Mechanical Behavior, Testing, and Manufacturing …testbank360.eu/sample/solution-manual-manufacturing-engineering... · Chapter 2 Mechanical Behavior, Testing, and Manufacturing

Chapter 2

Mechanical Behavior, Testing, andManufacturing Properties ofMaterials

QUALITATIVE PROBLEMS

2.16 Using the same scale for stress, the tensile true stress-true strain curve is higherthan the engineering stress-engineering strain curve. Explain whether this con-dition also holds for a compression test.

During a compression test, the cross-sectional area of the specimen increases as the load isincreased. Since true stress is defined as load divided by the instantaneous cross-sectionalarea of the specimen, the true stress in compression will be lower than the engineering stressfor a given load, providing that frictional forces (between the platens and the specimen) arenegligible.

2.17 With the aid of a simple sketch, explain whether it is necessary to use the offsetmethod to determine the yield stress, Y , of a material that has been highly coldworked.

As can be seen by reviewing Fig. 2.3 on p. 68, a highly cold-worked metal will have a distinctchange in slope on its stress-strain curve occurring at the yield point, so that the offset methodis not necessary. See also Fig. 2.6 on p. 72.

2.18 Explain why the difference between engineering strain and true strain becomeslarger as strain increases. Does this difference occur for both tensile and com-pressive strains? Explain.

The answer lies in the fact that the definitions of engineering strain and true strain aredifferent, the latter being based on the actual or instantaneous dimensions, as can be seen in

8

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Mechanical Behavior and Manufacturing Properties of Materials 9

Eqs. (2.2) and (2.7) on p. 67 and p. 70, respectively. In both cases of tension and compression,the difference increases as strain increases. This is shown quantitatively in Problem 2.51.

2.19 If a material does not have an endurance limit (for example, aluminum), howwould you estimate its fatigue life?

Materials without endurance limits have their fatigue life defined as a certain number ofcycles to failure at a given stress level. For engineering purposes, this definition allows foran estimate of the expected lifetime of a part. The part is then usually taken out of servicebefore its lifetime is reached. An alternative approach is to use nondestructive test techniques(Section 36.10 on p. 1132) to periodically measure the accumulated damage in a part, andthen use fracture mechanics approaches to estimate the remaining life.

2.20 Which hardness tests and scales would you use for very thin strips of metal, suchas aluminum foil? Why?

A hardness test that produces small indentations would have to be used; also, since aluminumfoil is relatively soft, a very light load would be required. Two scales that satisfy theserequirements are the Knoop microhardness (HK) and the Vickers hardness (HV) at verylight loads (see Fig. 2.12 on p. 80). An area of current research is the use of atomic forcemicroscopy and nanoindenters to obtain the hardness of very thin materials and coatings.The shape of the indenter used is not exactly the same as in Fig. 2.12, and the loads are inthe micro- to milli-Newton range.

2.21 Which of the two tests, tension or compression, requires higher capacity of testingmachine, and why?

The compression test requires a higher capacity machine since the cross-sectional area of thespecimen increases as the test progresses. The increase in area requires a load higher thanthat for the tension test to achieve the same stress level. Also, there is friction between theflat dies (platens) and the workpiece surfaces in a compression test (see Sections 2.3 and 14.2on p. 373) which results in higher pressures than in tension; this higher pressure then requireslarger forces for the same cross-sectional area. In addition, there is more redundant work incompression testing than in tension testing, so the material will work harden more (unlessthe test is conducted at elevated temperatures).

2.22 List and explain briefly the conditions that induce brittle fracture in an otherwiseductile metal.

Brittle fracture can be induced by high deformation rates, lower temperatures (particularlythose with bcc structure), the presence of stress concentration (notches and cracks), stateof stress, radiation damage, corrosion (including hydrogen embrittlement). In each case,the stress needed to cause yielding is raised above the stress needed to cause failure, or thestress needed for a crack to propagate is below the yield stress of the material (as with stressconcentration).

2.23 List the factors that you would consider in selecting a hardness test and in theninterpreting the results from this test.



Hardness tests mainly have three differences: (a) type of indenter, (b) applied load, and(c) method of indentation measurement, i.e., depth or surface area of indentation, or re-bound of indenter. The hardness test selected would depend on the estimated hardness ofthe workpiece, its size and thickness, and if average hardness or the hardness of individualmicrostructural components is desired. For instance, the scleroscope, which is portable, iscapable of measuring the hardness of large pieces that cannot be used for measurement byother techniques.

