Industrial Engineering Chapter 2 The Yield Criterion Used Does Not Matter State Plane Stress Thus The

2.66 A bar 1 m long is bent and then stress re-

lieved. The radius of curvature to the neutral

axis is 0.50 m. The bar is 30 mm thick and

is made of an elastic, perfectly plastic material

e=(0.030)

2(0.50) = 0.03

Since Y= 600 MPa and E= 200 GPa, we find

e= (2)(0.003) = 0.006

2.67 Assume that a material with a uniaxial yield

stress Yyields under a stress state of principal

(σ1−σ2)2+ (σ2−σ3)2+ (σ3−σ1)2= 2Y2

Now consider a new stress state where the prin-

cipal stresses are

+ [(σ2+p)−(σ3+p)]2

+ [(σ3+p)−(σ1+p)]2

2.68 Give two different and specific examples

in which the maximum-shear-stress and the

tension, simple compression, equal biaxial ten-

sion, and equal biaxial compression. Thus, ac-

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.

2.69 A thin-walled spherical shell with a yield stress

Yis subjected to an internal pressure p. With

criteria will give the same results. We will now

demonstrate this more rigorously. The princi-

find that

σ1−0 = Y

2.70 Show that, according to the distortion-energy

1.15Ywhere Yis the uniaxial yield stress of the

that σ2=σ1/2. Substituting these into the

distortion-energy criterion given by Eq. (2.37)

2.71 What would be the answer to Problem 2.70 if

the maximum-shear-stress criterion were used?

2.72 A closed-end, thin-walled cylinder of original

length l, thickness t, and internal radius ris

subjected to an internal pressure p. Using the

σ1=pr

2t;σ2=pr

t;σ3= 0

=pr

in order to produce a tensile membrane stress,

2.73 A round, thin-walled tube is subjected to ten-

sion in the elastic range. Show that both the

thickness and the diameter of the tube decrease

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.

2.74 Take a long cylindrical balloon and, with a thin

longitudinal direction, or (4) an ellipse? Per-

form this experiment and, based on your obser-

couraged to assign the students the task of pre-

dicting the shape numerically; an example of a

and l3. Assuming that the material is rigid and

perfectly plastic, show that volume constancy

requires that the following expression be satis-

From the definition of true strain given by

Eq. (2.9) on p. 35, ln l1

2.76 What is the diameter of an originally 30-mm-

diameter solid steel ball when it is subjected to

a hydrostatic pressure of 5 GPa?

Table 2.1 on p. 32 we take values for steel of

ν= 0.3 and E= 200 GPa, so that

2.77 Determine the effective stress and effective

strain in plane-strain compression according to

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.

2.78 (a) Calculate the work done in expanding a 2-

mm-thick spherical shell from a diameter of 100

mm to 140 mm, where the shell is made of a ma-

terial for which σ= 200+50ǫ0.5MPa. (b) Does

your answer depend on the particular yield cri-

terion used? Explain.

by

σ1=σ2=pt

2t

u= 2σ1ǫ1= 2Yln rf

ro

W= 8πY r2

otoZrf

ro

dr

r= 8πY r2

otoln rf

ro

which is the same expression obtained earlier.

To obtain a numerical answer to this prob-

lem, note that Yshould be replaced with an

average value ¯

Y. Also note that ǫ1=ǫ2=

ln(140/100) = 0.336. Thus,

1.5= 206 MPa

Hence the work done is

2.79 A cylindrical slug that has a diameter of 1

in. and is 1 in. high is placed at the center of

a 2-in.-diameter cavity in a rigid die. (See the

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.

Compressible

matrix

F

d

1″

2″

Obtain an expression for the force Fversus pis-

ton travel dup to d= 0.5 in.

The total force, F, on the piston will be

F=Fw+Fm,

where the subscript wdenotes the workpiece

and mthe matrix. As dincreases, the matrix

pressure increases, thus subjecting the slug to

transverse compressive stresses on its circum-

ference. Hence the slug will be subjected to tri-

axial compressive stresses, with σ2=σ3. Using

the maximum shear-stress criterion for simplic-

ity, we have

σ1=σ+σ2

where σ1is the required compressive stress on

the slug, σis the ﬂow stress of the slug mate-

rial corresponding to a given strain, and given

as σ= 15,000ǫ0.4, and σ2is the compressive

stress due to matrix pressure. Lets now deter-

mine the matrix pressure in terms of d.

volume of the cavity when d= 0 is π. Hence

the original volume of the matrix is Vom =3

4π.

