1. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-1-1
PROBLEM 1-1 Statement: It is often said, "Build a better mousetrap
and the world will beat a path to your door." Consider this problem
and write a goal statement and a set of at least 12 task
specifications that you would apply to its solution. Then suggest 3
possible concepts to achieve the goal. Make annotated, freehand
sketches of the concepts. Solution: Goal Statement: Create a
mouse-free environment. Task Specifications: 1. Cost less than
$1.00 per use or application. 2. Allow disposal without human
contact with mouse. 3. Be safe for other animals such as house
pets. 4. Provide no threat to children or adults in normal use. 5.
Be a humane method for the mouse. 6. Be environmentally friendly.
7. Have a shelf-life of at least 3 months. 8. Leave no residue. 9.
Create minimum audible noise in use. 10. Create no detectable odors
within 1 day of use. 11. Be biodegradable. 12. Be simple to use
with minimal written instructions necessary. Concepts and sketches
are left to the student. There are an infinity of possibilities.
P0101.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ.
All rights reserved. This publication is protected by Copyright and
written permission should be obtained from the publisher prior to
any prohibited reproduction, storage in a retrieval system, or
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Education, Inc., Upper Saddle River, NJ 07458.
2. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-2-1
PROBLEM 1-2 Statement: A bowling machine is desired to allow
quadriplegic youths, who can only move a joystick, to engage in the
sport of bowling at a conventional bowling alley. Consider the
factors involved, write a goal statement, and develop a set of at
least 12 task specifications that constrain this problem. Then
suggest 3 possible concepts to achieve the goal. Make annotated,
freehand sketches of the concepts. Solution: Goal Statement: Create
a means to allow a quadriplegic to bowl. Task Specifications: 1.
Cost no more than $2 000. 2. Portable by no more than two
able-bodied adults. 3. Fit through a standard doorway. 4. Provide
no threat of injury to user in normal use. 5. Operate from a 110 V,
60 Hz, 20 amp circuit. 6. Be visually unthreatening. 7. Be easily
positioned at bowling alley. 8. Have ball-aiming ability,
controllable by user. 9. Automatically reload returned balls. 10.
Require no more than 1 able-bodied adult for assistance in use. 11.
Ball release requires no more than a mouth stick-switch closure.
12. Be simple to use with minimal written instructions necessary.
Concepts and sketches are left to the student. There are an
infinity of possibilities. P0102.xmcd 2011 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This publication 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.
3. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-3-1
PROBLEM 1-3 Statement: A quadriplegic needs an automated page
turner to allow her to read books without assistance. Consider the
factors involved, write a goal statement, and develop a set of at
least 12 task specifications that constrain this problem. Then
suggest 3 possible concepts to achieve the goal. Make annotated,
freehand sketches of the concepts. Solution: Goal Statement: Create
a means to allow a quadriplegic to read standard books with minimum
assistance. Task Specifications: 1. Cost no more than $1 000. 2.
Useable in bed or from a seated position 3. Accept standard books
from 8.5 x 11 in to 4 x 6 in in planform and up to 1.5 in thick. 4.
Book may be placed, and device set up, by able-bodied person. 5.
Operate from a 110 V, 60 Hz, 15 amp circuit or by battery power. 6.
Be visually unthreatening and safe to use. 7. Require no more than
1 able-bodied adult for assistance in use. 8. Useable in absence of
assistant once set up. 9. Not damage books. 10. Timing controlled
by user. 11. Page turning requires no more than a mouth
stick-switch closure. 12. Be simple to use with minimal written
instructions necessary. Concepts and sketches are left to the
student. There are an infinity of possibilities. P0103.xmcd 2011
Pearson Education, Inc., Upper Saddle River, NJ. All rights
reserved. This publication 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.
4. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-4-1
PROBLEM 1-4 Statement: Convert a mass of 1 000 lbm to (a) lbf, (b)
slugs, (c) blobs, (d) kg. Units: blob lbf sec 2 in := Given: Mass M
1000 lb:= Solution: See Mathcad file P0104. 1. To determine the
weight of the given mass, multiply the mass value by the
acceleration due to gravity, g. W M g:= W 1000lbf= 2. Convert mass
units by assigning different units to the units place-holder when
displaying the mass value. Slugs M 31.081 slug= Blobs M 2.59 blob=
Kilograms M 453.592 kg= P0104.xmcd 2011 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This publication 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.
5. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-5-1
PROBLEM 1-5 Statement: A 250-lbm mass is accelerated at 40 in/sec2.
Find the force in lb needed for this acceleration. Given: Mass M
250 lb:= Acceleration a 40 in sec 2 := Solution: See Mathcad file
P0105. 1. To determine the force required, multiply the mass value,
in slugs, by the acceleration in feet per second squared. Convert
mass to slugs: M 7.770 slug= Convert acceleration to feet per
second squared: a 3.333s 2- ft= F M a:= F 25.9 lbf= P0105.xmcd 2011
Pearson Education, Inc., Upper Saddle River, NJ. All rights
reserved. This publication 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.
6. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-6-1
PROBLEM 1-6 Statement: Express a 100-kg mass in units of slugs,
blobs, and lbm. How much does this mass weigh? Units: blob lbf sec
2 in Given: M 100 kg Assumptions: The mass is at sea-level and the
gravitational acceleration is g 32.174 ft sec 2 or g 386.089 in sec
2 or g 9.807 m sec 2 Solution: See Mathcad file P0106. 1. Convert
mass units by assigning different units to the units place-holder
when displaying the mass value. The mass, in slugs, is M 6.85 slug
The mass, in blobs, is M 0.571 blob The mass, in lbm, is M 220.5 lb
Note: Mathcad uses lbf for pound-force, and lb for pound-mass. 2.
To determine the weight of the given mass, multiply the mass value
by the acceleration due to gravity, g. The weight, in lbf, is W M g
W 220.5 lbf The weight, in N, is W M g W 980.7 N 2011 Pearson
Education, Inc., Upper Saddle River, NJ. All rights reserved. This
publication 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.
7. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-7-1
PROBLEM 1-7 Statement: Prepare an interactive computer program
(using, for example, Excell, Mathcad, or TKSolver) from which the
cross-sectional properties for the shapes shown in the inside front
cover can be calculated. Arrange the program to deal with both ips
and SI unit systems and convert the results between those systems.
Solution: See the inside front cover and Mathcad file P0107. 1.
Rectangle, let: b 3 in h 4 in Area A b h A 12.000 in 2 A 7742 mm 2
Moment about x-axis Ix b h 3 12 Ix 16.000 in 4 Ix 6.660 10 6 mm 4
Moment about y-axis Iy h b 3 12 Iy 9.000 in 4 Iy 3.746 10 6 mm 4
Radius of gyration about x-axis kx Ix A kx 1.155 in kx 29.329 mm
Radius of gyration about y-axis ky Iy A ky 0.866 in ky 21.997 mm
Polar moment of inertia Jz Ix Iy Jz 25.000 in 4 Jz 1.041 10 7 mm 4
2. Solid circle, let: D 3 in Area A D 2 4 A 7.069 in 2 A 4560 mm 2
Moment about x-axis Ix D 4 64 Ix 3.976 in 4 Ix 1.655 10 6 mm 4
Moment about y-axis Iy D 4 64 Iy 3.976 in 4 Iy 1.655 10 6 mm 4
Radius of gyration about x-axis kx Ix A kx 0.750 in kx 19.05 mm
P0107.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ.
All rights reserved. This publication 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.
8. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-7-2
Radius of gyration about y-axis ky Iy A ky 0.750 in ky 19.05 mm
Polar moment of inertia Jz D 4 32 Jz 7.952 in 4 Jz 3.310 10 6 mm 4
3. Hollow circle, let: D 3 in d 1 in Area A 4 D 2 d 2 A 6.283 in 2
A 4054 mm 2 Moment about x-axis Ix 64 D 4 d 4 Ix 3.927 in 4 Ix
1.635 10 6 mm 4 Moment about y-axis Iy 64 D 4 d 4 Iy 3.927 in 4 Iy
1.635 10 6 mm 4 Radius of gyration about x-axis kx Ix A kx 0.791 in
kx 20.08 mm Radius of gyration about y-axis ky Iy A ky 0.791 in ky
20.08 mm Polar moment of inertia Jz 32 D 4 d 4 Jz 7.854 in 4 Jz
3.269 10 6 mm 4 4. Solid semicircle, let: D 3 in R 0.5 D R 1.5 in
Area A D 2 8 A 3.534 in 2 A 2280 mm 2 Moment about x-axis Ix 0.1098
R 4 Ix 0.556 in 4 Ix 2.314 10 5 mm 4 Moment about y-axis Iy R 4 8
Iy 1.988 in 4 Iy 8.275 10 5 mm 4 P0107.xmcd 2011 Pearson Education,
Inc., Upper Saddle River, NJ. All rights reserved. This publication
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.
9. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-7-3
Radius of gyration about x-axis kx Ix A kx 0.397 in kx 10.073 mm
Radius of gyration about y-axis ky Iy A ky 0.750 in ky 19.05 mm
Polar moment of inertia Jz Ix Iy Jz 2.544 in 4 Jz 1.059 10 6 mm 4
Distances to centroid a 0.4244 R a 0.637 in a 16.17 mm b 0.5756 R b
0.863 in b 21.93 mm 5. Right triangle, let: b 2 in h 1 in Area A b
h 2 A 1.000 in 2 A 645 mm 2 Moment about x-axis Ix b h 3 36 Ix
0.056 in 4 Ix 2.312 10 4 mm 4 Moment about y-axis Iy h b 3 36 Iy
0.222 in 4 Iy 9.250 10 4 mm 4 Radius of gyration about x-axis kx Ix
A kx 0.236 in kx 5.987 mm Radius of gyration about y-axis ky Iy A
ky 0.471 in ky 11.974 mm Polar moment of inertia Jz Ix Iy Jz 0.278
in 4 Jz 1.156 10 5 mm 4 P0107.xmcd 2011 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This publication 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.
10. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-8-1
PROBLEM 1-8 Statement: Prepare an interactive computer program
(using, for example, Excell, Mathcad, or TKSolver) from which the
mass properties for the solids shown in the page opposite the
inside front cover can be calculated. Arrange the program to deal
with both ips and SI unit systems and convert the results between
those systems. Units: blob lbf sec 2 in Solution: See the page
opposite the inside front cover and Mathcad file P0108. 1.
Rectangular prism, let: a 2 in b 3 in c 4 in 0.28 lbf in 3 Volume V
a b c V 24.000 in 3 V 393290 mm 3 Mass M V g M 0.017 blob M 3.048
kg Moment about x-axis Ix M a 2 b 2 12 Ix 0.019 blob in 2 Ix 2130.4
kg mm 2 Moment about y-axis Iy M a 2 c 2 12 Iy 0.029 blob in 2 Iy
3277.6 kg mm 2 Moment about z-axis Iz M b 2 c 2 12 Iz 0.036 blob in
2 Iz 4097.0 kg mm 2 Radius of gyration about x-axis kx Ix M kx
1.041 in kx 26.437 mm Radius of gyration about y-axis ky Iy M ky
1.291 in ky 32.791 mm Radius of gyration about z-axis kz Iz M kz
1.443 in kz 36.662 mm 2.Cylinder, let: r 2 in L 3 in 0.30 lbf in 3
Volume V r 2 L V 37.699 in 3 V 617778 mm 3 Mass M V g M 0.029 blob
M 5.13 kg 2011 Pearson Education, Inc., Upper Saddle River, NJ. All
rights reserved. This publication 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,
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permission(s), write to: Rights and Permissions Department, Pearson
Education, Inc., Upper Saddle River, NJ 07458.
11. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-8-2
Moment about x-axis Ix M r 2 2 Ix 0.059 blob in 2 Ix 6619.4 kg mm 2
Moment about y-axis Iy M 3 r 2 L 2 12 Iy 0.051 blob in 2 Iy 5791.9
kg mm 2 Moment about z-axis Iz M 3 r 2 L 2 12 Iz 0.051 blob in 2 Iz
5791.9 kg mm 2 Radius of gyration about x-axis kx Ix M kx 1.414 in
kx 35.921 mm Radius of gyration about y-axis ky Iy M ky 1.323 in ky
33.601 mm Radius of gyration about z-axis kz Iz M kz 1.323 in kz
33.601 mm 3. Hollow cylinder, let: a 2 in b 3 in L 4 in 0.28 lbf in
3 Volume V b 2 a 2 L V 62.832 in 3 V 1029630 mm 3 Mass M V g M
0.046 blob M 7.98 kg Moment about x-axis Ix M 2 a 2 b 2 Ix 0.296
blob in 2 Ix 3.3 10 4 kg mm 2 Moment about y-axis Iy M 12 3 a 2 3 b
2 L 2 Iy 0.209 blob in 2 Iy 2.4 10 4 kg mm 2 Moment about z-axis Iz
M 12 3 a 2 3 b 2 L 2 Iz 0.209 blob in 2 Iz 2.4 10 4 kg mm 2 Radius
of gyration about x-axis kx Ix M kx 2.550 in kx 64.758 mm Radius of
gyration about y-axis ky Iy M ky 2.141 in ky 54.378 mm 2011 Pearson
Education, Inc., Upper Saddle River, NJ. All rights reserved. This
publication 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.
12. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-8-3
Radius of gyration about z-axis kz Iz M kz 2.141 in kz 54.378 mm 4.
Right circular cone, let: r 2 in h 5 in 0.28 lbf in 3 Volume V r 2
h 3 V 20.944 in 3 V 343210 mm 3 Mass M V g M 0.015 blob M 2.66 kg
Moment about x-axis Ix 3 10 M r 2 Ix 0.018 blob in 2 Ix 2059.4 kg
mm 2 Moment about y-axis Iy M 12 r 2 3 h 2 80 Iy 0.023 blob in 2 Iy
2638.5 kg mm 2 Moment about z-axis Iz M 12 r 2 3 h 2 80 Iz 0.023
blob in 2 Iz 2638.5 kg mm 2 Radius of gyration about x-axis kx Ix M
kx 1.095 in kx 27.824 mm Radius of gyration about y-axis ky Iy M ky
1.240 in ky 31.495 mm Radius of gyration about z-axis kz Iz M kz
1.240 in kz 31.495 mm 5. Sphere, let: r 3 in Volume V 4 3 r 3 V
113.097 in 3 V 1853333 mm 3 Mass M V g M 0.082 blob M 14.364 kg
Moment about x-axis Ix 2 5 M r 2 Ix 0.295 blob in 2 Ix 33362 kg mm
2 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights
reserved. This publication 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.
13. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-8-4
Moment about y-axis Iy 2 5 M r 2 Iy 0.295 blob in 2 Iy 33362 kg mm
2 Moment about z-axis Iz 2 5 M r 2 Iz 0.295 blob in 2 Iz 33362 kg
mm 2 Radius of gyration about x-axis kx Ix M kx 1.897 in kx 48.193
mm Radius of gyration about y-axis ky Iy M ky 1.897 in ky 48.193 mm
Radius of gyration about z-axis kz Iz M kz 1.897 in kz 48.193 mm
2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights
reserved. This publication 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.
14. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-9-1
PROBLEM 1-9 Statement: Convert the template in Problem 1-7 to have
and use a set of functions or subroutines that can be called from
within any program in that language to solve for the
cross-sectional properties of the shapes shown on the inside front
cover. Solution: See inside front cover and Mathcad file P0109. 1.
Rectangle: Area A b h( ) b h Moment about x-axis Ix b h( ) b h 3 12
Moment about y-axis Iy b h( ) h b 3 12 2. Solid circle: Area A D( )
D 2 4 Moment about x-axis Ix D( ) D 4 64 Moment about y-axis Iy D(
) D 4 64 3. Hollow circle: Area A D d( ) 4 D 2 d 2 Moment about
x-axis Ix D d( ) 64 D 4 d 4 Moment about y-axis Iy D d( ) 64 D 4 d
4 4. Solid semicircle: Area A D( ) D 2 8 Moment about x-axis Ix R(
) 0.1098 R 4 Moment about y-axis Iy R( ) R 4 8 5. Right triangle:
Area A b h( ) b h 2 Moment about x-axis Ix b h( ) b h 3 36 Moment
about y-axis Iy b h( ) h b 3 36 2011 Pearson Education, Inc., Upper
Saddle River, NJ. All rights reserved. This publication 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.
15. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-10-1
PROBLEM 1-10 Statement: Convert the template in Problem 1-8 to have
and use a set of functions or subroutines that can be called from
within any program in that language to solve for the
cross-sectional properties of the shapes shown on the page opposite
the inside front cover. Solution: See the page opposite the inside
front cover and Mathcad file P0110. 1 Rectangular prism: Volume V a
b c( ) a b c Mass M a b c ( ) V a b c( ) g Moment about x-axis Ix a
b c ( ) M a b c ( ) a 2 b 2 12 Moment about y-axis Iy a b c ( ) M a
b c ( ) a 2 c 2 12 Moment about z-axis Iz a b c ( ) M a b c ( ) b 2
c 2 12 2. Cylinder: Volume V r L( ) r 2 L Mass M r L ( ) V r L( ) g
Moment about x-axis Ix r L ( ) M r L ( ) r 2 2 Moment about y-axis
Iy r L ( ) M r L ( ) 3 r 2 L 2 12 Moment about z-axis Iz r L ( ) M
r L ( ) 3 r 2 L 2 12 3. Hollow cylinder: Volume V a b L( ) b 2 a 2
L Mass M a b L ( ) V a b L( ) g Moment about x-axis Ix a b L ( ) M
a b L ( ) 2 a 2 b 2 Moment about y-axis Iy a b L ( ) M a b L ( ) 12
3 a 2 3 b 2 L 2 Moment about z-axis Iz a b L ( ) M a b L ( ) 12 3 a
2 3 b 2 L 2 2011 Pearson Education, Inc., Upper Saddle River, NJ.
