Industrial Engineering Chapter 2 When This Stress Relieved Such Shape Should Stiff For Better Con Removing

Chapter 2

Fundamentals of the Mechanical

Behavior of Materials

Questions

2.1 Can you calculate the percent elongation of ma-

terials based only on the information given in

Fig. 2.6? Explain.

on p. 37 only the necking strain (true and engi-

elongation of the specimen; also, note that the

2.2 Explain if it is possible for the curves in Fig. 2.4

to reach 0% elongation as the gage length is in-

creased further.

will decrease, but the total elongation will not

approach zero.

2.3 Explain why the difference between engineering

increases. Is this phenomenon true for both ten-

sile and compressive strains? Explain.

p. 35. This is true for both tensile and com-

2.4 Using the same scale for stress, we note that the

During a compression test, the cross-sectional

area of the specimen increases as the specimen

requires a higher capacity testing machine than

the other? Explain.

The compression test requires a higher capacity

This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited

reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or

likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

2.6 Explain how the modulus of resilience of a ma-

2.7 If you pull and break a tension-test specimen

rapidly, where would the temperature be the

highest? Explain why.

the energy dissipated per unit volume in plastic

deformation, is highest.

2.8 Comment on the temperature distribution if the

specimen in Question 2.7 is pulled very slowly.

2.9 In a tension test, the area under the true-stress-

the area under the load-elongation curve rep-

resents the work done on the specimen. If you

the same as the area under the true-stress-true-

strain curve? Explain. Will your answer be the

same for any value of strain? Explain.

2.10 The note at the bottom of Table 2.5 states that

as temperature increases, Cdecreases and m

strength of the material. The value of min-

creases with temperature because the material

becomes more strain-rate sensitive, due to the

2.11 You are given the Kand nvalues of two dif-

ferent materials. Is this information sufficient

to determine which material is tougher? If not,

what additional information do you need, and

why?

thus usually negligible with respect to the rest

2.12 Modify the curves in Fig. 2.7 to indicate the

effects of temperature. Explain the reasons for

Fig. 2.10 on p. 42.

2.13 Using a specific example, show why the defor-

mation rate, say in m/s, and the true strain rate

This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited

reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or

likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

2.14 It has been stated that the higher the value of

m, the more diffuse the neck is, and likewise,

strain hardening. However, the strain rate in

the necking region is also higher than in the rest

on p. 45). As expected, the elongation after

necking (postuniform elongation) also increases

2.15 Explain why materials with high mvalues (such

as hot glass and silly putty) when stretched

2.16 Assume that you are running four-point bend-

ing tests on a number of identical specimens of

the same length and cross-section, but with in-

creasing distance between the upper points of

loading (see Fig. 2.21b). What changes, if any,

would you expect in the test results? Explain.

distance increases.

2.17 Would Eq. (2.10) hold true in the elastic range?

2.18 Why have different types of hardness tests been

developed? How would you measure the hard-

ness of a very large object?

2.19 Which hardness tests and scales would you use

for very thin strips of material, such as alu-

minum foil? Why?

microhardness test such as Knoop or Vickers

and scale for a particular application.

Hardness tests mainly have three differences:

(a) type of indenter,

This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited

reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or

likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

2.21 In a Brinell hardness test, the resulting impres-

explanations for this phenomenon.

2.21 Referring to Fig. 2.22 on p. 52, note that the

material for indenters are either steel, tungsten

carbide, or diamond. Why isn’t diamond used

for all of the tests?

it would not, for example, be practical to make

and use a 10-mm diamond indenter because the

2.22 What effect, if any, does friction have in a hard-

ness test? Explain.

as thermoplastic components.

2.24 Referring to the two impact tests shown in

Fig. 2.31, explain how different the results

as stress raisers. Thus, cracks would not propa-

gate as they would when under tensile stresses.

Consequently, the specimens would basically

2.25 If you remove layer ad from the part shown in

Fig. 2.30d, such as by machining or grinding,

compression, tension, compression, and tension

springs.)

This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited

reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or

likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

2.26 Is it possible to completely remove residual

stresses in a piece of material by the technique

described in Fig. 2.32 if the material is elastic,

linearly strain hardening? Explain.

2.27 Referring to Fig. 2.32, would it be possible to

eliminate residual stresses by compression in-

stead of tension? Assume that the piece of ma-

2.28 List and explain the desirable mechanical prop-

erties for the following: (1) elevator cable, (2)

bandage, (3) shoe sole, (4) fish hook, (5) au-

dergo yielding as the load is increased.

These requirements thus call for a mate-

and outer surface resistant to environmen-

tal effects.

for comfort, with a high resilience. It

should be tough so that it absorbs shock

and should have high friction and wear re-

sistance.

