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weight can be expressed in terms of the initial
and instantaneous heights of the specimen:
KE = mv2
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=Y1 + tan α
Letting µ/ tan α=B, and using this relation-
σd=Y1 + B
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=Y1 + 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 αdx = 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
Dx2
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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