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# TEST BANK FOR Manufacturing Processes for Engineering Materials 5th Ed By Serope Kalpakjian

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2.1 Can you calculate the percent elongation of materials

based only on the information given in

Fig. 2.6? Explain.

Recall that the percent elongation is defined by

Eq. (2.6) on p. 33 and depends on the original

gage length (lo) of the specimen. From Fig. 2.6

on p. 37 only the necking strain (true and engineering)

and true fracture strain can be determined.

Thus, we cannot calculate the percent

elongation of the specimen; also, note that the

elongation is a function of gage length and increases

with gage length.

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

to reach 0% elongation as the gage length is increased

further.

The percent elongation of the specimen is a

function of the initial and final gage lengths.

When the specimen is being pulled, regardless

of the original gage length, it will elongate uniformly

(and permanently) until necking begins.

Therefore, the specimen will always have a certain

finite elongation. However, note that as the

specimen’s gage length is increased, the contribution

of localized elongation (that is, necking)

will decrease, but the total elongation will not

approach zero.

2.3 Explain why the difference between engineering

strain and true strain becomes larger as strain

increases. Is this phenomenon true for both tensile

and compressive strains? Explain.

The difference between the engineering and true

strains becomes larger because of the way the

strains are defined, respectively, as can be seen

by inspecting Eqs. (2.1) on p. 30 and (2.9) on

p. 35. This is true for both tensile and compressive

strains.

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

tensile true-stress-true-strain curve is higher

than the engineering stress-strain curve. Explain

whether this condition also holds for a

compression test.

During a compression test, the cross-sectional

area of the specimen increases as the specimen

height decreases (because of volume constancy)

as the load is increased. Since true stress is defined

as ratio of the load to the instantaneous

cross-sectional area of the specimen, the true

stress in compression will be lower than the engineering

stress for a given load, assuming that

friction between the platens and the specimen

is negligible.

2.5 Which of the two tests, tension or compression,

requires a higher capacity testing machine than

the other? Explain.

The compression test requires a higher capacity

machine because the cross-sectional area of the

1

© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.

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.

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specimen increases during the test, which is the

opposite of a tension test. The increase in area

requires a load higher than that for the tension

test to achieve the same stress level. Furthermore,

note that compression-test specimens

generally have a larger original cross-sectional

area than those for tension tests, thus requiring

higher forces.

2.6 Explain how the modulus of resilience of a material

changes, if at all, as it is strained: (1) for

an elastic, perfectly plastic material, and (2) for

an elastic, linearly strain-hardening material.

2.7 If you pull and break a tension-test specimen

rapidly, where would the temperature be the

highest? Explain why.

Since temperature rise is due to the work input,

the temperature will be highest in the necked

region because that is where the strain, hence

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.

If the specimen is pulled very slowly, the temperature

generated will be dissipated throughout

the specimen and to the environment.

Thus, there will be no appreciable temperature

rise anywhere, particularly with materials with

high thermal conductivity.

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

strain curve is the work done per unit volume

(the specific work). We also know that

the area under the load-elongation curve represents

the work done on the specimen. If you

divide this latter work by the volume of the

specimen between the gage marks, you will determine

the work done per unit volume (assuming

that all deformation is confined between

the gage marks). Will this specific work be

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

curve? Explain. Will your answer be the

same for any value of strain? Explain.

If we divide the work done by the total volume

of the specimen between the gage lengths, we

obtain the average specific work throughout the

specimen. However, the area under the true

stress-true strain curve represents the specific

work done at the necked (and fractured) region

in the specimen where the strain is a maximum.

Thus, the answers will be different. However,

up to the onset of necking (instability), the specific

work calculated will be the same. This is

because the strain is uniform throughout the

specimen until necking begins.

2.10 The note at the bottom of Table 2.5 states that

as temperature increases, C decreases and m

increases. Explain why.

The value of C in Table 2.5 on p. 43 decreases

with temperature because it is a measure of the

strength of the material. The value of m increases

with temperature because the material

becomes more strain-rate sensitive, due to the

fact that the higher the strain rate, the less time

the material has to recover and recrystallize,

hence its strength increases.

2.11 You are given the K and n values of two different

materials. Is this information sufficient

to determine which material is tougher? If not,

what additional information do you need, and

why?

