Industrial Engineering Chapter 8 For Long Plate The Cutting Time Then And Note That Plate Thickness

T=3.8Yf

ρc

3

K

Therefore, the temperature for the aluminum is

ρc

K=3.8(120)

2.6

97

8.112 In a dry cutting operation using a −5◦rake an-

gle, the measured forces were Fc= 1330 N and

Ft= 740 N. When a cutting fluid was used, these

forces were Fc= 1200 N and Ft= 710 N. What is

the change in the friction angle resulting from the

use of a cutting fluid?

tan β=740 + (1330) tan −5◦

1330 −740 tan −5◦= 0.447

Therefore, β= 24.1◦. With a cutting fluid,

8.113 In the dry machining of aluminum with a 10◦rake

angle tool, it is found that the shear angle is 25◦.

Determine the new shear angle if a cutting fluid is

8.114 Taking carbide as an example and using Eq. (8.30),

We begin with Eq. (8.32) which, for this case, can

tain

f2

f1

= 3−a/b

For carbide tools, approximate values are given on

= 0.83, or 83%.

8.115 With appropriate diagrams, show how the use of a

force, Ft. Consider the sketch given below. The

left sketch shows cutting without an effective cut-

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Chip

Tool

Workpiece

Fc

Ft

R

F

N

α

Chip

Tool

Workpiece

Fc

Fs

Ft

R

F

N

α

φ

β

β−α

8.116 An 8-in-diameter stainless-steel bar is being turned

on a lathe at 600 rpm and at a depth of cut, d= 0.1

in. If the power of the motor is 5 hp and has a me-

chanical efficiency of 80%, what is the maximum

feed that you can have at a spindle speed of 500

rpm before the motor stalls?

= 3.33 in3/min. Because the depth of cut is much

smaller than the workpiece diameter and referring

n= 0.3, calculate the percentage increase in tool

The Taylor equation for tool wear is given by

Eq. (8.31), which can be rewritten as

C=V T n

We can compare two cases as

V1Tn

1=V2Tn

2

or

V2

V1

=„T1

T2«n

solving for T1/T2,

T1

T2

=„V2

V1«1/n

(a) For the case where the speed is reduced by

30%, then V2= 0.7V1, and thus

(b) For a speed reduction of 60%, the new tool

life equation. (Assume a straight line relationship.)

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400 0.5 0.0014

4.0 0.0030

16.0 0.0082

54.0 0.0150

2.0 0.0035

8.0 0.0100

14 0.0160

2.0 0.0160

Tool life (min)

10

5

50

C= 1190. Therefore, the Taylor equation for

From Eq. (8.31) on p. 441 note that the value

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in rough turning a 1.5-m-long, annealed

aluminum-alloy round bar 75-mm in diameter,

or

t=L

fN =1.5

(0.002)(1018) = 0.74 min = 44 s

8.122 A 150-mm-long, 75-mm-diameter titanium-

alloy rod is being reduced in diameter to 65 mm

41.89 rad/s. The depth of cut can be calculated

from the information given as

= 2.2×105mm3/min

or MRR=3660 mm3/s. The actual time to cut

or 12.8 kW. The cutting force, Fc, is the tan-

gential force exerted by the tool. Since power

is the product of torque and rotational speed,

8.123 Calculate the same quantities as in Example 8.4

but for high-strength cast iron and at N= 500

ameter, Do, and is obtained from the expression

f=v

N=8 in./min

300 rpm = 0.0267 in/rev

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and since 1 hp = 396,000 in.-lb/min, the power

(300)2π= 38.8 in.-lb

is 0.005 in./rev, what is the material removal

rate? What is the MRR if the drill diameter is

tripled?

4(0.005)(300)

= 0.66 in3/min

2.25 in.), then the metal removal rate is

= 5.96 in3/min

It can be seen that this is a ninefold increase in

=π(15 mm)2

4(0.1 mm/rev)(500 rpm)

= 8840 mm3/min

Power is the product of the torque on the drill

and the rotational speed in radians per second,

D≫d.

d

x

lc



Referring to the figure given above, the hy-

2=x/2

R=x

2R

From the lower triangle,

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8.127 Calculate the chip depth of cut in Example 8.6.

The chip depth of cut, tc, is given by Eq. (8.42)

8.128 In Example 8.6, which of the quantities will be

affected when the spindle speed is increased to

200 rpm?

on a 20-in.-long, 6-in.-wide high-strength-steel

block at a feed of 0.01 in./tooth and a depth of

cut of 0.15 in. The cutter has a diameter of 2.5

in, has six straight cutting teeth, and rotates at

mm, w= 30 mm, l= 600 mm, d= 2 mm,

v= 1 mm/s, and N= 200 rpm. The cutter

has 10 inserts, and the workpiece material is

wd = (30)(2) = 60 mm2

Noting that the workpiece speed is v= 1 mm/s,

to Problem 8.126)

lc=√Dd =p(200)(2) = 20 mm

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8.131 Estimate the time required for face milling an

8-in.-long, 3-in.-wide brass block using a 8-in-

diameter cutter with 12 HSS teeth.