The volume of the matrix at any value of dis

σ2, is now given by

σ2=4(40,000)

3d=160,000

3d(psi)

The absolute value of the true strain in the slug

is given by

ǫ= ln 1

1−d,

with which we can determine the value of σfor

any d. The cross-sectional area of the workpiece

at any dis

Aw=π

4(1 −d)in2

and that of the matrix is

Am=π−π

4(1 −d)in2

The required compressive stress on the slug is

σ1=σ+σ2=σ+160,000

3d.

We may now write the total force on the piston

as

F=Awσ+160,000

3d+Am

160,000

3dlb.

The following data gives some numerical re-

sults:

d Awǫ σ F

(in.) (in2) (psi) (lb)

0.1 0.872 0.105 6089 22,070

0.2 0.98 0.223 8230 41,590

0.4 1.31 0.510 11,460 82,030

0.5 1.571 0.692 12,950 104,200

0

Displacement (in.)

0 0.1 0.2 0.3 0.4 0.5

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.

2.80 A specimen in the shape of a cube 20 mm on

each side is being compressed without friction

in a die cavity, as shown in Fig. 2.35d, where the

width of the groove is 15 mm. Assume that the

(20)(20)(20) = (h)(x)(x)

Fig. 2.35d on p. 67). The absolute value of the

true strain is

ǫ= ln 20

The cross-sectional area on which the force is

acting is

have σ1= 1.15Yf, or

2.81 Obtain expressions for the specific energy for

a material for each of the stress-strain curves

shown in Fig. 2.7, similar to those shown in

Section 2.12.

(a) For a perfectly-elastic material as shown in

Fig 2.7a on p. 40, this expression becomes

1

this is identical to an elastic material for

ǫ1< Y/E, and for ǫ1> Y /E it is

2Y

E2

E

=Y2

2E+Y ǫ1−Y2

E=Yǫ1−Y

2E

(e) For an elastic, linear strain hardening ma-

terial, the specific energy is identical to

an elastic material for ǫ1< Y/E and for

=Y1−Ep

Eǫ1+Epǫ2

1

2

jected to three principal (normal) stresses of σ1,

σ2= 0, and σ3=−σ1/2. What is the value of

σ1when the metal yields according to the von

Mises criterion? What if σ2=σ1/3?

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.

2.83 A steel plate has the dimensions 100 mm ×100

mm ×5 mm thick. It is subjected to biaxial

tension of σ1=σ2, with the stress in the thick-

ness direction of σ3= 0. What is the largest

possible change in volume at yielding, using the

von Mises criterion? What would this change

in volume be if the plate were made of copper?

2.85 An aluminum alloy yields at a stress of 50 MPa

in uniaxial tension. If this material is subjected

to the stresses σ1= 25 MPa, σ2= 15 MPa and

σ3=−26 MPa, will it yield? Explain.

According to the maximum shear-stress crite-

rion, the effective stress is given by Eq. (2.51)

on p. 69 as:

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.

V=πd2h

4=π(1)2(1)

4= 0.785 in3

Heat = Work = V u =VKǫn+1

1.5

(0.785)(30,000) = 4.20

Therefore, ǫ= 2.60. Using absolute values, we

2.87 A solid cylindrical specimen 100-mm high is

compressed to a final height of 40 mm in two

In the first step, we note that ho= 100 mm and

h1= 70 mm, so that from Eq. (2.1) on p. 30,

Note that if the operation were conducted in

one step, the following would result:

2.88 Assume that the specimen in Problem 2.87 has

for each step.

d1=dorho

h1

= 80r100

70 = 95.6 mm

4d2

A2=π

p. 37) so that Eq. (2.11) on p. 35 yields

σ1= 180(0.357)0.20 = 146.5 MPa

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.

2.89 Determine the specific energy and actual energy

expended for the entire process described in the

0.916, K= 180 MPa and n= 0.20, we have

2.90 A metal has a strain hardening exponent of

0.22. At a true strain of 0.2, the true stress

is 20,000 psi. (a) Determine the stress-strain

relationship for this material. (b) Determine

σ=Kǫn→20,000 = K(0.20)0.22

or K= 28,500 psi. Therefore, the stress-strain

relationship for this material is

ln Ao

Aneck =n= 0.22

Consequently,

Aneck =Aoe−0.22

and the maximum load is

P=σA =σultAneck.