Although the K and n values may give a good

estimate of toughness, the true fracture stress

and the true strain at fracture are required for

accurate calculation of toughness. The modulus

of elasticity and yield stress would provide

information about the area under the elastic region;

however, this region is very small and is

thus usually negligible with respect to the rest

of the stress-strain curve.

2.12 Modify the curves in Fig. 2.7 to indicate the

effects of temperature. Explain the reasons for

your changes.

These modifications can be made by lowering

the slope of the elastic region and lowering the

general height of the curves. See, for example,

Fig. 2.10 on p. 42.

2.13 Using a specific example, show why the deformation

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

are not the same.

The deformation rate is the quantity v in

Eqs. (2.14), (2.15), (2.17), and (2.18) on pp. 41-

46. Thus, when v is held constant during de-

2

© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.

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.

Name:Ghalib Thwapiah Email:[email protected] - [email protected] Work Phone:0041789044416

Address:Am glattbogen 112 - Zuerich - ch (Switzerland) - Zip Code:8050

formation (hence a constant deformation rate),

the true strain rate will vary, whereas the engineering

strain rate will remain constant. Hence,

the two quantities are not the same.

2.14 It has been stated that the higher the value of

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

the lower the value of m, the more localized the

neck is. Explain the reason for this behavior.

As discussed in Section 2.2.7 starting on p. 41,

with high m values, the material stretches to

a greater length before it fails; this behavior

is an indication that necking is delayed with

increasing m. When necking is about to begin,

the necking region’s strength with respect

to the rest of the specimen increases, due to

strain hardening. However, the strain rate in

the necking region is also higher than in the rest

of the specimen, because the material is elongating

faster there. Since the material in the

necked region becomes stronger as it is strained

at a higher rate, the region exhibits a greater resistance

to necking. The increase in resistance

to necking thus depends on the magnitude of

m. As the tension test progresses, necking becomes

more diffuse, and the specimen becomes

longer before fracture; hence, total elongation

increases with increasing values of m (Fig. 2.13

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

necking (postuniform elongation) also increases

with increasing m. It has been observed that

the value of m decreases with metals of increasing

strength.

2.15 Explain why materials with high m values (such

as hot glass and silly putty) when stretched

slowly, undergo large elongations before failure.

Consider events taking place in the necked region

of the specimen.

The answer is similar to Answer 2.14 above.

2.16 Assume that you are running four-point bending

tests on a number of identical specimens of

the same length and cross-section, but with increasing

distance between the upper points of

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

would you expect in the test results? Explain.

As the distance between the upper points of

loading in Fig. 2.21b on p. 51 increases, the

magnitude of the bending moment decreases.

However, the volume of material subjected to

the maximum bending moment (hence to maximum

stress) increases. Thus, the probability

of failure in the four-point test increases as this

distance increases.

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

Explain.

Note that this equation is based on volume constancy,

i.e., Aolo = Al. We know, however, that

because the Poisson’s ratio is less than 0.5 in

the elastic range, the volume is not constant in

a tension test; see Eq. (2.47) on p. 69. Therefore,

the expression is not valid in the elastic

range.

2.18 Why have different types of hardness tests been

developed? How would you measure the hardness

of a very large object?

There are several basic reasons: (a) The overall

hardness range of the materials; (b) the hardness

of their constituents; see Chapter 3; (c) the

thickness of the specimen, such as bulk versus

foil; (d) the size of the specimen with respect to

that of the indenter; and (e) the surface finish

of the part being tested.

2.19 Which hardness tests and scales would you use

for very thin strips of material, such as aluminum

foil? Why?

Because aluminum foil is very thin, the indentations

on the surface must be very small so as not

to affect test results. Suitable tests would be a

microhardness test such as Knoop or Vickers

under very light loads (see Fig. 2.22 on p. 52).

The accuracy of the test can be validated by observing

any changes in the surface appearance

opposite to the indented side.

2.20 List and explain the factors that you would consider

in selecting an appropriate hardness test

and scale for a particular application.

Hardness tests mainly have three differences:

(a) type of indenter,

(b) applied load, and

(c) method of indentation measurement

(depth or surface area of indentation, or

rebound of indenter).

## [Solved] TEST BANK FOR Manufacturing Processes for Engineering Materials 5th Ed By Serope Kalpakjian

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