Table 8.12 on p. 489), and the maximum feed

per tooth as 0.5 mm, or 0.02 in., The rotational

N=V

πD =59 in./s

π(8 in.) = 2.34 rev/s = 131 rpm

The workpiece speed can be obtained from

Therefore the cutting time is

8.132 A 12-in-long, 2-in-thick plate is being cut on a

band saw at 150 ft/min The saw has 12 teeth

per in. If the feed per tooth is 0.003 in., how

long will it take to saw the plate along its

length?

8.133 A single-thread hob is used to cut 40 teeth on a

If a single-threaded hob is used to cut forty

teeth, the hob and the blank must be geared so

that the hob makes forty revolutions while the

πD

Since the cutting speed is given as 200 ft/min

30.2/40 = 0.75 rpm.

8.134 In deriving Eq. (8.20) it was assumed that the

friction angle, β, was independent of the shear

angle, φ. Is this assumption valid? Explain.

8.135 An orthogonal cutting operation is being car-

thrust force = 200 N. Calculate the percent-

age of the total energy that is dissipated in the

shear plane during cutting.

The total power is

Eq. (8.1) on p. 420 as

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= 29.0◦

Note that because all necessary data is given,

we should not use the approximate shear-angle

or β= 36.7◦Also, Fsis calculated as

Fs=Rcos (φ+β−α)

or Pshear = 677 N-m/s. Hence the percentage is

p. 430.

8.136 An orthogonal cutting operation is being car-

ried out under the following conditions: depth

Pshear =FsVs

where

Fs=Rcos (φ+β−α)

and

= 16.7◦

Note that because all the necessary data is

given, we should not use the shear-angle rela-

= 36.9◦

= (250) cos (16.7◦+ 36.9◦−0◦)

= 148 lb

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(300)(0.020)(0.10)(12)=14.4 in3/min. Thus,

the heat content, Q, of the chip is

Qchip =cρV ∆T

8.137 It can be shown that the angle ψbetween the

shear plane and the direction of maximum grain

of the chip obtained from orthogonal cutting of

an annealed metal. The rake angle and cutting

Remembering that we only have a piece of the

chip and we do not know its relationship to the

between the angles ψand γ. Also from

Eq. (8.3) we have a relationship between

φand γ. Therefore, we can determine the

value of γ.

face. Based on observations such as those

given in Table 8.1, we may estimate this

8.138 A lathe is set up to machine a taper on a bar

stock 120 mm in diameter; the taper is 1 mm

per 10 mm. A cut is made with an initial depth

which is tapered. If the depth of cut were a con-

stant at 4 mm, the metal removal rate would be

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Cm=Tm(Lm+Bm)

=πLD

fV (Lm+Bm)

=πLD

V(Lm+Bm)1

f

The number of parts per tool grind is given as

8.140 Assuming that the coefficient of friction is 0.25,

calculate the maximum depth of cut for turning

a hard aluminum alloy on a 20-hp lathe (with

a mechanical efficiency of 80%) at a width of

cut of 0.25 in., rake angle of 0◦, and a cutting

speed of 300 ft/min. What is your estimate of

the material’s shear strength?

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measure the temperature in a cutting opera-

tion at a speed of 250 ft/min and feed of 0.0025

in./rev as 1200◦F. What would be the approx-

imate temperature if the cutting speed is in-

creased by 50%? What should the speed be to

lower the maximum temperature to 800◦F?

or

T=kV afb

where kis a constant. From Section 8.2.6, for

a carbide tool, a= 0.2 and b= 0.125. For

2fb

2

1

or T1

equation is problematic if either of the tem-

peratures T1or T2is zero or negative; there-

fore, an absolute temperature scale is required.

most, 2.0 hp.min/in3(see Table 8.3). There-

fore, the horsepower needed in the lathe motor,

corrected for 80% efficiency, is

P=2.0 hp-min/in3

36.5 in3/min = 0.05 hp

8.143 (a) A 6-in.-diameter aluminum bar with a

length of 12 in. is to have its diameter reduced

to 5 in. by turning. Estimate the machining

time if an uncoated carbide tool is used. (b)

What is the time for a TiN-coated tool?

d= 0.01 −0.35 in.

f= 0.003 −0.025 in.

V= 650 −2000 ft/min.

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d= 0.02 in. For the roughing cuts, the maxi-

mum allowable feed and speed can be used, so

that f= 0.025 in./rev and V= 2000 ft/min.

is 5.76 in., and 5.28 in. for the second cut. The

rotational speeds for first and second roughing

(0.02 in./rev)(60 rpm)

= 18.36 min

characteristics of the roughing and finishing

cuts selected. This answer will be based the

2= 5.76 in.