The Brinell hardness test leaves a fairly large indentation, thus providing a good measure ofaverage hardness, while the Knoop test leaves a small indentation that allows for determina-tion of the hardness of the individual phases in a two-phase alloy. The small indentation ofthe Knoop test also allows it to be useful in measuring the hardness of very thin layers orplated layers on parts. Note that the depth of indentation should be small relative to partthickness, and that any change in the appearance of the bottom surface the part will makethe test results invalid.

Figure 2.14 on p. 84 is a useful guide for determining which hardness test is valid for a classof material. Note that often numerous hardness tests are suitable for a material. In thesecases, the best hardness test is the one that has one or more of the following characteristics:

• The best hardness test is often one that can be performed quickly; thus, it may bedesirable to also select a hardness test based on available equipment.

• Hardness tests are often specified by customers as part of a quality control requirement.Whatever form of hardness test is specified by the customer is the appropriate one touse.

• A hardness test that is most commonly used in a plant may be the best choice sincetechnicians will be most familiar with the test protocol and the equipment is most likelyto be in good calibration.

• Experimental error can be minimized by selecting a hardness test that gives the largestpenetration or indentation size.

2.24 Using Fig. 2.6 only, explain why you cannot calculate the percent elongation ofthe materials listed.

Recall that elongation (total) is defined by Eq. (2.4) on p. 69 and depends on the originalgage length (lo) of the specimen. Note that if the gage length encompasses a necked regiononly, it will register a larger percent elongation than if the gage length is four or five times aslong as the necked region, for example. From Fig. 2.6 on p. 72, the true necking and fracturestrains can be determined. Since this is a true strain, it is not dependent on the gage length.That is, regardless of the gage length, the same true stress will be measured. Since there isno way of incorporating gage length, one cannot obtain the percent elongation from Fig. 2.6.(See also the answer to Problem 2.36.)

2.25 If you pull and break a tension-test specimen rapidly, where would the temper-ature be highest, and why?

Since temperature rise is due to work input, it is obvious that the temperature will be high-est in the necked region because that is where the strain is highest and, hence, the energydissipated per unit volume in plastic deformation is highest.



2.26 Will the disk test be applicable to a ductile material? Why or why not?

With a ductile material, a point load on a disk results in the circular disk being flattened atthe platens and attaining elliptical shape of the originally round specimen. The flatteningconverts the point load to a distributed load, completely changing the stress state in the piece.Therefore, Eq. (2.10) on p. 77 is not valid, and the usefulness of the test is compromised.

2.27 What hardness test is suitable for determining the hardness of a thin ceramiccoating?

For a thin ceramic coating, it is still important that the hardness of the coating and not thesubstrate be measured. Most ceramics have limited ductility (Section 8.3 on p. 224), so thatKnoop or Vickers tests are suitable, although the Mohs test can also be used to obtain aqualitative value. Because of the increasing importance of coatings, special microhardnesstests have been developed for their hardness measurement.

2.28 In a Brinell hardness test, the resulting impression is found to be an ellipse. Givepossible explanations for this result.

Two possible explanations for an elliptical impression after a Brinell test are: (a) An obviousreason is the possible presence of asymmetric residual stresses in the surface layers of thematerial before the test. (b) The material itself may be highly anisotropic, such as a fiber-reinforced composite material, or due to severe cold working.

2.29 Some coatings are extremely thin - some, as thin as a few nanometers. Explainwhy even the Knoop test is not able to obtain reasonable results for such coat-ings. Recent research has attempted to use highly polished diamonds (tip radiusaround 5 nanometers) to indent such coatings in atomic force microscopes. Whatconcerns would you have regarding the appropriateness of the results?

With a coating of thickness of 5 nm, the stressed volume has to be approximately one-tenththis depth, which begins to approach the size of individual atoms. Thus, a knoop indentorwould need to have a tip radius that was atomically sharp in order to get results. Even withhighly polished diamond tips in atomic force microscopes, this scale problem is unavoidable.However, there are additional concerns in that the diamond indenter may not be symmetric,there are large adhesive forces at the small scales, there are complicated elastic and viscoelasticrecovery at small length scales, there may be residual stresses at the surface, and the stressedvolume may or may not contain a dislocation (whereas with Knoop tests, there is always anumber of dislocations).

