasticmolding. Materials with lower melting points are usually easier to cast.

Specific heat The specific heat is the amount of heat required to change aunit mass of a substance by one degree in temperature. The amount of heat, Qneeded to change the temperature of a substance is calculated by

Q= mcT (6.27)

wherec is the specific heat of the metarial, m is the mass, and T is the tem-perature change. Specific heat is important because it determines how quickly asubstance will heat up or cool down. The smaller the specific heat of a substance,the quicker the substance heats up or cools down.

Thermal conductivity Thermal conductivity, is a measure of the abilityof a material to transfer heat. It is defined as the heat energy (Q) transferredthrough a unit thickness (L) in a direction normal to a surface of unit area(A) due to a unit temperature gradient (T). That is given by the followingequation

= Q LAT (6.28)

Materials with high values of thermal conductivity such as copper are goodconductors, and those with low values such as cork and polystyrene are goodinsulators. Some of the physical properties of the materials are given in TableB-1 in Appendix B.

6.1.3 Electrical Properties

General understanding of the electrical properties of the materials is necessaryto design electrical circuits. Electrical properties include: electrical resistance,electrical conductivity,

4David K. Hein, concrete coefficient of thermal expansion (cte) and its significance inmechanistic-empirical pavement design, Applied Research Associates, Inc., Toronto, Canada.


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Electrical resistivity

Electrical resistivity, r is the opposition of a material to the flow of electricalcurrent flow. The amount of resistance changes with respect to material types.Materials with low resistivity are good conductors of electricity and materialswith high resistivity are good insulators. The SI unit for electrical resistivity isthe ohm-meterand given by

r= RA

L (6.29)

where R is the resistance af the material, L is the length, and A is the crosssectional area of the wire.

Electrical conductivity

Electrical conductivity, is a measure of how well a material will allow themovement of an electric charge. The SI unit for electrical conductivity is theSiemens per meter(ohm-m)1 and calculated by the reciprocal of resistivity

=1

r (6.30)

Electrical conductivity knowledge is used for measuring the purity of water,sorting materials, checking for proper heat treatment of metals. Metals arethe best conductors and insulators and ceramics, wood, plastics are the poorconductors.

Permeability

Permeability is the measure of the ability of a material to hold the formation ofa magnetic field. It is the ratio of the flux density ( B) created within a materialto the magnetizing field (H) and is defined by the following equation:

= B

H (6.31)

A material with high carbon content will have low permeability and willretain more magnetic flux than a material with low carbon content.

Electrochemical corrosion

Corrosion can be defined as the deterioration of materials by an electrochemicalprocesses. Electrons to flow between the anodes and the cathodes is the drivingforce for corrosion to occur. Electrochemical corrosion occurring in aqueoussolutions at ambient temperatures is a major damaging process that results indestructive effects as the formation of rust and other corrosion products. It will

damage any product made by any metals except gold.Under a given set of environmental conditions, materials with low rate ofcorrosion tendency should be selected for specific applications. A crucial factorcontrolling the rate of corrosion is the existence of the phenomenon of passivity(conditions existing on a metal surface lower the rate of corrosion tendency) forcertain metals and alloys, such as stainless steels and titanium. Corrosion hasmany severe economic, health, safety, technological, and cultural consequencesto our society5.

5Jerome Kruger. electrochemistry of corrosion, http://electrochem.cwru.edu/encycl/, ac-cessed: August. 28, 2013


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6.1.4 Chemical Properties

A chemical property is basically whether or not the material will react withanother substance. For example, iron reacting with oxygen to form rust. Someof the material chemical properties include: Corrosion resistance and hygroscopy.

Corrosion resistanceSee electrochemical corrosion covered in Section 3.1.3.4.

hygroscopy

Hygroscopy is the ability of a material to attract and store water molecules fromthe surrounding environment through either absorption or adsorption. Materialwill change as a direct result of attracting and absorbing water. For example,the caking or hardening of sugar and salt after absorbing moisture from the air.The existence or absence of hygroscopic materials has a direct impact on indoorrelative humidity levels. For example, existence of hygroscopic materials willreduce the humidity level during periods of high humidity seasons by absorb-ing moisture from the environment and release the moisture during periods oflow humidity seasons and thus maintaining a relatively constant indoor relativehumidity level.

6.1.5 Environmental Properties

To find data on the environmental properties of materials, one can use databasespublished by companies and government organizations.

Embodied energy of materials

A materials embodied energy is the non-renewable energy that must be used toextract, process, transport, and process the material. A best way to reduce em-bodied energy is to specify recycled materials when designing a system to ensurethat those materials can be recovered at the products end of life. For example,using some of the recycled materials can cut embodied energy considerably.

Health impacts of materials

Materials can occasionally also have negative health impacts, and some materialsare regulated for this reason by national organizations. For example, materialssold in the United State need to meet the requirements of the government agen-cies such as Environmental Protection Agency (EPA) and National Institutefor Occupational Safety (NIOSH). For example, some building materials in thehousing sector that are harmful or potentially harmful to peoples health andthe environment. Risks to health usually result from producing harmful material

(toxic), use and disposal of such materials.

6.2 Processes Used to Alter the Properties of

Materials

Using various processes certain characteristics of metals and alloys can be changedin order to make them more suitable for a particular kind of application. Forexample, heat treatment can significantly influence mechanical properties suchas strength, hardness, ductility, toughness, and wear resistance of the materials.


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Figure 6.16: (a) Strain hardening, (b) Examples of cold work.

6.2.1 Strain Hardening

Strain hardeningalso calledwork hardening or cold workis a process to increasethe hardness and strength of the material by plastic deformation usually at am-bient temperature. Many ductile metals with a relatively high melting point aswell as several polymers can be strengthened in this way. This kind of hardeningis effective to increase the strength of metals that cannot be hardened by heattreatments.