=π(5.76)(0.24)(0.025)(110)

= 11.94 in3/min

MRR=11.94 in3/min and P= 3.28 hp. For

us=FsVs

wtoVand uf=F Vc

wtoV

Thus, their ratio becomes

ther by noting in the table for Problem 8.107

that the magnitudes of φand αare close to

each other. This expression can thus be ap-

we obtain us

= cot βcot α

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8.146 For a turning operation using a ceramic cutting

tool, if the cutting speed is increased by 50%,

by what factor must the feed rate be modified

to obtain a constant tool life? Let n= 0.5 and

y= 0.6.

Equation (8.33) will be used for this problem.

Since the tool life is constant, we can write the

so that f1

=1.521/1.2= 1.96

8.147 Using Eq. (8.35), select an appropriate feed for

R= 1 mm and a desired roughness of 1 µm.

How would you adjust this feed to allow for nose

f= 0.089 mm/rev

8.148 In a drilling operation, a 0.5-in. drill bit is be-

ing used in a low-carbon steel workpiece. The

The velocity of the drill into the workpiece

is v=fN = (0.010 in./rev)(700 rpm) = 7

in./min. Since the hole is to be tapped to a

depth of 1 in., it should be drilled deeper than

this distance. Note from Section 8.9.4 that the

point angle for steels ranges from 118◦to 135◦,

so that (using 118◦to get a larger number and

8.149 Assume that in the face-milling operation

shown in Fig. 8.54, the workpiece dimensions

0.005 in./tooth. Assume that the specific en-

ergy required for this material is 2 hp-min/in3

and that only 75% of the cutter diameter is en-

or MRR = 6.75 in3/min. Since the specific

8.150 Calculate the ranges of typical machining times

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0.015 in/tooth. The cutting time, as obtained

for 10 teeth in the cutter. is given below:

erance of 0.0001 in. in machining? A tolerance

of 0.001 in.? Assume that the spindle is made

that the temperatures involved are quite small,

8.152 In the production of a machined valve, the labor

rate is $19.00 per hour and the general overhead

Ψ = 1

45

60 (19 + 15) + 25+ 1(19 + 15)

lem 8.152 for maximum production.

8.154 Develop an equation for optimum cutting speed

in a face milling operation using a milling cutter

be acceptable, depending on the specific as-

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+lD

C1/nf mVn/(n−1)Ψ

8.156 Assume that you are an instructor covering the

topics in this chapter, and you are giving a quiz

on the quantitative aspects to test the under-

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8.157 Tool life could be greatly increased if an effec-

tive means of cooling and lubrication were de-

and students are encouraged to pursue creative

solutions. Methods of delivering fluid to the

cutting zone include (see also Section 8.7.1):

(a) Flooding or mist cooling of the cutting

zone, which has been the traditional ap-

proach.

(b) High-pressure coolant application.

8.158 Devise an experimental setup whereby you can

perform an orthogonal cutting operation on a

lathe using a short round tubular workpiece.

8.159 Cutting tools are sometimes designed so that

Fig. 8.7c). Explain the possible advantages of

By the student. The principal reason is that

by reducing the tool-chip contact, the friction

8.160 The accompanying illustration shows drawings

for a cast-steel valve body before (left) and af-

ter (right) machining. Identify the surfaces that

are to be machined (noting that not all sur-

faces are to be machined). . What type of

machine tool would be suitable to machine this

part? What type of machining operations are

involved, and what should be the sequence of

these operations?

Casting After machining

100 mm

By the student. Note that the dimensions of the

part suggest that most of these surfaces are produced

and (g) production rates achieved.

tions. Some examples of acceptable answers

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bar stock by placing the hex stock into a chuck

and machining the cylindrical shank of the bolt

the workpiece surface, and the associated dy-

namic stresses which, in turn, could lead to tool

area of the bolt also will have an influence on

8.163 Design appropriate fixtures and describe the

machining operations required to produce the

piston shown in Fig. 12.62.

By the student. Note that the piston has to be

greatly affected in a favorable way.

(b) Brazing or welding an insert is probably

withstand machining operation.

(b) It is likely that the tool will wear beyond

8.165 Describe your thought on whether chips pro-

istics and differences as compared to the same

products made by other manufacturing pro-

cesses. Which types of chips would be desirable

for this purpose? Explain.

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8.166 Experiments have shown that it is possible to

(0.003 in.) thick and 10 mm (4 in.) wide, which

would be similar to rolled sheet. Materials used

8.168 If expanded honeycomb panels (see Section

7.5.5) were to be machined in a form milling op-

you take to keep the sheet metal from buckling

due to cutting forces? Think of as many solu-

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7.282

4.625

4.093

0.439

0.625

0.591

1.741

1.156

0.125

0.75

0.460 0.500

0.375

1.207

0.813

0.500

0.062 R

90

30

130

30

60

30

0.38-24 UNF

Key seat width 0.096 x depth 0.151

Dimensions in inches

By the student. Note that the operations should be designed to incorporate, as appropriate, roughing

and finishing cuts and should minimize the need for tool changes or refixturing.

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