All rights reserved. This publication is protected by Copyright and
written permission should be obtained from the publisher prior to
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transmission in any form or by any means, electronic, mechanical,
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permission(s), write to: Rights and Permissions Department, Pearson
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16. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-10-2 4.
Right circular cone: Volume V r h( ) r 2 h 3 Mass M r h ( ) V r h(
) g Moment about x-axis Ix r h ( ) 3 10 M r h ( ) r 2 Moment about
y-axis Iy r h ( ) M r h ( ) 12 r 2 3 h 2 80 Moment about z-axis Iz
r h ( ) M r h ( ) 12 r 2 3 h 2 80 5. Sphere: Volume V r( ) 4 3 r 3
Mass M r ( ) V r( ) g Moment about x-axis Ix r ( ) 2 5 M r ( ) r 2
Moment about y-axis Iy r ( ) 2 5 M r ( ) r 2 Moment about z-axis Iz
r ( ) 2 5 M r ( ) r 2 2011 Pearson Education, Inc., Upper Saddle
River, NJ. All rights reserved. This publication 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.
17. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-1-1
PROBLEM 2-1 Statement: Figure P2-1 shows stress-strain curves for
three failed tensile-test specimens. All are plotted on the same
scale. (a) Characterize each material as brittle or ductile. (b)
Which is the stiffest? (c) Which has the highest ultimate strength?
(d) Which has the largest modulus of resilience? (e) Which has the
largest modulus of toughness? Solution: See Figure P2-1 and Mathcad
file P0201. 1. The material in Figure P2-1(a) has a moderate amount
of strain beyond the yield point, P2-1(b) has very little, and
P2-1(c) has considerably more than either of the other two. Based
on this observation, the material in Figure P2-1(a) is mildly
ductile, that in P2-1(b)is brittle, and that in P2-1(c) is ductile.
2. The stiffest material is the one with the grearesr slope in the
elastic range. Determine this by dividing the rise by the run of
the straight-line portion of each curve. The material in Figure
P2-1(c) has a slope of 5 stress units per strain unit, which is the
greatest of the three. Therefore, P2-1(c) is the stiffest. 3.
Ultimate strength corresponds to the highest stress that is
achieved by a material under test. The material in Figure P2-1(b)
has a maximum stress of 10 units, which is considerably more than
either of the other two. Therefore, P2-1(b) has the highest
ultimate strength. 4. The modulus of resilience is the area under
the elastic portion of the stress-starin curve. From observation of
the three graphs, the stress and strain values at the yield points
are: P2-1(a) ya 5:= ya 5:= P2-1(b) yb 9:= yb 2:= P2-1(c) yc 5:= yc
1:= Using equation (2.7), the modulus of resiliency for each
material is, approximately, P21a 1 2 ya ya:= P21a 12.5= P21b 1 2 yb
yb:= P21b 9= P21c 1 2 yc yc:= P21c 2.5= P2-1 (a) has the largest
modulus of resilience 5. The modulus of toughness is the area under
the stress-starin curve up to the point of fracture. By inspection,
P2-1 (c) has the largest area under the stress-strain curve
therefore, it has the largest modulus of toughness. P0201.xmcd 2011
Pearson Education, Inc., Upper Saddle River, NJ. All rights
reserved. This publication is protected by Copyright and written
permission should be obtained from the publisher prior to any
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transmission in any form or by any means, electronic, mechanical,
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18. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-2-1
PROBLEM 2-2 Statement: Determine an approximate ratio between the
yield strength and ultimate strength for each material shown in
Figure P2-1. Solution: See Figure P2-1 and Mathcad file P0202. 1.
The yield strength is the value of stress at which the
stress-strain curve begins to be nonlinear. The ultimate strength
is the maximum value of stress attained during the test. From the
figure, we have the following values of yield strength and tensile
strength: Figure P2-1(a) Sya 5:= Sua 6:= Figure P2-1(b) Syb 9:= Sub
10:= Figure P2-1(c) Syc 5:= Suc 8:= 2. The ratio of yield strength
to ultimate strength for each material is: Figure P2-1(a) ratioa
Sya Sua := ratioa 0.83= Figure P2-1(b) ratiob Syb Sub := ratiob
0.90= Figure P2-1(c) ratioc Syc Suc := ratioc 0.63= P0202.xmcd 2011
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reserved. This publication is protected by Copyright and written
permission should be obtained from the publisher prior to any
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19. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-3-1
PROBLEM 2-3 Statement: Which of the steel alloys shown in Figure
2-19 would you choose to obtain (a) Maximum strength (b) Maximum
modulus of resilience (c) Maximum modulus of toughness (d) Maximum
stiffness Given: Young's modulus for steel E 207 GPa Solution: See
Figure 2-19 and Mathcad file P0203. 1. Determine from the graph:
values for yield strength, ultimate strength and strain at fracture
for each material. Steel Yield Strength Ultimate Strength Fracture
Strain AISI 1020: Sy1020 300 MPa Sut1020 400 MPa f1020 0.365 AISI
1095: Sy1095 550 MPa Sut1095 1050 MPa f1095 0.11 AISI 4142: Sy4142
1600 MPa Sut4142 2430 MPa f4142 0.06 Note: The 0.2% offset method
was used to define a yield strength for the AISI 1095 and the 4142
steels. 2. From the values of Sut above it is clear that the AISI
4142 has maximum strength. 3. Using equation (2-7) and the data
above, determine the modulus of resilience. UR1020 1 2 Sy1020 2 E
UR1020 0.22 MN m m 3 UR1095 1 2 Sy1095 2 E UR1095 0.73 MN m m 3
UR4142 1 2 Sy4142 2 E UR4142 6.18 MN m m 3 Even though the data is
approximate, the AISI 4142 clearly has the largest modulus of
resilience. 4. Using equation (2-8) and the data above, determine
the modulus of toughness. UT1020 1 2 Sy1020 Sut1020 f1020 UT1020
128 MN m m 3 UT1095 1 2 Sy1095 Sut1095 f1095 UT1095 88 MN m m 3
UT4142 1 2 Sy4142 Sut4142 f4142 UT4142 121 MN m m 3 Since the data
is approximate, there is no significant difference between the 1020
and 4142 steels. Because of the wide difference in shape and
character of the curves, one should also determine the area under
the curves by graphical means. When this is done, the area under
the curve is about 62 square units for 1020 and 66 for 4142. Thus,
they seem to have about equal toughness, which is about 50% greater
than that for the 1095 steel. 5. All three materials are steel
therefore, the stiffnesses are the same. 2011 Pearson Education,
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20. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-4-1
PROBLEM 2-4 Statement: Which of the aluminum alloys shown in Figure
2-21 would you choose to obtain (a) Maximum strength (b) Maximum
modulus of resilience (c) Maximum modulus of toughness (d) Maximum
stiffness Given: Young's modulus for aluminum E 71.7 GPa Solution:
See Figure 2-21 and Mathcad file P0204. 1. Determine, from the
graph, values for yield strength, ultimate strength and strain at
fracture for each material. Alum Yield Strength Ultimate Strength
Fracture Strain 1100: Sy1100 120 MPa Sut1100 130 MPa f1100 0.170
2024-T351: Sy2024 330 MPa Sut2024 480 MPa f2024 0.195 7075-T6:
Sy7075 510 MPa Sut7075 560 MPa f7075 0.165 Note: The 0.2% offset
method was used to define a yield strength for all of the
aluminums. 2. From the values of Sut above it is clear that the
7075-T6 has maximum strength. 3. Using equation (2-7) and the data
above, determine the modulus of resilience. UR1100 1 2 Sy1100 2 E
UR1100 0.10 MN m m 3 UR2024 1 2 Sy2024 2 E UR2024 0.76 MN m m 3
UR7075 1 2 Sy7075 2 E UR7075 1.81 MN m m 3 Even though the data is
approximate, the 7075-T6 clearly has the largest modulus of
resilience. 4. Using equation (2-8) and the data above, determine
the modulus of toughness. UT1100 1 2 Sy1100 Sut1100 f1100 UT1100 21
MN m m 3 UT2024 1 2 Sy2024 Sut2024 f2024 UT2024 79 MN m m 3 UT7075
1 2 Sy7075 Sut7075 f7075 UT7075 88 MN m m 3 Even though the data is
approximate, the 7075-T6 has the largest modulus of toughness. 5.