When a material is subjected to a load, microscopic defects known as dis-location generation occur within the crystal structure of the material. Whenthe load continue to increase, these dislocations grow and interact with each

other, forming new internal structures that resist to slip deformation. Thesenew formations increase the materials yield strength, or ability to resist beingstressed.

As shown in Figure 6.16(a), when material is stressed, the strain increaseswith stress and a certain amount of deformation is put into the material (linefrom O to A). If the load is removed at this point, the material returns to azero stress at point B along a new elastic line AB with the same original slope,E. If the material reloaded immediately, the curve again rises from B to C andconsequently material ruptures if the loading is continued. Since, further increasein strength will cause reduction in ductility and formability, hardening processshould be stopped to anneal the metal. In other words, heat it up to remove theaccumulated dislocations before the sequence of hardening process continued. Asshown in the figure, by strain hardening yield strength of the metal significantlyincreased. The strain has two components recovered elastic strain, e and aninelastic strain (parmanent deformation), p. Then the total strain, T becomes

T =p+ e (6.32)


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EXAMPLE 6.7

If a steel solid rod is loaded to a stress of=480 MPa and then unloaded,what is the permanent strain? As shown in Figure 6.17, yield strength ofthe steel is 310 MPa corresponds to a strain of 0.0015 m/m and the totalstrain,T is approximately 0.005 m/m.

SOLUTIONThe original slope is

E=Sy

=

310

0.0015= 206.7 GPa

Figure 6.17:

Stress-strain curve.stress-strain curve.

Recovered elastic strain,eat 480 MPa can be calculated by using the originalelastic modulus, Eas

e=

E =

480

206.7= 2.32 103 m/m

The total strain, T is given by

T =p+ e

Then

p = T ep = 5 103 2.32 103 = 2.68 103 m/m

Strain Hardening and Residual Stresses

Figure 6.16(b) shows an example of strain hardening through metal-workingprocesses. This process is known as cold working or cold forming process. Dur-ing most metal forming operations, such as rolling and extrusion, the materialundergoes non-uniform plastic deformation resulting in a pattern of residualstresses throughout the materials cross section. Consider a metal sheet beingrolled as shown in Figure 6.16(b). During the rolling process, plastic deforma-tion occurs only near the surfaces that are in direct contact with the rollers. Thearising non-uniform deformation results in elongation of the surface fibers whilethe fibers near the center of the sheet stay unchanged. However, for the sheet

to remain in equilibrium as a continuous body, the center fibers tend to restrainthe surface fibers from stretching, whereas the surface fibers tend to elongatethe central fibers. This results in a pattern of residual stresses throughout thesheet with a compressive value at the surface and a tensile value at the centerof the sheet (see Figure 6.16(b-1).6 Note that the residual stresses shown inFigure 6.16(b-1) is simplified and approximated stress fields.

As seen from Figure 6.16(b-1), residual stresses are in the elastic region.In other words, they are below yield strength of the material. If a uniformly

6A. Ertas, and J. C. Jones, The Engineering Design Process, Second Addition, John Wiley& Sons, Inc., New York, 1996.


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distributed tensile stress (sufficiently over yield strength) is applied, first com-pressive stress will become zero then it exceed the level of yield strength andfinally stress distribution will become uniform as shown in Figure 6.16(b)-2.When the load is removed, elastic energy will be recovered and the material willbe free of residual stresses. Residual stresses can also be reduced or eliminatedby annealing.

Analysis of Strain Hardening

Although analysis of work hardening is difficult, there are several constitutivestress-strain relations have been formulated. Most commonly used formula isthe power law.

t = Knt (6.33)

Where t is the true stress, t is the true strain, K is the strength coefficient(numerically equal to the extrapolated value of true strain of 1.0), and n is thestrain hardening exponent. Taking log of both sides of Eq. 6.33 gives

log t = log K+ n log t (6.34)

Eq. 6.34 is a straight line with intercept, log Kand the slope, n. The truestress against true strain is plotted on a log-log scale is shown in Figure 6.18.

Figure 6.18:

Log/log plot of truestress-strain curve.

The values of strain-hardening exponent, n may change from n = 0 to n =1. Ifn=0, material is perfectly plastic and does not show any strain hardening.If n=1, material is elastic. As shown in Table 6.2, n values for most of thematerials changes between 0.10 and 0.5 (see Table 6.2).

If two points are known on the curve (not including the region after thenecking) shown in Figure 6.18, strain hardening exponent can be calculatedfrom Eq. 6.35.

Slope =n =

log t2

log t1log t2 log t1 (6.35)

EXAMPLE 6.8

A solid metal rod is stressed in tension beyond its yield stress. It had a gagelength of 50 mm and an area of 50 mm2 before the loading. At one pointduring the loading, the gage length L1becomes 60 mm and the correspondingengineering stress e1becomes 160 MPa. At another point during the loadingprior to necking, measured gage length, L2 is 80 mm and engineering stress,e2 is 190 MPa. Determine the strain-hardening exponent.

SOLUTION

Strain hardening exponent from Eq. 6.35 is

Slope =n = log t2 log t1

log t2 log t1Calculation of parameters for the first point is as follows:

Engineering strain, e1 is

e1 =L1 L0

L =

60 5050

= 0.2 mm/mm


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Table 6.2: Kand nvalues for selected metals and alloys.*

Strength Strain Hardening

Coefficient Exponent

Material K (MPa) n

Aluminum, 1100-O 180 0.202024-T4 690 0.16

5052-O 210 0.13

6061-O 205 0.20

6061-T6 410 0.05

7075-O 400 0.17

Brass, 7030, annealed 895 0.49

85-15, cold rolled 580 0.34

Bronze (phosphor), annealed 720 0.46

Cobalt-base alloy, heat treated 2070 0.50

Copper, annealed 315 0.54

Molybdenum, annealed 725 0.13Steel, low carbon, annealed 530 0.26

1045 hot rolled 965 0.14

1112 annealed 760 0.19

1112 cold rolled 760 0.08

4135 annealed 1015 0.17

4135 cold rolled 1100 0.14

4340 annealed 640 0.15

17- P-H, annealed 1200 0.05

52100, annealed 1450 0.07

304 stainless, annealed 1275 0.45

410 stainless, annealed 960 0.10

*Serope Kalpakjian and Steven R. Schmid, Manufacturing Processes forEngineering Materials, Printice Hall, 2008.