2.30 Select an appropriate hardness test for each of the following materials. Justifyyour answer.

a) Cubic boron nitride

b) Lead

c) Cold-drawn 0.5% C steel

d) Diamond

e) Caramel (candy)



f) Granite

Figure 2.14 on p. 84 is a useful guide for selecting hardness tests.

(a) Cubic boron nitride is very hard, and useful data can be obtained only from the Knoopand Mohs tests. The Mohs scale is qualitative and does not give numerical values forhardness, so the Knoop test is preferable.

(b) Lead. As shown in Fig. 2.14, lead is so soft that only the Brinell and Vickers tests yielduseful data. Recognizing that lead is very soft, the lightest loads in these tests shouldbe used. Consider the expected results in this test if a typical value of hardness is 4 HBor 4 HV. For the Brinell test, Fig. 2.12 suggests that the expected indentation for a 500kg load is:

HB =2P

(πD)(D −

√D2 − d2

)Therefore, solving for d,

d =

√D2 −

(D − 2P

(πD)(HB)

)2

=

√102 −

(10− 2(500)

[π(10)] (4)

)2

= 9.79 mm

Note that this dimension is almost the same as the diameter of the indentor, and makesthe usefulness of the test highly questionable. For the Vickers test, the expected inden-tation test, using the lowest allowable load of 1 kg, is:

HV =1.854P

L2→ L =

√1.854P

HX=

√1.854(1)

4= 0.68 mm

This is much more reasonable, suggesting that the Vickers test is the best alternativefor lead.

(c) Cold-drawn 0.5% steel. From Fig. 2.14, all of the hardness tests are suitable for thismaterial. As discussed in Problem 2.23, the best choice for this material will depend ona number of factors.

(d) Diamond. The hardness of diamond is difficult to obtain. The hardness of diamond isreally determined by extrapolating the hardness on the Mohs curve to another scale inFig. 2.14. The hardness of diamond is usually quoted as 8000 to 10,000 HK.

(e) Caramel (candy). This would be an interesting experiment to perform, but the resultwill be that none of the hardness tests can be used for this material because it is fartoo soft. Also, the hardness of caramel is strongly temperature-dependent and that itcreeps, so that hardness measurement may be meaningless.

(f) Granite. The hardness of granite varies according to the source, but it is approximatelyaround apatite on the Mohs scale. Thus, various hardness tests can give valuable infor-mation on granite. Note, however, that in inspecting granite surfaces, one can see variousregions within which there would be hardness variations. The particular hardness testselected will depend on various factors, as discussed in part (c) above.



QUANTITATIVE PROBLEMS

2.31 A paper clip is made of wire 1 mm in diameter. If the original material fromwhich the wire is made is a rod 18 mm in diameter, calculate the longitudinalengineering and true strains that the wire has undergone during processing.

Engineering strain is defined by Eq. (2.2) on p. 67. Thus, because of volume constancy inplastic deformation, we may write

e =lflo

=(

do

df

)2

=(

181.0

)2

= 324

Also, letting lo be unity, the engineering strain is (324 − 1)/1 = 323. True strain is definedby Eq. (2.7) on p. 70. Hence

ε = ln(

3231

)= 5.78

Note the large difference in strains, even though they both describe the same phenomenon.(See also the last paragraph in Sec. 2.2.3 on p. 70.)

2.32 A strip of metal is 250 mm long. It is stretched in two steps, first to 300 mm andthen to 400 mm. Show that the total true strain is the sum of the true strainsin each step - that is, that true strains are additive. Show that, in the case ofengineering strains, the strains cannot be added to obtain the total strain.

We first calculate the true strains for each step:

Step 1: ε1 = ln(300/250) = 0.182

Step 2: ε2 = ln(400/300) = 0.288

Thus the total true strain is

εtotal = 0.182 + 0.288 = 0.470

The total strain may also be calculated from the final and initial dimensions as

εtotal = ln(400/250) = 0.470

thus giving the same answer.

Engineering strains are calculated likewise. Thus,

Step 1: e1 = (300− 250)/250 = 0.200

Step 2: e2 = (400− 300)/300 = 0.333

Thus the total engineering strain is 0.200+0.333=0.533. However, when calculated from theinitial and final dimensions, the total engineering strain is

etotal = (400− 250)/250 = 0.600

which is higher than the sum of the engineering strains. Consequently, engineering strainsare not additive, whereas true strains are.



2.33 Identify the two materials in Fig. 2.6 that have the lowest and the highest uniformelongations. Calculate these quantities as percentages of the original gage lengths.