All three materials are aluminum therefore, the stiffnesses are the
same. 2011 Pearson Education, Inc., Upper Saddle River, NJ. All
rights reserved. This publication is protected by Copyright and
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21. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-5-1
PROBLEM 2-5 Statement: Which of the thermoplastic polymers shown in
Figure 2-22 would you choose to obtain (a) Maximum strength (b)
Maximum modulus of resilience (c) Maximum modulus of toughness (d)
Maximum stiffness Solution: See Figure 2-22 and Mathcad file P0205.
1. Determine, from the graph, values for yield strength, ultimate
strength, strain at fracture, and modulus of elasticity for each
material. Plastic Yield Strength Ultimate Strength Fracture Strain
Mod of Elasticity Nylon 101: SyNylon 63 MPa SutNylon 80 MPa fNylon
0.52 ENylon 1.1 GPa HDPE: SyHDPE 15 MPa SutHDPE 23 MPa fHDPE 3.0
EHDPE 0.7 GPa PTFE: SyPTFE 8.3 MPa SutPTFE 13 MPa fPTFE 0.51 EPTFE
0.8 GPa 2. From the values of Sut above it is clear that the Nylon
101 has maximum strength. 3. Using equation (2-7) and the data
above, determine the modulus of resilience. URNylon 1 2 SyNylon 2
ENylon URNylon 1.8 MN m m 3 URHDPE 1 2 SyHDPE 2 EHDPE URHDPE 0.16
MN m m 3 URPTFE 1 2 SyPTFE 2 EPTFE URPTFE 0.04 MN m m 3 Even though
the data is approximate, the Nylon 101 clearly has the largest
modulus of resilience. 4. Using equation (2-8) and the data above,
determine the modulus of toughness. UTNylon 1 2 SyNylon SutNylon
fNylon UTNylon 37 MN m m 3 UTHDPE 1 2 SyHDPE SutHDPE fHDPE UTHDPE
57 MN m m 3 UTPTFE 1 2 SyPTFE SutPTFE fPTFE UTPTFE 5 MN m m 3 Even
though the data is approximate, the HDPE has the largest modulus of
toughness. 5. The Nylon 101 has the steepest slope in the
(approximately) elastic range and is, therefore, the stiffest of
the three materials.. 2011 Pearson Education, Inc., Upper Saddle
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22. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-6-1
PROBLEM 2-6 Statement: A metal has a strength of 414 MPa at its
elastic limit and the strain at that point is 0.002. What is the
modulus of elasticity? What is the strain energy at the elastic
limit? Assume that the test speimen is 12.8-mm dia and has a 50-mm
gage length. Can you define the type of metal based on the given
data? Given: Elastic limit: Strength Sel 414 MPa Strain el 0.002
Test specimen: Diameter do 12.8 mm Length Lo 50 mm Solution: See
Mathcad file P0206. 1. The modulus of elasticity is the slope of
the stress-strain curve, which is a straight line, in the elastic
region. Since one end of this line is at the origin, the slope
(modulus of elasticity) is E Sel el E 207 GPa 2. The strain energy
per unit volume at the elastic limit is the area under the
stress-strain curve up to the elastic limit. Since the curve is a
straight line up to this limit, the area is one-half the base times
the height, or U'el 1 2 Sel el U'el 414 kN m m 3 The total strain
energy in the specimen is the strain energy per unit volume times
the volume, Uel U'el do 2 4 Lo Uel 2.7 N m 3. Based on the modulus
of elasticity and using Table C-1, the material is steel. 2011
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reserved. This publication is protected by Copyright and written
permission should be obtained from the publisher prior to any
prohibited reproduction, storage in a retrieval system, or
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23. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-7-1
PROBLEM 2-7 Statement: A metal has a strength of 41.2 kpsi (284
MPa) at its elastic limit and the strain at that point is 0.004.
What is the modulus of elasticity? What is the strain energy at the
elastic limit? Assume that the test speimen is 0.505-in dia and has
a 2-in gage length. Can you define the type of metal based on the
given data? Given: Elastic limit: Strength Sel 41.2 ksi Strain el
0.004 Sel 284 MPa Test specimen: Diameter do 0.505 in Length Lo
2.00 in Solution: See Mathcad file P0207. 1. The modulus of
elasticity is the slope of the stress-strain curve, which is a
straight line, in the elastic region. Since one end of this line is
at the origin, the slope (modulus of elasticity) is E Sel el E 10.3
10 6 psi E 71 GPa 2. The strain energy per unit volume at the
elastic limit is the area under the stress-strain curve up to the
elastic limit. Since the curve is a straight line up to this limit,
the area is one-half the base times the height, or U'el 1 2 Sel el
U'el 82.4 lbf in in 3 U'el 568 kN m m 3 The total strain energy in
the specimen is the strain energy per unit volume times the volume,
Uel U'el do 2 4 Lo Uel 33.0 in lbf 3. Based on the modulus of
elasticity and using Table C-1, the material is aluminum. 2011
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permission should be obtained from the publisher prior to any
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24. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-8-1
PROBLEM 2-8 Statement: A metal has a strength of 134 MPa at its
elastic limit and the strain at that point is 0.006. What is the
modulus of elasticity? What is the strain energy at the elastic
limit? Assume that the test speimen is 12.8-mm dia and has a 50-mm
gage length. Can you define the type of metal based on the given
data? Given: Elastic limit: Strength Sel 134 MPa Strain el 0.003
Test specimen: Diameter do 12.8 mm Length Lo 50 mm Solution: See
Mathcad file P0208. 1. The modulus of elasticity is the slope of
the stress-strain curve, which is a straight line, in the elastic
region. Since one end of this line is at the origin, the slope
(modulus of elasticity) is E Sel el E 45 GPa 2. The strain energy
per unit volume at the elastic limit is the area under the
stress-strain curve up to the elastic limit. Since the curve is a
straight line up to this limit, the area is one-half the base times
the height, or U'el 1 2 Sel el U'el 201 kN m m 3 The total strain
energy in the specimen is the strain energy per unit volume times
the volume, Uel U'el do 2 4 Lo Uel 1.3 N m 3. Based on the modulus
of elasticity and using Table C-1, the material is magnesium. 2011
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reserved. This publication is protected by Copyright and written
permission should be obtained from the publisher prior to any
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25. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-9-1
PROBLEM 2-9 Statement: A metal has a strength of 100 kpsi (689 MPa)
at its elastic limit and the strain at that point is 0.006. What is
the modulus of elasticity? What is the strain energy at the elastic
limit? Assume that the test speimen is 0.505-in dia and has a 2-in
gage length. Can you define the type of metal based on the given
data? Given: Elastic limit: Strength Sel 100 ksi Strain el 0.006
Sel 689 MPa Test specimen: Diameter do 0.505 in Length Lo 2.00 in
Solution: See Mathcad file P0209. 1. The modulus of elasticity is
the slope of the stress-strain curve, which is a straight line, in
the elastic region. Since one end of this line is at the origin,
the slope (modulus of elasticity) is E Sel el E 16.7 10 6 psi E 115
GPa 2. The strain energy per unit volume at the elastic limit is
the area under the stress-strain curve up to the elastic limit.
Since the curve is a straight line up to this limit, the area is
one-half the base times the height, or U'el 1 2 Sel el U'el 300 lbf
in in 3 U'el 2 10 3 kN m m 3 The total strain energy in the
specimen is the strain energy per unit volume times the volume, Uel
U'el do 2 4 Lo Uel 120.18 in lbf 3. Based on the modulus of
elasticity and using Table C-1, the material is titanium. 2011
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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,
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26. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-10-1
PROBLEM 2-10 Statement: A material has a yield strength of 689 MPa
at an offset of 0.6% strain. What is its modulus of resilience?