True stress, t1 is

t1 = e1(1 + e1) = 160(1 + 0.2) = 192 MPa

True strain, t1 is

t1 = lnL1L0

=ln60

50= 0.18 mm/mm

Calculation of parameters for the second point is as follows:

Engineering strain, e2 is

e2 =L2 L0

L0=

80 5050

= 0.6 mm/mm

True stress is

t2 = e2(1 + e2) = 190(1 + 0.6) = 304 MPa


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Figure 6.19:

Log-log scalestress-strain curve.

True strain, t2 is

t2 = lnL2L0

=ln80

50= 0.47 mm/mm

Using Figure 6.19, strain hardening exponent, n can be calculated as

n=log t2

log t1

log t2 log t1 = log(304)

log(192)

log(0.47) log(0.18)= 0.48

Material Instability in Tension

Now consider events that occur just before the necking (before the final fracture)of a specimen tested in uniaxial tension. When ultimate tensile strength isreached, necking will start and deformation will no longer be uniform. Finally,a point of plastic instability will be reached in the stress-strain curve. Althoughmaterial will get stronger with the increase of strain due to strain hardening,the cross-sectional area starts to decreases rapidly at some point along the gagesection of the specimen. Since the rate of decrease in cross-sectional area is

greater than the rate of increase in strength at the necking region, the strengthof the material will decrease considerably and eventually material will rupture.This condition is called as geometric softening. The true strain at the onsetof necking which corresponds to the maximum load is numerically equal to thestrain hardening exponent, n.7

EXAMPLE 6.9

Referring to Example 6.8, calculate the true ultimate tensile strength, S(ut)tand the engineering ultimate strength, S(ut)e.

SOLUTION

Using Figure 6.20, the slope, n is given by

n= 0.48 = y2 y1x2 x1

= log(192) log(K)log(0.18) log(1) =

2.283 logK0.745 0.0

Figure 6.20:

Log-log scalestress-strain curve.

Therefore,

logK= 2.64 K= 102.64 K= 437 MPa

The value ofKcan also be calculated directly from Eq. 6.32

K= t2nt2

= 3040.470.48

= 437 MPa

From Example 6.8 and Section 3.2.1.1, we have = n = 0.48at necking region.Then the true ultimate strength, S(ut)t in the region of necking is

S(ut)t= Knt = 437(0.48)

0.48 = 307 MPa

7Serope Kalpakjian and Steven R. Schmid, Manufacturing Processes for Engineering Ma-terials, Printice Hall, p. 38. 2008.


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The true strain at necking,(t)neck is

(t)neck = ln

A0Aneck

= n = 0.48

Above equation leads to

Aneck= A0e0.48

Necking strain corresponds to the maximum load, Pmax at which the materialreaches to its ultimate strength. That is,

(max)neck= PmaxAneck

where(max)neck is the true ultimate tensile strength, S(ut)t. Then the maximumload at necking is

P(max)neck= S(ut)t Aneck= 307(2.718)0.48A0 = 190A0

Finall, engineering ultimate strength, S(ut)e is

S(ut)e =Pmax

A0=

190A0A0

= 190 MPa

6.2.2 Forging

Forging is a manufacturing process of producing a metal component by hammer-ing or pressing to increase its strength. Forging improves the grain structure andalters physical properties of the metal. As the metal is shaped during the forgingprocess, grain structure deforms to follow the general shape of the component,thus giving rise to the metal component with improved strength characteristics.

6.2.3 Heat Treatment

Heat treatment is another process of altering the properties of the materials byheating and cooling operations. A few common heat-treating processes are givenbelow.

Aging (Age Hardening)

Age hardening is a type of heat treatment to strengthen metal alloys. It isalso called precipitation hardening, as it strengthens metal by creating solidimpurities, or precipitates, by heating it for a long time.

Annealing

Annealing is a heat treatment that alters a material to increase its ductility andto make it more workable by first heating and then slow cooling of a metal. Thisprocess removes residual stresses, gases, and improves cold working properties.


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Quenching

Quenching is the rapid cooling of a material by immersion in oil or water to alterthe material properties. Quenching is usually performed to maintain mechanicalproperties that would be vanished with slow cooling. It is usually applied tosteels, to increase the hardness.

Tempering

Tempering, is a process involving slow and moderate heating to increase thehardness and toughness of metals which was previously heat-treated. For exam-ple, quenched steel is tempered to reduce residual stresses while storing certainamount of ductility and improve toughness.

Case hardening (Surface Hardening)

Case hardening is the process of hardening the surface of a part so that it willhave tough and hard coating. This process is done by heating the material in

the presence of carbon. This process is the infusion of carbon into the outerlayer of the material to harden the surface. Case hardening is often done afterthe part has been formed into its final shape such as guns.

6.3 Classification of Materials

There are numerous kinds of materials used in the field of engineering. As shownin Table 6.3, materials can be classified into six groups. Namely, metals, ceramicand polymers, composites, semiconductors and new materials. These classes canbe further organized into various sub-groups based on their mechanical, physical,and chemical properties.