The magnitude of uniform elongation is directly related to the true strain at the onset ofnecking. As we see in Fig. 2.5c on p. 71, the necking strain on a true stress-true strain curvecorresponds to the beginning of the straight portion of the curve. Thus, from Fig. 2.6 onp. 72 we note that the lowest uniform elongation is for 1112 cold-rolled steel, with a neckingstrain of about 0.05. The highest is for 304 stainless steel, although 70-30 annealed brass isclose to it, with a necking strain of about 0.45.

To relate these values to a percentage of the gage length (lo), we convert true strain to percentelongation as follows:

ε = ln(l/lo)

Or, l/lo = eε. Hence,Elongation = (eε − 1)× 100

Thus, for 1112 cold-rolled steel

Elongation = (e0.05 − 1)× 100 = 5%

And for 70-30 brass, the specific value is

Elongation = (e0.45 − 1)× 100 = 57%

2.34 Plot the ultimate strength versus stiffness for the materials in Table 2.2, andprepare a three-dimensional plot for the metals in Table 2.2 where the third axisis maximum elongation in 50 mm.

The graphs are constructed using the minimum values in Table 2.2 on p. 67 when ranges ofvalues exist. They are as follows:

0 50 100 150 200 250 300 3500

100

200

300

400

500

600

700

800

UTS

(MPa

)

E (GPa)

Lead

Aluminum

Copper

Magnesium

Molybdenum

Nickel

SteelTitanium

Tungsten

Zinc

0 50 100 150 200 250 300 350 4000

500

1000

1500

2000

2500

3000

3500

Ceramics, glassSilicon carbide

Silicon nitride

ThermoplasticsThermoplastics, reinforced

Thermosets

Boron fibers

Carbon fibers

Glass fibers

Kevlar fibers

Spectra fibers

UTS

(MPa

)

E (GPa)



2.35 If you remove the layer of material ad from the part shown in Fig. 2.29d -for instance, by machining or grinding - which way will the specimen curve?(Hint: Assume that the part shown in sketch d in the figure is composed offour horizontal springs held at the ends. Thus, from the top down you havecompression, tension, compression, and tension springs.)

The part will bow downward, with the center of curvature below the part (that is, it will holdwater). The internal forces will have to achieve a state of static equilibrium. The top layerof material ad, which is under compression, has the tendency to bend the bar upward. Whenthis stress is relieved, such as by removing the top layer by machining or grinding, the barwill compensate by bending downward. Such residual stress systems can be modeled witha set of horizontal tension and compression springs. Thus, starting at the top, we have acompression, tension, compression, and tension spring.

2.36 Percent elongation is always described in terms of the original gage length, suchas 50 mm or 2 in. Explain how percent elongation varies as the gage length ofthe tensile specimen increases. (Hint: Recall that necking is a local phenomenon,and think of what happens to the elongation as the gage length becomes verysmall.)

As gage length increases, the percent elongation (total) will decrease to a value representinguniform elongation. For instance, consider taking a gage length that just encompasses thenecking region (thus approaching zero gage length); the percent elongation in this regionwill be very high because the effect of localized deformation will tend to greatly increase theelongation. As gage length increases, however, the localized deformation zone has less andless effect on the total elongation.

2.37 Make a sketch showing the nature and distribution of residual stresses in Fig.2.30a and b, before they were cut. (Hint: Assume that the split parts are freefrom any stresses, then force these parts back to the shape they had before theywere cut.)

As the problem states, we assume that the split parts shown in Fig. 2.30a and b on p. 96are free of stresses. After unbending the two halves (that is, closing the gap by pushing thetwo halves together), you will note that the inner (mating) surfaces will now be subjected tolongitudinal compressive stresses, while the outer surfaces are in tension. Consequently, thisrepresents the state of stress prior to slitting.

2.38 You are given the K and n values of two different metals. Is this information suf-ficient to determine which metal is tougher? If not, what additional informationdo you need?

Although the K and n values may give a good estimate of toughness, true fracture stressand true strain at fracture are also required for accurate calculation of the toughness of themetals. The modulus of elasticity and yield strength would provide information about thearea under the elastic region (known as modulus of resilience; see Problem 2.45), but thisregion is very small and is negligible compared to the rest of the curve.

2.39 A cable is made of two strands of different materials, A and B, and cross-sections



as follows:

For material A: K = 70, 000 psi, n = 0.5, Ao = 0.6 in2

For material B: K = 25, 000 psi, n = 0.5, Ao = 0.3 in2

Calculate the maximum tensile force that this cable can withstand prior to neck-ing.