Units: MJ 10 6 joule Given: Yield strength Sy 689 MPa Yield strain
y 0.006 Solution: See Mathcad file P0210. 1. The modulus of
resilience (strain energy per unit volume) is given by Equation
(2.7) and is approximately UR 1 2 Sy y UR 2.067 MJ m 3 UR 2.1 MPa
2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights
reserved. This publication 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
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27. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-11-1
PROBLEM 2-11 Statement: A material has a yield strength of 60 ksi
(414 MPa) at an offset of 0.2% strain. What is its modulus of
resilience? Units: MJ 10 6 joule Given: Yield strength Sy 60 ksi Sy
414 MPa Yield strain y 0.002 Solution: See Mathcad file P0211. 1.
The modulus of resilience (strain energy per unit volume) is given
by Equation (2.7) and is approximately UR 1 2 Sy y UR 60 in lbf in
3 UR 0.414 MJ m 3 UR 0.414 MPa 2011 Pearson Education, Inc., Upper
Saddle River, NJ. All rights reserved. This publication is
protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in
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28. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-12-1
PROBLEM 2-12 Statement: A steel has a yield strength of 414 MPa, an
ultimate tensile strength of 689 MPa, and an elongation at fracture
of 15%. What is its approximate modulus of toughness? What is the
approximate modulus of resilience? Given: Sy 414 MPa Sut 689 MPa f
0.15 Solution: See Mathcad file P0212. 1. Determine the modulus of
toughness using Equation (2.8). UT Sy Sut 2 f UT 82.7 MN m m 3 UT
82.7 MPa 2. Determine the modulus of resilience using Equation
(2.7) and Young's modulus for steel: E 207 GPa UR 1 2 Sy 2 E UR 414
kN m m 3 UR 0.41 MPa 2011 Pearson Education, Inc., Upper Saddle
River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the
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29. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-13-1
PROBLEM 2-13 Statement: The Brinell hardness of a steel specimen
was measured to be 250 HB. What is the material's approximate
tensile strength? What is the hardness on the Vickers scale? The
Rockwell scale? Given: Brinell hardness of specimen HB 250
Solution: See Mathcad file P0213. 1. Determine the approximate
tensile strength of the material from equations (2.10), not Table
2-3. Sut 0.5 HB ksi Sut 125 ksi Sut 862 MPa 2. From Table 2-3
(using linear interpolation) the hardness on the Vickers scale is
HV HB 241 277 241 292 253( ) 253 HV 263 3. From Table 2-3 (using
linear interpolation) the hardness on the Rockwell C scale is HRC
HB 241 277 241 28.8 22.8( ) 22.8 HRC 24.3 2011 Pearson Education,
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is protected by Copyright and written permission should be obtained
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30. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-14-1
PROBLEM 2-14 Statement: The Brinell hardness of a steel specimen
was measured to be 340 HB. What is the material's approximate
tensile strength? What is the hardness on the Vickers scale? The
Rockwell scale? Given: Brinell hardness of specimen HB 340
Solution: See Mathcad file P0214. 1. Determine the approximate
tensile strength of the material from equations (2.10), not Table
2-3. Sut 0.5 HB ksi Sut 170 ksi Sut 1172 MPa 2. From Table 2-3
(using linear interpolation) the hardness on the Vickers scale is
HV HB 311 341 311 360 328( ) 328 HV 359 3. From Table 2-3 (using
linear interpolation) the hardness on the Rockwell C scale is HRC
HB 311 341 311 36.6 33.1( ) 33.1 HRC 36.5 2011 Pearson Education,
Inc., Upper Saddle River, NJ. All rights reserved. This publication
is protected by Copyright and written permission should be obtained
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31. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-15-1
PROBLEM 2-15 Statement: What are the principal alloy elements of an
AISI 4340 steel? How much carbon does it have? Is it hardenable? By
what techniques? Solution: See Mathcad file P0215. 1. Determine the
principal alloying elements from Table 2-5 for 43xx steel.. 1.82%
Nickel 0.50 or 0.80% Chromium 0.25% Molybdenum 2. From "Steel
Numbering Systems" in Section 2.6, the carbon content is From the
last two digits, the carbon content is 0.40%. 3. Is it hardenable?
Yes, all of the alloying elements increase the hardenability. By
what techniques? It can be through hardened by heating, quenching
and tempering; and it can also be case hardened (See Section 2.4).
P0215.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ.
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any prohibited reproduction, storage in a retrieval system, or
transmission in any form or by any means, electronic, mechanical,
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32. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-16-1
PROBLEM 2-16 Statement: What are the principal alloy elements of an
AISI 1095 steel? How much carbon does it have? Is it hardenable? By
what techniques? Solution: See Mathcad file P0216. 1. Determine the
principal alloying elements from Table 2-5 for 10xx steel. Carbon
only, no alloying elements 2. From "Steel Numbering Systems" in
Section 2.6, the carbon content is From the last two digits, the
carbon content is 0.95%. 3. Is it hardenable? Yes, as a high-carbon
steel, it has sufficient carbon content for hardening. By what
techniques? It can be through hardened by heating, quenching and
tempering; and it can also be case hardened (See Section 2.4).
P0216.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ.
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any prohibited reproduction, storage in a retrieval system, or
transmission in any form or by any means, electronic, mechanical,
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permission(s), write to: Rights and Permissions Department, Pearson
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33. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-17-1
PROBLEM 2-17 Statement: What are the principal alloy elements of an
AISI 6180 steel? How much carbon does it have? Is it hardenable? By
what techniques? Solution: See Mathcad file P0217. 1. Determine the
principal alloying elements from Table 2-5 for 61xx steel.. 0.15%
Vanadium 0.60 to 0.95% Chromium 2. From "Steel Numbering Systems"
in Section 2.6, the carbon content is From the last two digits, the
carbon content is 0.80%. 3. Is it hardenable? Yes, all of the
alloying elements increase the hardenability. By what techniques?
It can be through hardened by heating, quenching and tempering; and
it can also be case hardened (See Section 2.4). P0217.xmcd 2011
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permission should be obtained from the publisher prior to any
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transmission in any form or by any means, electronic, mechanical,
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permission(s), write to: Rights and Permissions Department, Pearson
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34. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-18-1
PROBLEM 2-18 Statement: Which of the steels in Problems 2-15, 2-16,
and 2-17 is the stiffest? Solution: See Mathcad file P0218. 1.
None. All steel alloys have the same Young's modulus, which
determines stiffness. P0218.xmcd 2011 Pearson Education, Inc.,
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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,
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Permissions Department, Pearson Education, Inc., Upper Saddle
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35. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-19-1
PROBLEM 2-19 Statement: Calculate the specific strength and
specific stiffness of the following materials and pick one for use
in an aircraft wing spar. Given: Material Code Ultimate Strength
Young's Modulus Weight Density Steel st 0 Sut st 80 ksi E st 30 10
6 psi st 0.28 lbf in 3 Aluminum al 1 Sut al 60 ksi E al 10.4 10 6
psi al 0.10 lbf in 3 Titanium ti 2 Sut ti 90 ksi E ti 16.5 10 6 psi
ti 0.16 lbf in 3 Index i 0 1 2 Solution: See Mathcad file P0219. 1.
Specific strength is the ultimate tensile strength divided by the
weight density and specific stiffness is the modulus of elasticity
divided by the weight density. The text does not give a symbol to
these quantities. Specific strength Sut i i 1 in 328610 360010
356310 Specific stiffness E i i 1 in 610710 610410 610310 Steel
Aluminum Titanium 2. Based on the results above, all three
materials have the same specific stiffness but the aluminum has the
largest specific strength. Aluminum for the aircraft wing spar is
recommended. 2011 Pearson Education, Inc., Upper Saddle River, NJ.
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36. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-20-1
PROBLEM 2-20 Statement: If maximum impact resistance were desired
in a part, which material properties would you look for? Solution:
See Mathcad file P0220. 1. Ductility and a large modulus of
toughness (see "Impact Resistance" in Section 2.1). P0220.xmcd 2011
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37. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-21-1
PROBLEM 2-21 _____ Statement: Refer to the tables of material data
in Appendix A and determine the strength-to-weight ratios of the
following material alloys based on their tensile yield strengths:
heat-treated 2024 aluminum, SAE 1040 cold-rolled steel, Ti-75A
titanium, type 302 cold-rolled stainless steel. Given: Material
Yield Strength Specific Weight Mat 1 "2024 Aluminum, HT" Sy 1 290
MPa 1 0.10 lbf in 3 1 27.14 kN m 3 Mat 2 "1040 CR Steel" Sy 2 490
MPa 2 0.28 lbf in 3 2 76.01 kN m 3 Mat 3 "Ti-75A Titanium" Sy 3 517
MPa 3 0.16 lbf in 3 3 43.43 kN m 3 Mat 4 "Type 302 CR SS" Sy 4 1138
MPa 4 0.28 lbf in 3 4 76.01 kN m 3 i 1 2 4 Solution: See Mathcad
file P0221. 1. Calculate the strength-to-weight ratio for each
material as described in Section 2.1. SWR i Sy i i SWR i 10 4 m
1.068 0.645 1.190 1.497 Mat i "2024 Aluminum, HT" "1040 CR Steel"
"Ti-75A Titanium" "Type 302 CR SS" 2011 Pearson Education, Inc.,
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38. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-22-1
PROBLEM 2-22 _____ Statement: Refer to the tables of material data
in Appendix A and determine the strength-to-weight ratios of the
following material alloys based on their ultimate tensile
strengths: heat-treated 2024 aluminum SAE 1040 cold-rolled steel,
unfilled acetal plastic, Ti-75A titanium, type 302 cold-rolled
stainless steel. Given: Material Tensile Strength Specific Weight
Mat 1 "2024 Aluminum, HT" Sut 1 441 MPa 1 0.10 lbf in 3 1 27.14 kN
m 3 Mat 2 "1040 CR Steel" Sut 2 586 MPa 2 0.28 lbf in 3 2 76.01 kN
m 3 Mat 3 "Acetal, unfilled" Sut 3 60.7 MPa 3 0.051 lbf in 3 3
13.84 kN m 3 Mat 4 "Ti-75A Titanium" Sut 4 586 MPa 4 0.16 lbf in 3
4 43.43 kN m 3 Mat 5 "Type 302 CR SS" Sut 5 1310 MPa 5 0.28 lbf in
3 5 76.01 kN m 3 i 1 2 5 Solution: See Mathcad file P0222. 1.
