Industrial Engineering Chapter 6 The Neutral Point Will Shift Tion Thus Must High Enough Such That

weight can be expressed in terms of the initial

and instantaneous heights of the specimen:

KE = mv2

o−v2

2

where mis the mass of the falling body and

vois the velocity when the falling weight first

contacts the specimen. Equating the two ener-

gies, noting that Y,ho,V,m, and voare con-

stant, and simplifying, we find the relationship

between vand has

v∝√ln h+C

where Cis a constant. Inspection of this equa-

tion indicates that, qualitatively, the velocity

profile of the falling weight will be as shown in

the following figure:

6.81 Describe how would you go about estimating

the force acting on each die in a swaging oper-

ation.

First, an estimate has to be made of the con-

tact area between the die and the workpiece;

this can be done by studying the contact ge-

ometry. Then, a ﬂow stress, Yf, has to be de-

termined, which will depend on the workpiece

material, strain hardening exponent, n, and the

amount of strain the material is undergoing.

Also, as a first approximation, the hardness of

the workpiece material can be used (with appro-

priate units) since the deformation zone during

swaging is quite contrained, as in a hardness

test. Note also that the swaging process can be

assumed to be similar to hubbing, and conse-

quently, Eq. (6.23) on p. 281 may be used to

estimate the die force.

6.82 A mechanical press is powered by a 30-hp mo-

tor and operates at 40 strokes per minute. It

uses a ﬂywheel, so that the rotational speed of

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force is assumed later; actually, both the force

and velocity will vary. At 40 strokes per minute

p. 116). Therefore, using Eqs. (6.18) and (6.19)

on p. 272,

F= 24,750 lb

a 0.25-in-diameter head. Assume that the co-

0.125 in. in thickness.

Since we are asked for the force required to per-

= 27.88 ksi

The force is the product of pressure and area,

it as shown in Fig. 6.8. Balancing forces in the

radial direction,

dǫθ=2π dx

dǫr

=dǫθ

=dǫz

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6.85 In Example 6.4, calculate the velocity of the

strip leaving the rolls.

where hneutral is the thickness of the strip at the

neutral point and

0.8 in., respectively, let’s assume that hneutral =

0.85. Thus

Note that you are allowed to measure any quan-

tity other than torque or forces.

In this problem, we first measure the follow-

in Fig. (6.32), we can now determine φnfrom

which we obtain Hn, using Eq. (6.32) on p. 294.

To determine the coefficient of friction, we can

and front tension, σf, in rolling such that when

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6.89 Take an element at the center of the deforma-

tion zone in ﬂat rolling. Assuming that all

the stresses acting on this element are princi-

pal stresses, indicate the stresses qualitatively,

and state whether they are tensile or compres-

sive. Explain your reasoning. Is it possible for

all three principal stresses to be equal to each

other in magnitude? Explain.

p

p

the exit, hence the area under the entry-side

curve is larger than that for the exit-side curve.

(This is in order to supply energy through a net

frictional force during ordinary rolling.) Conse-

quently, a reduction in the height of the curve

by back tension (σb) has a greater effect than

that for the exit side by front tension.

6.91 It can be seen that in rolling a strip, the rolls

6.92 Derive Eq. (6.46).

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

Roll

/2

x/2

h0hf

R

z

Note that x=ho−hf, and is also called the

draft. For small α,z=Rsin α. Also, note that

for small angles,

x

2=zsin α

2

Therefore,

x=zsin α=Rtan2α

At small angles, the sine and tangent functions

are approximately equal; hence,

x=ho−hf=Rtan2α

Recall that the inclined-plane principle for fric-

tion states that α=tan−1µ, or µ= tan α.

Substituting, we have

6.93 In Steckel rolling, the rolls are idling, and thus

there is no net torque, assuming frictionless

bearings. Where, then, is the energy coming

from to supply the necessary work of defor-

mation in rolling? Explain with appropriate

6.94 Derive an expression for the tension required in

Steckel rolling of a ﬂat sheet, without friction,

In this process, the work done in rolling is sup-

plied by the front tension. Assuming a certain

reduction in thickness per pass, we first deter-

mine the absolute value of the true strain,

ǫ1= ln ho

hf

Since we know the behavior of the material as

σ=a+bǫ, we can determine the energy of

plastic deformation per unit volume, u, using

Eq. (2.59) on p. 71. We also know the cross-

sectional dimensions of the strip and the ve-

locities voand vf. The power dissipated is

the product of uand the volume rate of ﬂow

through the roll gap, which is given by the

quantity wohovo. This product is equal to the

power supplied by the front tension that acts

on the exiting cross-sectional area of the rolled

strip. Hence, assuming a plane-strain condi-

tion (that is, w= constant), we can write the

expression

uwhovo=σfwhfvf

from which the magnitude of the front tension

can be determined.