Note that necking will occur when ε = n = 0.5. At this point, the true stresses in each cableare (using σ = Kεn), respectively,

σA = 70, 000ε0.5 = 49, 500 psi

andσB = 25, 000ε0.5 = 17, 700 psi

The true areas at necking can be calculated as follows:

AA = 0.6e−0.5 = 0.364 in2

andAB = 0.3e−0.5 = 0.182 in2

Thus the total load that the cable can support is

Ptotal = (49, 500)(0.364) + (17, 700)(0.182) = 21, 200 lb

2.40 On the basis of the information given in Fig. 2.6, calculate the ultimate tensilestrength (engineering) of annealed copper.

Recall that the definition of engineering stress is σeng = P/Ao, and that for true stress it isσ = P/A. From Fig. 2.6, the true stress at necking for copper is found to be about 50,000psi, while the true strain is about 0.4. We also know that the ratio of the original to thenecked areas of the specimen is given by

ln(

Ao

Aneck

)= 0.4

orAneck

Ao= e−0.4 = 0.67

Thus, the engineering stress is calculated as

σeng = (50, 000)(0.67) = 33, 500 psi

2.41 In a disk test performed on a specimen 1.25 in. in diameter and 1/2 in. thick,the specimen fractures at a stress of 30,000 psi. What was the load on it?

For this problem, we use Eq. (2.10) on p. 77. Thus,

σ =2P

πdt; P =

σπdt

2=

(30, 000 psi)π(1.25 in.)(0.375 in.)2

= 22, 100 lb



2.42 A piece of steel has a hardness of 300 HB. Calculate the tensile strength in MPaand in psi.

Using Eq. (2.13) and (2.14) on pp. 82 and 83, respectively, we find that

UTS = 3.5(HB) = 3.5(300) = 1050 MPa

andUTS = 500(300) = 150, 000 psi

2.43 A material has the following properties: UTS = 50,000 psi and n = 0.25. Calcu-late its strength coefficient, K.

Let’s first note that the true UTS of this material is given by σ = Knn. We can then determinethe value of this stress from the UTS by following a procedure similar to the Example 2.1 onp. 73. Since n = 0.25, we can write

True UTS = UTS (Ao/Aneck)

where Ao/Aneck = en = 0.250.25 = 1.28. Hence, true UTS=50,000 (1.28)=64,200 psi. There-fore,

K = 64, 200/(0.25)0.25 = 90, 800 psi

2.44 A material has a strength coefficient K = 100,000 psi and n = 0.2. Assuming thata tensile-test specimen made from this material begins to neck at a true strainof 0.2, show that the ultimate tensile strength of this material is 59,340 psi.

The approach is very similar to that for Problem 2.43. The true UTS is given by

True UTS = Knn = 100, 000(0.2)0.2 = 72, 500 psi

The true UTS is also related to the UTS by:

True UTS = UTS(Ao/Aneck) = UTSen = 1.22 UTS

Hence, UTS=72,500/1.22=59,340 psi.

2.45 Modulus of resilience is defined as the area under the elastic region of the stress-strain curve of the material; it has the units of energy-per-unit volume. Derive anexpression for the modulus of resilience in terms of the yield stress and modulusof elasticity of the material.

The area under the elastic portion of the curve is simply

Area = (Yield stress)(Yield strain)/2

The modulus of resilience is therefore

MR = (Y )(Y/E)/2 = Y 2/2E



2.46 What is the modulus of resilience for a highly cold-worked piece of steel havinga hardness of 275 HB? For a piece of highly cold-worked copper with a hardnessof 100 HRB?

From the first paragraph in Section 2.6.2 on p. 82, we note that the yield strength of acold-worked metal is about three times the yield stress, Y . For the cold-worked steel with ahardness of 275 HB, the yield strength is then 825 MPa. From Table 2.2 on p. 67, Esteel = 190GPa. Therefore, using the expression derived in Problem 2.45,

MRsteel =Y 2

2E=

(825 MPa)2

2(190 GPa)= 1.79 MPa

Similarly, for the cold-worked copper, Y = 300 MPa, E = 105 GPa, and thus

MRcopper =Y 2

2E=

(300 MPa)2

2(105 GPa)= 0.43 MPa

2.47 Using only Fig. 2.6, calculate the maximum load in tension testing of a 304stainless steel specimen with an original diameter of 5 mm.