Calculate the strength-to-weight ratio for each material as
described in Section 2.1. SWR i Sut i i SWR i 10 4 m 1.625 0.771
0.438 1.349 1.724 Mat i "2024 Aluminum, HT" "1040 CR Steel"
"Acetal, unfilled" "Ti-75A Titanium" "Type 302 CR SS" 2011 Pearson
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39. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-23-1
PROBLEM 2-23 _____ Statement: Refer to the tables of material data
in Appendix A and calculate the specific stiffness of aluminum,
titanium, gray cast iron, ductile iron, bronze, carbon steel, and
stainless steel. Rank them in increasing order of this property and
discuss the engineering significance of these data. Units: Mg 10 3
kg Given: Material Modulus of Elasticity Density Mat 1 "Aluminum" E
1 71.7 GPa 1 2.8 Mg m 3 Mat 2 "Titanium" E 2 113.8 GPa 2 4.4 Mg m 3
Mat 3 "Gray cast iron" E 3 103.4 GPa 3 7.2 Mg m 3 Mat 4 "Ductile
iron" E 4 168.9 GPa 4 6.9 Mg m 3 Mat 5 "Bronze" E 5 110.3 GPa 5 8.6
Mg m 3 Mat 6 "Carbon steel" E 6 206.8 GPa 6 7.8 Mg m 3 Mat 7
"Stainless steel" E 7 189.6 GPa 7 7.8 Mg m 3 i 1 2 7 Solution: See
Mathcad file P0223. 1. Calculate the specific stiffness for each
material as described in Section 2.1. E' i E i i E' i 10 6 s 2 m 2
25.6 25.9 14.4 24.5 12.8 26.5 24.3 Mat i "Aluminum" "Titanium"
"Gray cast iron" "Ductile iron" "Bronze" "Carbon steel" "Stainless
steel" 2. Rank them in increasing order of specific stiffness. Mat
5 "Bronze" E' 5 10 6 s 2 m 2 12.8 Mat 3 "Gray cast iron" E' 3 10 6
s 2 m 2 14.4 Mat 7 "Stainless steel" E' 7 10 6 s 2 m 2 24.3 2011
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40. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-23-2 Mat
4 "Ductile iron" E' 4 10 6 s 2 m 2 24.5 Mat 1 "Aluminum" E' 1 10 6
s 2 m 2 25.6 Mat 2 "Titanium" E' 2 10 6 s 2 m 2 25.9 Mat 6 "Carbon
steel" E' 6 10 6 s 2 m 2 26.5 3. Bending and axial deflection are
inversely proportional to the modulus of elasticity. For the same
shape and dimensions, the material with the highest specific
stiffness will give the smallest deflection. Or, put another way,
for a given deflection, using the material with the highest
specific stiffness will result in the least weight. 2011 Pearson
Education, Inc., Upper Saddle River, NJ. All rights reserved. This
publication is protected by Copyright and written permission should
be obtained from the publisher prior to any prohibited
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41. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-24-1
PROBLEM 2-24 Statement: Call your local steel and aluminum
distributors (consult the Yellow Pages) and obtain current costs
per pound for round stock of consistent size in low-carbon (SAE
1020) steel, SAE 4340 steel, 2024-T4 aluminum, and 6061-T6
aluminum. Calculate a strength/dollar ratio and a stiffness/dollar
ratio for each alloy. Which would be your first choice on a
cost-efficiency basis for an axial-tension-loaded round rod (a) If
maximum strength were needed? (b) If maximum stiffness were needed?
Solution: Left to the student as data will vary with time and
location. 2011 Pearson Education, Inc., Upper Saddle River, NJ. All
rights reserved. This publication 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.
42. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-25-1
PROBLEM 2-25 Statement: Call your local plastic stock-shapes
distributors (consult the Yellow Pages) and obtain current costs
per pound for round rod or tubing of consistent size in plexiglass,
acetal, nylon 6/6, and PVC. Calculate a strength/dollar ratio and a
stiffness/dollar ratio for each alloy. Which would be your first
choice on a cost-efficiency basis for an axial-tension-loaded round
rod or tube of particular diameters. (a) If maximum strength were
needed? (b) If maximum stiffness were needed? Solution: Left to the
student as data will vary with time and location. 2011 Pearson
Education, Inc., Upper Saddle River, NJ. All rights reserved. This
publication 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.
43. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-26-1
PROBLEM 2-26 Statement: A part has been designed and its dimensions
cannot be changed. To minimize its deflections under the same
loading in all directions irrespective of stress levels, which
material woulod you choose among the following: aluminum, titanium,
steel, or stainless steel? Solution: See Mathcad file P0226. 1.
Choose the material with the highest modulus of elasticity because
deflection is inversely proportional to modulus of elasticity.
Thus, choose steel unless there is a corrosive atmosphere, in which
case, choose stainless steel. P0226.xmcd 2011 Pearson Education,
Inc., Upper Saddle River, NJ. All rights reserved. This publication
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,
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information regarding permission(s), write to: Rights and
Permissions Department, Pearson Education, Inc., Upper Saddle
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44. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-27-1
PROBLEM 2-27 Statement: Assuming that the mechanical properties
data given in Appendix Table A-9 for some carbon steels represents
mean values, what is the value of the tensile yield strength for
1050 steel quenched and tempered at 400F if a reliability of 99.9%
is required? Given: Mean yield strength Sy 117 ksi Sy 807 MPa
Solution: See Mathcad file P0227. 1. From Table 2-2 the reliability
factor for 99.9% is Re 0.753 . Applying this to the mean tensile
strength gives Sy99.9 Sy Re Sy99.9 88.1 ksi Sy99.9 607 MPa 2011
Pearson Education, Inc., Upper Saddle River, NJ. All rights
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permission should be obtained from the publisher prior to any
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permission(s), write to: Rights and Permissions Department, Pearson
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45. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-28-1
PROBLEM 2-28 Statement: Assuming that the mechanical properties
data given in Appendix Table A-9 for some carbon steels represents
mean values, what is the value of the ultimate tensile strength for
4340 steel quenched and tempered at 800F if a reliability of 99.99%
is required? Given: Mean ultimate tensile strength Sut 213 ksi Sut
1469 MPa Solution: See Mathcad file P0228. 1. From Table 2-2 the
reliability factor for 99.99% is Re 0.702 . Applying this to the
mean ultimate tensile strength gives Sut99.99 Sut Re Sut99.99 150
ksi Sut99.99 1031 MPa 2011 Pearson Education, Inc., Upper Saddle
River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the
publisher prior to any prohibited reproduction, storage in a
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46. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-29-1
PROBLEM 2-29 Statement: Assuming that the mechanical properties
data given in Appendix Table A-9 for some carbon steels represents
mean values, what is the value of the ultimate tensile strength for
4130 steel quenched and tempered at 400F if a reliability of 90% is
required? Given: Mean ultimate tensile strength Sut 236 ksi Sut
1627 MPa Solution: See Mathcad file P0229. 1. From Table 2-2 the
reliability factor for 90% is Re 0.897 . Applying this to the mean
ultimate tensile strength gives Sut99.99 Sut Re Sut99.99 212 ksi
Sut99.99 1460 MPa 2011 Pearson Education, Inc., Upper Saddle River,
NJ. All rights reserved. This publication 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,
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permission(s), write to: Rights and Permissions Department, Pearson
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47. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-30-1
PROBLEM 2-30 Statement: Assuming that the mechanical properties
data given in Appendix Table A-9 for some carbon steels represents
mean values, what is the value of the tensile yield strength for
4140 steel quenched and tempered at 800F if a reliability of
99.999% is required? Given: Mean yield strength Sy 165 ksi Sy 1138
MPa Solution: See Mathcad file P0230. 1. From Table 2-2 the
reliability factor for 99.999% is Re 0.659 . Applying this to the
mean tensile strength gives Sy99.9 Sy Re Sy99.9 109 ksi Sy99.9 750
MPa 2011 Pearson Education, Inc., Upper Saddle River, NJ. All
rights reserved. This publication 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,
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48. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-31-1
PROBLEM 2-31 Statement: A steel part is to be plated to give it
better corrosion resistance. Two materials are being considered:
cadmium and nickel. Considering only the problem of galvanic
action, which would you chose? Why? Solution: See Mathcad file
P0231. 1. From Table 2-4 we see that cadmium is closer to steel
than nickel. Therefore, from the standpoint of reduced galvanic
action, cadmium is the better choice. Also, since cadmium is less
noble than steel it will be the material that is consumed by the
galvanic action. If nickel were used the steel would be consumed by
galvanic action. P0231.xmcd 2011 Pearson Education, Inc., Upper
Saddle River, NJ. All rights reserved. This publication 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
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Permissions Department, Pearson Education, Inc., Upper Saddle
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49. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-32-1
PROBLEM 2-32 Statement: A steel part with many holes and sharp
corners is to be plated with nickel. Two processes are being
considered: electroplating and electroless plating. Which process
would you chose? Why? Solution: See Mathcad file P0232. 1.