6.3.1 Metals

The basic part of any element is the atom. Atoms are the basic building blocks ofall materials. A single atom contains of a positively charged nucleus, surroundedby a cloud of negatively charged particles called electrons. In an atom, theelectrical charges of the nucleus and electrons are equal, but opposite. It turnsout to be the overall electrical charge of an atom is neutral. The outmostelectrons in the atoms of metals are held loosely and they can travel easily fromatom to atom. The main attribute that distinguishes a metal from a nonmetalis the presence of these free electrons8.

Apure metalconsists of atoms of only one element and does not include anyother types of atoms. It has its own unique physical properties such as melting

point, boiling point, and thermal or heat conductivity. Examples of some puremetals are aluminum (Al), copper (Cu), zinc (Zn), and silver (Ag).

Analloyis a mixture of two or more elements into a solid, metallic solution.Therefore alloys have properties that are different from those of its constituentelements. Steel is the best known alloy which is made up mostly of iron andalso contains carbon. Inclusion of carbon makes the iron stronger increase intensile and/or shear strength.

8Materiasl Classification. ASME International, http://www.asminternational.org/content/ASM/StoreFiles/250491

ch.pdf, accessed: August 28, 2013


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Table 6.3: Classification of materials.

Material type Sub-groups

Metal

Ferrous metals and alloys (irons, carbon steels,

alloy steels, stainless steels, tool and die steels)

Nonferrous metals and alloys (aluminum, copper,magnesium, nickel, titanium, precious metals,

refractory metals, superalloys)

Ceramic

Glasses

Glass ceramics

Graphite

Diamond

Thermoplastics

Plastics Thermoset

Elastomers

Composite

Polymer-matrix composites

Metal-matrix compositesCeramic-matrix composites

Semiconductors Silicon

New Materials

Smart materials

Nano materials

Bio materials

Ferrous Metals and Alloys

Ferrous metals and alloys are iron-based materials that are used in many indus-

trial applications.IronsIron is the tenth most abundant element in the earth and it is the most essentialindustrial metal. It is shiny, ductile, malleable, and is the most used of all themetals.

Wrought iron Wrought iron is an iron alloy contains a very low carbon(less than 0.1%)and 12% slag. Wrought iron is a soft,ductile, easily worked,fibrous metal. Wrought iron is made by melting in a forge with charcoal, whichserved both as fuel and reducing agent. While it is still hot, it goes throughhammering and rolling operation to put the iron into a coherent mass. Itsresistance to corrosion is better than the steel.

Cast ironsCast iron is iron alloy with more than 2% carbon. Cast irons also contain from1 to 3% silicon which combined with the carbon provide excellent castability.Cast iron is usually made from pig iron (combination of iron ore, charcoal,and limestone melted together under extreme pressure). It is more brittle thanpure iron and steel, however it melts at a lower temperature. The meltingtemperature is ranging from 1,150 to 1,200 C (2,102 to 2,192 F), which is about300 C lower than the melting point of pure iron. Since it is more malleable itis good for industrial applications. Cast iron is used in pipes, machine parts,


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Table 6.4: Composition for typical cast irons.

Composition*

Type of iron Carbon (C) Silicon (Si) Manganese (Mn) Sulfur (S) Phosphorus (P)

Gray (FG) 2.5 4.0 1.0 3.0 0.2 1.0 0.02 0.25 0.002 1.0

Ductile (SG) 3.0 4.0 1.8 2.8 0.1 1.0 0.01 0.03 0.01 0.1Compacted 2.5 4.0 1.0 3.0 0.2 1.0 0.01 0.03 0.01 0.1

graphite (CG)

Malleable (cast 2.2 2.9 0.9 1.9 0.15 1.2 0.02 0.2 0.02 0.2

white) (TG)

White 1.8 3.6 0.5 1.9 0.25 0.8 0.06 0.2 0.06 0.2

*Reprented with permission of ASM International. All right reserved. www.engineershandbook.com.

automobile components and, skillets. Compositions of typical cast irons aregiven in Table 6.4.

White cast irons White cast irons are usually made by restricting thesilicon content to a maximum of 1.3 percent, so that no graphite is presentbut all of the carbon exists as cementite (Fe3C). White iron is hard and brittleand has a white crystalline fracture because it is essentially free of graphite.White irons are too hard to be machined.

Malleable iron The white iron castings become malleable iron through an-nealing process by heating at approximately 900 C. The casting is melted andthen kept at a controlled temperature for certain time. It is then allowed tocool very slowly over about 24 hours. The properties of malleable cast iron aresimilar to mild steel. Comparing with the gray cast iron, malleable cast ironhas better strength and ductility, particularly better impact resistance in low

temperatures.

Grey iron Gray Iron is a type of cast iron which includes carbon and siliconin addition to the iron, in various concentrations, with respect to different appli-cations. Grey iron is not very malleable or strong. It fractures easily. Therefore,it is not suitable for applications where high tensile strength is needed. Since itconducts heat very well, it is used to make cast iron pans. Gray iron containsmore silicon, thus less hard and more machinable than white iron. Properties ofgray cast iron is given in Tbale B-2 in Appendix B.

Ductile iron Ductile iron is a family of cast irons which has high strength,ductility and resistance to shock. It is heat treated by the process of annealing.Cast ductile iron can be put to different shapes without fracturing. The mechan-ical properties of ductile iron is better than either gray or malleable iron. DuctileIron castings is effective, lower cost alternative to malleable iron castings, steelforgings and steel fabrications. It is made of cast iron by adding magnesium init. This process causes the graphite to form in small spheroids which createsfewer discontinuities in the structure of the metal and produces a more strongerstructure.


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SteelsSteels are also mainly iron with carbon (C) as the main alloying element. Whilesteels contain less than 2% carbon, all cast irons contain more than 2% carbon.Properties of steels are given in Table B-3 in Appendix B.