(e) Automotive piston: This product must

have high strength at elevated tempera-

tures, high physical and thermal shock re-

erates at high temperatures (depending on

its location in the turbine); thus it should

should have high ductility to allow it to be

deformed without fracture, and also have

the split parts are free from any stresses. (Hint:

Force these parts back to the shape they were

in before they were cut.)

cross section, have a residual stress distribu-

tion of tension-compression-tension. Using the

Fig. 2.31b would have a similar residual stress

distribution prior to cutting.

2.30 It is possible to calculate the work of plastic

deformation by measuring the temperature rise

This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited

reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or

likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

ature by one degree. Consequently, the calcu-

specimen changes when the specimen is sub-

terms are positive, the product of these terms is

negative and, hence, there will be a decrease in

volume (This can also be deduced intuitively.)

For case (b), it will be noted that the volume

will increase.

2.32 We know that it is relatively easy to subject

a specimen to hydrostatic compression, such as

by using a chamber filled with a liquid. Devise a

means whereby the specimen (say, in the shape

of a cube or a thin round disk) can be subjected

to hydrostatic tension, or one approaching this

state of stress. (Note that a thin-walled, inter-

nally pressurized spherical shell is not a correct

answer, because it is subjected only to a state

of plane stress.)

Two possible answers are the following:

(a) A solid cube made of a soft metal has all its

six faces brazed to long square bars (of the

same cross section as the specimen); the

bars are made of a stronger metal. The six

arms are then subjected to equal tension

forces, thus subjecting the cube to equal

tensile stresses.

2.33 Referring to Fig. 2.19, make sketches of the

tube is externally pressurized. Assume that the

2.34 A penny-shaped piece of soft metal is brazed

to the ends of two ﬂat, round steel rods of the

same diameter as the piece. The assembly is

then subjected to uniaxial tension. What is the

state of stress to which the soft metal is sub-

jected? Explain.

The penny-shaped soft metal piece will tend

to contract radially due to the Poisson’s ratio;

however, the solid rods to which it attached will

prevent this from happening. Consequently, the

state of stress will tend to approach that of hy-

drostatic tension.

2.35 A circular disk of soft metal is being com-

pressed between two ﬂat, hardened circular

steel punches having the same diameter as the

disk. Assume that the disk material is perfectly

plastic and that there is no friction or any tem-

perature effects. Explain the change, if any, in

the magnitude of the punch force as the disk is

being compressed plastically to, say, a fraction

of its original thickness.

Note that as it is compressed plastically, the

disk will expand radially, because of volume

constancy. An approximately donut-shaped

This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited

reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or

likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

stress state σ1,σ2,σ3, where σ1> σ2> σ3.

Explain what happens if σ1is increased.

2.37 What is the dilatation of a material with a Pois-

son’s ratio of 0.5? Is it possible for a material to

have a Poisson’s ratio of 0.7? Give a rationale

It can be seen from Eq. (2.47) on p. 69 that the

(2.47) would then predict contraction under a

2.38 Can a material have a negative Poisson’s ratio?

Explain.

2.39 As clearly as possible, define plane stress and

plane strain.

Plane stress is the situation where the stresses

in one of the direction on an element are zero;

plane strain is the situation where the strains

in one of the direction are zero.

is less than this threshold, then one must ei-

ther rely on nontraditional hardness tests, or

2.41 List the advantages and limitations of the

stress-strain relationships given in Fig. 2.7.

possible. Two possible answers are: (1) there

and (2) some materials may be better suited for

2.42 Plot the data in Table 2.1 on a bar chart, show-

ing the range of values, and comment on the

results.

By the student. An example of a bar chart for

the elastic modulus is shown below.

This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited

reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or

likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

0 100 200 300 400 500

Aluminum

Copper

Lead

Magnesium

Molybdenum

Nickel

Steels

Stainless steels

Titanium

Tungsten

Elastic modulus (GPa)

Metallic materials

Carbon fibers

Glass fibers

Kevlar fibers

Spectra fibers

Non-metallic materials

Typical comments regarding such a chart are:

(a) There is a smaller range for metals than

(b) Thermoplastics, thermosets and rubbers

are orders of magnitude lower than met-

2.43 A hardness test is conducted on as-received

that the hardness is too high, thus the mate-

rial may not have sufficient ductility for the in-

tended application. The supplier is reluctant to

accept the return of the material, instead claim-

ing that the diamond cone used in the Rockwell

testing was worn and blunt, and hence the test

needed to be recalibrated. Is this explanation

plausible? Explain.