We note from the figure that the necking strain for 304 stainless steel is 0.45 and that thetrue stress at necking is about 900 MPa. The true area at necking is therefore

Aneck = Aoe−n =

[π

452

]e−0.45 = 12.5 mm2

Hence the maximum load is

Pmax = (900 MN/m2)(12.5 mm2) = 11, 270 N

2.48 Calculate the major and minor pyramid angles for a Knoop indenter and compareto Vickers and Rockwell A indenters.

Refer to Fig. 2.12 on p. 80 for the geometry of the indenters. For a Knoop indenter withb/t = 4, the pyramid’s minor angle is given by

β = tan−1(2) = 63◦

Also, since L/b = 7.11 and b/t = 4, then

L

t=

L

b

b

t= (7.11)(4) = 28.44

Therefore, the major pyramid angle is

α = tan−1(14.22) = 86◦

2.49 If a material has a target hardness of 350 HB, what is the expected indentationdiameter?



From Fig. 2.12 the indentation diameter depends on the load and the Brinell hardnessaccording to the equation

HB =2P

(πD)(D −

√D2 − d2

)where P is the load (either 500, 1500 or 3000 kg) and D is the diameter of the indenting ball(10 mm). This equation can be solved for d as:

d =

√D2 −

(D − 2P

πD(HB)

)2

Thus, (a) for a load of 500 kg, d is 1.34 mm, (b) a load of 1500 kg, d is 2.32 mm, and (c) fora load of 3000 kg, d is 3.25 mm.

2.50 A Rockwell A test was conducted on a material and a penetration depth of 0.1 mmwas recorded. What is the hardness of the material? What material will typicallygive such hardness values? If a Brinell hardness test were to be conducted on thismaterial, give an estimate of the indentation diameter if the load used is 1500 kg.

From Fig. 2.12, the Rockwell hardness is calculated from

HRA = 100− 500t = 100− 500(0.1) = 50 HRA

From Fig. 2.14 on p. 84, a material with a 50 HRA should have a Brinell hardness of 140HB. Therefore, from Fig. 2.12,

HB =2P

(πD)(D −

√D2 − d2

)or, solving for d,

d =

√D2 −

(D − 2P

(πD)(HB)

)=

√102 −

(10− 2(1500)

(π(10))(140)

)= 3.6 mm

2.51 A material is tested in tension. Over a 1-in. gage length, the engineering strainmeasurements are 0.01, 0.02, 0.03, 0.04, 0.05, 0.1, 0.15, 0.2, 0.5, and 1.0. Plotthe true strain versus engineering strain for these readings.

The engineering strain is easy to calculate from Eq. (2.2) on p. 67, and the true strain is givenby Eq. (2.7) on p. 70 as

ε = ln(

l

lo

)The following chart can then be made of the data points:

e 0.01 0.02 0.03 0.04 0.05 0.1 0.15 0.2 0.5 1.0ε 0.01 0.0198 0.0296 0.0392 0.0488 0.0953 0.1397 0.182 0.405 0.69

The plot is as follows:



Slope = 1

1.0

0.8

0.6

0.4

0.2

0

Tru

e st

rain

, ε

1.00.80.60.40.20Engineering strain, e

SYNTHESIS, DESIGN, AND PROJECTS

2.52 List and explain the desirable mechanical properties for (a) an elevator cable, (b)a paper clip, (c) a leaf spring for a truck, (d) a bracket for a bookshelf, (e) pianowire, (f) a wire coat hanger, (g) a gas-turbine blade, and (h) a staple.

(a) Elevator cable: The cable should not elongate elastically to a great extent or undergoyielding under the load applied. These requirements call for a material with a high elasticmodulus and yield stress. The cable should also be sufficiently flexible and ductile to bewrapped around the drums during its use.

(b) Paper clip: The clip material must possess high ductility to allow it to be formed withoutfracture. In its normal use, the clip should recover elastically when removed as well asapply sufficient force to hold papers together (thus should possess high elastic modulus).

(c) Leaf spring: The function of the leaf spring is to absorb energy elastically upon staticor dynamic loading. The energy should be absorbed in an elastic manner because afterabsorbing the energy, the spring should return to its original shape. This requires a highyield stress and high elastic modulus that maximizes the area under the elastic portionof the stress-strain curve (modulus of resilience).