Carbon steel Carbon steel is an alloy of iron and carbon in which thecarbon content may range from 0.015% to 2%. Adding this small amount ofcarbon produces a material which shows incredible strength, fatigue resistance,wear resistance, hardness and toughness. Carbon steel is one of the most widelyused materials in industrial applications. They are used broadly for automobilebodies, appliances, machinery, ships, containers, heavy wall pipe and tube, andbuilding industry.

Carbon Steel with a low carbon content has the same properties as iron. Itbecomes soft and easily formed. With the raising carbon content it becomesharder and stronger but less ductile and more difficult to weld. Usually, highercarbon content lowers steels melting point and its temperature resistance.

With respect to carbon content, grades of carbon steel covered by ASTM,ASME and API specifications are:

Mild (low carbon) steel has approximately 0.05% to 0.25% carbon contentwith up to 0.4% manganese content. Less strong but inexpensive.

Medium carbon steel has approximately 0.29% to 0.54% carbon contentwith 0.60 to 1.65% manganese content. Fairly good ductility and strengthand has good wear resistance.

High carbon steel has approximately 0.55% to 0.95% carbon content with0.30 to 0.90% manganese content. Very strong.

Very high carbon steel has approximately 0.96% to 2.1% carbon content,specially processed to produce specific atomic and molecular microstruc-tures.

Alloy steel Alloy steel is a steel in which one or more elements besidescarbon have been added to produce a preferred physical properties or character-istics. These elements are: molybdenum, boron, chromium and vanadium. Ingeneral, alloy steels are categorized into two groups high alloy and low alloysteels.

1. High alloy steels has more than eight percent of its weight is other elementsbesides iron and carbon.

2. Other elements in low alloy are considerably low compared with the amountof the entire steel.

Stainless steel Stainless steel also called inox steel is corrosion-resistant

metal alloy with a minimum of 10.5% chromium content by mass in it. Additionof this amount of chromium to low carbon steel provides stain resistance. Ad-vanced stainless steel may also contain other elements, such as nickel, niobium,molybdenum, and titanium. Stainless steel does not quickly corrode, rust orstain with water as ordinary steels do. It is used where both the properties ofsteel and resistance to corrosion are required. Stainless steel has different gradesand surface finishes going well with the environment and withstanding to highstresses.

The chromium in the steel come together with oxygen in the atmosphere andform an invisible passive thin film (a few atoms thick) layer. If the thin layer is


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cut or scratched, more oxide will quickly form and recover the exposed scratchedsurface, protecting it from oxidative corrosion. The passive film needs oxygento self-repair, so stainless steels have poor corrosion resistance in low-oxygenand poor circulation environments. It is extensively used in constructions aswell as in kitchens. It is ductile, ductile, long-lasting and much inexpensive ascompared to other metals. Properties of stainless steels are given in Table B-4

in Appendix B.There are three main types of stainless steels austenitic, martensitic, andferritic.

Austenitic stainless steels They are the most widespread and effec-tively used steels. Including other elements such as nickel, nitrogen and man-ganese, it contains 0.15% carbon and 16% chromium. It is ductile and alsohighly corrosion resistance.

Martensitic stainless steels They are low carbon steels with compo-sition of iron, 12% chromium, and 0.12% carbon. It has also nickel added (2%)for improved corrosion resistance. .Martensitic stainless steel are magnetic and

can be hardened by heat treating. The martensitic grades are primarily usedwhere hardness, strength, and wear resistance are necessary. Martensitic stain-less steel, containing the lowest alloy content of the three basic stainless steelswhich has low cost, broadly used and heat treatable stainless steel. Normally,it is used where corrosion is not severe (air, water, some chemicals, and foodacids). Typical applications include springs, cutlery, and highly stressed partsrequiring the combination of strength and corrosion resistance such as fasteners.

Ferritic stainless steels These types contain chromium in the rangeof 12% to 27% and whose structure consists largely of ferrite. Such steels havegood ductility and are easily worked but can not go through the processes ofhardening or tempering. Ferritic steel can be successfully used in sea water asit is highly resistant to corrosion, in washing machines, boilers, automotive trim

and architectural shield.

Tool and die steels They have carbon content between 0.7% and 1.5%and they are manufactured under carefully controlled conditions to produce thenecessary quality. Tool and die steels are high carbon steels have high hardness,strength and wear resistance. They are heat treatable. To increase hardnessand wear resistance of tool steels, chromium, tungsten, vanadium, manganese,and molybdenum are added to the composition.

Cast steels The differences between steel castings and its wrought steel aremainly in the method of production. In the case of wrought products (rolled orforged) are mechanically worked to produce sheet, bar, tube and other product

forms. On the other hand, steel castings are produced in the final product formwithout any intermediate mechanical working. Steel castings are almost netshape products. High carbon cast steels offer very good wear resistance. Thedifferences between cast steel and cast iron is, later has higher carbon contentof over 2% than cast steel. Both cast irons and cast steels have good physicalcharacteristics such as wear resistance, creep resistance, etc.

Cast steels and wrought steels of the same composition, have similar me-chanical properties. However, more complicated products can be made in onesingle step by using steel casting. Steel castings are used for extremly importantcomponents in the mining, railroad, truck, construction, military, and oil and


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gas industries. Typical products include, valves, pumps, tooling and industryparts. Cast steels with high yield strength in low temperature environment andgood weldability are used for offshore structures. Because of its creep resistant,cast steels play an important role in fossil fuel fired power plants.

Non-ferrous metals

They are any metals that do not contain iron in noticeable amounts. Althoughnon-ferrous metals are relatively more expensive they are used over ferrous met-als because of their low weight, higher strength, nonmagnetic properties, highermelting points, or resistance to chemical and atmospheric corrosion.

Some of the important non-ferrous metals include aluminum, copper, lead,nickel, tin, titanium and zinc, and alloys such as brass. Precious metals suchas gold, silver and platinum and exotic or rare metals such as cobalt, mercury,tungsten are also non-ferrous metals. Some of them are discussed below.