Refer to Fig. 2.22 on p. 52 and note that if an

indenter is blunt, then the penetration, t, un-

der a given load will be smaller than that using

a sharp indenter. This then translates into a

higher hardness. The explanation is plausible,

but in practice, hardness tests are fairly reliable

and measurements are consistent if the testing

equipment is properly calibrated and routinely

serviced.

2.44 Explain why a 0.2% offset is used to determine

the yield strength in a tension test.

The value of 0.2% is somewhat arbitrary and is

2.45 Referring to Question 2.44, would the off-

set method be necessary for a highly-strained-

hardened material? Explain.

The 0.2% offset is still advisable whenever it

can be used, because it is a standardized ap-

far more easily discernable than for the same

This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited

reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or

likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

2.46 A strip of metal is originally 1.5 m long. It is

stretched in three steps: first to a length of 1.75

m, then to 2.0 m, and finally to 3.0 m. Show

ǫ= ln l

1.75= 0.1335

Therefore the true strains are additive. Us-

e2= 0.1429, and e3= 0.5. The sum of these

strains is e1+e2+e3= 0.8096. The engineering

lo

1.5=1.5

1.5= 1

Note that this is not equal to the sum of the

2.47 A paper clip is made of wire 1.20-mm in di-

ameter. If the original material from which the

Assuming volume constancy, we may write

lf

=do

=15

= 156.25 ≈156

ed=1.2−15

15 =−0.92

The longitudinal true strain is given by

ǫd= ln 1.20

2.48 A material has the following properties: UTS =

Let us first note that the true UTS of this ma-

n= 0.25, we can write

= (50,000)(1.28) = 64,200 psi

This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited

reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or

likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

70-30 brass.

From Fig. 2.6 on p. 37, the true stress for an-

2.50 Calculate the ultimate tensile strength (engi-

neering) of a material whose strength coefficient

n= 0.20. Following the same procedure as in

Example 2.1, we find the true ultimate tensile

2.51 A cable is made of four parallel strands of dif-

ferent materials, all behaving according to the

equation σ=Kǫn, where n= 0.3 The materi-

Material B: K= 600 MPa, Ao= 2.5 mm2;

Material C: K= 300 MPa, Ao= 3 mm2;

Material D: K= 760 MPa, Ao= 2 mm2;

(b) Explain how you would arrive at an an-

swer if the nvalues of the three strands

were different from each other.

AD= (2)e−0.3= 1.48 mm2

+(209)(2.22) + (530)(1.48)

= 3650 N

2.52 Using only Fig. 2.6, calculate the maximum

load in tension testing of a 304 stainless-steel

round specimen with an original diameter of 0.5

in.

0.196 in2. Since n= 0.1, we follow a procedure

Ao

=e0.1= 1.1

This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited

reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or

likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Thus

1.1= 100,000 psi

2.53 Using the data given in Table 2.1, calculate the

values of the shear modulus Gfor the metals

listed in the table.

The important equation is Eq. (2.24) on p. 49

2.55 A cylindrical specimen made of a brittle mate-

It is found that fracture takes place at an angle

will be zero. The third principal stress acting

on this specimen is normal to the specimen and

its magnitude is

σ3=30,000

This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited

reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or

likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

the values given in the text; this is due to the

2.57 Calculate the work done in frictionless compres-

(2) annealed copper, (3) annealed 304 stainless

steel, and (4) 70-30 brass, annealed.

The work done is calculated from Eq. (2.62) on

Eq. (2.60). For example, for 1100-O aluminum,

where Kis 180 MPa and nis 0.20, uis calcu-

100,000 psi Assuming that a tensile-test spec-

maximum load and the necking strain for this

UTStrue = (100,000)(0.17)0.17 = 74,000 psi.

The cross-sectional area at the onset of necking

2.59 A tensile-test specimen is made of a material

represented by the equation σ=K(ǫ+n)n.

This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited

reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or

likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

ameter but different heights. Assume that both

specimens are compressed (frictionless) by the

and from volume constancy,

and

hsf

=Dso

2.61 A horizontal rigid bar c–cis subjecting specimen

(See the accompanying figure.) The force Fis

Hence, in terms of true stresses and instanta-

neous areas, we have

4.5KσaLb

La

4.5Lb

La√ǫa

Hence, for a deﬂection of x,

The true strain in specimen b is given by

4.5

This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited

reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or

likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

2.63 In a disk test performed on a specimen 40-mm

in diameter and 5 m thick, the specimen frac-

tures at a stress of 500 MPa. What was the

load on the disk at fracture?

tension, we algebraically add a uniform tensile

stress to this stress distribution. Note that the

change in the stresses is the same as that de-

picted in Fig. 2.32d, namely, the tensile stress

This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited

reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or

likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.