6.95 (a) Make a neat sketch of the roll-pressure dis-

bution, explaining your reasoning clearly. (c)

After completing part (b), further assume that

the roll bearings are becoming rusty and de-

prived of lubrication although rolling is still tak-

tion, the front tension must increase in order to

supply the additional work required to rotate

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6.96 Derive Eq. (6.28), based on the equation pre-

ceding it. Comment on how different the hval-

ues are as the angle φincreases.

h−hf=Rφ2

6.97 In Fig. 6.34, assume that L= 2L2. Is the roll

force, F, for Lnow twice or more than twice

the roll force for L2? Explain.

twice as high. This is due to the fact that

the roll-pressure distribution has the shape of

6.98 A ﬂat-rolling operation is being carried out

where h0= 0.2 in., hf= 0.15 in., w0= 10 in.,

R= 8 in., µ= 0.25, and the average ﬂow stress

of the material is 40,000 psi. Estimate the roll

force and the torque. Include the effects of roll

ﬂattening.

The roll force can be estimated from Eq. (6.40)

on p. 296, where the quantity Lis obtained from

Eq. (6.38). Therefore,

L=√R∆h=p(8)(0.20 −0.15) = 0.632 in.

= (0.632)(10)(40,000) 1 + (0.25)(0.632)

2(0.175) 

= 367,000 lb

We check for roll ﬂattening by using Eq. (6.48)

0.20 −0.15 

= 8.94 in.

predicts L= 0.671 and F= 397,000 lb, which

suggests a radius of R′= 9.02 in. Therefore,

6.99 A rolling operation takes place under the condi-

tions shown in the accompanying figure. What

is the position xnof the neutral point? Note

ened steel rolls; surface roughness of the rolls =

0.02 µm; rolling temperature = 210◦C.

R = 75 mm

x

V = 1.5 m/s 3 mm

5 mm V = 2 m/s

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(2.0)(3 mm)w=Vi(5 mm)w

Therefore, Vi= 1.20 m/s. At the neutral point,

(1.5)(h) = (2.0)(3) →h= 4.0 mm

Consider the sketch of the roll bite geometry

given below.

2

or θ= 6.62◦. Therefore,

L=√R∆h=p(200)(4) = 28.3 mm

and

ǫ= ln 10

6= 0.5108

The average yield stress can be obtained from

= 1.38 MN

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0.26

30

Stand 1 3 4 52

0.34

17.7 10.7 6.6 4.1 m/s

0.56 0.90 1.45 2.25 mm

Pay-off

reel

Take-up

reel

2.6 m/s

Stand 5

2.25 mm

1.45 mm

0.90 mm

Stand 4

In Section 6.3.1 starting on p. 290, the draft was

defined as ∆hfor a rolling operation. Therefore, the

answers are:

•Stand 5: 2.25 – 1.45 = 0.80 mm, or 36%.

•Stand 4: 1.45 – 0.90 = 0.55 mm, or 38%.

roll in Problem 6.101 in order to maintain a

Roll velocity

FS=0 FS=10%

Stand (m/s) (m/s)

1 30 27.3

2 17.7 16.1

4 6.6 6.0

5 2.6 2.36

The extrusion ratio is R= 62/22= 9, and

thus the true strain is ǫ= ln(9) = 2.20. For

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Consider the sketch below for an extrusion op-

eration.

x

6.105 Calculate the theoretical temperature rise in

the extruded material in Example 6.6, assum-

is

6.106 Using the same approach as that shown in Sec-

tion 6.5 for wire drawing, show that the extru-

sion pressure is given by the expression

p=Y1 + tan α

Letting µ/ tan α=B, and using this relation-

σd=Y1 + B

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=Y1 + tan α

6.107 Derive Eq. (6.56).

An estimate of the pressure pcan be obtained as

= 1.7Yln R

container interface. Assume that the frictional

stress is equal to the shear yield stress of the

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6.109 Estimate the force required in extruding 70-30

brass at 700◦C, if the billet diameter is 125 mm

6.110 Calculate the power required in Example 6.7 if

¯

Y=(895)(0.466)0.49

1.49 = 413 MPa

Fig. 6.63 for the following conditions: K= 100

MPa, n= 0.3, and µ= 0.04.