AluminumPure aluminum is a silvery-white metal with many attractive characteristics.

However, it is soft with low strength. With alloying addition of small amountsof copper, magnesium, silicon, manganese, and other elements to aluminumcreates aluminum alloys with desirable properties. Many different kinds of alu-minum alloys are available and their properties are given in Table B-5 in Ap-pendix B.

Aluminum is enormously available in the earths crust and substantially re-cyclable. However, it is not readly available in the nature, the Bayer process isused to refine aluminum from bauxite, an aluminum ore. Aluminum is excellentthermal and electrical conductor, highly corrosion resistant in air and water,non-toxic, non-magnetic, and it can be formed in almost any shape.

Aluminum alloys are used in building and construction industries, aeronau-tical, aviation and automotive industries, manufacture of electrical products,packaging and containers, and in many other applications.

CopperCopper provides a different range of properties good thermal and electrical con-ductivity, high resistance against corrosion, ease of forming and joining. Copperis non-magnetic, non-sparking and non-bacterial. However, copper and its al-loys have relatively low strength-to-weight ratios. Some copper alloys are alsosubject to stress-corrosion cracking unless they are stress relieved.

Copper alloys are metal alloys that have copper as their main component.Best known copper alloys are: the brasses (zinc is a significant addition) andbronzes (tin is a significant addition) are available in rod, plate, strip, sheet,tube shapes, forgings, and wire. With moderate strength and high fatigue re-sistance, copper alloys are main metals for electrical applications. Properties of

copper are given in Table B-6 and Table B-7 in Appendix B.

ZincZinc is inexpensive and enormously available metal in Earths crust with manyindustrial and biological uses. It has relatively low melting point (419.5C) andboiling point (907C). Its strength and hardness is considerably less than thatof aluminum or copper. Zinc is not a strong metal with a tensile strength lessthan half that of mild carbon steel. Therefore, it is generally not used in stressedapplications. One of the most important characteristics of zinc is its resistance toatmospheric corrosion. Fifty percent of its use is for the protection of steelwork.


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Table 6.5: Common magnesium alloys.*

Alloy Alloying Uses & Reasons

Designation Additivies Applications for Use

AZ91 9.0 % Al-0.7 % Zn-0.13 % Mn General casting Good castability, good mechanical

alloy properties at T


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Properties of nickel alloys are given in Table B-10 in Appendix B. As listedin this table, nickel 200 is used in food processing equipment, chemical ship-ping drums, piping, aerospace and missile components. Nickel 201 is neededover nickel 200 for applications involving exposure to temperature above 316C.Permanickel alloy 300 is used for magnetostriction devices, thermostat contactarms, solid-state capacitors, springs, clips, an fuel cells. Duranickel alloy 301is

an age- hardenable high nickel alloy which has high strength and hardness, goodcorrosion resistance, and good spring properties.Nickel-copper alloyshave high strength, good weldability, excellent corrosion

resistance, and toughness. Nickel-chromium-iron alloys, are developed for hightemperature oxidizing environments. Nickel-iron-chromium alloysare used inhigh temperature environments where resistance to oxidation or corrosion isrequired. Nickel-iron alloysare used for curing composite airframe components,for semiconductor lead frames and glass sealing applications. Nickel-chromium-molybdenum alloys preferred for severe corrosion environments. Nickel-powderalloyshave high creep-rupture strength and is also age hardenable that resultswith high strength at low temperatures up to 700C.9

Titanium

Titanium is widely found in the Earths crust. It is an extremely strong, lightmetal alloy. It is 60% heavier than aluminum, but it is twice as strong. Becauseof their unique properties, titanium alloys are crucial to the aerospace industry,medical, chemical and military applications. Pure titanium is a shiny whitemetal with low density, high strength, and high corrosion resistance. Titaniumcan be alloyed with iron, aluminum, vanadium, molybdenum, including otherelements, to produce strong lightweight alloys for many industrial applications.Properties of different grades of titanium are given in Table B-11 in AppendixB.

6.3.2 Ceramics

The word ceramic is from the Greek work keramicwhich meansmade of clay. In

the past, ceramic has been used as a material mainly for pottery and tableware.Ceramic is a non-organic material made from compounds formed between metal-lic and non-metallic elemnts. In general, they are brittle and porous. Recently,advanced developments have made ceramics to become tougher, more workable,and useful in wider range of fields.

Ceramic materials can be crystalline or non-crystalline. Non-crystalline ce-ramics, such as glasses, are formed from melts. If glass goes through heat-treatment it becomes partly crystalline. This resulting material is called glass-ceramic. Carbon ceramic materials are significantly strengthened by the in-clusion of carbon fibers. They are lightweight, remarkably strong and able tooperate at very high temperatures. For example, 1500C (2732F). Ceramicmaterials are very strong, stiff, brittle, and chemically inert. Although ceramicis widely used to make electrical insulators, some ceramic compounds are su-perconductors. Ceramic is a material for harsh environments that make it idealfor the paper processing industry. Although common uses for ceramic today arestill tableware, pottery, and tiles, ceramics are now used in bullet proof vests,cars, computers, toys, and repair material for dentures.

Diamond is a material with outstanding features such as extraordinarily hard,conducts heat well and is practically inert to chemical substances. Especiallyhigh-performance ceramics are also able to demonstrate special qualities such asthey are robust and withstands extreme temperatures. The composite material,

9Mechanical Engineers, handbook, 2nd addition, Wiley, New York, 1998.