Electroless plating is the better choice since it will give a
uniform coating thickness in the sharp corners and in the holes. It
also provides a relatively hard surface of about 43 HRC. P0232.xmcd
2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights
reserved. This publication is protected by Copyright and written
permission should be obtained from the publisher prior to any
prohibited reproduction, storage in a retrieval system, or
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50. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-33-1
PROBLEM 2-33 Statement: What is the common treatment used on
aluminum to prevent oxidation? What other metals can also be
treated with this method? What options are available with this
method? Solution: See Mathcad file P0233. 1. Aluminum is commonly
treated by anodizing, which creates a thin layer of aluminum oxide
on the surface. Titanium, magnesium, and zinc can also be anodized.
Common options include tinting to give various colors to the
surface and the use of "hard anodizing" to create a thicker, harder
surface. P0233.xmcd 2011 Pearson Education, Inc., Upper Saddle
River, NJ. All rights reserved. This publication 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
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Permissions Department, Pearson Education, Inc., Upper Saddle
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51. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-34-1
PROBLEM 2-34 Statement: Steel is often plated with a less nobel
metal that acts as a sacrificial anode that will corrode instead of
the steel. What metal is commonly used for this purpose (when the
finished product will not be exposed to saltwater), what is the
coating process called, and what are the common processes used to
obtain the finished product? Solution: See Mathcad file P0234. 1.
The most commonly used metal is zinc. The process is called
"galvanizing" and it is accomplished by electroplating or hot
dipping. P0234.xmcd 2011 Pearson Education, Inc., Upper Saddle
River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the
publisher prior to any prohibited reproduction, storage in a
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52. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-35-1
PROBLEM 2-35 Statement: A low-carbon steel part is to be
heat-treated to increase its strength. If an ultimate tensile
strength of approximately 550 MPa is required, what mean Brinell
hardness should the part have after treatment? What is the
equivalent hardness on the Rockwell scale? Given: Approximate
tensile strength Sut 550 MPa Solution: See Mathcad file P0235. 1.
Use equation (2.10), solving for the Brinell hardness, HB. Sut 3.45
HB= HB Sut 3.45 MPa HB 159 2. From Table 2-3, the equivalent
hardness on the Rocwell scale is 83.9HRB. 2011 Pearson Education,
Inc., Upper Saddle River, NJ. All rights reserved. This publication
is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in
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53. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-36-1
PROBLEM 2-36 Statement: A low-carbon steel part has been tested for
hardness using the Brinell method and is found to have a hardness
of 220 HB. What are the approximate lower and upper limits of the
ultimate tensile strength of this part in MPa? Given: Hardness HB
220 Solution: See Mathcad file P0236. 1. Use equation (2.10),
solving for ultimate tensile strength. Minimum: Sutmin 3.45 HB 0.2
HB( ) MPa Sutmin 715 MPa Maximum: Sutmax 3.45 HB 0.2 HB( ) MPa
Sutmax 803 MPa 2011 Pearson Education, Inc., Upper Saddle River,
NJ. All rights reserved. This publication 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,
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54. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-37-1
PROBLEM 2-37 Statement: Figure 2-24 shows "guide lines" for minimum
weight design when failure is the criterion. The guide line, or
index, for minimizing the weight of a beam in bending is f 2/3/,
where f is the yield strength of a material and is its mass
density. For a given cross-section shape the weight of a beam with
given loading will be minimized when this index is maximized. The
following materials are being considered for a beam application:
5052 aluminum, cold rolled; CA-170 beryllium copper, hard plus
aged; and 4130 steel, Q & T @ 1200F. The use of which of these
three materials will result in the least-weight beam? Units: Mg kg
3 Given: 5052 Aluminum Sya 255 MPa a 2.8 Mg m 3 CA-170 beryllium
copper Syb 1172 MPa b 8.3 Mg m 3 4130 steel Sys 703 MPa s 7.8 Mg m
3 Solution: See Mathcad file P0237. 1. The values for the mass
density are taken from Appendix Table A-1 and the values of yield
strength come from from Tables A-2, A-4, and A-9 for aluminum,
beryllium copper, and steel, respectively. 2. Calculate the index
value for each material. Index Sy Sy 0.667 Mg m 3 MPa 0.667
Aluminum Ia Index Sya a Ia 14.4 Beryllium copper Ib Index Syb b Ib
13.4 Steel Is Index Sys s Is 10.2 The 5052 aluminum has the highest
value of the index and would be the best choice to minimize weight.
2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights
reserved. This publication 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.
55. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-38-1
PROBLEM 2-38 Statement: Figure 2-24 shows "guide lines" for minimum
weight design when failure is the criterion. The guide line, or
index, for minimizing the weight of a member in tension is f/,
where f is the yield strength of a material and is its mass
density. The weight of a member with given loading will be
minimized when this index is maximized. For the three materials
given in Problem 2-37, which will result in the least weight
tension member? Units: Mg kg 3 Given: 5052 Aluminum Sya 255 MPa a
2.8 Mg m 3 CA-170 beryllium copper Syb 1172 MPa b 8.3 Mg m 3 4130
steel Sys 703 MPa s 7.8 Mg m 3 Solution: See Mathcad file P0238. 1.
The values for the mass density are taken from Appendix Table A-1
and the values of yield strength come from from Tables A-2, A-4,
and A-9 for aluminum, beryllium copper, and steel, respectively. 2.
Calculate the index value for each material. Index Sy Sy Mg m 3 MPa
Aluminum Ia Index Sya a Ia 91.1 Beryllium copper Ib Index Syb b Ib
141.2 Steel Is Index Sys s Is 90.1 The beryllium copper has the
highest value of the index and would be the best choice to minimize
weight. 2011 Pearson Education, Inc., Upper Saddle River, NJ. All
rights reserved. This publication 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.
56. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-39-1
PROBLEM 2-39 Statement: Figure 2-23 shows "guide lines" for minimum
weight design when stiffness is the criterion. The guide line, or
index, for minimizing the weight of a beam in bending is E1/2/,
where E is the modulus of elasticity of a material and is its mass
density. For a given cross-section shape the weight of a beam with
given stiffness will be minimized when this index is maximized. The
following materials are being considered for a beam application:
5052 aluminum, cold rolled; CA-170 beryllium copper, hard plus
aged; and 4130 steel, Q & T @ 1200F. The use of which of these
three materials will result in the least-weight beam? Units: Mg kg
3 Given: 5052 Aluminum Ea 71.7 GPa a 2.8 Mg m 3 CA-170 beryllium
copper Eb 127.6 GPa b 8.3 Mg m 3 4130 steel Es 206.8 GPa s 7.8 Mg m
3 Solution: See Mathcad file P0239. 1. The values for the mass
density and modulus are taken from Appendix Table A-1. 2. Calculate
the index value for each material. Index E ( ) E 0.5 Mg m 3 GPa 0.5
Aluminum Ia Index Ea a Ia 3.0 Beryllium copper Ib Index Eb b Ib 1.4
Steel Is Index Es s Is 1.8 The 5052 aluminum has the highest value
of the index and would be the best choice to minimize weight. 2011
Pearson Education, Inc., Upper Saddle River, NJ. All rights
reserved. This publication 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.
57. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-40-1
PROBLEM 2-40 Statement: Figure 2-24 shows "guide lines" for minimum
weight design when stiffness is the criterion. The guide line, or
index, for minimizing the weight of a member in tension is E/,
where E is the modulus of elasticity of a material and is its mass
density. The weight of a member with given stiffness will be
minimized when this index is maximized. For the three materials
given in Problem 2-39, which will result in the least weight
tension member? Units: Mg kg 3 Given: 5052 Aluminum Ea 71.7 GPa a
2.8 Mg m 3 CA-170 beryllium copper Eb 127.6 GPa b 8.3 Mg m 3 4130
steel Es 206.8 GPa s 7.8 Mg m 3 Solution: See Mathcad file P0240.