40

50

R

e

d

u

c

t

i

o

n

=

4

a ﬂat sheet or plate is given by the expression

For a plane-strain element, we can apply equi-

tan αdx = 0

From Eq. (2.36), and recognizing that positive

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6.113 Derive an analytical expression for the die pres-

sure in wire drawing, without friction or redun-

dant work, as a function of the instantaneous

Dx2

Consequently,

6.114 A linearly strain-hardening material with a

true-stress-true-strain curve σ= 5,000 +

25,000ǫpsi is being drawn into a wire. If the

original diameter of the wire is 0.25 in., what

is the minimum possible diameter at the exit

of the die? Assume that there is no redundant

Equating both expressions, as was done in

Eq. (6.71) on p. 322, we obtain

5,000+25,000ǫ1=σd= 1.15(5,000+12,500ǫ1)ǫ1

ual stress at the center of the rod is -80,000 psi.

Using the distortion-energy criterion, calculate

tive stress for the distortion-energy criterion,

given by Eq. (2.56) on p. 70. Four locations

are checked below.

(a) At the center, we have σL=−80,000 psi,

σR=−60,000 psi, and σT=−45,000 psi,

thus the effective stress is

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6.116 Derive an expression for the die-separating

force in frictionless wire drawing of a perfectly

plastic material. Use the same terminology as

in the text.

From the die-pressure curve such as that shown

in Fig. 6.62 on p. 322, which can be obtained

6.117 A material with a true-stress-true-strain curve

σ= 10,000ǫ0.3is used in wire drawing. As-

suming that the friction and redundant work

compose a total of 50% of the ideal work of de-

6.118 Derive an expression for the maximum reduc-

This problem requires the same approach as in

Problem 6.86 above. Thus, referring to Exam-

ple 6.8,

1.25ǫ1=n+ 1

6.119 Prove that the true-strain rate, ˙ǫ, in drawing or

extrusion in plane strain with a wedge-shaped

die is given by the expression

tan α=(ho−h)/2

x

or

h=ho−2xtan α.

˙ǫ=dǫ

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6.120 In drawing a strain-hardening material with

n= 0.25 what should be the percentage of

is 63%?

Referring to Example 6.8, we can write the

following expression to represent this situation

friction and redundant work is 25% of the ideal

6.121 A round wire made of a perfectly plastic ma-

be 0.1. Using both Eqs. (6.61) and (6.66), esti-

F=YavgAfln Ao

Af

= 82.3 lb

For µ= 0.1 and α= 15◦= 0.262 radians,

Eq. (6.66) on p. 321 yields

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

tive questions, and supply the answers.

and understanding on the part of the students,

6.123 Forging is one method of producing turbine

blades for jet engines. Study the design of such

blades and, referring to the relevant technical

ties that may be encountered in this operation.

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6.124 In comparing some forged parts with cast parts,

cons of each process, considering factors such

as part size, shape complexity, and design ﬂex-

6.125 Referring to Fig. 6.25, sketch the intermediate

steps you would recommend in the forging of a

wrench.

This is an open-ended problem, and there would

be several acceptable answers. Students should

be encouraged to describe the benefits of their

die layouts, including their limitations, if any.

If bar stock is the input material, an edging op-

eration is useful to distribute material to the

ends where the sockets will require extra ma-

terial. The blocking, finishing, and trimming

operations, as sketched in Fig. 6.25 on p. 285,

tailed list of the manufacturing steps involved

6.127 Figure 6.48a shows examples of products that

can be obtained by slicing long extruded sec-

axle that was made in this manner. Using the

Internet, the students should be able to give

6.128 Make an extensive list of products that either

are made of or have one or more components

and nails.

•Very thin wire as integrated circuit pack-

ages, communication cable (such as coax-

ial cable) shielding, and steel wool.

•Rods as axles, bolts and other fasteners,

reinforcing bars for concrete.

6.129 Although extruded products are typically

straight, it is possible to design dies whereby

the extrusion is curved, with a constant radius

of curvature. (a) What applications could you

think of for such products? (b) Describe your

sions. Students should be encouraged to de-

duced pre-bent, then the bending opera-

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6.130 Survey the technical literature, and describe the

design features of the various roll arrangements

shown in Fig. 6.41.

6.131 The beneficial effects of using ultrasonic vibra-

tion to reduce friction in some of the processes

were described in this chapter. Survey the tech-

6.132 In the Case Study at the end of this chapter,

it was stated that there was a significant cost

improvement using forgings when compared to

chining and finishing costs are significantly

reduced.

(c) Material costs may be very different; the

extruded alloy, for example, may be more

By the student. This is an open-ended problem,

and the students should consider, at a mini-

ual degradation in the surface produced.

•A polished die surface can produce a

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