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diamond-coated ceramics brings the best of both materials together. Placeswhere components and tools are subjected to heavy strain such as in pumps orforming and shaping dies, ceramic composites are used. It provides maximumwear-resistance with low values of friction. For more information see reference10

6.3.3 Polymers

Polymers are giant organic molecules composed of a long, repeating chain ofsmaller subunits called monomers. Polymers have the highest molecular weightamong any molecules, and may consist of billions of atoms. It is interesting toknow that human DNA is a polymer consist of over 20 billion atoms. Polymers,both natural and synthetic, are created via polymerization of many monomers.Many synthetic polymers are called plasticwhich comes from the Greek wordplastikos, suitable for molding or shaping. Plastics are divided into three maincategories which are themoplastics, thermosets, elatsomers.

Thermoplastics

When heated, thermoplasticsdo not undergo chemical change in their composi-tion. Thermoplastic materials can be formed into desired shapes under heat andpressure and returns to a solids upon cooling. If the same process is repeatedwith the same conditions of heat and pressure, they can be remolded again andagain. Five common types of thermoplastics are polyethylene, polypropylene,polystyrene, ABS, and acrylics.

Although polyethylenehas a wide variety of uses, it can be harmful to theenvironment. Its applications include gallon milk jugs, lawn chairs, garbagecans, and disposable gloves, etc.

In general, polypropylenehas certain advantages in improved strength, stiff-ness and higher temperature capability over polyethylene. Its application mayinclude dishes, automobile parts, and sterillizable hospital tools, etc.

Extrudedpolystyreneis lighter and has as much tensile strength as unalloyedaluminum. Polystyrene materials are used to make a variety of molded products

such as plastic tableware, CD cases and model cars, etc.ABS (combination of acrylonitrile, butadiene, and styrene monomers) hasgood impact strength compare to polystyrene. ABS is used for tough consumerproducts such as refrigigerator door liners, telephones, interior automative trimsetc.

One of the most common acrylic plastic is polymethyl methacrylate (PMMA),which is sold under the brand names of Plexiglas, Lucite, Perspex, and Crys-tallite. PMMA is a durable, highly transparent, and it has excellent resistanceto ultraviolet radiation and weathering. The applications are included: airplanewindshields, skylights, windows , automobile taillights, contact lenses, and out-door signs, etc.

Thermosets

Thermosetscan melt and take shape once under heat and pressure and cannotbe remolded after they have solidified, they stay solid. Cured thermoset ma-terials can be soften when heated, but do not melt or flow. Curing changes thematerial permanently. In general, thermoset materials are stronger than ther-moplastics due to polymer cross-linking and have a higher resistance to heat.Thermoset molding components have excellent dimensional stability, resist creepunder load at elevated temperatures, and provide excellent thermal properties.

10A powerful duo: diamond and ceramic, http://www.fraunhofer.de/en/press/research-news/2010/05/diamond-coated-ceramics.html, accessed: September 4, 2013.


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They are more brittle than thermoplastics and many cannot be recycled dueto irreversibility. Thermoset materials are very suitable for high-temperatureapplications.

Elastomer

Synthetic rubbers known aselastomer

is a polymer with low Youngs moduluscompared with other materials. Elastomers can be stretched to several timestheir original length, and can return back into their original shape without per-manent deformation. Elastomers are extensively used in automotive, aerospace,defence, and sport industries and many other applications such as dental prod-ucts, electrical products, paints and coatings, appliance parts, cement and con-crete adhesives, special wood protection coatings, medical and dental adhesives.

Properties of different types of plastics are given in Table B-12 in AppendixB.

6.3.4 Composites

In general, composites are made of two phases matrix and reinforcement. The

matrix is a less stiff and weaker material than the reinforcement and used tohold the reinforcement together to provide the main load-carrying capability ofthe composite.

Composite materials are usually used for buildings, bridges, swimming poolpanels, race car bodies, shower stalls, bathtubs, and storage tanks, imitationgranite and cultured marble sinks and counter tops. The fibers and matrix ofadvanced composites are used by the aerospace and automotive industry.

Different combinations of materials will create different kind of compositematerials. Three main catagories of composites are:

Polymer-Matrix Composites (PMC) Polymer-matrix composite is thematerial made up of a polymer (resin) matrix combined with a fibrousreinforcing dispersed phase. Polymer-matrix composite are very desirable

due to their low cost and simple fabrication process. Metal-Matrix Composites (MMC) Metal-matrix composites having high

specific stiffness and with almost zero coefficient of thermal expansion(CTE), have been developed for space applications.

Ceramic-Matrix Composites (CMC) Ceramic matrix composites, a ma-terial that combines the heat resistance of ceramics with the strength ofmetal. Important aviation applications of this material include combustorliners and blades.

6.3.5 Semiconductors

A semiconductor is a material which has electrical conductivity such as cop-per and is insulator such as glass. Semiconductors are the substance of solidstate electronics (transistors, solar cells, etc.) and digital and analog integratedcircuits.

6.3.6 New Materials

New materials are necessary for innovative new design development. Big spikesin design fields are often related with the innovation of new materials that impactnot only the feasible solutions but the very nature of design problems. The


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development and application of new materials may provide more safe, economicaland stringent design.

Material atoms can be combined in new ways to produce new materials withsmart properties. For example, a window glass changes color to control theroom temperature. New materials can be developed from existing materials byapplying high pressure, temperature, electric or magnetic fields. Some examples

of new materials are as following:

Smart Materials

Smart materials can considerably change their mechanical or other propertiesin a controllable manner. Based on input and output, the smart materials areclassified as follows:

Shape Memory Alloys The field of smart materials is growing rapidly,with one of the most fascinating areas being that of shape memory al-loys. A shape memory alloy can undergo significant plastic deformation,and then returns to its original shape by the application of heat. Nickel-titanium-copper, Gold-cadmium, and Nickel-titanium are good examples

for the shape memory alloys. Some of the examples of applications ofshape memory alloys are: Micro-actuators, thermostats, electrical circuitbreakers, fire dampers, and many medical applications.