1. The values for the mass density and modulus are taken from
Appendix Table A-1. 2. Calculate the index value for each material.
Index E ( ) E Mg m 3 GPa Aluminum Ia Index Ea a Ia 25.6 Beryllium
copper Ib Index Eb b Ib 15.4 Steel Is Index Es s Is 26.5 The steel
has the highest value of the index and would be the best choice to
minimize weight. 2011 Pearson Education, Inc., Upper Saddle River,
NJ. All rights reserved. This publication 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
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58. MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-1-1
PROBLEM 3-1 Statement: Which load class from Table 3-1 best suits
these systems? (a) Bicycle frame (b) Flag pole (c) Boat oar (d)
Diving board (e) Pipe wrench (f) Golf club. Solution: See Mathcad
file P0301. 1. Determine whether the system has stationary or
moving elements, and whether the there are constant or time-varying
loads. (a) Bicycle frame Class 4 (Moving element, time-varying
loads) (b) Flag pole Class 2 (Stationary element, time-varying
loads) (c) Boat oar Class 2 (Low acceleration element, time-varying
loads) (d) Diving board Class 2 (Stationary element, time-varying
loads) (e) Pipe wrench Class 2 (Low acceleration elements,
time-varying loads) (f) Golf club Class 4 (Moving element,
time-varying loads) P0301.xmcd 2011 Pearson Education, Inc., Upper
Saddle River, NJ. All rights reserved. This publication is
protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in
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59. MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-2a-1
PROBLEM 3-2a Statement: Draw free-body diagrams for the system of
Problem 3-1a (bicycle frame). Assumptions: 1. A two-dimensional
model is adequate. 2. The lower front-fork bearing at C takes all
of the thrust load from the front forks. 3. There are no
significant forces on the handle bars. Solution: See Mathcad file
P0302a. 1. A typical bicycle frame is shown in Figure 3-2a. There
are five points on the frame where external forces and moments are
present. The rider's seat is mounted through a tube at A. This is a
rigid connection, capable of transmitting two force components and
a moment. The handle bars and front-wheel forks are supported by
the frame through two bearings, located at B and C. These bearings
are capable of transmitting radial and axial loads. The pedal-arm
assembly is supported by bearings at D. These bearings are capable
of transmitting radial loads. The rear wheel-sprocket assembly is
supported by bearings mounted on an axle fixed to the frame at E.
dyF ex E F eyF dxF D eR axF ayF a aM A R dR bR Rc C ctbr B F F crF
FIGURE 3-2a Free Body Diagram for Problem 3-2a 2. The loads at B
and C can be determined by analyzing a FBD of the front wheel-front
forks assembly. The loads at D can be determined by analyzing a FBD
of the pedal-arm and front sprocket (see Problem 3-3), and the
loads at E can be determined by analyzing a FBD of the rear
wheel-sprocket assembly. 3. With the loads at B, C, D, and E known,
we can apply equations 3.3b to the FBD of the frame and solve for
Fax, Fay, and Ma. Fx : Fax Fbr cos ( ) Fcr cos ( )+ Fct sin ( ) Fdx
Fex+ 0= (1) Fy : Fay Fbr sin ( ) Fcr sin ( )+ Fct cos ( )+ Fdy Fey+
0= (2) Mz : Ma Rbx Fby Rby Fbx( )+ Rcx Fcy Rcy Fcx( )+ Rdx Fdy Rdy
Fdx( ) Rex Fey Rey Fex( )++ ... 0= (3) P0302a.xmcd 2011 Pearson
Education, Inc., Upper Saddle River, NJ. All rights reserved. This
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60. MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-2e-1
PROBLEM 3-2e Statement: Draw free-body diagrams for the system of
Problem 3-1e (pipe wrench). Assumptions: A two-dimensional model is
adequate. Solution: See Mathcad file P0302e. 1. A typical pipe
wrench with a pipe clamped in its jaw is shown in Figure 3-2e(a).
When a force Fhand is applied on the wrench, the piping system
provides an equal and opposite force and a resisting torque, Tpipe.
T F pipe hand hand a F (a) FBD of pipe wrench and pipe bd Fay Fax A
Fbt Fbn (b) FBD of pipe wrench only FIGURE 3-2e Free Body Diagrams
for Problem 3-2e 2. The pipe reacts with the wrench at the points
of contact A and B. The forces here will be directed along the
common normals and tangents. The jaws are slightly tapered and, as
a result, the action of Fhand tends to drive the wrench further
into the taper, increasing the normal forces. This, in turn, allows
for increasing tangential forces. It is the tangential forces that
produce the turning torque. 3. Applying equations 3.3b to the FBD
of the pipe wrench, Fx : Fax Fbn cos ( )+ Fbt sin ( ) 0= (1) Fy :
Fay Fbn sin ( )+ Fbt cos ( )+ Fhand 0= (2) MA : d Fbt cos ( ) Fbn
sin ( )+( ) d a+( ) Fhand 0= (3) 4. These equations can be solved
for the vertical forces if we assume is small so that sin() = 0 and
cos () = 1. P0302e.xmcd 2011 Pearson Education, Inc., Upper Saddle
River, NJ. All rights reserved. This publication 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.
61. MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-3-1
PROBLEM 3-3 Statement: Draw a free-body diagram of the pedal-arm
assembly from a bicycle with the pedal arms in the horizontal
position and dimensions as shown in Figure P3-1. (Consider the two
arms, pedals and pivot as one piece). Assuming a rider-applied
force of 1500 N at the pedal, determine the torque applied to the
chain sprocket and the maximum bending moment and torque in the
pedal arm. Given: a 170 mm b 60 mm Frider 1.5 kN Assumptions: The
pedal-arm assembly is supprted by bearings at A and at B. Solution:
See Figure 3-3 and Mathcad file P0303. 1. The free-body diagram
(FBD) of the pedal-arm assembly (including the sprocket) is shown
in Figure 3-3a. The rider-applied force is Frider and the force
applied by the chain (not shown) is Fchain. The radial bearing
reactions are Fax, Faz, Fbx, and Fbz. Thus, there are five unknowns
Fchain, Fax, Faz, Fbx, and Fbz. In general, we can write six
equilibrium equations for a three-dimensional force system, but in
this system there are no forces in the y-direction so five
equations are available to solve for the unknowns. Frider Pedal x b
Arm a Fax Fbx A B Arm (sectioned) y Sprocket Faz Fbz Fchain z (a)
FBD of complete pedal-arm assembly x Pedal b Arm Frider a cF Mc y z
Tc (b) FBD of pedal and arm with section through the origin FIGURE
3-3 Dimensions and Free Body Diagram of the pedal-arm assembly for
Problem 3-3 2. The torque available to turn the sprocket is found
by summing moments about the sprocket axis. From Figure 3-3a, it is
2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights
reserved. This publication 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.
62. MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-3-2
Ty-axis : a Frider r Fchain a Frider Tsprocket= 0= Tsprocket a
Frider Tsprocket 255 N m where r is the sprocket pitch radius. 3.
In order to determine the bending moment and twisting torque in the
pedal arm, we will cut the arm with a section plane that goes
through the origin and is parallel to the y-z plane, removing
everything beyond that plane and replacing it with the internal
forces and moments in the pedal arm at the section. The resulting
FBD is shown in Figure 3-3b. The internal force at section C is Fc
the internal bending moment is Mc, and the internal twisting moment
(torque) is Tc. We can write three equilibrium equations to solve
for these three unknowns: Shear force in pedal arm at section C Fz
: Fc Frider 0= Fc Frider Fc 1.5 kN Bending moment in pedal arm at
section C My-axis : a Frider Mc 0= Mc a Frider Mc 255 N m Twisting
moment in pedal arm at section C Mx-axis : b Frider Tc 0= Tc b
Frider Tc 90 N m 2011 Pearson Education, Inc., Upper Saddle River,
NJ. All rights reserved. This publication 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.
63. MACHINE DESIGN - An Integrated Approach,4th Ed. 3-4-1
PROBLEM 3-4 Statement: The trailer hitch from Figure 1-1 has loads
applied as shown in Figure P3-2. The tongue weight of 100 kg acts
downward and the pull force of 4905 N acts horizontally. Using the
dimensions of the ball bracket in Figure 1-5 (p. 15), draw a
free-body diagram of the ball bracket and find the tensile and
shear loads applied to the two bolts that attach the bracket to the
channel in Figure 1-1. Given: a 40 mm b 31 mm c 70 mm d 20 mm
Mtongue 100 kg Fpull 4.905 kN t 19 mm Assumptions: 1. The nuts are
just snug-tight (no pre-load), which is