(d) Bracket: Much like an elevator cable, but not as critical, a material with a high yieldstress and elastic modulus would be required. These properties would keep the shelffrom sagging excessively under load.

(e) Piano wire: A piano wire is under high tension (hence high stress) to achieve the desiredtone. The wire should be able to maintain this stress level for a period of time, orthe piano would become out of tune. Thus, its yield stress should be high in order tosustain the tension required to produce the tone, and should not be susceptible to stressrelaxation. Furthermore, it should be sufficiently ductile to be able to be wound aroundthe tightening mechanisms.



(f) Wire coat hanger: Much like the paper clip, the hanger requires a large amount ofdeformation in its manufacturing, so the material must possess high ductility. Thehanger should also be able to maintain its shape when clothes are hung on it, hence itshould possess sufficient yield strength and elastic modulus.

(g) Gas turbine blade: A gas turbine blade is required to operate at high temperatures(depending on its location or stage in the engine), so it should have high-temperaturestrength and be resistant to creep, as well as oxidation and corrosion due to combustionproducts during use.

(h) Staple: The properties should closely parallel a paper clip. The staple should have highductility to allow it to be deformed without fracture, and also a low yield stress so thestaple can be bent (as well as removed) easily without requiring the user to exert a greatamount of force.

2.53 When making a hamburger, have you observed the type of cracks shown in Fig.2.19d? What can you do to avoid such cracks? Note: Test hamburger pattiesby compressing them at different temperatures and observe the crack path (i.e.,through the fat particles, the meat particles, or their interface).

These cracks do occur, and they can be eliminated by making sure that the chunk of groundmeat is not too cold. In experiments, which can be performed by the student, it can beobserved that when cold, cracks propagate through the meat particles rather than throughthe fat. As temperature increases, cracks would propagate through the fat, which eventuallysoftens and melts as higher temperatures are reached (which is easily observed when fryingground meat in a pan to make, for example, meat sauce).

2.54 An inexpensive claylike material, called Silly Putty r©, is often available in storesthat sell toys and games. Obtain some and do the following experiments. (a)Shape it into a ball and drop it onto a flat surface. (b) Put the re-rounded ballon a table, and place a heavy book on it for five minutes. (c) Shape it intoa long cylindrical piece, and pull it apart - first slowly, then quickly. Describeyour observations and refer to the specific sections in this chapter where eachparticular observation is relevant.

Silly putty is an extremely strain-rate sensitive material. When shaped into a ball andbounced, it undergoes a high strain rate and acts as very rigid, elastic material; i.e., itbounces. When placed on a table and subjected to a static force such as placing a book init, it deforms as a very soft, plastic material and there is very little elastic recovery. Whenstretched slowly, it is very ductile; when stretched quickly, it can be seen that the ductility isdrastically reduced, and the stiffness and strength are also much higher.

2.55 Make individual sketches of the mechanisms of testing machines that, in youropinion, would be appropriate for tension, for torsion, and for compression testingof specimens at different rates of deformation. What modifications would youmake to include the effects of temperature on material properties?

By the student. The important features are that the jaws of the machine must move atconstant speed; this can be simply done using a power screw drive, with a load cell tomeasure force. More elaborate machines can use hydraulic power sources and control systems



to achieve any desired mode of loading. For testing at elevated temperatures, the machinemust be protected, and the specimen must be insulated from the environment to maintainconstant temperature during the test. This clearly requires a test chamber that is thermallyinsulated.

2.56 In Section 2.6.1, we described the Mohs hardness test. Obtain small pieces ofseveral different metallic and nonmetallic materials, including stones. Rub themagainst each other, observe the scratches made, and order them in a mannersimilar to the Mohs numbering system.

By the student. Some subtle points are that scratches may be difficult to detect and measurewhen the hardness of the specimen and that of the harder body are close to each other. Also,the materials to be tested may have hardness variations along their surfaces.

2.57 Demonstrate stress relaxation by tightly stretching thin plastic strings betweentwo nails on a long piece of wood. Pluck the strings frequently, to test thetension as a function of time and of temperature. (Note: Alter the temperatureby placing the fixture in an oven set on “low.”)

By the student. This experiment can be easily performed with a fishing line, but one mustbe careful with a wooden support in an oven. The string tone decreases in frequency overtime, demonstrating stress relaxation.

2.58 Demonstrate the impact toughness of a round piece of chalk by first using atriangular file to produce a V-notch, as shown in Fig. 2.18a, and then bendingthe chalk to break it.