Magnetostrictive Materials Magnetostrictive materials exhibit changein shape when exposed to a magnetic field. Some of the applications ofmagnetostrictive materials are: Actuators, transducers, magnetostrictivefilm applications, sensors and many other applications. Magnetostrictivematerials include nickel and alloys such as Fe-Ni, Co-Ni, and many others.

Piezoelectric Materials Piezoelectric materials produce an electric cur-rent if deformed by mechanical stress. The Piezoelectric process is also re-versible: creating deformation due to the application of an electrical field.Some of the applications of piezoelectric materials are: mechanical sensor

to pick up a mechanical deformation, used as an actuator, used for imag-ing, mostly in medicine, used as gas lighters. The most commonly knownpiezoelectric material is quartz Quartz (SiO2), berlinite (AlPO4), bar-ium titanate (BaTiO3), zinc oxide (ZnO), Aluminum Nitride (AlN).

Electro-Rheological Fluids Electro-Rheological (ER) fluids change theirphysical properties when exposed an electric field. They can be trans-formed from the liquid state into the solid state like gel in milliseconds byapplying an electric or a magnetic field. The change is reversible once theelectrical field is removed. Electro-Rheological (ER) fluids research areaproduced significant impacts on automobile industry, bridge and buildingconstruction, aerospace industry, defense industry, design of fast actuatedhydraulic devices, energy production and energy conservation.

Nanomaterials

Nanomaterials can be metals, ceramics, polymeric materials, or composite ma-terials. Their defining characteristic is containing nanoparticles, smaller than100 nanometres in at least one dimension. Nanomaterials properties are dif-ferent from those of the same materials with micron- or mm-scale dimensionsand intentionally produced and designed with very specific properties related toshape, size, surface properties and chemistry. Its relative-surface area is one ofthe main factors that increase its reactivity, strength and electrical properties.


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The applications of nanomaterials include, but are not limited to: better in-sulation and absorption materials, low-cost flat-panel displays, superior cuttingtools, high energy density batteries, high-sensitivity sensors, automobiles withbetter fuel efficiency, fuel cells, reliable satellites design, coatings, nano ballsbearings, nanoscale magnetic materials in data storage device, and nanostruc-tured membranes for water purification.

Although many potentially beneficial applications exist, in contrast, manyscientific institutions and government organizations across the world have under-scored the need to assess their possible health and environmental risks. Amongthem, worker protection during manufacture, use of nanomaterials, safe disposalof engineered nanomaterials, disposing of contaminated equipment, contamina-tion of soil, air, and water.

Bio-materials

Biomaterials are any natural or manmade materials that interacts with biologicalsystems. They are used to replace natural functions. Biomaterials are used to fillbony defects, an implant for young children that resorbs after the bodys boneshave healed, used to provide faster healing and reducing complications, used in

dental applications, surgery, and drug delivery. Some examples of biomaterialsare: a knee brace, contact lens, breast implant, heart transplant.

6.4 Machinability of Metals

Machinability of a material can be defined as the ease or difficulty with whichit can be machined. Machinability changes on the physical properties and thecutting conditions of the material. Materials with good machinability requireminimum power to machine, can be cut relatively fast, easily obtain a goodfinish, and do not wear the tooling as much.

There is no quantitative measurement on the machinability of a material. Ingeneral, machinability is expressed as a percentage or a normalized value. The

American Iron and Steel Institute (AISI) has determined AISI No. 1112 carbonsteel a machinability rating of 100%. Machinability of different materials basedon 100% machinability for AISI 1212 steel are given in Table B-13 in AppendixB.

6.5 Material Selection

Selecting proper materials and understanding the fabrication processes relatedwith design are two of the most critical responsibilities of a design engineer.In some cases, materials selection for a product design can be a challengingtask, especially when multi objectives and constraints such as minimum weight,flexible, strong, minimum cost, environmental impact, and recyclable etc., are

considered.A primary design requirement in the selection of a material for a product

design is that the material be capable of meeting the design service life require-ment at the least cost. For a defined service life, the designer normally narrowsavailable choices to a few candidate materials by using trade offs analysis. Thefinal selection of a particular material can be made from past experience withsimilar materials. Selecting materials based on past experience is still commonbecause the designer feels confident in using a tried and proven material.

For initial selection of candidate materials for a product design can also beperformed by using Ashby charts shown in Figures 6.21 and 6.22. These charts


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Figure 6.21: Ashby chart forYoungs modulus versus density. (Reprinted from: Ashby, MF. On the enginproperties of materials Acta Metallurgica, 37, p. 1278, 1989.)

provide physical insight into trade-offs (usually between product performanceand cost) by pairing properties which must normally both should be consideredat once. This process will help to prevent the need to work with many materialproperty tables. Ashby materials selection charts are a unique graphical way ofpresenting material property data. These charts provide the materials propertydata asballoonsin an easy way to compare and show the relative position for allof the materials being considered for a specific design application. This will helpto focus on possible candidates, and also helps to develop physical insight for therelative performance of materials. When data for a given material class such asmetals, ceramics, composites, and polymers are plotted, as shown in Figure 6.22


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Figure 6.22: Ashby chart for strength versus density. (Reprinted from: Ashby, MF. On theengineering properties of materials Acta Metallurgica, 37, p. 1280, 1989.)

they are grouped in an enclosed field balloon. These distinct balloons whichdefine the ranges of their properties may overlap and may have little balloonswithin their enclosed region.

6.5.1 Material Performance Requirements

Material properties and characteristics cover a broad range of parameters andplay an important role in satisfying the design requirements. In general, materialstrength has three distinct elements11:

11A. Ertas, and J. C. Jones, The Engineering Design Process, Second Addition, John Wiley& Sons, Inc., New York, 1996.


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1. Static st

materials for design - module 6

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