2.91 The area of each face of a metal cube is 400 m2,

and the metal has a shear yield stress, k, of 140

load applied to the z-direction to cause yield-

ing according to the Tresca criterion? Assume

stresses in the x– and y– directions are

σx=−40,000

400 =−100 MPa

σmax −σmin =Y= 2k= 280 MPa

It is stated that σ3is compressive, and is there-

2.92 A tensile force of 9 kN is applied to the ends of

a solid bar of 6.35 mm diameter. Under load,

the diameter reduces to 5.00 mm. Assuming

uniform deformation and volume constancy, (a)

determine the engineering stress and strain, (b)

determine the true stress and strain, (c) if the

original bar had been subjected to a true stress

of 345 MPa and the resulting diameter was 5.60

mm, what are the engineering stress and engi-

neering strain for this condition?

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.

from Eq. (2.1) on p. 30 as:

A=9000

4(5.00)2= 458 MPa

(c) If the final diameter is df= 5.60 mm, then

the final area is Af=π

4d2

f= 24.63 mm2.

Ao

π

4(6.35)2= 268 MPa

2.93 Two identical specimens 10-mm in diameter

and with test sections 25 mm long are made

of 1112 steel. One is in the as-received condi-

tion and the other is annealed. What will be

the true strain when necking begins, and what

will be the elongation of these samples at that

instant? What is the ultimate tensile strength

for these samples?

n= 0.19. At necking, ǫ=n, so that the strain

will be ǫ= 0.08 for the cold-rolled steel and

ǫ= 0.19 for the annealed steel. For the cold-

l=enlo=e0.08(25) = 27.08 mm

or 8.32 %. To calculate the ultimate strength,

we can write, for the cold-rolled steel,

UTS = P

Ao

=UTStrueAoe−n

Ao

= UTStruee−n

stress state σ1,σ2=σ1/3, σ3= 0. Sketch the

Mohr’s circle diagram for this stress state. De-

termine the stress σ1necessary to cause yielding

by the maximum shear stress and the von Mises

criteria.

For the stress state of σ1,σ1/3, 0 the following

figure the three-dimensional Mohr’s circle:

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.





1

2

3

32

3−02

9σ2

Solving for σ1gives σ1= 125 MPa. According

to the Tresca criterion, Eq. (2.36) on p. 64 on

2.95 Estimate the depth of penetration in a Brinell

hardness test using 500-kg load, when the sam-

worked aluminum with a yield stress of 200

kg/mm2. From Fig. 2.22 on p. 52, we can esti-

expression:

HB = 2P

(πD)(D−√D2−d2)

Because the radius is 5 mm and one-half the

penetration diameter is 1.5 mm, we can obtain

αas

α= sin−11.5

5= 17.5◦

The depth of penetration, t, can be obtained

from

t= 5 −5 cos α= 5 −5 cos 17.5◦= 0.23 mm

2.96 The following data are taken from a stainless

2500 0.02

3600 0.20

4500 0.60

4600 (max) 0.86

4586 (fracture) 0.98

(2.9), (2.10), and (2.8) on pp. 33-35:

0.02 2.02 0.00995 0.0554 45.1

0.2 2.2 0.0953 0.0509 70.7

0.6 2.6 0.262 0.0431 104

0.98 2.98 0.399 0.0376 120

The true stress-true strain curve is then plotted

25

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.

2.97 A metal is yielding plastically under the stress

state shown in the accompanying figure.

50 MPa

20 MPa

(a) Label the principal axes according to their

proper numerical convention (1, 2, 3).

(b) What is the yield stress using the Tresca

criterion?

(c) What if the von Mises criterion is used?

(d) The stress state causes measured strains

of ǫ1= 0.4 and ǫ2= 0.2, with ǫ3not being

measured. What is the value of ǫ3?

or

2Y2= 12,600

2.98 It has been proposed to modify the von Mises

yield criterion as:

(σ1−σ2)a+ (σ2−σ3)a+ (σ3−σ1)a=C

Fig. 2.36 on p. 67).

For plane stress, one of the stresses, say σ3, is

zero, and the other stresses are σAand σB. The

yield criterion is then

(σA−σB)a+ (σB)a+ (σA)a=C

For uniaxial tension, σA=Yand σB= 0 so

that C= 2Ya. These equations are difficult

to solve by hand; the following solution was

obtained using a mathematical programming

package:

Y

B

von Mises

a=12

a=4

2.99 Assume that you are asked to give a quiz to stu-

dents on the contents of this chapter. Prepare

three quantitative problems and three qualita-

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.

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.

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.