By the student. Chalk is relatively easy to break by hand. With a carefully filed, sharp notch,a dramatic reduction in the required force can be easily observed. With a simple fixture (suchas using a vise), the chalk can be cantilevered and small weights dropped from ever higherdistances to demonstrate the level of energy required to fracture the chalk.

2.59 Using a large rubber band and a set of weights, obtain the force-displacementcurve for the rubber band. How is this different from the stress-strain curvesshown in Fig. 2.5?

By the student. See also Fig. 7.14 on p. 214 and note that the behavior of rubber is highlynonlinear elastic. Two qualitative examples of load-elongation or stress-strain curves are givenbelow. Some rubber bands have a prestress, so that a non-zero force is needed to obtain asmall deflection, but these are rare. Most have section where the slope of the stress-straincurve is low, but the curve is very steep near fracture. This can be easily felt with a rubberband. Occasionally, there will be an initial stiff section as shown.



Load

or

stre

ss

Extension or strain

Finite load neededfor incremental strain

Initial high modulus may be absent

Low modulus extensionHigh modulusextension at high strain

Fracture

**

2.60 Find or prepare some solid circular pieces of brittle materials (such as chalk,ceramics, etc.) and subject them to the type of test shown in Fig. 2.9 by usingthe jaws of a simple vise. Describe your observations as to how the materialsfracture. Repeat the tests using ductile materials (such as clay, soft metals, etc.)and describe your observations.

By the student. When carefully performed, one can demonstrate the failure mode as describedin the disc test on p. 77, namely, that the brittle material fractures in tension along themidplane of the specimen. Note that most vises have a knurled, rough surface finish whichcan influence the results; a hard and flat spacer may need to be inserted into the jaws of thevise on opposite ends of the disk. Also, for a ductile material, the disk will flatten, and it maybe very difficult to actually cause fracture if a soft specimen is being used in the experiment.

2.61 Devise a simple fixture for the bend tests shown in Fig. 2.11 and test sticks ofbrittle materials by loading them with dead weights until they break. Verify thestatement that the specimens on the right in the figure will break sooner thanthe ones on the left.

By the student. This test can be performed using specially fabricated fixtures on a tension testmachine. However, it can also be done very simply by using a small-diameter rod, suspendedbetween two desks and the load applied by a basket with two strings. Three-point bendingcan be simulated by placing both strings at the same location, and four-point bending byspacing the strings along the rod. The load in the basket can be increased until the rod fails.Other alternatives are also possible.

2.62 By pushing a ball bearing against the top surfaces of various materials (such asclay, dough, etc.) observe with a magnifier the shape of the indentation, similarto those shown in Fig. 2.13a and b.

By the student. Note that with simple materials, the indentation shape (when viewed fromabove) will always be circular. An interesting specimen can be carefully prepared by reinforc-ing threads placed in the material so that a non-circular patch can be created. A side view of



a deformed zone can be obtained by laminating materials, such as using clays with differentcolors.

2.63 Embed a small steel ball in a soft block of material and compress it as shownin Fig. 2.23a. Then cut it carefully along the center plane and observe thedeformation of the material. Repeat the same experiment by embedding a smallround jelly bean in the material and deforming it.

By the student. The steel ball will not deform, just as is shown in the middle sketch of Fig.2.23b on p. 90. The jelly bean has to be soft in order to simulate the behavior shown in Fig.2.23a. An alternative is to use two colors of Play-Doh r© in a carefully prepared specimen,with one material chilled to increase its stiffness.

2.64 Devise a simple experiment and perform tests on materials commonly foundin a kitchen by bending them for a qualitative assessment of their transitiontemperature, as shown in Fig. 2.24.

By the student. This experiment can be performed fairly easily on materials such as butter,chocolate, or cheese. The temperature can be varied and controlled by putting the materialsin a freezer or increasing it by immersion into water heated to different temperatures. A qual-itative assessment of transition temperature can thus be done; also, a number of experimentscan be performed to assess ductility (such as shearing the material and observing the strainsthat can be achieved - shearing can include smearing with a knife).

2.65 Locate some solid and some tubular metal pieces, and cut them as shown in Fig.2.30 to determine if there are any residual stresses in the parts prior to cuttingthem.

By the student. The answers can vary widely depending on the materials and their processinghistory. Most tubes have been extruded and drawn and will have some residual stressesnotable. Some tubes may be annealed and therefore be residual stress free.


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